 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/6/a/ii/solution

On a branch with fixed signs of

u, v

and

uv \neq = 0

, use

u^{2} = x + y

and

v^{2} = y - x

. Differentiating these identities, or inverting the Jacobian matrix, gives

u_{x} = \frac{1}{2 u}, u_{y} = \frac{1}{2 u}, v_{x} = - \frac{1}{2 v}, v_{y} = \frac{1}{2 v} .

The inverse Jacobian matrix is therefore

\frac{\partial ( u , v )}{\partial ( x , y )} = (1/ (2 u) - 1/ (2 v) 1/ (2 u) 1/ (2 v)) .

These are local partial derivatives. Choosing the signs specifies an inverse function on

y > ∣ x ∣

; the original map has four preimages there and has no single global inverse. At

u = 0

v = 0

, the displayed inverse derivatives are not finite.

 New to topics? Read the docs here!