(almost) proof of TPID 13

[fivexthethird]

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]

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.