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Apr 21, 2014 at 20:47 comment added user44143 @DouglasZare, it does help for some purposes. The answer from Maple is more readable and understandable, but for large $n$ (and $L=2$), this will give you the answer more quickly and with better precision.
Apr 21, 2014 at 19:40 answer added user64494 timeline score: -5
Apr 21, 2014 at 17:57 comment added Douglas Zare When Mathematica produces answers like this, it often seems to be a restatement of the problem, and not really progress. Does this expression help if you want to approximate the value?
Apr 21, 2014 at 16:05 comment added Eckhard Mathematica also produces results for $N\neq n$ and larger values of $L$, albeit more complicated ones.
Apr 21, 2014 at 15:39 comment added user44143 For $L=2, N=n$, Mathematica gives an exact answer with the incomplete $\beta$ function: $-\frac{1}{2} i (1-p)^n \left(\frac{p}{p-1}\right)^{\left(\frac{1}{2}-\frac{i}{2}\right) n} \left((n-1) \left(\frac{p}{p-1}\right)^{i n} B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) n,n-1\right)-n \left(\frac{p}{p-1}\right)^{i n} B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) n,n\right)-(n-1) B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) n,n-1\right)+n B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) n,n\right)\right)$
Apr 21, 2014 at 15:31 review First posts
Apr 21, 2014 at 15:47
Apr 21, 2014 at 15:15 history asked Rony CC BY-SA 3.0