Find a good approximation of $r(x)$ for a proof

Question

Let $0<x\leq 0.5$ then we have :

$$x^{2(1-x)}+(1-x)^{2x}\leq 1$$

Sketch :

Using Weighted Ky-Fan inequality we have where $g(x)=\left(0.5\left(x\left(x-0.5\right)\left(1-x\right)\right)\right)^{2}$ :

$$\frac{x^{2\left(1-x\right)}}{\left(1-x\right)^{2x}}\leq h(x)=\frac{\left(x\cdot2\cdot\left(1-x\right)+\left(x+g\left(x\right)\right)\left(2x\right)\right)^{2}}{\left(\left(1-x\right)\left(2\left(1-x\right)\right)+\left(1-\left(x+g\left(x\right)\right)\right)\left(2x\right)\right)^{2}}\cdot\left(\frac{\left(x+g\left(x\right)\right)^{\left(2x\right)}}{\left(1-x\right)^{2\left(1-x\right)}}\right)^{-1}$$

Now I'm stuck because I want to find a good approximation of :

$$r(x)=\left(\left(x\right)^{\left(2x\right)}\right)^{-1}\leq q(x)$$

I denote by $t(x)=r(x)-q(x)$

Edit : I change the question because there is a big simplification !

Question :

How to find $q(x)$ on $(0,0.5]$ such that $t(x)\leq 10^{-3}$ and $t(0)=0=t(0.5)$?

Answer 1

This will a double sum as seen in this graph. I will use this exp(y) expansion:

$$\mathrm{\frac{(1-x)^{2(1-x)}}{x^{2x}}=e^{-2x ln(x)}e^{2ln(1-x)(1-x)}=exp\big(-4ln(x)ln(1-x)(1-x)\big)≈\quad\sum_{N=0}^t\sum_{n=0}^t\frac{(-1)^n2^{N+n}ln(1-x)^Nln^n(x)(1-x)^Nx^n}{N!n!}≈\sum_{n=0}^t\frac{[2(1-x)ln(1-x)-2xln(x)]^n}{n!}}$$

Simply choose small t for a small approximation and larger t for a better approximation. You can also use binomial theorem to split the two terms raised to the nth power, but this makes the expression more complicated. Using the simpler expansion, one can get, for t=4 a graphically indistinguishable plot:

x=$\frac14$,

t=$4$:error=$1.067…\,10^{-4}$

t=$5$:error=$4.625…\,10^{-7}$

t=$20$:error=-$1.11…\,10^{-16}$

…

For the simplified answer, it is simpler and you can just approximate using this other graph:

$$\mathrm{r(x)=x^{-2x}≈\sum_{n=0}^t\frac{(-1)^n2^nln^n(x) x^n}{n!}}$$

x=$\frac13$:

t=$4$: error=$1.995…\, 10^{-3}$

t=$5$: error=$2.39…\,10^{-4}$

t=$10$:error= $8.676…\, 10^{-10}$

$$\mathop\implies^?$$t=a: error order of magnitude=$10^{-a} \text{ or }10^{1-a}$

There are other approximations, but this one is intuitive. Please correct me and give me feedback!

Answer 2

This is not an answer.

It is not the first time that I notice that the plot of many of the functions appearing in your questions makes me thinking about the plots of the Gibbs excess energy for binary systems in vapor-liquid equilibria.

The oldest model, proposed by Margules in year $1895$, write $$\frac{G^{ex}}{RT}=x(1-x)(A+Bx)$$

I tried it for the function $$f(x)=x^{-2x}-1$$ using as only information that $x^{-2x}$ shows a maximum value of $e^{2/e}$ for $x=\frac 1e$.

We end with

$$x^{-2x}\sim g(x)=1+\frac{e^2 \left(e^{2/e}-1\right)}{(e-1)^2}x(1-x)\Big[ (2e-3)-(e-2)x\Big]$$ and $$\Phi=\int_0^1\Big[ x^{-2x}-g(x)\Big]^2 \,dx=8.33 \times 10^{-4}$$

Using it for $x=\frac 13$ as @Tyma Gaidash did, the approximate value is $2.07959$ while $3^{2/3}=2.08008$ (absolute error of $4.92 \times 10^{-4}$ ).