# Homework Help: Zwiebach problem 12.4

1. Oct 26, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
The first part of this problem makes no sense to me because n is a constant in gamma and when you multiply gamma by zeta the n is somehow supposed to get inside the zeta function...

2. Relevant equations

3. The attempt at a solution

2. Oct 27, 2007

### Jimmy Snyder

It's pretty straight forward. What does equation (1) look like after the replacement t -> nt?

3. Oct 27, 2007

### ehrenfest

I understand what he wants:

$$\Gamma (s) \zeta (s) = \sum_{n=1}^{\infty}\Gamma (s) n^{-s}$$

and it is easy from there.

But that equation is like saying that

$$(a x^2 + bx +n) (\sum_{n=1}^{\infty} n^2) = \sum_{n=1}^{\infty} (a x^2 + bx +n) n^2$$

My point is that there is very poor notation in this problem.

Zwiebach uses an n outside the summation which simply does not make sense.

4. Oct 27, 2007

### Jimmy Snyder

My bad. I should have written:

Replace t with nt in

$$\Gamma(s) = \int_0^{\infty}dte^{-t}t^{s-1}$$

after you put the gamma inside the sum.

Last edited: Oct 27, 2007
5. Oct 27, 2007

### ehrenfest

I see. I was stupidly replacing t with nt before I put the gamma inside the sum. I take back what I said about the poor notation.

6. Oct 28, 2007

### ehrenfest

How do you justify the last inequality?

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• ###### zwiebach_sol_125.jpg
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7. Oct 31, 2007

### ehrenfest

Sorry. I mean equality.

8. Oct 31, 2007

### Dick

For the last equality, they are expanding 1/(1+f(t))=1-f(t)+f(t)^2-f(t)^3+... and keeping only the first few terms.

9. Nov 1, 2007

### ehrenfest

Cool. I knew that 1/(1+t) = 1 -t + t^2 -t^3 +t^4. I didn't know it was true when you replaced t with an arbitrary function of t.

Is there a quick way to prove that?

10. Nov 1, 2007

### Jimmy Snyder

This is not a statement about the function f, it is a statement about the number f(t).

Last edited: Nov 1, 2007
11. Nov 1, 2007

### Dick

You can expand 1/(1+whatever) that way as long as |whatever|<1.

12. Nov 1, 2007

### ehrenfest

I see. Thanks.