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.
\( f_x(z) = (x+z)^{\frac{1}{x+z}} \)
\( g_x(w) = \sum_{n=0}^\infty f_x(n) \frac{w^n}{n!} \)
\( \sum_{n=0}^\infty (z)_n\frac{\Delta^nf_x(0)}{n!} = \frac{d^{z}}{dw^{z}}|_{w=0} g_x(w) =f_x(z) \)
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.
