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The Wallis Formula computes $\frac{\pi}{2}$ OEIS and the Pippenger Product computes $\frac{e}{2}$ OEIS. Here is a combined formula:

x = 0;  (*Wallis Formula (Pi/2)*)
Defer[
2^(1/2^x) Product[(1 - (2^j + 2 k + 1)^-2)^(2^x)^(-j - 1),
          {j,1, \[Infinity]}, {k, 0, 2^(j - 1) - 1}]]
N[%, 20]
N[Pi/4, 20]
x = 1;  (*Pippenger Product (E/2)*)
Defer[
2^(1/2^x) Product[(1 - (2^j + 2 k + 1)^-2)^(2^x)^(-j - 1),
          {j, 1, \[Infinity]}, {k, 0, 2^(j - 1) - 1}]]
N[%, 20]
N[E/(2 Sqrt[2]), 20]
x = 1; n = 20;  (*Pippenger Product (E/2) nth excludes Sqrt[2]*)
Defer[
Product[(1 - (2^j + 2 k + 1)^-2)^(2^x)^(-j - 1),
{j, n, n}, {k, 0, 2^(j - 1) - 1}]]
N[%, 20]

Edit Expanded formula for top two products:

$$2^{\frac{1}{2^x}} \prod _{j=1}^{\infty } \prod _{k=0}^{2^{j-1}-1} \left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right)^{\left(2^x\right)^{-j-1}}$$

Bounty Edit Below is boiler-plate to show the three symbolic results we need. The numerators are $\pi e,\pi,\text{ and } e.$ We need to modify Product to add some Patterns and Rules to achieve symbolic results. (i.e., to eliminate Solve and NProduct from the functions.) I have no idea how to do this!

Clear[m];
wallispippenger[] := 
Solve[2^(Log[2 Sqrt[2]]/Log[2]) NProduct[(1 - (2^j + 2 k + 1)^-2)^(
  1 + (2^1)^(-j - 1)), {j, 1, \[Infinity]}, {k, 0, 2^(j - 1) - 1},
   WorkingPrecision -> 100] == m/4, {m}]/4
wallispippenger[x_] := 
Solve[2^(1/2^x) NProduct[(1 - (2^j + 2 k + 1)^-2)^(2^x)^(-j - 1), {j,
    1, \[Infinity]}, {k, 0, 2^(j - 1) - 1}, 
  WorkingPrecision -> 100] == m/2, {m}]/2
wallis[] := wallispippenger[0]
pippenger[] := wallispippenger[1]
{wallispippenger[], wallis[], pippenger[]}

Edit The results before the LH power of two is applied: $$\frac{\pi e}{8 \sqrt{2}}, \frac{\pi}{4},\text{ and }\frac{e}{2 \sqrt{2}}.$$
These are mulitplied by: $$2\sqrt{2}, 2,\text{ and }\sqrt{2}.$$

share|improve this question
    
What did you mean by "symbolic output"? You wanted to see the fractions that comprise the terms of the infinite product? –  J. M. Jun 9 '13 at 19:50
    
@0x4A4D, I want $\frac{\pi}{2}$ and $\frac{e}{2}$ for the results of the first two calculations. –  Fred Kline Jun 9 '13 at 19:58
3  
In that case, I'd be shocked if Mathematica knew how to evaluate your double product symbolically... –  J. M. Jun 9 '13 at 20:01
1  
FullSimplify won't do the job (it takes quite long). –  Jacob Akkerboom Jun 9 '13 at 22:01
2  
Do you have any reason to believe that this is possible to evaluate symbolically? My best efforts result in complex summations over Gamma[] and Zeta[]. –  Corey Kelly Jun 12 '13 at 11:28
show 3 more comments

1 Answer

up vote 7 down vote accepted
+500

I'll start by saying that I don't have an answer, but I found this interesting and wanted to share some of the manipulations I noticed. I'm essentially "thinking out loud". Perhaps in recasting the question, somebody else will notice something. The following is a mixture of manipulations made by hand and some simplifications done in Mathematica.

I started by Loging both sides to write this as a summation identity instead, and doing some simplification

$$ f(x) = 2^{\frac{1}{2^x}} \prod _{j=1}^{\infty } \prod _{k=0}^{2^{j-1}-1} \left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right)^{(2^{x})^{-j-1}} $$ $$ \ln( f(x) ) = \frac{1}{2^x} \ln(2)+\sum _{j=1}^{\infty } \sum _{k=0}^{2^{j-1}-1} (2^{x})^{-j-1} \ln\left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right) $$ $$ \ln( f(x) ) = \frac{1}{2^x} \ln(2)+\sum _{j=1}^{\infty } (2^{x})^{-j-1} \sum _{k=0}^{2^{j-1}-1} \ln\left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right) $$

The inner sum can now be evaluated symbolically in Mathematica:

Sum[Log[1 - 1/(2^j + 2 k + 1)^2], {k, 0, 2^(j - 1) - 1}]
Out[] = (1/2)*Log[2*Pi] - Log[Gamma[2^(-1 + j)]] + \
Derivative[1, 0][Zeta][0, 2^j] - \
Derivative[1, 0][Zeta][0, 1 + 2^(-1 + j)] - \
2*Derivative[1, 0][Zeta][0, 1/2 + 2^j] + \
2*Derivative[1, 0][Zeta][0, (1/2)*(1 + 2^j)] +  \
Derivative[1, 0][Zeta][0, 1 + 2^j]

Mathematica won't simplify these derivatives, but we can, using $$ \frac{\partial}{\partial s} \zeta(0,a) = \ln(\Gamma(a))-\frac{1}{2}\ln(2\pi) $$

The factors of $\frac{1}{2}\ln(2\pi)$ all cancel, leaving

$$ \ln\Gamma(2^j)-\ln\Gamma(2^{j-1})+\ln\Gamma(1+2^j)-\ln\Gamma(1+2^{j-1})-2(\ln\Gamma(\frac{1}{2}+2^j)-\ln\Gamma(\frac{1}{2}+2^{j-1})) $$

and now Mathematica can help us out:

FullSimplify[
 Log[Gamma[2^j]/Gamma[2^(j - 1)]] + 
  Log[Gamma[1 + 2^j]/Gamma[1 + 2^(j - 1)]] - 
  2 Log[Gamma[1/2 + 2^j]/Gamma[1/2 + 2^(j - 1)]], {j >= 1}]

Out[] = Log[(2^(-1 + 2^(1 + j))*Gamma[(1/2)*(1 + 2^j)]^4)/(Pi*
Gamma[1/2 + 2^j]^2)]

Which we can rewrite

$$ \log \left(\frac{2^{2^{j+1}-1} \Gamma \left(\frac{1}{2} \left(1+2^j\right)\right)^4}{\pi \Gamma \left(\frac{1}{2}+2^j\right)^2}\right) = \left(2^{j+1}-1\right) \log (2)+2 \log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma \left(\frac{1}{2}+2^j\right)}\right)-\log (\pi ) $$

Our summation is now:

$$ \ln( f(x) ) = \frac{1}{2^x} \ln(2)+\sum _{j=1}^{\infty } (2^{x})^{-j-1} \left( \left(2^{j+1}-1\right) \log (2)+2 \log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma \left(\frac{1}{2}+2^j\right)}\right)-\log (\pi ) \right) $$

and the simpler sums can be evaluated and simplified in Mathematica:

$$ \sum _{j=1}^{\infty } \left(2^{j+1}-1\right) \log (2) \left(2^x\right)^{-j-1} = \frac{2^{-x} \left(3\ 2^x-2\right) \log (2)}{-3\ 2^x+4^x+2} $$

$$ \sum _{j=1}^{\infty } \log (\pi ) \left(2^x\right)^{-j-1} = \frac{\log (\pi )}{4^x-2^x} $$

leaving

$$ \ln( f(x) ) = \frac{1}{2^x} \ln(2) + \frac{2^{-x} \left(3\ 2^x-2\right) \log (2)}{-3\ 2^x+4^x+2} - \frac{\log (\pi )}{4^x-2^x} + 2 \sum _{j=1}^{\infty } (2^{x})^{-j-1} \log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma \left(\frac{1}{2}+2^j\right)} \right) $$

or:

$$ \ln( f(x) ) = \frac{\left(2^{1-x}-1\right) \log (\pi )+2^x \log (2)}{-3\ 2^x+4^x+2} + 2 \sum _{j=1}^{\infty } (2^{x})^{-j-1} \log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma \left(\frac{1}{2}+2^j\right)} \right) $$

At this point, I started playing around with manipulations of the $\Gamma$ functions, and had some success using identities for the half-integer values to rewrite in terms of factorials, but I noticed that my math must have gone awry somewhere, as the evaluated term has discontinuities at $x=0$ and $x=1$ and the summation seems to also converge at $x=0$. I think this is coming close to a closed form of $f(x)$ but something is slightly off.

EDIT:

A bit of further experimentation revealed that the equation

$$ \log \left(\frac{2^{2^{j+1}-1} \Gamma \left(\frac{1}{2} \left(1+2^j\right)\right)^4}{\pi \Gamma \left(\frac{1}{2}+2^j\right)^2}\right) $$

is where my trouble began. This form can be evaluated numerically as follows:

f[x_] := Exp[Log[2]/2^x + NSum[(2^x)^(-j - 1) Log[(2^(2^(j + 1) - 1) \
Gamma[1/2 + 2^(j - 1)]^4)/(Pi Gamma[1/2 + 2^j]^2)], {j, 1, Infinity}]]

{f[0],f[1]}
(* {1.5708, 1.35914} *)

but if you try to split the terms up, discontinuities start to creep in. Even just pulling out the $\ln(\pi)$ term "breaks" the evaluation at $x=0$:

NSum[(2^0)^(-j - 
 1) Log[(2^(2^(j + 1) - 1) Gamma[1/2 + 2^(j - 1)]^4)/(Gamma[
   1/2 + 2^j]^2)], {j, 1, Infinity}]

(*ComplexInfinity*)

If a closed form exists, it will likely involve the limiting behaviour of several discontinuous functions.

share|improve this answer
    
+1 for the great analysis. I'm reviving hamsters as I write this. The link to Pippenger in the OP has something similar for the $n$th steps. –  Fred Kline Jun 13 '13 at 20:00
    
Now if only I could figure out what I've done wrong. I think it's close to an analytical result. Might be that my substitutions for the Zeta[] derivatives are mistaken. –  Corey Kelly Jun 14 '13 at 10:57
1  
I was hoping to put a bit more effort into this before the bounty deadline expired, but I'm honestly not sure what else can be done. I think this can't be simplified to a problem that Mathematica is able to solve symbolically because of the discontinuities. My best guess would be to rewrite in terms of a contour integral, but my residue calculus is much too rusty to tackle a problem this complex. –  Corey Kelly Jun 17 '13 at 15:07
1  
The observation about the discontinuities is the answer. –  Fred Kline Jun 17 '13 at 21:19
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