 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/15/c/solution

Put

q (x) = π^{2} /16 + x

. The original PDF confirms that the

+ x

is inside the radical. For the composite cosine,

f^{'} (x) = - \frac{s i n q}{2 q}, f^{''} (x) = \frac{s i n q}{4 q ^{3}} - \frac{c o s q}{4 q ^{2}} .

x = 0

q = π /4

and

sin q = cos q = 2 /2

. Substituting into

f (0) + f^{'} (0) x + f^{''} (0) x^{2} /2

gives

f (x) = \frac{2}{2} - \frac{2}{π} x + \frac{2 ( 4 - π )}{π ^{3}} x^{2} + O (x^{3}) .

All three displayed coefficients are nonzero.

 New to topics? Read the docs here!