# Implicit => inverse function theorem ( due to exam, )

• yoyo_fen
In summary, the inverse function theorem states that there is an inverse function of f in a neighbourhood around the points a and b. This theorem is based on the implicit function theorem which states that if f(a,b)=0 then there is an F such that a=F(b) and likewise there exists a G such that b=G(a).
yoyo_fen

## Homework Statement

Prove the inverse function theorem, knowing the implicit function theorem.

## Homework Equations

The statements of both theorems... Can't think of much else.

## The Attempt at a Solution

Alright... I have a feeling that this is going to come up on my exam which is only two days from now, so please help me :(

I got a hint today: define g(x,y) = f(x)-y or g(x,y)=f(y)-x, the guy was not sure. I presume it should be the first because then the derivative with respect to the second variable y is 1 which is invertible and in the first case we are not sure whether it is.
I have a problem though because in the statement of the implicit function theorem I have, the function needs to be from R^n to R^(n+p) so that we can take the differential of the matrix w.r.t. the p+i variables (i goes from 1 to n). In the hint the function is from R^2 -> R.

Even if that was not a problem I don't know how to end up with the inverse function theorem.

I thank in advance for anyone who is going to help me out :)

Isn't this just bookwork?

What do you mean? Even if it is... I still cannot do it. :(

what does the implicit and inverse functon theorems say?

Inverse
f:R^n->R^n,
if f is C^1, f(a)=b and differential is invertible at a, then the inverse is well defined and C^1 on some open subsets U and V where a is in U and b is in V

Implicit
f:R^p+n-->R^n , f(a,b)=0, f is C^1 and the partials with respect to the variables from p onwards form and invertible matrix, then F(a)=b is well defined and C^1 on some open subsets U and V...

I know them... And I have tried... If I knew the answer I wasn't going to be asking right?

write g(a,b)=b-f(a)=0, what does the implicit function tell you now?

The implicit f(a)= b is C^1 and well defined in a neighbourhood around those points...?

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Or that there is an F such that F(b)=a...

How? I really don't see it... Is mine wrong?

Your's isn't wrong, but I used a different variable for mine.

The neighbourhoods are not the same, how did you just switch the letters a and b? And even so... Okay F(b) = b but how did f(a) become just a? What we want to show is that there is an inverse function of f in my previous post (f(a)=b), s.t. f^-1 (b)=a. This is why I don't get? How do we get that function. How are you able to write what you have written.

It didn't, remember the implicit function theorem says that if f(a,b)=0 which we have then there is an F such that a=F(b) but likewise there exists a G such that b=G(a), we are just applying the implicit function to each variable.

OH SNAP!
So I was wrong after all saying this "The implicit f(a)= b is C^1 and well defined in a neighbourhood around those points...?". This follows from b-f(a)= 0 as we've defined g(a,b) and the other result follows from implicit function theorem.

THANKS MAN! YOU ARE THE BEST!

There is more work to be done if you want to make it fully rigourous but what I said is the gist of the proof.

I help when I can.

## 1. What is the Implicit Inverse Function Theorem?

The Implicit Inverse Function Theorem is a mathematical theorem that relates to the existence and differentiability of inverse functions for a system of equations. It states that if a function satisfies certain conditions, then its inverse function also exists and is differentiable.

## 2. What are the conditions for the Implicit Inverse Function Theorem to hold?

The conditions for the Implicit Inverse Function Theorem to hold are that the function must be continuously differentiable and its Jacobian (partial derivatives) must be non-singular at the point in question.

## 3. How is the Implicit and Inverse Function Theorem different from each other?

The Implicit Inverse Function Theorem and the Inverse Function Theorem are closely related, but they have some important differences. The Implicit Inverse Function Theorem deals with a system of equations, while the Inverse Function Theorem deals with a single equation. Additionally, the Implicit Inverse Function Theorem only guarantees the existence and differentiability of the inverse function at a specific point, while the Inverse Function Theorem guarantees the existence and differentiability of the inverse function in a neighborhood around that point.

## 4. What are some applications of the Implicit Inverse Function Theorem?

The Implicit Inverse Function Theorem has various applications in mathematics, physics, and engineering. It is often used to solve systems of nonlinear equations, such as those that arise in optimization problems. It is also useful in studying dynamical systems and analyzing their stability. In physics, the theorem is used in the study of fluid dynamics and thermodynamics.

## 5. How is the Implicit Inverse Function Theorem proven?

The proof of the Implicit Inverse Function Theorem involves using the Inverse Function Theorem and the Implicit Function Theorem, both of which are fundamental theorems in calculus. The proof relies on the concept of the derivative and the chain rule to show that the inverse function is differentiable at the given point. It is a complex proof that requires a strong understanding of multivariable calculus and real analysis.

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