Question

Zadanie:
Wiadomo, że [tex]f(x)=3x+2[/tex] oraz [tex]\underbrace{g(g(g(...(x))))}_{ 100}=f(x)[/tex] . Wyznaczyć [tex]g(x)[/tex].

Answer 1

[tex]f(x)=3x+2\\\underbrace{g(g(g(\ldots(x))))}_{100}=f(x)\\\\\\g(x)=ax+b\\\\g(g(x))=a(ax+b)+b=a^2x+ab+b=a^2x+b(a+1)\\g(g(g(x)))=a(a^2x+ab+b)+b=a^3x+a^2b+ab+b=a^3x+b(a^2+a+1)\\g(g(g(g(x))))=a(a^3x+a^2b+ab+b)+b=a^4x+a^3b+a^2b+ab+b=a^4x+\\+b(a^3+a^2+a+1)\\\vdots\\\underbrace{g(g(g(\ldots(x))))}_{100}=a^{100}x+b(a^{99}+a^{98}+\ldots+1)[/tex]

To w nawiasie po prawej to suma ciągu geometrycznego.

[tex]a_1=a^{99}\\q=\dfrac{1}{a}\\n=100\\S_n=a_1\cdot\dfrac{1-q^n}{1-q}\\\\S_{100}=a^{99}\cdot\dfrac{1-\left(\dfrac{1}{a}\right)^{100}}{1-\dfrac{1}{a}}=a^{99}\cdot\dfrac{1-\dfrac{1}{a^{100}}}{\dfrac{a-1}{a}}=a^{99}\cdot \dfrac{a^{100}-1}{a^{100}}\cdot \dfrac{a}{a-1}=\dfrac{a^{100}-1}{a-1}[/tex]

[tex]\underbrace{g(g(g(\ldots(x))))}_{100}=a^{100}x+b\left(\dfrac{a^{100}-1}{a-1}\right)\\\\\\a^{100}x+b\left(\dfrac{a^{100}-1}{a-1}\right)=3x+2\\\\\\a^{100}=3 \wedge b\left(\dfrac{a^{100}-1}{a-1}\right)=2\\\\a=\sqrt[100]3 \vee a=-\sqrt[100]3\\\\b\left(\dfrac{3-1}{\sqrt[100]3-1}\right)=2\vee b\left(\dfrac{3-1}{-\sqrt[100]3-1}\right)=2\\\\\dfrac{2b}{\sqrt[100]3-1}=2\vee \dfrac{2b}{-\sqrt[100]3-1}=2\\\\2b=2 \sqrt[100]3-2\vee 2b=-2 \sqrt[100]3-2\\b=\sqrt[100]3-1\vee b=-\sqrt[100]3-1[/tex]

Zatem mamy dwie możliwe funkcje:

[tex]\boxed{\begin{aligned}&g(x)=\sqrt[100]3 x+\sqrt[100]3-1\\&g(x)=-\sqrt[100]3 x-\sqrt[100]3-1\end{aligned}}[/tex]