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Topic: Floor Summation (Read 8546 times) 

ThudnBlunder
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Floor Summation
« on: Nov 9^{th}, 2009, 6:00am » 
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Evaluate 81ntanh/10^{n} ^{n=1}


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Obob
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Re: Floor Summation
« Reply #1 on: Nov 9^{th}, 2009, 2:18pm » 
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It's roughly 1  2.413 * 10^{264}. Are you looking for an actual precise answer, or is the point just that tanh pi is close to 1?

« Last Edit: Nov 9^{th}, 2009, 2:50pm by Obob » 
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ThudnBlunder
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Re: Floor Summation
« Reply #2 on: Nov 11^{th}, 2009, 3:25am » 
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on Nov 9^{th}, 2009, 2:18pm, Obob wrote:It's roughly 1  2.413 * 10^{264}. Are you looking for an actual precise answer, or is the point just that tanh pi is close to 1? 
 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?


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Obob
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Re: Floor Summation
« Reply #3 on: Nov 11^{th}, 2009, 6:10am » 
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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.


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