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Question on isomorphism between addition and multiplication

  1. Nov 27, 2012 #1
    Hello,
    I want to find a family of functions [itex]\phi:\mathbb{R} \rightarrow \mathbb{C}[/itex] that have the property: [tex]\phi(x+y)=\phi(x)\phi(y)[/tex] where [itex]x,y\in \mathbb{R}[/itex].

    I know that any exponential function of the kind [itex]\phi(x)=a^x[/itex] with [itex]a\in\mathbb{C}[/itex] will satisfy this property.
    Is this the only choice, or are there other functions that I am missing that satisfy the above property?
     
  2. jcsd
  3. Nov 27, 2012 #2

    lavinia

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    Try playing around with the formula to answer this question for yourself.

    For instance the formula shows that [itex]\phi[/itex](0) = 1
     
  4. Nov 27, 2012 #3

    tiny-tim

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    hello mnb96! :smile:

    various ways, eg put ##\psi = ln\phi##, or what is ##\phi '(x+y)## ? :wink:
     
  5. Nov 27, 2012 #4

    micromass

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    Some remarks:

    1) I'm not sure how you define [itex]a^x[/itex] for [itex]a\in \mathbb{C}[/itex]. You got to be careful, because those exponents usually take on multiple values and you need to choose the correct one.

    2) You might want to add as an hypothesis that [itex]\varphi[/itex] is continuous. In that case, you will indeed be able to prove what you want. If [itex]\varphi[/itex] is not continuous, then there might be other functions which satisfy the equation, and those functions are very ill-behaved.
     
  6. Nov 27, 2012 #5

    lavinia

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    In the case of a real rather than complex valued function I would guess that the rule forces the function to be continuous.

    - the rule implies that f(0) = 1 and f(x) >0

    - the rule says that f(x) = f(x/n)^n so f(x/n) must approach 1 as n grows large. This indicates (but doesn't prove)continuity at zero. But if it is continuous at zero it is everyehere. If not, it is discontinuous everywhere.
     
  7. Nov 27, 2012 #6

    micromass

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    Well, here is a counterexample:
    We know that [itex]\mathbb{R}[/itex] is a [itex]\mathbb{Q}[/itex]-vector space, so it has an (infinite) basis E. Take a particular [itex]e\in E[/itex].
    Any element [itex]z\in \mathbb{R}[/itex] can be written as the finite sum

    [tex]z=\sum_{x\in E} \alpha_x x[/tex]

    Now define [itex]g(z)=\alpha_ee[/itex]. Then [itex]g:\mathbb{R}\rightarrow\mathbb{R}[/itex] satisfies [itex]g(x+y)=g(x)+g(y)[/itex] for all reals x and y. But it is not [itex]\mathbb{R}[/itex]-linear and thus not continuous.

    Now define [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] as [itex]f(z)=2^{g(z)}[/itex]. Then this satsifies [itex]f(x+y)=f(x)f(y)[/itex] but it is not continuous.
     
  8. Nov 27, 2012 #7

    lavinia

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    cool.

    so exponentiate any Q but not R linear map of the reals to the reals.

    So... the sequence x/n will have a constant coefficient divided by n with respect the the basis vector so that's why the function looks continuous on it.

    And this means that there is a number with a coefficient bounded away from zero in any interval around zero.
     
    Last edited: Nov 27, 2012
  9. Nov 27, 2012 #8
    Mmm...:confused: ... I am a bit confused.
    Let's stick for now with the case [itex]\phi:\mathbb{R} \rightarrow \mathbb{R}[/itex].
    Assuming [itex]\phi[/itex] is an isomorphism between (ℝ,+) and (ℝ+,×) that satisfies the property [itex]\phi(x+y)=\phi(x)\phi(y)[/itex], and that is continuous, we can say that:

    1) [itex]\phi(0)=\phi(x-x)=\phi(x)\phi(-x)[/itex] for all [itex]x\in \mathbb{R}[/itex], thus [itex]\phi(0)=1[/itex]

    2) from 1) we have that [itex]\phi(-x)=\frac{1}{\phi(x)}[/itex]

    3) [itex]\phi(x)=\phi(x/2+x/2)=\phi(x/2)^2 > 0[/itex], thus [itex]\phi(x)>0[/itex]

    4) [itex]\phi[/itex] must be bijective, thus [itex]\phi'(x)>0[/itex]

    5) [itex]\phi(n) = \phi(1+1+\ldots+1)=\phi(1)^n[/itex] for all [itex]n\in \mathbb{Z}[/itex]

    In conclusion [itex]\phi[/itex] must be a continuous positive monotonic increasing function passing through the point (0,1) and through the points [itex](n, \phi(1)^n)[/itex]. It seems clear that the only possibility is to choose [itex]\phi(x)=\phi(1)^x=a^x[/itex], although I don't know how to put it rigorously.

    Now the problem is, what if [itex]\phi:\mathbb{R}\rightarrow\mathbb{C}[/itex] maps the reals to a subset of the complex numbers?
     
    Last edited: Nov 27, 2012
  10. Nov 27, 2012 #9

    lavinia

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    [itex]\phi[/itex]( 0 + x) = [itex]\phi[/itex](0)[itex]\phi[/itex](x) so [itex]\phi(0)=1[/itex]

    you don't know that [itex]\phi[/itex] is differentiable just because it is continuous or bijective.


    You need to go from continuity to the conclusion
     
  11. Nov 28, 2012 #10

    tiny-tim

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    if ##\phi## is differentiable, there's a very quick proof

    i suspect that that proof can be adapted to the merely continuous case (but i haven't tried)
     
  12. Nov 28, 2012 #11
    Ok. But wouldn't it be possible to define a continuous (and monotonic increasing function) that still passes through the points [itex](n,a^n)[/itex] but does "strange things" between [itex](n,a^n)[/itex] and [itex](n+1,a^{n+1})[/itex] ?
     
  13. Nov 28, 2012 #12

    lavinia

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    The function e[itex]^{-x}[/itex] is bijective from the reals to the positive reals but its derivative is always negative.
     
  14. Nov 28, 2012 #13

    lavinia

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    a continuous monotonically increasing function may not be everywhere differentiable although it seems right that it can only have a discrete set of kinks. See if you can find some examples.
     
  15. Nov 28, 2012 #14

    lavinia

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    ln[itex]\phi[/itex] is linear over the rational numbers. If is is continuous it follows that it is linear over the reals. Easy proof.

    A continuous linear map of the reals to the reals is multiplication by a constant.
     
  16. Nov 28, 2012 #15

    lavinia

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    As micromass said for the complex case one need to wonder whether one can define a single branch of the logarithm on the values of [itex]\phi[/itex].

    If ln[itex]\phi[/itex] is single valued then its projections onto the x and y axis are linear.
     
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