 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/b/ii/solution

Put

h = x - 1

and

k = y - 1

. At

(1, 1)

, the value is

5 + 4 α

, the gradient is

(5 + 4 α, 4 + 2 α)

(8 2 α 2 α 2)

. Thus the quadratic Taylor expansion is

f (1 + h, 1 + k) = 5 + 4 α + (5 + 4 α) h + (4 + 2 α) k + 4 h^{2} + 2 α hk + k^{2} + O ((∣ h ∣ + ∣ k ∣)^{3}) .

The exact remainder is

h^{3}

, since the original polynomial has only the cubic term

x^{3}

beyond degree two.

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