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Does this function belong to an interesting class of functions?

 
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Dec2-11, 06:55 AM   #1
 

Does this function belong to an interesting class of functions?


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?
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Dec2-11, 07:45 AM   #2
 
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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).
Dec2-11, 08:52 AM   #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.
Dec2-11, 10:06 AM   #4
 
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Does this function belong to an interesting class of functions?


Quote by PhilDSP View Post
We could argue that the value becomes 1 when [itex]\phi = 1[/itex] couldn't we?
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.
Dec2-11, 02:04 PM   #5
 
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Quote by PhilDSP View Post
We could argue that the value becomes 1 when [itex]\phi = 1[/itex] couldn't we?
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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