Does this "factorial-like" product have a name?
$ (1 + t) \cdot (2 + t) \cdot (3 + t) \cdot \ldots \cdot (n + t) $
where $n \in \mathbb{N}$ and $0 < t < 1$?
So it's like a factorial in the sense that the factors differ by 1, but the factors are integers offset by a fixed real number $t$.
And is there a known way to numerically approximate the logarithm of such a product for large $n$ where an iterative approach would be too slow? (just like the logarithm of the Gamma function can be approximated well)
Best Answer
In terms of the gamma function $$P_n=\prod_{k=1}^n (k+t)=\frac{\Gamma (n+t+1)}{\Gamma (t+1)}$$ Taking logarithms $$\log(P_n)=\log (\Gamma (n+t+1))-\log (\Gamma (t+1))$$ For the first one, using Stirling approximation $$\log (\Gamma (n+t+1))=n (\log (n)-1)+(t+1) \log (n)+\frac{1}{2} \log \left(\frac{2 \pi }{n}\right)+\frac{6 t^2+6 t+1}{12 n}++O\left(\frac{1}{n^2}\right)$$ The second term is quite neglible since its maximum value is zero and its minimum value is $-0.121486$. So, ignore it.
Trying for $n=12345$ and $t=0.6789$ and , the above gives $$\log(P_{12345})=103969.35$$ while the exact value is $103969.45$
If you want an approximation
$$\log (\Gamma (t+1))\sim \frac 12 t(t-1)$$ which not too bad $$\int_0^1 \Bigg[\log (\Gamma (t+1))-\frac{1}{2} t(t-1) \Bigg]^2\,dt=3.36 \times 10^{-5}$$