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Homogeneous nonlinear function

  1. Jul 7, 2007 #1
    Could you give me an example of a function that satisfies scalar multiplication but not addition?

    more specifically, F: R[tex]^2[/tex] -> R such that F(av)=a F(v) but F(v1 + v2) != F(v1) + F(v2)

    The best thing I could come up with is F(x,y)= |x| . This obviously does not satisfy additivity, but satisfies sc. mul. for only a => 0.

    Help please.
     
  2. jcsd
  3. Jul 7, 2007 #2
    f(v) = 2, or any other constant function.
     
  4. Jul 7, 2007 #3
    This doesn't satisfy a multiplicative property like the original poster describes.

    If we have f(v)=2

    then af(v)=2a, but f(av)=2, so this is only true for a=1.
     
  5. Jul 7, 2007 #4

    matt grime

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    It is very important that you have R^2 mapping to R. Let's think of an example that fails, again. Suppose f is a simple poly in two variables, like f(x,y)=x^2+y^2, then f(kx,ky)=k^2f(x,y), so that won't do... but can we make it better? Next guess would be to take g(x) as sqrt(f(x)), but no, that gives |k|, not k. So can you think of a better examle? I don't know where you got the title of the thread from, but you should have thought about that.
     
  6. Jul 7, 2007 #5
    For one, we can't take sqrt because then F could send things to C.
    You said R^2 to R is significant...I've been playing around with x and y, but every nonlinear function seems to break both sc.mul and addition.
     
  7. Jul 7, 2007 #6
    I don't think that that was exactly what matt was getting at, if you work off of what he showed you should be able to come up with something that works with the property you want, I can think of a few that would work all based on the idea matt was trying to convey.
     
  8. Jul 7, 2007 #7
    Oh, I see. I found one!
    Thanks for the guidance. Took me hours, and would have taken hours more without your help.
     
  9. Jul 8, 2007 #8
    Here is one.

    f(x,y) = sqrt( x^2 + y^2) is one example. You can find several such functions.
     
  10. Jul 8, 2007 #9

    Dick

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    This function fails for a<0. Try again.
     
  11. Jul 8, 2007 #10
    No it does not. sqrt( x^2 + y^2) for (ax, ay) shall be

    (+-)a * sqrt(x^2 + y^2).

    For a < 0, you can find -a* sqrt(x^2+y^2) that satisfies the property

    f(av) = a*f(v).
     
    Last edited: Jul 9, 2007
  12. Jul 8, 2007 #11
    Alternatively, one can also have the function (x^3 + y^3) ^ (1/3).
     
  13. Jul 9, 2007 #12
    Just like cryptic26 did, put a square root to the whole function or in exponential form.

    F : R2 -> R

    suppose that R = f(x,y,z) = x3 + 5yz
    therefore, F : R2 = (x3 + 5yz)2
    so to satisfy our conditions, F(x,y,z) = ((x3 + 5yz)2)1/2

    or more generally
    F(g(x)) = R2 -> R = [F(g(x))]1/2

    I wonder if my notations and the way i wrote them is correct. :grumpy:
     
  14. Jul 9, 2007 #13

    matt grime

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    You are wrong. The square root operator always returns the positive square root. If you wish to artifically assert an 'if a>0 then, else if x<0 then...' fine, but what you wrote is still wrong.
     
  15. Jul 9, 2007 #14

    matt grime

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    this is acceptable and precisely the example i had in mind (which makes me wonder did you bother to read the posts ahead of you that actually answered the OPs questions?).

    thee cube root of a^3 is always a - the choice of principal branch takes care of that automatically.
     
  16. Jul 9, 2007 #15

    matt grime

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    I don't think you've understood the question at all. R^2 means the real plane. Not the square of a function R.
     
  17. Jul 9, 2007 #16
    You seem to be confused. Why would the square root operator return only the positive square root? Besides, x can take any value (both negative an positive). There are two solutions to that equation. It is not necessary for the function to be injective.
     
  18. Jul 9, 2007 #17

    Dick

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    A relation that associates multiple values with a given (x,y) is just that. A 'relation'. Not a function. +/-sqrt(x^2+y^2) defines TWO functions. Neither of which works for all a. Your example (x^3+y^3)^(1/3) works MUCH BETTER. Why not just go with that?
     
  19. Jul 9, 2007 #18
    I understand that (x^3+y^3)^(1/3) is a better example. But, even that can have multiple roots (only that in the real plane, there shall be just one root). If need to write for the exam, then I would chose this particular one. Thanks for your explanation.
     
  20. Jul 9, 2007 #19

    matt grime

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    1. the square root function by definition only returns one value (the principal root, usually).

    2. A function cannot return two possible values - that ceases to be a function

    The symbol [itex]\sqrt{y}[/itex] does not mean 'both solutions to the equation x^2=y'. It means the principal value, which for real numbers means the positive square root.

    your attempt at the function is not well defined: f(x,y)=sqrt(x^2+y^2) and then if we multiply (x,y) by a, we have to pull the correct square root out. This would require you to specify infinitely many things in a consistent way. For instance what of (3,-4) and (-3,4)? sqrt(x^2+y^2)=5 in both cases. Now which one do I choose to be 5 and which -5? Now what about (5,-12) and (-5,12)? Which of those do I choose to be 13 and which -13?
     
    Last edited: Jul 9, 2007
  21. Jul 9, 2007 #20

    matt grime

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    For the umpteenth time: x^{1/3} is a unique number, it is not a multiple choice. The 27^{1/3}=3, (-27)^{1/3}=-3. That is the definition of those symbols: x^1/3 does not mean the set of cube roots of x. It means _the_ value on the principal branch.
     
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