Power Series and Taylor Expansions

A structured guide to representing functions with power series, determining convergence, estimating approximation error, integrating series, and solving differential equations by coefficient matching.

The Basic Structure of

A centered at $a$ has the form

\sum_{n=0}^{\infty}c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots.

When $a=0$ , the series is a Maclaurin . For a fixed value of $x$ , the expression becomes an ordinary infinite series of numbers, so convergence depends on $x$ .

The basic model is the geometric series:

\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},\qquad |r|<1.

Replacing $r$ with $x-a$ gives

\sum_{n=0}^{\infty}(x-a)^n=\frac{1}{1-(x-a)},\qquad |x-a|<1.

This identity is the main starting point for constructing many other series.

Takeaway: Treat a as an infinite polynomial, but always determine where it converges before using it to represent a function.

Convergence and Endpoint Testing

For a series centered at $a$ , the $R$ describes the interior region where convergence is guaranteed:

|x-a|<R.

There are three possibilities:

$R=0$ : the series converges only at $x=a$ .
$R=\infty$ : the series converges for every real $x$ .
$0<R<\infty$ : the series converges inside the interval and diverges outside it.

To find $R$ , apply the ratio test to consecutive terms. For

\sum_{n=0}^{\infty}c_n(x-a)^n,

consider

L=\lim_{n\to\infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right|.

The ratio test gives convergence when $L<1$ and divergence when $L>1$ . Solve the resulting inequality, usually of the form $|x-a|<R$ , and then test $x=a-R$ and $x=a+R$ separately.

For example, consider

\sum_{n=0}^{\infty}\frac{(x-2)^n}{(n+1)3^n}.

The ratio limit is $|x-2|/3$ , so the interior condition is $|x-2|<3$ , or $-1<x<5$ . At $x=-1$ , the resulting alternating harmonic series converges; at $x=5$ , the harmonic series diverges. Therefore,

R=3,\qquad [-1,5).

Takeaway: The ratio test usually determines only the open interval. Endpoint tests determine whether the final interval uses parentheses or brackets.

Differentiating and Integrating Series

Inside the , a can be differentiated and integrated term by term. If

f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n,

then

f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}

and

\int f(x)\,dx=C+\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}.

The differentiated and integrated series have the same as the original series, although endpoint behavior can change.

Starting with

\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n, \qquad |x|<1,

differentiation gives

\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n,

while integration gives

-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}.

Similarly,

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

leads, after integration, to

\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}, \qquad -1<x\leq 1.

Takeaway: Differentiation creates factors involving the index, while integration increases powers and divides coefficients by the new exponent.

Taylor Expansions and Standard

The centered at $a$ is

\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n.

Its degree-
$n$ is

T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.

When $a=0$ , the expansion is a . Important examples include

e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},

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

and

\cos x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}.

The geometric and logarithmic expansions are also fundamental:

\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n, \qquad |x|<1,

and

\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}, \qquad -1<x\leq 1.

Having derivatives of every order does not automatically prove that the function equals its . One must establish that the remainder approaches zero on the interval under consideration.

Takeaway: A gives a finite approximation; the infinite represents the function only where its remainder vanishes.

Building New Series from Known Ones

Known series can be adapted through substitution, algebraic rearrangement, multiplication by powers, differentiation, and integration.

For a direct geometric-series construction,

\frac{1}{2+x}=\frac12\cdot\frac{1}{1+x/2} =\frac12\sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^n =\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^{n+1}}, \qquad |x|<2.

Substitution into a known series gives

e^{x^2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}

and

e^{-x^2}=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{n!}.

A nonzero center is useful when the approximation is intended near a point other than zero. For $f(x)=1/x$ centered at $a=1$ , write

\frac1x=\frac{1}{1+(x-1)} =\sum_{n=0}^{\infty}(-1)^n(x-1)^n, \qquad |x-1|<1.

Thus the is $(0,2)$ .

Takeaway: Rewrite the target function so that it matches a known template, then carry along the corresponding convergence condition.

Approximation and Error Control

If $T_n(x)$ approximates $f(x)$ , the remainder is

R_n(x)=f(x)-T_n(x).

Taylor’s theorem states that, for some $c$ between $a$ and $x$ ,

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

If $|f^{(n+1)}(t)|\leq M$ on the interval, then

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

For an alternating expansion with decreasing term magnitudes, the first omitted term provides a simpler estimate. For example, using

T_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}

to approximate $\sin(x)$ , the next term has magnitude $|x|^7/7!$ . At $x=\pi/18$ , this gives an error bound of approximately $9.8\times 10^{-10}$ .

A practical approximation procedure is:

Choose the center and the target value.
Select enough terms to meet the desired accuracy.
Compute the polynomial approximation.
Bound the remainder using a derivative bound or the first omitted alternating term.
Report the approximation together with its error guarantee.

Takeaway: The degree of the approximation should be justified by an explicit error estimate, not chosen only by appearance.

Series for Nonelementary Integrals

Term-by-term integration can evaluate definite integrals whose antiderivatives are not elementary. Since

e^{-t^2}=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{n!},

integrating from $0$ to $x$ gives

\int_0^x e^{-t^2}\,dt =\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)n!} =x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\cdots.

At $x=1$ , the first four terms give

1-\frac13+\frac1{10}-\frac1{42}.

Because the resulting series is alternating with decreasing term magnitudes, the error is at most the next term,

\frac{1}{9\cdot 4!}=\frac{1}{216}.

The general strategy is to expand the integrand, integrate each term over the desired limits, and then use a convergence or error test to control the approximation.

Takeaway: Series integration converts a difficult integral into a sequence of elementary power integrals while preserving a way to estimate the numerical error.

Power-Series Solutions of Differential Equations

To solve a differential equation by a , assume

y(x)=\sum_{n=0}^{\infty}a_nx^n.

Then differentiate term by term:

y'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}, \qquad y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}.

Substitute these expressions into the differential equation, rewrite all terms with matching powers of $x$ , and set each coefficient equal to zero.

For $y''+y=0$ , alignment of powers gives

(n+2)(n+1)a_{n+2}+a_n=0,

so the coefficient recurrence is

a_{n+2}=-\frac{a_n}{(n+2)(n+1)}.

The even and odd coefficients develop independently:

a_2=-\frac{a_0}{2!},\qquad a_4=\frac{a_0}{4!},\qquad a_6=-\frac{a_0}{6!},

and

a_3=-\frac{a_1}{3!},\qquad a_5=\frac{a_1}{5!},\qquad a_7=-\frac{a_1}{7!}.

Therefore,

y=a_0\cos x+a_1\sin x.

Initial conditions determine the starting coefficients: $a_0=y(0)$ and $a_1=y'(0)$ . For $y(0)=2$ and $y'(0)=-3$ , the solution is

y=2\cos x-3\sin x.

Takeaway: Power-series methods turn a differential equation into algebraic relations among coefficients; the recurrence and initial conditions determine the solution.

A Reliable Problem-Solving Workflow

Use this workflow when working with a power or Taylor expansion:

Identify the center and write all powers as $(x-a)^n$ .
Find the with the ratio test or root test.
Test both endpoints separately whenever the radius is finite.
Choose a construction method: geometric series, substitution, differentiation, integration, or Taylor’s formula.
For approximation, choose the smallest polynomial degree that meets the required accuracy.
Estimate the remainder with Taylor’s theorem or an alternating-series bound.
For differential equations, substitute the assumed series, align powers, and derive the coefficient recurrence.
State the interval on which the resulting series represents the function.

The central ideas are connected: convergence determines where a series is valid, term-by-term operations create new representations, Taylor polynomials provide local approximations, error bounds measure their reliability, and coefficient recurrences extend the method to differential equations.

The Basic Structure of Power Series