Inverse Function Theorem - Example

Example

Consider the vector-valued function F from R2 to R2 defined by

\mathbf{F}(x,y)= \begin{bmatrix} {e^x \cos y}\\ {e^x \sin y}\\ \end{bmatrix}.

Then the Jacobian matrix is

J_F(x,y)= \begin{bmatrix} {e^x \cos y} & {-e^x \sin y}\\ {e^x \sin y} & {e^x \cos y}\\ \end{bmatrix}

and the determinant is

\det J_F(x,y)= e^{2x} \cos^2 y + e^{2x} \sin^2 y= e^{2x}. \,\!

The determinant e2x is nonzero everywhere. By the theorem, for every point p in R2, there exists a neighborhood about p over which F is invertible. Note that this is different than saying F is invertible over its entire image. In this example, F is not invertible because it is not injective (because .)

Read more about this topic:  Inverse Function Theorem

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