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Topic:
Series Questions
Replies:
20
Last Post:
Apr 21, 1996 11:07 AM




Re: Series Questions
Posted:
Apr 11, 1996 5:15 PM


>sin(n)^n note that it is true that for almost all c sin(c+n)^n does >converge to 0 even though for the special case c = 0 there is no >convergence. This fact is not obvious but follows from probability >theory.
Can you post or email me a quick sketch of the reasoning for this? Naively, I tend to believe the opposite is true  that is diverges for almost all c  though I could easily believe I'm making an error. Here's my reasoning  though this is not a proof:
(I'm going to put my summation bounds BELOW the word "sum"  but just recognize that all bounds of integration are from 0 to 2pi):
 1) For any fixed N:
int sum (sin(c+n)^n) dc = sum int (sin(c+n)^n) dc n=0..N n=0..N
= sum int (sin theta)^n dtheta n=0..N 
2) int (sin theta)^n dtheta = 0 if n is odd but if n is even, this is 2pi*( product of odd numbers less than n)/product of even nos less than n)
 if my memory serves me, this should be close to 2*sqrt(pi/n) for even n  see material on "Wallis's product" in a good calculus book, like Spivak's  putting 1) and 2) together, we have
int sum (sin(c+n)^n) dc is close to sum 2sqrt(pi/n) n=0..N n = 2,4,6...N
which is itself close to 2*sqrt(pi*N)
so the AVERAGE value over c of the sum for the first N terms is... sqrt(N/pi)
Now it COULD be that the sum converges almost everywhere  perhaps it just doesn't converge to an integrable function  so this doesn't really prove anything  but this, along with the mixing that's going on (really, why should sin(n)^(n) be so special compared with sin(n+c)^n?)
another thing that makes me suspicious: look at the measure of the set (sin(theta))^n > 1/n for theta in 0 to 2pi (and n even, if it matters) i.e. sin(theta) > (1/n)^(1/n)
it sure seams to me like the measure of these sets for large n is more than 1/n  that plus the mixing makes me think that the sum ought to diverge almost everywhere.
I realize none of this is written very completely or even very carefully  but could you show me your argument why the sum should converge for almost all c?
Ted Alper alper@epgy.stanford.edu



