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        riddles >> putnam exam (pure math) >> Floor Summation
        
(Message started by: ThudanBlunder on Nov 9^th, 2009, 6:00am)

Title: Floor Summation
Post by ThudanBlunder on Nov 9^th, 2009, 6:00am

Evaluate
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif
81http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lfloor.gifntanhhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rfloor.gif/10ⁿ
ⁿ⁼¹

Title: Re: Floor Summation
Post by Obob on Nov 9^th, 2009, 2:18pm

It's roughly 1 - 2.413 * 10^-264. Are you looking for an actual precise answer, or is the point just that [hide]tanh pi is close to 1[/hide]?

Title: Re: Floor Summation
Post by ThudanBlunder on Nov 11^th, 2009, 3:25am

on 11/09/09 at 14:18:23, Obob wrote:

It's roughly 1 - 2.413 * 10^-264. Are you looking for an actual precise answer, or is the point just that [hide]tanh pi is close to 1[/hide]?

As I am not expecting an exact answer, perhaps I should have put this elsewhere.
The point is that the answer, a transcendental number, requires at least 239 decimal places before we can discover it does not equal 1. And even more if we want to consider rounding errors.

Do you normally sum series to such precision? :)

Title: Re: Floor Summation
Post by Obob on Nov 11^th, 2009, 6:10am

I just observed that the first time floor(n tanh pi) != n-1 occurs around n = 267 or so, summed the series 81(n-1)/10^n = 1, and then gave as a rough error estimate an approximation of the difference between the two series.

I guess I'm just saying that there is no reason to use tanh pi except for obfuscation; the real result lurking here is that

lim_x->1^- sum (floor(n x))/10^n = 1, and that the convergence occurs very quickly.