 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/15/c/i/solution

x

before taking their Maclaurin series:

\frac{s i n x}{s i n h x} = \frac{1 - x ^{2} /6 + x ^{4} /120 + O ( x ^{6} )}{1 + x ^{2} /6 + x ^{4} /120 + O ( x ^{6} )} .

For the reciprocal denominator, the geometric series gives

1 - x^{2} /6 + 7 x^{4} /360 + O (x^{6})

. Multiplying gives the Taylor expansion

\frac{sin x}{sinh x} = 1 - \frac{x ^{2}}{3} + \frac{x ^{4}}{18} + O (x^{6}) .

Strictly, the original quotient has a removable singularity at zero; this is the Taylor series of its continuous, analytic extension with value one there. The absence of odd powers also follows because the ratio of these two odd functions is even.

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