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Does this function belong to an interesting class of functions?

  1. Dec 2, 2011 #1
    Hello and thanks for your consideration,

    I'd like some insight into the function [itex]f(\phi) = \frac {1 - \phi}{\phi - 1}[/itex]

    Does this apply to any known modeling situations? Is it recognized as belonging to a more general class of functions that may have interesting or unique characteristics? Or can the function be transformed into a function that does?
     
  2. jcsd
  3. Dec 2, 2011 #2

    Mute

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    Your function, as written, is just equal to -1, except when [itex]\phi = 1[/itex], where there is a discontinuity because the denominator vanishes there.

    If you want an example of a function that has a similar form but isn't trivially some constant and has some applications, see Mobius transformation. (But note that the Mobius transformation is usually used with complex numbers. I don't know if it is used much in real number applications).
     
    Last edited: Dec 2, 2011
  4. Dec 2, 2011 #3
    Thanks, an association with the Mobius transformation does yield many interesting things to think about, especially since [itex]\phi[/itex] can be complex in the situation where the function popped up.

    We could argue that the value becomes 1 when [itex]\phi = 1[/itex] couldn't we? This almost sounds like a spinning sphere where the axis must be aligned parallel to a force acting on the sphere, but which can suddenly undergo a spin flip.
     
  5. Dec 2, 2011 #4

    Stephen Tashi

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    No, you can't define a function one way and then argue that it has a different definition. You can, however, define a function that is 1 when [itex] \phi = 1 [/itex] and equal to -1 elsewhere. You can argue that this definition applies to a certain practical situation. That would be an argument about physics.
     
  6. Dec 2, 2011 #5

    HallsofIvy

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    No. "f(x)= -1" and "g(x)= -1 if [itex]x\ne 1[/itex], and is not defined at [itex]x= 1[/itex]" are two different functions.
     
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