Hjilderda

Mijn Formule Test
Mijn Testpagina
Panorama
Numerieke Differentiatie Formules
Numerieke Integratie Formules

Closed Newton–Cotes Formulae

Degree	Common name	Formula	Error term
1	Trapezoid rule	${\frac {b-a}{2}}(f_{0}+f_{1})$	$-{\frac {(b-a)^{3}}{12}}\,f^{(2)}(\xi )$
2	Simpson's rule	${\frac {b-a}{6}}(f_{0}+4f_{1}+f_{2})$	$-{\frac {(b-a)^{5}}{2880}}\,f^{(4)}(\xi )$
3	Simpson's 3/8 rule	${\frac {b-a}{8}}(f_{0}+3f_{1}+3f_{2}+f_{3})$	$-{\frac {(b-a)^{5}}{6480}}\,f^{(4)}(\xi )$
4	Boole's rule	${\frac {b-a}{90}}(7f_{0}+32f_{1}+12f_{2}+32f_{3}+7f_{4})$	$-{\frac {(b-a)^{7}}{1935360}}\,f^{(6)}(\xi )$

Closed Newton–Cotes Formulae

Degree	Common name	Formula	Error term
1	Trapezoid rule	${\frac {h}{2}}(f_{0}+f_{1})$	$-{\frac {h^{3}}{12}}\,f^{(2)}(\xi )$
2	Simpson's rule	${\frac {h}{3}}(f_{0}+4f_{1}+f_{2})$	$-{\frac {h^{5}}{90}}\,f^{(4)}(\xi )$
3	Simpson's 3/8 rule	${\frac {3h}{8}}(f_{0}+3f_{1}+3f_{2}+f_{3})$	$-{\frac {3h^{5}}{80}}\,f^{(4)}(\xi )$
4	Boole's rule	${\frac {2h}{45}}(7f_{0}+32f_{1}+12f_{2}+32f_{3}+7f_{4})$	$-{\frac {8h^{7}}{945}}\,f^{(6)}(\xi )$

Closed Newton–Cotes Formulae

Degree	Common name	Formula	Error term
1	Trapezoid rule	${\frac {1}{2}}h(f_{0}+f_{1})$	$-{\frac {1}{12}}h^{3}\,f^{(2)}(\xi )$
2	Simpson's rule	${\frac {1}{3}}h(f_{0}+4f_{1}+f_{2})$	$-{\frac {1}{90}}h^{5}\,f^{(4)}(\xi )$
3	Simpson's 3/8 rule	${\frac {3}{8}}h(f_{0}+3f_{1}+3f_{2}+f_{3})$	$-{\frac {3}{80}}h^{5}\,f^{(4)}(\xi )$
4	Boole's rule	${\frac {2}{45}}h(7f_{0}+32f_{1}+12f_{2}+32f_{3}+7f_{4})$	$-{\frac {8}{945}}h^{7}\,f^{(6)}(\xi )$

$V={\begin{bmatrix}1&\alpha _{1}&\alpha _{1}^{2}&\dots &\alpha _{1}^{n-1}\\1&\alpha _{2}&\alpha _{2}^{2}&\dots &\alpha _{2}^{n-1}\\1&\alpha _{3}&\alpha _{3}^{2}&\dots &\alpha _{3}^{n-1}\\\vdots &\vdots &\vdots &\ddots &\vdots \\1&\alpha _{m}&\alpha _{m}^{2}&\dots &\alpha _{m}^{n-1}\end{bmatrix}}$

$V_{n}={\begin{bmatrix}x_{1}&x_{2}&\cdots &x_{n}\\x_{1}^{2}&x_{2}^{2}&\cdots &x_{n}^{2}\\\vdots &\vdots &\ddots &\vdots \\x_{1}^{n}&x_{2}^{n}&\cdots &x_{n}^{n}\end{bmatrix}}$

$V={\begin{bmatrix}0^{0}&0^{1}&0^{2}&\dots &0^{n}\\1^{0}&1^{1}&1^{2}&\dots &1^{n}\\2^{0}&2^{1}&2^{2}&\dots &2^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\n^{0}&n^{1}&n^{2}&\dots &n^{n}\end{bmatrix}}$

$f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(x_{0})}{n!}}(x-x_{0})^{n}.$

${\begin{bmatrix}f_{0}\\f_{1}\\f_{2}\\\vdots \\f_{n}\end{bmatrix}}={\begin{bmatrix}0^{0}&0^{1}&0^{2}&\dots &0^{n}\\1^{0}&1^{1}&1^{2}&\dots &1^{n}\\2^{0}&2^{1}&2^{2}&\dots &2^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\n^{0}&n^{1}&n^{2}&\dots &n^{n}\end{bmatrix}}{\begin{bmatrix}{\frac {f^{(0)}(x_{0})}{0!}}h^{0}\\{\frac {f^{(1)}(x_{0})}{1!}}h^{1}\\{\frac {f^{(2)}(x_{0})}{2!}}h^{2}\\\vdots \\{\frac {f^{(n)}(x_{0})}{n!}}h^{n}\end{bmatrix}}\Rightarrow {\begin{bmatrix}f^{(0)}(x_{0}){\frac {h^{0}}{0!}}\\f^{(1)}(x_{0}){\frac {h^{1}}{1!}}\\f^{(2)}(x_{0}){\frac {h^{2}}{2!}}\\\vdots \\f^{(n)}(x_{0}){\frac {h^{n}}{n!}}\end{bmatrix}}={\begin{bmatrix}0^{0}&0^{1}&0^{2}&\dots &0^{n}\\1^{0}&1^{1}&1^{2}&\dots &1^{n}\\2^{0}&2^{1}&2^{2}&\dots &2^{n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\n^{0}&n^{1}&n^{2}&\dots &n^{n}\end{bmatrix}}^{-1}{\begin{bmatrix}f_{0}\\f_{1}\\f_{2}\\\vdots \\f_{n}\end{bmatrix}}$

${\begin{bmatrix}{\frac {1}{1!}}&{\frac {1}{2!}}&{\frac {1}{3!}}&\dots &{\frac {1}{n!}}\\0&{\frac {1}{1!}}&{\frac {1}{2!}}&\dots &{\frac {1}{(n-1)!}}\\0&0&{\frac {1}{1!}}&\dots &{\frac {1}{(n-2)!}}\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\dots &{\frac {1}{1!}}\end{bmatrix}}^{-1}\!\!\!\!=\,{\begin{bmatrix}{\frac {B_{0}}{0!}}&{\frac {B_{1}}{1!}}&{\frac {B_{2}}{2!}}&\dots &{\frac {B_{n-1}}{(n-1)!}}\\0&{\frac {B_{0}}{0!}}&{\frac {B_{1}}{1!}}&\dots &{\frac {B_{n-2}}{(n-2)!}}\\0&0&{\frac {B_{0}}{0!}}&\dots &{\frac {B_{n-3}}{(n-3)!}}\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\dots &{\frac {B_{0}}{0!}}\end{bmatrix}}$

Constructing the interpolation polynomial

The red dots denote the data points (x_k,y_k), while the blue curve shows the interpolation polynomial.

Suppose that the interpolation polynomial is in the form

$p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{2}x^{2}+a_{1}x+a_{0}.\qquad (1)$

The statement that p interpolates the data points means that

$p(x_{i})=y_{i}\qquad {\mbox{for all }}i\in \left\{0,1,\dots ,n\right\}.$

If we substitute equation (1) in here, we get a system of linear equations in the coefficients $a_{k}$. The system in matrix-vector form reads

${\begin{bmatrix}x_{0}^{n}&x_{0}^{n-1}&x_{0}^{n-2}&\ldots &x_{0}&1\\x_{1}^{n}&x_{1}^{n-1}&x_{1}^{n-2}&\ldots &x_{1}&1\\\vdots &\vdots &\vdots &&\vdots &\vdots \\x_{n}^{n}&x_{n}^{n-1}&x_{n}^{n-2}&\ldots &x_{n}&1\end{bmatrix}}{\begin{bmatrix}a_{n}\\a_{n-1}\\\vdots \\a_{0}\end{bmatrix}}={\begin{bmatrix}y_{0}\\y_{1}\\\vdots \\y_{n}\end{bmatrix}}.$

We have to solve this system for $a_{k}$ to construct the interpolant $p(x).$ The matrix on the left is commonly referred to as a Vandermonde matrix.

$\det(V)=\prod _{i,j=0,i<j}^{n}(x_{i}-x_{j})=\prod _{0\leq i<j\leq n}(x_{i}-x_{j})$

$L(x):=\sum _{j=0}^{k}y_{j}\ell _{j}(x)$

of Lagrange basis polynomials

$\ell _{j}(x):=\prod _{\begin{smallmatrix}0\leq m\leq k\\m\neq j\end{smallmatrix}}{\frac {x-x_{m}}{x_{j}-x_{m}}}={\frac {(x-x_{0})}{(x_{j}-x_{0})}}\cdots {\frac {(x-x_{j-1})}{(x_{j}-x_{j-1})}}{\frac {(x-x_{j+1})}{(x_{j}-x_{j+1})}}\cdots {\frac {(x-x_{k})}{(x_{j}-x_{k})}}.$

My sandbox