text:
User:Egm6341.s10.Team4.andy/HW2

= Problem 1: Lagrange Basis Approximation =

Given
Refer Lecture slide 9-1 for problem statement.

From Lecture slide 8-3,

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

p_2(x) = c_2 x^2 + c_1 x + c_0 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)

 * }.
 * }.

Find
We need to approximate Equation (1) using,

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

p_2(x) = \sum_{i=0}^{2} l_i(x_j)f(x_i) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)

 * }.
 * }.

where,

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

l_i(x) = \prod_{j=0,i \ne j}^{2} \frac{x-x_j}{x_i-x_j} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)

 * }.
 * }.

and find $$\displaystyle c_0, c_1 \ and\ c_2 $$ in terms of the given points,

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

(x_i,f(x_i)), \ i = 0,1,2 $$
 * $$\displaystyle

 * }.
 * }.

Solution
Equation (3) is given by the following 3 Equations,

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

l_0(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)

 * }.
 * }.

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

l_1(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)

 * }.
 * }.

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

l_2(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (6)

 * }.
 * }.

Substituting Equations (4), (5) and (6) in (2),

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

p_2(x) = \frac{(x^2-(x_1+x_2)x+x_1x_2)}{(x_0-x_1)(x_0-x_2)} f(x_0) + \frac{(x^2-(x_0+x_2)x+x_0x_2)}{(x_1-x_0)(x_1-x_2)} f(x_1) + \frac{(x^2-(x_0+x_1)x+x_0x_1)}{(x_2-x_0)(x_2-x_1)} f(x_2) $$
 * $$\displaystyle

 * }.
 * }.

Rearranging,

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

\begin{align} \Rightarrow p_2(x) =  & x^2 \underbrace{\left [\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}  + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} \right]}_{c_2} +  \\ & x \underbrace{\left [\frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)}  + \frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)} \right]}_{c_1}+ \\ & \underbrace{\left [\frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)} \right]}_{c_0}
 * $$\displaystyle

\end{align} $$

 * }.
 * }.

So,

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow c_0 = \frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)}  + \frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)}

$$

 * style= |

 * }.
 * }.

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow c_1 = \frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)} + \frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)}

$$
 * style= |
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_2 = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

= Problem 7: Lagrange Error Term =

Given
Refer Lecture slide 12-3 for problem statement.

From Equation 4 of Lecture slide 10-1,

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

q_{n+1}(x) := (x - x_0)(x - x_1)(x - x_2)(x - x_3)\cdots(x - x_n) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)

 * }.
 * }.

Find
Show that the $$\displaystyle (n+1)^{th} $$ derivative of $$\displaystyle q_{n+1}(x) $$ is $$\displaystyle (n+1)! $$, i.e.,

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

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

 * }.
 * }.

Solution
Clearly from Equation (1),

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

q_{n+1}(x) \in \mathcal{P}_{n+1} $$
 * $$\displaystyle

 * }.
 * }.

where $$\displaystyle \mathcal{P}_{n+1} $$ is a set of polynomials of order $$\displaystyle n+1 $$.

So, Equation (1) can be written as,

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

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

 * }.
 * }.

Please note that $$\displaystyle a_{n+1} = 1 $$ in Equation (2).

On differentiating Equation (2), 1 time,

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

q_{n+1}^{1}(x) = (n+1) x^{n} + na_{n} x^{n-1} + (n-1)a_{n-1} x^{n-2} + \cdots + a_{1} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)

 * }.
 * }.

On differentiating Equation (2), 2 times,

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

q_{n+1}^{2}(x) = (n+1)(n) x^{n-1} + n(n-1)a_{n} x^{n-2} + (n-1)(n-2)a_{n-1} x^{n-3} + \cdots + 2a_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)

 * }.
 * }.

On differentiating Equation (2), n times,

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

q_{n+1}^{n}(x) = (n+1)(n)\cdots(2) x^{1} + n(n-1)\cdots(2)a_{n}x^0 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)

 * }.
 * }.

This can be realized by induction [in Equation (3), the constant term is $$\displaystyle a_1 $$ after differentiating 1 time; the constant term is $$\displaystyle 2a_2 $$ after differentiating 2 times;].

Differentiating Equation (5) one more time,

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

\begin{align} \Rightarrow q_{n+1}^{(n+1)}(x) &= (n+1)(n)\cdots(2)(1) \cancelto{1}{x^{1}} + n(n-1)\cdots(2)a_{n}\cancelto{0}{x^0} \\ &= (n+1)(n)\cdots(2)(1) \end{align} $$
 * $$\displaystyle

 * }.
 * }.

So,

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow q_{n+1}^{(n+1)}(x) = (n+1)!

$$

 * style= |

 * }.
 * }.

= Problem 13: Error in Integration =

Given
Refer Lecture slide 15-2 for problem statement

From Lecture slide 15-1,

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

G^{1}(t) = e^{1}(t) - 5t^4e(1) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)

 * }
 * }

where,

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

e(t) := \alpha(t) - \alpha_{2}(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)

 * }
 * }

where $$\displaystyle \alpha(t), \alpha_{2}(t) $$ are given by,

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

\alpha(t) := \int_{-t}^{t}F(t)dt $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)

 * }
 * }

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

\alpha^{2}(t) := \frac{t}{3} \left[F(-t) + 4F(0)+ F(t)\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)

 * }
 * }

and,

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

F(t) := f(x(t)) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)

 * }.
 * }.

Find
Show that,

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

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

 * }.
 * }.

Solution
On differentiating Equation (3),

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

\alpha^{1}(t) = F(-t) + F(t) $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (6)

.
 * }
 * }

Again differentiating Equation (6),

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

\alpha^{2}(t) = -F(-t) + F(t) $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (7)

.
 * }
 * }

On differentiating Equation (4),
 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow \alpha_{2}^{1}(t) &= \frac{1}{3} \left[ F(-t) + 4F(0) + F(t) \right] + \frac{t}{3} \left[-F^{1}(-t) + \cancelto{0}{4F^{1}(0)} + F^{1}(t)\right] \\ &= \frac{1}{3} \left[ F(-t) + 4F(0) + F(t) \right] + \frac{t}{3} \left[-F^{1}(-t) + F^{1}(t)\right] \end{align} $$
 * $$\displaystyle

 * }.
 * }.

Again on differentiating the above equation,

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

\begin{align} \Rightarrow \alpha_{2}^{2}(t) &= \frac{1}{3} \left[ -F^{1}(-t) + \cancelto{0}{4F^{1}(0)} + F^{1}(t) \right] + \frac{1}{3} \left[-F^{1}(-t) + F^{1}(t)\right] + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]\\ &= \frac{2}{3} \left[ -F^{1}(-t) + F^{1}(t) \right] + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right] \end{align} $$
 * $$\displaystyle

 * }.
 * }.

So,

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

 * $$\displaystyle

\Rightarrow \alpha_{2}^{2}(t) = -\frac{2}{3} F^{1}(-t) + \frac{2}{3} F^{1}(t) + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]

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

 * }.
 * }.

On differentiating Equation (1) we get,

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

G^{2}(t) = e^{2}(t) - 20t^3e(1) $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (9)

 * }.
 * }.

At point $$\displaystyle t = 0 $$,

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

G^{2}(0) = e^{2}(0) - \cancelto{0}{20(t)^3e(1)} \Rightarrow G^{2}(0) = e^{2}(0) $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (10)

 * }.
 * }.

$$\displaystyle e^{2}(t) $$ is given by differentiating Equation (2) twice,

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

e^{2}(t) = \alpha^{2}(t) - \alpha_{2}^{2}(t) $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (11)

 * }.
 * }.

Substituting Equations (7) and (8) in (11),

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

\Rightarrow e^{2}(t) = -F(-t) + F(t) +\frac{2}{3} F^{1}(-t) - \frac{2}{3} F^{1}(t) - \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right] $$
 * $$\displaystyle

 * }.
 * }.

At point $$\displaystyle t = 0 $$,

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

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

 * }.
 * }.

So,

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

 * $$\displaystyle

\Rightarrow e^{2}(0) = 0

$$

 * }.
 * }.

which means from Equation (10),

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow G^{2}(0) = 0

$$

 * style= |

 * }.
 * }.

title: User:Egm6341.s10.Team4.andy/HW2


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

title: File:Hwpic3.jpg


text:
User:EGM6341.s10.Team1.toddmock/HW2

= Problem 13-2: Error =

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

\begin{align} n=2, \\ & q_3(x)=(x-x_o)(x-x_1)(x-x_2)\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

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

\begin{align} \\ & x_o=a, x_2=b, x_1=\frac{a+b}{2} \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

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

\left | E_2 \right | = \frac{M_3}{3!}\int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right |\, dx = \frac{(b-a)^4}{192}M_3 $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

Solution
To capture positive region it is necessary to change the limits and multiply by two

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

\begin{align} \int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right |\, dx &= 2\int_{a}^{\frac{a+b}{2}}(x-a)\Big(x-\frac{a+b}{2}\Big)(x-b)\, dx\\ &=2\int_{a}^{\frac{a+b}{2}}x^3-\frac{a^2b}{2}-\frac{ab^2}{2}+\frac{a^2x}{2} + 2abx + \frac{b^2x}{2}-\frac{3ax^2}{2}-\frac{3bx^2}{2}, dx\\ &=\frac{1}{2}x^4-(a+b)x^3+\frac{1}{2}(a^2+4ab+b^2)x^2 - ab(a+b)x\, \Big |^{\frac{a+b}{2}}_a\\
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$
 * }.
 * }.

Evaluating at the limits gives

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

\begin{align} &=\frac{1}{2}\Bigg(\Big(\frac{a+b}{2}\Big)^4-a^4\Bigg)-(a+b)\Bigg(\Big(\frac{a+b}{2}\Big)^3-a^3\Bigg)+\frac{1}{2}(a^2+4ab+b^2)\Bigg(\Big(\frac{a+b}{2}\Big)^2-a^2\Bigg) - ab(a+b)\Bigg(\Big(\frac{a+b}{2}\Big)-a\Bigg)\
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$
 * }.
 * }.

Simplifying leads to

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

= \frac{1}{32}(b-a)^4 $$
 * $$\displaystyle
 * $$\displaystyle

 * style= |
 * }.
 * }.

Thus the solution becomes

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

$$\displaystyle \left | E_2 \right | = \frac{M_3}{3!}\Bigg(\frac{1}{32}(b-a)^4\Bigg) = \frac{(b-a)^4}{196}M_3 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

= Problem 15-2(b): Lagrange Error, e(t) =

Given
e(t) is given as:

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

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

 * }
 * }

Find
Determine the third derivative of e(t) which is known to be:

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

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

 * }
 * }

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

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

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

 * }
 * }

We must use Leibniz integral rule,

For a monovariant function $$g$$:

 * $$ {d\over dx}\, \int_{f_1(x)}^{f_2(x)} g(t) \,dt = g(f_2(x)) {f_2'(x)} - g(f_1(x)) {f_1'(x)} $$

Using a point k that lies somewhere in the interval [-t,t]:

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

Then evaluation of the first term is:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \alpha^{(1)}(t)=[F(k)+F(-t)]+[F(t)-F(k)]=F(-t)+F(t)
 * }
 * }

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

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

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

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

Evaluation of the second term:

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

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

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

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

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

All that remains is to add the two functions $$\alpha_1^{(3)}$$ and $$\alpha_2^{(3)}$$ :

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

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

$$ e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.

= Problem 9-3(a): Simple to Composite Trapezoidal and Simpson Rules =

Given
Given the simple Trapezoidal:

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

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

 * }
 * }

Find
Find the Composite Trapezoidal rule

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{b-a}{n}[\dfrac{1}{2}f_1+f_2+f_3+...+f_{n-1}+\dfrac{1}{2}f_n]

 * }
 * }

Where:

f_1=f(a), $$

f_n=f(b) $$

Solution
Simple trapezoidal rule uses only one trapezoid to calculate area underneath a function extending this principle to more than one trapezoid leads to the composite function.

The area of a trapezoid is:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Area=(b-a)\dfrac{h_1+h_2}{2}

 * }
 * }

Where $$h_1$$ and $$h_2$$ would be $$f(a)$$ and $$f(b)$$ respectively

Assuming five equal width trapezoids the base of each would be $$\dfrac{b-a}{5}$$:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \Rightarrow I_5=\dfrac{b-a}{5}(\dfrac{f_1+f_2}{2}+\dfrac{f_2+f_3}{2}+\dfrac{f_3+f_4}{2}+\dfrac{f_4+f_5}{2})

 * }
 * }

Collecting like terms

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_5=\dfrac{b-a}{5}(\dfrac{1}{2}f_1+f_2+f_3+f_4+\dfrac{1}{2}f_5)

 * }
 * }

Which displays a recognizable patter

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{b-a}{n}(\dfrac{1}{2}f_1+f_2+...+f_{n-1}+\dfrac{1}{2}f_n)

 * }.
 * }.

= Problem 9-3(b): Simple to Composite Trapezoidal and Simpson Rules =

Given
The simple Simpson's rule:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_2=\dfrac{h}{3}[f_0+4f_1+f_2]

 * }.
 * }.

where $$h=\dfrac{b-a}{n}$$ in this case $$n=2$$

Find
Find the composite Simpson's rule known to be

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_2=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+2f_4+...2f_{n-2}+4f_{n-1}+f_n]

 * }.
 * }.

where $$h=\dfrac{b-a}{n}$$

Solution
The simple Simpson's rule comes from calculation of the area underneath a polynomial using two sections

Repeating the pattern of the simple Simpson's rule will yield the composite rule. So if the same function from above is instead split into 5 equal sections then we have the following.

calculation for 2 sets of the sections $$[x_0,x_2]$$ and $$[x_2,x_4]$$ we will obtain

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I([x_0,x_2])=\dfrac{h}{3}[f_0+4f_1+f_2]

 * }
 * }

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I([x_2,x_4])=\dfrac{h}{3}[f_2+4f_3+f_4]
 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I([x_0,x_4])=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+f_4]
 * }
 * }

Thus producing the obvious pattern

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

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

= Problem 12-3(2): =

Given
Given the function:

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

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

 * }
 * }

and

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

$$ $$
 * $$\displaystyle t=2, x_0=3, x_1=4, ..., x_6=9
 * $$\displaystyle t=2, x_0=3, x_1=4, ..., x_6=9
 * $$\displaystyle

 * }
 * }

Find
Plot 3 figures like those on slide [[media:Egm6341.s10.mtg11.pdf|11-2]] for the functions

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

for $$\displaystyle i=3 (x_3=6)$$

and

$$\displaystyle x=5.5$$

Solution
The function f(x) and approximated function are defined as:

 * $$\displaystyle

f(x)=log(x) $$

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

Where the Lagrange equation for i=3:

l_{i,n}(x)=\prod_{i=1}^{n} \dfrac{x-x_j}{x_i-x_j} $$

therefore

l_{3,6}(x)=\Big(\frac{x-x_0}{x_3-x_0}\Big)\Big(\dfrac{x-x_1}{x_3-x_1}\Big)\Big(\dfrac{x-x_2}{x_3-x_2}\Big)\Big(\dfrac{x-x_4}{x_3-x_4}\Big)\Big(\dfrac{x-x_5}{x_3-x_5}\Big)\Big(\dfrac{x-x_6}{x_3-x_6}\Big) $$

 * $$\displaystyle

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

title: User:EGM6341.s10.Team1.toddmock/HW2


text:
User:Subbu.ase/HW2 pblm 15

Problem Statement
$$ -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{4}(\xi) $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|p.15-3]]

Solution
Use chain Rule

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

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

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

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

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

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

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

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

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

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

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

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

$$\displaystyle -\frac{1}{90}F^{(4)}(\zeta _{4})=-\frac{(b-a)^{4}}{1440}f^{4}(\xi) $$ where
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }
 * {| style="width:100%" border="0" align="left"

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

title: User:Subbu.ase/HW2 pblm 15


text:
User:Egm6341.s10.team4.anandankala/HW2-15

Prove that

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

 * $$\displaystyle e(1)= \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$

Deduce a relation between $$\displaystyle \xi and \zeta_4.$$
 * }.
 * }.

Solution
We have F(t)= f(x(t))= f(x). Let us consider

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

 * $$ \displaystyle x=\alpha + ht, where h= \frac{(b-a)}{2}$$.

 * }.
 * }.

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

 * $$\displaystyle F^'(t)= \frac{dF(t)}{dt}= \frac{df(x(t))}{dt}= \frac{df(x)}{dt}= \frac{df(x)}{dx}\cdot\frac{dx}{dt}= f^'(x)\cdot h$$

 * }.
 * }.

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

 * $$\displaystyle F^{}(t)= \frac{d}{dt}\cdot \frac{dF(t)}{dt}= \frac{d}{dt}\cdot(f^'(x)h) = h\cdot \frac{df^'(x)}{dt}= h\cdot \frac{df^'(x)}{dx}\cdot \frac{dx}{dt}= h^2\cdot f^{}(x)$$

 * }.
 * }.

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

 * $$\displaystyle F^{}(t)= \frac{d}{dt}\cdot (F^{}(t))= \frac{d}{dt}\cdot(h^2\cdot f^{}(x)) = h^2\cdot \frac{df^{}(x)}{dt}= h^2\cdot \frac{df^{}(x)}{dx}\cdot \frac{dx}{dt}= h^3\cdot f^{}(x)$$

 * }.
 * }.

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

 * $$\displaystyle F^{'}(t)= \frac{d}{dt}\cdot (F^{}(t))= \frac{d}{dt}\cdot(h^3\cdot f^{}(x)) = h^3\cdot \frac{df^{}(x)}{dt}= h^3\cdot \frac{df^{}(x)}{dx}\cdot \frac{dx}{dt}= h^4\cdot f^{'}(x)$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle F^4(t)=h^4\cdot f^4(x)$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle F^4(t)=\frac{(b-a)}{16} ^4\cdot f^4(x)$$

 * }.
 * }.

Substituting $$\displaystyle \xi$$ for $$\displaystyle x$$ and $$\displaystyle \zeta_4$$ for $$\displaystyle t$$

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

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

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$

 * }.
 * }.

Since $$ \displaystyle x=\alpha + ht$$, We have $$ \displaystyle \xi=\alpha + h\zeta_4$$

This is the required relation between $$\displaystyle \xi$$ and $$\displaystyle \zeta_4$$.

Hence proved.


text:
Going naked - Openism and freedom in academia

A 15-minute presentation about open academic philosophy and practice.

By James Neill - with responses by Leonard Low, Leigh Blackall.

Where: Hothouse, 1C32, University of Canberra

When: Friday 5th March, 2010 13:30

Now a solid(ish) draft - but still feel free to improve or comment

Summary: Going Naked - Openism and Freedom in Academia (Leigh Blackall blog)

Also recommended: Open education slides by David Wiley, 2010

The copyleft symbol mirrors the copyright symbol to represent freedom and openness in licensing intellectual property. See also free culture.

Abstract

Academia should be conducted in such a way as to benefit society. This means (among other things) that the processes and products of publically-funded academics' activities should, by default, be public (i.e., accessible and freely usable). It also means that academics should seek to use and promote tools (such as software) and materials (such as textbooks) which enable students emerging academics to utilise and foster public knowledge. The evolution towards open academia is a cultural challenge because closedness is the norm.

Academia is about sharing
If I give you a penny, you will be one penny richer and I’ll be one penny poorer. But if I give you an idea, you will have a new idea, but I shall still have it, too. - Albert Einstein

He who receives ideas from me, receives instruction himself without lessening mine; as he who lights his taper at mine receives light without darkening me. - Thomas Jefferson

Academia is about sharing - otherwise it is not academia. To not share is to retreat to an ivory tower. This is the "low game" of academia in which knowledge-development and knowledge-storing is approached as a competition (e.g., between staff, students, departments, institutions, sectors, countries etc.). The "higher game" in academia is to selflessly contribute to collective knowledge by freely disseminating one's knowledge and activities (for a deeper discussion of the "academic game", see De Ropp (1968) ).

This claim may be summarised as "academics are public servants and our work is public property".

The default is (becoming) open
The cultural standard for public institutions is increasingly moving towards a "default is open" philosophy and practice. Several business sectors are also participating in this evolution towards openness.

Openness-closedness is not a dichotomy - it is a continuum - arguably, an evolutionary continuum.

To not engage with the palpably shifting tide towards openness is risky for academic institutions (e.g., funding metrics are likely to become more accurate at measuring the social value generated by academic institutions). Even more importantly, democratic academic institutions should be leading the charge towards openness since innovatively fostering public knowledge is their raison d'etre.

The biggest barriers to openism in academia are not legal or technical, but rather cultural, organisational, and psychological (because closedness is the norm). Most academics, for example, use closed textbooks and websites for teaching, publish in closed rather than open journals, user proprietary software, and do not make their work openly available.

5 pillars of open academia
If you focus your mind on the freedom and community that you can build by staying firm, you will find the strength to do it." - Richard Stallman

In practice, going naked (free and open) can be a surprisingly complex and challenging endeavour - mainly because of the widespread cultural habits of closedness which need to be unlearnt. It can also be a liberating and empowering journey. By "going naked" and not carrying around the heavy weight of "closedness" much of one's energy and focus is freed.

Culturally, what's needed is to bulldoze the ivory tower and replace it with an open parthenon consisting of at least five pillars. These "five pillars of open academia" offer a practical, guiding framework for moving towards openism:

Expressed as a "creed", "'as an open academic I commit to providing open access to all my academic outputs (teaching, research and service) using open formats, open licensing and free software. I also commit to open management of my academic activities.'"

Open university
A closely related conceptualisation is provided by the Free Software Foundation (FSF; initiated by Students for Free Culture (SFC)). They define an open university as one in which:
 * 1) The research produced is open access;
 * 2) The course materials are open educational resources;
 * 3) The university embraces free software and open standards;
 * 4) The university’s patents are readily licensed for free software, essential medicine, and the public good;
 * 5) The university’s network reflects the open nature of the Internet.

The FSF goes further, to list specific criteria for each of these open university principles and university-rankings based on their degree of openness: Open University Report Cards.

Increasing transparency in higher education is currently being strongly by the Labour government in Australia. On the 3rd March, 2010 the government announced the creation of a "My University" federal website by 2012 (based on the "My School" site) which will provide transparent, comparative data about indicators of university quality (such as student satisfaction ratings). In response the chief executive of Universities Australia, Glenn Withers, welcomed the idea:

"Universities are fully committed to transparency. They are remarkably open already via their websites, public guides, reports to parliament, auditors general, ombudsmen and more. They fully welcome anything that can enhance this transparency through new, well-designed initiatives for universities and importantly, for all other tertiary providers." (Harrison, 2010 )

Making the change
If you want to accomplish something in the world, idealism is not enough - you need to choose a method that works to achieve the goal. - Richard Stallman

Two methods of moving towards greater openism in academia might be:
 * 1) Cold turkey (e.g., new institutions and projects and institutions undergoing major change may simply adopt an open charter from the outset)
 * 2) Iterative/progressive (e.g., for existing institutions, openist KPIs can be established and made part of their strategic plans and roadmaps can be developed (e.g., )
 * 3) Do a cost/benefit analysis measuring the costs of closed, including software, login caused inefficiencies and administration, copyright royalty costs, against the benefits such as free media, free software, skills, productivity, social capital, marketing benefits..

Activity
What are your thoughts about open academia? Helpful ways to structure responses could be in terms of barriers and benefits, and/or a SWOT analysis - just click edit, add your thoughts, and save :

Barriers and benefits
What barriers are there to open academia? What benefits might there be?

Barriers

 * Performance anxiety
 * Control and ownership
 * Ignorance
 * Time/prioritisation, full disclosure and informed choice

Benefits

 * Moral: Helps to fulfill the charter of public universities to contribute to the common good of society

Strengths

 * Addresses copyright ambiguity for educational use
 * Enables reusability
 * Liability is individualised

Weaknesses

 * Individually simple, organisationally difficult
 * Difficult to implement with so much of media commercially orientated, so formats and devices often don't record or play open formats
 * Skills shortage: Computing and IT sector has been trained to service commercially developed software, hardware, media (CISCO, Microsoft).

Opportunities

 * A principled stance against the questionable academic publishing norm
 * Increase the profile of UC at this time, as being a leader in the field, recognising the problems and hypocrisy, taking a principled stance
 * Good economic returns through marketing and profile gains, legal and administrative savings, infrastructural savings, efficiency and sustainability, and skills capacity building.

Threats

 * The status quo favours closedness
 * Commercial appropriation (and "enclosure") of freely available research
 * Cultural imperialism - forcing copyright restrictions on people with one hand, then swamping then with open educational media
 * Loosing or weakening whatever stake we in academia and education might have had with Fair Use/Dealings

Responses
Feel free to share your thoughts, ideas, resources, and experiences by clicking edit - or use the talk page.

Completely agree with the ethos of course. I'm interested in the assertion that other parts of society are becoming more open - I'm not convinced. Knowledge is horded by elites as a source of power, and as such is hidden by a variety of mechanisms, eg legal, technocratic language, by omission. It seems to me that universities pretty much invented the 'knowledge economy' - that is the commodity they trade in. They package it up into 'degrees', they market their 'unique' courses to prospective students, they create competition for knowledge by limiting access, and then by grading, rather than simply critiqing, students' work. The only point at which they "gift" knowledge is the publication of academic work - and they have managed to commodify that by linking publication to prestige (read market positioning of individuals and institutions)

University 'openism' is truly counter-cultural in that it means that universities get out of the business of simply re-creating elites. This would indeed be a revolution. So two lines of thought occur
 * 1) What then is the role of universities?... are they simply a trusted source of knowledge?  Does this not still link to elitism?  And who is funding this "public service" (180 degrees away from current corporatisation of unis)
 * 2)  Have we then delivered an egalitarian utopia... or do elites simply colonise another structure - and what would that be?

cheers, Jane

 * Thanks for finding the "edit" button and daring to share, Jane. Your comments have me pondering further, thank-you. Is the rest of society becoming more open? In making this suggestion, I'm thinking e.g., of the steps being made towards open government (making public data open by default), of p2p networks sharing (which no-one really can stop it seems), of the uptake of open source software (e.g., Firefox) etc. Maybe it is rose-coloured glasses from me - there is plenty of growth in closed/patented/copyrighted efforts too. But maybe both are true - maybe we are becoming simultaneously more open and more closed - that could be possible?

 * It seems to me, yes, as you say, universities have created (or at least fostered) a 'knowledge economy' to justify/maintain their existence - and lost focus on their social role/purpose - at least the university 'charters' don't seem to match the behaviour. And if universities can't/don't/won't do openism (which is their potential social value) then knowledge-brokering is in big trouble - how else can we expect openism in other facets of society to flourish if the institutions supposedly set up for knowledge-sharing are lapsing into small-minded protectionism? So the "egalitarian utopia" seems a long way off (but then, so do most utopias) - the multinationals have swallowed the textbook and journal industry and are busy clawing their fingernails into classrooms, learning management systems, etc. with academics seduced like doctors by pharmaceutical companies... but we can "subvert the dominant paradigm" - at least that was the graffiti I remember from my undergrad ways. -- Jtneill - Talk - c 13:38, 4 March 2010 (UTC)

 * I think we are becoming simultaneously more open and more closed. I think a 'knowledge war' of sorts exists. People feel knowledge is a right and simultaneously feel they have a right to profit from knowledge. I think polarization is pretty common in most things. When people cannot find common ground polarizations turns into extremes. I think detecting extreme polarizations is easy because terms like left, right, liberal, conservative. etc. tend to get thrown around, and people start identifying themselves and others by such labels. I won't be surprised when people start saying "I am a knowledge liberal" or "you are a right extremist on the knowledge issues". Just some thoughts, maybe this is not really on topic. -- dark lama  02:05, 5 March 2010 (UTC)

This is an interesting discussion. Based on the above thoughts, here's my new hypothesis. That individuals are becoming more open, due to the symbiotic relationship between technological advances and cultural changes to do with personal freedom and privacy issues. The individual is free to be altruistic and share knowledge. But this altruism does not translate well into the institutional/corporate sphere (at best partially, after a risk management process). The self-interest of the institution/corporation/professional groups (elites) is to maintain competitive advantage through ownership and control of knowledge resources (including education).

The university sector becomes the interesting case, as the institutional ideals clash with institutional self-interest. So I think your ideas around re-examining and reframing the institutional self-interest are practical... potentially 'subverting the dominant paradigm'??!! I guess the question is whether the university sector can be a master of its own destiny, or whether it is a tool of other dominant forces. cheers Jane

 * I tend to agree with some of the basic concepts in your new hypothesis. Knowledge is a natural resource like water. I think individuals can have both a personal self-interest and a professional self-interest in knowledge. The personal self-interest of individuals is to want free, open, and easy access to knowledge and the freedom to use and share knowledge. The professional self-interest of individuals is usually driven by a basic social need to have money to pay for other resources, like water. Personal self-interest looks for ways to make knowledge freely accessible and easily obtainable in order to have the freedom to use and share knowledge. Professional self-interest looks for ways to drive and/or control supply and demand, just like with water, in order to maintain itself and pay workers. I think transition between the personal and the profession (or visa versa) is naturally assumed to be incompatible because individuals often do not understand the driving forces or the motivations of the personal and professional parts of individuals. This is probably what leads to the idea that individuals or groups have bad intentions, and attempts to protect the personal and professional self-interests.

 * If you look and understand how the copyleft model manages to work, I think you may be able to see how both can work together. Universities can sell field experience and teachers as knowledge support, along the same lines as 'tech support' — in fact 'tech support' is a form of knowledge support because people seek technical knowledge on how to do or fix things. In this model knowledge is a resource that can be used and shared freely, and Universities offer experience and knowledge support as their services that people can buy. The change you talk about involves a move away from knowledge as an exploitable resource to a knowledge services support industry. I think a natural fear of change, the uncertainty and risk that comes from not knowing what would happen and how change would work, is usually what keeps individuals from pursuing change. I think professional change usually comes from necessity and from entrepreneurs. I think necessity comes into play when the majority of consumers refuse to buy the product being sold which in this case would be knowledge. When obtaining knowledge is no longer a major driving force in what University individuals decide to attend Universities will change by necessity. I think entrepreneurs will first provide alternative educational models that appeal to the majority of consumers before Universities catch up. Although there may be some intelligent individuals that see a trend and make changes to the University model before the majority of consumers want something else. -- dark lama  10:01, 5 March 2010 (UTC)

If history is any indicator, I think things will get worse before getting better. I can see a scenario where Universities become scared of their students undermining there business model and manage to control and limit access to knowledge by finding a way to make it illegal for students to share what they learned with others. -- dark lama  10:22, 5 March 2010 (UTC)

Some more random thoughts. In Abstract it is said that "Academia should be conducted in such a way as to benefit society." In Academia is about sharing it is said that "Academia is about sharing - otherwise it is not academia." I think an assumption is being made that these two statements are true, and everybody knows this to be true is taken for granted. I think anyone trying to learn from this page could benefit from understanding why academia should benefit society and why academia is about sharing and why if there is no sharing it is not academia. -- dark lama  02:20, 8 March 2010 (UTC)
 * Agreed - this is in need of expansion/development/argument. Curious, - what do you think of Academia - e.g., consistent or not so consistent with these claims - or itself in need of such development? -- Jtneill - Talk - c 02:32, 8 March 2010 (UTC)

 * I do not know enough about how things work at Wikipedia to comment on development there. I mostly just read Wikipedia, and I usually don't read an entire article because I think Wikipedia's articles usually suffer from information overload. I sometimes end up turning to Simple Wikipedia in order to get quicker answers to what I would like to know. What specifically about Academia should I be looking at to tell if its consistent or not so consistent with these statements? -- dark lama  02:48, 8 March 2010 (UTC)
 * No worries - I'll take a closer look - the quick gist I got was that it seemed at least loosely consistent with academia as characterised by sharing, but I need to read/think more in this area to articulate such a position. -- Jtneill - Talk - c 00:24, 9 March 2010 (UTC)

Some responses from Leonard Low and Leigh Blackall
Currently as noted by James Neill and Leigh Blackall
 * Leonard: There is an increasingly litigious intellectual property culture - therefore are we to play into the hands of commercial interests by releasing our material (for their use and gain/benefit)?
 * Leonard: Is it a university's duty to give back to society at large or to our students in particular? Leonard raises the issue that if a university makes all of its learning freely and accessibly available what advantage is there for paying students?
 * Leigh: A Share Alike restriction prevents exploitation of openism?
 * Leonard: If the knowledge is freely available, does this mean that a university is just a rubber-stamping, degree-granting institution?
 * Leonard: What's the advantage of going to university if the info is readily available?
 * Leigh: The value proposition of a university education is assessment and certification, not content.
 * James: A basic business model for an open university is that it:
 * provides training, support, guidance, teaching, assessment, feedback, and grants degrees etc. to paying students (funded publically or privately)
 * conducts research (funded publically or privately)
 * freely and openly shares its knowledge (teaching, research, and service) and activities (e.g., open management) - otherwise it is a private college, not a university
 * Leonard: Ivory tower is about engagement with society (means academics who do not engage with society) - one could still be in an ivory tower and open if the academic activity is not engaged with society's needs (i.e., relevance of content matters too - not just openness)
 * Leigh: Argues that the irrelevance of academic pursuits becoming "ivory towers" should be less of an issue through the Internet and can be addressed though popularisation - such as engaging in Wikimedia Foundation work, publishing video synopsis on Youtube, and networking online. See Using the Popular Internet in Teaching and Research
 * Leonard: On the balance of things, Leonard personally supports openness
 * Leigh: The risk of open education becoming a force for neo colonialism. While on the one hand our trade agreements impose copyright regulation on countries lacking resources to produce their own content, in the other we make our a selection of our own content free of use. At the moment, free content tends to be US centric for example. See New Colonialism in OER
 * Leigh: How much does UC pay Copyright Agency Limited (CAL) for generalised royalty fees for the copying of restricted content? How much could be saved if a Faculty were to declare only using free and open content?
 * Leonard: CAL fees are in the millions per year
 * Leigh: Regarding UC's copyright IP policy, and its recent review status. Leigh attempted to engage with UC leadership on the review but was advised that procedure requires only external agencies review UC policy. Leigh wants to know why this is and who the agency was, and what their advice was.

Proposed actions

 * The University of Canberra Teaching and Learning Centre to find out the annual fees paid to CAL, and if possible - a break down of that fee.
 * The University of Canberra Teaching and Learning Centre to find out why external agencies review UC policy, what agency reviewed UC's IP policy, and what their review was.
 * Leigh and James to run Workshops on how to source free content and reuse it in a free way
 * Leigh and James to propose the University of Canberra Faculty of Health lead the university and adopt a default open education and research position.


text: Category:User:Jtneill/Presentations

title: Category:User:Jtneill/Presentations


text:
18th century European scholarly societies and academies/Königliche Landwirthschaftsgesellschaft zu Celle

Summary

 * Society was founded March 18, 1764

 * Königliche Landwirthschaftsgesellschaft zu Celle translates Royal Agricultural Society in Celle

 * Founding City: Celle, Germany


text:
User:Egm6341.s10.Team4/HW2

= Problem 1: Lagrange Basis Approximation =

Given
Refer Lecture slide 9-1 for problem statement.

From Lecture slide 8-3,

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

p_2(x) = c_2 x^2 + c_1 x + c_0 $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

Find
We need to approximate Equation (1) using,

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

p_2(x) = \sum_{i=0}^{2} l_i(x_j)f(x_i) $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)$$
 * }.
 * }.

where,

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

l_i(x) = \prod_{j=0,i \ne j}^{2} \frac{x-x_j}{x_i-x_j} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)$$
 * }.
 * }.

and find $$\displaystyle c_0, c_1 \ and\ c_2 $$ in terms of the given points,

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

(x_i,f(x_i)), \ i = 0,1,2 $$
 * $$\displaystyle

 * }.
 * }.

Solution
Equation (3) is given by the following 3 Equations,

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

l_0(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)$$
 * }.
 * }.

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

l_1(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)$$
 * }.
 * }.

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

l_2(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (6)$$
 * }.
 * }.

Substituting Equations (4), (5) and (6) in (2),

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

p_2(x) = \frac{(x^2-(x_1+x_2)x+x_1x_2)}{(x_0-x_1)(x_0-x_2)} f(x_0) + \frac{(x^2-(x_0+x_2)x+x_0x_2)}{(x_1-x_0)(x_1-x_2)} f(x_1) + \frac{(x^2-(x_0+x_1)x+x_0x_1)}{(x_2-x_0)(x_2-x_1)} f(x_2) $$
 * $$\displaystyle

 * }.
 * }.

Rearranging,

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

\begin{align} \Rightarrow p_2(x) =  & x^2 \underbrace{\left [\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}  + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} \right]}_{c_2} +  \\ & x \underbrace{\left [\frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)}  + \frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)} \right]}_{c_1}+ \\ & \underbrace{\left [\frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)} \right]}_{c_0}
 * $$\displaystyle

\end{align} $$

 * }.
 * }.

So,

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow c_0 = \frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)}  + \frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)}

$$

 * style= |

 * }.
 * }.

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow c_1 = \frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)} + \frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}  + \frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)}

$$
 * style= |
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_2 = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}  + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 * 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.andy 21:55, 17 February 2010 (UTC) .

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

Given
Use equation (4) on slide [[media:Egm6341.s10.mtg8.pdf|8-3]], which is:

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

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

 * }.
 * }.

Find
to derive equation (simple simpson's rule Use following Integration (2) on slide [[media:Egm6341.s10.mtg7.pdf|7-2]], which is:

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

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

 * }.
 * }.

Solution
The primary equation is following polynomial:

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

From the lagrange equation we have:

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

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

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

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

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

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

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

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

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

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

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

Since,

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

And since,

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

We can rewrite the value of u as follows:

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

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

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

Similarly, for v and w we obtain:

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

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

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

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

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

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

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

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

Author
Solved and Typed by - Egm6341.s10.Team4.nimaa&m 02:37, 18 February 2010 (UTC) .

=Problem 3: Newton Cotes Application=

Given

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

f(x) = \frac{e^x - 1}{x}$$ on the interval [0,1].
 * $$\displaystyle

Consider n=0,1,2,4,8,16.
 * $$\displaystyle $$
 * }.
 * }.

Find

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

Construct $$ \displaystyle f_n(x)= \sum_{i=0}^n l_{i,n}(x)f(x_i)$$ a) Plot $$ \displaystyle f,f_n, n=1,2,4,8,16. $$ b) Compute $$ I_n= \int_{a}^{b}f_n(x)dx, n=1,2,4,8,16. $$ c) Plot $$ \displaystyle l_0, l_1, l_2, for n=4 $$
 * $$\displaystyle $$
 * }.
 * }.

Solution
From Newton Cotes formula, we have $$ \displaystyle f_n(x)= \sum_{i=0}^n l_{i,n}(x)f(x_i)$$                                                       $$\displaystyle  l_{i,n}(x)= \displaystyle\prod_{\underset{j\neq i}{j=0}}^{N}\frac{x-x_j}{x_i-x_j}$$

fn(x) has been constructed using matlab and its plots are shown below

Matlab Code: Plots: .

The function f(x) has been plotted and the plot is shown below: Plot of f(x) .

The value of In has been computed for different values of n in matlab and the details are tabulated below: $$\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c}\hline n & 1 & 2 & 4 & 8 & 16 \\ \hline I_n&1.3591&1.3180&1.3179&1.3179&1.3179\\ \hline \end{array} $$.

For n=4, the plots of $$l_0,l_1,l_2$$ are shown below including the matlab code: Matlab Code:

Plots: . . . Plot for $$\displaystyle l_3$$:

..

Plot for $$\displaystyle l_4$$:

.

The plots of $$\displaystyle l_3,l_4 $$ need not be looked at, since they are just the mirror images about $$\displaystyle x = 0.5$$, of the plots of $$\displaystyle l_1,l_0 $$ respectively.

Author
Solved and Typed by - Egm6341.s10.team4.anandankala 13:43, 19 February 2010 (UTC) .

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

Given
The simple Trapezoidal rule equation is written as follows:

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

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

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

 * }
 * }

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

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

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

 * }
 * }

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

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

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

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

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

By continuing this procedure for 4 intervals, we gain:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Author
Solved by ---Egm6341.s10.Team4.roni 05:11, 18 February 2010 (UTC)

Typed and reviewed by - Egm6341.s10.Team4.nimaa&m 02:38, 18 February 2010 (UTC) .

= Problem 5: Comparison of Error between Taylor series and Newton-Cotes formula =

Given
Refer Lecture slide 11-1 for problem statement
 * {| style="width:100%" border="0" align="left"

f(x) = \sin(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x \in [0,\pi] $$
 * $$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

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

\left | f^T_n\Big(\frac{7\pi}{8}\Big)-f\Big(\frac{7\pi}{8}\Big) \right | \le \underbrace{\left | f^L_4\Big(\frac{7\pi}{8}\Big)-f\Big(\frac{7\pi}{8}\Big) \right |}_{\left | E^L_4(\frac{7\pi}{8})\right |} \le \frac{\left | q_5(\frac{7\pi}{8}) \right |}{5!} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)$$
 * }.
 * }.

Solution
$$\displaystyle f^L_4\Big(\frac{7\pi}{8}\Big)$$, $$\displaystyle \left | E^L_4(\frac{7\pi}{8})\right |$$, $$\displaystyle \frac{\left | q_5(\frac{7\pi}{8}) \right |}{5!}$$, is claculated using Matlab.

 * $$\displaystyle f^L_4\Big(\frac{7\pi}{8}\Big) = \sum_{i=0}^4 l_{i,4}\Big(\frac{7\pi}{8}\Big)\sin(x_i)= 0.3812$$
 * $$\displaystyle \left | E^L_4(\frac{7\pi}{8})\right | = 0.0014808$$
 * $$\displaystyle \frac{\left | q_5(\frac{7\pi}{8}) \right |}{5!} = 0.0081716$$

At first, we can confirm that the seconde inequality of Equation(2)

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

\left | E^L_4(\frac{7\pi}{8})\right | = 0.0014808 \le \displaystyle \frac{\left | q_5(\frac{7\pi}{8}) \right |}{5!} = 0.0081716 $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)$$
 * }.
 * }.

Then, the Taylor series were calculated until the Error or taylor become smaller than that of Newton-Cotes formula at n=4. the result is like that

$$ \begin{array}{|c||c|c|} n & E^T_n(t) & E^T_n(t) - E^L_4(t) \\ \hline 1&1.7128&1.7113\\ 2&0.34976&0.3483\\	3&0.54236&0.5409\\ 4&0.10444&0.1030\\ 5&0.067533&0.0661\\	6&0.011256&0.0098\\	7&0.0045295&0.0030\\	8&0.00065513&-0.0008\\ \hline \end{array} $$

Matlab Code: .

Author
Solved and Typed by - Egm6341.s10.Team4.yunseok 02:03, 18 February 2010 (UTC) .

= Problem 6: Evaluation of (n+1)$$^{th}$$ Derivative of E(x) =

Given
We know that E(x) is defined as $$ \displaystyle\ E(x)=f(x)-f_n(x)$$.

Find
We are asked to show that the $$(n+1)^{th}$$ derivative of our error $$ \displaystyle\ E(x)=f(x)-f_n(x)$$ is $$\displaystyle\ E^{(n+1)}(x)=f^{(n+1)}(x)$$.

Solution
Assuming that $$ \displaystyle\ f_n(x)$$ is n+1 times differentiable, then $$ \displaystyle\ E_n^{(n+1)}(x)=f^{(n+1)}(x)-f_n^{(n+1)}(x)$$.

Let us assume because $$ \displaystyle\ f_n(x) \epsilon P_n$$, $$ \displaystyle\ f_n(x)$$ takes the form $$ \displaystyle\ f_n(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$.

We also know that $$ \displaystyle\ f_n(x) \epsilon P_n $$. Taking the $$n^{th}$$ derivative of $$ \displaystyle\ f_n(x)$$, we see that $$ \displaystyle\ f^{(n)}_n(x)=(n!)a_n$$, a constant. Taking the $$(n+1)^{th}$$ derivative of $$ \displaystyle\ f_n(x)$$, we see that $$ \displaystyle\ f^{(n+1)}_n(x)=0$$.

Substituting this back into our equation for $$ \displaystyle\ E^{(n+1)}_n(x)$$, we see that $$ \displaystyle\ E^{(n+1)}(x)=f^{(n+1)}-f^{(n+1)}_n(x)=f^{(n+1)}(x)-(0)=f^{(n+1)}(x)$$.

Author
Solved and Typed by - Egm6341.s10.Team4.riherd 14:56, 18 February 2010 (UTC) .

= Problem 7: Lagrange Error Term =

Given
Refer Lecture slide 12-3 for problem statement.

From Equation 4 of Lecture slide 10-1,

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

q_{n+1}(x) := (x - x_0)(x - x_1)(x - x_2)(x - x_3)\cdots(x - x_n) $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

Find
Show that the $$\displaystyle (n+1)^{th} $$ derivative of $$\displaystyle q_{n+1}(x) $$ is $$\displaystyle (n+1)! $$, i.e.,

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

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

 * }.
 * }.

Solution
Clearly from Equation (1),

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

q_{n+1}(x) \in \mathcal{P}_{n+1} $$
 * $$\displaystyle

 * }.
 * }.

where $$\displaystyle \mathcal{P}_{n+1} $$ is a set of polynomials of order $$\displaystyle \le n+1 $$.

So, Equation (1) can be written as,

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

q_{n+1}(x) = a_{n+1} x^{n+1} + a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{0} x^{0} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)$$
 * }.
 * }.

Please note that $$\displaystyle a_{n+1} = 1 $$ in Equation (2).

On differentiating Equation (2), 1 time,

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

q_{n+1}^{1}(x) = (n+1) x^{n} + na_{n} x^{n-1} + (n-1)a_{n-1} x^{n-2} + \cdots + a_{1} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)$$
 * }.
 * }.

On differentiating Equation (2), 2 times,

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

q_{n+1}^{2}(x) = (n+1)(n) x^{n-1} + n(n-1)a_{n} x^{n-2} + (n-1)(n-2)a_{n-1} x^{n-3} + \cdots + 2a_{2} $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)$$
 * }.
 * }.

On differentiating Equation (2), n times,

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

q_{n+1}^{n}(x) = (n+1)(n)\cdots(2) x^{1} + n(n-1)\cdots(2)a_{n}x^0 $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)$$
 * }.
 * }.

This can be realized by induction [in Equation (3), the constant term is $$\displaystyle a_1 $$ after differentiating 1 time; the constant term is $$\displaystyle 2a_2 $$ after differentiating 2 times;].

Differentiating Equation (5) one more time,

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

\begin{align} \Rightarrow q_{n+1}^{(n+1)}(x) &= (n+1)(n)\cdots(2)(1) \cancelto{1}{x^{1}} + n(n-1)\cdots(2)a_{n}\cancelto{0}{x^0} \\ &= (n+1)(n)\cdots(2)(1) \end{align} $$
 * $$\displaystyle

 * }.
 * }.

So,

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow q_{n+1}^{(n+1)}(x) = (n+1)!

$$

 * style= |

 * }.
 * }.

Author
Solved and Typed by - Egm6341.s10.Team4.andy 21:56, 17 February 2010 (UTC) .

= Problem 8: Lagrange Interpolation of the Natural Logarithm =

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

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

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

 * }
 * }

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

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

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

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

 * $$\displaystyle

f(x)=log(x) $$

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

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

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

 * $$\displaystyle

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

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

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

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

Matlab Code: .

Author
Solved and Typed by - Egm6341.s10.Team4.nimaa&m 02:39, 18 February 2010 (UTC) .

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

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

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

 * }.
 * }.

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

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

 * }.
 * }.

Solution

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

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

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

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

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

Author
Solved by --Egm6341.s10.Team4.roni 05:13, 18 February 2010 (UTC) and

Typed by - Egm6341.s10.Team4.nimaa&m 02:40, 18 February 2010 (UTC) .

= Problem 10: Proof of Error of Simple simpson's rule=

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

\begin{align} n=2, \\ & q_3(x)=(x-x_o)(x-x_1)(x-x_2)\\ & x_o=a, x_2=b, x_1=\frac{a+b}{2} \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

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

\left | E_2 \right | \le \frac{M_3}{3!}\int_{a}^{b}\left|(x-x_o)(x-x_1)(x-x_2) \right |\, dx = \frac{(b-a)^4}{192}M_3 $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

Solution

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

\begin{align} \int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right |\, dx &= 2\int_{a}^{\frac{a+b}{2}}(x-a)\Big(x-\frac{a+b}{2}\Big)(x-b)\,dx\\ &=2\int_{a}^{\frac{a+b}{2}}x^3-\frac{3(a+b)}{2}x^2+\frac{1}{2}(a^2+4ab+b^2)x - \frac{1}{2}ab(a+b)\, dx\\ &=\frac{1}{2}x^4-(a+b)x^3+\frac{1}{2}(a^2+4ab+b^2)x^2 - ab(a+b)x\, \Big |^{\frac{a+b}{2}}_a\\ &=\underbrace{\frac{1}{2}\Bigg(\Big(\frac{a+b}{2}\Big)^4-a^4\Bigg)}_{1st}-\underbrace{(a+b)\Bigg(\Big(\frac{a+b}{2}\Big)^3-a^3\Bigg)}_{2nd}+\underbrace{\frac{1}{2}(a^2+4ab+b^2)\Bigg(\Big(\frac{a+b}{2}\Big)^2-a^2\Bigg)}_{3rd} - \underbrace{ab(a+b)\Big(\frac{a+b}{2}-a\Big)}_{4th} \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

1st and 3rd term can be bundled by $$\displaystyle \frac{1}{2}\Bigg(\Big(\frac{a+b}{2}\Big)^2-a^2\Bigg)$$

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

\begin{align} &=\frac{1}{2}\Bigg(\Big(\frac{a+b}{2}\Big)^2-a^2\Bigg)\Bigg(\Big(\frac{a+b}{2}\Big)^2+a^2+(a^2+4ab+b^2)\Bigg)\\ &=\frac{1}{2}\Bigg(\Big(\frac{a+b}{2}\Big)+a\Bigg)\Bigg(\Big(\frac{a+b}{2}\Big)-a\Bigg)\Big(\frac{a^2+2ab+b^2}{4}+2a^2+4ab+b^2\Bigg)\\ &=\frac{1}{32}(3a+b)(b-a)(9a^2+18ab+5b^2)\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * {| style="width:１０0%" border="0" align="left"
 * {| style="width:１０0%" border="0" align="left"

$$\displaystyle =\frac{1}{32}(b-a)(27a^3+63a^2b+33ab^2+5b^3) $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (2)$$
 * style= |
 * }.
 * }.

2nd and 4th term can be bundled by $$\displaystyle -(a+b)\Big(\frac{a+b}{2}-a\Big)$$
 * {| style="width:100%" border="0" align="left"

\begin{align} &=-(a+b)\Big(\frac{a+b}{2}-a\Big)\Bigg(\Big(\frac{a+b}{2}\Big)^2+\frac{a+b}{2}a + a^2 + ab \Bigg)\\ &=-\frac{1}{8}(a+b)(b-a)(7a^2+8ab+b^2)\\ &=-\frac{1}{8}(b-a)(7a^3+15a^2b+9ab^2+b^3)\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * {| style="width:100%" border="0" align="left"

$$\displaystyle =-\frac{1}{32}(b-a)(28a^3+60a^2b+36ab^2+4b^3) $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (3)$$
 * style= |
 * }.
 * }.

Then, (2) + (3),
 * {| style="width:100%" border="0" align="left"

\begin{align} &\int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right |\, dx\\ &=\frac{1}{32}(b-a)(27a^3+63a^2b+33ab^2+5b^3)-\frac{1}{32}(b-a)(28a^3+60a^2b+36ab^2+4b^3)\\ &=\frac{1}{32}(b-a)\Big((27a^3+63a^2b+33ab^2+5b^3)-(28a^3+60a^2b+36ab^2+4b^3)\Big)\\ &=\frac{1}{32}(b-a)(-a^3+3a^2b-3ab^2+b^3)\\ &=\frac{1}{32}(b-a)(b-a)^3\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = \frac{1}{32}(b-a)^4 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (4)$$
 * style= |
 * }.
 * }.

So, plug in (4) to (1) equation.

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

$$\displaystyle = \left | E_2 \right | \le \frac{M_3}{3!}\int_{a}^{b}\left|(x-x_o)(x-x_1)(x-x_2) \right |\, dx = \frac{M_3}{3!}\Bigg(\frac{1}{32}(b-a)^4\Bigg) = \frac{(b-a)^4}{192}M_3 $$
 * 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.yunseok 02:04, 18 February 2010 (UTC) .

= Problem 11: Comparison of $$I $$ and $$I_2$$ of 3rd order polynomial =

Given
Refer Lecture slide 13-3 for problem statement
 * {| style="width:70%" border="0" align="left"

f(x)=6x^3-2x^2+8x+3 $$ x \in [0,1] $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

Find
Find $$\displaystyle I, I_2$$ (simpson's rule) and compare them.

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

\begin{align} &I=\int_{0}^{1} 6x^3-2x^2+8x+3,\ dx\\ &I_2=\frac{h}{3}\Big[f(x_0)+4f(x_1)+f(x_2)\Big] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Solution
$$\displaystyle I $$ is integrated like that
 * {| style="width:100%" border="0" align="left"

\begin{align} I&=\int_{0}^{1} 6x^3-2x^2+8x+3,\ dx\\ &= \frac{6}{4}x^4-\frac{2}{3}x^3+\frac{8}{2}x^2+3x \Big|^1_0 \\ &= \frac{3}{2}(1^4-0^4)-\frac{2}{3}(1^3-0^3)+4(1^2-0^2)+3(1-0)\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle I= \frac{3}{2}-\frac{2}{3}+4+3= \frac{47}{6} $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (1)$$
 * style= |
 * }.
 * }.

$$\displaystyle I_2 $$ is,
 * {| style="width:100%" border="0" align="left"

I_2=\frac{h}{3}\Big[f(x_0)+4f(x_1)+f(x_2)\Big] $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

\begin{align} &h=\frac{1}{2}\\ &f(x_0)=f(0)=3\\ &f(x_1)=f(\frac{1}{2})=6(\frac{1}{2})^3-2(\frac{1}{2})^2+8(\frac{1}{2})+3 = \frac{6}{8} - \frac{2}{4} + \frac{8}{2}+3 = \frac{29}{4}\\ &f(x_2)=f(1)=6(1)^3-2(1)^2+8(1)+3 =6-2+8+3 = 15\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

$$\displaystyle I_2=\frac{h}{3}\Big[f(x_0)+4f(x_1)+f(x_2)\Big] = \frac{(\frac{1}{2})}{3}\Big[3+4\frac{29}{4}+15\Big]=\frac{47}{6} $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (2)$$
 * style= |
 * }.
 * }.

in coclusion, From equation $$\displaystyle (1) ,\ (2)$$,

$$\displaystyle I$$ and $$\displaystyle I_2$$ is same for 3rd polynomial equation

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

$$\displaystyle I=I_2=\frac{47}{6} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and Typed by - Egm6341.s10.Team4.yunseok 02:10, 18 February 2010 (UTC) .

=Problem 12: Evaluation of $$ \alpha^{(1)}(t)$$=

Given
$$ \displaystyle\ \alpha(t)=\int_{-t}^{t} f(x(t'))dt'$$ where $$ \displaystyle\ f(x(t))=F(t)$$.

Find
Show that $$ \displaystyle\ \alpha^{(1)}(t)=F(-t)+F(t) $$.

Solution
We are asked to show that for $$ \displaystyle\ \alpha(t)=\int_{-t}^{t} f(x(t'))dt'=\int_{-t}^{t} F(t')dt',\alpha^{(1)}(t)=F(-t)+F(t)$$, where $$ \displaystyle\ f(x(t))=F(t)$$.

Using The Fundamental Theorem of Calculus (Second Part), we know that for some differentiable function $$ \displaystyle\ G(t)$$ with with derivative $$ \displaystyle\ g(t)=G'(t)$$, $$ \displaystyle\ \int_a^b g(t)dt=G(b)-G(a)$$.

Allowing $$ \displaystyle\ f(x(t))=F(t)$$ and $$ \displaystyle\ G'(t)=F(t)$$, then $$ \displaystyle\ \alpha(t)=\int_{-t}^{t} F(t')dt'=G(t)-G(-t)$$.

If the derivative of $$ \displaystyle \alpha $$ is now taken, the $$ \displaystyle\ \alpha^{(1)}(t)=\frac{d}{dt}(G(t)-G(-t))=G'(t)+G'(-t)$$.

Substituting in our definitions $$ \displaystyle\ F(t)=G'(t)$$ we see that

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

Author
Solved and Typed by - Egm6341.s10.Team4.riherd 14:55, 18 February 2010 (UTC) .

= Problem 13: Evaluation of $$ G^{2}(0) $$ =

Given
Refer Lecture slide 15-2 for problem statement

From Lecture slide 15-1,

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

G^{1}(t) = e^{1}(t) - 5t^4e(1) $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (1)$$
 * }
 * }

where,

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

e(t) := \alpha(t) - \alpha_{2}(t) $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (2)$$
 * }
 * }

where $$\displaystyle \alpha(t), \alpha_{2}(t) $$ are given by,

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

\alpha(t) := \int_{-t}^{t}F(t)dt $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (3)$$
 * }
 * }

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

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

and,

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

F(t) := f(x(t)) $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (5)$$
 * }.
 * }.

Find
Show that,

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

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

 * }.
 * }.

Solution
On differentiating Equation (3) [Please refer differentiating an integral and Problem 14 Solution below],

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

\alpha^{1}(t) = F(-t) + F(t) $$.
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (6)$$
 * }
 * }

Again differentiating Equation (6),

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

\alpha^{2}(t) = -F(-t) + F(t) $$.
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (7)$$
 * }
 * }

On differentiating Equation (4),
 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow \alpha_{2}^{1}(t) &= \frac{1}{3} \left[ F(-t) + 4F(0) + F(t) \right] + \frac{t}{3} \left[-F^{1}(-t) + \cancelto{0}{4F^{1}(0)} + F^{1}(t)\right] \\ &= \frac{1}{3} \left[ F(-t) + 4F(0) + F(t) \right] + \frac{t}{3} \left[-F^{1}(-t) + F^{1}(t)\right] \end{align} $$
 * $$\displaystyle

 * }.
 * }.

Again on differentiating the above equation,

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

\begin{align} \Rightarrow \alpha_{2}^{2}(t) &= \frac{1}{3} \left[ -F^{1}(-t) + \cancelto{0}{4F^{1}(0)} + F^{1}(t) \right] + \frac{1}{3} \left[-F^{1}(-t) + F^{1}(t)\right] + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]\\ &= \frac{2}{3} \left[ -F^{1}(-t) + F^{1}(t) \right] + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right] \end{align} $$
 * $$\displaystyle

 * }.
 * }.

So,

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

 * $$\displaystyle

\Rightarrow \alpha_{2}^{2}(t) = -\frac{2}{3} F^{1}(-t) + \frac{2}{3} F^{1}(t) + \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right]

$$
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (8)$$
 * }.
 * }.

On differentiating Equation (1) we get,

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

G^{2}(t) = e^{2}(t) - 20t^3e(1) $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (9)$$
 * }.
 * }.

At point $$\displaystyle t = 0 $$,

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

G^{2}(0) = e^{2}(0) - \cancelto{0}{20(t)^3e(1)} \Rightarrow G^{2}(0) = e^{2}(0) $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (10)$$
 * }.
 * }.

$$\displaystyle e^{2}(t) $$ is given by differentiating Equation (2) twice,

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

e^{2}(t) = \alpha^{2}(t) - \alpha_{2}^{2}(t) $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (11)$$
 * }.
 * }.

Substituting Equations (7) and (8) in (11),

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

\Rightarrow e^{2}(t) = -F(-t) + F(t) +\frac{2}{3} F^{1}(-t) - \frac{2}{3} F^{1}(t) - \frac{t}{3} \left[F^{2}(-t) + F^{2}(t)\right] $$
 * $$\displaystyle

 * }.
 * }.

At point $$\displaystyle t = 0 $$,

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

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

 * }.
 * }.

So,

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

 * $$\displaystyle

\Rightarrow e^{2}(0) = 0

$$

 * }.
 * }.

which means from Equation (10),

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

$$\displaystyle
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

\Rightarrow G^{2}(0) = 0

$$

 * style= |

 * }.
 * }.

Author
Solved and Typed by - Egm6341.s10.Team4.andy 21:56, 17 February 2010 (UTC) .

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

Given
If we define function e(t) as the following,

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

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

 * }
 * }

Find
Show that:

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

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

 * }
 * }

Solution
Let e(t) be,

 * $$\displaystyle

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Author
Solved and Typed by - Egm6341.s10.Team4.nimaa&m 02:42, 18 February 2010 (UTC) .

= Problem 15: Relationship between $$ \xi$$ and $$ \zeta_4$$=

Given

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

 * $$ \displaystyle x=\alpha + ht$$.

 * }.
 * }.

where,

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

 * $$ \displaystyle h= \frac{(b-a)}{2}$$.

 * }.
 * }.

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

 * $$\displaystyle \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$

 * }.
 * }.

and hence deduce a relationship between $$\displaystyle \xi$$ and $$\displaystyle \zeta_4$$

Solution
We have F(t)= f(x(t))= f(x). Given

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

 * $$ \displaystyle x=\alpha + ht$$.

 * }.
 * }.

where,

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

 * $$ \displaystyle h= \frac{(b-a)}{2}$$.

 * }.
 * }.

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

 * $$\displaystyle F^'(t)= \frac{dF(t)}{dt}= \frac{df(x(t))}{dt}= \frac{df(x)}{dt}= \frac{df(x)}{dx}\cdot\frac{dx}{dt}= f^'(x)\cdot h$$

 * }.
 * }.

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

 * $$\displaystyle F^{}(t)= \frac{d}{dt}\cdot \frac{dF(t)}{dt}= \frac{d}{dt}\cdot(f^'(x)h) = h\cdot \frac{df^'(x)}{dt}= h\cdot \frac{df^'(x)}{dx}\cdot \frac{dx}{dt}= h^2\cdot f^{}(x)$$

 * }.
 * }.

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

 * $$\displaystyle F^{}(t)= \frac{d}{dt}\cdot (F^{}(t))= \frac{d}{dt}\cdot(h^2\cdot f^{}(x)) = h^2\cdot \frac{df^{}(x)}{dt}= h^2\cdot \frac{df^{}(x)}{dx}\cdot \frac{dx}{dt}= h^3\cdot f^{}(x)$$

 * }.
 * }.

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

 * $$\displaystyle F^{'}(t)= \frac{d}{dt}\cdot (F^{}(t))= \frac{d}{dt}\cdot(h^3\cdot f^{}(x)) = h^3\cdot \frac{df^{}(x)}{dt}= h^3\cdot \frac{df^{}(x)}{dx}\cdot \frac{dx}{dt}= h^4\cdot f^{'}(x)$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle F^4(t)=h^4\cdot f^4(x)$$
 * }.
 * }.

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

 * $$\Rightarrow \displaystyle F^4(t)=\frac{(b-a)}{16} ^4\cdot f^4(x)$$
 * }.
 * }.

Substituting $$\displaystyle \xi$$ for $$\displaystyle x$$ and $$\displaystyle \zeta_4$$ for $$\displaystyle t$$

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

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

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

 * $$\Rightarrow \displaystyle \frac{-F^4(\zeta_4)}{90}= \frac{-(b-a)^4}{1440}f^4(\xi) $$
 * }.
 * }.

Since $$ \displaystyle x=\alpha + ht$$, We have $$ \displaystyle \xi=\alpha + h\zeta_4$$ This is the required relation between $$\displaystyle \xi$$ and $$\displaystyle \zeta_4$$. Hence proved.

Author
Solved and Typed by - Egm6341.s10.team4.anandankala 13:43, 19 February 2010 (UTC) .

= Problem 16: Runge phenomenon =

Given
Refer Lecture slide 16-1 for problem statement
 * {| style="width:100%" border="0" align="left"

I = \int^{5}_{-5} \frac{1}{1+x^2} dx $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

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

(2) Plot $$\displaystyle f, f_n $$ $$\mbox{,  }$$ for n=1, 2, 3, 8, 12

(3) Plot $$\displaystyle I_n$$ vs. $$\displaystyle n$$ and observe that $$\displaystyle I_n$$ does not converge as $$\displaystyle n \rightarrow \infty $$

(4) Observe weight $$\displaystyle W_{i,n}:=\int^{b}_{a} l_{i,n}(x),\ dx$$ are not all positive for n $$ \ge $$ 8. Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.

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

 * $$\displaystyle
 * $$\displaystyle

$$

 * }.
 * }.

Solution
(1) Find $$\displaystyle I_n$$ using Newton-Cotes for n=1, 2,. . ., 15

Newton-cotes formula $$\displaystyle P_n$$,

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

P_n = \sum_{i=0}^n l_{i,n}(x)f(x_i)$$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

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

l_{i,n}(x) = \prod_{j=0,j\ne i}^n \frac{x-x_j}{x_i-x_j}$$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

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

I_n = \int^{5}_{-5}P_n(x) dx = \sum_{i=0}^n \underbrace{\int^{5}_{-5}l_{i,n}(x) dx}_{W_i} f(x_i) = \sum_{i=0}^n w_i f(x_i) $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (2)$$
 * }.
 * }.

From the lecture slide 13-1,

Although this integration can be done numerically, there ia an another approach(analyitical solution) that derive the w(i) weight values directly.

the Error of integration of Newton-Cotes formula

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

\left | E_n(f(x)) \right | = \left | I-I_n \right | \le \int^{b}_{a} \left | f(x)-P_n(x)\right |  dx \le \frac {\int^{b}_{a}\left | q_{n+1}(x) \right| dx}{(n+1)!}\Big( max f^{(n+1)}(\xi), \xi \in [a,b] \Big)$$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (3)$$
 * }.
 * }.

From the equation (2),

We know the error is zero when the $$\displaystyle f(x) $$ is the polynomial of degree less than $$n$$

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

E_n(f(x)) = 0 $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

because, for the polynomial of degree less than $$\displaystyle n (\mathcal P_n)$$

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

if\quad f(x) \in \mathcal P_n, \qquad f^{(n+1)}(x)=0 $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

In order to find the $$\displaystyle w_i$$ in equation (2), we can use n equations and solve using matrix method like below,

for the $$\displaystyle f(x)=1,\; x, \; x^2, \; ...... \; x^n$$ (simplist and representative polynomial less than nth order), plug f(x) into equation(3)

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

E_n(f(x))= E_n(1)=E_n(x)=E_n(x^2)=......=E_n(x^n)=0 $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

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

\begin{align} &E_n(1)= \int^{5}_{-5}1 dx -[w_1(1)+w_2(1)+w_3(1)+......+w_n(1)] =0\\ &E_n(x)= \int^{5}_{-5}x dx -[w_1(x_0)+w_2(x_2)+w_3(x_3)+......+w_n(x_n)] =0\\ &E_n(x^2)= \int^{5}_{-5}x^2 dx -[w_1(x_0^2)+w_2(x_1^2)+w_3(x_3^2)+......+w_n(x_n^2)] =0\\ &E_n(x^3)= \int^{5}_{-5}x^3 dx -[w_1(x_0^3)+w_2(x_1^3)+w_3(x_3^3)+......+w_n(x_n^3)] =0\\ &.......................\\ &..............................\\ &......................................\\ &E_n(x^n)= \int^{5}_{-5}x^n dx -[w_1(x_0^n)+w_2(x_1^n)+w_3(x_3^n)+......+w_n(x_n^n)] =0 \end{align} $$
 * $$\displaystyle
 * }.
 * }.

Hence, the $$\displaystyle w_i$$ may be determined from

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

\underbrace{ \begin{bmatrix} 1 & 1 & 1 & 1 & ...... & 1\\ x_0 & x_1 & x_2 & x_3 & ...... & x_n\\ x_0^2 & x_1^2 & x_2^2 & x_3^2 & ...... & x_n^2\\ x_0^3 & x_1^3 & x_2^3 & x_3^3 & ...... & x_n^3\\ \vdots & \vdots& \vdots& \vdots& \vdots & \vdots \\ x_0^n & x_1^n & x_2^n & x_3^n & ...... & x_n^n\\ \end{bmatrix} }_{A} \underbrace{ \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\\ \vdots\\ w_n\\ \end{bmatrix} }_{W} = \underbrace{ \begin{bmatrix} \int^{5}_{-5}1 dx\\ \int^{5}_{-5}x dx\\ \int^{5}_{-5}x^2dx\\ \int^{5}_{-5}x^3dx\\ \vdots \\ \int^{5}_{-5}x^ndx\\ \end{bmatrix} = \begin{bmatrix} (b-a)\\ \frac{(b^2-a^2)}{2}\\ \frac{(b^3-a^3)}{3}\\ \frac{(b^4-a^4)}{4}\\ \vdots \\ \frac{(b^{n+1}-a^{n+1})}{n+1} \end{bmatrix} }_{B} $$
 * $$\displaystyle
 * }.
 * }.

$$\displaystyle W_i$$ can be obtained like this,
 * {| style="width:100%" border="0" align="left"

\begin{align} &AW=B\\ &W_i=A^{-1}B \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

plug obtained $$\displaystyle W_i$$ and \displaystyle f(x)=\frac{1}{1+x^2} into equation(2), $$\displaystyle I_n$$ can be calculated,
 * {| style="width:100%" border="0" align="left"

I_n = \sum_{i=0}^n w_i f(x_i) $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

$$ \begin{array}{|c||c|c|} n & I_n \\ \hline 1&0.384615\\ 2&6.794872\\	3&2.081448\\ 4&2.374005\\ 5&2.307692\\ 6&3.870449\\	7&2.898994\\	8&1.500489\\ 9&2.398618\\ 10&4.673301\\ 11&3.244773\\ 12&-0.31294\\ 13&1.919797\\ 14&7.899545\\ 15&4.155559\\ \hline \end{array} $$

Matlab Code:

(2) Plot $$\displaystyle f, f_n $$ $$\mbox{,  }$$ for n=1, 2, 3, 8, 12

$$\displaystyle f, f_n $$ were plotted by below matlab code.

Matlab Code:

(3) Plot $$\displaystyle I_n$$ vs. $$\displaystyle n$$ and observe that $$\displaystyle I_n$$ does not converge as $$\displaystyle n \rightarrow \infty $$

(4) Observe weight $$\displaystyle W_{i,n}:=\int^{b}_{a} l_{i,n}(x),\ dx$$ are not all positive for n $$ \ge $$ 8. Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.

This is the table that show the $$\displaystyle W_{i,n}:=\int^{b}_{a} l_{i,n}(x),\ dx$$ for n=8

 * Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.
 * Plot $$\displaystyle l_{i,n}(x)$$ for i=1, 2, ... ,8 & n=8.

 * Matlab Code:

.

Author
Solved - --Egm6341.s10.Team4.roni 05:15, 18 February 2010 (UTC)

Typed and reviewed by - Egm6341.s10.Team4.yunseok 02:06, 18 February 2010 (UTC) .

= Contributing Team Members =

Egm6341.s10.Team4.andy 21:57, 17 February 2010 (UTC) - Author of Problems:1,7,13 ; Proof Read: 2,3,4,10,14,15

Egm6341.s10.Team4.riherd 03:16, 10 February 2010 (UTC) - Author of Problems: 6,12

Egm6341.s10.Team4.nimaa&m 13:51, 19 February 2010 (UTC) 09:35, 10 February 2010 (UTC) - Author of Problems: 2, 8, 9, 14; Proof Read: 4

Egm6341.s10.team4.anandankala 3:25, 10 February 2010 (UTC)- Author of Problems: 3,15; Proof Read: 1,7,9.

Egm6341.s10.Team4.roni 05:54, 10 February 2010 (UTC) Author of Problems:4,9,16 ; Proof Read: 4,9,16

Egm6341.s10.Team4.yunseok 16:00, 10 February 2010 (UTC) Author of Problems:5,10,11 ; Proof Read: 16

title: User:Egm6341.s10.Team4/HW2