Is there a known transform to go from $\sum a_i*(x^{i+1}-1)/(x-1)$ to $\sum a_i*(x+1)^i$

Question

I have two functions $f:\mathbb{R} \rightarrow \mathbb{R}$ and $g:\mathbb{R} \rightarrow \mathbb{R}$:

$f(x) = a_0 + a_1*(x^2-1)/(x-1) + a_2*(x^3-1)/(x-1) + ...$

$g(x) = a_0 + a_1*(x+1) + a_2*(x+1)^2 + ...$

Is there a known transform to go from $f$ to $g$, or vice versa.

Answer 1

Not an answer proper, but too long for a comment (tl;dr not any "nice" closed form).

Expanding $\frac{x^{k+1}-1}{x-1}=\sum_{n=0}^{k} x^n$ and $(x+1)^k=\sum_{n=0}^{k} \binom{k}{n} x^n$, then regrouping:

$$f(x)=\sum_{n\ge 0}x^n \cdot \sum_{k \ge n}a_k$$

$$g(x)=\sum_{n\ge 0}x^n \cdot \sum_{k \ge n}\binom{k}{n} a_k$$

Taking the above as Taylor expansions at $x=0$:

$$ \sum_{k \ge n}a_k = \frac{f^{(n)}(0)}{n!} \quad \implies \quad a_n = \frac{f^{(n)}(0)}{n!} - \frac{f^{(n+1)}(0)}{(n+1)!} $$

$$ \begin{align} \sum_{k \ge n}\binom{k}{n} a_k = \frac{g^{(n)}(0)}{n!} \quad & \implies \quad g^{(n)}(0) = n! \;\sum_{k \ge n}\binom{k}{n}\left(\frac{f^{(k)}(0)}{k!} - \frac{f^{(k+1)}(0)}{(k+1)!}\right) \\ & \iff \quad g^{(n)}(0) = \sum_{k \ge n}\frac{1}{(k-n)!}\left(f^{(k)}(0) - \frac{f^{(k+1)}(0)}{k+1}\right) \\ & \iff \quad g^{(n)}(0) = f^{(n)}(0) +n \sum_{k \ge n+1}\frac{f^{(k)}(0)}{k\;(k-n)!} \end{align} $$

The latter gives a relationship between $f$ and $g$, though not a particularly "nice" one.