text:
User:Egm6341.s10.team3.sa/HW1

HW Problem Set 1

= Plotting a given function $$ f(x) $$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg2.pdf|p.2-2]]

Problem Statement
$$f(x) = \frac{e^x -1}{x} x \in [0,1] $$

(i) Find $$ \lim_{x\rightarrow 0} f(x)$$ (ii)Plot $$ \displaystyle f(x) $$

Solution
(i) By L'Hospital's rule,
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 * $$\lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0} \frac{e^x -1}{x}$$
 * }
 * }
 * }

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 * $$ = \lim_{x\rightarrow 0} \frac {\frac{d}{dx}(e^x -1)} {\frac {d}{dx}(x)}$$
 * }
 * }
 * }

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 * $$ = \lim_{x\rightarrow 0} {e^x}$$
 * }
 * }
 * }

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 * $$\displaystyle =1 $$
 * }
 * }
 * }

(ii) MATLAB Code to generate the plot

Plot :  $$\frac {e^x -1}{x}$$ vs $$ \displaystyle x $$

= Taylor Series Expansion of $$ f(x)=e^{x} $$= Ref: Lecture notes [[media:Egm6341.s10.mtg2.pdf|p.2-3]]

Problem Statement
$$\displaystyle f(x)=e^{x} $$

Find

(i)$$\displaystyle P_{n}(x) $$ and

(ii)$$\displaystyle R_{n+1}(x) $$

Solution
From [[media:Egm6341.s10.mtg2.pdf|p.2-2]] and [[media:Egm6341.s10.mtg2.pdf|p.2-3]] of the class note, $$\displaystyle P_{n}(x) $$ and $$\displaystyle R_{n+1}(x) $$ are defined by
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P_{n}(x)=f(x_{0})+\frac{(x-x_{0})}{1!}f^{(1)}(x_{0})+\cdots+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0}) $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
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R_{n+1}(x)=\frac{1}{n!}\int_{x_{0}}^{x}(x-t)^{n}f^{(n+1)}(t)dt=\frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}(\xi), \quad \xi\in[x_{0},x] $$ $$ respectively.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Also, note that the following property of $$\displaystyle e^{x} $$:
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\frac{d^{n}}{dx^{n}}\left(e^{x}\right)=e^{x}, \quad n=1, 2, \cdots. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

(i) Let's plug $$\displaystyle f(x)=e^{x} $$ into Eq. 1.
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P_{n}(x)=e^{x_{0}}+\frac{(x-x_{0})}{1!}\frac{d}{dx}\left(e^{x_{0}}\right)+\cdots+\frac{(x-x_{0})^{n}}{n!}\frac{d^{n}}{dx^{n}}\left(e^{x_{0}}\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Now, use Eq. 3 to simplify Eq. 4. Then,
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$$\displaystyle \Rightarrow P_{n}(x)=e^{x_{0}}+\frac{(x-x_{0})}{1!}e^{x_{0}}+\cdots+\frac{(x-x_{0})^{n}}{n!}e^{x_{0}} $$ $$
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 * $$\displaystyle (Eq. 5)
 * }
 * }

(ii) Let's plug $$\displaystyle f(x)=e^{x} $$ into Eq. 2.
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R_{n+1}(x)=\frac{1}{n!}\int_{x_{0}}^{x}(x-t)^{n}\frac{d^{n+1}}{dt^{n+1}}\left(e^{t}\right)dt=\frac{(x-x_{0})^{n+1}}{(n+1)!}\frac{d^{n+1}}{dx^{n+1}}\left(e^{x}\right), \quad x= \xi, \quad \xi\in[x_{0},x] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Now, use Eq. 3 to simplify Eq. 6. Then,
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$$\displaystyle \Rightarrow R_{n+1}(x)=\frac{1}{n!}\int_{x_{0}}^{x}(x-t)^{n}e^{t}dt=\frac{(x-x_{0})^{n+1}}{(n+1)!}e^{\xi}, \quad \xi\in[x_{0},x] $$ $$
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 * $$\displaystyle (Eq. 7)
 * }
 * }

Hence $$ \displaystyle f(x)$$ can be expressed as follows: $$\displaystyle f(x)=e^{x} $$ can be expressed in Taylor series expansion by
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$$\displaystyle f(x)=e^{x}=\underbrace{e^{x_{0}}\sum_{j=0}^{n}\frac{(x-x_{0})^{j}}{j!}}_{P_{n}(x)}+\underbrace{\frac{(x-x_{0})^{n+1}}{(n+1)!}e^{\xi}}_{R_{n+1}(x)}, \quad \xi\in[x_{0},x] $$ $$
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 * $$\displaystyle (Eq. 8)
 * }
 * }

= Plot Functions sin (x),-cos (x)and sin(x) - cos(x) =

Ref. Lecture notes [[media:Egm6341.s10.mtg3.pdf|p.3-3 ]]

Problem Statement
Plot f(x)= sin(x) and g(x) = - cos(x) $$ {\in} [0,{\pi}] $$

and also find $$||f(x)||_{\infin}, ||g(x)||_{\infin} and ||f(x)-g(x)||_{\infin}$$

Solution
Plot f(x)= sin(x) $$ {\in} [0,{\pi}] $$

Matlab code :

Plot :

f(x)=y= sin(x)

Plot g(x)= - cos(x) $$ {\in} [0,{\pi}] $$

Matlab code :

Plot :

g(x)= y = - cos(x)

Plot f(x)-g(x)= sin(x) + cos(x)

Matlab code :

$$||f(x)||_{\infin} = 1$$

$$||g(x)||_{\infin} = 1$$

$$||f(x)-g(x)||_{\infin} = \sqrt{2}$$

= Prove IMVT =

Ref: Lecture Notes [[media:Egm6341.s10.mtg5.pdf|p.5-1]]

Problem Statement
Prove :

(i) $$ \int_{a}^{b}w(x)\cdot f(x)dx = f(\xi )\int_{a}^{b}w(x)dx $$  for   $$ w(x)\geq 0 $$

(ii) Another version of IMVT $$w(x)<0$$ for all $$ x\in [a,b] $$

Solution
(i) $$ m\cdot w(x)\leq f(x)\cdot w(x)\leq M\cdot w(x)$$

where, m:=min f(x) M:=max f(x) m and M are constants

a) Integrating, $$ m\cdot \int_{a}^{b}w(x)dx\leq \int_{a}^{b}w(x)\cdot f(x)dx\leq M\cdot \int_{a}^{b}w(x)dx$$

b) Dividing by $$ \frac{1}{\int_{a}^{b}w(x)dx} $$ $$ m\leq \underbrace{\frac{1}{\int_{a}^{b}w(x)dx}\int_{a}^{b}w(x)\cdot f(x)dx}_{Z} \leq M $$ $$ Z=\frac{1}{\int_{a}^{b}w(x)dx}\int_{a}^{b}w(x)\cdot f(x) dx$$

c) By Intermediate Value Theorem, There exists $$ \xi \in [a,b] at   f(\xi)=Z $$

$$ \therefore f(\xi)=\frac{1}{\int_{a}^{b}w(x)dx}\int_{a}^{b}w(x)\cdot f(x)dx $$

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$$\displaystyle \therefore \int_{a}^{b}w(x)\cdot f(x)dx = f(\xi )\int_{a}^{b}w(x)dx \quad for \quad w(x)\geq 0 $$
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 * }
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(ii) $$ m\cdot \left | w(x) \right |\leq f(x)\cdot \left | w(x) \right |\leq M\cdot \left | w(x) \right | $$

where, m:=min f(x) M:=max f(x) m and M are constants

a) Integrating, $$ m\cdot \int_{a}^{b}\left | w(x) \right |dx\leq \int_{a}^{b}\left | w(x) \right |\cdot f(x)dx\leq M\cdot \int_{a}^{b}\left | w(x) \right |dx $$

b) Dividing by $$ \frac{1}{\int_{a}^{b}\left | w(x) \right |dx} $$ $$ m\leq \underbrace{\frac{1}{\int_{a}^{b}\left | w(x) \right |dx}\int_{a}^{b}\left | w(x) \right |\cdot f(x)dx}_{Z} \leq M $$ $$ Z=\frac{1}{\int_{a}^{b}\left | w(x) \right |dx}\int_{a}^{b}\left | w(x) \right |\cdot f(x)dx $$

c) By Intermediate Value Theorem, There exists $$ \xi \in [a,b] at   f(\xi)=Z $$

d) $$ w(x) $$ is strictly negative, so $$ f(\xi)=\frac{1}{\cancelto{}{(-1)}\cdot\int_{a}^{b}w(x)dx}\cdot\cancelto{}{(-1)}\cdot\int_{a}^{b}w(x)\cdot f(x)dx $$

$$ \therefore f(\xi)=\frac{1}{\int_{a}^{b}w(x)dx}\cdot\int_{a}^{b}w(x)\cdot f(x)dx $$

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$$\displaystyle \therefore \int_{a}^{b}w(x)\cdot f(x)dx = f(\xi )\int_{a}^{b}w(x)dx \quad for \quad w(x)< 0 $$
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= Proof : Taylor series by IMVT = Ref: Lecture Notes [[media:Egm6341.s10.mtg5.pdf|p.5-1]]

Problem Statement
See IMVT in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-3]]

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$$ with $$ w(x)\geqslant 0 \quad \forall \ x \in [a,b] $$
 * $$\displaystyle \int_{a}^{b} w(x)f(x)\ dt = f(\xi)\int_{a}^{b} w(x)\ dt
 * $$\displaystyle \int_{a}^{b} w(x)f(x)\ dt = f(\xi)\int_{a}^{b} w(x)\ dt
 * }

Use IMVT to show that (5) in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-2]] is equal to (1) in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-3]]

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= \frac{1}{n!} \frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi)
 * $$\displaystyle R_n+1(x) = \frac{1}{n!} \int_{x_0}^{x} (x-t)^n f^{(n+1)}(t)\ dt

$$
 * }

for $$\quad \xi \in [x_0,x]$$

Solution
Because $$t \in [x_0,x]$$, thus $$ (x-t)^n \geqslant 0 $$, according to IMVT we have following equations:

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= \frac{1}{n!} g(\xi)\int_{x_0}^{x} w(t)\ dt = \frac{1}{n!} f^{(n+1)}(\xi) \int_{x_0}^{x} (x-t)^n\ dt $$ $$ Int. $$ \int_{x_0}^{x} (x-t)^n\ dt $$, we have
 * $$ \displaystyle R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} \underbrace{(x-t)^n}_{w(t)} \underbrace{f^{(n+1)}(t)}_{g(t)}\ dt
 * $$ \displaystyle R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} \underbrace{(x-t)^n}_{w(t)} \underbrace{f^{(n+1)}(t)}_{g(t)}\ dt
 * $$\displaystyle (Eq. 1)
 * }
 * }

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= \cancelto{o}{-\frac{(x-x)^{n+1}}{n+1}} + \frac{(x-x_0)^{n+1}}{n+1} = \frac{(x-x_0)^{n+1}}{n+1} $$ $$
 * $$ \displaystyle \int_{x_0}^{x} (x-t)^n\ dt = \left [ -\frac{(x-t)^{n+1}}{n+1} \right ]_{t=x_0}^{t=x}
 * $$ \displaystyle \int_{x_0}^{x} (x-t)^n\ dt = \left [ -\frac{(x-t)^{n+1}}{n+1} \right ]_{t=x_0}^{t=x}
 * $$\displaystyle (Eq. 2)
 * }
 * }

Use Eq.1 & Eq.2 we have :

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$$ \displaystyle \Rightarrow  R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} (x-t)^n f^{(n+1)}(t)\ dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi)
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$$
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for $$\quad \xi \in [x_0,x]$$

= Taylor series - Generating higher order terms -Integration by parts = Ref: Lecture Notes [[media:Egm6341.s10.mtg5.pdf|p.5-3]]

Problem Statement

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$$ $$
 * $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\, dt
 * $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\, dt
 * $$\displaystyle (Eq. 1)
 * }
 * }

(i)Repeat int. by parts to reveal the following items with remainder

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 * $$\displaystyle \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0) $$

 * }

(ii)Assume (4)and (5) in [[media:Egm6341.s10.mtg2.pdf|Lecture p.2-2]] are true, do int. by parts one more time

Solution
(i)Int.$$\int_{x_0}^{x} (x-t){f^{(2)}(t)} dt $$ by parts in Eq 1:

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\int_{x_0}^{x} \underbrace{(x-t)}_{U^\prime} \underbrace{f^{(2)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^2}{2}\ f^{(2)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^2}{2}f^{(3)}(t)\, dt}_{W1} \\ & = \cancelto{o}{-\frac{(x-x)^2}{2}\ f^{(2)}(x)} + \frac{(x-x_0)^2}{2}\ f^{(2)}(x_0) + W1\\ & = \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + W1\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (Eq. 2)
 * }
 * }

Use Eq 1, Eq 2, we have:
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$$ $$
 * $$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{1}{2!}\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\, dt
 * $$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{1}{2!}\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\, dt
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Int. $$\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\,dt $$ by part one more time in Eq 3:
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\int_{x_0}^{x} \underbrace{(x-t)^2}_{U^\prime} \underbrace{f^{(3)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^3}{3}\ f^{(3)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^3}{3}f^{(4)}(t)\, dt}_{W2} \\ & = \cancelto{o}{-\frac{(x-x)^3}{3}\ f^{(3)}(x)} + \frac{(x-x_0)^3}{3}\ f^{(3)}(x_0) + W2\\ & = \frac{(x-x_0)^3}{3}\ f^{(3)}(x_0) + W2\\ \end{align} $$ $$ Use Eq 3, Eq 4, we have:
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 *
$$\displaystyle (Eq. 4)
 * }
 * }
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$$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0) + \underbrace{\frac{1}{3!}\int_{x_0}^{x} (x-t)^3f^{(4)}(t)\, dt}_{Remainder} $$
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 * }

(ii)Int. $$\int_{x_0}^{x} {(x-t)^n}f^{(n+1)}(t)\,dt $$ by part one more time:

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\int_{x_0}^{x} \underbrace{(x-t)^n}_{U^\prime} \underbrace{f^{(n+1)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^{n+1}}{n+1}\ f^{(n+1)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^{n+1}}{n+1}f^{(n+2)}(t)\, dt}_{W} \\ & = \cancelto{o}{-\frac{(x-x)^{n+1}}{n+1}\ f^{(n+1)}(x)} + \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0) + W\\ & = \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0) + W\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Use Eq 5, (4)& (5) in [[media:egm6341.s10.p2-2.png|Lecture p.2-2]], we have:
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$$\displaystyle \Rightarrow \ f(x) = \underbrace{f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \cdots + \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0)}_{P_{n+1}(x)} + \underbrace{\frac{1}{(n+1)!}\int_{x_0}^{x} (x-t)^{n+1}f^{(n+2)}(t)\, dt}_{R_{n+2}(x)} $$
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 * }

= Expressing a function $$ f(.)$$ using Taylor Series = Ref: Lecture Notes [[media:Egm6341.s10.mtg6.pdf|p.6-1]]

Problem Statement
$$ f(x)=sin x, x \in [0,\pi] $$ (i)Construct a Taylor series of $$f(x)$$ around $$x_0 = \frac{\pi}{4}$$ for $$\displaystyle n=0,1,2......,10.$$ (ii)Plot the series for each $$\displaystyle n $$ (iii)Estimate the maximum $$\displaystyle R(x)$$ at $$ x=\frac{\pi}{2}$$

Solution
(i) $$\displaystyle n^{th}$$ order polynomial is given by, $$\displaystyle P_n(x) = f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + ........+ \frac{(x-x_0)^n}{n!}f^n(x_0) $$

(a)$$\displaystyle P_n(x)$$ for $$\displaystyle n=0 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$

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 * $$ \displaystyle P_0(x)= f(x_0) $$
 * }
 * }
 * }

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$$\displaystyle P_0(x) = sin (\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$$
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 * }
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(b)$$\displaystyle P_n(x)$$ for $$\displaystyle n=1 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
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 * $$ \displaystyle P_1(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) $$
 * }
 * }
 * }

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$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) $$
 * }
 * }

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$$ \displaystyle P_1(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!}\Bigg] $$
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 * }
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(c)$$\displaystyle P_n(x)$$ for $$\displaystyle n=2 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
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 * $$ \displaystyle P_2(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) $$
 * }
 * }
 * }

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$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4})$$
 * }
 * }

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$$ \displaystyle P_2(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!}\Bigg] $$
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 * }
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(d)$$\displaystyle P_n(x)$$ for $$\displaystyle n=3 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
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 * $$ \displaystyle P_3(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0)$$
 * }
 * }
 * }

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$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) $$
 * }
 * }

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$$ \displaystyle P_3(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} \Bigg] $$
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(e)$$\displaystyle P_n(x)$$ for $$\displaystyle n=4 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
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 * $$ \displaystyle P_4(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) $$
 * }
 * }
 * }

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$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) $$
 * }
 * }

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$$ \displaystyle P_4(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} \Bigg] $$
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 * }
 * }

(f)$$\displaystyle P_n(x)$$ for $$\displaystyle n=5 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
 * {| style="width:100%" border="0" align="left"

 * $$ \displaystyle P_5(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0)$$
 * }
 * }
 * }

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

$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^5}{5!} cos(\frac{\pi}{4}) $$
 * }
 * }

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

$$ \displaystyle P_5(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} \Bigg] $$
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 * }
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(g)$$\displaystyle P_n(x)$$ for $$\displaystyle n=6 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
 * {| style="width:100%" border="0" align="left"

 * $$ \displaystyle P_6(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0)$$
 * }
 * }
 * }

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

$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^5}{5!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^6}{6!} sin(\frac{\pi}{4}) $$
 * }
 * }

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

$$ \displaystyle P_6(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} \Bigg] $$
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(h)$$\displaystyle P_n(x)$$ for $$\displaystyle n=7 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle + \frac{(x-x_0)^7}{7!} f^7(x_0)$$
 * $$ \displaystyle P_7(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * $$ \displaystyle P_7(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * }
 * }

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

$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^5}{5!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^6}{6!} sin(\frac{\pi}{4})$$ $$\displaystyle -\frac{(x-\frac{\pi}{4})^7}{7!} cos(\frac{\pi}{4}) $$
 * }
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 * {| style="width:100%" border="0" align="left"

$$ \displaystyle P_7(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} - \frac{(x-\frac{\pi}{4})^7}{7!}\Bigg] $$
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(i)$$\displaystyle P_n(x)$$ for $$\displaystyle n=8 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
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$$ \displaystyle + \frac{(x-x_0)^7}{7!} f^7(x_0) + \frac{(x-x_0)^8}{8!} f^8(x_0)$$
 * $$ \displaystyle P_8(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * $$ \displaystyle P_8(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * }
 * }

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

$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^5}{5!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^6}{6!} sin(\frac{\pi}{4})$$ $$\displaystyle -\frac{(x-\frac{\pi}{4})^7}{7!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^8}{8!} sin(\frac{\pi}{4}) $$
 * }
 * }

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

$$ \displaystyle P_8(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} - \frac{(x-\frac{\pi}{4})^7}{7!} + \frac{(x-\frac{\pi}{4})^8}{8!}\Bigg] $$
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(j)$$\displaystyle P_n(x)$$ for $$\displaystyle n=9 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle + \frac{(x-x_0)^7}{7!} f^7(x_0) + \frac{(x-x_0)^8}{8!} f^8(x_0) + \frac{(x-x_0)^9}{9!} f^9(x_0)$$
 * $$ \displaystyle P_9(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * $$ \displaystyle P_9(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * }
 * }

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

$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^5}{5!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^6}{6!} sin(\frac{\pi}{4})$$ $$\displaystyle -\frac{(x-\frac{\pi}{4})^7}{7!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^8}{8!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^9}{9!} cos(\frac{\pi}{4}) $$
 * }
 * }

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

$$ \displaystyle P_9(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} - \frac{(x-\frac{\pi}{4})^7}{7!} + \frac{(x-\frac{\pi}{4})^8}{8!} + \frac{(x-\frac{\pi}{4})^9}{9!} \Bigg] $$
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(k)$$\displaystyle P_n(x)$$ for $$\displaystyle n=10 $$ and $$\displaystyle x_0 = \frac{\pi}{4}$$
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$$ \displaystyle + \frac{(x-x_0)^7}{7!} f^7(x_0) + \frac{(x-x_0)^8}{8!} f^8(x_0) + \frac{(x-x_0)^9}{9!} f^9(x_0) + \frac{(x-x_0)^{10}}{10!} f^{10}(x_0)$$
 * $$ \displaystyle P_{10}(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * $$ \displaystyle P_{10}(x)= f(x_0) + \frac{(x-x_0)}{1!} f^1(x_0) + \frac{(x-x_0)^2}{2!} f^2(x_0) + \frac{(x-x_0)^3}{3!} f^3(x_0) +  \frac{(x-x_0)^4}{4!} f^4(x_0) + \frac{(x-x_0)^5}{5!} f^5(x_0) + \frac{(x-x_0)^6}{6!} f^6(x_0) $$
 * }
 * }

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

$$ \displaystyle = sin (\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})}{1!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^2}{2!} sin(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^3}{3!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^4}{4!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^5}{5!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^6}{6!} sin(\frac{\pi}{4})$$ $$\displaystyle -\frac{(x-\frac{\pi}{4})^7}{7!} cos(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^8}{8!} sin(\frac{\pi}{4}) + \frac{(x-\frac{\pi}{4})^9}{9!} cos(\frac{\pi}{4}) - \frac{(x-\frac{\pi}{4})^{10}}{10!} sin(\frac{\pi}{4}) $$
 * }
 * }

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

$$ \displaystyle P_{10}(x)= \frac{1}{\sqrt{2}}\Bigg[1 + \frac{(x-\frac{\pi}{4})}{1!} - \frac{(x-\frac{\pi}{4})^2}{2!} -\frac{(x-\frac{\pi}{4})^3}{3!} + \frac{(x-\frac{\pi}{4})^4}{4!} + \frac{(x-\frac{\pi}{4})^5}{5!} - \frac{(x-\frac{\pi}{4})^6}{6!} - \frac{(x-\frac{\pi}{4})^7}{7!} + \frac{(x-\frac{\pi}{4})^8}{8!} + \frac{(x-\frac{\pi}{4})^9}{9!} - \frac{(x-\frac{\pi}{4})^{10}}{10!}\Bigg] $$
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(ii) MATLAB Code for generating the plots for :     $$ \displaystyle n=0,1......,10 $$

Plot: $$\displaystyle f(x)=sin(x)$$ for $$\displaystyle n = 0, 1, 2, 3, 4 and 5$$

$$\displaystyle f(x)=sin(x)$$ for $$\displaystyle n = 6, 7, 8, 9 and 10$$

(iii)Error Estimation $$\displaystyle R_{n+1}(x) = \frac{(x-x_0)^{n+1}}{(n+1)!} f^{n+1}(\xi)$$ $$\displaystyle max R_{n+1}(x) = \frac{(x-x_0)^{n+1}}{(n+1)!} max \bigg| f^{n+1}(\xi)\bigg|$$

Since $$\displaystyle f(x) = sin (x)$$  $$\displaystyle max \bigg| f^{n+1}(\xi)\bigg| = 1  $$ for all $$\displaystyle n $$

(a)$$\displaystyle R_{n+1}(x)$$ for $$\displaystyle n=0 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{n+1}(x)_{max}= \frac{x-x_0}{1!} $$
 * }
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$$ \displaystyle = \frac{\frac{\pi}{2}-\frac{\pi}{4}}{1!} $$
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$$\displaystyle R_1(x)_{max} = 0.6168 $$
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(b)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=1 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{2}(x)_{max}= \frac{(x-x_0)^2}{2!}  $$
 * }
 * }
 * }

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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^2}{2!} $$
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$$\displaystyle R_2(x)_{max} = 0.3084 $$
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(c)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=2 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{3}(x)_{max}= \frac{(x-x_0)^3}{3!}  $$
 * }
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^3}{3!} $$
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$$\displaystyle R_3(x)_{max} = 0.0807 $$
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(d)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=3 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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 * $$ \displaystyle R_{4}(x)_{max}= \frac{(x-x_0)^4}{4!}  $$
 * }
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^4}{4!} $$
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$$\displaystyle R_4(x)_{max} = 0.0158 $$
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(e)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=4 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{5}(x)_{max}= \frac{(x-x_0)^5}{5!}  $$
 * }
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^5}{5!} $$
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$$\displaystyle R_5(x)_{max} = 2.4903 * 10^{-3} $$
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(f)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=5 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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 * $$ \displaystyle R_{6}(x)_{max}= \frac{(x-x_0)^6}{6!}  $$
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^6}{6!} $$
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$$\displaystyle R_6(x)_{max} = 3.2599 * 10^{-4}$$
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(g)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=6 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{7}(x)_{max}= \frac{(x-x_0)^7}{7!}  $$
 * }
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 * }

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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^7}{7!} $$
 * }
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$$\displaystyle R_7(x)_{max} = 3.6576 * 10^{-5} $$
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(h)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=7 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{8}(x)_{max}= \frac{(x-x_0)^8}{8!}  $$
 * }
 * }
 * }

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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^8}{8!} $$
 * }
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$$\displaystyle R_8(x)_{max} = 3.5908 * 10^{-6} $$
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(i)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=8 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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 * $$ \displaystyle R_{9}(x)_{max}= \frac{(x-x_0)^9}{9!}  $$
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^9}{9!} $$
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$$\displaystyle R_9(x)_{max} = 3.1336 * 10^{-7}$$
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(j)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=9 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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

 * $$ \displaystyle R_{10}(x)_{max}= \frac{(x-x_0)^{10}}{10!}  $$
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^{10}}{10!} $$
 * }
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$$\displaystyle R_{10}(x)_{max} = 2.4611 * 10^{-8} $$
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(k)$$\displaystyle R_{n+1}(x)_{max}$$ for $$\displaystyle n=10 $$ ,$$\displaystyle x_0 = \frac{\pi}{4} $$ and $$\displaystyle x = \frac{\pi}{2} $$

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 * $$ \displaystyle R_{11}(x)_{max}= \frac{(x-x_0)^{11}}{11!}  $$
 * }
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$$ \displaystyle = \frac{(\frac{\pi}{2}-\frac{\pi}{4})^{11}}{11!} $$
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$$\displaystyle R_{11}(x)_{max} = 1.7572 * 10^{-9} $$
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= Taylor, Trapezoidal and Simpson Rule =

Ref: Lecture Notes [[media:Egm6341.s10.mtg6.pdf|p.6-5]]

Problem Statement
$$ I=\int_{0}^{1}\frac{e^{x}-1}{x}dx $$ Use (i) Taylor series expansion fn (ii) Composite Trapezoidal rule (iii) Composite Simpson rule to find $$\displaystyle I_n $$ for n=2,4,8.......... until error is of order $$\displaystyle 10^{-6} $$

Solution
(i)Taylor series expansion The approximate value of the integral in Taylor series is given by, $$\displaystyle \sum_{j=1}^{n}\frac{1}{j! \cdot j}$$

The error range in Taylor series is calculated by, $$ \frac{1}{(n+1)!\cdot (n+1)}\leq I-I_{n}\leq \frac{e}{(n+1)!\cdot(n+1))}$$

a) $$I_{n}$$ for n=2, $$ I_{2}=\left ( \frac{1}{(1!\cdot 1)}+\frac{1}{(2!\cdot 2)} \right )=1.25 $$ a-1) Error for n=2, $$ \frac{1}{(2+1)!\cdot (2+1)}\leq I-I_{2}\leq \frac{e}{(2+1)!\cdot(2+1))}=0.055556\leq I-I_{2}\leq0.151016 $$

b) $$I_{n}$$ for n=4, $$ I_{4}=\left ( \frac{1}{(1!\cdot 1)}+\frac{1}{(2!\cdot 2)}+\frac{1}{(3!\cdot 3)}+\frac{1}{(4!\cdot 4)} \right )=1.3159722222223\approx 1.31597 $$ b-1) Error for n=4, $$\frac{1}{(4+1)!\cdot (4+1)}\leq I-I_{4}\leq \frac{e}{(4+1)!\cdot(4+1))}=0.001667\leq I-I_{4}\leq0.00453$$

c)$$I_{n}$$ for n=8, $$ I_{8}=\left ( \frac{1}{(1!\cdot 1)}+\frac{1}{(2!\cdot 2)}+\frac{1}{(3!\cdot 3)}+\frac{1}{(4!\cdot 4)}+\frac{1}{(5!\cdot 5)}+\frac{1}{(6!\cdot 6)}+\frac{1}{(7!\cdot 7)}+\frac{1}{(8!\cdot 8)}  \right )=1.3179018152401 \approx 1.3179 $$ c-1) Error for n=8, $$ \frac{1}{(8+1)!\cdot (8+1)}\leq I-I_{8}\leq \frac{e}{(8+1)!\cdot(8+1))}=3.06192\times 10^{-7}\leq I-I_{8}\leq8.32317\times 10^{-7} $$

$$\therefore I_n = 1.3179$$ Thus n=8 brings the error down to the order of $$10^{-6}$$

(ii) Composite Trapezoidal rule or the Corrected Trapezoidal rule It is hard to define the true value of I. Thus, MATLAB codes were used to generate the true value of I.

MATLAB Code to find the true value of I

True value of I = 1.317902151956861

a) The Composite Trapezoidal rule is given by $$ \int_{a}^{b}f(x)dx=\frac{b-a}{2n}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})] $$

b) The Corrected Trapezoidal rule is given by $$ \int_{a}^{b}f(x)dx=\frac{b-a}{2n}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})]-\frac{h^2}{12}*[f^{'}(b)-f^{'}(a)] $$

where
 $$ h=\frac{(b-a)}{n} $$

c) MATLAB Code for n=2

d) Results of Composite Trapezoidal rule

where $$\therefore I_{n} = 1.317904695 $$
En = The true value of I (=1.317902152) - In= -2.54263*10-6

Thus, the value of n=128 brings the error down to the order of 10-6 order.

(iii) Composite Simpson rule It is also hard to find the true value of I. Thus, MATLAB codes were used

a) The Composite Simpson rule is given by $$ \int_{a}^{b}f(x)dx=\frac{b-a}{3n}*[f(x_{0})+4f(x_{1})+2f(x_{2})+.........+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})] $$

b) MATLAB Code - Simpson's Rule

c) MATLAB Code for n=2

d) Results of Composite Simpson rule

where $$\therefore I_{n} = 1.317908917 $$
En = The true value of I (=1.317902152) - In= -6.76473*10-6 Thus, the value of n=4 brings the error down to the order of 10-6 order.

= Taylor Series Expansion - $$ f(x) = \frac{e^{x}-1}{x}=\frac{1}{x}\left[e^{x}-1\right] $$  with remainder = Ref: Lecture notes [[media:Egm6341.s10.mtg7.pdf|p.7-1]]

Problem Statement

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

f(x) = \frac{e^{x}-1}{x}=\frac{1}{x}\left[e^{x}-1\right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

(i)Expand $$\displaystyle e^{x} $$ in Taylor series with remainder,
 * {| style="width:100%" border="0" align="left"

R(x) = \frac{(x-0)^{n+1}}{(n+1)!}e^{\xi(x)}. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

(ii) Find Taylor series expansion and remainder of $$\displaystyle f(x) $$ to get (4) [[media:Egm6341.s10.mtg6.pdf|p.6-3]],
 * {| style="width:100%" border="0" align="left"

f(x)-f_{n}(x)=R_{n}(x)=\frac{(x-0)^{n}}{(n+1)!}e^{\xi(x)},\quad \xi\in[0,x]. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution
First, remember the Taylor series expansion for a function $$\displaystyle f(x) $$ around $$\displaystyle x=x_{0} $$:
 * {| style="width:100%" border="0" align="left"

f(x)=P_{n}(x)+R_{n+1}(x), $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_{n}(x)=f(x_{0})+\frac{(x-x_{0})}{1!}f^{(1)}(x_{0})+\cdots+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0}) $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

R_{n+1}(x)=\frac{1}{n!}\int_{x_{0}}^{x}(x-t)^{n}f^{(n+1)}(t)dt=\frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}(\xi), \quad \xi\in[x_{0},x] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

 (i) Expand $$\displaystyle f(x)=e^{x} $$ in Taylor series around $$\displaystyle x=x_{0}=0 $$. Recall that $$\displaystyle \frac{d^{n}}{dx^{n}}\left(e^{x}\right)=e^{x}, \quad n=1, 2, \cdots. $$
 * {| style="width:100%" border="0" align="left"

P_{n}(x)=e^{0}+\frac{(x-0)}{1!}e^{0}+\frac{(x-0)^{2}}{2!}e^{0}+\cdots+\frac{(x-0)^{n}}{n!}e^{0} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Rightarrow P_{n}(x)=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\cdots+\frac{x^{n}}{n!}=\sum_{j=0}^{n}\frac{x^{j}}{j!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

R_{n+1}(x)=\frac{(x-0)^{n+1}}{(n+1)!}e^{\xi}=\frac{x^{n+1}}{(n+1)!}e^{\xi} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

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

$$\displaystyle e^{x}=P_{n}(x)+R_{n+1}(x)=\sum_{j=0}^{n}\frac{x^{j}}{j!}+\underbrace{\frac{x^{n+1}}{(n+1)!}e^{\xi}}_{remainder} $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 *
$$\displaystyle (Eq. 6)
 * }
 * }

 (ii) Expand $$\displaystyle f(x)=\frac{e^{x}-1}{x}=\frac{1}{x}\left[e^{x}-1\right] $$ in Taylor series using Eq.6.
 * {| style="width:100%" border="0" align="left"

\frac{1}{x}\left[e^{x}-1\right]=\frac{1}{x}\left[\cancelto{0}{1}+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}e^{\xi}\cancelto{0}{-1}\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Rightarrow\frac{1}{x}\left[e^{x}-1\right]=\frac{1}{x}\left[\sum_{j=1}^{n}\frac{x^{j}}{j!}+\frac{x^{n+1}}{(n+1)!}e^{\xi}\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Rightarrow\frac{1}{x}\left[e^{x}-1\right]=\frac{1}{x}\sum_{j=1}^{n}\frac{x^{j}}{j!}+\frac{1}{x}\frac{x^{n+1}}{(n+1)!}e^{\xi} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Rightarrow\frac{1}{x}\left[e^{x}-1\right]=\sum_{j=1}^{n}\frac{x^{j-1}}{j!}+\frac{x^n}{(n+1)!}e^{\xi} $$ $$ Therefore,
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x)=\underbrace{\sum_{j=1}^{n}\frac{x^{j-1}}{j!}}_{f_{n}(x)=P_{n-1}(x)}+\underbrace{\frac{x^n}{(n+1)!}e^{\xi}}_{R_{n}(x)} $$ $$ Let's move the first term on RHS of Eq.11 to LHS. Then,
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle f(x)-f_{n}(x)=R_{n}(x)=\frac{x^{n}}{(n+1)!}e^{\xi} $$ $$ As Eq.12 shows, our final result is consistent with (4) [[media:Egm6341.s10.mtg6.pdf|p.6-3]] of our lecture note.
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 *
$$\displaystyle (Eq. 12)
 * }
 * }

= Prove Simple trapezoidal rule =

Ref. Lecture notes [[media:Egm6341.s10.mtg8.pdf|p.8-2]]

Problem Statement
Use (2) in Slide [[media:Egm6341.s10.mtg8.pdf|8-2]] to obtain Simple trapezoidal rule.

Solution
Given

$$I_1 =\Big[\int_a^b l_{\text{o}} (x)dx\Big] f(x_{\text{o}})$$ + $$\Big[\int_a^b l_1 (x)dx\Big] f(x_1)$$

where

$$l_{\text{o}}(x) = \frac{x - x_1} {(x_{\text{o}} - x_1)}$$,

$$l_1 (x) = \frac {x - x_{\text{o}}} {(x_1 - x_{\text{o}})}$$,

$$ x_{\text{o}} = a $$

$$ x_1 = b $$

Now $$  I_1 = \Bigg[\int_a^b \frac{x - x_1} {x_{\text{o}} - x_1} dx\Bigg] f(x_{\text{o}})$$ + $$\Bigg[\int_a^b \frac{x - x_{\text{o}}} {x_1 - x_{\text{o}}} dx\Bigg]  f(x_1)$$

= $$\Bigg[\frac{x^2 /2 - bx} {a - b}\Bigg]_a^b  f(a) + \Bigg[\frac{x^2 /2 - ax} {b - a}\Bigg]_a^b  f(b)$$

= $$\Bigg[\frac{(\frac{b^2}{2} - b^2)- (\frac{a^2} {2} - ab)} {(a - b)}\Bigg] f(a) + \Bigg[\frac{(\frac{b^2}{2} - ab)- (a^2 /2 - a^2)} {(b - a)}\Bigg] f(b)$$

= $$\Bigg[\frac{(b-a)^2}{2(b - a)}\Bigg] f(a)+ \Bigg[\frac{(b-a)^2}{2(b - a)}\Bigg] f(b)$$

= $$\Bigg[\frac{(b-a)}{2}\Bigg] (f(a) + f(b))$$

which is same as equation (1) Slide [[media:Egm6341.s10.mtg7.pdf|p.7-1]]

This completes the Proof of simple trapezoidal rule

= Expansion of Lagrange functions =

Ref: Lecture notes [[media:Egm6341.s10.mtg8.pdf|p.8-3]]

Problem Statement
Expand(4) from Slide [[media:Egm6341.s10.mtg8.pdf|8-3]] to obtain $$P_2 (x_j)= \sum_{i=0}^2 l_i(x_j) f(x_i) = f(x_j)$$

Solution
$$P_2 (x_j)= \sum_{i=0}^2 l_i(x_j) f(x_i)= l_{\text{o}}(x_j)f(x_{\text{o}})+ l_1(x_j)f(x_1)+ l_2 (x_j)f(x_2)$$

where  $$ j = 0 ,1 ,2 $$

but when $$ i=j $$ $$ l_i(x_j) = 1 $$

and when i$$ \ne  $$j $$ l_i(x_j) = 0 $$

then only surviving terms are given by

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

 * $$\displaystyle P_2 (x_j)= l_{\text{o}}(x_{\text{o}}) f(x_{\text{o}}) + l_1 (x_1)  f(x_1)+ l_2 (x_2)  f(x_2) = f(x_{\text{o}}) + f(x_1) + f(x_2)=f (x_j) $$
 * }

= Contributing Team Members = Subramanian Annamalai 06:21, 27 January 2010 (UTC)

Yong Nam Ahn 07:28, 27 January 2010 (UTC)

Min Zhong 10:31, 27 January 2010 (UTC)

Abhishek k singh 16:25, 27 January 2010 (UTC)

Heejun Chung 19:05, 27 January 2010 (UTC)

=References= 1. Introduction to Numerical Methods,2nd Edition by Kendall E. Atkinson, Tata Mc-Graw Hill Publications, 2008 2. Numerical Methods for Engineers,5th Edition by Steven C Chapra and Raymond P Canale, Wiley Publications,1989.

title: User:Egm6341.s10.team3.sa/HW1


text:
User:Danaemerritt/pagetitle

Character study Week 1 English Composition Danae Merritt

His original personality was that of a kind and simple man, and in general, that’s what he wanted. Possessing a sweet disposition, his only goal was to be special. But not just any kind special, he had to be predominant and famous. The problem was: it wasn’t his goal. His mother, more accurately, his adopted mother, had drummed into him that he had to be special because she was not. His mother was a highly religious agoraphobe, meaning she was afraid to leave her house. As his unbeknownst to him, adopted, father having walked out on him and his mother at a young age presumably, and his mother’s already fragile state of mind suffered for it. He was the one to reap the unfortunate benefits of his mother’s troubled psyche. Being as he was a young child, and therefore highly impressionable, his life lessons were not happy or pretty ones. A sad lesson it was, to believe that he was not good enough as is, so his drive to be more, to be special was given, or ingrained, at a young age. He, however, was content with being a simple mechanic in New York, following in the footsteps of his absent father. Although, he seemed to have an unnatural gift for his job, whereas his father’s skill was merely that of a trained man. By simply listening to a mechanic object, he could hear the faults in it instantly. On top of that, he could tell, from sound alone, how fast or slow the object was running. This was highly useful in his simple say to day life. What he did not yet know was how much he didn’t know. His mind had, on reflex, locked away memories, repressing them. Thos were the ones regarding his birth parents and just how much he remembered at an even younger age than his adopted father had left him. So, all in all, he was psychologically damaged from the start. While his mother knew he was adopted, as well she should, she never told him, preferring to make herself happy by keeping her special son to herself than tell him the truth. Her own twisted version of ‘happiness’ was what screwed with his future, binding him to her, perhaps out of guilt, but definitely pity. His thoughts regarding his mother were totally unconscious, his free will had been robbed, and while he thought independently, he didn’t know why. Make no mistake, he was not simple, retarded, or mentally challenged. Quite the opposite. He possessed a near genius mind that, granted in his current occupation went to waste, could astound many with his knowledge and memory capacity.

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

title: User:Danaemerritt/pagetitle


text:
User:Egm6341.s10.team3.heejun

= (4) Prove IMVT=

Problem Statement
(i) $$ \int_{a}^{b}w(x)\cdot f(x)dx = f(\xi )\int_{a}^{b}w(x)dx $$  for   $$ w(x)\geq 0 $$  (ii) Another version of IMVT $$w(x)<0$$ for all $$ x\in [a,b] $$

 Ref: Lecture Notes [[media:Egm6341.s10.mtg5.pdf|p.5-1]]

Solution
 Ref: Lecture Notes [[media:Egm6341.s10.mtg2.pdf|p.2-3]]  (i)

$$ m\cdot w(x)\leq f(x)\cdot w(x)\leq M\cdot w(x)$$

 where,

m:=min f(x)

M:=max f(x)

 a) Integrating,

$$ m\cdot \int_{a}^{b}w(x)dx\leq \int_{a}^{b}w(x)\cdot f(x)dx\leq M\cdot \int_{a}^{b}w(x)dx$$

 b) Dividing by $$ \frac{1}{\int_{a}^{b}w(x)dx} $$ $$ m\leq \underbrace{\frac{1}{\int_{a}^{b}w(x)dx}\int_{a}^{b}w(x)\cdot f(x)dx}_{Z} \leq M $$

$$ Z=\frac{1}{\int_{a}^{b}w(x)dx}\int_{a}^{b}w(x)\cdot f(x) dx$$

 c) By Interm. Value Them,

There exists $$ \xi \in [a,b] at   f(\xi)=Z $$

 $$ \therefore f(\xi)=\frac{1}{\int_{a}^{b}w(x)dx}\int_{a}^{b}w(x)\cdot f(x)dx $$

(ii)

$$ m\cdot \left | w(x) \right |\leq f(x)\cdot \left | w(x) \right |\leq M\cdot \left | w(x) \right | $$

 where,

m:=min f(x)

M:=max f(x)

<br\> a) Integrating, $$ m\cdot \int_{a}^{b}\left | w(x) \right |dx\leq \int_{a}^{b}\left | w(x) \right |\cdot f(x)dx\leq M\cdot \int_{a}^{b}\left | w(x) \right |dx $$

<br\> b) Dividing by $$ \frac{1}{\int_{a}^{b}\left | w(x) \right |dx} $$ $$ m\leq \underbrace{\frac{1}{\int_{a}^{b}\left | w(x) \right |dx}\int_{a}^{b}\left | w(x) \right |\cdot f(x)dx}_{Z} \leq M $$

$$ Z=\frac{1}{\int_{a}^{b}\left | w(x) \right |dx}\int_{a}^{b}\left | w(x) \right |\cdot f(x)dx $$

<br\> c) By Interm. Value Them,

There exists $$ \xi \in [a,b] at   f(\xi)=Z $$

<br\> d) $$ w(x) $$ is strictly negative, so

$$ f(\xi)=\frac{1}{\cancelto{}{(-1)}\cdot\int_{a}^{b}w(x)dx}\cdot\cancelto{}{(-1)}\cdot\int_{a}^{b}w(x)\cdot f(x)dx $$ <br\> $$ \therefore f(\xi)=\frac{1}{\int_{a}^{b}w(x)dx}\cdot\int_{a}^{b}w(x)\cdot f(x)dx $$ <br\>

= (8) Taylor, Trap and Simpson Rule=

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

(i) Taylor series exp. fn<br\> (ii) Comp. Trap. rule <br\> (iii) Comp. Simpson rule <br\> <br\> Ref: Lecture Notes [[media:Egm6341.s10.mtg6.pdf|p.6-5]] <br\>

Solution
(i)<br\> a) n=2,

$$ I_{2}=\left ( \frac{1}{(1!\cdot 1)}+\frac{1}{(2!\cdot 2)} \right )=1.25 $$

a-1) Error for n=2,

$$ \frac{1}{(2+1)!\cdot (2+1)}\leq I-I_{2}\leq \frac{e}{(2+1)!\cdot(2+1))}=0.055556\leq I-I_{2}\leq0.151016 $$

<br\> <br\> b) n=4,

$$ I_{4}=\left ( \frac{1}{(1!\cdot 1)}+\frac{1}{(2!\cdot 2)}+\frac{1}{(3!\cdot 3)}+\frac{1}{(4!\cdot 4)} \right )=1.3159722222223\approx 1.31597 $$

b-1) Error for n=4,

$$\frac{1}{(4+1)!\cdot (4+1)}\leq I-I_{2}\leq \frac{e}{(4+1)!\cdot(4+1))}=0.001667\leq I-I_{2}\leq0.00453$$

<br\> <br\> c) n=8,

$$ I_{4}=\left ( \frac{1}{(1!\cdot 1)}+\frac{1}{(2!\cdot 2)}+\frac{1}{(3!\cdot 3)}+\frac{1}{(4!\cdot 4)}+\frac{1}{(5!\cdot 5)}+\frac{1}{(6!\cdot 6)}+\frac{1}{(7!\cdot 7)}+\frac{1}{(8!\cdot 8)} \right )=1.3179018152401 \approx 1.3179 $$

c-1) Error for n=8,

$$ \frac{1}{(8+1)!\cdot (8+1)}\leq I-I_{2}\leq \frac{e}{(8+1)!\cdot(8+1))}=3.06192\times 10^{-7}\leq I-I_{2}\leq8.32317\times 10^{-7} $$

<br\>

(ii)<br\> This one can be calculated by '''the Comp. Trap. rule or the Corrected Trap. rule but it is hard to define the first term, X0'''.

Thus, some math-codes are required. <br\>

=Find the true value of I= MATLAB Code

<br\>

a) the Comp. Trap. rule

$$ \int_{a}^{b}f(x)dx=\frac{b-a}{2n}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})] $$

b) the Corrected Trap. rule

$$ \int_{a}^{b}f(x)dx=\frac{b-a}{2n}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})]-\frac{h^2}{12}*[f^{'}(b)-f^{'}(a)] $$

Where

<li> $$ h=\frac{(b-a)}{n} $$

<br\> <br\>

c) Results of Comp. Trap. rule

Where

<li>En = The true value of I (=1.317902152) - In

<br\> Finally, the value of n=128 is closer to the true value with about 10-6 order.

<br\> d) Example of MATLAB Code (n=2)

<br\>

(iii)<br\> This one can be calculated by '''the Comp. Simpson rule''' but it is also hard to define the first term, X0.

Thus, some math-codes are required.

<br\> <br\>

a) the Comp. Simpson rule

$$ \int_{a}^{b}f(x)dx=\frac{b-a}{3n}*[f(x_{0})+4f(x_{1})+2f(x_{2})+.........+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})] $$

<br\> <br\>

b) Results of Comp. Simpson rule

Where

<li> En = The true value of I (=1.317902152) - In

<br\> Finally, the value of n=4 is closer to the true value with about 10-6 order.

<br\> c) MATLAB Source Code <br\>

<br\> d) Example of MATLAB Code (Run, n=2) <br\>

--Heejun Chung 17:43, 27 January 2010 (UTC)

title: User:Egm6341.s10.team3.heejun


text:
User:Jbennett

I created this account to help with the Technical Writing pages. I've been a tech writer since 1987, and I hope I've picked up some knowledge to pass on.

The things I'd like to emphasize are simplicity, standards, and pride in the profession. Tech writers are important. We help satisfy customers by writing good instructions. We help businesses comply with government regulations. In many ways, we serve as the memory for a corporation or organization.

I'd also like to emphasize that we should serve as advocates for our readers.

John Bennett Wimberley, Texas

title: User:Jbennett


text:
User talk:Thejasonhaddadshow1

== ''Bold tex ==

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[Hide] [Help us with translations!] User talk:JasonHaddad From Wikimedia Commons, the free media repository Jump to: navigation, search Dear Rocket000, Here is Something for you Contents [hide] 1 Here is a New Inappropriate Adult Content or Language Movie (uncut) Show (edited) 2 Coming 3 yeah 4 Texting 5 Link title 6 CNN's HLNNews/FOX/Syndicated/Customized: Alvin and the Chipmunks Rated TV-MA (Redirected from FOX/Syndicated/Customized: Alvin and the Chipmunks Rated X) [edit] Here is a New Inappropriate Adult Content or Language Movie (uncut) Show (edited) "Kick My A×× and B×××s Sh×× Up Alvin"! Watch Out With For Adult Content: Who Sometimes E*clipit NOT FOR KIDS or 18+ Profanity/Language: Shut Up the 'S-H' Word Crap the 'C-R-A' Word Poop The 'P-O-O-P' Word B×lls the 'B' Word B*stard the 'B' Word and Lot of Bad Stuff Violence: THE ADULT ONES ARE BAD [edit] Coming on 2/14/2010. thank you Ass the 'A' Word and Assh*le the 'A-H' Word Can't is not a bad word but T*ts The 'T-I' Word Christmas is A GOOD WORD [edit] yeah these are the bad words and s*xy [edit] Texting Bold text This article is part of CNNews, your source for up-to-the-microsecond misinformation. was on Adult Channels, Heck No! it was 18 (21 in some areas and requires with wives or husbands). This article may not be suitable for all audiences. This article contains content not suitable for people with the brain of a person under 14 inches (35.5cm) tall or TransVestites. NOT SAFE FOR WORK!! The article you are looking at may not be work safe! ...Although, since you're already here it's kind of too late for this warning to actually be useful. If a boss or coworker sees this article, claim that it was spam and blame the IT guys. Otherwise, continue to read it until your lewd urges are satisfied.oldest rating is nc17 24 November 2007 TV-14-S (edited) TV-MA TV-ADULT (uncut) sometimes on dls. was on G×× Pride Pinky Dinky Doo and Tyler the one is younger guinea pig is youngest she is the oldest because of 7 year old girl. Shout Alvin and the Chipmunks have Beers, wines, Alcohal and Drug and Knife and Violent Refrences Lot of People gets the X rating and is no longer considered safe to 17- yr old kids' developing minds.UNDER 18 IS EXIT AND OVER 18 IS ENTER Why Alvin and The Chipmunks bad for kids was G all ages with cartoons and PG no cartoons but 9 and under needs a parent for PG. Breaking news! ALVIN, New York City, New York --The popular children's showChipmunks has been given an adults only|Rated R rating by the Board of Family Friendly Ratings Safety Commission. It has given the long-running show this new rating after an intense and through review, the Board has announced yesterday. FOX, Syndicated, Sesame Workshop, CNN's HLNNews http://images1.wikia.nocookie.net/uncyclopedia/images/c/c9/Big_bird_pothead.JPG The camera man offers to light up Big Bird's pot. Sesame Street bad for kids? We report, you decide!Its website, http://www.sesameworkshop.com (along with all its directories and pages) were also blocked at school and replaced with a shock site. A producer for the show, Jason Haddad, said that “Not until on Breaking R×tarded!” Not only has the aging of our target audience caused Sesame Street to change every year to allow new viewers. But this has also forced us to change the content of the show to meet society's standards. This new content contains the following: Violence: E×clipit Violence Alcohol and drug references Blood (lotsa lotsa blood!) Cookies Big Bird smoking pot and having delusions of a character named "Snuffleupagus" Sumo (NSFW) Oscar the Grouch's bipolar disorder Attmepts to eat disorders Cookie Monster's diabetes Smoking and eating pipes Ernie and Bert's h×mos×xuality Fatties (NSFW) Pedophilia (Especially an episode in which Bert is revealed to be an online predator) The Count's chronic masturbation Condoms Teenage pregnancy Urination fetishes Severe diarrhea "Fuzzy and Blue" supremacists: racism "No Dogs and Birds" allowed: discrimination "Girls can't play b××× and can't think too": s××ism Most popular episode #98765- Race war or Genocide. "The show could no longer suit the needs of today’s preschool child," said the producer "So we had to change the content of the show to meet the needs to modern3-7year-olds' needs."

Bert and Ernie are expected to come out this season.It was reported that the SesameWorkshopcharacters formed their own gang, putJamie Foxx out of business and possibly murdered him as well during the Battle of Chipmunks The first episode this season, entitled "Trapped in the Closet", will take place in Seattle, Wahington where Bert and Ernie take part in a g×× pride parade, revealing their h×mos×xuality. was 1°C and 0°F "We've been wanting to do that one for years," says Frank Oz, the longtime voice of Bert. "But SW and the parents would have sued our a××es a long time ago." This file has annotations. Move the mouse pointer over the image to see them. on Valintine's (M without Silence) Day it is the NEW "Kick My A×× and B×××s Sh×× Up" it's like meatballs or basketballs something like that instead of saying balls It has also been announced by the studio in charge of the production has also contacted "Fisher Price" to include profanity in the new "Kick My Ass Elmo" just in time for the Christmas holiday, and the popular toy will now contain phrases such as "go eat sh×× sl×××y" and "Dad's a ×××××××× q×eer and you know it!!" and "oh my g×d" the clean version was "Oh My Gosh" instead of God. "We hope this new show can meet the needs of modern society unlike our previously outdated piece-of-s*** show" said Henson. [edit] New Movie Coming Out Not only has Alvin and the Chipmunks been giving a rating not suitable for kids, but it has a new movie coming out called "Alvin and the Chipmunks: The Movie" is rated PG-13 or above that also has stuff not suitable for kids under 16-17 (18+ without), like blood, hate, farts and burps, h×mos×xuality and s×× It is unknown when this will be released. PG-13 was 13 and up off 15- iffy 16-17 on 18+ [edit] Sources Virginia Heffernan "Sweeping the Clouds Away". The New York Times, November 18, 2007 Retrieved from "http://commons.wikimedia.org/wiki/User_talk:JasonHaddad" all ages or 13+? or 16 or 17? 18 Views User page Discussion Edit +comment History Move Unwatch Gallery Orphans Untagged Personal tools Try Beta JasonHaddad My talk My preferences My watchlist My contributions Log out

Search Navigation Main Page Welcome Community portal Village pump Participate Upload file Recent changes Latest files Random file Help Contact us Donate Toolbox What links here Related changes User contributions Logs Upload file Special pages Printable version Permanent link Subpages This page was last modified on 24 January 2010, at 23:22. Text is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply. See Terms of Use for details. Privacy policy About Wikimedia Commons Disclaimers Cancel Edit Delete Preview revert Text of the note (may include Wiki markup) Could not save your note (edit conflict or other problem). Please copy the text in the edit box below and insert it manually by editing this page. Upon submitting the note will be published multi-licensed under the terms of the CC-BY-SA-3.0 license and of the GFDL, versions 1.2, 1.3, or any later version. See our terms of use for more details. Add a note Draw a rectangle onto the image above (press the left mouse button, then drag and release). This file has annotations. Move the mouse pointer over the image to see them. To edit the notes, visit page X. Why do you want to remove this note? translate Save To modify annotations, your browser needs to have the XMLHttpRequest object. Your browser does not have this object or does not allow it to be used (in Internet Explorer, it may be in a switched off ActiveX component), and thus you cannot modify annotations. We're sorry for the inconvenience. Adding image note$1 Changing image note$1 Removing image note$1 File:Untitled-2.png== Summary == Description	API English: Rated: TV-PG-S sometimes on DLS Date	1/24/2010 Source	HULU Author	Pinky Dinky Doo Permission (Reusing this file)	Me Other versions	NickJr Pinky is the oldest 2 is the younger and yougest was 3 almost 4. [edit] Licensing This media file is in the public domain in the United States. This applies to U.S. works where the copyright has expired, often because its first publication occurred prior to January 1, 1923. See this page for further explanation. Català | Česky | Deutsch | English | Español | Eesti | Suomi | Français | Gaeilge | Galego | עברית | Magyar | Italiano | Македонски | Malti | Plattdüütsch | Nederlands | Português | Română | Русский | ไทย | Vèneto | 中文 | 中文(简体) | 中文(繁體) | +/− This image might not be in the public domain outside of the United States; this especially applies in the countries and areas that do not apply the rule of the shorter term for US works, such as Canada, Mainland China (not Hong Kong or Macao), Germany, Mexico, and Switzerland. The creator and year of publication are essential information and must be provided. See Wikipedia:Public domain and Wikipedia:Copyrights for more details. 45 7 35 69 1130 599 TV PG S 223 438 100 47 1130 599 AY TV PG S AY

title: User talk:Thejasonhaddadshow1


text:
User:Ayushi~enwikiversity

HI EVERYBODY!!!! THIS IS AYUSHI

WELCOME TO MY USER PAGE

title: User:Ayushi~enwikiversity


text:
User talk:Egm6341.s10.team1.toddmock

 Hello Egm6341.s10.team1.toddmock, 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, 25 January 2010 (UTC)

uploads
Hi, I got your email. Yes, it takes 4 days before a new account becomes Autoconfirmed and able to upload images. It looks like your account was created 21 January 2010 so you should be able to upload within 24 hours. Sorry for the delay, but this is done to prevent spam accounts. Let me know if you have any trouble or if you need any help. You also may want to check out  This is one of our sister projects that was created to allow for easy file sharing. Any images there can be used here with just a simple link. --mikeu talk 18:03, 25 January 2010 (UTC)

Copyright problem with Image:Hwpic3.jpg

 * Read the previous five points. Thank you.Crochet.david 18:55, 15 February 2010 (UTC)

title: User talk:Egm6341.s10.team1.toddmock


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User talk:Egm6341.s10.team4.anandankala

 Hello Egm6341.s10.team4.anandankala, 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
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 * 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 08:18, 25 January 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 25 January 2010 (UTC)

title: User talk:Egm6341.s10.team4.anandankala