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Topic: Maximum of a sum(sin(2^n*x) / 2^n , n=0..inf)
Replies: 54   Last Post: Feb 1, 2009 1:19 PM

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 matt271829-news@yahoo.co.uk Posts: 2,136 Registered: 1/25/05
Re: Maximum of a sum(sin(2^n*x) / 2^n , n=0..inf)
Posted: Jan 31, 2009 7:37 PM

On Jan 25, 8:53 pm, Dmitry Shintyakov <shintya...@gmail.com> wrote:
> Everybody, thank you for your attention to my little conjecture.
>
> First of all, I am grateful to Matt and David Bernier for pointing out
> that the conjecture is wrong.
> Actually, it was very hard for me to believe it. I played with this
> problem for a quite long time and the
> perfect double-precision match between x_max and 36*pi/127 completely
> convinced me that the match is exact.
>
> First I thought that there can be some software glitch and tried to
> repeat your calculations for x = 36*Pi/127 - 1E-54,
> using 2000-bit (600 decimal digits) floats (quite an overkill huh).
> And I have found no mistake.
>
> After few minutes of calculation, I have made logarithmic graphs of
> differential quotient at very small
> scales of dx (from 1E-200 to 1E-20), and this became obvious.
> Here they are:http://dmishin.blogspot.com/2009/01/36127-conjecture-failed.html
>
> This graph helped me to find better counter-example:
>
> x = 36*Pi/127 - 1.998e-29
>
> This gives f(x) - f(36*Pi/127) \approx 3.06E-32. It seems to be best
> approximation to x_max for now.

Yes, I belatedly ran my program again at a "higher resolution" and it
came up with the same x-value as I posted elsewhere in this thread a
while ago (and was subsequently extended by Robert Israel), namely

x = 0.8905302010175791857059461558702704369...

which differs from 36*pi/127 by, as you say, about 1.998E-29. My
program is not absolutely guaranteed to find the global maximum, but
based on my results I'm, let's say, about 75% confident that this
really is the best that can be achieved (to the given precision).