# logarithmic trickery

Consider that since $\log(a)\log(b)=\log(a^{\log b})=\log(b^{\log{a}}$, and therefore, somewhat unexpectly: $a^{\log b} = b^{\log a}$
we can use it to construct new identities thusly

Let $A = \prod_{n=1}^{\infty} \sqrt[2]{1+\frac{1}{n^{2}}}$

Let $B = \prod_{m=1}^{\infty} \sqrt[e^{m}]{1+\frac{1}{m^{3}}}$

Therefore, since we know that $A^{\log(B)}=B^{\log(A)}$ we can use that
to explicitly calculate a new identity thusly:

1. Compute the logarithm of $A$, $\log(A)$:
$\log(A) = \frac{1}{2} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}\zeta(2v)}{v}$

2. Compute the logarithm of $B$, $\log(B)$:
$\log(B) = \sum_{s=1}^{\infty} \frac{(-1)^{s+1}\mathrm{Li}_{(3s)}(1/e)}{s}$

3. Combine them:

$\prod_{n=1}^{\infty} \prod_{s=1}^{\infty} \sqrt[s]{\left(1+\frac{1}{n^{2}}\right)^{(-1)^{s+1}\mathrm{Li}_{(3s)}(1/e)}} = \prod_{m=1}^{\infty} \prod_{v=1}^{\infty} \sqrt[ve^{m}]{\left(1+\frac{1}{m^{3}}\right)^{(-1)^{v+1}\zeta(2v)}}$

>>> A = fp.nprod(lambda n: sqrt(1+1/(n*n)), [1,inf])
>>>
>>> A
mpf(‘1.9109509100512501’)
>>> B = fp.nprod(lambda m: (1+1/(m**3))**(1/exp(m)), [1,inf])
>>>
>>> B
mpf(‘1.3140291251423164’)
>>> A**log(B)
mpf(‘1.1934623097049237’)
>>> B**log(A)
mpf(‘1.1934623097049237’)