text:
User:Egm6341.s10.team3.sa/HW2 updated

HW Problem Set 2

=Equivalence between an $$n^{th}$$  order polynomial and Lagrange interpolation polynomial= Ref: Lecture notes [[media:Egm6341.s10.mtg8.pdf|p.8-3]]

Problem Statement

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$$\displaystyle where\, c_0,\, c_1\,and\,  c_2\,  are\, constants.$$ $$\displaystyle P_2(x) = \sum_{i=0}^{2} l_i(x) f(x_i)$$ $$\displaystyle Find\, the\,  coefficients\,  c_i\, in\, terms\,  of\,  x_i\, and\,  f(x_i)\;\; i=0,1,2. $$
 * $$\displaystyle f_2(x) = P_2(x) = c_0 + c_1x + c_2x^2 $$
 * $$\displaystyle f_2(x) = P_2(x) = c_0 + c_1x + c_2x^2 $$
 * }
 * }

Solution

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 * $$\displaystyle P_2(x) = \sum_{i=0}^{2} l_i(x) f(x_i) $$
 * $$\displaystyle (Eq. 1) $$
 * }
 * }
 * }

We know from lecture slide Eq.2 [[media:Egm6341.s10.mtg7.pdf|p.7-3]],

Substitute i=0 and n=2 in Eq.(2),

Substitute i=1 and n=2 in Eq.(2),

Substitute i=2 and n=2 in Eq.(2),

Substituting Eq.(3) through Eq.(5) in Eq.(1),
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$$\displaystyle = \Big[\frac{x^2 -xx_2-xx_1+x_1x_2} {(x_0 - x_1)(x_0 - x_2)}\Big] f(x_0) + \Big[\frac{x^2 -xx_0-xx_2+x_0x_2} {(x_1 - x_0)(x_1 - x_2)}\Big] f(x_1) + \Big[\frac{x^2 -xx_0-xx_1+x_0x_1} {(x_2 - x_0)(x_2 - x_1)}\Big] f(x_2) $$ $$\displaystyle Grouping\, the\, x, x^2\, and\, constant\, terms\, $$ $$\displaystyle P_2(x)\quad= \bigg[\frac{x_1x_2} {(x_0 - x_1)(x_0 - x_2)} f(x_0)  +  \frac{x_0x_2} {(x_1 - x_0)(x_1 - x_2)} f(x_1) +  \frac{x_0x_1} {(x_2 - x_0)(x_2 - x_1)} f(x_2)\bigg]  $$ $$\displaystyle - \bigg[ \frac{x_1+x_2} {(x_0 - x_1)(x_0 - x_2)} f(x_0)  +  \frac{x_0+x_2} {(x_1 - x_0)(x_1 - x_2)} f(x_1) +  \frac{x_0+x_1} {(x_2 - x_0)(x_2 - x_1)} f(x_2)\bigg] x $$ $$\displaystyle + \bigg[ \frac{1} {(x_0 - x_1)(x_0 - x_2)} f(x_0)  +  \frac{1} {(x_1 - x_0)(x_1 - x_2)} f(x_1) +  \frac{1} {(x_2 - x_0)(x_2 - x_1)} f(x_2)\bigg] x^2 $$
 * $$\displaystyle P_2(x) = l_0(x) f(x_0) + l_1(x) f(x_1) + l_2(x) f(x_2) $$
 * $$\displaystyle P_2(x) = l_0(x) f(x_0) + l_1(x) f(x_1) + l_2(x) f(x_2) $$
 * $$\displaystyle (Eq. 6) $$
 * }
 * }

Comparing Eq.(6) with
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 * $$\displaystyle f_2(x) = c_0 + c_1x + c_2x^2 $$
 * }
 * }
 * }

we find that,
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$$\displaystyle c_0 = \frac{x_1x_2} {(x_0 - x_1)(x_0 - x_2)} f(x_0) +  \frac{x_0x_2} {(x_1 - x_0)(x_1 - x_2)} f(x_1) +  \frac{x_0x_1} {(x_2 - x_0)(x_2 - x_1)} f(x_2)  $$
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$$\displaystyle c_1 = -\Bigg[ \frac{x_1+x_2} {(x_0 - x_1)(x_0 - x_2)} f(x_0) +  \frac{x_0+x_2} {(x_1 - x_0)(x_1 - x_2)} f(x_1) +  \frac{x_0+x_1} {(x_2 - x_0)(x_2 - x_1)} f(x_2)\Bigg]  $$
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$$\displaystyle c_2 = \frac{1} {(x_0 - x_1)(x_0 - x_2)} f(x_0) +  \frac{1} {(x_1 - x_0)(x_1 - x_2)} f(x_1) +  \frac{1} {(x_2 - x_0)(x_2 - x_1)} f(x_2)  $$
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Subramanian Annamalai 21:37, 20 February 2010 (UTC)

= Derivation of the Simple Simpson's rule = Ref: Lecture Notes [[media:Egm6341.s10.mtg9.pdf|p.9-1]]

Problem Statement
Starting from $$\displaystyle p_2(x_j) = \sum_{i=0}^2 l_i(x_j) f(x_i)$$ derive the Simple Simpson's rule Use (4) in [[media:Egm6341.s10.mtg8.pdf|Lecture p.8-3]]to derive the Simple Simpson's rule (2) [[media:Egm6341.s10.mtg7.pdf|Lecture p.7-2]]

Solution
We use (4) in [[media:Egm6341.s10.mtg8.pdf|Lecture p.8-3]] to derive the left side of simple simpson's equation as follows:
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I_2 & = \int_{a}^{b} P_2(x)\ dx \\ & = \int_{a}^{b} \sum_{i=0}^{2}l_i(x)f(x_i)\ dx \\ & = \int_{a}^{b} l_0(x)f(x_0) + l_1(x)f(x_1) + l_2(x)f(x_2)\ dx \\ & = f(x_0)\int_{a}^{b} l_0(x)\ dx + f(x_1)\int_{a}^{b} l_1(x)\ dx + f(x_2)\int_{a}^{b} l_2(x)\ dx \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (Eq. 1)
 * }
 * }

Where $$ x_0 = a, x_1 = \frac{a+b}{2}, x_2 = b $$

Next is to do the integration of $$ {l_0(x), l_1(x),  l_2(x)} $$

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\int_{a}^{b} l_0(x)\ dx & = \int_{a}^{b}\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\ dx \\ & = \int_{a}^{b}\frac{(x-\frac{a+b}{2})(x-b)}{(a-\frac{a+b}{2})(a-b)}\ dx \\ & = \frac{2}{(b-a)^2}\left [ \frac{x^3}{3} - \frac{a+3b}{4}x^2 + \frac{b(a+b)}{2}x \right ]_{a}^b \\ & = \frac{2}{(b-a)^2} \left [ \frac{(b-a)(b^2 + ab + a^2)}{3} - \frac{a+3b}{4}(b-a)(b+a) + \frac{b(a+b)}{2}(b-a) \right ] \\ & = \frac{2}{(b-a)} \left [ \frac{(b^2 + ab + a^2)}{3} - \frac{a+3b}{4}(b+a) + \frac{b(a+b)}{2} \right ] \\ & = \frac{(b-a)^2}{6(b-a)} \\ & = \frac{b-a}{6}\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (Eq. 2)
 * }
 * }

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\int_{a}^{b} l_1(x)\ dx & = \int_{a}^{b}\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\ dx \\ & = \int_{a}^{b}\frac{(x-a)(x-b)}{(\frac{a+b}{2}-a)(\frac{a+b}{2}-b)}\ dx \\ & = \frac{4}{(b-a)(a-b)}\left [ \frac{x^3}{3} - \frac{a+b}{2}x^2 + abx \right ]_{a}^b \\ & = \frac{4}{(b-a)(a-b)}\left [ \frac{(b-a)(b^2+ab+a^2)}{3} - \frac{(a+b)(a+b)(a-b)}{2} + ab(b-a) \right ]_{a}^b \\ & = \frac{4}{(a-b)}\left [ \frac{(b^2+ab+a^2)}{3} - \frac{(a+b)(a+b)}{2} + ab \right ]_{a}^b \\ & = \frac{2(b-a)^2}{3(b-a)} \\ & = \frac{2(b-a)}{3}\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (Eq. 3)
 * }
 * }

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\int_{a}^{b} l_2(x) \ dx & = \int_{a}^{b}\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\ dx \\ & = \int_{a}^{b}\frac{(x-a)(x-\frac{a+b}{2})}{(b-a)(b-\frac{a+b}{2})}\ dx \\ & = \frac{2}{(b-a)^2}\left [ \frac{x^3}{3} - \frac{3a+b}{4}x^2 + \frac{a(a+b)}{2}x \right ]_{a}^b \\ & = \frac{2}{(b-a)^2} \left [ \frac{(b-a)(b^2 + ab + a^2)}{3} - \frac{3a+b}{4}(b-a)(b+a) + \frac{b(a+b)}{2}(b-a) \right ] \\ & = \frac{2}{(b-a)} \left [ \frac{(b^2 + ab + a^2)}{3} - \frac{3a+b}{4}(b+a) + \frac{b(a+b)}{2} \right ] \\ & = \frac{(b-a)^2}{6(b-a)} \\ & = \frac{b-a}{6}\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (Eq. 4)
 * }
 * }

Use Eq.1, Eq.2, Eq.3 and Eq.4, we have:

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$$ \displaystyle \begin{align} \Rightarrow \ I_2 & = \frac{b-a}{6}f(x_0)+\frac{2(b-a)}{3}f(x_1)+\frac{b-a}{6}f(x_2)\\ & = \frac{b-a}{6}[f(x_0)+4f(x_1)+f(x_2)]\\ & = \frac{h}{3}[f(x_0)+4f(x_1)+f(x_2)] \end{align} $$ Where $$h = \frac{b-a}{2}$$
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-- By Min Zhong 12:43, 22 February 2010 (UTC)

= Newton-Cotes Method - Constructing $$f_n(x)$$ for varying n = Ref: Lecture notes [[media:Egm6341.s10.mtg9.pdf|p.9-2]]

Problem Statement
Consider $$ f(x)=\frac{e^{x}-1}{x} $$ on $$ x\in[0,1], \quad x_{0}=a=0, \quad x_{n}=b=1 $$.

(i) Construct $$ f_{n}(x)= \sum_{i=0}^{n}l_{i,n}(x)f(x_{i}) $$ for n = 1, 2, 4, 8, 16.

(ii) Plot $$\displaystyle f(x) $$ and $$\displaystyle f_{n}(x) $$ for n = 1, 2, 4, 8, 16.

(iii) Compute $$ I_{n}=\int_{b}^{b}f_{n}(x)dx $$ for n = 1, 2, 4, 8 and compare to $$\displaystyle I $$.

(iv) For n = 4, plot $$\displaystyle l_{0}, l_{1}, l_{2} $$.

(v) Why don't we have to take a look at $$\displaystyle l_{3}, l_{4} $$ for n = 4?

Solution
Recall the formula
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l_{i,n}(x) = \prod_{j=0,i\neq j}^{n}\frac{x-x_{j}}{x_{i}-x_{j}} $$ $$ from [[media:Egm6341.s10.mtg7.pdf|p.7-3]] of lecture note.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

(i)

When n = 1 $$\displaystyle \rightarrow x_{0} = 0, x_{1} = 1 $$.

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f_{1}=\sum_{i=0}^{1}l_{i,1}(x)f(x_{i})=l_{0,1}(x)f(x_{0})+l_{1,1}(x)f(x_{1}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
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l_{0,1}=\prod_{j=0,j\neq 0}^{1}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{x-1}{0-1}, \quad f(x_{0})=f(0)=1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{1,1}=\prod_{j=0,j\neq 1}^{1}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{x-0}{1-0}, \quad f(x_{1})=f(1)=e-1. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

When n = 2 $$ \rightarrow x_{0} = 0, x_{1} = \frac{1}{2}, x_{2} = 1 $$.

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f_{2}=\sum_{i=0}^{2}l_{i,2}(x)f(x_{i})=l_{0,2}(x)f(x_{0})+l_{1,2}(x)f(x_{1})+l_{2,2}(x)f(x_{2}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
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l_{0,2}=\prod_{j=0,j\neq 0}^{2}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{2})(x-1)}{(0-\frac{1}{2}(0-1)}, \quad f(x_{0})=f(0)=1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{1,2}=\prod_{j=0,j\neq 1}^{2}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-1)}{(\frac{1}{2}-0)(\frac{1}{2}-1)}, \quad f(x_{1})=f(\frac{1}{2})=2\left(e^{\frac{1}{2}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{2,2}=\prod_{j=0,j\neq 2}^{2}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x-0)(x-\frac{1}{2})}{(1-0)(1-\frac{1}{2})}, \quad f(x_{2})=f(1)=e-1. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

When n = 4 $$ \rightarrow x_{0} = 0, x_{1} = \frac{1}{4}, x_{2} = \frac{2}{4}, x_{3} = \frac{3}{4}, x_{4} = 1 $$.

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f_{4}=\sum_{i=0}^{4}l_{i,4}(x)f(x_{i})=l_{0,4}(x)f(x_{0})+l_{1,4}(x)f(x_{1})+l_{2,4}(x)f(x_{2})+l_{3,4}(x)f(x_{3})+l_{4,4}(x)f(x_{4}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
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l_{0,4}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{4})(x-\frac{2}{4})(x-\frac{3}{4})(x-1)}{(0-\frac{1}{4})(0-\frac{2}{4})(0-\frac{3}{4})(0-1)}, \quad f(x_{0})=f(0)=1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{1,4}=\prod_{j=0,j\neq 1}^{4}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-\frac{2}{4})(x-\frac{3}{4})(x-1)}{(\frac{1}{4}-0)(\frac{1}{4}-\frac{2}{4})(\frac{1}{4}-\frac{3}{4})(\frac{1}{4}-1)}, \quad f(x_{1})=f(\frac{1}{4})=4\left(e^{\frac{1}{4}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{2,4}=\prod_{j=0,j\neq 2}^{4}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x-0)(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(\frac{2}{4}-0)(\frac{2}{4}-\frac{1}{4})(\frac{2}{4}-\frac{3}{4})(\frac{2}{4}-1)}, \quad f(x_{2})=f(\frac{2}{4})=2\left(e^{\frac{2}{4}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{3,4}=\prod_{j=0,j\neq 3}^{4}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x-0)(x-\frac{1}{4})(x-\frac{2}{4})(x-1)}{(\frac{3}{4}-0)(\frac{3}{4}-\frac{1}{4})(\frac{3}{4}-\frac{2}{4})(\frac{3}{4}-1)}, \quad f(x_{3})=f(\frac{3}{4})=\frac{4}{3}\left(e^{\frac{3}{4}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{4,4}=\prod_{j=0,j\neq 4}^{4}\frac{x-x_{j}}{x_{4}-x_{j}}=\frac{(x-0)(x-\frac{1}{4})(x-\frac{2}{4})(x-\frac{3}{4})}{(1-0)(1-\frac{1}{4})(1-\frac{2}{4})(1-\frac{3}{4})}, \quad f(x_{4})=f(1)=e-1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

When n = 8 $$ \rightarrow x_{0} = 0, x_{1} = \frac{1}{8}, x_{2} = \frac{2}{8}, x_{3} = \frac{3}{8}, x_{4} = \frac{4}{8}, x_{5} = \frac{5}{8}, x_{6} = \frac{6}{8}, x_{7} = \frac{7}{8}, x_{8} = 1 $$.

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f_{8}=\sum_{i=0}^{8}l_{i,8}(x)f(x_{i})=l_{0,8}(x)f(x_{0})+l_{1,8}(x)f(x_{1})+l_{2,8}(x)f(x_{2})+l_{3,8}(x)f(x_{3})+l_{4,8}(x)f(x_{4})+l_{5,8}(x)f(x_{5})+l_{6,8}(x)f(x_{6})+l_{7,8}(x)f(x_{7})+l_{8,8}(x)f(x_{8}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }
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l_{0,8}=\prod_{j=0,j\neq 0}^{8}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(0-\frac{1}{8})(0-\frac{2}{8})(0-\frac{3}{8})(0-\frac{4}{8})(0-\frac{5}{8})(0-\frac{6}{8})(0-\frac{7}{8})(0-1)}, \quad f(x_{0})=f(0)=1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{1,8}=\prod_{j=0,j\neq 1}^{8}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{1}{8}-0)(\frac{1}{8}-\frac{2}{8})(\frac{1}{8}-\frac{3}{8})(\frac{1}{8}-\frac{4}{8})(\frac{1}{8}-\frac{5}{8})(\frac{1}{8}-\frac{6}{8})(\frac{1}{8}-\frac{7}{8})(\frac{1}{8}-1)}, \quad f(x_{1})=f(\frac{1}{8})=8\left(e^{\frac{1}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{2,8}=\prod_{j=0,j\neq 2}^{8}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{2}{8}-0)(\frac{2}{8}-\frac{1}{8})(\frac{2}{8}-\frac{3}{8})(\frac{2}{8}-\frac{4}{8})(\frac{2}{8}-\frac{5}{8})(\frac{2}{8}-\frac{6}{8})(\frac{2}{8}-\frac{7}{8})(\frac{2}{8}-1)}, \quad f(x_{2})=f(\frac{2}{8})=\frac{8}{2}\left(e^{\frac{2}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{3,8}=\prod_{j=0,j\neq 3}^{8}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{3}{8}-0)(\frac{3}{8}-\frac{1}{8})(\frac{3}{8}-\frac{2}{8})(\frac{3}{8}-\frac{4}{8})(\frac{3}{8}-\frac{5}{8})(\frac{3}{8}-\frac{6}{8})(\frac{3}{8}-\frac{7}{8})(\frac{3}{8}-1)}, \quad f(x_{3})=f(\frac{3}{8})=\frac{8}{3}\left(e^{\frac{3}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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 * {| style="width:100%" border="0" align="left"

l_{4,8}=\prod_{j=0,j\neq 4}^{8}\frac{x-x_{j}}{x_{4}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{4}{8}-0)(\frac{4}{8}-\frac{1}{8})(\frac{4}{8}-\frac{2}{8})(\frac{4}{8}-\frac{3}{8})(\frac{4}{8}-\frac{5}{8})(\frac{4}{8}-\frac{6}{8})(\frac{4}{8}-\frac{7}{8})(\frac{4}{8}-1)}, \quad f(x_{4})=f(\frac{4}{8})=\frac{8}{4}\left(e^{\frac{4}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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l_{5,8}=\prod_{j=0,j\neq 5}^{8}\frac{x-x_{j}}{x_{5}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{5}{8}-0)(\frac{5}{8}-\frac{1}{8})(\frac{5}{8}-\frac{2}{8})(\frac{5}{8}-\frac{3}{8})(\frac{5}{8}-\frac{4}{8})(\frac{5}{8}-\frac{6}{8})(\frac{5}{8}-\frac{7}{8})(\frac{5}{8}-1)}, \quad f(x_{5})=f(\frac{5}{8})=\frac{8}{5}\left(e^{\frac{5}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{6,8}=\prod_{j=0,j\neq 6}^{8}\frac{x-x_{j}}{x_{6}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{7}{8})(x-1)}{(\frac{6}{8}-0)(\frac{6}{8}-\frac{1}{8})(\frac{6}{8}-\frac{2}{8})(\frac{6}{8}-\frac{3}{8})(\frac{6}{8}-\frac{4}{8})(\frac{6}{8}-\frac{5}{8})(\frac{6}{8}-\frac{7}{8})(\frac{6}{8}-1)}, \quad f(x_{6})=f(\frac{6}{8})=\frac{8}{6}\left(e^{\frac{6}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{7,8}=\prod_{j=0,j\neq 7}^{8}\frac{x-x_{j}}{x_{7}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-1)}{(\frac{7}{8}-0)(\frac{7}{8}-\frac{1}{8})(\frac{7}{8}-\frac{2}{8})(\frac{7}{8}-\frac{3}{8})(\frac{7}{8}-\frac{4}{8})(\frac{7}{8}-\frac{5}{8})(\frac{7}{8}-\frac{6}{8})(\frac{7}{8}-1)}, \quad f(x_{7})=f(\frac{7}{8})=\frac{8}{7}\left(e^{\frac{7}{8}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{8,8}=\prod_{j=0,j\neq 8}^{8}\frac{x-x_{j}}{x_{8}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})}{(1-0)(1-\frac{1}{8})(1-\frac{2}{8})(1-\frac{3}{8})(1-\frac{4}{8})(1-\frac{5}{8})(1-\frac{6}{8})(1-\frac{7}{8})}, \quad f(x_{8})=f(1)=e-1. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

When n = 16 $$ \rightarrow x_{0} = 0, x_{1} = \frac{1}{16}, x_{2} = \frac{2}{16}, x_{3} = \frac{3}{16}, x_{4} = \frac{4}{16}, x_{5} = \frac{5}{16}, x_{6} = \frac{6}{16}, x_{7} = \frac{7}{16}, x_{8} = \frac{8}{16}, x_{9} = \frac{9}{16}, x_{10} = \frac{10}{16}, x_{11} = \frac{11}{16}, x_{12} = \frac{12}{16}, x_{13} = \frac{13}{16}, x_{14} = \frac{14}{16}, x_{15} = \frac{15}{16}, x_{16} = 1 $$.

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

f_{16}=\sum_{i=0}^{16}l_{i,16}(x)f(x_{i})=l_{0,16}(x)f(x_{0})+l_{1,16}(x)f(x_{1})+l_{2,16}(x)f(x_{2})+l_{3,16}(x)f(x_{3})+l_{4,16}(x)f(x_{4})+l_{5,16}(x)f(x_{5})+l_{6,16}(x)f(x_{6})+l_{7,16}(x)f(x_{7})+l_{8,16}(x)f(x_{8})+l_{9,16}(x)f(x_{9})+l_{10,16}(x)f(x_{10})+l_{11,16}(x)f(x_{11})+l_{12,16}(x)f(x_{12})+l_{13,16}(x)f(x_{13})+l_{14,16}(x)f(x_{14})+l_{15,16}(x)f(x_{15})+l_{16,16}(x)f(x_{16}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

l_{0,16}=\prod_{j=0,j\neq 0}^{16}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{16})(x-\frac{2}{16})(x-\frac{3}{16})(x-\frac{4}{16})(x-\frac{5}{16})(x-\frac{6}{16})(x-\frac{7}{16})(x-\frac{8}{16})(x-\frac{9}{16})(x-\frac{10}{16})(x-\frac{11}{16})(x-\frac{12}{16})(x-\frac{13}{16})(x-\frac{14}{16})(x-\frac{15}{16})(x-1)}{(0-\frac{1}{16})(0-\frac{2}{16})(0-\frac{3}{16})(0-\frac{4}{16})(0-\frac{5}{16})(0-\frac{6}{16})(0-\frac{7}{16})(0-\frac{8}{16})(0-\frac{9}{16})(0-\frac{10}{16})(0-\frac{11}{16})(0-\frac{12}{16})(0-\frac{13}{16})(0-\frac{14}{16})(0-\frac{15}{16})(0-1)}, \quad f(x_{0})=f(0)=1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{1,16}=\prod_{j=0,j\neq 1}^{16}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-\frac{2}{16})(x-\frac{3}{16})(x-\frac{4}{16})(x-\frac{5}{16})(x-\frac{6}{16})(x-\frac{7}{16})(x-\frac{8}{16})(x-\frac{9}{16})(x-\frac{10}{16})(x-\frac{11}{16})(x-\frac{12}{16})(x-\frac{13}{16})(x-\frac{14}{16})(x-\frac{15}{16})(x-1)}{(\frac{1}{16}-0)(\frac{1}{16}-\frac{2}{16})(\frac{1}{16}-\frac{3}{16})(\frac{1}{16}-\frac{4}{16})(\frac{1}{16}-\frac{5}{16})(\frac{1}{16}-\frac{6}{16})(\frac{1}{16}-\frac{7}{16})(\frac{1}{16}-\frac{8}{16})(\frac{1}{16}-\frac{9}{16})(\frac{1}{16}-\frac{10}{16})(\frac{1}{16}-\frac{11}{16})(\frac{1}{16}-\frac{12}{16})(\frac{1}{16}-\frac{13}{16})(\frac{1}{16}-\frac{14}{16})(\frac{1}{16}-\frac{15}{16})(\frac{1}{16}-1)}, \quad f(x_{1})=f(\frac{1}{16})=16\left(e^{\frac{1}{16}}-1\right), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

$$\cdot$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$\cdot$$

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

l_{16,16}=\prod_{j=0,j\neq 16}^{16}\frac{x-x_{j}}{x_{16}-x_{j}}=\frac{(x-0)(x-\frac{1}{16})(x-\frac{2}{16})(x-\frac{3}{16})(x-\frac{4}{16})(x-\frac{5}{16})(x-\frac{6}{16})(x-\frac{7}{16})(x-\frac{8}{16})(x-\frac{9}{16})(x-\frac{10}{16})(x-\frac{11}{16})(x-\frac{12}{16})(x-\frac{13}{16})(x-\frac{14}{16})(x-\frac{15}{16})}{(1-0)(1-\frac{1}{16})(1-\frac{2}{16})(1-\frac{3}{16})(1-\frac{4}{16})(1-\frac{5}{16})(1-\frac{6}{16})(1-\frac{7}{16})(1-\frac{8}{16})(1-\frac{9}{16})(1-\frac{10}{16})(1-\frac{11}{16})(1-\frac{12}{16})(1-\frac{13}{16})(1-\frac{14}{16})(1-\frac{15}{16})}, \quad f(x_{16})=f(1)=e-1, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

(ii)

 Matlab code of this plots is provided in the end of this document

In order to compare the plots of different functions, it is definitely efficient to plot all functions in the same figure. In this case, however, the plots of functions $$\displaystyle f_{n}(x) $$ for n = 2, 4, 8, 16 almost overlap with $$\displaystyle f(x) $$ for $$\displaystyle x \in [0, 1] $$, and thus we cannot distinguish a specific plot from the others. Thus, we use separate figures for each plot of functions.

Plot $$\displaystyle f(x) = \frac{e^{x}-1}{x} $$

Plot $$\displaystyle f_{1}(x)$$

Plot $$\displaystyle f_{2}(x)$$

Plot $$\displaystyle f_{4}(x)$$

Plot $$\displaystyle f_{8}(x)$$

Plot $$\displaystyle f_{16}(x)$$

(iii)

 Matlab code of this calculations is provided in the end of this document

(iv)

 Matlab code of this plots is provided in the end of this document

Plots of $$\displaystyle l_{0}, l_{1}, l_{2} $$ when n = 4.

(v)

 Matlab code of this plots is provided in the end of this document

As the plots of $$\displaystyle l_{i} $$ for all $$\displaystyle i $$ in the figure below, the plots of $$\displaystyle l_{3} $$ and $$\displaystyle l_{4} $$ are the mirror images of $$\displaystyle l_{1} $$ and $$\displaystyle l_{0} $$ respectively with respect to $$\displaystyle x = x_{2}=0.5 $$. Therefore, we don't need to take a look at the plots of $$\displaystyle l_{3} $$ and $$\displaystyle l_{4} $$ if we already get the plots of $$\displaystyle l_{0} $$ and $$\displaystyle l_{1} $$.

Matlab code for (ii), (iii), (iv) and (v)

For this problem, one global code is much more efficient than the several short codes for each parts of problem. Therefore, we are providing a comprehensive code below.

Yong Nam Ahn 02:54, 21 February 2010 (UTC)

= Derivation:Composite trapezoidal rule and Composite simpson's rule from simple trapezoidal and simple simpson's rule =

Ref Lecture notes [[media:Egm6341.s10.mtg9.pdf|p.9-3]]

Problem Statement
Derive composite Trapezoidal rule[[media:Egm6341.s10.mtg7.pdf|p.7-1]] from Simple Trapezoidal rule [[media:Egm6341.s10.mtg7.pdf|p.7-1]]

and

Derive composite Simpson's rule [[media:Egm6341.s10.mtg7.pdf|p.7-2]] from simple Simpson's rule[[media:Egm6341.s10.mtg7.pdf|p.7-2]]

Solution
Composite Trapezoidal rule

From Simple trapezoidal rule we have ,

$$ I_1 = \frac{h}{2}[f_{\text{o}}+ f_1] $$

$$    = h [\frac{1}{2}f_{\text{o}}+\frac{1}{2} f_1]     $$

similarly we have

$$ I_2 = h [\frac{1}{2}f_1+\frac{1}{2} f_2] $$

$$ I_3 = h [\frac{1}{2}f_2+\frac{1}{2} f_3] $$ . . . . $$ I_{\text{n-1}} = h [\frac{1}{2}f_{\text{n-2}}+\frac{1}{2} f_{\text{n-1}}] $$

$$ I_{\text{n}} = h [\frac{1}{2}f_{\text{n-1}}+\frac{1}{2} f_{\text{n}}] $$

Summation of all of above expression gives

$$ I_{\text{n}} = h [\frac{1}{2}f_{\text{o}}+(\frac{1}{2} f_1 +\frac{1}{2} f_1)+(\frac{1}{2} f_2+\frac{1}{2} f_2)+.....+(\frac{1}{2} f_{\text{n-1}}+\frac{1}{2} f_{\text{n-1}})+ \frac{1}{2} f_{\text{n}} ] $$

$$ = h [\frac{1}{2}f_{\text{o}}+ f_1 + f_2 +.......+ f_{\text{n-1}} +\frac{1}{2} f_{\text{n}}]$$

Hence Proved..

Composite Simpson's Rule

From Simple Simpson's rule we obtain ,

$$ I_2 = \frac {h}{3} [f_{\text{o}}+4 f_1 +f_2] $$

where $$ h = \frac {b-a}{2}$$

Similarly

$$ I_4 = \frac {h}{3} [f_2+4 f_3 +f_4] $$

$$ I_6 = \frac {h}{3} [f_4+4 f_5 +f_6] $$ . . . . $$ I_{\text{n-2}} = \frac {h}{3} [f_{\text{n-4}}+4 f_{\text{n-3}} +f_{\text{n-2}}] $$

$$ I_{\text{n}} = \frac {h}{3} [f_{\text{n-2}}+4 f_{\text{n-1}} +f_{\text{n}}] $$

After Summation of above terms we obtain

$$ I_{\text{n}} = \frac {h}{3} [f_{\text{o}}+4 f_1+(f_2+f_2)+4f_{\text{3}}+(f_4+f_4)+4f_5+f_6.......+(f_{\text{n-2}}+f_{\text{n-2}})+4 f_{\text{n-1}} +f_{\text{n}}] $$

$$ I_{\text{n}} = \frac {h}{3} [f_{\text{o}}+4 f_1+2f_2+4f_{\text{3}}+2f_4+4f_5+f_6.......+2f_{\text{n-2}}+4 f_{\text{n-1}} +f_{\text{n}}] $$

where n = 2k  and k = 1,2,3,4.....

Hence Proved

Abhishekksingh 20:54, 21 February 2010 (UTC)

Find n such that error in Taylor series is less than the error from Lagrange series
Ref: Lecture Notes [[media:Egm6341.s10.mtg11.pdf|p.11-1]]

Problem Statement
Find n at $$ \displaystyle \left|{f}_{n}^{T}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq \left|{f}_{4}^{L}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq \left| \frac{{q}_{5}(t)}{5!} \right| $$  for  $$ \displaystyle f(x) = sin(x) $$ and $$ \displaystyle x_{0} = 0, x_{1} = \frac{\pi}{4}, x_{2} = \frac{\pi}{2}, x_{3} = \frac{3\pi}{4}, x_{4} = 1 $$.

Solution
Matlab codes for this problem is provided later.

First, construct Lagrange interpolating function $$ f_{4}^{L}(x) $$ and polynomial $$\displaystyle q_{5}(x) $$.

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

f_{4}^{L}(x)=P_{4}(x)= \sum_{i=0}^{4}l_{i,4}(x)f(x_{i}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

where

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

l_{i,4}(x)=\Pi_{j=0, j \neq 4}^{4}\frac{x-x_{j}}{x_{i}-x_{j}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Then, compare the constructed function $$\displaystyle f_{4}^{L}(x)$$ with the original function $$\displaystyle f(x)=sin(x)$$ by plotting them in the same figure.

Now, to check the fact $$\displaystyle \left|{f}_{4}^{L}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq \left| \frac{{q}_{5}(t)}{5!} \right| $$, compute $$\displaystyle f_{4}^{L}(\frac{7\pi}{8})$$, $$\displaystyle f(\frac{7\pi}{8})$$ and $$\displaystyle \frac{\left|q_{5}(\frac{7\pi}{8})\right|}{5!}$$.

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

f_{4}^{L}(\frac{7\pi}{8}) = 3.812026433338313e-001 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

f(\frac{7\pi}{8}) = sin(\frac{7\pi}{8}) = 3.826834260893118e-001 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\frac{\left|q_{5}(\frac{7\pi}{8})\right|}{5!} = 8.171616062246945e-003 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\left|E_{4}^{L}(\frac{7\pi}{8})\right|=\left|{f}_{4}^{L}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq \left| \frac{{q}_{5}(t)}{5!} \right| $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

$$\displaystyle \Rightarrow \left|E_{4}^{L}(\frac{7\pi}{8})\right| = 1.480782755480525e-003 \leq \frac{\left|q_{5}(\frac{7\pi}{8})\right|}{5!} = 8.171616062246945e-003 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

 Matlab code for above plot and calculations

Now, calculate $$ \displaystyle \left|{f}_{n}^{T}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| \leq 1.480782755480525e-003 $$ when $$ \displaystyle n=0, 1, 2, 3... n $$

Taylor Expansion of $$ \displaystyle f(x)=sin(x) $$ at $$ \displaystyle x_0 = \frac{\pi}{4} $$ Ref: Homework No. 1 [|Question 7]

a) Case 1 ($$\displaystyle n=0 $$) $$ \left|{f}_{0}^{T}(x) - f(x)\right| = \left| sin(x) - sin(x)\right| = 0 $$

b) Case 2 ($$\displaystyle n=1 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!}\Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!}\Bigg] -sin(\frac{7\pi}{8}) \right| = 1.712824266995947 $$

c) Case 3 ($$\displaystyle n=2 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 0.349764853003978 $$

d) Case 4 ($$\displaystyle n=3 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) + {f}_{3}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} - \frac{(x-\frac{\pi}{4})^3}{3!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^3}{3!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 0.542355447288433 $$

e) Case 5 ($$\displaystyle n=4 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) + {f}_{3}^{T}(x) + {f}_{4}^{T}(x)- f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} - \frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^3}{3!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^4}{4!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 0.104436918926408 $$

f) Case 6 ($$\displaystyle n=5 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) + {f}_{3}^{T}(x) + {f}_{4}^{T}(x)+ {f}_{5}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} - \frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^3}{3!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^4}{4!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^5}{5!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 0.067533285020216 $$

g) Case 7 ($$\displaystyle n=6 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) + {f}_{3}^{T}(x) + {f}_{4}^{T}(x) + {f}_{5}^{T}(x) + {f}_{6}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} - \frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^3}{3!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^4}{4!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^5}{5!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^6}{6!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 0.011256167379064 $$

h) Case 8 ($$\displaystyle n=7 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) + {f}_{3}^{T}(x) + {f}_{4}^{T}(x) + {f}_{5}^{T}(x) + {f}_{6}^{T}(x) + {f}_{7}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} - \frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} - \frac{(x-\frac{\pi}{4})^7}{7!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^3}{3!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^4}{4!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^5}{5!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^6}{6!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^7}{7!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 0.004529527205458 $$

i) Case 9 ($$\displaystyle n=8 $$) $$ \left|{f}_{0}^{T}(x) + {f}_{1}^{T}(x) + {f}_{2}^{T}(x) + {f}_{3}^{T}(x) + {f}_{4}^{T}(x) + {f}_{5}^{T}(x) + {f}_{6}^{T}(x) + {f}_{7}^{T}(x) + {f}_{8}^{T}(x) - f(x)\right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} - \frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} - \frac{(x-\frac{\pi}{4})^7}{7!} + \frac{(x-\frac{\pi}{4})^8}{8!} \Bigg] - sin(x) \right| $$ $$ = \left| \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})}{1!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^2}{2!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^3}{3!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^4}{4!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^5}{5!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^6}{6!} - \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^7}{7!} + \frac{(\frac{7\pi}{8}-\frac{\pi}{4})^8}{8!} \Bigg] -sin(\frac{7\pi}{8}) \right| = 6.551348508838095*10^{-4} $$

Compare with the value of $$ \left|{f}_{4}^{L}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| $$

Thus, $$ \left|{f}_{n}^{T}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| $$ is smaller than $$ \left|{f}_{4}^{L}(\frac{7\pi}{8}) - f(\frac{7\pi}{8})\right| $$ when n=8.

--Heejun Chung 03:24, 21 February 2010 (UTC)

=Lagrange Interpolation Error - $$(n+1)^{th} $$ derivative= Ref: Lecture Notes [[media:Egm6341.s10.mtg12.pdf|p.12-2]]

Problem Statement
Prove that
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle E^{n+1}(x) = f^{n+1}(x) $$
 * }
 * }
 * }

Solution
From lecture slide [[media:Egm6341.s10.mtg10.pdf|p.10-2]], we know that the Lagrange interpolation error is given by,
 * {| style="width:100%" border="0" align="left"

where,
 * $$\displaystyle E(x) := f(x) - f_n(x) $$
 * $$\displaystyle (Eq. 1) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle E(x) \; $$Interpolation error
 * }
 * }
 * }

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

 * $$\displaystyle f(x) \; $$ Actual function
 * }
 * }
 * }

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

 * $$\displaystyle f_n(x) \; $$ Inerpolating polynomial of order n
 * }
 * }
 * }

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

 * Now differentiating Eq (1) n+1 times,
 * }
 * }
 * }

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

 * $$\displaystyle E^{n+1}(x)= f^{n+1}(x) + \cancelto{0}{{f_n}^{n+1}(x)}$$
 * }
 * }
 * }

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

it's $$\displaystyle(n+1)^{th} $$ derivative is zero.
 * Recall that $$\displaystyle f_n(\cdot) $$ is a polynomial of order n ,
 * Recall that $$\displaystyle f_n(\cdot) $$ is a polynomial of order n ,
 * }
 * }

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

$$\displaystyle \therefore E^{n+1}(x)= f^{n+1}(x) $$
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style =|
 * }
 * }

Subramanian Annamalai 21:38, 20 February 2010 (UTC)

= Proof: $$ (n+1)^{th}  $$derivative  of $$q_{n+1}(x) = (n+1)! $$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg12.pdf|p.12-3]]

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

$$
 * $$\displaystyle q_{n+1}^{(n+1)}(x) = (n+1)!
 * $$\displaystyle q_{n+1}^{(n+1)}(x) = (n+1)!
 * }

Solution
From (4) in [[media:Egm6341.s10.mtg10.pdf|p.10-1]]we have:

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

$$ $$ q_{n+1} $$ is a n+1 polynomial, for simplicity, we express $$q_{n+1}(x)$$ as follows:
 * $$\displaystyle q_{n+1}(x) = (x-x_0)(x-x_1)\ldots (x-x_n)
 * $$\displaystyle q_{n+1}(x) = (x-x_0)(x-x_1)\ldots (x-x_n)
 * }
 * {| style="width:100%" border="0" align="left"

$$ Then the derivation of $$q_{n+1}$$ from (1) to (n+1) is shown as follows:
 * $$\displaystyle q_{n+1}(x) = x^{n+1}+c_n x^{n}+c_{n-1}x^{n-1}\ldots+ c_{1}x + c_0
 * $$\displaystyle q_{n+1}(x) = x^{n+1}+c_n x^{n}+c_{n-1}x^{n-1}\ldots+ c_{1}x + c_0
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle q_{n+1}^{(1)}(x) = (n+1)x^n + nc_nx^{n-1} + (n-1)c_{n-1}x^{n-2}\ldots+c_1+0
 * $$\displaystyle q_{n+1}^{(1)}(x) = (n+1)x^n + nc_nx^{n-1} + (n-1)c_{n-1}x^{n-2}\ldots+c_1+0
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle q_{n+1}^{(2)}(x) = (n+1)(n)x^{n-1} + n(n-1)c_nx^{n-2} + (n-1)(n-2)c_{n-1}x^{n-3}\ldots+0+0
 * $$\displaystyle q_{n+1}^{(2)}(x) = (n+1)(n)x^{n-1} + n(n-1)c_nx^{n-2} + (n-1)(n-2)c_{n-1}x^{n-3}\ldots+0+0
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \dots \dots$$
 * }
 * }

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

 * $$\displaystyle \dots \dots$$
 * }
 * }

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

 * $$\displaystyle \dots \dots$$
 * }
 * }

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

q_{n+1}^{(n)}(x) = (n+1)(n)(n-1)\ldots (2)x + 0 + \ldots+0+0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }

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

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

-- By Min Zhong 12:43, 22 February 2010 (UTC)

= Error analysis of function $$ f(x)=log(x) $$ = Ref: Lecture notes [[media:Egm6341.s10.mtg12.pdf|p.12-3]]

Problem Statement
Consider the function $$\displaystyle f(x)=log(x) $$ and set $$\displaystyle t=2, x_{0}=3, x_{1}=4, x_{2}=5, \cdot \cdot \cdot, x_{6}=9$$.

(i) Plot $$\displaystyle f(x) $$ and $$\displaystyle f_{n}(x)$$

(ii) Plot $$\displaystyle l_{i,n} $$ when $$\displaystyle n = 3 $$.

(iii)Plot $$\displaystyle q_{n+1}(x) $$.

(iv) Obtain $$\displaystyle G(x) $$ for $$\displaystyle x = 5.5 $$.

Solution
From the problem statement, we know that $$\displaystyle n = 6 $$.

(i)

Let's construct Lagrange interpolating function.

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

f_{6}(x)= \sum_{i=0}^{6}l_{i,6}(x)f(x_{i}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

 Matlab code of this plots is provided in the end of this document

Plot $$\displaystyle f(x) $$ and $$\displaystyle f_{6}(x) $$

(ii)

Let's calculate polynomial $$\displaystyle l_{3,6}(x) $$

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

l_{3,6}(x)=\Pi_{j=0,j \neq 3}^{6}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x-3)(x-4)(x-5)(x-7)(x-8)(x-9)}{(6-3)(6-4)(6-5)(6-7)(6-8)(6-9)} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 Matlab code of this plots is provided in the end of this document

Plot $$\displaystyle l_{3,6}(x) $$

In above figure, it is extremely hard to check the behavior of $$\displaystyle l_{3,6}(x) $$ around $$\displaystyle x = x_{i} $$. Thus, we shows the plot of the same function $$\displaystyle l_{3,6}(x) $$ for $$\displaystyle x \in [3,6] $$ below.

From this figure, we can easily see the fact that $$\displaystyle l_{3,6}(x_{i}) = 0 $$ for $$\displaystyle i = 0, 1, 2, 4, 5, 6 $$ and $$\displaystyle l_{3,6}(x_{3})=1 $$.

(iii)

Let's calculate polynomial $$\displaystyle q_{n+1}(x)=q_{7}(x) $$

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

q_{7}(x)=\Pi_{j=0}^{6}(x-x_{j})=(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 Matlab code of this plots is provided in the end of this document

Plot $$\displaystyle q_{7}(x) $$

In above figure, it is extremely hard to check the behavior of $$\displaystyle q_{7}(x) $$ around $$\displaystyle x = x_{i} $$. Thus, we shows the plot of the same function $$\displaystyle q_{7}(x) $$ for $$\displaystyle x \in [3,6] $$ below.

From this figure, we can easily see the fact that $$\displaystyle q_{7}(x_{i}) = 0 $$ for all $$\displaystyle i $$.

(iv)

 Matlab code of the calculations in this part is provided in the end of this document

Recall that $$\displaystyle G(x) $$ can be calculated by the following equation:

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

G(x)=E(x)-\frac{q_{7}(x)}{q_{7}(t)}E(t) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since $$\displaystyle t=2 $$ and $$\displaystyle x=5.5 $$,

$$\displaystyle E(x)=9.460393681548496e-006 $$, $$\displaystyle E(t)=-8.949684549653392e-003 $$,

$$\displaystyle q_{7}(x)=1.230468750000000e+001 $$ and $$\displaystyle q_{7}(t)=-5040 $$.

Therefore,

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

$$\displaystyle G(x) = -1.238942211350373e-005 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }
 * }

 Matlab code for (i), (ii), (iii) and (iv)

Yong Nam Ahn 02:55, 21 February 2010 (UTC)

Error calculation: Simple Trapezoidal rule
Ref: Lecture Notes [[media:Egm6341.s10.mtg13.pdf|p.13-2]]

Problem Statement
Show that $$ \left| {E}_{1}\right| \leq \frac{M_{2}}{2!}\int_{a}^{b}\left | q_{2}(x) \right | dx = \frac{{M}_{2}}{2!}\int_{a}^{b}(x-a)(b-x) dx = \frac{{(b-a)}^{3}}{12}{M}_{2} = \frac{h^{3}}{12}M_{2} $$

where $$ \displaystyle h:= b-a $$

Solution
Expansion $$ \displaystyle (x-a)(b-x) $$

$$ \frac{{M}_{2}}{2!}\int_{a}^{b}(x-a)(b-x) dx = -\frac{{M}_{2}}{2!}\int_{a}^{b}\left \{ x^{2}-(a+b)x+ab \right \}dx $$

Integrating $$ \displaystyle \left \{ x^{2}-(a+b)x+ab \right \} $$

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

$$ = -\frac{{M}_{2}}{2!6}(-b^{3}+3ab^{2}-3a^{2}b+a^{3}) $$

$$ = \frac{{M}_{2}}{2!6}(b^{3}-3ab^{2}+3a^{2}b-a^{3}) $$

$$ = \frac{M_{2}}{12}\underbrace {(b-a)^{3}}_{by Binomial Theorem} $$

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

$$\displaystyle \left| {E}_{1}\right| \leq \frac{M_{2}}{2!}\int_{a}^{b}\left | q_{2}(x) \right | dx = \frac{{M}_{2}}{2!}\int_{a}^{b}(x-a)(b-x) dx = \frac{{(b-a)}^{3}}{12}{M}_{2} = \frac{h^{3}}{12}M_{2} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

--Heejun Chung 03:25, 21 February 2010 (UTC)

Error calculation: Simple Simpson rule
Ref: Lecture Notes [[media:Egm6341.s10.mtg13.pdf|p.13-2]]

Problem Statement
Show that $$ \left| {E}_{n}\right| \leq \frac{{M}_{3}}{3!}\int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right| dx = \frac{{(b-a)}^{4}}{196}{M}_{3} = \frac{{2}^{4}{h}^{4}}{196}{M}_{3} $$

where $$ h:=\frac{b-a}{2} \Rightarrow b-a = 2h $$

Solution
Determination the range at $$ \displaystyle {(x-a)(x-\frac{(a+b)}{2})(x-b)} $$ always $$ \displaystyle \geq 0 $$

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

$$ = \frac{{M}_{3}}{3!}\left[\int_{a}^{\frac{(a+b)}{2}}(x-a)(\frac{(a+b)}{2} - x)(b-x)dx + \int_{\frac{(a+b)}{2}}^{b}(x-a)(x-\frac{(a+b)}{2})(b-x)dx \right] $$

$$ = \frac{{M}_{3}}{3!}\left[\underbrace {\int_{a}^{\frac{(a+b)}{2}}(x-a)(x-\frac{(a+b)}{2})(x-b)dx}_{call {P}_{1}} - \underbrace {\int_{\frac{(a+b)}{2}}^{b}(x-a)(x-\frac{(a+b)}{2})(x-b)dx}_{call {P}_{2}}\right] $$

Integrating $$ \displaystyle P_{1} $$

$$ \int_{a}^{\frac{(a+b)}{2}}(x-a)(x-\frac{(a+b)}{2})(x-b)dx $$

$$ = \frac{1}{4}(\frac{a+b}{2})(\frac{-a-b}{2})(\frac{{a}^{2}+2ab+{b}^{2}}{4}-\frac{{a}^{2}+2ab+{b}^{2}}{2}+2ab)-\frac{1}{4}a(-b)({a}^{2}-{a}^{2}-ab+2ab) $$

$$ = -\frac{1}{4}\times\frac{{(a+b)}^{2}}{4}\times \frac{{-(a-b)}^{2}+4ab}{4}+\frac{{a}^{2}{b}^{2}}{4} $$

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

Integrating $$ \displaystyle P_{2} $$

$$ \int_{\frac{(a+b)}{2}}^{b}(x-a)(x-\frac{(a+b)}{2})(x-b)dx $$

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

Computing $$ \displaystyle P_{1}-P_{2} $$

$$ \begin{align} P_{1}-P_{2} &= \frac{a^{2}b^{2}}{2}+\frac{(a+b)^{2}}{4^{3}}\left [ 2(a-b)^{2}-8ab \right ] \\ &=\frac{1}{32} \left [ (a^{4}+b^{4})-4ab(a^{2}+b^{2})+6a^{2}b^{2} \right ] \\ &=\frac{1}{32} \underbrace{(a^{4}-4a^{3}b+6a^{2}b^{2}-4ab^{3}+b^{4})}_{by Binomial Thereom}\\ &=\frac{(a-b)^{4}}{32} = \frac{(b-a)^{4}}{32}\\\end{align}\ $$

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

$$\displaystyle \displaystyle I=\frac{M_{3}}{3!}(P_{1}-P_{2}) = \frac{M_{3}}{3!} \frac{(b-a)^{4}}{32} = \frac{M_{3}}{192}(b-a)^{4} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

--Heejun Chung 03:26, 21 February 2010 (UTC)

=Proof: Simpson's rule integrates a $$3^{rd}$$ order polynomial exactly= Ref: Lecture notes [[media:Egm6341.s10.mtg13.pdf|p.13-3]]

Problem Statement

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

 * $$\displaystyle f_3(x) = P_3(x) = c_0 + c_1x + c_2x^2 + c_3x^3$$   $$\in [a,b] = [0,1] $$
 * }
 * }
 * }

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

 * $$\displaystyle c_0=3,$$ $$\displaystyle c_1=8,$$  $$\displaystyle c_2=-2, \, c_3 =6.$$
 * }
 * }
 * }

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

$$ \displaystyle Compare\, the\, results.$$
 * $$\displaystyle Find\, I\, and\, I_2 \,using\, Simpson's\, rule.$$
 * $$\displaystyle Find\, I\, and\, I_2 \,using\, Simpson's\, rule.$$
 * }
 * }

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

 * $$\displaystyle P_3(x) = 3+ 8x -2x^2 + 6x^3 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }

EXACT INTEGRAL Integrating Eq (1),

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

 * $$\displaystyle I = \int_{a}^{b} \Bigg(3+ 8x -2x^2 + 6x^3\Bigg)\,dx $$
 * }
 * }
 * }

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

$$\displaystyle = \int_{0}^{1} \Big(3+ 8x -2x^2 + 6x^3\Big)\,dx $$
 * }
 * }

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

$$\displaystyle =\Big[ 3x + 4x^2 - \frac{2x^3}{3} + \frac{3x^4}{2} \Big]_0^1 $$
 * }
 * }

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

$$\displaystyle = \Big[ 3 +4 - \frac{2}{3} + \frac{3}{2}\Big]$$
 * }
 * }

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

$$\displaystyle = \frac{47}{6} $$
 * }
 * }

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

$$\displaystyle \therefore I = 7.8333$$
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

SIMPSON'S RULE We know from lecture slide Eq(2) [[media:Egm6341.s10.mtg7.pdf|p.7-2]] that the Simple Simpson's Rule is given by,

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

 * $$\displaystyle I_2 = \frac{h}{3}\bigg[ f(x_0)+ 4f(x_1)+ f(x_2)\bigg]$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

where,

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

 * $$\displaystyle x_0\, = a\,=0 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3) $$
 * }
 * }
 * }

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

 * $$\displaystyle x_2\, = b\,=1 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4) $$
 * }
 * }
 * }

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

 * $$\displaystyle x_1\, = \frac{a+b}{2}\,= \frac{1}{2} $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5) $$
 * }
 * }
 * }

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

 * $$\displaystyle h\, = \frac{b-a}{2}\,= \frac{1}{2} $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6) $$
 * }
 * }
 * }

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

 * $$\displaystyle f(x_0)\, = \Big[ 3 + 4(0) - 2\cdot(0)^2 + 6\cdot(0)^3 \Big]\,= 3 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7) $$
 * }
 * }
 * }

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

 * $$\displaystyle f(x_1)\, = \Big[ 3 + 4(0.5) - 2\cdot(0.5)^2 + 6\cdot(0.5)^3 \Big]\,= 7.25 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8) $$
 * }
 * }
 * }

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

 * $$\displaystyle f(x_2)\, = \Big[ 3 + 4(1) - 2\cdot(1)^2 + 6\cdot(1)^3 \Big]\,= 15 $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9) $$
 * }
 * }
 * }

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

 * $$\displaystyle \therefore\, Substituting\, Eq (6)\, through\, Eq (9)\, in\, Eq(2)\, we\, have\,$$
 * }
 * }
 * }

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

 * $$\displaystyle I_2 = \frac{1}{6}\bigg[ 3+ 29 + 15 \bigg] =\, \frac{47}{6} $$
 * }
 * }
 * }

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

$$\displaystyle \therefore I_2 = 7.8333$$
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style =|
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle Since\, I = I_2\, it\, is\, very\, clear\, that\, the\, Simpson\, rule\, integrates\, a\, 3^{rd}\, order\, polynomial\, exactly $$
 * }
 * }
 * }

Subramanian Annamalai 21:38, 20 February 2010 (UTC)

= Derivation : $$\alpha^{(1)}(t) = F(-t) + F(t)$$ in Simple Simpson's rule= Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|p.15-1]]

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

$$
 * $$\displaystyle \alpha^{(1)}(t)= F(-t) + F(t)
 * $$\displaystyle \alpha^{(1)}(t)= F(-t) + F(t)
 * style = |
 * }

Solution
From (4) in [[media:Egm6341.s10.mtg14.pdf|p.14-2]], we can write down the expression of $$ \alpha(t) $$

$$ \alpha(t) = \int_{-t}^t f(x(t))\, dt $$

Given :

f(x(t)) = F(t)

Let there exists $$ g^1(t)= F(t) $$

$$ \alpha(t) = \int_{-t}^t f(x(t))\, dt

= \int_{-t}^t F(t))\, dt

= \int_{-t}^t g^1(t))\, dt

= g(t)- g(-t) $$

$$  \alpha^{(1)}(t) = \frac{d}{dt}[g(t)- g(-t)] $$

$$ \Rightarrow \alpha^{(1)}(t) = g^1(t)- (-g^1(-t)) $$

but $$ g^1(t)= F(t) $$

$$ \Rightarrow \alpha^{(1)}(t) = F(t)+F(-t)) $$

Hence Proved , Abhishekksingh 20:59, 21 February 2010 (UTC)

= Proof: $$ G^{(2)}(0)=0 $$ in Simpson's Rule = Ref: Lecture notes [[media:Egm6341.s10.mtg15.pdf|p.15-2]]

Problem Statement
For the Simpson's rule, prove that $$\displaystyle G^{(2)}(0)=0 $$.

Solution
From [[media:Egm6341.s10.mtg14.pdf|p.14-3]] of lecture note, $$\displaystyle G(t) $$ can be expressed by:

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

G(t) = e(t)-t^{5}e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

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

e(t)=\alpha(t)-\alpha_{2}(t)=\int_{-t}^{t}F(s)ds - \frac{t}{3}[F(-t)+4F(0)+F(t)] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

From [[media:Egm6341.s10.mtg15.pdf|p.15-1]] of lecture note, we already get the first derivatives of $$\displaystyle \alpha(x) $$ and $$\displaystyle \alpha_{2}(x) $$:

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

\alpha^{(1)}(t)=F(-t)+F(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

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

\alpha_{2}^{(1)}(t)=\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[F^{(1)}(-t)+F^{(1)}(t)] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

From the Eq.(3) and Eq.(4), we can obtain the second derivatives of $$\displaystyle \alpha(x) $$ and $$\displaystyle \alpha_{2}(x) $$:

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

\alpha^{(2)}(t)=F^{(1)}(-t)+F^{(1)}(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

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

\alpha_{2}^{(2)}(t)=\frac{2}{3}[F^{(1)}(-t)+F^{(1)}(t)]+\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

From the Eq.(1), the second derivative of $$\displaystyle G(t) $$ is

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

G^{(2)}(t) = e^{(2)}(t)-20t^{3}e(1) = \alpha^{(2)}(t)-\alpha_{2}^{(2)}(t)-20t^{3}e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Plug Eq.(5) and Eq.(6) into Eq.(7).

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

G^{(2)}(t) = \frac{1}{3}[F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]-20t^{3}e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Now, let's explicitly express $$\displaystyle F^{(1)}(t) $$ and $$\displaystyle F^{(1)}(t) $$ as follows.

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

F^{(1)}(t) = \frac{dF(t)}{dt} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

For the expression of the derivation of $$\displaystyle F^{(1)}(-t) $$, let $$\displaystyle -t=\tau $$ and use chain rule.

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

F^{(1)}(-t) = F^{(1)}(\tau)=\frac{dF(\tau)}{dt}=\frac{d\tau}{dt}\frac{dF(\tau)}{d\tau}=(-1)\frac{dF(\tau)}{d\tau} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Now, Plug Eq.(9) and Eq.(10) into Eq.(8). Then,

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

G^{(2)}(t) = \frac{1}{3}[(-1)\frac{dF(\tau)}{d\tau}+\frac{dF(t)}{dt}]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]-20t^{3}e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

To obtain $$\displaystyle G^{(2)}(0) $$, note that $$\displaystyle t=0 \rightarrow \tau=0 $$ and the second and the third terms of RHS in Eq.(11) are going to be zero when $$\displaystyle t=0 $$. Then,

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

G^{(2)}(0) = \frac{1}{3}[(-1)\left(\frac{dF(\tau)}{d\tau}\right)_{\tau=0}+\left(\frac{dF(t)}{dt}\right)_{t=0}] - 0 - 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

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

\Rightarrow G^{(2)}(0) = \frac{1}{3}[(-1)F^{(1)}(0)+F^{(1)}(0)] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }

Therefore,

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

$$\displaystyle G^{(2)}(0) = 0 $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

Yong Nam Ahn 02:55, 21 February 2010 (UTC)

= Proof:$$ e^3(t) = -(t/3) * [F^3(t)- F^3(-t)]$$ in Simple Simpson's rule =

Ref Lecture [[media:Egm6341.s10.mtg15.pdf|p.15-2]]

Problem Statement
Prove that

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

where

$$ e(t)= \int_{-t}^t f(x(t))\, dt - \frac {t}{3}[F(-t)+4 F(0)+F(t)] $$

Solution
$$ e(t)= \int_{-t}^t f(x(t))\, dt - \frac {t}{3}[F(-t)+4 F(0)+F(t)] $$

$$ e(t)= \int_{-t}^k f(x(t))\, dt +\int_{k}^t f(x(t))\, dt - \frac {t}{3}[F(-t)+4 F(0)+F(t)] $$

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

$$ e^{(2)}(t)= (-F^1(-t)+F^1(t)) - \frac {t}{3}[ F^2(-t)+F^2(t)] - \frac {1}{3}[-F^1(-t)+F^1(t)]-\frac {1}{3}[-F^1(-t)+F^1(t)] $$

$$ e^{(2)}(t)= (-F^1(-t)+F^1(t)) - \frac {t}{3}[ F^2(-t)+F^2(t)] - \frac {2}{3}[-F^1(-t)+F^1(t)] $$

$$ e^{(3)}(t)= (F^2(-t)+F^2(t)) - \frac {t}{3}[- F^3(-t)+F^3(t)] -\frac {1}{3}[F^2(-t)+F^2(t)]- \frac {2}{3}[F^2(-t)+F^2(t)] $$ $$ e^{(3)}(t)= (F^2(-t)+F^2(t)) - \frac {t}{3}[- F^3(-t)+F^3(t)] -(F^2(-t)+F^2(t)) $$

$$ e^{(3)}(t)= - \frac {t}{3}[- F^3(-t)+F^3(t)] $$

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

Hence Proved ..

Abhishekksingh 20:57, 21 February 2010 (UTC)

Relationship between $$ \xi $$ and $$ \zeta_{4} $$ in Simple Simpson's rule
Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|p.15-3]]

Problem Statement
Given that$$ -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{(4)}(\xi) $$ Find the relationship between $$\zeta _{4}$$ and $$ \xi $$

Solution
Use chain Rule

$$ \displaystyle F(t)=f(x) $$

$$ \displaystyle x(t)=x_{1}+ht $$

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

$$ \because \frac{d\left(x(t)\right)}{dt} = \frac{d(x_{1}+ht)}{dt} = h $$

$$ \displaystyle F^{(2)}(t) = \frac{d\left(F(t)^{(1)}\right)}{dt} = \frac{d\left(hf(x)^{(1)}\right)}{dx}\frac{dx}{dt} = hf^{(2)}(x) \times h = h^{2}f^{(1)}(x) $$

$$ \displaystyle F^{(3)}(t) = \frac{d\left(F(t)^{(2)}\right)}{dt} = \frac{d\left(h^{2}f(x)^{(2)}\right)}{dx}\frac{dx}{dt} = h^{2}f^{(3)}(x) \times h = h^{3}f^{(3)}(x) $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

Now, obtain the relationship between $$ \displaystyle \xi $$ and $$ \displaystyle \zeta_{4} $$

$$ \displaystyle x(\zeta_{4}) = x_{1} + h\zeta_{4} = \xi $$

And recall that $$ \displaystyle h=\frac{b-a}{2} $$ Then,

$$ \displaystyle -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{1}{90}\left(h^{4}f^{(4)}(\xi)\right)=-\frac{1}{90}\left(\frac{(b-a)^{4}}{2^{4}}f^{(4)}(\xi)\right) $$

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

$$\displaystyle -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{(4)}(\xi) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

where

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

$$\displaystyle \xi = x_{1}+h \zeta_{4} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

--Heejun Chung 03:27, 21 February 2010 (UTC)

=Runge Phenomenon - Motivation for composite Rules= Ref: Lecture notes [[media:Egm6341.s10.mtg16.pdf|p.16-1]]

Problem Statement

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

$$\displaystyle I = \int_{-5}^{5}f(x)\, dx $$
 * $$\displaystyle f(x) = \frac{1}{1+x^2}$$
 * $$\displaystyle f(x) = \frac{1}{1+x^2}$$
 * }
 * }

(i) Using Newton-Cotes obtain $$ I_n $$ for n==1,2,.......,15. (ii) Plot f(x) and$$\, f_n(x)$$ for n=1,2,3,8 and 12. (iii)Plot $$I_n\;vs\;n.$$. Comment on the pattern. (iv)Plot the Lagrange basis functions $$l_{i,n} $$ for n=8 and i=1,2......8.

Solution
EXACT INTEGRAL

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

$$\displaystyle = \int_{-5}^{5}\frac{1}{1+x^2}\, dx $$ $$\displaystyle = \Big[tan^{-1} (x)\Big]_{-5}^{5} =\;2.7468 $$
 * $$\displaystyle I = \int_{-5}^{5}f(x)\, dx $$
 * $$\displaystyle I = \int_{-5}^{5}f(x)\, dx $$
 * }
 * }

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

$$\displaystyle \therefore\, I = 2.7468 $$
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style =|
 * }
 * }

MATLAB Code that will generate the Lagrange Basis Functions, $$\displaystyle I_n\, and\, f_n(x)$$ for any given n

(i) $$\displaystyle I_n $$ obtained using Newton-Cotes formula for n=1,2,.....,15.

Results: Comparison of $$\displaystyle I_n\, vs\, n $$

(ii) $$\displaystyle Plot\,:\,I_n\, vs\, n$$

As it can be observed from the plot, we find that the value of $$I_n$$ never converges as n increases. This is called as the Runge phenomenon. And for this reason composite rules are preferred.

(iii) $$\displaystyle Plot\,:\,f(x)\, and\, f_n(x)\,for\, n=1,\,2,\, 3,\, 8,\, and \, 12.$$  f(x) vs x

$$\displaystyle Plot:\, f_1(x)\, vs\, x $$

$$\displaystyle Plot:\, f_2(x)\, vs\, x $$

$$\displaystyle Plot:\, f_3(x)\, vs\, x $$

$$\displaystyle Plot:\, f_8(x)\, vs\, x $$

$$\displaystyle Plot:\, f_12(x)\, vs\, x $$

(iii) $$\displaystyle Plot\,:\,l_{i,n}\, for\, n=8\, and\, i=\,0,\,1,\,.....\,8.$$

$$\displaystyle Plot\,:\,l_{i,n}\, for\, n=8\, and\, i=\,0,\,1,\,and\,2.$$

$$\displaystyle Plot\,:\,l_{i,n}\, for\, n=8\, and\, i=\,3,\,4,\,and\,5.$$

$$\displaystyle Plot\,:\,l_{i,n}\, for\, n=8\, and\, i=\,6,\,7,\,and\,8.$$

Subramanian Annamalai 21:39, 20 February 2010 (UTC)

= Contributing Team Members =

1.Subramanian Annamalai 21:41, 20 February 2010 (UTC) Authored:1,6,11,16  Proof-read:3,8 and 13 2. Yong Nam Ahn 02:57, 21 February 2010 (UTC) Authored: 3,8 and 13 Proof read: 5, 9, 10 and 15 3. Heejun Chung 03:30, 21 February 2010 (UTC) Authored: 5, 9, 10 and 15 Proof read: 1, 6, 11 and 16 4.Abhishekksingh 21:02, 21 February 2010 (UTC)Authored: 4,12 and 14 Proof read: 2 and 7 5. Min Zhong 03:30, 21 February 2010 (UTC) Authored: 2,7 and 12 Proof read: 4 and 14

=References= 1. Introduction to Numerical Methods,2nd Edition by Kendall E Atkinson Second Edition 2. Numerical Methods for Engineers,5th Edition by Steven C Chapra and Raymond P Canale

title: User:Egm6341.s10.team3.sa/HW2 updated


text:
Java Tutorial/Access modifiers

In Java, there are two types of access modifiers. They're  and. A private can call a public, but a public can't call a private. Example:

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

title: Java Tutorial/Access modifiers


text:
User:Egmontaz/monobook.js

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

title: User:Egmontaz/monobook.js


text:
User talk:TeqLImeNec

 Hello TeqLImeNec, 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, 21 February 2010 (UTC)

title: User talk:TeqLImeNec


text:
User talk:AndrewR~enwikiversity

 Hello AndrewR, 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, 21 February 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 AndrewR. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name AndrewR~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 23:27, 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) 06:02, 19 April 2015 (UTC)

title: User talk:AndrewR~enwikiversity


text:
User talk:LTravers

 Hello LTravers, 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, 21 February 2010 (UTC)

title: User talk:LTravers


text:
User:Troypheil

Hi.

I will be spending a little time looking around first. I have primarily come here to learn but hope to be able to contribute as well once I get acquainted with all the material and etc.

I have been on the internet for over ten years now and consider myself an independent web designer. Not so much developer as yet.

I do have my own web site -. I am 56 years old and self-taught. As I've said, one of the major reasons I have come here is to learn.

And ... that's about all for now.

peace

title: User:Troypheil


text:
User:EGM6321.f09.Team3.cgs/Misc

$$ Blocks$$

$$ \frac{1}{ab}\sum_ky_{..k}^2-\frac{y_{...}^2}{abn}$$

$$\sigma^2+ab\sigma_\delta^2$$

$$ A$$

$$\frac{1}{bn}\sum_iy_{i..}^2-\frac{y_{...}^2}{abn} $$

$$\sigma^2+\frac{bn\sum\tau_i^2}{a-1}$$

$$ \frac{MS_A}{MS_E}$$

$$ B\qquad $$

$$\frac{1}{an}\sum_jy_{.j.}^2-\frac{y_{...}^2}{abn}$$

$$\sigma^2+\frac{an\sum\beta_j^2}{b-1}$$ $$ \frac{MS_B}{MS_E} $$

$$ AB\qquad $$

$$\frac{1}{n}\sum_i\sum_jy_{ij.}^2-\frac{y_{...}^2}{abn}-SS_A-SS_B$$

$$\sigma^2+\frac{n\sum\sum(\tau\beta)_{ij}^2}{(a-1)(b-1)}$$

$$ \frac{MS_{AB}}{MS_E} $$

Error

Subtraction

$$ \sigma^2$$

$$ \sum_i\sum_j\sum_k y_{ijk}^2-\frac{y_{...}^2}{abn} $$

title: User:EGM6321.f09.Team3.cgs/Misc


text:
How Lesothosaurus defended itself

Lesothosaurus was an approx. 1 m long herbivore from the early Jurassic period. Its long legs, small arms, and slender tail all suggest that it was a fast runner. Given this fact, it defended itself by running away as fast as it could.

title: How Lesothosaurus defended itself


text: Template:Original research

title: Template:Original research


text:
User talk:Troypheil

 Hello Troypheil, 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 17:55, 21 February 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.

title: User talk:Troypheil


text:
User talk:Mohammad Al-Aggan

 Hello Mohammad Al-Aggan, 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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 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Mohammad Al-Aggan


text:
User:Abhishekksingh

hi

title: User:Abhishekksingh


text:
How to learn - asking questions finding answers

This is a learning activity for individuals or groups of people 13 years and over.

You will practice:
 * 1) posing a question to a question and answer forum space,
 * 2) extracting phrases and key words for cross checking with other sources
 * 3) documenting your work for review and assessment

You will need some computing and Internet ability, namely: using a keyboard; using a web browser such as clicking links, filling in online forms, creating user accounts with online services, and navigating multiple web pages; and being able to creating links to other web pages (see resources below).

You will need at least 1 hour to complete this activity.

Resources

 * A computer connected to the Internet
 * A pencil and paper
 * An idea for a good question you'd like answers for
 * How to ask a good question
 * How to make a hyperlink in Blogger

Yahoo! Answers
5 minutes - Create an account with Yahoo Answers (write down your user name and password)

5 minutes - Submit your question to Yahoo Answers. While you wait for answers spend some time editing your profile

Blogger
10 minutes - Create your own blog on Blogger (write down your user name and password, as well as the web address for your blog, e.g. www.yourname.blogspot.com)

20 minutes - Copy your question into your first blog post, and include copies of the best answers you have received so far. Include a link to your question page on Yahoo Answers

Cross referencing
20 minutes - Cross check those answers by finding information that verifies what is said.

Wikipedia
Search for Wikipedia articles relating to your question and answers, being sure to look at the discussion pages to see if there is any dispute, and refer to the sections in the articles called "see also" and "external links". Include links on your blog to articles and links you find that either support, confirm or contradict the answers you are verifying.

Youtube
Search Youtube based on words and phrases you have piced up about your question and answer so far. Include a link to your search result in your blog. Try to find one informative video related to your question and/or answers. Be sure to check the comments and related videos sections of the Youtube clips you find. Include a link in your blog to the best video you find.

Delicious
Search delicious.com based on keywords you should now have worked out about your question and answers. Include a link to your search result on your blog. Briefly survey the items in your search, and link to things of interest in your blog.

Twitter
Finally, do keyword searches on Twitter for any quick and up to the minute information relating to your question and answers. Include links to useful things you find in Twitter


text:
User:Brookeayers

About Me
M name is Brooke Ayers. I am currently enrolled in Appalachian State University's New Media and Global Education Masters Program. I love running and spending time with my family. I have lived in:
 * Lexington, NC
 * Salisbury, NC
 * Lumberton, NC

Education and Certifications
 * B.A. Graphic Design with a Minor in Graphic Design,| Emory and Henry College
 * Currently working on M.A in New Media and Global Education,| Appalachian State University

Experience
 * Assistant to Assistant Superintendent of Curriculum and Instruction, Davidson County Schools, Lexington, NC
 * Director, Design, Marketing and Customer Service, Granite Printing,Inc., Lexington, NC
 * Art Teacher and Yearbook Editor, Davidson Academy, Lexington, NC
 * Marketing Coordinator / Art Director, National Home Furnishings Association (NHFA), High Point, NC

title: User:Brookeayers


text:
User:Bob sears/Lutes: Lesson 1 The Intro

User: Bob Sears/Lutes:Lesson 1 The Intro


text:
How to use SPSS/Options

Editing SPSS defaults

 * 1) Edit - Options, e.g.,
 * 2) * Pivot Tables - TableLook

title: How to use SPSS/Options


text:
User:Amybedot1

Amy Brown, Director of eLearning, Guilford Technical Community College, NC

title: User:Amybedot1


text:
User:CraigJoseph/Political psychology

This is the seed page for the Political Psychology wiki.

title: User:CraigJoseph/Political psychology


text:
User talk:Charles Jeffrey Danoff

 Hello Charles Jeffrey Danoff, 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. --AFriedman (talk) 03:17, 22 February 2010 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

Yes! This is a haven for original research
I saw your post about original research. By all means, this is exactly the place to do what you've described. It fits well within the original research policy and you may want to contact other people, such as the people you've already communicated with, for feedback. --AFriedman (talk) 03:17, 22 February 2010 (UTC)
 * Thank you AFriedman. I have gotten a lot of support for this. I will begin. If you have time as I go on, I would love any feedback.--Charles Jeffrey Danoff 09:54, 22 February 2010 (UTC)

I probably won't be able to follow your research unless you tell me where on Wikiversity you put it. --AFriedman (talk) 05:13, 24 February 2010 (UTC)
 * Indeed, it would be difficult. The research is located at Reviewing My Semester with the FUN WITH ENGLISH 7A & 8A Textbooks. Thanks for being interested. --Charles Jeffrey Danoff 10:23, 24 February 2010 (UTC)

Hi, I've cleaned up the description of your research so it reads less like a diary and more like a proposal. What are your thoughts re: how this research will benefit the Wikiversity community? For example, I was thinking that this resource may be used to explore general principles re: which teaching methods are most effective (1) when teachers are attempting to communicate with students of different cultures, and (2) when teachers are attempting to expose students to other cultures. Were you and the students actually in China, and where do you come from originally? Also, you may want to link your main Userpage and relevant Topic pages (e.g. Topic:English language, Topic:Chinese) to your research page so your research is easier to find. On the page where you're doing your research, you may also want to add a link to whatever information is available about the textbook you used. --AFriedman (talk) 19:59, 24 February 2010 (UTC)
 * AFriedman, thanks for the ideas. I've added links to the publisher's pages for the textbooks, as well as information about where I taught, and a first draft of how this research will benefit the community. --Charles Jeffrey Danoff 01:08, 25 February 2010 (UTC)

Nice to see your work :). When are you going to begin the project, instead of simply describing it?  --AFriedman (talk) 02:47, 25 February 2010 (UTC)
 * AFriedman: Soon, I hope. If you'd like a preview, view my original attempt in handwritten html. I started there, then realized a wiki would be more appropriate, and then I found the original research on Wikiversity and I knew I'd found a home. --Charles Jeffrey Danoff 04:39, 25 February 2010 (UTC)

The question/answer format for background information may not make sense. My intent with providing questions was to provide examples of the sort of questions people reading your research might want to know in order to understand the details surrounding the research topic, as well as help provide a focus for you/researchers. The intent in including background information is to provide some answers. The only questions readers probably need to know about are the ones which are the focus of the research. The introduction section is a good start to the format I had in mind. I think any answers you want to include should be part of that introduction. If you want I could try to rewrite the questions you have answered so far in the format I had in mind. -- dark lama  14:02, 25 February 2010 (UTC)

 * DarkLama: Changing from a question/answer format to something more natural makes sense, especially because not all the questions are relevant for every kind of research. If you have time, please feel free to rewrite the questions so far in that format. This morning I added some more answers. If you'd be interested, please feel free also to add yourself to the list of people helping with the research. On a side note, I forgot to log-in before adding my answers. Is there some sort of protocol for how to note that I wrote them? I added a link on the talk page. Is that enough? --Charles Jeffrey Danoff 03:38, 26 February 2010 (UTC)
 * Something more natural is what I had in mind. I would like for Research process to reflect that as well while being useful for the first time researcher.
 * I believe common practice is to assume that anyone contributing under an IP address is an anonymous contributor who does not wish to receive any form of credit for the work they do. I wouldn't worry about it, unless credit for those revisions is important to you. What you wrote on the discussion page may be enough. You could also leave a message while logged out on the talk page confirming what you said while logged in. Custodians with access to checkuser tools can probably confirm what IP address(es) you have used when logged in. I don't know if using checkuser tools for this purpose though would be appropriate, it may violate wikimedia's checkuser or privacy policy. You may be able to find additional suggestions from the community if you asked at Colloquium. -- dark lama  10:20, 26 February 2010 (UTC)
 * I have edited the page twice. Which version do you prefer? [ A] or [ B]? -- dark lama  11:40, 26 February 2010 (UTC)
 * In certain ways I prefer B, but I think A is more appropriate and natural. B would be a good model for an abstract I may need to write later. In any event, thank you very much. You merged my words together better and quicker than I would have done. --Charles Jeffrey Danoff 02:27, 27 February 2010 (UTC)
 * Also, I asked the Colloquium for their advice. Finally, I agree that the Research process reflect something more natural while using the questions as a guide makes sense. Perhaps some journal examples of Background Info? And, or, should I finish my original research, mine? --Charles Jeffrey Danoff 02:43, 27 February 2010 (UTC)
 * What would you like to do? You are welcome to help revise and improve the research process too. It may be enough for you to document your experiences and what you have learned, and/or to actively discuss your experiences and what you learn as it happens with others, if that is what you rather do. I am likely to review and revise the research process over time, to reflect areas where people have struggled, and to document what works. Your own struggles, mistakes and successes with doing research on Wikiversity is likely to be enough to help me with that.
 * I think some kinks I still need to work out right now has to do with presentation and perspectives. I think a first-person narrative is a more natural frame of mind to be in while a research project is ongoing, but at some point revisions are probably needed to make the presentation more suitable for an audience. That probably doesn't need to be done until the research is finished.
 * I can't quiet put my finger on it, but I think maybe the background information section is missing some details. I think it has to do with the thing that is the subject/focus of your research. Like maybe more details about the textbooks and lessons you use are missing. How are teachers suppose to be able to know if your situation is similar to their situation if they don't know the details? How are other people suppose to be able to reproduce your research for themselves and see if they would reach similar conclusions or produce similar results if they don't know the details? Like what tools, materials, etc. you used, how you used them, or whatever? I hope you understand what I mean. -- dark lama  14:42, 27 February 2010 (UTC)
 * Darklama, I agree with you. I think I want to add the following questions:

In addition to answering these quesitons, a goal of this research is to create a legitimate web resource for any teachers using the FUN WITH ENGLISH listening textbooks. Previously I included my ideas about expanding the research to include what i'm doing right now, but upon further reflection, that can wait. Focus on getting this going first. --Charles Jeffrey Danoff 13:45, 2 March 2010 (UTC)
 * How can teachers archive old lesson plans for personal reference and other needs?
 * Is it possible to share lesson plans that are useful for others and if so, how?
 * What are effective teaching methods using the FUN WITH ENGLISH Listening texbooks?
 * What is the best way for teachers to review their own lesson plans to learn and improve?
 * How can teachers display their portfolio of archived teaching materials to show their personal evolution?
 * How can teachers share and use their culture teaching English in a foreign one?

 * Go for it. With further thought, I'm thinking you may want to add answers too for:
 * What is/are FUN WITH ENGLISH Listening textbooks?
 * Who is/are FUN WITH ENGLISH Listening textbooks suppose to teach?
 * Why is/are FUN WITH ENGLISH Listening textbooks used in your classroom?
 * How is/are FUN WITH ENGLISH Listening textbooks used in your classroom?
 * When is/are FUN WITH ENGLISH Listening textbooks used in your classroom?
 * -- dark lama  17:11, 2 March 2010 (UTC)
 * Will do, thanks for the support and ideas. --Charles Jeffrey Danoff 04:57, 4 March 2010 (UTC)

Wikimania returns
Hi, as I have seen your presentation, I would say there are more participants like you. E.g. this resource I have founded today looks good: Human Legacy Course/Early India.--Juan de Vojníkov 20:37, 2 August 2010 (UTC)
 * Juan glad to hear. I looked at the resource quickly and its intriguing. I hope I can read more later. How did you see my presentation? Thanks for getting in contact. --Charles Jeffrey Danoff 20:46, 2 August 2010 (UTC)

From the chair. Most of the facts of that blok I have known, so I had no opportunity to ask some questions.--Juan de Vojníkov 20:54, 2 August 2010 (UTC)

Category
Hi, could you put all the pages you have founded to a category? Thanks!--Juan de Vojníkov 18:14, 5 August 2010 (UTC)
 * That would be helpful. What would you suggest? Something unique like Mr. Danoff's FWE Lessons? And/or something more broad like TESOL [Teaching English as a Second or Other Language]? --Charles Jeffrey Danoff 21:46, 6 August 2010 (UTC)

It doesn't matter. If there are more pages of your course it can be category:Mr. Danoff's FWE Lessons. This category will be than in category:English as a Second Language.--Juan de Vojníkov 09:53, 8 August 2010 (UTC)
 * How about this Mr. Danoff's FWE 8B Lesson 2? Or, would it be better to make all the lesson plans a sub-page as part of an overall Mr. Danoff's FWE Lessons page like Nonlinear finite elements/Sets is a sub-page of Nonlinear finite elements? --Charles Jeffrey Danoff 19:03, 11 August 2010 (UTC)

The organization of Topic:TESOL material at Wikiversity with simple to use navigational elements will evolve as people submit pages on this topic and work with them. To start this, I put some category links into a template:TESOL that you can display on your user page or wherever you find it helpful. It is a box of links that floats right. To put it on a page, use this wiki text: as I have done here, to show you:  Cheers! --Rogerhc 04:36, 22 November 2010 (UTC)
 * Hi Roger, I think it'd be a helpful idea to add TESOL template to the TESOL resources I'm making. I tried here, only problem being it lines up to the left of the other templates I have. Any chance you could fix that?
 * Thanks for saying hi, we had some brief back and forths and wikigogy in the past, and I am glad to be back in contact again. --Charles Jeffrey Danoff 19:35, 22 November 2010 (UTC)
 * Fixed (added "clear:right;"). Sorry for the delay. Didn't see this till just now; don't frequent these parts often. --Rogerhc 04:25, 7 December 2010 (UTC)
 * thx, looks professional --Charles Jeffrey Danoff 05:40, 7 December 2010 (UTC)

Mr. Brown & Mr. Danoff's Halloween Lesson 2
This looks like a place where you took notes, on a subtopic.

May I suggest that this doesn't belong in mainspace, at the top level? If this is part of an existing resource, then placing it as a subpage underneath that resource would be appropriate. Or if it's your personal notes, in your user space, as subpages of User:Charles Jeffrey Danoff.

My guideline for what goes in mainspace, top level, is "Would someone be looking for a course on -- in this case -- "Mr. Brown & Mr. Danoff's Halloween Lesson 2"? Somehow I don't think so!

The question for me is not whether or not you can do on Wikiversity what you are doing. You can do it, it's welcome. The question is where and how.

You seem to be writing a book. Reviewing My Semester with the FUN WITH ENGLISH 7A & 8A Textbooks Yet that page is a collection of notes, not part of a book, per se. User space? Subspace of a Fun With English resource? The page I mention in the section header could be a subpage of a book page.

Many of your page titles have your name in them. I'm working on Cold fusion. I suppose I could start a page called Abd ul-Rahman Lomax's Experiences with Cold fusion. But that certainly would not be a collaborative project. If I were going to do that, I'd do it in my user space. as with User:Abd/My experiences with cold fusion.

What do you think? --Abd 21:35, 17 February 2011 (UTC)
 * Abd, thanks for your interest and comments. I agree that the organization is troublesome right now. Currently I am focusing on getting all the different components published this afternoon/evening, once they're all there I think the best way to organize will reveal itself to me. What I'm thinking now is:
 * TOP: Mr. Danoff's Anqing Teaching Portfolio
 * SUB: /7A Lessons
 * SUB: /1 (depending on lesson #)
 * what'd you think of that?--Charles Jeffrey Danoff 21:50, 17 February 2011 (UTC)
 * for the second semester stuff I took your advice and removed my name from the titles and put them all into so and so user spaces, then created a Wikiversity book out of it all: FUN WITH ENGLISH 7B & 8B: Unofficial Teacher's Guide --Charles Jeffrey Danoff 07:06, 18 February 2011 (UTC)

 * I suggest you use lesson names instead of lesson numbers. You can avoid a need to rename pages should the order in which the lessons are taught change. Other people will have an easer time finding the lesson they are after if they use the search function. Other people will also be able to navigate to the lesson they want to use or read quickly, instead of checking each page to see if this is the lesson they are after. -- dark lama  13:19, 18 February 2011 (UTC)
 * I like your ideas, I moved almost the 7A lessons under the sub-page and gave them new names. Is that what you meant? I still like having some sort of numbering ordering system, so I aligned them by chapter, a la this resource from Englipedia which was the most helpful thing I ever found as an EFL teacher. How's that? --Charles Jeffrey Danoff 05:39, 19 February 2011 (UTC)

 * I suggest not using your name in a top-level resource, i.e., in mainspace with no prefix, rather than being a subpage. If you are really writing only about yourself, and your personal experience, the page could be in your user space, but it may be just fine as a personal story under the general topic of Fun with English. I'm not sure about the "Books" as a top-level resource name! If it's a book, why not Wikibooks? --Abd 16:25, 18 February 2011 (UTC)
 * I agree that I shouldn't have my user name in there. I moved most of the the 7A lessons all into sub-pages under the main resource. Still need to move the the 8A lessons. The resource isn't a book, its designed to be a guide for teachers who are using the FUN WITH ENGLISH books. The idea is they'll type their book name into google and this'll come up ... does that make sense? Do you have a better idea for a title? --Charles Jeffrey Danoff 05:39, 19 February 2011 (UTC)

added link to you comment
Hi, I added a link to your messsage; I hope you dont mind. if you do mind, I don't mind if you remove it ;-) John Vandenberg 03:06, 23 February 2011 (UTC)
 * Silly me; I see now that you linked to it in your first message in that thread. John Vandenberg 03:16, 23 February 2011 (UTC)

United States English
Thanks for moving ENG 099. I see that the Wikipedia article for this subject is American English, but it does note that it is often called U.S. English. I'll add a redirect for our American English to the new title. There was a Dec. 2012 subpage. I've moved it under the current page. Let me know if you have any questions. -- Dave Braunschweig (discuss • contribs) 13:27, 21 November 2016 (UTC)

title: User talk:Charles Jeffrey Danoff


text:
User:JReid

Name: John R.

Background Macon, Georgia & Woodbridge, Virginia

Career Subject Matter Expert, Lockheed Martin 2007 Database Analyst, Global One Communications, 2000 Client Operations, Online Resources 2002 Technical Writer, IBM

Accomplishments Sports Editor, 1998 "Man On the Street" talk show, WVCU 1998 Two Worlds, One Love program 2010 Writer/Director "Brothers in Arms" 2009 Columnist, Bleacher Report 2007 Marketing Director, TCN television program 2007 Torchbearer Award for Academic Excellence 1995

title: User:JReid


text:
User talk:Wanderwits

 Hello Wanderwits, 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, 22 February 2010 (UTC)

title: User talk:Wanderwits