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

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Ted Alper

Posts: 51
Registered: 12/6/04
Re: Series Questions
Posted: Apr 11, 1996 5:15 PM
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>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

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

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

so the AVERAGE value over c of the sum for the first N terms is...

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

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