# Thread: f is a power series. Will f' and f'' have the same interval of convergence as f?

1. ## f is a power series. Will f' and f'' have the same interval of convergence as f?

Say I have the following funciton -

$f(x) = \sum_{n=1}^{+\infty}\frac{x^n}{n2}$

and I am asked to find the interval of convergence for f, f'' and f'''. What is the process for getting the IoC for f'' and f'''?

Here is what I think I should do, please correct me if Im wrong -

1. Find where f(x) converges using the Ratio Test, I get |x| < 1 for this.
2. Then check the function for convergence at endpoints -1 and 1. Neither of these converge for me.
3. So I have interval of convergence (-1, 1).
4. Here is where I am not sure what to do - Rather than doing the entire process again for f(x) I can just differentiate $\frac{x^n}{n2}$ and then check the endpoints against this new function? Because the radius of convergence is the same for f and f'' (and f'''), the interval of convergence will still be between -1 and 1 and I only need to check the endpoints? Is this correct?

2. ## Re: f is a power series. Will f' and f'' have the same interval of convergence as f?

Originally Posted by nukenuts
Say I have the following funciton -

$f(x) = \sum_{n=1}^{+\infty}\frac{x^n}{n2}$

and I am asked to find the interval of convergence for f, f'' and f'''. What is the process for getting the IoC for f'' and f'''?

Here is what I think I should do, please correct me if Im wrong -

1. Find where f(x) converges using the Ratio Test, I get |x| < 1 for this.
2. Then check the function for convergence at endpoints -1 and 1. Neither of these converge for me.
3. So I have interval of convergence (-1, 1).
4. Here is where I am not sure what to do - Rather than doing the entire process again for f(x) I can just differentiate $\frac{x^n}{n2}$ and then check the endpoints against this new function? Because the radius of convergence is the same for f and f'' (and f'''), the interval of convergence will still be between -1 and 1 and I only need to check the endpoints? Is this correct?
A basic theorem about power series extablishes that, given the power series...

$a_{0} + a_{1}\ z + a_{2}\ z^{2}+...+ a_{n}\ z^{n} + ...$ (1)

... that coverges inside a circle centered in z=0 and with radious r, then the series of derivatives...

$a_{1} + 2\ a_{2}\ z + 3\ a_{3}\ z^{2} ...+ n\ a_{n}\ z^{n-1} + ...$ (2)

... converges inside the same circle...

Kind regards

$\chi$ $\sigma$

3. ## Re: f is a power series. Will f' and f'' have the same interval of convergence as f?

Originally Posted by chisigma
A basic theorem about power series extablishes that, given the power series...

$a_{0} + a_{1}\ z + a_{2}\ z^{2}+...+ a_{n}\ z^{n} + ...$ (1)

... that coverges inside a circle centered in z=0 and with radious r, then the series of derivatives...

$a_{1} + 2\ a_{2}\ z + 3\ a_{3}\ z^{2} ...+ n\ a_{n}\ z^{n-1} + ...$ (2)

... converges inside the same circle...

Kind regards

$\chi$ $\sigma$
So in that case I dont even have to check the endpoints for f' and f''? The interval of convergence for f, f' and f'' is all exactly the same?

The reason I am unsure is because in Stewarts Calculus book it says the following
Although Theorem 2 says that the radius of convergence remains the same
when a power series is differentiated or integrated, this does not mean that the interval of
convergence remains the same. It may happen that the original series converges at an endpoint,
whereas the differentiated series diverges there.
So do I need to check the end points?

4. ## Re: f is a power series. Will f' and f'' have the same interval of convergence as f?

Originally Posted by nukenuts
So do I need to check the end points?
Absolutely (no pun intended!). BTW, is the 2 in your series

$\frac{1}{2n}$ or $\frac{1}{2^n}$?

5. ## Re: f is a power series. Will f' and f'' have the same interval of convergence as f?

Originally Posted by Danny
Absolutely (no pun intended!). BTW, is the 2 in your series

$\frac{1}{2n}$ or $\frac{1}{2^n}$?
Oops, its supposed to be n^2.

6. ## Re: f is a power series. Will f' and f'' have the same interval of convergence as f?

Yep - really makes a difference! Good question.