1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook