text:
Properties of quarks

Electric charge
Quarks have fractional electric charge values—either −$1/3$ or +$2/3$ times the elementary charge, depending on flavor. Up, charm, and top quarks (collectively referred to as up-type quarks) have a charge of +$2/3$, while down, strange, and bottom quarks (down-type quarks) have −$1/3$. Antiquarks have the opposite charge to their corresponding quarks; up-type antiquarks have charges of −$2/3$ and down-type antiquarks have charges of +$1/3$. Since the electric charge of a hadron is the sum of the charges of the constituent quarks, all hadrons have integer charges: the combination of three quarks (baryons), three antiquarks (antibaryons), or a quark and an antiquark (mesons) always results in integer charges. For example the hadron constituents of atomic nuclei, neutrons and protons, have charges of 0 and +1 respectively; the neutron is composed of two down quarks and one up quark, and the proton of two up quarks and one down quark.

Spin
Spin is an intrinsic property of elementary particles, and its direction is an important degree of freedom. It is sometimes visualized as the rotation of an object around its own axis (hence the name "spin"), though this notion is somewhat misguided at subatomic scales because elementary particles are believed to be point-like.

Spin can be represented by a vector whose length is measured in units of the reduced Planck constant ħ (pronounced "h bar"). For quarks, a measurement of the spin vector component along any axis can only yield the values +ħ/2 or −ħ/2; for this reason quarks are classified as spin-$1/2$ particles. The component of spin along a given axis—by convention the z axis—is often denoted by an up arrow ↑ for the value +$1/2$ and down arrow ↓ for the value −$1/2$, placed after the symbol for flavor. For example, an up quark with a spin of +$1/2$ along the z axis is denoted by u↑.

Weak interaction
A quark of one flavor can transform into a quark of another flavor only through the weak interaction, one of the four fundamental interactions in particle physics. By absorbing or emitting a W boson, any up-type quark (up, charm, and top quarks) can change into any down-type quark (down, strange, and bottom quarks) and vice versa. This flavor transformation mechanism causes the radioactive process of beta decay, in which a neutron "splits" into a proton, an electron  and an electron antineutrino  (see picture). This occurs when one of the down quarks in the neutron decays into an up quark by emitting a virtual  boson, transforming the neutron into a proton. The boson then decays into an electron and an electron antineutrino.

Both beta decay and the inverse process of inverse beta decay are routinely used in medical applications such as positron emission tomography (PET) and in high-energy experiments such as neutrino detection.

While the process of flavor transformation is the same for all quarks, each quark has a preference to transform into the quark of its own generation. The relative tendencies of all flavor transformations are described by a mathematical table, called the Cabibbo–Kobayashi–Maskawa matrix (CKM matrix). The approximate magnitudes of the entries of the CKM matrix are: where Vij represents the tendency of a quark of flavor i to change into a quark of flavor j (or vice versa).

There exists an equivalent weak interaction matrix for leptons (right side of the W boson on the above beta decay diagram), called the Pontecorvo–Maki–Nakagawa–Sakata matrix (PMNS matrix). Together, the CKM and PMNS matrices describe all flavor transformations, but the links between the two are not yet clear.

Strong interaction and color charge

Quarks possess a property called color charge. There are three types of color charge, arbitrarily labeled blue, green, and red. Each of them is complemented by an anticolor—antiblue, antigreen, and antired. Every quark carries a color, while every antiquark carries an anticolor.

The system of attraction and repulsion between quarks charged with different combinations of the three colors is called strong interaction, which is mediated by force carrying particles known as gluons; this is discussed at length below. The theory that describes strong interactions is called quantum chromodynamics (QCD). A quark charged with one color value can form a bound system with an antiquark carrying the corresponding anticolor; three (anti)quarks, one of each (anti)color, will similarly be bound together. The result of two attracting quarks will be color neutrality: a quark with color charge ξ plus an antiquark with color charge −ξ will result in a color charge of 0 (or "white" color) and the formation of a meson. Analogous to the additive color model in basic optics, the combination of three quarks or three antiquarks, each with different color charges, will result in the same "white" color charge and the formation of a baryon or antibaryon.

In modern particle physics, gauge symmetries—a kind of symmetry group—relate interactions between particles (see gauge theories). Color SU(3) (commonly abbreviated to SU(3)c) is the gauge symmetry that relates the color charge in quarks and is the defining symmetry for quantum chromodynamics. Just as the laws of physics are independent of which directions in space are designated x, y, and z, and remain unchanged if the coordinate axes are rotated to a new orientation, the physics of quantum chromodynamics is independent of which directions in three-dimensional color space are identified as blue, red, and green. SU(3)c color transformations correspond to "rotations" in color space (which, mathematically speaking, is a complex space). Every quark flavor f, each with subtypes fB, fG, fR corresponding to the quark colors, forms a triplet: a three-component quantum field which transforms under the fundamental representation of SU(3)c. The requirement that SU(3)c should be local—that is, that its transformations be allowed to vary with space and time—determines the properties of the strong interaction, in particular the existence of eight gluon types to act as its force carriers.

Mass
Two terms are used in referring to a quark's mass: current quark mass refers to the mass of a quark by itself, while constituent quark mass refers to the current quark mass plus the mass of the gluon particle field surrounding the quark. These masses typically have very different values. Most of a hadron's mass comes from the gluons that bind the constituent quarks together, rather than from the quarks themselves. While gluons are inherently massless, they possess energy—more specifically, quantum chromodynamics binding energy (QCBE)—and it is this that contributes so greatly to the overall mass of the hadron (see mass in special relativity). For example, a proton has a mass of approximately 938 MeV/c2, of which the rest mass of its three valence quarks only contributes about 11 MeV/c2; much of the remainder can be attributed to the gluons' QCBE.

The Standard Model posits that elementary particles derive their masses from the Higgs mechanism, which is related to the unobserved Higgs boson. Physicists hope that further research into the reasons for the top quark's large mass, which was found to be approximately equal to that of a gold nucleus (~171 GeV/c2), might reveal more about the origin of the mass of quarks and other elementary particles.

Table of properties
The following table summarizes the key properties of the six quarks. Flavor quantum numbers (isospin (I3), charm (C), strangeness (S, not to be confused with spin), topness (T), and bottomness (B′)) are assigned to certain quark flavors, and denote qualities of quark-based systems and hadrons. The baryon number (B) is +$9 °$ for all quarks, as baryons are made of three quarks. For antiquarks, the electric charge (Q) and all flavor quantum numbers (B, I3, C, S, T, and B′) are of opposite sign. Mass and total angular momentum (J; equal to spin for point particles) do not change sign for the antiquarks.

J = total angular momentum, B = baryon number, Q = electric charge, I3 = isospin, C = charm, S = strangeness, T = topness, B′ = bottomness. * Notation such as $1/3$ denotes measurement uncertainty. In the case of the top quark, the first uncertainty is statistical in nature, and the second is systematic.

title: Properties of quarks


text:
User:Eml5526.s11.team2.stewart/HW1

=Problem 1.1= = Given = $$  \displaystyle \frac{\partial }\left[ {E(x)A(x)\frac} \right] + f(x,t) = m(x)\frac $$ = Find = Prove the given equation using a balance of forces for the finite element shown from lecture slide. = Solution =

=Problem 1.2= = Given = $$  \displaystyle \frac{\partial }\left[ {E(x)A(x)\frac} \right] + f(x,t) = m(x)\frac $$ With a rectangular cross section as shown:

= Find = Prove the given equation using a balance of forces for the finite element shown from lecture slide. = Solution =

$$ \displaystyle \frac + f(x,t) = \underbrace {\rho \left( {x + \frac{2}} \right)A\left( {x + \frac{2}} \right)}_{m\left( {x + \frac{2}} \right)}\frac $$

$$ \displaystyle \mathop {\lim }\limits_{dx \to 0} \frac + f(x,t) = \mathop {\lim }\limits_{dx \to 0} \rho \left( {x + \frac{2}} \right)A\left( {x + \frac{2}} \right)\frac $$

$$ \displaystyle \partial \frac + f(x,t) = \rho (x)h(x)b\frac $$

$$ \displaystyle \sigma (x) = E(x)\frac $$

$$ \displaystyle \frac{\partial }\left[ {E(x)h(x)b\frac} \right] + f(x,t) = \rho (x)h(x)b\frac $$

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

title: User:Eml5526.s11.team2.stewart/HW1


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User:Eml5526.s11.team6

Team 06 - Members
Anand Tupsakhare

Shawn Gravois

Li Yongtao

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Nakul Deshpande

William Kurth

Shreyas Joglekar

title: User:Eml5526.s11.team6


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User:Eml5526.s11.team6.joglekar

Useful Links
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Homework #1

Homework #2

Homework #3

Homework #4

Homework #5

Homework #6

My Homework
Homework #2

Homework #3

Homework #4

Homework #5

Homework #6

My Team Members
Yongtao Li

Shreyas Joglekar

Nakul Deshpande

Anand Tupsakhare

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text: File:HW1-2 Fig1.jpg

title: File:HW1-2 Fig1.jpg


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User:EML5526.s11.team6.joglekar

Hi.This is Shreyas.

title: User:EML5526.s11.team6.joglekar


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User:Eml5526.s11.team6.kurth

EML 5526, Finite Element Method, Spring 2011 HW problem 2.8

HW problems 3.5 and 3.10

HW problem 4.7

HW problem 5.8

HW problem 6.2 and 6.9

Useful Links

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User talk:Ireoshang88

 Hello Ireoshang88, 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! --Leutha 12:10, 23 January 2011 (UTC)

title: User talk:Ireoshang88


text:
User talk:Reswan~enwikiversity

 Hello Reswan, 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! --Leutha 12:10, 23 January 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Reswan. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Reswan~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:50, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:19, 19 April 2015 (UTC)

title: User talk:Reswan~enwikiversity


text:
User talk:Gautam batra

 Hello Gautam batra, 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! --Leutha 12:11, 23 January 2011 (UTC)

title: User talk:Gautam batra


text:
User talk:Mhd.moaz

 Hello Mhd.moaz, 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! --Leutha 12:11, 23 January 2011 (UTC)

title: User talk:Mhd.moaz


text:
User talk:Ccmarinaro

 Hello Ccmarinaro, 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! --Leutha 12:12, 23 January 2011 (UTC)

title: User talk:Ccmarinaro


text:
User talk:Usmankhoso

 Hello Usmankhoso, 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! --Leutha 12:12, 23 January 2011 (UTC)

title: User talk:Usmankhoso


text:
User talk:Tangella

 Hello Tangella, 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! --Leutha 12:12, 23 January 2011 (UTC)

title: User talk:Tangella


text:
User talk:Shubhamkhg

 Hello Shubhamkhg, 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! --Leutha 12:13, 23 January 2011 (UTC)

title: User talk:Shubhamkhg


text:
User talk:Homi~enwikiversity

 Hello Homi, 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! --Leutha 12:13, 23 January 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Homi. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Homi~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:38, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:11, 19 April 2015 (UTC)

title: User talk:Homi~enwikiversity


text:
User talk:TheAstrobiologist

 Hello TheAstrobiologist, 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! --Leutha 12:14, 23 January 2011 (UTC)

title: User talk:TheAstrobiologist


text:
User talk:Hal 9000~enwikiversity

 Hello Hal 9000, 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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 * Visit a (kind of) random project.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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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! --Leutha 12:14, 23 January 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Hal 9000. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Hal 9000~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:33, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:10, 19 April 2015 (UTC)

title: User talk:Hal 9000~enwikiversity


text:
User talk:Ibn Alhassan

 Hello Ibn Alhassan, 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! --Leutha 12:15, 23 January 2011 (UTC)

title: User talk:Ibn Alhassan


text:
User talk:Leadtimpani

 Hello Leadtimpani, 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) 15:39, 23 January 2011 (UTC)

title: User talk:Leadtimpani


text:
User talk:Rambabu joshi

 Hello Rambabu joshi, 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) 15:39, 23 January 2011 (UTC)

title: User talk:Rambabu joshi


text:
User talk:Eml6341.s11.team4.kurth

 Hello Eml6341.s11.team4.kurth, 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) 15:40, 23 January 2011 (UTC)

title: User talk:Eml6341.s11.team4.kurth


text:
User talk:Eml5526.s11.team6.kurth

 Hello Eml5526.s11.team6.kurth, 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) 15:40, 23 January 2011 (UTC)

title: User talk:Eml5526.s11.team6.kurth


text:
User:Egm6341.s11.team4.yang

=Homework=

Homework-1

 * Homework 1.2

Homework-2

 * Homework 2.5, 2.6

Homework-3

 * Homework 3.5

Homework-4

 * Homework 4.2 4.9

Homework-5

 * Homework 5.4

Homework-5

 * Homework 6

=Links for this course=
 * EGM6341-NM1-Team4
 * EGM6341-NM1-Course Website
 * EGM6341-NM1-Course Lecture Plan

=The other course: EML5526-FEA=
 * EML5526-FEA-Team4-yang
 * EML5526-FEA-Team4
 * EML5526-FEA-Course Website
 * EML5526-FEA-Course Lecture Plan

title: User:Egm6341.s11.team4.yang


text: File:Lagrange.error.png

title: File:Lagrange.error.png


text: File:Lagrange.error.svg

title: File:Lagrange.error.svg


text:
User:Eml5526.s11.team7.jin

HW 1

HW 2

HW 3

HW 4

HW 5

HW 6

title: User:Eml5526.s11.team7.jin


text:
User talk:Egm6341.s11.team4.fields

HW1

title: User talk:Egm6341.s11.team4.fields


text:
User talk:Eml5526.s11.team7.jin

 Hello Eml5526.s11.team7.jin, 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) 17:25, 23 January 2011 (UTC)

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MaintenanceBot (discuss • contribs) 03:46, 11 November 2013 (UTC)

title: User talk:Eml5526.s11.team7.jin


text:
User talk:Egm6341.s11.team4.yang

 Hello Egm6341.s11.team4.yang, 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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title: User talk:Egm6341.s11.team4.yang


text:
TESOL/Point of view

The phrase point of view can mean the same as opinion, but from (someone's) point of view has a slightly different meaning.

Examples

 * That book represents its author's point of view/opinion.
 * A scientist must make decisions based on fact, not their point of view/opinion.
 * I agree with whatever point of view/opinion is most comfortable to me.

The phrase in (someone's) point of view has the same meaning.

 * In my point of view, the black bear is the coolest bear in the world.
 * He should be promoted in the boss's point of view.
 * In her point of view, the earth is flat.

Not equaling "opinion"
The phrase from (someone's) point of view has a slightly different meaning. It refers to someone's job, religion, age, gender, or other backgrounds.

Examples

 * From a child's point of view, driving a car can be very scary.
 * Young people seldom worry about health insurance but from a retiree's point of view, it is very serious.
 * Global warming is a real danger from a glacier's point of view.

Quiz
{In/from a doctor's point of view, going to a hospital is an everyday thing. - In + From
 * type=""}

{In/From her point of view, the president won the election because he was the tallest candidate. + In - From
 * type=""}

{A fly is delicious in/from a spider's point of view. - In + From
 * type=""}

title: TESOL/Point of view


text:
User:Egm6341.s11.team4.fields/HW1

=Problem 1.3 IMVT Proof=

Given
IMVT: $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Find
a.) Prove the Integral Mean Value Theorem (IMVT) for the non-negative function $$w(\cdot)$$

b.) Prove the IMVT for $$w(x) \neq 0 \,\,\,\, \forall \,x \in [a,b]$$

Solution
a.)

$$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Where w(x) and f(x) are continuous on [a,b], w(x) is non-negative and ξ is a point on [a,b].

For the case where w(x) = 0, both integrals = 0 and we are left with 0 = 0

The rest of this proof is for w(x) > 0

Since w(x) > 0, $$\int_{a}^{b}{w(x)dx}  >  0$$

Let $$\,m := min f(x) \,\,\forall x \in [a,b]$$

Let $$\,M := max f(x) \,\,\forall x \in [a,b]$$

It follows that:  $$\,m\,\le  \,f(x) \,\le\, M\,\, \forall x \in [a,b]$$

We now multiply through by w(x) to get $$\,m w(x)\,\le \,f(x) w(x) \,\le\, M w(x)\,\, \forall x \in [a,b]$$

Which leads to: $$\,m \int_{a}^{b}{w(x)dx}\,\le  \, \int_{a}^{b}{f(x)}{w(x)dx}\,\le\, M \int_{a}^{b}{w(x)dx}\,\, \forall x \in [a,b]$$

We can now divide through by $$\int_{a}^{b}{w(x)dx}$$

To get $$\,m \,\le \, \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}\,\le\, M \, \forall x \in [a,b]$$

Since f(x) is continuous, there must be at least one value ξ on the interval [a,b]; such that $$f(\xi) = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}$$

Plugging this value into the original statement $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx} = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\cancel{\int_{a}^{b}{w(x)dx}}}\cancel{\int_{a}^{b}{w(x)dx}}  =  \int_{a}^{b}{w(x)f(x)dx}$$

proves the theorem.

b.)

The proof for w(x) > 0 is contained in part a. above.

What follows is the proof for the IMVT with w(x) < 0.

$$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Where w(x) and f(x) are continuous on [a,b], w(x) < 0 and ξ is a point on [a,b].

Since w(x) < 0, $$\int_{a}^{b}{w(x)dx}  <  0$$

Let $$\,m := min f(x) \,\,\forall x \in [a,b]$$

Let $$\,M := max f(x) \,\,\forall x \in [a,b]$$

It follows that:  $$\,m\,\le  \,f(x) \,\le\, M\,\, \forall x \in [a,b]$$

We now multiply through by w(x) to get $$\,m w(x)\,\ge \,f(x) w(x) \,\ge\, M w(x)\,\, \forall x \in [a,b]$$

Which leads to: $$\,m \int_{a}^{b}{w(x)dx}\,\ge  \, \int_{a}^{b}{f(x)}{w(x)dx}\,\ge\, M \int_{a}^{b}{w(x)dx}\,\, \forall x \in [a,b]$$

We can now divide through by $$\int_{a}^{b}{w(x)dx}$$

To get $$\,m \,\le \, \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}\,\le\, M \, \forall x \in [a,b]$$

Since f(x) is continuous, there must be at least one value ξ on the interval [a,b]; such that $$f(\xi) = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}$$

Plugging this value into the original statement $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx} = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\cancel{\int_{a}^{b}{w(x)dx}}}\cancel{\int_{a}^{b}{w(x)dx}}  =  \int_{a}^{b}{w(x)f(x)dx}$$

proves the theorem.

title: User:Egm6341.s11.team4.fields/HW1


text:
User:Egm5526.s11.team2.sandhu

hw 5
User:Egm5526.s11.team2.sandhu/newpage =HW 4=

= Problem 4.3: Construct weak form of heat problem and find solution =

Given: Strong Form and Boundary Conditions
Statement:- Given the strong form for the heat conduction problem in a circular plate:
 * {| style="width:100%" border="0"

$$  \displaystyle k\frac{d}\left( {r\frac} \right) + rs = 0,0 < r \leqslant R $$ (4.3.1) natural boundary condition :$$ \frac\left( {r = 0} \right) = 0$$
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 *
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essential boundary condition :$$T\left( {r = R} \right) = 0$$

where R is the total radius of the plate, s is the heat source per unit length along the plate radius, T is the temperature and k is the conductivity. Assume that k, s and R are given:

a. Weak Form
Construct the weak form for the above strong form.

b. Solution using quadratic polynomial trial function
Use quadratic trial (candidate) solutions of the form $$T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2}$$ and weight functions of the same form to obtain a solution of the weak form.

c. Solve Differential equation
Solve the differential equation with the boundary conditions and showthat the temperature distribution along the radius is given by:

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$$  \displaystyle T = \frac{s}\left( {{R^2} - {r^2}} \right) $$     (4.3.2)
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 *
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a. Weak Form Derivation
Strong Form Can be written as:

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$$  \displaystyle p\left( T \right) = 0 $$
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 * }
 * }

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$$  \displaystyle p\left( T \right) = k\frac{d}\left( {r\frac} \right) + rs $$
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 * }
 * }

$$w\left( r \right)$$ is weight function Weighted Residual Form can be written as

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$$  \displaystyle \int\limits_0^R {w\left( r \right)p\left( T \right)dr = 0} $$     (4.3.3)
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 *
 * }

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$$  \displaystyle \int\limits_0^R {w\left( r \right)\left[ {k\frac{d}\left( {r\frac} \right) + rs} \right]dr} = 0 $$
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 * }
 * }

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$$  \displaystyle \int\limits_0^R {w\left( r \right)k\frac{d}\left( {r\frac} \right)dr + } \int\limits_0^R {w\left( r \right)rsdr} = 0 $$
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 * }
 * }

Write $$w\left( r \right) = w$$ Integrating the first term by parts.

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$$  \displaystyle \left[ {wkr\frac} \right]_0^R - \int\limits_0^R {\left( {\frackr\frac} \right)} dr + \int\limits_0^R {wrsdr} = 0 $$
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 * }
 * }

Assuming $$w\left( R \right) = 0$$ and using natural boundary condition $$\frac\left( {r = 0} \right) = 0$$

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

$$  \displaystyle 0 - 0 - \int\limits_0^R {\left( {\frackr\frac} \right)} dr + \int\limits_0^R {wrsdr} = 0 $$
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 * }
 * }

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$$  \displaystyle \int\limits_0^R {\left( {\frackr\frac} \right)} dr = \int\limits_0^R {wrsdr} $$     (4.3.4)
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 *
 * }

Writing in standard form:

 * }

b. Solution using quadratic polynomial trial function
let $$T$$ and $$w$$ are of the following
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$$  \displaystyle T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2} $$ and $$ \displaystyle w = {\beta _0} + {\beta _1}r + {\beta _2}{r^2} $$     (4.3.6)
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 *
 * }

Now using essential Boundary Condition i.e $$T\left( R \right) = 0$$

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$$  \displaystyle {\alpha _0} + {\alpha _1}R + {\alpha _2}{R^2} = 0 $$     (4.3.7)
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 *
 * }

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$$  \displaystyle {\alpha _0} = - \left( {{\alpha _1}R + {\alpha _2}{R^2}} \right) $$     (4.3.8)
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 *
 * }

Now using Natural boundary condition i.e. $$\frac\left( {r = 0} \right) = 0$$

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$$  \displaystyle {\alpha _1} + 2{\alpha _2}\left( 0 \right) = 0 $$     (4.3.9)
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 *
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$$  \displaystyle {\alpha _1} = 0 $$     (4.3.10)
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 *
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$$  \displaystyle {\alpha _0} = - {\alpha _2}{R^2} $$     (4.3.11)
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 *
 * }

Now using homogeneous boundary condition $$w\left( R \right) = 0$$
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$$  \displaystyle {\beta _0} + {\beta _1}R + {\beta _2}{R^2} = 0 $$     (4.3.12)
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 *
 * }

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$$  \displaystyle {\beta _0} = - \left( {{\beta _1}R + {\beta _2}{R^2}} \right) $$     (4.3.13)
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 *
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Also

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$$  \displaystyle \frac = {\alpha _1} + 2{\alpha _2}r $$     (4.3.14)
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 *
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As $${\alpha _1} = 0$$
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$$  \displaystyle \frac = 2{\alpha _2}r $$     (4.3.15)
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 *
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Now differentiating $$w$$ with respect to $$r$$
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$$  \displaystyle \frac = {\beta _1} + 2{\beta _2}r $$     (4.3.16)
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 *
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Putting values of $$w,\frac,\frac$$ in the standard weak form.

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$$  \displaystyle \int\limits_0^R {kr\left( {{\beta _1} + 2{\beta _2}r} \right)} \left( {2{\alpha _2}r} \right)dr = \int\limits_0^R {sr\left( {{\beta _0} + {\beta _1}r + {\beta _2}{r^2}} \right)} dr $$
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 * }
 * }

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$$  \displaystyle k\int\limits_0^R {\left( {2{\beta _1}{\alpha _2}{r^2} + 4{\beta _2}{\alpha _2}{r^3}} \right)dr} = s\int\limits_0^R {\left( {{\beta _0}r + {\beta _1}{r^2} + {\beta _2}{r^3}} \right)dr} $$
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 * }

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$$  \displaystyle \left[ {\frac{3} + k{\beta _2}{\alpha _2}{r^4}} \right]_0^R = \left[ {\frac{2} + \frac{3} + \frac{4}} \right]_0^R $$
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 * }

Putting value of $$$$ from equation

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$$  \displaystyle \frac{3} + k{\beta _2}{\alpha _2}{R^4} = \frac{2} + \frac{3} + \frac{4} $$
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 * }
 * }

Rearranging terms

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$$  \displaystyle {\beta _1}\left[ {\frac{3} + \frac{2} - \frac{3}} \right] + {\beta _2}\left[ {k{\alpha _2}{R^4} + \frac{2} - \frac{4}} \right] = 0 $$
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 * }
 * }

As $${\beta _1},{\beta _2}$$ are arbitrary they can't be zero, putting bracketed terms equal to zero both equation give same value of $${\alpha _2}$$.

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

$$  \displaystyle {\alpha _2} = \frac $$     (4.3.17)
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 *
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Using equation (4.3.10) $${\alpha _0}$$
 * {| style="width:100%" border="0"

$$  \displaystyle {\alpha _0} = - {\alpha _2}{R^2} = \frac $$     (4.3.18)
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 *

 * }

Putting these values in $$T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2}$$

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$$  \displaystyle T = \frac - \frac $$
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Now checking this solution by putting in given strong form given in equation.....

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$$  \displaystyle k\frac{d}\left( {r\frac{d}\left( {\frac{s}\left( {{R^2} - {r^2}} \right)} \right)} \right) + rs = 0 $$
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 * }

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$$  \displaystyle k\frac{d}\left( { - \frac} \right) + rs = 0 $$
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 * }
 * }

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$$  \displaystyle k\left( { - \frac{k}} \right) + rs = 0 $$
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$$  \displaystyle - \cancel{sr} + \cancel{sr} = 0 $$     (4.3.20)
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c. Exact solution for given differential equation
Starting from 4.3.1 and rearranging we get

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$$  \displaystyle d\left( {r\frac} \right) = - \frac{s}{k}rdr $$     (4.3.21) Integrating this we get
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 *

 * }
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$$  \displaystyle r\frac = - \frac{s}{r^2} + {c_1} $$     (4.3.22) Using the boundary condition and solving
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 *

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$$  \displaystyle \frac(r = 0) = 0 $$     (4.3.23)
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$$  \displaystyle \Rightarrow {c_1} = 0 $$     (4.3.24) Next we rearrange again and integrate
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 *

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$$  \displaystyle dT = - \frac{s}rdr $$     (4.3.25)
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 *

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

$$  \displaystyle T = - \frac{s}{r^2} + {c_2} $$     (4.3.26) Again using the boundary condition to solve for the constant of integration
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 *

 * }
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$$  \displaystyle T(R) = 0 $$     (4.3.27)
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 *

 * }
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$$  \displaystyle \Rightarrow {c_2} = \frac{s}{R^2} $$     (4.3.28) We arrive at the exact solution to the strong form which was found from the weak form
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 *

 * }
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$$  \displaystyle T(r) = \frac{s}\left( {{R^2} - {r^2}} \right) $$     (4.3.29)
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 *

 * }

=Problem 4.6: Elastic bar with a variable distributed spring force p(x)=

Given: bar of length l, cross-sectional area A(x), Young’s modulus E(x) with body force b(x)
Consider an elastic bar with a variable distributed spring p(x) along its length as shown in Figure 1. The distributed spring imposes an axial force on the bar in proportion to the displacement. Consider a bar of length l, cross-sectional area A(x), Young’s modulus E(x) with body force b(x) and boundary conditions as shown in Figure 1

a. Derivation of Strong Form
Let $$\displaystyle \vec{n}(x) $$ be defined as the unit vector having positive orientation in the direction of $$ \sigma $$ on the element we are considering. This means that the sign convention of our unit vectors will be

Which corresponds to the convention depicted in the free-body diagram shown above. Let us also define $$ \displaystyle \sigma(x) $$ as the force for a given area as,

Upon the rearrangement of terms we arrive at

Classical mechanics allows us to write the sum of the forces in static system is zero

Using the free-body diagram, let us now consider all of the forces acting on the elastic bar $$\displaystyle b(x),p(x), N(x), $$ and $$\displaystyle  N(x+dx) $$. Balancing the forces acting on our finite element results in the following relation,

Next we will substitute (1.5) and (1.6) into (1.10) to get

Recall that a Taylor's Series expansion (See:Taylor Series) about $$ a $$ near $$ x $$ has the following form:

Next we will need to determine a relationship for the displaced area and stress, namely $$\displaystyle \sigma(x+dx)$$ and $$ \displaystyle   A(x+dx)$$. To accomplish this we will performing a Taylor Series expansion choosing $$\displaystyle a=x+dx $$ and neglect all higher order terms.

Taking the product of (1.13) and (1.14) gives us

Substitute (1.15) into (1.11) results in

This can be simplified by canceling the like terms and dividing through by $$ \displaystyle dx $$ as shown below,

to get the following simplified relation,

Next we must apply the product rule in the reverse directions. Meaning we will consider (1.19) the expansion obtained by applying the product rule and recombine terms to get the unexpanded form shown here,

tensile stress according to Hooke's Law

Substituting (1.22) into our simplified and unexpanded form of our force balance shown in (1.20) we arrive at the solution,
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{d}[E(x)\cdot A(x)\frac] + b(x)-p(x) = 0 $$     (4.6.19) Standard form of the strong form can be written as
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 *

 * }

 * }

b. Derivation of weak form
Now strong form can be written as
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$$  \displaystyle P\left( u \right) = 0 $$     (4.6.21)
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 *

 * }

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

$$  \displaystyle p\left( u \right) = \frac{d}\left[ {E\left( x \right)A\left( x \right)\frac} \right] + \left( {b\left( x \right) - p\left( x \right)} \right) $$     (4.6.22)
 * style="width:95%" |
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 *

 * }

Let $${w\left( x \right)}$$ is a weight function, then writing Weighted Residual Form
 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {w\left( x \right)\left[ {P\left( u \right)} \right]dx = 0} $$     (4.6.23)
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 *

 * }

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

$$  \displaystyle \int\limits_0^l {w\left( x \right)\left[ {\frac{d}\left[ {E\left( x \right)A\left( x \right)\frac} \right] + \left( {b\left( x \right) - p\left( x \right)} \right)} \right]dx = 0} $$     (N)
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 *

 * }

Writing $$w\left( x \right) = w,E\left( x \right) = E,A\left( x \right) = A,b\left( x \right) = b,p\left( x \right) = p$$
 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {w\left[ {\frac{d}\left[ {EA\frac} \right] + \left( {b - p} \right)} \right]dx = 0} $$     (N)
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 *

 * }

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

$$  \displaystyle \int\limits_0^l {w\frac{d}\left[ {EA\frac} \right]dx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (N)
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 *

 * }

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

$$  \displaystyle \left[ {wEA\frac} \right]_0^l - \int\limits_0^l {EA\frac\fracdx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (N)
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 *

 * }

Now assuming $$w\left( 0 \right) = 0$$ and also from natural boundary condition $$E(l)\frac = \bar t$$ above integral will take form as
 * {| style="width:100%" border="0"

$$  \displaystyle w\left( l \right)A\bar t - 0 - \int\limits_0^l {EA\frac\fracdx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (N)
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 *

 * }

Rearranging terms

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$$  \displaystyle \int\limits_0^l {EA\frac\fracdx} = \int\limits_0^l {w\left( {b - p} \right)dx}  + w\left( l \right)A\bar t $$ (N)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now writing in standard format

 * }

title: User:Egm5526.s11.team2.sandhu


text:
User talk:Egm6341.s11.team3.sungsikkim

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MaintenanceBot (discuss • contribs) 03:45, 11 November 2013 (UTC)

title: User talk:Egm6341.s11.team3.sungsikkim


text:
User:Egm6341.s11.team4.KTH

Team4_Wiki

title: User:Egm6341.s11.team4.KTH


text:
User:Egm6341.s11.team2.DT

Homework sandbox =Problem 1.1=

Find
Plot f(x) and find:

 * $$ \lim_{x \to 0}f(x) $$

Given

 * $$ f(x) = \frac{x} $$ for $$ x \in [0,1] $$

Solution
Using l'Hôpital's rule,

 * $$ \lim_{x \to 0}\frac{x} = \lim_{x \to 0}\frac{1} = 1$$

 * Hw1aplot.jpg]

=Problem 2.4 - Comparision of Numerical Integration Techniques= Note: This problem was studied for $$ [a,b] = [0,1]$$ in s10.

''' This problem was solved without referring to S10 homework. '''

Given

 * $$ f(x) = \frac{x} $$ for $$ x \in [a,b] $$

where $$ [a,b] = [-1,1]$$

Find
Plot f(x), then approximate the integral of f(x) as
 * $$ I_{n}\approx I := \int_{a}^{b}f(x)dx$$

using the following techniques:

1) Taylor Series Expansion

2) Composite Trapezoidal Rule

3) Composite Simpson Rule

4) Gauss-Legendre Quadrature

The value of n should be selected as n = 2,4,8,... until $$I_{n}$$ has a maximum error on the order of $$10^{-6}$$.

Plot of f(x)

 * Egm6341.s11.hw2_4a.jpg]

2.4.1 Taylor Series
As shown in Problem 1.2, the Taylor series expansion of f(x) is Because this is a polynomial, each term in the summation can be integrated individually. Thus, the integral approximation can be found as

The maximum error of the Tayor series integration is shown in (Atkinson, 250) as The following table shows the results for integration by Taylor series expansion.

2.4.2 Composite Trapezoidal Rule
The formula for integration by composite trapezoidal rule is given by (Atkinson, 253)

The error in the composite trapezoidal integration is given by (Atkinson, 253)

Since we know that the error will achieve its maximum when Thus, the maximum error for integration by the composite trapezoidal rule is

The following table shows the results for integration by composite trapezoidal rule.

2.4.3 Composite Simpson Rule
The formula for integration by composite Simpson rule is given by (Atkinson, 257)

The error in the composite Simpson integration is given by (Atkinson, 257)

Like with the composite trapezoidal rule, since we know that the error will achieve its maximum when

Thus, the maximum error for integration by the composite Simpson rule is

The following table shows the results for integration by composite Simpson rule.

2.4.4 Gauss-Legendre Quadrature
The formula for Gaussian quadrature on the interval [-1,1] is given by (Atkinson, 276) The quadrature weights wj are given by (Atkinson, 276) and the nodes xj are the roots of the nth degree Legendre polynomial Pn(x) on the interval [-1,1]. The numerical values of xj and wj are listed in and verified in Problem 2.5.

The error using Gauss-Legendre quadrature is found as As in the composite trapezoidal and Simpson rules, the maximum value of f(2n)(η) is e. Thus, the maximum error using Gauss-Legendre quadrature is

The following table lists the results for integration by Gauss-Legendre quadrature.

Background
The maximum error of composite trapezoidal rule is ,

where $$\displaystyle {{M}_{2}}$$ is the maximum of $$\displaystyle \left| {{f}^{(2)}}(.) \right|$$

The maximum error of composite Simpson ‘ s rule is ,

where $$\displaystyle {{M}_{4}}$$ is the maximum of $$\displaystyle \left| {{f}^{(4)}}(.) \right|$$

Given
In the Problem 2.4, we have numerically calculate the order $$\displaystyle n$$ such that the following integration, with Taylor series expansion, composite trapezoidal rule and composite Simpson ‘ s rule, is with error less than $$\displaystyle Q={{10}^{-6}}$$.

where $$\displaystyle [a,b]=[-1,1]$$

And in Problem 2.3 we have the remainder of Taylor expansion of $$\displaystyle f(x)=\frac{{{e}^{x}}-1}{x}$$ is,

where $$\displaystyle \xi \in [0,x]$$

Find
(1)Use different error bound of Taylor series, composite trapezoidal rule and composite Simpson ‘ s rule to calculate the smallest order n needed such that the error of numerical integration of that in Problem 2.4 is less than $$\displaystyle Q={{10}^{-6}}$$. And then compare them with the actual numerical calculation.

Taylor series comparison
Using the same method in Mtg7 we can find the error bound of Taylor series,

Since we have set n to be even,

The error should be within $$\displaystyle Q$$

By WolframAlpha we have,

which coincide with the result in Problem 2.4.

Composite trapezoidal rule comparison
By 4.8.1 we have,

where $$\displaystyle {{M}_{2}}$$ can be calculated by WolframAlpha ,

Then we have,

Calculated by WolframAlpha we have,

While in Problem 2.4 we have $$\displaystyle n=512$$, the order produced by the error bound is larger than the numerical result.

Composite trapezoidal rule comparison
By 4.8.2 we have,

where $$\displaystyle {{M}_{4}}$$ can be calculated by WolframAlpha ,

Then we have,

Calculated by WolframAlpha we have,

In Problem 2.4 we calculated the case $$\displaystyle n=8$$ and $$\displaystyle n=16$$, the former case failed to satisfy the condition that $$\displaystyle \left| {{E}_{n}} \right|\le Q={{10}^{-6}}$$ while the latter succeed. This is a verification of the result above.

Conclusion
From the case of Taylor series and composite trapezoidal rule we can see the order of estimation given by the error bound is larger than or equal to that given by numerical calculation. This can be explained as: the error bound is stricter because it denotes the maximal error hence it is reasonably larger than or equal to the real error and thus requires higher order.

While in case of composite Simpson ‘ s rule we didn ‘ t calculate the case $$\displaystyle n=9$$, but the case $$\displaystyle n=8$$ and $$\displaystyle n=16$$ have demonstrate that if the error bound is satisfied with error limit then the numerical result must be satisfied.

Solution to part (2)
Figure 4.8-1 shows a plot of the logarithm of the Taylor series approximation for values of n = 2,4,8,16. This data was then fitted with a least square linear regression. The slope of this line is 16, however, this value increases as n is increased (and h is consequently decreased). Thus, the power of h is not a constant and the error using Taylor series shrinks at a faster rate than composite trapezoidal or Simpson's rule.

 * Egm6341.s11.team2.hw4_8_1.jpg

Figure 4.8-2 shows the logarithm of the composite trapezoidal rule error for varying values of log(h). The linear regression line has a slope of 2, Thus h appears as a quadratic term in finding the error.

 * Egm6341.s11.team2.hw4_8_2.jpg

Figure 4.8-3 shows the logarithm of the composite Simpson's rule error for varying values of log(h). The linear regression line has a slope of 4. Consequently, h appears as a 4th-order term in finding the integration error using the composite Simpson's rule.

 * Egm6341.s11.team2.hw4_8_3.jpg

=Problem 5.5: - Richardson Extrapolation and Romberg Integration applied to HW* 2.4=

Statement:
1) Modify the Matlab code from HW*2.4 (links) to make it more efficient, i.e. use Richardson extrapolation to compute higher order integral estimates starting from the trapezoidal rule.

2) Construct a Romberg table for results in 1). Compare to results from HW* 2.4.

Solution:
1) From problem 2.4, the composite trapezoidal rule was implemented for 2,4,8,16,32, and 64 node points. Equation 5.5.1 was then implemented to find the higher order integral estimates. The code is easily modified to include greater numbers of nodes or even higher order integral estimates.

The order of the error on each integral estimate can be found using Equation 5.5.2:

2) The following table presents the results of using Romberg integration on problem 2.4. The subscript on T corresponds to the iteration. Since the method starts with the composite trapezoidal rule, T0 corresponds to integration by the trapezoidal rule for various numbers of nodes.

Using Equation 5.5.2, the error of T3(4) is on the order of ~O(-7). Thus, using this method the third order estimate using only 4 nodes has O(-7) accuracy. However, this requires knowing the integral estimate using the composite trapezoidal rule for 2,4,8,16 and 32 nodes. For comparison, the composite trapezoidal rule alone only reaches O(-6) accuracy using 512 nodes, and the composite Simpson rule reaches O(-6) accuracy with 16 nodes. Legendre-Gauss quadrature reaches O(-7) accuracy with only 4 nodes. If the quadrature weights are readily available, Legendre-Gauss quadrature converges to an accurate solution with many fewer nodes required than the other methods. The Romberg integration scheme provides another useful method for numerically calculating the value of an integral, especially because it uses previous results to improve the integral estimate.

=Problem 6.6 - Understanding Kessler's Code=

Given
Kessler's code, given in his paper :

Find
In order to better understand Kessler's code and the HOTRE, improve upon the work of s10 and explain the code line by line. Then, run Kessler's code for $$i=1,2,3$$ to reproduce his results. The collective work of s10 is referenced here under HW5, Problem 10.

Solution
The following box presents Kessler's code commented line-by-line to describe the algorithm for using the HOTRE.

Running Kessler's code with n = 3 results in the following output:

These with the additional last line written into the code, the output matches the tables presented in Kessler's paper:

title: User:Egm6341.s11.team2.DT