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Topic: Sum over rationals (Read 1827 times) |
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Deedlit
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Sum over rationals
« on: Apr 11th, 2005, 10:51pm » |
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Here's a cute little summation I came up with.
Write each nonzero rational number r as a fraction pr/qr in lowest terms. Find the sum of 1 / (pr2 qr2) over all rationals r.
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| « Last Edit: Apr 12th, 2005, 8:38am by Deedlit » |
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Eigenray
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Re: Sum over rationals
« Reply #1 on: Apr 12th, 2005, 6:34am » |
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| hidden: | If q is prime,
[sum](a,q)=1 1/(q2+a2)
=[sum]a=1oo 1/(q2+a2) - [sum]k=1oo1/(q2+(kq)2)
>= 1/(q+1) - C/q2,
so just those terms make it diverge. |
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Deedlit
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Re: Sum over rationals
« Reply #2 on: Apr 12th, 2005, 8:16am » |
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Blech, I meant to multiply p and q, not add them!  I'll correct the OP.
Incidentally, how did you get 1/(q+1) ?
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Eigenray
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Re: Sum over rationals
« Reply #3 on: Apr 12th, 2005, 3:58pm » |
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on Apr 12th, 2005, 8:16am, Deedlit wrote:| Blech, I meant to multiply p and q, not add them! I'll correct the OP. |
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If you fix q, then summing f(p) over {(p,q)=1, p>0} is the same as summing mu(d)f(kd) over {d|q, k>0} (inclusion-exclusion). Thus:
[sum]q[sum](p,q)=1 1/(pq)2
= 2[sum]q[sum]d|q[sum]k>0 mu(d)1/(kdq)2
= 2 Zeta(2) [sum]q[sum]d|q mu(d)/(dq)2
= 2 Zeta(2)[sum]k,d, q=kd mu(d)/(d4k2)
= 2 Zeta(2)2 / Zeta(4)
= 5,
since [sum]k>0 mu(k)/ks = 1/Zeta(s), and Zeta(2)=[pi]2/6, Zeta(4)=[pi]4/90. Neat!
Quote:| Incidentally, how did you get 1/(q+1) ? |
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(q2+a2) < (q+a)2, and then integrate. Or you can integrate 1/(q2+x2) directly for 1/q ([pi]/2 - tan-1(1/q)) or something.
Maybe it's interesting to ask about the asymptotics of
[sum](p,q)=1, q<n 1/(p2+q2)?
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Deedlit
Senior Riddler
Posts: 476
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Re: Sum over rationals
« Reply #4 on: Apr 12th, 2005, 7:07pm » |
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Right. I like this problem a lot, but the reliance on the relatively unknown Zeta(4) bothered me a little - although I guess if there isn't a simple elementary solution for Zeta(4) (is there?) there can't be one for this problem.
on Apr 12th, 2005, 3:58pm, Eigenray wrote:
Maybe it's interesting to ask about the asymptotics of
[sum](p,q)=1, q<n 1/(p2+q2)?
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Of course, we have an upper bound of
[sum]q [pi]/(2q) ~ [pi]/2 (log n + gamma)
For a lower bound, we observe that
[sum]p=1[infty] 1/(p2+q2)
> [sum]k=1[infty] phi(q)/((kq)2+q2)
= phi(q)/q2 [sum]k=1[infty] 1/(k2+1)
> phi(q)/q2
> 1 / (2q log log q)
for q sufficiently large. Summing over q from 1 to n, we get a lower bound of
(log n + gamma) / (2 log log n)
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Barukh
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Re: Sum over rationals
« Reply #5 on: Apr 13th, 2005, 2:23am » |
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This one seemed very familiar to me. After some search, I have found the thread with almost the same name:
Sum Over the Rationals.
So, it was treated here almost a year ago... Note that the results of the two threads differ by a factor of 2; I believe this is because in another thread only rationals < 1 were considered.
I cannot believe I posted a different solution for this problem. What do you think about it, guys?
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Deedlit
Senior Riddler
Posts: 476
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Re: Sum over rationals
« Reply #6 on: Apr 15th, 2005, 3:30pm » |
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LOL... someone came up wit the same sum.  I guess it's not too surprising... (unless one of my friends submitted it, but it's probably just an independent creation.)
The different answer comes from restricting to positive rationals.
Neat proof, Barukh! 
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Grimbal
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Re: Sum over rationals
« Reply #7 on: Apr 16th, 2005, 4:16pm » |
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Actually, if you multiply
sum(1/p2q2)
by
sum(1/n4)
you get from the terms with gcd=1 all the terms with gcd=n, n=1, 2, ...
1/p2q2 * 1/n4 = 1/(np)2(nq)2
Therefore
sum[gcd(p,q)=1](1/p2q2)
= sum(1/p2q2) / sum(1/n4)
= sum(1/p2) * sum(1/q2) / sum(1/n4)
= (pi2/6) * (pi2/6) / (pi4/90)
= 90/36 = 5/2
[edit: typos]
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| « Last Edit: Apr 18th, 2005, 1:37am by Grimbal » |
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Deedlit
Senior Riddler
Posts: 476
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Re: Sum over rationals
« Reply #8 on: Apr 17th, 2005, 4:59am » |
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Yes, that was the proof I had in mind. 
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