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Math Help - finding local inverse of f: R^2->R^2

  1. #1
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    finding local inverse of f: R^2->R^2

    Could some kind soul look this over to see if I made any mistakes?
    Appreciate your comments on my problem areas (highlighted). thanks!


    Question.


    A function f: R^2->R^2 is defined by

    f(x,y)=(u,v)^T=(x^2+y, 2x-y)^T.

    Write down, where possible, the formula for the local inverse function at the points (x,y)=(1,0) and (-1,0), justifying your answers.

    For the local inverse function at one of the points above sketch the region in the (u,v)-plane on which it is defined and its image on (x,y)-plane. Find the derivative of the local inverse function at the point you have chosen previously.



    Answer.

    First, I'll find the critical points where the local inverse does not exist: det(J)=0

    |2x 1|
    |2 -1| is the Jacobian

    It's determinant det(J)=-2x-2=-2(x+1)=0 iff x=-1 which is a set of critical points, actually, a line: it divides the (x,y)-plane into two parts: J>0 to the left of the line x=-1, J<0 to the right of it.

    (does the Jacobian's sign have any meaning or use in the context of finding a local inverse?)


    So, of the two given points (1,0) and (-1,0), the local inverse only exists at (x,y)=(1,0).

    To find local inverse at this point, I will solve the equations

    u=x^2+y,v=2x-y for x and y.

    x=-1\pm\sqrt{1+u+v}

    I want x>0, therefore, x=-1+\sqrt{1+u+v}

    y=2x=2(-1+\sqrt{1+u+v})

    Therefore, the local inverse of f at (1,0) is described by the above equations and the point on the inverse correspoding to (x,y)=(1,0) is (u,v)=(1,2) (by using the definition of f).


    Next, the region on (u,v) where the local inverse is defined is limited by
    \sqrt{1+u+v}>1, 1+u+v>1, u+v>0, u>-v (since x>0) and I would draw a 45-degree line u=-v across II and IV quadrant of the (u,v)-plane, and the relevant region is the area above that line.

    I am not sure the above region is correct.


    For sketching coordinates of the inverse on (x,y)-plane, these are:y=1-x^2 (downward parabola crossing Y axis at y=1) and line y=2x-2. These two touch at a point (-1,0) - why is that? something to do with critical point or just a coincidence?


    Finally, the derivative of the inverse is the inverted Jacobian matrix calculated at the point (x,y)=(1,0) which I got to

    |1/4 1/4|
    |1/2 1/2|

    Thank you for your time!
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  2. #2
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    Quote Originally Posted by Volga View Post
    Could some kind soul look this over to see if I made any mistakes?
    Appreciate your comments on my problem areas (highlighted). thanks!


    Question.


    A function f: R^2->R^2 is defined by

    f(x,y)=(u,v)^T=(x^2+y, 2x-y)^T.

    Write down, where possible, the formula for the local inverse function at the points (x,y)=(1,0) and (-1,0), justifying your answers.

    For the local inverse function at one of the points above sketch the region in the (u,v)-plane on which it is defined and its image on (x,y)-plane. Find the derivative of the local inverse function at the point you have chosen previously.



    Answer.

    First, I'll find the critical points where the local inverse does not exist: det(J)=0

    |2x 1|
    |2 -1| is the Jacobian

    It's determinant det(J)=-2x-2=-2(x+1)=0 iff x=-1 which is a set of critical points, actually, a line: it divides the (x,y)-plane into two parts: J>0 to the left of the line x=-1, J<0 to the right of it.

    (does the Jacobian's sign have any meaning or use in the context of finding a local inverse?)
    If the sign is positive, the mapping can be achieved by a continuous deformation of the space (it is an isotopy, in the language of geometric topology). If it is negative, then the mapping involves a reflection in some sense.

    So, of the two given points (1,0) and (-1,0), the local inverse only exists at (x,y)=(1,0).

    To find local inverse at this point, I will solve the equations

    u=x^2+y,v=2x-y for x and y.

    x=-1\pm\sqrt{1+u+v}

    I want x>0, therefore, x=-1+\sqrt{1+u+v}

    y=2x=2(-1+\sqrt{1+u+v}) Shouldn't that be y = 2x – v ?

    Therefore, the local inverse of f at (1,0) is described by the above equations and the point on the inverse correspoding to (x,y)=(1,0) is (u,v)=(1,2) (by using the definition of f).


    Next, the region on (u,v) where the local inverse is defined is limited by
    \sqrt{1+u+v}>1, 1+u+v>1, u+v>0, u>-v (since x>0) and I would draw a 45-degree line u=-v across II and IV quadrant of the (u,v)-plane, and the relevant region is the area above that line.

    I am not sure the above region is correct.
    No. The condition that the square root should be >1 was needed to ensure that you used the branch of the inverse function that would give the correct inverse at the point (1,0). But for that branch of the inverse function to be defined, it is only necessary for the weaker condition 1+u+v > 0 to hold.

    For sketching coordinates of the inverse on (x,y)-plane, these are:y=1-x^2 (downward parabola crossing Y axis at y=1) and line y=2x-2. These two touch at a point (-1,0) - why is that? something to do with critical point or just a coincidence?
    If you take the domain of the inverse to be given by the condition 1+u+v > 0, I think you will find that it corresponds to the region x > –1 in the (x,y)-plane (which is what you might expect it to be).

    Finally, the derivative of the inverse is the inverted Jacobian matrix calculated at the point (x,y)=(1,0) which I got to

    |1/4 1/4|
    |1/2 1/2|

    Thank you for your time!
    ..
    Last edited by Opalg; March 28th 2011 at 04:20 AM.
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  3. #3
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    Thank you so much!

    Sorry, silly mistake in determining y...

    And I am much clearer now about the 'interval at which the inverse is defined'. As you can see, I was confused between requiring x>0 and requiring the expression under square root sign to be positive. So, to take another example, if the inverse is say sin^{-1}(x) then I would then require -1\leq{}x\leq{-1}, right?

    If you take the domain of the inverse to be given by the condition 1+u+v > 0, I think you will find that it corresponds to the region x > –1 in the (x,y)-plane (which is what you might expect it to be).
    So, I was thinking - why I would expect it to be there - and I couldn't see a simple answer:
    - the Jacobian is zero at x=-1, that brings about change of way of mapping (as per your comment above)...
    - if I look at gradient vectors of u and v, they are parallel at x=-1...
    - also, J<0 when x>-1...

    or am I overthinking and cannot see the obvious?
    Last edited by Volga; March 28th 2011 at 03:03 AM.
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  4. #4
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    Quote Originally Posted by Volga View Post
    So, I was thinking - why I would expect it to be there - and I couldn't see a simple answer:
    - the Jacobian is zero at x=-1, that brings about change of way of mapping (as per your comment above)...
    Yes, exactly that. The line x = 1 is where the Jacobian is zero, so that is where you ought to expect the inverse to go wrong in some way.
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