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Topic: Sum over the rationals (Read 1423 times) |
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Pietro K.C.
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Sum over the rationals
« on: Jun 7th, 2004, 9:28am » |
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I got this one from American Mathematical Monthly, section 'Problems and Solutions'. I think it was the latest issue (as of June 4th).
For x [in] [bbq], let a / b be the representation of x in its lowest terms, i.e.
x = a / b and gcd(a,b) = 1.
Define f : [bbq] [to] [bbz] by f(x) = ab. Show that
[sum] (f(x))-2 = 5/2.
x [in] [bbq]+
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Aryabhatta
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Re: Sum over the rationals
« Reply #1 on: Jun 7th, 2004, 7:03pm » |
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on Jun 7th, 2004, 9:28am, Pietro K.C. wrote:I got this one from American Mathematical Monthly, section 'Problems and Solutions'. I think it was the latest issue (as of June 4th).
For x [in] [bbq], let a / b be the representation of x in its lowest terms, i.e.
x = a / b and gcd(a,b) = 1.
Define f : [bbq] [to] [bbz] by f(x) = ab. Show that
[sum] (f(x))-2 = 5/2.
x [in] [bbq]+ |
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The value of this is [zeta][sup2](2)/[zeta](4) = 90/36 = 5/2
([zeta] is the riemann zeta function)
We use the fact that [sum]d [mu](d) d-n = 1/[zeta](n), [forall] n > 1, where [mu] is the moebius function.
The given sum = [sum]kk-2 [sum] {(n,k) = 1} n-2, k and n range from 1 to [infty]
For a given k, the inner sigma can be written as (using a standard theorem in number theory)
[sum] d|k [mu](d) [sum] m (md)-2 which is equal to
[sum] d|k [mu](d)d-2[sum] m m-2 =
[zeta](2)[sum] d|k [mu](d)d-2
so the total sum =
[zeta](2)([sum]kk-2[sum] d|k [mu](d)d-2)
Now,
[sum]kk-2[sum] d|k [mu](d)d-2 = [sum]k[sum] d|k [mu](d)d-2 k-2 =
[sum]d [sum]m[mu](d)d-2(md)-2 = [sum]d[mu](d)d-4[sum]mm-2 = [sum]d[zeta](2)[mu](d)d-4 = [zeta](2)/[zeta](4)
So the total sum = [zeta][sup2](2)/[zeta](4)
Also, in general, for the f which was defined in the problem,
[sum] (f(x))-t = [zeta][sup2](t)/[zeta](2t) where t > 1.
x [in] [bbq]+
Thanks for posting this, I have just started to learn about the moebius function and this turned out to be a good exercise! 
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| « Last Edit: Jun 7th, 2004, 7:48pm by Aryabhatta » |
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Pietro K.C.
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Re: Sum over the rationals
« Reply #2 on: Jun 8th, 2004, 4:49pm » |
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Well, you sure do know a lot of number theory for someone who's just starting to learn about the Möbius function! 
I don't know how it is abroad, but here in Brazil the Möbius function is introduced in the very first semester on number theory, long before zeta functions or even infinite sums!
I have two follow-ups now:
1. Let S(t) be the sum of my first post, but with the 2 replaced by t in the exponent. Find all integers t such that S(t) is rational.
2. Demonstrate the formula for 1 / [zeta](n) used by Aryabhatta. Generalize. (Intentionally open-ended, though I have a generalization myself.)
Note: I didn't know that formula for 1 / [zeta](n). Pretty!
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| « Last Edit: Jun 8th, 2004, 4:53pm by Pietro K.C. » |
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Barukh
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Re: Sum over the rationals
« Reply #3 on: Jun 9th, 2004, 5:37am » |
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on Jun 8th, 2004, 4:49pm, Pietro K.C. wrote:Well, you sure do know a lot of number theory for someone who's just starting to learn about the Möbius function!
I don't know how it is abroad, but here in Brazil the Möbius function is introduced in the very first semester on number theory, long before zeta functions or even infinite sums! |
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For somebody who hasn’t learned about Möbius function at any semester in the University (like myself), this thread is an excellent sourse of new treasures.
So, I would like to comment on some “well-known” results that were new for me. Let me begin with the definition of the Möbius function [mu]: [bbn] [to] (-1,0,1):
[mu](1) = 1; [mu](d) = (-1)r, if d is square-free and has r distinct prime factors; [mu](d) = 0, if d is not square-free.
on Jun 7th, 2004, 7:03pm, Aryabhatta wrote:For a given k …(using a standard theorem in number theory)
[sum] {(n,k) = 1} n-2 = [sum] d|k [mu](d) [sum] m (md)-2 |
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(Sorry, Aryabhatta, I changed your post). How do we show this? Use Inclusion-Exclusion Principle to find all the integers co-prime with k, that has r distinct prime factors p1, …, pn:
[bbn] = { [mu](1)m }
-p1[bbn] = { [mu](p1) p1m }
…
-pn[bbn] = { [mu](pn) pnm }
+p1p2[bbn] = { [mu](p1p2) p1 p2m }
…
(-1)rp1…pn[bbn] = { [mu](p1… pn) p1… pnm }
[smiley=blacksquare.gif]
I also encountered the following interesting identity involving Möbius function: [sum]d|n [mu](d) = 0 for n > 1. Can you see why?
Excellent stuff, Pietro and Aryabhatta! Will look at the follow-ups.
P.S. I hope I wasn’t too annoying…
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Barukh
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Re: Sum over the rationals
« Reply #4 on: Jun 10th, 2004, 2:28am » |
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on Jun 8th, 2004, 4:49pm, Pietro K.C. wrote:| 1. Let S(t) be the sum of my first post, but with the 2 replaced by t in the exponent. Find all integers t such that S(t) is rational. |
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All even integers satsify - using Aryabhatta’s generalization and the fact that [zeta](t) = [pi]t times rational number for even t.
Can’t say about odd integers, though… 
Quote:| 2. Demonstrate the formula for 1 / [zeta](n) used by Aryabhatta |
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Let M(n) = [sum]k[mu](k)k-n, [zeta](n) = [sum]kk-n. We have: M(n)[zeta](n) = [sum]kk-n[sum]d|k[mu](d). But the last inner sum equals 1 for k = 1, 0 otherwise (see my previous post).
Therefore, M(n)[zeta](n) = 1.
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Aryabhatta
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Re: Sum over the rationals
« Reply #5 on: Jun 11th, 2004, 1:19pm » |
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on Jun 9th, 2004, 5:37am, Barukh wrote:
I also encountered the following interesting identity involving Möbius function: [sum]d|n [mu](d) = 0 for n > 1. Can you see why?
P.S. I hope I wasn’t too annoying…
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No. You are not annoying 
The theorem which was used is this:
Given k,
[sum](n,k) = 1 f(n) = [sum]d|k[mu](d)[sum]m f(md)
The proof of this uses the fact that [sum]d|t [mu](d) = 0 for t > 1 and 1 for t =1. -(*)
Basically we rewrite [sum](n,k) = 1 f(n) as [sum]n f(n)g(n,k) where g(n,k) is 1 if (n,k) = 1 and 0 otherwise . From (*) we see that g(n,k) = [sum]d|(n,k)[mu](d)
I will leave it to you from here.
I knew the identity
[zeta](s)[sum]n[mu](n)n-s = 1 because i read it online somewhere (some course webpage) that this was the motivation for defining [mu].
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Barukh
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Re: Sum over the rationals
« Reply #6 on: Jun 14th, 2004, 6:04am » |
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Here’s another solution of the original problem.
[smiley=blacksquare.gif]
It’s clear that every natural number is a value of f(x), so I write the given sum S = [sum]ng(n)n-2, where g(n) is the number of times n appears in the sum. But g(n) is just a number of ways to write n as a product of two co-prime numbers, so g(n) = 2[omega](n), where [omega](n) is number of distinct prime factors of n; and g(1) = 0.
The next step is not obvious, but once written, it’s not difficult to get convinced:
S = [sum]n2[omega](n)n-2 = [prod] p(1+ 2p-2 + 2p-4 + 2p-6...),
where the product is taken over all prime numbers, and for every p, the sum is infinite. This formula follows from the Fundamental Theorem of Arithmetic (FTA): every natural number n may be written uniquely as a product of powers of [omega](n) prime factors.
Now, we have:
1+ 2p-2 + 2p-4 + 2p-6... = (1 + p-2) (1+ p-2 + p-4 + p-6...) = (1+ p-2)/(1- p-2) = (1- p-4)/(1- p-2)2,
so
S = [prod](1- p-4) / [prod](1- p-2)2 = ([sum]n-2)2 / ([sum]n-4) = [zeta]2(2)/[zeta](4).
This uses the identity (known already to Euler) [sum]n-s = [prod](1- p-s)-1, which is yet another consequence of FTA.
[smiley=blacksquare.gif]
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Aryabhatta
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Re: Sum over the rationals
« Reply #7 on: Jun 17th, 2004, 7:07am » |
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on Jun 14th, 2004, 6:04am, Barukh wrote:Here’s another solution of the original problem.
S = [sum]n2[omega](n)n-2 = [prod] p(1+ 2p-2 + 2p-4 + 2p-6...),
where the product is taken over all prime numbers, and for every p, the sum is infinite. |
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Nice!
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