Finding a closed form to a minimum of a function

Question

It's a try to find a closed form to the minimum of the function : Let $0<x<1$ then define : $$g(x)=x^{2(1-x)}+(1-x)^{2x}$$ Denotes $x_0$ the abscissa of the minimum .

Miraculously using Slater's inequality for convex function I have found that :

Define $f(x)=x^{2(1-x)}$ then :

$$\lim_{x\to x_0}\Bigg(0.5+\frac{(x-1)f'(x)-xf'(1-x)}{f'(x)+f'(1-x)}\Bigg)=0$$

And by the definition of the derivative :

$$\lim_{x\to x_0}g'(x)=0$$

For the first limit see here to compare with the second limit see here

My question :

With these two equations can we hope to find a nice closed form ?

Any helps is greatly appreciated

Thanks in advance because it's a hard nut .

Little update

Well I have got a down-vote there is no mistake if we use natural logarithm for the first limit .

Answer 1

Partial results since I am still working the problem.

Using the Inverse Symbolic Calculator, the point $x_*$ where the derivative cancels seems to be very close to $$x_*=10 \,(\gamma \, K_1(1)){}^{\Gamma \left(\frac{1}{4}\right)}$$ which numerically is $$x_*=0.216453828\qquad \implies g(x_*)=0.99066450008687554$$ while the exact results are $$x_*=0.216453839\qquad \implies g(x_*)=0.99066450008687550$$

A much better approximation seems to be $$x_*=\frac{97+800 e-307 e^2}{-263-205 e+113 e^2}$$ which is in an absolute error of $2.87 \times 10^{-19}$; for this value, $g(x_*)$ is in an absolute error of $2.76 \times 10^{-28}$.

Answer 2

Because $x_0$ the abscissa of the minimum we know that $g'(x_0)=0$ then you can calculate the function $g'(x)$: $$g'(x) = \bigr(x^{2(1−x)} \bigl)'+\bigr((1−x)^{2x}\bigl)' = \bigr(e^{\ln{(x^{2(1−x)})}} \bigl)'+\bigr(e^{\ln{((1−x)^{2x})}}\bigl)' = \bigr(e^{{2(1−x)} \ln{(x)}} \bigl)'+\bigr(e^{{2x} \ln{(1−x)}}\bigl)' = \bigl[ \bigr(e^{{2(1−x)} \ln{(x)}} \bigl)\bigr({{2(1−x)} \ln{(x)}} \bigl)' \bigl]+\bigl[ \bigr(e^{{2x} \ln{(1−x)}}\bigl)({2x} \ln{(1−x)})' \bigr] = $$ $$ = \bigl[x^{2(1−x)}\bigl(2(1-x){1\over x}-2\ln(x) \bigr) \bigr]+\bigl[(1-x)^{2x}\bigl(2\ln(1-x)+2{x\over (1-x)}\bigr)\bigr] = 2\bigl[x^{2(1−x)}({1\over x}-1-\ln(x))+(1-x)^{2x}(\ln(1-x)+{x\over (1-x)})\bigr]$$ Now because $g'(x_0)=0$: $$ 2\bigl[x_0^{2(1−x_0)}({1\over x_0}-1-\ln(x_0))+(1-x_0)^{2x_0}(\ln(1-x_0)+{x_0\over (1-x_0)})\bigr]=0$$ so we can get: $$x_0^{2(1−x_0)}({1\over x_0}-1-\ln(x_0))=-(1-x_0)^{2x_0}(\ln(1-x_0)+{x_0\over (1-x_0)})$$ $$$$ The solutions are: $x_0 = 0.216... \ , \ 0.783... \ \ and \ \ 0.5$, but you can easily find $x$ such as $g(x)<g(0.5)$ so $0.5$ is the max point and $x_0 = 0.216... \ , \ 0.783...$ are the minimum points. You can sea the solution here: https://www.wolframalpha.com/input/?i=-2+x%5E%281+-+2+x%29+%28x+%2B+x+log%28x%29+-+1%29+-+2+%281+-+x%29%5E%282+x+-+1%29+%28x+%2B+%28x+-+1%29+log%281+-+x%29%29+%3D0.