• Support PF! Buy your school textbooks, materials and every day products Here!

Calculate the Jacobian of this function

  • Thread starter mobe
  • Start date
20
0
1. Homework Statement
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. Homework 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:

Answers and Replies

743
1
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]
 
20
0
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:
20
0
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]
Yes, I believe that it is the dot product.
 
20
0
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.
Just to move it down here.
 
743
1
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:
20
0
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).
 

Related Threads for: Calculate the Jacobian of this function

  • Last Post
Replies
1
Views
8K
Replies
1
Views
2K
  • Last Post
Replies
0
Views
4K
Replies
3
Views
480
  • Last Post
Replies
2
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
819
  • Last Post
Replies
4
Views
1K
Top