Generalized Tent Map

(This is another mathematics blog post. Please expect mathematics on this blog post. If you dislike mathematics, we advise you to leave.)

Today, we are going to talk about the generalized tent map, namely

$${\displaystyle f\left(x\right)=a-\left|a\cdot2^{n}\left(x-\frac{1}{2}\right)^{n}\right|}$$, where ${\displaystyle 0\le a\le1}$, ${\displaystyle 0 \leq x \leq 1}$, and ${\displaystyle n = \frac{p}{q}}$ (${\displaystyle \operatorname{mod}\left(q,2\right)=1}$,${\displaystyle \gcd\left(p,q\right)=1}$, and ${\displaystyle n > 0}$)

Part I: Evaluating it with special values of ${\displaystyle x}$

Section I: ${\displaystyle x = 0}$

$${\displaystyle f\left(0\right)=a-\left|a\cdot2^{n}\left(0-\frac{1}{2}\right)^{n}\right|}$$

$${\displaystyle a=\left|a\right|+f\left(0\right)}$$

(However, since ${\displaystyle 0\le a\le1}$, ${\displaystyle a}$ is a real number, which entails${\displaystyle \left|a\right|=a}$.)

Therefore,${\displaystyle f\left(0\right)=0}$. ⬛

Section II: ${\displaystyle x=\frac{1}{2}}$

$${\displaystyle f\left(\frac{1}{2}\right)=a-\left|a\cdot2^{n}\left(\frac{1}{2}-\frac{1}{2}\right)^{n}\right|}$$

(However, since ${\displaystyle 0\le a\le1}$and ${\displaystyle n > 0}$, it implies ${\displaystyle a}$ and ${\displaystyle n}$are positive)

Therefore, we can rewrite the RHS in an alternative form, giving us:${\displaystyle f\left(\frac{1}{2}\right)=a}$. ⬛

Section III: ${\displaystyle x=1}$

$${\displaystyle f\left(1\right)=a-\left|a\cdot2^{n}\left(1-\frac{1}{2}\right)^{n}\right|}$$

$${\displaystyle a=\left|a\right|+f\left(1\right)}$$

(However, since ${\displaystyle 0\le a\le1}$, ${\displaystyle a}$ is a real number, which entails${\displaystyle \left|a\right|=a}$.)

Therefore,${\displaystyle f\left(1\right)=0}$. ⬛

Part II: Special values of ${\displaystyle n}$

Section I: ${\displaystyle n=0}$ (Zero)

$${\displaystyle f\left(x\right)=a-\left|a\cdot2^{0}\left(x-\frac{1}{2}\right)^{0}\right|=a-\left|a\right|}$$

(However, since ${\displaystyle 0\le a\le1}$, ${\displaystyle a}$ is a real number, which entails${\displaystyle \left|a\right|=a}$.)

Therefore,${\displaystyle f\left(x\right)=0}$. ⬛

Section II: ${\displaystyle n=1}$(Tent Map)

$${\displaystyle f\left(x\right)=a-\left|a\cdot2^{1}\left(x-\frac{1}{2}\right)^{1}\right|=a-2\left|a\left(x-\frac{1}{2}\right)\right|}$$

This can be shown to be equivalent to the tent map. The set of fixed points are: ${\displaystyle x=\left[0,\frac{2a}{2a+1}\right]}$(Under the constraint ${\displaystyle 0\le a\le1}$, the rightmost fixed point of the tent map can be found at ${\displaystyle x = \frac{2}{3}}$.)

${\displaystyle n=2}$(Logistic Map)

$${\displaystyle f\left(x\right)=a-\left|a\cdot2^{2}\left(x-\frac{1}{2}\right)^{2}\right|=4a\left(x\right)\left(1-x\right)}$$

This can be shown to be equivalent to the logistic map. The set of fixed points are: ${\displaystyle x=\left[0,\frac{4a-1}{4a}\right]}$(Under the constraint ${\displaystyle 0\le a\le1}$, the rightmost fixed point of the logistic map can be found at ${\displaystyle x=\frac{3}{4}}$.)

Part 3: Tablulation of rightmost fixed points vs integral ${\displaystyle n}$

Tablulation of rightmost fixed points vs integral ${\displaystyle n}$

$${\displaystyle n}$$	Rightmost fixed point
$${\displaystyle 0}$$	$${\displaystyle 0}$$
$${\displaystyle 1}$$	$${\displaystyle \frac{2}{3}}$$
$${\displaystyle 2}$$	$${\displaystyle \frac{3}{4}}$$
$${\displaystyle 3}$$	$${\displaystyle \frac{\frac{\sqrt[3]{\sqrt{87}+9}}{6^{\frac{2}{3}}}-\frac{1}{\sqrt[3]{6}\sqrt[3]{\sqrt{87}+9}}+1}{2}}$$
$${\displaystyle 4}$$	$${\displaystyle \frac{\sqrt[3]{6\sqrt{33}+26}}{12}-\frac{2^{\frac{2}{3}}}{3\sqrt[3]{3\sqrt{33}+13}}+\frac{2}{3}}$$

According to Galois theory, there is no known closed form for the rightmost fixed point of the generalized tent map for ${\displaystyle n \geq 5}$, as this involves solving a quintic or higher, and Galois theory says the roots of the resulting polynomial involved in the calculation would have no known closed forms. However, see below for a conjecture.

Part 4: Conjectures?

The rightmost fixed point of any generalized tent map is always less than or equal to ${\textstyle \frac{n+1}{n+2}}$. Any proof of this will be accepted.