text:
User:Egm6341.s10.Team4.roni/HW5/5.10

= Problem 10: Application of Engineering Orbital Mech. =

Given
Refer Lecture slide 30-4 for problem statement

Polar form relative to focus
 * {| style="width:100%" border="0" align="left"

 * [[Image:Ellipse Polar.svg|200px]]
 * }.
 * }.
 * }.

Eq (3) from Page [[media:Egm6341.s10.mtg30.djvu|30-3]]

$$\displaystyle \ \ r(\theta) = \frac{a(1-e^2)}{1-e \, cos(\theta)} $$

In our problem, a=1

where, eccentricity $$\displaystyle e = \sin(\frac{\pi}{12}) $$

First Method to compute the Arc Length for an elliptical curve is using Eq (4) from Page [[media:Egm6341.s10.mtg30.djvu|30-3]] Since for the Ellipse $$\displaystyle \theta$$ range is from 0 to 2 $$\displaystyle\pi $$

Circumference $$\displaystyle I( \theta) = \int_0^{2\pi} dl$$

Where Eq (1) $$\displaystyle dl= \sqrt{ {r}^{2}+ {\left(  \frac{dr}{d \theta}\right)}^{2}} d\theta $$.

The second method uses the elliptic integral of the second kind which is defined as:

$$E(e^2) = \int_0^{\pi/2}\sqrt {1-e^2 \sin^2\theta}\ d\theta\ $$    Eq(6)    in page     [[media:Egm6341.s10.mtg30.djvu|30-4]]

The circumference of an ellipse is $$\displaystyle C = 4 a E(e^2)$$    Eq(5)   in page   [[media:Egm6341.s10.mtg30.djvu|30-4]],

Giving the integral:

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

$$\displaystyle I= 4 E(e^2) = 4\int_0^{\pi/2}\sqrt {1-e^2 \sin^2\theta}\ d\theta $$

Find
Find Arc length of ellipse using the two Integrals and the integration methods below.

compute $$\displaystyle I_n$$ to the error $$\displaystyle 10^{-10}$$ ,

compute time using tic/toc matlab command and error estimate

1) Composit Trapozoidal rule

2) Romberg Table

3) Clencurt

Solution
The Integration result are summarized in the following table:

Note that checking both integral calculations against Ramanujan approximation 's:
 * $$C \approx \pi \left[3(a+b) - \sqrt{(3a+b)(a+3b)}\right]= \pi(3(a+b)-\sqrt{10ab+3(a^2+b^2)})$$

where, a = longest radius, b= shortest radius in ellipse

the results of our integration method and Ramanujan approximation were same. $$\displaystyle (c \simeq 6.176601987)$$

Discussion: One can clearly see that for these integrals the Trapezoidal method gives superior performance This is due to the fact that the Integrand is a periodic function. Clencurt has slightly better performance than Romberg. First method Integral gives lower quality on all methods and need more CPU time for similar performance.

(1) Composite Trapezoidal rule

 Matlab Code:

 subfunction ff5_10(x)

(2) Romberg Table

 Matlab Code:

(3) ClenCurt Integration)

 Matlab Code:

Author
Solved and typed by -

Proof read by -

title: User:Egm6341.s10.Team4.roni/HW5/5.10


text:
User:Egm6341.s10.Team4.andy/HW5

= Problem 1: Trapezoidal Rule Error Proof: Step 3b=

Given
Refer Lecture slide 26-3 for problem statement.

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

$$\displaystyle p_5(t)=\int p_4(t)dt = -\frac{t^5}{120} + \frac{t^3}{36} + c_5t + c_6 $$ $$
 * $$\displaystyle \longrightarrow (1)
 * }.
 * }.

Find
Selecting $$\displaystyle p_5(\pm 1)= 0 $$ and $$\displaystyle p_5(0)= 0 $$, find $$\displaystyle c_5 $$ and $$\displaystyle c_6 $$.

Solution
When $$\displaystyle p_5(0)= 0 $$ Equation (1),

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

$$\displaystyle p_5(0)= -\frac{0^5}{120} + \frac{0^3}{36} + c_5(0) + c_6 = 0 $$
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_6 = 0 $$
 * }.
 * }.

When $$\displaystyle p_5(1)= 0 $$ Equation (1),

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

$$\displaystyle p_5(1)= -\frac{1^5}{120} + \frac{1^3}{36} + c_5(1) = 0 $$
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_5 = \frac{1}{120} - \frac{1}{36} = -\frac{7}{360} $$
 * }.
 * }.

= Problem 6: Trapezoidal Rule Error by Canceling Odd Derivative Orders of $$ g(t)$$=

Given
Refer Lecture slide 28-2 for problem statement.

Higher Order Error for Trap. rule is given by [ Refer 21-1 ]:

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

E_n^1=\frac{h}{2} \sum_{k=0}^{n-1} \left[ \underbrace{\int_{-1}^{1}g_k(t)dt-(g_k(-1)+g_k(+1))}_{E}\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)
 * }.
 * }.

where $$\displaystyle E $$ can be written as,

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

E =\int_{-1}^{1}(-t)g_k^{(1)}(t)dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)
 * }.
 * }.

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

$$\displaystyle \begin{align} \Rightarrow E &=\int_{-1}^{1}\underbrace{(-t)}_{p_{1}(t)}g_k^{(1)}(t)dt \\ &= \left[p_2(t)g_k^{(1)}(t) \right]_{-1}^{+1} - \int_{-1}^{+1}p_2(t)g_k^{(2)}(t)dt \end{align} $$

where $$\displaystyle p_2(t) = \int p_1(t)dt$$.
 * }.
 * }.

After successive integration by parts,

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

$$\displaystyle

E = \left[p_2(t)g_k^{(1)}(t)-p_3(t)g_k^{(2)}(t)+p_4(t)g_k^{(3)}(t)-p_5(t)g_k^{(4)}(t)+p_6(t)g_k^{(5)}(t)+ \cdots +p_{\ell}(t)g_k^{(\ell-1)}(t)\right]_{-1}^{+1} - \int_{-1}^{1}p_{\ell}(t)g_k^{(\ell)}(t) dt

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

 * }.
 * }.

Find
Try to derive the Higher Order Trapezoidal Rule Error equation by canceling odd order derivative terms of $$\displaystyle g(t) $$, i.e., make terms $$\displaystyle p_{2}(t), p_{4}(t), p_{6}(t), \cdots, $$ to zero at $$\displaystyle t =  \pm 1$$.

Solution
We know,

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

$$\displaystyle p_{2}(t) = -\frac{t^2}{2} + c_3 $$
 * }.
 * }.

At $$\displaystyle t = \pm 1$$,

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

$$\displaystyle p_{2}(1) = -\frac{1^2}{2} + c_3 = 0; p_{2}(-1) = -\frac{(-1)^2}{2} + c_3 = 0; $$
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_3 = \frac{1}{2} $$
 * }.
 * }.

We also know,

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

$$\displaystyle p_{3}(t) = -\frac{t^3}{3!} + \frac{t}{2} + c_4 $$ And,
 * }.
 * }.
 * {| style="width:100%" border="0" align="left"

$$\displaystyle p_{4}(t) = -\frac{t^4}{4!} + \frac{t^2}{4} + c_4t + c_5 $$
 * }.
 * }.

At $$\displaystyle t = \pm 1$$,

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

$$\displaystyle p_{4}(1) = -\frac{(1)^4}{4!} + \frac{(1)^2}{4} + c_4(1) + c_5; p_{4}(-1) = -\frac{(-1)^4}{4!} + \frac{(-1)^2}{4} + c_4(-1) + c_5; $$

 * }.

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

$$\displaystyle \Rightarrow c_4 = 0; c_5 = \frac{-5}{24} $$
 * }.
 * }.

Functions $$\displaystyle p_{3}(t), p_{5}(t), p_{7}(t), \cdots, $$ are odd functions.

Since functions $$\displaystyle p_{2}(t), p_{4}(t), p_{6}(t), \cdots, $$ are made zero at $$\displaystyle t = \pm 1$$, Equation (3) becomes,

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

$$\displaystyle \begin{align} E & = \left[-p_3(t)g_k^{(2)}(t)-p_5(t)g_k^{(4)}(t)-p_7(t)g_k^{(6)}(t)\cdots\cdots-p_{(2\ell-1)}(t)g_k^{(2\ell-2)}(t)\right]_{-1}^{+1} + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt \\ & = \sum_{r=2}^{\ell} \left[ -p_{(2r-1)}g_k^{(2r-2)}\right]_{-1}^{+1} + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt \\ & = \sum_{r=2}^{\ell} \left[ -p_{(2r-1)}(1)g_k^{(2r-2)}(1) + p_{(2r-1)}(-1)g_k^{(2r-2)}(-1)\right] + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt \\ \end{align} $$

 * }.
 * }.

Since $$\displaystyle p_{(2r-1)}(t) $$ are odd functions, we know $$\displaystyle p_{(2r-1)}(1) + p_{(2r-1)}(-1) = 0 $$.

And so $$\displaystyle E $$ becomes,

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

$$\displaystyle

E = \sum_{r=2}^{\ell} p_{(2r-1)}(-1) \left[ g_k^{(2r-2)}(1) + g_k^{(2r-2)}(-1)\right] + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt

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

 * }.
 * }.

Using the below relationship

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

g_k^{(i)}(t)=\left(\frac{h}{2}\right)^if^{(i)}(x(t)); \ \ \ \ \ x \in [x_k,x_{k+1}] $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

In Equation (4) $$\displaystyle g_k(t)$$can be transferred to $$\displaystyle f(x_k)$$ as shown below:

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

\begin{align} &g_k^{(2r-2)}(+1)=\left(\frac{h}{2}\right)^{(2r-2)}f^{(2r-2)}(x_{k+1}),\\ &g_k^{(2r-2)}(-1)=\left(\frac{h}{2}\right)^{(2r-2)}f^{(2r-2)}(x_{k}),\\ &g_k^{(2\ell-1)}(t)=\left(\frac{h}{2}\right)^{(2\ell-1)}f^{(2\ell-1)}(x) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

And so, Equation (4) becomes,

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

$$\displaystyle

E =  \sum_{r=2}^{\ell}\left[ \left(\frac{h}{2}\right)^{(2r-2)} p_{(2r-1)}(-1) \Big[ f^{(2r-2)}(x_{k+1}) + f^{(2r-2)}(x_{k})\Big] \right] + \left(\frac{h}{2}\right)^{(2\ell-1)} \int_{x_k}^{x_{k+1}}p_{(2\ell-1)}(t_k(x))f^{(2\ell-1)}(x) dx

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

 * }.
 * }.

Substituting Equation (5) in Equation (1),

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

$$\displaystyle \begin{align} E_n^1 &= \sum_{r=2}^{\ell}\left[ \left(\frac{h}{2}\right)^{(2r-1)} p_{(2r-1)}(-1) \sum_{k=0}^{n}\Big[ f^{(2r-2)}(x_{k+1}) + f^{(2r-2)}(x_{k})\Big] \right] + \left(\frac{h}{2}\right)^{(2\ell)} \sum_{k=0}^{n} \left [\int_{-1}^{1}p_{(2\ell-1)}(t_k(x))f^{(2\ell-1)}(x) dx \right ] \\ & = \sum_{r=2}^{\ell}\left[ \left(\frac{h}{2}\right)^{(2r-1)} p_{(2r-1)}(-1) \Big[ \underbrace{f^{(2r-2)}(x_n) + f^{(2r-2)}(x_{n-1}) + \cdots \cdots + f^{(2r-2)}(x_{0})}_{\beta}\Big] \right] + \left(\frac{h}{2}\right)^{(2\ell)} \sum_{k=0}^{n} \left [\int_{-1}^{1}p_{(2\ell-1)}(t_k(x))f^{(2\ell-1)}(x) dx \right ] \end{align}

$$
 * }.
 * }.

In the above error equation $$\displaystyle \beta \neq f^{(2r-2)}(x_n) - f^{(2r-2)}(x_0)$$, but it contains terms that involve all the points from $$\displaystyle x_0 \ to\ x_n $$.

But instead of canceling out terms that involve odd order derivatives of $$\displaystyle g(t)$$, if we had canceled only terms that involve even order derivatives of $$\displaystyle g(t)$$ then the term $$\displaystyle \beta $$ would involve only first and last points, i.e., $$\displaystyle x_0 = a $$ and $$\displaystyle x_n = b $$, as we have got in Problem 5 above.

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

title: User:Egm6341.s10.Team4.andy/HW5


text:
User:Egm6341.s10.team3.ynahn81/HW5-11

= (11) Derivation of the formula to obtain arc length of ellipse from cosine raw = Ref: Lecture notes [[media:Egm6341.s10.mtg31.djvu|p.31-1]]

Problem Statement
Derive the following formula for arc length of ellipse from cosine raw.

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

Arclength(PQ) = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^{2}\right]^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution

Recall the cosine raw:

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

\overline{AB}^2=\overline{OA}^2+\overline{OB}^2-2\overline{OA}\,\overline{OB}\cos d\theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

To simply the derivation, let's use the following notations.

$$\displaystyle \overline{AB} = dl $$   $$\displaystyle \overline{OA} = r(\theta) = r $$   $$\displaystyle \overline{OB} = r(\theta+d\theta) \approx r + dr $$

Also, from the Taylor series expansion of $$\displaystyle \cos d\theta $$,
 * {| style="width:100%" border="0" align="left"

\cos d\theta = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}d\theta^{2n} = 1 - \frac{d\theta^2}{2!} + \frac{d\theta^4}{4!} - \cdots, $$  we can use the following approximated expression since the angle $$\displaystyle d\theta $$ is extremely small,
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\cos d\theta \approx 1 - \frac{d\theta^2}{2} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

As a consequence, (Eq.1) can be expressed as below,

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

\begin{align} dl^2 &= r(\theta)^2+r(\theta+d\theta)^2-2r(\theta)r(\theta+d\theta)\cos d\theta\\ &\approx r^2 + (r+dr)^2 -2r(r+dr)\left(1-\frac{1}{2}d\theta^2\right)\\ &=2r^2+2rdr+dr^2-(2r^2+2rdr)+(r^2+rdr)d\theta^2\\ &=dr^2+r^2d\theta^2+\underbrace{rdrd\theta^2}_{\approx 0}\\ &\approx d\theta^2\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

dl = d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Finally, the arc length PQ can be obtained by

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

$$\displaystyle Arclength(PQ)=\int_{\theta_{P}}^{\theta_{Q}}dl = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Above result is exactly same with (2) [[media:Egm6341.s10.mtg30.djvu|p.30-3]].

title: User:Egm6341.s10.team3.ynahn81/HW5-11


text:
University of Florida/Egm6341/s10.team3.aks/HW5

(5) Derive Trapezoidal Error expression
Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-1]]

Problem Statement
Derive equation 1 from Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Solution
We have

$$ E = \sum_{r=1}^l P_{(2r)}(+1) [g^{(2r-1)}- g^{(2r-1)}(-1)] - \int\limits_{-1}^{+1} P_{(2l)}(t) g^{(2l)}(t) dt]$$

and $$ g_k^i(t) = (\frac{h}{2})^i f^i(x(t)) $$

$$ g^{(2r-1)}(t) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (x(t)) $$

$$ g^{(2r-1)}(+1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (x(1)) $$

and $$ x(1) = b $$

$$ g^{(2r-1)}(+1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (b) $$

Similarly

$$ g^{(2r-1)}(-1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (a) $$

and $$ g^{(2l)} = (\frac{h}{2})^{(2l)} f^{(2l)} (x) $$

using above equations into first equation we obtain

$$ E = \sum_{r=1}^l P_{(2r)}(+1) \frac{h}{2} \frac{h}{2})^{(2r-1)} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - \sum_{k=1}^{n-1} \int\limits_{-1}^{+1} P_{(2l)}(t_k) (\frac{h}{2})^{2l} f^{(2l)}(x) dx]$$

$$ = \sum_{r=1}^l h^{2r} \frac{P_{(2r)}(+1)}{2^{2r}}[f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

but $$ \frac{P_{(2r)}(+1)}{2^{2r}} = \overline{d}_{2r}$$

So finally we obtain

$$ = \sum_{r=1}^l h^{2r} \overline{d}_{2r} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

(8) Obtain expressions for (P2,P3),(P4,P5),(P6,P7) Using recurrence formula
Ref: Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]]

Problem Statement
Obtain expressions for $$ (P_2,P_3),(P_4,P_5),(P_6,P_7) $$ using eq (6) Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-2]]

Solution
$$ (P_2,P_3)$$

$$ P_{2i}(t) = \sum_{j=0}^i C_{2j+1} \frac{t^{2(i-j)}}{(2(i-j))!} $$

$$ P_{2}(t) = \sum_{j=0}^i C_{2j+1} \frac{t^{2(1-j)}}{(2(1-j))!} $$

$$ P_{2}(t) = C_1 \frac{t^{2)}}{2!} + C_3 $$

Using eq (6) from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac {C_1}{3!}+ C_3 = 0 $$

$$ C_3 = - \frac {C_1}{3!} $$

$$    = - \frac{-1}{3!} = \frac{1}{6} $$

so $$ P_{2}(t) = - \frac{t^2}{2} + \frac{1}{6} $$

$$ P_3 = \sum_{j=0}^i C_{2j+1} \frac{t^{2(i-1)}}{(2(i-j))!} + C_{2i+2} $$

where $$ C_{2i+2} = 0  $$

$$ P_3 = C_1 \frac{t}{3!} + C_3 \frac{t}{1!}$$

where $$ C_1 = -1 and C_3 = \frac{1}{6} $$ from previous result

$$ P_3 = - \frac{t}{3!} + \frac{t}{6} = 0 $$

$$ P_4, P_5 $$

$$ P_4 = C_1 \frac{t^4}{4!} + C_3 \frac{t^2}{2!}+ C_5 $$

Using eq (6) from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac{C_1}{5!} + \frac {C_3}{3!}+\frac {C_5}{1!} = 0 $$

$$ C_5 = \frac {1}{5!} - \frac{1}{36} = \frac {-7}{360}$$

$$ P_4 = - \frac{t^4}{24} + \frac{t^2}{12} - \frac{7}{360} $$

$$ P_5 = C_1 \frac{t^3}{5!} + C_3 \frac{t^3}{3!}+ C_5 \frac{t^3}{1!} $$

$$ P_5 = (\frac {C_1}{5!} + \frac {C_3}{3!}+ \frac {C_5}{1!})t^3 $$

$$ P_5 = 0 $$

$$ P_6, P_7$$

$$ P_6 = C_1 \frac{t^6}{12!} + C_3 \frac{t^4}{10!} + C_5 \frac{t^2}{8!} + C_7 \frac{1}{6!} $$

By using eq 6 from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac{C_1}{7!} + \frac {C_3}{5!}+\frac {C_5}{3!} + C_7 = 0 $$

$$ C_7 = \frac{1}{7!} - \frac{1}{6!} - \frac{7}{2160}$$

$$   = \frac{11} {15120} $$

$$ P_6 = - \frac{t^6}{12!} + \frac{t^4}{6*10!}- \frac{7 t^2}{360 * 8!} + \frac {11}{15120} $$

$$ P_7 = C_1 \frac {t^5}{7!} + C_3 \frac {t^5}{5!} + C_5 \frac{t^5}{3!}+C_7 \frac{t^5}{1!} $$

$$ P_7 = (\frac {C_1}{7!} + \frac {C_3}{5!} + \frac {C_5}{3!}+ \frac {C_7}{1!}) t^5 $$

$$ P_7 = 0 * t^5 $$

$$ P_7 = 0 $$


text:
User:Egm6341.s10.team4.anandankala/HW5-3

Given

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

$$
 * $$\displaystyle x(t_k)= \frac{1}{2} (x_k+x_{k+1})+ t_k \frac{h}{2}

 * }.
 * }.

Find

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

 * Find tk in terms of x.
 * }.
 * }.

Solution

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

$$
 * $$\displaystyle x(t_k)= \frac{1}{2} (x_k+x_{k+1})+ t_k \frac{h}{2}

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k \frac{h}{2}=x-\frac{1}{2} (x_k+x_{k+1})

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x-\frac{1}{2} (x_k+x_{k+1}) )

 * }.
 * }.

Substituting x=xk.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x_k-\frac{1}{2} (x_k+x_{k+1}) )

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (\frac{1}{2} x_k- \frac{1}{2}x_{k+1})

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{-1}{h} (x_{k+1}-x_k)

 * }.
 * }.

We know that xk+1-xk=h.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{-1}{h} (h)

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =-1

 * }.
 * }.

Substituting x=xk+1.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x_{k+1}-\frac{1}{2} (x_k+x_{k+1}) )

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (\frac{1}{2}x_{k+1}-\frac{1}{2} x_k)

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{1}{h} (x_{k+1}-x_k)

 * }.
 * }.

We know that xk+1-xk=h.

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

$$
 * $$\Rightarrow \displaystyle t_k =\frac{1}{h} (h)

 * }.
 * }.

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

$$
 * $$\Rightarrow \displaystyle t_k =1

 * }.
 * }.


text: File:Table 5 10.png

title: File:Table 5 10.png


text: File:Table 5 10s.png

title: File:Table 5 10s.png


text:
User:Egm6341.s10.team3.heejun/HW5

=  Proof of Trapezoidal error =

Problem Statement
$$ \displaystyle i)\, Do\, steps\, 4a\, and\, 4b\, in\, order\, to\, determine\, P_{6}(t)\, and\, P_{7}(t)\, $$ Ref: Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

$$ \displaystyle ii)\, Steps\, 3a\, and\, 3b\, are\, given\, by\, Lecture\, 26\,$$

$$ E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_5(t)g^{(5)}(t)dt}_{Call\, D\,} $$

$$ \begin{align} P_4(t)&=-\frac{t^4}{4!}+\frac{t^2}{12}+\alpha \\ &=c_{1}\frac{t^4}{4!}+c_{3}\frac{t^2}{2!}+c_{5}\, (c_{5}=\alpha) \\ &=-\frac{t^4}{24}+\frac{t^2}{12}-\frac{7}{360} \\\end{align}\ $$

$$ \begin{align} P_5(t)&=-\frac{t^5}{120}+\frac{t^3}{36}+\alpha t + \beta \\ &=c_{1}\frac{t^5}{5!}+c_{3}\frac{t^3}{3!}+c_{5}t+c_{6}\, (c_{5}=\alpha\, and\, c_{6}=\beta) \\ &=-\frac{t^5}{120}+\frac{t^3}{36}-\frac{7t}{360} \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2&3]]

Solution
$$ \displaystyle i)\, Step\, 4a\, :\, Recall\, D\, and\, integrate\, by\, parts\, $$

$$ D=:\int_{-1}^{+1}\underbrace{P_5(t)}_{v'}\underbrace{g^{(5)}(t)}_{u}dt = [P_6(t)g^{(5)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}{P_6(t)g^{(6)}(t)}\, dt}_{Call\, E} $$

$$P_6(t) = c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7 = -\frac{t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha $$

$$ \displaystyle ii)\, Step\, 4b\, :\, Recall\, E\, and\, integrate\, by\, parts\, $$

$$ E=:\int_{-1}^{+1}\underbrace{P_6(t)}_{v'}\underbrace{g^{(6)}(t)}_{u}dt = [P_7(t)g^{(6)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}{P_7(t)g^{(7)}(t)}\, dt}_{Call\, F} $$

$$P_7(t) = c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(\frac{t}{1!})+c_8 = -\frac{t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta $$

$$ \displaystyle iii)\, Select\, points\, such\, that\, $$

$$ \displaystyle P_{7}(\pm 1)=0 $$ $$ \displaystyle P_{7}(0)=0 $$

$$\displaystyle Recall\, P_7(t)\, and\, plug\, in\, t=0\, and\, t=-1\,$$ $$P_7(t=0) = c_1(\frac{0}{7!})+c_3(\frac{0}{5!})+c_5(\frac{0}{3!})+c_7(\frac{0}{1!})+c_8 = -\frac{0}{5040}+\frac{0}{720}-\frac{0}{2160}+\alpha \times 0+\beta = 0 $$ $$\displaystyle Hence\, c_8\, =\, \beta\, =\, 0\,$$ $$P_7(t=-1) = c_1(\frac{(-1)^7}{7!})+c_3(\frac{(-1)^5}{5!})+c_5(\frac{(-1)^3}{3!})+c_7(\frac{(-1)}{1!}) = -\frac{(-1)}{5040}+\frac{(-1)}{720}-\frac{(-1)}{2160}+\alpha \times (-1) = 0 $$ $$\displaystyle Hence\, c_7\, =\, \alpha\, =\, \frac{31}{15120}\,$$

$$\displaystyle Therefore\, $$

$$ P_6(t)=-\frac{t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{15120} $$

$$ P_7(t)=-\frac{t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\frac{31}{15120} t $$

$$ E= [P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)+P_6(t)g^{(5)}(t)]_{-1}^{+1}-F $$

$$ \displaystyle where\, F=:\int_{-1}^{+1}{P_7(t)g^{(7)}(t)}\, dt $$

=  Compare the result of the given circumference formular with the complete elliptic integral of the second kind method =

Problem Statement
$$ \displaystyle i)\, The\, given\, circumference\, formular\, $$

$$ \int_{\theta=0}^{\theta=2\pi} dl = C $$

$$ \displaystyle Where\,\,\,\, \begin{align} dl &=d\theta [r^2+(\frac{dr}{d\theta})^2]^(\frac{1}{2})\\ r(\theta) &= \frac{1-e^2}{1-ecos(\theta)} \\ e &= sin(\frac{\pi}{12}) \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg25.djvu|p.25-2]] and [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle ii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle C = 4a E (e) $$

$$ \displaystyle E(e) = \int_{0}^{\pi /2} [1-e^2sin^2(\theta)]^{(\frac{1}{2})} d\theta $$

Ref: Team 4 in HW 4 [|problem_11] and Wikipedia[|Elliptic_integral]

Solution
$$ \displaystyle i)\, Define\, function\,\, dl\, $$

$$ \displaystyle ii)\, Integrate\,\, dl\, $$

$$ \displaystyle Therefore\,\,\, \int_{\theta=0}^{\theta=2\pi} dl = 6.176601938988738 $$

$$ \displaystyle iii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle Therefore\,\,\, C = 4aE(e) = 6.176601933080789\,\,\, when\, a=1\, $$

$$ \displaystyle iiii)\, The\, difference\, between\, 2)\, and\, 3)\, is\, 5.907948796846085\times 10^{-9}\, $$

$$ \displaystyle It\, is\, pretty\, much\, small\, so\, can\, be\, negligible.\, $$

$$ \displaystyle Therefore\, above\, two\, methods\, have\, the\, same\, results.\, $$

title: User:Egm6341.s10.team3.heejun/HW5


text:
User:Egm6341.s10.Team4/HW5

= Problem 1: Trapezoidal Rule Error Proof: Step 3b=

Given
Refer Lecture slide 26-3 for problem statement.

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

$$\displaystyle p_5(t)=\int p_4(t)dt = -\frac{t^5}{120} + \frac{t^3}{36} + c_5t + c_6$$
 * $$\displaystyle \longrightarrow (1) $$
 * }.
 * }.

Find
Selecting $$\displaystyle p_5(\pm 1)= 0 $$ and $$\displaystyle p_5(0)= 0 $$, find $$\displaystyle c_5 $$ and $$\displaystyle c_6 $$.

Solution
When $$\displaystyle p_5(0)= 0 $$ Equation (1),

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

$$\displaystyle p_5(0)= -\frac{0^5}{120} + \frac{0^3}{36} + c_5(0) + c_6 = 0$$
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_6 = 0$$
 * }.
 * }.

When $$\displaystyle p_5(1)= 0 $$ Equation (1),

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

$$\displaystyle p_5(1)= -\frac{1^5}{120} + \frac{1^3}{36} + c_5(1) = 0$$
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_5 = \frac{1}{120} - \frac{1}{36} = -\frac{7}{360}$$
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.andy 08:18, 24 March 2010 (UTC).

Proofed by - Egm6341.s10.Team4.riherd 14:14, 24 March 2010 (UTC) and --Egm6341.s10.Team4.roni 18:35, 24 March 2010 (UTC)

= Problem 2: Steps 4a and 4b in the proof of higher order derivation of Trap. error =

Given
Envisage the acquired expression for E at the end of step 3 on [[media:Egm6341.s10.mtg26.djvu|26-3]] as following:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle$$
 * $$\displaystyle E=[p_2(t).g^{(1)}(t)+p_4(t).g^{(3)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt$$
 * $$\displaystyle E=[p_2(t).g^{(1)}(t)+p_4(t).g^{(3)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt$$

 * }
 * }

Find
What are the expressions for $$\displaystyle p_6(t) $$, and $$\displaystyle p_7(t) $$ ?

Solution
At the end of step 3 of this procedure, we have obtained:

 * $$\displaystyle

p_4(t)=-\dfrac{t^4}{24}+\dfrac{t^2}{12}-\dfrac{7}{360}$$

 * $$\displaystyle

p_5(t)=-\dfrac{t^5}{120}+\dfrac{t^3}{36}-\dfrac{7}{360}t$$

Step 4a:

E=[p_2(t).g^{(1)}(t)+p_4(t).g^{(3)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt$$

If we call the second term (integration term) as:

D:=\int\limits_{-1}^{+1} p_5(t).g^{(5)}(t)dt$$

By integrating by parts we have:

dv=p_5(t)dt\Rightarrow v=p_6(t)=\int p_5(t)dt$$

u=g^{(5)}(t)\Rightarrow du=g^{(6)}(t)dt$$

\Rightarrow D=[uv]_{-1}^{+1}-\int\limits_{-1}^{+1} vdu=[p_6(t).g^{(5)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} p_6(t).g^{(6)}(t)dt$$

p_6(t)=\int p_5(t)dt=\int (-\dfrac{t^5}{120}+\dfrac{t^3}{36}-\dfrac{7}{360}t)dt=-\dfrac{t^6}{6!}+\dfrac{t^4}{144}-\dfrac{7}{720}t^2+\alpha$$

Step 4b:

Then by defining F as:

F:=\int\limits_{-1}^{+1} p_6(t).g^{(6)}(t)dt$$

We repeat integration by parts method;

dv=p_6(t)dt\Rightarrow v=p_7(t)=\int p_6(t)dt$$

u=g^{(6)}(t)\Rightarrow du=g^{(7)}(t)dt$$

\Rightarrow F=[p_7(t).g^{(6)}]_{-1}^{+1}-\int\limits_{-1}^{+1} p_7(t).g^{(7)}(t)dt$$

p_7(t)=-\dfrac{t^7}{7!}+\dfrac{t^5}{720}-\dfrac{7t^3}{2160}+\alpha t+\beta$$

We want to make $$\displaystyle [p_7(t).g^{(6)}(t)]_{-1}^{+1} $$ to be equal to zero. So, for an odd function like $$\displaystyle p_7(t) $$, we have:

p_7(0)=0\Rightarrow \beta=0$$

p_7(\pm 1)=0$$

\Rightarrow p_7(1)=-\dfrac{1}{7!}+\dfrac{1}{720}-\dfrac{7}{2160}+\alpha=0$$

\Rightarrow \alpha=\dfrac{3!}{7!\times 3}=\dfrac{31}{15120}$$

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

$$\displaystyle p_6(t)=-\dfrac{t^6}{6!}+\dfrac{t^4}{144}-\dfrac{7t^2}{720}+\dfrac{31}{15120}$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * {| style="width:１０0%" border="0" align="left"
 * {| style="width:１０0%" border="0" align="left"

$$\displaystyle p_7(t)=-\dfrac{t^7}{7!}+\dfrac{t^5}{720}-\dfrac{7t^3}{2160}+\dfrac{31}{15120}t$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&m 23:25, 18 March 2010 (UTC)

Proofed by - Egm6341.s10.Team4.riherd 14:19, 24 March 2010 (UTC) .

= Problem 3: Deducing a relation between x and t in proof of trapezoidal error =

Given

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

 * $$\displaystyle x(t_k)= \frac{1}{2} (x_k+x_{k+1})+ t_k \frac{h}{2}$$

 * }.
 * }.

Find

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

 * Find tk in terms of x.
 * }.
 * }.

Solution

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

 * $$\displaystyle x(t_k)= \frac{1}{2} (x_k+x_{k+1})+ t_k \frac{h}{2}$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k \frac{h}{2}=x-\frac{1}{2} (x_k+x_{k+1})$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x-\frac{1}{2} (x_k+x_{k+1}) )$$

 * }.
 * }.

Substituting x=xk.

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

 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x_k-\frac{1}{2} (x_k+x_{k+1}) )$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (\frac{1}{2} x_k- \frac{1}{2}x_{k+1})$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =\frac{-1}{h} (x_{k+1}-x_k)$$

 * }.
 * }.

We know that xk+1-xk=h.

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

 * $$\Rightarrow \displaystyle t_k =\frac{-1}{h} (h)$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =-1$$

 * }.
 * }.

Substituting x=xk+1.

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

 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (x_{k+1}-\frac{1}{2} (x_k+x_{k+1}) )$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =\frac{2}{h} (\frac{1}{2}x_{k+1}-\frac{1}{2} x_k)$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =\frac{1}{h} (x_{k+1}-x_k)$$

 * }.
 * }.

We know that xk+1-xk=h.

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

 * $$\Rightarrow \displaystyle t_k =\frac{1}{h} (h)$$

 * }.
 * }.

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

 * $$\Rightarrow \displaystyle t_k =1$$

 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.team4.anandankala 12:50, 24 March 2010 (UTC). Reviewed by - Egm6341.s10.Team4.andy 19:25, 24 March 2010 (UTC)

= Problem 4: Comparing Richarson Extrap. Coeff. to Coeff. from Trap. Error Estimate =

Given
Richardson Extrapolation Coefficients on slide [[media:Egm6341.s10.mtg19.djvu|19-3]], which is:

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

 * $$\displaystyle$$
 * $$\displaystyle a_i=d_i*(f \overset{(2i-1)}{(b)} -f \overset{(2i-1)}{(a)})$$
 * style= |
 * }
 * }
 * }

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

 * $$\displaystyle$$
 * $$\displaystyle d_i=- \frac{{B}_{2i}}{(2i)!}$$
 * style= |
 * }
 * }
 * }

$$\displaystyle$$ $$\displaystyle d_1=- \frac{1}{12} \ \ d_2= \frac{1}{720} \ \  d_3=- \frac{1}{30240}$$

Find
That these coefficients are equal to the following: equation (3) on slide [[media:Egm6341.s10.mtg27.djvu|27-1]]

$$\displaystyle$$ $$\displaystyle \bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^{2r})!}$$

$$\displaystyle$$ $$\displaystyle d_1=\bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^{2r})!} \ \ where  \ \  r=1$$

$$\displaystyle$$ $$\displaystyle d_2=\bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^{2r})!} \ \ where  \ \  r=2$$

$$\displaystyle$$ $$\displaystyle d_3=\bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^{2r})!} \ \ where  \ \  r=3$$

Solution
Recalling:

equation (2) on slide [[media:Egm6341.s10.mtg21.djvu|21-2]]
 * $$\displaystyle

P_1(t)=-t$$

equation (1) on slide [[media:Egm6341.s10.mtg21.djvu|21-3]]

P_2(t)=- \frac{t^2}{2}+\frac{1}{6}$$

equation (4) on slide [[media:Egm6341.s10.mtg21.djvu|21-3]]

P_3(t)=-\frac{t^3}{6}+\frac{t}{6}$$

on slide [[media:Egm6341.s10.mtg21.djvu|26-3]]

P_4(t)=- \frac{t^4}{24}- \frac{t^2}{12}- \frac{72}{360}$$

In General, from the Paper by Patch Kessler we have:

P_{2k}(t)=c1 \frac{t^{2k}}{(2k)!} + c_2 \frac{t^{2k-2}}{(2k-2)!} + c_5 \frac{t^{2k-4}}{(2k-4)!} + c_{2k-3} \frac{t^{4}}{(4)!} + c_{2k-1} \frac{t^{2}}{(2)!} + c_{2k+1} \frac{t^{0}}{(0)!}$$

Giving (also from HW 5.2 and it's solution) [[media:Egm6341.s10.mtg27.djvu|27-1]]

P_6(t)= -\frac{t^{6}}{(720)}+\frac{t^{4}}{(144)}-\frac{7t^{2}}{(720)}+\frac{(31)}{(15120)}$$

We only need P1, P4 and P6

Giving,

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

$$\displaystyle d_1=\bar{d}_{2}= \frac{{P}_{2} (t=1)}{(4)} \ \ =-1/12$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

 * style= |
 * }
 * }

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

$$\displaystyle d_2=\bar{d}_{4}= \frac{{P}_{4} (t=1)}{(16)} \ \ = 1/16*(-1/241/12-7/360)=1/720$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

 * style= |
 * }
 * }

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

$$\displaystyle d_3=\bar{d}_{6}= \frac{{P}_{6} (t=1)}{(64)} \ \ = 1/64*(-1/720+1/144-7/720+31/15120)=-1/30240$$
 * 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.roni 18:23, 24 March 2010 (UTC)

= Problem 5: Derivation of Compsit Trap. Rule Error =

Given
Refer Lecture slide 28-3 for problem statement

The Error of Composit Trap. Rule

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

E^1_n= \sum_{k=0}^{n-1} \left[  \int_{x_{k}}^{x_{k+1}}f(x)dx-\frac{h}{2} \left\{ f(x_k)+f(x_{k+1}) \right\}  \right]$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Find
Derive the Compsit Trap.Rule Error equation.
 * {| style="width:100%" border="0" align="left"

E=\sum_{r=1}^{\ell} 2 \, \overline{d}_{2r} \, h^{2r-1}\Bigg[f^{(2r-1)}(b)-f^{(2r-1)}(a)\Bigg]-\frac{h^{2\ell}}{2^{2\ell}} \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} \, p_{2\ell}(t_{k}(x)) \, f^{(2\ell)}(x)dx$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

where,

 * $$\displaystyle \overline{d}_{2r} = \frac{p_{2r}(+1)}{2^{2r}} $$
 * $$\displaystyle x \in [a,b] $$

Solution
The Error of Composit Trap. Rule

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

E^1_n= \sum_{k=0}^{n-1} \left[  \int_{x_{k}}^{x_{k+1}}f(x)dx-\frac{h}{2} \left\{ f(x_k)+f(x_{k+1}) \right\}  \right]$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Transfer int.$$\displaystyle [x_k, x_{k+1}]$$ to $$\displaystyle [1,+1]$$

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

x(t):=t\frac{h}{2}+\frac{x_k+x_{k+1}}{2}, \, \, \, \, \, \, \, t \in [-1,1]\, \, \,\, \, \,, h:=(b-a)/n$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

g_k(t) = f(x(t)) \,\,\,\,\,\,such \,\,\,that\,\,\,\,\,x \in [x_k, x_{k+1}]$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

E^1_n= \frac{h}{2}\sum_{k=0}^{n-1} \left[  \int_{-1}^{+1}g_k(t)dt- \left\{ g_k(-1)+g_k(+1) \right\}  \right]$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

through the successive integration by parts, the below equation can be obtained21-2,326-3

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

\begin{align} E^1_n & = \frac{h}{2} \left[ \sum_{k=0}^{n-1}\Big[p_2(t)g_k^{(1)}(t)+p_4(t)g_k^{(3)}(t_k)+p_6(t)g_k^{(5)}(t)+......+p_{2\ell}(t)g_k^{(2\ell-1)}(t)\Big]_{-1}^{+1}- \sum_{k=0}^{n-1} \left[ \int_{-1}^{1}p_{2\ell}(t)g_k^{(2\ell)}(t) dt \right]\right]\\ &= \frac{h}{2} \left[ \sum_{r=1}^{\ell}\sum_{k=0}^{n-1}\Big[p_{2r}(t)g_k^{(2r-1)}(t)\Big]_{-1}^{+1} - \sum_{k=0}^{n-1} \left[    \int_{-1}^{1}p_{2\ell}(t)g_k^{(2\ell)}(t) dt \right]\right] \end{align}$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Extend the first summation term.

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

\begin{align} E^1_n &= \frac{h}{2} \left[ \sum_{r=1}^{\ell}\sum_{k=0}^{n-1}\Big[p_{2r}(+1)g_k^{(2r-1)}(+1)-p_{2r}(-1)g_k^{(2r-1)}(-1)\Big]- \sum_{k=0}^{n-1} \left[   \int_{-1}^{1}p_{2l}(t)g_k^{(2\ell)}(t) dt \right]\right]\\ & because \,\,\,\,\,\,p_{2r}(+1)=p_{2r}(-1),\\ &= \frac{h}{2} \left[ \sum_{r=1}^{\ell}\sum_{k=0}^{n-1} p_{2r}(+1) \Big[g_k^{(2r-1)}(+1)-g_k^{(2r-1)}(-1)\Big]- \sum_{k=0}^{n-1} \left[   \int_{-1}^{1}p_{2l}(t)g_k^{(2\ell)}(t) dt\right]\right] \end{align}$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Using below relationship,

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

g_k^{(i)}(t)=\left(\frac{h}{2}\right)^if^{(i)}(x(t)),,\,\,\,\,\ x \in [x_k,x_{k+1}]$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

$$\displaystyle g_k(t)$$can be changed to $$\displaystyle f_(x_k)$$ and below was obtained.
 * {| style="width:100%" border="0" align="left"

\begin{align} &g_k^{(2r-1)}(+1)=\left(\frac{h}{2}\right)^{2r-1}f^{(2r-1)}(x_{k+1}),\\ &g_k^{(2r-1)}(-1)=\left(\frac{h}{2}\right)^{2r-1}f^{(2r-1)}(x_{k}),\\ &g_k^{(2\ell)}(t)=\left(\frac{h}{2}\right)^{2\ell}f^{(2\ell)}(x_{t}) \end{align}$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Plug in thess relationships to the equation $$\displaystyle E_n^1$$
 * {| style="width:100%" border="0" align="left"

\begin{align} E^1_n & = \left(\frac{h}{2}\right) \left[ \sum_{r=1}^{\ell} p_{2r}(+1) \left(\frac{h}{2}\right)^{2r-1} \sum_{k=0}^{n-1} \Big[f^{(2r-1)}(x_{k+1})-f^{(2r-1)}(x_k)\Big]- \sum_{k=0}^{n-1} \left[   \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))\left(\frac{h}{2}\right)^{2\ell}f^{(2\ell)}(x) dx \right] \right]\\ &= \sum_{r=1}^{\ell} h^{2r} \frac{p_{2r}(+1)}{2^{2r}} \Big[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)\Big]-  \left(\frac{h}{2}\right)^{2\ell+1} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right] \end{align}$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

by the definition of $$\displaystyle \overline{d}_{2r}$$
 * $$\displaystyle \overline{d}_{2r} = \frac{p_{2r}(+1)}{2^{2r}} $$
 * $$\displaystyle x_n=b, x_0=a $$

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

E^1_n = \sum_{r=1}^{\ell} h^{2r} \overline{d}_{2r} \Big[f^{(2r-1)}(b)-f^{(2r-1)}(a)\Big]- \underbrace{\left(\frac{h}{2}\right)^{2\ell+1}}_{(?)} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right]$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

This solution is little bit different from the given solution of the lecture.

the coefficient of second summation term that marked $$\displaystyle (?) $$

we got $$\displaystyle \left(\frac{h}{2}\right)^{2\ell+1} $$ instead of $$\displaystyle \left(\frac{h}{2}\right)^{2\ell} $$.

Author
Solved and typed by - Egm6341.s10.Team4.yunseok 16:59, 24 March 2010 (UTC) Reviewed by - Egm6341.s10.Team4.andy 18:41, 24 March 2010 (UTC)

= Problem 6 - Evaluation of d2i =

Given
The hyperbolic cotangent is defined as $$ \displaystyle\ coth(x)=\frac{cosh(x)}{sin(x)}=\frac{e^x+e^{-x}}{e^x-e^{-x}}. $$.

The function $$ \displaystyle\ f(x)=xcosh(x)=\sum_{i=0}^{\infty}\frac{2^{2i}B_{2i}}{(2i)!}x^{2i}$$, where Bi are the Bernoulli numbers.

Find
We are asked to verify the values of di for i=2,4, and 6, where di is the constant $$ \displaystyle\ d_{2i}=\frac{-B_{2i}}{(2i)!}$$. This is done previously given the values d2=-1/12, d4=1/720 and d6=-1/30240.

We are also asked to determine the values of d8 and d10.

Additionally, we are asked to compare these values of di to another relationship for di used in our error of the trapezoidal rule, where $$ \displaystyle\ d_{2i}=\frac{p_{2i}(1)}{2^{2i}}$$.

Solution
Expanding cosh(x) and sinh(x) in terms of the exponential function and then expanding those exponential functions as series around the point x=0, we see that

$$ \displaystyle\ f(x)=xcosh(x)=\sum_{i=1}^{\infty}-2^{2i}d_{2i}x^{2i}$$

$$ \displaystyle\ cosh(x)=1+\frac{1}{2}x^2-\frac{1}{24}x^4+\frac{1}{720}x^6+\frac{1}{40320}x^8+\frac{1}{3628800}x^{10}+...$$

$$ \displaystyle\ sinh(x)=x+\frac{1}{6}x^3+\frac{1}{120}x^5+\frac{1}{5040}x^7+\frac{1}{362880}x^9+\frac{1}{39916800}x^{11}+...$$

Substituting all of this into the function $$ \displaystyle\ f(x)=xcosh(x)$$, we see that

$$ \displaystyle\ f(x)=\frac{x+\frac{1}{2}x^3-\frac{1}{24}x^5+\frac{1}{720}x^7+\frac{1}{40320}x^9+\frac{1}{3628800}x^{11}...}{x+\frac{1}{6}x^3+\frac{1}{120}x^5+\frac{1}{5040}x^7+\frac{1}{362880}x^9+\frac{1}{39916800}x^{11}+...}$$

Performing the appropriate polynomial division (which will not be shown here due to length and no clear way to show long division in Latex), we see that

$$ \displaystyle\ f(x)=xcosh(x)=1+\frac{1}{3}x^2-\frac{1}{45}x^4+\frac{2}{975}x^6-\frac{1}{4725}x^8+\frac{2}{93555}x^{10}-... $$

This series expansion of xcoth(x) in hand, we are able to determine the values of $$ \displaystyle\ 2^{2i}d_{2i} $$ to be -

Additionally, we are asked to calculate d2i using a definition given by the polynomial functions defined in our trapezoidal error analysis, where $$ \displaystyle\ d_{2i}=\frac{p_{2i}(1)}{2^{2i}}$$.

From analysis done in class and coefficients as given by Kessler, P. (Trapezoidal Rule Error, http://www.mechanicaldust.com/UCB/math128a/extrap.pdf, 2008), we know that

$$ \displaystyle\ p_2(t)=-\frac{t^2}{2}+\frac{1}{6}, p_2(1)=-\frac{1}{3}$$

$$ \displaystyle\ p_4(t)=-\frac{t^4}{24}+\frac{t^2}{12}-\frac{7}{360}, p_4(1)=\frac{1}{45}$$

$$ \displaystyle\ p_6(t)=-\frac{t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{15120}, p_6(1)=-\frac{2}{945}$$

$$ \displaystyle\ p_8(t)=-\frac{t^8}{40320}+\frac{t^6}{4320}-\frac{7t^4}{8640}+\frac{31t^2}{30240}-\frac{127}{604800}, p_8(1)=\frac{1}{4725} $$

$$ \displaystyle\ p_{10}=-\frac{t^10}{3628800}+\frac{t^8}{241920}-\frac{7t^6}{259200}+\frac{31t^4}{362880}-\frac{127t^2}{1209600}+\frac{73}{3421440}, p_{10}(t)=\frac{-2}{93555} $$

With these coefficients in hand, we are able to calculate the values for d2i using the formula as given above relating p2i(1) and d2i.

Comparing these values, we see that the formulas given to find d2i are consistent with each other.

Author
Solved and Written by Egm6341.s10.Team4.riherd 07:13, 24 March 2010 (UTC)

= Problem 7: Trapezoidal Rule Error by Canceling Odd Derivative Orders of $$ g(t)$$=

Given
Refer Lecture slide 28-2 for problem statement.

Higher Order Error for Trap. rule is given by [ Refer 21-1 ]:

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

E_n^1=\frac{h}{2} \sum_{k=0}^{n-1} \left[ \underbrace{\int_{-1}^{1}g_k(t)dt-(g_k(-1)+g_k(+1))}_{E}\right]$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1) $$
 * }.
 * }.

where $$\displaystyle E $$ can be written as,

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

E =\int_{-1}^{1}(-t)g_k^{(1)}(t)dt$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2) $$
 * }.
 * }.

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

$$\displaystyle \begin{align} \Rightarrow E &=\int_{-1}^{1}\underbrace{(-t)}_{p_{1}(t)}g_k^{(1)}(t)dt \\ &= \left[p_2(t)g_k^{(1)}(t) \right]_{-1}^{+1} - \int_{-1}^{+1}p_2(t)g_k^{(2)}(t)dt \end{align}$$

where $$\displaystyle p_2(t) = \int p_1(t)dt$$.
 * }.
 * }.

After successive integration by parts,

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

$$\displaystyle

E = \left[p_2(t)g_k^{(1)}(t)-p_3(t)g_k^{(2)}(t)+p_4(t)g_k^{(3)}(t)-p_5(t)g_k^{(4)}(t)+p_6(t)g_k^{(5)}(t)+ \cdots +p_{\ell}(t)g_k^{(\ell-1)}(t)\right]_{-1}^{+1} - \int_{-1}^{1}p_{\ell}(t)g_k^{(\ell)}(t) dt $$
 * $$\displaystyle \longrightarrow (3) $$

 * }.
 * }.

Find
Try to derive the Higher Order Trapezoidal Rule Error equation by canceling odd order derivative terms of $$\displaystyle g(t) $$, i.e., make terms $$\displaystyle p_{2}(t), p_{4}(t), p_{6}(t), \cdots, $$ to zero at $$\displaystyle t =  \pm 1$$.

Solution
We know,

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

$$\displaystyle p_{2}(t) = -\frac{t^2}{2} + c_3$$
 * }.
 * }.

At $$\displaystyle t = \pm 1$$,

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

$$\displaystyle p_{2}(1) = -\frac{1^2}{2} + c_3 = 0; p_{2}(-1) = -\frac{(-1)^2}{2} + c_3 = 0;$$
 * }.
 * }.

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

$$\displaystyle \Rightarrow c_3 = \frac{1}{2}$$
 * }.
 * }.

We also know,

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

$$\displaystyle p_{3}(t) = -\frac{t^3}{3!} + \frac{t}{2} + c_4$$ And,
 * }.
 * }.
 * {| style="width:100%" border="0" align="left"

$$\displaystyle p_{4}(t) = -\frac{t^4}{4!} + \frac{t^2}{4} + c_4t + c_5$$
 * }.
 * }.

At $$\displaystyle t = \pm 1$$,

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

$$\displaystyle p_{4}(1) = -\frac{(1)^4}{4!} + \frac{(1)^2}{4} + c_4(1) + c_5; p_{4}(-1) = -\frac{(-1)^4}{4!} + \frac{(-1)^2}{4} + c_4(-1) + c_5;$$

 * }.

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

$$\displaystyle \Rightarrow c_4 = 0; c_5 = \frac{-5}{24}$$
 * }.
 * }.

Functions $$\displaystyle p_{3}(t), p_{5}(t), p_{7}(t), \cdots, $$ are odd functions.

Since functions $$\displaystyle p_{2}(t), p_{4}(t), p_{6}(t), \cdots, $$ are made zero at $$\displaystyle t = \pm 1$$, Equation (3) becomes,

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

$$\displaystyle \begin{align} E & = \left[-p_3(t)g_k^{(2)}(t)-p_5(t)g_k^{(4)}(t)-p_7(t)g_k^{(6)}(t)\cdots\cdots-p_{(2\ell-1)}(t)g_k^{(2\ell-2)}(t)\right]_{-1}^{+1} + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt \\ & = \sum_{r=2}^{\ell} \left[ -p_{(2r-1)}g_k^{(2r-2)}\right]_{-1}^{+1} + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt \\ & = \sum_{r=2}^{\ell} \left[ -p_{(2r-1)}(1)g_k^{(2r-2)}(1) + p_{(2r-1)}(-1)g_k^{(2r-2)}(-1)\right] + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt \\ \end{align}$$

 * }.
 * }.

Since $$\displaystyle p_{(2r-1)}(t) $$ are odd functions, we know $$\displaystyle p_{(2r-1)}(1) + p_{(2r-1)}(-1) = 0 $$.

And so $$\displaystyle E $$ becomes,

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

$$\displaystyle

E = \sum_{r=2}^{\ell} p_{(2r-1)}(-1) \left[ g_k^{(2r-2)}(1) + g_k^{(2r-2)}(-1)\right] + \int_{-1}^{1}p_{(2\ell-1)}(t)g_k^{(2\ell-1)}(t) dt $$
 * $$\displaystyle \longrightarrow (4) $$

 * }.
 * }.

Using the below relationship

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

g_k^{(i)}(t)=\left(\frac{h}{2}\right)^if^{(i)}(x(t)); \ \ \ \ \ x \in [x_k,x_{k+1}]$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

In Equation (4) $$\displaystyle g_k(t)$$can be transferred to $$\displaystyle f(x_k)$$ as shown below:

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

\begin{align} &g_k^{(2r-2)}(+1)=\left(\frac{h}{2}\right)^{(2r-2)}f^{(2r-2)}(x_{k+1}),\\ &g_k^{(2r-2)}(-1)=\left(\frac{h}{2}\right)^{(2r-2)}f^{(2r-2)}(x_{k}),\\ &g_k^{(2\ell-1)}(t)=\left(\frac{h}{2}\right)^{(2\ell-1)}f^{(2\ell-1)}(x) \end{align}$$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

And so, Equation (4) becomes,

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

$$\displaystyle

E =  \sum_{r=2}^{\ell}\left[ \left(\frac{h}{2}\right)^{(2r-2)} p_{(2r-1)}(-1) \Big[ f^{(2r-2)}(x_{k+1}) + f^{(2r-2)}(x_{k})\Big] \right] + \left(\frac{h}{2}\right)^{(2\ell-1)} \int_{x_k}^{x_{k+1}}p_{(2\ell-1)}(t_k(x))f^{(2\ell-1)}(x) dx $$
 * $$\displaystyle \longrightarrow (5) $$

 * }.
 * }.

Substituting Equation (5) in Equation (1),

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

$$\displaystyle \begin{align} E_n^1 &= \sum_{r=2}^{\ell}\left[ \left(\frac{h}{2}\right)^{(2r-1)} p_{(2r-1)}(-1) \sum_{k=0}^{n}\Big[ f^{(2r-2)}(x_{k+1}) + f^{(2r-2)}(x_{k})\Big] \right] + \left(\frac{h}{2}\right)^{(2\ell)} \sum_{k=0}^{n} \left [\int_{-1}^{1}p_{(2\ell-1)}(t_k(x))f^{(2\ell-1)}(x) dx \right ] \\ & = \sum_{r=2}^{\ell}\left[ \left(\frac{h}{2}\right)^{(2r-1)} p_{(2r-1)}(-1) \Big[ \underbrace{f^{(2r-2)}(x_n) + f^{(2r-2)}(x_{n-1}) + \cdots \cdots + f^{(2r-2)}(x_{0})}_{\beta}\Big] \right] + \left(\frac{h}{2}\right)^{(2\ell)} \sum_{k=0}^{n} \left [\int_{-1}^{1}p_{(2\ell-1)}(t_k(x))f^{(2\ell-1)}(x) dx \right ] \end{align} $$
 * }.
 * }.

In the above error equation $$\displaystyle \beta \neq f^{(2r-2)}(x_n) - f^{(2r-2)}(x_0)$$, but it contains terms that involve all the points from $$\displaystyle x_0 \ to\ x_n $$.

But instead of canceling out terms that involve odd order derivatives of $$\displaystyle g(t)$$, if we had canceled only terms that involve even order derivatives of $$\displaystyle g(t)$$ then the term $$\displaystyle \beta $$ would involve only first and last points, i.e., $$\displaystyle x_0 = a $$ and $$\displaystyle x_n = b $$, as we have got in Problem 5 above.

Author
Solved and typed by - Egm6341.s10.Team4.andy 08:21, 24 March 2010 (UTC) Reviewed by- Egm6341.s10.team4.anandankala 12:55, 24 March 2010 (UTC).

= Problem 8: Using Recurrence formula to find $$\displaystyle c_i $$ =

Given
Consider the Recurrence formula on slide [[media:Egm6341.s10.mtg29.djvu|29-2]] as follows:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle$$
 * $$\displaystyle c_{2i+1}=-\dfrac{c_1}{(2i+1)!}-\dfrac{c_3}{(2i-1)!}-\dfrac{c_5}{(2i-3)!}-...-\dfrac{c_{2i-1}}{3!}$$
 * $$\displaystyle c_{2i+1}=-\dfrac{c_1}{(2i+1)!}-\dfrac{c_3}{(2i-1)!}-\dfrac{c_5}{(2i-3)!}-...-\dfrac{c_{2i-1}}{3!}$$

 * }
 * }

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

 * $$\displaystyle$$
 * $$\displaystyle p_1(t)=-t\Rightarrow c_1=-1$$
 * $$\displaystyle p_1(t)=-t\Rightarrow c_1=-1$$

 * }
 * }

Find
Obtain $$\displaystyle (p_2,p_3) $$, $$\displaystyle (p_4,p_5) $$ and $$\displaystyle (p_6,p_7) $$:

Solution

 * $$\displaystyle

p_1(t)=-t\Rightarrow p_2(t)=-\frac{t^2}{2}+c_3\Rightarrow p_3(t)=-\frac{t^3}{3!}$$

For $$\displaystyle (p_2,p_3) $$ we can get $$\displaystyle i=1 $$. Thus,

 * $$\displaystyle

c_{(2\times 1+1)}=c_3=-\frac{c_1}{(2\times 1+1)!}$$

 * $$\displaystyle

\Rightarrow c_3=\frac{1}{3!}$$

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

$$\displaystyle p_2(t)=-\frac{t^2}{2!}+\dfrac{1}{6}$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * {| style="width:１０0%" border="0" align="left"
 * {| style="width:１０0%" border="0" align="left"

$$\displaystyle p_3(t)=-\frac{t^3}{3!}+\dfrac{1}{6}t$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Likewise, for $$\displaystyle (p_4,p_5) $$, we can write:

p_4(t)=\int p_3(t)dt=\int (-\frac{t^3}{3!}+c_3t)dt=\int (-\frac{t^3}{3!}+\frac {t}{6})dt=-\frac{t^4}{4!}+\frac{t^2}{12}+c_5$$

p_5(t)=\int p_4(t)dt=\int (-\frac{t^4}{4!}+\frac{t^2}{12}+c_5)dt=-\frac{t^5}{5!}+\frac{t^3}{36}+c_5t$$

By setting $$\displaystyle i=2 $$ in Recurrence equation:

c_{2\times 2+1}=c_5=-\frac{-1}{(2\times 2+1)!}-\frac{\frac{1}{6}}{(2\times 2-1)!}=\frac{1}{5!}-\frac{1}{36}=-\frac{7}{360}$$

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

$$\displaystyle p_4(t)=-\frac{t^4}{4!}+\frac{t^2}{12}-\frac{7}{360}$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * {| style="width:１０0%" border="0" align="left"
 * {| style="width:１０0%" border="0" align="left"

$$\displaystyle p_5(t)=-\frac{t^5}{5!}+\frac{t^3}{36}-\frac{7}{360}t$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Ultimately, for $$\displaystyle (p_6,p_7) $$, we have:

p_6(t)=\int p_5(t)dt=\int (-\frac{t^5}{5!}+\frac{t^3}{36}-\frac{7}{360}t)dt=-\frac{t^6}{6!}+\frac{t^4}{144}-\frac{7t^2}{720}+c_7$$

p_7(t)=\int p_6(t)dt=\int (-\frac{t^6}{6!}+\frac{t^4}{144}-\frac{7t^2}{720}+c_7)dt=-\frac{t^7}{7!}+\frac{t^5}{720}-\frac{7t^3}{2160}+c_7t$$

By setting $$\displaystyle i=3 $$ in Recurrence equation:

c_{2\times 3+1}=c_7=-\frac{-1}{(2\times 3+1)!}-\frac{\frac{1}{6}}{(2\times 3-1)!}-\frac{\frac{-7}{360}}{(2\times 3-3)!}=\frac{1}{7!}-\frac{1}{720}+\frac{7}{2160}=\frac{31}{15120}$$

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

$$\displaystyle p_6(t)=-\dfrac{t^6}{6!}+\dfrac{t^4}{144}-\dfrac{7t^2}{720}+\dfrac{31}{15120}$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

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

$$\displaystyle p_7(t)=-\dfrac{t^7}{7!}+\dfrac{t^5}{720}-\dfrac{7t^3}{2160}+\dfrac{31}{15120}t$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&m 00:44, 19 March 2010 (UTC) Reviewed by- Egm6341.s10.team4.anandankala 13:05, 24 March 2010 (UTC). .

= Problem 9: Kessler's Code =

Given

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

 * Given below is the Kessler's code

 * }.
 * }.

Find

 * {| style="width:85%" border="0" align="center"

Use (p2,p3),(p4,p5)(p6,p7) to understand the Kessler's code                       $$\displaystyle $$
 * }.
 * }.

Solution

 * {| style="width:85%" border="0" align="center"

The above code has been used to compute the coefficients p(2k) and the constants c(k) where k varies from 1 to n. In the above code a function fracsum has been defined and called accordingly in order to compute c(k) and p(2k). The various commands used in the code are gcd, lcm and round. Gcd is used to find the greatest common divisor between two numbers, lcm is used to find the least common multiple of the two numbers and round is used to round off a number to its nearest intezer. We know that $$\displaystyle p_2(t)= -\frac {t^2}{2!} + \frac {1}{6}$$

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

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

$$\displaystyle \Rightarrow p_2(1)= -\frac {1}{2!} + \frac {1}{6}= -\frac {1}{3}$$ $$\displaystyle $$
 * }.
 * }.

 * {| style="width:76%" border="0" align="center"

$$\displaystyle p_4(t)= -\frac {t^4}{24} + \frac {t^2}{12} - \frac {7}{360}$$ $$\displaystyle $$
 * }.
 * }.

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

$$\displaystyle \Rightarrow p_4(1)= -\frac {1}{24} + \frac {1}{12} - \frac {7}{360}= \frac {1}{45}$$ $$\displaystyle $$
 * }.
 * }.

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

$$\displaystyle c1=-1, c3=\frac{1}{6}, c5=\frac{-7}{360}. $$ $$\displaystyle $$
 * }.
 * }.

 * {| style="width:85%" border="0" align="center"

In the code, the value of c1 is taken to be -1. The value of k ranges from 1 to n, the value of n being user defined. Considering the case where k=1, we get the value of c3 and p2(1). We have $$\displaystyle f=f\cdot g\cdot (g+1)$$ and cn=-1, cd=1. The initial values of f and g are 1,2. The final value of f is equal to 6. The variables for the function fracsum are $$\displaystyle(-1\cdot cn)$$ and $$\displaystyle(cd\cdot f)$$, which give the new values of c after every iteration. The array size of the input variables for the function increases by a unit size after every iteration. The funtion inputs in the first iteration are (1,6) which are equivalent to (n,d). The function returns the output values of these as (1,6). The cn array has now been updated to [-1 1] and the cd array to [1,6], where cn and cd are the numerator and denominator of constant c(k+2). To compute the value of p2(1), the following steps are done in the program. f array has been changed to [f;1] and g array to [2+g(1);g]. For k=1, we get it as f=[6 1] and g= [4 2]. Now these input variables are returned to the function as $$\displaystyle(g-1)\cdot cn $$and $$\displaystyle f\cdot cd$$ where cn= [-1 1] and cd= [1 6].

In this case we get arrays with two elements each and the commands gcd and lcm are implemented independently on each of those elements. we get the values of newpn and newpd as -1 and 3, where pn and pd are the numerator and denominator of p2k(1). We get p=[pn pd]=[-1 3]

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

 * {| style="width:85%" border="0" align="center"

In the next iteration the value of k is increased to 2. $$\displaystyle f=f\cdot g\cdot (g+1)$$. We get f= [120 6], cn= [-1 1] and cd= [1 6]. From the function, we get the values of newcn and newcd as -7 and 360.

For computing p4(1), f is again changed to [120 6 1] and g to [6 4 2]. The function is called again with input variables as $$\displaystyle(g-1)\cdot cn $$and $$\displaystyle f\cdot cd$$. We get the values of newpn and newpd as 1 and 45. pn=[-1 1] and pd=[3 45].

similar cycle is continued for all the higher iterations and we get the higher p and c.

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

Author
Solved and typed by -Egm6341.s10.team4.anandankala 13:15, 24 March 2010 (UTC).

.

= Problem 10: Application of Engineering Orbital Mech. =

Given
Refer Lecture slide 30-4 for problem statement

Polar form relative to focus
 * {| style="width:100%" border="0" align="left"

 * [[Image:Ellipse Polar.svg|200px]]
 * }.
 * }.
 * }.

Eq (3) from Page [[media:Egm6341.s10.mtg30.djvu|30-3]]

$$\displaystyle \ \ r(\theta) = \frac{a(1-e^2)}{1-e \, cos(\theta)}$$

In our problem, a=1

where, eccentricity $$\displaystyle e = \sin(\frac{\pi}{12}) $$

First Method to compute the Arc Length for an elliptical curve is using Eq (4) from Page [[media:Egm6341.s10.mtg30.djvu|30-3]] Since for the Ellipse $$\displaystyle \theta$$ range is from 0 to 2 $$\displaystyle\pi $$

Circumference $$\displaystyle I( \theta) = \int_0^{2\pi} dl$$

Where Eq (1) $$\displaystyle dl= \sqrt{ {r}^{2}+ {\left(  \frac{dr}{d \theta}\right)}^{2}} d\theta $$.

The second method uses the elliptic integral of the second kind which is defined as:

$$E(e^2) = \int_0^{\pi/2}\sqrt {1-e^2 \sin^2\theta}\ d\theta\ $$    Eq(6)    in page     [[media:Egm6341.s10.mtg30.djvu|30-4]]

The circumference of an ellipse is $$\displaystyle C = 4 a E(e^2)$$    Eq(5)   in page   [[media:Egm6341.s10.mtg30.djvu|30-4]],

Giving the integral:

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

$$\displaystyle I= 4 E(e^2) = 4\int_0^{\pi/2}\sqrt {1-e^2 \sin^2\theta}\ d\theta $$
 * }

Find
Find Arc length of ellipse using the two Integrals and the integration methods below.

compute $$\displaystyle I_n$$ to the error $$\displaystyle 10^{-10}$$ ,

compute time using tic/toc matlab command and error estimate

1) Composit Trapozoidal rule

2) Romberg Table

3) Clencurt

Solution
The Integration result are summarized in the following table:

Note that checking both integral calculations against Ramanujan approximation 's:
 * $$C \approx \pi \left[3(a+b) - \sqrt{(3a+b)(a+3b)}\right]= \pi(3(a+b)-\sqrt{10ab+3(a^2+b^2)})$$

where, a = longest radius, b= shortest radius in ellipse

the results of our integration method and Ramanujan approximation were same. $$\displaystyle (c \simeq 6.176601987)$$

Discussion: One can clearly see that for these integrals the Trapezoidal method gives superior performance This is due to the fact that the Integrand is a periodic function. Clencurt has slightly better performance than Romberg. First method Integral gives lower quality on all methods and need more CPU time for similar performance.

(1) Composite Trapezoidal rule

Matlab Code:

subfunction ff5_10(x)

(2) Romberg Table

Matlab Code:

(3) ClenCurt Integration)

Matlab Code:

Author
Solved and typed by - --Egm6341.s10.Team4.roni 18:25, 24 March 2010 (UTC)

= Problem 11: Derivation of the equation of arclength using the cosine law =

Given
Refer Lecture slide for problem statement

[Figure 1]

 * $$\displaystyle

r(\theta) = \frac{1-e^2}{1-e \, cos(\theta)}$$

where, eccentricity $$\displaystyle e = \sin(\frac{\pi}{12}) $$

Find
Derive the equation of arclength using triangle OAB and cosine law.

 * $$d\ell=d\theta \left[ r^2+ \left( \frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}}$$

Solution

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

[reference]
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

the law of cosine
 * $$c^2 = a^2 + b^2 - 2ab\cos(\gamma)\,$$

 * style= |
 * }.
 * }.

Apply the cosine law to triangle OAB in the Figure1 (See Figure 1)

 * $$\displaystyle dl^2=\overline{AB}^2 = \overline{OA}^2 + \overline{OB}^2 - 2\overline{OA}\,\,\, \overline{OB}\cos(d\theta)\, $$
 * $$\displaystyle dl = \sqrt{\overline{OA}^2 + \overline{OB}^2 - 2\overline{OA}\,\,\, \overline{OB}\cos(d\theta)}\, $$
 * $$\displaystyle dl = \sqrt{r(\theta)^2 + r(\theta+d\theta)^2 - 2r(\theta)r(\theta+d\theta)\cos(d\theta)}\, $$
 * $$\displaystyle dl = \sqrt{\underbrace{\Big[r(\theta) - r(\theta+d\theta)\Big]^2}_{(a)} + 2r(\theta)r(\theta+d\theta)\underbrace{(1-\cos(d\theta))}_{(b)}}\, $$

Part$$\displaystyle (a) $$ is approximately
 * $$\displaystyle (r(\theta) - r(\theta+d\theta))^2 \simeq (dr)^2 \, $$

For part$$\displaystyle (b) $$

Using the other cosine summation law

 * $$\displaystyle cos(\alpha + \beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) \, $$

 * $$\displaystyle cos(d\theta)=cos\left(\frac{d\theta}{2}+\frac{d\theta}{2}\right) =cos\left(\frac{d\theta}{2}\right)cos\left(\frac{d\theta}{2}\right)-sin\left(\frac{d\theta}{2}\right)sin\left(\frac{d\theta}{2}\right)= cos^2\left(\frac{d\theta}{2}\right)-sin^2\left(\frac{d\theta}{2}\right) \, $$

because, $$\displaystyle cos^2\left(\frac{d\theta}{2}\right)= 1-sin^2\left(\frac{d\theta}{2}\right)\, $$

 * $$\displaystyle \Rightarrow cos(d\theta)= \left(1-sin^2\left(\frac{d\theta}{2}\right)\right)-sin^2\left(\frac{d\theta}{2}\right) = 1-2sin^2\left(\frac{d\theta}{2}\right)\ \, $$

In small $$\displaystyle d\theta$$ approximately,
 * $$\displaystyle sin(\frac{d\theta}{2}) \simeq \frac{d\theta}{2} \, $$

using this fact,

 * $$\displaystyle 1-cos(d\theta)= 1-\left(1-2sin^2\left(\frac{d\theta}{2}\right)\right) = 2\left(\frac{d\theta}{2}\right)^2 =\frac{(d\theta)^2}{2} $$

So, Plug in part(a) and Part(b)

 * $$\displaystyle dl = \sqrt{(r(\theta) - r(\theta+d\theta))^2 + 2r(\theta)r(\theta+d\theta)(1-\cos(d\theta))}\, $$
 * $$\displaystyle = \sqrt{(dr)^2 + 2r(\theta)r(\theta+d\theta)(\frac{(d\theta)^2}{2})} = \sqrt{(dr)^2 + r(\theta)r(\theta+d\theta)(d\theta)^2} \, $$
 * $$\displaystyle = d\theta \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r(\theta)r(\theta+d\theta)} \, $$

when $$\displaystyle d\theta$$ is very small, approximately, $$\displaystyle r(\theta)r(\theta+d\theta) \simeq r(\theta)^2 \, $$

 * $$\displaystyle \simeq d\theta \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r(\theta)^2} \, $$

Author
Solved and typed by - Egm6341.s10.Team4.yunseok 17:01, 24 March 2010 (UTC)

Proof read by - Egm6341.s10.Team4.nimaa&m 16:06, 24 March 2010 (UTC) and --Egm6341.s10.Team4.roni 18:26, 24 March 2010 (UTC)

= Contributing Members = Egm6341.s10.Team4.andy 08:28, 24 March 2010 (UTC) - Author of Problems: 1, 7 ; Proof Read: 5, 3

Egm6341.s10.Team4.riherd 14:19, 24 March 2010 (UTC) - Author of Problem: 6 ; Proof Read: 1, 2

Egm6341.s10.Team4.nimaa&m 16:06, 24 March 2010 (UTC) - Author of Problems: 2, 8 ; Proof Read: 11

Egm6341.s10.Team4.yunseok 17:02, 24 March 2010 (UTC) - Author of Problems: 5, 11

Egm6341.s10.team4.anandankala 12:50, 24 March 2010 (UTC)- Author of problems: 3,9; Proofread:7, 8.

Egm6341.s10.Team4.roni 18:28, 24 March 2010 (UTC) - Author of problems: 4,10; Proofread: 1,11.

title: User:Egm6341.s10.Team4/HW5


text: File:6341 5 1.jpg

title: File:6341 5 1.jpg


text: File:6341 5 11.JPG

title: File:6341 5 11.JPG


text: Category:Experimental physics

title: Category:Experimental physics


text: Category:Science experiments

title: Category:Science experiments


text:
Category:Scientific method

A scientific method is a sequence or collection of processes that are considered characteristic of scientific investigation and the acquisition of new scientific knowledge based upon physical evidence.

title: Category:Scientific method


text:
BucketSort

In BucketSort we have an input list and a number of buckets. In order to distribute the data records to the buckets we need a function which tells, into which bucket an actual record has to go. And this function must comply with the following relation: Having m buckets, the buckets are numbered and the function returns an integer between 1 and m. Normally, after a sort, in every bucket will be records with different key values. Therefor BucketSort in itself is normally not a complete sorting algorithm; with one exception: there are as many buckets as there are possible key values. If we want to sort by unsigned 16-bit-integers and we put up 65.536 buckets, then the sort can be done in one go. Any key is inspected only once; thus the time for a sort increases in a linear fashion with the number of records.

The algorithm is stable and the time for a sort does not depend on the key values.

Timing
Normally, after one distribution, there will be records with different key values in the same bucket; it is necessary to sort them internally. For this internal sort BucketSort can be used recursively, but then a different function for selecting the buckets is necessary (different key values!). Depending on this function, the behavior of BucketSort changes completely.

If we are sorting by a positive 8-bit number, put up 256 buckets, and take the value of the key as the number of the destination bucket, the algorithm is identical with ExtraDix. Hence the timing is

 * T = E * n * lg256 (N) = E * n

where E is a constant reflecting the implementation. The time increases in a linear fashion with the number of records.

We use only two buckets and a pivot for the distribution function. If the key value is smaller or equal to the pivot, the record goes into the first bucket or else into the second bucket. Then we call the function recursively. The result is that BucketSort converts into Quicksort. Hence the timing is

 * T = C * n * lg2 (n)

where C is a constant reflecting the implementation.

We use ten buckets. The function to determine the bucket number converts the key value into a decimal number and depending on the repetitions the corresponding digit (starting with the most significant digit) is selected; the digit gives directly the number of the bucket where the record shall go. BucketSort behaves like an inverted RadixSort and the timing is

 * T = D * n * lg10 (N)

where D is a constant reflecting the implementation.

We use ten buckets. The function to determine the bucket number converts the key value into a decimal number and depending on the repetitions the corresponding digit (starting with the least significant digit) is selected; the digit gives directly the number of the bucket where the record shall go. After distributing all records to the buckets, the records are recollected and the sort is repeated for the next digit position. BucketSort now behaves exactly like RadixSort and the timing is

 * T = E * n * lg10 (N)

where E is a constant reflecting the implementation.

The normal way how BucketSort is used recursively is to use a fixed number m of buckets on any level. The keys of all records are checked in order to find the smallest value and the biggest one. Then the difference between the smallest and biggest value is divided evenly into m intervals and for each interval a start value is calculated. The distribution function is just a nested sequence of if-statements; if the actual key value is bigger than x put the record into bucket m, if the actual key value is bigger than y put the record into bucket m-1 and so on. Now the content of a bucket is recursively distributet to m sub-buckets; therefor the timing is

 * T = F * n * lgm (N)

where F is a constant reflecting the implementation.

Example
For the demonstration we use the variant simulating Quicksort with two buckets. The number of necessary division levels is calculated dynamically by help of the average of the key values. All records with a value smaller or equal to the average go into the first bucket and all others into the second bucket. Buckets with sorted content are displayed in orange.

Input List: average = 385

Bucket 1: average = 148

Bucket 2: average = 622

We continue with bucket 1. The average is 148; therefor all records with a value smaller or equal to 148 go into bucket three and all others into bucket four.

Bucket 3: average = 15

Bucket 4: average = 237

We continue with bucket 3. The average is 15; therefor all records with a value smaller or equal to 15 go into bucket five and all others into bucket six.

Bucket 5:

Bucket 6:

We now recombine Bucket 5 and 6 to form the sorted bucket 3:

Sorted Bucket 3:

We continue with bucket 4. The average is 237; therefor all records with a value smaller or equal to 237 go into bucket seven and all others into bucket eight.

Bucket 7: average = 200

Bucket 8:

We continue with bucket 7. The average is 200; therefor all records with a value smaller or equal to 200 go into bucket nine and all others into bucket ten.

Bucket 9:

Bucket 10:

We can recombine them to form the sorted bucket 7:

Sorted Bucket 7:

Bucket 8 has only one element and is sorted by definition. We can recombine it with bucket 7 to build sorted bucket 4:

Sorted Bucket 4:

We can now recombine bucket 3 and bucket 4 to form the sorted bucket 1:

Sorted Bucket 1:

We now make a shortcut and get directly the sorted bucket 2:

Sorted Bucket 2:

The Sorted List is build by recombining bucket 1 and 2:

It is obvious from the relative position of the buckets to each other that BucketSort can easily be realized as an in place algorithm.

title: BucketSort


text:
DITeLearning

IMPORTANT: The content from Wikiresources page was copied to this new page to change the title and adjust to a research project.

Home | Learning Theories | Instructional Design | ePortfolio | Storyboard | Wiki Analysis | Communication Tools | Resources Analysis | Group Work | Teaching Philosophy | Accessibility | Evaluation | Glossary

Introduction
This Wikiversity project is part of one of the requirements and learning outcomes of the MSc. in Applied eLearning at the Dublin Institute of Technology, specifically in the Instructional Design and eAuthoring module, as part of which students are required to work as a group to design and develop an activity/eLearning object that can be used as a tool for learning.

The genesis of this Wiki was the desire to provide a structure for students to identify and assimilate their knowledge, ideas and conclusions about the different digital technologies available to eLearning. The choice of Wikiversity as a platform for this project arose from the fact that, as an education project, it operates on an academic level and addresses a university community as its audience. The increasing sophistication in technology, along with advances in research in the eLearning environment, may in time render the information in this wiki inaccurate or even obsolete; therefore this wiki is not to be viewed as a definitive work, since users may continuously update it. Rather it is conceived as a dynamic expansion of resources and elements that in themselves constitute its continued usefulness. Future students might use it as a learning resource, and they may include or change concepts, links, images, videos, or add other elearning objects based on their studies and conceptual conclusions.

Learning Objectives
Although this project is not intended to be definitive, we hope that it will meet the following objectives:

 * 1) To motivate learners to explore, evaluate and record knowledge about diverse resources relevant to eLearning
 * 2) 	To operate as a digital laboratory wherein participants may experience and apply models and methods available for instructional design
 * 3) 	To serve as a guide to the various tools introduced on the course
 * 4) 	To provide a model that encourages new investigators wishing to explore eLearning
 * 5) 	To promote the skills needed to design instructional eLearning objects
 * 6) 	To support collaborative action and knowledge sharing

Instructional Design and eAuthoring Module
This section allows you to see the requirements on the Instructional Design and eAuthoring module in the MSc. Applied eLearning.

In this section you will find information about learning theories, models and techniques applied to instructional design. In addition, information about essential reading and module assessment is included.

You can jump to the Instructional Design and eAuthoring section by clicking on this blue arrow.

Learning Theories

This section presents a concise account of the learning theories and theorists most applicable to this eLearning project. For a project such as this to be of use to other users it is fundamental that it be based on sound pedagogical principles and each building block reflects this.

Go to the Learning Theories section by clicking on this blue arrow.

Storyboard

The first and perhaps most important phase in creating an elearning object is its storyboard. In this section you will discover the principles behind a storyboard, its creation and content, and its importance as an elearning project. Moreover you can view the storyboard created by the Wikiresources Group for this elearning object along with an interesting alternative experiment in storyboarding, called Storybird, which uses collaborative storytelling.

Go to the Storyboard and enjoy this section by clicking on this blue arrow.

Wiki Analysis
We believe that the initial step for our project should be based on a lucid examination of the tool used to develop the resource. Therefore we can begin by examining the pros and cons of our tool. The Wiki Analysis section will demonstrate how you can determine the best kind of platform or tool to employ according your learning objectives. Consequently the Wikiresources Group decided to examine a PMI analysis (Plus, Minus and Interesting) and a SWOT (Strength, Weaknesses, Opportunities and Threats) analysis of the wiki. An initial PMI analysis will afford valuable information about Wikis as a tool in our group elearning experience, in which we discuss and highlight why a wiki is suitable as an elearning resource. The entire discussion then is based on a fundamental question: Why this tool?

If you want to know the answer go to the Wiki Analysis section by clicking on this blue arrow.

ePortfolio

Part of this project is creating an ePortfolio, and identifying its significance and support for ideas and reflections. We created a Wikiresources Group website to support the creation of the ePortfolio. The contents of this website should be presented in greater detail to clarify our analysis of the course and also to explain why we consider a website as the best model. Below you will find a link to the Wikiresources Group website, which will provide more information on ePortfolios.

Go to the ePortfolio section by clicking on this blue arrow.

Resources Analysis

An eLearning environment affords a significant number of technological resources; yet these are continually replaced by more efficient and attractive devices. In this section we attempt to present the characteristics and ways that these artifacts can be used in eLearning, and to identify the most recent resources available. We have listed primarily the technological resources that are free of charge to the user. This does not however exclude paid options.

For information on various eLearning resources, go to Resources Analysis section by clicking on this blue arrow.

Communication Tools

This section will provides an analysis of communication tools that facilitate the flow of information between individuals and groups. It demonstrates how a tool can be applied in the eLearning field. Since there are many tools available that create efficient communication, we will present in this section some tools used by our group for our daily communication as we construct our eLearning project. You can go to the Communication Tools section by clicking on this blue arrow.

Accessibility
In this section you will have access to relevant information in relation to accessibility. The importance of the interface usability between the user and the object of learning.

Accessibility assumes that any learner can use the same means to full fill the requirements: identical if possible, equivalent when not.

An elearning object should be designed to be useful and accessible by learners with diverse abilities and respectful of diversity. They also should feel the learning object a fair and safe learning environment.

The Accessibility section is available by clicking on this blue arrow.

Group Work

A group is an interaction between a collection of people for the purpose of achieving a common objective. There are many factors and variables that contribute to a group dynamic and to a successful and productive outcome.

The Group Work section is available by clicking on this blue arrow.

Teaching Philosophy

To participate in any recurring activity usually involves adopting a methodology that underlies how that activity is structured. So too is the case in educational activities, where an underlying teaching philosophy guides what we do and how we do it. Such a philosophy is central to the planning, design and development of our courses. What we achieve in any course is linked to the clarity of our aspirations and the practical implementation of a sound and productive philosophy. In order to achieve these results, it is important to understand the principles behind teaching philosophies.

For the reasons why a Teaching Philosophy is desirable, click on the blue arrow.

Evaluation

Attempts have been made to create a resource that is as user-friendly as possible through the use of clear navigational direction, links, colour combinations, graphics, etc. Evaluation of the resource will be carried out by our current peers on the MSc course. Access to an evaluation questionnaire will be placed on the home page of the wiki for their use.

Go to the Evaluation section by clicking on this blue arrow.

Glossary

You may find some unfamiliar words in this wiki project. If so please consult the Glossary section for the meaning of these words by clicking on this blue arrow.

Feedback

Our group would appreciate if you give us your feedback about this project. If you have any query, suggestion or information, please contact us sending a message to our email: wikiresourcesgroup@gmail.com

title: DITeLearning


text:
DITeLearning/Learning Theories

Home | Learning Theories | Instructional Design | ePortfolio | Storyboard | Wiki Analysis | Communication Tools | Resources Analysis | Group Work | Teaching Philosophy | Accessibility | Evaluation | Glossary

Introduction
This wiki project can be considered as a laboratory for several learning theories, especially those most relevant to adult learners. The Learning Theories section will give you a short brief review of these theories and summarize the characteristics of each to allow you to reflect on the way that learning theory has been an essential underpinning for development of this project.

Learning Outcomes
At the end of this chapter the learner will be able to:
 * 1) Understand which learning theories are most relevant to adult learners.
 * 2) Understand what the main characteristics of each of these theories.
 * 3) Understand why the knowledge of the learning theories is essential for development of a learning object.
 * 4) Understand which learning theories that most influenced the instructional design concept.

We would appreciate if you give us your contribution in this project. If you have any query, suggestion or feedback, don't hesitate to contact us sending a message to our email: wikiresourcesgroup@gmail.com

Overview
The major theme of Vygotsky's theoretical framework is that social interaction plays a fundamental role in the development of cognition. Vygotsky (1978) states:
 * "Every function in the child's cultural development appears twice: first, on the social level, and later, on the individual level; first, between people (inter-psychological) and then inside the child (intra-psychological). This applies equally to voluntary attention, to logical memory, and to the formation of concepts. All the higher functions originate as actual relationships between individuals." (Vogotsky, 1962).

A second aspect of Vygotsky's theory is the idea that the potential for cognitive development depends upon the "zone of proximal development" (ZPD): a level of development attained when children engage in social behaviour. Full development of the ZPD depends upon full social interaction. The range of skill that can be developed with adult guidance or peer collaboration exceeds what can be attained alone.

Vygotsky's theory was an attempt to explain consciousness as the end product of socialization. For example, in the learning of language, our first utterances with peers or adults are for the purpose of communication but once mastered they become internalized and allow "inner speech".

Scope/Application
This is a general theory of cognitive development. Most of the original work was done in the context of language learning in children (Vygotsky, 1962), although later applications of the framework have been broader (see Wertsch, 1985).

 * Example:
 * Vygotsky (1978, p56) provides the example of pointing a finger. Initially, this behavior begins as a meaningless grasping motion; however, as people react to the gesture, it becomes a movement that has meaning. In particular, the pointing gesture represents an interpersonal connection between individuals.

Principles

 * Cognitive development is limited to a certain range at any given age.
 * Full cognitive development requires social interaction.

Overview
This theory stipulates that there are several different types or levels of learning. The significance of these classifications is that each different type requires different types of instruction. Gagne identifies five major categories of learning: verbal information, intellectual skills, cognitive strategies, motor skills and attitudes. Different internal and external conditions are necessary for each type of learning. For example, for cognitive strategies to be learned, there must be a chance to practice developing new solutions to problems; to learn attitudes, the learner must be exposed to a credible role model or persuasive arguments.

Gagne suggests that learning tasks for intellectual skills can be organized in a hierarchy according to complexity: stimulus recognition, response generation, procedure following, use of terminology, discriminations, concept formation, rule application, and problem solving. The primary significance of the hierarchy is to identify prerequisites that should be completed to facilitate learning at each level. Prerequisites are identified by doing a task analysis of a learning/training task. Learning hierarchies provide a basis for the sequencing of instruction.

In addition, the theory outlines nine instructional events and corresponding cognitive processes. These events should satisfy or provide the necessary conditions for learning and serve as the basis for designing instruction and selecting appropriate media (Gagne, Briggs & Wager, 1992).

Scope/Application
While the nine instructional events covers all aspects of learning, the focus of the theory is on intellectual skills. The theory has been applied to the design of instruction in all domains (Gagne & Driscoll, 1988). In its original formulation (Gagne, 1962), special attention was given to military training settings. Gagne (1987) addresses the role of instructional technology in learning.

 * Example:
 * The following example illustrates a teaching sequence corresponding to the nine instructional events for the objective, Recognize an equilateral triangle:
 * Gain attention - show variety of computer generated triangles
 * Identify objective - pose question: "What is an equilateral triangle?"
 * Recall prior learning - review definitions of triangles
 * Present stimulus - give definition of equilateral triangle
 * Guide learning- show example of how to create equilateral
 * Elicit performance - ask students to create 5 different examples
 * Provide feedback - check all examples as correct/incorrect
 * Assess performance- provide scores and remediation
 * Enhance retention/transfer - show pictures of objects and ask students to identify equilaterals (1985, chapter 12) provides examples of events for each category of learning outcomes.

Principles

 * Different instruction is required for different learning outcomes.
 * Events of learning operate on the learner in ways that constitute the conditions of learning.
 * The specific operations that constitute instructional events are different for each different type of learning outcome.
 * Learning hierarchies define what intellectual skills are to be learned and a sequence of instruction.

Overview
A major theme in the theoretical framework of Bruner is that learning is an active process in which learners construct new ideas or concepts based upon their current/past knowledge. The learner selects and transforms information, constructs hypotheses, and makes decisions, relying on a cognitive structure to do so. Cognitive structure (i.e., schema, mental models) provides meaning and organization to experiences and allows the individual to "go beyond the information given".

As far as instruction is concerned, the instructor should try and encourage students to discover principles by themselves. The instructor and student should engage in an active dialogue (i.e., Socratic learning). The task of the instructor is to translate information to be learned into a format appropriate to the learner's current state of understanding. Curriculum should be organized in a spiral manner so that the student continually builds upon what they have already learned.

Bruner (1966) states that a theory of instruction should address four major aspects:
 * Predisposition towards learning
 * The ways in which a body of knowledge can be structured so that it can be most readily grasped by the learner
 * The most effective sequences in which to present material
 * The nature and pacing of rewards and punishments. Good methods for structuring knowledge should result in simplifying, generating new propositions, and increasing the manipulation of information.

In his more recent work, Bruner (1986, 1990, 1996) has expanded his theoretical framework to encompass the social and cultural aspects of learning as well as the practice of law.

Scope/Application
Bruner's constructivist theory is a general framework for instruction based upon the study of cognition. Much of the theory is linked to child development research (especially Piaget). The ideas outlined in Bruner (1960) originated from a conference focused on science and maths learning. Bruner illustrated his theory in the context of mathematics and social science programs for young children (see Bruner, 1973). The original development of the framework for reasoning processes is described in Bruner, Goodnow & Austin (1951). Bruner (1983) focuses on language learning in young children. Note that Constructivism is a very broad conceptual framework in philosophy and science and Bruner's theory represents one particular perspective.

 * Example:
 * This example is taken from Bruner (1973):
 * "The concept of prime numbers appears to be more readily grasped when the child, through construction, discovers that certain handfuls of beans cannot be laid out in completed rows and columns. Such quantities have either to be laid out in a single file or in an incomplete row-column design in which there is always one extra or one too few to fill the pattern. These patterns, the child learns, happen to be called prime. It is easy for the child to go from this step to the recognition that a multiple table, so called, is a record sheet of quantities in completed multiple rows and columns. Here is factoring, multiplication and primes in a construction that can be visualized."

Principles

 * Instruction must be concerned with the experiences and contexts that make the student willing and able to learn (readiness).
 * Instruction must be structured so that it can be easily grasped by the student (spiral organization).
 * Instruction should be designed to facilitate extrapolation and or fill in the gaps (going beyond the information given).

Overview
The social learning theory of Bandura emphasizes the importance of observing and modelling the behaviours, attitudes, and emotional reactions of others. Bandura (1977) states: "Learning would be exceedingly laborious, not to mention hazardous, if people had to rely solely on the effects of their own actions to inform them what to do. Fortunately, most human behaviour is learned observationally through modelling: from observing others one forms an idea of how new behaviours are performed, and on later occasions this coded information serves as a guide for action." (p22). Social learning theory explains human behaviour in terms of continuous reciprocal interaction between cognitive, behavioural, and environmental influences.

The component processes underlying observational learning are:
 * Attention, including modelled events (distinctiveness, affective valence, complexity, prevalence, functional value) and observer characteristics (sensory capacities, arousal level, perceptual set, past reinforcement)
 * Retention, including symbolic coding, cognitive organization, symbolic rehearsal, motor rehearsal)
 * Motor Reproduction, including physical capabilities, self-observation of reproduction, accuracy of feedback
 * Motivation, including external, vicarious and self reinforcement.

Because it encompasses attention, memory and motivation, social learning theory spans both cognitive and behavioural frameworks. Bandura's theory improves upon the strictly behavioural interpretation of modelling provided by Miller & Dollard (1941). Bandura’s work is related to the theories of Vygotsky and Lave which also emphasize the central role of social learning.

Scope/Application
Social learning theory has been applied extensively to the understanding of aggression (Bandura, 1973) and psychological disorders, particularly in the context of behaviouur modification (Bandura, 1969). It is also the theoretical foundation for the technique of behaviour modelling which is widely used in training programs. In recent years, Bandura has focused his work on the concept of self-efficacy in a variety of contexts (e.g., Bandura, 1997).

 * Example:
 * The most common (and pervasive) examples of social learning situations are television commercials. Commercials suggest that drinking a certain beverage or using a particular hair shampoo will make us popular and win the admiration of attractive people. Depending upon the component processes involved (such as attention or motivation), we may model the behaviour shown in the commercial and buy the product being advertised.

Principles

 * The highest level of observational learning is achieved by first organizing and rehearsing the modelled behaviour symbolically and then enacting it overtly. Coding modelled behaviour into words, labels or images results in better retention than simply observing.
 * Individuals are more likely to adopt a modelled behaviour if it results in outcomes they value.
 * Individuals are more likely to adopt a modelled behaviour if the model is similar to the observer and has admired status and the behavior has functional value.

Overview
This theory suggests that learning happens best under conditions that are aligned with human cognitive architecture. The structure of human cognitive architecture, while not known precisely, is discernible through the results of experimental research. Recognizing George Miller's research showing that short term memory is limited in the number of elements it can contain simultaneously, Sweller builds a theory that treats schemas, or combinations of elements, as the cognitive structures that make up an individual's knowledge base. (Sweller, 1988)

The contents of long term memory are "sophisticated structures that permit us to perceive, think, and solve problems," rather than a group of rote learned facts. These structures, known as schemas, are what permit us to treat multiple elements as a single element. They are the cognitive structures that make up the knowledge base (Sweller, 1988). Schemas are acquired over a lifetime of learning, and may have other schemas contained within themselves.

The difference between an expert and a novice is that a novice hasn't acquired the schemas of an expert. Learning requires a change in the schematic structures of long term memory and is demonstrated by performance that progresses from clumsy, error-prone, slow and difficult to smooth and effortless. The change in performance occurs because as the learner becomes increasingly familiar with the material, the cognitive characteristics associated with the material are altered so that it can be handled more efficiently by working memory.

From an instructional perspective, information contained in instructional material must first be processed by working memory. For schema acquisition to occur, instruction should be designed to reduce working memory load. Cognitive load theory is concerned with techniques for reducing working memory load in order to facilitate the changes in long term memory associated with schema acquisition.

Scope/Application
Sweller's theories are best applied in the area of instructional design of cognitively complex or technically challenging material. His concentration is on the reasons that people have difficulty learning material of this nature. Cognitive load theory has many implications in the design of learning materials which must, if they are to be effective, keep cognitive load of learners at a minimum during the learning process. While in the past the theory has been applied primarily to technical areas, it is now being applied to more language-based discursive areas.

 * Example:
 * In combining an illustration of blood flow through the heart with text and labels, the separation of the text from the illustration forces the learner to look back and forth between the specified parts of the illustration and the text. If the diagram is self-explanatory, research data indicates that processing the text unnecessarily increases working memory load. If the information could be replaced with numbered arrows in the labeled illustration, the learner could concentrate better on learning the content from the illustration alone. Alternatively, if the text is essential to intelligibility, placing it on the diagram rather than separated will reduce cognitive load associated with searching for relations between the text and the diagram (Sweller, 1999).

Principles
Specific recommendations relative to the design of instructional material include:

 * Change problem solving methods to avoid means-ends approaches that impose a heavy working memory load, by using goal-free problems or worked examples.
 * Eliminate the working memory load associated with having to mentally integrate several sources of information by physically integrating those sources of information.
 * Eliminate the working memory load associated with unnecessarily processing repetitive information by reducing redundancy.
 * Increase working memory capacity by using auditory as well as visual information under conditions where both sources of information are essential (i.e. non-redundant) to understanding.

Overview
When preparing eLearning materials, you should consider how users like to take in and process information (learning styles). Learning style inventories are related to personality indicators derived by Jung. The Myers-Briggs Type Indicator (MBTI) assessment is a psychometric questionnaire developed during WWII by designed to measure psychological preferences in how people perceive the world and make decisions. The initial questionnaire grew into the Myers-Briggs Type Indicator, which was first published in 1962. The classic learning styles inventories are based on Kolb’s Learning Inventory which flows from Kolb’s experiential learning cycle. These essentially have two components, an active experimentation (AC) - reflective observation (RO) continuum and a concrete experience (CE) - abstract conceptualisation (AC) continuum leading to four poles of learning preferences.
 * AC-RO Assimilating
 * AC-AE Converging
 * CE –AE Accommodating
 * CE-RO Diverging

title: DITeLearning/Learning Theories