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′′ 
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$$  \displaystyle \boldsymbol{f_2(x)=p_2(x)=c_2 x^2+c_1 x+c_0} $$     (1 p8-3)
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$$  \displaystyle \boldsymbol{p_2(x)=\sum_{i=0}^{2}(l_i(x)f(x_i))} $$     (3 p8-3)
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Solution
We have the general formula for the Lagrange basis function $$\boldsymbol{l_i(x)}$$ as


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$$  \displaystyle \boldsymbol{l_i(x)=\prod_{j=0;j\neq i}^{n} \left (\frac{x-x_j}{x_i-x_j} \right )} $$     (2 p7-3) for the case of Simple Simpson's Rule, n =2 i.e i=0,1,2. For the given interval $$[a,b]$$
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$$\boldsymbol{x_0=a}$$

$$\boldsymbol{x_2=b}$$

$$\boldsymbol{x_1=\frac{a+b}{2}}$$

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 $$\boldsymbol{p_2(x)=\left ( \frac{x^2-(x_2+x_1)x+(x_1x_2)}{(x_0-x_1)(x_0-x_2)} \right )f(x_0)+\left ( \frac{x^2-(x_2+x_0)x+(x_0x_2)}{(x_1-x_0)(x_1-x_2)} \right )f(x_1)+\left ( \frac{x^2-(x_0+x_1)x+(x_1x_0)}{(x_2-x_0)(x_2-x_1)} \right )f(x_2)}$$

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_1x_2]}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)[x_0x_2]}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)[x_0x_1]}{(x_2-x_0)(x_2-x_1)} \right \}}$$

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


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$$  \displaystyle \boldsymbol{c_2=\left \{ \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} \right \}}
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$$     (1)
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$$  \displaystyle \boldsymbol{c_1=-\left \{\frac{f(x_0)[x_2+x_1]}{(x_0-x_1)(x_0-x_2)}+ \frac{f(x_1)[x_2+x_0]}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)[x_1+x_0]}{(x_2-x_0)(x_2-x_1)} \right \}} $$     (2)
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$$  \displaystyle \boldsymbol{c_0=\left \{ \frac{f(x_0)[x_1x_2]}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)[x_0x_2]}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)[x_0x_1]}{(x_2-x_0)(x_2-x_1)} \right \}} $$     (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.
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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
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:

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

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

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.

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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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:


 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{E^{(n+1)}(x)=f^{(n+1)}(x)-0} $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

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


 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{E^{n+1}(x)=f^{n+1}(x)-0} $$     (1)
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

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

$$\boldsymbol{I=[3x+4x^2-\frac{2}{3}x^3+\frac{3}{2}x^4]_0^1}$$

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


 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{I=7.833} $$     (1)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

Case B: Using Simple Simpson's rule
We have the Simple Simpson's rule as
 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{I_2=\frac{h}{3}\left \{ f(x_0)+4f(x_1)+f(x_2) \right \}} $$     (2 p7-2) where $$\boldsymbol{h=\frac{a+b}{2}=\frac{0+1}{2}=0.5}$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

$$\boldsymbol{I_2=\frac{0.5}{3}\left \{ f(0)+4f(0.5)+f(1) \right \}}$$ we know $$\boldsymbol{f(0)=3;f(0.5)=7.25;f(1)=15}$$ substituting we get

$$\boldsymbol{I_2=\frac{0.5}{3}\left \{ 3+4(7.25)+15 \right \}=\frac{0.5}{3}\left \{ 47 \right \}}$$


 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{I_2=7.833} $$     (2)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

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 $$e\left ( t \right )$$

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:

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

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:

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

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: $$I_{1},I_{2},\cdots I_{15}$$

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)