Showing that $\frac{(j+a)!(j-a)!}{a^2}$, for $a=1, \ldots, j$, reaches its minimum at $a\approx \sqrt{j}$

Question

Let $j$ be a positive integer, and $a=1,\dots,j$. From numerical result, I get that $$\frac{(j+a)!(j-a)!}{a^2}$$ reaches its minimum at $a\approx\sqrt{j}$. But how to prove this?

More general form: $\alpha \neq 2$ case

Answer 1

Consider $$f(a)=\frac{(j+a)!\,(j-a)!}{a^2}\qquad \text{and} \qquad g(a)=\log[f(a)]$$ $$g'(a)=\frac {f'(a)}{f(a)}$$ Simplify to show that $$f'(a)=0 \implies H_{j+a}-H_{j-a}-\frac{2}{a}=0$$ Suppose that $j \gg a$ and use the asymptotics of harmonic numbers; it gives $$H_{j+a}-H_{j-a}-\frac{2}{a}=-\frac{2}{a}+\frac{2 a}{j}-\frac{a}{j^2}+O\left(\frac{1}{j^3}\right)$$ If you use the expansion to $O\left(\frac{1}{j^2}\right)$, it gives your result. Using the next term would give, as better approximation $$a \sim \frac{j\sqrt{2} }{\sqrt{2 j-1}}$$

Using more terms and series reversion, $$a=\sqrt j \left(1+\frac{1}{12 j}+\frac{71}{1440 j^2}+\frac{2951}{34560 j^3}+O\left(\frac{1}{j^4}\right)\right)\tag 1$$

Trying for $j=16$, the above truncated series gives $$a=\frac{28465051}{7077888}\approx 4.021687$$ while the solution given by Newton method is $4.021616$

Edit

Using $(1)$, we could easily show that $$H_{j+a}-H_{j-a}-\frac{2}{a}=\frac{3463}{12096\,j^{7/2}}+O\left(\frac{1}{j^{9/2}}\right)$$ Knowing the possible form of $a$, let $$a=\sqrt j \,\,\sum _{i=0}^3 \frac {b_i} { j^{i}}$$ This leads to $$a=\sqrt j \left(1+\frac{1}{12 j}+\frac{71}{1440 j^2}+\frac{491}{5760j^3}+O\left(\frac{1}{j^4}\right)\right)\tag 2$$ This changed very little coefficient $b_3$

$$\frac{2951}{34560}-\frac{491}{5760}=\frac{1}{6912}$$ but this is sufficient to make $$H_{j+a}-H_{j-a}-\frac{2}{a}=-\frac{4429051}{1036800\,j^{9/2}}+O\left(\frac{1}{j^{11/2}}\right)$$

Answer 2

Approximation of the factorial

Observe that $$\log N!=\sum_{n=1}^{N}\log n\approx \int\limits_{1}^{N}\log x\,\mathrm{d}x = N\log N-N+1\approx N\log N-N.$$ Raising both sides to $e$ yields $$N!\approx\left(\frac{N}{e}\right)^N.$$ One may even use Stirlings formula but at the end we'll see that this approximation is more than enough.

Replacing

If we replace the factorial in your formula with this approximation we have $$\frac{1}{a^2}\left(\frac{j+a}{e}\right)^{j+a}\left(\frac{j-a}{e}\right)^{j-a}=\frac{1}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}.$$ The domain of this function is $\{x\in\mathbb{R}\mid j>a.\}$

Setting up the equation

In order to find the minimum we have to compute the derivative of this function and set it equal to zero $$-\frac{2}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}+\frac{1}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}(\log(j-a)+1)+\frac{2}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}(\log(j+a)+1)=0.$$ Solving the equation

1. Write the left hand side as a single fraction. $$\frac{1}{a^2}e^{-2j}(j+a)^{j+a}(j-a)^{j-a}(\log(j+a)+\log(j-a))=0$$
2. Multiply both sides by $a^2e^{2j}(j-a)^a$. $$(j+a)^{j+a}(j-a)^{j}(\log(j+a)+\log(j-a))=0$$
3. Now we have $3$ different possibilities:

P1: $(j-a)^j=0$. Violates $j>a$.

P2: $(j+a)^{j+a}=0$. Base has to be $0$ but $0^0$ is undefined.

P3: $\log(j+a)+\log(j-a)=0$. Is solvable.

Solving P3

Combine logarithms $$\log((j+a)(j-a))=0 \Rightarrow (j+a)(j-a)=1.$$ Expand out the terms and add $a^2$ on both sides to find the minimum. $$j^2=a^2+1\Rightarrow j=\sqrt{a^2+1}.$$ The equation above does not have $2$ solutions since our domain was $\{x\in\mathbb{R}\mid j>a\}$ (=no negative solution). Equivalently $$a=\sqrt{j^2-1}.$$ This is as good as $a=j$, since for big values of $j$ we may even say $j>6$, the difference is very small. For $j\to\infty$ there is no difference.