Wikipedia:Reference desk/Archives/Mathematics/2011 February 14

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February 14

Heuristic justification for Goldbach's conjecture

In the excellent wikipedia article about Goldbach's conjecture, the section Goldbach's conjecture#Heuristic_justification mentions an asymptotic approximation to the Goldbach partition counting function as being n/[2ln^2(n)]. Though I am by no means an expert with asymptotic expansions, I come up with n/[2ln^2(n/2)]. From a pedestrian point of view, it seems that the denominator in the summand, [ln(m)ln(n-m)] flattens out as n tends to infinity. If we approximate the summation with a definite integral, then for sufficiently large n, the integral dt/[ln(t)ln(n-t)] ~ t/[ln(t)ln(n-t)] + c, again since the denominator is relatively constant. Evaluating this quasi anti-derivative from 3 to n/2, we have n/[2ln^2(n/2)] - 3/[ln3ln(n-3)], with the highest order term being n/[2ln^2(n/2)]. I also got the same result by keeping only the highest order term after differentiating: d[t^a/ln^2(t)]~at^(a-1)/ln^2(t)dt and then 'reverse guessing' the anti-derivative. Let us not loose sight of the fact that this does not affect the way the article reads one way or the other.

None the less, I must be missing something here. Would it be possible for someone to point me in the right direction? I am truly curious about Goldbach and am hungry to learn more about asymptotic expansions. I am new here and please forgive my igornant lack of TeX insertions (I hope to figure that out soon). Thank you in advance for your time and consideration. Mathup (talk) 03:41, 14 February 2011 (UTC)

P.S.: Is it just me, or is the prime addend order not important? That is, are p+q and q+p the same partition? I think they are the same because of the way the partitions are counted in the article. Perhaps the phrase, "the sum of two odd primes" needs a parenthetical caveat such as, "the sum of two odd primes (without regard to order)" in order to clarify the situation for picky literal fools like me. If you agree, I would like to dip my toe into the water of wiki edits and do my first mod.

First note that the difference between ${\frac {n}{2\ln ^{2}n}}$ and ${\frac {n}{2\ln ^{2}(n/2)}}$ is very small - even if the derivation does indeed lead to the latter being more accurate, the author of that piece might have just discarded the extra detail since it has no bearing on the argument in question.

At the resolution in which the two expressions differ, I'm not sure either is very accurate. While the summand is flat when $n\approx \infty ,\ m\approx n/2$, it is not so flat around $m\approx 3$, and to get higher accuracy you need to pay more attention to that region - with both estimating the sum as an integral and estimating that integral.

In general, to estimate $\sum _{k=a}^{b}f(k)$, it's more accurate to use $\int _{a-1/2}^{b+1/2}f(k)\ dk$ than $\int _{a}^{b}f(k)\ dk$.

Note that the term asymptotic expansion usually refers to something distinct than what is discussed here.

Help:Displaying a formula is a useful resource for using LaTeX on Wikipedia.

As far as the conjecture itself is concerned, saying "the sum of two odd primes (without regard to order)" doesn't make much sense. For counting how many partitions there are, it makes sense to clarify how we're counting but doesn't matter much if we do it one way or the other, as long as we're consistent and don't switch definitions mid-argument.

It is recommended to be bold with editing Wikipedia, though in this particular case I don't personally think there's a problem with how the article is currently. -- Meni Rosenfeld (talk) 10:50, 14 February 2011 (UTC)

I've done some numerical investigation and it looks like the correct expression is ${\frac {n}{2\ln ^{2}n-4\ln n-c}}$, where c tends to some number around 0.7 (maybe it's $\ln 2$). -- Meni Rosenfeld (talk) 11:30, 14 February 2011 (UTC)

I have un unrelated problem with the heuristic justification section. It correctly argues: "Since this quantity goes to infinity as n increases, we expect that every large even integer has not just one representation as the sum of two primes, but in fact has very many such representations."

But I don't think it's sufficient justification for the whole conjecture. We would need more such as summing the chance of n being a counter example, for all n above the computational search limit. If this sum is close to 0 (it is but we don't show it), then we have a real argument for the conjecture.

To show the need for this sum argument, suppose the heuristics were different and each n had chance 1/n of having no representation. The search limit is above 10¹⁸ so every new n would only have a tiny chance below 10^-18 and tending to 0. But the harmonic series diverges, and so does the sum of 1/n for all even n above a given limit, so with that hypothetical heuristic we would expect infinitely many numbers with no representation. PrimeHunter (talk) 13:27, 14 February 2011 (UTC)

I agree, this should be mentioned, but I find the fact that the sum is close to 0 is too trivial to justify an explicit demonstration. Keeping the heuristic that primes are random, the number of partitions of n is distributed Poissonly. An expectation of ${\frac {n}{\ln ^{2}n}}\approx n$ means a probability of roughly $e^{-n}$ of no partition, and of course the sum of this from some large number to $\infty$ is very close to 0. -- Meni Rosenfeld (talk) 13:59, 14 February 2011 (UTC)

Great answers. So, I was absolutely right about me missing something here. I thought it was an excellent article to begin with and now, am absolutely certain of it. As I slowly plod through the advanced information you so kindly provided, I truly appreciate your collective indulgence in this matter of my extreme curiosity. I guess my concern about '(without regard to order)' applies to only one sentence, "Nevertheless, if one pursues this heuristic, one might expect the total number of ways to write a large even integer n as the sum of two odd primes (without regard to order) to be roughly..." Everywhere else, "the sum of two odd primes" is quite okay without the caveat. Sorry, I did not present my thoughts more clearly. Thanks guys.Mathup (talk) 00:47, 15 February 2011 (UTC)

Separation of variables problem

Hi. I've been trying to solve a PDE for the last few hours and I've gotten stuck at one point and I'm not sure how to continue approaching the problem. The governing equation is

${\frac {\partial ^{2}F}{\partial x\partial y}}+F-k=0$

If I assume that $F$ is separable, defined between $0\leq \{x,y\}\leq 1$ and that $F=u(x)v(y)$, I can get to

${\frac {\partial u}{\partial x}}{\frac {\partial v}{\partial y}}+uv=k$

but from that point on, I am unable to fully separate the equation into two systems. I know it is too early to try to apply boundary conditions to try to clean up the problem further. Is there something obvious I'm missing? Should I solve the homogeneous system first? Titoxd^{(?!? - cool stuff)} 05:44, 14 February 2011 (UTC)

Alternatively, the problem reduces to the homogeneous system if I assume that $F=u(x)v(y)+k$. Is it mathematically valid to do that? Titoxd^{(?!? - cool stuff)} 05:49, 14 February 2011 (UTC)

It IS a way to find a valid solution, though it may not yield the most general solution, and depending on the boundary conditions, it may not solve your particular problem.--Leon (talk) 12:01, 14 February 2011 (UTC)

It just corresponds to noticing that the constant function F(x,y)=k solves the original equation, so any solution will be k plus a solution to the homogeneous equation. The latter are easier to work with, because the solutions to the homogeneous equation form a vector space.

To proceed from here I'd suggest diagonalizing the differential operator by setting p=x-y, q=x+y. You'll get something formally equivalent to the Klein-Gordon equation whose known solutions you can then exploit. –Henning Makholm (talk) 12:32, 14 February 2011 (UTC)

It suffices to solve the homogeneous problem. First look for solutions of the form $Ce^{i(ax+by)}$. There will be a relation on a and b. Then the general solution can be witten as a (integral) superposition of these plane wave solutions. Sławomir Biały (talk) 13:25, 14 February 2011 (UTC)

• Yep, adding $k$ to the separation of variables statement fits in nicely with the boundary conditions, and the whole thing reduces to a very simple linear problem. Thanks! Titoxd^{(?!? - cool stuff)} 18:54, 14 February 2011 (UTC)