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        riddles >> putnam exam (pure math) >> Sum over rationals
        
(Message started by: Deedlit on Apr 11th, 2005, 10:51pm)
Title: Sum over rationals
Post by Deedlit on Apr 11th, 2005, 10:51pm
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.
Title: Re: Sum over rationals
Post by Eigenray on Apr 12th, 2005, 6:34am
[hideb]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.[/hideb]
Title: Re: Sum over rationals
Post by Deedlit on Apr 12th, 2005, 8:16am
Blech, I meant to multiply p and q, not add them!   :-[  I'll correct the OP.

Incidentally, how did you get 1/(q+1) ?
Title: Re: Sum over rationals
Post by Eigenray on Apr 12th, 2005, 3:58pm

on 04/12/05 at 08:16:51, Deedlit wrote:
Blech, I meant to multiply p and q, not add them!   :-[  I'll correct the OP.

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) ?

(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)?
Title: Re: Sum over rationals
Post by Deedlit on Apr 12th, 2005, 7:07pm
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 04/12/05 at 15:58:55, Eigenray wrote:
Maybe it's interesting to ask about the asymptotics of
[sum](p,q)=1, q<n 1/(p2+q2)?


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)
Title: Re: Sum over rationals
Post by Barukh on Apr 13th, 2005, 2:23am
This one seemed very familiar to me. After some search, I have found the thread with almost the same name:

Sum Over the Rationals. (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1086625708;start=4)

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?
Title: Re: Sum over rationals
Post by Deedlit on Apr 15th, 2005, 3:30pm
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!  ;D
Title: Re: Sum over rationals
Post by Grimbal on Apr 16th, 2005, 4:16pm
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]
Title: Re: Sum over rationals
Post by Deedlit on Apr 17th, 2005, 4:59am
Yes, that was the proof I had in mind.  8)


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