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 (almost) proof of TPID 13 fivexthethird Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2013 05/06/2016, 11:50 AM Actually, the statement I'm proving is more general: Theorem: Let $f(z)$ be holomorphic and bounded on the right half-plane $\Re(z) > c$ for some $c < 0$. Then $f(x)$ is equal to its newton series starting at 0 on that half-plane, We need the following very simple lemma: Lemma: Let $\mathcal{M}\{f(x)\}(s) = \int_1^\infty x^{s-1}f(x) dx + \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n! (n+s)}$ be the analytic continuation of the mellin transform. Then $\mathcal{M}\{\sum_{k=0}^\infty f_k(x) \}(s) = \sum_{k=0}^\infty \mathcal{M}\{f_k(x)\}(s)$ if 1. The sum is absolutely convergent for all x 2. The $f_k$ are all holomorphic. 3. The derivative of the sum at 0 is equal to its term-wise derivative at 0 Proof: The sum and the integral are trivially interchanged. The other term is just $\sum_{n=0}^{\infty} \sum_{k=0}^\infty \frac{f_k^{(n)}(0)}{n!(n+s)}$ The inner sum is clearly absolutely convergent, so we can interchange the sums. Then we can add the two sums of the transform term-wise to get the result. A more general result is most likely well-known but I haven't found any proof of it. Now, $f$ satisfies the conditions for Ramanujan's master theorem to hold, so we have : $f(s) = \mathcal{M}\{\frac{1}{\Gamma(-s)} \sum_{k=0}^{\infty} (-x)^k \frac{f(k)}{k!}\}(-s) [tex] = \mathcal{M}\{\frac{e^{-x}}{\Gamma(-s)} \sum_{k=0}^{\infty} (-x)^k \frac{\Delta^k f(0)}{k!}\}(-s)$ $=\sum_{k=0}^{\infty} \frac{\mathcal{M}\{e^{-x}(-x)^k\}(-s)}{\Gamma(-s)} \frac{\Delta^k f(0)}{ k!} =\sum_{k=0}^{\infty} \frac{(-1)^k \Gamma(k-s)}{\Gamma(-s) } \frac{\Delta^k f(0)}{k!}$ $=\sum_{k=0}^{\infty} (s)_k\frac{\Delta^k f(0)}{k!}$ As the Mellin transform will converge when $\Re(s) > c$, the result follows. Of course, this isn't quite what TPID 13 actually wants: this proves convergence of the newton series of $n^{\frac{1}{n}}$ starting at every $n>0$, but not starting at the desired $n=0$. 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 »

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