 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/5/a/ii/solution

e^{u} = 1 + u + u^{2} /2 + \dots

with

u = x^{3} - x^{5}

. Since

u

begins at cubic order and

u^{2}

begins at sixth order, through cubic terms the Taylor series is

f (x) = 1 + x^{3} + O (x^{5}) .

The linear and quadratic coefficients are zero. The stronger remainder

O (x^{5})

, in big O notation, follows because the fourth-order term is also zero; the next term is

- x^{5}

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