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

  1. Apr 6, 2004 #1

    matt grime

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    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. jcsd
  3. Apr 6, 2004 #2
    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.
     
  4. Apr 6, 2004 #3

    matt grime

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    I'd certainly like to know if I've missed something.
     
  5. Apr 7, 2004 #4
    I hope this comment wasn't directed at me because of my question in the other thread... ;)
     
  6. Apr 7, 2004 #5

    matt grime

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