User talk:Peter Luschny/ComputationAndAsymptoticsOfBernoulliNumbers

On 22:43, June 17, 2012 Charles R Greathouse IV wrote (see history page):

> (fix incorrect formula -- that is not the series for B, which is

> 1/12 * n^-1 - 1/360 * n^-3 + 1/1260 * n^-5 - 1/1680 * n^-7 + O(n^-9))

Charles, you confuses me.

Writing for your expression R(n) Stirling's series gives

ln(n!) ~ n ln(n) - n + ln(2 Pi n)/2 + R(n).

Ok, but there is nothing explicitly related to the factorial here,

neither in the B0 approximation (which is http://dlmf.nist.gov/24.11#E2 )

nor in B1 or B2, which are extensions of B0.

So how can their be an "incorrect formula"?

Now LogB2(n) is an asymptotic approximation to the logarithm of the

absolute value of the Bernoulli number B(n) as are LogB0(n) and LogB1(n).

Thus exp(LogB2(n)) is an asymptotic approximation to the absolute value

of the Bernoulli number B(n).

So no index 2 in the final formula. I will undo your change

until you can convince me. Peter Luschny 14:16, 18 June 2012 (UTC)

As it happens I must leave on a trip soon, so I must be somewhat brief. I will return a week from Friday if you have further questions.

I don't know why you discuss the factorial, which I had not mentioned. I ignore it below.

We are agreed, I take it, that

$\mid {\text{B}}(n)\mid \ ={\sqrt {8\pi n}}\ \left({\frac {n}{2\pi e}}\right)^{n}\exp \left({\frac {1}{12}}\,{n}^{-1}-{\frac {1}{360}}\,{n}^{-3}+kn^{-5}+O\left({n}^{-6}\right)\right)$

so define a function f such that

$\mid {\text{B}}(n)\mid \ ={\sqrt {8\pi n}}\ \left({\frac {n}{2\pi e}}\right)^{n}\exp \left({\frac {1}{12}}\,{n}^{-1}-{\frac {1}{360}}\,{n}^{-3}+f(n)n^{-5}\right).$

Your claim, then, is that $f(n)=2857/3600000+O(1/n)$ while mine is that $f(n)=1/1260+\Theta (1/n^{2}).$

It is simple to verify that f(10^6) = 0.000793650793650198412..., f(10^7) = 0.000793650793650787698..., f(10^10) = 0.000793650793650793650..., f(10^15) = 0.000793650793650793650....

• Now if my asymptotic is correct, the errors are 6 * 10^-16, 6 * 10^18, 6 * 10^-24, 6 * 10^-34: very small, and gaining two digits of precision each time I multiply the argument by 10, just as you'd expect for a function with the next therm $\Theta (1/n^{2}).$
• The error terms in your case are -3.96825390873015873... * 10^-8, -3.96825396765873015... * 10^-8, -3.96825396765873015... * 10^-8, and -3.96825396765873015... * 10^-8: basically constant (all displayed digits identical in the last three). Just what you'd expect if your $n^{-5}$ term was off by about -3.96825396765873015... * 10^-8, which happens to be very close to 2857/3600000 - 1/1260.

So it's clear numerically that my formula is correct for B and your is incorrect. I have no time for an analytic proof at the moment, but I hope this will convince you.

As to the issue of the subscript: I assumed that your formula, though wrong for B, was right for exp(LogB2(n)). Perhaps not: it might be wrong for both. I have not checked that your series works for that approximation, neither numerically nor analytically.

Charles R Greathouse IV 22:03, 21 June 2012 (UTC)