Homework Help: Is there a differnt way to write this expression?

1. Mar 7, 2010

Firepanda

f(z) = SUM [(-1)n

Where SUM = sum from n=0 to infinity

Looks to me like there isn't but there must be a simpler way to write that.

Thanks

Last edited: Mar 7, 2010
2. Mar 7, 2010

Dick

It's divergent, isn't? The nth term doesn't approach 0.

3. Mar 7, 2010

Firepanda

The question is to write this power series as a function

Last edited: Mar 7, 2010
4. Mar 7, 2010

Dick

z^(2n)/n! isn't equal to e^(z^2). The SUM of the series z^(2n)/n! is e^(z^2). That would be fine if the (-1)^n weren't there. Can't you think of a function that's a lot like e^(z^2) that will give you the alternating sign in the series?

5. Mar 7, 2010

Firepanda

I just keep getting into the routine of trying to simply it like before. Do you mean replace e^(z^2) or multiply in some function f(z)/f(z)?

6. Mar 7, 2010

Dick

(-1)^n*z^(2n). Can you think of a way to write that as r^n for some number r? What would r be?

7. Mar 7, 2010

Firepanda

ahh yes

Last edited: Mar 7, 2010
8. Mar 7, 2010

Firepanda

For the 2nd part I have it down to

z-2 SUM (-1)n.(z2/4)n+1

Is there anyway I can progress from here or should I go back and try a different way?

9. Mar 7, 2010

Dick

It's a geometric series.

10. Mar 7, 2010

Firepanda

(z2/4)n+1 starting at n=0

= 1/(1-z2/4) - 1

=> Series = z-2 SUM (-1)n/(1-z2/4) + (-1)n+1

11. Mar 7, 2010

Dick

You just did the same wrong thing you did in the first problem. You replaced a term is the series with the sum of a series. And you missed the cue to do the same thing you did with the (-1)^n in the first problem. You series has the form a*r^n. r just might have a (-1) in it.

12. Mar 7, 2010

Firepanda

ah yes, i have my r now thanks, was simple after that!

Thanks alot