The integral form of the <Taylor theorem> about the left endpoint is
$$
f(x)=q_{k-1}(x)+\frac1{(k-1)!}
\int_a^b(x-t)_+^{k-1}f^{(k)}(t)\,dt,
$$
where $q_{k-1}\in\mathcal P_{k-1}$. Apply the order-$k$ <divided difference> at $x_i,\ldots,x_{i+k}$. The polynomial term vanishes, and linearity permits interchange with the integral:
$$
f[x_i,\ldots,x_{i+k}]
=\frac1{(k-1)!}\int_a^b
[x_i,\ldots,x_{i+k}](\mathord\cdot-t)_+^{k-1}
f^{(k)}(t)\,dt.
$$
By the definition $M_i(t)=k[x_i,\ldots,x_{i+k}](\mathord\cdot-t)_+^{k-1}$,
$$
\boxed{f[x_i,\ldots,x_{i+k}]
=\frac1{k!}\int_a^bM_i(t)f^{(k)}(t)\,dt}.
$$
This is the <Peano kernel theorem> for the divided-difference functional.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 5 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-318/5/2/a/solution