University of Florida/Egm6341/s10.Team2/HW2

Problem-1: Expressions for co-efficients of second degree Newton-Cotes integration polynomial

P. 9-1 of Lecture 9 Notes

Problem Statement

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

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

(1 p8-3)

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

(3 p8-3)

Solution

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

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

(2 p7-3)

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

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

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

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

Expanding equation 3 p8-3 we get:

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

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

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

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

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

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

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

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

(1)

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

(2)

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

(3)

Inference: This solution essentially shows the equivalence of the two methods for the Simpson's rule i.e. cubic polynomial and using Lagrange basis functions.

Solution for problem 1:Egm6341.s10.team2.niki 00:18, 8 February 2010 (UTC)

Proofread problem 1: Srikanth Madala 07:04, 11 February 2010 (UTC)

Problem-2: Application of Newton-Cotes method to derive the simple Simpson's rule

P. 9-1 of Lecture 9 Notes

Problem Statement

Derive the simple Simpson's rule using Eq.(4) on slide 8-3 Lecture-8 notes i.e. the second degree polynomial of Newton-Cotes method: $P_{2}\left(x\right)=\sum _{i=0}^{2}l_{i}\left(x\right)f\left(x_{i}\right)$

Solution

$I_{2}\left(f\right)$	$=\int _{x_{0}}^{x_{2}}f_{2}\left(x\right).dx=\int _{x_{0}}^{x_{2}}P_{2}\left(x\right).dx$
	$=\int _{x_{0}}^{x_{2}}\sum _{i=0}^{2}l_{i}\left(x\right)\cdot f\left(x_{i}\right)\cdot dx$
	$=\int _{x_{0}}^{x_{2}}\left[l_{0}\left(x\right)\cdot f\left(x_{0}\right)+l_{1}\left(x\right)\cdot f\left(x_{1}\right)+l_{2}\left(x\right)\cdot f\left(x_{2}\right)\right]\cdot dx$

But we know that $l_{i}\left(x\right)=\prod _{j=0\;\;j\neq i}^{j=2}{\frac {x-x_{j}}{x_{i}-x_{j}}}$

So we get the values of $l_{0}\left(x\right),l_{1}\left(x\right)$ and $l_{2}\left(x\right)$ as follows:

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

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

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

But we know that $x_{1}={\frac {\left(x_{0}+x_{2}\right)}{2}}$ or $2x_{1}=\left(x_{0}+x_{2}\right)$

$\therefore l_{0}\left(x\right)={\frac {\left(2x-x_{0}-x_{2}\right)}{\left(2x_{0}-x_{0}-x_{2}\right)}}{\frac {\left(x-x_{2}\right)}{\left(x_{0}-x_{2}\right)}}={\frac {2x^{2}-\left(3x_{2}+x_{0}\right)x+\left(x_{2}^{2}+x_{2}x_{0}\right)}{\left(x_{0}-x_{2}\right)^{2}}}$

Similarily by substituting the value of $x_{1}$ in the equations of $l_{1}\left(x\right)$ and $l_{2}\left(x\right)$, we get:

$l_{1}\left(x\right)={\frac {-4\left(x^{2}-\left(x_{0}+x_{2}\right)x+x_{0}x_{2}\right)}{\left(x_{0}-x_{2}\right)^{2}}}$

$l_{2}\left(x\right)={\frac {\left(2x-x_{0}-x_{2}\right)}{\left(2x_{2}-x_{0}-x_{2}\right)}}{\frac {\left(x-x_{0}\right)}{\left(x_{2}-x_{0}\right)}}={\frac {2x^{2}-\left(3x_{0}+x_{2}\right)x+\left(x_{0}^{2}+x_{2}x_{0}\right)}{\left(x_{0}-x_{2}\right)^{2}}}$

Now plugging back the values of $l_{0}\left(x\right),l_{1}\left(x\right)$ and $l_{2}\left(x\right)$ in the Equation for $I_{2}\left(f\right)$ , we get:

$I_{2}\left(f\right)=\int _{x_{0}}^{x_{2}}\left[\left({\frac {2x^{2}-\left(3x_{2}+x_{0}\right)x+\left(x_{2}^{2}+x_{2}x_{0}\right)}{\left(x_{0}-x_{2}\right)^{2}}}\right)\cdot f\left(x_{0}\right)+\left({\frac {-4\left(x^{2}-\left(x_{0}+x_{2}\right)x+x_{0}x_{2}\right)}{\left(x_{0}-x_{2}\right)^{2}}}\right)\cdot f\left(x_{1}\right)+\left({\frac {2x^{2}-\left(3x_{0}+x_{2}\right)x+\left(x_{0}^{2}+x_{2}x_{0}\right)}{\left(x_{0}-x_{2}\right)^{2}}}\right)\cdot f\left(x_{2}\right)\right]\cdot dx$

Now isolating the $x^{2}$ terms, $x$ terms and constant terms, we get three consolidated terms as follows:

Term-1: $x^{2}$ -term

${\frac {x^{2}}{\left(x_{0}-x_{2}\right)^{2}}}\left[2f\left(x_{0}\right)-4f\left(x_{1}\right)+2f\left(x_{2}\right)\right]$

Integrating this term between limits $x_{0}$ and $x_{2}$, we get:

${\frac {x_{2}^{3}-x_{0}^{3}}{3\left(x_{0}-x_{2}\right)^{2}}}\left[2f\left(x_{0}\right)-4f\left(x_{1}\right)+2f\left(x_{2}\right)\right]={\frac {\left(x_{2}^{2}+x_{0}^{2}+x_{2}x_{0}\right)}{3\left(x_{0}-x_{2}\right)}}\left[2f\left(x_{0}\right)-4f\left(x_{1}\right)+2f\left(x_{2}\right)\right]$

Term-2: $x$ -term

${\frac {-x}{\left(x_{0}-x_{2}\right)^{2}}}\left[\left(3x_{2}+x_{0}\right)f\left(x_{0}\right)-4\left(x_{2}+x_{0}\right)f\left(x_{1}\right)+\left(3x_{0}+x_{2}\right)f\left(x_{2}\right)\right]$

Integrating this term between limits $x_{0}$ and $x_{2}$, we get:

${\frac {x_{2}^{2}-x_{0}^{2}}{2\left(x_{0}-x_{2}\right)^{2}}}\left[\left(3x_{2}+x_{0}\right)f\left(x_{0}\right)-4\left(x_{2}+x_{0}\right)f\left(x_{1}\right)+\left(3x_{0}+x_{2}\right)f\left(x_{2}\right)\right]={\frac {x_{2}+x_{0}}{2\left(x_{2}-x_{0}\right)}}\left[\left(3x_{2}+x_{0}\right)f\left(x_{0}\right)-4\left(x_{2}+x_{0}\right)f\left(x_{1}\right)+\left(3x_{0}+x_{2}\right)f\left(x_{2}\right)\right]$

Term-3: Constant term

${\frac {1}{\left(x_{0}-x_{2}\right)^{2}}}\left[\left(x_{2}^{2}+x_{2}x_{0}\right)f\left(x_{0}\right)-4\left(x_{2}x_{0}\right)f\left(x_{1}\right)+\left(x_{0}^{2}+x_{2}x_{0}\right)f\left(x_{2}\right)\right]$

Integrating this term between limits $x_{0}$ and $x_{2}$, we get:

${\frac {x_{2}-x_{0}}{\left(x_{2}-x_{0}\right)^{2}}}\left[\left(x_{2}^{2}+x_{2}x_{0}\right)f\left(x_{0}\right)-4\left(x_{2}x_{0}\right)f\left(x_{1}\right)+\left(x_{0}^{2}+x_{2}x_{0}\right)f\left(x_{2}\right)\right]={\frac {1}{\left(x_{2}-x_{0}\right)}}\left[\left(x_{2}^{2}+x_{2}x_{0}\right)f\left(x_{0}\right)-4\left(x_{2}x_{0}\right)f\left(x_{1}\right)+\left(x_{0}^{2}+x_{2}x_{0}\right)f\left(x_{2}\right)\right]$

Note that $I_{2}\left(f\right)=Term\;1+Term\;2+Term\;3$

Now consolidating the coefficients of $f\left(x_{0}\right),f\left(x_{1}\right)and\;f\left(x_{2}\right)$ in all the three terms, we get:

Consolidated Coefficient of $f\left(x_{0}\right)$ as:

$={\frac {1}{x_{2}-x_{0}}}\left[{\frac {2}{3}}x_{2}^{2}+{\frac {2}{3}}x_{0}^{2}+{\frac {2}{3}}x_{2}x_{0}-{\frac {3}{2}}x_{2}^{2}-{\frac {1}{2}}x_{2}x_{0}-{\frac {3}{2}}x_{2}x_{0}-{\frac {1}{2}}x_{0}^{2}+x_{2}^{2}+x_{2}x_{0}\right]$
$={\frac {1}{x_{2}-x_{0}}}\left[\left(1+{\frac {2}{3}}-{\frac {3}{2}}\right)x_{2}^{2}+\left({\frac {2}{3}}-{\frac {1}{2}}\right)x_{0}^{2}+\left({\frac {2}{3}}-{\frac {1}{2}}-{\frac {3}{2}}+1\right)x_{0}x_{2}\right]$
$={\frac {1}{6\left(x_{2}-x_{0}\right)}}\left[\left(x_{2}-x_{0}\right)^{2}\right]={\frac {\left(x_{2}-x_{0}\right)}{6}}$

Consolidated Coefficient of $f\left(x_{1}\right)$ as:

$={\frac {1}{x_{2}-x_{0}}}\left[{\frac {-4}{3}}x_{2}^{2}-{\frac {4}{3}}x_{0}^{2}-{\frac {4}{3}}x_{2}x_{0}+{\frac {4}{2}}x_{2}^{2}+{\frac {4}{2}}x_{0}^{2}+\left({\frac {4}{2}}\right)2x_{2}x_{0}-4x_{2}x_{0}\right]$
$={\frac {4}{6\left(x_{2}-x_{0}\right)}}\left[\left(x_{2}-x_{0}\right)^{2}\right]={\frac {4\left(x_{2}-x_{0}\right)}{6}}$

Consolidated Coefficient of $f\left(x_{2}\right)$ as:

$={\frac {1}{x_{2}-x_{0}}}\left[{\frac {2}{3}}x_{2}^{2}+{\frac {2}{3}}x_{0}^{2}+{\frac {2}{3}}x_{2}x_{0}-{\frac {3}{2}}x_{2}x_{0}-{\frac {3}{2}}x_{0}^{2}-{\frac {1}{2}}x_{2}^{2}-{\frac {1}{2}}x_{2}x_{0}+x_{0}^{2}+x_{2}x_{0}\right]$
$={\frac {1}{x_{2}-x_{0}}}\left[\left({\frac {2}{3}}-{\frac {1}{2}}\right)x_{2}^{2}+\left({\frac {2}{3}}-{\frac {3}{2}}+1\right)x_{0}^{2}+\left({\frac {2}{3}}-{\frac {1}{2}}-{\frac {3}{2}}+1\right)x_{0}x_{2}\right]$
$={\frac {1}{6\left(x_{2}-x_{0}\right)}}\left[\left(x_{2}-x_{0}\right)^{2}\right]={\frac {\left(x_{2}-x_{0}\right)}{6}}$

Now the value of $I_{2}\left(f\right)$ is the sum of all the consolidated coefficient multiplied by their respective $f\left(x_{i}\right)$ term. i.e.

$I_{2}\left(f\right)$	$={\frac {\left(x_{2}-x_{0}\right)}{6}}\left[f\left(x_{0}\right)\right]+{\frac {4\left(x_{2}-x_{0}\right)}{6}}\left[f\left(x_{1}\right)\right]+{\frac {\left(x_{2}-x_{0}\right)}{6}}\left[f\left(x_{2}\right)\right]$
	$={\frac {\left(x_{2}-x_{0}\right)}{6}}\left[f\left(x_{0}\right)+4f\left(x_{1}\right)+f\left(x_{2}\right)\right]={\frac {h}{3}}\left[f\left(x_{0}\right)+4f\left(x_{1}\right)+f\left(x_{2}\right)\right]$
	$\displaystyle {\boldsymbol {={\frac {h}{3}}\left[f\left(x_{0}\right)+4f\left(x_{1}\right)+f\left(x_{2}\right)\right]}}$

Where $h:={\frac {\left(x_{2}-x_{0}\right)}{2}}$

Thus, the simple Simpson's rule is derived using the Lagrange polynomial of second degree.

Solution for problem 2: Srikanth Madala 07:04, 11 February 2010 (UTC)

Proofread problem 2:

Problem-3: Study of Newton-Cotes method of numerical integration with the help of an example

Problem Statement

P.9-2 of Lecture 9 Notes

For the function:

$F(x)={\frac {\exp ^{x}-1}{x}}$ on the interval between 0 and 1 and letting $\qquad x_{0}=a\quad and\quad x_{n}=b$

Consider: n=1,2,4,8,16.

1) Construct $F_{n}(x)=\sum _{i}^{n}l_{i,n}(x)f(x_{i})$

Plot Fn for n=1,2,4,8,16

2) Compute the integral for n=1,2,4,8 and compare to I=1.3179022

$\int _{a}^{b}F_{n}(x)dx$

3) for n=4 plot:

$l_{0}\quad ,\quad l_{1},\quad l_{2}$

1: Calculation of Function

$F_{n}(x)=\sum _{i}^{n}L_{i,n}(x)F(x_{i})$

For n=1

$x_{0}=0\quad x_{1}=1$

$\;F_{n}(x)=xf(x_{1})-xf(x_{0})+f(x_{0})$

For n=2

$x_{0}=0\quad x_{1}=0.5\quad x_{2}=1$

$\;F_{n}(x)=(2x^{2}-3x+1)f(x_{0})+(-4x^{2}+4x)f(x_{1})+(2x^{2}-x)f(x_{2})$

For n=4

$\;x_{0}=0\quad x_{1}=0.25\quad x_{2}=0.5\quad x_{3}=0.75\quad x_{4}=1$

$\;F_{n}(x)=L_{0}(x)F(x_{0})+L_{1}(x)F(x_{1})+L_{2}(x)F(x_{2})+L_{3}(x)F(x_{3})+L_{4}(x)F(x_{4})$

$\;L_{0}=10.66667x^{4}-26.66667x^{3}+23.33333x^{2}-8.33333x+1$

$\;L_{1}=-42.66667x^{4}+96x^{3}-69.33333x^{2}+16x$

$\;L_{2}=64x^{4}-128x^{3}+76x^{2}-12x$

$\;L_{3}=-42.66667x^{4}+74.66667x^{3}-37.33333x^{2}+5.33333x$

$\;L_{4}=10.66667x^{4}-16x^{3}+7.33333x^{2}-x$

For n=8

$\;x_{0}=0\quad x_{1}={\frac {1}{8}}\quad x_{2}={\frac {2}{8}}\quad x_{3}={\frac {3}{8}}\quad x_{4}={\frac {4}{8}}\quad x_{5}={\frac {5}{8}}\quad x_{6}={\frac {6}{8}}\quad x_{7}={\frac {7}{8}}\quad x_{8}={\frac {8}{8}}$

$\;F_{n}(x)=L_{0}(x)F(x_{0})+L_{1}(x)F(x_{1})+L_{2}(x)F(x_{2})+L_{3}(x)F(x_{3})+L_{4}(x)F(x_{4})+L_{5}(x)F(x_{5})+L_{6}(x)F(x_{6})+L_{7}(x)F(x_{7})+L_{8}(x)F(x_{8})$

$\;L_{0}=416.10158x^{8}-1872.45714x^{7}+3549.86667x^{6}-3686.40x^{5}+2280.53334x^{4}-854.40x^{3}+187.49841x^{2}-21.74286x+1$

$\;L_{1}=-3328.81269x^{8}+14563.55555x^{7}-26578.48889x^{6}+26168.88889x^{5}-14973.15556x^{4}+4963.55556x^{3}-879.54286x^{2}+64x$

$\;L_{2}=11650.84445x^{8}-49516.08889x^{7}+87017.24445x^{6}-81464.88889x^{5}+43488.71111x^{4}-13051.02222x^{3}+1987.2x^{2}-112x$

$\;L_{3}=-23301.688889x^{8}+96119.46667x^{7}-162747.73333x^{6}+145408x^{5}-73181.86667x^{4}+20403.2x^{3}-2848.71111x^{2}+149.33333x$

$\;L_{4}=29127.11111x^{8}-116508.44444x^{7}+190236.44444x^{6}-162929.77778x^{5}+78172.44444x^{4}-20721.77778x^{3}+2764x^{2}-140x$

$\;L_{5}=-23301.68889x^{8}+90294.04444x^{7}-142358.75556x^{6}+117464.17778x^{5}-54294.75556x^{4}+13912.17778x^{3}-1804.8x^{2}+89.6x$

$\;L_{6}=11650.84445x^{8}-43690.66666x^{7}+66628.26667x^{6}-53248.0x^{5}+23918.93333x^{4}-5984x^{3}+761.95556x^{2}-37.33333x$

$\;L_{7}=-3328.81267x^{8}+12066.94603x^{7}-17840.35555x^{6}+13880.88889x^{5}-6098.48889x^{4}+1499.02222x^{3}-188.34286x^{2}+9.14286x$

$\;L_{8}=416.10159x^{8}-1456.35556x^{7}+2093.51111x^{6}-1592.88889x^{5}+687.64444x^{4}-166.75556x^{3}+20.74285x^{2}-x$

For n=16

$\;L_{0}=881657.95157x^{16}-7494092.58831x^{15}+29273799.17310x^{14}-69671642.03199x^{13}+112925052.51958x^{12}-131967887.58325x^{11}+114825387.35065x^{10}-75729669.30819x^{9}+38171908.26140x^{8}-14714527.588281x^{7}+04309632.78652x^{6}-945274.82454x^{5}+151495.79946x^{4}-17046.98375x^{3}+1260.15769x^{2}-54.09166x+1$

$\;L_{1}=-14106527.22506x^{16}+119023823.46147x^{15}-460941797.80334x^{14}+1085937410.14913x^{13}-1738929752.17911x^{12}+2002803091.82081x^{11}-1712031004.37163x^{10}+1104672771.15791x^{9}-541708483.98513x^{8}+201575661.16345x^{7}-56355645.76161x^{6}+11602169.33258x^{5}-1698797.20815x^{4}+166576.91462x^{3}-9751.46595x^{2}+256x$

$\;L_{2}=105798954.18798x^{16}-0886066241.32429x^{15}+3402097620.60708x^{14}-7935334841.26295x^{13}+12559089447.19286x^{12}-14266260329.09093x^{11}+11995763940.94187x^{10}-7588089824.36582x^{9}+3632117763.32303x^{8}-1311728590.17858x^{7}+353189860.61015x^{6}-69284246.36891x^{5}+9518965.13969x^{4}-855767.40852x^{3}+44247.99734x^{2}-960x$

$\;L_{3}=-493728452.87722x^{16}+4104117764.54188x^{15}-15623805456.08729x^{14}+36086656014.89820x^{13}-56471781408.17665x^{12}+63313558032.58708x^{11}-52430924785.25504x^{10}+32577816415.35461x^{9}-15267928048.50856x^{8}+5377398940.34311x^{7}-1405132059.13719x^{6}+265891640.65593x^{5}-34982965.07744x^{4}+2987004.95254x^{3}-145624.88060x^{2}+2986.66667x$

$\;L_{4}=1604617471.8510x^{16}-13238094142.7704x^{15}+49968790959.3588x^{14}-114310190758.3826x^{13}+176946047896.0571x^{12}-195945043427.5014x^{11}+159995944121.3113x^{10}-97829012110.5894x^{9}+45015620008.1122x^{8}-15526535208.6470x^{7}+3961897869.3057x^{6}-729925713.3421x^{5}+93240926.6909x^{4}-7715278.7671x^{3}+364667.3131x^{2}-7280.0000x$

$\;L_{5}=-3851081932.4423x^{16}+31530733321.8714x^{15}-118014600625.0386x^{14}+267446169700.4090x^{13}-409679701374.1887x^{12}+448410826284.2035x^{11}-361428908733.8343x^{10}+217840661558.8782x^{9}-98659688548.6735x^{8}+33441903834.1596x^{7}-8373881063.3471x^{6}+1512122601.3085x^{5}-189195339.1544x^{4}+15337681.5698x^{3}-711343.3212x^{2}+13977.6x$

$\;L_{6}=7060316876.1442x^{16}-57365074618.6718x^{15}+212912680796.2242x^{14}-478088254093.5941x^{13}+725020725291.7742x^{12}-784916071782.2535x^{11}+625178174985.6762x^{10}-372001376200.4074x^{9}+166180125282.1892x^{8}-55516389946.1491x^{7}+13692893124.6509x^{6}-2434925873.1974x^{5}+300081159.6671x^{4}-23981811.0600x^{3}+1098163.6741x^{2}-21354.6667x$

$\;L_{7}=-10086166965.920x^{16}+81319721162.733x^{15}-299314884531.628x^{14}+666093322863.382x^{13}+1000446772071.373x^{12}+1072017171171.158x^{11}-844594918904.077x^{10}+496837139865.232x^{9}-219320381819.450x^{8}+72381528563.948x^{7}-17635280331.068x^{6}+3098508847.931x^{5}-377514324.910x^{4}+29854977.045x^{3}-1354651.574x^{2}+26148.571x$

$\;L_{8}=11346937836.660x^{16}-90775502693.283x^{15}+331366044011.222x^{14}-730991010946.104x^{13}+1087849920454.064x^{12}-1154501752969.400x^{11}+900551858718.191x^{10}-524364914637.391x^{9}+229090002005.604x^{8}-74830969058.400x^{7}+18049489433.320x^{6}-3140942275.210x^{5}+379279801.510x^{4}-29754780.212x^{3}+1340839.429x^{2}-25740x$

$\;L_{9}=-10086166965.9203x^{16}+80058950291.9925x^{15}-289859103001.0773x^{14}+633997839407.8629x^{13}-935238816157.1799x^{12}+983640799863.9704x^{11}-760304481367.9756x^{10}+438676146116.2794x^{9}-189931298320.0814x^{8}+61497840304.9114x^{7}-14709663906.2828x^{6}+2539758045.4894x^{5}-304498045.2923x^{4}+23737343.7162x^{3}-1063948.1905x^{2}+20337.7778x$

$\;L_{10}=7060316876.1442x^{16}-55599995399.6357x^{15}+199674586653.4538x^{14}-433133892733.7696x^{13}+633595137618.2660x^{12}-660801882755.2526x^{11}+506520525181.9884x^{10}-289867863581.2933x^{9}+124513226619.0337x^{8}-40013170290.0740x^{7}+9503307923.1606x^{6}-1630193342.9661x^{5}+194307522.7623x^{4}-15070044.2088x^{3}+672565.1911x^{2}-12812.8x$

$\;L_{11}=-3851081932.4423x^{16}+30086577597.2055x^{15}-107183432690.0446x^{14}+230637122421.3279x^{13}-334693607740.9036x^{12}+346333877641.7664x^{11}-263452751068.9336x^{10}+149663429178.3096x^{9}-63841287725.7351x^{8}+20382171194.1772x^{7}-4811733315.5251x^{6}+820893779.1809x^{5}-97369178.8764x^{4}+7519914.5780x^{3}-334427.5394x^{2}+6353.4545x$

$\;L_{12}=1604617471.8510x^{16}-12435785406.8449x^{15}+43951475439.9177x^{14}-93838781918.2841x^{13}+135144509146.9397x^{12}-138823173541.3109x^{11}+104864824822.2035x^{10}-59179379524.2647x^{9}+25088338392.5610x^{8}-7964186416.2542x^{7}+1870391859.2768x^{6}-317606286.2109x^{5}+37517640.3682x^{4}-2887280.1637x^{3}+128026.8822x^{2}-2426.6667x$

$\;L_{13}=-493728452.87722x^{16}+3795537481.49362x^{15}-13309453333.22533x^{14}+28202188704.66898x^{13}-40323751088.42652x^{12}+41138969287.27365x^{11}-30876804370.45528x^{10}+17321211261.59502x^{9}-7302784476.34160x^{8}+2306623062.41091x^{7}-539263122.24266x^{6}+91202614.92199x^{5}-10735523.07630x^{4}+823698.40506x^{3}-36433.35509x^{2}+689.23077x$

$\;L_{14}=105798954.18797x^{16}-806717025.68331x^{15}+2806978503.29972x^{14}-5904490853.45158x^{13}+8384576805.58060x^{12}-8499641805.10702x^{11}+6341859902.60960x^{10}-3538432902.20016x^{9}+1484500201.94362x^{8}-466805633.89329x^{7}+108701004.725845x^{6}-18319599.810063x^{5}+2149846.10200x^{4}-164522.71173x^{3}+7261.55064x^{2}-137.14286x$

$\;L_{15}=-14106527.22506x^{16}+106680612.13954x^{15}-368367712.88886x^{14}+769401541.678536x^{13}-1085486894.98980x^{12}+1093842237.27907x^{11}-811729100.16131x^{10}+450678677.52992x^{9}-188239271.99821x^{8}+58958123.91421x^{7}-13680883.41476x^{6}+2298568.99135x^{5}-269024.36205x^{4}+20541.40071x^{3}-904.95995x^{2}+17.06667x$

$\;L_{16}=881657.95157x^{16}-6612434.63675x^{15}+22661364.53636x^{14}-47010277.49563x^{13}+65914775.02395x^{12}-66053112.55929x^{11}+48772274.79135x^{10}-26957394.51683x^{9}+11214513.74456x^{8}-3500013.84371x^{7}+809618.94280x^{6}-135655.88173x^{5}+15839.91772x^{4}-1207.06603x^{3}+53.09166x^{2}-x$

2:Integral Comparison

Integrals using Newton Cotes
	Actual Value	1.3179022
n	Estimated Value	Percent Difference
n=1	1.35915	3.12981%
n=2	1.318008	.00802791%
n=4	1.318009	.00810379%
n=8	1.31790215	3.7939E-6

3: Lagrange Polynomial Plots

The plots for l3 and l4 (3rd and 4th Lagrange Polynomial) are not necessary to be plotted as their graph are symmetrical to one another (a mirror image of the l0 and l1 polynomials).

Solution for problem 3: Guillermo Varela

Problem-4: Simple to composite Simpson's rule

Solution for problem 4: Egm6341.s10.team2.lee 18:48, 7 February 2010 (UTC)

Proofread problem 4:

Problem-5: Error Bound

Refer to P. 11-1 on Lecture-11 Notes

Problem Statement

Find $n\!$ if:

$\left|f_{n}^{T}({\frac {7\pi }{8}})-f({\frac {7\pi }{8}})\right|\leq \left|f_{4}^{L}({\frac {7\pi }{8}})-f({\frac {7\pi }{8}})\right|\leq {\frac {\left|q_{4+1}(t)\right|}{(4+1)!}}$

For $f(x)=sin(x)\!$ and $x_{0}=0,x_{1}={\frac {\pi }{4}},x_{2}={\frac {\pi }{2}},x_{3}={\frac {3\pi }{4}},x_{4}=1$

Solution

Constructing the Lagrangian Interpolating Function:

$f_{4}^{L}(x)=P_{4}(x)=\sum _{i=0}^{4}l_{i,4}(x)f(x_{i})$

Where

$l_{i,4}(x)=\prod _{j=0,j\neq 4}^{4}{\frac {x-x_{j}}{x_{i}-x_{j}}}$

The Lagrangian Error

$E_{4}^{L}({\frac {7\pi }{8}})=\left|f_{4}^{L}({\frac {7\pi }{8}})-f({\frac {7\pi }{8}})\right|\leq {\frac {\left|q_{5}({\frac {7\pi }{8}})\right|}{5!}}$

By Computing

$f_{4}^{L}({\frac {7\pi }{8}}),f({\frac {7\pi }{8}}),{\frac {q_{5}({\frac {7\pi }{8}})}{5!}}$

Must plot the functions to compute

$E_{4}^{L}({\frac {7\pi }{8}})=1.4808e-3$

The Taylor Series Error

$\left|f_{n}^{T}({\frac {7\pi }{8}})-f({\frac {7\pi }{8}})\right|$

For $n=0,1,2,3...\!$

n=0

$\left|f_{0}^{T}(x)-f(x)\right|=\left|sin(x)-sin(x)\right|=0$

n=1

$=\left|{\frac {1}{\sqrt {2}}}\left[1+{\frac {{\frac {7\pi }{8}}-{\frac {\pi }{4}}}{1!}}\right]-sin({\frac {7\pi }{8}})\right|$

$=1.7128\!$

...n=9

$\left|f_{n}^{T}({\frac {7\pi }{8}})-f({\frac {7\pi }{8}})\right|=6.31E-4\leq E_{4}^{L}({\frac {7\pi }{8}})=1.4808e-3$

Solution for problem 5:Egm6341.s10.team2.patodon 21:43, 10 February 2010 (UTC)

Proofread problem 5:

Problem 6: (n+1)^th derivative of Lagrange Interpolation Error

P. 12-2 reference: Lecture-12 Notes

Problem Statement

For the Lagrange Interpolation Error verify the following:

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

(1)

Solution

We can write the Lagrange Interpolation error as

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

differentiating the above expression once we get

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

differentiating the expression (n+1) times we get

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

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

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

(1)

Solution for problem 6: Egm6341.s10.team2.niki 00:20, 8 February 2010 (UTC)
Proofread problem 6:

Problem-7: Intermediate step in the proof of Error in Newton Cotes method

P. 12-3 (top); reference: Lecture-12 Notes

Problem Statement

To Prove that the $\left(n+1\right)^{th}$ derivative of $q_{n+1}\left(x\right)$ is $\left(n+1\right)!$

Solution

We know that by definition $q_{n+1}\left(x\right)$ is:

$q_{n+1}\left(x\right)=\prod _{j=0}^{n}\left(x-x_{j}\right)$

Expanding the above terms in the product, we get:

$q_{n+1}\left(x\right)=\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{2}\right).....\left(x-x_{n}\right)$

We see that the above expression is a $\left(n+1\right)^{th}$ degree polynomial in $x$

Let it be expressed as:

$q_{n+1}\left(x\right)=x^{n+1}+C_{0}x^{n}+C_{1}x^{n-1}+C_{2}x^{n-2}+.....+C_{n}x^{0}$

Note that here the coefficient of $\;x^{n+1}$ is 1. As we successively differentiate the $q_{n+1}\left(x\right)$ term $\;\left(n+1\right)$ times, all the lower degree terms that are less than $\;\left(n+1\right)$ vanish. Therefore, we are concerned only about the $\;\left(n+1\right)^{th}$ degree term i.e. $\;x^{n+1}$ as we successively differentiate the equation.

${\frac {\mathrm {d} }{\mathrm {d} x}}\left[q_{n+1}\left(x\right)\right]=\left(n+1\right)x^{\left(n+1\right)-1}+other\;terms$

${\frac {\mathrm {d^{2}} }{\mathrm {d} x^{2}}}\left[q_{n+1}\left(x\right)\right]=\left(n+1\right)\left(\left(n+1\right)-1\right)x^{\left(n+1\right)-2}+other\;terms$

$\vdots \qquad \qquad \qquad \vdots \qquad \qquad \qquad \vdots \qquad \qquad \qquad \vdots$
$\vdots \qquad \qquad \qquad \vdots \qquad \qquad \qquad \vdots \qquad \qquad \qquad \vdots$

${\begin{matrix}{\frac {\mathrm {d} ^{n+1}}{\mathrm {d} x^{n+1}}}\left[q_{n+1}\left(x\right)\right]&=&\left(n+1\right)\left(\left(n+1\right)-1\right)\cdots \cdots \left(\left(n+1\right)-n\right)x^{\left(n+1\right)-\left(n+1\right)}\\\\\ &=&\left(n+1\right)\left(\left(n+1\right)-1\right)\cdots \cdots \left(\left(n+1\right)-n\right)\\\\\ &=&\left(n+1\right)\cdot \left(n\right)\cdots \cdots 2\;\cdot 1\\\\\ &=&\left(n+1\right)!\end{matrix}}$

Thus, we proved that the $\left(n+1\right)^{th}$ derivative of $q_{n+1}\left(x\right)$ is $\left(n+1\right)!$

Solution for problem 7: Egm6341.s10.team2.madala 11:58, 8 February 2010 (UTC)

Proofread problem 7:

Problem-8: Geometric Interpretation of G(x) Error for Log(x)

P.12-3

Problem Statement

Let $\;F(x)=\log(x)\quad and\quad t=2\quad x_{0}=3\quad x_{1}=4\quad ...\quad x_{6}=9$

Plot:

1) A graph of $\;F(x)$ and $\;F_{n}(x)$

2) A plot of the Lagrange Polynomial when i=3

3) A plot of $\;q_{n+1}(t)$ when n=6

Part 1

The Lagrange Interpolation Function is as follows:

$\;F_{n}=-4.92641E-6x^{6}+.000212^{5}+-.0039x^{4}+.039876x^{3}-.25286x^{2}+1.114899x-.77941$

The plot is as follows:

Part 2

The Lagrange Polynomial is as follows:

$\;L_{3}(x)=-0.027777778x{^{6}}+x^{5}-14.61111111x^{4}+110.6666667x^{3}-457.3611111x^{2}+976.3333333x-840$

The plot as follows:

Part 3

$\;q_{7}(x)=x^{7}-42x^{6}+742x^{5}-7140x^{4}+40369x^{3}-133938x^{2}+241128x-181440$

The Plot as Follows:

Solution for problem 8:

Proofread problem 8: Srikanth Madala 07:04, 11 February 2010 (UTC)

Problem-9: Error in simple Trapezoidal rule

Solution for problem 9: Egm6341.s10.team2.lee 20:48, 9 February 2010 (UTC)

Proofread problem 9:

Problem 10: Simple Simpson's Rule Error

P. 13-2

Problem Statement

Show that the Error for the simple simpson's rule is:

$n=2\qquad q_{3}(x)=(x-x_{0})(x-x_{1})(x-x_{2})$

$\left|E_{2}\right|={\frac {(b-a)^{4}}{192}}M_{3}={\frac {2^{4}h^{4}M_{3}}{192}}$

$h={\frac {a+b}{2}}$

Solution

The error can be written:

${\frac {M_{3}}{3!}}\int _{a}^{b}\left|(x-a)(x-{\frac {a+b}{2}})(x-b)\right|dx$

with $x_{0}=a\quad x_{1}={\frac {a+b}{2}}\quad x_{2}=b$

The Integral can be evaluated as follows:

$\int _{a}^{b}\left|(x-a)(x-{\frac {a+b}{2}})(x-b)\right|dx=\int _{a}^{b}(x-a)(x-{\frac {a+b}{2}})(b-x)dx$

$\int _{a}^{b}(xb-x^{2}-ab+xa)(x-{\frac {a+b}{2}}dx$

$\int _{a}^{b}-x^{3}+x^{2}((b+a)+({\frac {a+b}{2}})+x(-ab-b{\frac {a+b}{2}}-a({\frac {b+a}{2}}))+ab({\frac {a+b}{2}})dx$

After algebraic manipulation and simplification the following is obtained:

${\frac {1}{4}}\left[-b^{4}-a^{4}+(2a+2b)(b^{3}-a{3})+(a^{2}+b^{2})(b^{2}-a^{2})+(a+b)(2ab^{2}-2a^{2}b)\right]$

which reduces to:

${\frac {1}{32}}(b-a)^{4}$

This result is used in the error function to yield:

${\frac {(b-a)^{4}}{32}}{\frac {M_{3}}{3!}}={\frac {(b-a)^{4}}{192}}M_{3}={\frac {h^{4}16}{192}}M_{3}$

Solution for problem 10: Guillermo Varela

Problem 11:To show that Simpson's rule can be used to integrate a cubic polynomial exactly

P. 13-3, reference: Lecture-13 Notes

Problem Statement

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

Case A: Determination of Exact Integral

${\boldsymbol {I=\int _{0}^{1}(3+8x-2x^{2}+6x^{3})dx}}$

${\boldsymbol {I=[3x+4x^{2}-{\frac {2}{3}}x^{3}+{\frac {3}{2}}x^{4}]_{0}^{1}}}$

${\boldsymbol {I=3+4-{\frac {2}{3}}+1.5}}$

$\displaystyle {\boldsymbol {I=7.833}}$

(1)

Case B: Using Simple Simpson's rule

We have the Simple Simpson's rule as

$\displaystyle {\boldsymbol {I_{2}={\frac {h}{3}}\left\{f(x_{0})+4f(x_{1})+f(x_{2})\right\}}}$

(2 p7-2)

where ${\boldsymbol {h={\frac {a+b}{2}}={\frac {0+1}{2}}=0.5}}$

${\boldsymbol {I_{2}={\frac {0.5}{3}}\left\{f(0)+4f(0.5)+f(1)\right\}}}$ we know ${\boldsymbol {f(0)=3;f(0.5)=7.25;f(1)=15}}$ substituting we get

${\boldsymbol {I_{2}={\frac {0.5}{3}}\left\{3+4(7.25)+15\right\}={\frac {0.5}{3}}\left\{47\right\}}}$

$\displaystyle {\boldsymbol {I_{2}=7.833}}$

(2)

Conclusion

From the above we see that ${\boldsymbol {I=I_{2}=7.833}}$ which proves that the Simpsons rule can integrate a cubic polynomial exactly

Solution for problem 11: Egm6341.s10.team2.niki 00:23, 8 February 2010 (UTC)

Proofread problem 11:

Problem-12: Differentiation of a definite integral

Solution for problem 12:Egm6341.s10.team2.lee 20:54, 9 February 2010 (UTC)

Proofread problem 12:

Problem-13: Intermediate step in proving the tight error bound of the Simpson's rule

Solution for problem 13: Egm6341.s10.team2.lee 20:47, 9 February 2010 (UTC)

Proofread problem 13: Srikanth Madala 07:04, 11 February 2010 (UTC)

Problem-14: An intermediate step in the proof of tight error bound of the simple Simpson's rule

P. 15-2 (bottom), Refer: Lecture-15 Notes

Title

To Prove that: $e^{(3)}\left(t\right)={\frac {-t}{3}}\left[F^{(3)}\left(t\right)-F^{(3)}\left(-t\right)\right]$ which is an intermediate step in the proof of tight error bound of simple Simpson's rule

Solution

We know by the definition $e\left(t\right)$

$e\left(t\right)$	$=\left[\int _{-t}^{t}F\left(\tau \right).d\tau \right]-\left({\frac {t}{3}}\right)\left[F\left(-t\right)+4F\left(0\right)+F\left(t\right)\right]$
	$=\left[\int _{-t}^{k}F\left(\tau \right).d\tau +\int _{k}^{t}F\left(\tau \right).d\tau \right]-\left({\frac {t}{3}}\right)\left[F\left(-t\right)+4F\left(0\right)+F\left(t\right)\right]$

Differentiating with respect to 't', we get:

$e^{(1)}\left(t\right)=\left[F\left(-t\right)+F\left(t\right)\right]-\left({\frac {1}{3}}\right)\left[F\left(-t\right)+4F\left(0\right)+F\left(t\right)\right]-\left({\frac {t}{3}}\right)\left[-{F}'\left(-t\right)+{F}'\left(t\right)\right]$

Again Differentiating with respect to 't', we get:

$e^{(2)}\left(t\right)$	$=\left[-{F}'\left(-t\right)+{F}'\left(t\right)\right]-\left({\frac {1}{3}}\right)\left[-{F}'\left(-t\right)+{F}'\left(t\right)\right]-\left({\frac {1}{3}}\right)\left[-{F}'\left(-t\right)+{F}'\left(t\right)\right]-\left({\frac {t}{3}}\right)\left[{F}''\left(-t\right)+{F}''\left(t\right)\right]$
	$=\left({\frac {1}{3}}\right)\left[-{F}'\left(-t\right)+{F}'\left(t\right)\right]-\left({\frac {t}{3}}\right)\left[{F}''\left(-t\right)+{F}''\left(t\right)\right]$

Again Differentiating with respect to 't', we get:

$e^{(3)}\left(t\right)$	$=\left({\frac {1}{3}}\right)\left[{F}''\left(-t\right)+{F}''\left(t\right)\right]-\left({\frac {1}{3}}\right)\left[{F}''\left(-t\right)+{F}''\left(t\right)\right]-\left({\frac {t}{3}}\right)\left[-{F}'''\left(-t\right)+{F}'''\left(t\right)\right]$
	$={\frac {-t}{3}}\left[{F}'''\left(t\right)-{F}'''\left(-t\right)\right]={\frac {-t}{3}}\left[F^{(3)}\left(t\right)-F^{(3)}\left(-t\right)\right]$

Thus Proved.

Solution for problem 14: Srikanth Madala 07:04, 11 February 2010 (UTC)

Proofread problem 14:

Problem-15: Relationship between ζ and ζ ₄

P. 15-3 on Lecture-15 Notes

Problem Statement

Proof:

${\frac {-F^{(4)}(\zeta _{4})}{90}}={\frac {-(b-a)^{4}f^{(4)}(\zeta )}{1440}}$

Solution

Given:

$x(t)=x_{1}+ht\qquad and\qquad h={\frac {b-a}{2}}$

${\frac {dx(t)}{dt}}=h$

$\;F(t)=F(x(t))$

$F^{1}(t)={\frac {df(x(t))}{dt}}={\frac {df(x)}{dx}}{\frac {dx}{dt}}=f^{1}(x)h$

$F^{2}(t)={\frac {df^{1}(x(t))}{dt}}=h{\frac {df^{1}(x)}{dx}}{\frac {dx}{dt}}=f^{2}(x)h^{2}$

$F^{3}(t)={\frac {df^{2}(x(t))}{dt}}=h^{2}{\frac {df^{2}(x)}{dx}}{\frac {dx}{dt}}=f^{3}(x)h^{3}$

$F^{4}(t)={\frac {df^{3}(x(t))}{dt}}=h^{3}{\frac {df^{3}(x)}{dx}}{\frac {dx}{dt}}=f^{4}(x)h^{4}$

The next step is to substitute h, and manipulate the equations as follows:

$F^{4}(t)=f^{4}(x){\frac {(b-a)^{4}}{16}}$

$-F^{4}(\zeta _{4})={\frac {-(b-a)^{4}}{16}}f^{4}(x)$

${\frac {-F^{4}(\zeta _{4})}{90}}={\frac {-(b-a)^{4}}{1440}}F^{4}(x)$

It is then seen that the relation between $\zeta \quad and\quad \zeta _{4}$ must be:

$\;\zeta =x1+h\zeta _{4}$

Solution for problem 15:

Proofread problem 15:

Problem 16: Illustration of Runge Phenomenon

P. 16-1 on Lecture-16 Notes

Problem Statement

Given data: $I=\int _{-5}^{5}\left({\frac {1}{1+x^{2}}}\right)\cdot dx$
To find:

1. Using Newton-Cotes for $n=1,2,3\cdots \cdots 15$ find the numerical integral $\;I_{n}$, and find Exact Integral $\;I$ also for comparison
2. Plot $f,\;f_{n}$ for selected values of $\;n=1,2,3,8,12$
3. Plot $\;I_{n}$ vs $\;n$ and prove that the value of $\;I_{n}$ does not converge with increasing value of $\;n$
4. Prove that the weights $W_{i,n}:=\int _{a}^{b}l_{i,n}\left(x\right)\cdot dx$ for $n\geq 8$ are not all positive and plot the $l_{i,n}\left(x\right)$ for $i=1,2,3\cdots 8$ with $\;n=8$

Solution

1. From the below Matlab code, we generate the values of $\;I$ and various values of $\;I_{n}$ as tabulated below:

Integrals using Newton Cotes
	Exact Integral value (I i.e 'IE' in Matlab code)	2 Arctan(5)=2.7468
n	Numerical Integral Value (I_{n} i.e 'I' in Matlab code)	Percent Difference
n=1	0.3846	614.1684
n=2	6.7949	-59.5754
n=3	2.0814	31.9659
n=4	2.3740	15.7033
n=5	2.3077	19.0281
n=6	3.8704	-29.0314
n=7	2.8990	-5.2499
n=8	1.5005	83.0604
n=9	2.3986	14.5160
n=10	4.6733	-41.2235
n=11	3.2448	-15.3469
n=12	-0.3129	-977.7504
n=13	1.9198	43.0777
n=14	7.8995	-65.2284
n=15	4.1556	-33.9006

2. We already know that:

$f\left(x\right)=\left({\frac {1}{1+x^{2}}}\right)$

And from Matlab code below we know that:

$f_{1}(x)={\frac {1}{26}}$

$\;f_{2}(x)=-1/130*y*(-1/10*y+1/2)+(1/5*y+1)*(-1/5*y+1)+1/130*(1/10*y+1/2)*y$

$\;f_{3}(x)=1/26*(-3/10*y-1/2)*(-3/20*y+1/4)*(-1/10*y+1/2)+9/34*(3/10*y+3/2)*(-3/10*y+1/2)*(-3/20*y+3/4)+9/34*(3/20*y+3/4)*(3/10*y+1/2)*(-3/10*y+3/2)+1/26*(1/10*y+1/2)*(3/20*y+1/4)*(3/10*y-1/2)$

$\;f_{8}(x)=-1/130*(-4/5*y-3)*(-2/5*y-1)*(-4/15*y-1/3)*y*(-4/25*y+1/5)*(-2/15*y+1/3)*(-4/35*y+3/7)*(-1/10*y+1/2)-64/3615*(4/5*y+4)*(-4/5*y-2)*(-2/5*y-1/2)*y*(-1/5*y+1/4)*(-4/25*y+2/5)*(-2/15*y+1/2)*(-4/35*y+4/7)-8/145*(2/5*y+2)*(4/5*y+3)*(-4/5*y-1)*y*(-4/15*y+1/3)*(-1/5*y+1/2)*(-4/25*y+3/5)*(-2/15*y+2/3)-64/205*(4/15*y+4/3)*(2/5*y+3/2)*(4/5*y+2)*y*(-2/5*y+1/2)*(-4/15*y+2/3)*(-1/5*y+3/4)*(-4/25*y+4/5)+(1/5*y+1)*(4/15*y+1)*(2/5*y+1)*(4/5*y+1)*(-4/5*y+1)*(-2/5*y+1)*(-4/15*y+1)*(-1/5*y+1)+64/205*(4/25*y+4/5)*(1/5*y+3/4)*(4/15*y+2/3)*(2/5*y+1/2)*y*(-4/5*y+2)*(-2/5*y+3/2)*(-4/15*y+4/3)+8/145*(2/15*y+2/3)*(4/25*y+3/5)*(1/5*y+1/2)*(4/15*y+1/3)*y*(4/5*y-1)*(-4/5*y+3)*(-2/5*y+2)+64/3615*(4/35*y+4/7)*(2/15*y+1/2)*(4/25*y+2/5)*(1/5*y+1/4)*y*(2/5*y-1/2)*(4/5*y-2)*(-4/5*y+4)+1/130*(1/10*y+1/2)*(4/35*y+3/7)*(2/15*y+1/3)*(4/25*y+1/5)*y*(4/15*y-1/3)*(2/5*y-1)*(4/5*y-3)$

$\;f_{12}(x)=-1/130*(-6/5*y-5)*(-3/5*y-2)*(-2/5*y-1)*(-3/10*y-1/2)*(-6/25*y-1/5)*y*(-6/35*y+1/7)*(-3/20*y+1/4)*(-2/15*y+1/3)*(-3/25*y+2/5)*(-6/55*y+5/11)*(-1/10*y+1/2)-216/16525*(6/5*y+6)*(-6/5*y-4)*(-3/5*y-3/2)*(-2/5*y-2/3)*(-3/10*y-1/4)*y*(-1/5*y+1/6)*(-6/35*y+2/7)*(-3/20*y+3/8)*(-2/15*y+4/9)*(-3/25*y+1/2)*(-6/55*y+6/11)-27/1090*(3/5*y+3)*(6/5*y+5)*(-6/5*y-3)*(-3/5*y-1)*(-2/5*y-1/3)*y*(-6/25*y+1/5)*(-1/5*y+1/3)*(-6/35*y+3/7)*(-3/20*y+1/2)*(-2/15*y+5/9)*(-3/25*y+3/5)-8/145*(2/5*y+2)*(3/5*y+5/2)*(6/5*y+4)*(-6/5*y-2)*(-3/5*y-1/2)*y*(-3/10*y+1/4)*(-6/25*y+2/5)*(-1/5*y+1/2)*(-6/35*y+4/7)*(-3/20*y+5/8)*(-2/15*y+2/3)-27/170*(3/10*y+3/2)*(2/5*y+5/3)*(3/5*y+2)*(6/5*y+3)*(-6/5*y-1)*y*(-2/5*y+1/3)*(-3/10*y+1/2)*(-6/25*y+3/5)*(-1/5*y+2/3)*(-6/35*y+5/7)*(-3/20*y+3/4)-216/305*(6/25*y+6/5)*(3/10*y+5/4)*(2/5*y+4/3)*(3/5*y+3/2)*(6/5*y+2)*y*(-3/5*y+1/2)*(-2/5*y+2/3)*(-3/10*y+3/4)*(-6/25*y+4/5)*(-1/5*y+5/6)*(-6/35*y+6/7)+(1/5*y+1)*(6/25*y+1)*(3/10*y+1)*(2/5*y+1)*(3/5*y+1)*(6/5*y+1)*(-6/5*y+1)*(-3/5*y+1)*(-2/5*y+1)*(-3/10*y+1)*(-6/25*y+1)*(-1/5*y+1)+216/305*(6/35*y+6/7)*(1/5*y+5/6)*(6/25*y+4/5)*(3/10*y+3/4)*(2/5*y+2/3)*(3/5*y+1/2)*y*(-6/5*y+2)*(-3/5*y+3/2)*(-2/5*y+4/3)*(-3/10*y+5/4)*(-6/25*y+6/5)+27/170*(3/20*y+3/4)*(6/35*y+5/7)*(1/5*y+2/3)*(6/25*y+3/5)*(3/10*y+1/2)*(2/5*y+1/3)*y*(6/5*y-1)*(-6/5*y+3)*(-3/5*y+2)*(-2/5*y+5/3)*(-3/10*y+3/2)+8/145*(2/15*y+2/3)*(3/20*y+5/8)*(6/35*y+4/7)*(1/5*y+1/2)*(6/25*y+2/5)*(3/10*y+1/4)*y*(3/5*y-1/2)*(6/5*y-2)*(-6/5*y+4)*(-3/5*y+5/2)*(-2/5*y+2)+27/1090*(3/25*y+3/5)*(2/15*y+5/9)*(3/20*y+1/2)*(6/35*y+3/7)*(1/5*y+1/3)*(6/25*y+1/5)*y*(2/5*y-1/3)*(3/5*y-1)*(6/5*y-3)*(-6/5*y+5)*(-3/5*y+3)+216/16525*(6/55*y+6/11)*(3/25*y+1/2)*(2/15*y+4/9)*(3/20*y+3/8)*(6/35*y+2/7)*(1/5*y+1/6)*y*(3/10*y-1/4)*(2/5*y-2/3)*(3/5*y-3/2)*(6/5*y-4)*(-6/5*y+6)+1/130*(1/10*y+1/2)*(6/55*y+5/11)*(3/25*y+2/5)*(2/15*y+1/3)*(3/20*y+1/4)*(6/35*y+1/7)*y*(6/25*y-1/5)*(3/10*y-1/2)*(2/5*y-1)*(3/5*y-2)*(6/5*y-5)$

Using the above equations, we generate plots (as show in the picture below) of the function $f\left(x\right),\;f_{n}\left(x\right)$ for selected values of $\;n=1,2,3,8,12$

3. The following picture is generated using the Matlab code below and from the graph we can observe that the value of $\;I_{n}$ does not converge with increasing value of $\;n$
For each projected dotted line from X-axis on to the blue curve, we have a distinct numerical value of $\;I_{n}$ on the Y-axis. The Exact Integral- $\;I$ is the green straight line on the plot. We can clearly see that there is no convergence in the value of $\;I_{n}$(on Y-axis) as $\;n$ (on X-axis) value increases.

4. The Values of weights are evaluated by integrated the Lagrange polynomials within the integration limits and they are presented in the table below. It can be noted that the Bold weights in the table are negative

Weights of the function

n-value

W0 Value

W1 Value

W2 Value

W3 Value

W4 Value

W5 Value

W6 Value

W7 Value

W8 Value

W9 Value

W10 Value

W11 Value

W12 Value

W13 Value

W14 Value

W15 Value

8

989/2835

5888/2835

-928/2835

10496/2835

-908/567

10496/2835

-928/2835

5888/2835

989/2835

9

2857/8960

15741/8960

27/224

1209/560

2889/4480

2889/4480

1209/560

27/224

15741/8960

2857/8960

10

80335/299376

132875/74844

-80875/99792

28375/6237

-24125/5544

89035/12474

-24125/5544

28375/6237

-80875/99792

132875/74844

80335/299376

11

434293/1741824

4495513/2903040

-3237113/8709120

560593/193536

-1599257/1451520

2582261/1451520

2582261/1451520

-1599257/1451520

560593/193536

-3237113/8709120

4495513/2903040

434293/1741824

12

1364651/6306300

25008/15925

-210774/175175

1786256/315315

-1144251/140140

2431008/175175

-1045204/75075

2431008/175175

-1144251/140140

1786256/315315

-210774/175175

25008/15925

1364651/6306300

13

8181904909/40236134400

56280729661/40236134400

-1737125143/2235340800

11148172711/2874009600

-6066382933/1609445376

22964826443/4470681600

-3592666051/3353011200

-3592666051/3353011200

22964826443/4470681600

-6066382933/1609445376

11148172711/2874009600

-1737125143/2235340800

56280729661/40236134400

8181904909/40236134400

14

90241897/500385600

44436679/31274100

-770720657/500385600

109420087/15637050

-6625093363/500385600

789382601/31274100

-5600756791/166795200

101741867/2606175

-5600756791/166795200

789382601/31274100

-6625093363/500385600

109420087/15637050

-770720657/500385600

44436679/31274100

90241897/500385600

15

25221445/147603456

147529925/114802688

-129408925/114802688

1746295975/344408064

-124034975/16400384

1367713705/114802688

-10001664025/1033224192

566004225/114802688

566004225/114802688

-10001664025/1033224192

1367713705/114802688

-124034975/16400384

1746295975/344408064

-129408925/114802688

147529925/114802688

25221445/147603456

The plot of $l_{i,n}\left(x\right)$ for $i=1,2,3\cdots 8$ with $\;n=8$

MATLAB Code:

To generate the values of numerical Intergrals: $I_{1},I_{2},\cdots I_{15}$

%To generate Lagrange polynomials l(x)%

for n=1:15
    for i=1:n+1
        x(n,i) = -5+(i-1)*10/n;
        f(n,i) = 1/(1+x(n,i)^2);
    end
end
syms y;
for n=1:15
    for i=1:n+1
        pr=1;
        for j=1:n+1
            if j~=i
                pr=pr*((y-x(n,j))/(x(n,i)-x(n,j)));
            end
        end
        l(n,i)=pr;
    end
end

%To generate Fn(x)for numerically integrating%

for n=1:15
    for i=1:n+1
        F(n,i)=l(n,i)*f(n,i);
    end
end
for n=1:15
    sum=0;
    for i=1:n+1
        sum=sum+F(n,i);
    end
    G(n)=sum;
end
for n=1:15
    T(n,1)=G(n);
end

%for calculating Exact Integral and Numerical Integral%

Ie=double(int(1/(1+y^2),y,-5,5));
for n=1:15
I(n,1)=int(T(n,1),y,-5,5);
IE(n,1)=Ie;
end
n=linspace(1,15,15);
I=double(I);
plot(n,I,n,IE);
legend('I=Numerical Integral(I_n)','IE=Exact Integral(I)');
xlabel('n-Value (Degree of the Newton-Cotes appoximating polynomial)');
ylabel('Numerical Value of Exact Integral (I) and Numerical Integrals (I_n)');

%code for calculating percentage error%
for n=1:15
PE(n,1)=(((IE(n,1)-I(n,1))/I(n,1))*100);
end

Te=(1/(1+y^2));
%code for plotting f,f1,f2,f3,f8 and f12 that are represented by Te, T(1,1), T(2,1), T(3,1), T(8,1), T(12,1) respectively in this code%
figure (2);
EZPLOT(Te);
hold on;
EZPLOT(T(1,1))
hold on;
EZPLOT(T(2,1))
hold on;
EZPLOT(T(3,1))
hold on;
EZPLOT(T(8,1))
hold on;
EZPLOT(T(12,1))
hold on;
xlabel('x- Value ');
ylabel('f(x),f1(x),f2(x),f3(x),f8(x) and f12(x)');

%code for calculating the Weights by integrating the lagrange polnomial%
for n=1:15
    for i=1:n+1
W(n,i)=int(l(n,i),y,-5,5);
    end
end
%code for plotting L0,L1,L2,L3,L4...L7 and L8 for n=8%
figure (3);
ezplot(l(8,1));
hold on;
ezplot(l(8,2));
hold on;
ezplot(l(8,3));
hold on;
ezplot(l(8,4));
hold on;
ezplot(l(8,5));
hold on;
ezplot(l(8,6));
hold on;
ezplot(l(8,7));
hold on;
ezplot(l(8,8));
hold on;
ezplot(l(8,9));
hold on;
xlabel('x- Value ');
ylabel('L0(x),L1(x),L2(x),L3(x)....L7(x) and L8(x)');

Solution for problem 16: Srikanth Madala 07:04, 11 February 2010 (UTC)

Proofread problem 16:

Contributing Authors

Solutions for problem 5 and proofread problems: 3,6,8,and 12--Egm6341.s10.team2.patodon 16:42, 10 February 2010 (UTC)

--Egm6341.s10.Team2.GV 21:55, 10 February 2010 (UTC)

Solutions for problems 1,6,11 and proofread 5,9,14--Niki Nachappa Chenanda Ganapathy

Solutions for problems 4,9,12,13 and proofread 2,10,11--Pengxiang Jiang

Solutions for problems 2,7,14,16 and proofread 1,8,13--Srikanth Madala 07:04, 11 February 2010 (UTC)

Problem Assignments
Problem	Solution	Proofread
Problem 1	NN	SM
Problem 2	SM	JP
Problem 3	GV	PO
Problem 4	JP	GV
Problem 5	PO	NN
Problem 6	NN	PO
Problem 7	SM	GV
Problem 8	PO	SM
Problem 9	JP	NN
Problem 10	GV	JP
Problem 11	NN	JP
Problem 12	GV	PO
Problem 13	JP	SM
Problem 14	SM	NN
Problem 15	PO	GV
Problem 16	SM	PO