Part III · Differential Calculus and Local Approximation

Chapter 9 — Inverse Functions, Differentials, and Taylor Approximation

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 x² becomes invertible.

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

A restricted one-to-one curve and its inverse are reflections in y equals 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

Reflection in y equals x exchanges a steep forward slope with a shallow inverse slope: reciprocal slopes.

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

The tangent T₁ is the first Taylor overlay; its gap from the curve is the local linearisation error.

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

Choose the expansion centre a.
Match the value and successive derivatives at a.
Use a remainder estimate or local scale before trusting the approximation away from a.

Ordered Taylor overlays T₀, T₂, and T₆ match increasingly many local changes of cosine near zero.

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 $$1$$ .

[Figure: A one-to-one curve and its inverse reflected in

\(y=x\)

; tangent-line and quadratic Taylor approximations compared with a curved function near

\(a\)

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

Reciprocal slopes explain inverse sensitivity, while the labelled local tangent keeps differential error distinct from an exact finite change.

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 $0.980392\ldots$ .

[Figure: A smooth curve with

\(T_0\)

, tangent

\(T_1\)

, and curved

\(T_2\)

near

\(a\)

; show approximation error increasing away from the expansion point.]

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}

The ordered cosine overlays include T₆(0.2), with its numerical error below 10⁻⁷ at the labelled local input.

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

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