let f be a function that satisfies the statement

f(u+v)=f(u)+f(v)+1-cos(uv)

find f(0)

show work

good luck and thanks i am helpless on this one

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- Oct 16th 2007, 11:39 AM #17thPathGuest

- Oct 16th 2007, 12:39 PM #2
not exactly sure what you are after here. but we have f(0) when u + v = 0 => u = -v, so we could write,

f(0) = f(-v) + f(v) + 1 - cos(-v^2)

but that doesn't really get us anywhere, i think. so let's try something else. what if i solved for f(u) first, and then plug in u = 0?

f(u) = f(u + v) - f(v) + cos(uv) - 1

if u = 0, we get:

f(0) = f(v) - f(v) + cos(0) - 1 = 0

so f(0) = 0

the same thing happens if we solve for f(v) and let v = 0

so i suppose f(0) = 0 is actually the answer

- Oct 16th 2007, 01:03 PM #3