Chapter 11

Rigorous

Taylor Series Advanced

00 · Symbol Glossary

$T_n(x)$T sub n of x — nth-degree Taylor polynomial

The finite partial sum of a Taylor series (Chapter 10) through the $x^n$ term: $T_n(x) = \displaystyle\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k$ . An ordinary polynomial, unlike the full infinite Taylor series.

$R_n(x)$R sub n of x — remainder

The error: $R_n(x) = f(x) - T_n(x)$ , the gap between the true function value and the $n$ -th degree Taylor polynomial's approximation of it.

$c$c — the unnamed point in the Lagrange remainder

A specific (but generally unknown) number strictly between $a$ and $x$ , guaranteed to exist by the Lagrange remainder formula in Section 01. Its exact value is rarely known — only its existence and the resulting bound matter in practice.

$M$M — a bound on the (n+1)th derivative

Any number satisfying $\lvert f^{(n+1)}(t)\rvert \leq M$ for every $t$ between $a$ and $x$ . Used to convert the Lagrange remainder formula (which involves the unknown $c$ ) into a concrete numerical bound on the error.

01 · Taylor's Theorem and the Lagrange Remainder

Chapter 10 built the infinite Taylor series of a function. In practice, only a finite number of terms are ever actually computed — so the natural question is: how far off is that finite approximation?

Definition — Taylor's Theorem with Lagrange Remainder

If $f$ has $n+1$ derivatives on an interval containing $a$ and $x$ , then

f(x) = T_n(x) + R_n(x), \qquad R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

for some (generally unknown) number $c$ strictly between $a$ and $x$ .

Reading the remainder formula

$R_n(x)$ has exactly the same shape as the next term that would appear in the Taylor series ( $n+1$ derivatives, divided by $(n+1)!$ , times $(x-a)^{n+1}$ ) — the only difference is that the derivative is evaluated at the mystery point $c$ instead of at $a$ itself. The proof of why such a $c$ must exist is a calculus result in its own right and is not derived here; the formula is used as a given tool for bounding error.

02 · Bounding the Error

Since $c$ is unknown, $R_n(x)$ itself cannot usually be computed exactly. But if a bound $M$ on $\lvert f^{(n+1)}\rvert$ can be found over the relevant interval, the remainder can be bounded without ever knowing $c$ .

Definition — Taylor Error Bound

If $\lvert f^{(n+1)}(t)\rvert \leq M$ for every $t$ between $a$ and $x$ , then

\lvert R_n(x)\rvert \leq \frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}

Step-by-step — Bound the error of $T_3(x)$ approximating $e^x$ at $x=0.5$

Recall the Maclaurin series of $e^x$ (Chapter 10, Section 03):

T_3(x) = 1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}

(degree-3 partial sum,

a=0

Identify $n$ and the derivative needed:

n=3

, so the remainder involves

f^{(4)}(c) = e^c

(Chapter 02, Section 10: every derivative of

e^x

e^x

Bound $e^c$ over the relevant interval:

c

lies between

a=0

and

x=0.5

, so

0<c<0.5

. Since

e^t

is increasing (Chapter 02, Section 10's derivative

e^x>0

everywhere means

e^x

is always increasing, Chapter 04, Section 01), the largest value of

e^c

on this interval is bounded by

e^{0.5}<e^1<3

(using

e\approx2.718<3

). Take

M=3

Apply the error bound formula:

\lvert R_3(0.5)\rvert \leq \dfrac{M}{(n+1)!}\lvert x-a\rvert^{n+1} = \dfrac{3}{4!}(0.5)^4 = \dfrac{3}{24}\times0.0625

Compute the numeric bound:

\dfrac{3}{24}=0.125

;

0.125\times0.0625 = 0.0078125

. So the degree-3 Taylor polynomial approximates

e^{0.5}

with error at most about

0.0078

— without ever computing

e^{0.5}

exactly first.

Verify against the known value:

T_3(0.5) = 1+0.5+0.125+0.0208\overline{3} = 1.6458\overline{3}

; the true value

e^{0.5}\approx1.64872

. Actual error

\approx0.00289

, comfortably within the bound of

0.0078

from step 5 — the bound is a guarantee, not a tight estimate.

❌ Using a bound M that only holds at one point, not the whole interval

On the same problem, a common mistake: setting $M = f^{(4)}(0) = e^0 = 1$ — evaluating the bounding derivative only at the center $a=0$ instead of over the whole interval from $a$ to $x$ .

Why it breaks: the error bound formula requires $M$ to bound $\lvert f^{(n+1)}(t)\rvert$ for every $t$ between $a$ and $x$ , not just at the endpoint $a$ . Here $e^t$ is increasing, so its largest value on $[0,0.5]$ occurs at $t=0.5$ , not $t=0$ — using $M=1$ understates the true maximum of $e^c$ on the interval.

Consequence: the resulting "bound" $\dfrac{1}{24}(0.5)^4 \approx 0.0026$ is smaller than the actual error computed in step 6 above ( $\approx0.00289$ ) — an invalid bound that the true error already exceeds. $M$ must be a valid upper bound over the entire interval between $a$ and $x$ , which in this case meant using the right endpoint, not the center.

03 · Common Maclaurin Series, Radius of Convergence, and Approximation

Definition — Reference Table of Maclaurin Series

e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \qquad \sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} \qquad \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}

\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} \qquad \frac{1}{1-x} = \sum_{n=0}^\infty x^n

The first three are built from Chapter 02's derivatives of $e^x$ , $\sin x$ , $\cos x$ (Sections 09–10) following Chapter 10, Section 03's construction; $\ln(1+x)$ was built in Chapter 10, Exercise 10.3; $\dfrac{1}{1-x}$ is Chapter 10, Section 01's geometric series.

Step-by-step — Apply the ratio test (Chapter 10, Section 02) to confirm $e^x$'s series converges for every real x

Set up the ratio test on the terms $a_n=\dfrac{x^n}{n!}$ :

\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert = \left\lvert\dfrac{x^{n+1}}{(n+1)!}\cdot\dfrac{n!}{x^n}\right\rvert = \dfrac{\lvert x\rvert}{n+1}

(using

\dfrac{(n+1)!}{n!}=n+1

Take the limit as $n\to\infty$ for any fixed $x$ :

\dfrac{\lvert x\rvert}{n+1}\to0

, since the numerator is a fixed constant and the denominator grows without bound.

Conclude:

\rho=0<1

for every real

x

, so by the ratio test the series converges for all

x

— radius of convergence

R=\infty

. This holds for

\sin x

and

\cos x

's series too, by the identical argument (their coefficients also carry a factorial in the denominator).

❌ Assuming ln(1+x)'s series converges everywhere, like e^x's does

A tempting generalization from Section 03's step-block: assuming every Maclaurin series converges for all real $x$ , since $e^x$ 's did.

Why it breaks: applying the ratio test to $\ln(1+x)=\sum\frac{(-1)^{n+1}x^n}{n}$ : $\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert = \left\lvert\dfrac{x^{n+1}}{n+1}\cdot\dfrac{n}{x^n}\right\rvert = \lvert x\rvert\cdot\dfrac{n}{n+1} \to \lvert x\rvert$ as $n\to\infty$ (no factorial in the denominator to force the ratio to $0$ , unlike $e^x$ 's series). The ratio test then requires $\lvert x\rvert<1$ for convergence — a finite radius of convergence $R=1$ , not all of $\mathbb{R}$ .

Consequence: the series for $\ln(1+x)$ is only valid for $-1<x<1$ (with the right endpoint $x=1$ separately checkable); plugging in, say, $x=5$ into the series gives a divergent sum of numbers with no relationship to the actual value $\ln6$ . Each Taylor series carries its own radius of convergence — it must be checked individually, never assumed from a different function's series.

04 · Exercises

EXERCISE 11.1

Identify $n$ , find the $(n+1)$ -th derivative, bound it over the interval from $a$ to $x$ , then apply the error bound formula.

Approximating $\cos(0.3)$ with $T_2(x)=1-\dfrac{x^2}{2}$ ( $n=2$ , $a=0$ ): the third derivative of $\cos x$ is $\sin x$ (Chapter 02, Section 09: $\cos x\to-\sin x\to-\cos x\to\sin x$ ), and $\lvert\sin t\rvert\leq1$ for all $t$ , so $M=1$ . Bound: $\lvert R_2(0.3)\rvert \leq \dfrac{1}{3!}(0.3)^3 = \dfrac{0.027}{6}=0.0045$ .

Bound the error when $T_2(x)=1-\dfrac{x^2}{2}$ is used to approximate $\cos(0.3)$ .

EXERCISE 11.2

Compute $T_2$ directly from the coefficients, then compare to the true value using a calculator-level approximation of $\cos(0.3)$ .

$T_2(0.3) = 1-\dfrac{(0.3)^2}{2} = 1-\dfrac{0.09}{2}=1-0.045=0.955$ . True value $\cos(0.3)\approx0.95534$ . Actual error $\approx0.00034$ , well within the bound of $0.0045$ from Exercise 11.1.

Compute $T_2(0.3)$ from Exercise 11.1 and compare to the true value of $\cos(0.3)\approx0.95534$ .

EXERCISE 11.3

Apply the ratio test to the general term of the series, exactly as done for $e^x$ 's series in Section 03.

$\sin x = \sum\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ : ratio of consecutive terms' absolute values involves $\dfrac{\lvert x\rvert^{2n+3}}{(2n+3)!}\cdot\dfrac{(2n+1)!}{\lvert x\rvert^{2n+1}} = \dfrac{\lvert x\rvert^2}{(2n+3)(2n+2)}\to0$ as $n\to\infty$ for any fixed $x$ (fixed numerator, denominator grows without bound). So $\rho=0<1$ always — converges for every real $x$ , $R=\infty$ .

Show that the Maclaurin series for $\sin x$ converges for every real number $x$ , using the ratio test.

EXERCISE 11.4

Recall which of the series in Section 03 was shown to have a finite radius of convergence, and by how much its ratio-test limit failed to vanish.

$\ln(1+x)$ 's series (Section 03's FailBlock) has radius of convergence $R=1$ , unlike $e^x$ , $\sin x$ , and $\cos x$ 's series, which all converge for every real $x$ ( $R=\infty$ ). The difference traces back to the denominator of the coefficients: the factorial $n!$ in $e^x$ , $\sin x$ , $\cos x$ 's coefficients forces the ratio-test limit to $0$ regardless of $x$ , while $\ln(1+x)$ 's coefficients have only a plain $n$ in the denominator, leaving a nonzero $\lvert x\rvert$ in the ratio-test limit.

Which Maclaurin series from Section 03 has a finite radius of convergence rather than $R=\infty$ , and what feature of its coefficients causes the difference?

EXERCISE 11.5

Determine how many terms are needed by requiring the error bound formula to be less than the target precision, solving for the smallest valid $n$ .

Approximate $e^{0.2}$ to within $0.0001$ using $T_n$ centered at $a=0$ . Bound: $\lvert R_n(0.2)\rvert \leq \dfrac{M}{(n+1)!}(0.2)^{n+1}$ with $M$ bounding $e^c$ for $0<c<0.2$ ; since $e^{0.2}<e<3$ , take $M=3$ . Try $n=2$ : $\dfrac{3}{3!}(0.2)^3 = \dfrac{3}{6}(0.008)=0.004$ — too large. Try $n=3$ : $\dfrac{3}{4!}(0.2)^4 = \dfrac{3}{24}(0.0016)=0.0002$ — still too large. Try $n=4$ : $\dfrac{3}{5!}(0.2)^5=\dfrac{3}{120}(0.00032)=0.000008$ — within $0.0001$ . So $n=4$ terms (through $T_4$ ) suffice.

How many terms of the Maclaurin series for $e^x$ are needed to approximate $e^{0.2}$ to within $0.0001$ ? Use the error bound formula to determine the smallest sufficient $n$ .

05 · Chapter Summary

Concept	Formula / Rule
Taylor polynomial	$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$
Remainder	$R_n(x) = f(x)-T_n(x)$
Lagrange remainder	$R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ , $c$ between $a,x$
Error bound	$\lvert R_n(x)\rvert \leq \dfrac{M}{(n+1)!}\lvert x-a\rvert^{n+1}$ where $\lvert f^{(n+1)}\rvert\leq M$ on $[a,x]$
$e^x$ , $\sin x$ , $\cos x$ series	Converge for all real $x$ ( $R=\infty$ )
$\ln(1+x)$ series	Converges only for $-1<x<1$ ( $R=1$ )

Next: Chapter 12 — Convergence Tests, which builds out the full toolkit for deciding whether a series converges, beyond the ratio test introduced here.