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.5 Change of variables: the two legal bookkeeping paths
For
\[\int_a^b f(g(x))g'(x)\,dx,\]
there are two safe paths.
Path 1: change bounds. Set \(u=g(x)\), replace \(g'(x)dx\) by \(du\), and replace \(a,b\) with \(g(a),g(b)\). Finish entirely in \(u\).
Path 2: return to the original variable. Find an antiderivative in \(u\), substitute \(u=g(x)\) back, and only then use \(x=a,b\).
The unsafe hybrid changes to \(u\) but keeps \(x\)-bounds. A simple dimensional check—asking “which variable does this bound describe?”—often exposes the error.
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/trig | matching 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 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.