Compute the infinite product
[sin(x) cos(x/2)]^1/2 · [sin(x/2) cos(x/4)]^1/4 · [sin(x/4) cos(x/8)]^1/8 · ... ,
where 0 ≤ x ≤ 2π
I was able to prove it this morning but didn't have time to type it up. Here's something of a proof that the expression given is equivalent to sin(x)/2.
f(x) = [sin(x) cos(x/2)]^1/2 · [sin(x/2) cos(x/4)]^1/4 · [sin(x/4) cos(x/8)]^1/8 · ...,
Apply the double angle identity for sine in every term of f(x) where sine appears.
In the first term sin(x)cos(x/2) = 2sin(x/2)cos²(x/2)
likewise in all the other terms a 2 is introduced, the angle in the sine is halved and the cosine gets squared. If we reshuffle by root we get:
f(x) = cos(x/2) · [2sin(x/2) cos(x/4)]^1/2 · [2sin(x/4) cos(x/8)]^1/4 · [2sin(x/8) cos(x/16)]^1/8 · ... ,
now multiply both sides by sin(x) and divide by all the roots of 2
f(x)*sin(x)/[2^(1/2)·2^(1/4) ·2^(1/8)... ] = sin(x)cos(x/2) · [sin(x/2) cos(x/4)]^1/2 · [sin(x/4) cos(x/8)]^1/4 · ...,
But this expression is exactly equal for f²(x).
So we have
f(x)*sin(x)/[2^(1/2)·2^(1/4) ·2^(1/8)... ] = f²(x)
or
f(x) = sin(x)/[2^(1/2)·2^(1/4) ·2^(1/8)... ]
But that denominator is just equal to 2 (If you square it it doubles) so the final simplified version is
f(x) = sin(x)/2

Posted by Jer
on 20131229 19:10:25 