Re: Where do I begin?
- From: Don Coppersmith <dcopper@xxxxxxxxxx>
- Date: Mon, 02 May 2005 08:24:48 EDT
> Don Coppersmith wrote:
> >>>Consider the related series
> >>>
> >>>sum ((-1)^n)/(ln n + cos ( 2 \pi n p/q ) )
> >>>
> >>>where p/q is a fraction in lowest terms.
> >>>If q is odd this converges.
> >>>If q is even it seems to diverge;
> >>>known to be true for q=2, 4, 6, 8, but I haven't
> >>>proven
> >>>it for general even q.
> >>>
> >>>Don Coppersmith
> >>
> >>Don, thank you so much, I think you are right that
> it
> >>has something to do with this.
> >>Can you explain how you know that it converges for
> >>q=2, 4, 6, 8 etc..
> >
> >
> > Let q be even; we will show that it DIverges.
> > Pick some large m, a multiple of q.
> > Consider the terms for n=m+1,m+2,...,m+q.
> > log n differs from log m by O(1/m),
> > which we will be able to ignore.
> > For shorthand set x=log m.
> > In that range, consider
> > sum (-1)^n / ( log n + cos ( 2 \pi n p/q ) ).
> > Incur an error O(1/(m log m)) in each term
> > by replacing log n by x.
> > The sum of these q terms is then
> > sum[k=1..q] (-1)^k /(x+cos(2 \pi k p/q)) +O(q^2/m
> log m).
> > Without the error term,
> > that sum is a nonzero fraction in x;
> > its absolute value is at least
> > 1/x^{1 + q/2} = 1/(log m)^{1 + q/2}.
>
> I chose q=16, p=1, and your x is my M; then:
>
> f(M) := sum[k=0..15] (-1)^k /(M+cos(2 \pi k/16))
> seems to be O(1/M^9),
> (consistent with your lower bound)
>
> according to my numerical experiments, but I have no
> proof.
The sum involves terms c/(x+cos(*)) for nonzero
numerators c and for 1+q/2 different values of cos(*),
so it is a nonzero fraction whose denominator is the
product of these 1+q/2 different (x+cos(*)), so a
polynomial of degree 1+q/2. The numerator is probably
a constant, due to cancellation of the coefficients
of x^i for i>0.
> An idea (to do with the original series in the
> thread)
> would be to look at the rational approximations p/q
> of 1/(2Pi) from the continued fraction expansion,
> with q even such as 1/6.
The trouble is that unless it's an exceedingly close
rational approximation, its error will imply a
gradual phase shift, and the positive bias won't have
a chance to accumulate before the phase shift makes
it a negative bias.
> David Bernier
>
> > This exceeds the error term.
> > It provides consistent bias in one direction;
> > for every q terms of the original series, we get a
> > bias of about 1/(log m)^(1+q/2).
> > The sum of those biases diverges.
> >
> > Do you also see why it CONverges for odd q?
> >
> > Don Coppersmith
.
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- Re: Where do I begin?
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