Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 8

Multiply the Taylor expansions $\ln (1+x)=x-x^2/2+x^3/3+O(x^4)$ and $(1-x{)}^{-1}=1+x+x^2+O(x^3)$. Collecting the first three nonzero coefficients givesThe cubic coefficient is $1-1/2+1/3=5/6$.

Use the geometric series $(1+u{)}^{-1}=1-u+u^2+O(u^3)$ with $u=\sin x=x+O(x^3)$. The constant term counts as the first nonzero term, so the requested expansion is

Write $\cos x=1+u$ with $u=-x^2/2+x^4/24-x^6/720+O(x^8)$. Substitution into $\ln (1+u)=u-u^2/2+u^3/3+O(u^4)$ givesThus the first three nonzero terms of this even Taylor series areThe real logarithm is well defined near zero because $\cos x>0$ there.

Multiply the two power series and collect equal powers of $x$. The Cauchy product givesOnly terms whose two exponents add to the target exponent contribute.

In the Cauchy product, $a_j{x}^j$ combines with $b_{i-j}{x}^{i-j}$ to give a term of degree $i$. HenceThis coefficient identity holds for formal power series. For convergent Taylor series it also gives the function product throughout the common open interval of convergence, where absolute convergence justifies regrouping.