Expert: Paul Klarreich - 10/9/2008

Question
Hello,

Thank you for replying.  I didn't know how to edit the last question, so I just re-ask it here.

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SUM [d(n)*(z^n)] = SUM [z^n/(1-z^n)] for lzl<1 (both sums from 0 to infinity)

where d(n) denotes the number of divisors of n.

My complex analysis class just finished proving the following theorems:

Holomorphic functions are analytic (complex differentiable functions at a point have a convergent power series expansion at that point).

Lioville: If f is entire and bounded, then f is constant.

Corollary to Lioville (Fundamental Theorem of Algebra): Every non-constant polynomial with complex coefficients has a root in C.

Morera: If f is continuous and the path integral over any closed toy contours equals zero, then f is holomorphic.



I have no idea of how to relate d(n) with any thing else in the problem... I listed of terms and counted the number of divisors... please help.

Answer
Questioner:   John
Category:  Advanced Math
Private:  No
 
Subject:  Complex Analysis
Question:  Hello,

Thank you for replying.  I didn't know how to edit the last question, so I just re-ask it here.

Show
inf
SUM [d(n)*(z^n)] = SUM [z^n/(1-z^n)] for abs(z) < 1
n=0
where d(n) denotes the number of divisors of n.

>> I THINK YOU WANT   n = 1 to infinity, not zero.  If n = 0, then two things are wrong:

z^0/(1 - z^0) = 1/0, also undefined.
The expansion (see below) does not work.

................................

My complex analysis class just finished proving the following theorems:

Holomorphic functions are analytic (complex differentiable functions at a point have a convergent power series expansion at that point).

Lioville: If f is entire and bounded, then f is constant.

Corollary to Lioville (Fundamental Theorem of Algebra): Every non-constant polynomial with complex coefficients has a root in C.

Morera: If f is continuous and the path integral over any closed toy contours equals zero, then f is holomorphic.



I have no idea of how to relate d(n) with any thing else in the problem... I listed of terms and counted the number of divisors... please help.
 
Hi, John,

Thank you for resending the question.  I am not sure of the answer here, but I have started some stuff and decided to send you what I have so far.  Perhaps you will be able to finish up.

The following expression has a well known expansion (it's actually the binomial expansion, or you can just do grade-school long division):
  1
------ = 1 + x + x^2 + x^3 + ..., for abs(x) < 1, of course.
1 - x

So your 'fraction' on the right can be developed:
 1
------ = 1 + z^n + z^2n + z^3n + ...
1-z^n

and multiplying by z^n:

z^n
------ = z^n(1 + z^n + z^2n + z^3n + ...)
1-z^n

=        z^n + z^2n + z^3n + z^4n + ...

Now the right side can be written:

                inf   inf
Sum (fraction) = SUM [ SUM(z^k + z^2k + z^3k + z^4k +...) ]
n=1              n=1   k=1


Now I think you can do something with the double sum on the right, along these lines:  It looks like this:

k = 1: z^1 +  z^2 +  z^3 +  z^4 +...
k = 2: z^2 +  z^4 +  z^6 +  z^8 +...
k = 3: z^3  + z^6  + z^9  + z^12 +...
...
------------------------------------ SUM IT ALL UP

Now how many times would you have a term like, say, z^42?

[42 is a nice number, right?]

I mean, how many of those  'k = something' lines would say + k^42 ?

It would appear in k = 1, in k = 2, in k = 3, k = 6,7,14,21,42.  Exactly the number of divisors of 42.

Maybe you can take it from there.