09/17/2010, 10:52 AM
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.).
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.).