# Invertible function

1. Mar 18, 2016

### mr.tea

1. The problem statement, all variables and given/known data
prove, using the definition, that the mapping
$\mathbf{u}=\mathbf{u}(u_1(x_1,x_2),u_2(x_1,x_2))$ where
$u_1=\tan(x_1)+x_2$
$u_2=x_2^3$

is a bijection from the strip $-\frac{pi}{2}<x_1<\frac{pi}{2}$ in the $x_1x_2$-plane onto the entire $u_1u_2$-plane, and find the inverse.

2. Relevant equations

3. The attempt at a solution
Finding the inverse is not the problem. My problem is that when I am trying to apply the inverse function theorem, I get that at $x_2=0$ the Jacobian is 0. The Jacobian that I have found is:
$\frac{\partial (u_1,u_2)}{\partial (x_1,x_2)}= \begin{vmatrix} \frac{1}{cos^2(x_1)} & 1\\ 0& 3x_2^2\\ \end{vmatrix}$

What am I missing here?

Thank you.

2. Mar 18, 2016

### Samy_A

Hint:
Let $f: \mathbb R \to \mathbb R \ :\ x \mapsto x³$.
Is $f$ a bijection?
Is the derivative of $f$ different from 0 in every point?

What does this tell us about the inverse function theorem?

3. Mar 20, 2016

### mr.tea

the answers are "yes"(but the derivative of f is 0 at 0).
Does this tell us that the inverse function theorem doesn't apply to that point(0 in your example)? Does it mean that we need to prove for that point without using the inverse function theorem?

Thank you.

4. Mar 21, 2016

### Samy_A

Yes.

Let's look at the precise statement of the inverse function theorem:

Let f : ℝn → ℝn be continuously diﬀerentiable on some open set containing a, and suppose that Df(a) is not singular. Then there is some open set V containing a and an open W containing f(a) such that f : V → W has a continuous inverse f−1 : W → V which is diﬀerentiable for all y ∈ W.

The Jacobian not being singular is a sufficient condition for a function to have an inverse (locally), not a necessary one.

5. Mar 22, 2016

### mr.tea

OK, great. I think I have got it.
By the way, do you know a good book for self learning multivariable analysis?(standard course for mathematicians)

Thank you again!

6. Mar 22, 2016

### Samy_A

This insight from @micromass may help.