- #1
Fractal20
- 74
- 1
I am trying to understand the role of the Jacobian in the Implicit Function Theorem. However, I have had a hard time finding any discussions that use the Jacobian and are accessible for my level. http://mathworld.wolfram.com/ImplicitFunctionTheorem.html has been the best thing I have found.
More specifically I am unsure about maintaining that the Jacobian be square. So for the theorem to apply book is it true that given m equations with n variables (m < n) the number of dependent variables needs to be equal to the number of equations? If so, then I think I see why the Jacobian will be square (maybe it shows my ignorance to even suggest that something not square could be Jacobian?).
Also, in the case of something like f(x,y) = 0 and we are interested in finding y as a function of x, does the Jacobian still play a role? I want to say the Jacobian is a 1x1: [itex]\frac{\deltad}{\deltay}[/itex]. Is it okay to have 1x1 Jacobians and is the determinant of a 1x1 matrix simply the single element? Furthermore, if in such a case the partial deriv with respect to x is 0, does the theorem still apply? Or does it just mean dy/dx = 0?
Hopefully this wasn't too rambling and unspecific. Any insight would be appreciated! Thanks!
More specifically I am unsure about maintaining that the Jacobian be square. So for the theorem to apply book is it true that given m equations with n variables (m < n) the number of dependent variables needs to be equal to the number of equations? If so, then I think I see why the Jacobian will be square (maybe it shows my ignorance to even suggest that something not square could be Jacobian?).
Also, in the case of something like f(x,y) = 0 and we are interested in finding y as a function of x, does the Jacobian still play a role? I want to say the Jacobian is a 1x1: [itex]\frac{\deltad}{\deltay}[/itex]. Is it okay to have 1x1 Jacobians and is the determinant of a 1x1 matrix simply the single element? Furthermore, if in such a case the partial deriv with respect to x is 0, does the theorem still apply? Or does it just mean dy/dx = 0?
Hopefully this wasn't too rambling and unspecific. Any insight would be appreciated! Thanks!