LEC ANA1 32

Real analytic functions

Let $\sum c_n{x}^n$ converge for $|x|<R$. Define

$$\begin{array}{rl}& f:\{x\in \mathbb{R}:|x|<R\}\to \mathbb{R},\\ & f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n.\end{array}$$E1

$R$ has to be less than or equal to the radius of convergence. Functions which are locally (i.e, on some interval) defined by a convergent power series are called analytic functions. $f$ is a special kind of analytic function, since its power series at a single point converges on its entire domain, and hence $f$ is defined on its entire domain by the power series at that point, namely $0$. In this lecture, we will only be working with such functions. In general, however, such a point may not exist. Here’s the general definition:

Definition 154.1(Definition).

A function $g:(a,b)\to \mathbb{R}$ is analytic if for each $x\in (a,b)$, there exists a power series

$$\sum a_r{h}^r$$

and a $\delta >0$ such that if $|h|<\delta$, then the series converges and

$$g(x+h)=\sum _{r=0}^{\infty }{a}_r{h}^r.$$

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We will soon show, in Thm 6, that functions like $f$ satisfy Def 1.

Let $f$ be as in (E1). Natural questions:

1. Is $f$ continuous? Differentiable? Integrable?
2. What functions are real analytic? In the above example, we started with a convergent series and defined a function. Can we go the other way around, i.e, start with a function, and express it as a convergent power series centered at any point in its domain?

More questions

At this point, we could just make the definitions

$$\begin{array}{r}{e}^x\equiv 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots ,\\ \sin x\equiv x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots ,\\ \cos x\equiv 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\dots .\end{array}$$

We have the tools to say that the above series converge for all $x$, so $\sin x$ and $\cos x$ are defined for all $x\in \mathbb{R}$. We can ask the above questions in particular for these functions.

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Theorem 154.2(Rudin (1976) 8.1).

Suppose the series ${\sum }_{n=0}^{\infty }{c}_n{x}^n$ converges for $|x|<R$. This allows us to define

$$f(x)\equiv \sum _{n=0}^{\infty }{c}_n{x}^n(|x|<R),$$

which is the limit ${\lim }_{n\to \infty }{f}_n$, where $f_n={\sum }_{i=0}^n{c}_i{x}^i$.

Then,

1. $(f_n)$ converges uniformly on $[-R+ϵ,R-ϵ]$ for all $ϵ>0$.
2. $f$ is continuous and differentiable on $(-R,R)$, and

$$f^{\prime }(x)=\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}(|x|<R).$$

Since $f^{\prime }(x)$ is also an analytic function, we can turn the crank again to get $f^{\prime \prime }(x)$, which is also analytic, and so on. Thus, any analytic function is infinitely differentiable. This yields an explicit formula for the $n$th derivative of $f$:

Corollary 154.3(Corollary).

If

$$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$

on $(-R,R)$, then $f^{(k)}(x)$ exists for all $k\ge 0$, and is given by

$$\begin{array}{rlr}{f}^{(k)}(x)=\sum _{n=k}^{\infty }n(n-1)\dots (n-k+1)c_n{x}^{n-k}& & x\in (-R,R).\end{array}$$

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The derivatives at $0$ are of particular interest:

$$\begin{array}{r}{f}^{(k)}(0)=k!c_k\end{array}$$

$$⟹c_k=\frac{f^{(k)}(0)}{k!}.$$E2

Remark 154.4(Remark).

This tells us that if a function is analytic, it must have a unique power series representation! The series

$$\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{h}^n$$

Is called the Taylor series for $f$ at $0$. Furthermore, the series $\sum a_r{h}^r$ from Def 1 must be the Taylor series for $g$ at $x$:

$$\sum _{n=0}^{\infty }\frac{g^{(n)}(x)}{n!}{h}^n.$$

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Additionally, we only need the derivatives at a single point to construct this power series about that point.

Remark 154.5(Remark).

Although a function $f$ may have derivatives of all orders (i.e, be smooth), the series $\sum c_n{x}^n$, where $c_n$ is computed by (E2), need not converge to $f(x)$ for any $x\ne 0$. Since if $f$ had a power series representation, it would have to be $\sum c_n{x}^n$, this tells us that $f$ is not analytic on any neighborhood of $0$. For example, consider

$$\begin{array}{r}f(x)=\{\begin{array}{ll}{e}^{-1/x^2}& x\ne 0\\ 0& x⩽0.\end{array}\end{array}$$

$f^{(n)}(0)=0$ for all $n\in \mathbb{N}$. However, $f$ is not constant in any neighborhood of $0$.

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Having shown that an analytic function $f$ on $(-R,R)$ is infinitely differentiable on $(-R,R)$, we can calculate all derivatives $f^{(0)}(t),f^{(1)}(t),f^{(2)}(t),\dots$ at any point $t\in (-R,R)$. So, we can calculate the Taylor polynomial of $f$ at $t$ to any degree. But do these converge? If they do, do they converge to $f$? We will see that this is indeed the case.

Theorem 154.6(Taylor, Rudin (1976) 8.4).

Suppose

$$\begin{array}{rlr}f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n& & x\in (-R,R).\end{array}$$

If $a\in (-R,R)$, then $f$ can be expressed as a power series about the point $x=a$ which converges in $|x-a|<R-|a|$, and

$$\begin{array}{rlr}f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a{)}^n& & |x-a|<R-|a|.\end{array}$$

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Misc

Suppose ${\sum }_{n=0}^{\infty }{c}_n{x}^n$ converges for $x\in (-R,R)$. If the series also converges at an endpoint, say $x=R$, then $f:={\sum }_{n=0}^{\infty }{c}_n{x}^n$ is continuous not only in $(-R,R)$, but also at $x=R$. We take $R=1$ for simplicity:

Proposition 154.7(Rudin (1976) 8.2).

Suppose $\sum c_n$ converges. Put

$$\begin{array}{rlr}f(x):=\sum _{n=0}^{\infty }{c}_n{x}^n& & -1⩽x⩽1.\end{array}$$

Then,

$$\underset{x\to 1}{\lim }f(x)=\sum _{n=0}^{\infty }{c}_n.$$

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References

Rudin, W. (1976). Principles of Mathematical Analysis (3d ed). McGraw-Hill.