 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/15/c/i/solution

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

- 1/2

gives

(1 + u)^{- 1/2} = 1 - u /2 + 3 u^{2} /8 + O (u^{3})

. Substituting

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

yields

(cosh x)^{- 1/2} = 1 - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{x ^{4}}{24}) + \frac{3}{8} (\frac{x ^{2}}{2})^{2} + O (x^{6}) = 1 - \frac{x ^{2}}{4} + \frac{7 x ^{4}}{96} + O (x^{6}) .

These are the first three nonzero terms; the remainder is written in Big O notation.

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