 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/12/a/ii/solution

ℓ = ln 2

2^{x} - 2^{- x} = e^{ℓ x} - e^{- ℓ x} = 2 ℓ x + \frac{ℓ ^{3}}{3} x^{3} + O (x^{5}), sin x = x - \frac{x ^{3}}{6} + O (x^{5}) .

Multiply the two power series, retaining every contribution through degree four:

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

Since

0 < ln 2 < 1

x^{4}

is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.

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