About the proof of Zeta Transform
I have to prove $$Z[k^n]=(-1)^nD^n\left(\frac{z}{z-1}\right)$$ where $$D=z\frac{d}{dz}$$ and $n$ varies over the set $\mathbb{Z}$ My book doesn't give me any advice; how can I go further?
$\endgroup$ 21 Answer
$\begingroup$Define:
$$f_n(z)=Z[k^n]=\sum_{k=0}^\infty k^n z^{-k}$$
Prove it by induction on $n$.
If $n=0$, then $$f_0(z)=\sum_{k=0}^\infty z^{-k} = \frac{1}{1-z^{-1}}=\frac{z}{z-1} = (-1)^0D^0\left(\frac{z}{z-1}\right)$$
Next, show that:
$$f_{n+1}(z)=(-1)D[f_n(z)]$$
by:
$$\begin{align}(-1)D[f_n(z)]&=(-1)D\left[\sum_{k=0}^\infty k^n z^{-k}\right] \\&= (-1)z\sum_{k=0}^\infty (-1)k^{n+1} z^{-k-1}\\&=f_{n+1}(z) \end{align}$$
More generally:
$$Z[ka_k] = (-1)D\left(Z[a_k]\right)$$
This is the property called differentiation on the wikipedia page for the Z-transform.
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