How do I find the Inverse of $f(x)$?

Question

Given $f(x+1)+f(x-1)=x^2$

I have subtituted $(a=x+1)$ and $(a=x-1)$ and got $$f(x)+f(x-2)=(x-1)^2 \text{ and } f(x+2)+f(x)=(x+1)^2$$ Combining those equations, I got

$$f(x+2)-f(x-2)=4x$$

I could not even find $f(x)$

Please help me

Answer 1

Note that $f(x)=1$ leads to $f(x+1)+f(x-1)=2$, $f(x)=x$ leads to $f(x+1)+f(x-1)=2x$, $f(x)=x^2$ leads to $f(x+1)+f(x-1)=2x^2+2$. In order to obtain $x^2$ on the right hand side, we might therefore combine $$ f(x)=\frac12x^2-\frac12$$ and verify that this indeed makes $$ f(x+1)+f(x-1)=x^2.$$ However, $f$ is not unique. For example, $$ f(x)=\frac12x^2-\frac12+\sin\frac{\pi x}2$$ is another solution

Answer 2

HINT:

WLOG let $$f(x)=g(x)+\sum_{r=0}^na_rx_r$$

$$x^2=g(x-1)+g(x+1)+2a_0+2a_1x+a_2((x+1)^2+(x-1)^2)+\cdots$$

Set $a_r=0$ for $r\ge3$

$$x^2=g(x-1)+g(x+1)+2a_0+2a_1x+2a_2(x^2+1)$$

Compare the coefficients of $x$ to find $a_1=0$

that of $x^2,1=2a_2$

those of constants, $0=a_2+a_0\iff a_2=-a_0$ so that

$$f(x)=g(x)+\dfrac{x^2-1}2\text{ and } g(x-1)+g(x+1)=0\iff g(x+1)=-g(x-1)=\cdots=g(x+3)$$

So, $g(x)$ could be any periodic function with period $=4$