Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 3

Since the derivative of the arctangent is $1/(1+x^2)=1-x^2+O(x^4)$ near zero, termwise integration givesThe integration constant is zero because $\arctan 0=0$.

The expression in the PDF is $1/a^x-1$, with the subtraction outside the fraction. Using the exponential series with $a^{-x}=e^{-x\ln a}$ givesFor $a>0$, $a\ne 1$, these are the first two nonzero terms. If $a=1$, the function is identically zero, so there are no nonvanishing terms.

The square is on the cosine: near zero, $\ln ({\cos }^{2}x)=2\ln (\cos x)$. Substituting $\cos x=1-x^2/2+x^4/24+O(x^6)$ into the logarithmic series givesTherefore

Since $g(0)=0$, only the first three positive powers of $g$ contribute through degree three. The composition coefficients of power series follow by using $g^2=g_1^2{x}^2+2g_1{g}_2{x}^3+O(x^4)$ and $g^3=g_1^3{x}^3+O(x^4)$ in $f_0+f_{1}g+f_2{g}^2+f_3{g}^3+\cdots$. HenceHere $f_m$ and $g_m$ are coefficients of the power series, rather than raw derivatives; their definitions already contain any factorial factors.