If f is a function from R to R such that f(x)f(y)=f(x+y)+f(z-y) and f(1)=3, then what is the value of f(7)?

Question

asked Apr 12, 2023 8.1k views

1 Answer

Jakub · Answer 1 · 2023-04-16T13:24:47+0000

I suppose you mean

f(x) f(y) = f(x + y) + f(x - y)

Let
x=1 and
y=0 . Then

$f(1) f(0) = f(1 + 0) + f(1 - 0) \implies f(1) f(0) = 2 f(1) \implies f(0) = 2$

Now if we fix
y=1 , the functional equation reduces to the recurrence relation

$f(x) f(1) = f(x + 1) + f(x - 1) \implies f(x+1) - 3f(x) + f(x-1) = 0$

and we can find
f(7) with a few rounds of substitution.

$x=1 \implies f(2) - 3f(1) + f(0) = 0 \implies f(2) = 3f(1) - f(0) = 7$

$x=2 \implies f(3) - 3f(2) + f(1) = 0 \implies f(3) = 3f(2) - f(1) = 18$

$x=3 \implies f(4) - 3f(3) + f(2) = 0 \implies f(4) = 3f(3) - f(2) = 47$

$x=4 \implies f(5) - 3f(4) + f(3) = 0 \implies f(5) = 3f(4) - f(3) = 123$

$x=5 \implies f(6) - 3f(5) + f(4) = 0 \implies f(6) = 3f(5) - f(4) = 322$

$x=6 \implies f(7) - 3f(6) + f(5) = 0 \implies f(7) = 3f(6) - f(5) = \boxed{843}$