Function that sends $\sum_{i=0}^n x^i$ to $\sum_{i=0}^n x^{-i}$?

Question

Is there a known function $f(S,x)$ that takes

$$S=1+x+x^2+x^3+x^4+\dots + x^n$$

and $x$ as its input and returns

$$S'=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\dots+\frac{1}{x^n} \ $$ for $x \in$ R, $x>0$ and $n>0$?

Edit: Simply evaluating $S(\frac{1}{x})$ is disallowed, as mentioned in the comments. I am looking for a more elegant formula, which does not require knowledge of $n$.

Answer 1

The following formula inverts the terms of the geometric series $\displaystyle{S=\sum_{i=0}^nx^i}$ independent of n: $$S'=\frac{x}{\frac{1}{S}+x-1}.$$

As should be clear, $$S'=\sum_{i=0}^nx^{-i}=\frac{S}{x^n}.$$ Since $S$ is a geometric series, $S=\frac{1-x^n}{1-x}$ and $$ \begin{align} 1-x^n &= (1-x)S \\ x^n &= 1-S+xS \\ n &= \log_x(1-S+xS). \end{align} $$ Therefore, $$ \begin{align} S' &=\frac{S}{x^n} \\ &=\frac{S}{x^{\log_x(1-S+xS)}} \\ &= \frac{S}{1-S+xS} \\ &= \frac{x}{\frac{1}{S}+x-1} \end{align} $$ for $x \in$ R and $x>0, n\geq 0$.

Answer 2

If I understand the question, the answer is "no." The function can't be well-defined. For instance:

$$f(7) = f(1+2+4) = 1+\frac{1}{2}+\frac{1}{4} = \frac{7}{4}.$$

But

$$f(7) = f(1+6) = 1 +\frac{1}{6} = \frac{7}{6}.$$