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You are not logged in. #2 20071212 03:44:40
Re: James Stirling Formulasweetness! A logarithm is just a misspelled algorithm. #3 20071212 03:56:04
Re: James Stirling Formula
Very much so, if you think an error of is ok.#4 20071212 04:18:01
Re: James Stirling FormulaI just like how it contains both pi and e. A logarithm is just a misspelled algorithm. #5 20071212 04:31:17
Re: James Stirling Formula
Which comes out to a relative error of 0.008%. I'll take that. Wrap it in bacon #6 20071212 04:36:09
Re: James Stirling Formula.. not exactly big! EDIT: Aah post collison Last edited by Daniel123 (20071212 04:37:01) #7 20071212 05:17:37
Re: James Stirling Formuladepends on what your priorities are i guess. A logarithm is just a misspelled algorithm. #8 20071212 05:19:17
Re: James Stirling Formula
So does this actually converge on the factorial value as n goes to infinity? #9 20071212 05:22:16
Re: James Stirling Formulawell both n! and the approximation both diverge to infinity as n goes to infinity The Beginning Of All Things To End. The End Of All Things To Come. #10 20071212 05:35:06
Re: James Stirling FormulaYes. Wrap it in bacon #11 20071212 05:35:32
Re: James Stirling Formulai think he meant, does Last edited by mikau (20071212 05:36:33) A logarithm is just a misspelled algorithm. #12 20071212 06:39:54
Re: James Stirling FormulaI agree that percentage difference is more important than absolute difference. , but this is true: Why did the vector cross the road? It wanted to be normal. #13 20071212 09:01:51
Re: James Stirling Formulaisnt that identical? the only time that that would converge to 1, is if the first converged to 0? The Beginning Of All Things To End. The End Of All Things To Come. #14 20071212 10:18:16
Re: James Stirling FormulaBecause the error may grow, but not as fast as n! grows. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #15 20071212 12:19:35
Re: James Stirling Formulayeah. Note but Last edited by mikau (20071212 12:20:49) A logarithm is just a misspelled algorithm. #16 20071212 13:35:00
Re: James Stirling FormulaWhat additional requirement can we impose so that Luca's statement holds? "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #17 20071213 01:05:42
Re: James Stirling FormulaEquality? That is, instead of just getting arbitrarily close to 1, it actually has to get there.Why did the vector cross the road? It wanted to be normal. #18 20071213 02:53:20
Re: James Stirling FormulaCertainly you can come up with a restriction far less restricting than that. Remember, this restriction can't apply to Mikau's example. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #19 20071213 04:07:28
Re: James Stirling Formulaif the two limits each converge to the same finite number? Last edited by mikau (20071213 04:08:17) A logarithm is just a misspelled algorithm. #20 20071213 04:21:43
Re: James Stirling FormulaBingo. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #21 20071213 04:32:25
Re: James Stirling FormulaIt is also possible for the absolute error to approach 0 while the limits themselves diverge to infinity. As a trivial example, let f(x) = x^2 and g(x) = x^2 + 1/x. Then and but Last edited by TheDude (20071213 07:09:28) Wrap it in bacon #22 20071213 06:55:02
Re: James Stirling Formula
awesome! But are there any other restrictions that would do it? A logarithm is just a misspelled algorithm. 