Derivative of a Power Series

Can any one help me out on these questions? Not too sure what to do...

Define a function f by,

$f(x) := \displaystyle\sum_{i=0}^{\infty} \frac{3^n x^n}{(n+1)!}$

for those$x \in \mathbb{R}$ for which the series converge, find f'

Thanks for any help!

Well, assuming you've identified the values of x for which the series converges, you should know a general result that tells you how to differentiate a power series within its radius of convergence

That is correct, however it's the justification of the derivative that is important.

You don't know (from the question) that the derivative f' converges, thus use Spandy's simple method to rewrite the summand and show convergence.

The question is only difficult because you have to prove the convergence of the derivative as well as calculating it. The derivative you have calculated above will be valid for those x in the reals where the original series f converges.

Hope that's clear!

That is correct, however it's the justification of the derivative that is important.

You don't know (from the question) that the derivative f' converges, thus use Spandy's simple method to rewrite the summand and show convergence.

The question is only difficult because you have to prove the convergence of the derivative as well as calculating it. The derivative you have calculated above will be valid for those x in the reals where the original series f converges.

Hope that's clear!

Not if you've already been given the standard result that a power series is differentiable within its radius of convergence!

I'm not sure, but I suspect the point of this question is that you can in fact express the derivative as a well-known function of x by doing a bit of rearrangement and taking out a couple of factors, but I haven't tried it myself!

Ok in that case you're safe with f being differentiable everywhere in R.

However, you need to be careful with derivatives. It's derivative will CONVERGE on all R, otherwise f would not be differentiable on R. So you're right there (incidentally, the power series is differentiable because it is just a sum of positive powers of x; convergence means you can 'switch the derivative and sum' signs without any trouble).

You claimed that f' would be differentiable everywhere. I don't think this is true, take, for example f' = |x| and f = sgn(x) x^2 / 2

Then f is differentiable on R but f' is not (not differentiable at zero).

If you have any questions on the subtlety let me know

You claimed that f' would be differentiable everywhere. I don't think this is true,

take, for example f' = |x| and f = sgn(x) x^2 / 2

Then f is differentiable on R but f' is not (not differentiable at zero).

There is a bit of subtlety in this question because we're talking about power series. In summary, for a power series:

- It is differentiable (term-by-term) within its radius of convergence; but the derivative may not be convergent at all points

This is sort of true, but it's quite a lot more subtle than you seem to think. I say "sort of true" because you need the derivative to converge as well. You certainly can have "trouble" where it doesn't. For example consider

$\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n x^n}{2n+1} = \log(1+x)$ for (-1 < x <= 1).

The sum converges when x = 1, but differentiating the power series term by term gives a divergent series when x = 1, even though you can differentiate log(1+x) at that point.

If the power series for f converges everywhere, this is true. More generally:

If you have a power series $f(x) = \sum_0^\infty a_n x^n$ and it converges for any particular value $x=x_0$, then the series converges whenever $|x|<|x_0|$ , and for such values of x differentiating term by term gives a convergent series that converges to f'(x). You can repeat the argument to show all the derivatives exist inside the radius of convergence.

This is all standard bookwork - albeit bookwork that you're often not expected to be able to prove.

This is true, but you can't express f as a power series so it isn't a counterexample to what was claimed.

Power series are special.

Im pretty sure thats not the power series expansion of log(1+x). Coefficient of x^n should be [(-1)^{n-1}]/n

Power series are special.