text:
User:Egm6341.s10.team3.Min/HW6

2 Show $$ dp_{\overline{y}}=mVdr$$
Ref: Lecture Notes [[media:Egm6341.s10.mtg33.djvu|p.33-2]]

Problem Statement
At t+dt, consider V+dV Show that $$ dp_{\overline{y}}=mVdr$$

Solution

At t+dt, V+dV,see the above figure
 * {| style="width:100%" border="0" align="left"

$$ By negnecting higher oder terms, mdVdr, We have: At t+dt, V+dV,see the above figure
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ End the proof
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * }
 * }

-- By Min Zhong 12:00, 7 Arip 2010 (UTC)

7 Show that $$ Z(\varsigma= \frac{1}{2})= \frac{1}{2}(Z_i-Z_{i+1})+ \frac{h}{8}(f_i-f_{i+1}) $$
Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that $$ Z(\varsigma=2)= \frac{1}{2}(Z_i-Z_{i+1})+ \frac{h}{8}(f_i-f_{i+1}) $$

In Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2]](2), We have:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle (Eq. 7-1)
 * }
 * }

In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.35-4]](1), we have:

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

\begin{Bmatrix} C_0 \\ C_1 \\ C_2 \\ C_3 \\ \end{Bmatrix}= \begin{bmatrix} 1 & 0 & 0&0 \\ 0 & 1 & 0&0 \\  -3 & -2 & 3&-1 \\  2 & 1 & -2&1\\ \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^\prime \\ Z_{i+1} \\ Z_{i+1}^\prime \\ \end{Bmatrix}
 * $$\displaystyle
 * $$\displaystyle

$$ $$ In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.36-1]](1), we have:
 * $$\displaystyle (Eq. 7-2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle (Eq. 7-3)
 * }
 * }

Solution
We start the proof from Equation 1.
 * {| style="width:100%" border="0" align="left"

\begin{align} Z(\varsigma= \frac{1}{2})& = \sum_{n=0}^{3}C_i{ \frac{1}{2}}^i\\ &=C_0+ \frac{1}{2}C_1+ \frac{1}{4}C_2+ \frac{1}{8}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-4)
 * }
 * }

Next, we will find the expression of Ci by Zi and fi From the matrix Eq. 7-2, we have:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle (Eq. 7-5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle (Eq. 7-6)
 * }
 * }

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

$$ $$
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-7)
 * }
 * }

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

$$ $$
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-8)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.7-4,We can get:
 * {| style="width:100%" border="0" align="left"

\begin{align} Z(\varsigma= \frac{1}{2})& =Z_i+ \frac{1}{2}Z_i^\prime + \frac{1}{4}(-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime)+ \frac{1}{8}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime)\\ &=\frac{1}{2}Z_i+\frac{1}{2}Z_{i+1}+\frac{1}{8}Z_i^\prime-\frac{1}{8}Z_{i+1}^\prime\\ &=\frac{1}{2}(Z_i+Z_{i+1})+\frac{1}{8}(Z_i^\prime-Z_{i+1}^\prime)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-9)
 * }
 * }

Plug Eq 3 into above equation, we have

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

$$\displaystyle Z(\varsigma=\frac{1}{2})=\frac{1}{2}(Z_i+Z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

-- By Min Zhong 12:00, 7 Apri 2010 (UTC)

8 Show $$ Z(\varsigma=\frac{1}{2})^\prime=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime-Z_{i+1}^\prime)$$
Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that Z(\varsigma=\frac{1}{2})^\prime=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime-Z_{i+1}^\prime) From Lecture Notes [[media:Egm6341.s10.mtg56.djvu|p.35-1]](2)

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

$$ $$
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle (Eq. 8-1)
 * }
 * }

Solution
We start the proof from Eq.8-1,

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

$$ $$
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle (Eq. 8-2)
 * }
 * }

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

\begin{align} Z(\varsigma={\frac{1}{2}}^\prime)&=C_1+ 2C_2\frac{1}{2}+ 3C_3{\frac{1}{2}}^2\\ &=C_1+ C_2 +\frac{3}{4}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8-3)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.8-3, we get:

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

\begin{align} Z(\varsigma=\frac{1}{2})^\prime & = Z_i^\prime-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime+\frac{1}{2}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime) \\ &=-\frac{3}{2}Z_i+\frac{3}{2}Z_{i+1}-\frac{1}{4}Z_i^\prime-\frac{1}{4}Z_{i+1}^\prime\\ &=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime+Z_{i+1}^\prime)\\ \end{align}
 * $$\displaystyle
 * $$\displaystyle

$$ -- By Min Zhong 12:00, 7 Apri 2010 (UTC)
 * }
 * }

9 Show $$ Z_{i+1}=Z_i+\frac{h/2 }{3}[f_i+4f_{i+\frac{1}{2}}+f_{i+1}]$$
Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]]

Problem Statement
Proof that $$ Z_{i+1}=Z_i+\frac{h/2 }{3}[f_i+4f_{i+\frac{1}{2}}+f_{i+1}]$$

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]](1)

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

$$ $$
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * $$\displaystyle (Eq. 9-1)
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

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

\dot Z_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9-2)
 * }
 * }

Solution
When $$\triangle=0$$, from Eq.9-1, we have :
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * $$\displaystyle (Eq. 9-3)
 * }
 * }

Plug Eq. 9-3 into Eq, 9-2, we can get:
 * {| style="width:100%" border="0" align="left"

f_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9-4)
 * }
 * }

Rearrange Eq. 9-4 we have:

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

\begin{align} Z_{i+1}&=Z_i+ \frac{h}{6}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ &=Z_i+ \frac{h/2}{3}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-2)
 * }
 * }

title: User:Egm6341.s10.team3.Min/HW6


text:
University of Florida/Egm6341/s10.team3.aks/HW6

(3) Evaluate the rest of the coefficient of matrix
Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-3]]

Problem Statement
Evaluate the remaining coefficient of Matrix by using degrees of Freedom

Solution
We have

$$ Z(s)=\sum_{i=0}^{3}c_is^i=\sum_{i=0}^{3}N_i(s)d_i $$

such that

$$ d_1=Z_i$$

$$ d_2=\dot{Z_i}$$

$$ d_3=Z_{i+1}$$

$$ d_4={\dot{Z}_{i+1}}$$

where

$$ \dot{Z}=\frac{dZ}{dt}={\frac{dZ}{ds}} \frac{ds}{dt} $$

such that $$ \frac{ds}{dt} = \frac {1}{h} $$

We know the coefficient of matrix for first two rows from lecture notes   [[media:Egm6341.s10.mtg32.djvu|p.35-3]]

$$\begin{bmatrix} 1 &0 &0  &0 \\  0&1  &0  &0 \end{bmatrix}$$

Using the equations above we have

$$  d_3=Z_{i+1}=Z(s=1)=c_0+c_1+c_2+c_3 $$

$$  d_4=\dot{Z}_{i+1}=\frac{1}{h}Z'(s=1)=c_1+2c_2+3c_3 $$

Putting the results in matrix form we obtain

$$\begin{bmatrix} 1 &0 &0  &0 \\ 0 &1  &0  &0 \\ 1 &1  &1  &1 \\ 0 &1  &2  &3 \end{bmatrix}\begin{Bmatrix} c_0\\ c_1\\ c_2\\

c_3\end{Bmatrix}= \begin{Bmatrix} Z_i\\ Z'_i\\ Z_{i+1}\\ Z'_{i+1} \end{Bmatrix}$$

(4) Verify the inverse of matrix using Matlab
Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Problem Statement
Find the inverse of given Matrix

A =

1    0     0     0     0     1     0     0     1     1     1     1     0     1     2     3

Solution
which is same as the one given on [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Hence Verified

(5) Identify basis functions and plot them
Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Problem Statement
Identify the basis functions

$$ N_i (s) $$

where $$ i = 1,2,3,4$$

Solution
We have

$$ z(s) = \sum_{i=0}^3 c^i s^i = \sum_{i=1}^4 N_i(s) d_i $$

Expanding above we obtain

$$ N_1 d_1 + N_2 d_2 + N_3 d_3 + N_4 d_4 = C_0 s^0 + C_1 s^1 + C_2 s^2 + C_3 s^3$$

$$                                      = C_0  + C_1 s^1 + C_2 s^2 + C_3 s^3$$

$$d_1 = z_i = z(s=0) = C_0$$

$$d_2 = \dot z_i = \dot z_i (s=0) = C_1$$

$$d_3 = z_{i+1}= z_{i+1} (s=1) = C_0 + C_1 + C_2  + C_3 $$

$$ d_4 = \dot z_{i+1} =C_1 + 2 C_2 + 3 C_3 $$

Inserting above values in first eq we obtain

$$(N_1 + N_3) C_0 + (N_2+N_3+N_4) C_1 + (N_3+2 N_4)C_2 + (N_3 + 3 N_4) C_3 = C_0 + C_1 s + C_2 s^2 + C_3 s^3$$

Comparing both LHS and RHS we obtain

$$N_1 + N_3 = 1$$

$$N_2+N_3+N_4 = s$$

$$N_3+2 N_4=s^2$$

$$N_3 + 3 N_4 = s^3$$

Solving above we obtain basis functions

$$N_1 = 1- 3 s^2 +2 s^3$$

$$N_2 = s^3 - 2 s^2 + s $$

$$N_3 = 3 s^2 - 2 s^3$$

Below is the plot of above basis functions

$$ N_4 = s^3 - s^2$$

= (6) Show that s = s(t) =

Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.36-1]]

Problem Statement
We have to show that s is the function of t (s = s(t) )

Solution
We have (from [[media:Egm6341.s10.mtg32.djvu|p.35-1]] eq (1))

$$ t(s)= (1-s) t_i + s t_{i+1} $$

$$    = t_i + s (t_{i+1} - t_i ) $$

$$ but \frac {ds}{dt} = \frac{1}{h} $$

$$ or \frac {dt}{ds} = h $$

so $$ t_{i+1} - t_i = \frac{t_{i+1} - t_i}{1-0}= h $$

$$ t(s) = t_i +h s $$ $$ h s = t - t_i $$ $$ s = \frac {(t - t_i)} {h} $$ $$ s = s(t) $$

Hence Proved

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


text:
Category:Inexplicable physics pages

__EXPECTUNUSEDCATEGORY__ '''This category remains for historical purposes, it should not be used for existing pages. Rather, see Requests for Deletion/Archives/10 for the ultimate disposition of 24 pages so tagged (plus another, next on the RfD archive page). --Abd (discuss • contribs) 21:01, 21 December 2014 (UTC)'''

This category was created, 06:54, 7 April 2010, by User:Sj to apply to pages created by IP and the following is the original research or opinion of Sj even though it was unsigned. This category will be emptied, because it is POV. The comments here may be moved elsewhere after the cleanup is complete. --Abd (discuss • contribs) 15:24, 27 September 2014 (UTC)

Most of these inexplicable physics pages are either copies of Wikipedia articles with original research subtlely interleaved, or articles that were deleted from or rejected from Wikipedia.

These pages are all a mixture of real physics, fringe presentations and interpretations of fundamental physical principles, and original research. They are combined in such a way that it is difficult to separate simple and well-known observations from novel links or ideas which may or may not be correct. The language is a mixture of perfect English (which inevitably comes from another source -- often Wikipedia, from current or past-and-deleted articles, but sometimes possibly a copyvio) and broken English (with quirks such as '80-ties' for the decade of the 80s, and ungrammatical sentences).

Many of these pages were initially posted as Articles for Creation requests on Wikipedia, then deleted as fringe material that only linked to itself and did not sufficiently identify its fringe status (despite its interesting and motely constellation of references).

Characteristics of inexplicable physics pages

 * Sections devoted to 'resonators' related to whatever the nominal topic is
 * Sections related to tiny and gigantic scales, quantum scales, &c
 * Extended references to work by physicists from the end of the 19th century, especially Heaviside.

Most of the pages include material from Wikipedia pages (which is all well sourced and referenced) and interlaces inexplicable statements and original research. Some of them are more clearly OR than others, and the 'research' in question isn't presented as research but as known fact and equations (which don't always make sense).

Deletion discussion
The initial en:wp deletion discussion and review is here: Wikipedia:Articles for deletion/Selfconsistent gravidynamic constants.

Science conspiracies
Related discussions by the author about science conspiracies include this interesting diff.

title: Category:Inexplicable physics pages


text:
User:Egm6341.s10.Team4.andy/HW6

= Problem 1: Arc Length Calculation =

Given
Refer Lecture slide 32-1 for problem statement

The value of $$\displaystyle r(\theta) $$ from Page [[media:Egm6341.s10.mtg30.djvu|30-3]] is given by,

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

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

Here, $$\displaystyle a=1 $$ and $$\displaystyle e = \sin(\pi/12) $$

Arc Length for an elliptical curve with $$\displaystyle \theta$$ varying from 0 to $$\displaystyle \frac{\pi}{3} $$ is given by,

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

$$\displaystyle I = \int_0^{\frac{\pi}{3}} \sqrt{ {r}^{2}+ {\left( \frac{dr}{d \theta}\right)}^{2}} d\theta $$ $$
 * $$\displaystyle \longrightarrow (2)
 * }.
 * }.

The elliptic integral of the second kind is given as:

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

$$\displaystyle E(e^2)=I = \int_0^{\pi/3}\sqrt {1-e^2 \sin^2\theta}\ d\theta\ $$ $$
 * $$\displaystyle \longrightarrow (3)
 * }.
 * }.

Find
1. Use error estimate for Composite Trapezoidal Rule to find n, such that the error is to the order of $$\displaystyle O(10^{-10}) $$.

Error estimate is given by,

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

$$ $$
 * $$\displaystyle
 * E_{n}| \leq \frac{(b-a)^3}{12n^2}M_{2}
 * E_{n}| \leq \frac{(b-a)^3}{12n^2}M_{2}
 * $$\displaystyle \longrightarrow (4)
 * }.
 * }.

2. Use successive numerical Integration results as a stopping criterion, i.e,

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

$$ $$
 * $$\displaystyle
 * I_{2n} - I_{n}| < 10^{-10}
 * I_{2n} - I_{n}| < 10^{-10}
 * $$\displaystyle \longrightarrow (5)
 * }.
 * }.

3. Verify results of Romberg Integration, Clencurt Integration, Chebfun sum Integration with that of Composite Trapezoidal Rule.

Solution
1.

Using the error estimate of Composite Trapezoidal Rule given by Equation (4) for Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

n \le 1.675553588956571 \times 10^{4} \approx 16756 $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Using the error estimate of Composite Trapezoidal Rule given by Equation (4) for Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

n \le 8.006607416431191 \times 10^{3} \approx 8007
 * $$\displaystyle
 * $$\displaystyle

$$  Matlab Code:
 * }.
 * }.

2. Using Successive Numerical Integration Results to obtain error tolerance of order of $$\displaystyle O(10^{-10}) $$.

For Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 16384 \\ E &= 2.685143218883468 \times 10^{-10} \\ I_n &= 1.26369048548985 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

From above we get $$\displaystyle n < 16756 $$.

For Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 4096 \\ E &= 4.863667246723935 \times 10^{-10} \\ I_n &= 1.03682664319432 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

From above we get $$\displaystyle n < 8007 $$.

 Matlab Code:  . 3. Verifying results with Romberg Integration, Clencurt Integration and Chebfun Sum Integration.

Using Romberg table, the integration results are given by:

For Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 64 \\ E &= 3.917310920087402 \times 10^{-11} \\ I_n &= 1.26368461998287 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

For Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 32 \\ E &= 3.602149689641010 \times 10^{-10} \\ I_n &= 1.03682398743891 \\ \end{align} $$  Matlab Code:  .
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Using Clencurt, the integration results are given by:

For Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 8 \\ E &= 1.575350960791866 \times 10^{-10} \\ I_n &= 1.26369048600540 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

For Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 8 \\ E &= 2.562394740834861 \times 10^{-13} \\ I_n &= 1.03682664384306 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:  .

Using Chebfun sum, the integration results are given by:

For Equation (2), an error tolerance $$\displaystyle E $$ is obtained as,

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

\begin{align} E &= 5.107025913275720 \times 10^{-15} \\ I_n &= 1.263690485847865 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

For Equation (3), an error tolerance $$\displaystyle E $$ is obtained as,

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

\begin{align} E &= 2.220446049250313 \times 10^{-16} \\ I_n &= 1.036826643842808 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:

title: User:Egm6341.s10.Team4.andy/HW6


text:
User:Berny68

Hi,

I'm Berny

Subpages

 * interesting

title: User:Berny68


text:
User:Berny68/interesting

Wikipedia

 * Inside Wikipedia

title: User:Berny68/interesting


text:
User talk:Berny68

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title: User talk:Berny68


text:
Survey research and design in psychology/Tutorials/Correlation

Contents

 * 1) /General steps/
 * 2) /Types of correlation and level of measurement/
 * 3) /Types of correlations - Exercises/
 * 4) /Scatterplot correlation guess/
 * 5) /Effects of outliers/
 * 6) /Correlations and non-linear relations/

Readings

 * 1) Howitt & Cramer (2014)
 * 2) Part 2: Descriptive statistics (particularly Chapters 8 to 11)
 * 3) Allen & Bennett (2010)
 * 4) Chapter 12 "Correlation"
 * 5) Chapter 16 "Non-parametric procedures"
 * 6) 16.3 Chi-square for contingencies
 * 7) 16.10 Cramer's V
 * 8) 16.11 Spearman's Rho and Kendall's Tau-B
 * 9) Francis (2007)
 * 10) Section 3.1 "Relations Between Metric Variables"

Recordings

 * 1) Virtual tutorial recording, 2018

 * 1) Screencasts


text:
Survey research and design in psychology/Tutorials/Multiple linear regression/Exercises/Deterrence theory

== Deterrence theory and compliance with the law == Allen & Bennett Ch 13 Data file: data_13_2.sav

Deterrence theory suggests that compliance with the law is shaped by perceived certainty of punishment and perceived severity of punishment. When the perceived certainty (i.e., likelihood) and severity (i.e., personal costs) of punishment for breaking the law are high, law-breaking behaviour is predicted to be low. Conversely, when perceived punishment certainty and severity decrease, non-compliance with the law should increase.

 * Standard

To test deterrence theory, 62 participants were asked to complete a three-part questionnaire:
 * 1) Frequency of participants' everyday law breaking behaviours (e.g., littering, speeding, driving while intoxicated, shop-lifting etc.)
 * 2) Perceived certainty (likelihood) of being punished
 * 3) Personal severity (costs) of punishment.

Thus, the multiple linear regression model is:

Results: R = .20, R2 = .04, F (2, 59) = 1.23, p = .30
 * DV = (likelihood of) compliance (with the law).
 * IV1 = (perceived) certainty (of being caught).
 * IV2 = (perceived) severity (of punishment).

Regression equation: Compliance = 2.969 + .154xCertainty + .015xSeverity
 * Hierarchical

For the second analysis, use IV1 and IV2 in Step 1 of a hierarchical MLR, with Step 2 consisting of:
 * IV3 = (perceived) legitimacy (of the law / law-abidingness) (used in the hierarchical analysis).


text:
User:Prof. Loc Vu-Quoc

Prof. Loc Vu-Quoc of the University of Florida has had students in each of the following classes organize their work here on Wikiversity, since 2008, using a different teacher account for each class. Below are a few courses; for a complete list, see Other courses.

 * Fall 2008
 * User:Eas4200c.f08 - EAS 4200 Aerospace Structures
 * User:Eml4500.f08 - EML 4500 Finite Element Analysis and Design, Fall 2008

 * Spring 2009
 * User:Cis4914.s09 - CIS 4914 Senior Design
 * User:Egm6322.s09 - EGM 6322 Principles of Engineering Analysis 2
 * User:Egm6936.s09 - EGM 6936 Seminar

 * Fall 2009
 * User:Egm6321.f09 - EGM 6321 Principles of Engineering Analysis 1
 * User:Egm6936.f09 - EGM 6936 Seminar

 * Spring 2010
 * User:Egm6341.s10 - EGM 6341 Numerical Methods for Engineering Analysis 1

title: User:Prof. Loc Vu-Quoc


text:
Talk:Inside Wikipedia

This is a fabulous start. Kudos to the editors. –SJ + > 08:04, 7 April 2010 (UTC)

 * Thanks. :-) Cormaggio talk 09:16, 8 April 2010 (UTC)

title: Talk:Inside Wikipedia


text: Template:Barnstar Scholar

title: Template:Barnstar Scholar


text: File:Basis function plot.jpg

title: File:Basis function plot.jpg


text:
Multiple linear regression/Assumptions

Level of measurement

 * DV: A normally distributed interval or ratio variable
 * 1) IVs: Two or more normally distributed interval or ratio variables or dichotomous variables. Note that it may be necessary to recode non-normal interval or ratio IVs or multichotomous categorical or ordinal IVs into dichotomous variables or a series of dummy variables).

Sample size
Enough data is needed to provide reliable estimates of the correlations. As the number of IVs increases, more inferential tests are being conducted, therefore more data is needed, otherwise the estimates of the regression line are probably unstable and are unlikely to replicate if the study is repeated.

Some rules of thumb:
 * 1) Use at least 50 cases plus at least 10 to 20 as many cases as there are IVs.
 * 2) Green (1991) and Tabachnick and Fidell (2007):
 * 3)  50 + 8(k) for testing an overall regression model and
 * 4) 104 + k when testing individual predictors (where k is the number of IVs)
 * 5) Based on detecting a medium effect size (β >= .20), with critical α <= .05, with power of 80%.

To be more accurate, study-specific power and sample size calculations should be conducted (e.g., use A-priori sample Size calculator for multiple regression; note that this calculator uses f2 for the anticipated effect size - see the Formulas link for how to convert R2 to to f2).

Normality

 * 1) Check the univariate descriptive statistics (M, SD, skewness and kurtosis). As a general guide, skewness and kurtosis should be between -1 and +1.
 * 2) Check the histograms with a normal curve imposed.
 * 3) Note: Be wary (i.e., avoid!) using inferential tests of normality (e.g., the Shapiro–Wilk test - they are notoriously overly sensitive for the purposes/needs of regression).
 * 4) Normally distributed variables will enhance the MLR solution. Estimates of correlations will be more reliable and stable when the variables are normally distributed, but regression will be reasonably robust to minor to moderate deviations from non-normal data when moderate to large sample sizes are involved. Also examine scatterplots for bivariate outliers because non-normal univariate data may make bivariate and multivariate outliers more likely.
 * 5) Further information:
 * 6) * Does your data violate linear regression assumptions? (quality-control-plan.com)
 * 7) * Regression when the OLS residuals are not normally distributed (StackExchange)
 * 8) * How do I perform a regression on non-normal data which remain non-normal when transformed? (StackExchange)

Linearity
Check scatterplots between the DV (Y) and each of the IVs (Xs) to determine linearity:
 * 1) Are there any bivariate outliers? If so, consider removing the outliers.
 * 2) Are there any non-linear relationships? If so, consider using a more appropriate type of regression.

Homoscedasticity
Based on the scatterplots between the IVs and the DV:
 * 1) Are the bivariate distributions reasonably evenly spread about the line of best fit?
 * 2) Also can be checked via the normality of the residuals.

Multicollinearity
IVs should not be overly correlated with one another. Ways to check:
 * 1) Steps are shown in this screencast:
 * 2) Examine bivariate correlations and scatterplots between each of the IVs (i.e., are the predictors overly correlated - above ~.7?).
 * 3) Check the collinearity statistics in the coefficients table:
 * 4) Various recommendations for acceptable levels of VIF and Tolerance have been published.
 * 5) Variance Inflation Factor (VIF) should be low (< 3 to 10) or
 * 6) Tolerance should be high (> .1 to .3)
 * 7) Note that VIF and Tolerance have a reciprocal relationship (i.e., TOL=1/VIF), so only one of the indicators needs to be used.
 * 8) For more information, see

Multivariate outliers
Check whether there are influential MVOs using Mahalanobis' Distance (MD) and/or Cook’s D (CD):
 * 1) Steps are shown in these screencasts:
 * 2) SPSS: Linear Regression - Save - Mahalanobis (can also include Cook's D)
 * 3) After execution, new variables called mah_1 (and coo_1) will be added to the data file.
 * 4) In the output, check the Residuals Statistics table for the maximum MD and CD.
 * 5) The maximum MD should not exceed the critical chi-square value with degrees of freedom (df) equal to number of predictors, with critical alpha =.001. CD should not be greater than 1.
 * 6) If outliers are detected:
 * 7) Go to the data file, sort the data in descending order by mah_1, identify the cases with mah_1 distances above the critical value, and consider why these cases have been flagged (these cases will each have an unusual combination of responses for the variables in the analysis, so check their responses).
 * 8) Remove these cases and re-run the MLR.
 * 9) If the results are very similar (e.g., similar R2 and coefficients for each of the predictors), then it is best to use the original results (i.e., including the multivariate outliers).
 * 10) If the results are different when the MVOs are not included, then these cases probably have had undue influence and it is best to report the results without these cases.

Normality of residuals
The residuals should be normally distributed around 0.
 * 1) Residuals are more likely to be normally distributed if each of the variables normally distributed, so check normality first.
 * 2) There are three ways of visualising residuals. In SPSS - Analyze - Regression - Linear - Plots:
 * 3) Scatterplot: ZPRED on the X-axis and ZRESID on the Y-axis
 * 4) Histogram: Check on
 * 5) Normal probability plot: Check on
 * 6) Scatterplot should have no pattern (i.e. be a "blob").
 * 7) Histogram should be normally distributed
 * 8) Normal probability plot should fall along the diagonal line
 * 9) If residuals are not normally distributed, there is probably something wrong with the distribution of one or more variables - re-check

title: Multiple linear regression/Assumptions


text:
Survey research and design in psychology/Lectures/Multiple linear regression I

Outline
Introduces and explains the use of linear regression and multiple linear regression, a multivariate correlational statistical technique, in the context of psychology.

More specifically, this lecture covers:
 * 1) Correlation (Review)
 * 2) Simple linear regression
 * 3) Multiple linear regression
 * 4) General steps
 * 5) Assumptions
 * 6) R, coefficients
 * 7) Equation
 * 8) Types
 * 9) Summary
 * 10) MLR I Quiz - Practice questions

Prerequisite

 * 1) Correlation (Lecture)

Readings

 * 1) Howitt and Cramer (2014a):
 * 2) Chapter 09: Regression: Prediction with precision (pp. 120-132)
 * 3) Chapter 32: Multiple regression and multiple correlation (pp. 444-459)
 * 4) StatSoft (2016). How to find relationship between variables, multiple regression. StatSoft Electronic Statistics Handbook.
 * 5) Tabachnick, B. G., & Fidell, L. S. (2013) (6th ed. - International ed.). Multiple regression (Ch. 5) [Includes sample write-ups] . Using multivariate statistics (6th ed., pp. 117-196). Boston, MA: Pearson. Includes example write-ups.  UCLearn reading list (Psychometric instrument development)

Handouts

 * Lecture slides (Google Slides)
 * 2018 handouts:
 * [[Media:SRDP Lecture07Handout MultipleLinearRegression 6slidesperpage.pdf|Download 6 slides to a page]]: SRDP Lecture07Handout MultipleLinearRegression 6slidesperpage.pdf
 * [[Media:SRDP Lecture07Handout MultipleLinearRegression 3slidesperpage.pdf|Download 3 slides to a page]]: SRDP Lecture07Handout MultipleLinearRegression 3slidesperpage.pdf


text:
Algebraic Properties of Equality

The equal sign that depicts the fact that both sides of it are equal is a very strange symbol with many properties. It tells you various traits of each side, and it allows you to manipulate each side in specific ways. Here are the different properties of that sign:

title: Algebraic Properties of Equality


text:
Talk:Algebraic Properties of Equality

Attribution
This page was created with content imported from Wikibooks:Algebra/Solving Equations on 7 April 2010, which see for specific details of attribution. A complete list of contributors of that content is as follows. --Moonriddengirl 13:22, 7 April 2010 (UTC)
 * Wikibooks:User:Andre anckaert
 * Wikibooks:User:Thenub314
 * IP contributor 71.8.83.44
 * Wikibooks:User:Aztecvic
 * IP contributor 69.54.131.78
 * IP contributor 76.110.240.64
 * IP contributor 71.198.35.231
 * Wikibooks:User:Darklama
 * Wikibooks:User:Hoogli
 * Wikibooks:User:Whiteknight
 * IP contributor 202.3.130.17
 * IP contributor 203.217.78.14
 * IP contributor 24.21.17.1
 * IP contributor 81.236.19.142
 * Wikibooks:User:Doctormatt
 * Wikibooks:User:Phoenix7477
 * Wikibooks:User:NoSpin
 * Wikibooks:User:Jguk
 * IP contributor 69.34.97.8

title: Talk:Algebraic Properties of Equality


text:
Template:Help me learn

Please help me learn by including some useful and verifiable facts as part of a discussion or analysis of my idea, theory, or argument.

title: Template:Help me learn


text: File:Exampleyun.png

title: File:Exampleyun.png


title: File:Grap2.png


text: File:Girl with poster 2.jpg

title: File:Girl with poster 2.jpg


text:
User:Egm6341.s10.Team4/HW6

= Problem 1: Arc Length Calculation =

Given
Refer Lecture slide 32-1 for problem statement

The value of $$\displaystyle r(\theta) $$ from Page [[media:Egm6341.s10.mtg30.djvu|30-3]] is given by,

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

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

Here, $$\displaystyle a=1 $$ and $$\displaystyle e = \sin(\pi/12) $$

Arc Length for an elliptical curve with $$\displaystyle \theta$$ varying from 0 to $$\displaystyle \frac{\pi}{3} $$ is given by,

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

$$\displaystyle I = \int_0^{\frac{\pi}{3}} \sqrt{ {r}^{2}+ {\left( \frac{dr}{d \theta}\right)}^{2}} d\theta $$
 * $$\displaystyle \longrightarrow (2)$$
 * }.
 * }.

The elliptic integral of the second kind is given as:

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

$$\displaystyle E(e^2)=I = \int_0^{\pi/3}\sqrt {1-e^2 \sin^2\theta}\ d\theta\ $$
 * $$\displaystyle \longrightarrow (3)$$
 * }.
 * }.

Find
1. Use error estimate for Composite Trapezoidal Rule to find n, such that the error is to the order of $$\displaystyle O(10^{-10}) $$.

Error estimate is given by,

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

$$
 * $$\displaystyle
 * E_{n}| \leq \frac{(b-a)^3}{12n^2}M_{2}
 * E_{n}| \leq \frac{(b-a)^3}{12n^2}M_{2}
 * $$\displaystyle \longrightarrow (4)$$
 * }.
 * }.

2. Use successive numerical Integration results as a stopping criterion, i.e,

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

$$
 * $$\displaystyle
 * I_{2n} - I_{n}| < 10^{-10}
 * I_{2n} - I_{n}| < 10^{-10}
 * $$\displaystyle \longrightarrow (5)$$
 * }.
 * }.

3. Verify results of Romberg Integration, Clencurt Integration, Chebfun sum Integration with that of Composite Trapezoidal Rule.

Solution
1.

Using the error estimate of Composite Trapezoidal Rule given by Equation (4) for Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

n \le 1.675553588956571 \times 10^{4} \approx 16756 $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Using the error estimate of Composite Trapezoidal Rule given by Equation (4) for Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

n \le 8.006607416431191 \times 10^{3} \approx 8007
 * $$\displaystyle
 * $$\displaystyle

$$  Matlab Code:
 * }.
 * }.

2. Using Successive Numerical Integration Results to obtain error tolerance of order of $$\displaystyle O(10^{-10}) $$.

For Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 16384 \\ E &= 2.685143218883468 \times 10^{-10} \\ I_n &= 1.26369048548985 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

From above we get $$\displaystyle n < 16756 $$.

For Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 4096 \\ E &= 4.863667246723935 \times 10^{-10} \\ I_n &= 1.03682664319432 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

From above we get $$\displaystyle n < 8007 $$.

 Matlab Code:  .

3. Verifying results with Romberg Integration, Clencurt Integration and Chebfun Sum Integration.

Using Romberg table, the integration results are given by:

For Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 32 \\ E &= 3.970122008922772 \times 10^{-10} \\ I_n &= 1.26366702185545 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

For Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 32 \\ E &= 9.005374224102525 \times 10^{-11} \\ I_n &= 1.03682398743891 \\ \end{align} $$  Matlab Code:  .
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Using Clencurt, the integration results are given by:

For Equation (2), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 8 \\ E &= 1.575350960791866 \times 10^{-10} \\ I_n &= 1.26369048600540 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

For Equation (3), an error tolerance of the order of $$\displaystyle O(10^{-10}) $$ is obtained for,

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

\begin{align} n &= 8 \\ E &= 2.562394740834861 \times 10^{-13} \\ I_n &= 1.03682664384306 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:  .

Using Chebfun sum, the integration results are given by:

For Equation (2), an error tolerance $$\displaystyle E $$ is obtained as,

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

\begin{align} E &= 5.107025913275720 \times 10^{-15} \\ I_n &= 1.263690485847865 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

For Equation (3), an error tolerance $$\displaystyle E $$ is obtained as,

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

\begin{align} E &= 2.220446049250313 \times 10^{-16} \\ I_n &= 1.036826643842808 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:  .

Author
Solved and typed by - Egm6341.s10.Team4.andy 14:18, 7 April 2010 (UTC)

Reviewed By - Egm6341.s10.Team4.roni 18:31, 7 April 2010 (UTC)

= Problem 2: Rate of momentum change for optimal control problem =

Given
Envisage the below figure as free body diagram of an aircraft:

.

likewise consider the shown axes and vectors, for $$\displaystyle v+dv $$ at $$\displaystyle t+dt $$:

.

Find
Show that $$\displaystyle dp_{\bar y}=mvd\gamma $$.

Solution
According to the above figure, we can survey these two cases at $$\displaystyle t $$ and $$\displaystyle t+dt $$, the velocity of the aircraft after $$\displaystyle dt $$ will reach to $$\displaystyle v+dv $$ and the angle between the aircraft and horizontal axis will reach to the $$\displaystyle \gamma+d\gamma $$. Thus, regarding $$\displaystyle d\gamma $$ generated angle between two velocity vectors, we can write:

 * $$\displaystyle

d\gamma\approxeq sin(d\gamma)=\frac{dv_{\bar y}}{v+dv} $$

 * $$\displaystyle

\Rightarrow dv_{\bar y}=v.d\gamma+dv.d\gamma $$

The amount of $$\displaystyle dv.d\gamma $$ can be neglected in front of $$\displaystyle v.d\gamma $$.

 * $$\displaystyle

\Rightarrow dv_{\bar y}=v.d\gamma $$

On the other hand, momentum is defined as $$\displaystyle p=mv $$. So, we have:

dp_{\bar y}=d(m.v_{\bar y})=v_{\bar y}.dm+m.dv_{\bar y} $$

Assuming the amount of $$\displaystyle dm $$ to be negligible in front of changes in velocity;

dp_{\bar y}=d(m.v_{\bar y})=m.dv_{\bar y} $$

Finally, we can summarize the answer as:

dp_{\bar y}=m.dv_{\bar y}=m.v.d\gamma $$

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

$$\displaystyle dp_{\bar y}=m.v.d\gamma $$
 * 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 03:08, 3 April 2010 (UTC) Reviewed by- Egm6341.s10.team4.anandankala 13:45, 7 April 2010 (UTC).

= Problem 3: Row verification for the matrix=

Given

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

 * $$\underbrace{\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix}.}_{A} \underbrace{\begin{bmatrix}

C_0 \\ C_1 \\ C_2 \\ C_3 \end{bmatrix}}_{C} = \begin{bmatrix}

z_i \\ z^'_i \\ z_{i+1} \\ z^'_{i+1} \end{bmatrix} $$
 * }.
 * }.

Find

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

 * Arrive at the third and fourth rows of the matrix A using the equations,

$$\displaystyle d_3=z_{i+1}=z(s=1).$$

$$\displaystyle d_4=z^'_{i+1}=z^'(s=1).$$

 * }.
 * }.

Solution

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

We have $$\displaystyle z(s)= \sum_{i=0}^{3} C_i s^i= C_0+ C_1s+ C_2s^2+ C_3s^3$$
 * $$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

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

$$\displaystyle z^'(s)= \sum_{i=1}^{3} iC_i s^{i-1}= C_1+ 2C_2s+ 3C_3s^2$$
 * $$\displaystyle \longrightarrow (2)$$
 * }.
 * }.

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

We get the product of the third row of the matrix A and the column vector C by substituting s=1 in the equation(1). $$\displaystyle \Rightarrow z_{i+1}= z(s=1)= C_0+ C_1+ C_2+ C_3$$
 * $$\displaystyle \longrightarrow (3)$$
 * }.
 * }.

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

We get the product of the fourth row of the matrix A and the column vector C by substituting s=1 in the equation(2). $$\displaystyle \Rightarrow z^'_{i+1}= z^'(s=1)= C_1+ 2C_2+ 3C_3$$
 * $$\displaystyle \longrightarrow (4)$$

 * }.
 * }.

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

By looking at the equations (3) and (4), one can say that the third and fourth rows of the matrix A are [1 1 1 1] and [0 1 2 3].
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.team4.anandankala 13:57, 7 April 2010 (UTC) Reviewed by - Egm6341.s10.Team4.roni 18:33, 7 April 2010 (UTC) .

= Problem 4: Matrix Inversion for Hermite - Simpson Algorithm =

Given
Matrix A [[media:Egm6341.s10.mtg35.djvu|35-3]], which is:

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

$$
 * $$\displaystyle
 * $$\displaystyle

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 1 & 1 & 1 & 1\\ 0 & 1 & 2 & 3 \end{bmatrix}
 * $$\displaystyle A= \begin{bmatrix}

$$
 * style= |
 * }
 * }

Find
Show that the Inversion of this Matrix A Gives the following Matrix: on slide [[media:Egm6341.s10.mtg35.djvu|35-4]]

$$\displaystyle $$

$$\displaystyle Inv( A ) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ -3 & -2 & 3 & -1\\ 2 & 1 & -2 & 1 \end{bmatrix}

$$

Solution
Using MatLab:

 Matlab Code:

Giving Result,

A_Inv =

1    0     0     0     0     1     0     0    -3    -2     3    -1     2     1    -2     1

Author
Solved and typed by - --Egm6341.s10.Team4.roni 01:15, 7 April 2010 (UTC) Reviewed by -

= Problem 5: Identification of basis function $$\displaystyle \overline{N_i}$$ =

Given
Refer Lecture slide 35-2,4 for problem statement

The relationship between $$\displaystyle Z(s)$$ and $$\displaystyle d_i$$(degree of freedom)

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

Z(s)=\sum_{i=0}^{3}c_i s^i = \sum_{i=1}^{4}N_i(s) d_i = \sum_{i=1}^{4} \overline{N_i}(s) \, \overline{d_i} $$ where,
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)$$
 * }.
 * }.
 * $$\displaystyle N_i(s)$$ = basis function
 * $$\displaystyle d_i(s)$$ = degree of freedom

$$\displaystyle d_1=Z_i $$ $$\displaystyle \,\,\,\,\, ,d_2=\dot{Z_i} $$ $$\displaystyle \,\,\,\,\, ,d_3=Z_{i+1} $$ $$\displaystyle \,\,\,\,\, ,d_4=\dot{Z}_{i+1} $$

Find
Identify the the basis function $$\displaystyle \left\{ N_i(s) \right\}\,\,\, ,i=1,2,3,4 $$, and plot $$\displaystyle \overline{N_i}(s)$$'s

Solution
From the equation (1) of 35-4(lecture slide)

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

\begin{Bmatrix} c_0 \\ c _1 \\ c _2 \\ c_3 \end{Bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^' \\ Z_{i+1} \\ Z_{i+1}^' \end{Bmatrix} =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} \overline{d}_1 \\ \overline{d}_2 \\ \overline{d}_3 \\ \overline{d}_4 \\ \end{Bmatrix} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

where,

$$\displaystyle \overline{d}_2= h \, d_2$$ $$\displaystyle, \,\,\,\,\, \overline{d}_4= h \, d_4$$ $$\displaystyle, \,\,\,\,\, \overline{d}_1=d_1$$ $$\displaystyle, \,\,\,\,\, \overline{d}_3=d_3$$

Expand the matrix equation.

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

\begin{align} &c_0= \overline{d}_1 \\ &c_1= \overline{d}_2 \\ &c_2= -3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4 \\ &c_3= 2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4 \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

plug in $$\displaystyle c_i$$ to equation (1)

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

\begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i = c_0 + c_1 s^1 + c_2 s^2 + c_3 s^3 \\ &= \overline{d}_1 + \overline{d}_2 s^1 + (-3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4) s^2 + (2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4) s^3 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

arrange in terms of $$\displaystyle \overline{d}_i $$

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

\begin{align} & = (2s^3-3s^2+1)\overline{d}_1+(s^3-2s^2+s)\overline{d}_2+(-2s^3+3s^2)\overline{d}_3+(s^3-s^2)\overline{d}_4 \\ &=\overline{N}_1(s) \, \overline{d}_1+\overline{N}_2(s) \, \overline{d}_2+\overline{N}_3(s) \, \overline{d}_3+\overline{N}_4(s) \, \overline{d}_4 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

\begin{align} & \overline{N}_1= 2s^3-3s^2+1 \\ & \overline{N}_2= s^3-2s^2+s \Rightarrow N_2= h(s^3-2s^2+s)\\ & \overline{N}_3= -2s^3+3s^2 \\ & \overline{N}_4= s^3-s^2 \Rightarrow N_4= h(s^3-s^2) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Plot of $$\displaystyle N_i $$
 * [[File:basis_function.png]]

.

Author
Solved and typed by - Egm6341.s10.Team4.yunseok 14:55, 7 April 2010 (UTC) Reviewed by - Egm6341.s10.Team4.riherd 14:46, 7 April 2010 (UTC)

= Problem 6 - Inversion of Local Domain Transformation=

Given
In using our Hermit interpolation we use the transformation from $$ \displaystyle t \epsilon [t_i,t_{i+1}]$$ to $$ \displaystyle s  \epsilon [0,1]$$ using the transformation $$ \displaystyle t(s)=(1-s)t_i+st_{i+1}$$

Find
We are asked to invert this transformation from the form of $$ \displaystyle t=t(s)$$ to $$ \displaystyle s=s(t)$$

Solution
Inverting this transformation to find s in terms of t is a simple algebra problem.

$$ \displaystyle t=(1-s)t_i+st_{i+1} $$

$$ \displaystyle t=t_i+s(t_{i+1}-t_{i}) $$

$$ \displaystyle s=\frac{t-t_{i}}{t_{i+1}-t_{i}} $$

Author
Solved and typed by - Egm6341.s10.Team4.riherd 14:45, 7 April 2010 (UTC) Reviewed by - Egm6341.s10.team4.anandankala 15:57, 7 April 2010 (UTC)

= Problem 7: Expression for Hermitian interpolation at $$ t_{i+\frac{1}{2}} $$ =

Given
Consider the Hermitian interpolation by the following equation (on slide [[media:Egm6341.s10.mtg35.djvu|35-2]]):
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^3 c_is^i

 * }
 * }

Find
Show the following expression can be obtained for $$\displaystyle z'_{i+\frac{1}{2}} $$:

 * $$\displaystyle

z_{i+\frac{1}{2}}=z(s=\frac{1}{2})=\frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$

Solution
By differentiating from the equation for $$\displaystyle z(s) $$, we will attain:

 * $$\displaystyle

z'(s)=\sum_{i=1}^3 ic_is^{i-1} $$

Now, we can compute the followings:

 * $$\displaystyle

z_i=z(s=0)=\sum_{i=0}^3 c_i(0)^i=c_0+0+0+0=c_0 $$

 * $$\displaystyle

z_{i+1}=z(s=1)=\sum_{i=0}^3 c_i(1)^i=c_0+c_1+c_2+c_3 $$

 * $$\displaystyle

z'_i=z'(s=0)=\sum_{i=1}^3 ic_i(0)^{i-1}=c_1+0+0=c_1 $$

 * $$\displaystyle

z'_{i+1}=z'(s=1)=\sum_{i=1}^3 ic_i(1)^{i-1}=c_1+2c_2+3c_3 $$

 * $$\displaystyle

\dot{z}=\frac{dz}{dt}=\frac{dz}{ds}\times\frac{ds}{dt}=z'\times\frac{1}{h} $$

 * $$\displaystyle

\Rightarrow h(f_i-f_{i+1})=h(\dot{z}_i-\dot{z}_{i+1})=(z'_i-z'_{i+1}) $$

 * $$\displaystyle

\Rightarrow \frac{h}{8}(f_i-f_{i+1})=\frac{h}{8}(\dot{z}_i-\dot{z}_{i+1})=\frac{1}{8}(z'_i-z'_{i+1}) $$

 * $$\displaystyle

\Rightarrow\begin{cases} z_i+z_{i+1}=2c_0+c_1+c_2+c_3 \\ f_i-f_{i+1}=-2c_2-3c_3 \end{cases} $$

 * $$\displaystyle

\Rightarrow\begin{cases} \frac{1}{2}(z_i+z_{i+1})=c_0+\frac{1}{2}c_1+\frac{1}{2}c_2+\frac{1}{2}c_3 \\ \frac{h}{8}(f_i-f_{i+1})=-\frac{1}{4}c_2-\frac{3}{8}c_3 \end{cases} $$

 * $$\displaystyle

\Rightarrow\frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1})=c_0+\frac{1}{2}c_1+\frac{1}{2}c_2+\frac{1}{2}c_3-\frac{1}{4}c_2-\frac{3}{8}c_3=c_0+\frac{1}{2}c_1+\frac{1}{4}c_2+\frac{1}{8}c_3 $$

The acquired foregoing equation is equal to RHS of the expression. Now, we can compute the LHS of it as:

 * $$\displaystyle

LHS=z_{i+\frac{1}{2}}=z(s=\frac{1}{2})=\sum_{i=0}^3 c_i(\frac{1}{2})^i=c_0\times 1+c_1\times\frac{1}{2}+c_2\times\frac{1}{4}+c_3\times\frac{1}{8}=c_0+\frac{1}{2}c_1+\frac{1}{4}c_2+\frac{1}{8}c_3 $$

 * $$\displaystyle

\Rightarrow LHS=RHS $$

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

$$\displaystyle z_{i+\frac{1}{2}}=z(s=\frac{1}{2})=\frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&m 04:15, 3 April 2010 (UTC) .

= Problem 8: Expression for derivative of Hermitian interpolation at $$\displaystyle t_{i+\frac{1}{2}} $$ =

Given
Consider the Hermitian interpolation by the following equation (on slide [[media:Egm6341.s10.mtg35.djvu|35-2]]):
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^3 c_is^i

 * }
 * }

Find
Show the following expression can be obtained for $$\displaystyle z'_{i+\frac{1}{2}} $$:

 * $$\displaystyle

z'_{i+\frac{1}{2}}=z'(s=\frac{1}{2})=-\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z'_i+z'_{i+1}) $$

Solution
By differentiating from the equation for $$\displaystyle z(s) $$, we will attain:

 * $$\displaystyle

z'(s)=\sum_{i=1}^3 ic_is^{i-1} $$

Now, we can compute the followings:

 * $$\displaystyle

z_i=z(s=0)=\sum_{i=0}^3 c_i(0)^i=c_0+0+0+0=c_0 $$

 * $$\displaystyle

z_{i+1}=z(s=1)=\sum_{i=0}^3 c_i(1)^i=c_0+c_1+c_2+c_3 $$

 * $$\displaystyle

z'_i=z'(s=0)=\sum_{i=1}^3 ic_i(0)^{i-1}=c_1+0+0=c_1 $$

 * $$\displaystyle

z'_{i+1}=z'(s=1)=\sum_{i=1}^3 ic_i(1)^{i-1}=c_1+2c_2+3c_3 $$

 * $$\displaystyle

\Rightarrow\begin{cases} z_i-z_{i+1}=c_0-(c_0+c_1+c_2+c_3)=-c_1-c_2-c_3 \\ z'_i+z'_{i+1}=2c_1+2c_2+3c_3 \end{cases} $$

 * $$\displaystyle

\Rightarrow\begin{cases} -\frac{3}{2}(z_i-z_{i+1})=\frac{3}{2}c_1+\frac{3}{2}c_2+\frac{3}{2}c_3\\ -\frac{1}{4}(z'_i+z'_{i+1})=-\frac{1}{2}c_1-\frac{1}{2}c_2-\frac{3}{4}c_3 \end{cases} $$

 * $$\displaystyle

\Rightarrow -\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z'_i+z'_{i+1})=(\frac{3}{2}-\frac{1}{2})c_1+(\frac{3}{2}-\frac{1}{2})c_2+(\frac{3}{2}-\frac{3}{4})c_3=c_1+c_2+\frac{3}{4}c_3 $$

The acquired foregoing equation is equal to RHS of the expression. Now, we can compute the LHS of it as:

 * $$\displaystyle

LHS=z'_{i+\frac{1}{2}}=z'(s=\frac{1}{2})=\sum_{i=1}^3 ic_i(\frac{1}{2})^{i-1}=1\times c_1\times (\frac{1}{2})^{1-1}+2\times c_2\times (\frac{1}{2})^{2-1}+3\times c_3\times (\frac{1}{2})^{3-1}=c_1+c_2+\frac {3}{4}c_3 $$

 * $$\displaystyle

\Rightarrow LHS=RHS $$

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

$$\displaystyle z'_{i+\frac{1}{2}}=z'(s=\frac{1}{2})=-\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z'_i+z'_{i+1}) $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&m 04:23, 3 April 2010 (UTC) .

=Problem 9: Expression for Hermitian interpolation at $$\displaystyle t_{i+1}$$ =

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

 * $$z_{i+1}=z_i+ \dfrac {h/2}{3}[f_i+ 4f_{i+1/2}+ f_{i+1}]$$

 * }.
 * }.

Solution

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

 * $$z^'_{i+1/2}= z^'(s=1/2)= \dfrac {-3}{2}(z_i-z_{i+1})- \dfrac{1}{4}(z^'_i +z^'_{i+1})$$

 * }.
 * }.

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

 * We have $$\displaystyle z^'= hz^.$$

 * }.
 * }.

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

 * Applying collocation at $$\displaystyle t_i$$ and $$\displaystyle t_{i+1}$$, we have $$z^._i= f_i$$ and $$z^._{i+1}=f_{i+1}$$

 * }.
 * }.

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

 * $$\displaystyle \Rightarrow z^._{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1}) $$

 * }.
 * }.

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

 * $$\displaystyle \bigtriangleup := z^._{i+1/2}- f_{i+1/2} $$

 * }.
 * }.

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

 * Applying collocation at $$\displaystyle t_{i+1/2}$$, we have $$\displaystyle \bigtriangleup= 0$$ and $$\displaystyle z^._{i+1/2}= f_{i+1/2}$$

 * }.
 * }.

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

 * $$\displaystyle \Rightarrow f_{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1}) $$

 * }.
 * }.

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

 * $$\displaystyle \Rightarrow \dfrac {1}{4}(f_i+ 4f_{i+1/2}+ f_{i+1})= -\dfrac {3}{2h}(z_i-z_{i+1})= \dfrac {3}{2h}(z_{i+1}-z_i) $$

 * }.
 * }.

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

 * $$\displaystyle \Rightarrow z_{i+1}-z_i= \dfrac {h}{6} (f_i+ 4f_{i+1/2}+ f_{i+1}) $$

 * }.
 * }.

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

$$
 * $$\displaystyle \Rightarrow z_{i+1}= z_i+ \underbrace{\dfrac {(h/2)}{3} (f_i+ 4f_{i+1/2}+ f_{i+1})}_{Simpson's Rule}

 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.team4.anandankala 13:57, 7 April 2010 (UTC) .

= Problem 10: Trapezoidal Rule Error Computation =

Given
Kessler's Paper, MatLab code and Result Table.

Patch Kessler Code for Trapezoidal Rule Error 5-9 problem given in Page [[media:Egm6341.s10.mtg30.djvu|30-2]]

 Matlab Code:

Find
Run Kessler's Code line by line to reproduce his table. Complete HW 5-9 Refer Lecture slide [[media:Egm6341.s10.mtg30.djvu|30-2]] Produce (P2,P3) (P4,P5) (P6,P7)and put comments in his code

Solution
Table 1 shows the results from running Kessler's Code

Table 2 shows the results from running Team4 Code

Team4 Trapezoidal rule Error (alternative to to Kessler code)

 Matlab Code:

Discussion:

Under Kessler's code, the program computes P(2k) for n=1 to any n number selected. From the paper we can see that (P3=P5=P7=0 at t=1). So we can use this equality to compute the C coefficients.

Kessler's code results for C coefficients and P terms are shown in Table 1. Team4 algorithm is based on decimal calculation of the C coefficients and therefore all coefficients are accurate to C9. Team4 algorithm allows the user to convert the decimal to fraction after the computation of the coefficients. The loss of accuracy in the fraction calculation from C7 in Team4 code is due to numerical errors of using single precision variables and functions.

Kessler's code in the paper took 0.102152 seconds to compute coefficients for n=1 to 8. The code developed by Team4 is much faster.021921 seconds for n=2 to 9 and does the work accurately for the decimal terms. Fraction terms are accurate up to C6 and P10. Those are shown in table 2 and could be improved if Double Precision is used to find the C coefficients and the factorial terms.

Author
Solved and typed by - --Egm6341.s10.Team4.roni 04:08, 7 April 2010 (UTC)

= Problem 11: Verification and comaprision of Problem 10 of HW5 =

Given
The given condition of No 10 problem of HW5

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

Refer Lecture slide 30-4 for problem statement
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |

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 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

 * style= |
 * }.
 * }.

Find
verify the result of pb10 of HW5(Team3) and compare with the own code.

Solution
first method $$\displaystyle I( \theta) = \int_0^{2\pi} \sqrt{ {r}^{2}+ {\left( \frac{dr}{d \theta}\right)}^{2}} d\theta$$

When run the Team 3's clencult code, they does'nt porvide answer(while loop continuously run.) the reason is that they use the I_exact value as Quad's solution(6.17660193898874),

but this value is not correct. when they calculate the Quad solution, they didn't designate the Toll error level(e.g. $$\displaystyle 10^{-10}$$). However their clencult code itself is correct,so when I change the I_exact value correctly, they provide correct answer.

Team3

Own work

Romberg Table is correct.

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

Team 3 only calculated the quad method.

 Own work

In conclusion, Team 3's work is almost correct except for

1. quad function -- the error order$$\displaystyle (10^{-8}) $$ was bigger than criterion $$\displaystyle 10^{-10} $$ because of not designating error tolerance level in the function

2. seconde method was not calculated except for quad function.

Also, when compare the result bwtween first and second mehtod, the results were same. and in the Romberg table, the second method was faster than first method.(see the upper table)

 Matlab Code:  method 1-Quad

 Matlab Code:  method 1-Composit Trap.Rule

 subfunction ff5_10

 Matlab Code:  method 1 - clencult

 Matlab Code:  method 1 - Romberg Table

 subfuntion ff:

 Matlab Code:  method 2 - Quad

 Matlab Code:  method 2 - Compsit Trap. rule

 subfunction ff

 Matlab Code:  method 2 - Clencurt

 Matlab Code:  method 2 - RombergTable .

Author
Solved and typed by - Egm6341.s10.Team4.yunseok 14:50, 7 April 2010 (UTC)

= Contributing Members = Egm6341.s10.Team4.andy 14:39, 7 April 2010 (UTC) - Author of Problems: 1 ; Proof read: 2,3,4.

Egm6341.s10.Team4.riherd 14:47, 7 April 2010 (UTC) Author of Problems: 6 ; Proof read: 5.

Egm6341.s10.Team4.nimaa&m 14:54, 7 April 2010 (UTC) - Author of Problems: 2, 7, 8 ; Proof read: 5.

Egm6341.s10.Team4.yunseok 14:57, 7 April 2010 (UTC) - Author of Problems: 5, 11 ;

Egm6341.s10.team4.anandankala 13:57, 7 April 2010 (UTC)- Author of problems: 3,9; Proof read: 2,6

Egm6341.s10.Team4.roni 18:35, 7 April 2010 (UTC) - Author of problems 4,10; Proof read: 1,3

title: User:Egm6341.s10.Team4/HW6


title: File:Bvp2.ogv


text:
User talk:Norc

 Hello Norc, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Norc


title: File:Egm6341.s10.Team2.Prob2hw6.jpg


text:
User:EGM6341.s10.Team1.toddmock/HW6

= Problem 3: Determine 3rd and 4th rows of the Matrix =

Given
The matrix is given to be:
 * {| style="width:70%" border="0" align="center"

 * $$\underbrace{\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix}.}_{A} \underbrace{\begin{bmatrix}

C_0 \\ C_1 \\ C_2 \\ C_3 \end{bmatrix}}_{C_i} = \begin{bmatrix}

z_i \\ z'_i \\ z_{i+1} \\ z'_{i+1} \end{bmatrix} $$
 * }.
 * }.

Find
Need to determine the 3rd and 4th row of the matrix using

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \bar{d_3}=z_{i+1}=z(s=1)
 * }.
 * }.

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \bar{d_4}=z'_{i+1}=z'(s=1)
 * }.
 * }.

Solution
We know that

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^{3} C_i s^i= C_0+ C_1s+ C_2s^2+ C_3s^3
 * }.
 * }.

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z'(s)= \sum_{i=1}^{3} iC_i s^{i-1}= C_1+ 2C_2s+ 3C_3s^2
 * }.
 * }.

thus using the above relations where $$s=1$$ in both cases we then have

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}= C_0+ C_1+ C_2+ C_3
 * }.
 * }.

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z'_{i+1}= C_1+ 2C_2+ 3C_3
 * }.
 * }.

Therefore from here it is easy to surmise that the 3rd and 4th rows are $$[1 1 1 1]$$ and $$[0 1 2 3]$$ respectively.

Author --Egm6341.s10.team1.toddmock 19:13, 7 April 2010 (UTC)

= Problem 4: verify the inverse of the Matrix =

Given
From the matrix

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

 * $$A=\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix} $$
 * }.
 * }.

Find
Use Matlab to show that the inverse is

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

 * $$A^{-1}=\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} $$
 * }.
 * }.

Solution
This yields the result

Which is exactly what we were hoping for.

Author --Egm6341.s10.team1.toddmock 19:13, 7 April 2010 (UTC)

= Problem 9: Proof of $$z_{i+1}$$ =

Given
We know that

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dot{z}_{i+1/2}=\dfrac {-3}{2h}(z_i-z_{i+1})-\dfrac {1}{4}(f_i+f_{i+1})
 * }.
 * }.

and we want to use collocation at $$t_{i+1/2}$$ which gives the relation

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \triangle = 0 = \dot{z}_{i+1/2}-f_{i+1/2}
 * }.
 * }.

Find
The mission here is to find a relation for $$z_{i+1}$$

Solution
Therefore using the collocation we have
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f_{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1})
 * }.
 * }.

which can be rearranged to give

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{1}{4}f_i+\dfrac{1}{4}f_{i+1}+f_{i+1/2}=\dfrac{3}{2h}(z_{i+1}-z_i)
 * }.
 * }.

or

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{1}{4}(f_i+4f_{i+1/2}+f_{i+1})=\dfrac{3}{2h}(z_{i+1}-z_i)
 * }.
 * }.

rearranging this gives

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}-z_i=\dfrac{h}{6}(f_i+4f_{i+1/2}+f_{i+1})
 * }.
 * }.

here it should be noted that the right hand side is Simpson's rule, proceeding further we obtain the expression

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}=z_i+\dfrac{h/2}{3}(f_i+4f_{i+1/2}+f_{i+1})
 * }.
 * }.

Author --Egm6341.s10.team1.toddmock 19:14, 7 April 2010 (UTC)

title: User:EGM6341.s10.Team1.toddmock/HW6


title: File:KesslerTable2.png


text:
Portal:Agape World Fellowship/Tasks/Archive

 * Develop Christian Leadership Program
 * Finish Gospel of John
 * Begin Mere Christianity
 * Finish Mere Christianity


text: File:Prob5 hw6 NM.jpg

title: File:Prob5 hw6 NM.jpg


text: File:Problem setup.ogv

title: File:Problem setup.ogv


text:
User:EGM6341.s10.Team1.Kumanchik/HW6

=Problem 1: Compute the arc length of an ellipse on $$[0^{\circ},60^{\circ}]$$=

Statement
An arc is denoted $$\widehat{PQ}$$ and lies on an ellipse defined by,
 * $$r(\theta)=\frac{a(1-e^2)}{1-e \cos{(\theta)}}$$

where $$a=1$$ and $$e=\sin{(\frac{\pi}{12})}$$. The equation for the arc length is,
 * $$PQ=\int_a^bdl=\int_{\theta_1}^{\theta_2}d\theta \sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}$$

1.) Estimate the number of nodes required for the composite trapezoidal estimate of the integral to have an error $$O(10^{-10})$$ 2.)  Using successive numerical integrations, find the actual number of nodes required for the trapezoidal technique by considering the stopping criterion $$\left |I_{2n}-I_n \right |<10^{-10}$$ 3.) How many nodes are required for the Romberg table (verify vs trapezoidal), clencurt.m (verify with a similar stopping criterion), and sum (chebfun - compare to trapezoidal).

Solution
The integrating function is,
 * $$F=\sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}=\frac{1-e^2}{(1-e\cos{\theta})^2}\sqrt{1-2e\cos{\theta}+e^2}$$

1.) An estimate for the error in the composite trapezoidal technique is,
 * $$\left | E_n \right |\leq \frac{(b-a)^3}{12n^2}M_2$$

Rearranging for the number of nodes, $$n$$, gives,
 * $$n=\sqrt{\frac{(b-a)^3}{12(10^{-10})}M_2}$$

where $$M_2=max\left | F^{(2)}(\zeta) \right |, \zeta\in [a,b]$$ and $$[a,b]=[0,60\frac{\pi}{180}]$$. The following MATLAB code solves for $$M_2$$ and determines $$n<16,755$$

2.) Numerically integrating $$F$$ using the trapezoidal rule detemines that at $$n=16,384$$ we get $$\left | I_{16,384}-I_{8,192} \right | < 10^{-10}$$. The following MATLAB code is used.  Notice that in step sizes of 2n, the previous summation can be utilized as shown in the code.

3.) Numerically integrate using the Romberg, clencurt, and sum (chebfun).

Romberg gives $$O(10^{-10})$$ compared to itself at $$n=2^7=128$$, see code.

For the sum (chebfun) the code is shown next. Comparing this to the composite trapezoidal result shows a difference $$O(10^{-11})$$ so is considered accurate enough.

And finally for clencurt.m is shown last. The result is compared to itself and at n=10 has an error less than $$O(10^{-10})$$

=Problem 10: Use Kessler's code to generate his table=

Statement
Use Kessler's code to generate his table and then follow the code line-by-line to get $$(P_2,P_3)(P_4,P_5)(P_6,P_6)$$. Comment on the code.

Solution
Kessler's table was generated with the code below for n=8:

The code computes the Trapezoidal error polynomials evaluated at t=1. The polynomial equation is,
 * $$p_{2k}(t)=c_1\frac{t^{2k}}{(2k)!}+c_3\frac{t^{2k-2}}{(2k-2)!}+c_5\frac{t^{2k-4}}{(2k-4)!}+...+c_{2k-3}\frac{t^4}{4!}+c_{2k-1}\frac{t^2}{2!}+c_{2k+1}\frac{t^0}{0!}$$

When evaluated at t=1, the equation is,
 * $$p_{2k}(1)=\frac{c_1}{(2k)!}+\frac{c_3}{(2k-2)!}+\frac{c_5}{(2k-4)!}+...+\frac{c_{2k-3}}{4!}+\frac{c_{2k-1}}{2!}+\frac{c_{2k+1}}{0!}$$

The coefficients, $$c$$, are computed by solving the below equation using $$c_1=-1$$ as a starting point,
 * $$\frac{c_1}{(2k-1)!}+\frac{c_3}{(2k-3)!}+\frac{c_5}{(2k-5)!}+...+\frac{c_{2k-5}}{5!}+\frac{c_{2k-3}}{3!}+\frac{c_{2k-1}}{1!}=0$$

For illustration, the first few coefficients and polynomial equations are shown below.

Notice that the factorials in the denominators of the polynomials differ by 1 from those in the coefficients. Also, to solve for the next coefficient, the preceding coefficients only need to be summed and then moved to the other side of the equal sign, i.e $$c_7=-(\frac{c_1}{7!}+\frac{c_3}{5!}+\frac{c_5}{3!})$$.

The following code will be commented line-by-line to explain the algorithm of Kessler:

Finally, the code will be followed to generate $$(P_2,P_3)(P_4,P_5)(P_6,P_6)$$: The polynomials are written as,
 * $$p_2(t)=c_1\frac{t^2}{2!}+c_3$$
 * $$p_3(t)=c_1\frac{t^3}{3!}+c_3t$$
 * $$p_4(t)=c_1\frac{t^4}{4!}+c_3\frac{t^2}{2!}+c_5$$
 * $$p_5(t)=c_1\frac{t^5}{5!}+c_3\frac{t^3}{3!}+c_5t$$
 * $$p_6(t)=c_1\frac{t^6}{6!}+c_3\frac{t^4}{4!}+c_5\frac{t^2}{2!}+c_7$$
 * $$p_7(t)=c_1\frac{t^7}{7!}+c_3\frac{t^5}{5!}+c_5\frac{t^3}{3!}+c_7t$$

Following the code to get $$c_{1-7}$$, n=3 because 3 loops will generate all 7 coefficients f=1, g=2, cn=-1, cd=1 entering loop which runs 3 times f=1*2*3 (3! = 6) enter fracsum with (1,6) and will sum the fractions w/o leaving a decimal div = gcd(1,6) = 1 n = 1 d = 6 dsum = 1 enter loop (runs once) dsum = 6 leave loop nsum = 6*(1/6)= 1 div = gcd (1,6) = 1 nsum = 1 dsum = 6 leave fracsum with (1,6) cn = [-1; 1] cd = [1; 6] f = [6;1] g = [4;2] enter fracsum with ([-3;1],[6;6]) div = [3;1] n=[-1;1] d=[2;6] enter loop dsum = 2 then 6 leave loop nsum=6*-1/2+6*1/6=-3+1=-2; div=2 nsum=-1; dsum=3; leave fracsum with (-1,3) pn(1,1)=-1; pd(1,1)=3; move on to next loop k=2

At this point it is apparent that fracsum will sum up fractions without reducing them to decimals Also, the second fracsum call does not affect cn so we will skip calculating it And, cn=[-1;1] cd=[1;6] so we have c1 and c3

now k=2 f=[6;1].*[4;2].*[5;3] = [24;2].*[5;3] = [120;6] enter fracsum with (-[-1;1],[1;6].*[120;6])=([1;-1],[120;36]) this is like 1/120-1/36 = -7/360 leave fracsum cn = [-1;1;-7]; cd = [1;6;360]; f = [120;6;1] g = [2+4;[4;2]] = [6;4;2] skip to k=3 f=[120;6;1].*[6;4;2].*[7;5;3] = [720;24;2].*[7;5;3] = [5040;120;6] enter fracsum with ([1;-1;7],[1;6;360].*[5040;120;6]) = ([1;-1;7],[5040;720;2160]) this is 1/5040-1/720+6/2160 = 31/15120 leave fracsum cn = [-1;1;-7;31]; cd = [1;6;360;15120]

The result is,
 * $$c_1=-1 \!$$
 * $$c_3=-\frac{1}{6}$$
 * $$c_5=-\frac{-7}{360}$$
 * $$c_7=-\frac{31}{15120}$$

Plugging into the expressions for $$p_1(t) -> p_7(t)$$ generates the correct result.

title: User:EGM6341.s10.Team1.Kumanchik/HW6


text:
Digital Libraries/Image Retrieval

 * Older versions of the draft developed by UNC/VT Project Team (2009-12-09 PDF WORD)

Module name
Image Retrieval

Scope
The module covers basic explanation of Image Retrieval, various techniques used, and its working in existing systems.

Learning objectives:
At the end of the module, students will be able to:

 * a. Explain the basics of Image Retrieval and its needs in digital libraries.
 * b. Explain the different Image Retrieval types.
 * c. Classify Content Based Image Retrieval based on various features and differentiate between the techniques being used.
 * d. Explain the various Image Retrieval systems in existence and evaluation strategies.

5S characteristics of the module:

 * a. Streams: Text and visual queries which are inputs and images that serve as the output of an Image Retrieval system. These are sequences of characters or bits.
 * b. Structures: Images are stored in dynamic organizational structures such as R Trees and R* Trees aiding in easy retrieval.
 * c. Spaces: Feature spaces for images that are indexed. The screen of an interface, images in 2D spaces.
 * d. Society: End users of the system such as librarians, journalists, and curator of image collections and software agents who process the information.
 * e. Scenarios: Situations where users submit images, create collections, search and retrieve images, invoke relevance feedback and submit queries by sample images.

Level of effort required:

 * a. Prior to class: 2 hours of reading
 * b. In class: 3 hours
 * i. 2 hours for learning the concepts of image retrieval
 * ii. 1 hour for class discussions and activities

Relationships with other modules:

 * 6-a: Info needs, relevance explains how information needs and the state of people’s minds evolve when looking for information, browsing and retrieving information. It is related to 7-a (1) where the user searches for images based on their information needs.
 * 6-b: Online info seeking and search strategy covers the theories, models, and practices related to online information seeking behaviors in different settings of digital libraries. This is related to 7-a (1) where it details the search strategies for images.

Prerequisite knowledge required:

 * (completion optional)
 * Basic knowledge of internet, images and computer systems is required.

Introductory remedial instruction:

 * a. None

I. Image Retrieval and Its Needs

 * The below architecture explains the architecture of an image retrieval system. The user uses the query interface to submit the query which is processed and browses the image collection to extract the visual features or the texts. This is based on the type of the image retrieval system being used.

 * a. Need for image retrieval systems
 * The need for a desired image is by groups such as artists, designers, teachers, historians, advertising agencies, photographers, engineers and journalists. The requirement for images and corresponding use varies considerably among the groups mentioned above. Before the advent of digitized images, librarians and archivists were providing access to images through text descriptors and classification codes manually.
 * b. Challenges faced by Image Retrieval technologies
 * The basic features that users look for in images include color, shape or texture [2]. With the huge amount of information present over the internet, it needs to be organized efficiently for effective browsing search and retrieval. The knowledge on how the systems interact with the visual information is required to further develop an understanding of Image Retrieval systems.

II. Basic Image Retrieval Types

 * The image retrieval techniques focus on two aspects of image research. They are text based and content based image retrieval techniques. These topics are discussed in detail in the following sub-sections.

 * a. Text based Image Retrieval
 * i. How Text based Image Retrieval works
 * Text based image retrieval has been employed from the 1970s. Collections of images are annotated. The annotation of images employs keywords describing the image, caption of an image, text surrounding/embedded in it, complete text of the containing page, and its filename. These systems use text based Database Management systems to retrieve them. These solutions assume a high degree of relevance of the text with the image.

 * ii. Textual descriptors
 * The annotations are based on semantic reasoning that is done in two ways, 1) made by human manually with the aid of tools and 2) made by automatically relating the semantics of the terms with relevant descriptions. Another approach followed in adding text to images is to tag images based on its metadata. Metadata (data about data) contains attributes of the image such as image creation date, image creator, format and simple descriptions of the digital object.

 * iii. Images in databases
 * After annotation, the various categories and the images are indexed for easy retrieval. For example, Getty’s Art and Architecture Thesaurus maintained over 150,000 terms under topics such as history, architecture and cultural objects and images. Similar collections were used as databases to retrieve images. Running a search query on these databases fetches images that are related to query terms.

 * iv. Problems faced in Text-based Image Retrieval
 * Manually adding annotations and textual attributes is difficult and the amount of man-hours it consumes is huge. Also, the probability of disagreement between indexers is high and this adds to the existing problems in text based image retrieval. For example, an image could be interpreted in many ways and that could lead to conflicts in indexing. Though there have been examples of automatic assignment of attributes, lack of captions or accompanying text accounts as a disadvantage for the text based system. Studies show that user’s needs for images is predominantly related to visual features and hence the need arises for an image retrieval system based on visual features such as shape, color and texture arises.

 * The following sub-sections describe the various content based techniques in existence.

 * b. Content-based Image Retrieval
 * The queries for CBIR systems include image attributes or visual examples such as a sketch, texture palette, etc. Following are some of the common image needs and patterns of use such as searches for a particular image, pictures that are related to a story or an article, searches for images that illustrate a document, similarity searches, etc. The primitive features in images that are of importance to image retrieval are color, shape and texture [2]. Some of the other features include spatial and face recognition feature extractions [10].
 * i. Color feature based retrieval
 * Image retrieval systems extract the color feature and similarity measures from images.
 * o Color Histogram
 * Color Histogram is a commonly used color feature representation. It basically calculates the intensity levels of the three color channels (Red, Green and Blue) and denotes their joint probability. Histograms can be represented as one 3-D histogram or three separate 1-D histograms. These color representations holds irrespective of the rotation and translation variants of the image. Using the mathematical representations of these histograms, the similarity of the queried image and the stored images in the database is calculated based on two methods.

 * o Histogram Intersection
 * This method calculates the normalized color histograms of each image in the database with the normalized color histograms of the queried image. The similarity value that ranges between 0 and 1 determines the extent of the relationship between the images.

 * o Distance Method
 * Calculation of the Euclidean distance between the feature vectors calculated from the histograms can also be used as a similarity measure. The lower the value, the higher the similarity between the images.

 * o Color Moments and Color Sets
 * The lower order moments such as first moment (mean) and the second and third moments (variance and skewness) are extracted from the images and checked for similarity. Additionally, binary feature vectors such as Color sets which are created by the transforming the color space into uniformed space such as HSV (Hue-Saturation Value) and then quantizing the color space into bins called Color sets. These color sets are binary vectors and binary tree search can be used for faster execution.

 * ii. Shape based image retrieval
 * Shape is an additional feature that should be extracted if the colors of the images are the same throughout. Shape features are based on two methods, Boundary based and Region based. Boundary based methods uses the outer boundary characteristics of the image/objects while the region-based involve the entire region. Shape features may also be local or global. A shape feature is local if it is derived from some proper subpart of an object, while it is global if it is derived from the entire object.

 * o Histogram method
 * Identification of features such as boundaries, lines, circularities and edge detections are primary areas concerning boundary based methods. Similar to color based techniques, these methods use histograms depicting the edge directions. A histogram intersection technique matches the parts of the images and is used to calculate the similarity values.

 * c. Texture based image retrieval
 * Texture based techniques are classified into three categories; probabilistic, spectral and structural. In probabilistic, texture patterns are treated as certain random fields; features are extracted from these fields. Spectral methods involve decomposing of images into channels and analyzing them to extract relevant features. Structural techniques model texture features based on heuristic rules of primitive image elements that mimics human perception of textural patterns. Visual textual features such as coarseness, contrast and roughness are used in quantifying the textual patterns in an image.

 * i. Image segmentation
 * Images containing textures usually contain multiple objects. Hence, images are segmented into regions of similarity using image segmentation algorithms or tools. Each region represents a set of pixels coherent in their local texture properties and its corresponding texture descriptor is defined. These descriptors are used to determine the similarity between the regions. Image segmentation is also used in Color based schemes.

III. Indexing in Image Retrieval

 * Indexing techniques are useful when images are stored in large databases and size of the images stored grows exponentially. Images are represented as feature vectors in image retrieval systems.

 * a. Multi-dimensional indexing
 * Multi-dimensional feature vectors are stored in trees that contain special properties. They are represented in structures that contain aspects such as:

 * i. Leaves at the same level contain entries of the form (reference, feature vector) where reference is the pointer to the image in the database.
 * ii. Presence of intermediate nodes including area representation.
 * iii. Every node of the tree has a specification of the capacity it can hold.
 * o R-Trees are dynamic organizational tree structures that conform to the above aspects such that each node has a variable number of entries and each non-leaf node contains a child node and a bounding box.
 * o R* Trees are a variant of R-Trees where they contain better storage efficiency by means of margin minimization and better spatial distribution of the tree entries. QBIC is an example Image Retrieval system using this structure.
 * o SS-Tree is a dynamic organization where the area field of the node entry is indicated with a centroid and radius, thus accounting for less space with more area information.
 * o M-Tree or metric tree is a data structure that suits indexing of generic metric spaces.
 * o Feature vectors also use Incremental Clustering Techniques for dynamic information retrieval. They have a dynamic structure which is capable of handling high dimensional data and deals with Non-Euclidean similarity measures.
 * Tools such as Self Organization Map (SOM) are used to construct the tree indexing structure that aids in efficient retrieval. SOM supports dynamic clustering and similarity measures.

IV. Image Retrieval Systems in Existence

 * Among the different systems, some commercial and research prototypes such as QBIC [6], VisualSEEK [3, 9], Virage [3], MARS [15] that encompass the above said features are discussed below.

 * a. QBIC (Query by Image Content)
 * QBIC is a content based image retrieval system by IBM [6]. The queries used in the system include example sketches, images, drawings and color patterns with textures. It uses RGB color features. The texture retrieval features take into consideration, characteristics such as coarseness, contrast and directionality. Image segmentation is employed to identify interesting objects in the images and retrieve them. QBIC incorporates high dimensional feature indexing and R* Trees is the dynamic organization data structure being used. Another feature is that it integrates both text based keyword search and content based similarity search in its interface.

 * b. VisualSeek
 * VisualSEEK is a content based image query system integrating feature based image indexing and spatial query methods [3]. It employs color sets which is useful in automated region extraction through color set-back projection. Another advantage of using color sets is it is easily indexed and retrieval costs are low. VisualSEEK improves image retrieval efficiency by automating the extraction of localized regions and its features and by supporting feature extraction of compressed data. VisualSEEK tools can be ported to an application aiding in the search of images that is specific to the application [9]. The retrieval process of the images is made fast by use of binary tree based indexing algorithms such as R Trees and M Trees.

 * c. VIRAGE
 * Virage developed by Virage Inc. is a content based image search engine supporting visual queries [3]. It uses features like color, composition, layout, boundary information and texture. The difference between Virage and other systems lie in its ability to provide weights and to lets users adjust the weights based on their preference. It supports both general features such as color, shape, and textures and domain specific features like face recognition and cell detection.

 * d. MARS
 * MARS stands for Multimedia Analysis and Retrieval System. It focuses on organizing the various visual features of the image rather than concentrating one particular visual feature such as color, shape, or texture by seamlessly integrating DBMS and ranked retrieval technologies. It follows an integrated Relevance Feedback Architecture [15].
 * The system achieves relevance feedback at feature representation and similarity measure levels. The user queries are refined based on the relevance feedback provided by the user.
 * MARS provides options to retrieve images based on a specific feature. For example, the user can retrieve images that follow a color space (HSV, YIQ, and LUV) or a particular color feature (color moments, color histogram), texture features or image scales.

V. Trends and Improvements

 * a. Linguistic indexing
 * Associating pictures with textual descriptions can be further developed using automatic image annotation or linguistic indexing. Images that lack reliable metadata may require linguistic indexing. Content based Image retrieval focuses more on the features and fails when user needs an image related to a particular content. Hence, automatic annotations facilitates content based image search using text.
 * b. Stories, Aesthetics in pictures
 * Certain concepts described in an article or stories are mapped with relevant pictures. An average image retrieval system fails images are not stored in proper indexing schemes. Hence, the representative pictures needs to be indexed based on a ranking scheme that discriminates the relevant pictures for a given text.
 * In aesthetics, image retrieval system developers need to consider the most desirable image to a query. This aspect of an image is calculated by measuring its quality. Quality is perceived at levels such as size, aspect ratio and color depth.

 * c. Human Interaction
 * Interaction and feedback in image retrieval has caused major changes to the way the image retrieval systems have been developed. Techniques such as user interactive segmentation and dynamic feature vector re-computation are based on relevance feedback provided by the users.
 * d. Web Oriented Engines
 * With the advent of the World Wide Web, the ability to store images has increased manifold. The current image search engines are predominantly text based and subject browsing based. The major concerns regarding web oriented content based engines are due to indexing features and the concern of interoperability.

Resources

 * Reading list for Students:

 * Yong Rui, Thomas S. Huang, Shih-fu Chang (1997), Image Retrieval: Past, Present, And Future, Journal of Visual Communication and Image Representation, Page 12-17, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.40.5149

 * Abby. A. Goodrum (2000), Image Information Retrieval: An Overview of Current Research, Informing Science - Special Issue on Information Science Research, Vol. 3 No 2, Page 1-5, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.26.8589

 * Yong Rui, Thomas S. Huang, Shih-fu Chang (1999), Image Retrieval: Current Techniques, Promising Directions and Open Issues, Journal of Visual Communication and Image Representation, Pages 39-62, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.32.7819

 * Reading list for Instructors:
 * Ritendra Datta, Dhiraj Joshi, Jia Li, and James Z. Wang (2006), Image Retrieval: Ideas, Influences, and Trends of the New Age, ACM Computing Survey, Volume 39, Pages 1-60, http://citeseerx.ist.psu.edu/viewdoc/versions?doi=10.1.1.91.7203

 * Anil K. Jain, Aditya Vailaya (1996), Image Retrieval using Color and Shape, Patter Recognition, Volume 29, Issue 8, Pages 1233-1244, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.47.9699

 * M. Flickner, H.Sawhney, W.Niblack, J. Ashley, Q. Huang, D. Steele (1995), Query by Image and Video Content: The QBIC System, Computer, Vol. 28, No. 9, Pages 23-32, http://doi.ieeecomputersociety.org/10.1109/2.410146

 * J.Chen, T. Pappas, B. Rogowitz (2002), Adaptive Image Segmentation Based On Color and Texture (2002), International Conference on Image Processing (ICIP) Volume 2, Pages 789-792, http://citeseerx.ist.psu.edu/viewdoc/versions?doi=10.1.1.14.2102

 * S.B.Chad, C. Carson, H.Greenspan, J.Malik (1998), Color and Texture-Based Image Segmentation Using EM and Its Application to Content-Based Image Retrieval, Proceedings of the Sixth International Conference on Computer Vision, Pages 675-682, http://portal.acm.org/citation.cfm?id=939161

 * VisualSEEK Image retrieval system, http://www.ee.columbia.edu/ln/dvmm/researchProjects/MultimediaIndexing/VisualSEEk/VisualSEEk.htm

 * Grosky, W.I. (n.d.). Image retrieval - existing techniques, content-based (cbir) systems. Retrieved from http://encyclopedia.jrank.org/articles/pages/6763/Image-Retrieval.html

 * Zhuge, H. (n.d.). Semantic-based web image retrieval. Retrieved from http://www2003.org/cdrom/papers/poster/p172/p172-zhuge/p172-zhuge.htm

 * A Content Based Image Search Engine, http://www.tiltomo.com/

 * Anaktisi – Content based Image Search, http://orpheus.ee.duth.gr/anaktisi/

 * Celentano, A., & Sabbadin, S. (n.d.). Multiple features indexing in image retrieval systems. Retrieved from http://www.dsi.unive.it/~auce/docs/celentano_cbmi99.pdf

 * MARS Image Retrieval System, http://www.ifp.illinois.edu/~qitian/MARS.html

Concept map
(created by students)
 * After studying the material in this module, students will create a concept map, which represents the concepts in the module and their relationships with one another. By transforming the knowledge in their mind into a graphical representation, students will have a 'clearer picture' of the content.

 * Students might create concept maps not only for the content in the body of knowledge section, but also for the learning activities section. For example, students may show the steps to search, browse, add, delete, import or export an item in a concept map.  Or, they can list features of different DL application software and compare them to promote critical thinking. Even a concept map (or multiple concept maps) can be created for the semester-long DL development project, showing several phases of the project such as preparation step, actual installation and configuration of the software, content selection, collection development, etc.

 * Note: IHMC Cmap Tools is an open source client tool to create concept maps. CmapServer enables the users to collaborate and share concept maps anywhere on the internet.  Both software can be downloaded freely for educational purposes from http://cmap.ihmc.us/download/index.php

Exercises / Learning activities

 * In-Class exercise 1 (30 minutes)

 * i. Split the class into groups of 4 members each.

 * ii. Have them use the following web based image retrieval search engines and make them perform the following activities.
 * The web sites are:
 * 1. http://www.tiltomo.com/ “Tiltomo” - A Content Based Image Search Engine
 * 2. http://orpheus.ee.duth.gr/anaktisi/ “Anaktisi – Content based Image Search”

 * iii. Provide textual/visual queries to each of the tools.

 * iv. Identify which content based image retrieval scheme is being used (mentioned in the curriculum)

 * v. analyze the advantages and disadvantages of each scheme and document the results.

 * In-Class exercise 2 (15 minutes)

 * i. Use Google ‘Similar Images’ feature at http://images.google.com and identify how accurate is the similarity between the retrieved results.

 * ii. Provide three examples of queries that return at least one non-similar result for an image.

 * iii. Each of the features, color, shape and texture, should be represented in the examples.

 * iv. Analyze the dissimilarity of the retrieved result and explain what could be the reason for the behavior.

 * In-Class exercise 3 (15 minutes)

 * i. The below link contains demo to a sample R-Tree structure
 * o http://donar.umiacs.umd.edu/quadtree/points/rtrees.html

 * ii. Using the demo, study how insertion and deletion of nodes work.

 * iii. Document the steps and submit.

 * iv. The idea of this learning activity is to help students understand the intricacies of these structures.

Evaluation of learning outcomes

 * a. Do students know the fundamental concepts, and techniques of Image Retrieval?

 * b. Do students effectively distinguish the features of various content based image retrieval techniques?

Glossary

 * Collections A group of items, often documents.
 * Feature Information extracted from an object and used during query processing.
 * Index A data structure built for the images to speed up searching.
 * Information Retrieval Part of computer science which studies retrieval of information (not data) from a collection of written documents.
 * Metadata Attributes of a data or a document.
 * Query The expression of the user information need in the input language provided by the information system.
 * Repository A physical or digital place where objects are stored for a period of time, from which individual objects can be obtained if they are requested.
 * Relevance feedback An interactive process of obtaining information from the user about the relevance and the non-relevance of retrieved documents.
 * Feature vector An n-dimensional vector of numerical features that represent some object.
 * Feature space The vector space associated with feature vectors.
 * Color Histogram It is the representation of distribution of colors in an image.
 * Euclidean Distance It is the ordinary distance between two points in a vector space.

Contributors

 * a. Authors
 * Nagarajan Kuppuswami
 * Edward A. Fox
 * b. Reviewers
 * Seungwon Yang
 * Tarek Kanan
 * Venkatasubramaniam Ganesan
 * John Ewers

title: Digital Libraries/Image Retrieval


text: File:Setup w(x,t) v(x).ogv

title: File:Setup w(x,t) v(x).ogv


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Mere Christianity (CBL)/Week 1

Assignments

 * 1) Read pg. 3-12
 * 2) Write a well developed response that expands upon a theme or concept from this week's reading.
 * 3) Answer weekly question set
 * 4) Study for the Book I Quiz next week

Discussion
This week's discussion will be held at TIME on DATE in the AWF virtual universe, Agape World at the AWF School facility.There is a section on the talk page where you can post topics and questions for us to go over. If you are unable to make it, your questions can also be posted in a separate section to the talk page, where I will do my best to answer them shortly.

Final Essay
Each week there will be a section similar to this one covering the Final Essay. You will be able to submit your essay from the beginning of Week 7 to the end of Week 10. Essays submitted in Weeks 7 and 8 will be returned during Week 9 and you will be able to revise and resubmit for a higher grade.

Handouts

 * Literary Response Essay - This will be your final for this course

Questions
After reading pages 3-12 and writing a response, please answer the following 10 questions in complete sentences. Each question will be worth one point, for a total of ten points possible.
 * 1) According to Lewis, is morality individual or societal? How does he justify his position?
 * 2) What does Lewis call the Law of Human Nature? Why?
 * 3) Do different civilizations and ages have different moralities? How do we know this is the case?
 * 4) What form of hypocrisy, or double standard can be seen in the mindset of moral relativists?
 * 5) What is the foundation of all clear thinking about ourselves and the universe we live in?
 * 6) Is the Moral Law a herd instinct? Why or why not?
 * 7) What does Lewis say about impulses?
 * 8) What tune does the Moral Law make? How?
 * 9) What does the book differentiate from the Law of Human Nature on page 10? Why is this important and how does it relate to Question 3?
 * 10) What's so important about the Moral Law? Why do we need it?

Quiz
The quiz on this unit will open next week. It will consist of 25 free response questions similar to the weekly questions. Any material from Book I is fair game and you should begin studying now to maximize your score. Each question is worth 1 point and will be graded based upon completeness and comprehension.

title: Mere Christianity (CBL)/Week 1