# Derivative of a Power Series

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#2

I’m not quite sure- but this is how I believe it will workout-

The derivative will be n*(3^n)*(x^[n-1])/(n+1)!

This further reduces to {(3^n)*(x^[n-1])/(n)!}- {(3^n)*(x^[n-1])/(n+1)!},which converges.

The derivative will be n*(3^n)*(x^[n-1])/(n+1)!

This further reduces to {(3^n)*(x^[n-1])/(n)!}- {(3^n)*(x^[n-1])/(n+1)!},which converges.

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#3

(Original post by

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

Define a function f by,

for those for which the series converge, find f'

Thanks for any help!

**Blue7195**)Can any one help me out on these questions? Not too sure what to do...

Define a function f by,

for those for which the series converge, find f'

Thanks for any help!

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(Original post by

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

**davros**)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

?

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#5

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!

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!

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(Original post by

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!

**nathanturnerspc**)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!

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#7

**nathanturnerspc**)

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!

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!

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#8

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

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

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(Original post by

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!

**davros**)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!

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(Original post by

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

**nathanturnerspc**)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

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#11

(Original post by

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).

**nathanturnerspc**)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).

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.

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

**is**true. More generally:

If you have a power series and it converges for any particular value , then the series converges whenever , 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.

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).

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

Power series are

*.*

**special**
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#12

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

- The derivative of the series will be differentiable at all points by virtue of it being a series of powers of x

Just FYI the series you have for log(1+x) isn't quite right, although I agree the derivative series does not converge at x=1. But no matter for the question we're looking at here.

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

- The derivative of the series will be differentiable at all points by virtue of it being a series of powers of x

Just FYI the series you have for log(1+x) isn't quite right, although I agree the derivative series does not converge at x=1. But no matter for the question we're looking at here.

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#13

(Original post by

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

**nathanturnerspc**)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

See also https://gowers.wordpress.com/2014/02...-power-series/

- The derivative of the series will be differentiable at all points by virtue of it being a series of powers of x

Just FYI the series you have for log(1+x) isn't quite right, although I agree the derivative series does not converge at x=1. But no matter for the question we're looking at here.

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#14

(Original post by

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

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

If you have a power series and it converges for any particular value , then the series converges whenever , 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

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

**DFranklin**)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

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 and it converges for any particular value , then the series converges whenever , 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**
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#15

(Original post by

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

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

Edit: I should perhaps have acknowledged the error in the previous post but was posting from my phone and having enough trouble cut/pasting a link! Having lost the post twice I just wanted to get it out there.

Point still holds though: the power series for log(1+x) converges for -1<x<=1 but the derivative diverges at 1, despite the fact that log(1+x) is differentiable there.

The arctan series is actually a more interesting example in some ways (hopefully got it right this time!):

The RHS only converges for -1 < x <= 1 and the (term-by-term) derivative diverges at x=1.

But the LHS is infinitely differentiable for all x.

The reason being that arctan z (z complex) has singularities at z=+/- i. Of course, I've done the complex analysis work to understand why this all happens, but I still find it kind of amazing how a function's behaviour well away from the real line can still have very observable effects on its behaviour when restricted to the real line.

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