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E,ln and differentiation.

  1. Sep 26, 2006 #1
    i have three questions which go like this:
    the function f(x) satisfies the equation f(x+y)=f(x)f(y)
    1)if f(x) is differentiable, either f(x)=0 or f(x)=e^ax
    2)if f(x) is continuous, either f(x)=0 or f(x)=e^ax

    3)if a differentiable function f(x) satisfies the equation f(xy)=f(x)+f(y)
    then f(x)=a*log(x)

    for the first question i followed the definition:
    f'(x)=lim (f(x+h)-f(x))/h=lim (f(x)(f(h)-1)/h)=f(x)*f'(0)
    there's a theorem which states that if y'=ay then the y=ce^ax
    so i think because f'(0) is a constant, it follows from this that f(x)=e^ax
    about the second question im not sure, perhaps i should use the intermediate theorem here, but im not given any interval, and i thought perhaps using the definition of conintuity (at least one of them) lim (f(x+h)-f(x))=0 as h appraoches zero, but it didnt get me anywhere.

    about the third question, here what i did:
    if we put g(x)=e^f(x) then we have this equality:
    g(xy)=g(x)g(y) obviously we have here a function of the form x^a, but how do i prove it, i started using the defintion of derivative but this also havent got me anywhere.

    your help is appreciated.
     
  2. jcsd
  3. Sep 26, 2006 #2

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    For the second, you can easily show that for integers y, f(y)=f(1)^y. You can then extend this to rational numbers, and finally to all real numbers by continuity.

    For the third, try taking a derivative of the equation f(xy)=f(x)+f(y).
     
  4. Sep 26, 2006 #3
    let me see, if i understand:
    f(x)=f(x)*f(0)
    if f(x) doenst equal 0 then f(0)=1
    f(1)=f(f(0))
    f(2)=f(1+1)=f(1)f(1)=f(1)^2
    f(3)=f(2)f(1)=f(1)^3
    i.e by induction we have f(x)=f(1)^x
    and then i extend it as you said to the reals, but shouldn't i be showing that f(1)=e^a to finalise the proof?

    for the third if i take the derivative i get:
    f'(xy)=f'(x)+f'(y)
    after that i should use the fact that g(xy)=e^f(xy) that g'(xy)=f'(xy)e^f(xy)=(f'(x)+f'(y))e^(f(x)+f(y)) but is this enough to show that g(x)=x^a, i feel that something is missing?
     
  5. Sep 26, 2006 #4

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    For the first, e^a can be made to be any number by the right choice of a (or if you want to restrict f(x) to be real, argue why f(1) needs to be positive). For the second, you need to differentiate with respect to either x or y.
     
    Last edited: Sep 26, 2006
  6. Sep 26, 2006 #5
    so i should take either x or y as a constant and the other as a variable:
    so it should be yf'(xy)=f'(x)?
    and yg'(xy)=(f'(x)/y)*e^f(xy), but how from here i get that g(x)=x^a?

    p.s
    im not sure my differentiation is correct, i differntiated wrt x.
     
  7. Sep 26, 2006 #6

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    Forget g! Play around with that first equation in your last post. I can't give you any more hints without giving it away.
     
  8. Sep 27, 2006 #7
    just some queries, for the second question in order to prove that f(1) i have assumed that it's negative and got a contradiction to the fact that it's continuous, is it the right idea here?

    for the third question i got this far:
    af'(ax)=f'(x)
    so if we divide by f(ax) when it's different than 0, then af'(ax)/f(ax)=f'(x)/f(x) the lhs is (ln(f(ax)))', i tried to integrate this equation, but i didnt succeded in arrived at the equation i need to derive.
     
  9. Sep 27, 2006 #8

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    f(1) could be negative and f could be continuous if f is complex, so I don't know how this would work. You can show what you've done, but there's a simpler proof that doesn't use continuity.

    For the third, remember you're trying to show f(x)=a*log(x). What should f'(x) look like?
     
  10. Sep 28, 2006 #9
    it should be a/x, so i should divide by x?
    af'(ax)/x=f'(x)/x
    (aln(x))'f'(ax)=f'(x)/x
    (aln(x))'=f'(x)/f'(ax)x
    now i need to show that f'(x)/xf'(ax)=f'(x), how do i do that?

    btw, for the other question, what is the other approach?
    im not sure if my approach is correct but i assumed that f(1)<0 and thus because f(2)=f(1)^2>0 then for a suitable d, we have |1-0|<d |f(1)-f(0)|<e and for d' |2-1|<d' |f(2)-f(1)|<e', so we have for s=min(d,d') that |2-0|<s but because s is the smaller between them then it could be less than 2.
    but as i said i dont think it's correct.
     
  11. Sep 28, 2006 #10

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    Your problem is with f'(ax). Differential equations usually don't involve things like y'(ax), only functions of single variables, like y'(x) (or y'(a) (hint)).

    ?? How could you write this and then keep going for another paragraph? Just modify this slightly.
     
    Last edited: Sep 28, 2006
  12. Sep 28, 2006 #11
    about your last remark, obviously the function isn't monotone if f(1)<0 but does it mean that it's not continuous? i can't find in my book a theorem that states so.

    anyway, for the other question, a is a constant so we are still dealing here with one variable,x.
    for x=1 we have f'(a)=f'(1)/a but how do i use it on af'(ax)/x=f'(x)/x?
     
  13. Sep 28, 2006 #12

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    You just showed f(2)>0 without using any assumptions of continuity, monotonicity, etc. Can you modify this to show f(1)>0?

    Who says a is a constant? That equation is true for all a, so just integrate it with respect to a to get f(a), and thus f(x). (note a here is not the same as a in a*log(x). I think you plugged in this a originally for a different reason, but go back and see that you can replace a with x, y, or whatever letter you want.)
     
  14. Sep 28, 2006 #13
    f(2)=f(1)^2
    f(1)=f(2)/f(1)
    obviously we have here a geometric sequence, but i dont see how can i conclude that f(1) must be positive.
    if f(1)<0 then f(1)<f(0)=1
    f(f(1))<f(1)<f(0)
    f(f(1))<f(2)/f(1)<f(0)
    f(1)<f(2)<f(1)f(f(1))=f(1+f(1))=f(1)^(1+f(1))
    but i dont how to procceed from here.
     
  15. Sep 29, 2006 #14

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    You're overthinking this. f(1)=f(1/2+1/2)=f(1/2)^2>0.
     
  16. Sep 29, 2006 #15
    foolish me, forgot about this.
    thanks, status.
     
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