ANA1_L32

Real analytic functions

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

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

$R$ is has to be less than or equal to the radius of convergence, as defined in the previous lecture. Functions which are locally(i.e, on some interval) defined by a convergent power series are called analytic functions. The one above is a real analytic function defined locally around $0$ by a convergent power series.

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.

To answer these questions, we will need to understand uniform convergence.

---

A look ahead

Let us black box “uniform convergence” for the moment and have a look at the results it affords us.

Theorem 1(Theorem Rudin 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

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

tldr; If a function is analytic, then the derivative exists and is given by (that expression). Note that this implicitly asserts that the sequence defining $f^{\prime }(x)$ converges for $|x|<R$.

Additionally, 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. Any analytic function is infinitely differentiable, and applying the theorem on $f(x)$ also yields an expected formula for the nth derivative of $f$:

Theorem 2(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

$$f^{(k)}(x)=\sum _{n=k}^{\infty }n(n-1)\dots (n-k+1)c_n{x}^{n-k}.$$

The derivatives at $0$ are of particular interest:

$$\begin{array}{r}{f}^{(k)}(0)=k(k-1)(k-2)\dots (1)c_n\end{array}$$ $$⟹c_n=\frac{f^{(k)}(0)}{k!}$$

This tells us that if a function is analytic, it must have a unique power series representation! Additionally, we only need the derivatives at a single point to construct this power series about that point.

This raises a question. We have shown that an analytic function on $(-R,R)$ is infinitely differentiable on $(-R,R)$. This allows us to 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 the function at $t$ to any degree. But do these converge? If they do, do they converge to $f$?

$$g(x)=\sum _{n=0}^{\infty }\left(\frac{f^{(n)}(t)}{n!}\right)(x-t{)}^n.$$

In other words, is $g=f$? We will see that this is indeed the case. If a function is analytic on $(-R,R)$, is is also analytic on any subinterval of $(-R,R)$.