[CALCULUS] Power series

Determine where an infinite polynomial converges and use calculus inside its radius.

A power series starts with coefficients and asks where their infinite polynomial converges. This reverses the Taylor-series viewpoint: a Taylor series starts with a function, derives coefficients and asks where the series equals that function.

From coefficients to a function

A power series defines a function only where it converges, so find that set before studying its algebra or calculus. Convergence has its partial-sum meaning and the convergence tests find where it holds.

[DEFINITION]

Infinite series

The series $\sum_{n=1}^{\infty}a_n$ converges to $L$ precisely when its partial sums satisfy

\lim_{k\to\infty}s_k=L, \qquad s_k=\sum_{n=1}^{k}a_n.

In that case,

\sum_{n=1}^{\infty}a_n=L.

[DEFINITION]

Power series

A power series centred at $a$ is a series of the form

\sum_{k=0}^{\infty}c_k(x-a)^k.

The series itself may define a new function at every value of $x$ where it converges.

[DEFINITION]

Radius and interval of convergence

The radius of convergence $R$ of

\sum_{k=0}^{\infty}c_k(x-a)^k

is the number such that the series converges absolutely when $|x-a|<R$ and diverges when $|x-a|>R$ .

The interval of convergence is the open interval $(a-R,a+R)$ together with whichever endpoints converge.

With $R=0$ the series converges only at its centre, while $R=\infty$ gives convergence for every real $x$ . Here $R$ denotes a radius; the Taylor remainder is written separately as $R_{n+1}(x)$ .

[1]:Power series
[2]:Convergence has a central interval

Find the radius and test the endpoints

The radius gives the open interval where the power series converges, but it does not settle the boundary. At each endpoint the power series becomes an ordinary series, to be tested with the convergence tests.

[THEOREM]

Convergence has a central interval

Every real power series has a radius $R\in[0,\infty]$ with the following behaviour:

absolute convergence for $|x-a|<R$ ;
divergence for $|x-a|>R$ ;
separate endpoint behaviour when $0<R<\infty$ .

To see why the region has this shape, suppose the series converges at some $x_0\ne a$ . Then its terms satisfy $|c_k(x_0-a)^k|\le M$ for some $M>0$ . At any point with $|x-a|<|x_0-a|$ ,

|c_k(x-a)^k| \le M\left|\frac{x-a}{x_0-a}\right|^k.

The right-hand side is geometric with ratio below $1$ , so convergence at one point forces absolute convergence at every point closer to the centre.

[METHOD]

Find the interval of convergence

For a power series centred at $a$ :

apply the ratio or root test when its limit exists and is decisive; otherwise estimate the coefficients by another convergence test;
rewrite that condition as $|x-a|<R$ ;
record the open interval $(a-R,a+R)$ ;
if $0<R<\infty$ , substitute $x=a-R$ and $x=a+R$ into the original series;
apply the appropriate ordinary series test at each finite endpoint.

The endpoint series may be geometric, harmonic, alternating or something else and the two endpoints need not behave alike. If $R=0$ there are no endpoints to test; if $R=\infty$ every real $x$ is already inside the radius.

[EXAMPLE]

A power series with two divergent endpoints

Consider

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

Writing its term as $a_k$ , the ratio test gives

\left|\frac{a_{k+1}}{a_k}\right| =|x-3|\frac{3^k+2}{3^{k+1}+2} \longrightarrow\frac{|x-3|}{3}.

Therefore the series converges when $|x-3|<3$ , giving the possible interval $(0,6)$ .

At $x=0$ ,

a_k=\frac{(-3)^k}{3^k+2}, \qquad |a_k|\longrightarrow1,

so the terms do not tend to zero. At $x=6$ ,

a_k=\frac{3^k}{3^k+2}\longrightarrow1,

so the terms again fail the $n$ -th term test. The interval of convergence is

{(0,6)}.

[exam]

[1]:Find the interval of convergence

[EXAMPLE]

A power series with one convergent endpoint

Consider

\sum_{k=1}^{\infty}\frac{4^k}{k}(x-2)^k.

The ratio test gives

\left|\frac{a_{k+1}}{a_k}\right| =4|x-2|\frac{k}{k+1} \longrightarrow4|x-2|.

Thus $R=1/4$ , with possible interval $(7/4,9/4)$ .

At $x=7/4$ ,

\frac{4^k}{k}\left(-\frac14\right)^k =\frac{(-1)^k}{k},

which converges by the alternating-series test. At $x=9/4$ ,

\frac{4^k}{k}\left(\frac14\right)^k =\frac1k,

which diverges. Hence

{\left[\frac74,\frac94\right)}.

[exam]

[1]:Find the interval of convergence

Algebra inside a common radius

Inside a common interval of absolute convergence, power series can be added and multiplied like polynomials.

[THEOREM]

Algebra of power series

Inside a common interval where the power series converge absolutely, if

f(x)=\sum_{k=0}^{\infty}a_k(x-a)^k, \qquad g(x)=\sum_{k=0}^{\infty}b_k(x-a)^k,

then

f(x)\pm g(x) =\sum_{k=0}^{\infty}(a_k\pm b_k)(x-a)^k,

and

f(x)g(x) =\sum_{k=0}^{\infty} \left(\sum_{j=0}^{k}a_jb_{k-j}\right)(x-a)^k.

The second formula is the Cauchy product. The coefficient of $(x-a)^k$ collects every pair of terms whose powers add to $k$ . Products at endpoints require separate justification when endpoint convergence is only conditional.

[1]:Power series

Calculus inside the radius

Inside the open interval of convergence, a power series can be differentiated and integrated term by term.

[THEOREM]

Differentiate and integrate term by term

Inside the radius of convergence,

\frac{d}{dx}\sum_{k=0}^{\infty}a_k(x-a)^k =\sum_{k=1}^{\infty}ka_k(x-a)^{k-1},

and

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

Both resulting power series have the same open radius of convergence as the original one.

On every closed interval $[a-r,a+r]$ with $0<r<R$ , the original series and its differentiated series converge uniformly. This local uniform control justifies the termwise operations; pointwise convergence alone would not.

Uniform convergence on a set means that for every tolerance $\varepsilon>0$ , there is one index $N$ such that $|S_n(x)-f(x)|<\varepsilon$ for every $n>N$ and every input $x$ in the set. Pointwise convergence may require a different $N$ at each input.

The theorem concerns only the open interval. Uniform convergence at either endpoint is a separate question and endpoint behaviour may change after differentiating or integrating.

f(x)=\sum_{k=0}^{\infty}a_k(x-a)^k,

then evaluating derivatives at the centre gives

f^{(j)}(a)=j!a_j.

Consequently, the power series is the Taylor series of the function it defines.

[EXAMPLE]

Recovering $\cos x$ from $\sin x$

Termwise differentiation of

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

gives

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

[routine]

[1]:Differentiate and integrate term by term

[EXAMPLE]

Find $\sum_{k=0}^{\infty}k^2x^k$

Start with the geometric series:

\frac1{1-x}=\sum_{k=0}^{\infty}x^k, \qquad |x|<1.

Differentiate and multiply by $x$ :

\frac{x}{(1-x)^2}=\sum_{k=0}^{\infty}kx^k.

Differentiate once more and multiply by $x$ :

\frac{x(1+x)}{(1-x)^3} =\sum_{k=0}^{\infty}k^2x^k, \qquad |x|<1.

Therefore

{ \sum_{k=0}^{\infty}k^2x^k =\frac{x(1+x)}{(1-x)^3}, \qquad |x|<1. }

[exam]

[1]:Differentiate and integrate term by term

Why pointwise convergence is not enough

The termwise calculus theorem is special to power series. For a general sequence of functions, a limit at each fixed input does not control the derivatives.

[PROPOSITION]

Pointwise convergence is not enough

Convergence at each fixed $x$ (called pointwise convergence) does not, by itself, justify differentiating the limit term by term.

Consider

f_n(x)=\frac{\sin(n^2x)}n.

For every fixed $x$ ,

|f_n(x)|\le\frac1n\longrightarrow0,

so $f_n(x)\to f(x)=0$ . However,

f_n'(x)=n\cos(n^2x), \qquad f_n'(0)=n,

which does not converge to $f'(0)=0$ . In general,

\lim_{n\to\infty}\lim_{h\to0} \frac{f_n(x+h)-f_n(x)}h \quad\text{and}\quad \lim_{h\to0}\lim_{n\to\infty} \frac{f_n(x+h)-f_n(x)}h

need not agree. Power-series theorems supply the extra control inside the radius of convergence.

[1]:Differentiate and integrate term by term

Because a power series is the Taylor series of its sum, its partial sums are the Taylor polynomials of that sum. The remainder in Taylor's theorem controls their error and the Taylor series page decides when a Taylor series built from a given function equals it. The next chapter carries Taylor approximation into several variables.

Power series

Infinite series

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