452 pi

[tommy1729]

Continuum product has lowest period \( 2\pi \), so since 452 is a multiple of 2, yes.

Note that

\( \prod_{n=0}^{x-1} sin(n) + 5/4 = \exp\left(\sum_{n=0}^{x-1} \log(\sin(n) + 5/4)\right) \).

Now we consider the Fourier expansion of \( \log(\sin(x) + 5/4) \), from which we find the continuum sum. The zeroth-order Fourier coefficient is 0 (though I don't have a proof, rather was done via numerical integration -- an explicit anti-derivative requires the poly-logarithm and is horrifically complicated, at least according to Wolfram's integrator), so the continuum sum will be periodic and have period equal to the original function, i.e. \( 2 \pi \) (since there is no constant term in the Fourier series and thus the continuum sum reduces to a simple coefficient transformation which does not alter the period), and thus the continuum product will have the same period (if \( g \) is periodic with period \( P \), then \( f \circ g \) is periodic with the same period.).

thanks.

i wonder why i got the larger 452 , maybe i need to ajust my precision for estimations. ( i was afraid of conj 2pi since some numerical data indicated otherwise so i took a multiple for certainty that did fit )

at least we are working on the same method.

the idea that periodicity is not altered is intresting. ( in general , maybe in particular for tetration and real iterations of arbitrary functions ? )

im still thinking about an algoritm to find the period of a function ( taylor series )

and the continuum sum might be an important step ...

dont know if you ever considered that ...

btw giving the maximum value of the above continuum product in closed form might also be intresting.

i need more time to study all this ...