MA1101ContentsClose ×18 chapters

Static textbook · Complete edition

MA1101 — Mathematics for Electronics IMathematics for Electronics I

Read mathematics as the engineering language for modelling, analysis, and design.

Begin with Chapter 01

A Standalone Textbook for Electronics Students

Mathematics is one of the ways engineers make a physical idea precise. A circuit description such as “the current becomes smaller as the resistor becomes larger” is useful intuition; the relationship \(V=IR\) lets us calculate, predict, test, and redesign. This book develops the mathematics needed to do that well.

The emphasis is not on memorising a pile of formulas. For each major idea we will ask four questions: what does it mean, why is it needed, how is it derived or used, and where does it become useful in an electronic system? The course is about mathematics first. Circuit, signal, and measurement examples appear when they make the mathematics easier to understand.

How to Use This Book

Read a derivation with paper nearby. Before looking at a worked solution, try to identify the unknowns, the known quantities, and the relationship between them. For exercises, do not merely compare a final number: compare your choice of method, assumptions, units, and checks. A calculator is valuable for arithmetic; it cannot decide whether a denominator may be zero or whether a graph represents the intended physical quantity.

Notation Used Throughout

Unless stated otherwise, variables take real values. We use \(t\) for time, \(x\) for a generic independent variable, and \(n\) for a positive-integer index. A function \(f\) sends an input \(x\) to an output \(f(x)\). The symbol \(i\) satisfies \(i^2=-1\); in AC-circuit contexts engineers often write \(j\) for the same number so that \(i\) can remain available for current. An interval such as \([a,b]\) contains its endpoints, while \((a,b)\) does not.


Part I

Part I — Mathematical Language, Functions, and Algebraic Models

Mathematical modelling begins by choosing quantities, assigning symbols, and writing relationships that are true under clearly stated assumptions. The first Part develops the language for doing that, then uses it for lines, quadratics, and systems of equations.

Part I · Mathematical Language, Functions, and Algebraic Models

01

Chapter 1 — Mathematics as an Engineering Language

Open chapter view →

Why This Matters

An electronic product is specified by quantities: supply voltage, current, resistance, frequency, power, gain, timing error, and noise level. Numbers alone are not enough. Engineering requires relationships between quantities, together with units and assumptions. Mathematics gives those relationships a form that can be checked and used.

1.1 Quantities, numbers, and units

A physical quantity is a measurable property together with a unit. For example, \(5\) is only a number; \(5\ \mathrm{V}\) is a voltage. The same care is needed when writing equations: both sides of a physically meaningful equation must have the same units.

Real numbers are the main number system of this course. They include natural numbers such as \(0,1,2,\ldots\), integers such as \(-2,-1,0,1,2\), rational numbers that can be written as \(p/q\) with \(q\ne0\), and irrational numbers such as \(\sqrt2\) and \(\pi\). Visualise the real numbers as points on a number line: values increase as we move to the right, decrease as we move to the left, and there is no largest or smallest real number.

Scientific notation keeps very large and very small quantities readable:

\[4700\ \Omega = 4.7\times10^3\ \Omega, \qquad 0.000,002\ \mathrm{F}=2\times10^{-6}\ \mathrm{F}.\]

The prefixes kilo-, milli-, micro-, and nano- are shorthand for powers of ten. Always estimate the order of magnitude before trusting a numerical output. A calculated capacitor voltage of \(3.3\times10^9\ \mathrm{V}\) in a \(5\ \mathrm{V}\) laboratory circuit is usually a model or unit error, not an exciting discovery.

From first principles. A unit travels through an equation in the same way as a numerical factor. This makes dimensional checking a calculation: multiplying volts by amperes gives watts, so a claimed power value must have the unit V A, which is the watt.

\[1\,\mathrm{W}=1\,\mathrm{V}\times1\,\mathrm{A}\]

1.2 Variables and equations

A variable is a symbol representing a quantity that may vary or whose value is currently unknown. An equation asserts that two expressions have the same value. In the resistor model

\[V=IR,\]

\(V\) is voltage, \(I\) is current, and \(R\) is resistance. If \(R\) is constant and \(V\) changes, then \(I\) changes as well. The equation makes that relationship testable.

A disciplined modelling workflow is:

  1. State the situation and the assumptions.
  2. Name each relevant quantity and include units.
  3. Write the basic relationships.
  4. Solve only for quantities that the model actually determines.
  5. Check the result using units, signs, magnitude, and limiting cases.

Example — Turning words into equations

Two components cost ₹12 and ₹20 each. A purchase contains \(x\) of the first component and \(y\) of the second, has \(8\) items in total, and costs ₹136.

The count and cost statements become

\[x+y=8, \qquad 12x+20y=136.\]

The mathematics has not yet solved the problem, but it has separated two pieces of information that can now be combined systematically in Chapter 4.

Keep the statement balanced. An equation says that two expressions name the same quantity. Applying the same reversible operation to both sides preserves that statement; applying it to only one side changes the problem. This is the reason an unknown may be isolated without changing the physical relation it represents.

  1. Start with the relation containing the unknown.
  2. Undo additions or subtractions on both sides.
  3. Undo multiplication or division by a non-zero quantity on both sides, then check the unit of the result.

1.3 Expressions and algebraic equivalence

An expression is a meaningful combination of numbers, variables, and operations, such as \(3x^2-4x+1\). Two expressions are equivalent if they have the same value for every input where both are defined. For example,

\[\frac{x^2-9}{x-3}=x+3\]

is true only for \(x\ne3\). The left-hand expression is undefined at \(x=3\), while the right-hand expression is defined there. Algebraic cancellation simplifies an expression; it does not silently repair a forbidden input.

Example — Unit-aware rearrangement

From \(V=IR\), solving for resistance gives

\[R=\frac{V}{I}, \qquad I\ne0.\]

If \(V=9\ \mathrm{V}\) and \(I=3\ \mathrm{mA}=3\times10^{-3}\ \mathrm A\), then

\[R=\frac{9}{3\times10^{-3}}\ \Omega=3\times10^3\ \Omega=3\ \mathrm{k}\Omega.\]

The unit calculation confirms that volt per ampere is an ohm.

Equivalence has a condition. Two expressions may produce the same value wherever both are defined, yet one may conceal a restriction of the other. Factoring and cancelling are useful because they expose structure, but the original denominator or square-root condition still controls the permissible inputs.

  1. Record the original domain before simplifying.
  2. Apply a reversible algebraic transformation and state its condition.
  3. Use the simplified form for calculation without silently enlarging the original domain.

1.4 Geometry as a model-building tool

Equations often encode geometry. A rectangle of base \(b\) and perpendicular height \(h\) has area \(A=bh\). A parallelogram with the same base and perpendicular height has the same area: cut a triangular piece from one side and move it to the other side to form a rectangle. This is why its area is also \(bh\).

The word perpendicular matters. A slanted side of a parallelogram is not automatically its height. The area of a triangle with base \(b\) and perpendicular height \(h\) is half the matching parallelogram area:

\[A=\frac12 bh.\]

This kind of rearrangement argument is a useful habit: replace a difficult shape or equation by an equivalent one whose meaning is easier to see.

Area is controlled by perpendicular height. A triangle can be paired with a matching copy to form a parallelogram, and a sloping side can be cut and shifted to make a rectangle. These rearrangements preserve area because they preserve the same base and perpendicular height; they do not turn an arbitrary side length into a height.

\[A_{\triangle}=\frac12 bh,\qquad A_{\text{parallelogram}}=bh\]
  1. Choose a base and draw its perpendicular height.
  2. Rearrange the matching pieces into a rectangle or parallelogram with that base and height.
  3. Apply the half-area relation only after identifying the perpendicular height.

Common Mistakes

  • Treating a unitless number as a completed engineering answer.
  • Cancelling a factor without recording the value at which that factor is zero.
  • Calling a slanted length a height when it is not perpendicular to the base.
  • Using an equation outside the assumptions that produced it, such as applying the ideal-resistor law to a component whose resistance changes strongly with temperature without saying so.

Chapter Summary

Engineering mathematics starts with quantities, units, variables, relationships, and checks. Real numbers provide the main numerical setting. Equations express relationships; algebra changes their form without changing their permitted meaning. Units and domain restrictions are part of the mathematics, not an afterthought.

Key Formulas

\[V=IR,\qquad A_{\text{rectangle}}=bh,\qquad A_{\text{triangle}}=\frac12bh.\]

Exercises

Conceptual Questions

  1. Explain why \(\frac{x^2-1}{x-1}=x+1\) is not an equality of functions on all real numbers.
  2. State the unit of \(V/I\) and the restriction needed to compute it.
  3. A parallelogram has side lengths \(8\) cm and \(5\) cm. Why is its area not necessarily \(40\ \mathrm{cm}^2\)?

Basic Problems

  1. Write \(0.000,047\ \mathrm F\) in scientific notation and convenient engineering-prefix form.
  2. A \(12\ \mathrm V\) source drives \(4\ \mathrm{mA}\) through a resistor. Find its resistance.
  3. A triangular sensor region has base \(18\) mm and perpendicular height \(7\) mm. Find its area.

Part I · Mathematical Language, Functions, and Algebraic Models

02

Chapter 2 — Coordinates, Graphs, and Functions

Open chapter view →

Why This Matters

When a quantity depends on another quantity, a graph lets us see the relationship before we calculate with it. Voltage may depend on time, current may depend on voltage, and a measured output may depend on a control setting. Functions make such dependencies precise.

2.1 Cartesian coordinates

The Cartesian plane has a horizontal \(x\)-axis and a vertical \(y\)-axis. A point \((x,y)\) means move \(x\) units horizontally from the origin and then \(y\) units vertically. The order matters: \((2,5)\) and \((5,2)\) are different points.

To visualise a graph of \(y=f(x)\), imagine choosing an input \(x\), calculating the associated output, and plotting the point \((x,f(x))\). Repeating this process creates a curve or collection of points.

2.2 What a function is

A function assigns exactly one output to each allowed input. We write

\[f:D\to R,\]

where \(D\) is the domain of permitted inputs and \(R\) is a set containing the outputs. The range is the set of outputs actually attained.

For \(f(x)=x^2\), if the domain is all real numbers, then the range is \([0,\infty)\). The value \(f(-3)=9\) is a number. The function \(f\) is the whole rule, not the particular output \(9\).

The vertical-line test gives a graphical version of the definition: if a vertical line meets a graph more than once, some input has two outputs, so the graph does not represent a single-valued function of \(x\).

Example — Domain before calculation

Consider

\[g(x)=\frac{\sqrt{x-1}}{x-4}.\]

The square root requires \(x-1\ge0\), so \(x\ge1\). The denominator requires \(x\ne4\). Therefore

\[\operatorname{dom}(g)=[1,4)\cup(4,\infty).\]

Only after finding this domain should we substitute values or sketch the graph.

Read the arrow before calculating. A function is the complete assignment from permitted inputs to outputs, not one value produced by it. A domain restriction is therefore part of the rule: it tells us which plotted horizontal positions may be read, while the range records the vertical positions actually reached.

\[f:D\to R,\qquad f(-3)=9\text{ for }f(x)=x^2\]

2.3 Function algebra

Given functions \(f\) and \(g\), we can define

\[(f+g)(x)=f(x)+g(x),\quad (fg)(x)=f(x)g(x),\quad \left(\frac f g\right)(x)=\frac{f(x)}{g(x)}.\]

The quotient is defined only where \(g(x)\ne0\). Function algebra is useful for building models from simpler pieces. For example, a sinusoidal input plus a constant offset has the form \(v(t)=V_0+A\sin(\omega t)\).

Combine both values and both restrictions. Function addition and multiplication use the two values at one common input. A quotient has one further condition: the denominator function must not be zero there. Algebra builds a new rule, but it cannot make either original input restriction disappear.

\[\left(\frac fg\right)(x)=\frac{f(x)}{g(x)},\qquad g(x)\ne0\]
  1. List inputs permitted by each component function.
  2. Apply the selected operation to their values at a shared permitted input.
  3. For a quotient, exclude every input that makes the denominator zero.

2.4 Graphs as solution tools

To solve an equation \(f(x)=g(x)\) graphically, plot \(y=f(x)\) and \(y=g(x)\) on the same axes. Their intersection points have equal \(y\)-coordinates, so their \(x\)-coordinates solve the equation. A graph provides insight and estimates; it does not replace an exact algebraic method when close roots or scale distortions matter.

Example — A graphical model with a check

Suppose current through a resistor is modeled by \(I=V/1000\) amperes and a load requires \(I=0.003\) A. The intersection of

\[y=\frac{x}{1000},\qquad y=0.003\]

occurs at \(x=3\). Algebra confirms the graph:

\[\frac V{1000}=0.003 \implies V=3\ \mathrm V.\]

Read an intersection as a simultaneous statement. At a crossing, both graph rules assign the same output to the same input. The horizontal coordinate solves f(x)=g(x); the vertical coordinate is the shared value. A graph is therefore a geometric way to audit an algebraic solution.

\[f(x)=g(x)\quad\Longleftrightarrow\quad (x,f(x))\text{ lies on both graphs}\]

2.5 Transformations and interpretation

Starting from \(y=f(x)\):

  • \(y=f(x)+k\) shifts the graph upward by \(k\).
  • \(y=f(x-h)\) shifts it right by \(h\).
  • \(y=af(x)\) scales output values vertically.
  • \(y=f(-x)\) reflects the graph in the \(y\)-axis.

These are not rules to memorise without meaning. For instance, replacing \(x\) by \(x-h\) asks the original function to receive the value \(x-h\). To produce the old feature at input \(0\), the new graph needs \(x-h=0\), hence \(x=h\): the feature moves right.

Why this works. The expression inside a function changes the input before the function acts. Thus f(x-h) asks the old graph for a value h units earlier, so the visible graph shifts right by h. The minus sign is therefore an input correction, not a leftward motion.

\[g(x)=a f(x-h)+k\]

Common Mistakes

  • Confusing \(f\) with \(f(x)\).
  • Assuming every algebraic expression is defined for every real number.
  • Reading a graph scale carelessly, especially when one axis is compressed.
  • Treating graph intersections as exact values when the plot is only an estimate.

2.6 Domain, range, and transformations as graph-reading tools

The domain answers “which inputs make sense?”; the range answers “which outputs can the rule actually produce?” To find a domain, exclude zero denominators, negative even-root inputs, and invalid logarithms. To find a range, use a graph or rewrite the expression to expose a lowest, highest, or excluded value.

Example — range from a transformed parabola

Let \(h(x)=-2(x-3)^2+5\). Since \((x-3)^2\ge0\),

\[-2(x-3)^2\le0,\qquad h(x)\le5.\]

The maximum \(5\) is achieved at \(x=3\), so the range is \((-\infty,5]\) while the domain is all real numbers. The negative factor reflects the basic parabola, the \(-3\) inside moves its vertex right, and \(+5\) moves it upward.

Example — composition carries an input restriction

Let \(f(x)=\sqrt{x}\) and \(g(x)=4-x\). Then

\[(f\circ g)(x)=\sqrt{4-x}.\]

Although \(g\) accepts every real input, the composite requires \(4-x\ge0\). Therefore its domain is \((-\infty,4]\). Composition is not multiplication: it asks one function to supply the input to the other.

Example — inverse intuition from calibration

If a sensor model is \(V=0.02T\), then its inverse is \(T=50V\). The inverse swaps the roles of known input and inferred output. Graphically, inverse graphs are reflections in \(y=x\) once compatible axis scales are understood.

Keep a domain ledger while transforming. A horizontal shift changes which input reaches a restricted inner rule. A vertical shift changes outputs but cannot repair an input that was never permitted. Reading the inside first is a reliable way to carry exclusions through a transformed graph.

Chapter Summary

Coordinates locate points; a graph displays input-output pairs. A function gives one output for each permitted input. Domain and range are part of the definition. Function algebra and graph transformations build useful models, while graph intersections give a visual interpretation of equations.

2.7 Reading a graph as a mathematical statement

A graph simultaneously communicates domain, range, zeros, intercepts, increasing and decreasing portions, symmetry, and limiting behaviour. Read its axes before reading a numerical value: a plotted “3” may mean \(3\) V, \(3\) ms, or \(3\times 10^{-3}\) depending on the scale. Algebraic form often predicts the graph before any points are plotted.

Example — rational form predicts a graph

For

\[r(x)=\frac{x+1}{x-2},\]

the denominator excludes \(x=2\), so there is a vertical asymptote there. Division or rearrangement gives

\[r(x)=1+\frac{3}{x-2}.\]

Thus the graph is the reciprocal curve \(1/x\), shifted right by \(2\) and up by \(1\), with horizontal asymptote \(y=1\). Its range excludes \(1\), because \(3/(x-2)\) cannot be zero. This conclusion comes from structure, not a dense table of points.

Example — graphical solutions are estimates with a meaning

To solve \(\sqrt{x+1}=3-x\), first require \(x+1\ge0\) and \(3-x\ge0\) because a square root is nonnegative. A graph of the two sides suggests an intersection near \(x=2\). Squaring is then safe under those restrictions:

\[x+1=(3-x)^2=x^2-6x+9,\]

so \(x^2-7x+8=0\). Of its two algebraic roots, only \((7-\sqrt{17})/2\) satisfies the original equation. The graph locates the solution and alerts us to the one-intersection expectation; substitution rejects the extraneous root created by squaring.

Example — a horizontal-line test is a range test in disguise

For \(y=(x-1)^2+2\), a horizontal line at \(y=1\) does not meet the graph, so \(1\) is not in the range. Every horizontal line \(y=k\) with \(k>2\) meets twice, so the all-real function is not one-to-one. Restricting to \(x\ge1\) keeps one branch and makes an inverse possible.

Read structure before points. In r(x)=(x+1)/(x-2)=1+3/(x-2), the denominator records the excluded input x=2, and the separated constant records the horizontal asymptote y=1. The two branches approach those lines without supplying a value at the excluded input.

\[r(x)=\frac{x+1}{x-2}=1+\frac3{x-2},\qquad x\ne2\]
  1. Find excluded inputs from the unsimplified denominator.
  2. Rewrite the rational expression to expose its shifted reciprocal form.
  3. Use the exclusion and the constant term to identify the vertical and horizontal asymptotes before estimating values from the graph.

Exercises

  1. Find the domain and range of \(f(x)=\sqrt{9-x^2}\).
  2. Is the relation \(x^2+y^2=1\) a function of \(x\)? Explain using the vertical-line test.
  3. Describe the transformations taking \(y=x^2\) to \(y=-2(x-3)^2+5\).
  4. A sensor has output \(v(t)=2+0.5\sin(100\pi t)\). State its offset, amplitude, and period.
  5. Find the domain and range of \(q(x)=1/(x-2)^2\) and describe its asymptotes.
  6. Given \(f(x)=x^2+1\) and \(g(x)=3-x\), find \(f\circ g\) and \(g\circ f\).
  7. Explain why \(y=x^2\) has no all-real inverse, then give a restriction that does.
  8. Without plotting points, state the domain, range, and asymptotes of \((x-4)/(x+1)\) by rewriting it.
  9. Solve \(\sqrt{x+2}=4-x\), checking all algebraic candidates in the original equation.

Part I · Mathematical Language, Functions, and Algebraic Models

03

Chapter 3 — Lines, Linear Models, and Quadratics

Open chapter view →

Why This Matters

Linear relationships describe proportional change and local approximations. Quadratic relationships describe curvature, extrema, and roots. Together they form the algebraic vocabulary needed for circuit equations, optimisation, and later calculus.

3.1 Straight lines and slope

A non-vertical straight line has equation

\[y=mx+b.\]

The constant \(m\) is the slope: change in \(y\) per unit change in \(x\). The constant \(b\) is the \(y\)-intercept, because \(x=0\) gives \(y=b\).

For two distinct points \((x_1,y_1)\) and \((x_2,y_2)\) on a non-vertical line,

\[m=\frac{y_2-y_1}{x_2-x_1},\qquad x_2\ne x_1.\]

Why does this work? The line equation gives \(y_1=mx_1+b\) and \(y_2=mx_2+b\). Subtracting eliminates \(b\):

\[y_2-y_1=m(x_2-x_1).\]

Dividing by the nonzero horizontal change gives the slope formula. A vertical line has \(x_2-x_1=0\), so its slope is undefined and its equation is \(x=c\).

Slope is a ratio with direction. Moving right by a chosen run and measuring the signed rise gives the same quotient anywhere on a straight line. A negative slope does not mean a negative length; it records that the output decreases as the input increases.

\[m=\frac{\Delta y}{\Delta x}\]

3.2 Point-slope and two-point forms

Starting with \(y=mx+b\) and a point \((x_1,y_1)\) on the line,

\[y-y_1=m(x-x_1).\]

This point-slope form is often the quickest way to write a line. Substituting the two-point slope gives

\[\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1},\]

provided the line is not vertical and \(x\ne x_1\) during the displayed division. The undivided point-slope form remains safer in symbolic work.

Example — Constructing a calibration line

A sensor gives \(0.4\ \mathrm V\) at \(20^\circ\mathrm C\) and \(2.0\ \mathrm V\) at \(100^\circ\mathrm C\). If the response is linear, then

\[m=\frac{2.0-0.4}{100-20}=0.02\ \mathrm{V}/^\circ\mathrm C.\]

Using \((20,0.4)\),

\[V-0.4=0.02(T-20),\]

so \(V=0.02T\). The zero intercept is a feature of this particular data set, not a general property of sensors.

Subtract first so the intercept disappears. Two points on one non-vertical line share the same intercept, so subtracting their equations leaves only the rate of change. Once that slope is known, point-slope form rebuilds the whole line from either measured point without requiring a separately guessed intercept.

\[m=\frac{11-2}{4-1}=3,\qquad y-2=3(x-1)\]
  1. Subtract the y-values and the non-zero x-values in the same order.
  2. Divide the signed rise by the signed run to obtain the slope.
  3. Substitute that slope and either original point into y minus y one equals m times x minus x one.

3.3 Intersections and solution cases

The intersection of \(y=m_1x+b_1\) and \(y=m_2x+b_2\) satisfies both equations. Equating them gives

\[m_1x+b_1=m_2x+b_2.\]

If \(m_1\ne m_2\), then

\[x=\frac{b_2-b_1}{m_1-m_2}\]

and there is one intersection. If \(m_1=m_2\) but \(b_1\ne b_2\), the lines are parallel and there is no solution. If both slope and intercept agree, the two equations describe the same line and there are infinitely many solutions.

Three pictures, three algebraic outcomes. Distinct non-parallel lines meet once; distinct parallel lines never share a point; coincident lines share every point. Before manipulating symbols, compare slopes and intercepts: that comparison predicts whether a unique answer is even possible.

3.4 Quadratic functions and their shape

A quadratic function has the form

\[f(x)=ax^2+bx+c,\qquad a\ne0.\]

Its graph is a parabola. The sign of \(a\) controls opening: \(a>0\) gives a bowl opening upward; \(a<0\) gives a bowl opening downward. To see the centre and extreme value clearly, rewrite the quadratic in vertex form.

Start by factoring \(a\) from the first two terms:

\[ax^2+bx+c=a\left(x^2+\frac ba x\right)+c.\]

Inside the bracket, add and subtract \(\left(\frac{b}{2a}\right)^2\):

\[f(x)=a\left[x^2+\frac ba x+\left(\frac{b}{2a}\right)^2 -\left(\frac{b}{2a}\right)^2\right]+c.\]

The first three terms form a square, hence

\[f(x)=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.\]

The vertex is therefore

\[\left(-\frac b{2a},\ c-\frac{b^2}{4a}\right).\]

Example — Completing the square

Find the vertex and minimum of \(f(x)=2x^2-8x+3\).

\[\begin{aligned} f(x)&=2(x^2-4x)+3\\ &=2[(x-2)^2-4]+3\\ &=2(x-2)^2-5. \end{aligned}\]

Because \(2(x-2)^2\ge0\), the minimum is \(-5\) at \(x=2\).

Derivation. Completing the square manufactures the square that reveals a parabola's vertex. The added and subtracted quantity is the same, so the function is not changed; only its geometry becomes visible.

\[x^2-6x+4=(x-3)^2-5\]

3.5 Roots and the quadratic formula

Roots solve \(ax^2+bx+c=0\). Using the completed-square expression,

From the completed-square form, move the constant term carefully:

\[a\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a}-c.\]

Multiply by \(4a\):

\[4a^2\left(x+\frac{b}{2a}\right)^2=b^2-4ac.\]

Taking square roots and solving for \(x\) gives

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\]

The quantity \(\Delta=b^2-4ac\) is the discriminant. If \(\Delta>0\), there are two distinct real roots; if \(\Delta=0\), one repeated real root; and if \(\Delta<0\), no real roots. Complex roots are studied in Chapter 14.

The discriminant is a geometric count. After completing the square, the square root in the quadratic formula is real twice, once, or not at all according as the discriminant is positive, zero, or negative. This is exactly the count of x-axis intersections of a parabola, not a separate graphical rule.

\[x^2-6x+4=(x-3)^2-5,\qquad \Delta\gtrless0\Longleftrightarrow 2,1,0\text{ real roots}\]
  1. Compute b squared minus four a c before taking a square root.
  2. Classify its sign as positive, zero, or negative.
  3. Use the matching root count to check the expected number of graph intersections.

3.6 Lines and parabolas as geometric models

Slope is a rate with units. If voltage is plotted against time, slope has units V/s; an intercept has voltage units. A negative slope says the output decreases as input increases. A line through two measurements is an interpolation model, not proof that the same rule remains valid far outside the measurements.

Example — parallel and perpendicular lines

The line \(2x-3y=6\) is \(y=(2/3)x-2\). A parallel line through \((3,1)\) is

\[y-1=\frac23(x-3).\]

A perpendicular non-vertical line has slope \(-3/2\), because the product of slopes is \(-1\), so through the same point it is

\[y-1=-\frac32(x-3).\]

Vertical and horizontal lines are the special case: one is perpendicular to the other, but a vertical line has no finite slope to invert.

Example — discriminant as a graphical test

For \(2x^2+4x+5=0\),

\[\Delta=4^2-4(2)(5)=-24.\]

There are no real roots, so the upward-opening parabola never meets the \(x\)-axis. Completing the square confirms this:

\[2x^2+4x+5=2(x+1)^2+3>0.\]

Engineering Example — a line is a local approximation

Near an operating current \(I_0\), a non-linear device curve can be approximated by \(V(I_0)+m(I-I_0)\). The slope \(m\) is a local resistance-like rate \(\Delta V/\Delta I\). Chapter 7 makes that local slope precise as a derivative.

An intersection makes both models true. Replacing the line's output by the parabola's output produces one quadratic equation in the shared input. Its discriminant predicts whether the line crosses twice, is tangent once, or misses the parabola; substitution then returns each corresponding vertical coordinate.

\[ax^2+bx+c=mx+d\Longrightarrow ax^2+(b-m)x+(c-d)=0\]

Problem-Solving Strategy

For a line, choose slope-intercept form when the slope/intercept are known and point-slope form when a point and slope are known. For a quadratic, use vertex form to study shape and extrema, factorisation when factors are visible, and the quadratic formula when a general reliable method is needed.

Exercises

  1. Find the equation of the line through \((2,5)\) and \((6,17)\).
  2. Determine whether \(3x-2y=7\) and \(6x-4y=14\) have one, none, or infinitely many solutions.
  3. Complete the square for \(x^2+6x-7\) and state its vertex.
  4. Solve \(3x^2-5x-2=0\) and verify both roots by substitution.
  5. A power model is \(P(R)=\frac{36R}{(R+3)^2}\) watts for \(R>0\). Sketch its qualitative behaviour and identify why it is not a linear model.
  6. Find a line through \((1,4)\) perpendicular to \(3x+2y=7\).
  7. Use discriminant and vertex form to explain why \(x^2-6x+13=0\) has no real root.
  8. A calibration line passes through \((10,0.8)\) and \((70,2.6)\). Find its model and state the units of its slope.

Part I · Mathematical Language, Functions, and Algebraic Models

04

Chapter 4 — Linear Systems and Matrix Methods

Open chapter view →

Why This Matters

One equation describes one relationship. A circuit node, a network of constraints, or a polynomial fitted to data usually produces several relationships at once. A linear system lets us solve them together and, just as importantly, tells us when the data are inconsistent or leave freedom in the answer.

4.1 Linear systems as models

A linear equation in variables \(x_1,\ldots,x_n\) has the form

\[a_1x_1+a_2x_2+\cdots+a_nx_n=b.\]

The variables occur only to the first power and are not multiplied by each other. A linear system consists of several such equations.

Engineering Connection — node voltages

At a circuit node, Kirchhoff’s current law says that the algebraic sum of currents is zero. If each branch current is written with Ohm’s law, unknown node voltages often appear linearly. The mathematics does not replace circuit reasoning: it organises the relationships produced by it.

Keep every constraint in view. A simultaneous solution is one ordered list of unknowns that makes every equation true at once. Each equation can be read as a line or plane of allowed points; solving a system finds their shared set rather than solving each relation independently.

\[x+y=3,\qquad2x-y=0\]
  1. Write one equation for each independent constraint.
  2. Use an operation that preserves the full common solution set.
  3. Substitute the candidate back into every original equation, not only the last simplified row.

4.2 Elimination and back-substitution

Follow one row operation at a time. The augmented matrix below represents the same system throughout. Advance the stage only after reading the operation: the last column is transformed with the coefficient columns because it belongs to the equation, not because it is an answer column.

Consider

\[\begin{aligned} 2x+y&=7,\\ x-y&=2. \end{aligned}\]

The \(y\)-coefficients are already opposites, so add the two equations directly:

\[(2x+y)+(x-y)=7+2.\]

The \(y\) terms cancel, leaving \(3x=9\), hence \(x=3\). Substituting into the second equation gives \(3-y=2\), so \(y=1\). The cancellation was deliberate: we chose an operation that removes one unknown while preserving the solution set.

The guiding idea is to replace equations by equivalent equations that remove one variable. Valid row operations are:

  1. Exchange two equations.
  2. Multiply one equation by a nonzero constant.
  3. Add a multiple of one equation to another.

They preserve the solution set because each new equation is a logical consequence of the old system and the operation can be reversed.

From first principles. Replacing one equation by a non-zero multiple, or adding a multiple of one equation to another, does not change the pair of values that satisfies both equations. Elimination is therefore a solution-preserving rewrite, not a numerical trick.

  1. Keep the first equation unchanged.
  2. Replace the second equation by the second equation minus twice the first.
  3. Solve the resulting one-variable equation, then substitute back.

4.3 Augmented matrices

The coefficients and right-hand sides can be recorded compactly as an augmented matrix:

\[\left[\begin{array}{cc|c} 2&1&7\\ 1&-1&2 \end{array}\right].\]

The vertical bar separates the coefficient matrix from the constants. Matrix notation reduces bookkeeping; it does not change the underlying equations.

The bar keeps constants attached to their equations. An augmented matrix is a compact transcription of a system, not a new problem. Every row operation changes all entries in its row, including the right-hand side, because the row continues to stand for one complete equation.

\[\left[\begin{array}{cc|c}1&1&3\\2&-1&0\end{array}\right]\]
  1. Align coefficients in one fixed variable order.
  2. Place the constants after the augmented bar.
  3. Apply any row operation to the complete row, including its final entry.

4.4 Row-echelon form, pivots, and rank

After elimination, a system is easiest to read in row-echelon form: each successive nonzero row begins farther to the right. The first nonzero entry in a nonzero row is a pivot. The number of pivots in the coefficient matrix is its rank, denoted \(r\).

For a consistent system with \(n\) unknowns:

  • \(r=n\) means every variable is determined: there is one solution.
  • \(r<n\) means \(n-r\) variables are free: there are infinitely many solutions.

An augmented row of the form

\[[0\quad0\quad\cdots\quad0\mid b],\qquad b\ne0,\]

means \(0=b\), a contradiction. The system is inconsistent and has no solution.

Example — Free variable and vector form

Solve

\[\begin{aligned} x+2y-z&=4,\\ 2x+4y-2z&=8. \end{aligned}\]

The second equation is twice the first, so it adds no new restriction. Let \(y=s\) and \(z=t\) be free. Then

\[x=4-2s+t.\]

The complete solution is

\[\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}4\\0\\0\end{bmatrix} +s\begin{bmatrix}-2\\1\\0\end{bmatrix} +t\begin{bmatrix}1\\0\\1\end{bmatrix}.\]

The first vector is one particular solution. The other vectors describe allowed directions of change that keep the equations true.

What a pivot records. A pivot is not merely a leading number. It marks a variable direction that the equations constrain independently. A row of zeros contributes no new constraint; a row with only a non-zero right-hand side records an impossibility. Rank counts the independent constraint directions that remain.

  1. Move a non-zero candidate into the next pivot position.
  2. Use reversible row operations to clear entries below it.
  3. Compare the pivots in the coefficient and augmented columns before declaring a solution type.

4.5 Matrix-vector form

For

\[\begin{aligned} a_{11}x_1+a_{12}x_2&=b_1,\\ a_{21}x_1+a_{22}x_2&=b_2, \end{aligned}\]

we write

\[\underbrace{\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}_{A} \underbrace{\begin{bmatrix}x_1\\x_2\end{bmatrix}}_{\mathbf x} =\underbrace{\begin{bmatrix}b_1\\b_2\end{bmatrix}}_{\mathbf b}.\]

Thus \(A\mathbf x=\mathbf b\). Matrix multiplication means that \(A\mathbf x\) is a linear combination of the columns of \(A\). Solving the system asks whether \(\mathbf b\) can be made from those columns, and if so, in how many ways.

Columns describe available directions. In A x=b, multiplying by the unknown vector forms a weighted combination of the columns of A. A solution exists precisely when the target vector can be assembled from those columns; multiple solutions occur when different weights assemble the same target.

\[A\mathbf{x}=\mathbf{b}\]

Common Mistakes

  • Swapping or scaling coefficients without doing the same operation to the right-hand side.
  • Dividing by a pivot before checking that it is nonzero; swap rows first when possible.
  • Saying “rank \(<n\) means infinite solutions” without first checking consistency.
  • Calling a parameter an unknown to be solved. A free parameter labels a family of valid solutions.

4.6 Elimination is reversible bookkeeping

Replacing equation 2 by equation 2 minus three times equation 1 preserves every solution: any solution of the old equations satisfies the new equation, and adding three times equation 1 back reverses the operation. Scaling is reversible only by a nonzero number, which is why multiplying a row by zero is not allowed.

Example — complete elimination with a row swap

Solve

\[\begin{aligned} y+z&=3,\\ x-y+z&=1,\\ 2x+y-z&=4. \end{aligned}\]

Use the second equation as pivot row:

\[\left[\begin{array}{ccc|c} 1&-1&1&1\\ 0&1&1&3\\ 2&1&-1&4 \end{array}\right] \xrightarrow{R_3-2R_1} \left[\begin{array}{ccc|c} 1&-1&1&1\\ 0&1&1&3\\ 0&3&-3&2 \end{array}\right].\]

Then \(R_3\leftarrow R_3-3R_2\) gives \(-6z=-7\), so \(z=7/6\), \(y=11/6\), and \(x=5/3\). Substitution into the original system is the final check, not optional decoration.

Example — rank distinguishes none from infinitely many

The forms

\[\left[\begin{array}{cc|c}1&0&2\\0&1&-1\end{array}\right],\quad \left[\begin{array}{cc|c}1&2&3\\0&0&0\end{array}\right],\quad \left[\begin{array}{cc|c}1&2&3\\0&0&1\end{array}\right]\]

mean, respectively, one solution, infinitely many solutions, and no solution. In the middle case rank is \(1<2\) and the system is consistent, so one variable is free. In the last case the augmented column creates \(0=1\).

Each row operation has an undo. From x+y=3 and 2x-y=0, replace the second row by the second minus twice the first, scale its non-zero pivot, then clear the first row. Back-substitution is visible in the resulting identity matrix: the same solution satisfies the starting equations because every rewrite was reversible.

\[\left[\begin{array}{cc|c}1&1&3\\2&-1&0\end{array}\right]\to\left[\begin{array}{cc|c}1&1&3\\0&-3&-6\end{array}\right]\to\left[\begin{array}{cc|c}1&0&1\\0&1&2\end{array}\right]\]
  1. R two becomes R two minus two R one, giving negative three y equals negative six.
  2. Scale R two by negative one third, giving y equals two.
  3. Replace R one by R one minus R two, giving x equals one, then verify both original equations.

Chapter Summary

Linear systems model simultaneous constraints. Elimination produces a triangular or echelon form from which solutions can be read. Pivots reveal rank, consistency, and the number of free variables. Matrix-vector notation makes the geometry and bookkeeping of larger systems manageable.

Exercises

  1. Solve by elimination: \(3x+2y=12\), \(x-y=1\).
  2. Reduce the augmented matrix \(\left[\begin{smallmatrix}1&2&-1&|&3\\2&4&1&|&8\end{smallmatrix}\right]\) and classify its solution set.
  3. Write a two-node circuit model using node voltages \(v_1,v_2\) if currents through resistors are specified by voltage differences divided by resistance.
  4. A polynomial \(p(x)=a+bx+cx^2\) passes through \((0,1)\), \((1,2)\), and \((2,5)\). Form and solve the linear system for \(a,b,c\).
  5. Solve the system in Section 4.6 by augmented-matrix operations.
  6. Give an augmented matrix representing a \(2\times2\) system with no solution.
  7. A node at voltage \(v\) connects through \(1\ \mathrm{k}\Omega\) to \(5\) V and through \(2\ \mathrm{k}\Omega\) to ground. Form KCL and solve for \(v\).

Part II

Part II — Limits, Continuity, and the Calculus Foundation

Part II · Limits, Continuity, and the Calculus Foundation

05

Chapter 5 — Sequences and Infinite Processes

Open chapter view →

Why This Matters

Many engineering ideas are iterative: a sampled signal is a sequence, a numerical algorithm produces successive approximations, and a transient circuit approaches a steady state. Limits give a precise language for “approaches.”

5.1 Sequences

A sequence is a function whose inputs are positive integers. We write

\[(a_n)_{n=1}^{\infty}=a_1,a_2,a_3,\ldots\]

For example, \(a_n=1/n\) gives \(1,1/2,1/3,\ldots\). Unlike a continuous graph, visualise a sequence as isolated dots above the integer positions \(1,2,3,\ldots\).

A sequence may be explicit, such as \(a_n=3n-1\), or recursive, such as

\[a_{n+1}=\frac12\left(a_n+\frac2{a_n}\right),\qquad a_1>0.\]

The recursive rule above is Newton’s iteration for \(\sqrt2\).

A sequence is discrete data. Its input is the positive integer index, so a plotted sequence is a collection of dots rather than a continuous curve. The notation a_n names one term; it does not mean that every real horizontal position is an allowed input.

\[(a_n)_{n=1}^{\infty}=a_1,a_2,\ldots,\qquad a_n=\frac1n\]

5.2 Limits of sequences

We say \(a_n\) converges to \(L\), written \(a_n\to L\), if its terms can be made as close to \(L\) as desired by going sufficiently far along the sequence. Formally, for every \(\varepsilon>0\) there is an integer \(N\) such that

\[n\ge N\implies |a_n-L|<\varepsilon.\]

The value \(L\) need not appear as a sequence term. For \(a_n=1/n\), no term is zero, but terms eventually fit inside every band \((-\varepsilon,\varepsilon)\) around zero.

Example — An \(\varepsilon\)\(N\) proof

Prove \(1/n\to0\). Given \(\varepsilon>0\), choose an integer \(N>1/\varepsilon\). For \(n\ge N\),

\[\left|\frac1n-0\right|=\frac1n\le\frac1N<\varepsilon.\]

This proves convergence. The proof works because it tells us exactly how far we must go for any requested accuracy.

The index depends on the requested accuracy. For 1/n, choose an integer beyond 1/ε. Then every later denominator is still larger, so every later term lies in the same ε-band; this proves an eventual statement, not merely a pattern among early dots.

\[N>\frac1{\varepsilon},\ n\ge N\Longrightarrow\left|\frac1n-0\right|\le\frac1N<\varepsilon\]
  1. State a positive tolerance ε.
  2. Choose an integer N greater than 1/ε.
  3. Use n ≥ N to bound 1/n by 1/N, then by ε.

5.3 Boundedness and monotonicity

A sequence is bounded above by \(M\) if \(a_n\le M\) for all \(n\), and bounded below by \(m\) if \(a_n\ge m\) for all \(n\). It is non-decreasing if \(a_{n+1}\ge a_n\) and non-increasing if \(a_{n+1}\le a_n\).

The theorem guarantees existence; it does not by itself find the value of the limit. For a recursive sequence, first prove the appropriate bounds and monotonicity, then take limits on both sides of the recursion.

Bounded monotonic motion has somewhere to settle. A non-decreasing real sequence with an upper bound converges, and the reversed statement holds for a non-increasing sequence with a lower bound. This theorem guarantees existence only; a separate limit equation or estimate is needed to identify the value.

\[a_{n+1}\ge a_n,\ a_n\le M\Longrightarrow(a_n)\text{ converges}\]
  1. Prove the proposed bound holds for every term.
  2. Compare consecutive terms to establish monotonicity.
  3. Use the theorem for existence before solving a fixed-point equation for the limit.

5.4 Series and partial sums

Given a sequence \((a_n)\), the expression

\[\sum_{n=1}^{\infty}a_n\]

is a series. Its meaning is the limit of partial sums

\[S_N=\sum_{n=1}^{N}a_n.\]

Do not confuse \(a_n\to0\) with convergence of the series. It is necessary for a series to converge, but it is not sufficient: the harmonic series \(\sum_{n=1}^{\infty}1/n\) diverges even though \(1/n\to0\).

A series is defined through a new sequence. The individual terms a_n and the partial sums S_N answer different questions. Term convergence to zero is necessary, but only convergence of the partial-sum sequence makes the infinite addition finite.

\[S_N=\sum_{n=1}^{N}a_n,\qquad a_n\to0\not\Longrightarrow S_N\text{ converges}\]

5.5 Sequence limit laws

If \(a_n\to A\) and \(b_n\to B\), then sums, products, and scalar multiples behave as expected:

\[a_n+b_n\to A+B,\qquad a_nb_n\to AB, \qquad \frac{a_n}{b_n}\to\frac AB\quad(B\ne0).\]

For rational expressions in \(n\), divide numerator and denominator by the highest power of \(n\) that occurs.

Some sequence limits are not rational expressions. A standard one is

\[\lim_{n\to\infty}a^{1/n}=1\qquad(a>0).\]

It says that repeatedly taking a root pulls every fixed positive number toward \(1\). One way to see the structure is to write \(a^{1/n}=e^{(\ln a)/n}\): the exponent tends to \(0\), and continuity of the exponential gives the result. This is not a substitution of infinity into an exponent; it is a composition of two ordinary limits.

Example — Dominant powers

\[\lim_{n\to\infty}\frac{5n^2-3n+1}{2n^2+7} =\lim_{n\to\infty}\frac{5-3/n+1/n^2}{2+7/n^2}=\frac52.\]

We did not substitute “\(\infty\)” for \(n\). Infinity is not an ordinary real number; the division exposes terms whose limits are known.

Limit laws apply to known component limits. Dividing a rational expression by its highest power exposes constants and reciprocal powers whose limits are controlled. It is not substitution of infinity, and a quotient law still requires a non-zero limiting denominator.

\[\frac{5n^2-3n+1}{2n^2+7}=\frac{5-3/n+1/n^2}{2+7/n^2}\longrightarrow\frac52\]
  1. Identify the highest power occurring in numerator and denominator.
  2. Divide every term by that power.
  3. Apply the sum and quotient laws only after checking the limiting denominator is non-zero.

Common Mistakes

  • Treating a sequence as a continuous curve rather than a discrete list.
  • Assuming a limit must equal some term.
  • Calling \(0/0\) an answer instead of recognising an indeterminate form.
  • Applying a quotient law when the limiting denominator is zero.

5.6 Tails, oscillation, and partial sums

Convergence concerns the tail of a sequence: changing finitely many early terms does not change its limit. Picture a horizontal \(\varepsilon\)-band around \(L\). Once every later dot lies in that band, the definition is met. Oscillation alone does not prevent convergence; what matters is whether the distance to one value shrinks.

Example — an oscillating sequence that converges

For \(a_n=(-1)^n/n\),

\[|a_n-0|=\frac1n.\]

Given \(\varepsilon>0\), choose \(N>1/\varepsilon\). Every \(n\ge N\) satisfies \(|a_n|<\varepsilon\), independent of its sign. Thus \(a_n\to0\).

Example — a fixed point needs an existence argument

Let \(a_{n+1}=(a_n+4)/2\) with \(a_1=0\). A possible limit satisfies \(L=(L+4)/2\), giving \(L=4\). To justify taking that limit, observe \(a_n\le4\) and

\[a_{n+1}-a_n=\frac{4-a_n}{2}\ge0.\]

The sequence is increasing and bounded above, so it converges; only then does the fixed-point calculation determine its limit.

Example — terms tend to zero but partial sums grow

Group the harmonic series:

\[1+\frac12+\left(\frac13+\frac14\right) +\left(\frac15+\cdots+\frac18\right)+\cdots.\]

Every group after the first is at least \(1/2\). The partial sums consequently exceed \(1+m/2\) after \(m\) such groups and are unbounded. Thus \(1/n\to0\) does not imply \(\sum1/n\) converges.

Oscillation is harmless when its size shrinks. The signs of (−1)^n/n alternate, but its distance from zero is exactly 1/n, so its tail enters every epsilon band. In contrast, harmonic partial sums grow because grouped positive terms keep adding a fixed minimum amount.

\[\left|\frac{(-1)^n}{n}\right|=\frac1n\to0,\qquad H_N=\sum_{n=1}^{N}\frac1n\to\infty\]

5.7 A sequence is not the series built from it

The sequence \(a_n\) is a list of individual contributions. The series \(\sum a_n\) is a new object whose values are the partial sums \(S_N=a_1+\cdots+a_N\). A convergent sequence can therefore produce either a convergent or divergent series. The term test only provides a one-way decision: if \(a_n\) fails to approach zero, the series diverges; if it does approach zero, another test is required.

Example — geometric partial sums approach a finite total

For \(a_n=(1/2)^{n-1}\), the terms approach \(0\). More importantly,

\[S_N=1+\frac12+\cdots+\left(\frac12\right)^{N-1} =2\left(1-\left(\frac12\right)^N\right).\]

Since \((1/2)^N\to0\), \(S_N\to2\). The series converges because the partial sums have a limit, not merely because individual contributions shrink.

Example — a necessary condition that fails immediately

For \(\sum_{n=1}^{\infty}(n+1)/(n+2)\),

\[\lim_{n\to\infty}\frac{n+1}{n+2}=1\ne0.\]

The terms do not even become small, so adding infinitely many of them cannot settle to a finite value. No ratio, comparison, or more elaborate test is needed.

Example — recursive convergence and an error estimate

If \(a_{n+1}=(a_n+4)/2\) has limit \(4\), subtract \(4\) from both sides:

\[a_{n+1}-4=\frac12(a_n-4).\]

Hence \(|a_{n+1}-4|=\tfrac12|a_n-4|\). Every iteration halves the current error. This connects convergence to a practical numerical stopping rule: after \(k\) further steps, the error is at most \(2^{-k}\) times the current error.

Derive the infinite result from a finite identity. For a geometric sum, subtracting half the sum cancels every middle term. Only after obtaining the formula for S_N may we take its limit; for ratio one half, the third partial sum is 1.75 and the limiting total is two.

\[S_N=1+\frac12+\cdots+\left(\frac12\right)^{N-1}=2\left(1-\left(\frac12\right)^N\right)\to2\]
  1. Write the finite sum S sub N.
  2. Subtract one half S sub N so the middle terms cancel.
  3. Take the limit only after the finite identity is established.

Exercises

  1. Decide whether \(a_n=(-1)^n\) converges. Explain geometrically.
  2. Prove that \(3/(2n+1)\to0\) using the definition.
  3. Find \(\lim_{n\to\infty}(7n^3-n)/(2n^3+5n)\).
  4. Find the first five terms of \(a_{n+1}=\frac12(a_n+3/a_n)\) with \(a_1=2\).
  5. Explain why \(a_n\to0\) is necessary but not sufficient for \(\sum a_n\) to converge.
  6. Prove \((-1)^n/n\to0\) using an \(\varepsilon\)\(N\) argument.
  7. Find a suitable \(N\) so \((3n-1)/(n+2)\) is within \(0.01\) of its limit.
  8. Establish convergence of \(a_{n+1}=(a_n+4)/2\), \(a_1=0\), before finding its limit.
  9. For \(a_n=(1/2)^{n-1}\), write \(S_N\) and show \(\lim S_N=2\).
  10. Use the term test to decide whether \(\sum(n+1)/(n+2)\) can converge.

Part II · Limits, Continuity, and the Calculus Foundation

06

Chapter 6 — Powers, Exponentials, Limits, and Continuity

Open chapter view →

Why This Matters

Calculus relies on understanding what happens near a point, not merely at it. Limits formalise this nearby behaviour. Continuity tells us when small input changes produce small output changes. Exponentials provide the model for growth, decay, and later circuit transients.

6.1 Real powers and exponential functions

For integer exponents, \(a^n\) means repeated multiplication. For arbitrary real exponents, the clean real-valued theory takes \(a>0\). Rational powers use roots, for example \(a^{m/n}=\sqrt[n]{a^m}\) when this is real and defined. Irrational exponents are defined consistently through limits of rational approximations.

The functions \(x^a\) and \(a^x\) are different. In \(x^a\), the base varies; in \(a^x\), the exponent varies. For \(a>1\), \(a^x\) passes through \((0,1)\), increases strictly, tends to \(0\) as \(x\to-\infty\), and grows without bound as \(x\to\infty\). For \(0<a<1\), the direction is reversed.

Real powers retain domain conditions. The usual real-valued definition of a^x for arbitrary real exponents begins with a>0. It also distinguishes the variable-base function x^a from the variable-exponent function a^x; their domains and long-run behaviours need not match.

\[a>0,\qquad a^x=e^{x\ln a}\]

6.2 Limits of functions

We write

\[\lim_{x\to c}f(x)=L\]

when \(f(x)\) becomes arbitrarily close to \(L\) as \(x\) becomes sufficiently close to \(c\), without requiring \(x=c\). The value \(f(c)\) may be different from \(L\) or may not even be defined.

The left and right limits must agree for a two-sided limit to exist:

\[\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L.\]

Example — Removable discontinuity

For \(x\ne0\),

\[\frac{2^x-1}{x}\]

is undefined at \(0\). Its limit as \(x\to0\) is \(\ln2\), a fact derived later from the derivative of \(2^x\). Assigning the value \(\ln2\) at \(x=0\) creates a continuous extension. The original expression is not already defined there.

A limit ignores the centre while it is approached. Nearby inputs determine the limiting value, so the function value at the centre may be different or absent. A two-sided limit requires the left and right approaches to agree on one value.

\[\lim_{x\to c}f(x)=L\Longleftrightarrow\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L\]

6.3 Continuity

A function is continuous at \(c\) when all three conditions hold:

  1. \(f(c)\) is defined.
  2. \(\lim_{x\to c}f(x)\) exists.
  3. The limit equals \(f(c)\).

Polynomials are continuous on all real numbers. Rational functions are continuous wherever their denominators are nonzero. Root functions require their real-domain restrictions. These facts are useful because they allow direct substitution only after the domain has been checked.

Continuity requires three linked facts. The point value must exist, the nearby limit must exist, and those two values must agree. Direct substitution is therefore justified for a rational expression only after its denominator restriction has been checked.

\[f(c)\text{ defined},\quad\lim_{x\to c}f(x)\text{ exists},\quad\lim_{x\to c}f(x)=f(c)\]
  1. Check that the input is in the domain.
  2. Find the nearby two-sided limit.
  3. Compare it with the actual point value before declaring continuity.

6.4 One-sided limits and discontinuities

A switch can change a circuit model abruptly. At the switching instant, a piecewise function may have a left limit and a different right limit. This is a jump discontinuity. A rational function may have an infinite discontinuity or vertical asymptote where its denominator approaches zero. A missing point with a finite surrounding limit is a removable discontinuity.

A point assignment cannot repair unequal approaches. A jump has distinct left and right limits, whereas a removable discontinuity has one finite surrounding limit and a missing or mismatched point. A vertical asymptote records unbounded nearby behaviour rather than an ordinary finite value.

\[\lim_{x\to0^-}f(x)=1\ne3=\lim_{x\to0^+}f(x)\]

6.5 Limits at infinity

Limits at infinity describe long-run behaviour. For example,

\[\lim_{x\to\infty}\frac{3x^2+1}{x^2-4}=3.\]

The leading powers control the result. Exponentials eventually dominate any fixed polynomial power: for every fixed nonnegative integer \(m\),

\[\lim_{x\to\infty}x^m e^{-x}=0.\]

This comparison later explains why exponential transients decay despite polynomial factors that may appear in intermediate calculations.

Divide by the dominant scale. In a rational function, the largest power controls the ratio at large magnitude. Equal degrees give the ratio of leading coefficients; a smaller numerator degree gives zero; a larger numerator degree has no finite horizontal asymptote.

\[\frac{3x^2+1}{x^2-4}=\frac{3+1/x^2}{1-4/x^2}\longrightarrow3\]

Problem-Solving Strategy

Before evaluating a limit, check the domain and try direct substitution. If it produces an indeterminate form such as \(0/0\), simplify algebraically by factorisation, rationalisation, or dominant-term division. Use one-sided limits for piecewise definitions and denominator sign changes. Do not perform algebra on the symbol \(\infty\) as though it were a real number.

6.6 Limits describe nearby behaviour, not a point calculation

Direct substitution is a useful first test, not the definition of a limit. A \(0/0\) result says numerator and denominator both become small; it does not say the quotient is zero. Factorisation or rationalisation can expose a simpler nearby expression, provided the excluded input is still recorded.

Example — factor, cancel, then take the limit

\[\lim_{x\to2}\frac{x^2-4}{x-2} =\lim_{x\to2}\frac{(x-2)(x+2)}{x-2} =\lim_{x\to2}(x+2)=4.\]

Cancellation is valid only for \(x\ne2\), which is exactly the punctured neighbourhood used by the limit. It does not make the original function defined at \(2\).

Example — one-sided limits determine a jump

Let \(f(x)=x+1\) for \(x<0\) and \(f(x)=x^2\) for \(x\ge0\). Then the left limit at \(0\) is \(1\) and the right limit is \(0\). The two-sided limit does not exist even though \(f(0)\) is defined; one point value cannot repair unequal approaches.

Example — limits at infinity use dominant terms

\[\lim_{x\to\infty}\frac{5x^3-2x}{-x^3+7} =\lim_{x\to\infty}\frac{5-2/x^2}{-1+7/x^3}=-5.\]

Dividing by the highest power exposes the horizontal asymptote. If numerator degree is smaller, the limit is \(0\); if larger, no finite horizontal asymptote results.

Cancellation simplifies nearby values, not the original domain. Factoring x²−4 exposes x+2 for x≠2, so the limit is four. The exclusion remains part of the original quotient; the hole is not filled by algebra alone.

\[\frac{x^2-4}{x-2}=x+2\ (x\ne2),\qquad\lim_{x\to2}\frac{x^2-4}{x-2}=4\]
  1. Record the input excluded by the original denominator.
  2. Factor and simplify only on the punctured neighbourhood.
  3. Evaluate the simplified nearby expression for the limit without assigning the original function a value there.

Exercises

  1. State the domain of \(x^{2/3}\) and of \(x^{\sqrt2}\) in the usual real-valued convention.
  2. Evaluate \(\lim_{x\to3}(x^2-9)/(x-3)\) and state why substitution initially fails.
  3. Determine the one-sided limits of \(f(x)=|x|/x\) at \(0\).
  4. Find the horizontal asymptote of \((4x^3-x)/(2x^3+7)\).
  5. Sketch a piecewise function with a removable discontinuity at \(x=1\) and a jump discontinuity at \(x=3\).
  6. Evaluate \(\lim_{x\to3}(\sqrt{x+1}-2)/(x-3)\) by rationalising.
  7. Give a piecewise rule continuous at \(0\) but not differentiable there.
  8. Find horizontal asymptotes of \((2x+1)/(x^2+4)\) and \((x^3+1)/(2x^2-1)\).

Part III

Part III — Differential Calculus and Local Approximation

Part III · Differential Calculus and Local Approximation

07

Chapter 7 — Derivatives from First Principles

Open chapter view →

Why This Matters

An average speed over ten seconds does not tell us the speed at one instant. In the same way, the ratio \(\Delta V/\Delta t\) describes an average voltage change over an interval, while a circuit or control problem often needs the instantaneous rate of change. The derivative is the limiting version of that average rate.

7.1 Average rate of change and secant slope

For a function \(y=f(x)\), the average change in output between \(x\) and \(x+h\) is

\[\frac{f(x+h)-f(x)}{h},\qquad h\ne0.\]

On a graph, the two points \((x,f(x))\) and \((x+h,f(x+h))\) determine a secant line. Its slope is precisely the displayed quotient. In a physical model where \(x=t\) is time, it is average rate of change over a time interval of length \(h\).

As \(h\) becomes smaller, the second point moves towards the first. If the secant slopes approach one definite number, that number is the slope of the tangent line and the instantaneous rate of change.

Average rate is a line through two measured points. Moving from x to x+h changes both input and output, so the quotient is the secant slope. In a time model it is a rate over an interval, not yet an instantaneous rate.

\[\frac{f(x+h)-f(x)}h,\qquad h\ne0\]

7.2 Definition of derivative

The derivative of \(f\) at \(x\) is

\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h},\]

provided the limit exists. It is a new function when it can be computed at many inputs. Other notations include \(\frac{df}{dx}\) and, when \(y=f(x)\), \(\frac{dy}{dx}\).

Example — Derivative of a square from the definition

Let \(f(x)=x^2\). Start with the difference quotient:

\[\frac{f(x+h)-f(x)}h=\frac{(x+h)^2-x^2}{h}.\]

Expand the square before cancelling:

\[\frac{x^2+2xh+h^2-x^2}{h}=\frac{h(2x+h)}h=2x+h.\]

The cancellation is valid because the quotient is considered for \(h\ne0\). Now take the limit:

\[f'(x)=\lim_{h\to0}(2x+h)=2x.\]

At \(x=3\), the tangent slope is \(6\). This result is not guessed from the graph; the graph is an interpretation of the limiting calculation.

Expand before taking the limit. For a square, the non-zero offset permits cancellation only after expansion produces a factor of h. The simplified secant slope 2x+h then has a limit of 2x, which is the tangent slope.

\[\frac{(x+h)^2-x^2}{h}=2x+h\longrightarrow2x\]
  1. Keep h non-zero while forming the difference quotient.
  2. Expand and factor h before cancelling it.
  3. Only then let h tend to zero to obtain the derivative.

7.3 When a derivative does not exist

Differentiability is stronger than continuity. A differentiable function is continuous, but a continuous function need not be differentiable.

For \(f(x)=|x|\), the left-hand slopes at \(0\) are \(-1\) and the right-hand slopes are \(1\). Because these do not agree, \(f'(0)\) does not exist. A jump discontinuity also prevents differentiability. A graph may additionally have a vertical tangent, where the ordinary finite derivative fails to exist.

Continuity does not settle the slope question. At the corner of |x|, left slopes are negative one and right slopes are positive one, so no single derivative exists. Jumps and vertical tangents likewise fail to supply an ordinary finite tangent slope.

\[\lim_{h\to0^-}\frac{|h|}{h}=-1\ne1=\lim_{h\to0^+}\frac{|h|}{h}\]

7.4 Linearity rules

The derivative respects addition and constant scaling:

\[\frac d{dx}[cf(x)]=cf'(x), \qquad \frac d{dx}[f(x)+g(x)]=f'(x)+g'(x).\]

These follow from the corresponding limit laws. The derivative of a constant is \(0\), because its output does not change, and \(\frac d{dx}x=1\).

Local increments add and scale before their limit is taken. The difference quotient of a sum splits into two quotients, and a constant factor stays attached to its changing output. Constants differentiate to zero because their local increment is zero.

\[(cf+g)'=cf'+g',\qquad\frac d{dx}C=0\]

7.5 Product and quotient rules

For a product, both factors may change. Add and subtract the intermediate term \(f(x+h)g(x)\):

\[\begin{aligned} &f(x+h)g(x+h)-f(x)g(x)\\ ={}&f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)]. \end{aligned}\]

Divide by \(h\) and take the limit. Continuity of differentiable functions lets \(f(x+h)\to f(x)\), yielding

\[(fg)'=f'g+fg'.\]

For \(g(x)\ne0\), write \(f/g=f\,g^{-1}\) and differentiate, or derive directly:

\[\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}.\]

Example — Why the product rule needs two terms

For \(P(t)=v(t)i(t)\), electrical power changes because voltage can change, current can change, or both can change:

\[\frac{dP}{dt}=v\frac{di}{dt}+i\frac{dv}{dt}.\]

Keeping only one term would assume, without stating it, that the other factor is constant.

Two changing factors create two contributions. Add and subtract the intermediate product f(x+h)g(x) to split the total change into a change of g and a change of f. The quotient rule inherits the additional condition that the denominator remains non-zero.

\[(fg)'=f'g+fg',\qquad\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}\ (g\ne0)\]
  1. Insert and subtract f of x plus h times g of x.
  2. Group the two resulting local increments and divide by h.
  3. Take the limit while retaining the quotient domain restriction where needed.

7.6 Chain rule

Suppose \(y=f(u)\) and \(u=g(x)\). A small change in \(x\) first changes \(u\), which then changes \(y\). The derivative is

\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(g(x))g'(x).\]

This is the chain rule. It is a statement about composition, not a trick about parentheses.

Example — A nested power

Differentiate \(y=(3x^2-1)^5\). The outside function is \(u^5\) and the inside is \(u=3x^2-1\). Therefore

\[\frac{dy}{dx}=5(3x^2-1)^4\cdot6x=30x(3x^2-1)^4.\]

A local change travels through each layer. A small change in x first changes the inner quantity and then changes the outer output. The chain rule multiplies those local rates in their physical order; it is not a rule for merely removing parentheses.

\[\frac{dy}{dx}=f'(g(x))g'(x)\]
  1. Name the inner quantity u equals g of x.
  2. Differentiate the outer rule with respect to u.
  3. Multiply by du over dx to return the rate to x.

7.7 Power, polynomial, and rational functions

For a real exponent \(a\), the power rule

\[\frac d{dx}x^a=ax^{a-1}\]

is used on intervals where the function is real and differentiable. For arbitrary real \(a\), the standard safe domain is \(x>0\). Integer powers have their usual larger domains; selected rational powers may have additional real inputs.

Term-by-term differentiation gives, for example,

\[\frac d{dx}(4x^3-5x^2+7x-9)=12x^2-10x+7.\]

For rational functions, use the quotient rule or first simplify carefully while retaining all domain restrictions.

Rules inherit the original function's domain. The power rule applies where the real-valued power is defined and differentiable; arbitrary real exponents are safely handled on x>0. Simplifying a rational expression never grants a derivative at an input excluded by its original denominator.

\[\frac d{dx}x^a=ax^{a-1}\quad\text{on a real differentiable domain}\]

7.8 Higher derivatives

The second derivative is the derivative of the derivative:

\[f''(x)=\frac{d}{dx}f'(x).\]

In motion it measures acceleration; in a graph it records how the slope changes. For a voltage signal, \(v'(t)\) and \(v''(t)\) describe increasingly rapid aspects of variation. Higher derivatives later provide Taylor approximations and determine the form of second-order differential equations.

The next derivative tracks change in the current rate. If f′ is slope or instantaneous rate, then f″ measures how that slope itself changes. This connects acceleration, curvature, and the higher local information used later in Taylor approximation.

\[f''(x)=\frac d{dx}f'(x)\]

Common Mistakes

  • Cancelling \(h\) before expanding a difference quotient or forgetting that it is nonzero before the limit is taken.
  • Using the product rule as \((fg)'=f'g'\).
  • Forgetting the derivative of the inside function in a chain-rule calculation.
  • Applying \(x^a\) formulas at points outside the real domain of the original function.
  • Inferring differentiability solely from a graph that has not been examined at the relevant scale.

Chapter Summary

The derivative begins as average change over an interval and becomes instantaneous change through a limit. Difference quotients justify the derivative definition. Linearity, product, quotient, chain, and power rules make derivatives practical, but their domain conditions remain part of every calculation.

7.9 Derivative rules as controlled local changes

The product rule has two terms because both factors can change. The chain rule has an inside derivative because the outer function receives a changing input. Before differentiating, identify the outermost operation and the independent variable; this prevents applying rules mechanically.

Example — quotient and domain together

For \(f(x)=(x^2+1)/(x-1)\), \(x\ne1\). The quotient rule gives

\[f'(x)=\frac{(2x)(x-1)-(x^2+1)}{(x-1)^2} =\frac{x^2-2x-1}{(x-1)^2}.\]

The derivative cannot be used to claim anything about \(x=1\), where the original function is not defined.

Example — higher derivatives describe changing slope

For \(s(t)=t^3-6t^2+9t\),

\[s'(t)=3t^2-12t+9,\qquad s''(t)=6t-12.\]

The first derivative gives instantaneous rate; the second says whether that rate is increasing or decreasing. At \(t=2\), \(s''(2)=0\), so the slope changes its trend there. Later, this same information supports curvature and Taylor approximations.

Name the changing pieces before selecting a rule. A product has two independently changing factors, a composition has an inner scale, and a quotient retains an excluded denominator. Higher derivatives then give a second layer of rate information rather than a new unrelated calculation.

\[s(t)=t^3-6t^2+9t\Longrightarrow s'(t)=3t^2-12t+9,\ s''(t)=6t-12\]

Exercises

  1. Use the limit definition to find the derivative of \(f(x)=3x-2\).
  2. Explain why \(|x|\) is continuous but not differentiable at \(0\).
  3. Differentiate \((x^2+1)(x^3-4)\).
  4. Differentiate \(\sqrt{1+4x^2}\) on its real domain.
  5. If \(q(t)=C v(t)\) for a capacitor with constant capacitance \(C\), derive a formula for \(dq/dt\).
  6. Differentiate \((2x-1)^4/(x+1)\) and state its domain.
  7. A signal is \(v(t)=t^3-6t^2+9t\). Find \(v'\) and \(v''\), then locate points where its slope is zero.

Part III · Differential Calculus and Local Approximation

08

Chapter 8 — Derivative Theorems and Function Behaviour

Open chapter view →

Why This Matters

The derivative becomes most useful when it reveals global behaviour: whether a function rises or falls, where a design reaches a maximum, and how a difference between two values is controlled by a slope in between. These conclusions require theorems with hypotheses, not pictures alone.

8.1 Rolle’s theorem and the Mean Value Theorem

If a smooth graph begins and ends at the same height, it must have a horizontal tangent somewhere in between. The endpoint and differentiability assumptions are essential; a sharp corner can defeat the conclusion.

The average slope of a secant line is therefore achieved by a tangent somewhere inside the interval. The MVT is the rigorous bridge from local derivative signs to interval-wide trends.

The conclusion comes after the hypotheses. Continuity on the closed interval and differentiability on its interior let the Mean Value Theorem guarantee an interior tangent parallel to the endpoint secant. Without the whole-interval conditions, the parallel tangent is not promised.

\[f'(c)=\frac{f(b)-f(a)}{b-a},\qquad c\in(a,b)\]

8.2 Monotonicity

If \(f'(x)>0\) throughout an interval, then \(f\) is strictly increasing there; if \(f'(x)<0\), it is strictly decreasing. By the MVT,

\[f(y)-f(x)=f'(c)(y-x)\]

for some point \(c\) between \(x\) and \(y\). When \(y>x\), the sign of the difference is therefore the sign of \(f'(c)\).

Example — Sign chart

Let \(f(x)=x^3-3x\). Then

\[f'(x)=3x^2-3=3(x-1)(x+1).\]

The derivative is positive for \(x<-1\), negative for \(-1<x<1\), and positive for \(x>1\). Thus \(f\) rises, falls, then rises. The derivative zeros identify changes to investigate; their signs explain the behaviour.

Derivative sign controls an interval through MVT. For y>x, the difference f(y)−f(x) has the sign of f′(c) at a point between them. Thus a positive derivative throughout gives increase and a negative derivative throughout gives decrease.

\[f(y)-f(x)=f'(c)(y-x)\]

8.3 Extrema and optimisation

An interior local maximum or minimum of a differentiable function must occur at a point where \(f'(c)=0\). Such points are critical points, but they are only candidates. A local extremum can also occur where the derivative does not exist, and an absolute extremum on a closed interval may occur at an endpoint.

Engineering Example — maximum power transfer

Let a source with internal resistance \(R_s>0\) drive a load \(R>0\). Up to a positive constant, load power is

\[P(R)=\frac{R}{(R_s+R)^2}.\]

Differentiate:

\[P'(R)=\frac{(R_s+R)^2-2R(R_s+R)}{(R_s+R)^4} =\frac{R_s-R}{(R_s+R)^3}.\]

The denominator is positive. Thus \(P'(R)>0\) for \(R<R_s\) and \(P'(R)<0\) for \(R>R_s\). Power rises then falls, so the maximum occurs at

\[R=R_s.\]

The conclusion comes from the sign change, not merely from setting a derivative to zero.

Critical points are candidates, not verdicts. On a closed interval, compare every valid interior critical point with both endpoints. A sign change from positive to negative identifies a local maximum, while the reverse identifies a local minimum; a zero derivative alone identifies neither.

\[P'(R)=\frac{R_s-R}{(R_s+R)^3},\qquad R=R_s\text{ gives the maximum}\]
  1. List derivative-zero points and points where the derivative fails to exist.
  2. Add both endpoints for a closed-interval absolute-extrema problem.
  3. Compare function values or use a sign change before making the classification.

8.4 L’Hôpital’s rule

The rule is not a licence to differentiate any expression in a limit. First verify the quotient form and its hypotheses. Products, differences, and powers must be transformed before the rule can be considered.

Example — repeated use with conditions checked

\[\lim_{x\to0}\frac{e^x-1-x}{x^2}.\]

Direct substitution gives \(0/0\). Differentiate numerator and denominator:

\[\lim_{x\to0}\frac{e^x-1}{2x},\]

which is again \(0/0\). Differentiate once more:

\[\lim_{x\to0}\frac{e^x}{2}=\frac12.\]

Taylor expansion in Chapter 9 will give the same result and explain the local shape behind it.

Differentiate a quotient only after its form and hypotheses qualify. L'Hôpital's rule begins with a 0/0 or ∞/∞ quotient, differentiability on a punctured interval, non-zero derivative denominator there, and a derivative-quotient limit. A product such as x sin(1/x) is not initially an eligible quotient.

\[\frac{e^x-1-x}{x^2}\xrightarrow{0/0}\frac{e^x-1}{2x}\xrightarrow{0/0}\frac{e^x}{2}\to\frac12\]
  1. Check that the expression is an eligible quotient form.
  2. Verify the punctured-neighbourhood derivative conditions.
  3. Differentiate numerator and denominator together, then recheck the form before repeating.

Common Mistakes

  • Calling every derivative-zero point a maximum or minimum.
  • Forgetting interval endpoints in an absolute-extrema problem.
  • Applying MVT without checking continuity on the closed interval and differentiability on the open interval.
  • Applying L’Hôpital’s rule to a non-indeterminate quotient.

8.5 From derivative signs to reliable conclusions

Critical points are candidates, not automatic extrema. Make a sign chart by splitting the domain at derivative zeros and points where the derivative fails to exist; then test the sign on each interval. For absolute extrema on a closed interval, compare function values at every valid critical point and both endpoints.

Example — complete closed-interval extremum check

Find absolute extrema of \(f(x)=x^3-3x\) on \([-2,2]\). Since

\[f'(x)=3(x^2-1),\]

the interior candidates are \(x=-1,1\). Evaluate all required points:

\[f(-2)=-2,\quad f(-1)=2,\quad f(1)=-2,\quad f(2)=2.\]

Thus both \(2\) and \(-2\) occur as absolute extrema. Omitting endpoints would lose valid answers even though the derivative calculation itself is correct.

Example — when L'Hôpital may not start

For \(\lim_{x\to0}x\sin(1/x)\), direct substitution is not a quotient \(0/0\) or \(\infty/\infty\), so L'Hôpital's rule is not the first method. Instead,

\[|x\sin(1/x)|\le|x|\to0,\]

so the squeeze theorem gives limit \(0\). Method selection begins with the form, not with a preferred rule.

A sign chart becomes reliable when its domain is complete. Split at derivative zeros and derivative failures, then include both endpoints before comparing absolute values. This separates an interval-wide conclusion from the tempting but insufficient step of solving only f′=0.

\[f(-2)=-2,\ f(-1)=2,\ f(1)=-2,\ f(2)=2\]

8.6 Theorems give conclusions only under their hypotheses

The Mean Value Theorem is a guarantee about a smooth unbroken graph on a whole interval. It does not say every secant slope occurs as a tangent slope if the graph jumps, has a corner, or is undefined at an endpoint. Derivative signs use the same hypothesis through the MVT: a positive derivative on an interval gives increase only when the function is differentiable throughout its interior.

Example — MVT applied with every condition visible

For \(f(x)=x^2\) on \([1,4]\), polynomials are continuous and differentiable everywhere. The secant slope is

\[\frac{f(4)-f(1)}{4-1}=\frac{16-1}{3}=5.\]

Set \(f'(c)=2c=5\), giving \(c=5/2\), which lies in \((1,4)\) as required. The theorem does not identify a unique point in general; it guarantees at least one.

Example — a hypothesis failure matters

Let \(f(x)=|x|\) on \([-1,1]\). Its endpoint values agree, and it is continuous, but it is not differentiable at \(0\). Rolle's theorem cannot be applied. Indeed, \(f'(x)\) is \(-1\) or \(1\) where defined and is never \(0\). This is a counterexample to the incorrect claim that equal endpoint values alone force a horizontal tangent.

Example — L'Hôpital after form, not before

Evaluate \(\lim_{x\to0}(1-\cos x)/x^2\). It has the allowed \(0/0\) quotient form. One application gives \(\sin x/(2x)\), still \(0/0\); a second gives \(\cos x/2\to1/2\). By contrast, \(\lim_{x\to0}(1-\cos x)/x\) may use the same rule once, but an expression such as \(x\ln x\) is not initially a quotient and must be rewritten or handled by another argument.

Equal endpoint values alone do not give Rolle's conclusion. The graph of |x| on [−1,1] is continuous but has a corner at zero, so differentiability fails and no horizontal tangent is guaranteed. The theorem's conditions are working assumptions, not formal decoration.

\[f(-1)=f(1),\qquad f'(x)=-1\text{ or }1\text{ for }x\ne0\]

Exercises

  1. Verify the hypotheses of the MVT for \(f(x)=x^2\) on \([1,4]\) and find the point(s) \(c\) it guarantees.
  2. Find and classify the critical points of \(x^3-6x^2+9x\).
  3. Optimise the area of a rectangle with perimeter \(40\) m.
  4. Evaluate \(\lim_{x\to0}(\sin x-x)/x^3\) using L’Hôpital’s rule.
  5. Explain why the result of Exercise 4 cannot be obtained by simply replacing \(\sin x\) with \(x\) before examining the error.
  6. Find absolute extrema of \(x^3-3x\) on \([-2,2]\).
  7. Evaluate \(\lim_{x\to0}x\sin(1/x)\) without L'Hôpital's rule.
  8. Apply MVT to \(x^2\) on \([1,4]\) and identify every hypothesis used.
  9. Explain precisely why Rolle's theorem cannot be used for \(|x|\) on \([-1,1]\).
  10. Evaluate \((1-\cos x)/x^2\) as \(x\to0\) using L'Hôpital's rule and state why it is permitted.

Part III · Differential Calculus and Local Approximation

09

Chapter 9 — Inverse Functions, Differentials, and Taylor Approximation

Open chapter view →

Why This Matters

Engineers repeatedly reverse relationships: infer temperature from a sensor voltage, infer resistance from current and voltage, or infer a phase from a sinusoidal measurement. Local linear approximations then tell us how errors travel through those inversions and how a complicated function can be approximated near an operating point.

9.1 Inverse functions

A function \(f:D\to R\) has an inverse on its range when it is one-to-one: no two inputs in \(D\) have the same output. Its inverse reverses the mapping:

\[f^{-1}(f(x))=x\quad(x\in D),\qquad f(f^{-1}(y))=y\quad(y\in f(D)).\]

The notation \(f^{-1}(x)\) means inverse function, not reciprocal \(1/f(x)\). A horizontal-line test expresses one-to-one behaviour graphically. The graph of an inverse is the reflection of the original graph in the line \(y=x\).

For \(f(x)=x^2\), all-real input is not one-to-one because \(f(2)=f(-2)\). Restricting to \(x\ge0\) gives an inverse \(f^{-1}(y)=\sqrt y\) for \(y\ge0\).

An inverse reverses a one-to-one assignment. Its graph reflects across y=x, exchanging inputs and outputs. The notation f⁻¹ is not a reciprocal, and a domain restriction may be necessary before a familiar curve such as becomes invertible.

\[f^{-1}(f(x))=x,\qquad (x,y)\mapsto(y,x)\]

9.2 Derivative of an inverse

Suppose \(y=f^{-1}(x)\), so \(f(y)=x\). Differentiate with respect to \(x\):

\[f'(y)\frac{dy}{dx}=1.\]

Therefore

\[(f^{-1})'(x)=\frac1{f'(f^{-1}(x))},\]

provided the required derivatives exist and the denominator is nonzero. An inverse may exist at a point where its derivative is not finite; the nonzero condition must not be omitted.

Reflection turns a slope into its reciprocal. From f(y)=x, implicit differentiation gives f′(y)dy/dx=1. The original derivative must be non-zero: otherwise the reciprocal slope is not a finite derivative of the inverse.

\[(f^{-1})'(x)=\frac1{f'(f^{-1}(x))},\qquad f'(f^{-1}(x))\ne0\]

9.3 Linearisation and differentials

Near \(x=a\), a differentiable function is approximated by its tangent line:

\[f(x)\approx f(a)+f'(a)(x-a).\]

Writing \(dx\) for a small input change and \(dy\) for the corresponding linearised output change gives

\[dy=f'(a)\,dx.\]

The differential \(dy\) is an approximation to the actual finite change \(\Delta y=f(a+\Delta x)-f(a)\). The approximation improves as \(\Delta x\) becomes smaller, assuming the local curvature is controlled.

Engineering Example — propagated measurement error

The volume of a sphere is \(V=\frac43\pi r^3\). Solving for radius gives

\[r(V)=\left(\frac{3V}{4\pi}\right)^{1/3}.\]

For a small volume error \(dV\), the radius error is approximately

\[dr=r'(V)\,dV.\]

The derivative depends on the operating volume, so equal absolute errors in volume do not produce equal errors in radius. This is why relative error and sensitivity matter in measurement design.

The tangent is the first local model. Near the operating input a, the tangent line supplies the linear approximation and its change dy. It approximates the finite change Δy; controlled curvature and a smaller input increment make the two closer, not automatically identical.

\[f(x)\approx f(a)+f'(a)(x-a),\qquad dy=f'(a)\,dx\]

9.4 Taylor polynomials

The best polynomial approximation near \(x=a\) matches the value and successive derivatives of \(f\) at \(a\):

\[T_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+ \frac{f^{(n)}(a)}{n!}(x-a)^n.\]

\(T_1\) is the tangent-line approximation. Higher terms capture curvature and successive changes of curvature.

The theorem gives both an approximation and an error mechanism. A Taylor series is not automatically equal to the function merely because its terms can be written down; equality requires a remainder that tends to zero.

Example — local approximation of a reciprocal

Approximate \(1/1.1\) around \(a=1\) for \(f(x)=1/x\). We have

\[f(1)=1,\quad f'(1)=-1,\quad f''(1)=2.\]

With \(x-1=0.1\),

\[T_2(1.1)=1-0.1+\frac{2}{2}(0.1)^2=0.91.\]

The exact value is approximately \(0.90909\), so the quadratic approximation is already close. Its accuracy is local: using the same polynomial at \(x=5\) would not be sensible.

Each coefficient is forced by one derivative match. Matching value fixes the constant term, matching slope fixes the linear coefficient, and matching curvature fixes the quadratic coefficient after factorial scaling. Taylor's remainder describes why this local agreement does not promise accuracy far from the centre.

\[T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k\]
  1. Choose the expansion centre a.
  2. Match the value and successive derivatives at a.
  3. Use a remainder estimate or local scale before trusting the approximation away from a.

9.5 Inverse slopes, differentials, and approximation error

At corresponding points on inverse graphs, slopes are reciprocals: a steep forward graph has a shallow inverse graph. This is why the denominator in the inverse-derivative formula must be nonzero.

Example — inverse derivative at a measured output

Let \(f(x)=x^3+x\). Since \(f'(x)=3x^2+1>0\), it is one-to-one on the reals. At \(y=2\), the inverse input is \(x=1\), so

\[(f^{-1})'(2)=\frac1{f'(1)}=\frac14.\]

A small output change near \(2\) therefore corresponds to an input change about one quarter as large.

Engineering Example — resistance uncertainty and power

For fixed voltage \(P=V^2/R\), differentiation gives

\[dP=-\frac{V^2}{R^2}\,dR=-\frac{P}{R}\,dR.\]

Hence \(dP/P\approx-dR/R\). A \(1\%\) positive resistance error produces roughly a \(1\%\) negative power error. The sign is useful: it tells the direction of the change, while the absolute value gives the uncertainty magnitude.

Example — Taylor approximation with a size check

For \(\sqrt{1.04}\), use \(f(x)=\sqrt{x}\) at \(a=1\):

\[T_2(x)=1+\frac12(x-1)-\frac18(x-1)^2.\]

Thus \(T_2(1.04)=1+0.02-0.0002=1.0198\). The exact value is about \(1.019804\). The small error is plausible because \(0.04\) is close to the expansion point; the same polynomial should not be trusted unchanged far from \(1\).

Differentials report local sensitivity with direction. For fixed voltage, dP/P≈−dR/R says a positive resistance error gives an approximately equal negative relative power error. A Taylor approximation likewise needs a size check: its error is governed by distance from the centre and the remaining curvature.

\[dP=-\frac{V^2}{R^2}dR=-\frac PRdR\]

9.6 Taylor polynomials: matching local behaviour term by term

At \(x=a\), \(T_0\) matches only the function value, \(T_1\) also matches the tangent slope, and \(T_2\) also matches curvature. The factorials ensure that differentiating a Taylor term the required number of times leaves the matching derivative coefficient. This is why Taylor polynomials fit a function near the expansion point more faithfully as order rises.

Example — increasing order improves a local cosine approximation

At \(a=0\),

\[T_0(x)=1,\qquad T_2(x)=1-\frac{x^2}{2},\qquad T_4(x)=1-\frac{x^2}{2}+\frac{x^4}{24}.\]

At \(x=0.5\), these give \(1\), \(0.875\), and approximately \(0.877604\), while \(\cos(0.5)\approx0.877583\). The fourth-order term improves the result because \(0.5\) is close enough to the centre that the next powers are small. Increasing order does not automatically help far from the centre without an error analysis.

Example — remainder as an engineering tolerance

Approximate \(e^{0.1}\) by \(T_2=1+x+x^2/2\), giving \(1.105\). On \([0,0.1]\), the third derivative of \(e^x\) is at most \(e^{0.1}\), so

\[|R_2(0.1)|\le\frac{e^{0.1}}{3!}(0.1)^3<0.000185.\]

The approximation is therefore justified to three decimal places. A tolerance requirement can determine how many Taylor terms are needed.

Example — linearisation is the first Taylor polynomial

For \(f(x)=1/(1+x)\) at \(a=0\), \(T_1(x)=1-x\). Thus a \(2\%\) fractional increase \(x=0.02\) gives \(1/(1.02)\approx0.98\). Adding the quadratic term gives \(1-x+x^2=0.9804\), closer to the exact value \(0.980392\ldots\).

Order helps locally because the next powers are small. For cosine near zero, even orders add the next matched curvature behaviour. At x=0.2, the sixth-order polynomial differs from cosine by less than 10⁻⁷; farther away, order alone is not an error guarantee.

\[T_6(0.2)=1-\frac{0.2^2}{2!}+\frac{0.2^4}{4!}-\frac{0.2^6}{6!},\qquad |T_6(0.2)-\cos(0.2)|<10^{-7}\]

Common Mistakes

  • Confusing inverse and reciprocal notation.
  • Inverting a function without first restricting it to a one-to-one domain.
  • Treating \(dy\) as exactly equal to a finite change rather than a local approximation.
  • Using a Taylor polynomial far from its expansion point without estimating error.

Exercises

  1. Find the inverse of \(f(x)=3x-5\) and verify both compositions.
  2. Restrict \(f(x)=x^2-4\) to a domain that makes it invertible, then find the inverse.
  3. Use linearisation to approximate \(\sqrt{24.8}\) around a convenient point.
  4. Find \(T_3(x)\) for \(e^x\) about \(0\).
  5. A resistance \(R\) is measured with a small error \(dR\). For \(P=V^2/R\) with fixed \(V\), derive the differential approximation for \(dP\).
  6. For \(f(x)=x^3+x\), find \((f^{-1})'(2)\).
  7. Approximate \(\sqrt{1.04}\) using a second-degree Taylor polynomial about \(1\).
  8. Compare \(T_0,T_2,T_4\) for \(\cos x\) at \(x=0.5\).
  9. Use a Taylor remainder bound to justify the accuracy of \(1+x+x^2/2\) for approximating \(e^{0.1}\).

Part IV

Part IV — Series, Exponentials, and Integral Calculus

Part IV · Series, Exponentials, and Integral Calculus

10

Chapter 10 — Infinite Series and Convergence Tests

Open chapter view →

Why This Matters

Why the ratio test works — and when it does not. The ratio test checks whether the tail of a series eventually behaves like a convergent geometric series by measuring the limiting ratio between consecutive terms.

The ratio test as geometric comparison

Starting point. Let \(\sum a_n\) be a series with non-zero terms, and let \(L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|\).

  1. \[|a_{n+1}| \approx L |a_n| \quad \text{for large } n\]
    For large \(n\), consecutive terms scale by a factor close to \(L\). The tail approximates a geometric progression with ratio \(L\).reversible
  2. \[|a_{N+k}| \approx L^k |a_N|\]
    After \(k\) additional steps past index \(N\), the term magnitude scales as \(L^k |a_N|\).reversible
  3. \[L < 1 \implies \text{converges}, \quad L > 1 \implies \text{diverges}\]
    When \(L < 1\), the geometric series \(\sum L^k\) converges, bounding the tail. When \(L > 1\), terms grow without bound.reversible

Conclusion. The ratio test compares the series tail to a geometric series. When \(L < 1\), terms decay exponentially fast; when \(L > 1\), terms do not approach zero.

An infinite expression is meaningful only when its partial results settle to a definite value. Series make it possible to define and approximate functions that finite polynomials cannot express exactly, including the exponential function that governs capacitor transients.

10.1 Geometric series

The finite geometric sum is

\[S_N=1+r+r^2+\cdots+r^N.\]

Multiply by \(r\) and subtract:

\[\begin{aligned} S_N-rS_N&=(1+r+\cdots+r^N)-(r+r^2+\cdots+r^{N+1})\\ &=1-r^{N+1}. \end{aligned}\]

Therefore, for \(r\ne1\),

\[S_N=\frac{1-r^{N+1}}{1-r}.\]

If \(|r|<1\), then \(r^{N+1}\to0\) and

\[\sum_{n=0}^{\infty}r^n=\frac1{1-r}.\]

If \(|r|\ge1\), the terms do not tend to zero in the required way, so the series diverges. The condition is part of the formula.

Engineering Connection — repeated attenuation

If each stage passes a fraction \(r\) of a quantity and \(|r|<1\), accumulated contributions can have the form \(A(1+r+r^2+\cdots)=A/(1-r)\). The model is valid only when the repeated effect really is linear and each stage has the same ratio.

Derivation. Multiply a finite geometric sum by its ratio and subtract. Every middle term cancels, leaving only the first term and one tail term. Infinity enters only after this finite identity is established.

\[S_n=a\frac{1-r^n}{1-r}\quad(r\ne1)\]

10.2 Positive-term comparison and \(p\)-series

For nonnegative terms, comparison turns a difficult series into a known one. If

\[0\le a_n\le b_n\]

eventually and \(\sum b_n\) converges, then \(\sum a_n\) converges. If \(a_n\ge b_n\ge0\) eventually and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

The \(p\)-series gives a central benchmark:

\[\sum_{n=1}^{\infty}\frac1{n^p} \begin{cases} \text{converges},&p>1,\\ \text{diverges},&p\le1. \end{cases}\]

The harmonic series is the \(p=1\) case. Its terms approach zero but its partial sums do not settle, demonstrating again that term limit zero is not sufficient.

Why \(p = 1\) is the boundary for convergence. The \(p\)-series \(\sum 1/n^p\) is the central reference benchmark for comparison tests. The threshold \(p = 1\) reflects the balance point where the continuous improper integral transitions from finite to infinite area.

Integral test for p-series

Starting point. Let \(f(x) = \frac{1}{x^p}\) be positive and decreasing on \([1, \infty)\).

  1. \[\int_1^{n+1} f(x)\,dx \le \sum_{k=1}^n f(k) \le f(1) + \int_1^n f(x)\,dx\]
    The sum of rectangle areas is sandwiched between two shifted continuous integrals.reversible
  2. \[\int_1^\infty \frac{dx}{x^p} < \infty \iff p > 1\]
    The improper integral converges if and only if \(p > 1\), since \(\lim_{b\to\infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b\) is finite only when \(1-p < 0\).reversible
  3. \[\sum_{n=1}^\infty \frac{1}{n^p} < \infty \iff p > 1\]
    Since the partial sums are bounded by the convergent integral, the \(p\)-series converges exactly when \(p > 1\).reversible

Conclusion. The \(p\)-series \(\sum_{n=1}^\infty \frac{1}{n^p}\) converges if and only if \(p > 1\). The harmonic series (\(p = 1\)) sits right on the boundary and diverges.

  1. Identify the series as a \(p\)-series \(\sum \frac{1}{n^p}\) or compare it to one using the Limit Comparison Test.
  2. If \(p > 1\): the series converges (useful for establishing upper bounds).
  3. If \(p \le 1\): the series diverges (useful for establishing lower bounds to prove divergence).
  4. Remember: \(p = 1.0001\) converges, while \(p = 0.9999\) diverges. The boundary at \(p = 1\) is sharp.

10.3 Ratio and root tests

For a series \(\sum a_n\), the ratio test examines

\[L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.\]

If \(L<1\), the series converges absolutely; if \(L>1\) or \(L=\infty\), it diverges; if \(L=1\), the test says nothing. The root test has the corresponding quantity

\[L=\lim_{n\to\infty}|a_n|^{1/n}.\]

It has the same conclusions. Factorials often make ratios simple; powers raised to the \(n\)th power often make roots simple.

Example — Factorials favour the ratio test

Study \(\sum_{n=1}^{\infty}\frac{3^n}{n!}\). Then

\[\left|\frac{a_{n+1}}{a_n}\right| =\frac{3^{n+1}/(n+1)!}{3^n/n!}=\frac3{n+1}\to0.\]

Because \(0<1\), the series converges absolutely.

The ratio test is an eventual comparison. Once consecutive absolute terms are consistently reduced by a factor below one, the remaining tail is dominated by a convergent geometric tail. When the limiting ratio is one, that comparison becomes neutral; it says nothing about the series.

10.4 Power series and radius of convergence

A power series centred at \(0\) has form

\[\sum_{n=0}^{\infty}a_nx^n.\]

There is a radius \(R\in[0,\infty]\) such that the series converges absolutely for \(|x|<R\) and diverges for \(|x|>R\). At \(|x|=R\), each endpoint must be tested separately. When the relevant root limit exists,

\[\frac1R=\lim_{n\to\infty}|a_n|^{1/n}.\]

For a series centred at \(a\), replace \(x\) by \(x-a\) throughout.

Why endpoints require their own work. The ratio test describes an open interval where the terms shrink decisively and an exterior where they do not. At the boundary its diagnostic often becomes exactly one, which gives no conclusion. Each endpoint is a new series and must be tested on its own terms.

10.5 Functions defined by power series

Within its radius of convergence, a power series is continuous and may be differentiated term by term. The derivative has the same radius of convergence:

\[\frac d{dx}\sum_{n=0}^{\infty}a_nx^n =\sum_{n=1}^{\infty}na_nx^{n-1}.\]

This theorem is powerful precisely because it is restricted to the open interval of convergence. It is not an invitation to differentiate a divergent expression.

Why power series always converge on symmetric intervals. The behaviour of \(\sum a_n (x-c)^n\) depends on \(|x-c|\) — the absolute geometric distance from the centre \(c\) — rather than on the sign of \(x-c\).

Computing R from the ratio test

From
  • Apply the ratio test: \(\lim_{n\to\infty} \left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = |x-c| \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = |x-c| L\).
  • Convergence requires \(|x-c| L < 1\).
  1. Solving for distance: \(|x-c| < \frac{1}{L}\).
  2. Define the radius of convergence \(R = \frac{1}{L}\) (with \(R = \infty\) if \(L = 0\), and \(R = 0\) if \(L = \infty\)).

Result. The series converges absolutely for \(|x-c| < R\) and diverges for \(|x-c| > R\), forming an interval centred at \(c\).

Limitation. The endpoints \(x = c - R\) and \(x = c + R\) must be tested separately using numerical series convergence tests.

Common Mistakes

  • Treating a sequence and its series as the same object.
  • Concluding a series converges solely because its terms tend to zero.
  • Reporting “ratio test equals one” as convergence or divergence.
  • Forgetting that the radius test does not decide the endpoints.

Fundamental distinction — sequence versus series. A sequence \(a_1, a_2, a_3, \ldots\) is an ordered list of numbers; each term exists independently. A series \(\sum a_n\) adds those numbers into a single accumulated value. Confusing the two is like confusing a shopping list with the total bill. The sequence \(1/n\) approaches \(0\), but the harmonic series \(\sum 1/n\) diverges — terms shrink, yet there are infinitely many of them.

Partial sums define the series

Starting point. Let \(S_N = \sum_{n=1}^N a_n\) be the \(N\)-th partial sum.

  1. \[S_1 = a_1\]
    The first partial sum is just the first term.reversible
  2. \[S_2 = a_1 + a_2\]
    The second partial sum adds the second term.reversible
  3. \[S_N = S_{N-1} + a_N\]
    Each new partial sum adds one more term to the previous accumulation.reversible
  4. \[\sum_{n=1}^\infty a_n = \lim_{N\to\infty} S_N\]
    The infinite series is defined as the limit of the partial-sum sequence \((S_N)\).reversible

Conclusion. Convergence of a series means the sequence of partial sums \((S_N)\) settles to a finite number \(S\). It is not merely about individual terms \(a_n\) shrinking.

  1. First check the \(n\)-th term: if \(\lim_{n\to\infty} a_n \neq 0\), the series diverges immediately (Divergence Test).
  2. If \(\lim_{n\to\infty} a_n = 0\), the test is inconclusive — you must use a convergence test (Comparison, Integral, Ratio, Root, or Alternating Series).
  3. Never conclude a series converges just because \(a_n \to 0\).

10.6 Partial sums, test selection, and endpoint discipline

A series converges only when its partial sums settle. The term test is a fast necessary-condition check: if \(a_n\) does not tend to zero, \(\sum a_n\) diverges immediately. Passing this check proves nothing further. Choose comparison for positive rational-looking terms, ratio for factorials or exponentials, and root for an expression naturally raised to the \(n\)th power.

Example — finite geometric sum before the infinite limit

For \(1+1/2+\cdots+(1/2)^4\), the finite formula gives

\[S_4=\frac{1-(1/2)^5}{1-1/2}=\frac{31}{16}.\]

The infinite sum is \(2\), not because five terms equal it, but because the remainder \((1/2)^{N+1}/(1-1/2)\) tends to zero as \(N\) grows.

Example — comparison with a \(p\)-series

For \(n\ge1\),

\[0<\frac1{n^2+n}\le\frac1{n^2}.\]

Since \(\sum1/n^2\) converges, \(\sum1/(n^2+n)\) converges. The inequality direction matters: a smaller nonnegative series inherits convergence from a larger known convergent series, not the other way around.

Example — radius is not an interval

For \(\sum_{n=1}^{\infty}(x-1)^n/n\), the ratio test gives \(|x-1|<1\). At \(x=2\), the series is harmonic and diverges; at \(x=0\), it is alternating harmonic and converges. The interval is \([0,2)\), not simply “radius one.”

A systematic approach to choosing convergence tests. With several convergence tests available, the essential skill is matching the test to the algebraic structure of the summand.

  1. <strong>Divergence Test first:</strong> if \(\lim_{n\to\infty} a_n \neq 0\), the series diverges immediately. Done.
  2. <strong>Factorials (\(n!\)) or products:</strong> use the Ratio Test, since factorials simplify cleanly in ratios \(\frac{(n+1)!}{n!} = n+1\).
  3. <strong>\(n\)-th powers (\(a_n = (b_n)^n\)):</strong> use the Root Test to eliminate the outer exponent.
  4. <strong>Rational or algebraic functions of \(n\):</strong> use the Limit Comparison Test with a \(p\)-series \(\sum 1/n^p\).
  5. <strong>Alternating signs with \(|a_n|\) decreasing to \(0\):</strong> use the Alternating Series Test.
  6. <strong>Continuous, positive, decreasing function \(f(n) = a_n\):</strong> use the Integral Test.
  7. <strong>Power series:</strong> use the Ratio or Root Test to find the radius \(R\), then test each endpoint \(x = c \pm R\) separately.

Endpoint testing is independent. The Ratio or Root Test yields the open interval \((c-R, c+R)\). At \(x = c+R\) and \(x = c-R\), substitute the numerical value and test each resulting series on its own merits. One endpoint may converge while the other diverges.

10.7 Choosing a convergence test and representing functions

Tests are tools with limited conclusions. A ratio or root limit equal to \(1\) is not a near-convergence result; it is no result at all from that test. For a power series, first find the open radius interval, then test each endpoint in the original series. Endpoint behaviour can differ because signs or known benchmark series change there.

Example — factorials signal the ratio test

For

\[\sum_{n=1}^{\infty}\frac{n!}{4^n},\]

the term ratio is

\[\left|\frac{a_{n+1}}{a_n}\right| =\frac{(n+1)!/4^{n+1}}{n!/4^n} =\frac{n+1}{4}\to\infty.\]

The terms eventually grow rather than approach zero, so the series diverges. The ratio test was chosen because factorial cancellation reveals the behaviour in one step.

Example — an \(n\)th power signals the root test

For

\[\sum_{n=1}^{\infty}\left(\frac{2n+1}{3n+4}\right)^n,\]

the root test gives

\[\lim_{n\to\infty}\left|a_n\right|^{1/n} =\lim_{n\to\infty}\frac{2n+1}{3n+4}=\frac23<1.\]

Therefore the series converges absolutely. A ratio test could also work but would require more algebra; the root test matches the expression's structure.

Example — power-series endpoints are separate problems

For

\[\sum_{n=1}^{\infty}\frac{x^n}{n\,2^n},\]

the ratio test gives \(|x|/2<1\), so \(|x|<2\). At \(x=2\) the series is harmonic and diverges. At \(x=-2\) it is alternating harmonic and converges. Thus the convergence interval is \([-2,2)\).

Example — a power series represents a local function

Within \(|x|<1\),

\[\frac1{1-x}=1+x+x^2+x^3+\cdots.\]

At \(x=0.1\), the first four terms give \(1.111\), while the exact value is \(1/0.9=1.111\overline1\). The omitted tail is geometric and has size \(0.1^4/(1-0.1)\), so its smallness is quantified. At \(x=1.1\), the rational function exists but this series diverges: a function and one particular power series representation need not share the same domain.

Why the geometric series formula holds. The sum formula is not an approximation — it comes from exact algebraic cancellation (a telescoping identity) followed by a limit.

Algebraic derivation of the geometric sum

Starting point. Let \(S_n = \sum_{k=0}^{n-1} a r^k = a + ar + ar^2 + \cdots + ar^{n-1}\).

  1. \[S_n = a + ar + ar^2 + \cdots + ar^{n-1}\]
    Write out the finite sum explicitly.reversible
  2. \[r S_n = ar + ar^2 + \cdots + ar^n\]
    Multiply every term by the common ratio \(r\).reversible
  3. \[S_n - r S_n = a - ar^n\]
    Subtract the second line from the first: intermediate terms \(ar, ar^2, \dots, ar^{n-1}\) cancel in pairs.reversible
  4. \[(1-r) S_n = a(1 - r^n)\]
    Factor out \(S_n\) on the left and \(a\) on the right.reversible
  5. \[S_n = \frac{a(1-r^n)}{1-r} \quad (r \neq 1)\]
    Divide by \(1-r\). For \(|r| < 1\), as \(n \to \infty\), \(r^n \to 0\), giving \(S_\infty = \frac{a}{1-r}\).reversible

Conclusion. The finite geometric sum is \(S_n = \frac{a(1-r^n)}{1-r}\). When \(|r| < 1\), the infinite sum converges to \(S = \frac{a}{1-r}\).

The r = 1 edge case

From
  • The derivation divides by \(1-r\), which is zero when \(r = 1\).
  • When \(r = 1\), every term in the sum equals \(a\).
  1. If \(r = 1\): \(S_n = a + a + \cdots + a = n a\).
  2. As \(n \to \infty\), \(n a \to \infty\) (for \(a \neq 0\)).

Result. The geometric series with \(r = 1\) diverges to infinity. The edge case requires separate arithmetic because division by \(1-r\) is undefined.

Limitation. When \(r = 1\), multiplying by \(r\) changes nothing, so subtraction yields \(0 = 0\) rather than a formula.

Exercises

  1. Find \(\sum_{n=0}^{\infty}(1/3)^n\).
  2. Decide whether \(\sum_{n=1}^{\infty}1/(n^2+1)\) converges by comparison.
  3. Use the ratio test for \(\sum n!/5^n\).
  4. Find the radius and test endpoints for \(\sum_{n=1}^{\infty}x^n/n\).
  5. Explain why \(\sum x^n=1/(1-x)\) cannot be used at \(x=2\).
  6. Find the sum of the first six terms of \(1+1/3+1/9+\cdots\) and its infinite sum.
  7. Decide convergence of \(\sum1/(n^2+n)\) by comparison.
  8. Find the convergence interval of \(\sum(x-1)^n/n\).
  9. Choose and apply an appropriate test to \(\sum n!/4^n\).
  10. Choose and apply an appropriate test to \(\sum((2n+1)/(3n+4))^n\).
  11. Find the interval of convergence of \(\sum x^n/(n\,2^n)\), testing both endpoints.

Part IV · Series, Exponentials, and Integral Calculus

11

Chapter 11 — The Exponential and Logarithmic Functions

Open chapter view →

Why This Matters

The equation \(y'=y\) describes a quantity whose rate of change is proportional to its current value. It appears in ideal growth and decay models and, after scaling, in first-order circuit responses. The exponential function is the function that solves it. The logarithm reverses exponential relationships.

\(\exp(x)\) is defined by its derivative, not by repeated multiplication. The exponential function generalises exponentiation to all real inputs. Its defining property is \(\frac{d}{dx}e^x = e^x\) — the function equals its own rate of change.

Why e^x equals its own derivative

Starting point. Define \(f(x) = e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\).

  1. \[f'(x) = \frac{d}{dx}\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right)\]
    Differentiate the power series term-by-term (valid for all real \(x\)).reversible
  2. \[f'(x) = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \frac{4x^3}{4!} + \cdots\]
    Apply the power rule \(\frac{d}{dx} x^n = n x^{n-1}\) to each individual term.reversible
  3. \[f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = f(x)\]
    Simplify coefficients \(\frac{n}{n!} = \frac{1}{(n-1)!}\): the original series reappears identically.reversible

Conclusion. The function \(f(x) = e^x\) is its own derivative: \(f'(x) = f(x)\) with initial condition \(f(0) = 1\). It is the unique function satisfying this property.

The base e as a natural limit

From
  • Continuous compounding: compound interest with rate \(r = 1\) split into \(n\) intervals yields \(\left(1 + \frac{1}{n}\right)^n\).
  • The derivative definition requires \(\lim_{h\to0} \frac{e^h - 1}{h} = 1\).
  1. Letting \(n = 1/h\), the limit becomes \(\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = e \approx 2.71828\).
  2. The number \(e\) is not an arbitrary constant: it is the exact base that makes the derivative of \(b^x\) have scale factor \(1\) (since \(\frac{d}{dx}b^x = b^x \ln b\)).

Result. \(e = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828\). The natural base simplifies all calculus derivatives by eliminating extra \(\ln b\) scale factors.

11.1 Defining \(\exp(x)\)

Exponential models in engineering. Whenever a quantity changes at a rate proportional to its current size, the response is exponential: growth (\(k > 0\)) or decay (\(k < 0\)).

Deriving the time constant

Starting point. For a first-order system with decay rate \(k = 1/\tau\): \(y(t) = A e^{-t/\tau}\).

  1. \[y(0) = A\]
    At \(t = 0\), the initial value is \(A\).reversible
  2. \[y(\tau) = A e^{-1} \approx 0.368 A\]
    After one time constant \(t = \tau\), the quantity falls to \(1/e \approx 36.8\%\) of its initial value (a \(63.2\%\) drop).reversible
  3. \[y(3\tau) = A e^{-3} \approx 0.0498 A\]
    After three time constants \(t = 3\tau\), under \(5\%\) of the initial value remains.reversible
  4. \[y(5\tau) = A e^{-5} \approx 0.0067 A\]
    After five time constants \(t = 5\tau\), less than \(1\%\) remains — practically settled in engineering circuits.reversible

Conclusion. The time constant \(\tau = 1/k\) sets the natural timescale: \(1\tau \to 36.8\%\), \(3\tau \to 5\%\), \(5\tau \to <1\%\).

  1. Identify the initial value \(A = y(0)\).
  2. Determine the rate constant \(k\) or time constant \(\tau = 1/k\) from the physical circuit parameters (e.g. \(\tau = RC\)).
  3. Write the model: \(y(t) = A e^{-t/\tau}\) (decay) or \(y(t) = A(1 - e^{-t/\tau})\) (step response).
  4. Check units: the exponent \(-t/\tau\) must be dimensionless.

The exponential function can be defined by its power series

\[\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.\]

The ratio test shows convergence for every real \(x\). Differentiating term by term returns the same series:

\[\frac d{dx}\exp(x)=\exp(x).\]

Let \(e=\exp(1)\). The addition law

\[\exp(x+y)=\exp(x)\exp(y)\]

implies \(\exp(x)=e^x\) for real \(x\). It also gives \(e^{-x}=1/e^x\), so \(e^x>0\) for all real \(x\).

Why the exponential is unusually stable. Its power-series coefficients are chosen so that differentiating shifts every term into the same series. That self-reproduction is the analytic reason the exponential models processes whose rate is proportional to their current amount.

\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\]

11.2 Exponential growth and decay

The function \(Ae^{kt}\) has initial value \(A\). If \(k>0\), it grows; if \(k<0\), it decays. Its derivative is

\[\frac d{dt}(Ae^{kt})=kAe^{kt}.\]

The time constant \(\tau>0\) is often used in decay as \(e^{-t/\tau}\). After one time constant the remaining fraction is \(e^{-1}\approx0.368\); after five it is about \(0.0067\). These are model-based landmarks, not abrupt physical cutoffs.

Engineering consequence. Exponential change appears when the instantaneous rate is proportional to the amount currently present. Equal time intervals then multiply the amount by the same factor rather than adding the same amount.

\[\frac{dy}{dt}=ky\;\Longrightarrow\;y(t)=Ae^{kt}\]

11.3 Logarithms as inverses

The natural logarithm \(\ln(x)\) undoes the exponential. Because \(e^x > 0\) for all real \(x\), its inverse \(\ln(x)\) is defined strictly for \(x > 0\).

Derivative of ln(x) via inverse function rule

Starting point. Let \(y = \ln x\), so \(x = e^y\).

  1. \[\frac{d}{dx}(x) = \frac{d}{dx}(e^y)\]
    Differentiate both sides with respect to \(x\) using the chain rule.reversible
  2. \[1 = e^y \frac{dy}{dx}\]
    The derivative of \(e^y\) with respect to \(x\) is \(e^y \frac{dy}{dx}\).reversible
  3. \[\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\]
    Solve for \(\frac{dy}{dx}\) and substitute back \(e^y = x\).reversible

Conclusion. \(\frac{d}{dx}\ln x = \frac{1}{x}\) for all \(x > 0\). This fills the missing power rule gap: \(\int x^{-1} dx = \ln|x| + C\).

Because \(e^x\) is strictly increasing and positive, it has an inverse \(\ln x\) on \(x>0\):

\[\ln(e^x)=x,\qquad e^{\ln x}=x\quad(x>0).\]

The addition law for exponentials becomes logarithm laws:

\[\ln(ab)=\ln a+\ln b,\qquad \ln(a^r)=r\ln a,\]

when the real expressions are defined. Differentiating \(e^{\ln x}=x\) gives

\[e^{\ln x}\frac d{dx}\ln x=1,\]

so \(\frac d{dx}\ln x=1/x\) for \(x>0\).

For \(u(x)\ne0\) on an interval,

\[\frac d{dx}\ln|u(x)|=\frac{u'(x)}{u(x)}, \qquad \int\frac{u'(x)}{u(x)}\,dx=\ln|u(x)|+C.\]

Undoing a scale factor in the exponent. The logarithm answers the exponent question: which power of the base gives this positive number? Because an exponential with positive base is one-to-one, its logarithm reverses it. This also explains why logarithm laws come from exponent laws rather than from memorised symbol patterns.

\[\ln(e^x)=x,\qquad e^{\ln y}=y\;(y>0)\]

11.4 General bases

For \(a>0\), \(a\ne1\),

\[\log_a x=\frac{\ln x}{\ln a},\qquad x>0.\]

This gives

\[\frac d{dx}a^x=a^x\ln a.\]

Example — solving an exponential equation

Solve \(5e^{-t/2}=0.2\). Divide by \(5\):

\[e^{-t/2}=0.04.\]

Take natural logs, which is legal because both sides are positive:

\[-\frac t2=\ln(0.04),\qquad t=-2\ln(0.04).\]

The positive answer is expected: a positive decaying exponential reaches a smaller positive threshold at a later time.

Converting between general bases and base \(e\). Every exponential \(b^x\) and logarithm \(\log_b x\) can be rewritten in base \(e\) using natural logarithms.

Base conversion formulas

Starting point. Let \(b > 0, b \neq 1\). Use the identity \(b = e^{\ln b}\).

  1. \[b^x = (e^{\ln b})^x = e^{x \ln b}\]
    Rewrite general base \(b^x\) in natural exponential form.reversible
  2. \[\frac{d}{dx}(b^x) = \frac{d}{dx}(e^{x \ln b}) = (\ln b) e^{x \ln b} = (\ln b) b^x\]
    Differentiate using the chain rule: an extra factor of \(\ln b\) appears.reversible
  3. \[y = \log_b x \iff b^y = x \iff y \ln b = \ln x \iff y = \frac{\ln x}{\ln b}\]
    Apply natural log to both sides of \(b^y = x\) to derive the change-of-base formula.reversible

Conclusion. \(b^x = e^{x \ln b}\) and \(\log_b x = \frac{\ln x}{\ln b}\). In calculus, always convert general bases to base \(e\) before differentiating or integrating.

  1. To differentiate \(b^x\): write \(b^x = e^{x \ln b}\), then differentiate to get \((\ln b) b^x\).
  2. To differentiate \(\log_b x\): write \(\log_b x = \frac{\ln x}{\ln b}\), then differentiate to get \(\frac{1}{x \ln b}\).
  3. Notice that when \(b = e\), \(\ln e = 1\), recovering the simpler natural formulas.

11.5 Exponentials, logarithms, and model checks

The equation \(e^{u}=e^{v}\) permits \(u=v\) because the exponential is one-to-one. Taking \(\ln\) of an equation is valid only when both sides are positive real numbers. A logarithm can turn a product into a sum and an exponential unknown into a linear unknown, but it cannot repair a negative or zero argument.

Example — solve and interpret a decay threshold

If \(Q(t)=12e^{-t/5}\) mC, find when \(Q=3\) mC:

\[3=12e^{-t/5},\qquad \frac14=e^{-t/5}.\]

Taking natural logarithms gives

\[\ln\frac14=-\frac{t}{5},\qquad t=5\ln4.\]

The positive time agrees with decay. A negative answer would signal an algebraic sign error or a threshold above the initial quantity.

Example — logarithmic differentiation of a variable power

For \(y=x^x\) on \(x>0\), take logs before differentiating:

\[\ln y=x\ln x.\]

Then \(y'/y=\ln x+1\), hence

\[\frac{d}{dx}x^x=x^x(\ln x+1).\]

Neither the ordinary power rule nor the \(a^x\) rule alone applies because both base and exponent vary.

Engineering Example — time constant from two measurements

If a decaying voltage follows \(v(t)=V_0e^{-t/\tau}\), then

\[\ln v=\ln V_0-\frac{t}{\tau}.\]

A plot of \(\ln v\) against \(t\) is a straight line of slope \(-1/\tau\), provided the measurements are positive and the single-exponential model is appropriate. This is a mathematical linearisation of an exponential model.

Sanity checks for exponential models. Always check asymptotic limits, dimensions, and signs when setting up exponential equations.

  1. <strong>Initial value:</strong> at \(t = 0\), does \(y(0)\) match the known initial condition? (For \(y = A e^{-kt}\), \(y(0) = A\). For \(y = A(1 - e^{-kt})\), \(y(0) = 0\).)
  2. <strong>Long-term limit:</strong> as \(t \to \infty\), does \(y(t)\) approach the correct physical equilibrium? (Decay \(\to 0\), charging \(\to V_s\).)
  3. <strong>Exponent units:</strong> the argument of an exponential must be dimensionless (\(kt\) or \(t/\tau\) has unit \(1\)).
  4. <strong>Monotonicity:</strong> check whether the physical quantity should be strictly increasing or decreasing.

Charging capacitor model check

From
  • Model: \(v(t) = V_s (1 - e^{-t/RC})\).
  • Circuit physically starts uncharged and charges toward supply voltage \(V_s\).
  1. Check \(t = 0\): \(v(0) = V_s (1 - e^0) = V_s (1 - 1) = 0\text{ V}\) (correct: uncharged capacitor).
  2. Check \(t \to \infty\): \(v(\infty) = V_s (1 - 0) = V_s\) (correct: fully charged to supply).
  3. Check derivative: \(v'(t) = \frac{V_s}{RC} e^{-t/RC} > 0\) (correct: monotonically increasing voltage).

Result. The model passes all three boundary and derivative sanity checks.

Exercises

  1. Differentiate \(7e^{-3t}\).
  2. Solve \(2^x=15\) using logarithms.
  3. State the domain of \(\ln(4-x)\).
  4. Evaluate \(\int 6x/(3x^2+1)\,dx\).
  5. A quantity decays as \(Q(t)=Q_0e^{-t/RC}\). Explain the roles of \(Q_0\), \(R\), \(C\), and \(t\) without solving an equation.
  6. Solve \(8e^{-t/4}=1\) and interpret the sign of the answer.
  7. Differentiate \(x^x\) on its real positive domain by logarithmic differentiation.
  8. Explain how a straight-line plot can be used to estimate an exponential time constant from positive data.

Part IV · Series, Exponentials, and Integral Calculus

12

Chapter 12 — Definite Integrals and the Fundamental Theorem

Open chapter view →

Why This Matters

Derivatives describe local rate. Integration reverses the question: if a rate is known, how much has accumulated? Current is rate of charge flow, so accumulated charge is an integral of current. Area is another familiar accumulation model.

The Fundamental Theorem connects rate of change to accumulation. Differentiation computes instantaneous rates; integration accumulates those rates over an interval. They are exact inverses.

Fundamental Theorem of Calculus (Part 1 and 2)

Starting point. Let \(f\) be continuous on \([a, b]\), and define the accumulator \(g(x) = \int_a^x f(t) dt\).

  1. \[g'(x) = \lim_{h\to0} \frac{g(x+h) - g(x)}{h} = \lim_{h\to0} \frac{1}{h} \int_x^{x+h} f(t) dt\]
    Write the derivative of the accumulation function by definition.reversible
  2. \[\lim_{h\to0} \frac{1}{h} \left[ f(c) \cdot h \right] = \lim_{h\to0} f(c) = f(x)\]
    By the Mean Value Theorem for Integrals, \(\int_x^{x+h} f(t) dt = f(c) h\) for some \(c \in [x, x+h]\). As \(h \to 0\), \(c \to x\).reversible
  3. \[\int_a^b f(t) dt = F(b) - F(a)\]
    Since \(g(x)\) is an antiderivative of \(f(x)\), any antiderivative \(F\) satisfies \(\int_a^b f(t)dt = F(b) - F(a)\).reversible

Conclusion. FTC Part 1: \(\frac{d}{dx} \int_a^x f(t) dt = f(x)\). FTC Part 2: \(\int_a^b f(x) dx = F(b) - F(a)\).

12.1 Riemann sums and signed area

For a function \(f\) on \([a,b]\), split the interval into narrow pieces of width \(\Delta x\). A rectangle of height \(f(x_k^*)\) has signed area \(f(x_k^*)\Delta x\). Adding the rectangles and taking the limit defines

\[\int_a^b f(x)\,dx.\]

Area above the axis contributes positively and area below contributes negatively. The integral is therefore signed accumulation, not automatically total geometric area.

Geometric picture. A definite integral is built from ordinary rectangle areas. Narrower rectangles reduce the mismatch between the curve and each rectangle; portions below the axis contribute negative signed area.

\[\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_k^*)\Delta x\]

12.2 Antiderivatives and the FTC

An antiderivative of \(f\) is a function \(F\) with \(F'=f\). Antiderivatives differ by a constant because constants differentiate to zero.

The theorem makes exact integration practical. It also says that the accumulation function \(A(x)=\int_a^x f(t)\,dt\) has derivative \(A'(x)=f(x)\) when \(f\) is continuous.

Example — accumulated charge

If \(i(t)=(2t+1)\) mA, with \(t\) measured in milliseconds and \(0\le t\le3\), then the charge is

\[q=\int_0^3(2t+1)\ \mathrm{mA\,ms} =\left[t^2+t\right]_0^3\ \mathrm{\mu C} =12\ \mathrm{\mu C}.\]

The unit conversion is built into \(\mathrm{mA\,ms}=10^{-6}\ \mathrm C\). Writing the time unit explicitly also prevents a common factor-of-\(1000\) error.

Accumulation produces rate. Let A(x) be the signed area accumulated from a fixed starting point to x. Extending the endpoint by a tiny amount adds a very thin strip whose area is approximately its height f(x) times its width. Dividing by that width and taking the limit gives A'(x)=f(x).

\[A(x)=\int_a^x f(t)\,dt\quad\Longrightarrow\quad A'(x)=f(x)\]

12.3 Basic antiderivatives

For \(a\ne-1\),

\[\int x^a\,dx=\frac{x^{a+1}}{a+1}+C.\]

The excluded exponent has its own rule:

\[\int\frac1x\,dx=\ln|x|+C\]

on intervals not crossing zero. Also,

\[\int e^{ax}\,dx=\frac1a e^{ax}+C\quad(a\ne0).\]

Antidifferentiation is the reverse of differentiation. If \(F'(x) = f(x)\), then \(\int f(x) dx = F(x) + C\). The constant \(+C\) represents a family of vertically shifted parallel curves all sharing the same slope at every \(x\).

Why the constant of integration is necessary

Starting point. Let \(F(x)\) and \(G(x)\) be two antiderivatives of \(f(x)\) on an interval.

  1. \[\frac{d}{dx}[F(x) - G(x)] = F'(x) - G'(x) = f(x) - f(x) = 0\]
    The derivative of their difference is identically zero.reversible
  2. \[F(x) - G(x) = C\]
    By the Mean Value Theorem, a function whose derivative is everywhere zero on an interval must be a constant.reversible
  3. \[F(x) = G(x) + C\]
    Any two antiderivatives differ by at most an additive constant.reversible

Conclusion. The indefinite integral \(\int f(x)dx = F(x) + C\) captures all possible antiderivatives. Omitting \(+C\) loses solutions to differential equations.

  1. Power rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) for all \(n \neq -1\).
  2. Reciprocal rule: \(\int \frac{1}{x} dx = \ln|x| + C\).
  3. Exponential rule: \(\int e^{kx} dx = \frac{1}{k} e^{kx} + C\).
  4. Always add \(+C\) to indefinite integrals.

12.4 Substitution

Substitution reverses the chain rule. If \(u=g(x)\) and \(du=g'(x)\,dx\), then

\[\int f(g(x))g'(x)\,dx=\int f(u)\,du.\]

For a definite integral, either change the limits to \(u\)-limits or return to \(x\) before substituting the original limits.

Example — substitution with changed bounds

\[\int_0^1 2x e^{x^2}\,dx.\]

Let \(u=x^2\), so \(du=2x\,dx\). The bounds become \(u=0\) and \(u=1\):

\[\int_0^1e^u\,du=e-1.\]

Substitution names the changing inner quantity. When an inner expression changes by a known derivative factor, replacing it by a new variable turns a composite accumulation into a simpler one-variable accumulation. For a definite integral, the limits must change too because they describe values of the new variable.

  1. Choose the repeated inner expression as u.
  2. Write du with its differential factor and account for any constant multiplier.
  3. For definite bounds, convert each original endpoint to its u-value before evaluating.

12.5 Building a definite integral from rectangles

The symbol \(\int_a^b f(x)\,dx\) is compact, but it represents a very concrete limiting measurement. Suppose \(f(x)=x^2\) on \([0,1]\). Divide the interval into \(n\) equal pieces. Each has width \(\Delta x=1/n\). If right endpoints are used, the \(k\)th height is \((k/n)^2\), so the right-rectangle total is

\[S_n=\sum_{k=1}^n\left(\frac{k}{n}\right)^2\frac1n =\frac1{n^3}\sum_{k=1}^n k^2.\]

The finite sum formula \(\sum_{k=1}^n k^2=n(n+1)(2n+1)/6\) gives

\[S_n=\frac{n(n+1)(2n+1)}{6n^3} =\frac{2n^3+3n^2+n}{6n^3}.\]

As \(n\to\infty\), the lower-order terms disappear and \(S_n\to1/3\). Thus

\[\int_0^1x^2\,dx=\frac13.\]

This calculation is not a replacement for the Fundamental Theorem of Calculus; it explains what that theorem evaluates efficiently. A graph should be imagined here: the curve \(y=x^2\) rises from \((0,0)\) to \((1,1)\), while the narrow rectangles leave ever smaller gaps above or below the curve.

\(n=4\), then a second panel with many thin rectangles. Label \(\Delta x\) and a sample height \(f(x_k^*)\).]

Example — signed accumulation is not total area

Evaluate

\[\int_{-2}^{2}x\,dx.\]

An antiderivative is \(x^2/2\), so

\[\left[\frac{x^2}{2}\right]_{-2}^{2}=2-2=0.\]

The result follows also from symmetry: the triangular region below the axis on \([-2,0]\) has the same magnitude as the triangular region above it on \([0,2]\). If the question asks for geometric area instead, split at the zero and reverse the sign on the negative part:

\[\int_{-2}^{0}(-x)\,dx+\int_0^2x\,dx=2+2=4.\]

Always decide which quantity the model asks for. Net charge can be signed; the total distance travelled cannot be obtained from velocity by allowing opposite directions to cancel.

Riemann sums construct the exact integral. The limit of Riemann sums as \(\Delta x \to 0\) (\(n \to \infty\)) defines the integral regardless of the sample point choice (left, right, or midpoint).

Definite integral as Riemann limit

Starting point. Divide \([a, b]\) into \(n\) equal subintervals of width \(\Delta x = \frac{b-a}{n}\) with grid points \(x_k = a + k \Delta x\).

  1. \[R_n = \sum_{k=1}^n f(x_k) \Delta x\]
    Right Riemann sum: sample heights at right endpoints \(x_k\).reversible
  2. \[L_n = \sum_{k=0}^{n-1} f(x_k) \Delta x\]
    Left Riemann sum: sample heights at left endpoints \(x_k\).reversible
  3. \[\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{k=1}^n f(x_k^*) \Delta x\]
    As \(n \to \infty\) and \(\Delta x \to 0\), all Riemann sums converge to the same unique real number for any continuous \(f\).reversible

Conclusion. \(\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{k=1}^n f(x_k^*) \Delta x\). The integral represents the net signed area bounded by \(y = f(x)\) and the \(x\)-axis.

12.6 Why the Fundamental Theorem works

There are two complementary parts to the theorem. The first says that an accumulation function has the original rate as its derivative. Define

\[A(x)=\int_a^x f(t)\,dt.\]

The letter \(t\) is a dummy variable: changing it to \(u\) does not change the meaning. For a small change \(h\),

\[\frac{A(x+h)-A(x)}{h} =\frac1h\int_x^{x+h}f(t)\,dt.\]

The right-hand side is the average value of \(f\) over a short interval. When \(f\) is continuous, every value in that short interval becomes close to \(f(x)\) as \(h\to0\). Therefore \(A'(x)=f(x)\). This is the accumulation analogue of the secant line becoming a tangent line.

The second part says that if \(F'=f\), then the total accumulation from \(a\) to \(b\) is \(F(b)-F(a)\). Since \(A'=F'\), the functions \(A\) and \(F\) differ by a constant. Because \(A(a)=0\), that constant is \(-F(a)\), and hence

\[A(b)=F(b)-F(a).\]

The continuity hypothesis gives a dependable ordinary Riemann integral and makes the average-value argument valid. More advanced mathematics extends the theorem further, but this course uses the stated continuous-function setting.

Example — an accumulation function before evaluating it

Let

\[A(x)=\int_1^x\left(3t^2-4\right)\,dt.\]

Without first performing any integration, the Fundamental Theorem gives

\[A'(x)=3x^2-4.\]

To find the function itself, use \(F(t)=t^3-4t\):

\[A(x)=(x^3-4x)-(1-4)=x^3-4x+3.\]

Checking \(A(1)=0\) is essential. An accumulation from a point to itself must be zero; this catches a lost lower-limit term.

Why differentiation and integration undo each other. The Fundamental Theorem links the local slope of an area function directly to the height of the integrand.

Proof of FTC Part 1

Assumptions
  • \(f\) is continuous on \([a, b]\).
  • Define \(A(x) = \int_a^x f(t) dt\).
  1. \(A(x+h) - A(x) = \int_x^{x+h} f(t) dt\).By interval additivity of definite integrals.
  2. \(\int_x^{x+h} f(t) dt = f(c) h\) for some \(c \in [x, x+h]\).By the Mean Value Theorem for Integrals (since \(f\) is continuous).
  3. \(A'(x) = \lim_{h\to0} \frac{A(x+h) - A(x)}{h} = \lim_{h\to0} f(c) = f(x)\).As \(h \to 0\), the squeeze \(x \le c \le x+h\) forces \(c \to x\), and by continuity \(f(c) \to f(x)\).

Conclusion. \(\frac{d}{dx} \int_a^x f(t) dt = f(x)\). The rate of area accumulation equals the instantaneous curve height.

Scope. Requires \(f\) to be continuous at \(x\).

12.7 Substitution as a change of scale

The chain rule says

\[\frac d{dx}F(g(x))=F'(g(x))g'(x).\]

Substitution reads this equation backwards. The inner expression \(g(x)\) is a new coordinate, and \(g'(x)\,dx\) tells us how a small change in \(x\) changes that coordinate. It is not enough for an integrand to contain something that looks like an inner expression: the accompanying differential must match, possibly after extracting a constant.

Example — recognise and repair the missing factor

Find

\[\int x\cos(x^2)\,dx.\]

The inner expression is \(x^2\) and its derivative is \(2x\). Write the available factor as \(x\,dx=\tfrac12(2x\,dx)\), then set \(u=x^2\):

\[\int x\cos(x^2)\,dx =\frac12\int\cos u\,du =\frac12\sin u+C =\frac12\sin(x^2)+C.\]

Differentiation confirms the result: the outer factor \(1/2\) and chain-rule factor \(2x\) combine to recover \(x\cos(x^2)\).

Example — a definite integral with two safe routes

Evaluate

\[\int_1^2\frac{2x}{x^2+3}\,dx.\]

With \(u=x^2+3\), \(du=2x\,dx\). Changing bounds gives \(u=4\) at \(x=1\) and \(u=7\) at \(x=2\), so

\[\int_4^7\frac1u\,du=[\ln u]_4^7=\ln\frac74.\]

Equally, one may return to \(x\) first and calculate \([\ln(x^2+3)]_1^2=\ln7-\ln4\). The unsafe hybrid is to use \(u\) but retain the old bounds \(1\) and \(2\); those numbers name \(x\)-positions, not \(u\)-values.

\(u\)-substitution scales the coordinate axis. Substituting \(u = g(x)\) transforms the integration variable; the differential relation \(du = g'(x) dx\) accounts for the local stretching or compression of coordinates.

12.8 A reliable integral-solving routine

Before calculating, identify the quantity being accumulated and attach units. Then inspect the integrand: a sum invites term-by-term integration; a composite expression with its inner derivative invites substitution; a product may need integration by parts in Chapter 13. For a definite integral, sketch signs, zeros, and any discontinuities. Finally differentiate an indefinite answer or check an exact definite result by a size estimate.

Engineering Example — charge and average current

A current is \(i(t)=4-0.5t\) mA for \(0\le t\le6\) ms. The charge delivered is

\[Q=\int_0^6(4-0.5t)\,dt\ \mathrm{mA\,ms} =\left[4t-0.25t^2\right]_0^6\ \mu\mathrm C =15\ \mu\mathrm C.\]

The average current over the interval is total charge divided by elapsed time:

\[i_{\rm avg}=\frac{Q}{6\ \mathrm{ms}}=2.5\ \mathrm{mA}.\]

This is also the average height of the graph. Since current falls linearly from \(4\) mA to \(1\) mA, the same answer is the mean of the endpoint values. That geometric check is available only because the graph is a straight line.

\(u\)-substitution is the chain rule in reverse. If an integral contains an inner function \(g(x)\) alongside a multiple of its derivative \(g'(x)\), substitution converts it into a simpler integral in terms of \(u\).

Substitution formula

Starting point. By the chain rule, \(\frac{d}{dx}[F(g(x))] = F'(g(x)) g'(x) = f(g(x)) g'(x)\).

  1. \[\int f(g(x)) g'(x) dx = F(g(x)) + C\]
    Integrate both sides with respect to \(x\).reversible
  2. \[\text{Let } u = g(x) \implies du = g'(x) dx\]
    Substitute variable \(u\) and differential \(du\).reversible
  3. \[\int f(u) du = F(u) + C = F(g(x)) + C\]
    Evaluate the simpler integral in \(u\), then substitute back \(u = g(x)\).reversible

Conclusion. \(\int f(g(x)) g'(x) dx = \int f(u) du\). For definite integrals, transform the integration limits: \(\int_a^b f(g(x))g'(x)dx = \int_{g(a)}^{g(b)} f(u)du\).

  1. Identify the 'inner' function \(u = g(x)\) whose derivative \(g'(x)\) appears as a multiplying factor.
  2. Calculate the differential: \(du = g'(x) dx \implies dx = \frac{du}{g'(x)}\).
  3. Substitute \(u\) and \(du\), ensuring all instances of \(x\) cancel completely.
  4. For definite integrals, update limits: \(u_1 = g(a)\) and \(u_2 = g(b)\) (no need to back-substitute).
  5. For indefinite integrals, evaluate in \(u\), then back-substitute \(u = g(x)\).

Exercises

  1. Interpret \(\int_0^4 v(t)\,dt\) when \(v\) is a velocity.
  2. Evaluate \(\int_0^2(3x^2-4x+1)\,dx\).
  3. Find \(\int(5x-1)^7\,dx\) by substitution.
  4. Explain why \(\int_{-1}^1x\,dx=0\) does not mean the geometric area is zero.
  5. A current is \(i(t)=I_0e^{-t/\tau}\). Find total charge delivered from \(0\) to \(T\).
  6. Use a right Riemann sum to derive \(\int_0^1x\,dx=1/2\).
  7. Let \(A(x)=\int_0^x\sin t\,dt\). Find \(A'(x)\) and \(A(\pi)\).
  8. Evaluate \(\int_0^1 3x^2/(1+x^3)\,dx\), showing the changed limits.
  9. A velocity is \(v(t)=t^2-4t+3\) for \(0\le t\le3\). Find displacement and total distance; explain why they differ.

Part IV · Series, Exponentials, and Integral Calculus

13

Chapter 13 — Integration Techniques and Geometric Applications

Open chapter view →

Why This Matters

Once an integral represents an accumulation, the remaining challenge is choosing a method that exposes an antiderivative or a geometry. Substitution handles a hidden chain rule; integration by parts handles a product. Applications require careful intervals and diagrams, not formula substitution alone.

13.1 Integration by parts

The product rule says \((uv)'=u'v+uv'\). Rearranging and integrating gives

\[\int u\,dv=uv-\int v\,du.\]

Choose \(u\) so differentiation simplifies it, and choose \(dv\) so integration is available.

Example — polynomial times exponential

\[\int xe^x\,dx.\]

Take \(u=x\), \(dv=e^x\,dx\). Then \(du=dx\), \(v=e^x\), so

\[\int xe^x\,dx=xe^x-\int e^x\,dx=e^x(x-1)+C.\]

Differentiate the final answer to check it.

Integration by parts is the product rule read backwards. Differentiate a product, rearrange the two derivative terms, and integrate. The useful choice is the one that leaves an integral simpler than the one you started with; the formula itself does not decide that choice.

\[\int u\,dv=uv-\int v\,du\]
  1. Choose u so differentiating it simplifies it.
  2. Choose dv so integrating it is direct.
  3. Write du and v explicitly before substituting into the formula.

13.2 Area between curves

If \(f(x)\ge g(x)\) on \([a,b]\), area between the curves is

\[\int_a^b[f(x)-g(x)]\,dx.\]

If the curves cross, first solve \(f(x)=g(x)\) and split the interval at each intersection. Otherwise positive and negative pieces cancel and yield signed, not physical, area.

Why this works. A thin vertical strip has height equal to top curve minus bottom curve. Adding strips across the interval gives area only after the intersection points establish where that ordering is true.

\[A=\int_a^b(\text{top}-\text{bottom})\,dx\]

13.3 Arc length and volumes

For a differentiable curve \(y=f(x)\), approximate a small arc by a straight segment:

\[ds=\sqrt{dx^2+dy^2}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.\]

Thus

\[L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx.\]

When a region is rotated about an axis, thin disks or washers produce volume. For rotation about the \(x\)-axis with outer radius \(R(x)\) and inner radius \(r(x)\),

\[V=\pi\int_a^b\left[R(x)^2-r(x)^2\right]\,dx.\]

Draw the region and a representative slice before choosing bounds.

Length begins with local straightness. Over a tiny horizontal change, a smooth curve is nearly its tangent. The Pythagorean length of that tiny segment combines horizontal and vertical changes; adding the segments and taking the limiting refinement produces the arc-length integral.

13.4 Improper integrals

An integral over an infinite interval or with an unbounded integrand is defined by a limit. For example,

\[\int_1^\infty\frac1{x^p}\,dx =\lim_{b\to\infty}\int_1^b\frac1{x^p}\,dx.\]

It converges precisely when \(p>1\). The infinity symbol is a limiting process, not a permitted endpoint at which an antiderivative is directly evaluated.

Circle formulas as an integration check

The upper unit semicircle is \(y=\sqrt{1-x^2}\) for \(-1\le x\le1\). Its derivative is \(y'=-x/\sqrt{1-x^2}\), so the arc-length integrand becomes

\[\sqrt{1+(y')^2}=\frac1{\sqrt{1-x^2}}.\]

The endpoints are unbounded, so the semicircle length is an improper integral, not an ordinary substitution at \(x=\pm1\):

\[\pi=\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}} =2\int_0^1\frac{dx}{\sqrt{1-x^2}}.\]

Scaling every length by a radius \(R\) gives circumference \(2\pi R\). Likewise, the unit-circle area is four equal quarter-circle areas,

\[\pi=4\int_0^1\sqrt{1-x^2}\,dx,\]

so a radius-\(R\) circle has area \(\pi R^2\). These formulas are familiar, but their derivation is a useful lesson in distinguishing a convergent endpoint singularity from an invalid direct evaluation.

Improper integrals are evaluated strictly as limits. Infinite bounds or unbounded integrands cannot be evaluated by direct substitution — a limit must be taken.

Improper integral of 1/x^2 on [1, infinity)

Starting point. Evaluate \(\int_1^\infty \frac{1}{x^2} dx\).

  1. \[\int_1^\infty \frac{dx}{x^2} = \lim_{b\to\infty} \int_1^b x^{-2} dx\]
    Replace the infinite upper bound with a finite parameter \(b\) and take a limit.reversible
  2. \[\int_1^b x^{-2} dx = \left[ -\frac{1}{x} \right]_1^b = -\frac{1}{b} - (-1) = 1 - \frac{1}{b}\]
    Evaluate the definite integral using the Fundamental Theorem.reversible
  3. \[\lim_{b\to\infty} \left(1 - \frac{1}{b}\right) = 1 - 0 = 1\]
    Take the limit as \(b \to \infty\).reversible

Conclusion. \(\int_1^\infty \frac{dx}{x^2} = 1\) (converges). In contrast, \(\int_1^\infty \frac{dx}{x} = \lim_{b\to\infty} \ln b = \infty\) (diverges).

  1. Identify the source of impropriety: infinite bound (Type I) or vertical asymptote / discontinuity (Type II).
  2. Replace each improper point with a limit variable (e.g. \(\lim_{b\to\infty} \int_a^b\) or \(\lim_{c\to 0^+} \int_c^b\)).
  3. Evaluate the definite integral in terms of the parameter.
  4. Take the limit: if the limit is a finite real number, the integral <strong>converges</strong>; if the limit is \(\pm\infty\) or does not exist, it <strong>diverges</strong>.

13.5 Choosing integration by parts deliberately

Integration by parts transfers a derivative from one factor to the other. It is most useful when that transfer makes the remaining integral simpler. A practical preference order for choosing \(u\) is often: logarithmic functions, inverse trigonometric functions, algebraic polynomials, trigonometric functions, then exponentials. This is a guide, not a theorem. The decisive question is whether \(\int v\,du\) is easier than the original integral.

Example — a logarithm needs an implicit second factor

Find

\[\int\ln x\,dx \qquad (x>0).\]

There is no visible product, but write \(\ln x\cdot1\). Choose

\[u=\ln x,\quad dv=dx, \qquad du=\frac{dx}{x},\quad v=x.\]

Then

\[\int\ln x\,dx=x\ln x-\int x\frac{dx}{x} =x\ln x-\int1\,dx =x\ln x-x+C.\]

The restriction \(x>0\) comes from the real logarithm. On a negative interval one would instead work with \(\ln|x|\).

Example — repeated integration by parts

Evaluate \(\int e^x\sin x\,dx\). Let this integral be \(I\). First choose \(u=\sin x\), \(dv=e^x\,dx\):

\[I=e^x\sin x-\int e^x\cos x\,dx.\]

Call the remaining integral \(J\). Applying integration by parts to \(J\) with \(u=\cos x\) gives

\[J=e^x\cos x+\int e^x\sin x\,dx=e^x\cos x+I.\]

Substitute this into the first equation:

\[I=e^x\sin x-e^x\cos x-I.\]

Therefore \(2I=e^x(\sin x-\cos x)\) and

\[\int e^x\sin x\,dx=\frac{e^x}{2}(\sin x-\cos x)+C.\]

The reappearance of the original integral is not failure; it creates a linear equation for that integral.

Strategic choice of \(u\) and \(dv\). The goal of integration by parts is to make \(\int v\,du\) strictly simpler than \(\int u\,dv\).

  1. <strong>Polynomial \(\times\) Exponential / Trig:</strong> set \(u = \text{polynomial}\) (so \(du\) reduces in degree) and \(dv = e^{ax} dx\) or \(\cos(ax) dx\).
  2. <strong>Logarithmic or Inverse Trig factors:</strong> set \(u = \ln x\) or \(u = \arctan x\) (so \(du\) becomes an algebraic fraction) and \(dv = dx\) or \(x^n dx\).
  3. <strong>Repeated parts:</strong> for \(x^2 e^x\), apply parts twice to eliminate powers of \(x\).
  4. <strong>Cyclic parts:</strong> for \(e^{ax}\cos(bx)\), applying parts twice reproduces the original integral; solve for the integral algebraically.

Why LIATE works

From
  • Functions higher in LIATE (L, I) simplify dramatically when differentiated (logs become rational; inverse trig becomes algebraic).
  • Functions lower in LIATE (T, E) do not become more complicated when integrated (\(e^{ax}\) and \(\sin(ax)\) retain their form).
  1. Differentiating \(u = \ln x\) gives \(du = \frac{1}{x} dx\) (simplifies).
  2. Integrating \(dv = x^2 dx\) gives \(v = \frac{x^3}{3}\) (manageable).
  3. The product \(v\,du = \frac{x^2}{3} dx\) is a simple monomial.

Result. LIATE maximizes algebraic simplification in the resulting integral \(\int v\,du\).

13.6 From a drawing to an area or volume integral

An application integral begins with geometry, not an antiderivative. Mark all intersections, choose vertical or horizontal slices, label the slice dimensions, and only then write the integral. For area with vertical slices, the height is “top minus bottom.” For a washer about the \(x\)-axis, the cross-sectional area is \(\pi(R^2-r^2)\), not \(\pi(R-r)^2\).

slice labelled top \(x\), bottom \(x^2\), height \(x-x^2\). A companion washer diagram labels outer radius \(R(x)\) and inner radius \(r(x)\).]

Example — area after finding the intersections

Find the area enclosed by \(y=2x\) and \(y=x^2\). Solve

\[x^2=2x\quad\Longrightarrow\quad x(x-2)=0,\]

so the intersections are at \(x=0\) and \(x=2\). A test value such as \(x=1\) gives \(2x=2\) and \(x^2=1\), so the line is above the parabola. Hence

\[A=\int_0^2(2x-x^2)\,dx =\left[x^2-\frac{x^3}{3}\right]_0^2 =4-\frac83=\frac43.\]

The limits came from intersections, not from an arbitrary visible portion of a sketch.

Example — a disk volume

Rotate the region under \(y=\sqrt{x}\) from \(x=0\) to \(x=4\) about the \(x\)-axis. At position \(x\), the cross-section is a disk with radius \(\sqrt{x}\), so its area is \(\pi(\sqrt{x})^2=\pi x\). Therefore

\[V=\pi\int_0^4x\,dx =\pi\left[\frac{x^2}{2}\right]_0^4=8\pi.\]

Squaring the radius is why the radical disappears. A volume must have cubic units; the factor \(dx\) supplies the slice thickness.

Integration by parts is the product rule in reverse. It trades an intractable integral \(\int u\,dv\) for a manageable integral \(\int v\,du\).

Derivation of integration by parts

Starting point. By the product rule for differentiation: \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

  1. \[u(x)v(x) = \int u'(x)v(x) dx + \int u(x)v'(x) dx\]
    Integrate both sides with respect to \(x\).reversible
  2. \[u v = \int v\,du + \int u\,dv\]
    Rewrite using differential notation \(du = u'(x)dx\) and \(dv = v'(x)dx\).reversible
  3. \[\int u\,dv = u v - \int v\,du\]
    Rearrange to isolate \(\int u\,dv\).reversible

Conclusion. \(\int u\,dv = u v - \int v\,du\). For definite integrals: \(\int_a^b u\,dv = [u v]_a^b - \int_a^b v\,du\).

  1. Choose \(u\) and \(dv\) using the <strong>LIATE</strong> priority rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
  2. Differentiate \(u\) to get \(du = u' dx\); integrate \(dv\) to get \(v = \int dv\) (choose \(C = 0\)).
  3. Apply formula: \(\int u\,dv = u v - \int v\,du\).
  4. Check that the new integral \(\int v\,du\) is simpler than the original.

13.7 Arc length: local straightness, global curvature

Over a very short horizontal displacement \(dx\), a smooth curve changes vertically by approximately \(dy=f'(x)dx\). Pythagoras gives the short length

\[ds\approx\sqrt{(dx)^2+(dy)^2} =\sqrt{1+[f'(x)]^2}\,dx.\]

Adding these short lengths and taking a limit gives the arc-length formula from Section 13.3. The formula is not “height plus width”: a curved path is the hypotenuse of many infinitesimal right triangles.

Example — arc length of a straight line verifies the formula

For \(y=3x\) on \([0,2]\), \(f'(x)=3\). Thus

\[L=\int_0^2\sqrt{1+9}\,dx=2\sqrt{10}.\]

This agrees with distance between endpoints \((0,0)\) and \((2,6)\): \(\sqrt{2^2+6^2}=2\sqrt{10}\). A general formula earns trust when it returns a known geometric result in a simple case.

Arc length sums infinitesimal Pythagorean hypotenuses. A smooth curve is locally straight; the differential element of arc length is \(ds = \sqrt{dx^2 + dy^2}\).

Arc length formula derivation

Starting point. Consider a curve segment with endpoints \((x, y)\) and \((x+dx, y+dy)\).

  1. \[(ds)^2 = (dx)^2 + (dy)^2\]
    Apply the Pythagorean theorem to an infinitesimal step.reversible
  2. \[ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx\]
    Factor out \(dx = \sqrt{(dx)^2}\) from under the radical.reversible
  3. \[L = \int_a^b ds = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx\]
    Integrate all infinitesimal arc length elements from \(x = a\) to \(x = b\).reversible

Conclusion. The arc length of \(y = f(x)\) on \([a, b]\) is \(L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx\). Because \(\sqrt{1 + (f')^2} \ge 1\), \(L \ge b-a\) (the straight-line minimum).

13.8 Improper integrals and convergence checks

When an endpoint is infinite, replace it by a finite variable before integrating. For \(p\ne1\),

\[\int_1^b x^{-p}\,dx=\frac{b^{1-p}-1}{1-p}.\]

If \(p>1\), then \(1-p<0\) and \(b^{1-p}\to0\), so the limiting integral is \(1/(p-1)\). If \(p<1\), the power grows without bound. At \(p=1\) the integral is \(\ln b\), which also grows without bound. This establishes the threshold \(p>1\) rather than merely memorising it.

Example — a finite integral with an infinite endpoint value

Consider

\[\int_0^1\frac{dx}{\sqrt{x}}.\]

The integrand is unbounded at zero, so write

\[\lim_{a\to0^+}\int_a^1x^{-1/2}\,dx =\lim_{a\to0^+}\left[2\sqrt{x}\right]_a^1 =2.\]

An unbounded height does not by itself imply infinite accumulated area. The limit decides; directly inserting \(x=0\) into \(1/\sqrt{x}\) would not.

Area between curves requires finding intersections first. Area is always non-negative: \(\int (\text{upper} - \text{lower}) dx\).

Area between y = x and y = x^2

Starting point. Find the area bounded by \(y_1 = x\) and \(y_2 = x^2\).

  1. \[x = x^2 \implies x(x-1) = 0 \implies x = 0, \quad x = 1\]
    Find intersection points to determine integration limits \([0, 1]\).reversible
  2. \[\text{On } [0, 1]: \quad x \ge x^2 \implies \text{upper} = x, \quad \text{lower} = x^2\]
    Test a point in \((0, 1)\) (e.g. \(x = 0.5\): \(0.5 > 0.25\)) to confirm curve ordering.reversible
  3. \[A = \int_0^1 (x - x^2) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]
    Integrate \(\text{upper} - \text{lower}\) over \([0, 1]\).reversible

Conclusion. The bounded area is exactly \(\frac{1}{6}\). If curves cross multiple times, split the integral at each intersection so the integrand is always \(|y_1 - y_2|\).

  1. Set \(f(x) = g(x)\) and solve for all intersection points \(x = a, b, c, \dots\)
  2. Determine which function is on top on each subinterval \([a, b]\) by testing a sample point.
  3. Set up \(\int_a^b (f(x) - g(x)) dx\) where \(f \ge g\).
  4. Evaluate and sum the areas of all enclosed regions.

Common Mistakes

  • Choosing \(u\) and \(dv\) for integration by parts without checking that the new integral is simpler.
  • Mixing \(u\)-limits with an antiderivative returned to \(x\).
  • Subtracting curves in the wrong order or failing to split at crossings.
  • Treating an improper integral as ordinary finite-bound evaluation.

Exercises

  1. Evaluate \(\int x\cos x\,dx\).
  2. Find the area between \(y=x\) and \(y=x^2\) on \([0,1]\).
  3. Find the volume generated by rotating \(y=x\) on \([0,2]\) about the \(x\)-axis.
  4. Determine whether \(\int_1^\infty1/x^{3/2}\,dx\) converges.
  5. Derive the circumference of a circle from a parametric or geometric argument and explain why radians are natural for arc length.
  6. Evaluate \(\int x\ln x\,dx\) by integration by parts.
  7. Show all steps in evaluating \(\int e^x\cos x\,dx\).
  8. Find the area enclosed by \(y=4-x^2\) and the \(x\)-axis.
  9. Find the volume obtained by rotating the region under \(y=2x\) on \([0,1]\) about the \(x\)-axis.
  10. Determine whether \(\int_0^1x^{-2/3}\,dx\) converges and evaluate it if it does.

Part V

Part V — Complex Numbers, Trigonometry, and Advanced Functions

Part V · Complex Numbers, Trigonometry, and Advanced Functions

14

Chapter 14 — Complex Numbers and Polar Representation

Open chapter view →

Why This Matters

The equation \(x^2+1=0\) has no real solution, yet such equations arise naturally when we analyse oscillation. Complex numbers enlarge the number system in a controlled way. Their geometry also provides the cleanest language for amplitude and phase in sinusoidal steady-state analysis.

Complex numbers turn AC circuit trigonometry into algebra. In AC circuits, voltages and currents oscillate at frequency \(\omega\). Phasors represent these sinusoidal signals as complex numbers \(V = |V| e^{i\phi}\), turning differential equations into simple Ohm's-law-like algebraic relations.

Algebraic rules of the imaginary unit

Starting point. Define \(i\) by \(i^2 = -1\) (or \(j^2 = -1\) in electronics notation).

  1. \[i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1\]
    Powers of \(i\) form a 4-cycle representing repeated \(90^\circ\) rotations in the complex plane.reversible
  2. \[(a + bi)(c + di) = (ac - bd) + (ad + bc)i\]
    Multiply algebraically and use \(i^2 = -1\) to group real and imaginary parts.reversible
  3. \[\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}\]
    Multiply numerator and denominator by the complex conjugate \(c - di\) to make the denominator real.reversible

Conclusion. The complex plane extends the real numbers into a 2D field. Every non-zero complex number has an exact multiplicative inverse: \(\frac{1}{z} = \frac{\bar{z}}{|z|^2}\).

  1. To add/subtract: \((a+bi) \pm (c+di) = (a\pm c) + (b\pm d)i\) (componentwise).
  2. To multiply: expand with FOIL and substitute \(i^2 = -1\).
  3. To divide: multiply top and bottom by the complex conjugate \(\bar{z} = c - di\).
  4. Modulus: \(|a+bi| = \sqrt{a^2 + b^2}\). Conjugate: \(\overline{a+bi} = a - bi\).

14.1 The complex number system

Define \(i\) by \(i^2=-1\). A complex number is

\[z=x+iy,\]

where \(x\) and \(y\) are real. The real and imaginary parts are \(\operatorname{Re}z=x\) and \(\operatorname{Im}z=y\). Two complex numbers are equal only when both corresponding parts are equal.

Addition and multiplication follow algebra, with \(i^2=-1\):

\[(a+ib)(c+id)=(ac-bd)+i(ad+bc).\]

The conjugate of \(z=x+iy\) is \(\bar z=x-iy\). Multiplying conjugates removes the imaginary cross terms:

\[z\bar z=x^2+y^2.\]

This motivates the modulus

\[|z|=\sqrt{z\bar z}=\sqrt{x^2+y^2}.\]

For \(z\ne0\),

\[\frac1z=\frac{\bar z}{|z|^2}.\]

Example — division by a complex number

\[\frac{3+4i}{1-2i}=\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)} =\frac{-5+10i}{5}=-1+2i.\]

The conjugate is chosen because it makes the denominator real and positive.

Complex arithmetic is pair arithmetic with one new rule. Write a complex number as an ordered pair of real and imaginary coordinates, then impose i²=-1. Addition combines coordinates; multiplication follows expansion and collects real and imaginary parts. The plane picture is a consequence of this algebra, not an unrelated diagram.

14.2 Complex plane, arguments, and radians

Plot \(z=x+iy\) as the point \((x,y)\) in a plane: horizontal coordinate is real part, vertical coordinate is imaginary part. Its distance from the origin is \(|z|\). An angle from the positive real axis to the point is an argument.

Arguments are not unique: if \(\theta\) is an argument, so is \(\theta+2\pi k\) for every integer \(k\). A chosen principal argument \(\operatorname{Arg}z\) uses one stated interval, commonly \((-\pi,\pi]\).

Radians are defined by arc length divided by radius. On a unit circle, an angle \(\theta\) has arc length \(\theta\). This is why derivative and power-series formulas for sine and cosine use radians rather than degrees.

Cartesian versus polar representation. A complex number \(z = a + bi\) specifies coordinates \((a, b)\) on the real and imaginary axes; polar form \(z = r e^{i\theta}\) specifies distance \(r = |z|\) from the origin and counterclockwise angle \(\theta = \operatorname{Arg}(z)\) from the positive real axis.

Converting between Cartesian and polar forms

Starting point. Let \(z = x + iy\) with \(x, y \in \mathbb{R}\).

  1. \[r = |z| = \sqrt{x^2 + y^2}\]
    The modulus \(r\) is the Euclidean distance from \((0, 0)\) to \((x, y)\).reversible
  2. \[x = r \cos\theta, \quad y = r \sin\theta\]
    Project polar vector \((r, \theta)\) onto the real and imaginary axes.reversible
  3. \[z = r(\cos\theta + i\sin\theta) = r e^{i\theta}\]
    Combine using Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\).reversible

Conclusion. Cartesian form \(x + iy\) is ideal for addition and subtraction. Polar form \(r e^{i\theta}\) is ideal for multiplication, division, and powers.

  1. Compute modulus: \(r = \sqrt{x^2 + y^2}\).
  2. Compute reference angle: \(\alpha = \arctan(|y|/|x|)\).
  3. Determine quadrant and set \(\theta = \operatorname{Arg}(z) \in (-\pi, \pi]\):
  4. Quadrant I: \(\theta = \alpha\); Quadrant II: \(\theta = \pi - \alpha\); Quadrant III: \(\theta = -\pi + \alpha\); Quadrant IV: \(\theta = -\alpha\).

14.3 Polar coordinates: converting points and checking quadrants

The same geometry describes an ordinary Cartesian point \((x,y)\) and a complex number \(z=x+iy\). Its polar coordinates are a radius \(r\ge0\) and an angle \(\theta\) measured from the positive real axis:

\[x=r\cos\theta,\qquad y=r\sin\theta,\qquad r=\sqrt{x^2+y^2}.\]

For \(r>0\), the two equations determine the quadrant as well as the reference angle. A calculator command based only on \(\arctan(y/x)\) can be misleading, because the same ratio occurs in opposite quadrants and because it is undefined when \(x=0\). Use a quadrant-aware angle function such as \(\operatorname{atan2}(y,x)\) when available, or use the signs of both \(x\) and \(y\) to select the correct argument.

Example — rectangular to polar form

For \(z=-\sqrt3+i\), the modulus is

\[r=\sqrt{(-\sqrt3)^2+1^2}=2.\]

The reference angle is \(\pi/6\), but the point lies in quadrant II. Therefore a principal argument is \(5\pi/6\), and

\[z=2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right).\]

The value \(-\pi/6\) has the same tangent but is in the wrong quadrant, so it is not an argument of this point.

The \(n\) roots of a complex number form a regular polygon. Solving \(z^n = w\) produces exactly \(n\) distinct complex roots evenly distributed on a circle of radius \(|w|^{1/n}\).

Finding the n-th roots

Starting point. Solve \(z^n = w = r e^{i\theta}\). Write \(z = R e^{i\phi}\).

  1. \[R^n e^{in\phi} = r e^{i(\theta + 2\pi k)} \quad (k = 0, 1, \dots, n-1)\]
    Account for \(2\pi k\) periodicity of angles in the complex plane.reversible
  2. \[R = r^{1/n}, \quad \phi_k = \frac{\theta + 2\pi k}{n}\]
    Equate moduli and divide angles by \(n\).reversible
  3. \[z_k = r^{1/n} e^{i(\theta + 2\pi k)/n} = r^{1/n} \left( \cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n} \right)\]
    List the \(n\) distinct solutions for \(k = 0, 1, \dots, n-1\).reversible

Conclusion. The \(n\) roots have equal modulus \(r^{1/n}\) and are spaced at equal angle intervals \(\Delta\phi = \frac{2\pi}{n}\), forming the vertices of a regular \(n\)-gon centred at the origin.

  1. Express \(w\) in polar form: \(w = r e^{i\theta}\).
  2. Modulus of roots: \(R = r^{1/n}\).
  3. Angles of roots: \(\phi_k = \frac{\theta + 2\pi k}{n}\) for \(k = 0, 1, 2, \dots, n-1\).
  4. Convert each \(z_k = R(\cos\phi_k + i\sin\phi_k)\) back to Cartesian form if required.

14.4 Polar form and multiplication

Every nonzero complex number can be written as

\[z=r(\cos\theta+i\sin\theta),\qquad r=|z|\ge0.\]

Later Euler’s formula writes this more compactly as \(z=re^{i\theta}\). Multiplying polar forms multiplies magnitudes and adds arguments:

\[r_1e^{i\theta_1}r_2e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}.\]

Geometrically, multiplication performs a scale and a rotation. Division divides magnitudes and subtracts angles.

Geometric picture. In polar form, multiplication combines two independent actions: moduli multiply, so lengths scale; arguments add, so directions rotate. This is why polar multiplication is especially natural for phasors.

\[r_1e^{i\theta_1}r_2e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}\]

14.5 Arithmetic, conjugates, and geometric checks

Complex arithmetic is ordinary polynomial algebra with the one replacement \(i^2=-1\). Keeping real and imaginary parts collected is important: a complex number is zero only if both parts are zero. The conjugate changes the sign of the imaginary coordinate, so it is a reflection in the real axis. Its product with the original number is a squared distance:

\[(x+iy)(x-iy)=x^2+y^2=|z|^2.\]

That fact explains division rather than supplying a trick to memorise. A complex denominator is made real by multiplying numerator and denominator by its conjugate; this multiplies the fraction by \(1\) and changes no value.

Example — multiplication with an error check

Compute \((2-3i)(4+i)\):

\[\begin{aligned} (2-3i)(4+i) &=8+2i-12i-3i^2\\ &=8-10i+3\\ &=11-10i. \end{aligned}\]

The final \(+3\) comes from \(-3i^2=-3(-1)\); losing this sign is one of the most common complex-number errors. As a quick magnitude check,

\[|2-3i|\,|4+i|=\sqrt{13}\sqrt{17}=\sqrt{221},\]

while \(|11-10i|=\sqrt{121+100}=\sqrt{221}\), as required.

Example — division from first principles

Write \((5-i)/(2+3i)\) in rectangular form:

\[\frac{5-i}{2+3i} =\frac{(5-i)(2-3i)}{(2+3i)(2-3i)} =\frac{10-15i-2i+3i^2}{4+9}.\]

Since \(3i^2=-3\), this becomes

\[\frac{7-17i}{13}=\frac7{13}-\frac{17}{13}i.\]

The denominator is positive and real. If an \(i\) remains below the fraction bar, the rationalisation step is incomplete.

Multiplication scales moduli and adds arguments. In polar coordinates, complex multiplication decomposes into two independent geometric operations: radial scaling and angular rotation.

Product of two polar complex numbers

Starting point. Let \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\).

  1. \[z_1 z_2 = (r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = (r_1 r_2) e^{i(\theta_1 + \theta_2)}\]
    Use exponent addition rule \(e^a e^b = e^{a+b}\).reversible
  2. \[|z_1 z_2| = |z_1| \cdot |z_2| = r_1 r_2\]
    The length of the product is the product of the lengths.reversible
  3. \[\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)\]
    The angle of the product is the sum of the angles.reversible

Conclusion. \(z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}\) and \(\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}\). Moduli multiply/divide; angles add/subtract.

14.6 Polar form is scale plus rotation

In rectangular coordinates, multiplication requires four products. In polar coordinates, a number has length \(r\) and direction \(\theta\). Multiplying by \(r_2e^{i\theta_2}\) scales every distance by \(r_2\) and rotates every direction by \(\theta_2\). This gives the rules

\[|z_1z_2|=|z_1||z_2|, \qquad \arg(z_1z_2)=\arg z_1+\arg z_2\pmod {2\pi}.\]

For division, \(z_2\) must be nonzero:

\[\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}, \qquad \arg\left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2\pmod {2\pi}.\]

Example — powers without repeated rectangular multiplication

Let \(z=\sqrt3+i\). It has modulus \(2\) and argument \(\pi/6\), so

\[z^5=2^5e^{i5\pi/6} =32\left(-\frac{\sqrt3}{2}+\frac{i}{2}\right) =-16\sqrt3+16i.\]

This is De Moivre's formula,

\[\left[r(\cos\theta+i\sin\theta)\right]^n =r^n(\cos n\theta+i\sin n\theta),\]

for integer \(n\). It follows directly by multiplying magnitudes and adding arguments repeatedly. It is especially efficient for powers and roots, not usually for addition: addition is still simpler in rectangular form.

De Moivre's Theorem powers complex numbers. Repeated multiplication in polar form leads to an elegant formula for integer powers and roots.

De Moivre's Theorem

Starting point. Let \(z = r(\cos\theta + i\sin\theta) = r e^{i\theta}\), and let \(n\) be an integer.

  1. \[z^n = (r e^{i\theta})^n = r^n e^{in\theta}\]
    Apply power rule to the exponential form.reversible
  2. \[z^n = r^n(\cos(n\theta) + i\sin(n\theta))\]
    Expand back into trigonometric components using Euler's formula.reversible

Conclusion. \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))\). Computing \(z^n\) in polar form takes \(O(1)\) arithmetic, avoiding tedious binomial expansion.

14.7 Roots and the multivalued argument

To solve \(w^n=z=re^{i\theta}\) with \(r>0\), the argument of \(z\) may be \(\theta+2\pi k\) for any integer \(k\). Dividing by \(n\) gives the \(n\) distinct roots

\[w_k=r^{1/n}e^{i(\theta+2\pi k)/n}, \qquad k=0,1,\ldots,n-1.\]

Example — cube roots have three distinct directions

Solve \(w^3=-8\). Write \(-8=8e^{i(\pi+2\pi k)}\). Its cube roots have modulus \(2\) and arguments

\[\frac{\pi}{3},\qquad \pi,\qquad \frac{5\pi}{3}.\]

Thus the roots are \(1+\sqrt3i\), \(-2\), and \(1-\sqrt3i\). They form an equilateral triangle centered at the origin. Reporting only the principal root would miss two algebraic solutions.

Why roots are evenly spaced. Taking an nth root divides an argument by n, but the original angle may differ by any whole turn. Those possible whole turns become angle steps of 2π/n, placing every root on one circle at equal angular intervals.

\[z_k=r^{1/n}e^{i(\theta+2\pi k)/n},\quad k=0,\ldots,n-1\]

Common Mistakes

  • Using \(\tan^{-1}(y/x)\) without correcting its quadrant.
  • Treating \(\operatorname{Arg}z\) as defined at \(z=0\); the zero vector has no direction.
  • Writing \(\sqrt{x^2+y^2}=x+y\) or omitting the nonnegative square root.
  • Adding polar magnitudes and arguments. The simple polar rules are for multiplication and division, not addition.

Exercises

  1. Compute \((2-3i)(4+i)\).
  2. Write \((1+i)/(1-i)\) in \(a+ib\) form.
  3. Find modulus and a principal argument of \(-1+\sqrt3i\).
  4. Convert \(4(\cos\frac\pi6+i\sin\frac\pi6)\) to rectangular form.
  5. Explain the geometric effect of multiplying a complex number by \(i\).
  6. Compute \((3+2i)(1-4i)\) and check the result using moduli.
  7. Write \((4+3i)/(1-i)\) in rectangular form.
  8. Express \((1-i)^8\) using polar form.
  9. Find all cube roots of \(8i\), giving both polar and rectangular forms.

Part V · Complex Numbers, Trigonometry, and Advanced Functions

15

Chapter 15 — Trigonometric, Sinusoidal, and Complex Exponential Functions

Open chapter view →

Why This Matters

Sinusoids model periodic signals. Their amplitude, frequency, phase, and rate of change all have geometric meaning on the unit circle. Complex exponentials turn trigonometric identities and phase shifts into ordinary algebra.

15.1 Sine and cosine from the unit circle

On the unit circle, the point at radian angle \(\theta\) has coordinates

\[(\cos\theta,\sin\theta).\]

Therefore

\[\sin^2\theta+\cos^2\theta=1.\]

Both functions are \(2\pi\)-periodic. Sine is odd, \(\sin(-\theta)=-\sin\theta\); cosine is even, \(\cos(-\theta)=\cos\theta\).

The signal

\[v(t)=V_0+A\cos(\omega t+\phi)\]

has offset \(V_0\), amplitude \(|A|\), angular frequency \(\omega\) radians per second, period \(T=2\pi/\omega\) when \(\omega>0\), and phase \(\phi\). A phase shift moves a wave horizontally; it is not a change in amplitude.

Coordinates, not triangle mnemonics. A point rotating once around the unit circle has horizontal coordinate cos θ and vertical coordinate sin θ. Signs, periodicity, and the identity sin²θ+cos²θ=1 follow immediately from the circle equation, including in quadrants where a right-triangle picture is awkward.

\[(\cos\theta)^2+(\sin\theta)^2=1\]

15.3 Derivatives and series

With radian measure,

\[\frac d{dx}\sin x=\cos x,\qquad \frac d{dx}\cos x=-\sin x.\]

Repeated differentiation cycles every four derivatives. Taylor’s theorem then gives

\[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,\]
\[\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots.\]

These series converge for every real \(x\). In particular, \(\lim_{x\to0}\sin x/x=1\), a central small-angle fact.

The expression \(\operatorname{sinc}x=\sin x/x\) is therefore assigned the continuous value \(\operatorname{sinc}(0)=1\). This normalized sinc function is common in signal processing; its value at zero comes from a limit, not division by zero.

Trigonometric functions define circular motion. On the unit circle \(x^2 + y^2 = 1\), a point at angle \(\theta\) has coordinates \((\cos\theta, \sin\theta)\).

Pythagorean identity and angle-sum formulas

Starting point. Let \((x, y) = (\cos\theta, \sin\theta)\) lie on the unit circle \(x^2 + y^2 = 1\).

  1. \[\cos^2\theta + \sin^2\theta = 1\]
    Substitute coordinates directly into the circle equation \(x^2 + y^2 = 1\).reversible
  2. \[1 + \tan^2\theta = \sec^2\theta\]
    Divide by \(\cos^2\theta\) (for \(\cos\theta \neq 0\)).reversible
  3. \[\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\]
    Double-angle formula from the angle-addition identity \(\cos(\alpha + \beta)\) with \(\alpha = \beta = \theta\).reversible
  4. \[\sin(2\theta) = 2\sin\theta\cos\theta\]
    Double-angle formula for sine.reversible

Conclusion. Trigonometric identities are geometric facts about circular projection. In electronics, double-angle formulas explain frequency doubling in non-linear mixers.

  1. Power-reduction: \(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\), \(\sin^2\theta = \frac{1 - \cos(2\theta)}{2}\) (essential for AC power integrals).
  2. Product-to-sum: \(\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]\) (essential for amplitude modulation).
  3. Derivatives: \((\sin x)' = \cos x\), \((\cos x)' = -\sin x\), \((\tan x)' = \sec^2 x\).

15.4 Euler’s formula

Insert \(ix\) into the power series for \(e^x\). Even powers of \(i\) alternate between \(1\) and \(-1\); odd powers provide a factor \(i\). Grouping real and imaginary terms gives

\[e^{i\theta}=\cos\theta+i\sin\theta.\]

This is Euler’s formula. It does not replace real sine and cosine with “imaginary tricks”; it packages their linked behaviour into one expression.

Adding angles now follows from multiplication:

\[e^{i(a+b)}=e^{ia}e^{ib}.\]

Use Euler’s formula on both sides:

\[\cos(a+b)+i\sin(a+b) =(\cos a+i\sin a)(\cos b+i\sin b).\]

Multiplication gives a real part and an imaginary part,

\[(\cos a\cos b-\sin a\sin b) +i(\sin a\cos b+\cos a\sin b).\]

Equality of complex numbers requires equality of the corresponding parts, so

\[\cos(a+b)=\cos a\cos b-\sin a\sin b,\]
\[\sin(a+b)=\sin a\cos b+\cos a\sin b.\]

Putting \(b=-\pi/2\) shows \(\cos(x-\pi/2)=\sin x\): sine and cosine are the same shape with a quarter-cycle phase shift. This derivation is often more reliable than memorising several separate identities.

Engineering Connection — phasor intuition

The real part of \(Ae^{i(\omega t+\phi)}\) is \(A\cos(\omega t+\phi)\). For linear steady-state sinusoidal calculations, the common time factor can be handled with complex amplitudes, then a real physical voltage or current is recovered at the end. The method has conditions; it is not a replacement for transient analysis.

One compact expression carries two coordinates. Euler's formula places cosine in the real coordinate and sine in the imaginary coordinate of a unit complex rotation. Multiplying such expressions therefore adds angles while multiplying magnitudes, which links circular motion, phasors, and trigonometric identities.

\[e^{i\theta}=\cos\theta+i\sin\theta\]

15.5 Roots of unity

The \(n\)th roots of unity solve \(z^n=1\). In polar form, their distinct values are

\[z_k=e^{2\pi ik/n},\qquad k=0,1,\ldots,n-1.\]

They are equally spaced around the unit circle. Complex conjugate roots occur in pairs when polynomial coefficients are real; paired factors can combine into real quadratic factors. For example, roots \(e^{i\theta}\) and \(e^{-i\theta}\) produce

\[(z-e^{i\theta})(z-e^{-i\theta}) =z^2-2(\cos\theta)z+1,\]

which has real coefficients. The imaginary components cancel because the roots are conjugates.

Euler's formula unites trigonometry and exponential growth. \(e^{i\theta} = \cos\theta + i\sin\theta\) connects circular rotation in the complex plane to exponential arithmetic.

Derivation of Euler's formula via Taylor series

Starting point. Expand \(e^z\), \(\cos\theta\), and \(\sin\theta\) using their Taylor series at \(0\).

  1. \[e^{i\theta} = \sum_{n=0}^\infty \frac{(i\theta)^n}{n!} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \cdots\]
    Substitute \(z = i\theta\) into the exponential series.reversible
  2. \[e^{i\theta} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)\]
    Group even powers (\(i^2 = -1, i^4 = 1\)) as real part and odd powers (\(i^1 = i, i^3 = -i\)) as imaginary part.reversible
  3. \[e^{i\theta} = \cos\theta + i\sin\theta\]
    Recognise the real series as \(\cos\theta\) and the imaginary series as \(\sin\theta\).reversible

Conclusion. \(e^{i\theta} = \cos\theta + i\sin\theta\). Consequently, \(\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\) and \(\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}\).

Euler's identity: the most beautiful equation

From
  • Euler's formula: \(e^{i\theta} = \cos\theta + i\sin\theta\).
  • Set \(\theta = \pi\).
  1. \(\cos\pi = -1, \quad \sin\pi = 0\).
  2. \(e^{i\pi} = -1 + 0i = -1\).
  3. \(e^{i\pi} + 1 = 0\).

Result. \(e^{i\pi} + 1 = 0\). It links the five fundamental constants of mathematics: \(e, i, \pi, 1, 0\).

15.6 Reading and constructing a sinusoid

A sinusoid is best understood as a moving unit-circle coordinate. In \(A\cos(\omega t+\phi)\), the input angle advances at \(\omega\) radians each second. One complete turn requires \(2\pi\) radians, which explains

\[T=\frac{2\pi}{|\omega|},\qquad f=\frac1T=\frac{|\omega|}{2\pi}.\]

The amplitude is \(|A|\), not necessarily \(A\): a negative coefficient can be absorbed as a phase shift because \(-\cos\theta=\cos(\theta+\pi)\). The horizontal shift is found by setting the internal angle to a landmark value. For example, \(\cos(\omega t+\phi)\) has a maximum when \(\omega t+\phi=2\pi k\).

Example — decode a voltage waveform

For

\[v(t)=2+3\sin(100\pi t-\pi/6)\ \mathrm V,\]

the midline is \(2\) V and the amplitude is \(3\) V. The angular frequency is \(100\pi\) rad/s, hence

\[f=\frac{100\pi}{2\pi}=50\ \mathrm{Hz},\qquad T=0.020\ \mathrm s.\]

The phase \(-\pi/6\) means the sine waveform is delayed relative to \(3\sin(100\pi t)\) by

\[\frac{\pi/6}{100\pi}=\frac1{600}\ \mathrm s.\]

It is incorrect to call \(-\pi/6\) “a delay of \(\pi/6\) seconds”: phase is an angle and becomes a time shift only after division by angular frequency.

From first principles. A sinusoid records one coordinate of uniform circular motion. Amplitude sets the vertical radius, angular frequency sets how quickly the point turns, and phase states where the rotation begins.

\[v(t)=V_0+A\cos(\omega t+\phi)\]

15.7 Identities as geometry and algebra

The Pythagorean identity comes from the unit-circle equation \(x^2+y^2=1\). Dividing it by \(\cos^2x\), where \(\cos x\ne0\), gives

\[1+\tan^2x=\sec^2x.\]

Dividing by \(\sin^2x\), where \(\sin x\ne0\), gives

\[1+\cot^2x=\csc^2x.\]

The restrictions matter because division by a zero coordinate is not allowed. The addition formulas in Section 15.4 generate useful double-angle formulas:

\[\sin2x=2\sin x\cos x,\qquad \cos2x=\cos^2x-\sin^2x.\]

Using \(\sin^2x+\cos^2x=1\), cosine also has the alternatives \(\cos2x=1-2\sin^2x=2\cos^2x-1\). Choose the version that contains the quantity already present in a problem.

Example — simplify without changing the domain

For inputs where \(\cos x\ne0\),

\[\frac{1-\sin^2x}{\cos x} =\frac{\cos^2x}{\cos x} =\cos x.\]

The original expression is undefined at \(\cos x=0\), whereas \(\cos x\) itself is defined there. Thus this is a simplification on the original domain, not an unqualified equality of functions on all real numbers.

Radians make calculus formulas natural. In calculus, all trigonometric derivatives require angles to be measured in radians. The identity \(\lim_{\theta\to0} \frac{\sin\theta}{\theta} = 1\) is true only in radians.

Why radians are the natural angle unit

Starting point. By definition of radian measure on a circle of radius \(r\): arc length \(s = r\theta\).

  1. \[\text{For } r = 1: \quad s = \theta\]
    On the unit circle, angle in radians equals arc length along the circumference.reversible
  2. \[\lim_{\theta\to0} \frac{\sin\theta}{\theta} = 1\]
    The chord length \(2\sin(\theta/2)\) approaches the arc length \(\theta\) as \(\theta \to 0\).reversible
  3. \[\frac{d}{d\theta}\sin\theta = \lim_{h\to0} \frac{\sin(\theta+h) - \sin\theta}{h} = \cos\theta\]
    The standard derivative formula \(\frac{d}{d\theta}\sin\theta = \cos\theta\) holds without extra conversion constants only in radians.reversible

Conclusion. If degrees were used, \(\frac{d}{d\theta}\sin\theta^\circ = \frac{\pi}{180} \cos\theta^\circ\). Radians eliminate the awkward \(\frac{\pi}{180}\) scale factor throughout physics and engineering.

  1. Conversion: \(\pi\text{ radians} = 180^\circ \implies 1\text{ rad} = \frac{180^\circ}{\pi} \approx 57.3^\circ\).
  2. Degrees to radians: multiply by \(\frac{\pi}{180^\circ}\).
  3. Radians to degrees: multiply by \(\frac{180^\circ}{\pi}\).

15.8 Differentiation, rates, and small-angle reasoning

The derivative formula \((\sin x)'=\cos x\) depends on radian measure. If an angle is measured in degrees, an extra constant conversion factor appears. This is why calculus and engineering signal models conventionally use radians.

Example — differentiate an AC waveform

If

\[v(t)=12\cos(500t+\pi/4)\ \mathrm V,\]

then the chain rule gives

\[\frac{dv}{dt}=-12(500)\sin(500t+\pi/4) =-6000\sin(500t+\pi/4)\ \mathrm{V/s}.\]

The factor \(500\) is not optional; it records how rapidly the input angle changes. At a voltage peak, the sine factor is zero, so the graph has horizontal tangent. At a midline crossing, the magnitude of the rate is largest.

Engineering Example — capacitor current leads voltage rate

For an ideal capacitor, \(i_C=C\,dv_C/dt\). If \(v_C(t)=V_m\cos(\omega t)\), then

\[i_C(t)=-\omega CV_m\sin(\omega t) =\omega CV_m\cos(\omega t+\pi/2).\]

The current amplitude is \(\omega CV_m\) and the current is one quarter-cycle ahead of voltage in this cosine convention. This conclusion comes from differentiation; it should not be extended blindly to non-sinusoidal sources or non-ideal components.

Common Mistakes

  • Confusing angular frequency \(\omega\) (rad/s) with ordinary frequency \(f\) (Hz).
  • Applying an identity after division without retaining its domain restriction.
  • Forgetting the chain-rule multiplier when differentiating \(\sin(\omega t+\phi)\).
  • Treating a phase angle as a time interval without using \(\omega\).

Exercises

  1. Find amplitude, period, and phase of \(3\sin(4t-\pi/2)\).
  2. Verify \(\sin(a+b)\) using Euler’s formula.
  3. Find all fourth roots of unity and plot their locations conceptually.
  4. Differentiate \(5\cos(3t+\pi/4)\).
  5. Explain why a phase angle must be stated with a reference convention.
  6. Find \(\frac{d}{dt}\tan(2t)\) and state the values of \(t\) at which the original function is undefined.
  7. A camera is \(10\) m from a road. At an instant \(x=10\) m and \(dx/dt=15\) m/s. Use \(\tan\theta=x/10\) to find \(d\theta/dt\) at that instant.
  8. For \(v(t)=4-2\cos(20t)\) V, identify offset, amplitude, frequency, and range.
  9. Prove \(1+\tan^2x=\sec^2x\) and state where the proof applies.
  10. Differentiate \(7\sin(3t-\pi/5)\) and interpret the maximum possible rate.
  11. A capacitor of \(10\ \mu\)F has \(v(t)=5\cos(1000t)\) V. Find \(i_C(t)\).

Part V · Complex Numbers, Trigonometry, and Advanced Functions

16

Chapter 16 — Rational, Inverse-Trigonometric, and Hyperbolic Functions

Open chapter view →

Why This Matters

Rational expressions appear in transfer-like functions and partial fractions. Inverse trigonometric functions recover angles from ratios. Hyperbolic functions arise naturally from exponentials and describe a geometry parallel to, but different from, circle trigonometry.

16.1 Function characteristics and the sigmoid

When a mathematical model is designed from requirements, begin with its desired characteristics: domain, range, symmetry, monotonicity, smoothness, limiting values, roots, and rate of change. The sigmoid

\[s(x)=\frac1{1+e^{-x}}\]

is a useful example. It is defined for every real \(x\), lies strictly between \(0\) and \(1\), approaches \(0\) as \(x\to-\infty\), approaches \(1\) as \(x\to\infty\), and is strictly increasing because

\[s'(x)=\frac{e^{-x}}{(1+e^{-x})^2}>0.\]

It demonstrates a general design lesson: a function need not be chosen because it looks familiar. Its properties can be checked against the requested behaviour.

Invertibility requires a one-to-one mapping. A function \(f\) has an inverse \(f^{-1}\) on a domain if and only if it passes the Horizontal Line Test: each output corresponds to exactly one input.

Reflective symmetry across y = x

Starting point. Let \((a, b)\) lie on the graph of \(y = f(x)\), so \(b = f(a)\).

  1. \[b = f(a) \iff a = f^{-1}(b)\]
    By definition of the inverse function.reversible
  2. \[(a, b) \in \text{graph}(f) \iff (b, a) \in \text{graph}(f^{-1})\]
    Swapping inputs and outputs reflects every point across the diagonal line \(y = x\).reversible
  3. \[(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}\]
    The slope of the reflected tangent line is the reciprocal of the original slope.reversible

Conclusion. \(\text{Domain}(f^{-1}) = \text{Range}(f)\) and \(\text{Range}(f^{-1}) = \text{Domain}(f)\). The slopes are reciprocals: \((f^{-1})'(y) = \frac{1}{f'(x)}\) where \(y = f(x)\).

  1. Check that \(f\) is strictly monotonic (increasing or decreasing) on the specified interval.
  2. Set \(y = f(x)\) and solve algebraically for \(x\) in terms of \(y\).
  3. Swap variables: write \(y = f^{-1}(x)\).
  4. State domain and range restrictions explicitly.

16.2 Rational functions and partial fractions

A rational function is \(N(x)/D(x)\) where \(N\) and \(D\) are polynomials and \(D(x)\ne0\). Zeros come from roots of \(N\) not cancelled by \(D\); roots of \(D\) give excluded inputs and may give poles. If the numerator degree is at least the denominator degree, perform polynomial division first.

For distinct linear factors,

\[\frac{P(x)}{(x-a)(x-b)}=\frac A{x-a}+\frac B{x-b}.\]

Multiplying through by the denominator and choosing convenient \(x\) values finds the coefficients. Repeated factors and irreducible quadratics require expanded templates; do not use the distinct-linear template blindly.

Example — partial fractions and a local power-series view

Decompose

\[\frac1{x^2-1}=\frac A{x-1}+\frac B{x+1}.\]

Multiplication by \((x-1)(x+1)\) gives

\[1=A(x+1)+B(x-1).\]

Putting \(x=1\) gives \(A=1/2\); putting \(x=-1\) gives \(B=-1/2\). Hence

\[\frac1{x^2-1}=\frac1{2(x-1)}-\frac1{2(x+1)}.\]

Partial fractions reveal both poles, \(x=\pm1\). They can also prepare a power series. For example, when \(|x|<1\),

\[\frac1{1-x}=1+x+x^2+\cdots,\]

because it is the geometric series with ratio \(x\). The inequality \(|x|<1\) is part of the statement; a rational expression exists at many values where this particular series does not converge.

Decomposition reverses recombination. A rational expression with factored denominator can be written as simpler fractions whose common-denominator sum reproduces the original numerator. Solving for the constants is coefficient matching: equality for every permitted input forces matching polynomial coefficients.

  1. Factor the denominator and write the required fraction shape.
  2. Clear denominators to obtain a polynomial identity.
  3. Choose convenient inputs or match coefficients, then verify by recombining.

16.3 Inverse trigonometric functions

Trigonometric functions are periodic, so they must be restricted before they can have inverse functions. Standard principal branches are

\[\arcsin:[-1,1]\to[-\pi/2,\pi/2],\quad \arccos:[-1,1]\to[0,\pi],\]
\[\arctan:\mathbb R\to(-\pi/2,\pi/2).\]

The notation \(\arcsin x\) avoids confusing an inverse with \(1/\sin x\).

Implicit differentiation of \(\sin y=x\) gives

\[\cos y\frac{dy}{dx}=1.\]

On the arcsine range, \(\cos y\ge0\), and \(\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}\). Thus

\[\frac d{dx}\arcsin x=\frac1{\sqrt{1-x^2}},\qquad -1<x<1.\]

Similarly,

\[\frac d{dx}\arccos x=-\frac1{\sqrt{1-x^2}},\qquad -1<x<1,\]
\[\frac d{dx}\arctan x=\frac1{1+x^2}\qquad(x\in\mathbb R).\]

For \(a>0\),

\[\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+C, \qquad \int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+C.\]

The reciprocal inverse functions require especially careful absolute values. On their standard principal branches,

\[\frac{d}{dx}\operatorname{arcsec}x= \frac{1}{|x|\sqrt{x^2-1}},\qquad \frac{d}{dx}\operatorname{arccsc}x= -\frac{1}{|x|\sqrt{x^2-1}},\qquad |x|>1.\]

The \(|x|\) is not optional: \(\sqrt{x^2}=|x|\), not generally \(x\). Different texts make different endpoint choices for these inverse functions; the displayed derivatives are asserted only on intervals where a consistent branch has been chosen.

Why a branch is necessary. Sine repeats values, so it has no inverse on all real inputs. Restricting it to one interval makes each allowed output identify exactly one angle; the inverse notation records that chosen branch.

16.4 Hyperbolic functions

Define

\[\cosh x=\frac{e^x+e^{-x}}2,\qquad \sinh x=\frac{e^x-e^{-x}}2.\]

\(\cosh x\) is even and \(\sinh x\) is odd. Their identity is

\[\cosh^2x-\sinh^2x=1,\]

so \((\cosh t,\sinh t)\) traces the hyperbola \(X^2-Y^2=1\), unlike the circle identity for sine and cosine. Their derivatives are

\[(\cosh x)'=\sinh x,\qquad(\sinh x)'=\cosh x.\]

The inverse functions have logarithmic forms on their principal real domains:

\[\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right),\]
\[\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right),\qquad x\ge1.\]

The quotient

\[\tanh x=\frac{\sinh x}{\cosh x} =\frac{e^x-e^{-x}}{e^x+e^{-x}}\]

is defined for every real \(x\), is odd, and satisfies \(-1<\tanh x<1\). The identity \(\cosh^2x-\sinh^2x=1\) gives

\[(\tanh x)'=\frac{\cosh^2x-\sinh^2x}{\cosh^2x} =\operatorname{sech}^2x, \qquad \operatorname{sech}x=\frac1{\cosh x}.\]

Its inverse has the real domain \(-1<x<1\):

\[\operatorname{artanh}x=\frac12\ln\left(\frac{1+x}{1-x}\right), \qquad \frac{d}{dx}\operatorname{artanh}x=\frac1{1-x^2}.\]

The remaining useful inverse-hyperbolic derivatives are

\[\frac{d}{dx}\operatorname{arsinh}x=\frac1{\sqrt{1+x^2}} \quad(x\in\mathbb R), \qquad \frac{d}{dx}\operatorname{arcosh}x=\frac1{\sqrt{x^2-1}} \quad(x>1).\]

Consequently, for \(a>0\),

\[\int\frac{dx}{\sqrt{a^2+x^2}} =\operatorname{arsinh}\frac xa+C, \qquad \int\frac{dx}{a^2-x^2} =\frac1a\operatorname{artanh}\frac xa+C\]

on intervals where the integrands are defined. The latter statement is local: its denominator excludes \(x=\pm a\).

Build the hyperbolic functions from exponentials. The symmetric combination keeps the even part of the exponential, while the antisymmetric combination keeps the odd part. Adding and subtracting their definitions reconstructs e^x and e^{-x}, which is the quickest route to many identities.

\[\cosh x=\frac{e^x+e^{-x}}2,\qquad\sinh x=\frac{e^x-e^{-x}}2\]

16.5 Inverse functions: branch first, formula second

An inverse function must return one output for each permitted input. Because \(\sin\theta=1/2\) at infinitely many angles, \(\arcsin(1/2)\) cannot mean “all solutions.” It means the one angle in the chosen interval \([-\pi/2,\pi/2]\), namely \(\pi/6\). Solving a trigonometric equation is a different task: after finding a reference angle, all periodic solutions must be listed if the problem requests them.

Example — inverse value versus all angles

The statement

\[\arccos(-1/2)=\frac{2\pi}{3}\]

uses the range \([0,\pi]\). In contrast, the equation \(\cos\theta=-1/2\) has

\[\theta=\frac{2\pi}{3}+2\pi k \quad\hbox{or}\quad \theta=\frac{4\pi}{3}+2\pi k,\qquad k\in\mathbb Z.\]

Mixing these two answers is a branch error, not an arithmetic error.

Example — derive the arctangent derivative

Let \(y=\arctan x\), so \(\tan y=x\) with \(-\pi/2<y<\pi/2\). Differentiate:

\[\sec^2y\frac{dy}{dx}=1.\]

The identity \(\sec^2y=1+\tan^2y=1+x^2\) gives

\[\frac{dy}{dx}=\frac1{1+x^2}.\]

Unlike arcsine, no square-root sign choice is needed because the tangent identity already produces a positive denominator. The formula is valid for all real \(x\).

Calculus of hyperbolic and inverse hyperbolic functions. Hyperbolic derivatives mirror trigonometric derivatives, but with key sign changes due to \(\cosh^2 x - \sinh^2 x = 1\).

Hyperbolic derivatives

Starting point. Differentiate the exponential definitions of \(\sinh x\) and \(\cosh x\).

  1. \[\frac{d}{dx}\sinh x = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x\]
    The derivative of \(\sinh x\) is \(+\cosh x\).reversible
  2. \[\frac{d}{dx}\cosh x = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh x\]
    The derivative of \(\cosh x\) is \(+\sinh x\) (positive sign, unlike \(-\sin x\)).reversible
  3. \[\frac{d}{dx}\tanh x = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x\]
    Apply the quotient rule and use \(\cosh^2 x - \sinh^2 x = 1\).reversible

Conclusion. \((\sinh x)' = \cosh x\), \((\cosh x)' = \sinh x\), \((\tanh x)' = \operatorname{sech}^2 x\). Inverse hyperbolic formulas: \(\operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1})\), \(\operatorname{arcosh} x = \ln(x + \sqrt{x^2 - 1})\) (\(x \ge 1\)).

  1. \(\int \sinh x\,dx = \cosh x + C\), \(\int \cosh x\,dx = \sinh x + C\).
  2. \(\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh}\left(\frac{x}{a}\right) + C = \ln\left(x + \sqrt{x^2 + a^2}\right) + C\).
  3. \(\int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh}\left(\frac{x}{a}\right) + C = \ln\left(x + \sqrt{x^2 - a^2}\right) + C\) (\(x > a > 0\)).

16.6 Inverse-trigonometric integrals by pattern recognition

The expressions \(a^2-x^2\) and \(a^2+x^2\) signal different geometries. Before using a formula, factor constants until the denominator exactly matches one of the standard patterns.

Example — an arcsine pattern

Evaluate

\[\int\frac{dx}{\sqrt{25-4x^2}}.\]

Factor \(25\) inside the square root:

\[\sqrt{25-4x^2}=5\sqrt{1-\left(\frac{2x}{5}\right)^2}.\]

Let \(u=2x/5\), so \(dx=(5/2)\,du\). Then

\[\int\frac{dx}{\sqrt{25-4x^2}} =\frac12\int\frac{du}{\sqrt{1-u^2}} =\frac12\arcsin u+C =\frac12\arcsin\frac{2x}{5}+C.\]

For real values, the original integrand requires \(|x|<5/2\) on an interval; at the endpoints the derivative form is unbounded.

Example — an arctangent pattern

Evaluate

\[\int\frac{3\,dx}{x^2+9}.\]

Since \(x^2+9=x^2+3^2\),

\[\int\frac{3\,dx}{x^2+9} =3\left(\frac13\arctan\frac{x}{3}\right)+C =\arctan\frac{x}{3}+C.\]

Differentiating \(\arctan(x/3)\) gives \((1/3)/(1+x^2/9)=3/(x^2+9)\), which checks both the scale and the result.

Inverse trigonometric functions require restricted principal branches. Because \(\sin\), \(\cos\), and \(\tan\) are periodic, their domains must be restricted to establish one-to-one branches.

Derivatives of inverse trigonometric functions

Starting point. Let \(y = \arcsin x\) for \(x \in [-1, 1]\) and \(y \in [-\pi/2, \pi/2]\).

  1. \[\sin y = x \implies \frac{d}{dx}(\sin y) = \frac{d}{dx}(x) \implies \cos y \frac{dy}{dx} = 1\]
    Differentiate implicitly with respect to \(x\).reversible
  2. \[\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}\]
    Since \(y \in [-\pi/2, \pi/2]\), \(\cos y \ge 0\), so \(\cos y = +\sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}\).reversible

Conclusion. \(\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}\), \(\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}\), \(\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}\).

  1. \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C\) (\(a > 0\)).
  2. \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C\) (\(a > 0\)).
  3. Look for quadratic denominators: complete the square to fit these standard inverse-trig forms.

16.7 Hyperbolic functions are exponential combinations

Circular functions use a plus sign in \(\sin^2x+\cos^2x=1\) and are periodic. Hyperbolic functions use a minus sign in \(\cosh^2x-\sinh^2x=1\) and are not periodic. From their exponential definitions,

\[\cosh x+\sinh x=e^x,\qquad \cosh x-\sinh x=e^{-x}.\]

Adding and subtracting those equations recovers the definitions. This makes hyperbolic functions natural whenever a calculation already contains both growing and decaying exponentials.

Example — a hyperbolic simplification

Show that \(\tanh x\) approaches \(1\) as \(x\to\infty\). Divide numerator and denominator by \(e^x\):

\[\tanh x=\frac{1-e^{-2x}}{1+e^{-2x}}\longrightarrow\frac{1-0}{1+0}=1.\]

Similarly it approaches \(-1\) as \(x\to-\infty\). The bounded S-shape contrasts with \(\tan x\), which has vertical asymptotes.

Example — inverse-hyperbolic integral

Evaluate

\[\int\frac{dx}{\sqrt{x^2+16}}.\]

The form is \(a^2+x^2\), but it is inside a square root rather than an ordinary quadratic denominator. With \(a=4\),

\[\int\frac{dx}{\sqrt{x^2+16}} =\operatorname{arsinh}\frac{x}{4}+C =\ln\left(\frac{x+\sqrt{x^2+16}}{4}\right)+C.\]

The two antiderivatives differ from \(\ln(x+\sqrt{x^2+16})\) only by the constant \(-\ln4\), so all are equivalent indefinite answers.

Hyperbolic functions parametrize hyperbolas. Just as \((\cos t, \sin t)\) parametrizes the circle \(x^2 + y^2 = 1\), the hyperbolic pair \((\cosh t, \sinh t)\) parametrizes the hyperbola \(x^2 - y^2 = 1\).

Definitions and fundamental hyperbolic identity

Starting point. Define \(\cosh x = \frac{e^x + e^{-x}}{2}\) (even part of \(e^x\)) and \(\sinh x = \frac{e^x - e^{-x}}{2}\) (odd part of \(e^x\)).

  1. \[\cosh^2 x = \frac{(e^x + e^{-x})^2}{4} = \frac{e^{2x} + 2 + e^{-2x}}{4}\]
    Square \(\cosh x\) algebraically.reversible
  2. \[\sinh^2 x = \frac{(e^x - e^{-x})^2}{4} = \frac{e^{2x} - 2 + e^{-2x}}{4}\]
    Square \(\sinh x\) algebraically.reversible
  3. \[\cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1\]
    Subtract: the exponential terms cancel, leaving identically \(1\).reversible

Conclusion. \(\cosh^2 x - \sinh^2 x = 1\) for all real \(x\). Also: \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}\).

Common Mistakes

  • Cancelling a rational factor and forgetting its excluded input.
  • Using inverse trig notation without a specified principal range.
  • Losing the positive square-root sign in the arcsine derivative.
  • Replacing the hyperbolic identity with the circular identity; its sign differs.

Exercises

  1. Decompose \(1/(x^2-1)\) into partial fractions.
  2. Find \(\arcsin(\sqrt3/2)\) and explain the principal-value choice.
  3. Differentiate \(\arctan(2x)\).
  4. Verify \(\cosh^2x-\sinh^2x=1\) from exponential definitions.
  5. Sketch the qualitative difference between \(\tanh x\) and \(\tan x\).
  6. Differentiate \(\operatorname{arccos}(x)\) and explain why its sign differs from that of \(\operatorname{arcsin}(x)\).
  7. Evaluate \(\int dx/\sqrt{9+x^2}\) and state why an inverse hyperbolic function is appropriate.
  8. Distinguish \(\arcsin(1/2)\) from all solutions of \(\sin\theta=1/2\).
  9. Evaluate \(\int dx/\sqrt{16-9x^2}\) and state the real domain of the integrand.
  10. Evaluate \(\int 5\,dx/(x^2+25)\).
  11. Prove from definitions that \(\cosh x+\sinh x=e^x\).
  12. Sketch the key qualitative differences between \(\tan x\) and \(\tanh x\).

Part VI

Part VI — Differential Equations for Dynamic Systems

Part VI · Differential Equations for Dynamic Systems

17

Chapter 17 — First-Order Linear ODEs and RC Transients

Open chapter view →

Why This Matters

An algebraic equation relates quantities at one instant. A differential equation relates a quantity to how it changes. That is the natural language of a capacitor, whose current depends on the time rate of change of voltage, and of any system with memory.

17.1 What an ODE is

An ordinary differential equation (ODE) contains an unknown function of one variable and one or more derivatives. Its order is the highest derivative present. For example,

\[\frac{dy}{dt}+ky=q(t)\]

is first order. A solution is a function \(y(t)\) that makes the equation true on the interval of interest.

An initial condition such as \(y(0)=y_0\) selects one physical response from a family of mathematical solutions. Without enough initial conditions, an ODE usually has arbitrary constants and therefore does not yet describe one unique trajectory.

An ODE constrains a whole curve. An algebraic equation selects numbers; a differential equation relates a value to its permitted local rate of change. An initial condition chooses one curve from the family that satisfies the same rate law, which is why circuit state matters at the switching instant.

17.2 Linear first-order equations and integrating factors

The standard linear form is

\[y'(t)+P(t)y(t)=Q(t).\]

Define the integrating factor

\[\mu(t)=e^{\int P(t)\,dt}.\]

Because \(\mu'=P\mu\),

\[\frac d{dt}[\mu(t)y(t)]=\mu(t)y'(t)+\mu(t)P(t)y(t)=\mu(t)Q(t).\]

Integrating gives

\[\mu y=\int\mu Q\,dt+C,\]

and then divide by the nonzero exponential \(\mu\). For constant \(P=k\), the integrating factor is simply \(e^{kt}\).

Example — constant forcing

Solve

\[y'+2y=6,\qquad y(0)=1.\]

Here \(\mu=e^{2t}\), so

\[\frac d{dt}(e^{2t}y)=6e^{2t}.\]

Integrating,

\[e^{2t}y=3e^{2t}+C,\]

so \(y=3+Ce^{-2t}\). The initial condition gives \(1=3+C\), hence

\[y(t)=3-2e^{-2t}.\]

The constant \(3\) is the steady-state value; the exponential term is the transient that decays to zero.

Choose the multiplier that creates a product derivative. For y'+p(t)y=q(t), seek a non-zero function μ such that (μy)'=μy'+μpy. Matching coefficients requires μ'=pμ, so μ is an exponential of an integral of p. The method is a constructed product rule, not a separate rule to memorise.

\[\mu=e^{\int p(t)\,dt},\qquad (\mu y)'=\mu q\]
  1. Put the equation in standard linear form.
  2. Form the integrating factor from the coefficient of y.
  3. Multiply every term, integrate the resulting product derivative, then apply the initial condition.

17.3 RC circuit response

For a series resistor-capacitor circuit driven by a constant source \(V_s\), KVL and the capacitor relation \(i=C\,dv_C/dt\) lead to

\[RC\frac{dv_C}{dt}+v_C=V_s.\]

With an initially uncharged capacitor, \(v_C(0)=0\), the solution is

\[v_C(t)=V_s\left(1-e^{-t/(RC)}\right).\]

The current is

\[i(t)=\frac{V_s}{R}e^{-t/(RC)}.\]

Every symbol has a role: \(V_s\) is the applied step voltage, \(R>0\) and \(C>0\) set the time constant \(\tau=RC\), and the initial condition determines the transient coefficient. A different initial capacitor voltage produces a different solution.

Engineering consequence. The time constant is a scale, not a finish line. After one time constant a charging response has completed about 63.2 percent of its total change; each further time constant removes the same fraction of the remaining gap.

\[v_C(t)=V_f+(V_i-V_f)e^{-t/\tau}\]

17.4 Sinusoidal forcing and complex method

An AC-forced first-order equation often contains integrals such as \(\int e^{pt}\cos(\omega t)\,dt\). Euler’s formula lets us compute the real part of \(\int e^{(p+i\omega)t}\,dt\), then take the real component. This method is efficient because differentiation and integration of exponentials preserve their form. The completed response separates into a decaying transient and a persistent sinusoidal steady state.

First-order differential equations model dynamic states. In an electric circuit, inductors and capacitors store energy; the rates of change \(\frac{di}{dt}\) and \(\frac{dv}{dt}\) generate first-order ODEs governed by Kirchhoff's laws.

Standard form of a first-order linear ODE

Starting point. Consider the linear ODE \(\frac{dy}{dt} + P(t) y = Q(t)\).

  1. \[\frac{dy}{dt} + P(t) y = 0\]
    Homogeneous equation (\(Q(t) = 0\)): models natural unforced decay.reversible
  2. \[\frac{dy}{y} = -P(t) dt \implies \ln|y| = -\int P(t) dt + C_1 \implies y_h(t) = A e^{-\int P(t) dt}\]
    Separate variables to find the homogeneous solution \(y_h(t)\).reversible
  3. \[y(t) = y_h(t) + y_p(t)\]
    The general solution is the sum of the natural response \(y_h\) and any particular forced response \(y_p\).reversible

Conclusion. Every linear first-order ODE has a general solution composed of the natural response (initial condition decay) plus the forced response (steady-state driving).

  1. Put equation in standard form: \(y' + P(t)y = Q(t)\).
  2. Classify: separable or linear?
  3. If separable: separate \(\frac{dy}{g(y)} = f(t) dt\) and integrate both sides.
  4. If linear: find integrating factor \(\mu(t) = e^{\int P(t) dt}\).
  5. Apply initial condition \(y(t_0) = y_0\) to determine the constant \(C\).

17.5 Separation, linearity, and what an initial condition selects

Some first-order ODEs are separable, meaning they can be rearranged into a function of \(y\) times \(dy\) equal to a function of \(t\) times \(dt\). For example,

\[\frac{dy}{dt}=ky\]

has, where \(y\ne0\),

\[\frac{dy}{y}=k\,dt.\]

Integrating gives \(\ln|y|=kt+C\), hence \(y=Ae^{kt}\). The missing zero solution is also contained by choosing \(A=0\). Separation is efficient when available, but a linear equation such as \(y'+P(t)y=Q(t)\) need not be separable. The integrating factor method works for every equation in that standard linear form.

Example — exponential decay with units and an initial condition

Suppose a stored quantity satisfies \(dq/dt=-0.4q\), with \(t\) in seconds and \(q(0)=10\) mC. Separation gives

\[\frac{dq}{q}=-0.4\,dt,\qquad \ln|q|=-0.4t+C.\]

Exponentiating and using the initial value gives

\[q(t)=10e^{-0.4t}\ \mathrm{mC}.\]

The coefficient \(0.4\) has units \(\mathrm{s}^{-1}\), because its product with \(t\) must be dimensionless inside an exponential. This unit check detects a common model-writing mistake.

Separation of variables splits the differentials. If an ODE can be factored as \(\frac{dy}{dt} = f(t) g(y)\), group all \(y\) terms on the left and all \(t\) terms on the right.

Separation of variables procedure

Starting point. Let \(\frac{dy}{dt} = f(t) g(y)\) with \(g(y) \neq 0\) on the domain of interest.

  1. \[\frac{1}{g(y)} dy = f(t) dt\]
    Divide by \(g(y)\) and multiply by \(dt\) to separate variables.reversible
  2. \[\int \frac{1}{g(y)} dy = \int f(t) dt + C\]
    Integrate each side with respect to its own independent variable.reversible
  3. \[H(y) = F(t) + C \implies y(t) = H^{-1}(F(t) + C)\]
    Solve algebraically for \(y\) whenever an explicit inverse \(H^{-1}\) exists.reversible

Conclusion. Separation of variables converts a differential equation into two independent antiderivatives. Remember that division by \(g(y)\) assumes \(g(y) \neq 0\); equilibrium solutions \(y = y_0\) where \(g(y_0) = 0\) must be checked separately.

  1. Check for constant equilibrium solutions: solve \(g(y) = 0\).
  2. Separate variables: \(\frac{dy}{g(y)} = f(t) dt\).
  3. Integrate both sides: \(\int \frac{dy}{g(y)} = \int f(t) dt + C\).
  4. Solve for \(y(t)\) explicitly and apply the initial condition to determine \(C\).

17.6 Integrating factors, derived as a product rule

The integrating factor is sometimes taught as a recipe. Its purpose is to make the left side of a linear ODE into one derivative. Start with

\[y'+P(t)y=Q(t).\]

We seek a nonzero function \(\mu(t)\) such that

\[(\mu y)'=\mu y'+\mu' y\]

equals \(\mu y'+\mu P y\). This requires \(\mu'=P\mu\). Dividing by \(\mu\) and integrating gives

\[\frac{\mu'}{\mu}=P,\qquad \ln|\mu|=\int P(t)\,dt, \qquad \mu=e^{\int P(t)\,dt}.\]

Any nonzero constant multiple would work and cancels from the final solution.

Example — variable coefficient, all steps shown

Solve

\[y'+\frac{2}{t}y=t^2,\qquad t>0,\qquad y(1)=3.\]

Here \(P(t)=2/t\), so

\[\mu(t)=e^{\int 2/t\,dt}=e^{2\ln t}=t^2\]

on the specified positive interval. Multiply the ODE by \(t^2\):

\[t^2y'+2ty=t^4.\]

The left side is \((t^2y)'\), so integration gives

\[t^2y=\frac{t^5}{5}+C,\qquad y=\frac{t^3}{5}+\frac{C}{t^2}.\]

Apply \(y(1)=3\): \(3=1/5+C\), hence \(C=14/5\). Thus

\[y(t)=\frac{t^3}{5}+\frac{14}{5t^2},\qquad t>0.\]

The interval matters. The coefficient \(2/t\) is undefined at \(t=0\), so a solution on \(t>0\) is not automatically a solution across zero.

The integrating factor is derived directly from the product rule. The requirement that \((\mu y)' = \mu y' + \mu P y\) uniquely determines \(\mu(t) = e^{\int P(t) dt}\).

Why the integrating factor is unique

Assumptions
  • The ODE is linear: \(y' + P(x) y = Q(x)\) with continuous \(P, Q\).
  • We seek a non-zero differentiable multiplier \(\mu(x)\).
  1. Multiplying by \(\mu(x) = e^{\int P(x) dx}\) makes the left side equal to \((\mu y)'\).By the chain rule: \(\mu'(x) = \mu(x) P(x)\), so \(\mu y' + \mu P y = \mu y' + \mu' y = (\mu y)'\).
  2. Integrating \((\mu y)' = \mu Q\) yields \(\mu y = \int \mu Q\,dx + C\).By the Fundamental Theorem of Calculus: one integration produces exactly one constant of integration \(C\).
  3. The solution \(y(x) = \frac{1}{\mu(x)} \int \mu Q\,dx + \frac{C}{\mu(x)}\) is uniquely determined by \(y(x_0) = y_0\).Because \(\mu(x) = e^{\int P dx} > 0\) is never zero, division by \(\mu(x)\) is always defined and invertible.

Conclusion. The integrating factor method yields the complete, unique solution to any first-order linear initial value problem.

Scope. Covers all first-order linear ODEs on intervals where \(P(x)\) and \(Q(x)\) are continuous.

17.7 RC response from the circuit law

For a series resistor and capacitor with source \(V_s\), Kirchhoff's voltage law is

\[V_s=v_R+v_C.\]

The resistor voltage is \(v_R=Ri\), and the capacitor current is \(i=C\,dv_C/dt\) under the passive sign convention. Substitution gives

\[V_s=RC\frac{dv_C}{dt}+v_C.\]

Dividing by \(RC\) puts the equation in standard form:

\[\frac{dv_C}{dt}+\frac1{RC}v_C=\frac{V_s}{RC}.\]

The integrating factor is \(e^{t/(RC)}\). Carrying out the method gives the general step response

\[v_C(t)=V_s+(V_0-V_s)e^{-t/(RC)},\]

where \(V_0=v_C(0)\). It passes two physical checks:

\[v_C(0)=V_0,\qquad \lim_{t\to\infty}v_C(t)=V_s.\]

Engineering Example — numerical RC charging

Let \(R=2\ \mathrm{k}\Omega\), \(C=50\ \mu\mathrm F\), \(V_s=10\) V, and \(V_0=0\). Then

\[\tau=RC=(2000)(50\times10^{-6})=0.10\ \mathrm s.\]

Therefore

\[v_C(t)=10(1-e^{-t/0.10})\ \mathrm V.\]

At one time constant, \(v_C(0.10)=10(1-e^{-1})\approx6.32\) V. The capacitor has not “finished charging”; it has completed about \(63.2\%\) of the initial-to-final change. Its current is

\[i(t)=\frac{10}{2000}e^{-t/0.10}=5e^{-10t}\ \mathrm{mA}.\]

RC circuit differential equation. Applying Kirchhoff's Voltage Law to a series RC circuit with step voltage \(V_s\) gives \(R C \frac{dv_C}{dt} + v_C = V_s\).

Step response of an RC circuit

Starting point. Solve \(\frac{dv_C}{dt} + \frac{1}{RC} v_C = \frac{V_s}{RC}\) with initial condition \(v_C(0) = V_0\).

  1. \[P(t) = \frac{1}{RC} \implies \mu(t) = e^{\int (1/RC) dt} = e^{t/RC}\]
    Find the integrating factor with time constant \(\tau = RC\).reversible
  2. \[\frac{d}{dt}[e^{t/RC} v_C] = \frac{V_s}{RC} e^{t/RC}\]
    Multiply the ODE by \(\mu(t) = e^{t/RC}\).reversible
  3. \[e^{t/RC} v_C(t) = \int \frac{V_s}{RC} e^{t/RC} dt = V_s e^{t/RC} + C\]
    Integrate both sides.reversible
  4. \[v_C(t) = V_s + C e^{-t/RC}\]
    Divide by \(e^{t/RC}\).reversible
  5. \[v_C(0) = V_0 \implies V_0 = V_s + C \implies C = V_0 - V_s\]
    Apply the initial condition at \(t = 0\).reversible

Conclusion. \(v_C(t) = V_s + (V_0 - V_s)e^{-t/RC} = V_{\text{final}} + (V_{\text{initial}} - V_{\text{final}})e^{-t/\tau}\). This universal formula applies to all first-order RC/RL step responses.

  1. Identify initial voltage \(V_i = v_C(0)\) and final asymptotic voltage \(V_f = \lim_{t\to\infty} v_C(t)\).
  2. Compute time constant \(\tau = R_{\text{Th}} C\) using the Thévenin resistance seen by the capacitor.
  3. Apply the master first-order formula: \(v(t) = V_f + (V_i - V_f)e^{-t/\tau}\).
  4. Current through capacitor: \(i_C(t) = C \frac{dv_C}{dt} = \frac{V_f - V_i}{R} e^{-t/\tau}\).

17.8 First-order problem-solving strategy

  1. Name the independent variable and the unknown function; state an interval.
  2. Classify the equation: separable, linear, or neither of these standard forms.
  3. For linear equations, identify \(P(t)\) and \(Q(t)\) before forming \(\mu\).
  4. Keep the arbitrary constant until every integration is complete.
  5. Apply initial data, then substitute the result back into the ODE and inspect \(t=0\) and long-time behaviour.

For a stable constant-coefficient equation \(y'+ky=q\) with \(k>0\), the homogeneous part \(Ce^{-kt}\) decays. The particular constant \(q/k\) is the steady state. This separation is a mathematical result, not an assumption that every real circuit is ideal for all times.

A systematic diagnostic workflow for first-order ODEs. Follow this roadmap to select the optimal solution method.

  1. 1. Write the equation in standard form: \(y' = F(t, y)\) or \(y' + P(t)y = Q(t)\).
  2. 2. Check if separable: can you write \(F(t, y) = f(t) g(y)\)? If yes, separate \(\frac{dy}{g(y)} = f(t) dt\).
  3. 3. Check if linear: is the equation in the form \(y' + P(t)y = Q(t)\)? If yes, compute \(\mu(t) = e^{\int P(t) dt}\).
  4. 4. Integrate both sides and include the constant of integration \(+C\).
  5. 5. Solve for \(y(t)\) explicitly if possible.
  6. 6. Apply the initial condition \(y(t_0) = y_0\) to determine \(C\), and verify by differentiating your solution back into the original ODE.

Verification by substitution

From
  • Proposed solution: \(y(t) = 5 + 3 e^{-2t}\).
  • Original ODE: \(y' + 2y = 10\) with \(y(0) = 8\).
  1. Check initial condition: \(y(0) = 5 + 3 e^0 = 5 + 3 = 8\) (matches).
  2. Compute derivative: \(y'(t) = -6 e^{-2t}\).
  3. Substitute into left side: \(y' + 2y = -6e^{-2t} + 2(5 + 3e^{-2t}) = -6e^{-2t} + 10 + 6e^{-2t} = 10\).
  4. Matches right side \(Q(t) = 10\) for all \(t\).

Result. The solution is certified correct.

Common Mistakes

  • Calling an equation “linear” merely because it contains derivatives; \(yy'\) is nonlinear in \(y\).
  • Forgetting the initial condition after finding a general solution.
  • Using \(e^{\int Pdt}\) without the integral in a variable-coefficient equation.
  • Writing a numerical RC formula such as \(5e^{-t/RC}/R\) without stating the source voltage and initial condition that justify the \(5\).

The integrating factor turns the left side into a product rule derivative. Multiplying \(y' + P(t)y = Q(t)\) by \(\mu(t) = e^{\int P(t)dt}\) allows direct integration.

Derivation of the integrating factor

Starting point. We seek a multiplier \(\mu(t)\) such that \(\mu(t)y' + \mu(t)P(t)y = \frac{d}{dt}[\mu(t) y]\).

  1. \[\frac{d}{dt}[\mu y] = \mu y' + \mu' y\]
    Apply the product rule to the right side.reversible
  2. \[\mu y' + \mu P y = \mu y' + \mu' y \implies \mu' = P \mu\]
    Equate coefficients of \(y\): this yields a separable ODE for \(\mu\).reversible
  3. \[\frac{d\mu}{\mu} = P(t) dt \implies \ln|\mu| = \int P(t) dt \implies \mu(t) = e^{\int P(t) dt}\]
    Integrate to find the integrating factor (set \(C = 0\) as any non-zero multiplier works).reversible
  4. \[\frac{d}{dt}[\mu(t) y] = \mu(t) Q(t) \implies \mu(t) y = \int \mu(t) Q(t) dt + C\]
    Multiply the original ODE by \(\mu(t)\) and integrate both sides.reversible
  5. \[y(t) = \frac{1}{\mu(t)} \int \mu(t) Q(t) dt + \frac{C}{\mu(t)}\]
    Divide by \(\mu(t)\) to obtain the explicit general solution.reversible

Conclusion. \(y(t) = e^{-\int P(t)dt} \left[ \int e^{\int P(t)dt} Q(t) dt + C \right]\). The constant \(C\) is inside the brackets and must be divided by \(\mu(t)\).

Exercises

  1. Solve \(y'-3y=0\) with \(y(0)=4\).
  2. Solve \(y'+y=t\) using an integrating factor.
  3. Derive the current in an RC charging circuit from the capacitor-voltage result.
  4. State transient and steady-state parts of \(y(t)=7+2e^{-t/5}\).
  5. A capacitor initially has voltage \(V_0\). Write the expected qualitative change to the RC charging formula if the source is \(V_s\).
  6. Solve \(dy/dt=-3y\) with \(y(0)=7\), and state the units required for \(3\) if time is measured in seconds.
  7. Solve \(y'+(2/t)y=t^2\) for \(t>0\) with \(y(1)=1\).
  8. Starting from \(V_s=v_R+v_C\), derive the first-order RC equation.
  9. A \(1\ \mathrm{k}\Omega\) resistor and \(100\ \mu\)F capacitor are charged from \(5\) V with \(v_C(0)=1\) V. Find \(v_C(t)\) and \(v_C(0.1)\).

Part VI · Differential Equations for Dynamic Systems

18

Chapter 18 — Second-Order Linear ODEs and RLC Responses

Open chapter view →

Why This Matters

Two energy-storage elements create richer dynamics. Inductors and capacitors can exchange energy, producing oscillation; resistance removes energy, producing damping. Second-order ODEs classify these behaviours through the roots of one quadratic equation.

18.1 Homogeneous constant-coefficient equations

Consider

\[ay''+by'+cy=0,\qquad a\ne0.\]

Try \(y=e^{st}\). Since \(y'=se^{st}\) and \(y''=s^2e^{st}\),

\[(as^2+bs+c)e^{st}=0.\]

The exponential never equals zero, so \(s\) must satisfy the characteristic equation

\[as^2+bs+c=0.\]

The nature of its roots determines the solution.

  • Distinct real roots \(s_1,s_2\): \(y=C_1e^{s_1t}+C_2e^{s_2t}.\)
  • Repeated real root \(s\): \(y=(C_1+C_2t)e^{st}.\)
  • Complex roots \(\alpha\pm i\beta\): \(y=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t).\)

Two initial conditions, typically \(y(0)\) and \(y'(0)\), determine \(C_1\) and \(C_2\).

Why exponentials are the trial shape. Differentiating e^{rt} only multiplies it by r, so every derivative in a constant-coefficient equation remains the same shape. Substitution turns the differential equation into an algebraic condition on r; the available roots determine the available response shapes.

\[y=e^{rt}\quad\Longrightarrow\quad ar^2+br+c=0\]

18.2 RLC damping

For a series RLC natural response, a standard current equation can be written

\[i''+2\alpha i'+\omega_0^2i=0,\]

where

\[\alpha=\frac{R}{2L},\qquad \omega_0=\frac1{\sqrt{LC}}.\]

The characteristic roots are

\[s=-\alpha\pm\sqrt{\alpha^2-\omega_0^2}.\]

If \(\alpha>\omega_0\), roots are distinct real values: the response is over-damped, non-oscillatory. If \(\alpha=\omega_0\), there is a repeated root: critical damping, the fastest non-oscillatory return under this model. If \(\alpha<\omega_0\), roots are complex and the response is under-damped:

\[i(t)=e^{-\alpha t}\left(C_1\cos\omega_dt+C_2\sin\omega_dt\right), \qquad \omega_d=\sqrt{\omega_0^2-\alpha^2}.\]

\(\omega_d\) is an oscillation frequency only in the under-damped case. It should not be defined with an absolute value to imply oscillation in an over-damped circuit.

Why the discriminant matters. The characteristic roots determine the ingredients available to the response. Two real roots give two decays, a repeated root is the boundary case, and complex conjugate roots combine decay with oscillation.

\[r^2+2\zeta\omega_n r+\omega_n^2=0\]

18.3 Non-homogeneous equations

For

\[ay''+by'+cy=q(t),\]

the general solution is

\[y=y_h+y_p,\]

where \(y_h\) solves the homogeneous equation and \(y_p\) is one particular solution. For polynomial, exponential, sine, or cosine forcing, the method of undetermined coefficients proposes a trial form whose derivatives have the same family.

If the proposed trial duplicates a term already in \(y_h\), multiply the trial by a sufficient power of \(t\) to make it independent. This resonance correction is essential; otherwise substituted coefficients cannot determine a new solution.

Example — constant forcing

Solve a particular solution for

\[y''+3y'+2y=10.\]

Try \(y_p=A\). Its derivatives are zero, so \(2A=10\) and \(A=5\). The homogeneous characteristic equation is \((s+1)(s+2)=0\), hence

\[y=C_1e^{-t}+C_2e^{-2t}+5.\]

Initial conditions would determine \(C_1,C_2\).

Separate what the system does from what drives it. The homogeneous solution is the natural response allowed by the stored energy and damping. A particular solution accounts for the forcing. Their sum is required because differentiation is linear: substituting each contribution explains a different part of the equation.

\[y=y_h+y_p\]

18.4 Physical interpretation

The homogeneous part contains natural modes determined by stored energy and losses. In a stable damped circuit, it decays and becomes the transient. The particular part is driven by the source and becomes the steady-state response. With \(R=0\), the ideal LC equation has undamped sinusoidal solutions; calling it an LC oscillator equation is more precise here than confusing it with the spatial wave-equation PDE.

The RLC circuit is the physical embodiment of a second-order ODE. Each coefficient has a direct electrical correspondence: \(a = L\) (inductance, opposes current changes), \(b = R\) (resistance, dissipates energy as heat), and \(c = 1/C\) (reciprocal capacitance, stores electrostatic energy).

Characteristic roots as system poles

From
  • Characteristic equation: \(L r^2 + R r + \frac{1}{C} = 0\).
  • Roots with negative real parts (\(\operatorname{Re}(r) < 0\)) correspond to stable, decaying physical responses.
  1. Real negative roots (\(r_1, r_2 < 0\)): overdamped — non-oscillatory exponential return.
  2. Repeated real root (\(r = -R/(2L)\)): critically damped — fastest decay without overshoot.
  3. Complex conjugate roots (\(r = -\alpha \pm i\omega_d\)): underdamped — decaying sinusoidal oscillation at damped frequency \(\omega_d = \sqrt{\omega_0^2 - \alpha^2}\).

Result. The pole positions in the complex \(s\)-plane completely dictate the transient waveforms and stability of the electrical circuit.

Limitation. Applies to linear time-invariant systems. Non-linear components (such as saturating inductors or semiconductor diodes) require non-linear phase-plane analysis.

18.5 Why the characteristic equation supplies all three cases

Trying \(y=e^{st}\) is useful because differentiation only multiplies this trial by \(s\) or \(s^2\). Substitution into a constant-coefficient homogeneous equation therefore reduces a function problem to the algebraic quadratic

\[as^2+bs+c=0.\]

If the roots are distinct, \(e^{s_1t}\) and \(e^{s_2t}\) are independent solutions, so their linear combination contains two arbitrary constants. If the root is repeated, simply writing the same exponential twice does not give two independent solutions. The second solution is \(te^{st}\), producing

\[y=(C_1+C_2t)e^{st}.\]

If roots are \(\alpha\pm i\beta\), Euler's formula turns their exponential solutions into real functions:

\[e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos\beta t+i\sin\beta t).\]

Adding and subtracting conjugate solutions gives the real basis \(e^{\alpha t}\cos\beta t\) and \(e^{\alpha t}\sin\beta t\). The envelope \(e^{\alpha t}\) must remain; it describes growth or decay of the oscillation.

Example — distinct roots and initial conditions

Solve

\[y''-5y'+6y=0,\qquad y(0)=1,\qquad y'(0)=0.\]

The characteristic equation factors:

\[s^2-5s+6=(s-2)(s-3)=0.\]

Thus \(y=C_1e^{2t}+C_2e^{3t}\). The first initial condition gives

\[C_1+C_2=1.\]

Differentiate:

\[y'=2C_1e^{2t}+3C_2e^{3t},\]

so \(2C_1+3C_2=0\) at \(t=0\). Subtract twice the first equation from the second: \(C_2=-2\), hence \(C_1=3\). Therefore

\[y(t)=3e^{2t}-2e^{3t}.\]

The growth is mathematically correct even though it would signal an unstable physical model if \(y\) represented a passive circuit state.

Example — repeated root without losing a solution

For \(y''+4y'+4y=0\), the characteristic polynomial is

\[s^2+4s+4=(s+2)^2.\]

The correct family is

\[y=(C_1+C_2t)e^{-2t},\]

not merely \(Ce^{-2t}\). A second-order ODE needs two independently selectable constants before two independent initial conditions can generally be met.

Example — damped oscillation

Solve \(y''+6y'+25y=0\). The roots are

\[s=\frac{-6\pm\sqrt{36-100}}2=-3\pm4i.\]

Hence

\[y=e^{-3t}(C_1\cos4t+C_2\sin4t).\]

The visible oscillation frequency is \(4\) rad/s, while \(e^{-3t}\) reduces its amplitude. A graph should show crossings continuing while the envelope shrinks, not a sinusoid of fixed height.

Why substituting \(y = e^{rt}\) produces the characteristic equation. The exponential function is the unique eigenfunction of differentiation: each derivative simply multiplies the function by the constant scalar \(r\), converting differential operators into polynomials.

Why the exponential ansatz is complete

Assumptions
  • The differential equation is linear, homogeneous, with constant coefficients: \(a y'' + b y' + c y = 0\).
  • The solution vector space has dimension 2 (two initial conditions specify a unique trajectory).
  1. \(e^{rt}\) satisfies the ODE if and only if \(a r^2 + b r + c = 0\).Substituting \(y = e^{rt}\) yields \((a r^2 + b r + c)e^{rt} = 0\). Since \(e^{rt} \neq 0\), \(a r^2 + b r + c = 0\) must hold.
  2. For distinct roots (\(r_1 \neq r_2\)), \(e^{r_1 t}\) and \(e^{r_2 t}\) are linearly independent.The Wronskian determinant \(W(e^{r_1 t}, e^{r_2 t}) = (r_2 - r_1)e^{(r_1 + r_2)t} \neq 0\) is non-zero for all \(t\).
  3. For a repeated root (\(r_1 = r_2 = r\)), \(e^{rt}\) and \(t e^{rt}\) are linearly independent.Reduction of order confirms \(t e^{rt}\) satisfies the ODE, with non-vanishing Wronskian \(W = e^{2rt} \neq 0\).
  4. For complex roots \(\alpha \pm i\beta\), \(e^{\alpha t}\cos(\beta t)\) and \(e^{\alpha t}\sin(\beta t)\) form an independent real basis.They are the real and imaginary parts of the complex solution \(e^{(\alpha + i\beta)t}\), both satisfying the real-coefficient ODE.

Conclusion. In all cases, the characteristic quadratic equation produces two linearly independent solutions that span the complete 2D solution space.

Scope. Covers all second-order constant-coefficient linear homogeneous ODEs.

18.6 RLC classification from the discriminant

For the RLC characteristic equation

\[s^2+2\alpha s+\omega_0^2=0,\]

the discriminant is \(4(\alpha^2-\omega_0^2)\). Its sign creates the three physical response classes:

ComparisonRootsNatural response
\(\alpha>\omega_0\)distinct real negative for passive componentsover-damped, two decays
\(\alpha=\omega_0\)repeated negativecritically damped
\(\alpha<\omega_0\)\(-\alpha\pm i\omega_d\)under-damped, decaying oscillation

Engineering Example — classify before calculating constants

Let \(R=4\ \Omega\), \(L=1\ \mathrm H\), and \(C=1/9\ \mathrm F\). Then

\[\alpha=\frac{R}{2L}=2\ \mathrm{s}^{-1}, \qquad \omega_0=\frac1{\sqrt{LC}}=3\ \mathrm{rad/s}.\]

Because \(\alpha<\omega_0\), the response is under-damped and

\[\omega_d=\sqrt{3^2-2^2}=\sqrt5\ \mathrm{rad/s}.\]

Only after this classification should one apply initial current and initial derivative conditions to find \(C_1\) and \(C_2\). Calling \(\omega_0\) the observed oscillation frequency would be inaccurate when damping is present.

The discriminant classifies physical damping regimes. For \(a y'' + b y' + c y = 0\), the discriminant \(D = b^2 - 4ac\) dictates whether the circuit response is overdamped, critically damped, or underdamped.

Roots and solution forms from the quadratic formula

Starting point. For \(a r^2 + b r + c = 0\), the characteristic roots are \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

  1. \[D > 0 \implies r_1, r_2 \in \mathbb{R}, \quad y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\]
    Two distinct real exponential rates: pure decay without oscillation (overdamped).reversible
  2. \[D = 0 \implies r = -\frac{b}{2a}, \quad y_h(t) = (C_1 + C_2 t) e^{-(b/2a)t}\]
    Repeated real root: the factor \(t\) provides the second linearly independent mode (critically damped).reversible
  3. \[D < 0 \implies r = \alpha \pm i\beta, \quad y_h(t) = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t)\]
    Complex conjugate roots with real part \(\alpha = -\frac{b}{2a}\) and frequency \(\beta = \frac{\sqrt{4ac - b^2}}{2a}\) (underdamped).reversible

Conclusion. The three mathematical cases correspond to three distinct physical dynamics. The sign of \(D = b^2 - 4ac\) uniquely selects the regime.

18.7 Non-homogeneous equations: choose, test, correct

For \(Ly=q(t)\), first solve \(Ly=0\) to obtain \(y_h\). Then choose a trial \(y_p\) whose derivatives remain in the same family as \(q(t)\). Substitute the entire trial and equate coefficients. If the trial overlaps \(y_h\), multiply it by \(t\) once for each overlap order; otherwise it cannot create a genuinely new solution.

Example — polynomial forcing

Solve a particular solution for

\[y''-y'=6t.\]

A degree-one polynomial forcing needs a degree-one trial, but its second derivative may remove terms; use \(y_p=At^2+Bt\) so that \(y_p''-y_p'\) can still produce a linear term. Then

\[y_p'=2At+B,\qquad y_p''=2A,\]

and substitution gives

\[2A-(2At+B)=6t.\]

Matching the \(t\) coefficient gives \(-2A=6\), so \(A=-3\). Matching constants gives \(2A-B=0\), so \(B=-6\). Thus \(y_p=-3t^2-6t\). The homogeneous solution should be added afterwards.

Example — resonance correction, fully checked

Find a particular solution of

\[y''+y=3\cos t.\]

Here \(y_h=C_1\cos t+C_2\sin t\), so a usual trial \(A\cos t+B\sin t\) overlaps. Multiply by \(t\) and try \(y_p=At\sin t\). Then

\[y_p'=A\sin t+At\cos t,\]
\[y_p''=2A\cos t-At\sin t.\]

Therefore

\[y_p''+y_p=2A\cos t.\]

Matching \(3\cos t\) gives \(A=3/2\), hence

\[y_p=\frac32t\sin t.\]

The factor \(t\) explains the increasing amplitude in the ideal undamped resonance model; it is not an arbitrary decoration.

Trial forms for undetermined coefficients. The trial function \(y_p(t)\) must include all derivatives of the forcing function \(q(t)\).

  1. If \(q(t) = \text{polynomial of degree } n\): guess \(y_p(t) = A_n t^n + A_{n-1} t^{n-1} + \cdots + A_1 t + A_0\) (complete polynomial).
  2. If \(q(t) = C e^{kt}\): guess \(y_p(t) = A e^{kt}\). If \(k\) matches a characteristic root, multiply by \(t\); if a repeated root, multiply by \(t^2\).
  3. If \(q(t) = C\cos(\omega t)\) or \(D\sin(\omega t)\): guess \(y_p(t) = A\cos(\omega t) + B\sin(\omega t)\) (always include both sine and cosine terms). Multiply by \(t\) if \(\pm i\omega\) matches a characteristic root (resonance).
  4. If \(q(t)\) is a sum of terms: guess the sum of the respective individual trial forms (superposition principle).
  5. If \(q(t)\) is a product (e.g. \(t e^{2t}\cos 3t\)): guess the product of the corresponding trial forms.
  6. After formulating \(y_p\), substitute into the ODE, collect like terms, and solve the algebraic system for the unknown coefficients.

18.8 A second-order solution checklist

  1. Put the ODE in standard form and identify whether it is homogeneous.
  2. Form and solve the characteristic equation before writing a solution shape.
  3. Select the correct distinct, repeated, or complex-root family.
  4. If forced, find one particular solution independent of the homogeneous family.
  5. Apply both initial conditions and substitute back into the original equation.
  6. Interpret signs of real parts: negative means decay, zero means persistent oscillation, positive means growth.

Second-order linear ODEs govern oscillatory and resonant systems. The equation \(a y'' + b y' + c y = q(t)\) models mass-spring-damper systems and RLC circuits.

The characteristic equation

Starting point. Substitute trial solution \(y = e^{rt}\) into the homogeneous equation \(a y'' + b y' + c y = 0\).

  1. \[y = e^{rt} \implies y' = r e^{rt}, \quad y'' = r^2 e^{rt}\]
    Differentiate the exponential ansatz.reversible
  2. \[a(r^2 e^{rt}) + b(r e^{rt}) + c(e^{rt}) = 0\]
    Substitute derivatives into the differential equation.reversible
  3. \[(a r^2 + b r + c) e^{rt} = 0\]
    Factor out \(e^{rt}\) (which is strictly positive and never zero).reversible
  4. \[a r^2 + b r + c = 0\]
    The characteristic algebraic quadratic equation must equal zero.reversible

Conclusion. The characteristic equation \(a r^2 + b r + c = 0\) is obtained by replacing \(y'' \to r^2\), \(y' \to r\), and \(y \to 1\). Its roots \(r_1, r_2\) define the exponential modes of the system.

  1. Write characteristic equation: \(a r^2 + b r + c = 0\).
  2. Compute discriminant: \(D = b^2 - 4ac\).
  3. Case 1 (\(D > 0\), distinct real roots \(r_1 \neq r_2\)): \(y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\) (overdamped).
  4. Case 2 (\(D = 0\), repeated real root \(r\)): \(y_h(t) = (C_1 + C_2 t) e^{rt}\) (critically damped).
  5. Case 3 (\(D < 0\), complex roots \(\alpha \pm i\beta\)): \(y_h(t) = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t)\) (underdamped).
  6. Apply two initial conditions \(y(0) = y_0\) and \(y'(0) = v_0\) to find \(C_1\) and \(C_2\).

Common Mistakes

  • Using the distinct-root formula when the characteristic roots repeat.
  • Omitting the exponential envelope in an under-damped response.
  • Calling every non-homogeneous solution a steady state even when the forcing is not persistent or the system is unstable.
  • Choosing a particular trial that overlaps the homogeneous solution and failing to multiply by \(t\).

The general solution to a non-homogeneous ODE is \(y = y_h + y_p\). The homogeneous solution \(y_h\) captures the natural response (determined by the circuit's internal components). The particular solution \(y_p\) captures the forced steady-state response (driven by the source \(q(t)\)).

Method of undetermined coefficients

Starting point. For constant-coefficient ODE \(a y'' + b y' + c y = q(t)\), guess a trial form \(y_p(t)\) matching \(q(t)\).

  1. \[q(t) = P_n(t) e^{\gamma t} \implies y_p(t) = t^s Q_n(t) e^{\gamma t}\]
    Match the functional form of \(q(t)\). The multiplier \(t^s\) (\(s = 0, 1, 2\)) accounts for overlap with the homogeneous solution, where \(s\) is the multiplicity of \(\gamma\) as a characteristic root.reversible
  2. \[a y_p'' + b y_p' + c y_p = q(t)\]
    Substitute \(y_p\) into the differential equation and equate coefficients of matching terms (\(t^k\), \(\cos\omega t\), \(\sin\omega t\), \(e^{kt}\)) to solve for the unknown constants.reversible

Conclusion. The particular solution mirrors the driving function \(q(t)\), modified by \(t^s\) if there is duplication with the homogeneous natural modes.

  1. 1. Solve the homogeneous equation: find \(y_h(t)\) from the characteristic roots.
  2. 2. Inspect the driving term \(q(t)\): polynomial, exponential, or sinusoid?
  3. 3. Formulate the trial guess \(y_p(t)\) with undetermined coefficients.
  4. 4. If any term in the guess appears in \(y_h(t)\), multiply the entire guess by \(t\) (or \(t^2\) for double roots).
  5. 5. Substitute \(y_p\) into the ODE and solve the linear system for the unknown coefficients.
  6. 6. Write the complete general solution: \(y(t) = y_h(t) + y_p(t)\).
  7. 7. <strong>Crucial rule:</strong> apply initial conditions to the FULL solution \(y(t) = y_h(t) + y_p(t)\), never to \(y_h\) alone!

Chapter Summary

Second-order constant-coefficient ODEs reduce to a quadratic characteristic equation. Root types determine exponential, repeated-root, or damped-sinusoidal forms. In RLC systems, damping compares \(\alpha\) with \(\omega_0\). Forced responses combine a homogeneous transient with a particular driven response.

Exercises

  1. Solve \(y''-5y'+6y=0\).
  2. Solve \(y''+4y'+4y=0\) with \(y(0)=1\), \(y'(0)=0\).
  3. Classify an RLC response with \(\alpha=5\ \mathrm{s}^{-1}\) and \(\omega_0=12\ \mathrm{s}^{-1}\).
  4. Find a particular solution of \(y''+y=3\cos t\), explaining the required trial modification.
  5. Explain, in physical terms, why resistance changes an ideal LC oscillator into a decaying response.
  6. Solve \(y''-5y'+6y=0\) with \(y(0)=1\) and \(y'(0)=0\).
  7. Solve \(y''+6y'+25y=0\) and identify its envelope and oscillation frequency.
  8. For \(R=4\ \Omega\), \(L=1\ \mathrm H\), \(C=1/9\ \mathrm F\), classify the natural response and find \(\omega_d\) where applicable.
  9. Find a particular solution of \(y''-y'=6t\).
  10. Solve for a particular solution of \(y''+y=3\cos t\) and explain why a non-resonant trial fails.

Reference appendices

Appendices

Appendix A — Essential Algebra and Notation Reference

Algebraic identities

\[(a+b)^2=a^2+2ab+b^2,qquad(a-b)^2=a^2-2ab+b^2,\]
\[a^2-b^2=(a-b)(a+b),qquad (a+b)(a-b)=a^2-b^2.\]

Essential derivative and integral conditions

\[\frac d{dx}x^n=nx^{n-1},\quad \frac d{dx}e^{ax}=ae^{ax},\quad \frac d{dx}\ln x=\frac1x\;(x>0),\]
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\;(n\ne-1),\quad \int\frac1x\,dx=\ln|x|+C.\]

Trigonometric and hyperbolic identities

\[\sin^2x+\cos^2x=1,qquad e^{ix}=\cos x+i\sin x,\]
\[\cosh^2x-\sinh^2x=1.\]

Appendix B — Selected Hints and Answers

These are checks, not substitutes for working through the exercises.

  • Chapter 3 Exercise 2: the second equation is twice the first, so the two lines coincide.
  • Chapter 5 Exercise 1: the sequence alternates between \(-1\) and \(1\); use an \(\varepsilon\) smaller than \(1\) to see why no single limit works.
  • Chapter 8 Exercise 4: three applications of L’Hôpital’s rule give \(-1/6\).
  • Chapter 10 Exercise 4: the radius is \(1\); one endpoint diverges and the other converges conditionally.
  • Chapter 12 Exercise 5: use \(u=t/\tau\) or integrate the exponential directly.
  • Chapter 18 Exercise 3: it is under-damped and \(\omega_d=\sqrt{12^2-5^2}\ \mathrm{s}^{-1}\).

Appendix C — Source-Coverage Ledger

All 88 source lectures have been synthesised into the textbook rather than reproduced as lecture units:

Source lecturesTextbook chapters
L01–L021
L03–L05, L142, 7
L06–L073
L08–L134
L15–L215
L22–L286, 11
L29–L357
L36–L448–9
L45–L5010
L51–L53, L64–L6611
L54–L5712
L58–L6313
L67–L69, L7114
L70, L72–L74, L77–L7815
L75–L76, L79–L8216
L83–L8517
L86–L8818

Every source lecture is represented at least once. Repeated lecture treatments are deliberately merged where they develop the same mathematical concept.

Appendix D — Extended Worked Methods

This appendix gives longer examples for methods that tend to need more than one exposure. It is not a replacement for the chapters: each example identifies the same assumptions and checks that should become habits in new problems.

D.1 A complete Gaussian-elimination example

Solve

\[\begin{aligned} x+2y-z&=3,\\ 2x-y+3z&=7,\\ 3x+y+2z&=8. \end{aligned}\]

Use the first row as a pivot. Replace row 2 by row 2 minus twice row 1, and row 3 by row 3 minus three times row 1:

\[\left[\begin{array}{ccc|c} 1&2&-1&3\\ 0&-5&5&1\\ 0&-5&5&-1 \end{array}\right].\]

Now replace row 3 by row 3 minus row 2:

\[\left[\begin{array}{ccc|c} 1&2&-1&3\\ 0&-5&5&1\\ 0&0&0&-2 \end{array}\right].\]

The final row says \(0=-2\). It is a contradiction, so the original three planes have no common point and the system has no solution. Continuing with back-substitution would be a mistake: that procedure applies only after consistency has been established.

D.2 A recursive sequence: existence before value

Let

\[a_{n+1}=\frac12\left(a_n+\frac2{a_n}\right),\qquad a_1=2.\]

The intended limit appears to be \(\sqrt2\), but guessing it is not a proof that a limit exists. First note \(a_n>0\). For any positive \(a_n\), the arithmetic-geometric mean inequality gives

\[\frac12\left(a_n+\frac2{a_n}\right)\ge\sqrt{a_n\frac2{a_n}}=\sqrt2.\]

Thus \(a_{n+1}\ge\sqrt2\). One then proves that once a term is at least \(\sqrt2\), the next term does not exceed it. Indeed,

\[a_{n+1}-a_n=\frac{2-a_n^2}{2a_n}\le0\]

when \(a_n\ge\sqrt2\). The sequence is non-increasing from its initial value and bounded below by \(\sqrt2\), so it converges. If its limit is \(L>0\), taking limits in the recursion gives

\[L=\frac12\left(L+\frac2L\right),\]

so \(L^2=2\). Positivity selects \(L=\sqrt2\). The order is important: establish convergence, then solve the limiting equation, then select the root compatible with the sequence.

D.3 A derivative-based optimisation checklist

An open box is made by cutting squares of side \(x\) from the corners of a \(30\text{ cm}\times20\text{ cm}\) sheet and folding up the sides. The physical domain is not all real \(x\):

\[0<x<10.\]

The volume is

\[V(x)=x(30-2x)(20-2x)=600x-100x^2+4x^3.\]

Differentiate:

\[V'(x)=600-200x+12x^2.\]

Set this equal to zero and divide by \(4\):

\[3x^2-50x+150=0.\]

The quadratic formula gives candidate values. Only values inside \((0,10)\) are physically admissible. Finally compare the derivative sign or the volumes near the valid critical point and the limiting endpoint behaviour. This is more secure than reporting every algebraic root as a proposed box size.

D.4 Taylor approximation with an error bound

Approximate \(e^{0.2}\) by the cubic Maclaurin polynomial:

\[T_3(0.2)=1+0.2+\frac{0.2^2}{2}+\frac{0.2^3}{6}=1.221\overline{3}.\]

For \(e^x\), every derivative equals \(e^x\). On the interval from \(0\) to \(0.2\), \(e^\xi\le e^{0.2}\). Taylor’s remainder therefore satisfies

\[|R_3(0.2)|\le\frac{e^{0.2}}{4!}(0.2)^4.\]

The error estimate explains why three terms beyond the constant are already effective at this small input. It also explains why the same approximation should not be trusted automatically at a much larger input.

D.5 Choosing a convergence test

Consider

\[\sum_{n=1}^{\infty}\frac{n^3}{2^n}.\]

Factorials are absent, but the exponential \(2^n\) suggests a ratio. With \(a_n=n^3/2^n\),

\[\left|\frac{a_{n+1}}{a_n}\right| =\frac{(n+1)^3}{2n^3} =\frac12\left(1+\frac1n\right)^3\to\frac12<1.\]

The series converges absolutely. In contrast, for \(\sum 1/(n^2+4n)\), comparison with \(1/n^2\) or partial fractions is more natural than a ratio test that approaches \(1\) and says nothing.

D.6 Power series interval, not just radius

Find the convergence interval of

\[\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}.\]

The ratio test gives \(|x-2|<1\), so the radius is \(1\). At \(x=3\), the series is \(\sum1/n\), which diverges. At \(x=1\), it is

\[\sum_{n=1}^{\infty}\frac{(-1)^n}{n},\]

which converges by the alternating-series criterion. Hence the interval is \([1,3)\), not simply “radius one.” A radius describes distance from the centre; the interval records endpoint decisions.

D.7 Area requires a region analysis

Find the area between \(y=x\) and \(y=x^2\) from \(x=0\) to \(x=1\). On this interval, \(x\ge x^2\), so

\[A=\int_0^1(x-x^2)\,dx =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\frac16.\]

If the interval were \([-1,1]\), the relative position changes: \(x^2\) lies above \(x\) on \([-1,0]\), while \(x\) lies above \(x^2\) on \([0,1]\). One integral with one subtraction order would produce signed cancellation, not the desired total area.

D.8 Phase and complex multiplication

Let \(z_1=2e^{i\pi/6}\) and \(z_2=3e^{-i\pi/4}\). Then

\[z_1z_2=6e^{i(\pi/6-\pi/4)}=6e^{-i\pi/12}.\]

No rectangular multiplication was required because multiplication is naturally expressed in polar form. The magnitude becomes \(6\) and the phase becomes \(-\pi/12\). In an AC calculation, this is the reason gains multiply while phase shifts add.

D.9 A first-order RC initial-condition check

For an RC circuit with constant source \(V_s\) and initial capacitor voltage \(V_0\), the general response is

\[v_C(t)=V_s+(V_0-V_s)e^{-t/(RC)}.\]

At \(t=0\), it gives \(v_C(0)=V_0\). As \(t\to\infty\), it gives \(V_s\). Both checks are quick ways to catch a sign error. Differentiating gives

\[\frac{dv_C}{dt}=-\frac{V_0-V_s}{RC}e^{-t/(RC)},\]

which has the expected sign: if \(V_0<V_s\), voltage initially increases.

D.10 Second-order resonance correction

Solve a particular solution for

\[y''+y=4\cos t.\]

The homogeneous solution already contains \(\cos t\) and \(\sin t\). A trial \(A\cos t+B\sin t\) would reproduce the homogeneous family and fail to generate the forcing. Multiply by \(t\) and try

\[y_p=t(A\sin t+B\cos t).\]

After differentiating and substituting, coefficients determine the required values \(A=2\) and \(B=0\), so \(y_p=2t\sin t\). The additional factor \(t\) reflects resonance: forcing at a natural frequency builds a response whose amplitude is not a bounded steady sinusoid in the ideal undamped model.

Appendix E — Theorem Conditions and Further Connections

E.1 A proof pattern behind the Mean Value Theorem

The Mean Value Theorem is often remembered as a formula, but its proof explains why the conclusion is plausible. Let the secant slope between \((a,f(a))\) and \((b,f(b))\) be

\[m=\frac{f(b)-f(a)}{b-a}.\]

Construct the line through \((a,f(a))\) with that slope:

\[\ell(x)=f(a)+m(x-a).\]

Now define \(g(x)=f(x)-\ell(x)\). The two functions agree at \(a\) and \(b\), so \(g(a)=g(b)=0\). If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then so is \(g\). Rolle’s theorem supplies a \(c\) with \(g'(c)=0\). Since

\[g'(c)=f'(c)-m,\]

we get \(f'(c)=m\). The line is not an artificial trick: subtracting it removes the overall rise so that the remaining graph begins and ends at the same height.

E.2 Why a local extremum has zero derivative

Suppose \(f\) has a local maximum at an interior point \(c\) and is differentiable there. For small positive \(h\),

\[f(c+h)-f(c)\le0,\]

so dividing by positive \(h\) gives a right difference quotient no greater than zero. For small negative \(h\), the numerator is still no greater than zero but division reverses the inequality, so the left quotient is no less than zero. If the two-sided derivative exists, it must be simultaneously at most and at least zero. Hence \(f'(c)=0\).

This result does not say every derivative-zero point is an extremum: \(f(x)=x^3\) has \(f'(0)=0\) but increases through \(0\). It also does not include a sharp corner, where a maximum can occur even though no derivative exists.

E.3 From geometric series to a rational function

Within \(|x|<1\),

\[1+x+x^2+\cdots=\frac1{1-x}.\]

Differentiate both sides within the common radius of convergence:

\[1+2x+3x^2+\cdots=\frac1{(1-x)^2}.\]

The equality is not a formal cancellation of an infinite string. It means every partial polynomial has a limit there, the derivative series has a limit there, and the power-series theorem justifies exchanging limiting and differentiating. At \(x=1\), neither series converges, so neither identity is valid as a numerical sum at that endpoint.

E.4 The fundamental theorem as a rate-of-accumulation statement

Let

\[A(x)=\int_a^x f(t)\,dt.\]

For a small increment \(h\), the change is

\[A(x+h)-A(x)=\int_x^{x+h}f(t)\,dt.\]

Divide by \(h\). The right side is the average value of \(f\) over the tiny interval from \(x\) to \(x+h\). When \(f\) is continuous, that average approaches \(f(x)\) as the interval shrinks. Thus \(A'(x)=f(x)\). This is the accumulated-quantity counterpart of the secant-to-tangent idea behind derivatives.

E.6 Branches and inverse functions

An inverse must return one value. That is why \(\arcsin(1/2)=\pi/6\), not an unlimited list of angles such as \(5\pi/6\) or \(\pi/6+2\pi k\). Those other angles solve \(\sin\theta=1/2\), but they are not values of the chosen inverse-function branch. The same distinction appears for complex arguments: \(\arg z\) can denote all angles differing by \(2\pi\), while \(\operatorname{Arg}z\) denotes one chosen principal value.

E.7 A compact particular-solution trial guide

For a linear constant-coefficient ODE, choose a trial whose derivatives stay in the same family:

Forcing \(q(t)\)First trial for \(y_p\)
Constant \(K\)\(A\)
Polynomial degree \(m\)polynomial degree \(m\)
\(e^{\lambda t}\)\(Ae^{\lambda t}\)
\(\cos\omega t\) or \(\sin\omega t\)\(A\cos\omega t+B\sin\omega t\)
\(e^{\lambda t}\) times polynomial/trigmatching product form

If any trial component appears in \(y_h\), multiply the whole trial by \(t\). If it still overlaps because the characteristic root is repeated, multiply by the needed higher power of \(t\). This rule is a linear-independence requirement, not a memorisation exception.

Appendix F — Cumulative Review Problems

The problems below deliberately mix topics. They are useful after completing a Part because practical models seldom announce which mathematical tool is wanted.

Part I Review

  1. A device is described by \(V=12-0.5I\), with \(V\) in volts and \(I\) in amperes. State the slope and intercept, and explain their units.
  2. Find the line through \((1,-2)\) perpendicular to \(2x-3y=6\).
  3. A quadratic has roots \(1\) and \(4\) and leading coefficient \(2\). Construct it, then find its vertex.
  4. Determine all values of \(k\) for which the system \(x+y=1\), \(2x+2y=k\) is consistent.
  5. Explain geometrically why a consistent \(2\times3\) system can have a line of solutions but a consistent \(3\times2\) system may still have only one point.

Part II Review

  1. Prove that \((2n+1)/(n+3)\to2\) without treating infinity as a number.
  2. Find each one-sided limit of \((x-1)/|x-1|\) at \(x=1\).
  3. Decide whether \(f(x)=\sqrt{x^2}\) is differentiable at \(0\) and explain why the expression must first be recognised as \(|x|\) over the reals.
  4. Construct a continuous piecewise function that is not differentiable at \(0\).
  5. Compare the long-run behaviour of \(x^5\) and \(e^{0.1x}\); explain why a graph over a short interval can make the wrong one appear larger.

Part III Review

  1. Find the tangent line to \(y=\ln x\) at \(x=1\) and use it to approximate \(\ln(1.03)\).
  2. A measured quantity is \(A=\pi r^2\). Derive both absolute and relative differential error approximations in terms of \(dr\).
  3. Find intervals of increase and decrease for \(f(x)=x/(x^2+1)\).
  4. Determine the absolute maximum and minimum of \(x^3-3x\) on \([-2,2]\).
  5. Use a second-degree Taylor polynomial to approximate \(\cos(0.1)\) and state why the omitted cubic term vanishes.

Part IV Review

  1. Decide whether \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\) converges absolutely, conditionally, or diverges. State the test used at each stage.
  2. Find the first four nonzero terms of the power series for \(1/(1+x)\) and its interval of convergence.
  3. Compute \(\int_0^1 x\ln x\,dx\) by integration by parts and explain the limiting interpretation at \(x=0\).
  4. A current is \(i(t)=I_0(1-e^{-t/\tau})\). Find charge from \(0\) to \(T\) and check the small-\(T\) limit qualitatively.
  5. Find the area enclosed by \(y=x^2\) and \(y=2x\); identify the intersections before integrating.

Part V–VI Review

  1. Convert \(3-3i\) to polar form and compute its cube using De Moivre’s rule.
  2. Show that \(e^{i\pi}+1=0\) follows immediately from Euler’s formula; state why this is a special identity rather than a definition of \(e\) or \(i\).
  3. Solve \(y'+4y=8\sin t\) by an integrating factor or a complex-exponential particular solution.
  4. For \(y''+2y'+5y=0\), classify the roots and write the real general solution.
  5. An RLC response has \(\alpha=\omega_0\). Explain the characteristic-root form and the physical meaning of critical damping.

Appendix G — Selected Worked Solutions and Hints

These solutions model complete method choice, not just answer production. Try a problem first. Then compare your diagram, stated domain, method, algebra, units, and final check with the solution. A short hint is given for a few problems whose main benefit is the first decision rather than lengthy arithmetic.

Example — G.1 Quadratic construction and vertex (Appendix F, Problem 3)

A quadratic with roots \(1\) and \(4\) has factors \((x-1)\) and \((x-4)\). A leading coefficient of \(2\) gives

\[f(x)=2(x-1)(x-4)=2x^2-10x+8.\]

For the vertex, complete the square:

\[\begin{aligned} f(x)&=2(x^2-5x)+8\\ &=2\left[x^2-5x+\left(\frac52\right)^2-\left(\frac52\right)^2\right]+8\\ &=2\left(x-\frac52\right)^2-\frac{25}{2}+8\\ &=2\left(x-\frac52\right)^2-\frac92. \end{aligned}\]

The vertex is \((5/2,-9/2)\). The coefficient \(2>0\) means the parabola opens upward, so this vertex is a minimum. Factored form was best for roots; completed square form was best for the vertex. Keeping both forms is often more useful than expanding once and discarding the structure.

Example — G.2 Consistency in a parameterised system (Appendix F, Problem 4)

The system is

\[x+y=1,\qquad 2x+2y=k.\]

Doubling the first equation gives \(2x+2y=2\). Thus the second equation is consistent exactly when \(k=2\). If \(k=2\), the two equations represent the same line and there are infinitely many solutions. If \(k\ne2\), they represent parallel distinct lines and there is no solution. There is never one unique solution because the left sides are dependent.

Example — G.3 A one-sided limit and its graph (Appendix F, Problem 7)

For \(x>1\), \(|x-1|=x-1\), so

\[\frac{x-1}{|x-1|}=1.\]

For \(x<1\), \(|x-1|=-(x-1)\), so the quotient is \(-1\). Therefore

\[\lim_{x\to1^-}\frac{x-1}{|x-1|}=-1, \qquad \lim_{x\to1^+}\frac{x-1}{|x-1|}=1.\]

The two-sided limit does not exist because the one-sided limits differ. The function itself is also undefined at \(x=1\). Visualise two horizontal rays, one at height \(-1\) to the left of \(1\) and one at height \(1\) to its right, each with an open circle at \(x=1\).

Example — G.4 Tangent-line approximation (Appendix F, Problem 11)

For \(f(x)=\ln x\), the point \(x=1\) is convenient because

\[f(1)=0,\qquad f'(x)=\frac1x,\qquad f'(1)=1.\]

The tangent line is

\[L(x)=f(1)+f'(1)(x-1)=x-1.\]

Thus

\[\ln(1.03)\approx L(1.03)=0.03.\]

Because \(\ln x\) is concave down for \(x>0\) (\(f''(x)=-1/x^2<0\)), its tangent line lies above the graph nearby. Therefore \(0.03\) is a slight overestimate, an interpretive check that is unavailable from calculator arithmetic alone.

Example — G.5 Differential error in an area (Appendix F, Problem 12)

The area is \(A=\pi r^2\). Differentiate:

\[dA=2\pi r\,dr.\]

For a small measurement error \(dr\), this is the absolute-error approximation \(\Delta A\approx2\pi r\,\Delta r\). Divide by \(A=\pi r^2\) (with \(r\ne0\)):

\[\frac{dA}{A}=\frac{2\pi r\,dr}{\pi r^2}=2\frac{dr}{r}.\]

Hence the relative area error is approximately twice the relative radius error. This result is dimensionless, as every relative error must be.

Example — G.6 Alternating series: conditional versus absolute convergence

Consider

\[\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n.\]

For absolute convergence, remove signs:

\[\sum_{n=1}^{\infty}\left|\frac{(-1)^{n+1}}n\right| =\sum_{n=1}^{\infty}\frac1n.\]

This harmonic series diverges. So the original series is not absolutely convergent. Its positive magnitudes \(1/n\) decrease to \(0\), and the signs alternate; the alternating-series test therefore proves convergence. It is conditionally convergent. “Terms go to zero” is only one required condition; it is not itself a convergence test.

Example — G.7 Integration by parts with an improper endpoint (Appendix F, Problem 18)

The integral

\[\int_0^1x\ln x\,dx\]

is improper because \(\ln x\) is unbounded at \(0\). First replace zero by \(a>0\):

\[\int_a^1x\ln x\,dx.\]

Choose \(u=\ln x\), \(dv=x\,dx\). Then \(du=dx/x\) and \(v=x^2/2\), so

\[\int x\ln x\,dx=\frac{x^2}{2}\ln x-\int\frac{x}{2}\,dx =\frac{x^2}{2}\ln x-\frac{x^2}{4}+C.\]

Therefore

\[\int_0^1x\ln x\,dx =\lim_{a\to0^+} \left[\frac{x^2}{2}\ln x-\frac{x^2}{4}\right]_a^1.\]

At \(x=1\) the value is \(-1/4\). The limit \(a^2\ln a\) is \(0\) (for example, use \(a=e^{-u}\) and note \(-ue^{-2u}\to0\)), so the lower endpoint contributes zero. The answer is \(-1/4\). The sign is plausible because \(x\ln x<0\) on \((0,1)\).

Example — G.8 A complete first-order forced response (Appendix F, Problem 23)

Solve

\[y'+4y=8\sin t.\]

The integrating factor is \(\mu=e^{4t}\). Multiplication gives

\[(e^{4t}y)'=8e^{4t}\sin t.\]

Use the standard result, obtainable by two integrations by parts,

\[\int e^{4t}\sin t\,dt =\frac{e^{4t}}{17}(4\sin t-\cos t)+C.\]

Thus

\[e^{4t}y=\frac{8e^{4t}}{17}(4\sin t-\cos t)+C,\]

and

\[y(t)=\frac{32}{17}\sin t-\frac8{17}\cos t+Ce^{-4t}.\]

The trigonometric terms are the persistent forced response; \(Ce^{-4t}\) is the transient. An initial condition would determine \(C\).

Example — G.9 Complex power in polar form (Appendix F, Problem 21)

For \(z=3-3i\),

\[|z|=\sqrt{3^2+(-3)^2}=3\sqrt2, \qquad \operatorname{Arg}z=-\frac{\pi}{4}.\]

Hence

\[z=3\sqrt2\,e^{-i\pi/4}.\]

De Moivre's formula gives

\[z^3=(3\sqrt2)^3e^{-3i\pi/4} =54\sqrt2\left(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}\right) =-54-54i.\]

The magnitude check is \(|z^3|=|z|^3=54\sqrt2\), which agrees with \(|-54-54i|=54\sqrt2\).

Example — G.10 Critical damping (Appendix F, Problem 25)

When \(\alpha=\omega_0\), the characteristic equation

\[s^2+2\alpha s+\omega_0^2=0\]

becomes

\[s^2+2\alpha s+\alpha^2=(s+\alpha)^2=0.\]

There is one repeated root \(s=-\alpha\), so the response is

\[i(t)=(C_1+C_2t)e^{-\alpha t}.\]

It does not oscillate because no imaginary root part remains. It tends to zero for \(\alpha>0\), and it returns without the slower two-mode behaviour associated with over-damping. The phrase “fastest non-oscillatory return” describes this family under the ideal linear model; actual component tolerances can shift a constructed circuit away from exact critical damping.

Hints for additional practice

  • Chapter 12 Exercise 9: find the zeros of velocity first; distance integrates \(|v(t)|\), so split the interval where the sign changes.
  • Chapter 13 Exercise 8: solve the curve-axis intersections before deciding top and bottom.
  • Chapter 14 Exercise 9: write \(8i=8e^{i(\pi/2+2\pi k)}\) and divide arguments by three.
  • Chapter 15 Exercise 11: differentiate voltage, then multiply by capacitance; convert microfarads to farads.
  • Chapter 17 Exercise 9: use \(v_C=V_s+(V_0-V_s)e^{-t/(RC)}\) and check both endpoint behaviours.
  • Chapter 18 Exercise 10: identify the overlap with the homogeneous \(\cos t\) term before choosing the trial.