 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/9/b/solution

h = x - x_{s}

and

k = y - y_{s}

, the second-order multivariable Taylor expansion is

z = z_{s} + (z_{x})_{s} h + (z_{y})_{s} k + \frac{1}{2} [(z_{xx})_{s} h^{2} + 2 (z_{x y})_{s} hk + (z_{yy})_{s} k^{2}] + o (h^{2} + k^{2}) .

(x_{s}, 0)

sin x_{s} = - 2 x_{s} cos x_{s}

and

z_{s} = cos x_{s} e^{- x_{s}^{2}}

(z_{xx})_{s} = - (4 x_{s}^{2} + 3) z_{s}, (z_{x y})_{s} = 0, (z_{yy})_{s} = - 2 z_{s} .

Thus, since this particular function is analytic, the expansion can be written

z (x, y) = z_{s} [1 - \frac{4 x _{s}^{2} + 3}{2} (x - x_{s})^{2} - y^{2}] + O (((x - x_{s})^{2} + y^{2})^{3/2}) .

There are no linear terms because the gradient vanishes, and no mixed quadratic term because the mixed derivative vanishes on the axis.

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