# How can you absolutely prove that a function has an infinite power series representation?

I am beginning to grasp onto the idea of a power series and the way that we can "hunt" for the constants in a maclaurin series expansion. However, what I don't yet understand is how we know that there is an infinite representation in the first place.

• we don't always. There are examples of smooth functions which do not admit a Taylor series at a particular point. Fortunately, most of your favorite functions appear as solutions to differential equations at an ordinary point hence, they are analytic, meaning there is a power series for them. For some discussion, about nonanalytics, see math.stackexchange.com/q/189841/36530 – James S. Cook Oct 8 '14 at 5:25
• This is a very important question, and it has no simple answer! Much of the modern study of analysis began as an attempt to understand when functions have power series expansions and under what conditions those power series converge. – MJD Oct 8 '14 at 14:07
• It becomes a little clearer when you get into complex numbers and functions. Having a convergent power series on some open interval of the reals means that the function can be extended to an "analytic" complex function, which is relatively easy to define. – Thomas Andrews Oct 11 '14 at 16:22
• This is only loosely related to your question, but a lot of power series representations can be proven using the binomial theorem. For example, expanding the following gives their power series: $e^x=\lim\limits_{n\to\infty}(1+\frac xn)^n$, $\ln(1+x)=\lim\limits_{\epsilon\to0}\dfrac{(1+x)^\epsilon-1}\epsilon$, $\cos(x)+i\sin=\lim\limits_{n\to\infty}(1+i\frac xn)^n$. (For the last one, you probably need to use some combination of De Moivre and L'Hôpital to prove it.) – Akiva Weinberger Oct 15 '14 at 16:04
• More related to the question—doesn't Taylor's theorem involve some sort of error term? If the error term goes to zero, than the power series given by Taylor's theorem converges to the function... I think. – Akiva Weinberger Oct 15 '14 at 16:07

Taken from the first chapter of Markushevich's Complex Analysis, Vol 1:

A function $f(x)$ defined in a neighborhood of $x_0$ is analytic there if and only if

1. $f(x) \in C^{\infty}$ at a neighborhood of $x_0$
2. There exists positive numbers $\delta$ and $M^{'}$, such that for any $x \in (x_0 - \delta, x_0 + \delta)$ the inequality: $$|f^k(x)| \leq M^{'} \dfrac{k!}{\delta^k}$$

Holds for any $k \geq 0$.

As I cannot see the inside of the book to link from google books to link, I'll just leave the proof here.

Proof: Supose $f$ is analytic at $x_0$, that means one some neighborhood of $(x_0 - \varepsilon, x_0 + \varepsilon)$ we have$$f(x) = a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + \dots$$

We also know that it can be differentiated term by term getting $$f^{(k)}(x) = k!a_k + \dfrac{(k+1)!a_{k+1}}{1!}(x-x_0) + \dfrac{(k+2)!a_{k+2}}{2!}(x-x_0)^2 + \dots$$

Now, let $0 < 2\delta < \varepsilon$, if $|x-x_0| < 2\delta$, then the series converges at $x$. This means that $\lim\limits_{n\to \infty} a_n(2\delta)^n = 0$ (Why?).

Or in other words the set $\{a_n(2\delta)^n\}$ is bounded, that means we have an $M > 0$, such that $M > |a_n(2\delta)^n|$ for all n .

But now, if we are in the interval $(x_0 - \delta, x_0 + \delta)$: $$|f^{(k)}(x)| \leq k!|a_k|+ \dfrac{(k+1)!|a_{k+1}|}{1!}\delta + \dfrac{(k+2)!|a_{k+2}|}{2!}\delta^2 + \dots$$

$$\leq k!\dfrac{M}{2\delta^k}+ \dfrac{(k+1)!}{1!}\dfrac{M}{(2\delta)^{k+1}} \delta + \dfrac{(k+2)!}{2!} \dfrac{M}{(2\delta)^{k+2}} \delta^2 + \dots$$

$$= \dfrac{k!M}{(2 \delta)^k} \bigg[1 + \dfrac{k+1}{1!} \dfrac{1}{2} + \dfrac{(k+1)(k+2)}{2!} \dfrac{1}{2^2}+ \dots \bigg]$$

$$= \dfrac{k!M}{2^k \delta^k} \bigg(1-\dfrac{1}{2}\bigg)^{-(k+1)} \quad \text{(Why?)}$$

$$= 2M\dfrac{k!}{\delta^k}$$

To finish Just Put $M^{'} = 2M$

Now, we need to prove the other way, supose $f$ is infinitely differentiable in $(x_0 - \delta, x_0 + \delta)$ and the inequality is satisfied for all the derivatives of $f$ in that interval. Expanding in a Taylor series with Lagrange remainder we get:

$$f(x) = f(x_0) + \dfrac{f^{(1)}(x_0)(x-x_0)}{1!} + \dfrac{f^{(2)}(x_0)(x-x_0)^2}{2!} + \dots + \dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}$$

With $0 < |\xi - x_0| < |x-x_0|$. Applying 2 we get that:

$$\bigg\lvert \dfrac{f^{(n+1)}(\xi)}{(n+1)!} \bigg\vert |x-x_0|^{(n+1)} \leq \dfrac{M(n+1)!}{\delta^{n+1}(n+1)!}|x-x_0|^{n+1} = M \bigg( \dfrac{|x-x_0|}{\delta} \bigg)^{n+1}$$

But this mean that if $|x-x_0| < \delta$ when $n \rightarrow \infty$ the error term goes to $0$. Meaning $f$ is analytic there.

Both conditions are necessary, if you only had ($1$) for example, wouldnt suffice, as there are functions that are infinitely differentiable ($C^{\infty}$) but which dont have an infinite taylor series representation at any point! This can be proved using the Baire Category Theorem, the proper statement would be:

The set of functions in $C^{\infty}$ that are nowhere analytic is of the second category.

This means that nowhere analytic functions are a dense set in the set of infinite differentiable functions. A proof can be found on Dugundji's Topology p.$301$.

• For instance, $f : x-> exp(\frac{-1}{x^2})$ is $C^∞$ on ]0, +∞[ , but you can prove quite easily that such a function doesn't have a series expansion (using the property $a_n =\frac{f^{(k)}(0)}{k!}$) – mvggz Oct 16 '14 at 8:09

On an intuitive level, we think something like this "I think I understand polynomials, but all those other functions are super confusing. Is there a way I could think of those other things as like 'big' polynomials or something?"

The answer is a bit complicated, but the short of it is this: you can certainly try, sometimes this 'big' polynomial approach is cool, sometimes it doesn't work out so well. Let's see why (This is Taylor series, for MacLaurin let a=0).

If we write $$f(x) = a_0 + a_1 (x-a) + a_2 (x-a)^2 + \dots$$ then if we plug in $a$ we find out that $a_0 = f(a)$. If we differentiate both sides we see $$f'(x) = a_1 + 2a_2 (x-a) + 3a_3 (x-a)^2 + \dots$$ then if we plug in $a$ we see that $a_1 = f'(a)$. Continuing this process we find that $$a_n = \frac{f^{(n)}(a)}{n!}.$$ We can always think about writing a 'big' polynomial (Taylor series) so long as all the derivatives we need exist. (This is the part about requiring that $f$ have infinitely many continuous derivatives.)

The problem comes when we try to use this series for something that's not $a$. At $a$ we know the series converges since it's just $f(a) +0$. Now we need to ask a hard question: How far can we go from $a$ before things break down? The answer is the radius of convergence of the series. Usually we find this with the ratio test.

If the radius is 0, which happens sometimes, we have to abandon the series approach. If the radius of convergence is non-zero we call the function analytic at $a$ and can (carefully!!) treat it a little bit like a really long polynomial.