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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


 * $$ R(x) = f[x_0,\ldots,x_k,x] \ell(x) = \ell(x) \frac{f^{(k+1)}(\xi)}{(k+1)!},$$

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 === 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 $$F^{(k+1)}(\xi)$$:


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


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


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

The equation can be rearranged as


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

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