3
$\begingroup$

I'm trying to reproduce the results from a certain infamous paper that has been moving around the web for the last few days. The details are irrelevant.

This paper claims to have a closed-form expression for the so-called fine-structure constant $Ж:=\alpha^{-1}=137.035\dots$, to wit, $$ Ж=\lim_{\substack{n\to\infty\\j\to\infty}}2^{-2n}B^n_{k(j)}\tag{8.11} $$ where $k(0):=0$ and $k(j+1):=2^{k(j)}$, and $B^n_k$ are the Bernoulli polynomials.

A very naïve implementation of this formula is as follows:

Clear[k]
k[0] = 0;
k[j_?NumericQ] := 2^k[j - 1];
Table[N[2^(-2 n) BernoulliB[n, k[n]], 50], {n, 1, 10}]

which diverges (and actually overflows), suggesting that the limit does not exist. Is this correct, or is the apparent divergence just an issue of numerical instability? If so, what is the correct value for the limit above? Is $Ж=137.035\dots$ correct?

The paper offers alternative formulas, such as $$ Ж=\frac{\pi}{\gamma}Ч\tag{1.1} $$ where $\gamma$ is the Euler constant, and $$ Ч:=\frac12\lim_{n\to\infty}\sum_{j=1}^n2^{-j}\left(1-\int_{1/j}^j\log_2(x)\mathrm dx\right)\tag{7.1} $$

Unlike before, this alternative formula seems to converge, but not to the claimed value. Indeed,

π/(2 EulerGamma) NSum[2^-j (1 - Integrate[Log[2, x], {x, 1/j, j}]), {j, 1, ∞}]

evaluates to 0.16026, nowhere near $137$. Again, is this numerical calculation reliable, or is the disagreement due to numerical error? If so, what is the correct value of this sum?

$\endgroup$
  • 3
    $\begingroup$ I am often very wary of these speculative closed form proposals. That being said, did you check if the definition of Bernoulli polynomials that person was using is the same as Mathematica's definition? $\endgroup$ – J. M. will be back soon Sep 25 '18 at 14:39
  • $\begingroup$ @J.M.issomewhatokay. Good point. The author does not explicitly define these polynomials, but he writes down a couple of formulas for them. I'll try and check if these formulas are satisfied by MMA's BernoulliB. Thanks! $\endgroup$ – AccidentalFourierTransform Sep 25 '18 at 14:46
  • 2
    $\begingroup$ The formula for Ч can be evaluated in closed form: $$\frac1{2\log 2}\left(\text{Li}_{-1}^{(1,0)}\left(\frac{1}{2}\right)+\text{Li}_1^{(1,0)}\left(\frac{1}{2}\right)+2\right)$$, or in Mathematica format: (2 + Derivative[1, 0][PolyLog][-1, 1/2] + Derivative[1, 0][PolyLog][1, 1/2])/(2 Log[2]) $\endgroup$ – J. M. will be back soon Sep 25 '18 at 14:47
  • 2
    $\begingroup$ That would be NorlundB[] in Mathematica, if that's the definition he refers to. I haven't worked on those in a while, but I remember that the asymptotics for those will not let the proposed limit converge. $\endgroup$ – J. M. will be back soon Sep 25 '18 at 15:03
  • 1
    $\begingroup$ There are going to be many comments, and this needs a chatroom. $\endgroup$ – Szabolcs Sep 25 '18 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.