Proof inverse f'n th'mimplicit f'n th'm?

  • Thread starter Thread starter calvino
  • Start date Start date
  • Tags Tags
    Inverse Proof
Click For Summary
SUMMARY

The discussion focuses on applying the Inverse Function Theorem to demonstrate that if the function g: [0,1] x [0,1] -> R^3 is continuously differentiable and the determinant of its Jacobian matrix, denoted as det[Dg], is non-zero at a point (s', t'), then there exists a neighborhood around (s', t') such that the image of g forms the graph of a function. The Inverse Function Theorem states that for a continuously differentiable function F: U -> R^n with an invertible Jacobian at a point x', there exists a neighborhood V where F has a continuously differentiable inverse. This theorem is crucial for establishing the local behavior of g in the vicinity of (s', t').

PREREQUISITES
  • Understanding of the Inverse Function Theorem in multivariable calculus
  • Knowledge of continuously differentiable functions (C1 functions)
  • Familiarity with Jacobian matrices and determinants
  • Basic concepts of neighborhoods in topology
NEXT STEPS
  • Study the Inverse Function Theorem in detail, focusing on its applications in multivariable calculus
  • Explore examples of continuously differentiable functions and their Jacobians
  • Learn about the implications of non-zero determinants in the context of function graphs
  • Investigate the concept of neighborhoods in R^n and their significance in analysis
USEFUL FOR

Mathematics students, particularly those studying multivariable calculus, researchers in mathematical analysis, and educators looking to deepen their understanding of the Inverse Function Theorem and its applications.

calvino
Messages
108
Reaction score
0
Show that if g: [0,1] x [0,1]-> R^3 is continuously differentiable with
det [Dg] (s',t') not equal to 0, then there exists S>0 such that
{g(s,t): (s,t) are elements of the ball, radius S, centred at (s',t')} is the graph of some function.
(s',t') is just some point.

I'm just not sure how to start. The prof said that it might be best to use inverse function theorem, but yah...any help would be great.
 
Physics news on Phys.org
what does the inverse function thm say
 
The inverse function theorem says:

Suppose U, a proper subset of R^n, is open, and x' is an element of U. F: U -> R^n is C1, and DF(x') is invertible. Then there is a neighbourhood V, a proper subset of U, of x' on which F has a C1 inverse function. So there are neighbourhoods V of x' and W of F(x'), and a function g: W-> V so that

F(g(y))=y for all y in W, and g(F(x))=x for all x in V.

Moreover, if F(x)=y, then

Dg(y)= (DF(x))^-1

[note: we're taking everything as vector-valued, although it should't really matter]
 

Similar threads

Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K