Quddusaliquddus stop reading if you don't want the solution.(adsbygoogle = window.adsbygoogle || []).push({});

OK?

Stopped?

Right, to find all cont, functions from R to R satisfying f(x+y)f(x-y) = (f(x)f(y))^2

put x=y=0 to see f(0)^2=f(0)^4 ie f(0) =0 1 or -1

also x=y shows f(0)f(2x)=f(x)^4

so f(0)=0 implies f=0

if f(0)=1 then we see the relation

f(2x)=f(x)^4

if you're stupid you get mixed up and conclude that only the constant solution f=1 works. Then you realize that there are obviously non-constant solutions, duh! such as 2^{x^2} so you figure out why these ones are the only kind:

so, let f(1) = k

then f(2)=k^4, f(4)=k^16, hmm, f(3) can be got from f(3)f(1)=(f(2)f(1))^2

and pretty soon you realize that for every integer n, f(n)=k^{n^2}

now you go on to think, but we can do it for all rationals with powers of 2 in the denominator, and it works there, and as they're dense in the reals you see that actually f_k(x) = k^{x^2} forms a complete set of solutions (well, there's the solutions g_k = -f_k but whose counting).

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Solution to quddusaliquddus's cont func. questions if you want it

**Physics Forums | Science Articles, Homework Help, Discussion**