If a function satisfies f(x+1)+f(x-1)=root(2).f(x),....

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SUMMARY

The function f(x) satisfies the equation f(x+1) + f(x-1) = √2 · f(x), leading to the conclusion that the period of f(x) is 4. By manipulating the original equation, the user derived f(x+2) + f(x-2) = 0, which indicates that f(x+4) = -f(x) and ultimately leads to f(x+4) = f(x). This confirms that the function is periodic with a period of 4.

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Titan97
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Homework Statement


If a function satisfies f(x+1)+f(x-1)=√2.f(x), the the period of f(x)=_____

Homework Equations


none

The Attempt at a Solution


I first tried to get rid of the irrational term.
f(x+2)+f(x)= √2.f(x+1)
f(x)+f(x-2) = √2.f(x-1)

adding the above equation, and substituting the given equation,

f(x+2)+f(x-2)+2f(x)=√2.(√2.f(x))=2f(x)
f(x+2)+f(x-2)=0
f(x+4)+f(x)=0

Now instead of getting f(a-x)=f(x) equation, i am getting f(a-x)=-f(x). How can I find its period?
 
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If f(x+a) = -f(x), what is f(x+2a)?
 
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f(x+2a)=f((x+a)+a)=-f(x+a)=f(x):headbang: I did not think it was this easy.
 

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