Chapter 10

Medium

Power Series and Taylor Series

00 · Symbol Glossary

$\displaystyle\sum_{n=0}^{\infty} c_n(x-a)^n$power series centered at a

A series (Chapter 09) whose terms depend on a variable $x$ , each term a constant $c_n$ times a power of $(x-a)$ . Unlike Chapter 09's series of fixed numbers, a power series can converge for some values of $x$ and diverge for others — it defines a function of $x$ wherever it converges.

$c_n$c sub n — coefficient

The constant multiplying $(x-a)^n$ in a power series. Distinct from a sequence term $a_n$ (Chapter 09) — here $c_n$ does not depend on $x$ at all, only on the index $n$ .

$R$R — radius of convergence

A number (possibly $0$ or $\infty$ ) such that the power series converges whenever $\lvert x-a\rvert < R$ and diverges whenever $\lvert x-a\rvert > R$ . The behavior exactly at $\lvert x-a\rvert=R$ must be checked separately in each case.

$\displaystyle\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$ratio test quantity

The limit of the absolute value of consecutive-term ratios in a series $\sum a_n$ . Section 02 introduces this as a tool needed to find $R$ ; Chapter 12 revisits it as one of a full toolkit of convergence tests.

$f^{(n)}(a)$f sup n of a — nth derivative at a

The $n$ -th derivative of $f$ (Chapter 02, Section 08), evaluated at the specific point $x=a$ — a single number, not a function. The building block of the Taylor series in Section 03.

$n!$n factorial

The product $n\times(n-1)\times(n-2)\times\cdots\times2\times1$ , with $0!=1$ by convention. Appears in the denominator of each Taylor series term.

01 · Power Series

A power series looks like a polynomial with infinitely many terms. Where it converges, it defines a genuine function — one built entirely out of powers of $(x-a)$ .

Definition — Power Series

A power series centered at $a$ is $\displaystyle\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots$

For each fixed value of $x$ , this is an ordinary series of numbers (Chapter 09) — it may converge or diverge depending on which $x$ is plugged in.

Example — A power series that is secretly a geometric series

$\displaystyle\sum_{n=0}^{\infty} x^n$ is a power series centered at $a=0$ with every $c_n=1$ . For a fixed $x$ , this is exactly Chapter 09's geometric series with first term $1$ and common ratio $r=x$ . By Chapter 09, Section 03, it converges precisely when $\lvert x\rvert<1$ , to the value $\dfrac{1}{1-x}$ .

This single power series therefore represents the function $f(x)=\dfrac{1}{1-x}$ , but only on the interval $-1<x<1$ — outside that interval the series diverges even though $f(x)=\frac{1}{1-x}$ itself is defined for every $x\neq1$ .

02 · Radius of Convergence and the Ratio Test

Section 01's example showed a power series can converge on an interval centered at $a$ and diverge outside it. Finding exactly how wide that interval is requires a new tool: comparing the size of consecutive terms.

Definition — The Ratio Test

For a series $\displaystyle\sum a_n$ with nonzero terms, let $\rho = \displaystyle\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$ .

If $\rho<1$ , the series converges absolutely (Chapter 12 defines this term precisely; for now, read it as "converges").
If $\rho>1$ (including $\rho=\infty$ ), the series diverges.
If $\rho=1$ , the test is inconclusive.

Why the ratio test works — geometric intuition

If consecutive terms shrink by a roughly constant factor $\rho<1$ once $n$ is large, the tail of the series behaves like a geometric series with ratio $\rho$ (Chapter 09, Section 03) — and geometric series with $\lvert r\rvert<1$ converge. If $\rho>1$ , consecutive terms are eventually growing, so the terms cannot shrink to $0$ , and the test for divergence (Chapter 09, Section 04) applies. A full proof is outside this chapter's scope, but the geometric-series comparison is the reason the test works.

Step-by-step — Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(x-2)^n}{n+1}$

Set up the ratio test on the terms $a_n = \dfrac{(x-2)^n}{n+1}$ : form

\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert = \left\lvert\dfrac{(x-2)^{n+1}}{n+2}\cdot\dfrac{n+1}{(x-2)^n}\right\rvert

Cancel the common power of $(x-2)$ :

\dfrac{(x-2)^{n+1}}{(x-2)^n} = (x-2)

, leaving

\left\lvert(x-2)\cdot\dfrac{n+1}{n+2}\right\rvert = \lvert x-2\rvert\cdot\dfrac{n+1}{n+2}

Take the limit as $n\to\infty$ :

\dfrac{n+1}{n+2}\to1

(divide numerator and denominator by

n

, Chapter 09 Exercise 9.1 style), so

\rho = \lvert x-2\rvert\cdot1 = \lvert x-2\rvert

Apply the ratio test's convergence condition $\rho<1$ :

\lvert x-2\rvert<1

— the series converges when

x

is within distance

1

of the center

a=2

Read off the radius:

R=1

. (Checking the two boundary points

x=1

and

x=3

separately, using Chapter 09's tools, would complete the exact interval of convergence — the ratio test alone only settles the interior.)

❌ Concluding convergence when the ratio test gives exactly 1

For $\displaystyle\sum_{n=1}^{\infty}\frac1n(x-0)^n$ evaluated at the boundary point $x=1$ : the ratio test on the series $\sum\frac1n$ itself gives $\rho = \lim_{n\to\infty}\left\lvert\dfrac{1/(n+1)}{1/n}\right\rvert = \lim_{n\to\infty}\dfrac{n}{n+1}=1$ .

Why it breaks: $\rho=1$ is explicitly the inconclusive case in the ratio test's definition — the test simply does not decide convergence or divergence here, in either direction.

Consequence: a separate check is required: at $x=1$ , the series is exactly $\sum\frac1n$ , the harmonic series, which Chapter 09, Section 05 proved diverges. The ratio test's silence at $\rho=1$ is not evidence of convergence — every boundary point of an interval of convergence must be checked by an independent method, never assumed.

03 · Taylor Series

Section 01 showed some functions happen to equal a power series. The Taylor series is a general recipe for building a power series out of any function's derivatives — designed so that the series matches the function's value and every derivative's value at a single point $a$ .

Definition — Taylor Series

The Taylor series of $f$ centered at $a$ is

f(x) \approx \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

When $a=0$ , this is called a Maclaurin series: $\displaystyle\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$ .

Why divide by n! — matching derivatives at a, term by term

Each coefficient is built so that differentiating the series $n$ times and evaluating at $x=a$ reproduces $f^{(n)}(a)$ exactly: differentiating $(x-a)^n$ a total of $n$ times gives the constant $n!$ (each of the $n$ derivatives brings down one factor: $n, n-1, n-2,\ldots,1$ , multiplying to $n!$ ), and all lower-power terms vanish after $n$ differentiations while all higher-power terms still contain a factor of $(x-a)$ and vanish at $x=a$ . Dividing by $n!$ exactly cancels that leftover factor, leaving $f^{(n)}(a)$ — the series is constructed to agree with $f$ and all its derivatives at the single point $a$ .

Step-by-step — Build the Maclaurin series for $f(x)=e^x$

Recall the derivative of $e^x$ (Chapter 02, Section 10):

\dfrac{d}{dx}[e^x]=e^x

— every derivative of

e^x

e^x

itself, so

f^{(n)}(x)=e^x

for every

n

Evaluate every derivative at $a=0$ :

f^{(n)}(0) = e^0 = 1

for every

n=0,1,2,\ldots

Substitute into the Taylor series formula with $a=0$ :

\displaystyle\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{1}{n!}x^n

— every

f^{(n)}(0)=1

from step 2.

Write out the first few terms:

1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots = 1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\cdots

Numerically check at $x=1$ : the partial sum through the

x^4

term gives

1+1+0.5+0.1\overline{6}+0.041\overline{6}\approx2.708\overline{3}

, already close to

e^1\approx2.71828

— each additional term tightens the approximation.

04 · Exercises

EXERCISE 10.1

Set up the ratio test on the general term, simplify the ratio of consecutive terms, then take the limit as $n\to\infty$ .

$\displaystyle\sum_{n=0}^{\infty}\frac{(x+1)^n}{2^n}$ : $a_n=\dfrac{(x+1)^n}{2^n}$ . $\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert = \left\lvert\dfrac{(x+1)^{n+1}}{2^{n+1}}\cdot\dfrac{2^n}{(x+1)^n}\right\rvert = \dfrac{\lvert x+1\rvert}{2}$ (independent of $n$ , so the limit is itself). Convergence requires $\dfrac{\lvert x+1\rvert}{2}<1 \implies \lvert x+1\rvert<2$ . Radius of convergence: $R=2$ .

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\dfrac{(x+1)^n}{2^n}$ using the ratio test.

EXERCISE 10.2

Use $\dfrac{d}{dx}[\sin x]=\cos x$ and $\dfrac{d}{dx}[\cos x]=-\sin x$ (Chapter 02, Section 09) to compute successive derivatives, then evaluate each at $x=0$ .

$f(x)=\sin x$ : $f'(x)=\cos x$ , $f''(x)=-\sin x$ , $f'''(x)=-\cos x$ , $f^{(4)}(x)=\sin x$ (the pattern repeats every 4 derivatives). At $x=0$ : $f(0)=0$ , $f'(0)=1$ , $f''(0)=0$ , $f'''(0)=-1$ , $f^{(4)}(0)=0$ . Maclaurin series: $0+1\cdot x+\dfrac{0}{2!}x^2+\dfrac{-1}{3!}x^3+\dfrac{0}{4!}x^4+\cdots = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots$ (only odd powers survive, alternating sign).

Build the Maclaurin series for $f(x)=\sin x$ through the $x^5$ term, using the derivatives established in Chapter 02.

EXERCISE 10.3

Compute successive derivatives of $\ln(1+x)$ directly (using Chapter 02's derivative of $\ln x$ combined with the chain rule from Chapter 03), then evaluate at $x=0$ .

$f(x)=\ln(1+x)$ : $f'(x)=\dfrac{1}{1+x}$ , $f''(x)=-\dfrac{1}{(1+x)^2}$ , $f'''(x)=\dfrac{2}{(1+x)^3}$ . At $x=0$ : $f(0)=0$ , $f'(0)=1$ , $f''(0)=-1$ , $f'''(0)=2$ . Series: $0+1\cdot x+\dfrac{-1}{2!}x^2+\dfrac{2}{3!}x^3+\cdots = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots$ .

Build the Maclaurin series for $f(x)=\ln(1+x)$ through the $x^3$ term.

EXERCISE 10.4

The boundary points of an interval of convergence are never decided by the ratio test itself — check them as ordinary numerical series using Chapter 09's tools.

From Exercise 10.1, the series converges for $\lvert x+1\rvert<2$ , i.e. $-3<x<1$ , with boundary points $x=-3$ and $x=1$ requiring separate checks (not answered by the ratio test, since $\rho=1$ exactly there) — for instance at $x=1$ : terms become $\frac{2^n}{2^n}=1$ for every $n$ , which does not shrink to $0$ , so by the test for divergence (Chapter 09) the series diverges at that endpoint.

Explain why the ratio test alone cannot determine whether the series from Exercise 10.1 converges at its two boundary points $x=-3$ and $x=1$ , and check convergence at $x=1$ directly.

EXERCISE 10.5

Substitute $-x$ for $x$ directly into the known Maclaurin series for $e^x$ from Section 03 — no new derivative computation is required.

From Section 03, $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ . Substituting $x\to-x$ : $e^{-x} = \sum_{n=0}^\infty\frac{(-x)^n}{n!} = \sum_{n=0}^\infty\frac{(-1)^n x^n}{n!} = 1-x+\dfrac{x^2}{2}-\dfrac{x^3}{6}+\cdots$ .

Using the Maclaurin series for $e^x$ from Section 03, find the Maclaurin series for $e^{-x}$ by substitution (no new derivatives needed).

05 · Chapter Summary

Concept	Formula / Rule
Power series	$\sum c_n(x-a)^n$ ; converges for some $x$ , diverges for others
Ratio test	$\rho=\lim\lvert a_{n+1}/a_n\rvert$ : $\rho<1$ converges, $\rho>1$ diverges, $\rho=1$ inconclusive
Radius of convergence	$R$ : converges for $\lvert x-a\rvert<R$ ; endpoints checked separately
Taylor series	$\sum \dfrac{f^{(n)}(a)}{n!}(x-a)^n$
Maclaurin series	Taylor series centered at $a=0$
Maclaurin series for $e^x$	$\sum \dfrac{x^n}{n!} = 1+x+\dfrac{x^2}{2}+\cdots$

Next: Chapter 11 — Taylor Series Advanced, where the gap between a Taylor series' partial sum and the true function value is measured and bounded.