Is there a known transform to go from ∑ai∗(xi+1−1)/(x−1)\sum a_i*(x^{i+1}-1)/(x-1) to ∑ai∗(x+1)i\sum a_i*(x+1)^i

I have two functions f:R→Rf:\mathbb{R} \rightarrow \mathbb{R} and g:R→Rg:\mathbb{R} \rightarrow \mathbb{R}:

f(x)=a0+a1∗(x2−1)/(x−1)+a2∗(x3−1)/(x−1)+…f(x) = a_0 + a_1*(x^2-1)/(x-1) + a_2*(x^3-1)/(x-1) + …

g(x)=a0+a1∗(x+1)+a2∗(x+1)2+…g(x) = a_0 + a_1*(x+1) + a_2*(x+1)^2 + …

Is there a known transform to go from ff to gg, or vice versa.

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

Expanding xk+1−1x−1=∑kn=0xn\frac{x^{k+1}-1}{x-1}=\sum_{n=0}^{k} x^n and (x+1)k=∑kn=0(kn)xn(x+1)^k=\sum_{n=0}^{k} \binom{k}{n} x^n, then regrouping:

f(x)=∑n≥0xn⋅∑k≥nakf(x)=\sum_{n\ge 0}x^n \cdot \sum_{k \ge n}a_k

g(x)=∑n≥0xn⋅∑k≥n(kn)akg(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=0x=0:

∑k≥nak=f(n)(0)n!⟹an=f(n)(0)n!−f(n+1)(0)(n+1)!
\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)!}

∑k≥n(kn)ak=g(n)(0)n!⟹g(n)(0)=n!∑k≥n(kn)(f(k)(0)k!−f(k+1)(0)(k+1)!)⟺g(n)(0)=∑k≥n1(k−n)!(f(k)(0)−f(k+1)(0)k+1)⟺g(n)(0)=f(n)(0)+n∑k≥n+1f(k)(0)k(k−n)!
\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 ff and gg, though not a particularly “nice” one.