# Finding the series representation of complex function

• Nov 11th 2009, 12:00 PM
douber
Finding the series representation of complex function
the funtion is cosh(z)/(1 -z) about z=0

to find the series representation would i not just multiply the series of cosh(z) with that of 1/(1-z) about z=0

After that is there any simplifying i could do? or will it just be a multiplication of 2 sums?

am i missing any tricks? seems too easy to do the way i'm doing it.

thanks for any help!
• Nov 11th 2009, 02:55 PM
TKHunny
Easy to SAY. How long will it take you to compute 100 terms?
• Nov 11th 2009, 07:12 PM
Drexel28
Quote:

Originally Posted by douber
the funtion is cosh(z)/(1 -z) about z=0

to find the series representation would i not just multiply the series of cosh(z) with that of 1/(1-z) about z=0

After that is there any simplifying i could do? or will it just be a multiplication of 2 sums?

am i missing any tricks? seems too easy to do the way i'm doing it.

thanks for any help!

This depends on exactly how you define series. I mean would $\sum_{n=0}^{\infty}\frac{z^{2n}}{(1-z)(2n)!}$ work?

If this isn't satisfactory notice that $\frac{1}{1-z}=\sum_{m=0}^{\infty}z^m$ so that the above becomes $\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{z^{2n+ m}}{(2n)!}$ which is a compact double sum. Of course, this depends on the fact that $|z|<1$

As TKHunny pointed out computing $\left(\sum_{n=0}^{\infty}\frac{z^{2n}}{(2n)!}\righ t)\left(\sum_{n=0}^{\infty}z^n\right)$ is no trivial task.
• Nov 11th 2009, 07:35 PM
douber
Thanks guys!

I did have that compact double sum as my answer so now i can be confident about it. that makes a lot of sense that it would be crazy to compute otherwise.

thanks for everything!!
• Nov 11th 2009, 08:18 PM
TKHunny
There are a few things like this. The first time I encountered it, I learned the term "Essentially Impossible". This was further defined as "Sure, you could do it, but who would want to?" Really, it's an existential argument. It is enough to say that it can be done. We don't actually have to do it.

I am delighted if you have learned today that math can cause a smile or two, even at this level. (Nod)
• Nov 12th 2009, 10:17 AM
douber
hahaha

yea smile is good every now and then lol