Power series and real analyticity

A Taylor polynomial approximates a function to finite order. An analytic representation makes an infinite series equal the function on a neighborhood. Smoothness alone does not guarantee that equality. Power-series convergence and remainder control explain the difference.

1 Radius of convergence

A power series centered at $a$ is $\sum_{n=0}^\infty c_n(x-a)^n$. Put $L=\limsup_{n\to\infty}|c_n|^{1/n}$. Its radius is $R=1/L$, with conventions $1/0=\infty$ and $1/\infty=0$.

Radius theorem. The series converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$. On every closed smaller interval $|x-a|\le r<R$, convergence is uniform and absolute.

Proof. If $r<R$, choose $s$ with $r<s<R$. The root test gives convergence of $\sum|c_n|s^n$, and $|c_n(x-a)^n|\le|c_n|r^n\le|c_n|s^n$. The M-test yields uniform absolute convergence. If $|x-a|>R$, the limsup of the term's $n$th root is greater than 1, so its terms cannot tend to zero. Boundary points are not decided by this argument. $\square$

For $\sum x^n/n$, $n\ge1$, the radius is 1. At $x=1$ it diverges; at $x=-1$ it converges conditionally. The radius alone does not settle either endpoint.

2 Differentiating and integrating inside the radius

Inside the convergence interval,

$$\frac d{dx}\sum c_n(x-a)^n=\sum_{n\ge1}nc_n(x-a)^{n-1}.$$

To justify this, choose $r<s<R$. Convergence at $s$ bounds $|c_n|s^n$ by a constant $M$. For $|x-a|\le r$ and $r>0$,

$$|nc_n(x-a)^{n-1}|\le\frac Ms n(r/s)^{n-1},$$

whose scalar sum is finite. Thus the derivative series converges uniformly on the smaller interval. Apply the derivative theorem to the partial sums, whose values at $a$ equal $c_0$. Repeating gives all higher derivatives and

$$c_n=\frac{f^{(n)}(a)}{n!}.$$

Uniform convergence also permits termwise integration on every closed smaller interval. Endpoint formulas require their own convergence arguments; they are not obtained simply by substituting a boundary value into an interior theorem.

Example. The geometric series gives $1/(1-x)=\sum_{n\ge0}x^n$ for $|x|<1$. Integrating from zero to $x$ gives

$$-\log(1-x)=\sum_{n\ge1}\frac{x^n}{n}.$$

Differentiating again gives $1/(1-x)^2=\sum_{n\ge0}(n+1)x^n$. Each manipulation is valid on smaller closed intervals inside $(-1,1)$ and hence at every interior point.

3 Analytic versus infinitely differentiable

A function is real analytic at $a$ if it equals a convergent power series on some neighborhood of $a$. The coefficient formula makes that representation its Taylor series. Being analytic at every domain point does not mean that one center's series represents the function over the entire domain.

Define

$$f(x)=\begin{cases}e^{-1/x^2},&x\ne0,\\0,&x=0.\end{cases}$$

Here are all steps proving smoothness and failure of analyticity. For $x\ne0$, put $t=1/x$ and define polynomials recursively by

$$P_0(t)=1,\qquad P_{j+1}(t)=2t^3P_j(t)-t^2P_j'(t).$$

Since $dt/dx=-t^2$, the product and chain rules show that the derivative of $P_j(1/x)e^{-1/x^2}$ is $P_{j+1}(1/x)e^{-1/x^2}$. To control the origin, for every fixed power $p\ge0$ and integer $N>p$ the nonnegative exponential series gives $e^u\ge u^N/N!$ for $u>0$, whence $u^pe^{-u}\le N!u^{p-N}\to0$ as $u\to\infty$. Thus every polynomial in $1/x$ times $e^{-1/x^2}$ tends to zero; multiplying by one further $1/x$ still has that property.

Let $g_j(x)=P_j(1/x)e^{-1/x^2}$ off zero and $g_j(0)=0$. Each $g_j$ is continuous at zero. Its difference quotient there is

$$\frac{g_j(x)-g_j(0)}x=\frac1xP_j(1/x)e^{-1/x^2}\longrightarrow0.$$

Away from zero its derivative is $g_{j+1}$; at zero the quotient just proved gives the same derivative value $g_{j+1}(0)=0$. Starting with $g_0=f$, induction proves $f\in C^\infty(\mathbb R)$ and $f^{(j)}(0)=0$ for every $j$. Every Taylor coefficient vanishes, but $f(x)>0$ at every nonzero $x$. No neighborhood representation by its zero Taylor series is possible. The exponential identities and positive-series bound used here are constructed in the elementary-functions chapter independently of this counterexample.

For comparison, suppose a $C^\infty$ function has fixed constants $M,K>0$ such that $|f^{(n+1)}(t)|\le M K^{n+1}$ for every integer $n\ge0$ and every $t$ in one common neighborhood of $a$. The one-variable Taylor remainder then has bound $M K^{n+1}|x-a|^{n+1}/(n+1)!$, which tends to zero: the ratio of successive bounds is $K|x-a|/(n+2)\to0$. This proves equality to the Taylor series at each sufficiently near point. Bounds for just finitely many orders, or only at $a$, cannot replace this common-neighborhood hypothesis.

4 A bounded multivariable extension

For a multi-index $\alpha=(\alpha_1,\ldots,\alpha_d)$, write $|\alpha|=\sum\alpha_i$, $\alpha!=\prod\alpha_i!$ and $h^\alpha=\prod h_i^{\alpha_i}$. A local real-analytic expansion has form $\sum_\alpha c_\alpha h^\alpha$. If for some $s>0$, $\sum_\alpha|c_\alpha|s^{|\alpha|}<\infty$, the series converges absolutely and uniformly on every smaller closed cube $|h_i|\le r<s$. Derivative series are also uniformly convergent there: the additional factor $\alpha_j$ is controlled by a convergent polynomial-times-geometric bound between $r$ and $s$. Hence $c_\alpha=\partial^\alpha f(a)/\alpha!$.

For $1/(1-x-y)$, the geometric expansion and binomial theorem give

$$\frac1{1-x-y}=\sum_{k=0}^\infty\sum_{i=0}^k\binom{k}{i}x^iy^{k-i},\qquad |x|+|y|<1.$$

The absolute sum is bounded by $\sum_k(|x|+|y|)^k$. The displayed region is a sufficient domain of absolute convergence. It is not obtained by treating $x$ and $y$ as unrelated one-dimensional series.

For the following converse, take a positive integer $n$, a center $a\in\mathbb R^n$, $s>0$ and real coefficients $c_\alpha$. Define $f(a+h)=\sum_\alpha c_\alpha h^\alpha$ on $|h_i|<s$ under the stated finite absolute coefficient bound. This defines the function whose regularity is to be proved; $C^\infty$ regularity is not assumed.

Converse: absolute coefficient control permits differentiation. Suppose for some $s>0$,

$$A=\sum_{\alpha\in\mathbb N_0^n}|c_\alpha|s^{|\alpha|}<\infty.$$

Interpret this absolute sum as the supremum of finite subsums, or enumerate multi-indices in increasing total degree. On every closed cube $|h_i|\le r<s$, the original series is uniformly convergent by the M-test. For one partial derivative and $m=|\alpha|\ge1$, the term is zero when $\alpha_j=0$; otherwise, for $0<r<s$,

$$|\alpha_jc_\alpha h^{\alpha-e_j}|\le\frac{|c_\alpha|s^m}{s}\,m(r/s)^{m-1}.$$

The factor $m q^{m-1}$ is uniformly bounded for $q=r/s<1$, for instance by the convergent scalar sum $\sum_{m\ge1}m q^{m-1}=1/(1-q)^2$. Thus the differentiated terms have a summable coefficient majorant and converge uniformly on the cube.

Here is why convergence really gives a derivative. Let $P_N$ be the finite sum of monomials of total degree at most $N$. Fix all coordinates except the jth, and use that coordinate as a variable on $[-r,r]$. Both $P_N$ and $\partial_jP_N$ converge uniformly; the values at coordinate zero converge as well. The one-variable uniform-derivative theorem proves that the limit has that partial derivative. The uniform limit of the derivative polynomials is continuous on the whole cube. Doing this for every coordinate proves C1 regularity on its interior by the continuous-partials differentiability criterion.

For a prescribed higher derivative order $\ell$, the falling-factorial coefficient is at most $m^\ell$, and its terms are bounded on $|h_i|\le r$ by $|c_\alpha|s^m s^{-\ell}m^\ell(r/s)^{m-\ell}$, with zero terms when the relevant exponents are too small. This polynomial-times-geometric factor is uniformly bounded; apply the same argument successively on nested cubes strictly between $r$ and $s$. Every derivative order exists and is continuous. Evaluating at $h=0$ leaves just the matching monomial, proving $\partial^\alpha f(a)=\alpha!c_\alpha$. This argument supplies the termwise rule rather than assuming it from formal notation.

Exercises with complete solutions

E1 Foundation. Find the radius and endpoint behavior of $\sum_{n\ge1}x^n/n^2$.

S1. The root limsup is 1, so $R=1$. At both $x=1$ and $x=-1$ the absolute series is $\sum1/n^2$, which converges. Thus the interval of convergence is $[-1,1]$; uniform convergence on this entire interval also follows separately by the M-test.

E2 Series manipulation. Obtain a power series for $1/(2-x)$ about zero and for $1/(2-x)^2$, including the common radius.

S2. $1/(2-x)=\frac12\sum_{n\ge0}(x/2)^n=\sum x^n/2^{n+1}$ for $|x|<2$. Differentiating gives $1/(2-x)^2=\sum_{n\ge0}(n+1)x^n/2^{n+2}$, also with radius 2.

E3 Uniqueness. Prove that two convergent power series about the same center representing the same function near that center have identical coefficients.

S3. Differentiate the equal functions $n$ times at the center using the justified termwise rule. Both $n$th derivatives are $n!$ times the corresponding coefficient, so the coefficients agree for every $n$.

E4 Several variables. Find the terms of total degree at most two in the series for $e^{x+y}$.

S4. Expand $1+(x+y)+(x+y)^2/2+\cdots$. The terms are $1+x+y+x^2/2+xy+y^2/2$. The coefficient of $xy$ is 1, consistent with division by $1!1!$, not by $2!$ without the binomial factor.

Summary

Power series converge uniformly on smaller regions inside their radius, allowing justified termwise calculus. Analyticity means equality to that local series. Infinite differentiability alone is weaker; a Taylor remainder or another convergence argument must establish equality.