Almost everyone is familiar with the famous Taylor Series:
$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n $
which, if it converges at more than one point, will converge in some interval about $a$. Has anyone considered the "Reverse" Taylor Series:
$ g(x) = \sum_{n=0}^\infty \frac{f^{(n)}(x-a)}{n!} a^n $
I call it reverse, because its what you get for symbolically taking $a \rightarrow (x-a)$. You might be saying that there is absolutely no reason to believe that this series should converge to $f(x)$, but there are two big examples where it does.
For $e^x$:
$g(x) = \sum_{n=0}^\infty \frac{e^{(x-a)}}{n!} a^n = \frac{e^x}{e^a} \sum_{n=0}^\infty \frac{a^n}{n!} = e^x $
For $x^k$:
$g(x) = \sum_{n=0}^k \frac{k(k-1)...(k-n+1)(x-a)^{k-n} a^n}{n!} = \sum_{n=0}^k {k \choose n} (x-a)^{k-n} a^n = ((x-a) + a)^k $
I have yet to find a counter-example for when the Reverse Taylor Series does not give back the original function for an analytic function, and I also have yet to think of a way to prove that the Reverse Taylor Series should converge for a given function. Does anyone have any ideas?