16
$\begingroup$

This is a direct followup to the post Possible new series for $\pi$ by Timothy Chow and its numerous answers and comments.

Using another formula in the same string theory paper by Saha and Sinha one can obtain analogous but much simpler formulas for $\pi$. For instance: define $$S(z)=\sum_{k\ge1}\dfrac{1}{k!}\left(1+\dfrac{1}{4k}\right)_{k-1}\dfrac{1}{k+z/2}\;.$$ Then $$S(z)+S(1/z)-2S(1)=\pi-\Gamma(z/2)\Gamma(1/(2z))/\Gamma(z/2+1/(2z)+1)$$ For instance $$S(-1)-S(1)=\pi/2,\ \ \ \ S(0)-2S(1)=\pi-4$$ Several questions:

  1. Can H.~Rosengren's proof be generalized to this case ?
  2. I am unable to check the numerical validity of these formulas to more than 5 or 6 decimals. Is there a way to accelerate the convergence ?
  3. Since $S(1)$ seems to be fundamental here, is there a "closed formula" in a suitable sense apart from the definition ?

EDIT: Thanks to Jorge Zuniga, Q2 is answered, and for those who want to play with number recognition, $$S(1)=1.28604460098720412334031787206457048830225866881745983306306...$$

Improve this question
$\endgroup$
2
  • $\begingroup$ Side observation: we have \begin{align}1-\frac\pi4&=\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k\frac{2k-1}{2k+1}\\1-\frac\pi2\operatorname{csch}\frac\pi2&=\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k\frac{(2k)^2-1}{(2k)^2+1}\end{align} I wonder if there is a similar closed form for $$?=\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k\frac{(2k)^3-1}{(2k)^3+1}$$ $\endgroup$
    – TheSimpliFire
    Commented Jul 27 at 23:13
  • $\begingroup$ $$1-\frac\pi{12}-\frac{\left|\Gamma(-e^{i\pi/3}/2)\right|^2}{6\sqrt\pi}=\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k\frac{(2k)^3-1}{(2k)^3+1}$$ It is possible to extend this to any integer exponent using your functional equation repeatedly, matching conjugate roots of unity. $\endgroup$
    – TheSimpliFire
    Commented Jul 29 at 11:27

2 Answers 2

Reset to default
12
$\begingroup$

This could be a simple comment, but it answers the second question. The variant $\Delta$-Euler-MacLaurin is a summation method by Henri Cohen, "Numerical Recipes for Number Theory", Université de Bordeaux (2010) Section 2.7 pp.110-115. It accelerates the convergence of this sum pretty fine.

Pi sum high precision computing

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Thanks! Quite surprising, I tested newer implementations of this which did not work at all, while the 2010 one does. I will check what is wrong with the newer ones. Can you tell me from where you obtained the sumnumdelta script (and others), it took me some time to find in my archives, and also what is the "Numerical recipes for number theory" that you mention since it is supposed to be superseded by my book with Karim Belabas ? $\endgroup$
    – Henri Cohen
    Commented Jul 22 at 8:51
  • 1
    $\begingroup$ @Henri_Cohen. I do not remember where I downloaded this pdf containing your scripts several years ago (2012-2014). I guess it comes from your old Home Page or some Site from the Université de Bordeaux. Convergence Acceleration is a topic I was working in deep in those years. I can send you the document if you want. For me sumnumdelta is the best general convergence acceleration method for monótone series. $\endgroup$
    – Jorge Zuniga
    Commented Jul 22 at 13:23
  • $\begingroup$ Sorry for the noise. Found everything you mention, thanks. $\endgroup$
    – Henri Cohen
    Commented Jul 25 at 21:34
2
$\begingroup$

It is unlikely $S(1)$ has a closed form.

We have \begin{align}S(1)&=4\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k\frac{2k}{2k+1}\end{align} and (substituting $\lambda=0$ in the original Saha-Sinha formula for $\pi$) \begin{align}4-\pi&=4\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k\frac{2k-1}{2k+1}\end{align} so $$S(1)=2T+2-\frac\pi2$$ where \begin{align}T=\sum_{k\ge1}(-1)^k\binom{-1/(4k)}k.\end{align} I initially thought that $T=3/7$ (see here: Is it true that $\sum\limits_{n\ge1}\binom{n+\frac1{4n}-1}n=\frac37$?) but it isn't the case; otherwise, $S(1)$ would equal $1.28635\cdots$ rather than $1.28604\cdots$

Note that $T$ can also be written as $$\frac1{2\pi i}\sum_{k\ge1}\int_Cz^{1-k}(1-z)^{-1/(4k)}\,dz$$ over the unit circle and I don't think this is tractable either.

Improve this answer
$\endgroup$
2

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .