text:
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)


text:
User:Pmlineditor/monobook.js

mw.loader.load('https://meta.wikimedia.org/w/index.php?title=User:Pmlineditor/global.js&action=raw&ctype=text/javascript');

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Pmlineditor/monobook.js


title: File:Prob5.JPG


text:
User:Amit Bhise

Hey..

Lets get down to work as of now.

title: User:Amit Bhise


text:
User:Prince Kassad

m:wikt:User:Prince Kassad

title: User:Prince Kassad


text:
User:Trum5770

Topic:Turing_Programming_Language

title: User:Trum5770


text:
User talk:Egm6341.s2010.team1

 Hello Egm6341.s2010.team1, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 29 January 2010 (UTC)

title: User talk:Egm6341.s2010.team1


text:
User talk:Morris~enwikiversity

 Hello Morris, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 29 January 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Morris. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Morris~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:44, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:16, 19 April 2015 (UTC)

title: User talk:Morris~enwikiversity


text:
User talk:Richard ndonga

 Hello Richard ndonga, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 29 January 2010 (UTC)

title: User talk:Richard ndonga


text:
User:Csell

Spring 2010 Courses
EML5526

EMA6136

title: User:Csell


title: User:Csell/eml5526


text:
User:Csell/eml5526.HW1

1. In the 3-node triangle, field quantity Ø can be written as $$\Phi=a_1+a_2x+a_3y$$

title: User:Csell/eml5526.HW1


title: File:Egm6341.s10.mtg15.pdf


title: File:Egm6341.s10.mtg16.pdf


text:
R (programming language)/Tutorials/Connecting Fortran and R

This is to briefly describe the process of connecting the programming languages: Fortran and R. At least basic knowledge of both programming languages is assumed here.

Basic requirements

 * R - R should be installed on your computer. In this tutorial R version 2.10.1 (2009-12-14) was used on a Windows XP machine. For information about setup visit the installation how-to.

 * The Fortran program - You have to have a program source code, that you want to use in R. Please note, that R is able to call only subroutines. In case of functions a so called "wrapper subroutine" should be used. You find more about this here.

Sample program
Copy the program below to a text file, and name it: multiply.f

c sample subroutine to include for R      subroutine multiply(ax,bx,cx) implicit none

integer ax,bx,cx

cx=ax*bx

end

Making a dll
In order to connect these two programming languages you need to compile the Fortran code. You will need a Fortran compiler for this one. The following example assumes a gfortran compiler.

To make a dll (Dynamic-link library) from the source code, open the command prompt, navigate to the directory with the source code and type:

gfortran -shared -o multiply.dll multiply.f

Where multiply.f is the name of the file which contains the source code.

In Lahey Fortran compiler you should use the -dll switch instead of -shared.

Inside R

 * Open R. The simplest case is to move the newly created dll file to the working directory of R. (Note: type getwd to find out your working directory.)
 * You have to load the dll file to R. For this use dyn.load like this:

dyn.load("d:/work/R_work_files/general/multiply.dll")

Where d:/work/R_work_files/general/multiply is the path to the dll file.

is.loaded("multiply")
 * To test if it was loaded correctly, type:

If you get "[1] TRUE", you are ok. If not, find out the cause of it (some hints are here).

 * The last thing you have to do is call the compiled Fortran subroutine from R using ".Fortran" like this:

a=5 b=2 .Fortran("multiply",as.integer(a),as.integer(b),c=integer(1))

and you will get:

1 [1] 5

2 [1] 2

$c [1] 10

Where $c equals 10, which is exactly 5 times 2.

Move to R function
The process below is optional, but it makes the usage of the "final product" much more simple and user friendly.

Source code

 * Open a new script, paste the following code there and save it as fortranConnection.r

star = function(a,b){ # res holds the results for the Fortran call res =.Fortran("multiply",as.integer(a),as.integer(b),c=integer(1)) return(res$c) }
 * 1) script to test the link between Fortran and R

Note: As an outcome you are interested just in the value of c, so the return value will be res$c and not the whole res.

 * In R open Files> Source R code..., find fortranConnection.r and source it in.

 * Use your brand new R function as:

star(5,2)

Which leads to not-so-surprising result:

[1] 10

Special cases
is.loaded("multiply_")
 * Some compilers leave a trailing underscore after the subroutine name inside the dll file. If the is.loaded test above gives a FALSE for you, try this as one of the possible solutions:

DLL_EXPORT mySubroutine
 * Also other compiler specific issues might come in, such as for Lahey the following line could be inserted into the Fortran source code:

Where mySubroutine is the subroutine name to be exported to a dll file. More information here.


text: FORTRAN

title: FORTRAN


text:
Human anatomy quizzes/Shoulder muscles

Muscles of the shoulder joint
The muscles of the shoulder joint below include, amongst others, the rotator cuff, composed of supraspinatus, infraspinatus, teres minor, and subscapularis.

{ Based on the image above, answer the following:  Name muscle A. { Supraspinatus (i) } Name muscle B. { Infraspinatus (i) } Name muscle C. { Teres minor (i) } Name muscle D. { Lateral head (i) |Lateral head triceps (i) |Lateral head of triceps (i) |Lateral tricep (i) } What is muscle E called? { Long head of triceps (i) } What is muscle F called? { Latissimus dorsi (i) } What is muscle G called? { Teres major (i) } What is the main function of muscle G? { Adducts and medially rotates humerus (i)| Adduction and medial rotation of humerus (i) }</li>
 * type="{}"}

</ol>

title: Human anatomy quizzes/Shoulder muscles


text:
Human anatomy quizzes

Anatomy quizzes for medical students
This page contains anatomy learning resources designed to be used as a revision tool in preparation for medical school and anatomy spot tests.

Shoulder and pectoral girdle

 * /Pectoral girdle/
 * /Scapula/
 * /Humerus/
 * /Shoulder joint/
 * /Sternoclavicular joint/
 * /Shoulder capsule/

Forearm and Hand

 * /Radius and ulna/
 * /Elbow joint ligaments/
 * /Elbow joint bursa/
 * /Hand/
 * /Lateral aspect of the hand/

Shoulder and Pectoral Girdle

 * /Superficial muscles of pectoral region/
 * /Deep muscles of the pectoral region/
 * /Shoulder muscles/
 * /Back muscles/

Forearm

 * /Superficial muscles of the anterior compartment/

Neurovascular Supply of the Upper Limb

 * /Brachial plexus/
 * /Brachial plexus in situ/

Picture bank
 * http://commons.wikimedia.org/wiki/Category:Gray%27s_Anatomy_plates

title: Human anatomy quizzes


text:
User talk:Dworn21

 Hello Dworn21, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Gbaor 16:00, 29 January 2010 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

names
Hi! Just want to say that spaces in article/quiz titles are also allowed. --Gbaor 16:00, 29 January 2010 (UTC)

reason why I added the categories
About : could I explain why I did this? So other participants (in the future) find the content faster and can use your content. I added at pages some more until I saw the edit. If this is disturbing you, please tell. I will stop for now. Erkan Yilmaz uses the Chat (try) 18:16, 25 March 2010 (UTC)

title: User talk:Dworn21


text:
User talk:Raf db

 Hello Raf db, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Spelling - region
I fixed the spelling of Bones of the shoulder regoin to Bones of the shoulder region. I also fixed the spelling for 2, 3, and 4. Ottava Rima (talk) 16:21, 29 January 2010 (UTC)

title: User talk:Raf db


text:
Anatomy/Bones of the shoulder region

Bones and Ligaments of the Pectoral Girdle

{ Label the picture  A: { Acromioclavicular joint (i) } B: { Sternoclavicular joint (i) } C: { Acromion (i) } D: { Shoulder joint (i) } E: { Coracoid process (i) } F: { Scapula (i) } G: { Clavicle (i) } H: { Humerus (i) }</li>
 * type="{}"}

</ol>

title: Anatomy/Bones of the shoulder region


text:
Anatomy/Bones of the shoulder region 2

The Sternoclavicular Joint

{ Label the picture  A: { Clavicle (i) } B: { Sternoclavicular joint (i) } C: { Intra-articular disk (i) } D: { Costoclavicular ligament (i) } E: { Primary cartillagenous joint (i) } F: { Manubrium (i)|Sternum (i) } G: { Rib (i)|1st rib (i)| first rib (i) }  (note: Label E refers to the type of joint in question, not it's name)
 * type="{}"}

title: Anatomy/Bones of the shoulder region 2


text:
Anatomy/Bones of the shoulder region 4

The Shoulder Capsule

{ Label the picture  A: { Transverse Ligament(i) } B: { Synovial Membrane (i) } C: { Tendon of long head of biceps (i) } D: { Suprascapular ligament(i) } E: { Sholder capsule(i) }
 * type="{}"}

title: Anatomy/Bones of the shoulder region 4


text:
Anatomy/Bones of the shoulder region 3

The Shoulder Joint

{ Label the picture  A: { Acromioclavicular ligament (i) } B: { Acromion (i) } C: { Coracoacromial arch (i) } D: { Coracohumeral ligament(i) } E: { Tendon of long head of biceps(i) } F: { Lesser tubercle(i) } G: { Shoulder joint(i) } H: { Coracoid process(i) }

I: { Coracoclavicular ligament(i) }

J: { Clavicle(i) }

K: { Greater tubercle(i) }

 * type="{}"}

title: Anatomy/Bones of the shoulder region 3


text:
Wikiversity:Research process

Wikiversity was largely created for participants to engage in research. New participants are encouraged to be bold and be creative when creating or contributing to research, but what is research, and how can you do research? This page explores questions new researchers may have and a way that a research process might work.

What is research?
Research is an examination or investigation process that attempts to answer questions or check hypotheses. Any time you search for answers or test hypotheses you are doing research. You can search for answers and test hypotheses many ways, and consequently there are many forms of research. Research is also done as exploration, it is not essential that there be a hypothesis being tested. Rather, the first levels of research simply study a topic, collect data, and, for the researcher, create a background of understanding that may then allow the formation of hypotheses grounded in fact.

Primary research
Primary research is about obtaining an abundance of quality data or information. Good primary research collects a large, representative pool of data or information to examine. More data or information means any assessment of that data or information is more likely to be accurate. Conversely less data or information means less accuracy.

Examples of primary research:

 * A government agency does primary research when collecting lots of information about their country's population through a census.
 * A student does primary research when asking a bunch of fellow students what type of pizza they like.

 * May I suggest that while the above examples are good ones the further back in time one goes the more complex the issue of accuracy can become. Quantity of data does not necessarily equate to accuracy to data. It is quite possible for a false fact to be repeated so often in different documents that it is wrongly accepted as true. Evaluation of accuracy may require some expertise in the knowledge of context. (2A00:23C5:EC0A:8200:79BD:CC9C:FA62:EC48 (discuss) 14:07, 7 June 2018 (UTC)BethANZ)

Secondary research
Secondary research is about assessing existing knowledge. Good secondary research can often lead to new ways of looking at things and progress towards the creation of new knowledge. Awareness of past efforts can help save you time that might otherwise have been wasted recreating the wheel. A common way for Wikiversity participants to learn is to participate in secondary research projects that review existing literature, and compose summarizations of what was learned, which can in turn help other participants learn.

Examples of secondary research:

 * Students do secondary research to write papers. Reading a large number of books about a subject, and taking notes.
 * Authors of nonfiction books do secondary research so that their books will be correct.

For emerging academics, there is a natural progression from secondary research to primary (original) research.

Original research
Original research is about discovering new knowledge. Good original research is motivated by curiosity. Original research can be done anywhere, like in the workplace or in the field (called field work), but is often done in an academic or laboratory setting. You are doing original research anytime you attempt to discover something new that has never been published and peer reviewed.

Examples of original research:

 * Scientists do original research when carrying out new experiments
 * Anthropologists do original research when examining newly unearthed bones
 * Wikiversitans do original research when they collect and analyze data. Ideally, the collection itself is not original research, but collection can be incomplete or biased. The original research, then, exists in decisions of what to collect, in the choices made in collection, if any, and, then, in analysis, if that analysis is not published elsewhere.

What are research ethics?
Research ethics try to answer what actions are good, and what actions are bad. The key issues for research published on Wikiversity are:

 * Verifiability: Fully document what methods are intended to be used and what methods were used. Explain any changes in methods used.
 * Transparency: Allow participants to draw their own independent conclusions by clearly explaining every decision and endeavor undertaken in the course of the research in a open and transparent manner.
 * Honesty: Report without omissions, even if portions are undesirable, don't fit predetermined beliefs, or initial hypotheses. Participants are not lead into drawing specific conclusions.
 * Objectivity: Strive to conduct and report research in a way that allows experts to independently test, verify and confirm the validity of the research.
 *  Subjectivity: You are free to express your views and opinions, but they should be clearly identified as such.
 * Sources: When sources of previous knowledge, data or other information is relied on cite your sources. This allows examination of those works too. Be clear to distinguish previous knowledge from new knowledge.
 * Safety: Do no harm by conducting research in a safe and lawful manner.
 * Consensual: Inform and ask for consent before involving others in your research. People, groups, and organizations must be willing participants.
 * Disclosures: Report any held opinions and beliefs about the topic and any research subjects, anyone or anything involved in or backing the research, and the motivations for doing the research.
 * Review process: Ask Wikiversity participants to review the research. This is an essential step for quality control and to ascertain the standing of a research project.

Following these basic principles will usually foster trust, and help the community to better understand your point of view.

How is research reported?
Research can be reported in many different ways. Sharing your research with other people can often feel overwhelming. This section explores ways you can share you research with Wikiversity participants.

Stages
Research has a beginning, middle, and end. In the beginning stage you define your goals, time constraints, and tools needed to complete research.

Background information
Background information is what knowledge, data, or facts are known about a topic. It is usually obtained by reading or studying any existing material on the subject matter. It can help you to understand what the focus of your research should be, and can keep you from becoming overwhelmed. By including background information you can help your audience and yourself to better understand the effects of your research on existing knowledge. Good research reports should include background information that answers some questions readers might have. Table 1.1 includes some example questions that might help you when you are looking to include background information.

Table 1.1

What else should I know?

 * Research (Wikipedia)
 * Research (Simple Wikipedia)
 * Original research (Wikiversity)
 * Original research (Wikipedia)
 * Human subject research (Wikipedia)
 * Informed consent in research (Wikipedia)
 * Research guidelines at Beta Wikiversity
 * Getting Background Information at USG
 * Conducting Research (Purdue OWL)
 * Scientific Reports (University of North Carolina)

title: Wikiversity:Research process


text:
User:Ninetyone

hello there. unfortunately, i do not edit here. if you are looking for me, you will find me on the english-language wikipedia.

title: User:Ninetyone


title: User:Joejill67


text:
User:Egm6341.s10.Team4.nimaa&m/HW2

= Problem 2: Proof of simple simpson's rule =

Given
Use equation (4) on slide [[media:Egm6341.s10.mtg8.pdf|8-3]], which is:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle P_2(x_j)=\sum_{k=0}^2 l_i(x_j)f(x_i)=f(x_j)

 * }
 * }

Find
to derive equation (simple simpson's rule Use following Integration (2) on slide [[media:Egm6341.s10.mtg7.pdf|7-2]], which is:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]              ;             h=\dfrac{b-a}{2}

 * }
 * }

Solution
The primary equation is following polynomial:

P_2(x)=\sum_{i=0}^2 l_i(x)f(x_i)=f_n(x) $$

From the lagrange equation we have:

l_i(x)=\prod_{i=1}^{N} \dfrac{x-x_j}{x_i-x_j} \Rightarrow l_0(x)=\dfrac{x-x_1}{x_0-x_1}\times\dfrac{x-x_2}{x_0-x_2} $$

\Rightarrow I_2=\int\limits_{x_0}^{x_2} P_2(x)\, dx=\int\limits_{x_0}^{x_2} \sum_{i=0}^2 l_i(x)f(x_i)\, dx=\int\limits_{x_0}^{x_2} [l_0(x)f(x_0)+l_1(x)f(x_1)+l_2(x)f(x_2)]\, dx $$

Similarly for other $$ l_i(x) $$ we can write:

l_1(x)=\dfrac{x-x_0}{x_1-x_0}\times\dfrac{x-x_2}{x_1-x_2} $$

l_2(x)=\dfrac{x-x_0}{x_2-x_0}\times\dfrac{x-x_1}{x_2-x_1} $$

\Rightarrow I_2=\int\limits_{x_0}^{x_2} [\dfrac{x-x_1}{x_0-x_1}\times\dfrac{x-x_2}{x_0-x_2}f(x_0)+\dfrac{x-x_0}{x_1-x_0}\times\dfrac{x-x_2}{x_1-x_2}f(x_1)+\dfrac{x-x_0}{x_2-x_0}\times\dfrac{x-x_1}{x_2-x_1}f(x_2)]\, dx $$

=\int\limits_{x_0}^{x_2} \dfrac{x-x_1}{x_0-x_1}\times\dfrac{x-x_2}{x_0-x_2}f(x_0)\, dx+\int\limits_{x_0}^{x_2} \dfrac{x-x_0}{x_1-x_0}\times\dfrac{x-x_2}{x_1-x_2}f(x_1)\, dx+\int\limits_{x_0}^{x_2} \dfrac{x-x_0}{x_2-x_0}\times\dfrac{x-x_1}{x_1-x_2}f(x_2)\, dx= $$

=\dfrac{f(x_0)}{(x_0-x_1)(x_0-x_2)}\int\limits_{x_0}^{x_2} (x^2-x_2x-x_1x+x_1x_2)\, dx+\dfrac{f(x_1)}{(x_1-x_0)(x_1-x_2)}\int\limits_{x_0}^{x_2} (x^2-x_2x-x_0x+x_0x_2)\, dx+\dfrac{f(x_2)}{(x_2-x_0)(x_2-x_1)}\int\limits_{x_0}^{x_2} (x^2-x_0x-x_1x+x_0x_1)\, dx= $$

=\dfrac{f(x_0)}{(x_0-x_1)(x_0-x_2)}[\dfrac{x^3}{3}-x_2\dfrac{x^2}{2}-x_1\dfrac{x^2}{2}+x_1x_2x]+\dfrac{f(x_1)}{(x_1-x_0)(x_1-x_2)}[\dfrac{x^3}{3}-x_2\dfrac{x^2}{2}-x_0\dfrac{x^2}{2}+x_0x_2x]+\dfrac{f(x_2)}{(x_2-x_0)(x_2-x_1)}[\dfrac{x^3}{3}-x_0\dfrac{x^2}{2}-x_1\dfrac{x^2}{2}+x_0x_1x] $$

If we name these three terms by u, v and w, respectively:

u=\dfrac{f(x_0)}{(x_0-x_1)(x_0-x_2)}[\dfrac{x_2^3}{3}-\dfrac{x_2^3}{2}-x_1\dfrac{x_2^2}{2}+x_1x_2^2-\dfrac{x_0^3}{3}+x_2\dfrac{x_0^2}{2}+x_1\dfrac{x_0^2}{2}-x_0x_1x_2] $$

Since,

x_0=a; x_1=\dfrac{a+b}{2}; x_2=b $$

And since,

(x_0-x_1)=-h;(x_0-x_2)=-2h; $$

We can rewrite the value of u as follows:

u=\dfrac{f(x_0)}{2h^2}[\dfrac{-b^3}{6}+(\dfrac{b+a}{4})b^2-\dfrac{a^3}{3}+\dfrac{a^2}{2}(\dfrac{3b+a}{2})-ab(\dfrac{b+a}{2})]= $$

u=\dfrac{f(x_0)}{2h^2}[\dfrac{b^3}{12}-\dfrac{a^3}{12}-\dfrac{ab^2}{4}+\dfrac{ba^2}{4}]=\dfrac{f(x_0)}{24h^2}(b^3-a^3-3ab^2+3ba^2)=\dfrac{f(x_0)}{24h^2}(b-a)^3=\dfrac{f(x_0)}{24h^2}\times(2h)^3=\dfrac{f(x_0)}{24h^2}\times8h^3 $$

\Rightarrow u=\dfrac{h\times f(x_0)}{3} $$

Similarly, for v and w we obtain:

v=\dfrac{f(x_1)}{-h^2}[\dfrac{-b^3}{6}+\dfrac{ab^2}{2}+\dfrac{a^3}{6}-\dfrac{ba^2}{2}]=\dfrac{f(x_1)}{6h^2}(b-a)^3=\dfrac{f(x_1)}{6h^2}(2h)^3=\dfrac{f(x_1)}{6h^2}8h^3 $$

\Rightarrow v=\dfrac{h\times 4f(x_1)}{3} $$

w=\dfrac{f(x_2)}{2h^2}[\dfrac{b^3}{3}-a\dfrac{b^2}{2}-\dfrac{a+b}{2}(\dfrac{b^2}{2})+a(\dfrac{a+b}{2})b-\dfrac{a^3}{3}+\dfrac{a^3}{2}+\dfrac{a+b}{2}(\dfrac{a^2}{2})-\dfrac{a+b}{2}a^2]= $$

w=\dfrac{f(x_2)}{2h^2}[\dfrac{b^3}{12}-\dfrac{a^3}{12}-\dfrac{ab^2}{4}+\dfrac{ba^2}{4}]=\dfrac{f(x_2)}{24h^2}(b^3-a^3-3ab^2+3ba^2)=\dfrac{f(x_0)}{24h^2}(b-a)^3=\dfrac{f(x_2)}{24h^2}\times(2h)^3=\dfrac{f(x_2)}{24h^2}\times8h^3 $$

\Rightarrow u=\dfrac{h\times f(x_2)}{3} $$

Eventually, by adding u, v and w, we gain the entire equation:

 * {| style="width:60%" border="0"

$$ I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)] $$
 * style="width:30%; padding:10px; border:1px solid #8888aa" |
 * style="width:30%; padding:10px; border:1px solid #8888aa" |
 * style= |
 * }

= Problem 4: From simple Trap. rule and simple Simpson's rule to the composite types =

Given
The simple Trapezoidal rule equation is written as follows:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle I_1=\dfrac{b-a}{2}[f(a)+f(b)]
 * $$\displaystyle I_1=\dfrac{b-a}{2}[f(a)+f(b)]

Also, the simple Simpson's rule equation is written as: $$
 * $$\displaystyle I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]
 * $$\displaystyle I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]

 * }
 * }

Find
Show that from these two simple forms of equations we can develop composite forms of them respectively, as below:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle I_n=h[\dfrac{1}{2}f_0+f_1+...+\dfrac{1}{2}f_{n-1}]
 * $$\displaystyle I_n=h[\dfrac{1}{2}f_0+f_1+...+\dfrac{1}{2}f_{n-1}]

$$
 * $$\displaystyle I_n=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+2f_4+...+2f_{n-2}+4f_{n-1}+f_n]
 * $$\displaystyle I_n=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+2f_4+...+2f_{n-2}+4f_{n-1}+f_n]

 * }
 * }

Solution
Part 1: For Trapezoidal rule, for one interval (n=1) and h equal to width of the interval (a is start point and b is the end point), we have:

I_1=\dfrac{x_1-x_0}{2}(f(x_0)+f(x_1)) $$

When we take two intervals (n=2) and h equal to width of each interval:

I_2=\dfrac{x_1-x_0}{2}(f(x_0)+f(x_1))+\dfrac{x_2-x_1}{2}(f(x_1)+f(x_2)) $$

\Rightarrow I_2=\dfrac{h}{2}(f(x_0)+2f(x_1)+f(x_2)) $$

By continuing this procedure for 4 intervals, we gain:

I_4=\dfrac{x_1-x_0}{2}(f(x_0)+f(x_1))+\dfrac{x_2-x_1}{2}(f(x_1)+f(x_2))+\dfrac{x_3-x_2}{2}(f(x_2)+f(x_3))+\dfrac{x_4-x_3}{2}(f(x_3)+f(x_4)) $$

\Rightarrow I_4=\dfrac{h}{2}(f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)) $$

Thus, when n intervals are selected, this procedure can be resulted in:

\Rightarrow I_n=h(\dfrac{1}{2}f(x_0)+f(x_1)+f(x_2)+...+f(x_{n-1})+\dfrac{1}{2}f(x_{n})) $$

Part 2: Similarly, when we have 2 intervals for Simpson's rule (trivially h is the width of each one), we can write following from the simple case:

I_2=\dfrac{h}{3}(f(x_0)+4f(x_1)+f(x_2)) $$

For 4 intervals, we can extend the above equation to:

I_4=\dfrac{h}{3}(f(x_0)+4f(x_1)+f(x_2))+\dfrac{h}{3}(f(x_2)+4f(x_3)+f(x_4)) $$

\Rightarrow I_4=\dfrac{h}{3}(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)) $$

For 6 intervals, we can extend the above equation to:

I_6=\dfrac{h}{3}(f(x_0)+4f(x_1)+f(x_2))+\dfrac{h}{3}(f(x_2)+4f(x_3)+f(x_4))+\dfrac{h}{3}(f(x_4)+4f(x_5)+f(x_6)) $$

\Rightarrow I_6=\dfrac{h}{3}(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)) $$

Thus, when n intervals are selected, this pattern can be translated in:

I_n=\dfrac{h}{3}(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)) $$

= Problem 8 =

Given
Let function $$\displaystyle f(x)$$ and $$\displaystyle t=2, x_0=3, x_1=4, ..., x_6=9$$

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

$$ $$
 * $$\displaystyle f(x)=log(x)
 * $$\displaystyle f(x)=log(x)
 * $$\displaystyle

 * }
 * }

Find
Plot 3 figures similar to plotted graphs on slide [[media:Egm6341.s10.mtg11.pdf|11-2]] showing following functions,

$$\displaystyle f(x), f_n(x), l_{i,n}(x), q_{n+1}(x)$$

for $$\displaystyle i=3 (x_3=6)$$ and $$\displaystyle x=5.5$$

Solution
The primary function f(x) and estimated function by Newton-Cotes formula will be defined as:

 * $$\displaystyle

f(x)=log(x) $$

f_n(x)=\sum_{i=0}^6 l_{i,n}(x)f(x_i)=\sum_{i=0}^6 l_{i,n}(x)log(x_i) $$

Where the Lagrange equation is introduced with below equation for i=3:

l_{i,n}(x)=\prod_{i=1}^{n} \dfrac{x-x_j}{x_i-x_j} \Rightarrow l_{3,6}(x)=\dfrac{x-x_0}{x_3-x_0}\times\dfrac{x-x_1}{x_3-x_1}\times\dfrac{x-x_2}{x_3-x_2}\times\dfrac{x-x_4}{x_3-x_4}\times\dfrac{x-x_5}{x_3-x_5}\times\dfrac{x-x_6}{x_3-x_6} $$

 * $$\displaystyle

q_{n+1}(t)=(t-x_0)(t-x_1)(t-x_2)...(t-x_n) $$

Constructing these fucntions by the Matlab codes, facilitate us to attain following graphs:

Ultimately, the error of interpolation error theorem can be attained as following for this problems for at t=2 and x=5.5 by n=1000.

$$ \begin{array}{|c||c|c|} x & q_{n+1}(t) & f(t) - f_n(t) \\ \hline t=2&-5040&0.003887\\ x=5.5&12.30469&-4.1086E^{-6}\\ \hline \end{array} $$

 Matlab Code:

= Problem 9: Proof of error for simple trapezoidal rule =

Given
Refer to Lecture slide 13-2 for problem statement
 * {| style="width:100%" border="0" align="left"

\begin{align} n=1, \\ & q_2(x)=(x-x_o)(x-x_1)\\ & x_o=a, x_1=b, \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

Find
Show that
 * {| style="width:100%" border="0" align="left"

\left | E_1 \right | = \frac{M_2}{2!}\int_{a}^{b}\left|q_2(x) \right |\, dx = \frac{M_2}{2!}\int_{a}^{b}\left|(x-x_o)(x-x_1) \right |\frac{(b-a)^3}{12}M_2 $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

Solution

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

\begin{align} \dfrac {M_2}{2!}\int\limits_{a}^{b} (x-a)(b-x)\, dx=\dfrac{M_2}{2!}\int\limits_{a}^{b} (bx-x^2-ab+ax)\, dx=\dfrac{M_2}{2!}[\dfrac{bx^2}{2}-\dfrac{x^3}{3}-abx+\dfrac{ax^2}{2}]\, \Big |^b_a\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

\dfrac{M_2}{2!}[\dfrac{b^3}{2}-\dfrac{b^3}{3}-ab^2+\dfrac{ab^2}{2}-\dfrac{ba^2}{2}+\dfrac{a^3}{3}+a^2b-\dfrac{a^3}{2}]=\dfrac{M_2}{2!}[\dfrac{b^3}{6}-\dfrac{a^3}{6}-\dfrac{ab^2}{2}+\dfrac{a^2b}{2}]=\dfrac{M_2}{12}[b^3-a^3-3ab^2+3a^2b]=\dfrac{M_2}{12}(b-a)^3=\dfrac{M^2}{12}h^3 $$

 * {| style="width:30%" border="0"

$$ \Rightarrow \left | E_1 \right | = \frac{M_2}{2!}\int_{a}^{b}\left|q_2(x) \right |\, dx=\dfrac{M^2}{12}h^3 $$
 * style="width:10%; padding:10px; border:1px solid #8888aa" |
 * style="width:10%; padding:10px; border:1px solid #8888aa" |
 * style= |
 * }

= Problem 14: Third derivative of e(t) in Lagrange Intp. Error =

Given
If we define function e(t) as the following,
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e(t)=\int\limits_{-t}^{t} F(t)\, dt-\dfrac{t}{3}[F(-t)+4F(0)+F(t)]

 * }
 * }

Find
Show that:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)]

 * }
 * }

Solution
Let e(t) is the summation of these two terms:

 * $$\displaystyle

e(t)=\alpha(t)-\alpha_2(t) $$

Basically, we have the following in the calculus for differentiating an integral:

({\int\limits_{g(x)}^{h(x)} f(x)\, dx})'=\dfrac{d}{dx}\int\limits_{g(x)}^{h(x)} f(x)\, dx=f(h(x)).h'(x)-f(g(x)).g'(x) $$ Assuming k a point in the [-t,t] interval:

\alpha (t)=\int\limits_{-t}^{k} F(t)\, dt+\int\limits_{k}^{t} F(t)\, dt $$

\Rightarrow \alpha^{(1)}(t)=[F(k)\times0-F(-t)\times(-1)]+[F(t)\times 1-F(k)\times 0]=F(-t)+F(t) $$

\Rightarrow \alpha^{(2)}(t)=[F^{(1)}(-t)\times(-1)+F^{(1)}(t)\times 1]=F^{(1)}(t)-F^{(1)}(-t) $$

\Rightarrow \alpha^{(3)}(t)=F^{(2)}(t)+F^{(2)}(-t) $$

\alpha_2^{(1)}(t)=\dfrac{1}{3}[F(-t)+4F(0)+F(t)]-\dfrac{t}{3}F^{(1)}(-t)+0+\dfrac{t}{3}F^{(1)}(t) $$

\Rightarrow \alpha_2^{(2)}(t)=-\dfrac{1}{3}F^{(1)}(-t)+\dfrac{1}{3}F^{(1)}(t)-\dfrac{1}{3}F^{(1)}(-t)+\dfrac{t}{3}F^{(2)}(-t)+\dfrac{1}{3}F^{(1)}(t)+\dfrac{t}{3}F^{(2)}(t) $$

=-\dfrac{2}{3}F^{(1)}(-t)+\dfrac{2}{3}F^{(1)}(t)+\dfrac{t}{3}F^{(2)}(-t)+\dfrac{t}{3}F^{(2)}(t) $$

\Rightarrow \alpha_2^{(3)}(t)=\dfrac{2}{3}F^{(2)}(-t)+\dfrac{2}{3}F^{(2)}(t)+\dfrac{1}{3}F^{(2)}(-t)-\dfrac{t}{3}F^{(3)}(-t)+\dfrac{1}{3}F^{(2)}(t)+\dfrac{t}{3}F^{(3)}(t) $$

=F^{(2)}(-t)+F^{(2)}(t)+\dfrac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$

\Rightarrow e^{(3)}(t)=\alpha^{(3)}(t)-\alpha_2^{(3)}(t)=F^{(2)}(t)+F^{(2)}(-t)-F^{(2)}(-t)-F^{(2)}(t)+\dfrac{t}{3}[F^{(3)}(-t)-F^{(3)}(t)]= $$

=\dfrac{t}{3}[F^{(3)}(-t)-F^{(3)}(t)]=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$

 * {| style="width:60%" border="0"

$$ e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$
 * style="width:30%; padding:10px; border:1px solid #8888aa" |
 * style="width:30%; padding:10px; border:1px solid #8888aa" |
 * style= |
 * }

title: User:Egm6341.s10.Team4.nimaa&m/HW2


text: File:HW 2 9 1.pdf

title: File:HW 2 9 1.pdf


text:
Talk:Python Concepts/Libraries

Capitalization
Wow. The capitalization on this page is going to give me a seizure.

This Page Needs Fixing But I'm Too Lazy.
 * I guess it was fixed. Devourer09  ( t · c ) 00:57, 9 May 2011 (UTC)

Libraries vs. Resources
It seems to me that libraries and resources serve different purposes. Libraries seems specific to code libraries, while resources appears to be a collection of resources that help you learn about Python. -- Dave Braunschweig (discuss • contribs) 00:07, 15 April 2015 (UTC)

 * After some consideration, I've decided to expand it and add it under the appendix section. --I8086 (discuss • contribs) 14:39, 15 April 2015 (UTC)

 * One of the libraries that should certainly be mentioned here is Pywikibot. -- Dave Braunschweig (discuss • contribs) 14:59, 15 April 2015 (UTC)

 * Indeed, something like this should be on there. I'll add it on within the week. --I8086 (discuss • contribs) 15:31, 15 April 2015 (UTC)

title: Talk:Python Concepts/Libraries


text:
User talk:H.L.Jolly

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 30 January 2010 (UTC)

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User talk:Nday~enwikiversity

hello i want to learn dutch language to english and i want to heard how to prononce,

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Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Nday. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Nday~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:45, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:17, 19 April 2015 (UTC)

title: User talk:Nday~enwikiversity


text:
User talk:Andrew c

 Hello Andrew c, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. SB_Johnny talk 12:45, 30 January 2010 (UTC)

title: User talk:Andrew c


text:
Wikiversity:Activity unit template

Guidelines for the writing of activity units

 * 1) Read carefully the introduction to, and rationale for, the course, paying particular attention to the curriculum.
 * 2) Look at and study the existing "Activity units" to see how they are constructed and where there are gaps in the series which need to be filled.
 * 3) The use of I.T., computation, web research and organisational skills are given due emphasis in each Activity unit. Some assignments involve downloading and working with PDF format worksheets. So the Adobe Acrobat Reader is needed. Open source software such as Open Office and Inkscape can be employed and students are encouraged to used their acquired skills and knowledge to make contributions to Wiki projects.
 * 4) List what you consider to be the gaps in the course.
 * 5) Draft an Activity unit, modelled on the previously compiled ones, to fill a gap that you have identified.
 * 6) If the ideal support material does not exist online then produce it.(Wikipedia article, Wikimedia images, Wikibooks)
 * 7) Carry out the activities that you have suggested on your draft Activity unit.
 * 8) Make any modifications needed to the Activity unit and support materials in the light of your experience in carrying out the activities.
 * 9) Choose a suitable title for your Activity unit page. (Usually a verb and a noun will be sufficient.)
 * 10) Put you suggested Activity unit "live" online, list it at the appropriate place in the course index by copying just the grey index box, and link it to the programme of study.(The grey box in the template automatically repeats the name of the page on which it is used so must be altered when used in the course index.)
 * 11) Wait for, evaluate and be prepared to act on feedback from other writers and students.

Activity unit template
{| cellpadding="10" cellspacing="5" style="width: 900px; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #969696; border: 2px solid black; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Module:
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Techniques:
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Skills:
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Food knowledge:
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Equipment & appliances:
 * }
 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#7F462C solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #FFE7CC; border: 4px solid #C47451; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Objectives
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#7F462C solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #FFE7CC; border: 4px solid #C47451; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Assignment
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#7F462C solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #FFE7CC; border: 4px solid #C47451; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Essential reading
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#7F462C solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #FFE7CC; border: 4px solid #C47451; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Vocabulary
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Preparation skills
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Weights and measures:
Insert text here.

Methods and techniques
Insert text here.

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Knowledge of:
Insert text here.

Experience with:
Insert text here.

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Knowledge of:
Insert text here.

Experience with:
Insert text here.

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Health and safety
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Purchasing and budgeting
Insert text here

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Presentation

Search various sources for presentation ideas and record your findings on a search worksheet. Use the ideas gained to present your food attractively.

Comment:

Google images and Wikimedia Commons are useful sources of images.
 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#827839 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #ECE5B6; border: 4px solid #ADA96E; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Recording and evaluation

Keep your completed worksheets in a file as a record of your completion of this Activity unit. You can also photograph your cooking as it progresses and when it is completed ready for serving.

 * }

{| border=1 cellspacing=0 cellpadding=7 width="900px" style="border:#1a3947 solid 4px; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; text-align: left;"
 * style="width: 40%; background-color: #b7ccd6; border: 4px solid #87AFC7; vertical-align: top; -moz-border-radius-topleft: 3px; -moz-border-radius-bottomleft: 3px; -moz-border-radius-topright: 3px; -moz-border-radius-bottomright: 3px"|

Further and background reading:
Insert text here.

More experience:
Insert text here.

Extension activities:
Insert text here.

See also:

 * }

title: Wikiversity:Activity unit template