# Find all functions satisfying $f(x+1)=\frac{f(x)-5}{f(x)-3}$

Find all functions satisfying $$f(x+1)=\frac{f(x)-5}{f(x)-3}$$

My try:

We have $$f(x+1)=1-\frac{2}{f(x)-3}$$

Letting $$g(x) =f(x+1)-3$$

We get $$g(x+1)=-2-\frac{2}{g(x)}$$

Any clue here?

• Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens. – Crostul Jan 2 at 9:35
• And observe that you can select the value of $f(t)$, $t\in[0,1)$ any which way you want. Only if you also require continuity is there something to worry. – Jyrki Lahtonen Jan 2 at 9:58
• Why did you tag polynomials ? – Claude Leibovici Jan 2 at 10:10
• Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks – Ekaveera Kumar Sharma Jan 2 at 10:17

The hint.

Prove that: $$f(x+4)=f(x).$$

• But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question . – Kavi Rama Murthy Jan 2 at 12:20
• @Kavi Rama Murthy It was the hint. $f(x)\notin\{0,1,2,3\}$ of course. If you want, you can write a full solution. – Michael Rozenberg Jan 2 at 12:22
• Yes one of the functions that satisfy is $f(x)=\tan\left(\frac{\pi x}{4}\right)+2$ where $x \ne 4p-2$ ,$\forall$, $p \in \mathbb{Z}$. Obviously Cot also satisfies. I will be happy if there are any other functions – Umesh shankar Mar 4 at 16:49

Let $$f(x)=g(x)+3$$ ,

Then $$g(x+1)+3=\dfrac{g(x)+3-5}{g(x)+3-3}$$

$$g(x+1)+3=\dfrac{g(x)-2}{g(x)}$$

$$g(x+1)+3=1-\dfrac{2}{g(x)}$$

$$g(x+1)=-2-\dfrac{2}{g(x)}$$

Let $$g(x)=\dfrac{h(x+1)}{h(x)}$$ ,

Then $$\dfrac{h(x+2)}{h(x+1)}=-2-\dfrac{2h(x)}{h(x+1)}$$

$$\dfrac{h(x+2)}{h(x+1)}=-\dfrac{2h(x+1)+2h(x)}{h(x+1)}$$

$$h(x+2)+2h(x+1)+2h(x)=0$$

$$h(x)=\theta_1(x)(-1+i)^x+\theta_2(x)(-1-i)^x$$ , where $$\theta_1(x)$$ and $$\theta_2(x)$$ are arbitrary periodic functions with unit period

$$h(x)=\theta_1(x)e^{x\ln(-1+i)}+\theta_2(x)e^{x\ln(-1-i)}$$ , where $$\theta_1(x)$$ and $$\theta_2(x)$$ are arbitrary periodic functions with unit period

$$h(x)=\theta_1(x)e^{\frac{x\ln2}{2}+\frac{3i\pi x}{4}}+\theta_2(x)e^{\frac{x\ln2}{2}-\frac{3i\pi x}{4}}$$ , where $$\theta_1(x)$$ and $$\theta_2(x)$$ are arbitrary periodic functions with unit period

$$h(x)=\Theta_1(x)2^\frac{x}{2}\sin\dfrac{3\pi x}{4}+\Theta_2(x)2^\frac{x}{2}\cos\dfrac{3\pi x}{4}$$ , where $$\Theta_1(x)$$ and $$\Theta_2(x)$$ are arbitrary periodic functions with unit period

$$\therefore f(x)=\dfrac{\Theta_1(x+1)2^\frac{x+1}{2}\sin\dfrac{3\pi(x+1)}{4}+\Theta_2(x+1)2^\frac{x+1}{2}\cos\dfrac{3\pi(x+1)}{4}}{\Theta_1(x)2^\frac{x}{2}\sin\dfrac{3\pi x}{4}+\Theta_2(x)2^\frac{x}{2}\cos\dfrac{3\pi x}{4}}+3$$ , where $$\Theta_1(x)$$ and $$\Theta_2(x)$$ are arbitrary periodic functions with unit period

$$f(x)=\dfrac{\sqrt2\sin\dfrac{3\pi(x+1)}{4}+\Theta(x)\sqrt2\cos\dfrac{3\pi(x+1)}{4}}{\sin\dfrac{3\pi x}{4}+\Theta(x)\cos\dfrac{3\pi x}{4}}+3$$ , where $$\Theta(x)$$ is an arbitrary periodic function with unit period