Product of global and local maxima

Let $L= L(x) = \lim\limits_{n\to\infty} \left( \dfrac{n^n(x+n)\left( x+\dfrac{n}{2}\right)\left( x+\dfrac{n}{3}\right)... \left( x+\dfrac{n}{n}\right)}{n!(x^2+n^2)\left( x^2+\dfrac{n^2}{4}\right)\left( x^2+\dfrac{n^2}{9}\right)...\left( x^2+\dfrac{n^2}{n^2}\right)}\right)^{\dfrac{x}{n}} \newline$

We know for sure $L(x) > 0$ because it is in the form of an exponent and exponents are always positive .

$L = \lim\limits_{n\to\infty} \prod\limits_{r=1}^n \left( \dfrac{n}{r} \dfrac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}} \right)^{\frac{x}{n}} \newline$

Justification of the line above is given in the end (conversion to product form).

$\ln(L) = \lim\limits_{n\to\infty} \dfrac{x}{n}ln\left( \prod\limits_{r=1}^n \dfrac{n}{r} \dfrac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}} \right)$

$\ln(L) = \lim\limits_{n\to\infty} \dfrac{x}{n} \sum\limits_{r=1}^n ln\left( \dfrac{n}{r} \dfrac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}} \right)$

$\ln(L) = \lim\limits_{n\to\infty} \dfrac{x}{n} \sum\limits_{r=1}^n ln\left( \dfrac{1}{r/n} \dfrac{x+\frac{1}{r/n}}{x^2+\frac{1}{(r/n)^2}} \right)$

Converting a Riemann Sum to a Definite Integral

$\ln(L) = \int_{0}^{1} x ln\left( \dfrac{1}{p} \dfrac{x+\frac{1}{p}}{x^2+\frac{1}{p^2}} \right)dp$

$\ln(L) = \int_{0}^{1} x ln\left( \dfrac{1}{\cancel{p}} \dfrac{xp+1}{\cancel{p}}\dfrac{\cancel{p^2}}{x^2p^2+1} \right)dp = \int_{0}^{1} x ln\left( \dfrac{1+xp}{1+x^2p^2} \right)dp$

Let $xp = t; xdp = dt$

$\ln(L) = \cancel{x} \int_{0}^{x} ln\left( \dfrac{1+t}{1+t^2} \right)\dfrac{dt}{\cancel{x}}$

$\color{#D61F06} \implies \ln(L) = \int_{0}^{x} ln\left( \dfrac{1+t}{1+t^2} \right)dt$

Differentiate both sides:

$\dfrac{L'(x)}{L(x)} = ln\left( \dfrac{1+x}{1+x^2} \right)$

Since L(x)>0 (proved in the start), the sign of L'(x) depends solely the RHS of the above expression. The sign of L'(x) will give us information about the function L(x) (regions where it is increasing and decreasing).

$\ln(\dfrac{1+x}{1+x^2})<0 \implies 0<\dfrac{1+x}{1+x^2}<1 \implies -1<x<0, x>1 \newline \ln(\dfrac{1+x}{1+x^2})\geq 0 \implies \dfrac{1+x}{1+x^2}\geq1 \implies 0\leq x \leq 1$

Now we know the behavior of L(x). L(x) is defined only for x>-1 as the argument of a logarithm must be greater than 0. Also L(x) decreases from (-1,0], then increases from (0,1] and then decreases from (1, $\infty$ ). Hence candidates for the maximum values of L(x) are: $L(1) \text{and }\lim\limits_{x\to-1} L(x)$ .

Hence $M_L = f(1)$ and $M_G= \max (f(1), f(-1))$ .

Now we will find L(x) using the equation from before (by solving the definite integral on the RHS):

$\color{#D61F06} \ln(L) = \int_{0}^{x} ln\left( \dfrac{1+t}{1+t^2} \right)dt = \ln(x+1) +xln\left( \dfrac{1+x}{1+x^2} \right) -2arctan(x) + x$

$L(x) = e^{ \ln(x+1) +xln\left( \dfrac{1+x}{1+x^2} \right) -2arctan(x) + x} = \dfrac{(x+1)^{x+1}} {(x^2+1)^x} e^{-2arctan(x)+x}$

Given the fact that f(x) is continuous and the findings from before:

$f(x) = \begin{cases} \dfrac{(x+1)^{x+1}} {(x^2+1)^x} e^{-2arctan(x)+x} & \text{if }x>-1 \\ a(x+2): a(-1+2) = \lim\limits_{x\to-1} \dfrac{(x+1)^{x+1}} {(x^2+1)^x} e^{-2arctan(x)+x}& \text{if }x\leq-1 \end{cases}$

We know $f(1)= M_L = \dfrac{(1+1)^{1+1}} {(1^2+1)^1} e^{-2arctan(1)+1} = 2e^{1-\frac{\pi}{2}}$ and $M_G = \max\left( \lim\limits_{x\to-1} \dfrac{(x+1)^{x+1}} {(x^2+1)^x} e^{-2arctan(x)+x}, f(1) \right)$ . Now let us find this limit to see who wins... f(1) or f(-1):

$\lim\limits_{x\to-1} \dfrac{(x+1)^{x+1}} {(x^2+1)^x} e^{-2arctan(x)+x} = \lim\limits_{x\to-1} \dfrac{(x+1)^{x+1}} {(1^2+1)^{-1}} e^{-2arctan(-1)-1} = 2e^{\frac{\pi}{2} -1}\lim\limits_{x\to-1} (x+1)^{x+1} = 2e^{\frac{\pi}{2} -1}$ .

$f(-1) > f(1) \implies M_G = f(-1) = 2e^{\frac{\pi}{2} -1}$

$\therefore M_GM_L = 4$

$f(x) = \begin{cases} \dfrac{(x+1)^{x+1}} {(x^2+1)^x} e^{-2arctan(x)+x} & \text{if }x>-1 \\ 2e^{\frac{\pi}{2} -1}(x+2) & \text{if }x\leq-1 \end{cases}$

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$\text{1. Justification of Line 2 (conversion to product)}$ $\newline \frac{n^n}{n!} = \prod\limits_{r=1}^n \frac{r}{n} \newline (x+n)\left( x+\dfrac{n}{2}\right)\left( x+\dfrac{n}{3}\right)... \left( x+\dfrac{n}{n}\right) = \prod\limits_{r=1}^n \left(x+\dfrac{n}{r}\right) \newline (x^2+n^2)\left( x^2+\dfrac{n^2}{4}\right)\left( x^2+\dfrac{n^2}{9}\right)...\left( x^2+\dfrac{n^2}{n^2}\right) = \prod\limits_{r=1}^n \left(x+\left(\dfrac{n}{r}\right)^2\right) \newline$

$\text{2. }\log(\prod\limits_{x=1}^n x) = \sum\limits_{x=1}^n \log(x)$

$\text{3. }$ Converting a Riemann Sum to a Definite Integral