A question based on finding the periodicity of function.

Question

Find the periodicity of the function $f(x+1)+f(x-1)=\sqrt2 f(x)$. I tried to solve the problem by replacing $x$ with $(x+1)$ then $(x+2)$ and $(x+3)$. But could not get any answer please help me out as well as tell me the procedure as to how to go about these types of questions.

Answer 1

That's not a function, it's an equation. You want the period of a solution $f(x)$ of the equation.

Hint: Express $f(x)$ in terms of $f(x-1)$ and $f(x-2)$, then in terms of $f(x-2)$ and $f(x-3)$ (by using the expression for $f(x-1)$), then $f(x-3)$ and $f(x-4)$. This should tell you something useful.

Answer 2

Applying the given equation with $x$ changed to $x+1$ in the first term and $x$ change to $x-1$ in the second term we get$$[f(x+2)+f(x)]+[f(x)+f(x-2)]$$ $$=\sqrt 2 f(x+1)+\sqrt 2 f(x-1)=\sqrt2 \sqrt 2f(x)=2f(x).$$ Hence $[f(x+2)+f(x-2)]=0$ or $f(x+2)=-f(x-2)$. Applying this twice you get $f(x+4)=f(x-4)$ or $f(x+8)=f(x)$.