05/06/2016, 11:50 AM
Actually, the statement I'm proving is more general:
Theorem: Let
be holomorphic and bounded on the right half-plane
for some
. Then
is equal to its newton series starting at 0 on that half-plane,
We need the following very simple lemma:
Lemma: Let
be the analytic continuation of the mellin transform. Then
if
1. The sum is absolutely convergent for all x
2. The
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
}(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,
satisfies the conditions for Ramanujan's master theorem to hold, so we have :
 = \mathcal{M}\{\frac{1}{\Gamma(-s)} \sum_{k=0}^{\infty} (-x)^k \frac{f(k)}{k!}\}(-s)<br />
[tex] = \mathcal{M}\{\frac{e^{-x}}{\Gamma(-s)} \sum_{k=0}^{\infty} (-x)^k \frac{\Delta^k f(0)}{k!}\}(-s))
^k\}(-s)}{\Gamma(-s)} \frac{\Delta^k f(0)}{ k!}<br />
=\sum_{k=0}^{\infty} \frac{(-1)^k \Gamma(k-s)}{\Gamma(-s) } \frac{\Delta^k f(0)}{k!})
_k\frac{\Delta^k f(0)}{k!})
As the Mellin transform will converge when
, the result follows.
Of course, this isn't quite what TPID 13 actually wants: this proves convergence of the newton series of
starting at every
, but not starting at the desired
.
Theorem: Let
We need the following very simple lemma:
Lemma: Let
1. The sum is absolutely convergent for all x
2. The
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
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,
As the Mellin transform will converge when
Of course, this isn't quite what TPID 13 actually wants: this proves convergence of the newton series of