 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/15/c/ii/solution

e^{- x^{3}} = 1 - x^{3} + O (x^{6})

and

cosh x = 1 + x^{2} /2 + x^{4} /24 + O (x^{6})

. Inverting the latter using the geometric series gives

\frac{1}{c o s h x} = 1 - \frac{x ^{2}}{2} + (\frac{1}{4} - \frac{1}{24}) x^{4} + O (x^{6}) = 1 - \frac{x ^{2}}{2} + \frac{5}{24} x^{4} + O (x^{6}) .

Multiplying the two expansions yields

\frac{e ^{- x^{3}}}{cosh x} = 1 - \frac{x ^{2}}{2} - x^{3} + \frac{5}{24} x^{4} + O (x^{5}) .

O (x^{5})

because the product already generates an

x^{5}

term from

(- x^{3}) (- x^{2} /2)

 New to topics? Read the docs here!