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Math Help - [SOLVED] Jacobian derivative matrix

  1. #1
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    [SOLVED] Jacobian derivative matrix

    Given that G(x,y) = (x^2+1, y^2) and F(u,v) = (u + v, v^2). compute the Jacobian derivative matrix of F(G(x,y)) at the point (x,y) = (1,1).

    A hint to get started will suffice, thanks!
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  2. #2
    ynj
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    Quote Originally Posted by garymarkhov View Post
    Given that G(x,y) = (x^2+1, y^2) and F(u,v) = (u + v, v^2). compute the Jacobian derivative matrix of F(G(x,y)) at the point (x,y) = (1,1).

    A hint to get started will suffice, thanks!
    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)),where (u0,v0)=G(x,y) but you regard it as a constant vector.
    J(A) means the jacobian of A
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  3. #3
    Member garymarkhov's Avatar
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    Quote Originally Posted by ynj View Post
    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)),where (u0,v0)=G(x,y) but you regard it as a constant vector.
    J(A) means the jacobian of A
    Hmm, I'm going to need more of a push than that. I guess I really have no idea how to do this question.
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  4. #4
    ynj
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    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))is the basic theorem of Jacobian Matrix
    Let we define a series of functions
    f1(x1,x2,...xn)
    f2(x1,x2....xn)
    ...
    fm(x1,x2...,xn)
    this set of functions will be written as F,where
    F(x1,...,xn)=(y1,...ym)
    yi=fi(x1,...xn)
    then J(F) will be
    (f1'x1,f1'x2.....f1'xn
    f2'x1,f2'x2.....f2'xn
    ....
    fm'x1,fm'x2,......fm'xn)
    it is a n*m matrix
    so J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))=
    (1,0
    0,2v)
    *
    (2x,0
    0,2y)
    =(2x,0
    0,4vy)
    =(2x,0
    0,4y^3)
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  5. #5
    Member garymarkhov's Avatar
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    Quote Originally Posted by ynj View Post
    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))is the basic theorem of Jacobian Matrix
    Let we define a series of functions
    f1(x1,x2,...xn)
    f2(x1,x2....xn)
    ...
    fm(x1,x2...,xn)
    this set of functions will be written as F,where
    F(x1,...,xn)=(y1,...ym)
    yi=fi(x1,...xn)
    then J(F) will be
    (f1'x1,f1'x2.....f1'xn
    f2'x1,f2'x2.....f2'xn
    ....
    fm'x1,fm'x2,......fm'xn)
    it is a n*m matrix
    so J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))=
    (1,0
    0,2v)
    *
    (2x,0
    0,2y)
    =(2x,0
    0,4vy)
    =(2x,0
    0,4y^3)
    Hmm, my textbook gives an answer of:

    <br />
\left[\begin{array}{cc}2&2 \\ 0&4 \end{array}\right]<br />

    Can you explain the steps you took to get:

    <br />
\left[\begin{array}{cc}1&0 \\ 0&2v \end{array}\right]\left[\begin{array}{cc}2x&0 \\ 0&2y \end{array}\right]<br />
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  6. #6
    ynj
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    Quote Originally Posted by garymarkhov View Post
    Hmm, my textbook gives an answer of:

    <br />
\left[\begin{array}{cc}2&2 \\ 0&4 \end{array}\right]<br />

    Can you explain the steps you took to get:

    <br />
\left[\begin{array}{cc}1&0 \\ 0&2v \end{array}\right]\left[\begin{array}{cc}2x&0 \\ 0&2y \end{array}\right]<br />
    oh,there is a small mistake...
    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))=
    (1,1-----------mistake
    0,2v)
    *
    (2x,0
    0,2y)
    =(2x,0
    0,4vy)
    =(2x,0
    0,4y^3)
    (1,1
    0,2v)means
    ((u+v)'u,(u+v)'v
    (v^2)'u,(v^2)'v)
    (2x,0
    0,2y)means
    ((x^2+1)'x,(x^2+1)'y
    (y^2)'x,(y^2 )'y)
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  7. #7
    ynj
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    Quote Originally Posted by garymarkhov View Post
    Hmm, my textbook gives an answer of:

    <br />
\left[\begin{array}{cc}2&2 \\ 0&4 \end{array}\right]<br />

    Can you explain the steps you took to get:

    <br />
\left[\begin{array}{cc}1&0 \\ 0&2v \end{array}\right]\left[\begin{array}{cc}2x&0 \\ 0&2y \end{array}\right]<br />
    oh,there is a small mistake...
    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))=
    (1,1-----------mistake
    0,2v)
    *
    (2x,0
    0,2y)
    =(2x,2y
    0,4vy)
    =(2x,2y
    0,4y^3)
    (1,1
    0,2v)means
    ((u+v)'u,(u+v)'v
    (v^2)'u,(v^2)'v)
    (2x,0
    0,2y)means
    ((x^2+1)'x,(x^2+1)'y
    (y^2)'x,(y^2 )'y)
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  8. #8
    Member garymarkhov's Avatar
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    Quote Originally Posted by ynj View Post
    oh,there is a small mistake...
    J(F(G(x,y)))=J(F(u0,v0))J(G(x,y))=
    (1,1-----------mistake
    0,2v)
    *
    (2x,0
    0,2y)
    =(2x,2y
    0,4vy)
    =(2x,2y
    0,4y^3)
    (1,1
    0,2v)means
    ((u+v)'u,(u+v)'v
    (v^2)'u,(v^2)'v)
    (2x,0
    0,2y)means
    ((x^2+1)'x,(x^2+1)'y
    (y^2)'x,(y^2 )'y)
    Okay, I'm starting to get the hang of this. But where did you get 4y^3 from? I get 4vy in that position.
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  9. #9
    ynj
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    v=y^2
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  10. #10
    Member garymarkhov's Avatar
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    Quote Originally Posted by ynj View Post
    v=y^2
    ynj, I know I'm the densest person you've ever had the pain of corresponding with, but it would really help if you would add a short explanation to what you are writing. I can already tell that you think v=y^2 based on what you had written earlier - I want to know why!
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  11. #11
    ynj
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    Quote Originally Posted by garymarkhov View Post
    ynj, I know I'm the densest person you've ever had the pain of corresponding with, but it would really help if you would add a short explanation to what you are writing. I can already tell that you think v=y^2 based on what you had written earlier - I want to know why!
    I do not exactly know what you are confusing about....
    you may understand J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)) like
    (fg)'(x)=f'(g(x))g'(x)
    the function G actually put (x,y) to (x^2+1,y^2)
    and F put (u,v) to (u+v,v^2)
    For the format reason, we may define f(x,y)'x to be the partial deriviation of x, say the lim (f(x+dx,y)-f(x,y))/dx.

    if you want to know (FG)(x,y)'x or (FG)(x,y)'y, you may first know F(u,v)'v or F(u,v)'v or G(x,y)'x or G(x,y)'y because it will be easier.
    The Jacobian's theorem actually provides you a perfect way to solve the problem.

    Let we define a series of functions
    f1(x1,x2,...xn)
    f2(x1,x2....xn)
    ...
    fm(x1,x2...,xn)
    this set of functions will be written as F,where
    F(x1,...,xn)=(y1,...ym)
    yi=fi(x1,...xn)
    then J(F) will be
    (f1'x1,f1'x2.....f1'xn
    f2'x1,f2'x2.....f2'xn
    ....
    fm'x1,fm'x2,......fm'xn)
    it is a n*m matrix
    I hope you can understand the definition of Jacobian, although it will be a little abstract..
    To get the answer, you may first write (FG)(x,y) as F(g1(x,y),g2(x,y)), where g1(x,y)=x^2+1,g2(x,y)=y^2
    In this case, we can let u0=g1(x,y),v0=g2(x,y).
    Then we can use our equation J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)).
    The first part J(F(u0,v0)) tells you that you can ignore the existence of x,y, just looking at u0 and v0.After calculating, you just let u0=g1(x,y),v0=g2(x,y), then everything is completed.
    so J(F(u0,v0))=
    ((u0+v0)'u0,(u0+v0)'v0
    (v0^2)'u0,(v0^2)'v0)
    =(1,1
    0,2v0)
    The second part J(G(x,y)) tells you where the x,y will appear.
    so J(G(x,y))=
    ((x^2+1)'x,(x^2+1)'y
    (y^2)'x,(y^2 )'y)
    =(2x,0
    0,2y)
    Then we calculate J(F(u0,v0))J(G(x,y))
    it will be
    (2x,2y
    0,4v0y)
    then v0=g2(x,y)=y^2
    so the answer will be
    (2x,2y
    0,4y^3)
    =(2,2
    0,4)
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    Super Member Random Variable's Avatar
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     Df(u,v) = \begin{bmatrix} 1&1\\0&2v\end{bmatrix}

     Dg(x,y) = \begin{bmatrix} 2x&0\\0&2y\end{bmatrix}


    when  (x,y)=(1,1) \ , \ (u,v) = (2,1)


     D f\big(g(1,1)\big) = Df(2,1)Dg(1,1) = \begin{bmatrix} 1&1\\0&2\end{bmatrix}\begin{bmatrix} 2&0\\0&2\end{bmatrix}  = \begin{bmatrix} 2&2\\0&4\end{bmatrix}
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  13. #13
    Member garymarkhov's Avatar
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    Quote Originally Posted by ynj View Post
    I do not exactly know what you are confusing about....
    you may understand J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)) like
    (fg)'(x)=f'(g(x))g'(x)
    the function G actually put (x,y) to (x^2+1,y^2)
    and F put (u,v) to (u+v,v^2)
    For the format reason, we may define f(x,y)'x to be the partial deriviation of x, say the lim (f(x+dx,y)-f(x,y))/dx.

    if you want to know (FG)(x,y)'x or (FG)(x,y)'y, you may first know F(u,v)'v or F(u,v)'v or G(x,y)'x or G(x,y)'y because it will be easier.
    The Jacobian's theorem actually provides you a perfect way to solve the problem.

    Let we define a series of functions
    f1(x1,x2,...xn)
    f2(x1,x2....xn)
    ...
    fm(x1,x2...,xn)
    this set of functions will be written as F,where
    F(x1,...,xn)=(y1,...ym)
    yi=fi(x1,...xn)
    then J(F) will be
    (f1'x1,f1'x2.....f1'xn
    f2'x1,f2'x2.....f2'xn
    ....
    fm'x1,fm'x2,......fm'xn)
    it is a n*m matrix
    I hope you can understand the definition of Jacobian, although it will be a little abstract..
    To get the answer, you may first write (FG)(x,y) as F(g1(x,y),g2(x,y)), where g1(x,y)=x^2+1,g2(x,y)=y^2
    In this case, we can let u0=g1(x,y),v0=g2(x,y).
    Then we can use our equation J(F(G(x,y)))=J(F(u0,v0))J(G(x,y)).
    The first part J(F(u0,v0)) tells you that you can ignore the existence of x,y, just looking at u0 and v0.After calculating, you just let u0=g1(x,y),v0=g2(x,y), then everything is completed.
    so J(F(u0,v0))=
    ((u0+v0)'u0,(u0+v0)'v0
    (v0^2)'u0,(v0^2)'v0)
    =(1,1
    0,2v0)
    The second part J(G(x,y)) tells you where the x,y will appear.
    so J(G(x,y))=
    ((x^2+1)'x,(x^2+1)'y
    (y^2)'x,(y^2 )'y)
    =(2x,0
    0,2y)
    Then we calculate J(F(u0,v0))J(G(x,y))
    it will be
    (2x,2y
    0,4v0y)
    then v0=g2(x,y)=y^2
    so the answer will be
    (2x,2y
    0,4y^3)
    =(2,2
    0,4)
    That was actually extremely helpful.

    I only have this one question to learn what a Jacobian matrix is, so I'm desperate to try some other examples. Do you have some?

    Also, what is the Jacobian matrix good for? What's the intuition behind it? Judging by how little material I can find on the web, this is probably a difficult question that not many will be able to answer.
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  14. #14
    ynj
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    yeah,Jacobian Matrix is very useful especially when the function is very complex. And it will be used in some important theorems multivariable deriviation.(But I do not know its English name..)
    If you use the Jacobian, you can get several answer at the same time, and the process will be more easier and clearer.
    you may want some exercises??I have some..
    1.
    f(x,y,z)=(x^2+y+z,2x+y+z^2,0),g(u,v,w)=(uv^2w^2,(w ^2sinv),u^2e^v)
    Calculate J(F(G(u,v,w)))
    2.
    if f(x,y,z)=F(u,v,w), x^2=vw,y^2=wu,z^2=uv
    prove:
    xf'x+yf'y+zf'z=uF'u+vF'v+wF'w
    I hope these two will be helpful....
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  15. #15
    Member garymarkhov's Avatar
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    Quote Originally Posted by ynj View Post
    yeah,Jacobian Matrix is very useful especially when the function is very complex. And it will be used in some important theorems multivariable deriviation.(But I do not know its English name..)
    If you use the Jacobian, you can get several answer at the same time, and the process will be more easier and clearer.
    you may want some exercises??I have some..
    1.
    f(x,y,z)=(x^2+y+z,2x+y+z^2,0),g(u,v,w)=(uv^2w^2,(w ^2sinv),u^2e^v)
    Calculate J(F(G(u,v,w)))
    2.
    if f(x,y,z)=F(u,v,w), x^2=vw,y^2=wu,z^2=uv
    prove:
    xf'x+yf'y+zf'z=uF'u+vF'v+wF'w
    I hope these two will be helpful....

    Okay for the first problem, I get:

    <br /> <br />
\left[\begin{array}{ccc}2x&1&1 \\ 2&1&2z \\ 0&0&0 \end{array}\right]*\left[\begin{array}{ccc}v^2w^2&2uvw^2&2uv^2w \\ 0&w^2cosv&2wsinv \\ 2ue^v&u^2e^v&0 \end{array}\right] = \left[\begin{array}{ccc}\end{array}\right]<br />

    Which should be converted to:

    <br /> <br />
\left[\begin{array}{ccc}2uv^2w^2&1&1 \\ 2&1&2u^2e^v \\ 0&0&0 \end{array}\right]*\left[\begin{array}{ccc}v^2w^2&2uvw^2&2uv^2w \\ 0&w^2cosv&2wsinv \\ 2ue^v&u^2e^v&0 \end{array}\right]<br />

    =

    <br /> <br />
\left[\begin{array}{ccc}2uv^4w^4+2ue^v&4u^2v^3w^4+w^2cos  v+u^2e^v&4u^2v^4w^3+2wsinv \\ 2v^2w^2+4u^3e^{2v}&4uvw^2+w^2cosv+2u^4e^{2v}&4uv^2  w+2wsinv \\ 0&0&0 \end{array}\right]<br /> <br />

    Is it correct?
    Last edited by garymarkhov; August 14th 2009 at 11:10 PM. Reason: Figured out how to put 2v as exponent
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