Understanding the Proof of the Jacobian

  • Thread starter Thread starter leospyder
  • Start date Start date
  • Tags Tags
    Jacobian Proof
leospyder
Messages
9
Reaction score
0
Can someone explain to me this part of the proof of the jacobian?

Idea of the Proof

As usual, we cut S up into tiny rectangles so that the image under T of each rectangle is a parallelogram.



We need to find the area of the parallelogram. Considering differentials, we have

T(u + Du,v) @ T(u,v) + (xuDu,yuDu)

T(u,v + Dv) @ T(u,v) + (xvDv,yvDv)


Thus the two vectors that make the parallelogram are

P = guDu i + huDu j

Q = gvDv i + hvDv j

I don't know what they're talking about...I can follow the rest (the cross product bla bla bla bla bla) but I don't know how they're getting these two vectors...I figured it has something to do with partial differentials but I am still confused. If anyone could provide any insight Id be appreciative. :eek:
 
Physics news on Phys.org
The notation is a bit screwy, but here's what I think they're doing.

So suppose the surface is parametrised by \vec T(u,v).
Take a small rectangle in the domain with dimensions \Delta u, \Delta v, the bottom left corner being the point (u_0,v_0).
The image of this rectangle is a patch of area, which can be approximated by the parallelogram formed by the vectors:

\vec T(u_0+\Delta u,v_0)-\vec T(u_0,v_0)
and
\vec T(u_,v_0+\Delta v)-\vec T(u_0,v_0) (a picture helps here).

These vectors are in turn approximated by
\frac{\partial}{\partial u}\vec T(u_0,v_0)\Delta u
and
\frac{\partial}{\partial v}\vec T(u_0,v_0)\Delta v
respectively.

So area patch is about |\vec T_u \times \vec T_v|\Delta u \Delta v and you can figure out the rest.

Hope that helps.
 
Last edited:
It might help (me, anyways) if you would say what you're trying to prove. The Jacobian is a number associated with a matrix; it doesn't make any more sense to ask about a proof of the Jacobian than it does to ask about a proof of the number 2.
 
Galileo said:
The notation is a bit screwy, but here's what I think they're doing.
So suppose the surface is parametrised by \vec T(u,v).
Take a small rectangle in the domain with dimensions \Delta u, \Delta v, the bottom left corner being the point (u_0,v_0).
The image of this rectangle is a patch of area, which can be approximated by the parallelogram formed by the vectors:
\vec T(u_0+\Delta u,v_0)-\vec T(u_0,v_0)
and
\vec T(u_,v_0+\Delta v)-\vec T(u_0,v_0) (a picture helps here).
These vectors are in turn approximated by
\frac{\partial}{\partial u}\vec T(u_0,v_0)\Delta u
and
\frac{\partial}{\partial v}\vec T(u_0,v_0)\Delta v
respectively.
So area patch is about |\vec T_u \times \vec T_v|\Delta u \Delta v and you can figure out the rest.
Hope that helps.

Oh, I see it better now. THanks a lot, just wanted to say that before I go to bed. If i need further clarification Ill post the fool proof. THakns a lot guys
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

Similar threads

Back
Top