Curious result related to the function $f(x)=\exp\Big(\frac{x-1}{x}\ln(3)\Big)$

Question

Let me defines somethings :

Let $0<x<1$ let $f(x)$ be the function : $$f(x)=\exp\Big(\frac{x-1}{x}\ln(3)\Big)$$ And : $$g(x)=f(1-x)$$ Denote by : $$\min_{x\in(0,1)}(f(x)+g(x))=\frac{2}{3}$$ Define $f^n(x)$ by ($n\geq2$ a natural number): $$f^n(x)=\underbrace{f(f(f(f(\cdots(x)\cdots)}_{n \quad \text{times}}$$ And $g^n(x)$ by : $$g^n(x)=g(\underbrace{g(1-g(1-g(1-g(\cdots g(1-x))\cdots)}_{(n-1) \quad \text{times}}$$ Then we have $$\min_{x\in(0,1)}(f^n(x)+g^n(x))=\frac{2}{3}$$

Let me show $$\min_{x\in(0,1)}(f(x)+g(x))=\frac{2}{3}$$

The derivative is :

$$f'(x)+g'(x)= 3^{\frac{(x - 1)}{x}} \Big(\frac{1}{x} - \frac{(x - 1)}{x^2}\Big) \log(3) + 3^{\frac{-x}{1-x}} \Big(-\frac{x}{(1 - x)^2} - \frac{1}{(1 - x)}\Big) \log(3)$$

Or :

$$f'(x)+g'(x)=\frac{\Big(3^{\frac{(x - 1)}{x}} (x - 1)^2 - 3^{\frac{-x}{1-x}} x^2\Big) \log(3)}{((x - 1)^2 x^2)}$$

Now it's not hard to show that the derivative vanishes at $x=0.5$ and get the desired result using the closed interval method .

For the other cases I believe that there is a trick or somethings like that .

Thanks a lot for your time and patience .

Ps:I think furthermore that we can replace the value $\ln(3)$ be something more general $\ln(\alpha)$ by example with $\alpha\geq 2$ .

Answer 1

In the following I will use a subscript notation $f_n, g_n$ instead of the given superscript one as it might lead to confusion regarding the exponentiation.

At first one needs to note that the direct definitions of $f_n, g_n$ can be rewritten in terms of recursive definitions. For $f_n$ it is rather obvious $f_n(x) = f(f_{n-1}(x))$ whereas for $g_n$ it might help to write down the first couple of terms \begin{align*} g_2(x) &:= g(g(1-x)) = g(f(x)) \\ &= g(f_1(x)) \\ g_3(x) &:= g(g(1-g(1-x)))= g(g(1-f(x)))= g(f(f(x))) \\ &= g(f_2(x)) \\ g_4(x) &:= g(g(1-g(1-g(1-x)))) \\ &= g(f_3(x)) \end{align*} which gives the general recursive formula $$ g_n(x) = g(f_{n-1}(x)) \,. $$ (To be exact I definied $f_1 := f$.)

Next we need to differentiate $f_n+g_n$ to find its extrema \begin{align*} (f_n(x)+g_n(x))' &= f_n'(x)+g_n'(x) \\ &= (f(f_{n-1}(x)))' + (g(f_{n-1}(x)))' \\ &= f'(f_{n-1}(x)) \cdot f_{n-1}'(x) + g'(f_{n-1}(x)) \cdot f_{n-1}'(x) \\ &= f_{n-1}'(x) \left[ f'(f_{n-1}(x)) + g'(f_{n-1}(x)) \right] \\ &= f_{n-1}'(x) \left[ f'(y) + g'(y) \right] \,, \qquad y := f_{n-1}(x) \\ &\overset{!}{=} 0 \end{align*} where I introduced the definition of $y$ to make it more obvious that both terms in the sum depend on the same variable.

As $f_{n-1}'(x) \neq 0$ we need to solve $$ 0 = f'(y) + g'(y) $$ and the solution $y=y_0=\frac{1}{2}$ has been shown in the first part.

Compute the extremal value ($x_0$ refers to $y_0=f_{n-1}(x_0)$) \begin{align*} f_n(x_0)+g_n(x_0) &= f(y_0)+g(y_0) = \frac{2}{3} \end{align*} where the summation value follows from the first part.

For the argument to be complete note that $f$ is one-to-one. The same goes for $f_n$.

Is it still a minima (we only showed that its derivative is zero at $x_0$)?
Note that $f_n+g_n$ is continuous and the limit for $x\rightarrow0^+$ is easily computed to be 1. Thus, it it still a minima.

Answer 2

We have

$\forall x\in (0,1), f(x)+g(x)=3^{-\frac{1-x}{x}}+3^{-\frac{x}{1-x}}$

The function $\varphi:x\mapsto \frac{1-x}{x}$ is a bijection from (0,1) to $(0,+\infty)$

Moreover,

$\forall x\in (0,1), f(x)+g(x)=3^{-\varphi(x)}+3^{-\frac{1}{\varphi(x)}}$

We consider the function $h:x\mapsto 3^{-x}+3^{-\frac{1}{x}}$ on $(0,+\infty)$

The derivative is given by: $h'(x)=\frac{3^{-1/x}\ln(3)-3^{-x}\ln(3)x^2}{x^2}$

$\forall x\in(0,+\infty), h'(x)>0\Leftrightarrow 3^{-1/x}>3^{-x}x^2$

$\Leftrightarrow -\frac{1}{x}\ln(3)>-x\ln(3)+2\ln(x)$

$\Leftrightarrow \frac{1}{x}\ln(3)-x\ln(3)+2\ln(x)<0$

Study the function $\psi:x\mapsto 2\ln(x)-x\ln(3)+\frac{\ln(3)}{x}$

The dérivative $\psi'(x)=-\frac{x^2\ln(3)-2x+\ln(3)}{x^2}$ is negative on $(0,+\infty)$

So $\psi$ is strictly decreasing and $\psi(1)=0$

$\forall x\in (0,+\infty), h'(x)>0\Leftrightarrow \psi(x)<0\Leftrightarrow x>1$

$\forall x\in (0,+\infty), h'(x)=0\Leftrightarrow \psi(x)=0\Leftrightarrow x=1$

Thus h is strictly decreasing on (0,1) and striclty increasing on $(1,+\infty)$. Moreover h is continuous, so $h$ admits a minimum at 1.

Hence $f+g$ admits a minimum at $\varphi^{-1}(1)=\frac12$.

For the second part I resume the work of @jack

We have

$\forall x\in(0,1), f^n(x)+g_n(x)=f(f^{n-1}(x))+g(f^{n-1}(x))$

We remark that $f((0,1))\subseteq (0,1)$ so $f^{n-1}((0,1))\subseteq (0,1)$

We deduce that

$\forall x\in(0,1), f(f^{n-1}(x))+g(f^{n-1}(x))\geq \min_{x\in(0,1)} (f(x)+g(x))$

Then

$\forall x\in(0,1), f^n(x)+g_n(x)\geq \frac{2}{3}$

Thus $\inf_{x\in(0,1)} (f^n(x)+g_n(x))\geq \frac{2}{3}$

However the function $f$ is a bijection from (0,1) to (0,1)

By considering $a=f^{-1}\left(f^{-1}\left(...f^{-1}\left(\frac12\right)...\right)\right)$ (n-1 times), we have $a\in(0,1)$ and

$f^n(a)+g_n(a)=f(f^{n-1}(a))+g(f^{n-1}(a))=f\left(\frac12\right)+g\left(\frac12\right)=\frac32$

Hence the infimum is reached and it is a minimum

$\min_{x\in(0,1)} (f^n(x)+g_n(x))=\frac{2}{3}$