1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculate the Jacobian of this function

  1. Mar 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Could you please help me with this problem?

    Let f(x) = (f_1(x), f_2(x)) map R[tex]^{2}[/tex] into itself where f_1, f_2 have continuous 1st/ 2nd partial derivatives in each variable. Assume that f maps origin to itself and that J_f(x)(Jacobian matrix) is an invertible 2x2 matrix for all x. Put g(x) = x - f'(x)[tex]^{-1}[/tex].f(x)

    (i) Explicitly compute J_g(x) using relation J_f[tex]^{-1}[/tex]. J_f = Identity matrix I[tex]_{2}[/tex]


    Thanks in advance!!!


    2. Relevant equations


    3. The attempt at a solution
    What are f'(x)[tex]^{-1}[/tex] and f'(x)[tex]^{-1}[/tex].f(x)
    ? I am just having trouble with the notations.
    Can you give me hints?
    ( For some reason, I can't you tex right???)[tex]^{}[/tex]
     
    Last edited: Mar 18, 2008
  2. jcsd
  3. Mar 18, 2008 #2
    Well firstly, I don't think you've entered your question correctly. If [itex] f\in C^2(\mathbb{R}^2), \quad f:\mathbb{R}^2\rightarrow \mathbb{R}^2 [/itex] then it's more likely that [itex] f(x,y) = \displaystyle \left( f_1 (x,y), f_2 (x,y) \right) [/itex]. If f were indeed simply a function of x, then your Jacobian matrix would be singular.

    Now your Jacobian matrix for f will look like

    [tex] Jf = \begin{pmatrix} \partial_x f_1 & \partial_y f_1 \\ \partial_x f_2 & \partial_y f_2 \end{pmatrix} [/tex]

    Then you don't need to explicitly calculate [itex] f^{-1} (x,y) [/tex] since you only need it's Jacobian matrix.

    [tex] Jf^{-1} = \frac{1}{\partial_x f_1 \partial_y f_2 - \partial_x f_2 \partial_y f_1} \begin{pmatrix} \partial_y f_2 & -\partial__y f_1 \\ -\partial_x f_2 & \partial_x f_1 \end{pmatrix} [/tex].

    Is [itex] f^\prime (x) ^{-1} \cdot f(x) [/itex] the dot product? If it is then [itex] g:\mathbb{R}^2 \rightarrow \mathbb{R} \text{ and } Jg \in M_{1\times 2} (\mathbb{R} ) [/itex]
     
  4. Mar 18, 2008 #3
    Thanks for the reply!
    I got up to g = x - f'(x)Click to see the LaTeX code for this image.f(x) in term of J_f and J_f^-1. I want to compute J_g but continue doing the same thing ( taking partial derivatives) would make J_g look awfully ugly. I am curious if there is another way to get J_g? I got a hint: (J_f)^(-1).J_f = Identity matrix I_2 but I do not know how to use it.
     
    Last edited: Mar 18, 2008
  5. Mar 18, 2008 #4
    Yes, I believe that it is the dot product.
     
  6. Mar 18, 2008 #5
    Just to move it down here.
     
  7. Mar 18, 2008 #6
    I'm pretty sure you should be able to decompose g(x) over it's sums. That is

    [tex] g(x) &=& x - \frac{df^{-1}}{dx} f(x) \\[/tex]
    [tex] =x-f_1 \frac{df_1^{-1}}{dx} - f_2 \frac{df_2^{-1}}{dx} [/tex]

    Now I'm not 100% sure about this, but I think that in your case, since [itex] g:\mathbb{R}^2 \rightarrow \mathbb{R} \text{ and } Jg \in M_{1\times 2} (\mathbb{R} ) [/itex]

    That you can say [itex] J(f+g) = J(f) + J(g) \text{ and } J(fg) = (Jf)g+f(Jg) [/tex] and just substitute f and g with the appropriate functions. Play around with it and see what happens.

    Edit: I'm not sure if this will hold in higher dimensions, but I think it holds in [itex]M_{1\times 2} (\mathbb{R} )[/itex] since the Jacobian just ends up being the gradient of each scalar function. Not sure if that helps you at all.
     
    Last edited: Mar 18, 2008
  8. Mar 18, 2008 #7
    I think [itex]
    g:\mathbb{R}^2 \rightarrow \mathbb{R} \text{ and } Jg \in M_{1\times 2} (\mathbb{R} )
    [/itex] is right. I will use your suggestion and see where it gets me. Thanks! ( I may post more question).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculate the Jacobian of this function
  1. Calculate the jacobian (Replies: 1)

Loading...