Solving an infinite alternating product

Question

Idea is to alternately multiply infinite many values from $1$ to $x$, with const distance $h$.

Reweriting it to math:

$$ h = \frac{x-1}{n} $$ $$ f(x) = \lim_{n \to \infty} \left( \prod_{m=0}^{n} (x - mh)^{(-1)^{m}} \right) $$

I have played with the idea in Python and it seems for me that: $$ f(x) = \sqrt{x} $$ $$ x \gt 0 $$

I see some similarities with function for calculating the length $L$ of some curve $f(x)$ but there are two main differences:

1. $\prod$ instead of $\sum$
2. alterantion $(-1)^m$

How I can prove it ?

Answer 1

The function $f$ can be rescribed as $$f(x)=\lim_{n\to+\infty}\prod_{m=0}^n(x-m(x-1)/n)^{(-1)^m}=\lim_{n\to+\infty}e^{g_n(x)}$$ where $$g_n(x)=\sum_{m=0}^n(-1)^m\log(x-m(x-1)/n)$$ For $x>0$ we have: \begin{align} g_{2n+1}(x) &=\sum_{m=0}^{2n+1}(-1)^m\log(x-m(x-1)/n)\\ &=\sum_{k=0}^n[(-1)^{2k}\log(x-2k(x-1)/n)+(-1)^{2k+1}\log(x-(2k+1)(x-1)/n)]\\ &=-\sum_{k=0}^n\log\left(1-\frac{(x-1)/n}{x-(x-1)2k/n}\right)\\ &=\sum_{k=0}^n\frac{(x-1)/n}{x-(x-1)2k/n}+O(1/n)\\ &\xrightarrow{n\to+\infty}\frac 12\int_0^1\frac{x-1}{x-(x-1)t}\mathrm dt\\ &=\frac 12\log(x) \end{align}

Answer 2

$$P_n=\prod_{m=0}^n\left(x-\frac{m }{n}(x-1)\right)^{(-1)^m}$$ $$P_n=\frac{n x \left(\frac{2(1- x)}{n}+x\right) \left(\frac{2(1-x)}{n}\right)^{\left\lceil \frac{1-n}{2}\right\rceil +\left\lfloor \frac{n}{2}\right\rfloor -1} \left(\frac{(n-4) x+4}{2(1- x)}\right)_{\left\lfloor \frac{n}{2}\right\rfloor -1}}{((n-1) x+1) \left(\frac{(n-3) x+3}{2(1- x)}\right)_{\left\lfloor \frac{n-1}{2}\right\rfloor }}$$ is not very pleasant to work with.

However $$P_{2n+1}=\frac{\Gamma \left(\frac{n x}{2 (1-x)}+\frac{1}{2}\right) \Gamma \left(\frac{nx }{2 (1-x)}+n+1\right)}{\Gamma \left(\frac{n x}{2 (1-x)}\right) \Gamma \left(\frac{nx }{2 (1-x)}+n+\frac{3}{2}\right)}$$

Taking logarithms, using four times Stirling approximation and continuing with Taylor series for large values of $n$

$$\log(P_{2n+1})=\frac{1}{2} \log \left(\frac{x}{2-x}\right)-\frac{1-x^2}{2 (2-x) x}\frac 1n+O\left(\frac{1}{n^2}\right)$$ So, for large values of $n$ $$\color{blue}{P_{2n+1}\sim \sqrt{\frac{x}{2-x}}}$$

On the other hand $$P_{2n}=\frac{x ((n-2) x+2) \left(\frac{(n-4) x+4}{2-2 x}\right)_{n-1}}{((n-1) x+1) \left(\frac{(n-3) x+3}{2-2 x}\right)_{n-1}}$$

$$\log(P_{2n})=\frac{1}{2} \log (x(2-x) )-\frac{(x-1)^2}{2 (2-x) x}\frac 1n+O\left(\frac{1}{n^3}\right)$$

So, for large values of $n$ $$\color{blue}{P_{2n}\sim \sqrt{(2-x)x}}$$

Combining the two series $$\log(P_{2n}\,P_{2n+1})=\log(x)+\frac{x-1}{4 n x}-\frac{x^2-1}{16 n^2 x}+O\left(\frac{1}{n^3}\right)$$

$$\large\color{red}{\sqrt{P_{2n} \,P_{2n+1}}\sim \sqrt{x}\exp\Big[\frac{x-1}{8 n x} \Big] \quad \to \quad \sqrt{x}}$$

Trying for $x=\frac 12$ and $n=100$, the exact value is

$$\sqrt{P_{200} \,P_{201}}=\frac 12\sqrt{ \frac{401}{804}}=0.7062267$$ while $$\frac{1}{\sqrt{2} \sqrt[800]{e}}=0.7062234$$