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With the help of wolfram alpha we got very long closed form for $\pi$ in terms of algebraic numbers, logarithms of algebraic numbers and $cot^{-1},coth^{-1}$ which can be expressed as logarithms.

From mathworld

$$\pi=\sum_{k=0}^\infty \frac{1}{16^k}\left( \frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6} \right) \qquad (1)$$

Rearranging (1) we get:

$$ \pi= \sum_{n=0}^\infty(1/16^n*(4/(8*n+1)-2/(8*n+4)-1/(8*n+5)) )+\sum_{n=0}^\infty(-1/16^n*(1/(8*n+6)))\qquad (2)$$

Plugging (2) in Wolfram alpha gives closed form in terms of logarithms and algebraic numbers. The form is about one screen long:

-2 coth^(-1)(4) + 1/5 ((5 (-1)^(3/4) (-log(2)/2 - (i π)/4))/sqrt(2) - (5 (-1)^(1/4) (-log(2)/2 + (i π)/4))/sqrt(2) + (5 log(1 - 1/sqrt(2)))/sqrt(2) - (5 log(1 + 1/sqrt(2)))/sqrt(2) - (5 i (1/2 log(3/2) - i cot^(-1)(sqrt(2))))/sqrt(2) + (5 i (1/2 log(3/2) + i cot^(-1)(sqrt(2))))/sqrt(2) + (5 (-1)^(1/4) (1/2 log(5/2) - i cot^(-1)(3)))/sqrt(2) - (5 (-1)^(3/4) (1/2 log(5/2) + i cot^(-1)(3)))/sqrt(2)) + 4 (((-1)^(3/4) (-log(2)/2 - (i π)/4))/(4 sqrt(2)) - ((-1)^(1/4) (-log(2)/2 + (i π)/4))/(4 sqrt(2)) - log(1 - 1/sqrt(2))/(4 sqrt(2)) + log(1 + 1/sqrt(2))/(4 sqrt(2)) + (i (1/2 log(3/2) - i cot^(-1)(sqrt(2))))/(4 sqrt(2)) - (i (1/2 log(3/2) + i cot^(-1)(sqrt(2))))/(4 sqrt(2)) + ((-1)^(1/4) (1/2 log(5/2) - i cot^(-1)(3)))/(4 sqrt(2)) - ((-1)^(3/4) (1/2 log(5/2) + i cot^(-1)(3)))/(4 sqrt(2))) + 1/6 (12 cot^(-1)(2) - 12 coth^(-1)(2))

Q1 Is this closed form for $\pi$ trivial?

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    $\begingroup$ In your closed form there is $\pi$ itself appearing. $\endgroup$
    – Francesco Polizzi
    Commented May 16, 2022 at 5:31
  • $\begingroup$ @FrancescoPolizzi Indeed, thanks. Looks like we can solve for $\pi$ if it is only in linear terms. $\endgroup$
    – joro
    Commented May 16, 2022 at 5:44
  • $\begingroup$ Even if $\pi$ is in the LHS, this might give algebraic dependency of $\pi$ and the logarithms. $\endgroup$
    – joro
    Commented May 16, 2022 at 5:55
  • $\begingroup$ $i\pi$ is a logarithm of -1 itself, and I bet there is no essentially different algebraic dependence of $\pi$ and the logarithms $\endgroup$
    – Fedor Petrov
    Commented May 16, 2022 at 6:01
  • $\begingroup$ @FedorPetrov there are definitely other algebraic dependencies of pi and logarithms, but the logarithms are linearly dependent over the rationals. $\endgroup$
    – joro
    Commented May 16, 2022 at 6:34

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Here is what Maple says about it... image

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  • $\begingroup$ Thanks. Wouldn't sufficiently good CAS solve $\pi=RHS$ by computing if the RHS is $\pi$? Is $\pi=4 \arctan(1)$ trivial too? $\endgroup$
    – joro
    Commented May 18, 2022 at 14:58

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