 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2013/f2/a/iv/solution

h

, direct expansion and the chain rule give

D F_{x} (h) = A h, D g_{x} (h) = 2 x^{T} h, D k_{x} (h) = 2 (A x)^{T} A h = 2 x^{T} A^{T} A h .

For example,

∥ x + h ∥^{2} - ∥ x ∥^{2} = 2 x^{T} h + ∥ h ∥^{2}

, so the quadratic remainder divided by

∥ h ∥

tends to zero. The corresponding gradients are

\nabla g (x) = 2 x

and

\nabla k (x) = 2 A^{T} A x

F

is the constant matrix

A

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