 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/15/c/solution

For

x > 0

, the definition of hyperbolic cosine simplifies the composite function exactly:

cosh (ln x) = \frac{e ^{l n x} + e ^{- l n x}}{2} = \frac{1}{2} (x + \frac{1}{x}) .

Let

h = x - 2

\frac{1}{2 + h} = \frac{1}{2} (1 - \frac{h}{2} + \frac{h ^{2}}{4} - \frac{h ^{3}}{8} + O (h^{4})) .

Adding

2 + h

and dividing by two gives the requested Taylor expansion:

cosh (ln x) = \frac{5}{4} + \frac{3}{8} (x - 2) + \frac{1}{16} (x - 2)^{2} - \frac{1}{32} (x - 2)^{3} + O ((x - 2)^{4}) .

The underlying infinite geometric expansion has

∣ x - 2∣ < 2

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