My motivation is that I am looking for some given number $y$ in Pascal's triangle by searching the diagonals (essentially iterating through $k$, omitting division by $k!$. Currently, I am taking advantage of the fact that the diagonals are monotonic, so I can take an upper and lower bound, evaluate at the middle and readjust the bounds as needed (binary search).
This works fine, but if I can directly computer the function inverse, that would be a lot faster.
I tried applying Newton's method as well, which would be much faster than binary search, but it seems like computing the derivatives is non-trivial (given arbitrary $k$).
So my question is, is there an easy way to find $x$, given $y=x(x+1)(x+2)\cdots (x+k-1)$?
(Sorry for any errors, on mobile).
Assuming $x>0$ and taking logarithms this becomes finding $x$ such that $\log y = \sum_{i=1}^{k}\log(x+i-1)$ or $$f(x) = \sum_{i=1}^k \log(x+i-1)-\log y=0.$$The derivative $$f'(x) = \sum_{i=1}^k \frac{1}{x+i-1}$$ so we can find such an $x$ by iterating Newton's method with an initial guess $x_0>0$: $$x_{n+1} = x_n - \left(\sum_{i=1}^k \frac{1}{x+i-1}\right)^{-1}\left(\sum_{i=1}^k \log(x+i-1)-\log y\right).$$
For assistance finding an initial guess, note that for large $x\gg k$ $$\log y = \sum_{i=1}^k\log(x+i-1) \approx k\log(x+k-1)$$ so, solving for $x$, $$ x \approx y^{1/k}+1-k\text{ and we can take }x_0 = \max\{0, y^{1/k} + 1 - k\}. $$