If a fn. satisfies f(x+1)+f(x-1)=√2 f(x)
then find the period of f(x).
62f(x+1)+f(x-1)=√2 f(x)
substitute x= x+1
f(x+2)+f(x)=√2 f(x+1)
x=x-1, f(x)+f(x-2)=√2 f(x-1)
add these,
f(x+2)+2f(x)+f(x-2)=√2 {f(x-1)+f(x+1)}
substitute RHS from the first equation
f(x+2)+2f(x)+f(x-2)=√2 √2 f(x) = 2f(x)
thus f(x+2)+f(x-2) = 0
now substitute x=x+4
f(x+6)+f(x+2)=0
from the last 2 equations,
f(x+6)=f(x-2)
thus the period is 8
1The purpose is to remove √2. See how this can be done
first replace x by x +1
f(x+2) + f(x)= √2 f(x+1) -----------(1)
replace x by x-1
f(x) + f(x-2) = √2f(x-1)---------------(2)
add both
f(x+2) +2* f(x) + f(x-2) = √2[f(x+1) + f(x-1)]
= √2[√2[f(x)] ..see parent fn equation and see the relation
hence f(x+2) + f(x-2) = 0
