3
$\begingroup$

Let $\gamma$ be Euler constant and $W$ Lambert W function.

One can show:

$$-2/3\,{\frac {\gamma+\ln \left( \pi \right) }{W \left( -1/3\,{\frac { \left( \gamma+\ln \left( \pi \right) \right) { {\rm e}^{-1/3\,\gamma}}}{\sqrt [3]{\pi }}} \right) }} = 2 \qquad (1) $$

This means at least one of $ \left( -1/3\,{\frac { \left( \gamma+\ln \left( \pi \right) \right) { {\rm e}^{-1/3\,\gamma}}}{\sqrt [3]{\pi }}} \right)$ and $\gamma+\ln \left( \pi \right)$ is transcendental.

Q1 Is this known?

Q2 Can one solve (1) for $\gamma$ or $\pi$? (sage, maple and Wolfram Alpha couldn't).

In machine readable form:

 def eulerpi():
      import mpmath
      from mpmath import lambertw,euler,pi,log,exp
      mpmath.mp.dps=10**4
      a= -2*(euler + log(pi))/lambertw(-1/3*( (euler+log(pi))*exp(-euler/3))/pi**(1/3))/3-2
      print mpmath.chop(a)
$\endgroup$
5
$\begingroup$

By the definition of the Lambert function $W\left(-\frac13\alpha e^{-\alpha/3}\right)=-\frac\alpha3\,$ for any $\alpha$, so $$ -2/3\frac\alpha{W\left(-\frac13\alpha e^{-\alpha/3}\right)}=2. $$ Putting here $\alpha=\gamma+\ln\pi$ gives your formula.

$\endgroup$
3
  • $\begingroup$ I don't get numerical support for your second formula for alpha=4 and alpha=5, in both sage and Maple. Is it a CAS bug? (Though it works for gamma + log(pi)) $\endgroup$ – joro Dec 1 '13 at 15:35
  • $\begingroup$ @joro for some intervals the Lambert function is multiwalued en.wikipedia.org/wiki/Lambert_W_function. May be the program computes another branch. $\endgroup$ – Andrew Dec 1 '13 at 16:16
  • $\begingroup$ Btw, I tried to check if the stupid question is a trivial tautology, but branches of $W$ fooled me. $\endgroup$ – joro Dec 12 '13 at 15:20

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.