Zwiebach Problem 12.4: Struggling to Understand

[ehrenfest]

Homework Statement

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...

Homework Equations

The Attempt at a Solution

 

[Jimmy Snyder]

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...

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

 

[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.

 

[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.

 

[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.

 

[ehrenfest]

How do you justify the last inequality?

 

[ehrenfest]

Sorry. I mean equality.

 

[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.

 

[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?

 

[Jimmy Snyder]

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?

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

 

[Dick]

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

 

[ehrenfest]

I see. Thanks.