Power series $\sum n^3a_nz^n$ 
If $f(z)=\sum a_nz^n$, what is $\sum n^3a_nz^n$?

The desired sum is $a_1z+8a_2z^2+27a_3z^3+\cdots$. I can't see how to write the desired sum in terms of $f$. For example, I could substitute $kz$ for $z$ to get $f(kz)=\sum k^na_nz^n$, but that doesn't help me get the coefficient $n^3$.
 A: Hint: Differentiate, multiply by $z$, differentiate, $\dots$.
A: If $f(z)=\sum a_nz^n$, then $zf'(z) = \sum n a_n z^n$. Applying this thrice you would get-
$\sum n^3 a_n z^n = z \left(f'(z) + 3z f''(z) + z^2 f'''(z)\right)$
A: Find the coordinates $(\alpha_0,\alpha_1,\alpha_2,\alpha_3$) of $x^3$ in the basis $(1,x,x(x-1),x(x-1)(x-2))$ of $\mathbb R_3[x]$
so 
$$\sum n^3a_n z^n=\alpha_0f(z)+\alpha_1 z f'(z)+\alpha_2 z^2 f''(z)+\alpha_3z^3f'''(z)$$
A: Hint: $$z(a_0 + a_1 z + a_2 z^2 + \ldots)' = a_1 z + 2 a_2 z^2 + \ldots$$
A: For the general case,
you will encounter the
Stirling numbers
of the second kind.
In this case,
you are given
$f(z)=\sum_{n=0}^{\infty} a_nz^n$
and want to find
$f_k(z)
=\sum_{n=0}^{\infty} n^ka_nz^n
$.
By repeated differentiation,
$f^{(k)}(z)=\sum_{n=k}^{\infty} (n)_k a_nz^{n-k}$
where
$(n)_k
=n(n-1)...(n-k+1)
$
is the Pochhammer symbol
denoting the falling factorial.
Therefore
$\sum_{n=0}^{\infty} (n)_k a_nz^{n}
=z^kf^{(k)}(z)
$.
We now use the definition
of the Stirling numbers
of the second kind:
$\sum_{j=0}^k S(k, j) (n)_j
=n^k
$.
Note: The Stirling numbers of the first kind
go the other way,
expressing
$(n)_k 
=\sum_{j=0}^k s(k, j) n^j
$
These make is easy to get a formula
for $f_k(z)$:
$\begin{align}
f_k(z)
&=\sum_{n=0}^{\infty} n^ka_nz^n\\
&=\sum_{n=0}^{\infty} a_nz^n\sum_{j=0}^k S(k, j) (n)_j\\
&=\sum_{j=0}^k\sum_{n=0}^{\infty} a_nz^n S(k, j) (n)_j\\
&=\sum_{j=0}^k S(k, j) \sum_{n=0}^{\infty} a_nz^n  (n)_j\\
&=\sum_{j=0}^k S(k, j) z^k f^{(k)}(z)\\
\end{align}
$
For $k=3$,
$S(3, 1..3) = [1, 3, 1]$,
so
$\sum_{n=0}^{\infty} n^3a_nz^n
=z^3 f'''(z)+3z^2f''(z)+zf'(z)
$.
For $k=4$,
$S(4, 1..4) = [1, 7, 6, 1]$,
so
$\sum_{n=0}^{\infty} n^4a_nz^n
=z^4 f^{(4)}(z)+7z^3 f'''(z)+6z^2f''(z)+zf'(z)
$.
