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Homework Help: Multivariate inverses

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    If I am asked to find the inverse functions of two functions:

    a=f(x, y)
    b=g(x,y)

    Does this mean find the two equations:

    x=m(a, b)
    y=n(a,b)

    If so, how do m, and n compare to f and g? Inverses of single variables mirror their inverse on a graph. How does this work with multiple variables?

    Thanks!
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 15, 2010 #2

    Mark44

    Staff: Mentor

    Since f and g map R2 to R, their inverses would have to map R to R2. IOW, you would have (x, y) = f-1(a). Same for g-1.

    EDIT: On second thought, I don't think it even makes sense to talk about the inverse of a function from R2 to R.
     
    Last edited: Feb 15, 2010
  4. Feb 15, 2010 #3
    Hmm, not sure exactly what you mean.

    I'm trying to get the inverse functions of the first two functions of x and y, f and g, which my text states to simply manipulate the equations and solve for x and y as functions of a and b.

    Supposedly these new equations are the inverse functions.

    Is this correct?

    I'm also wondering, when dealing with a single variable function, we end up interchanging variables which is something not done above. Why are they inverses then?
     
  5. Feb 15, 2010 #4

    Mark44

    Staff: Mentor

    How about giving us the exact problem?

    BTW, the whole business of switching variable names is highly overrated, IMO. If you have a one-to-one function f, such that y = f(x), then the inverse relationship is x = f-1(y). The graphs of y = f(x) and x = f-1(y) are exactly the same.

    For example, log functions in a particular base and exponential functions in the same base are related as described above. If y = log x, then x = 10y, and both these equations have the same graph. The two equations are merely two ways of saying the exact same thing. The business of switching variables so that you get two different graphs obscures what is IMO the most important concept of functions and their inverses -- providing a way to take an equation in the form y = f(x) and solve it for x in terms of y.
     
    Last edited: Feb 15, 2010
  6. Feb 15, 2010 #5
    I'm asked to find the inverses of:

    u=f(x, y)=excos y
    v=g(x,y)= exsin y

    The inverses should be:

    x=f-1(u, v)
    y=g-1(u, v)

    So far I've gotten:

    y = arctan (v/u)

    x = Ln[tex]\frac{u}{cos (arctan \frac{v}{u})}[/tex]

    I'll be taking the Jacobian of this, so the more I can simplify it the better.
     
    Last edited: Feb 15, 2010
  7. Feb 15, 2010 #6

    vela

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    Try adding the squares of u and v together to solve for x. You'll get a nicer expression for x.
     
  8. Feb 15, 2010 #7
    Hi Vela,

    I'm an idiot though, it was sin/cos y as updated above. Thanks though.
     
  9. Feb 15, 2010 #8

    vela

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    Yeah, I figured as much, but my hint still works.
     
  10. Feb 15, 2010 #9
    Wow, thanks so much for that!
     
  11. Feb 15, 2010 #10

    vela

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    It's better to think of it as

    [tex]\begin{pmatrix}u\\v\end{pmatrix}=f\begin{pmatrix}x\\y\end{pmatrix}[/tex]

    and

    [tex]\begin{pmatrix}x\\y\end{pmatrix}=f^{-1}\begin{pmatrix}u\\v\end{pmatrix}[/tex]

    As Mark44 noted above, with the individual functions, you can't find inverses because they're not one-to-one. Generally, curves in R2 will map to a single point in R, so the inverse relation isn't a function.
     
  12. Feb 15, 2010 #11
    I'm not sure I follow... Is that combinatorial notation?
     
  13. Feb 15, 2010 #12

    Mark44

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    No, it's vector notation.
     
  14. Feb 15, 2010 #13

    vela

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    Yeah, it's vector notation. You can think of it as one mapping from R2 to R2, rather than two individual mappings from R2 to R. The mapping you have for this problem, for instance, maps vertical lines in the xy-plane to circles in the uv-plane.
     
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