If a function $f$ satisfies $f(x+1)+f(x-1)=\sqrt{2} f(x)$, then $f(x+2)+f(x-2)=$

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Aicharya · Accepted Answer

Given, $f(x+1)+f(x-1)=\sqrt{2} f(x)$ ...(i) Replace $x$ by $x+1$ $f(x+2)+f(x)=\sqrt{2} f(x+1)$ ...(ii) Replace $x$ by $x-1$ in Eq. (i), $f(x)+f(x-2)=\sqrt{2} f(x-1)$ ...(iii) On adding Eqs. (ii) and (iii), $\begin{aligned} f(x+2)+f(x-2) & +2 f(x) \\ = & \sqrt{2}[f(x+1)+f(x-1)]\end{aligned}$

$\therefore f(x+2)+f(x-2)=0$

If a function $f$ satisfies $f(x+1)+f(x-1)=\sqrt{2} f(x)$, then $f(x+2)+f(x-2)=$

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