Implicit function theorem part 2

Click For Summary
SUMMARY

The discussion centers on the application of the Implicit Function Theorem, specifically verifying the conditions required for its use. The conditions outlined include that F(x_0, z_0) must equal 0 and the partial derivative ∂F/∂z at the point (x_0, z_0) must be non-zero. In this case, with (x_0, z_0) = (1, 2) and ∂F/∂z = -1, both conditions are satisfied. Consequently, there exists a function g such that x = g(z_1) in the neighborhood of (z_0 - ε, z_0 + ε).

PREREQUISITES
  • Understanding of the Implicit Function Theorem
  • Knowledge of partial derivatives
  • Familiarity with function notation and neighborhoods
  • Basic calculus concepts
NEXT STEPS
  • Study the implications of the Implicit Function Theorem in multivariable calculus
  • Explore examples of functions that satisfy the conditions of the theorem
  • Learn about the continuity and differentiability of implicit functions
  • Investigate applications of the theorem in solving real-world problems
USEFUL FOR

Mathematics students, educators, and professionals in fields requiring advanced calculus, particularly those focusing on multivariable analysis and differential equations.

Lambda96
Messages
233
Reaction score
77
Homework Statement
see Post
Relevant Equations
Implicit function theorem
Hi,

I'm not sure if I've solved the problem correctly

Bildschirmfoto 2024-07-06 um 16.56.27.png


In order for the Implicit function theorem to be applied, the following two properties must hold ##F(x_0,z_0)=0## and ##\frac{\partial F(x_0,z_0)}{\partial z} \neq 0##. ##(x_0,z_0)=(1,2)## is a zero and ##\frac{\partial F(x_0,z_0)}{\partial z} =-x^2=-1## so both properties are fulfilled


According to the Implicit function theorem, there now exists a function ##g## for which the following relation ##x=g(z_1)## is valid in the neighborhood ##(z_0 - \epsilon, z_0 + \epsilon)##.


Then the following applies ##F(x,z_1)=F(g(z_1),z_1)##
 
Physics news on Phys.org
You're doing good! keep going. You could end up with a value of some ##\epsilon>0## in the end, because that's the the problem is asking for. Then, you could show that the statement in the problem is true using the value of ##\epsilon## you found and the implicit function theorem.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
3K
Replies
7
Views
2K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 0 ·
Replies
0
Views
499