| The limiting radius is prod(n=2->inf) cos(pi/(2^n)), or
(1) r = cos(pi/4) * cos(pi/8) * cos(pi/16) * ...
From elementary trigonometry,
(2) sin(a) = 2*cos(a/2)*sin(a/2).
Letting a = pi/4 and substituting (2) into (1) (since cos(pi/4) = sin(pi/4)),
(3) r = 2 * cos^2(pi/8) * sin(pi/8) * cos(pi/16) * cos(pi/32) * ...
Let a = pi/8 and do it again,
(4) r = 2^2 * cos^2(pi/8) * cos^2(pi/16) * sin(pi/16) * cos(pi/32) * ...
If you keep doing it successively with a = pi/2^n, you get:
(5) r = lim(n->inf)(2^(n-2) * sin(pi*2^-n)) * prod(n=3->inf)(cos^2(pi/(2^n))
Now, = sin(pi*2^-n) = pi*2^-n + (higher terms), so the limit is equal to pi/4,
and the product in (5) is the square of the original product without the first
term, so
(6) r = pi/4 * (r / cos(pi/4))^2 = pi/2 * r^2 (since cos^2(pi/4) = 1/2).
Solving for r,
(7) r = 2 / pi.
As advertised, an elegant result.
|
| >A generalization of the previous result is:
>
> Prod(n=0 to inf)(cos(x/(2^n))) = sin(2x) / 2x.
Can we generalize further?
Prod(n=0 to inf)(cos(x/(K^n))) = ?,
for K = 3,4,5...
Andrew
|
| re .3
> Can we generalize further?
>
> Prod(n=0 to inf)(cos(x/(K^n))) = ?,
> for K = 3,4,5...
The obvious generalisation of the method of .1 doesn't answer your question but
here are my scratchings anyway.
Let sin Kx = f (x) K sin x
K
where f (x) = P (cos x)
K K
where P is a polynomial of degree K-1
K
Then
oo
+---+
| |
| | f (x/K^n) = (sin Kx)/Kx
| | K
n=0
K=2: f (x) = cos x and this is the result from .2.
2
K=3: f (x) = (4cos�x - 1)/3 but the resulting product doesn't look
3
particularly useful.
K=4 looks even messier but in fact gives the same product as K=2.
Today's trivia: P (1) = 1 for all K
K
Now some easy observations on the question asked in .3 ...
Define g (x) as the required infinite product.
K
For any x in [0,pi/2] and K>=2, from the monotone convergence theorem it can be
seen that g (x) is defined. Considered as a sequence in K, g (x) is
K K
monotonically increasing, strictly except at x=0 and x=pi/2.
sin 2x
g (x) = ------ from .2
2 2x
g (x) . g (x/2) = g (x) Is this solvable for g (x)?
4 4 2 4
We can do similar things for other non-prime K but I don't get anywhere.
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