Let $e_m$ denote the $m$th <elementary symmetric polynomial> of $t_{i+1},\ldots,t_{i+k-1}$. Expanding both sides of the <Marsden identity> in powers of $x$ and equating the coefficient of $x^{k-1-m}$ gives
$$
\boxed{a_{m,i}=
\frac{e_m(t_{i+1},\ldots,t_{i+k-1})}{\binom{k-1}{m}}},
\qquad0\leq m\leq k-1.
$$
Thus $a_{0,i}=1$, while $a_{1,i}$ and $a_{2,i}$ are respectively the means of the interior knots and of their pairwise products.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-318/4/b/solution