For ε sufficiently small

Here is an interesting functional equation problem. Can you solve it? Find all functions   so that a)   for all   and b)   for all  Here is how I solved it. Substitute u for -u and v for -v. Then our functional equation  becomes F(-2u)= F(u+v)F(v-u)+F(u-v)F(-u-v)=F(2u). Let 2u=x, then we get F(x)=F(-x). Now let u=0. So we get F(0)=F(v)^2+ F(v)F(-v)= 2F(v)^2. If we let v=0, we get that F(0)=2F(0)^2. Solving for F(0), we get that F(0)= 0 or F(0)= 1/2. If F(0)=0, then F(x)=0 for all x. But if F(0)= 1/2, then F(x)= 1/2 or -1/2 for all x. However, F(x)≥0, so the only solutions are F(x)=0 or F(x)= 1/2.

Hello everyone! This is my blog about Maths. Posts will be daily with probability ε, for ε sufficiently small ;) I will also talk about interesting Maths resources and websites.