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Extensive functions

  1. Jan 10, 2012 #1
    Are logarithms the only functions for which f(xy) = f(x) + f(y)?
     
  2. jcsd
  3. Jan 10, 2012 #2

    CompuChip

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    Depending on your definition of the logarithm, yes.

    I managed to show for example, that [itex]f'(x) \propto \frac{1}{x}[/itex].

    The trick was to look at
    [tex]\frac{\mathrm d}{\mathrm dx} f(x) = \lim_{h \downarrow 0} \frac{f(x + h)}{h}[/tex]
    where
    [tex]f(x + h) = f(x(1 + h/x)) = f(x) + f(1 + h/x)[/tex]

    You can throw in an expansion:
    [tex]f\left(1 + \frac{h}{x}\right) = f(1) + c \frac{h}{x} + \mathcal O(h^2)[/tex]
    The second order terms in h divided by h will be O(h) and will vanish. You can show that f(1) = 0 by looking at f(x) = f(1x) = f(1) + f(x).

    Combining everything will give you
    [tex]\frac{\mathrm d}{\mathrm dx} f(x) = c \cdot \frac{1}{x}[/tex]

    Now it follows that f(x) must be a logarithm, because you can appropriately choose c to get any [itex]\log_a{x} = c \ln x[/itex] (c = 1/ln(a)).

    If you prefer the definition that ln is the function satisfying [itex]e^{\ln x} = x[/itex] with ln(1) = 0, you can use the chain rule to show that this is equivalent to the solution of the differential equation.
     
  4. Jan 10, 2012 #3
    Isn't this a much simpler way to prove that [itex]f'(x) \propto \dfrac{1}{x}[/itex]?
    If you suppose [itex]f[/itex] differentiable on its domain, and if you derivate the equation with respect to y, you get that f satisfies:
    [tex]\forall x, y \in \mathbb R, \hspace{10pt} xf'(xy) = f'(y)[/tex]
    By chosing y = 1:
    [tex]\forall x \in \mathbb R, \hspace{10pt} xf'(x) = f'(1)[/tex]
    Thus, forall nonzero x:
    [tex]f'(x) = \dfrac{f'(1)}{x}[/tex]
     
  5. Jan 10, 2012 #4

    Stephen Tashi

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    Those answers assume differntiability. Perhaps the technical answer to the question:
    is No unless you add some condition like continuity or differentiability.
     
  6. Jan 11, 2012 #5

    morphism

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    Indeed. Consider the function g(x)=f(e^x). It satisfies the functional equation g(x+y) = g(x) + g(y) (i.e., g is additive) and determines f on the positive reals, hence on all the reals (why?). It's a pretty well-known fact that simple additional requirements, such as continuity, will force an additive function to be linear (i.e. to be of the form g(x)=cx for some constant c). However, in the absence of such requirements, one can in fact find http://planetmath.org/encyclopedia/ThereExistAdditiveFunctionsWhichAreNotLinear.html [Broken]. These functions are fairly pathological. In particular, they aren't continuous. So, if we take one such example for our g, then the resulting f won't be a logarithm.
     
    Last edited by a moderator: May 5, 2017
  7. Jan 11, 2012 #6

    Stephen Tashi

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    I just noticed that the differentiable constant function f(x) = 0 also works.
     
  8. Jan 11, 2012 #7

    morphism

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    Presumably by "logarithms" the OP meant functions of the form f(x) = c log|x| for some constant c. (This is at least consistent with CompuChip and sachav's posts.) You get the zero function when you take c=0.

    Edit: It should be pointed out that one must exclude 0 from the domain of f in order to get anything interesting. If you insist on defining f at 0, then f(0) has to be zero (because f(0)=f(0*0)=f(0)+f(0)), and from this it follows that f(x)=0 for all x. (Proof: f(x)=f(x)+f(0)=f(x*0)=f(0)=0.)
     
    Last edited: Jan 11, 2012
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