User:Ngfio/tests

Remainder in Lagrange interpolation formula

When interpolating a given function f by a polynomial of degree $k$ at the nodes $x_{0},...,x_{k}$ we get the remainder $R(x)=f(x)-L(x)$ . If $f(x)$ is $k+1$ -times differentiable, then for any $x\in [x_{0},x_{k}]$ this remainder can be expressed as^[1]

R(x)=f[x_{0},\ldots ,x_{k},x]\ell (x)=\ell (x){\frac {f^{(k+1)}(\xi )}{(k+1)!}},

where $\xi =\xi (x)\in (x_{0},x_{k})$ and $f[x_{0},\ldots ,x_{k},x]$ is the notation for divided differences. Alternatively, the remainder can be expressed as a contour integral in complex domain as

R(z)={\frac {\ell (z)}{2\pi i}}\int _{C}{\frac {f(t)}{(t-z)(t-z_{0})\cdots (t-z_{k})}}dt={\frac {\ell (z)}{2\pi i}}\int _{C}{\frac {f(t)}{(t-z)\ell (t)}}dt.

The remainder can be bound as

|R(x)|\leq {\frac {(x_{k}-x_{0})^{k+1}}{(k+1)!}}\max _{x_{0}\leq \xi \leq x_{k}}|f^{(k+1)}(\xi )|.

Derivation^[2]

Clearly, $R(x)$ is zero at nodes. Fix a point ${\hat {x}}$ in the interval $[x_{0},x_{k}]$ , different from any of the nodes $x_{0},x_{1},\ldots ,x_{k}$ . We want to find $R({\hat {x}})$ . To this end, define a new function $F(x)=f(x)-L(x)-C\cdot \prod _{i=0}^{k}(x-x_{i})$ , $C$ is chosen so that $R({\hat {x}})=C\cdot \prod _{i=0}^{k}({\hat {x}}-x_{i})$ or equivalently $F({\hat {x}})=0$ . Now $F(x)$ has $k+2$ zeroes (at all nodes and ${\hat {x}}$ ) between $x_{0}$ and $x_{k}$ (including endpoints). Assuming that $f(x)$ is $k+1$ -times differentiable, $L(x)$ and $C\cdot \prod _{i=0}^{k}(x-x_{i})$ are polynomials, and therefore, are infinitely differentiable. By Rolle's theorem, $F^{(1)}(x)$ has $k+1$ zeroes, $F^{(2)}(x)$ has $k$ zeroes, ..., $F^{(k+1)}$ has 1 zero, say $\xi ,\,x_{0}<\xi <x_{k}$ . Explicitly writing $F^{(k+1)}(\xi )$ :