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
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