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

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

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

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

The remainder can be bound as

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

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

${\displaystyle F^{(k+1)}(\xi )=f^{(k+1)}(\xi )-L^{(k+1)}(\xi )-R^{(k+1)}(\xi )}$

${\displaystyle L^{(k+1)}=0,R^{(k+1)}=C\cdot (k+1)!}$ (Because the highest power of ${\displaystyle x}$ in ${\displaystyle R(x)}$ is ${\displaystyle k+1}$

${\displaystyle 0=f^{(k+1)}(\xi )-C\cdot (k+1)!}$

The equation can be rearranged as

${\displaystyle C={\frac {f^{(k+1)}(\xi )}{(k+1)!}}}$

^[3][4][5][6]

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