Expert: Sherman D. - 3/13/2004

Question
I want to work out all the continuous functions f, such that x and y are reals and

f(x+y)f(x-y) = {f(x)f(y)}^2

I'm especially confused about the braces squared!

Any help appreciated, THANKS!

Answer
Here is what i can tell you

the brackets or braces, just means that the f(x) * f(y) is being squared.

According to http://math.chapman.edu/cgi-bin/structures.pl?Function_rings

f(x + y)f(x - y) = {f(x)f(y)}^2
becomes
{f(x)f(y)}^2 = (f(x) + f(y))*(f(x) - f(y))
{f(x)f(y)}^2 = f(x)^2 - f(x)f(y) + f(x)f(y) - f(y)^2
{f(x)f(y)}^2 = f(x)^2 - f(y)^2

divide both sides by f(x)^2 and you get

1 - [{f(y)^2}/{f(x)^2}] = f(y)^2

divide both sides by f(y)^2 and you get

[1/{f(y)^2}] - [1/{f(x)^2}] = 1

or

{f(x)^2 - f(y)^2}/[{f(x)f(y)}^2]

Sorry, but that is about as far as i got, hopefully that is all you needed, or you are able to do the rest.

as for the [ and }, the difference is, which ever is inside, is done first, then which ever is outside is done last. In short, [ and { are just like (, so you don't get confused, they are used instead of just more parenthesis.