Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 13

For $|t|<1$, the geometric series gives $(1+t{)}^{-1}={\sum }_{n\ge 0}(-t{)}^n$. Its uniform convergence on every closed interval strictly inside $(-1,1)$ permits termwise integration from zero to $x$. ThereforeSubtracting cancels every even power and doubles every odd power, giving the Taylor series

Use the binomial series and exponential Maclaurin series:The Cauchy product isAdding $\log (1+x)=x-x^2/2+x^3/3-x^4/4+O(x^5)$ cancels the first three powers. Hence the first nonzero term is

Put $d=a-b$. The rational factor has Taylor expansion $1+dx-bd{x}^2+b^{2}d{x}^3+O(x^4)$. Multiplication by the logarithmic power series makes the coefficients of $x^2,x^3$ equal to $d-1/2$ and $1/3-d/2-bd$. Setting both to zero gives $d=1/2$ and $b=1/6$, henceThe coefficient of $x^4$ is $-1/4+d/3+bd/2+b^{2}d=-1/36$. Thus the resulting expansion begins $x-x^4/36+O(x^5)$.