Question

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

• A. \(2f(x)\)
• B. \(f(x+1)-f(x-1)\)
• C. \(4f(x)\)
• D. \(0\)

Hint:

In functional equations involving \(f(x+1)\) and \(f(x-1)\), substitute \(x+1\) and \(x-1\) separately to obtain expressions involving \(f(x+2)\) and \(f(x-2)\).

Answer 1

The Correct Option is D

Step 1: Find the period by repeated substitution.
Shift x by 1 in the given relation \ (f(x+1)+f(x-1)=\sqrt{2}f(x)\) to get \(f(x+2)+f(x)=\sqrt{2}f(x+1)\). Shift again: \(f(x+3)+f(x+1)=\sqrt{2}f(x+2)\). Continue until you see the pattern repeat. Adding the relation for x and x+2 eliminates middle terms and gives \(f(x+2)+f(x-2)=\sqrt{2}f(x+1)-f(x)+\sqrt{2}f(x-1)-f(x)\).

Step 2: Use telescoping with the two shifted equations.
From x shifted by +1: \(f(x+2)=\sqrt{2}f(x+1)-f(x)\). From x shifted by -1: \(f(x-2)=\sqrt{2}f(x-1)-f(x)\). So \(f(x+2)+f(x-2)=\sqrt{2}(f(x+1)+f(x-1))-2f(x)=\sqrt{2}\cdot\sqrt{2}f(x)-2f(x)=2f(x)-2f(x)=0\). \ \[ \boxed{0} \]