

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/10^{n} ^{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:
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) != n1 occurs around n = 267 or so, summed the series 81(n1)/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. 

Powered by YaBB 1 Gold  SP 1.4! Forum software copyright © 20002004 Yet another Bulletin Board 