Is every power series infinitely differentiable everywhere?

Question

I have found various sources on the internet that say that power series are infinitely differentiable on their interval of convergence:

Wikipedia:

Once a function $f(x)$ is given as a power series as above, it is differentiable on the interior of the domain of convergence.

Northwestern University:

[...] power series are (infinitely) differentiable on their intervals of convergence [...]

But isn't every power series (infinitely) differentiable everywhere?

After all, a power series is just an infinite polynomial and a polynomial of degree $n$ is differentiable $n+1$ times. Source

Doesn't this imply, that a polynomial of "degree $\infty$" is differentiable $\infty$ times?

"a power series is just an infinite polynomial". Following that logic, $\sum_{i=1}^{\infty} 1$ is just a number. — Mathematician 42, Jan 10 '20 at 09:59
It‘s rather about where the power series is even well-defined. — Qi Zhu, Jan 10 '20 at 10:01
You are right that a formal power series is just an infinite degree polynomial. This is a purely algebraic viewpoint that is sometimes useful; see for example the book "Generatingfunctionology", by Herbert Wilf. However, the analytic aspects of the manipulation of power series depend on its convergence properties; differentiability is one such aspect. — Giuseppe Negro, Jan 10 '20 at 10:01

score 6 · Answer 1 · answered Jan 10 '20 at 09:58

6

There is lot of difference between polynomials and power series.

You cannot define $\sum z^{n}$ for $|z| >1$. So there is no question of differentiability of this on $\{z: |z| >1\}$.

answered Jan 10 '20 at 09:58

Kavi Rama Murthy

359,332

Thank you for your answer. My issue was that I did not understand what it means for a derivative to exist. Can you take a look at my own answer and check that my new understanding is correct? – user734796 Jan 10 '20 at 11:26

user734796 · Accepted Answer · 2020-01-10T12:04:56.067

I think this was an issue of language and my understanding of what it means for a derivative to exist.

It is of course possible to "derive" (e.g. blindly following the rules of differentiation) term-by-term like this: $$ f(x) = \sum_{n=0}^\infty a_n(x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + a_3(x-c)^3 + ... $$ $$ f'(x) = \sum_{n=1}^\infty na_n(x-c)^{n-1} = a_1 + 2a_2(x-c) + 3a_3(x-c)^2 + ... $$

However, an infinite sum only exists, if it converges!
Therefore, if the infinite sum does not converge (which it might, depending on $x$), it does not exist, and therefore the derivative itself does not exist for those values of $x$.

The radius of convergence of the power series and its derivatives is the same: Source 1, Source 2.

Edit
I would like to highlight this statement by Kavi Rama Murthy from the comments, because it complements this answer:

[The terms of the derived series] make sense for any $x$ but the infinite sum makes sense only within the circle of convergence.

To be more explicit let me make the following statements: The series $\sum n a_n (x-c)^{n-1}$ can be called the 'derived series'. Its terms make sense for any $x$ but the infinite sum makes sense only within the circle of convergence. And in this circle $f'(x)$ exists and it is given by the infinite sum. — Kavi Rama Murthy, Jan 10 '20 at 11:51

Is every power series infinitely differentiable everywhere?

2 Answers2