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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




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 doubleprecision 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  1E54, > using 2000bit (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 1E200 to 1E20), and this became obvious. > Here they are:http://dmishin.blogspot.com/2009/01/36127conjecturefailed.html > > This graph helped me to find better counterexample: > > x = 36*Pi/127  1.998e29 > > This gives f(x)  f(36*Pi/127) \approx 3.06E32. 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 xvalue 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 1.998E29. 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).



