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title: User talk:Nebmatyc


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User:Egm6341.s10.team4.anandankala/HW3-3

=Problem 3=

Given

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

$$I=\int\limits_{0}^{1} f(x)\, dx$$

$$where f(x)=\frac{(e^x-1)}{x}.$$
 * $$\displaystyle $$
 * }.
 * }.

Find

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

 * 1) Find n such that En=I-In is of the order of 10-6 using the error estimates for the Taylor's series, composite Trapezoidal rule, composite Simpson's rule and compare to numerical results.
 * 2) Numerically find the power of h in the error. Plot error versus h on log graph and measure slope.
 * 1) Find n such that En=I-In is of the order of 10-6 using the error estimates for the Taylor's series, composite Trapezoidal rule, composite Simpson's rule and compare to numerical results.
 * 2) Numerically find the power of h in the error. Plot error versus h on log graph and measure slope.

 * $$\displaystyle $$
 * }.
 * }.

Solution
The error for Taylor series is analytically given by $$\displaystyle \frac{1}{(n+1)!(n+1)} \leq E_n \leq \frac{e}{(n+1)!(n+1)}$$

The maximum value of En is equal to $$\displaystyle \frac{e}{(n+1)!(n+1)} $$.

The value of the error given in the problem is equal to 10-6.

 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle \frac{e}{(n+1)!(n+1)}= 10^{-6} $$
 * }.
 * }.

 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle (n+1)!(n+1)= e \cdot10^6 $$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n=8(approximately). $$
 * }.
 * }.

The error for composite Trapezoidal rule is analytically given by $$\displaystyle |E^1_n| \leq \frac{(b-a)^3}{12n^2}M_2$$

where M2 is the maximum of $$\displaystyle f^{''}(x).$$ in the interval [0,1]. $$\Rightarrow a=0,b=1.$$

 * {| style="width:70%" border="0" align="center"

 * Given $$\displaystyle f(x)=\frac{(e^x-1)}{x}.$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow f^'(x)=\frac {xe^x-e^x+1}{x^2}.$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow f^{''}(x)=\frac {x^2e^x-2xe^x+2e^x-2}{x^3}$$

Plot:
 * }.
 * }.

.

The above function $$\displaystyle f^{''}(x)$$ is plotted with respect to x in the interval [0,1] and a maximum of 0.72 is obtained at x=1.

 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle |E^1_n|=\frac{(b-a)^3}{12n^2}M_2=\frac{M_2}{12n^2}$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle \frac{M_2}{12n^2}=10^{-6}$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n^2=\frac{0.72*10^6}{12}$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n=245$$(approximately).

 * }.
 * }.

The error for composite Simpson's rule is analytically given by $$\displaystyle |E^2_n| \leq \frac{(b-a)^5}{180n^4}M_4.$$

where M4 is the maximum of $$\displaystyle f^4(x).$$ in the interval [0,1].

 * {| style="width:70%" border="0" align="center"

 * Given $$\displaystyle f(x)=\frac{(e^x-1)}{x}.$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow f^4(x)=\frac {x^4e^x-4x^3e^x+12x^2e^x-24xe^x+24e^x-24}{x^5}.$$

 * }.
 * }.

Plot:

.

The above function $$\displaystyle f^4(x)$$ is plotted with respect to x in the interval [0,1] and a maximum of 0.465 is obtained at x=1.

 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle |E^2_n|=\frac{(b-a)^5}{180n^4}M_4=\frac{M_4}{180n^4}.$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle \frac{M_4}{180n^4}=10^{-6}$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n^4=\frac{0.465*10^6}{180}$$

 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n=8$$(approximately).

 * }.
 * }.

The above analytical values of n are compared to the numerical values of n for an error of 10-6 which in turn are obtained from the matlab code.

 Matlab Code: for $$ E_{Taylors series} $$

 Matlab Code: for $$ E_{n,Trapezoidal} $$

 Matlab Code: for $$ E_{n,simpson} $$

The values of n obtained numerically for an error of 10-6 are 8,200,8 for Taylors series,trapezoidal rule and Simpson's rule which are almost equal to the values 8,245,8 found analytically.

In the next part we are required to find the convergence of the error and power of h in the error numerically. Hence we plot the values of the error and h on a log plot or alternatively plot the values of log(Error) and log(h) in excel plot using the numerical values.

Semilog Plot:

.

Semilog Plot:

.

Logarithmic Plot:

.

The slope of the line in the above graph is obtained to be equal to 2. This is the required power of h in the error formula for trapezoidal rule.

Semilog Plot:

.

Logarithmic Plot:

.

The slope of the line in the above graph is obtained to be equal to 4. This is the required power of h in the error formula for simpsons rule.

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


text:
User:Egm6341.s10.Team4.andy/HW3

= Problem 1: Proof of Tighter Error Bound for Simpson's rule [Simple]=

Given
Refer Lecture slide [[media:Egm6341.s10.mtg17.djvu|17-1]] for problem statement.

From Lecture slide [[media:Egm6341.s10.mtg14.pdf|14-1]], Error for Simple Simpson's rule is given by,

 * {| style="width:90%" border="0" align="center"

\begin{align} E_{2} & = -\frac{(b-a)^5}{2880}f^{(4)}(\xi)\ \, \xi \in [a,b] \\ & = -\frac{h^5}{90}f^{(4)}(\xi)\ \, h := \frac{b-a}{2} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)

 * }.
 * }.

In order to prove the above Error for Simple Simpson's rule [ [[media:Egm6341.s10.mtg15.pdf|proof]] ], we defined

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^5 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (2)

 * }.
 * }.

where,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

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

$$ $$
 * $$\displaystyle \longrightarrow (3)

 * }.
 * }.

Find
 Part I.  Try to replicate the proof, by using the below equations instead of Equation (2) and point out where the proof breaks down.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^4 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (4)

 * }.
 * }.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^6 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (5)

 * }.
 * }.

''' Part II. ''' Using Equation(2) find $$\displaystyle G^{(3)}(0) $$ and follow the same steps in the proof. Observer what happens.

Solution
 Part I. Using Equation (4):

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^4 \underbrace{e(1)}_{Constant}

$$

 * }.
 * }.

On differentiating,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(1)}(t) = e^{(1)}(t) - 4t^3 e(1)

$$

 * }.
 * }.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(2)}(t) = e^{(2)}(t) - 12t^2 e(1)

$$

 * }.
 * }.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(3)}(t) = e^{(3)}(t) - 24t e(1)

$$ $$
 * $$\displaystyle \longrightarrow (6)

 * }.
 * }.

From, HW2-Problem-14-Solution and from the lecture slide [[media:Egm6341.s10.mtg15.pdf|15-2]] we know,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

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

$$ $$
 * $$\displaystyle \longrightarrow (7)

 * }.
 * }.

Applying Rolle's theorem thrice 15-2], we have shown that

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\exists \ \zeta_3 \in \ (0,\zeta_2) \ such\ that\ G^{(3)}(\zeta_3) = 0,

$$ $$
 * $$\displaystyle \longrightarrow (8)

 * }.
 * }.

Substituting Equations (7) and (8) in (6),

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow G^{(3)}(\zeta_3) &= e^{(3)}(\zeta_3) - 24\zeta_3 e(1) \\ &= -\frac{\zeta_3}{3} \left[ F^{(3)}(\zeta_3) - F^{(3)}(-\zeta_3)\right] - 24\zeta_3 e(1) \\ &= 0 \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (9)

 * }.
 * }.

Using DMVT [[media:Egm6341.s10.mtg12.pdf|Derivative Mean Value Theorem, 12-1]], we get

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow \left[ F^{(3)}(\zeta_3) - F^{(3)}(-\zeta_3) \right] &= \left[ \zeta_3 - (-\zeta_3) \right] F^{(4)}(\zeta_4) \\ &= 2\zeta_3 F^{(4)}(\zeta_4) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (10)

 * }.
 * }.

Using Equation (10) in (9),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow 0 = -\frac{2\zeta_{3}^{2}}{3} F^{(4)}(\zeta_4) - 24\zeta_3 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (11)

 * }.
 * }.

Solving for $$\displaystyle e(1) $$,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{\zeta_{3}}{36} F^{(4)}(\zeta_4)

$$ $$
 * $$\displaystyle \longrightarrow (12)
 * }.
 * }.

We know from Lecture slide [[media:Egm6341.s10.mtg14.pdf|14-2]],

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1)

$$ $$
 * $$\displaystyle \longrightarrow (13)
 * }.
 * }.

On substituting Equation (12) in (13) gives us,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = - h \frac{\zeta_{3}}{36} F^{(4)}(\zeta_4)

$$ $$
 * $$\displaystyle \longrightarrow (14)
 * }.
 * }.

From the relationship between $\displaystyle \xi$ and $\displaystyle \zeta_4$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow \displaystyle F^4(\zeta_4)=\frac{(b-a)}{16} ^4\cdot f^4(\xi)

$$ $$
 * $$\displaystyle \longrightarrow (15)
 * }.
 * }.

Equation (14) then becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = - \frac{\zeta_{3}(b-a)^5}{1152} f^{(4)}(\xi)

$$ $$
 * $$\displaystyle \longrightarrow (16)
 * }.
 * }.

This error equation has an extra term $$\displaystyle \zeta_{3} $$ on the numerator as opposed to Equation (1). The proof breaks down in Equation (11), where we expect $$\displaystyle \zeta_{3}^2 $$ to cancel out, only if we had $$\displaystyle -24\zeta_{3}^2 e(1) $$ instead of $$\displaystyle -24\zeta_{3} e(1) $$.

 Part I. Using Equation (5):

On differentiating Equation (5),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(3)}(t) = e^{(3)}(t) - 120t^3 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (17)

 * }.
 * }.

In this case, Equation (11) becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow 0 = -\frac{2\zeta_{3}^{2}}{3} F^{(4)}(\zeta_4) - 120\zeta_3^3 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (18)

 * }.
 * }.

Solving for $$\displaystyle e(1) $$,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{1}{180\zeta_{3}} F^{(4)}(\zeta_4)

$$ $$
 *
$$\displaystyle \longrightarrow (19)
 * }.
 * }.

Substituting in Equation (13),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = - h \frac{1}{180\zeta_{3}} F^{(4)}(\zeta_4)

$$ $$
 *
$$\displaystyle \longrightarrow (20)
 * }.
 * }.

Using Equation (15),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = -  \frac{(b-a)^5}{5760\zeta_{3}} f^{(4)}(\xi)

$$ $$
 *
$$\displaystyle \longrightarrow (21)
 * }.
 * }.

Here also an extra term $$\displaystyle \zeta_{3} $$ is on the denominator as opposed to Equation (1). The proof breaks down in Equation (18), where we expect $$\displaystyle \zeta_{3}^2 $$ to cancel out, only if we had $$\displaystyle -120\zeta_{3}^2 e(1) $$ instead of $$\displaystyle -120\zeta_{3}^3 e(1) $$.

So in order to have a tighter bound on the Error Equation, we choose $$\displaystyle G(t) $$ to be $$\displaystyle e(t)-t^5 e(1) $$, in which case Equation (11) becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow 0 = -\frac{2\zeta_{3}^{2}}{3} F^{(4)}(\zeta_4) - 60\zeta_3^2 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (22)

 * }.
 * }.

Here $$\displaystyle \zeta_{3} $$ term cancels out to give,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{1}{90} F^{(4)}(\zeta_4)

$$ $$
 *
$$\displaystyle \longrightarrow (23)
 * }.
 * }.

and the Error Equation becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = -  \frac{(b-a)^5}{2880} f^{(4)}(\xi)

$$ $$
 *
$$\displaystyle \longrightarrow (24)
 * }.
 * }.

''' Part II. '''

Using Equation (7) in the third derivative of $$\displaystyle G(t) = e(t)-t^5 e(1) $$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G^{(3)}(t) = -\frac{t}{3} \left[ F^{(3)}(t) - F^{(3)}(-t)\right] - 60t^2 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (25)

 * }.
 * }.

At $$\displaystyle t = 0$$,

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow G^{(3)}(0) &= -\frac{0}{3} \left[ F^{(3)}(0) - F^{(3)}(-0)\right] - 60(0)^2 e(1) \\ &= 0
 * $$\displaystyle

\end{align} $$ $$
 *
$$\displaystyle \longrightarrow (26)

 * }.
 * }.

Since $$\displaystyle G^{(3)}(t)$$ is zero at $$\displaystyle t = 0$$ and $$\displaystyle t = \zeta_3$$ [from Equation (8)],

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\exists \ \zeta_4 \in \ (0,\zeta_3) \ such\ that\ G^{(4)}(\zeta_4) = 0,

$$ $$
 *
$$\displaystyle \longrightarrow (27)

 * }.
 * }.

Please note that, $$\displaystyle (0,\zeta_3) $$ is an open interval and $$\displaystyle \zeta_4 \ne 0 $$ or $$\displaystyle \zeta_4 \ne \zeta_3 $$.

Now on differentiating Equation (25),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G^{(4)}(t) = -\frac{1}{3} \left[ F^{(3)}(t) - F^{(3)}(-t)\right] -\frac{t}{3} \left[ F^{(4)}(t) + F^{(4)}(-t)\right] - 120t e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (28)

 * }.
 * }.

At $$\displaystyle \zeta_4 $$,

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow G^{(4)}(\zeta_4) &= -\frac{1}{3} \underbrace{\left[ F^{(3)}(\zeta_4) - F^{(3)}(-\zeta_4)\right]}_{2 \zeta_4 F^{(4)}(\zeta_5)} -\frac{\zeta_4}{3} \left[ F^{(4)}(\zeta_4) + F^{(4)}(-\zeta_4)\right] - 120\zeta_4 e(1) \\ &= 0 \end{align} $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (29)

 * }.
 * }.

where $$\displaystyle \zeta_5 \in [-\zeta_4,\zeta_4]$$.

Solving for $$\displaystyle e(1)$$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{1}{180}  F^{(4)}(\zeta_5) - \frac{1}{360} \left[ F^{(4)}(\zeta_4) + F^{(4)}(-\zeta_4)\right]

$$ $$
 *
$$\displaystyle \longrightarrow (30)

 * }.
 * }.

Using the relationship between $\displaystyle \xi$ and $\displaystyle \zeta_4$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{(b-a)^4}{2880}  f^{(4)}(\xi_2) - \frac{(b-a)^4}{5760} \left[ f^{(4)}(\xi_1) + f^{(4)}(-\xi_1)\right]

$$ $$
 *
$$\displaystyle \longrightarrow (31)

 * }.
 * }.

where $$\displaystyle \xi_1 = x_1 + h \zeta_4 $$ and $$\displaystyle \xi_2 = x_1 + h\zeta_5 $$, as we know $$\displaystyle x = x_1 + ht $$ from [[media:Egm6341.s10.mtg14.pdf|14-2]].

So, as long as $$\displaystyle f(x) \in \mathcal{P}_{3}$$ we know $$\displaystyle f^{4}(x) = 0 $$ from [[media:Egm6341.s10.mtg14.pdf|14-1]].

And so, $$\displaystyle e(1) = 0 \Rightarrow E_{2} = 0 $$.

Author
Review and Correction - Egm6341.s10.Team4.andy 16:27, 17 February 2010 (UTC) .

= Problem 2: Error for composite Simpson's rule =

Given
Use composite Simpson's equation (3) on slide [[media:Egm6341.s10.mtg7.pdf|7-2]], which is:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{h_1}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]

 * }
 * }

Find
To demonstrate error of composite Simpson's rule is equal to the following:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle 	\left | {E_n}^2\right |\leqslant\dfrac{(b-a)^5}{2880n^4}M_4=\dfrac{(b-a)h^4}{2880}M_4

M_4:=max\left |f^{(4)}(\xi)\right | $$

\xi\in[a,b] $$
 * style= |
 * }.
 * }.

Solution
The error definition can be done as:

{E_n}^2=I-I_n=\int\limits_{a}^{b} f(x)dx-I_n $$

From the composite form of Simpson's rule, we have:

I_n=\dfrac{h_1}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)] $$

\Rightarrow {E_n}^2=\int\limits_{a}^{b} f(x)dx-\dfrac{h_1}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)] $$

=\sum_{i=1}^{n} [\int\limits_{x_{i-1}}^{x_{i}} f(x)dx-\dfrac{h_1}{3}[f(x_{i-1})+4f(\dfrac{x_{i-1} + x_{i}}{2})+f(x_{i})]] $$

Here, $$\displaystyle h_1 = \frac{x_i - x_{i-1}}{2} $$

According to Lagrange interpolation error for [[media:Egm6341.s10.mtg14.pdf|simple Simpson's rule]] for $$\displaystyle [x_{i-1},x_{i}] $$ ;

\left |{E_n}^2 \right | \le \sum_{i=1}^{n} max \left | \dfrac {(x_{i}-x_{i-1})^5}{2880} f^{(4)}(\xi)\right | =\dfrac{h^5}{2880}\times \sum_{i=1}^{n} max \left | f^{(4)}(\xi)\right | $$

h=x_{i}-x_{i-1} =\dfrac {b-a}{n} $$

M_4:=max\left | f^{(4)}(\xi) \right | $$

\xi\in [x_{i-1},x_{i}] $$

\overline{M_4}:=\sum_{i=1}^{n} max\left |f^{(4)}(\xi) \right | $$

\overline{M_4} = n .M_4 $$

\Rightarrow\left |{E_n}^2\right |\le\dfrac{h^5}{2880}\times n.M_4=\dfrac{h^5}{2880}\times \dfrac{b-a}{h}.M_4 $$

 * {| style="width:20%" border="0"

$$\displaystyle \Rightarrow\left |{E_n}^2\right | \le \dfrac{(b-a)h^4}{2880} M_4 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

 * style= |

.
 * }
 * }

Author
Review and Correction - Egm6341.s10.Team4.andy 16:27, 17 February 2010 (UTC) .

= Problem 2: Error for composite Simpson's rule [Nima's Solution]=

Given
Use composite Simpson's equation (3) on slide [[media:Egm6341.s10.mtg7.pdf|7-2]], which is:

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

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

 * }
 * }

Find
To demonstrate error of composite Simpson's rule is equal to the following:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle 	\left | {E_n}^2\right |\leqslant\dfrac{(b-a)^5}{2880n^4}M_4=\dfrac{(b-a)h^4}{2880}M_4

M_4:=max\left |f^{(4)}(\xi)\right | $$

\xi\in[a,b] $$

 * style= |
 * }.
 * }.

Solution
The error definition can be done as:

{E_n}^2=I-I_n=\int\limits_{a}^{b} f(x)dx-I_n $$

From the composite form of Simpson's rule, we have:

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

\Rightarrow {E_n}^2=\int\limits_{a}^{b} f(x)dx-\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]= $$

=\sum_{i=1}^{n/2} [\int\limits_{x_{2i-2}}^{x_{2i}} f(x)dx-\dfrac{h}{3}[f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i})]] $$

According to Lagrange interpolation error for simple Simpson's rule (refer to 14-1) for $$\displaystyle [x_{i-1},x_{i+1}] $$ ;

E_2=-\dfrac {x_{i+1}}{x_{i-1}} f^{(4)}(\xi) $$

\xi\in [x_{i-1},x_{i+1}] $$

\Rightarrow \left |{E_n}^2 \right | \le \sum_{i=1}^{n/2} max \left | \dfrac {(x_{i+1}-x_{i-1})^5}{90\times 2^5} f^{(4)}(\xi)\right | =\dfrac{h^5}{90}\times \sum_{i=1}^{n/2} max \left | f^{(4)}(\xi)\right | $$

h=\dfrac {x_{i+1}-x_{i-1}}{2}=\dfrac {b-a}{n} $$

M_4:=max\left | f^{(4)}(\xi) \right | $$

\xi\in [x_{i-1},x_{i+1}] $$

\overline{M_4}:=\sum_{i=1}^{n/2} max\left |f^{(4)}(\xi) \right | $$

\overline{M_4}\le max\left | \dfrac {n}{2} .M_4 \right | $$

\Rightarrow\left |{E_n}^2\right |\le\dfrac{h^5}{90}\times \dfrac {n}{2}.M_4=\dfrac{h^5}{90}\times \dfrac {\dfrac{b-a}{h}}{2}.M_4 $$

 * {| style="width:20%" border="0"

$$\displaystyle \Rightarrow\left |{E_n}^2\right | \le \dfrac{(b-a)h^4}{180} M_4 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

 * style= |

.
 * }
 * }

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&m 15:31, 17 February 2010 (UTC) .

title: User:Egm6341.s10.Team4.andy/HW3


text:
User:Rjorjorjo/HW3

=Problem 17-3(1).(2) ~  18-1 (3)~(6)=

Statement
Reproduce Table 5.1, Table 5.3, Table 5.4, Table 5.5, Table 5.6 and Table 5.7 in text book(Atkinson)

Consider Trapezoidal and Simpson integration

Solution

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg7.pdf|P7-1]] [[media:Egm6341.s10.mtg7.pdf|P7-2]]
 * }
 * }

i Table 5.1

ii Table 5.3

iii Table 5.4

iv Table 5.5

v Table 5.6

vi Table 5.7

MATLAB
i    Table 5.1
 * For In

 * For En

 * For Ratio

 * For E'n

ii    Table 5.3
 * For In

 * For En

 * For Ratio

 * For E'n

iii    Table 5.4
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

iv    Table 5.5
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

v    Table 5.6
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

vi    Table 5.7
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

EGM6341.s10.Team1.Ya-Chiao Chang 06:54, 17 February 2010 (UTC)

title: User:Rjorjorjo/HW3


text:
18th century European scholarly societies and academies/Deutsche Gesellschaft in Leipzig

Summary
Name of Society:Deutsche Gesellschaft in Leipzig

Alternate Name(s):Görlitzische Poetische Gesellschaft, Teutschübende Poetische Gesellschaft,

Country:Germany

City: Leipzig, Saxony

Active Dates: 1697-1827

The Deutsch Gesellschaft was created in 1697 to further the importance of the German language and sought to be like the Académie Française which held the same ideals but in France instead of Germany. The Gesellschaft has been known by three different names Görlitzische Poetische Gesellschaft, Teutschübende Poetische Gesellschaft, and finally Deutsche Gesellschaft. The Deutsche Gesellschaft existed on its own until 1827 when it merged with an academy named Verein zur Erforschung vaterländischer Geschichte und Alterthümer and the combination of the two was to be named Deutsche Gesellschaft zur Erforschung vaterländischer Sprache und Alterthümer.


text:
University of Florida/Egm6341/s10.team3.aks/HW3

Problem Statement
Use Error estimate for Taylor series, composite trapezoidal and composite Simpson's rule to find n such that $$ E_n = I - I_n = O(10^{-6}) $$ and compare to numerical results.

Solution

 * $$I=\int_{0}^{1}\frac{e^{x}-1}{x}dx$$

Taylor Series

 * $$e^{x}=\sum _{j=0}^{\infty }\frac{x^j}{j!}\Rightarrow \frac{e^{x}-1}{x}=\sum _{j=1}^{\infty }\frac{x^{j-1}}{j!}$$
 * $$I_n=\int_{0}^{1}f_n(x)dx=\int_{0}^{1}\sum _{j=1}^{\infty} \frac{x^{j-1}}{j!}dx=\sum _{j=1}^{n}\frac{x^j}{j!j}|_0^1=\sum _{j=1}^n \frac{1}{j!j}$$
 * $$f(x)-f_n(x)=R_n(x)=\frac{(x-x_0)^n}{(n+1)!}f^{(n+1)}(\xi)$$ with $$\xi \in [0,x] \,$$ and $$x_0=0 \,$$
 * $$\Rightarrow f(x)-f_n(x)=\frac{x^n}{(n+1)!}f^{(n+1)}(\xi)$$
 * $$E_n=I-I_n=\int_{0}^{1}\left [ f(x)-f_n(x) \right ]dx=\int_{0}^{1}\underbrace{\frac{x^n}{(n+1)!}}_{w(x)}\underbrace{f^{(n+1)}\left (\xi(x) \right )}_{g(x)}dx$$
 * $$\Rightarrow g(\alpha)\int_{0}^{1}w(x)dx$$ for $$\alpha \in [0,1]$$
 * $$E_n =\,$$
 * $$\Rightarrow max = \frac{g(\alpha)}{(n+1)!(n+1)}$$, $$\alpha= \frac{e}{(n+1)!(n+1)} \,$$
 * $$\Rightarrow min = \frac{g(\alpha)}{(n+1)!(n+1)}$$, $$\alpha= \frac{1}{(n+1)!(n+1)}\,$$
 * $$n=2, E_2 \leq .151016 \,$$
 * $$n=4, E_5 \leq .00453 \,$$
 * $$ n=6, E_6 \leq 6.2792 X 10^{-4} \,$$
 * $$ n=8, E_8 \leq 8.42 X 10^{-6} \,$$

Below are the values from Numerical Analysis of Taylor series from HW_1

 * $$n=2, E_2 \leq .151016 \,$$
 * $$n=4, E_5 \leq .00453 \,$$
 * $$ n=6, E_6 \leq 6.2792 X 10^{-4} \,$$
 * $$ n=8, E_8 \leq 8.42 X 10^{-6} \,$$

We are getting same values from both analysis for Taylor series

Trapazoidal Rule

Error for Composite Trapazoidal rule is given by

$$  \displaystyle \left | E_n^1 \right |\leq \frac{(b-a)^3}{12n^2}M_2 $$

where $$M_2 = max \left | f^{(2)} (\zeta )\right |$$ for $$\zeta \varepsilon [a,b]$$

For the given function $$f(x)= \frac{e^x - 1}{x}$$, we have

$$  \displaystyle f^{(2)}(x)= \frac{e^x [x^2-2x+2]-2}{x^3} $$

For the given interval [0,1] the maximum value of function $$f^{(2)}(x)$$ is achieved at $$x=1$$

$$  \displaystyle M_2= f^{(2)}(x=1)=\frac{e[1+2-2]-2}{1}=0.71828182 $$

$$ E_n^1 \leq \frac{(1-0)^3}{12n^2}X 0.71828182 $$

$$ E_2^1 \leq .014964204 $$

$$ E_4^1 \leq 3.74105 X 10^{-3} $$

$$ E_6^1 \leq 1.662689 X 10^{-3} $$

$$ E_8^1 \leq 9.3526278 X 10^{-4}$$

$$ E_{16}^1 \leq 2.3381569 X 10 ^{-4}$$

$$ E_{32}^1 \leq 5.845392 X 10 ^{-5}$$

$$ E_{64}^1 \leq 1.4613481 X 10^{-5}$$

$$ E_{128}^1 \leq 3.65337026 X 10^{-6} $$

$$ E_{256}^1 \leq 9.1334256 X 10^{-7}$$

Below are the results from Numerical analysis from HW 1

$$ E_2^1 \leq 0.010389576$$

$$ E_4^1 \leq 0.002602468 $$

$$ E_8^1 \leq 0.650935 X 10^{-3}$$

$$ E_{16}^1 \leq 162753 X 10^{-4}$$

$$ E_{32}^1 \leq 4.06892 X 10 ^{-5}$$

$$ E_{64}^1 \leq 1.0172 X 10^{-5}$$

$$ E_{128}^1 \leq 2.54263 X 10^{-6} $$

$$ E_{256}^1 \leq 6.3528 X 10^{-7}$$

We can see that we are getting order $$ 10^{-6}$$ at n=128 from both the analysis.

Composite Simpsons Rule

The error estimate of the Composite Simpson's rule is given as

$$   E_n^2 \leq \frac{(b-a)^5}{2880n^4}M_4 $$

where $$M_4 = max \left | f^{(4)} (\zeta )\right |$$ for $$\zeta \varepsilon [a,b]$$

For the given function $$f(x)= \frac{e^x - 1}{x}$$, we have

$$  \displaystyle f^{(4)}(x)= \frac{e^x [x^4-4x^3+12x^2-24x]-24}{x^5} $$

The function $$f^{4}(x)$$ has maximum value at $$x=1$$

$$  \displaystyle M_4= \frac{e[1-4+12-24+24]-24}{1}=0.46453645

$$

$$ E_n^1 \leq \frac{(1-0)^5}{2880n^4}X 0.46453645 $$

$$ E_2^2 \leq 1.00810 X 10^{-5} $$

$$ E_4^2 \leq 6.300678 X 10^{-7} $$

Below are the results from Numerical analysis from HW 1

$$ E_2^2 \leq 1.06514 X 10^{-4} $$

$$ E_4^2 \leq 6.76473 X 10^{-6} $$

We can see that we are getting order $$ 10^{-6}$$ at n=4 from both the analysis.


text:
18th century European scholarly societies and academies/Reale Accademia delle Scienze e Belle-Lettere di Napoli

Summary
Name of Society: Reale Accademia delle Scienze e Belle-Lettere di Napoli

Country: Italy

City: Naples

Years Active: 1778-1788

Reale Accademia delle Scienze e Belle-Lettere di Napoli or in English Royal Academy of Science and Humanities of Naples was a short lived Academy that was conceived in 1778 and was legitimized by King Ferdinado IV in 1780 and disbanded in 1788, however there is some debate over whether the disbanding of the society happened in 1788 or 1787. This academy is considered a partial ancestor of the present day Societa' Nazionale Di Scienze, Lettere E Arti in Napoli.

Resources
Scholarly Societies Project


text:
Heritage Language

Definition
The term heritage language is used to identify any languages other than the dominant language(s) in a given social context Other researchers defined the term heritage language as the language associated with one's cultural background and it may or may not be spoken in the home Van Deusen-Scholl (2003) states that the term heritage language cannot be identified in a static, definitive sense; rather, it is more a metaphor in relation to the structuring of attitudes, beliefs, and assumptions that enable or constrain linguistic practices.

There have been critical views on the use of the term heritage language. Despite the term heritage language, heritage language speaker, and heritage language learner are gaining currency, the concept remains ill defined and is sensitive to a variety of interpretations within a social, political, regional, and national contexts

Using the label heritage itself has been challenged by many and has been regarded as extremely negative -even offensive- by some and counterproductive by others. Baker and Jones state that the term heritage language points more to the past and less to the future, to traditions rather to the contemporary so that the danger is that the term becomes associated with ancient cultures, past traditions and more "primitive times" and so may fail to give the impression of a modern language that is of value in a technological society

Alternative terms for heritage language proposed by researchers and used or preferred in different countries and contexts are: Though some of these terms are equally problematic as heritage language. For example, native language is also problematic because we cannot assume that the language does not necessarily students' native language. Those students may have a limited knowledge of the language or simply may not speak it at all. Ancestral language also conveys a negative connotation that it belongs to the distant past or a previous generation which does not have a present or future affiliation.
 * native language
 * minority language
 * ancestral language
 * immigrant minority language
 * allochthonous language
 * home language
 * language of origin
 * community language

Heritage Language Learners
Valdés provided three main criteria to identify heritage language learners. According to the proposal, heritage language learners are individuals who:

 * are raised in homes where a non-English language is spoken
 * speak or merely understand the heritage language
 * are to some degree bilingual in English and the heritage language

We have to keep in mind that heritage language learners are not a monolithic group so that we cannot assume or overgeneralize their proficiency. Rather, degree of heritage language fluency, proficiency, cultural background, connection or knowledge are widely varied among students. For example, heritage language learners include students of fluent native speakers to nonspeakers who may be generations removed but who may feel culturally connected to a language (Van Deusen-Scholl, 2003).

title: Heritage Language


text: File:Niki-trap-semilog.png

title: File:Niki-trap-semilog.png


text: File:Niki-trap-log.png

title: File:Niki-trap-log.png


text: File:Niki-CSR-semilog.png

title: File:Niki-CSR-semilog.png


text: File:Niki-CSR-log.png

title: File:Niki-CSR-log.png


text:
User talk:Davidhsjenkins

 Hello Davidhsjenkins, 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. --Mu301Bot 08:17, 17 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:17, 17 February 2010 (UTC)

title: User talk:Davidhsjenkins


text:
User:Uyenbui

=Humanities I/The Commons=

title: User:Uyenbui


text:
User:EGM6341.s10.Team1.Kumanchik/HW3

=Homework 3=

Find
Where does the proof of bounding the error for the simple Simpson's rule break down?

Given
1) $$G(t):= e(t)-t^4e(1) \,$$

2) $$G(t):= e(t)-t^6e(1) \,$$

3) $$G(t):= e(t)-t^5e(1) \,$$, where for this case continue differentiating to get $$G^{(3)}(0) \,$$

Solution
Start by performing derivatives of $$e(t) \,$$,

 * $$e^{(0)}(t)=\int_{-t}^{t}F(t)dt - \frac{t}{3}[F(-t)+4F(0)+F(t)]$$
 * $$e^{(1)}(t)=F(-t)+F(t)-\frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]$$
 * $$e^{(2)}(t)=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]$$
 * $$e^{(3)}(t)=-\frac{t}{3}[-F^{(3)}(-t)+F^{(3)}(t)]$$
 * $$e^{(4)}(t)=-\frac{1}{3}[-F^{(3)}(-t)+F^{(3)}(t)]-\frac{t}{3}[F^{(4)}(-t)+F^{(4)}(t)]$$

Note that $$e^{(n)}(0)=0 \,$$ for $$n=0,1,2,3,4 \,$$.

Case 1:

 * $$G^{(2)}(t)=e^{(2)}(t)-12t^2e(1) \,$$

and,
 * $$G^{(2)}(0)=0 \,$$

Plugging in $$\zeta_2 \,$$,
 * $$G^{(2)}(\zeta_2)=\frac{1}{3}\underbrace{[-F^{(1)}(-\zeta_2)+F^{(1)}(\zeta_2)]}_{2\zeta_2F^{(2)}(\zeta_3)}-\frac{\zeta_2}{2}[F^{(2)}(-\zeta_2)+F^{(2)}(\zeta_2)]-12\zeta_2^2e(1)=0$$

Solving for $$e(1) \,$$,

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

e(1)=\frac{1}{18\zeta_2}F^{(2)}(\zeta_3)-\frac{1}{36}F^{(2)}\underbrace{(-\zeta_2)}_{problem}-\frac{1}{36}F^{(2)}(\zeta_2) $$ This variable is not on the interval between 0 and $$\zeta_2 \,$$ $$
 * $$\displaystyle
 * }
 * }

Case 2:

 * $$G^{(4)}(t)=e^{(4)}(t)-360t^2e(1) \,$$

and,
 * $$G^{(4)}(0)=0 \,$$

Plugging in $$\zeta_4 \,$$
 * $$G^{(4)}(\zeta_4)=-\frac{1}{3}[-F^{(3)}(-\zeta_4)+F^{(3)}(\zeta_4)]-\frac{\zeta_4}{3}[F^{(4)}(-\zeta_4)+F^{(4)}(\zeta_4)]-360(\zeta_4^2)e(1)=0 \,$$

Solving for $$e(1) \,$$,

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

e(1)=\frac{-1}{540\zeta_4}F^{(4)}(\zeta_5)-\frac{1}{1080\zeta_4}[F^{(4)}(\underbrace{-\zeta_4)}_{problem}+F^{(4)}(\zeta_4)] $$ This variable is not on the interval between 0 and $$\zeta_4 \,$$ $$
 * $$\displaystyle
 * }
 * }

Case 3:

 * $$G^{(3)}(t)=e^{(3)}(t)-60t^2e(1) \,$$

and,
 * $$G^{(3)}(0)=0 \,$$

We continue taking derivatives of $$G \,$$
 * $$G^{(4)}(t)=e^{(4)}(t)-120te(1) \,$$

to get,
 * $$G^{(4)}(\zeta_4)=-\frac{1}{3}[-F^{(3)}(-\zeta_4)+F^{(3)}(\zeta_4)]-\frac{\zeta_4}{3}[F^{(4)}(-\zeta_4)+F^{(4)}(\zeta_4)]-120(\zeta_4^2)e(1)=0 \,$$

Solving for $$e(1) \,$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

e(1)=\frac{-3}{360}F^{(4)}(\zeta_4)-\frac{1}{360}F^{(4)}\underbrace{(-\zeta_4)}_{problem} $$ This variable is not on the interval between 0 and $$\zeta_4 \,$$ $$
 * $$\displaystyle
 * }
 * }

title: User:EGM6341.s10.Team1.Kumanchik/HW3


text:
Java server faces

Java Server Faces is a view technology for web applications developed using java technology

Developed through the Java Community Process under JSR - 314, JavaServer Faces technology establishes the standard for building server-side user interfaces. With the contributions of the expert group, the JavaServer Faces APIs are being designed so that they can be leveraged by tools that will make web application development even easier. Several respected tools vendors were members of the JSR-314 expert group, which developed the JavaServer Faces 1.0 specification. These vendors are committed to supporting the JavaServer Faces technology in their tools, thus promoting the adoption of the JavaServer Faces technology standard.

title: Java server faces


text:
User talk:Mattisse

Welcome
Welcome to Wikiversity, Mattisse. If you have any questions or concerns, feel free to contact me. :) Ottava Rima (talk) 14:56, 17 February 2010 (UTC)
 * Thanks Ottava. It's nice to see you. Mattisse 14:59, 17 February 2010 (UTC)
 * Nice to see you too. You'll notice that projects like Wikiversity, Source, etc, tend to have more of a "small town feel" - a little slower, a lot more peaceful, and more straight forward but laid back. If you have any project areas or any topics you would want to work on, there are members in a large range of topics that would definitely help out if you need anything. My user page has a list of different areas that I frequent with Colloquium being the central hub for any concerns or ideas that don't seem to fit anywhere else. Ottava Rima (talk) 15:17, 17 February 2010 (UTC)

I haven't seen you in a long time. Good to see that you are alive and active. :) Ottava Rima (talk) 19:01, 20 February 2011 (UTC)

 * Hi! I have been very active on Wikisource since last July, which I really enjoy. But I might find some things here to do also. I have to look around and see.  Best wishes, Mattisse 19:05, 20 February 2011 (UTC)
 * I stopped being active over there when Poetlister started using Wikisource to email me. Blah. If you need any help with anything, feel free to drop a line. Ottava Rima (talk) 19:11, 20 February 2011 (UTC)

 * I haven't seen that name around. Maybe he is no longer there on Wikisource. Mattisse 19:17, 20 February 2011 (UTC)
 * Ottava Rima (talk) 19:34, 20 February 2011 (UTC)
 * Well, I admit I don't understand. Apparently that account is block but he continues to edit under another name. I have experienced nothing but helpfulness and pleasantness at Wikisource, so I don't know why a disruptive editor is tolerated. Mattisse 19:42, 20 February 2011 (UTC)

title: User talk:Mattisse


text:
Romanticism and Revolution

The Romantic movement of 19th century art and literature was influenced by revolutionary events such as the French and American revolutions.

The 18th century Romantic poets were influenced by many outside influences but chief among them was the revolution occurring in France. Their poetry reflects the social turmoil raging across Europe and their own dreams and worries. The link between the French Revolution and poetry is best shown by examining three English Romantics: Wordsworth, Shelley, and Coleridge, and their political interests and activities in the French Revolution, and how that was reflected in their poems.

Wordsworth
Wordsworth’s politics and his poem “Tintern Abbey” demonstrate how his work and social life was influenced by the revolution. Politically, Wordsworth didn’t become involved in the French situation until 1791 when he revisited France, became friends with Beaupuis, and came across a poor girl who was starving because of the way France was being run. As Hancock puts it, this “experience sank deep into Wordsworth’s heart. It made him an advocate of the cause of rebellious France." After this, Wordsworth’s ideas on politics and government changed completely. He took an eager interest in the revolution and emitted his enthusiasm and energy, but he “thought little about the abstract principles of human rights”. He became interested in politics and government as he had never been in England, since he felt a connection to the French cause and people. However, as the Revolution continued it became increasingly dark in the optimistic and visionary eyes of Wordsworth and it ended with him becoming disillusioned and depressed as the Revolution’s outcome fell short of his goals. “Tintern Abbey” echoes William Wordsworth’s anxieties about the French Revolution. This poem was written towards the end of the French Revolution, so by this time, Wordsworth has experienced the effects of Revolution. “Tintern Abbey” is a poem about the loss of innocence and childhood as well as the loss of dreams. For Wordsworth, one of these dreams was the ideas brought forth by the Revolution that fell short of his expectations. As Yu Liu says, “he was utterly disenchanted with his own activism, but was unable to change his perception of the French Revolution as what should or ought to succeed in theory, and”… “he was utterly disillusioned with the turns of events in France." “Tintern Abbey” has a melancholy tone, full of reminiscence and loss. In lines 83-85 this longing for the past is exhibited “time is past and all its aching joys are now no more, and all its dizzy raptures." These lines and the ones before it are referring to growing up and the ending of dreams, including Wordsworth’s interest in the French Revolution. This poem also shows Wordsworth’s love of nature, and his use of it to express the thought and ideas inside him. “While he was still under the spell of the Revolution, he could boldly result to natural metaphor." However, when he lost faith in the Revolution he sank deeper into nature instead of making a “sacrifice of himself for what had already proven to be a lost or losing cause, his response in the event, as it so richly and fruitfully worked itself out in the triumphant publication of the Lyrical Ballads in 1798, was an instinctive and analogical movement toward nature." Through many of Wordsworth’s poems, one can see the effect the French Revolution had on his mind and his poetry, from his initial hopes and joys to his devastated dreams.

Percy Bysshe Shelley
Percy Bysshe Shelley is another well known Romantic poet who was affected by the French Revolution. He embodied the spirit of the Revolution shown “in an unrestrained denunciation of the past with its tyrannical government of priests and kings, and an unshakeable faith for a future with its perfected humanity." Shelley had a lot of ideas about the ideal future that he thought the Revolution could bring about. His excitement for change was shown in his poetry by using such emotions as anger to demonstrate his point. Stauffer points out that “Shelley was attracted to anger precisely because of its renovating force." His poems are distinct by the emotion he puts into them and an example of this, combined with his political feelings, is the poem “Queen Mab.”

“Queen Mab” is an interesting poem, especially since it is “Shelley’s first poem of importance, broadly professing his radical ideas”. The part to focus on when applying it to Shelley’s stance on the Revolution is the final two parts. These describe his vision of the perfect world or “Shelley’s picture of the Golden Age”; one that he hopes can be achieved with the help of the Revolution. In canto eight, Shelley describes a fantastic vision of peace and harmony overtaking the world from “Hope was seen beaming through the mists of fear; Earth was no longer a hell” to such frivolous images of “Fruits”… [being] “ever ripe, flowers ever fair”. This shows how this new, heavenly earth will be perfect in all things, from Earth’s peacefulness as a planet to the little things. An important line to consider is one found a few lines later predicting “the lion now forgets to thirst for blood”. The lion in this line demonstrates that, if a natural predator like a lion no longer needs to kill, neither will human beings have to kill each other, and there will be world peace. An interesting perspective is Stauffer’s where he says “Shelley’s poetry almost invariably contravenes its own wrath with more conciliatory and hopeful imaginings”. It is these “hopeful imaginings” that make up his beliefs and stance towards the government and politics. After this Utopian view for the future, the next canto goes one step further. “Then, that sweet bondage which is freedom’s self, and rivets with sensation’s softest tie the kindred sympathies of human souls, needed no fetters of tyrannic law”. These lines show that to achieve this unending and unlimited peace, “tyrannic law” needs to be gone. In this poem, Shelley’s vision and politics come through. His hopes for a peaceful future are defined by the absent of government and his excitement for the French Revolution embodies this.

Samuel Taylor Coleridge
Samuel Taylor Coleridge is the third example of an eighteenth century poet who was affected by the French Revolution. However, he wasn’t necessarily physically involved in the Revolution, but both he and Wordsworth “were thinkers, but of the emotional and imaginative, not of the ratiocinative, order”. Both Wordsworth and Coleridge are often grouped together but Coleridge “was more open, sensitive, more receptive to the multitudinous impressions of a world of human affairs”. Coleridge exhibited as much emotion and enthusiasm for change as Wordsworth but Coleridge often looked through a religious lens. As Dowden puts it, “the idea of God possessed him,” and this was reflected in his writings and in the way he viewed the world. Coleridge’s poem “Destruction of the Bastille” shows his emotional devotion and interest in the French Revolution.

In Coleridge’s “Destruction of the Bastille,” his love for freedom and the French Revolution become incredibly apparent. This poem, written shortly after the Bastille fell, contains the proud tone of a victor. The speaker declares his victory “Go, Tyranny! Beneath some barbarous sky thy terrors lost, and ruin’d power deplore!" This is an exhilarated cry taunting “Tyranny” and it’s loss of power over the French people. This victorious and proud tone continues into the later stanzas. “Yes! Liberty the soul of Life shall reign, shall throb in every pulse, shall flow thro’ every vein!” Now, the speaker isn’t just talking about a victory over “Tyranny,” he is talking about hope for the future though “Liberty.” A final aspect of this poem to consider are the questions in stanza six “Shall France alone a Despot spurn? Shall she alone, O Freedom boast thy care?” The lines after these expand upon this idea of universal revolution and political upheaval. This poem ends considering the chance that Europe is presented right now, the chance for a loss of “Tyranny,” a gain of “Liberty” and a new political order. Everyone in Europe was affected in some way by the French Revolution, the question is to what extend did this influence affect their lives. By examining a few eighteenth century poets, it is clear that the revolution was on the mind, even in a different country. Being a Romantic writer means reacting to the surrounding world, and part of that was the French Revolution. As Butler put it “the French Revolution was new because it involved more of the people more crucially in national affairs, ‘big’ politics, than had ever been the case in European, or perhaps any politics before." Duff phrases it a little differently by saying that “the French Revolutionaries complained of everything." But the Revolution was about change, and it couldn’t happen without complaining, talking and writing. Writing and poetry was a way to express this change in political and emotional relations. Writing was as important to the French Revolution as the French Revolution was to eighteenth century writers. “Words, discourse, the media of communication did matter in the French Revolution” and is did “indeed appear to have mattered to an unprecedented degree."

title: Romanticism and Revolution


title: File:Niki1.png


title: File:Niki2.png


text:
User:EGM6341.s10.Team1.toddmock/HW3

= Problem : Error on Numerical Integrals =

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

Find
Using three numerical integration methods (Taylor series expansion, Simpson's rule, trapezoidal rule) find at what iteration (n) the error approaches an order of 1e-6

Solution
In general there error of a numerical integration is the exact integral minus the numerical integral:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n=I-I_n
 * }
 * }

For the Taylor series error is just the integral of the remainder term which is

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle R_n(x)=e^{\xi }\frac{x^n}{(n+1)!}
 * }
 * }

Then from p.7-1:

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

$$  \displaystyle \int_{0}^{1}[f(x)-f_n(x)]dx=e^{\xi } \int_{0}^{1}\frac{x^n}{(n+1)!}dx $$    (3) Completing the integration gives
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$ $$ Choosing $$\xi=0,1$$ then the error is bounded
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e^{\xi }\frac{1}{(n+1)!(n+1)}
 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \frac{1}{(n+1)!(n+1)} 16-3:
 * {| style="width:100%" border="0" align="left"

$$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \left | E_n^1 \right |\leq \frac{(b-a)}{12n^2}M_2
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ for
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle M_2 = max | f^{(2)} (\xi )|
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ Here
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \xi \varepsilon [a,b]
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ the max occurs at $$\xi = 1$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f^{(2)}=\frac{(x^2-2x+2)e^x-2}{x^3}
 * }
 * }

Finally for the Simpson's rule:
 * {| style="width:100%" border="0" align="left"

$$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \left | E_n^2 \right |\leq \frac{(b-a)^5}{2880n^4}M_4
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ for
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle M_4 = max | f^{(4)} (\xi )|
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \xi \varepsilon [a,b]
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f^{(4)}=\frac{(x^4-4x^3-24x+24)e^x-24}{x^5}
 * }
 * }

Using these three MATLAB codes:

the results below were obtained:

$$ \begin{array}{|c||c|c|c|} n & E_n^{Trap} & E_n^{Taylor} & E_n^{Simpson} \\ \hline 1&0.059900000&0.679600000&0.000161300\\ 2&0.007500000&0.151000000&0.000010081\\ 3&0.002200000&0.028300000&0.000001991\\ 4&0.000935260&0.004500000&--\\	5&0.000478850&0.000629230&--\\	6&0.000277110&0.000077049&--\\	7&0.000174510&0.000008427&--\\	8&0.000116910&--&--\\		9&0.000082108&--&--\\		10&0.000059857&--&--\\		11&0.000044971&--&--\\	12&0.000034639&--&--\\		13&0.000027245&--&--\\	14&0.000021814&--&--\\		15&0.000017735&--&--\\		16&0.000014613&--&--\\		17&0.000012183&--&--\\		18&0.000010264&--&--\\		19&0.000008727&--&--\\ \hline \end{array} $$

From this table it can be seen that the Simpson's rule for numerical integration seems to be the best in that it gives an accurate answer in minimal spatial steps

= Problem :Error on Numerical Integrals Cont'd =

Find
Plot the step size (h) against the integration error in such a way as to determine a power relation between the two

Solution
Plotting the Log of the step size (h) against the log of the error is most appropriate here because the plot will be a straight line. From the fit equation:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Log(E_n)=a[Log(h)]+b

 * }
 * }

it can then be shown that the slope (a) becomes the power term on h.

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle 10^{Log(E_n)}=10^{a(log(h))+b}

 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle\Rightarrow E_n=10^bh^a

Thus it can be seen in the plots that for the Trapezoidal rule
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n\propto h^3

 * }
 * }

and for the Simpson's rule

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n\propto h^4

 * }
 * }

$$ \begin{array}{|c||c|c|} n & Log(h) & Log(E_n) \\ \hline 1&0.000000&1.2228864\\ 2&0.301030&2.1259764\\ 3&0.477121&2.6542501\\ 4&0.602060&3.0290663\\ 5&0.698970&3.3197964\\ 6&0.778151&3.5573401\\ 7&0.845098&3.7581805\\ 8&0.903090&3.9321563\\ 9&0.954243&4.0856139\\ 10&1.000000&4.2228864\\ \hline \end{array} $$

$$ \begin{array}{|c||c|c|} n & Log(h) & Log(E_n) \\ \hline 1&0.00000&3.792372685\\ 2&0.30103&4.996492668\\ 3&0.47712&5.700857704\\ 4&0.60206&6.200612651\\ 5&0.69897&6.588252703\\ 6&0.77815&6.904977687\\ 7&0.84510&7.172764845\\ 8&0.90309&7.404732633\\ 9&0.95424&7.609342723\\ 10&1.00000&7.792372685\\ \hline \end{array} $$

title: User:EGM6341.s10.Team1.toddmock/HW3


text: File:Trapezoidal.jpg

title: File:Trapezoidal.jpg


text: File:Simrule.jpg

title: File:Simrule.jpg


text: File:Romberg.JPG

title: File:Romberg.JPG


text:
User:LTravers

This is my user page with Wikiversity.

title: User:LTravers


text:
User:Egm6341.s2010.Team1/HW3

=Problem 17-1=

Find
Where does the proof for bounding the error of the simple Simpson's rule break down?

Given
1) $$G(t):= e(t)-t^4e(1) \,$$

2) $$G(t):= e(t)-t^6e(1) \,$$

3) $$G(t):= e(t)-t^5e(1) \,$$, where for this case continue differentiating to get $$G^{(3)}(0) \,$$

Solution
Start by performing derivatives of $$e(t) \,$$,

 * $$e^{(0)}(t)=\int_{-t}^{t}F(t)dt - \frac{t}{3}[F(-t)+4F(0)+F(t)]$$
 * $$e^{(1)}(t)=F(-t)+F(t)-\frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]$$
 * $$e^{(2)}(t)=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]$$
 * $$e^{(3)}(t)=-\frac{t}{3}[-F^{(3)}(-t)+F^{(3)}(t)]$$
 * $$e^{(4)}(t)=-\frac{1}{3}[-F^{(3)}(-t)+F^{(3)}(t)]-\frac{t}{3}[F^{(4)}(-t)+F^{(4)}(t)]$$

Note that $$e^{(n)}(0)=0 \,$$ for $$n=0,1,2,3,4 \,$$.

Case 1:

 * $$G^{(2)}(t)=e^{(2)}(t)-12t^2e(1) \,$$

and,
 * $$G^{(2)}(0)=0 \,$$

Plugging in $$\zeta_2 \,$$,
 * $$G^{(2)}(\zeta_2)=\frac{1}{3}\underbrace{[-F^{(1)}(-\zeta_2)+F^{(1)}(\zeta_2)]}_{2\zeta_2F^{(2)}(\zeta_3)}-\frac{\zeta_2}{2}[F^{(2)}(-\zeta_2)+F^{(2)}(\zeta_2)]-12\zeta_2^2e(1)=0$$

Solving for $$e(1) \,$$,

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

e(1)=\frac{1}{18\zeta_2}F^{(2)}(\zeta_3)-\frac{1}{36}F^{(2)}\underbrace{(-\zeta_2)}_{problem}-\frac{1}{36}F^{(2)}(\zeta_2) $$ This variable is not on the interval between 0 and $$\zeta_2 \,$$ $$
 * $$\displaystyle
 * }
 * }

Case 2:

 * $$G^{(4)}(t)=e^{(4)}(t)-360t^2e(1) \,$$

and,
 * $$G^{(4)}(0)=0 \,$$

Plugging in $$\zeta_4 \,$$
 * $$G^{(4)}(\zeta_4)=-\frac{1}{3}[-F^{(3)}(-\zeta_4)+F^{(3)}(\zeta_4)]-\frac{\zeta_4}{3}[F^{(4)}(-\zeta_4)+F^{(4)}(\zeta_4)]-360(\zeta_4^2)e(1)=0 \,$$

Solving for $$e(1) \,$$,

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

e(1)=\frac{-1}{540\zeta_4}F^{(4)}(\zeta_5)-\frac{1}{1080\zeta_4}[F^{(4)}(\underbrace{-\zeta_4)}_{problem}+F^{(4)}(\zeta_4)] $$ This variable is not on the interval between 0 and $$\zeta_4 \,$$ $$
 * $$\displaystyle
 * }
 * }

Case 3:

 * $$G^{(3)}(t)=e^{(3)}(t)-60t^2e(1) \,$$

and,
 * $$G^{(3)}(0)=0 \,$$

We continue taking derivatives of $$G \,$$
 * $$G^{(4)}(t)=e^{(4)}(t)-120te(1) \,$$

to get,
 * $$G^{(4)}(\zeta_4)=-\frac{1}{3}[-F^{(3)}(-\zeta_4)+F^{(3)}(\zeta_4)]-\frac{\zeta_4}{3}[F^{(4)}(-\zeta_4)+F^{(4)}(\zeta_4)]-120(\zeta_4^2)e(1)=0 \,$$

Solving for $$e(1) \,$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

e(1)=\frac{-3}{360}F^{(4)}(\zeta_4)-\frac{1}{360}F^{(4)}\underbrace{(-\zeta_4)}_{problem} $$ This variable is not on the interval between 0 and $$\zeta_4 \,$$ $$ EGM6341.s10.Team1.Kumanchik 19:30, 17 February 2010 (UTC)
 * $$\displaystyle
 * }
 * }

=Problem 17-2(1)=

=Problem 17-2(2)=

Find
Using three numerical integration methods (Taylor series expansion, Simpson's rule, trapezoidal rule) find at what iteration (n) the error approaches an order of 1e-6

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

Solution
In general there error of a numerical integration is the exact integral minus the numerical integral:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n=I-I_n
 * }
 * }

For the Taylor series error is just the integral of the remainder term which is

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle R_n(x)=e^{\xi }\frac{x^n}{(n+1)!}
 * }
 * }

Then from p.7-1:

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

$$  \displaystyle \int_{0}^{1}[f(x)-f_n(x)]dx=e^{\xi } \int_{0}^{1}\frac{x^n}{(n+1)!}dx $$    (3) Completing the integration gives
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$ $$ Choosing $$\xi=0,1$$ then the error is bounded
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e^{\xi }\frac{1}{(n+1)!(n+1)}
 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \frac{1}{(n+1)!(n+1)} 16-3:
 * {| style="width:100%" border="0" align="left"

$$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \left | E_n^1 \right |\leq \frac{(b-a)}{12n^2}M_2
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ for
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle M_2 = max | f^{(2)} (\xi )|
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ Here
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \xi \varepsilon [a,b]
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ the max occurs at $$\xi = 1$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f^{(2)}=\frac{(x^2-2x+2)e^x-2}{x^3}
 * }
 * }

Finally for the Simpson's rule:
 * {| style="width:100%" border="0" align="left"

$$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \left | E_n^2 \right |\leq \frac{(b-a)^5}{2880n^4}M_4
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ for
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle M_4 = max | f^{(4)} (\xi )|
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \xi \varepsilon [a,b]
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f^{(4)}=\frac{(x^4-4x^3-24x+24)e^x-24}{x^5}
 * }
 * }

Using these three MATLAB codes:

the results below were obtained:

$$ \begin{array}{|c||c|c|c|} n & E_n^{Trap} & E_n^{Taylor} & E_n^{Simpson} \\ \hline 1&0.059900000&0.679600000&0.000161300\\ 2&0.007500000&0.151000000&0.000010081\\ 3&0.002200000&0.028300000&0.000001991\\ 4&0.000935260&0.004500000&--\\	5&0.000478850&0.000629230&--\\	6&0.000277110&0.000077049&--\\	7&0.000174510&0.000008427&--\\	8&0.000116910&--&--\\		9&0.000082108&--&--\\		10&0.000059857&--&--\\		11&0.000044971&--&--\\	12&0.000034639&--&--\\		13&0.000027245&--&--\\	14&0.000021814&--&--\\		15&0.000017735&--&--\\		16&0.000014613&--&--\\		17&0.000012183&--&--\\		18&0.000010264&--&--\\		19&0.000008727&--&--\\ \hline \end{array} $$

From this table it can be seen that the Simpson's rule for numerical integration seems to be the best in that it gives an accurate answer in minimal spatial steps --Egm6341.s10.team1.toddmock 19:50, 17 February 2010 (UTC)

=Problem 17-2(3)=

Find
Plot the step size (h) against the integration error in such a way as to determine a power relation between the two

Solution
Plotting the Log of the step size (h) against the log of the error is most appropriate here because the plot will be a straight line. From the fit equation:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Log(E_n)=a[Log(h)]+b

 * }
 * }

it can then be shown that the slope (a) becomes the power term on h.

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle 10^{Log(E_n)}=10^{a(log(h))+b}

 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle\Rightarrow E_n=10^bh^a

Thus it can be seen in the plots that for the Trapezoidal rule
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n\propto h^3

 * }
 * }

and for the Simpson's rule

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle E_n\propto h^4

 * }
 * }

$$ \begin{array}{|c||c|c|} n & Log(h) & Log(E_n) \\ \hline 1&0.000000&1.2228864\\ 2&0.301030&2.1259764\\ 3&0.477121&2.6542501\\ 4&0.602060&3.0290663\\ 5&0.698970&3.3197964\\ 6&0.778151&3.5573401\\ 7&0.845098&3.7581805\\ 8&0.903090&3.9321563\\ 9&0.954243&4.0856139\\ 10&1.000000&4.2228864\\ \hline \end{array} $$

$$ \begin{array}{|c||c|c|} n & Log(h) & Log(E_n) \\ \hline 1&0.00000&3.792372685\\ 2&0.30103&4.996492668\\ 3&0.47712&5.700857704\\ 4&0.60206&6.200612651\\ 5&0.69897&6.588252703\\ 6&0.77815&6.904977687\\ 7&0.84510&7.172764845\\ 8&0.90309&7.404732633\\ 9&0.95424&7.609342723\\ 10&1.00000&7.792372685\\ \hline \end{array} $$

--Egm6341.s10.team1.toddmock 19:50, 17 February 2010 (UTC)

=Problem 17-3(1-2) 18-1(3-6)=

Find
Reproduce Table 5.1, Table 5.3, Table 5.4, Table 5.5, Table 5.6 and Table 5.7 in text book(Atkinson)

Consider Trapezoidal and Simpson integration

Solution

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg7.pdf|P7-1]] [[media:Egm6341.s10.mtg7.pdf|P7-2]]
 * }
 * }

i Table 5.1

ii Table 5.3

iii Table 5.4

iv Table 5.5

v Table 5.6

vi Table 5.7

MATLAB
i    Table 5.1
 * For In

 * For En

 * For Ratio

 * For E'n

ii    Table 5.3
 * For In

 * For En

 * For Ratio

 * For E'n

iii    Table 5.4
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

iv    Table 5.5
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

v    Table 5.6
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

vi    Table 5.7
 * For Trapezoidal Rule Error

 * For Trapezoidal Rule Ratio

 * For Simpson's Rule Error

 * For Simpson's Rule Ratio

EGM6341.s10.Team1.Ya-Chiao Chang 06:54, 17 February 2010 (UTC)

=Problem 19-1=

Find
a) Modify matlab code to make the comp. of $$ T_{0}\left ( 2^j \right )\,$$ efficient, i.e,$$ T_{0}\left ( 2^j \right )=T_{0}\left (2^{(j-1)}\right )+...\,$$

b) Romberg table, compare to previous results

Given
Cont'd  [[media:Egm6341.s10.mtg6.pdf|Lec. 6-5]]
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle f(x)= \frac{e^x-1}{x} $$
 * $$\displaystyle I= \int\limits_{0}^{1}\frac{e^x-1}{x}\, dx $$
 * }
 * }
 * }

Solution
a) From

 * $$ T(2n)\,$$ comp. based on $$ T(n)\,$$, and $$ f(x_i), i=1,3,5,...\,$$

 * $$ T(n)=h [1/2f_0+f_1+f_2+...+f_{n-1}+1/2f_n], h=(b-a)/n \,$$

 * $$ T(2n)=\underbrace{\left (h [1/2f_0+f_2+f_4...+f_{2(n-1)}+1/2f_n]\right)}_{T(n)}+h [f_1+f_3+f_5+...+f_{2n-1}], h=(b-a)/2n \,$$

 * $$\Rightarrow T_{0}\left ( 2^j \right )= T_{0}\left (2^{(j-1)}\right )+ (b-a)/2n [f_1+f_3+f_5+...+f_{2(j-1)}] \,$$

Based on this method,modify matlab code to make the comp.of $$ T_{0}\left ( 2^j \right )\,$$ efficient.

Modified Matlab Code:

b) Romberg table display the results as follows:

Previous results:

From the comparation, for a given value of n,the most accurate value is for the largest value of k.

Matlab Code:

Egm6341.s10.team1.lei 16:06, 17 February 2010 (UTC)

=Contributing members= EGM6341.s10.Team1.Kumanchik 19:34, 17 February 2010 (UTC) Author p17-1

Egm6341.s10.team1.toddmock 19:51, 17 February 2010 (UTC) Author p17-2(2 & 3)

Egm6341.s10.team1.andrewdugan 20:17, 17 February 2010 (UTC) Author p17-2(1)

Egm6341.s10.team1.lei 20:31, 17 February 2010 (UTC)

Egm6341.s10.team1.Ya-Chiao Chang 21:23, 17 February 2010 (UTC) Auhor p19-2

title: User:Egm6341.s2010.Team1/HW3


text:
User:AndrewR~enwikiversity

=Humanities I/The Commons=

title: User:AndrewR~enwikiversity


title: File:Egm6341.s10.mtg11.djvu


title: File:Egm6341.s10.mtg12.djvu


title: File:Egm6341.s10.mtg20.djvu


title: File:Egm6341.s10.mtg21.djvu


text:
User talk:KadenceEnglehardt

 * Kadence,

I received your email, but my reply bounced, so I'll try leaving a note here. Your idea for a project sounds great. Take a look at the Wikipedia page on Mail Art for some inspiring ideas.

I hope you're feeling better,
 * Stevenarntson

Noncommercial licensing
The pciture referenced above is currently listed as noncommercial. This is not an appropriate license for Wikiversity, as it doesn't count as a free license. You will need to relicense the picture under a free license like one of the CC ones or the picture will be deleted. Geoff Plourde 18:03, 4 August 2010 (UTC)

title: User talk:KadenceEnglehardt


text:
User:Calusarul

Hello.

I hope a Romanian Wikiversitate will be soon created. I support any open source learning resources.

title: User:Calusarul


text:
User talk:Calusarul

Hello.

Modify this page if you want to say anything to me. And don't forget to leave a signature. Calusarul 22:21, 17 February 2010 (UTC)

 Hello Calusarul, 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) 16:11, 19 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.

Romanian Wikiversity
Re: your posts about this topic, Most of the people here don't speak Romanian. I don't, for example. However, it's great that you have so much enthusiasm about creating another language's Wikiversity. You may be able to recruit Romanian speakers from Romanian Wikipedia. Have you tried that yet? --AFriedman (talk) 16:10, 19 February 2010 (UTC)

title: User talk:Calusarul


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Teamwork through World of Warcraft

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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Grade Level: High School/College
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: Humanities Course
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: The Benefits of Teamwork in a Society
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 3-4 days
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: World of Warcraft
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Introduction/Specifications
The World of War Craft (a.k.a. WoW) is a massively multiplayer online role-playing game which takes place within the Warcraft world of Azeroth.

For this lesson, students will be asked to create an avatar, explore their surroundings, fight various monsters, complete quests, and interact with other players. It is important to be sure that each student is apart of the same realm so that they can interact within the same world.

To create a new character,players must choose between the opposing factions of Alliance or Horde. Characters from the opposing factions can perform minor communication, but only members of the same faction can speak, email, group, and share guilds. For the sake of this particular lesson, half of the students will join the Alliance faction and the other half will join the Horde faction. The player selects the new character's race (species), and a class for the character, with choices such as mages, warriors and priests.

As characters become more developed through quests and other adventures, they gain various talents and skills, that will require students to further define the abilities of that character (i.e. blacksmithing, mining, cooking, fishing, and first aid).

Students will focus on questing for this several day lesson since the central focus is on teamwork and playing your role in a society and seeing how your contributions are apart of a greater whole. Quests usually reward the player with experience points, items, and/or in-game money.

While a character can be played on its own, players can also group up with others in order to tackle more challenging content. In this way, character classes are used in specific roles within a group.[28][30] World of Warcraft uses a "rested bonus" system, increasing the rate that a character can gain experience points after the player has spent time away from the game.[26] When a character dies, it becomes a ghost (or wisp for Night Elf characters) at a nearby graveyard.[31] Characters can be resurrected by other characters that have the ability, or can self-resurrect by moving from the graveyard to the place where they died. When a character dies, the items equipped by the character degrade, requiring in-game money and a specialist NPC to repair them. Items that have degraded heavily become unusable until they are repaired. If the location of the character's body is unreachable, they can use a special NPC known as a spirit healer to resurrect at the graveyard. When the spirit healer revives a character, items equipped by the character at that time suffer increased degradation, and the character is significantly weakened for ten minutes. This "Resurrection Sickness" does not occur and item degradation is less severe if the character revives by locating its body, or is resurrected by another player, through special items or spells.[32][33]

World of Warcraft contains a variety of mechanisms for player-versus-player (PvP) play. Some realms allow player-versus-player combat almost anywhere in the game world, at any time. In these environments, members of opposing factions can attack each other at almost any time or location. Player-versus-environment (PvE) servers, by contrast, allows a player to choose whether or not to engage in combat against other players. On both server types, there are special areas of the world where free-for-all combat is permitted. Players on PvE servers can also opt to "flag" themselves, making them attackable targets to players of the opposite faction.[34] Battlegrounds, for example, are similar to dungeons: only a set number of characters can enter a single battleground, but additional copies of the battleground can be made to accommodate additional players.[35] Each battleground has a set objective, such as capturing a flag or defeating an opposing general, that must be completed in order to win the battleground. Competing in battlegrounds rewards the character with tokens and honor points that can be used to buy armour and weapons.[34]

Setting World of Warcraft is a massively multiplayer online role-playing game. In addition to sharing the "Warcraft" name with the real-time strategy games in the Warcraft series, it is set in the world of Azeroth and has similar art direction.[21]

World of Warcraft takes place in a 3D representation of the Warcraft universe that players can interact with through their characters. The game world initially features two continents of Azeroth (Kalimdor and the Eastern Kingdoms), with two separate expansions adding the realm of Outlands and the continent of Northrend adding to the playable area. In this game world, players use their characters to explore locations, defeat creatures, and complete quests, among other activities. Doing these, characters gain experience that allows them to attain higher levels, gain access to new skills and abilities, explore new areas, and attempt new quests.[31] As a player explores new locations, different routes and means of transportion become available. Players can access "flight masters" in newly discovered locations to fly to previously discovered locations in other parts of the world.[36] Players can also use boats, zeppelins, or portals to move from one continent to another. Although the game world remains reasonably similar from day to day, seasonal events reflecting real world events such as Halloween,[37] Christmas, Children's Week,[34] Easter and Midsummer have been added. Locations also have variable weather including, among other things, rain, snow, and dust storms.[36]

A number of facilities are available for characters while in towns and cities. In each major city characters can access a bank in order to deposit items, such as treasure or crafted items. Each character has access to personal bank storage with the option to purchase additional storage space using in-game gold.[38] Additionally, guild banks are available for use by members of a guild with restrictions being set by the guild leader.[39] Auction houses are also available for players to buy and sell items to others in a similar way to online auction sites such as eBay.[40] Players can also use mailboxes, which can be found in almost every town. The mailbox can be used to collect items won at auction and also to send messages, items and even in-game money to other characters.[26]

Some of the harder challenges in World of Warcraft require players to group together to defeat them. These usually take place in dungeons, also known as instances, that a group of characters can enter together. The term comes from each group or party having a separate copy or instance of the dungeon, complete with their own enemies to defeat and their own treasure or rewards.[41] This allows players to explore areas and complete quests without other players outside the group interfering. Dungeons are spread over the game world and are designed for characters of varying progression. A typical dungeon will allow up to five characters to enter as part of a group. Some dungeons require more players to group together and form a raid of limited size (up to forty players) to face some of the most difficult challenges.[42] As well as dungeon-based raid challenges, several creatures exist in the normal game environment that are designed for raids to attack

The focus of this lesson will involve quests

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Ideas/Examples
Students will download their 10-day trial of the game at the following site listed below:

https://us.battle.net/account/creation/wow/signup/

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Software
)

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Articles that Support this Lesson
Dewey, J. (1916). Demogracy and education: An introduction to the philosophy of education(pp.49-62). New York, NY: The Macmillan Company.

Stigler, J.W., & Hiebert, J. (2009). The teaching gap: Best ideas from the world's teachers for improving education in the classroom (pp.25-54). New York, NY: Free Press.

Gentner, D., & Loewenstein, J., & Thompson, L. (2003). Learning and transfer: A general role for analogical encoding. Journal of Educational Psychology, 95(2), 393-405.

Barab, S., Scott, B., Siyahhan, S., Goldstone, R., Ingram-Goble, A., Zuiker, S., et al. (2009). Transformational play as a cirricular scaffold: Using videogames to support science education. Journal of Science and Educational Technology, 18(4) 305-320.

Gee, J. (2008). Games for learning institute. Retrieved from http://vimeo.com/4513412

LINKS:

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title: Teamwork through World of Warcraft


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American Life vs. Haitian Life

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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Grade Level: High School
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: Sociology Course
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: Comparing and Contrasting American Life vs. Haitian Life
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 1 week
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: WordPress
 * }
 * }

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Workshop Overview

 * California State Standard Addressed:

Historical and Social Sciences Analysis Skills ''The intellectual skills noted below are to be learned through, and applied to, the content standards for grades nine through twelve. They are to be assessed only in conjunction with the content standards in grades nine through twelve. In addition to the standards for grades nine through twelve, students demonstrate the following intellectual, reasoning, reflection, and research skills.'' Chronological and Spatial Thinking

Students relate current events to the physical and human characteristics of places and regions.

Students will carry out this task by comparing and contrasting American life vs. Haitian life. This will include relating current events such as the earthquake in Haiti and the struggling economy in the America. These types of current events effect how Americans and Haitians live their lives.

For this lesson, students will complete the following tasks:

 * 1) Students will accept a WordPress invite from their teacher.
 * 2) Students will address the first question posted on the blog: "In which ways does your American life differ from Haitian life?"
 * 3) Student will share their findings with their classmates.
 * 4) Third, students will address the second question posted on the blog: "In what ways is your American life similar to Haitian Life?"
 * 5) Next, students will compare and contrast their findings in a roundtable discussion.
 * 6) Students will create a class blog and correspond with a Haitian partner school.
 * 7) Students will reflect on the similarities and differences of their lives in their own country and Haiti.

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Before You Get Started
WordPress has grown to be the largest self-hosted blogging tool in the world, used on millions of sites and seen by tens of millions of people every day. WordPress is an open source project, which means there are hundreds of people all over the world working on it. It also means you are free to use it without paying anyone a license fee and a number of other important freedoms. It started as just a blogging system, but has evolved to be used as a full content management system and so much more through the thousands of plugins, widgets, and themes.

For an in-depth tutorial on Wordpress visit this YouTube video:

Wordpress Tutorial

To visit the WordPress website click on the link below:

Visit Wordpress

For an example of a WordPress blog visit the link below:

Example Blog

For information on how to embed a video into Wordpress, follow the link below:

EMBED A VIDEO IN WORDPRESS

For further information on WordPress, visit the following website:

Further Information

''For this assignment, it would be advisable to let the students have access to your admin account, that way only your class can make alterations to the class blog. Those who are not able to log into the admin account can only leave comments. They will not be able to alter any of the content that you post on there as a class.''

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Workshop Plan/Teacher Suggestions

 * It is important to scaffold some lessons on American life and Haitian life in order for students to have some prior knowledge to complete the tasks below. Although they will conduct some research on their own, it is important to provide them with some background information before expecting them to complete all of these tasks.

 * 1) Students will accept a Wordpress invite via email from their teacher. Be sure to let students have access to the admin login and password so that the students can add blogs to the post.  This will ensure that only your students will be allowed to make changes to the actual blogs and not any other outside parties.
 * 2) Students will address the first question posted on the blog: "How does your American life differ from Haitian life?" This can be done by clicking on the comment button beneath the question.  Students will be required to conduct some research. This may be an appropriate time to teach students how to conduct research so that they are finding credible sources.  For assistance on how to navigate the internet to know the difference between credible and not so credible sources, click the following link: Truth  Be sure to have students provide the link for their findings when they respond to the question.  If students need more guidance give them suggestions for what they can talk about. This can include home life, school life, culture, religion, traditions, etc. (i.e. What dining traditions do you have? How is this the same from Haitian dining traditions? How is it different?).  OR... You can have students list 10 things about their everyday lives; that way they can have an idea of where to go in terms of their research.
 * 3) Once the above task is complete, pick 3 or 4 student responses from the blog to share out with the class. Try to pick responses that can spur class discussion and expose students to a variety of perspectives.
 * 4) Next, students will address the second question posted on the blog: "How is your American life similar to Haitian Life?" This question will also require some research.  Links supporting ones findings are also encouraged.  Examples to provide students could be the following: How does my religion guide me through daily life? Does it affect my daily life?  Do religious Haitians follow a religion that affects their daily lives? If so, how is their life similar to mine in religious terms?  Students can also embed videos to support their answers as well (review the Before You Get Started section to learn how to embed videos).
 * 5) Next, students will compare and contrast their findings from their research in a roundtable discussion. Provide students with a list of possible questions that will be asked in order to drive the conversation.  Think socratic seminar format: Socratic Seminar. Although discussion is not around a piece of literature, possible questions may include: What struck you most about your findings?  Does your life differ as much from Haitian life as you expected? If so, how? If not, how? If you lived in Haiti, which part of their everyday life be most difficult for you to adapt to based on your research?
 * 6) As a class, students will create a class blog and correspond with another classroom of students in Haiti to compare and contrast their lives and the lives of Haitian students. The compare and contrast process is done again to see if the research you found is similar or different from what your dialogue with the school is.
 * 7) Reflection; have students address the following sections in a write-up: In what ways was your research on Haitian life different from your findings from your dialogue with the partner school? In your opinion, what is the biggest difference between Haitian life and American life?  Before beginning this lesson, did you anticipate that your life was more different or more similar to Haitian life?

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Articles that Support this Lesson/Additional Resources
Dewey, J. (1916). Demogracy and education: An introduction to the philosophy of education(pp.49-62). New York, NY: The Macmillan Company.

This article supports this lesson in a sense that it gives students to chance to grow on their own and to be reflective. This lesson also connects to this article because it promotes interdependence in a social context. For this lesson students must respond and react to one another's responses on WordPress. WordPress and being given the time to truly reflect, compare, and contrast might be new to most students and will allow them the potential for growth that is described in the article. The lesson also attempts to eliminate habituation, a common habit that occurs in education.

Stigler, J.W., & Hiebert, J. (2009). The teaching gap: Best ideas from the world's teachers for improving education in the classroom (pp.25-54). New York, NY: Free Press.

This article described teaching in Germany, United States, and Japan as the following: "developing advanced procedures" "learning terms and practicing procedures" and "structured problem". This article relates to this lesson in a few ways. This particular attempts to allow students some breathing room to compare and contrast their own way, and to also block out the noise that can occur in the classroom if this assignment were to be carried out in the classroom without the use of WordPress. This article also brought up the importance of structure. Not only is structure important for students in the classroom, it is also important in an online environment.

Gentner, D., & Loewenstein, J., & Thompson, L. (2003). Learning and transfer: A general role for analogical encoding. Journal of Educational Psychology, 95(2), 393-405.

The article tested three studies that claimed that analogical encoding- comparing two instances of a to-be learned principle- is a powerful ways to promote learning. This relates to this particular lesson because this lesson attempts to engage students by having them compare and contrast their own lives to that of others in a real way. By learning how to compare and contrast in this instance, it is possible that students will be able to transfer this compare-contrast schema to another situation that calls for it.

Barab, S., Scott, B., Siyahhan, S., Goldstone, R., Ingram-Goble, A., Zuiker, S., et al. (2009). Transformational play as a cirricular scaffold: Using videogames to support science education. Journal of Science and Educational Technology, 18(4) 305-320.

As educators, we want our students to be able to take the knowledge they gain in the classroom and apply it to the real world. This article mentions that the more control and structure a teacher places on a particular task, the less room there is to learn for the students. This brings up the fine line between too much breathing room and stifling a student's creativity. This lesson attempts to place students in a safe and somewhat controlled environment, but does not dictate them and leave no room for expressing their own opinions.

Gee, J. (2008). Games for learning institute. Retrieved from http://vimeo.com/4513412

This video discussed intentional problems in games that promote intrinsic learning and 21st century skills. This connects to this lesson in the way that this lesson is organized, to illicit deep conceptual understanding and critical thinking skills, which are both important 21st century skills.

http://en.wikiversity.org/wiki/Educational technology

http://en.wikiversity.org/wiki/Fostering Classroom Norms: Using Wikispaces

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title: American Life vs. Haitian Life


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South African Law/Persons/Unborn child

Under Ethiopian Law a child merly concieved may be considered born for certain legal purposes provided it's interest so requires and is born alive and viable(Art. 2 of the 1960 Civil code). The requirments that the merly conceived child's interest requires and it be born alive and viable are cummulative all the time. Its interest is said to be involved if for example, the father of the baby dies while still conceived and the succession of the deceased is opened and other heirs are to take it with out the conceived child's participation. In this case the child is said to be a double loser in a sense he lost his father and, if he is to be excluded from the succession only be cause he in to be a person yet. So to protect the interest of the conceived child from the regours of the principle that only birth gives rise to personality, this exception has be devised by the law. The other requirments are concomitant to one an other: namely birth and viability. For the purpose of birth the child must be born alive and of course it is a prerequisite for the requirment of viability for we cannot talk about viability of a child born with out life. Article 4 of the Civil Code provides the meaning of what viability for the purpose of Art. 2 of the Civil Code. Accordingly, viability is the ability of the child to sustain life for 48 hours after birth. The law says a child is viable if it lived for 48 hours after birth(Art.4/1). And viability of a child who lived for 48 hours is irribattable, meaning no contrary proof is allowed. This even so if the life of the child was supported with an external mechanism with out which it couldn't survive for more than 4 hours, let say due to some desease it has contracted while in the womb. On the other hand if a child dies due to a cause other than its internal constitution, ie due to external factors example the child suffered a head fructure due to an error by the nurse, or it was saffocated due to lack of ventlation or other causes that it might suffer in the hands of any one or causes not attributable to any desease he might have contracted while in the womb.

title: South African Law/Persons/Unborn child


text:
User talk:Mokoia

 Hello Mokoia, 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. --Mu301Bot 08:17, 18 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:17, 18 February 2010 (UTC)

title: User talk:Mokoia