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 (almost) proof of TPID 13 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/06/2016, 04:12 PM Cant we just take the limit as $n \to 0$? Namely $f_x(z) = (x+z)^{\frac{1}{x+z}}$ $g_x(w) = \sum_{n=0}^\infty f_x(n) \frac{w^n}{n!}$ $\sum_{n=0}^\infty (z)_n\frac{\Delta^nf_x(0)}{n!} = \frac{d^{z}}{dw^{z}}|_{w=0} g_x(w) =f_x(z)$ And therefore $z^{1/z} = \lim_{x\to 0} f_x(z) = \lim_{x\to 0} \sum_{n=0}^\infty \frac{(z)_n}{n!} \sum_{j=0}^n \binom{n}{j}(-1)^{n-j}(j+x)^{\frac{1}{j+x}} = \sum_{n=0}^\infty \frac{(z)_n}{n!} \sum_{j=0}^n \binom{n}{j}(-1)^{n-j}j^{\frac{1}{j}}$ Granted showing the limit can be pulled through is trivial. Maybe I'm missing something though. « Next Oldest | Next Newest »

 Messages In This Thread (almost) proof of TPID 13 - by fivexthethird - 05/06/2016, 11:50 AM RE: (almost) proof of TPID 13 - by JmsNxn - 05/06/2016, 04:12 PM

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