 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/15/c/solution

cosh x = 1 + \frac{x ^{2}}{2} + \frac{x ^{4}}{24} + O (x^{6}), (1 + x^{2})^{- 1/2} = 1 - \frac{x ^{2}}{2} + \frac{3 x ^{4}}{8} + O (x^{6}) .

The quadratic terms cancel, while the quartic coefficient is

1/24 - 1/4 + 3/8 = 1/6

. Hence

\frac{cosh x}{1 + x ^{2}} = 1 + \frac{x ^{4}}{6} + O (x^{6}) .

The absence of odd powers follows from the function's even symmetry.

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