text:
TESOL/Talking about your job

Practice
{Do you - your job? + enjoy - find - kind - job
 * type=""}

{Where do you - ? - enjoy - job + work - like
 * type=""}

{What do you - ? - work + do - job - office
 * type=""}

{No, - really. + not - no - kind of - a
 * type=""}

{It's - .? - enjoy - like + interesting - interested
 * type=""}

title: TESOL/Talking about your job


text:
A Translation of the Bible/Genesis

Genesis

Chapter 1
The Week of Creation

1 In the beginning of God's creation of the heaven and the earth, 2 when the earth was unformed and void, and darkness was upon the face of the deep; and a mighty wind blew over the face of the waters, 3 then God said: 'Let there be light.' And there was light. 4 And God saw the light, that it was good; and God divided the light from the darkness. 5 And God called the light Day, and the darkness He called Night. And there was evening and there was morning, one day.

6 And God said: 'Let there be an expanse in the middle of the waters, and let it split up the waters.' 7 And God made the expanse, and split the waters which were under the expanse from the waters which were above the expanse; and it was so. 8 And God called the expanse Heaven. And there was evening and there was morning, a second day.

9 And God said: 'Let the waters under the heaven be gathered together into one place, and let the dry land appear.' And it was so. 10 And God called the dry land Earth, and He called the gathering together of the waters seas. And God saw that it was good. 11 And God said: 'Let the Earth sprout grass, plants yielding seeds, and fruit-trees bearing fruit after their kinds, in which is their seeds, upon the Earth.' And it was so. 12 And the earth sprouted grass, plants yielding seeds after their kinds, and trees bearing fruit, in which is their seeds, after their kinds; and God saw that it was good. 13 And there was evening and there was morning, a third day.

14 And God said: 'Let there be lights in the expanse of the heaven to distinguish the day from the night; and let them be for signs, and for festivals, and for days and years; 15 and let them be for lights in the expanse of the heaven to give light upon the earth.' And it was so. 16 And God made the two great lights: the greater light to rule the day, and the lesser light to rule the night; and He made the stars. 17 And God set them in the expanse of the heaven to give light upon the earth, 18 and to rule over the day and over the night, and to distinguish the light from the darkness; and God saw that it was good. 19 And there was evening and there was morning, a fourth day.

20 And God said: 'Let the waters swarm with swarms of living creatures, and let birds fly above the earth in the open expanse of heaven.' 21 And God created the great sea-monsters, and every living creature that creeps, with which the waters swarmed, after its kind, and every winged bird after its kind; and God saw that it was good. 22 And God blessed them, saying: 'Be fruitful, and multiply, and fill the waters in the seas, and let birds multiply on the earth.' 23 And there was evening and there was morning, a fifth day.

24 And God said: 'Let the Earth bring forth living creatures after their kinds, cattle, and creeping thing, and beasts of the earth after their kinds.' And it was so. 25 And God made the beasts of the earth after their kinds, and the cattle after their kind, and everything that creeps on the ground after its kind; and God saw that it was good. 26 And God said: 'Let us make man in our image, after our likeness; and let them have dominion over the fish of the sea, and over the birds of the air, and over the cattle, and over all the Earth, and over every creeping thing that creeps on the earth.' 27 And God created man in His own image - in the image of God He created him; he created them, male and female. 28 And God blessed them; and God said to them: 'Be fruitful, and multiply, and replenish the earth, and subdue it; and have dominion over the fish of the sea, and over the birds of the air, and over every living thing that creeps on the earth.' 29 And God said: 'Behold, I have given you every plant yielding seed, which is on the face of all the Earth, and every tree, in which is the fruit of a tree yielding seed; to you it shall be for food. 30 And to every beast of the earth, and to every bird of the air, and to everything that creeps on the earth, in which there is a living soul, every green plant shall be for food.' And it was so. 31 And God saw everything that He had made, and, behold, it was very good. And there was evening and there was morning, the sixth day.

Chapter 2
1 And the Heaven and the Earth were finished, and all that they contained. 2 And on the seventh day God had finished His work which He had made; and He rested on the seventh day from all His work which He had made. 3 And God blessed the seventh day, and made it holy, because on it He rested from all His work which God in creating had made.

The Garden of Eden

4 This is the story of the Heaven and of the Earth when they were created, in the day that the LORD God made Earth and Heaven. 5 No shrub of the field was yet on the Earth, and no plant of the field had yet sprung up; for the LORD God had not caused it to rain upon the Earth, and there was no man to till the ground. 6 But a mist went up from the Earth, and watered the whole face of the ground. 7Then the LORD God formed man out of the dust of the ground, and breathed the breath of life into his nostrils; and man became a living soul. 8 And the LORD God planted a garden in the East, in Eden; and there He put the man whom He had formed. 9 And the LORD God made every tree that is pleasant to the sight, and good for food, grow out of the ground; the tree of life also in the middle of the garden, and the tree of the knowledge of good and evil. 10 And a river went out of Eden to water the garden; and from there it was split, and became four headwaters. 11 The name of the first is Pishon; that is the one which encircles the whole land of Havilah, where there is gold; 12 and the gold of that land is good; there is aromatic gum and the onyx stone. 13 And the name of the second river is Gihon; that is the one which encircles the whole land of Cush. 14 And the name of the third river is Tigris; that is the one which goes toward the east of Assyria. And the fourth river is the Euphrates. 15 And the LORD God took the man, and put him in the garden of Eden to tend it and to keep it. 16 And the LORD God commanded the man, saying: 'You may freely eat of every tree in the garden; 17 but you shall not eat any of the tree of the knowledge of good and evil; for on the day that you eat any of it you shall surely die.

18 And the LORD God said: 'It is not good that the man should be alone; I will make a helper suitable for him.' 19 And the LORD God formed every beast of the field and every bird of the air out of the ground, and brought them unto the man to see what he would call them; and whatsoever the man called every living creature, that was to be its name. 20 And the man gave names to all cattle, and to the birds of the air, and to every beast of the field; but for Adam there was not found a helper suitable for him. 21 And the LORD God caused a deep sleep to fall upon the man, and he slept; and He took one of his ribs, and closed up the place with flesh instead. 22 And the rib, which the LORD God had taken from the man, He made into a woman, and brought her to the man. 23 And the man said: 'This is at last bone of my bones, and flesh of my flesh; she shall be called Woman, because she was taken out of Man.' 24 Therefore a man shall leave his father and his mother, and shall cling to his wife, and they shall be like one person. 25 And they were both nude, the man and his wife, and were not embarrassed.

Chapter 3
Expulsion from the Garden of Eden

1 And the serpent was cunning, more than any beast of the field which the LORD God had made. And he said to the woman: "But has God said: You shall not eat of any tree of the garden?" 2 And the woman said to the serpent: "We may eat the fruit of the trees of the garden; 3 but of the fruit of the tree which is in the middle of the garden, God has said: You shall not eat it, nor shall you touch it, in case you die." 4 And the serpent said to the woman: "No, you shall not die; 5 for God knows that in the day you eat of it, then your eyes shall be opened, and you shall be like God, knowing good and evil." 6 And when the woman saw that the tree was good for food, and that it was a delight for the eyes, and that the tree was to be desired to make one wise, she took some of its fruit, and ate it; and she also gave it to her husband with her, and he ate it. 7 And the eyes of both of them were opened, and they knew that they were naked; and they sewed fig-leaves together, and made themselves aprons. 8 And they heard the voice of the LORD God walking in the garden toward the evening ; and the man and his wife hid themselves from the presence of the LORD God among the trees of the garden. 9 And the LORD God called to the man, and said to him: "Where are you?" 10And he said: "I heard your voice in the garden, and I was afraid, because I was naked; and I hid myself." 11 And He said: "Who told you that you were naked? Have you eaten of the tree, of which I commanded you that you should not eat?" 12 And the man said: "The woman whom You gave to be with me, she gave me of the tree, and I ate it." 13 And the LORD God said to the woman: "What is this that you have done?" And the woman said: "The serpent tricked me, and I ate." 14 And the LORD God said to the serpent: "Because you have done this, you are cursed from among all cattle, and from among all beasts of the field; you shall go on your belly, and you shall eat dust all the days of your life. 15 And I will put hatred between you and the woman, and between your descendants and her descendants; they will bruise your head, and you shall bruise their heel." 16 He said to the woman: 'I will greatly increase you pain and your labour; you will give birth to children in pain; and your desire shall be towards your husband, and he shall rule over you." 17 And He said to Adam: "Because you have heeded the voice of your wife, and have eaten of the tree, of which I commanded you, saying: You shall not eat of it, the ground is cursed for your sake; you shall eat of it in toil all the days of your life. 18 It will bring forth thorns and thistles for you, and you shall eat the plants of the field. 19 You shall eat bread with the sweat of your face, until you return to the ground; for out of it you were taken; for you are dust, and you will return to dust." 20 And the man gave his wife the name Eve, because she was the mother of all living. 21 And the LORD God made garments of skin for Adam and for his wife, and clothed them.

22 And the LORD God said: "Behold, the man has become like one of us, to know good and evil; and now, in case he reaches out his hand, and takes also from the tree of life, and eats, and lives for ever." 23 Therefore the LORD God expelled him from the garden of Eden, to till the ground from whence he was taken. 24 So He drove out the man; and He placed the cherubim at the east of the garden of Eden, and the flaming sword which turned every way, to defend the way to the tree of life.

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

title: A Translation of the Bible/Genesis


text:
User:EML5526.S11.Team3.risher/homework2/solutions

Problem 2.1
Energy balance for a control volume can be written as:

heat generated + heat flow in - heat flow out = 0

$$ s(x+\Delta x/2)\Delta x + q(x)A(x) - q(x+\Delta x)(A(x+\Delta x) $$

Balance of heat: heat flow into a control volume (H1) + heat generated (H2) = heat due to change in temperature (H3)

$$H1+H2=H3$$

Heat due to change in temperature: $$H3 = A(x) \rho(x) c\frac{\partial u}{\partial t}$$

Rearranging H1 and H2 we get:

$$ \frac{ q(x+\Delta x)(A(x+\Delta x)-q(x)A(x)}{\Delta x} + s(x+\Delta x/2) = A(x)\rho(x) c\frac{\partial u}{\partial t}$$

If we take the limit of as $$\Delta x$$ → 0, then we get:

$$ \frac{\partial (qA)}{\partial x}+s =A(x)\rho(x) c\frac{\partial u}{\partial t}$$

Using Fourier's Law,

$$ q = k \frac{\partial u}{\partial x} $$

This becomes:

$$ \frac{\partial}{\partial x}\left[ A(x)k(x) \frac{\partial u}{\partial x}\right]+f(x,t)=-A(x)\rho(x) c\frac{\partial u}{\partial t} $$

Problem 2.2
(1)

$$ det\left[ b_{jk} \right]= det\left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 &6 \\ \end{matrix}\right]= -(-1*1*3)-(1*3*2)-(1*2*6)+(-1*1*6)+(1*3*3)+(1*2*2) = -8 $$

$$ det\left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 &6 \\ \end{matrix}\right]=-8 $$

(2)

$$ \underline{b}_{j} = b_{jk}\underline{a}_{k} $$

$$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K} = K_{ij}=\underline{b}_{i}.\underline{b}_{j}$$

$$ \underline{b}_{i}.\underline{b}_{j}=\underline{b}_{i}.\underline{b}_{i}^T = \left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & 6 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]$$

$$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K} = \left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]$$

$$det \underline{ \Gamma }= 64 $$

(3)

$$\underline{F}=\underline{b}_{i}*\underline{v}$$

$$\underline{F}= \left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & 6 \\ \end{matrix}\right]$$ $$ \left[\begin{matrix} 5 \\ -7  \\ -4  \\ \end{matrix}\right]$$

$$\underline{F}= \left[\begin{matrix} -6 \\ 5 \\ -23 \\ \end{matrix}\right]$$

(4)

$$ \underline{K} \underline{d} = \underline{F} $$

$$ \underline{d} = \underline{K}^{-1} \underline{F} $$

$$\underline{d}= \left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]^{-1}$$ $$\left[\begin{matrix} -6 \\ 5 \\ -23 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

$$\underline{d}=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

(5)

Using $$\underline{w}_{i}=\underline{a}_{i}$$ where i = 1,2,...,n and $$\underline{a}_{i}$$ is orthonormal basis:

$$\underline{a}_{i}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$

Looking at

$$ \bar{\underline{K}} \underline{d} = \bar{\underline{F}} $$

We can express this as:

$$ \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]\underline{d}$$ $$=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

Which is simplified to:

$$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]\underline{d}$$ $$=\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

Where:

$$\bar{\underline{K}}=\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]$$

$$\bar{\underline{F}}= \left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

(5)

$$ \bar{\underline{K}} \underline{d} = \bar{\underline{F}} $$

$$ \underline{d} = \bar{\underline{K}}^{-1} \bar{\underline{F}} $$

$$\underline{d}=\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]^{-1}$$ $$\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

$$\underline{d}=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

Both $$\underline{d}$$ for each case are equal, which makes since.

(7)

$$\underline{K}$$ is always symmetric while $$\bar{\underline{K}}$$ is not. Due to this symmetry, the Bubnov-Galerkin method may be easier to solve, while the effort for finding $$\underline{K}$$ may be a little more labor intensive. The lack of symmetry in $$\bar{\underline{K}}$$ for the Petrov-Galerkin method may be difficult to solve, but it is very easy to find $$\bar{\underline{K}}$$.

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text:
User:Egm6341.s11.team4.hylon/hw1

=HW 1.2 problem statement=

Find $${{p}_{n}}(x)$$ and $${{R}_{n+1}}(x)$$ of

1).$${{e}^{x}}$$

2).$${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$

at point $${{x}_{0}}=0$$.

=Solution:=

According to the Taylor series from the lecture [[media:nm1.s11.mtg3.djvu|Mtg 3 (d)]], the Taylor polynomials is as following:

$${{p}_{n}}(x)=f({{x}_{0}})+\frac{(x-{{x}_{0}})}{1!}{{f}^{(1)}}({{x}_{0}})+\frac{2!}{{f}^{(2)}}({{x}_{0}})....+\frac{n!}{{f}^{(n)}}({{x}_{0}})$$

$${{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_^{x}{{{(x-t)}^{n}}{{f}^{(n+1)}}(t)}dt=\frac{(n+1)!}{{f}^{(n+1)}}(\xi )\begin{matrix} , & \xi \in \left[ {{x}_{0}},x \right] \\ \end{matrix}$$

1). For $$f(x)={{e}^{x}}$$, take the i-th order derivative:

$$ {{f}^{(1)}}(x)={{f}^{(2)}}(x)=...={{f}^{(n)}}(x)={{e}^{x}}$$.

At point $${{x}_{0}}=0$$,

$$ {{f}^{(1)}}(0)={{f}^{(2)}}(0)=...={{f}^{(n)}}(0)=1$$.

Thus, the n-th order polynomial and residual is:
 * {| style="width:70%" border="0" align="center"

$$\displaystyle {{p}_{n}}(x)=1+\frac{x}{1!}+\frac{2!}+...+\frac{n!} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

$$\displaystyle {{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_{0}^{x}dt=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

2). For $${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$, based on the result of 1): $${{f}_{1}}(x)={{p}_{n}}(x)+{{R}_{n+1}}(x)$$ , we obtain:

$$\begin{align} & {{f}_{2}}^ – (x)=\frac{{{f}_{1}}^ – (x)-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & =\frac{{{p}_{n}}(x)+{{R}_{n+1}}(x)-1}{x} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1+\frac{x}{1!}+\frac{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }}-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }} \\ \end{align}$$

Thus, At point $${{x}_{0}}=0$$ ,the n-th order polynomial and residual for $${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$ is:

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

$$\displaystyle {{p}^{'}}_{n}(x)=\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

$$\displaystyle {{R}^{'}}_{n+1}(x)=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

title: User:Egm6341.s11.team4.hylon/hw1


title: File:Fe1.s11.team6.HW1.fig2.jpg


text:
User:Bhawani Gautam Rhk

Please see my userpage on the Nepli Wikipedia and respond at my Nepali talk page. Thank you.

title: User:Bhawani Gautam Rhk


text:
User:Eml5526.s11.team3.hylon/Homework 1

= Homework 1 = Formulation of Partial Differential Equation (PDE) for a 1- Dimensional Problem.

Refer to lecture notes 6.1

--

Problem Statement
1.1 To find the 1-Dimensional Partial Differential Equation(PDE) of an elastic bar of length L and varying cross-section, subjected to a body force or distributed force f(x,t) along the axis of the bar.

1.2 Now if the bar has a rectangular cross-section where the breadth is constant, perform the balance of forces to formulate the 1-D PDE for this case.

Solution 1.1
Consider an element of length $$d\left(x\right)$$ of the elastic bar, the free body diagram of the bar is shown as below.

In the above figure $$A\left(x\right)$$, $$ \sigma\left (x\right)$$ and  $$n\left(x\right)$$ represent the area, stress and normal to the plane on the left cross-section of the bar and

$$A\left(x+dx\right)$$, $$\sigma\left(x+dx\right)$$ and   $$n\left(x+dx\right)$$ represent the area, stress and normal to the plane on the right cross-section of the bar.

For the left cross-section $$n\left(x\right)=-1$$ and for the right cross-section $$n\left(x+dx\right)=1$$

For the element of the elastic bar shown, we have $$m\left(x\right)=\rho(x) A(x)dx$$

By using Newton's 2nd law of motion, we can have

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

|$$\displaystyle N(x+dx)-N(x)+f(x)dx= \rho(x) A(x)dx\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 1)

 * }
 * }

where $$N\left(x+dx\right)$$ is the net force due to the stress $$\sigma\left(x+dx\right)$$ which is given by

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

|$$\displaystyle N\left(x+dx\right)=\sigma(x+dx)A(x+dx)n(x+dx)$$ $$
 * $$\displaystyle (Eq. 2)

 * }
 * }

$$N\left(x\right)$$ is the net force due to the stress $$\sigma\left(x\right)$$ which is given by $$N\left(x\right)=\sigma(x)A(x)n(x)$$

Substituting for $$N\left(x+dx\right)$$,$$N\left(x\right)$$,$$n\left(x+dx\right)$$,$$n\left(x\right)$$, we get

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

|$$\displaystyle \sigma (x+dx)A(x+dx)-\sigma (x)A(x)+f(x,t)dx=\rho (x)A(x)dx\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 3)

 * }
 * }

Dividing the above equation by $$d\left(x\right)$$, we get

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

|$$\displaystyle \frac{\sigma (x+dx)A(x+dx)-\sigma (x)A(x)}{dx}+f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 4)

 * }
 * }

Applying the limit as $${dx}{\to0}$$ in the above equation,by the definition of partial derivative, as we know that

$$ \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}= \frac{\mathrm{d} f(x)}{\mathrm{d} x} $$

the first term on the LHS reduces to $$ \frac{\partial\left[\sigma (x)A(x)\right]}{\partial x} $$, the second term on the LHS remains the same as there is no $$d\left(x\right)$$ term. the RHS also remains the same.

Equation (4) reduces to

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

\frac{\partial\left[\sigma (x)A(x)\right]}{\partial x} +f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)

 * }
 * }

The stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain as $$\sigma (x)=E(x)\frac{\partial u}{\partial x}$$

Hence the PDE for 1 Dimensional elastic bar is

$$ \frac{\partial\left[E(x)A(x)\frac{\partial u}{\partial x}\right]}{\partial x}+f(x,t)= \rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$ $$ $$
 *  $$ \displaystyle
 *  $$ \displaystyle (Eq. 6)

Solution 1.2
The Partial Differential Equation (PDE) of an elastic bar subjected to an axial load derived in the section (1.1) is

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

|$$\displaystyle \frac{\partial}{\partial x}[E(x) A(x))\frac{\partial u}{\partial x}] + f(x,t) = \rho(x) A(x) \frac{\partial^2 u}{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 1)

 * }
 * }

Now, a particular case is taken into consideration, where the bar has a rectangular cross-section. The breadth of the bar is a constant and is equal to 'b'. But the height of the bar varies along the length of the bar and hence it is function of x, which is 'h(x).' (For figure, refer page 6-1 of meeting 6).

Let us consider the inertial force at the center of the thin strip of the elastic bar. Therefore the mass at that location is given by,

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

|$$\displaystyle m(x + \frac{dx}{2}) = \rho(x + \frac{dx}{2}) b h(x + \frac{dx}{2}) $$ $$
 * $$\displaystyle (Eq. 2)

 * }
 * }

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

|$$\displaystyle m(x + \frac{dx}{2}) = \rho(x + \frac{dx}{2}) b \frac{1}{2} [h(x) + h(x+dx)] $$ $$
 * $$\displaystyle (Eq. 3)

 * }
 * }

Performing the balance of forces similar to part 1.1 and rearranging the terms we get

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

|$$\displaystyle \frac{\sigma (x+dx)bh(x+dx)-\sigma (x)bh(x)}{dx}+f(x,t)=\rho(x + \frac{dx}{2}) b \frac{1}{2} [h(x) + h(x+dx)]\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 4)

 * }
 * }

Applying the limit as $${\frac{dx}{2}\to0}$$ in the above equation,by the definition of partial derivative

$$\displaystyle \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}= \frac{\mathrm{d} f(x)}{\mathrm{d} x} $$

we get,

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

|$$\displaystyle \frac{\partial \left[\sigma (x)bh(x+0)\right]}{\partial x}+f(x,t)=\rho(x + 0) b \frac{1}{2} [h(x) + h(x+0)]\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 5)

 * }
 * }

By Hooke's Law, the stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain $$\frac{\partial u}{\partial x}$$ as

$$\displaystyle \sigma (x)=E(x)\frac{\partial u}{\partial x} $$

Hence the PDE for 1 Dimensional elastic bar is

$$ \frac{\partial \left[E(x)bh(x)\frac{\partial u}{\partial x}\right]}{\partial x}+f(x,t)= \rho (x)bh(x)\frac{\partial^2u }{\partial t^2}$$ $$
 *  $$ \displaystyle (Eq. 6)

$$\frac{\partial\left[E(x)h(x)\frac{\partial u}{\partial x}\right]}{\partial x}+\frac{1}{b}f(x,t)= \rho (x)h(x)\frac{\partial^2u }{\partial t^2}$$ $$
 *  $$ \displaystyle (Eq. 7)

Reference
1. Newton's 2nd law of motion 2. Partial derivative 3. Hooke's Law

title: User:Eml5526.s11.team3.hylon/Homework 1


text:
User:Eml5526.s11.team3/Homework 2

 Homework 2

=--Problem 2.1--=

Problem Statment
From notes 6-2: Derive heat problem in 1-D, i.e, (1) p.4-2

Where $$m = \rho A$$

Solution
Energy balance for a control volume can be written as: heat generated + heat flow in - heat flow out = 0

Balance of heat: heat flow into a control volume (H1) + heat generated (H2) = heat due to change in temperature (H3)

Heat due to change in temperature: $$H3 = A(x) \rho(x) c\frac{\partial u}{\partial t}$$

Rearranging H1 and H2 we get:

If we take the limit as $$\Delta x$$ → 0, then we get:

Using Fourier's Law,

and also knowing that $$s$$ is a function of $$x$$ and $$t$$:

This becomes:

Problem solved by EML5526.S11.Team3.risher and Eml5526.s11.team3.perry.

Reference
Used wolframalph to do matrix math.

Solution
From balance of heat, we have following:

Where $${{H}_{1}}$$ represent the flow in heat minus the flow out heat, $${{H}_{2}}$$ stands for the heat generated inside the volume, $${{H}_{3}}$$ is the heat due to the change of temperature.

Using Taylor Series to expand above terms, we can get

Hence, we have

Neglect the higher order term, we hence have

By Fourier's Law, We have

Finally, we get the PDE of heat problem in $$ 1-D$$ case.

Problem solved by Hailong Chen.

Reference
1.Taylor Series 2.Fourier's Law

=--Problem 2.2--=

Problem Statement
From notes 7-3 and 7-4:

1.) Find $$det \left[ b_{jk} \right]$$

2.) Find $$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K}, det \underline{ \Gamma }$$

3.) $$ \underline{F} = \left[ F_{i} \right] = \left[ \underline{b}_{i} * \underline{v} \right]   $$

4.) Solve (5) on p. 7-2 for $$ \underline{d} = \left[ v_{j} \right] $$

(5) on p. 7-2 : $$ \underline{K} \underline{d} = \underline{F} $$

5.) Use $$ \underline{w}_{i} * \underline{P} (\underline{v}) = 0 $$, $$ \forall_{i} = 1, ..., n $$ to find $$ \bar { \underline{K}} \underline{d} = \bar{ \underline{F}} $$

What is $$ \bar{ \underline{K} } $$ and $$ \bar{ \underline{F} } $$ ? Note: $$ \underline{d} = \left[ v_{j} \right] $$

6.) Solve for $$ \underline{d} $$; compare to $$ \underline{d} $$ in part 4.)

7.) Observe symetric properties of $$ \underline{K} $$ and $$ \bar{ \underline{K} } $$

Discuss pros and cons of the 2 methods

Note: (5) p. 7-2 → Bubnov-Galerkin

$$ \underline{K} \underline{d} = \underline{F} $$

Note: (2) p. 7-4 → Petrov-Galerkin

$$ \bar { \underline{K}} \underline{d} = \bar{ \underline{F}} $$

Solution
(1)

(2)

$$ \underline{b}_{j} = b_{jk}\underline{a}_{k} $$

$$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K} = K_{ij}=\underline{b}_{i}.\underline{b}_{j}$$

$$ \underline{b}_{i}.\underline{b}_{j}=\underline{b}_{i}.\underline{b}_{i}^T = \left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & 6 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]$$

$$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K} = \left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]$$

(3)

$$\underline{F}=\underline{b}_{i}*\underline{v}$$

$$\underline{F}= \left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & 6 \\ \end{matrix}\right]$$ $$ \left[\begin{matrix} 5 \\ -7  \\ -4  \\ \end{matrix}\right]$$

(4)

$$ \underline{K} \underline{d} = \underline{F} $$

$$ \underline{d} = \underline{K}^{-1} \underline{F} $$

$$\underline{d}= \left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]^{-1}$$ $$\left[\begin{matrix} -6 \\ 5 \\ -23 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

(5)

Using $$\underline{w}_{i}=\underline{a}_{i}$$ where i = 1,2,...,n and $$\underline{a}_{i}$$ is orthonormal basis:

$$\underline{a}_{i}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$

Looking at

$$ \bar{\underline{K}} \underline{d} = \bar{\underline{F}} $$

We can express this as:

$$ \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]\underline{d}$$ $$=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

Which is simplified to:

$$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]\underline{d}$$ $$=\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

Where:

(6)

$$ \bar{\underline{K}} \underline{d} = \bar{\underline{F}} $$

$$ \underline{d} = \bar{\underline{K}}^{-1} \bar{\underline{F}} $$

$$\underline{d}=\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]^{-1}$$ $$\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

Both $$\underline{d}$$ for each case are equal, which makes since.

(7)

$$\underline{K}$$ is always symmetric while $$\bar{\underline{K}}$$ is not. Due to this symmetry, the Bubnov-Galerkin method may be easier to solve, while the effort for finding $$\underline{K}$$ may be a little more labor intensive. The lack of symmetry in $$\bar{\underline{K}}$$ for the Petrov-Galerkin method may be difficult to solve, but it is very easy to find $$\bar{\underline{K}}$$.

Problem solved by EML5526.S11.Team3.risher and Eml5526.s11.team3.perry.

Reference
Used wolframalph to do matrix math.

=--Problem 2.3--=

Problem Statement
To show that $$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ is equivalent to:

$$ \underline{a}_{i} * \underline{P} (\underline{v}) = 0 $$ $$ \forall_{i} = 1, ..., n $$

where {$$\underline{a_{i}}$$} is an orthonormal basis vector.

and $$ \underline{w} = \Sigma_{i} \beta_{i} \underline{a}_{i} $$ and $$ \forall \left[ \beta_{1}, ..., \beta_{n} \right] \in \mathbb{R}^n$$

Solution
It is given that $$\underline{\omega} = \sum_{i=1}^{n} \beta _{i}\underline{a_{i}}$$, where {$$ \underline{a_{i}} $$} is an orthonormal basis vector.

Then (1) p. 8-1 is equivalent to:
 * {| style="width:100%" border="0" align="left"

 * $$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ $$ \forall $$ $$ \underline{w} = \Sigma_{i} \beta_{i} \underline{a}_{i} $$ $$\in \mathbb{R}^n$$
 * $$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ $$ \forall $$ $$ \underline{w} = \Sigma_{i} \beta_{i} \underline{a}_{i} $$ $$\in \mathbb{R}^n$$

 *

$$\displaystyle (Eq. 1) $$

 * }
 * }

From Eq.(1) substituting for $$\underline{\omega}$$, we get

$$\sum_{i=1}^{n} \beta _{i}\underline{a_{i}}$$* $$ \underline{P} (\underline{v})$$ = 0

Expanding we can write,
 * {| style="width:100%" border="0" align="left"

 * ($$ \beta _{1}\underline{a_1}+\beta _{2}\underline{a_2}+\beta _{3}\underline{a_3}+....\beta _{n}\underline{a_n}$$)*$$ \underline{P} (\underline{v})$$ = 0
 * ($$ \beta _{1}\underline{a_1}+\beta _{2}\underline{a_2}+\beta _{3}\underline{a_3}+....\beta _{n}\underline{a_n}$$)*$$ \underline{P} (\underline{v})$$ = 0

 *

$$\displaystyle (Eq. 2) $$

 * }
 * }

Choice 1

Let $$\beta_1 = 1$$ and  $$\beta _2 = \beta _3=...\beta _n=0$$

Therefore, equation (1) can be written as-

$$\underline{a_1}.\underline{P}(\underline{v})=0$$

Choice 2

Let $$\beta_2 = 1$$ and  $$\beta _1 = \beta _3=...\beta _n=0$$

Therefore, equation (1) can be written as-

$$\underline{a_2}.\underline{P}(\underline{v})=0$$

.

.

Choice n

Let $$\beta_n = 1$$ and  $$\beta _1 = \beta _2=...\beta _{n-1}=0$$

Therefore, equation (1) can be written as-

$$\underline{a_n}.\underline{P}(\underline{v})=0$$

Hence we can state that :

$$ \underline{a}_{i} * \underline{P} (\underline{v}) = 0 $$  for $$ \forall_{i} = 1, ..., n $$

Problem solved by ushnish12.

=--Problem 2.4--=

Problem Statement
To solve for a G1DM1.0/D1 given in (3) in P 9-2

Given

An elastic bar of unit length is applied with a distributed load of $$f(x)=5x$$  along the axis of the bar, and linear variation of product of elastic modulus and area of the bar is given as  $$a_{2}(x)=2+3x$$

Boundary conditions

1) Essential boundary condition $$\gamma _{g}={1}, g=4$$ , which implies  $$ u(1)=4$$

2) Natural boundary condition $$\gamma _{h}={1}, h=6$$ , which implies  $$-\frac{\partial u(x=0)}{\partial x} =6 $$

Hence the above problem reduces to finding exact solution to the differential equation given by,

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \forall x \in ] 0,1 [ $$

to solve the above differential equation we first solve the following integrals

1) $$\int lnxdx$$ 2) $$\int xlnxdx$$ 3) $$\int \frac{x^2}{1+cx}$$ 4) $$\int \frac{x^2}{a+bx}$$ 5) Find the exact solution of

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \forall x \in ] 0,1 [ $$

$$ u(1)=4, -\frac{\partial u(x=0)}{\partial x} =6 $$

6) Plot u(x) 7) Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

1
$$\int lnxdx$$ To the solve the above integral we use Integration by parts method: {| style="width:100%" border="0" \int{u}dv =uv-\int{v}du $$ {| style="width:100%" border="0" where,
 * style="width:100%"
 * style="width:100%"
 * style="width:100%" |
 * style="width:100%" |

2
$$\int xlnxdx$$ we again use Integration by parts

here,

3
$$\int \frac{x^2}{1+cx}$$ here,

hence,

$$\int \frac{x^2}{1+cx}= \frac{x^2}{c}ln(1+cx)-\int \frac{2}{c}ln(1+cx)xdx$$

for, $$\int ln(1+cx)xdx$$

hence,

$$= \frac{x^2}{c}ln(1+cx)-\frac{2}{c}[[\frac{x}{c}(1+cx)ln(1+cx)-\frac{x}{c}(1+cx)]-\int \frac{((1+cx)ln(1+cx)-(1+cx))}{c}dx]$$

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{2}{c^2}\int (1+cx)ln(1+cx)dx-\frac{2}{c^2}\int (1+cx)dx$$

for, $$\int (1+cx)ln(1+cx)dx$$

hence,

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{2}{c^2}[\frac{(1+cx)^2ln(1+cx)}{2c}-\int \frac{(1+cx)^2cdx}{2c(1+cx)}]-\frac{2}{c^2}\int (1+cx)dx$$

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{(1+cx)^2}{c^3}ln(1+cx)-\frac{1}{c^2}\int (1+cx)dx-\frac{2}{c^2}\int (1+cx)dx$$

$$=ln(1+cx)[\frac{x^2}{c}-\frac{2x}{c^2}(1+cx)+\frac{(1+cx)^2}{c^3}]+\frac{2x}{c^2}(1+cx)-\frac{3}{c^2}\int (1+cx)dx$$

$$=ln(1+cx)[\frac{x^2}{c}-\frac{2x}{c^2}-\frac{2x^2c}{c^2}+\frac{1}{c^3}+\frac{c^2x^2}{c^3}+\frac{2cx}{c^3}]+\frac{2x}{c^2}(1+cx)-\frac{3}{c^2}[\frac{(1+cx)^2}{2c}]$$

$$=ln(1+cx)[\frac{1}{c^3}]+\frac{2x}{c^2}+\frac{2x^2}{c}-\frac{3}{2c^3}[1+c^2x^2+2cx]$$

$$=ln(1+cx)[\frac{1}{c^3}]+\frac{2x}{c^2}+\frac{2x^2}{c}-\frac{3}{2c^3}-\frac{3x^2}{2c}-\frac{3x}{c^2}$$

$$=\frac{1}{c^3}ln(1+cx)+\frac{x^2}{2c}-\frac{x}{c^2}-\frac{3}{2c^3}$$

$$=\frac{2ln(1+cx)+x^2c^2-2cx}{2c^3}+constant$$

4
$$\int \frac{x^2}{a+bx}$$ here,

hence,

$$\int \frac{x^2}{a+bx}= \frac{x^2}{b}ln(a+bx)-\int \frac{2}{b}ln(a+bx)xdx$$

for, $$\int ln(a+bx)xdx$$

hence,

$$= \frac{x^2}{b}ln(a+bx)-\frac{2}{b}[[\frac{x}{b}(a+bx)ln(a+bx)-\frac{x}{b}(a+bx)]-\int \frac{((a+bx)ln(a+bx)-(a+bx))}{b}dx]$$

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{2}{b^2}\int (a+bx)ln(a+bx)dx-\frac{2}{b^2}\int (a+bx)dx$$

for, $$\int (a+bx)ln(a+bx)dx$$

hence,

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{2}{b^2}[\frac{(a+bx)^2ln(a+bx)}{2b}-\int \frac{(a+bx)^2bdx}{2b(a+bx)}]-\frac{2}{b^2}\int (a+bx)dx$$

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{(a+bx)^2}{b^3}ln(a+bx)-\frac{1}{b^2}\int (a+bx)dx-\frac{2}{b^2}\int (a+bx)dx$$

$$=ln(a+bx)[\frac{x^2}{b}-\frac{2x}{b^2}(a+bx)+\frac{(a+bx)^2}{b^3}]+\frac{2x}{b^2}(a+bx)-\frac{3}{b^2}\int (a+bx)dx$$

$$=ln(a+bx)[\frac{x^2}{b}-\frac{2ax}{b}-\frac{2x^2b}{b^2}+\frac{a^2}{b^3}+\frac{b^2x^2}{b^3}+\frac{2abx}{b^3}]+\frac{2x}{b^2}(a+bx)-\frac{3}{b^2}[\frac{(a+bx)^2}{2b}]$$

$$=ln(a+bx)[\frac{a^2}{b^3}]+\frac{2ax}{b^2}+\frac{2x^2}{b}-\frac{3}{2b^3}[a^2+b^2x^2+2abx]$$

$$=ln(a+bx)[\frac{a^2}{b^3}]+\frac{2ax}{b^2}+\frac{2x^2}{b}-\frac{3a^2}{2b^3}-\frac{3x^2}{2b}-\frac{3ax}{b^2}$$

$$=\frac{a^2}{b^3}ln(a+bx)+\frac{x^2}{2b}-\frac{ax}{b^2}-\frac{3a^2}{2b^3}$$

$$=\frac{2a^2ln(a+bx)+x^2b^2-2bax}{2b^3}+constant$$

5
Find the exact solution of

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}=-5x $$

Integrating the above expression we get,

$$ (2+3x)\frac{\partial u}{\partial x}=\frac{-5x^2}{2}+k1$$

$$\frac{\partial u}{\partial x}=\frac{-5x^2}{2(2+3x)}+\frac{k1}{2+3x}$$

given,

$$-\frac{\partial u(x=0)}{\partial x} =6$$

$$-6= 0+ \frac{k1}{2}$$

hence $$k1=-12$$

$$\frac{\partial u}{\partial x}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x}$$

$$\partial u=\frac{-5x^2}{2(2+3x)}\partial x+\frac{k1}{2+3x}\partial x$$

Integrating the above equation

$$u=\frac{-5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x}$$

from (Eq.2) $$\int \frac{x^2}{a+bx}dx=\frac{2a^2ln(a+bx)+bx(bx-2a)}{2b^3}$$

hence $$\int \frac{x^2}{2+3x}dx=\frac{8ln(2+3x)+3x(3x-4)}{54}$$

$$u=\frac{-5}{2}[\frac{8ln(2+3x)+3x(3x-4)}{54}]-12[\frac{ln(2+3x)}{3}]+k2$$

$$u=\frac{-118}{27}ln(2+3x)-\frac{5x}{36}(3x-4)+k2$$

given $$ u(1)=4$$

$$4=\frac{-118}{27}ln(5)-\frac{5}{36}(-1)+k2$$

$$k2=10.89495$$

hence the exact solution is

6
Plot of u(x)

7
Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

The above integral is a special case if (Eq.2) where  $$a=b=1$$

hence,

$$\int \frac{x^2}{1+x}= \frac{2*1^2ln(1+x)+x(x-2)}{2*1^3}+constant$$

Problem Solved by akj.

=--Problem 2.5--=

Problem Statement
To show that

$$ \underline{b}_{i} * \underline{P} (\underline{v}) = 0 $$ where i=1, ..., n

is equivalent to

$$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ where $$ \underline{w} = \Sigma_{i} \alpha_{i} \underline{b}_{i} $$

and $$ \forall \left[ \alpha_{1}, ..., \alpha_{n} \right] \in \mathbb{R}^n$$

Solution
It is given that-
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \underline{b}_{i} * \underline{P} (\underline{v}) = 0 $$  where   i = 1, ..., n $$
 * $$\displaystyle (Eq. 1)

 * }
 * }

Choice 1

Taking i=1, we get

$$ \underline{b}_{1} * \underline{P} (\underline{v}) = 0 $$

Multiplying both sides by $$\alpha _{1}$$ , we get-

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

|$$\displaystyle \alpha _{1}\underline{b}_{1} * \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 2)

 * }
 * }

Choice 2

Taking i=2, we get

$$ \underline{b}_{2} * \underline{P} (\underline{v}) = 0 $$

Multiplying both sides by $$\alpha _{2}$$ , we get-

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

|$$\displaystyle \alpha _{2}\underline{b}_{2} * \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 3)

 * }
 * }

......

......

......

Choice n

Taking i=n, we get

$$ \underline{b}_{n} * \underline{P} (\underline{v}) = 0 $$

Multiplying both sides by $$\alpha _{n}$$ , we get-

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

|$$\displaystyle \alpha _{n}\underline{b}_{n} * \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 4)

 * }
 * }

Adding Equations 2 to 4, we get

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

|$$\displaystyle(\alpha _{1}\underline{b}_{1} + \alpha _{2}\underline{b}_{2} +\alpha _{n}\underline{b}_{n}$$) $$ *   \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 5)

 * }
 * }

or we can write Eq(5) as -

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

|$$\displaystyle \sum_{i=1}^{n} \alpha _{i}\underline{b_{i}}$$* $$ \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 6)

 * }
 * }

Now by definition,

$$\underline{\omega} = \sum_{i=1}^{n} \alpha _{i}\underline{b_{i}}$$ where $$ \forall \left[ \alpha_{1}, ..., \alpha_{n} \right] \in \mathbb{R}^n$$ and {$$\underline{b_{i}}$$} is an orthonormal basis vector.

So Eq.(6) can be written as -

$$ \underline{w} * \underline{P} (\underline{v}) = 0 $$

Problem solved by ushnish12.

=--Problem 2.6--=

Problem Statement
Given

Consider family of functions $$\displaystyle F=\left \{ 1,cosiwx,siniwx \right \} $$ on interval $$\displaystyle \Omega \left [ 0,T \right ]$$ for i =1,2

So

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

\displaystyle F=\left \{ b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x) \right \} $$     (Eq 6.1)
 * style="width:95%"
 * style="width:95%"
 *
 * }

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

\displaystyle F=\left \{ 1,coswx,cos2wx,sinwx,sin2wx \right \} $$     (Eq 6.2)
 * style="width:95%"
 * style="width:95%"
 *
 * }

Find
6.1 Construct $$\displaystyle \underline{\Gamma }(F)$$ and observe properties of $$\displaystyle \underline{\Gamma }(F)$$

6.2 Find the determinant of  $$\displaystyle \underline{\Gamma }(F)$$

6.3 Conclude F is orthogonal basis.

6.1

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

\displaystyle \underline{\Gamma }={{[{{\Gamma }_{ij}}]}_{5*5}}=\left( \begin{matrix}  {{\Gamma }_{11}} & \ldots  & {{\Gamma }_{15}}  \\   \vdots  & \ddots  & \vdots   \\   {{\Gamma }_{51}} & \cdots  & {{\Gamma }_{55}}  \\ \end{matrix} \right) $$     (Eq 6.3)
 * style="width:95%"
 * style="width:95%"
 *
 * }

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

\displaystyle \Gamma _{ij}==\int_{\Omega }b_{i}(x)b_{j}(x)dx $$     (Eq 6.4)
 * style="width:95%"
 * style="width:95%"
 *
 * }

We need to integrate on interval $$ \Omega \left [ 0,T \right ] $$ and $$T = \frac{w} $$ to find each elements of Gram matrix

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

\displaystyle \Gamma _{11}=\int_{0}^{T}1dx=\frac{2\Pi }{w} $$     (Eq 6.5)
 * style="width:95%"
 * style="width:95%"
 *
 * }

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

\displaystyle \Gamma _{21}=\Gamma _{12}=\int_{0}^{T}coswxdx=0 $$     (Eq 6.6)
 * style="width:95%"
 * style="width:95%"
 *
 * }
 * {| style="width:100%" border="0"

\displaystyle \Gamma _{31}=\Gamma _{13}=\int_{0}^{T}cos2wxdx=0 $$     (Eq 6.7)
 * style="width:95%"
 * style="width:95%"
 *

 * }

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

\displaystyle \Gamma _{41}=\Gamma _{14}=\int_{0}^{T}sinwxdx=0 $$     (Eq 6.8)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{51}=\Gamma _{15}=\int_{0}^{T}sin2wxdx=0 $$     (Eq 6.9)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{22}=\int_{0}^{T}coswx.coswxdx=\frac{\Pi }{w} $$     (Eq 6.10)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{23}=\Gamma _{32}=\int_{0}^{T}coswx.cos2wxdx=0 $$     (Eq 6.11)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{24}=\Gamma _{42}=\int_{0}^{T}coswx.sinwxdx=0 $$     (Eq 6.12)
 * style="width:95%"
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 *

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

\displaystyle \Gamma _{25}=\Gamma _{52}=\int_{0}^{T}coswx.sin2wxdx=0 $$     (Eq 6.13)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{33}=\int_{0}^{T}cos2wx.cos2wxdx=\frac{\Pi }{w} $$     (Eq 6.14)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{34}=\Gamma _{43}=\int_{0}^{T}cos2wx.sinwxdx=0 $$     (Eq 6.15)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{35}=\Gamma _{53}=\int_{0}^{T}cos2wx.sin2wxdx=0 $$     (Eq 6.16)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{44}=\int_{0}^{T}sinwx.sinwxdx=\frac{\Pi }{w} $$     (Eq 6.17)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{45}=\Gamma _{45}=\int_{0}^{T}sinwx.sin2wxdx=0 $$     (Eq 6.18)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \Gamma _{55}=\int_{0}^{T}sin2wx.sin2wxdx=\frac{\Pi }{w} $$     (Eq 6.19)
 * style="width:95%"
 * style="width:95%"
 *

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

\displaystyle \underline{\Gamma }(F)=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \frac{2\Pi }{w} & 0 \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & \frac{\Pi }{w} \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & \frac{\Pi }{w} & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & \frac{\Pi }{w} & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & 0 & \frac{\Pi }{w}  \\ \end{matrix} \\ \end{matrix} \right]
 * style="width:95%"
 * style="width:95%"

$$     (Eq 6.20)
 *

 * }

It is shown that gram matrix is diagonal matrix.

6.2
Determinant of gram matrix has been calculated by wolframalpha.

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

\displaystyle \det \underline{\Gamma }(F)=\frac{2{{\Pi }^{5}}} $$     (Eq 6.21) According to theorem on lecture note meeting 10-2
 * style="width:95%"
 * style="width:95%"
 *

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

 * style="width:95%" |
 * style="width:95%" |

It is $$\displaystyle \det \underline{\Gamma }(F)\neq 0$$, so the family of functions are linearly independent. Also they are basis functions.
 * }

6.3
From the lecture note Eq.(2) meeting 7-1

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

\displaystyle {{\delta }_{ij}}=\left\{ \begin{matrix} \begin{matrix} 1 & for & i=j \\ \end{matrix} \\ \begin{matrix} 0 & for & i\ne j \\ \end{matrix} \\ \end{matrix} \right.
 * style="width:95%"
 * style="width:95%"

$$     (Eq 6.22) and
 *

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

\displaystyle {{\Gamma }_{ij}}=<{{b}_{i}},{{b}_{j}}>={{\delta }_{ij}} $$     (Eq 6.23)
 * style="width:95%"
 * style="width:95%"
 *

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

For $$\ i = j$$   Kronecker $$\ {\delta } = 1 $$      (diagonal elements of Gram matrix in Eq 6.20)
 * style="width:95%" |
 * style="width:95%" |

For $$ i \ne j$$  Kronecker $$\ {\delta } = 0 $$      (off-diagonal elements of Gram matrix in Eq 6.20)
 * }

As it is seen on above, both elements of Gram matrix are satisfying the orthogonal condition. Therefore, $$\displaystyle F $$ is a family of orthogonal basis functions.

Reference
Used wolframalpha to do integrates.

Problem solved byEml5526.s11.team3.sahin.

=--Problem 2.7--=

Problem Statement
Consider F = {$$1,x,x^{2},x^{3},x^{4}$$} in the interval $$ \Omega=[0,1]$$ (2.7.1) Construct $$\underline{\Gamma }$$(F) (2.7.2) Find the determinant of $$\underline{\Gamma }$$(F) (2.7.3) Is F Orthogonal Family?

Solution
2.7.1 F = {$$1,x,x^2,x^3,x^4$$} = {$$b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x)$$} $$\Gamma = \begin{bmatrix} \Gamma_{11}&\Gamma_{12}&\Gamma_{13} &\Gamma_{14}  &\Gamma_{15} \\ \Gamma_{21}&\Gamma_{22}&\Gamma_{23} &\Gamma_{24}  &\Gamma_{25} \\ \Gamma_{31}&\Gamma_{32}&\Gamma_{33} &\Gamma_{34}  &\Gamma_{35} \\ \Gamma_{41}&\Gamma_{42}&\Gamma_{43} &\Gamma_{44}  &\Gamma_{45} \\ \Gamma_{51}&\Gamma_{52}&\Gamma_{53} &\Gamma_{54}  &\Gamma_{55} \end{bmatrix}$$

$$\;\Gamma_{ij}=\int_{\Omega}b_{i}(x)b_{j}(x)dx$$ $$\Gamma_{11} = <b_{1}(x),b_{1}(x)> = \int_{0}^{1}1 dx = 1$$ $$\Gamma_{12} = <b_{1}(x),b_{2}(x)> = \int_{0}^{1}1.x dx =\frac{1}{2} $$ $$\Gamma_{13} = <b_{1}(x),b_{3}(x)> = \int_{0}^{1}1.x^2 dx =\frac{1}{3} $$ $$\Gamma_{14} = <b_{1}(x),b_{4}(x)> = \int_{0}^{1}1.x^3 dx =\frac{1}{4} $$ $$\Gamma_{15} = <b_{1}(x),b_{5}(x)> = \int_{0}^{1}1.x^4 dx =\frac{1}{5} $$ $$\Gamma_{21} = <b_{2}(x),b_{1}(x)> = \int_{0}^{1}x.1 dx =\frac{1}{2} $$ $$\Gamma_{22} = <b_{2}(x),b_{2}(x)> = \int_{0}^{1}x.x dx =\frac{1}{3} $$ $$\Gamma_{23} = <b_{2}(x),b_{3}(x)> = \int_{0}^{1}x.x^2 dx =\frac{1}{4}$$ $$\Gamma_{24} = <b_{2}(x),b_{4}(x)> = \int_{0}^{1}x.x^3dx =\frac{1}{5} $$ $$\Gamma_{25} = <b_{2}(x),b_{5}(x)> = \int_{0}^{1}x.x^4dx =\frac{1}{6} $$ $$\Gamma_{31} = <b_{3}(x),b_{1}(x)> = \int_{0}^{1}x^2.1 dx =\frac{1}{3}$$ $$\Gamma_{32} = <b_{3}(x),b_{2}(x)> = \int_{0}^{1}x^2.x dx =\frac{1}{4}$$ $$\Gamma_{33} = <b_{3}(x),b_{3}(x)> = \int_{0}^{1}x^2.x^2 dx =\frac{1}{5} $$ $$\Gamma_{34} = <b_{3}(x),b_{4}(x)> = \int_{0}^{1}x^2.x^3 dx =\frac{1}{6} $$ $$\Gamma_{35} = <b_{3}(x),b_{5}(x)> = \int_{0}^{1}x^2.x^4 dx =\frac{1}{7} $$ $$\Gamma_{41} = <b_{4}(x),b_{1}(x)> = \int_{0}^{1}x^3.1 dx =\frac{1}{4} $$ $$\Gamma_{42} = <b_{4}(x),b_{2}(x)> = \int_{0}^{1}x^3.x dx =\frac{1}{5} $$ $$\Gamma_{43} = <b_{4}(x),b_{3}(x)> = \int_{0}^{1}x^3.x^2 dx =\frac{1}{6}$$ $$\Gamma_{44} = <b_{4}(x),b_{4}(x)> = \int_{0}^{1}x^3.x^3 dx =\frac{1}{7} $$ $$\Gamma_{45} = <b_{4}(x),b_{5}(x)> = \int_{0}^{1}x^3.x^4 dx =\frac{1}{8} $$ $$\Gamma_{51} = <b_{5}(x),b_{1}(x)> = \int_{0}^{1}x^4.1 dx =\frac{1}{5} $$ $$\Gamma_{52} = <b_{5}(x),b_{2}(x)> = \int_{0}^{1}x^4.x dx =\frac{1}{6} $$ $$\Gamma_{53} = <b_{5}(x),b_{3}(x)> = \int_{0}^{1}x^4.x^2 dx =\frac{1}{7} $$ $$\Gamma_{54} = <b_{5}(x),b_{4}(x)> = \int_{0}^{1}x^4.x^3 dx =\frac{1}{8} $$ $$\Gamma_{55} = <b_{5}(x),b_{5}(x)> = \int_{0}^{1}x^4.x^4 dx =\frac{1}{9} $$

$$ \underline{\Gamma}(F) = \begin{bmatrix}

1 &\frac{1}{2} & \frac{1}{3}  & \frac{1}{4}  &\frac{1}{5} \\

\frac{1}{2} & \frac{1}{3} & \frac{1}{4}  & \frac{1}{5}  & \frac{1}{6} \\

\frac{1}{3} &\frac{1}{4} &\frac{1}{5}  &\frac{1}{6}  &\frac{1}{7} \\

\frac{1}{4} & \frac{1}{5} &\frac{1}{6} &\frac{1}{7}  &\frac{1}{8} \\

\frac{1}{5} &\frac{1}{6} &\frac{1}{7}  &\frac{1}{8}  &\frac{1}{9} \end{bmatrix}$$

2.7.2 The determinant of $$\underline{\Gamma}(F)\;\;$$ is

$$\left | \underline{\Gamma}(F) \right | = 3.74\times 10^{-12}$$

2.7.3 Consider any two basis functions. Say, $$ b_{i}$$ and $$b_{j}$$. If $$\Gamma_{ij} = <b_{i},b_{j}> = \delta_{ij} $$, then F is a family of orthogonal basis of functions. In other words, if $$\;\;\;\delta_{ij} = 1 \;for\; i=j\;$$  or if $$\;\;\; \delta_{ij} = 0\; for\; i\neq j$$, then F is a family of orthogonal basis of functions.

But in the $$\underline{\Gamma}(F)$$ matrix of Solution (2.7.1) , we find none of the elements satisfying the above condition. Hence, F is not a family of orthogonal basis of functions. Problem solved by vnarayanan.

=--Problem 2.8--=
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\; w^{h}(x)\;\;$$ where $$\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)\;$$ $$
 *
$$\displaystyle (Eq. 1)

 * }
 * }

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

|$$\displaystyle \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 *
$$\displaystyle (Eq. 2)

 * }
 * }

Show that Eq.1 $$\;\Leftrightarrow\; $$ Eq.2

Solution
Part 1

Eq.1 $$\;\Rightarrow\;\;\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\;\; w^{h}(x)$$ where $$\;\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)$$

Case (1) : $$\;\; c_{1} = 1, c_{2}=c_{3}=...=c_{n} = 0$$

$$\therefore\;\; w^{h}(x) = b_{1}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{1}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 3)

 * }
 * }

Case (2) : $$\;\; c_{2}=1,c_{1} = c_{3}=...=c_{n} = 0$$

$$\therefore\;\; w^{h}(x) = b_{2}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{2}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 4)

 * }
 * }

Case (3) : $$\;\; c_{1} =c_{2}= c_{3}=...=c_{n-1} = 0,c_{n}=1,$$

$$\therefore\;\; w^{h}(x) = b_{n}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{n}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 5)

 * }
 * }

Hence, from Eq.(3), Eq.(4) and Eq.(5) it can be concluded that

Part 2

Eq.(2) $$\; \Rightarrow\;\int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n $$

Multiply $$\;\;c_{i}\;\;$$ on both sides of the Eq.(2). We get, $$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$

For, $$\; i = 1\;$$ $$\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx = 0 $$

For, $$\; i = 2\;$$ $$\;\int_{\Omega}c_{2}b_{2}(x)P(u^{h}(x))dx = 0 $$

For, $$\; i = n\;$$ $$\;\int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0 $$

Since, $$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n\;\;$$ individually, the sum of all terms should also be equal to 0 $$\therefore\;\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx + \int_{\Omega}c_{2}b_{2}(x)P(u^{h}(x))dx + .\; .\;. + \int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0$$

$$\Rightarrow\;\;\int_{\Omega}\left [ (c_{1}b_{1}(x)P(u^{h}(x))) + (c_{2}b_{2}(x)P(u^{h}(x)))+ .\; .\;. + (c_{n}b_{n}(x)P(u^{h}(x)))\right ]dx = 0$$

$$\Rightarrow\;\;\int_{\Omega}\left [ c_{1}b_{1}(x) + c_{2}b_{2}(x)+ .\; .\;. + c_{n}b_{n}(x)\right ]P(u^{h}(x))dx = 0$$
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \Rightarrow\;\;\int_{\Omega}\sum_{i=1}^{n}c_{i}b_{i}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 7)

 * }
 * }

We know that, $$\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)\;$$

$$\therefore\; Eq.(7) $$ becomes,

Problem solved by vnarayanan.

=--Problem2.9--=

Problem Statement
Given $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$. $$1).$$ Find two equations that enforce boundary conditions for $${{u}^{h}}(x)=\sum\limits_{j=0}^{n}{{{d}_{j}}{{b}_{j}}(x)}$$. Natural boundary condition: $$-\frac{d{{u}^{h}}(x=0)}{dx}=4$$; essential boundary condition: $${{u}^{h}}(x=1)=0$$ . $$2).$$ Find one more equation to solve for $$\underline{d}=\left\{ {{d}_{j}} \right\} (j=0,1,2)$$by project the residue $$P({{u}^{h}})$$ on a basis function $${{b}_{k}}(x)$$ with $$k=0,1,2 $$, such that the additional equation is linear independent from the above two equations in $$(1)$$. $$3).$$ Display three equations in matrix form $$\underline{K}\underline{d}=\underline{F}$$, observe symmetric property of $$\underline{K}$$. $$4).$$ Solve for $$\underline{d}$$. $$5).$$ Construct $$u_{2}^{h}(x)$$ and plot $$u_{2}^{h}(x)$$ vs. $$u_{2}^ – (x)$$. $$6).$$Compute $$u_{2}^{h}(x=0.5)$$ and Error $${{e}_{2}}(x=0.5)=u(x=0.5)-{{u}^{h}}(x=0.5)$$. $$7).$$ Repeat $$1)-6)$$ for $$n=4,n=6$$. $$8).$$ Plot $${{e}_{2}}(x=0.5)$$ vs. $$n$$ Refer to lecture slide [[media:fe1.s11.mtg12.djvu|12-1]] for more information.

Solution
For $$n=2$$ : $$1).$$ Since $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$, then $${{b}_{0}}(x)=\cos (\phi ),{{b}_{1}}(x)=\cos (x+\phi ),{{b}_{2}}(x)=\cos (2x+\phi )$$  . In order to make sure that $$b_{i}^{'}(x=0)\ne 0$$,we choose $$\phi =\frac{\pi }{4}$$. Hence we have From natural boundary condition: $$-\frac{d{{u}^{h}}(x=0)}{dx}=4$$, we can obtain Hence we have

From essential boundary condition: $${{u}^{h}}(x=1)=0$$, we can have We also have Hence we can get the second equation, that

$$2).$$ Projecting the residue $$P({{u}^{h}})$$ on the $${{b}_{i}}(x)$$,$$i=0,1,2$$, we have For static form, we know that $$P({{u}^{h}})=\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)$$. And it’s already given that $${{a}_{2}}(x)=2, f(x)=3$$. Thus, So we can get one more equation

$$3).$$ Till now, we obtain three equations as follow: Display them in matrix form, yields

Hence,

non-symmetric. $$4).$$ Using Matlab, we can easily get

$$5).$$

Since $$2\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+3=0$$, $$u(x=1)=0$$ and $${{u}^{'}}(x=0)=-4$$, we can obtain

Plot

Matlab Code: $$6).$$ We have already got $$u_{2}^{h}(x)=3.803\cos (\frac{\pi }{4})-0.1495\cos (x+\frac{\pi }{4})+2.9032\cos (2x+\frac{\pi }{4})$$ and $$u(x)=-\frac{3}{4}{{x}^{2}}-4x+\frac{19}{4}$$. Hence

For $$n=4$$ : $$1).$$ We also choose $$\phi =\frac{\pi }{4}$$ here. Since $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$, then Hence we have

From natural boundary condition:$$ -\frac{d{{u}^{h}}(x=0)}{dx}=4$$, we can obtain

Hence, we have

From essential boundary condition: $${{u}^{h}}(x=1)=0$$, we can have We get that Hence we can get the second equation, that

$$2).$$ Projecting the residue $$P({{u}^{h}})$$ on the $${{b}_{i}}(x)$$,$$i=0,1,2,3,4$$, we have $$\int\limits_{0}^{1}{{{b}_{i}}(x)P({{u}^{h}})}dx=0$$. For static form, we know that $$P({{u}^{h}})=\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)$$. And it’s already given that $${{a}_{2}}(x)=2,f(x)=3$$. Thus, $$\int\limits_{0}^{1}{{{b}_{i}}(x)\{\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)}\}dx=\int\limits_{0}^{1}{{{b}_{i}}(x)[2{{d}_{j}}{{b}_{j}}^{(2)}(x)+3}]dx=0$$. We can have three more equations here,

$$3).$$ Till now, we obtain five equations as follow: Display them in matrix form, yields

Hence,

non-symmetric. $$4).$$ Using Matlab, we can easily get

$$5).$$

We have already obtained

Plot

Matlab Code: $$6).$$ We have already got $$u_{4}^{h}(x)=14.1214\cos (\frac{\pi }{4})-22.8534\cos (x+\frac{\pi }{4})+26.5637\cos (2x+\frac{\pi }{4})-12.2778\cos (3x+\frac{\pi }{4})+3.0541\cos (4x+\frac{\pi }{4})$$ and $$u(x)=-\frac{3}{4}{{x}^{2}}-4x+\frac{19}{4}$$. Hence

For $$n=6$$ : $$1).$$ We also choose $$\phi =\frac{\pi }{4}$$ here. Since $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$, then

Hence we have From natural boundary condition: $$-\frac{d{{u}^{h}}(x=0)}{dx}=4$$, we can obtain

Hence, we have

From essential boundary condition: $${{u}^{h}}(x=1)=0$$, we can have

We get that

Hence we can get the second equation, that

$$2).$$ Projecting the residue $$P({{u}^{h}})$$ on the $${{b}_{i}}(x)$$,$$i=0,1,2,3,4,5,6$$, we have For static form, we know that $$P({{u}^{h}})=\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)$$. And it’s already given that $${{a}_{2}}(x)=2,f(x)=3$$. Thus,

We can have five more equations here,

$$3).$$ Till now, we obtain seven equations as follow:

Display them in matrix form, yields

Hence,

is non-symmetric. $$4).$$ Using Matlab, we can easily get

$$5).$$

We have already obtained

Plot

Matlab Code: $$6).$$ We have already got $$u_{6}^{h}(x)=0.6678\cos (\frac{\pi }{4})+5.2048\cos (x+\frac{\pi }{4})+6.9757\cos (2x+\frac{\pi }{4})-10.7753\cos (3x+\frac{\pi }{4})+7.5430\cos (4x+\frac{\pi }{4})-2.2711\cos (5x+\frac{\pi }{4})+0.0017\cos (6x+\frac{\pi }{4}) $$ and $$u(x)=-\frac{3}{4}{{x}^{2}}-4x+\frac{19}{4}$$. Hence

Plot $${{e}_{n}}(x=0.5)$$ vs. $$n$$:

Problem solved by Hailong Chen.

Reference
All matrix operations and integrations are solved using Matlab.

title: User:Eml5526.s11.team3/Homework 2


text:
User:EGM6341.S11.team5.reiss

HW1 HW2 HW3 HW4 HW5 HW6 HW7

title: User:EGM6341.S11.team5.reiss


text:
TESOL/Using hear and listen

The words hear and listen both mean that we perceive something with our ears, but the meanings differ.

Concentration
We use listen when we are concentrating harder on a sound. Hear may happen by accident, but listen is never an accident.

 * I thought I heard someone say my name. I thought I listened to someone say my name.
 * Suddenly I heard a dog bark. Suddenly I listened to a dog bark.
 * If you hear my cell phone ring, let me know. If you listen to my cell phone right, let me know.

Progressive
Hear is used as a progressive less often than most verbs. Instead, "can hear" is used.

 * I can hear the rain fall. I am hearing the rain fall.
 * Can you hear him breathing? Are you hearing him breathing?
 * I could hear man shouting when the car crashed. I was hearing a man shouting when the car crashed.

"Hearing" may also be an infinitive, which is spelled the same way as a progressive. In that case, it's fine.
 * I hate hearing his excuses.
 * Hearing the train come makes me excited.
 * Hearing you say that makes me happy.

Listen and listen to
Listen is used without "to" if there is no object. If there is an object, we have to use listen to.

 * Listen! Listen me!
 * You should listen to your parents. You should listen your parents.

Quiz
{Your voice is too soft, so I can't _____ you. + hear - listen - listen to
 * type=""}

{I can _____ a strange noise in the bathroom. - listen - listen to + hear
 * type=""}

{I _____ carefully to remember. - heard - listened to + listened
 * type=""}

{I love ______ the sound of the bell so I can go home. + hearing - listening to - listening
 * type=""}

{I was _______ my favorite music on my music player when you came in. - listening + listening to - hearing
 * type=""}

{I wish my husband would _____ me. - listen + listen to - hear
 * type=""}

title: TESOL/Using hear and listen


text:
File:PPMCycle.jpg

A diagram looking at the lifecycle of managing projects within IT. The framework represented by the diagram is known as Project Portfolio Management.

title: File:PPMCycle.jpg


text:
User talk:RhondaM

 Hello RhondaM, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 19:47, 26 January 2011 (UTC)

title: User talk:RhondaM


text:
User talk:Eml5526.s11.team5.smith

 Hello Eml5526.s11.team5.smith, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twiter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 19:47, 26 January 2011 (UTC)

title: User talk:Eml5526.s11.team5.smith


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User talk:Eml5526.s11.team5.mcdaniel

 Hello Eml5526.s11.team5.mcdaniel, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 20:15, 26 January 2011 (UTC)

title: User talk:Eml5526.s11.team5.mcdaniel


title: File:Fe1.s11.team1.hw1.fig2..jpg


title: File:Nm1.s11.team5.HW1.fig2.jpg


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Communication and Identities in Institutional Arenas - Part I/Friday 18 March

Week 11 Friday 18 March Sal F 133 09.30-14.30

Theme - Identities & communities of practices

We will discuss the following:

Adjustments to our course learning resource on wikiversity

Membership issues in communities of practices

Emperical fields of study (as communities of practices) & intersectionality

Literature for discussion:

 * Background - literature from 25 January & 14 february
 * Davies, Alan (2003): The Native Speaker: Myth and Reality. 2nd ed, 237 s. Clevedon: Multilingual matters (chapters 5, 6, 7, 8, 9 & 10)
 * Hornberger, Nancy H ed. (2003): Continua of Biliteracy. An ecological framework for educational policy, research and practice in multilingual settings, Clevedon: Multilingual matters (parts 1, 3, 5 & 6)
 * Wenger, Etienne (1998): Communities of Practice. Learning, Meaning and Identity. Learning in Doing Series – Social, cognitive and computational perspectives, Cambridge: Cambridge University Press

Participants are required to present their literature logs on the KOIIA-I learning resource page for each of the studied texts latest by Wednesday 16 March noon. Particpants are subsequently required to study the logs presented by their course colleagues before our third campus meeting on Friday 18 March.


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Communication and Identities in Institutional Arenas - Part I/Monday-Tuesday 16-17 May

Week 20

Monday 16 May Sal F 135 10.00-14.00 & 15.15-17.00

Tuesday 17 May Sal F 136 09.30-14.00

Theme - Analysis of Linguistic-Landscapes & Commentaries of KOIIA-I resources

Guest teacher at this final campus based two-day meeting will be Dr. Francis Hult. In addition to presenting an analytical point of departure for the analysis of multilingual landscapes Dr. Hult will also comment the KOIIA-I resources that have been developed during the spring of 2011 by participants in the course.

Dr. Francis Hult will also present an open seminar on Monday 16 May. The open seminar is part of the KOIIA-I course and course participants are required to attend.


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User:Eml5526.s11.team6.tupsakhare

Useful Links
Team 06 Home Page

My Team Members
Shawn Gravois Yongtao Li Chris Vork Nakul Deshpande William Kurth Shreyas Joglekar

title: User:Eml5526.s11.team6.tupsakhare


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User:EML5526.S11.Team3.risher

Team Page
Team 3 Page

Homework
Homework #2 - Risher

Homework #3 - Risher

Homework #5 - Risher

title: User:EML5526.S11.Team3.risher


title: File:Nm1.s11.mtg12.djvu


title: File:Fe1.s11.mtg12.djvu


text:
Paideia High School/Common Sense

Common Sense by Thomas Paine

Common Sense

title: Paideia High School/Common Sense