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

1. Apr 6, 2004

matt grime

Quddusaliquddus stop reading if you don't want the solution.

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).

2. Apr 6, 2004

quddusaliquddus

Lol...I must say you have unbounded enthusiasm for maths! Shall i post the answer I got from someone else? It looks slightly different, so you might be interested.

3. Apr 6, 2004

matt grime

I'd certainly like to know if I've missed something.

4. Apr 7, 2004

Muzza

I hope this comment wasn't directed at me because of my question in the other thread... ;)

5. Apr 7, 2004

matt grime

nope, it was directed at me, sorry if you thought it was meant for someone else.