Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 15

sinh t = t + t^{3} /3! + t^{5} /5! + O (t^{7})

, applied with

t = x^{2}

and then multiplied by

x

, gives

x sinh (x^{2}) = x^{3} + \frac{x ^{7}}{6} + \frac{x ^{11}}{120} + O (x^{15}) .

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

b

Put

u = ln (1 + x) = x - x^{2} /2 + x^{3} /3 + O (x^{4})

. Since

u = O (x)

, the outer Taylor series requires

u^{2} = x^{2} - x^{3} + O (x^{4})

and

u^{3} = x^{3} + O (x^{4})

. Therefore

ln (1 + ln (1 + x)) = u - \frac{u ^{2}}{2} + \frac{u ^{3}}{3} + O (u^{4}) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{1}{2} (x^{2} - x^{3}) + \frac{x ^{3}}{3} + O (x^{4}) = x - x^{2} + \frac{7 x ^{3}}{6} + O (x^{4}) .

Both logarithms are defined on a neighbourhood of zero, so this composition of power series is valid there.

Factor the leading term out of the Maclaurin series for the sine:

sin x = x (1 - \frac{x ^{2}}{6} + \frac{x ^{4}}{120} + O (x^{6})) .

The binomial theorem gives, through degree four inside the bracket,

(1 - \frac{x ^{2}}{6} + \frac{x ^{4}}{120} + O (x^{6}))^{6} = 1 + 6 (- \frac{x ^{2}}{6} + \frac{x ^{4}}{120}) + (2 6) \frac{x ^{4}}{36} + O (x^{6}) .

x^{4}

1/20 + 5/12 = 7/15

. Multiplying back by

x^{6}

yields

sin^{6} x = x^{6} - x^{8} + \frac{7}{15} x^{10} + O (x^{12}) .

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