University of Florida/Egm6341/s10.team2.niki/HW2

Using the following equations find the expressions for ${\displaystyle {\boldsymbol {c_{i}}}}$ in terms of ${\displaystyle {\boldsymbol {x_{i}}}}$ and ${\displaystyle {\boldsymbol {f(x_{i})}}}$ where i=0,1,2

${\displaystyle \displaystyle {\boldsymbol {f_{2}(x)=p_{2}(x)=c_{2}x^{2}+c_{1}x+c_{0}}}}$

(1 p8-3)

${\displaystyle \displaystyle {\boldsymbol {p_{2}(x)=\sum _{i=0}^{2}(l_{i}(x)f(x_{i}))}}}$

(3 p8-3)

We have the general formula for the Lagrange basis function ${\displaystyle {\boldsymbol {l_{i}(x)}}}$ as

${\displaystyle \displaystyle {\boldsymbol {l_{i}(x)=\prod _{j=0;j\neq i}^{n}\left({\frac {x-x_{j}}{x_{i}-x_{j}}}\right)}}}$

(2 p7-3)

for the case of Simple Simpson's Rule, n =2 i.e i=0,1,2. For the given interval ${\displaystyle [a,b]}$

${\displaystyle {\boldsymbol {x_{0}=a}}}$

${\displaystyle {\boldsymbol {x_{2}=b}}}$

${\displaystyle {\boldsymbol {x_{1}={\frac {a+b}{2}}}}}$

Expanding equation 3 p8-3 we get:

${\displaystyle {\boldsymbol {p_{2}(x)=l_{0}(x)f(x_{0})+l_{1}(x)f(x_{1})+l_{2}(x)f(x_{2})}}}$ where,

${\displaystyle {\boldsymbol {l_{0}(x)=\left({\frac {x-x_{1}}{x_{0}-x_{1}}}\right)\left({\frac {x-x_{2}}{x_{0}-x_{2}}}\right)}}}$;

${\displaystyle {\boldsymbol {l_{1}(x)=\left({\frac {x-x_{0}}{x_{1}-x_{0}}}\right)\left({\frac {x-x_{2}}{x_{1}-x_{2}}}\right)}}}$;

${\displaystyle {\boldsymbol {l_{2}(x)=\left({\frac {x-x_{0}}{x_{2}-x_{0}}}\right)\left({\frac {x-x_{1}}{x_{2}-x_{1}}}\right)}}}$

Thus we have the polynomial as ${\displaystyle {\boldsymbol {p_{2}(x)=\left({\frac {x^{2}-(x_{2}+x_{1})x+(x_{1}x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2}))}}\right)f(x_{0})+\left({\frac {x^{2}-(x_{2}+x_{0})x+(x_{0}x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2}))}}\right)f(x_{1})+\left({\frac {x^{2}-(x_{0}+x_{1})x+(x_{1}x_{0})}{(x_{2}-x_{0})(x_{2}-x_{1}))}}\right)f(x_{2})}}}$

Grouping coefficients of ${\displaystyle {\boldsymbol {x^{2},x^{1},x^{0}}}}$, ${\displaystyle {\boldsymbol {p_{2}(x)=\left\{{\frac {f(x_{0})}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {f(x_{2})}{(x_{2}-x_{0})(x_{2}-x_{1})}}\right\}x^{2}-\left\{{\frac {f(x_{0})[x_{2}+x_{1}]}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {f(x_{1})[x_{2}+x_{0}]}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {f(x_{2})[x_{1}+x_{0}]}{(x_{2}-x_{0})(x_{2}-x_{1})}}\right\}x+\left\{{\frac {f(x_{0})[x_{1}x_{2}]}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {f(x_{1})[x_{0}x_{2}]}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {f(x_{2})[x_{0}x_{1}]}{(x_{2}-x_{0})(x_{2}-x_{1})}}\right\}}}}$

Comparing this equation with eqn 1p8-3 we see that

${\displaystyle \displaystyle {\boldsymbol {c_{2}=\left\{{\frac {f(x_{0})}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {f(x_{2})}{(x_{2}-x_{0})(x_{2}-x_{1})}}\right\}}}}$

(1)

${\displaystyle \displaystyle {\boldsymbol {c_{1}=-\left\{{\frac {f(x_{0})[x_{2}+x_{1}]}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {f(x_{1})[x_{2}+x_{0}]}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {f(x_{2})[x_{1}+x_{0}]}{(x_{2}-x_{0})(x_{2}-x_{1})}}\right\}}}}$

(2)

${\displaystyle \displaystyle {\boldsymbol {c_{0}=\left\{{\frac {f(x_{0})[x_{1}x_{2}]}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {f(x_{1})[x_{0}x_{2}]}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {f(x_{2})[x_{0}x_{1}]}{(x_{2}-x_{0})(x_{2}-x_{1})}}\right\}}}}$

(3)

For the Lagrange Interpolation Error verify the following:

${\displaystyle \displaystyle {\boldsymbol {E^{(n+1)}(x)=f^{(n+1)}(x)-0}}}$

(1)

We can write the Lagrange Interpolation error as

${\displaystyle {\boldsymbol {E(x)=f(x)-f_{n}(x)}}}$

differentiating the above expression once we get

${\displaystyle {\boldsymbol {E^{1}(x)=f^{1}(x)-f_{n}^{1}(x)}}}$

differentiating the expression (n+1) times we get

${\displaystyle {\boldsymbol {E^{n+1}(x)=f^{n+1}(x)-f_{n}^{n+1}(x)}}}$

But since ${\displaystyle {\boldsymbol {f_{n}(x)}}}$ is a polynomial of degree n the (n+1)th derivative is zero

${\displaystyle \displaystyle {\boldsymbol {E^{n+1}(x)=f^{n+1}(x)-0}}}$

(1)

Given the polynomial ${\displaystyle {\boldsymbol {f_{3}(x)=P_{3}(x)=3+8x^{1}-2x^{2}+6x^{3}}}}$ where ${\displaystyle {\boldsymbol {x\epsilon \left[0,1\right]}}}$ determine the exact integral ${\displaystyle {\boldsymbol {I}}}$ and the integral using Simpson's Rule ${\displaystyle {\boldsymbol {I_{n}}}}$