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
(In reply to
computer solution by Charlie)
I
have tried to solve it analytically and therefore compiled a table similar to
yours,
but
did not find a formula as requested.
After seeing
your surmise, based on the fact that the infinite
product column and the "sin(x)/2 column" match it is relatively easy to show the correctness
of your conclusion.<o:p></o:p>
If
f(x)=(sin(x))/2 then f(x/2)= (sin(x/2))/2
Squaring
the infinite expression we get :
f2(x)= sin(x)*cos(x/2)*(sin(x/2))/2 which can be rewritten
as
=1/4*( sin(x)*2*cos(x/2)*(sin(x/2))==(sin(x))2/4
.So f(x) as
calculated from the LHS equals the
RHS i.e. (sin(x/2))/2
The above as
presented does not constitute deriving an unknown formula, but turning a
conjecture into fact.
If anyone
wants to translate it into a "formal" proof, go ahead.
re: computer solution ---second the motion
