Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 1

The Taylor series $\sin x=x-x^3/6+O(x^5)$ gives ${\sin }^{2}x=x^2(1-x^2/3+O(x^4))$. Taking its reciprocal and cancelling the singular term givesThus the apparent singularity at zero is removable, and the first nonzero term is the constant $1/3$.

The Taylor series of the sine and arctangent give $\sin x-\arctan x=x^3/6+O(x^5)$, while $x^2\ln (1+x)=x^3(1-x/2+O(x^2))$. Dividing the two series givesThe first nonzero term after cancellation is the constant $1/6$.

Factor the numerator as $(e^x-1{)}^2=x^2+O(x^3)$. The Taylor series of the cosine gives $\cos 3x-2\cos 2x+\cos x=(-9/2+4-1/2)x^2+O(x^4)=-x^2+O(x^4)$. HenceThe first nonzero term is the constant $-1$.