Editing Lagrange polynomial (section)

==Remainder in Lagrange interpolation formula==
When interpolating a given function ''f'' by a polynomial of degree {{mvar|k}} at the nodes <math>x_0,...,x_k</math> we get the remainder <math>R(x) = f(x) - L(x)</math> which can be expressed as<ref>{{AS ref|25, eqn 25.2.3|878}}</ref>

:<math> R(x) = f[x_0,\ldots,x_k,x] \ell(x) = \ell(x) \frac{f^{(k+1)}(\xi)}{(k+1)!},  \quad \quad x_0 < \xi < x_k,</math>

where <math>f[x_0,\ldots,x_k,x]</math> is the notation for [[divided differences]]. Alternatively, the remainder can be expressed as a contour integral in complex domain as

:<math>R(x) = \frac{\ell(x)}{2\pi i} \int_C \frac{f(t)}{(t-x)(t-x_0) \cdots (t-x_k)} dt = \frac{\ell(x)}{2\pi i} \int_C \frac{f(t)}{(t-x)\ell(t)} dt.</math>

The remainder can be bound as

:<math>|R(x)| \leq \frac{(x_k-x_0)^{k+1}}{(k+1)!}\max_{x_0 \leq \xi \leq x_k} |f^{(k+1)}(\xi)|. </math>

=== Derivation===
Clearly, <math>R(x) </math> is zero at nodes. To find <math>R(x)</math> at a point <math>x_p  </math>, define a new function <math>F(x)=R(x)-\tilde{R}(x)=f(x)-L(x)-\tilde{R}(x)</math> and choose <math display="inline">\tilde{R}(x)=C\cdot\prod_{i=0}^k(x-x_i)</math>  where <math>C</math> is the constant we are required to determine for a given <math>x_p</math>. We choose <math>C</math> so that <math>F(x)</math> has <math>k+2</math> zeroes (at all nodes and <math>x_p</math>) between <math>x_0</math> and <math>x_k</math> (including endpoints). Assuming that <math>f(x)</math> is <math>k+1</math>-times differentiable, since <math>L(x)</math> and <math>\tilde{R}(x)</math> are polynomials, and therefore, are infinitely differentiable, <math>F(x)</math> will be <math>k+1</math>-times differentiable. By [[Rolle's theorem]], <math>F^{(1)}(x)</math> has <math>k+1</math> zeroes, <math>F^{(2)}(x)</math> has <math>k</math> zeroes... <math>F^{(k+1)}</math> has 1 zero, say <math>\xi,\, x_0<\xi<x_k</math>. Explicitly writing <math>F^{(k+1)}(\xi)</math>:

:<math>F^{(k+1)}(\xi)=f^{(k+1)}(\xi)-L^{(k+1)}(\xi)-\tilde{R}^{(k+1)}(\xi)</math>

:<math>L^{(k+1)}=0,\tilde{R}^{(k+1)}=C\cdot(k+1)!</math> (Because the highest power of <math>x</math> in <math>\tilde{R}(x)</math> is <math>k+1</math>)

:<math>0=f^{(k+1)}(\xi)-C\cdot(k+1)!</math>

The equation can be rearranged as<ref>{{Cite web|url=https://sam.nitk.ac.in/sites/default/Numerical_Methods/Interpolation/interpolation.pdf|title=Interpolation|pages=12–15|archive-url=https://web.archive.org/web/20170215114507/http://www.sam.nitk.ac.in/sites/default/Numerical_Methods/Interpolation/interpolation.pdf|archive-date=2017-02-15}}</ref>

:<math>C=\frac{f^{(k+1)}(\xi)}{(k+1)!}</math>
Since <math>F(x_p) = 0</math> we have <math>R(x_p)=\tilde{R}(x_p) = \frac{f^{k+1}(\xi)}{(k+1)!}\prod_{i=0}^k(x_p-x_i)</math>