## Another prime mess

Posted by Chris on April 1, 2011 – 1:33 pm

Evaluate Product((p² + 1)/(p² – 1)) where the product is over all the primes, p.

The answer is about 10% less than e.

The primes are: 2,3,5,7,11,13,…

I don’t know how to do it (but I haven’t really tried yet). It may require advanced mathematics (but I really don’t know if that’s the case). It was discovered by one of the most brilliant mathematicians that ever walked this planet.

April 1st, 2011 at 3:29 pm

Hi Chris,

Not sure if I’m reading this right.

All terms would be negative, so the product would alternate between positive and negative values as you go through the primes.

And it looks like it would approach +/- 1.

April 1st, 2011 at 3:57 pm

Hi, Wiz. I believe Chris made a mistake and the product is (p

^{2}+1)/(p^{2}-1)=1+2/(p^{2}-1). Written in this form it is clear that the product is actually finite, but to say anything more than that is, in my opinion, very very difficult. One definitely should use at least the law for the (asymptotic) distribution of the prime numbers and this already is a far too enormous spoon for our mouths…April 1st, 2011 at 4:29 pm

Hi Slavy,

If this is the correct statement of the problem, then the product would tend towards 1, not 10% less then e as Chris states.

April 1st, 2011 at 4:45 pm

Hi. Slavy’s correction is right. I’ve fixed it. I’m far from convinced that the prime number theorem is relevant – especially as the answer is exact (in fact it’s rational) and the PNT is approximate.

The factors tends to 1 as p -> infinty, but the first few are > 1.

p = 2 => 5/3 alone. I’m at my brother’s place for the weekend (using his rubbish computer), but for p = 2-17 I get approx 2.435 using the Windows calculator.

April 1st, 2011 at 5:01 pm

I haven’t thought on the problem, so there is a good chance that I am wrong with my first comment. However, I have serious work for the weekend and I really hope I will not contribute with any solutions here

April 1st, 2011 at 5:39 pm

I’ve spent a few minutes on it now, but courtesy of some rather nice Australian Shiraz, I cannot (LOL) get much further tonight.

As it’ll do no harm, the value we are seeking is 2.5 and Ramanujan is the chap who worked it out. He thrived on infinite products and sums. G. H. Hardy remarked that every number is a personal friend of Ramanujan. GHH was a truly brilliant mathematician, but he said his greatest contribution to mathematics was discovering Ramanujan.

April 1st, 2011 at 10:00 pm

Too much cheap local shiraz led me to think that the product tended to 1. Of course, it’s only the terms that tend to 1. I put the series on an Excel spreadsheet for primes up to 149 and the product at that stage reaches 2.494217 so it certainly looks like it gets to 2.5 for all primes.

Proving it looks like another matter, no matter what I drink.

April 2nd, 2011 at 6:46 am

Gutted. I can’t think of a way to do it.

My excuse for posting it was that I thought I’d look up the most beautiful equations in mathematics. This one caught my eye. Definitely beautiful and definitely beyond my abilities

April 3rd, 2011 at 6:02 am

Eureka. I’ve cracked it

Product[(p² + 1)/(p² – 1)] = Product[(1 + 1/p²)/(1 - 1/p²)]

Multiply each factor, top and bottom, by (1 – 1/p²)

= Product[(1 + 1/p²)(1 - 1/p²)/(1 - 1/p²)(1 - 1/p²)]

courtesy of the difference of two squares identity, we get

= Product[(1 - 1/p

^{4})/(1 - 1/p²)²]= Product]1 – 1/p

^{4})] / Product[1 - 1/p²)]²= ζ(2)² / ζ(4) = (Π²/6)² / Π

^{4}/90) = 90/36 = 5/2ζ(s) is the Riemann Zeta function. The two particular values ζ(2) and ζ(4) were found by Euler. The first is known as the Basel problem. Euler found the ζ(4), ζ(6) (and others? while he was at it) – ninety years before Riemann was born.

At last, I’ve introduced the Riemann Zeta function non-gratuitously

April 3rd, 2011 at 6:59 am

That came to me in a flash whilst drinking my first coffee of the day. I had been really stuck because I knew that ζ(s) = Product[1/(1 - 1/p^s)] = 1 + 1/2

^{s}+ 1/3^{s}+ 1/4^{s}+…, but couldn’t see a really nice way to change the – to a +.I had written out some pretty evil looking expansions (which invloved those + signs sometimes being – signs) and was beginning to think that the signs were the same as for the Mobius function (but I now think that isn’t going to be the case).

Corresponding to Product[1/(1 - 1/p

^{s})] = ζ(s), I now realise thatProduct[1/(1 + 1/p

^{s})] = ζ(2s) / ζ(s). I bet I can find that somewhere on Google now that I know what to search for.For the posted problem we then have (ζ(2)/ζ(4))ζ(2) = 5/2

April 3rd, 2011 at 7:35 am

Hi Chris, this may lead to something, but I think you are still far away from a complete proof. So far you just confirmed my opinion that using the Prime number theorem is inevitable. What you maybe need is the link between the Riemann zeta function and the Chebyshev function, see http://en.wikipedia.org/wiki/Prime_number_theorem which is a highly non-trivial result and is in the core of the proof of the Prime number theorem. Plus I don’t know that if it is enough for you. Anyway, I still stay behind my words that this problem is too complicated

April 3rd, 2011 at 9:37 am

Hi slavy. I’m not going to prove the Basel problem (I do realise that Euler’s proof was dreadful by today’s standards). However, I’m taking

ζ(2) and ζ(4) as “well-known” and that the manipulations are with convergent products. I believe that my solution is good (but not rigorous) and is independent of the PNT. As both 2 and 4 are > 1, we don’t even need Riemann’s analytic continuation of Euler’s result. I think that’s the only loose end. But I’ve only learnt engineering maths, and am quite unable to produce a modern day idea of a rigorous proof of anything.

I certainly don’t see the PNT as being required or helpful. I accept that the non-trivial zeroes of ζ(s) being inside the critical strip i.e. 0 < s < 1 was required to prove the PNT. But I don't see that has anything to do with the posted problem.

Personally I think that Euler's form of the zeta function should be in the top two most beautiful mathematical equations. When I first saw it, about two or three years ago, I was in total awe of it for many hours – I still am occasionally. I don't think Euler's identity hit me in quite the same way – but it is very beautiful.

Thanks for that link. I’ve just taken a quick glance at it. Strangely, I don’t think I’ve seen that particular wiki before.

April 3rd, 2011 at 12:23 pm

Hi again slavy. I’m not sure why you say my solution is incomplete, did you read it too quickly (or is there a real problem with it)?

April 4th, 2011 at 1:05 am

Hi Chris. Yes, I read it too quickly and noticed too much heavy artillery used that I don’t know by heart, so I missed the main point. I really don’t have time to study the solution in detail, but to some extend you convinced me that it may be complete. However, I really don’t like such kind of proofs, where one just states some big results and that is it. Of course, I may be angry at myself here, because I didn’t know/remember your formulas, and this might be the reason to still be so harsh on the topic

April 6th, 2011 at 12:25 pm

is it pie or cake because i dont know it mire be cupcake

April 13th, 2011 at 9:32 am

Prime number is discovered by ariabhata