8
$\begingroup$

This question is related to Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$? by Tito Piezas III.

Andrew Fraker (2014) found an almost-integer which is equivalent to the following approximation $$e^\pi\approx\frac{5^4}{3^3}$$ https://en.wikipedia.org/wiki/Mathematical_coincidence#Containing_both_.CF.80_and_e

However, $e^\pi<\frac{5^4}{3^3}$, so it cannot be obtained by truncating of the following series of rational terms, because the sequence of partial sums is monotonically increasing.

$$ e^\pi=e^{6asin\left(\frac{1}{2}\right)}=\sum_{k=0}^\infty\frac{3\left(e^{3\pi}-\left(-1\right)^ke^{-3\pi}\right)\Gamma\left(\frac{k}{2}+3i\right)\Gamma\left(\frac{k}{2}-3i\right)}{2 \pi k!} $$

This formula is built from the integer sequence http://oeis.org/A166748 with contributions by Povolotsky, Hasler, Mathar and Kotesovec.

A similar series with slower convergence is obtained from Kotesovec closed form for http://oeis.org/A166741

$$ e^\pi=e^{2asin\left(1\right)}=\sum_{k=0}^\infty\frac{2^{k-1}\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{\pi k!} $$

Using $\frac{1}{2}$ instead of $1$ as the argument for $asin$, the third root of $e^\pi$ is obtained.

$$ e^\frac{\pi}{3}=e^{2asin\left(\frac{1}{2}\right)}=\sum_{k=0}^\infty\frac{\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{2\pi k!} $$

For the sixth root of $e^{\pi}$ the argument of the gamma function becomes a single fraction.

$$e^{\frac{\pi}{6}}=e^{asin\left(\frac{1}{2}\right)}=\sum_{k=0}^{\infty}\frac{\left(e^{\frac{\pi}{2}} - (-1)^k e^{-\frac{\pi}{2}}\right)\Gamma\left(\frac{k+i}{2}\right)\Gamma\left(\frac{k-i}{2}\right)}{4\pi k!}$$

Q1: How can Fraker's approximation be derived by truncating a series for a fractional power of $e^\pi$?

Q2: How can integrals related to these series be obtained?

$\endgroup$
  • $\begingroup$ Imho, the first reason (and of course I don't mean the only reason) to have a series and truncate it is to find an approximation to some number. And so these series for numbers like $e^\pi$ which require that one use $e^\pi$ while still summing lose that utility. $\endgroup$ – alex.jordan Jan 21 '16 at 22:23
  • $\begingroup$ In fact one does not require to use $e^\pi$ nor $\pi$. It should be noted that $e^\pi$ and $\pi$ do appear in this closed form, but when the fractions are computed recursively, only integers are used. Please follow the links to the OEIS for the recursive definitions of the numerator divided by $\pi$. $\endgroup$ – Jaume Oliver Lafont Jan 21 '16 at 22:29
  • $\begingroup$ @alex.jordan In this application, the series is actually truncated to get a rational approximation (although this may not be obvious either...) math.stackexchange.com/a/1599014/134791 $\endgroup$ – Jaume Oliver Lafont Jan 21 '16 at 22:47
  • 1
    $\begingroup$ OK, I did not see that the output from $\Gamma$ "rationalizes" the factors with $e^\pi$ and $\pi$. $\endgroup$ – alex.jordan Jan 21 '16 at 23:57
  • $\begingroup$ That's pretty awesome how we get rational series with this kind of general term, especially for such a cool number as $e^\pi$. I wonder if this series can be used for an alternative proof of irrationality of this number, not relying on Schneider-Gelfond $\endgroup$ – Yuriy S Jul 2 '19 at 6:20
1
$\begingroup$

I'll try to obtain a more clear form of the series first, then maybe something about the integral.

Let's work with the series:

$$e^\pi=\sum_{k=0}^\infty\frac{2^{k-1}\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{\pi k!}$$

First we explicitly separate even and odd terms for clarity:

$$e^\pi=\sum_{n=0}^\infty\frac{2^{2n} \sinh (\pi) \Gamma\left(n+i\right)\Gamma\left(n-i\right)}{\pi (2n)!}+ \\ +\sum_{n=0}^\infty\frac{2^{2n+1} \cosh (\pi) \Gamma\left(n+\frac{1}{2}+i\right)\Gamma\left(n+\frac{1}{2}-i\right)}{\pi (2n+1)!}=S_1+S_2$$

Using information from Wikipedia, we can write for $n \geq 1$:

$$\Gamma\left(n+i\right)\Gamma\left(n-i\right)=\frac{\pi}{\sinh \pi} (i)_n (-i)_n=\frac{\pi}{\sinh \pi} \prod_{j=0}^{n-1} \left(j^2+1 \right)$$

$$\Gamma\left(n+\frac{1}{2}+i\right)\Gamma\left(n+\frac{1}{2}-i\right)=\frac{\pi}{\cosh \pi} \left(\frac{1}{2}+i \right)_n \left(\frac{1}{2}-i \right)_n= \\ =\frac{\pi}{\cosh \pi} \prod_{j=0}^{n-1} \left(\left(j+\frac{1}{2} \right)^2+1 \right)$$

Now we obtained explicitly rational series:

$$S_1=1+\sum_{n=1}^\infty\frac{2^{2n} }{(2n)!} \prod_{j=0}^{n-1} \left(j^2+1 \right)$$

$$S_2=2+\sum_{n=1}^\infty\frac{2^{2n+1} }{(2n+1)!} \prod_{j=0}^{n-1} \left(\left(j+\frac{1}{2} \right)^2+1 \right)$$

Explicit form of these series is (easy to obtain by writing the ratio of adjacent terms):

$$S_1={_2 F_1} \left(i,-i; \frac{1}{2};1 \right)$$

$$S_2=2~{_2 F_1} \left(\frac{1}{2}+i,\frac{1}{2}-i; \frac{3}{2};1 \right)$$

Numerical check:

Hypergeometric2F1[I,-I,1/2,1]+2 Hypergeometric2F1[1/2+I,1/2-I,3/2,1]-Exp[Pi]

Wolfram Alpha gives $0$.

We can use integral representations of the Hypergeometric functions to obtain the desired integral form, though it would contain complex exponentials.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I used the word "explicit" three times, that requires some talent $\endgroup$ – Yuriy S Jul 2 '19 at 10:16

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.