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Differential of the exp function prove

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f: R->R be a function which satisfies the condition:
    [tex]f(x+y) = f(x) \cdot f(y)[/tex]
    [tex]\forall x,y \in R[/tex]

    a)
    Show that if f is everywhere differntial it satisfies the equation:
    [tex]f'(x) = f'(0) \cdot f(x)[/tex]

    Conclude that either f is the zero function or there exists a number c so that
    [tex]f(x) = exp(c \cdot x)[/tex]
    [tex]\forall x \in R[/tex]


    b)
    Show that if f is differntial in 0, then f is the zero function or there exists a number c so that:
    [tex]f(x) = exp(c \cdot x)[/tex]
    [tex]\forall x \in R[/tex]


    2. Relevant equations
    [tex]\frac{d}{dx} ln|f(x)| = \frac{f'(x)}{f(x)}[/tex]
    [tex]\frac{(f \cdot g)'}{f \cdot g} = \frac{f'}{f} + \frac{g'}{g}[/tex]

    As well as we might need some ln and exp rules, the most important one being:
    [tex] a^{x} = exp(x \cdot ln(a))[/tex]


    3. The attempt at a solution
    Here's what i've done so far:

    If we diffrientiate the ln (natural log) both sides we get.

    Left side:
    [tex](ln|f(x+y)|)' = \frac{f'(x+y)}{f(x+y)}[/tex]

    Right side:
    [tex](ln|f(x) \cdot f(y)|)' = \frac{f'(x)}{f(x)} + \frac{f'(y)}{f(y)}[/tex]

    Next, I isolated f'(x) but I couldn't see that new equation helping me anything.

    Could someone please point me in the right direction, I've really no idea how to approach this problem.


    Thanks in advance,
    Sennap
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2
    Should this be [tex]f(x+y) = f(x) f(y)[/tex]?
     
  4. Nov 9, 2009 #3
    Yes, sorry about that. It's fixed now.

    - Sennap
     
  5. Nov 9, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    How about using the definition of derivative?
    [tex]f'(x)= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}= \lim_{h\to 0}\frac{f(x)f(h)- f(x)}{h}= f(x)lim_{h\to 0}\frac{f(h)- 1}{h}[/tex]
    The fact that f is differentiable tells you that that limit exists. But
    [tex]f'(0)= \lim_{h\to 0}\frac{f(h)- f(0)}{h}[/tex]
    and f(x+0)= f(x)f(0) tells you that f(0)= 1.
     
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