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Does this function belong to an interesting class of functions?

by PhilDSP
Tags: belong, class, function, functions, interesting
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PhilDSP
#1
Dec2-11, 06:55 AM
P: 606
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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Mute
#2
Dec2-11, 07:45 AM
HW Helper
P: 1,391
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).
PhilDSP
#3
Dec2-11, 08:52 AM
P: 606
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.

Stephen Tashi
#4
Dec2-11, 10:06 AM
Sci Advisor
P: 3,256
Does this function belong to an interesting class of functions?

Quote 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.
HallsofIvy
#5
Dec2-11, 02:04 PM
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Emeritus
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Thanks
PF Gold
P: 39,357
Quote 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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