 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/6/b/solution

Put

z = (x, y)^{T}

r^{2} = z^{T} z

and

f = 1 + r^{2}

. The chain rule yields

\nabla f = \frac{z}{f}, H_{f} = \frac{I}{f} - \frac{z z ^{T}}{f ^{3}} = \frac{1}{f ^{3}} (1 + y^{2} - x y - x y 1 + x^{2}) .

For every nonzero vector

v

v^{T} H_{f} v = \frac{( 1 + r ^{2} ) ∥ v ∥ ^{2} - ( z ^{T} v ) ^{2}}{f ^{3}} \geq \frac{∥ v ∥ ^{2}}{f ^{3}} > 0.

Thus the Hessian is a positive-definite matrix everywhere. The Hessian criterion for convexity implies that $f$ is strictly convex on $R^{2}$ . Its unique global minimum is at

z = 0

, where

f = 1

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