Put $m=k-1$, so the <Marsden identity> is
$$
(x-t)^m=m!\sum_{i=1}^n\psi_i(x)N_i(t).
$$
Differentiating $m-j$ times with respect to $x$ gives
$$
\frac{(x-t)^j}{j!}
=\sum_{i=1}^n\psi_i^{(m-j)}(x)N_i(t),
\qquad 0\leq j\leq m.
$$
The exact <Taylor formula for a polynomial> is
$$
p(t)=\sum_{j=0}^m\frac{p^{(j)}(x)}{j!}(t-x)^j.
$$
Substitution of the differentiated Marsden identities yields
$$
\boxed{p(t)=\sum_{i=1}^n\lambda_i(p,x)N_i(t)},
$$
where
$$
\boxed{\lambda_i(p,x)=
\sum_{j=0}^{k-1}(-1)^j
\psi_i^{(k-1-j)}(x)p^{(j)}(x)}.
$$

Differentiating this expression for $\lambda_i$ produces two sums whose adjacent terms cancel. The uncancelled endpoints contain $\psi_i^{(k)}$ and $p^{(k)}$, both zero because the two functions have degree at most $k-1$. Thus
$$
\frac d{dx}\lambda_i(p,x)=0,
$$
so the <Marsden dual functional> $\lambda_i(p)$ is independent of the auxiliary point $x$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-318/4/a/solution