[THEOREMS] Power series

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.

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

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.

Theorems from Power series

Convergence has a central interval

Algebra of power series

Differentiate and integrate term by term

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