Tassus Posted August 3, 2012 Posted August 3, 2012 Can anyone help me to find this limit? See the attached picture. Thank you.
P. G. Luan Posted August 3, 2012 Posted August 3, 2012 I guess you can just apply the Stirling formula listed below to get the desired result. [Remember that x!=Gamma(x+1)]
Tassus Posted August 3, 2012 Author Posted August 3, 2012 So, the application of this approximation to the Gamma function is independent of the number set that the argument of Gamma belongs to? i.e. wether the argument is natural or real number etc Thank you!!! I guess you can just apply the Stirling formula listed below to get the desired result. [Remember that x!=Gamma(x+1)]
Severian Posted August 3, 2012 Posted August 3, 2012 (edited) I don't think it has a wel defined limit. For large v, it will grow like v! (that is a factorial, not an exclamation). Edit: Ack! I take that back. I missed the s^(-v/2) at the front, which is exponential decay. The exponential trumps the factorial, so I think the limit is 0. Edit 2: Err... no, I think factorial trumps exponential, so now I am back to thinking it is infinite. Let's just check with mathematica..... and yes indeed it is infinite. Edited August 3, 2012 by Severian
P. G. Luan Posted August 4, 2012 Posted August 4, 2012 I found that the result depends on the value of s. 1. For 0<s<1, the result is infinity. 2. For s>1, the result is 0. I found that the result depends on the value of s. 1. For 0<s<1, the result is infinity. 2. For s>1, the result is 0. If s=1, the result is also 0.
Tassus Posted August 5, 2012 Author Posted August 5, 2012 Thank you all for your help! So according to your comments, I sum up: see the attached picture.
Tassus Posted August 6, 2012 Author Posted August 6, 2012 A Correction: After the 5th line (at the previous picture), replace v->+infinity with y->+infinity Thank you all for your help! So according to your comments, I sum up: see the attached picture.
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