Chapter 20

Medium

Taylor Expansions and Approximation in Applied Contexts

00 · Symbol Glossary

$L(x)$L of x — linearization

The degree- $1$ Taylor polynomial $T_1(x)$ (Chapter 11), renamed $L(x)$ here to emphasize its role as the best straight-line approximation to $f$ near a point.

$\Delta x$Delta x — input perturbation

A small change applied to the input of a function, distinct from $dx$ in that it need not be infinitesimally small — used throughout this chapter to represent a realistic-sized error or shift.

$\Delta f \approx f'(a)\,\Delta x$propagated change (linear approximation)

The estimated change in $f$ 's output caused by a small change $\Delta x$ in its input, using only the first-derivative term of the Taylor polynomial (Chapter 11).

01 · Linearization

Chapter 11 built the Taylor polynomial $T_n(x)$ and bounded its error with the Lagrange remainder. The simplest, most-used case in practice is $n=1$ : approximating a function by its tangent line.

Definition — Linearization

The linearization of $f$ at $a$ is $T_1(x)$ from Chapter 11, given a new name to emphasize its use:

L(x) = f(a) + f'(a)(x-a)

For $x$ near $a$ : $f(x) \approx L(x)$ , with error bounded by Chapter 11's formula (with $n=1$ ): $\lvert R_1(x)\rvert \leq \dfrac{M}{2!}\lvert x-a\rvert^2$ , where $M$ bounds $\lvert f''\rvert$ between $a$ and $x$ .

Step-by-step — Linearize $f(x)=\sqrt{x}$ at $a=25$ and estimate $\sqrt{26}$

Compute $f(a)$ and $f'(a)$ :

f(25)=\sqrt{25}=5

f'(x)=\dfrac12x^{-1/2}

(Chapter 02 power rule), so

f'(25) = \dfrac{1}{2\sqrt{25}} = \dfrac{1}{10}

Build the linearization:

L(x) = 5+\dfrac{1}{10}(x-25)

Substitute $x=26$ :

L(26) = 5+\dfrac{1}{10}(1) = 5.1

Bound the error using Chapter 11's remainder formula:

f''(x) = -\dfrac14x^{-3/2}

, so

\lvert f''(x)\rvert = \dfrac{1}{4x^{3/2}}

, decreasing for larger

x

; on

[25,26]

its largest value is at

x=25

M=\dfrac{1}{4(125)}=\dfrac{1}{500}=0.002

Apply the bound with $n=1$ , $\lvert x-a\rvert=1$ :

\lvert R_1(26)\rvert \leq \dfrac{0.002}{2!}(1)^2 = 0.001

Check against the true value:

\sqrt{26}\approx5.0990

; the estimate

5.1

has actual error

\approx0.0010

, matching the bound from step 5 almost exactly.

02 · Quadratic Approximation

Including one more term from the Taylor polynomial (Chapter 10, Section 03) improves accuracy, at the cost of a slightly more complex formula.

Definition — Quadratic Approximation

Q(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2

This is $T_2(x)$ from Chapter 11, with error bounded by $\lvert R_2(x)\rvert \leq \dfrac{M}{3!}\lvert x-a\rvert^3$ where $M$ bounds $\lvert f'''\rvert$ .

Step-by-step — Quadratic approximation of $\sqrt{x}$ at $a=25$, re-estimating $\sqrt{26}$

**Reuse

f(25)=5

f'(25)=0.1

from Section 01.

Compute $f''(25)$ :

f''(x)=-\dfrac14x^{-3/2}

(Section 01, step 4), so

f''(25) = -\dfrac{1}{4(125)} = -0.002

Build $Q(x)$ :

Q(x) = 5+0.1(x-25) - \dfrac{0.002}{2}(x-25)^2 = 5+0.1(x-25)-0.001(x-25)^2

Substitute $x=26$ :

Q(26) = 5+0.1(1)-0.001(1)^2 = 5+0.1-0.001 = 5.099

Compare to the true value $\sqrt{26}\approx5.09902$ : the quadratic estimate's error is about

0.00002

, roughly

50\times

smaller than the linear estimate's error from Section 01 — one extra term meaningfully tightens the approximation.

03 · Error Propagation

Linearization has a second, equally important use: estimating how an error or uncertainty in an input propagates into an error in the output of a function applied to it.

Definition — Propagated Error (Linear Approximation)

If $x$ carries a small error or perturbation $\Delta x$ , the resulting change in $f(x)$ is approximately

\Delta f \approx f'(a)\,\Delta x

— exactly Section 01's linearization formula, read as "change in output $\approx$ slope $\times$ change in input."

Step-by-step — Propagate a pricing input error through $f(x) = x^3$

Set up the scenario: a quantity

x=10

is measured with a possible error of

\Delta x = 0.2

; find the approximate resulting error in

f(x)=x^3

Differentiate:

f'(x) = 3x^2

(Chapter 02 power rule), so

f'(10) = 300

Apply the propagated-error formula:

\Delta f \approx f'(10)\cdot\Delta x = 300(0.2) = 60

Check against the exact change:

f(10.2)-f(10) = 10.2^3-10^3 = 1061.208-1000=61.208

. The linear estimate (

60

) is close to the exact change (

61.208

) but not identical — the gap (

1.208

) is exactly the higher-order remainder terms the linear approximation ignores.

❌ Trusting the linear approximation far from the center point

Applying the same linearization from Section 01 ( $L(x)=5+0.1(x-25)$ for $\sqrt x$ ) to estimate $\sqrt{100}$ instead of $\sqrt{26}$ : $L(100) = 5+0.1(75) = 5+7.5=12.5$ .

Why it breaks: the true value is $\sqrt{100}=10$ , so the linear estimate is off by $2.5$ — far worse than Section 01's error of about $0.001$ at $x=26$ . Chapter 11's error bound formula explicitly grows with $\lvert x-a\rvert^{n+1}$ ; here $\lvert x-a\rvert = \lvert100-25\rvert=75$ , dramatically larger than the $\lvert x-a\rvert=1$ used validly in Section 01.

Consequence: a linearization (or any fixed-order Taylor approximation) is only trustworthy for $x$ close to the center $a$ — the error bound's dependence on $\lvert x-a\rvert^{n+1}$ means accuracy degrades rapidly, not gracefully, as $x$ moves away from $a$ . Always check that the intended input is close to the center before relying on a linear or quadratic approximation.

04 · Quant Application — Price Sensitivity to a Changing Input

A quantity's sensitivity to a changing driver is a direct linear/quadratic approximation problem, exactly as in Sections 01–02.

Example — Approximating a price change from first and second derivatives

A bond's price as a function of the prevailing yield $y$ is $P(y)$ , currently at $y=0.04$ (i.e. $4\%$ ) with $P(0.04)=1000$ dollars. Suppose $P'(0.04) = -4200$ and $P''(0.04)=38{,}000$ (values obtained from the pricing model, not derived here). Estimate the price if the yield rises to $y=0.045$ (a $\Delta y=0.005$ increase), using the quadratic approximation from Section 02:

\Delta P \approx P'(0.04)\,\Delta y + \frac{P''(0.04)}{2}(\Delta y)^2

$\Delta P \approx (-4200)(0.005) + \dfrac{38{,}000}{2}(0.005)^2 = -21 + 19{,}000(0.000025) = -21+0.475 = -20.525$ .

The linear term alone ( $-21$ ) already captures most of the price move (bond prices fall as yields rise, consistent with $P'(0.04)<0$ ); the quadratic term ( $+0.475$ ) is a smaller correction that slightly offsets the linear estimate — in finance, these two terms are traditionally named duration (the linear, first-derivative sensitivity) and convexity (the quadratic, second-derivative correction), but the underlying mathematics is exactly Section 02's Taylor approximation, nothing more.

05 · Small-Input Approximations

Certain Taylor approximations are common enough, for small enough inputs, to be used as standing rules of thumb.

Definition — Small-Angle Approximations

For $x$ near $0$ (in radians — Chapter 02's trig derivatives were built on limit facts stated in radians, and these approximations inherit that requirement):

\sin x \approx x \qquad \cos x \approx 1-\frac{x^2}{2}

Both follow directly from truncating the Maclaurin series (Chapter 10, 11) after the first nonzero term.

Step-by-step — Bound the error of sin(x) ≈ x at x = 0.1

Identify this as a Taylor approximation with $a=0$ , $n=1$ :

\sin x \approx x

T_1(x)

for

f(x)=\sin x

(since

f(0)=0

f'(0)=\cos0=1

, giving

T_1(x)=0+1\cdot x=x

, Chapter 02, Section 09).

Apply Chapter 11's error bound with $M$ bounding $\lvert f''\rvert=\lvert-\sin x\rvert\leq1$ always:

\lvert R_1(0.1)\rvert \leq \dfrac{1}{2!}(0.1)^2 = \dfrac{0.01}{2}=0.005

Check against the true value:

\sin(0.1)\approx0.09983

; the approximation

0.1

has actual error

\approx0.00017

, comfortably inside the bound of

0.005

❌ Applying the small-angle approximation with degrees instead of radians

Estimating $\sin(10°)$ using $\sin x\approx x$ by plugging in $x=10$ directly: $\sin(10°)\overset{?}{\approx}10$ .

Why it breaks: Chapter 02, Section 09 built $\dfrac{d}{dx}[\sin x]=\cos x$ using the limit fact $\lim_{h\to0}\dfrac{\sin h}{h}=1$ , which is only true when $h$ is measured in radians — in degrees, that limit evaluates to a different constant entirely (related to $\pi/180$ ), so the entire derivative (and hence the whole Taylor series built from it) assumed radian input from the start.

Consequence: $10°$ in radians is $10\times\dfrac{\pi}{180}\approx0.1745$ , and $\sin(10°)=\sin(0.1745\text{ rad})\approx0.1736$ — reasonably close to $0.1745$ (small-angle approximation applied correctly, in radians). But treating " $10$ " as if it were already in radians and writing $\sin(10°)\approx10$ is wrong by roughly a factor of $57$ — the small-angle approximation is only valid once the angle is converted to radians first.

06 · Exercises

EXERCISE 20.1

Compute $f(a)$ and $f'(a)$ at a nearby convenient point, then build and evaluate the linearization.

$f(x)=\sqrt[3]{x}$ near $a=8$ : $f(8)=2$ , $f'(x)=\frac13x^{-2/3}$ , $f'(8)=\frac13(4)^{-1}=\frac{1}{12}$ . $L(x)=2+\frac{1}{12}(x-8)$ . Estimate $\sqrt[3]{9}$ : $L(9)=2+\frac{1}{12}(1)=2+0.0833=2.0833$ . (True value $\sqrt[3]9\approx2.0801$ .)

Use linearization to estimate $\sqrt[3]{9}$ , centering at $a=8$ .

EXERCISE 20.2

Add the quadratic term to Exercise 20.1's linearization, using the second derivative evaluated at the same center point.

$f''(x) = -\frac29x^{-5/3}$ , $f''(8) = -\frac29(32)^{-1} = -\frac{2}{288}=-\frac{1}{144}$ . $Q(x) = 2+\frac{1}{12}(x-8) - \frac{1}{288}(x-8)^2$ . At $x=9$ : $Q(9) = 2+0.0833-\frac{1}{288}(1) = 2.0833-0.00347\approx2.0799$ , closer to the true value $2.0801$ than the linear estimate.

Use the quadratic approximation to re-estimate $\sqrt[3]{9}$ from Exercise 20.1, and compare accuracy.

EXERCISE 20.3

Differentiate to find the sensitivity (rate of change), then multiply by the given input perturbation.

$f(x) = x^2$ , $x=5$ , $\Delta x=0.1$ . $f'(x)=2x$ , $f'(5)=10$ . $\Delta f \approx 10(0.1)=1$ . Exact: $f(5.1)-f(5)=26.01-25=1.01$ — the linear estimate is close, off by $0.01$ .

Estimate the propagated change in $f(x)=x^2$ at $x=5$ when $\Delta x=0.1$ , and compare to the exact change.

EXERCISE 20.4

Recall which factor in the Taylor error bound formula grows as the input moves away from the center, and by how much.

The error bound $\lvert R_n(x)\rvert\leq\frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}$ (Chapter 11) grows with the $(n+1)$ -th power of the distance from the center $a$ . So doubling the distance $\lvert x-a\rvert$ increases the worst-case error bound by a factor of $2^{n+1}$ — for a linearization ( $n=1$ ), that's a factor of $4$ ; moving far from the center degrades accuracy rapidly, not proportionally.

Explain, using Chapter 11's error bound formula, why a linear approximation becomes unreliable "rapidly" (not just gradually) as the input moves away from the center point.

EXERCISE 20.5

This is the small-angle cosine approximation from Section 05 — apply it directly, making sure the input is in radians.

Estimate $\cos(0.2)$ (radians) using $\cos x\approx1-\frac{x^2}{2}$ : $1-\frac{(0.2)^2}{2} = 1-\frac{0.04}{2}=1-0.02=0.98$ . True value $\cos(0.2)\approx0.98007$ — accurate to about $0.00007$ .

Use the small-angle approximation to estimate $\cos(0.2)$ (in radians), and compare to the true value $\cos(0.2)\approx0.98007$ .

EXERCISE 20.6

Apply the same linear-plus-quadratic sensitivity formula from Section 04, using the given first and second derivative values.

A portfolio's value $V(r)$ as a function of an interest rate $r$ has $V(0.03)=50{,}000$ , $V'(0.03)=-180{,}000$ , $V''(0.03)=900{,}000$ . Estimate $\Delta V$ for $\Delta r=0.002$ : $\Delta V \approx V'(0.03)\Delta r + \frac{V''(0.03)}{2}(\Delta r)^2 = (-180{,}000)(0.002) + \frac{900{,}000}{2}(0.002)^2 = -360+450(0.000004) = -360+0.0018 \approx -359.998$ . The linear term dominates almost entirely here; the quadratic correction is negligible at this small $\Delta r$ .

A portfolio's value $V(r)$ has $V(0.03)=50{,}000$ , $V'(0.03)=-180{,}000$ , $V''(0.03)=900{,}000$ . Estimate the change in value $\Delta V$ if the rate rises by $\Delta r=0.002$ .

07 · Chapter Summary

Concept	Formula / Rule
Linearization	$L(x)=f(a)+f'(a)(x-a)$ ; error $\leq\frac{M}{2!}\lvert x-a\rvert^2$
Quadratic approximation	$Q(x)=L(x)+\frac{f''(a)}{2}(x-a)^2$ ; error $\leq\frac{M}{3!}\lvert x-a\rvert^3$
Propagated error	$\Delta f \approx f'(a)\Delta x$
Accuracy vs. distance from center	Error bound grows with $\lvert x-a\rvert^{n+1}$ — degrades rapidly, not gradually
Price sensitivity (linear + quadratic)	$\Delta P \approx P'(y)\Delta y + \frac{P''(y)}{2}(\Delta y)^2$
Small-angle approximations	$\sin x\approx x$ , $\cos x\approx1-\frac{x^2}{2}$ for small $x$ in radians

This closes the full 20-chapter Calculus sequence: limits and derivatives (01–05), integrals (06–08), series and Taylor approximation (09–12, 20), curves and multivariable calculus (13–17), and differential equations (18–19) — the complete single- and multi-variable toolkit, from first principles through applied approximation.