The required <Taylor series> give
$$
(1+\beta)-(1+\beta)^{-2}=3\beta-3\beta^2+O(\beta^3),
$$
$$
(2+\beta)-(2+\beta)^{-2}
=\frac74+\frac54\beta-\frac3{16}\beta^2+O(\beta^3).
$$
Seek $\beta=a\epsilon+b\epsilon^2+O(\epsilon^3)$. Substitution in the exact equation from part (iii) gives successively
$$
3a+\frac74=0,
\qquad
3b-3a^2+\frac54a=0.
$$
Thus $a=-7/12$ and $b=7/12$, so
$$
\boxed{\beta=-\frac7{12}\frac{\mu_2}{\mu_1}
+\frac7{12}\left(\frac{\mu_2}{\mu_1}\right)^2
+O\left((\mu_2/\mu_1)^3\right)
=\alpha+\frac7{12}\left(\frac{\mu_2}{\mu_1}\right)^2+\cdots}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 3 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-316/3/iv/solution