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School:Electrical engineering/Biography ELE

Michael Faraday (*September 22, 1791 – †August 25, 1867) Although Faraday received little formal education and knew little of higher mathematics, such as calculus, he was one of the most influential scientists in history. Some historians of science refer to him as the best experimentalist in the history of science. The SI unit of capacitance, the farad, is named after him, as is the Faraday constant, the charge on a mole of electrons (about 96,485 coulombs). "Nothing is too wonderful to be true if it be consistent with the laws of nature, and in such things as these, experiment is the best test of such consistency."

 * read more...


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School:Electrical engineering/Biography ELE-4

Joseph Henry (December 17, 1797 – May 13, 1878) was an American scientist who served as the first Secretary of the Smithsonian Institution, as well as a founding member of the National Institute for the Promotion of Science, a precursor of the Smithsonian Institution. During his lifetime, he was highly regarded. While building electromagnets, Henry discovered the electromagnetic phenomenon of self-inductance. He also discovered mutual inductance independently of Michael Faraday, though Faraday was the first to publish his results. The SI unit of inductance, the henry, is named in his honor. "Seeds of great discoveries are constantly floating around us, but they only take root in minds well prepared to receive them."

 * read more...

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


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School:Electrical engineering/Biography ELE-2

Thomas Alva Edison (February 11, 1847 – October 18, 1931) was an American inventor and businessman. He developed many devices that greatly influenced life around the world, including the phonograph, the motion picture camera, and a long-lasting, practical electric light bulb. Dubbed "The Wizard of Menlo Park" (now Edison, New Jersey) by a newspaper reporter, he was one of the first inventors to apply the principles of mass production and large teamwork to the process of invention, and therefore is often credited with the creation of the first industrial research laboratory. "Restlessness is discontent and discontent is the first necessity of progress. Show me a thoroughly satisfied man and I will show you a failure."

 * read more...


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User:Abhiabhi~enwikiversity

Abhikant Singh

title: User:Abhiabhi~enwikiversity


text:
User:Mikehicks

/enes100/

title: User:Mikehicks


title: File:Problem2Rep3.jpg


title: File:S4.png


text:
File:R3.1PLOT.jpg

Licensing
for school project

title: File:R3.1PLOT.jpg


text:
Talk:3D Scan/Kinect Makerbot/Howard Community College/Spring2012/P1501msn

check out this processing library for face recognition

download processing, install the library below and begin playing with the examples.

processing reference

1sfoerster 19:00, 22 February 2012 (UTC)

project grading ... 1sfoerster (talk) 19:53, 20 March 2012 (UTC)
 * 100 points ... great job !!! + 100 points for the decision matrix


text:
User:Egm4313.s12.team8.kyle/R4.1

The general form of (2) p.7b-13 is shown on p.7c-20 and is as follows:
 * $$\sum_{j=0}^{n-2}[c_{j+2} (j+2) (j+1) + a c_{j+1} (j+1) + bc_j]x^j

+ac_nnx^{n-1} + b[c_{n-1}x^{n-1} + c_nx^n] = \sum_{j=0}^{n} d_jx^j $$

When j=0:
 * $$2c_{2}+ac_{1}+bc_{0}=d_{0}$$

When j=1:
 * $$6c_{3}+2ac_{2}+bc_{1}=d_{1}$$

When j=2:
 * $$12c_{4}+3ac_{3}+bc_{2}=d_{2}$$

When j=(n-2):
 * $$[c_{n}(n)(n-1)+ac_{n-1}(n-1)+bc_{n-2}]=d_{n-2}$$

When j=(n-1):
 * $$ac_{n}n+bc_{n-1}=d_{n-1}$$

When j=n:
 * $$bc_{n}=d_{n}$$

Where $$c=[c_0, c_1, ..., c_{n-1}, c_n]^{T}$$ and $$d=[d_0, d_1, ..., d_{n-1}, d_n]^{T}$$

and $$Ac=d$$

The general form of matrix A can be created from the coefficients shown. $$A= \begin{bmatrix} b &     a &  2     &      0 &       0 &       0 &       0 \\ 0 &     b & 2a     &      6 &       0 &       0 &       0 \\ 0 &     0 &  b     &     3a &      12 &       0 &       0 \\ 0 &     0 &      0 & \ddots & \ddots  &  \ddots &       0 \\ 0 &     0 &      0 &      0 &       b & a (n-1) & n (n-1) \\ 0 &     0 &      0 &      0 &       0 &       b & a n     \\ 0 &     0 &      0 &      0 &       0 &       0 &   b \end{bmatrix} $$

title: User:Egm4313.s12.team8.kyle/R4.1


text:
User:Nkansahrexford

Please read my profile here

title: User:Nkansahrexford


text:
User:Itsme

Please leave comments on my Dutch Talk page.

title: User:Itsme


text:
Exercise and metabolic disease/prostate

This is a resource created by exercise physiology students about prostate cancer related androgen deprivation therapy (ADT) and exercise. It provides an overview of prostate cancer and ADT, in relation to the following topics:

 * Diagnosis
 * Prevalence
 * Pathophysiology
 * Medical and Clinical Considerations

ADT and exercise are then discussed on the following pages:

 * Exercise Training
 * Exercise Prescription

title: Exercise and metabolic disease/prostate


text:
Dominant group/Materials science

"Materials science is studying the substances and their properties that make them useful in structures, machines, tools, devices, or products.

Dominant group

 * 1) Accident hypothesis: dominant group is an accident of whatever processes are operating.
 * 2) Artifact hypothesis: dominant group may be an artifact of human endeavor or may have preceded humanity.
 * 3) Association hypothesis: dominant group is associated in some way with the original research.
 * 4) Bad group hypothesis: dominant group is the group that engages in discrimination, abuse, punishment, and additional criminal activity against other groups. It often has an unfair advantage and uses it to express monopolistic practices.
 * 5) Control group hypothesis: there is a control group that can be used to study dominant group.
 * 6) Entity hypothesis: dominant group is an entity within each field where a primary author of original research uses the term.
 * 7) Evolution hypothesis: dominant group is a product of evolutionary processes, such groups are the evolutionary process, produce evolutionary processes, or are independent of evolutionary processes.
 * 8) Identifier hypothesis: dominant group is an identifier used by primary source authors of original research to identify an observation in the process of analysis.
 * 9) Importance hypothesis: dominant group signifies original research results that usually need to be explained by theory and interpretation of experiments.
 * 10) Indicator hypothesis: dominant group may be an indicator of something as yet not understood by the primary author of original research.
 * 11) Influence hypothesis: dominant group is included in a primary source article containing original research to indicate influence or an influential phenomenon.
 * 12) Interest hypothesis: dominant group is a theoretical entity used by scholarly authors of primary sources for phenomena of interest.
 * 13) Metadefinition hypothesis: all uses of dominant group by all primary source authors of original research are included in the metadefinition for dominant group.
 * 14) Null hypothesis: there is no significant or special meaning of dominant group in any sentence or figure caption in any refereed journal article.
 * 15) Object hypothesis: dominant group is an object within each field where a primary author of original research uses the term.
 * 16) Obvious hypothesis: the only meaning of dominant group is the one found in Mosby's Medical Dictionary.
 * 17) Original research hypothesis: dominant group is included in a primary source article by the author to indicate that the article contains original research.
 * 18) Primordial hypothesis: dominant group is a primordial concept inherent to humans such that every language or other form of communication no matter how old or whether extinct, on the verge of extinction, or not, has at least a synonym for dominant group.
 * 19) Purpose hypothesis: dominant group is written into articles by authors for a purpose.
 * 20) Regional hypothesis: dominant group, when it occurs, is only a manifestation of the limitations within a region. Variation of those limitations may result in the loss of a dominant group with the eventual appearance of a new one or none at all.
 * 21) Source hypothesis: dominant group is a source within each field where a primary author of original research uses the term.
 * 22) Term hypothesis: dominant group is a significant term that may require a 'rigorous definition' or application and verification of an empirical definition.

Examples from primary sources are to be used to prove or disprove each hypothesis. These can be collected per subject or in general.

Chemicals
On the right is an image of a specimen of the tektosilicate reedmergnerite from Darai-Pioz Glacier, Alayskiy Range, Tien Shan Mountains, Tajikistan. Its formula is NaBSi3O8. It has an albite structural type of a plagioclase feldspar with boron replacing aluminum.

For the "reedmergnerite unit, BSi4O10 ... The spectra showed small amounts of it, and we saw no evidence of fragmentation (i.e., no corresponding smaller fragments that would appear to be constituents). Thus, it does not appear to be the dominant group in the glass network. ... The borate association remains dominant, however, even as we increase the soda content."

Alloys
"The dominant group of Al–Si foundry alloys contain between 5 and 25 wt.% Si, with Mg, Ni and Cu additions."

"There is in reality a distribution of magnetic cluster sizes although the smallest magnetic clusters will constitute the dominant group."

"From the above results we can see that the industry group metal alloys (33) is the dominant group in terms of number of employees, fixed capital and value added."

"The dominant group is (SiGe)SiH, at T, = 190 "C and is replaced by (Ge,)SiH, at T, = 250 "C."

Materials
Notation: let PE stand for polyethylene plastics.

There "is a dominant group (PE) that is still growing on the market. ... Pilot-scale composting tests are also suitable instruments for investigating any negative effects of the test materials on the composting process if sufficient test material is introduced."

Materials sciences
In materials science, the cultural or technical term dominant group may be an artifact, an evolutionary process, or only an entity.

Def. "[m]atter which may be shaped or manipulated" is called a material.

Def. "[a] region within a material having a single crystal structure or direction" is called a grain.

Def. the "study of materials; their internal structures, properties and modes of failure" is called materials science.

Def. an understanding of the nature of substances whose properties make them useful in structures, machines, devices, or products, leading to theories or descriptions that explain how structure relates to composition, properties, and behavior is called materials science.

Ceramics
Def. any nonmetallic solid that remains hard when heated is called a ceramic.

"The typology of the lustre ceramics shows that there is a small group (G1 = MZ2 + MZ6 + MZ12) with red lustre decoration on the interior and green-yellow lustre decoration on the exterior and a dominant group (G2) with yellowish lustre on both sides (Table 1)."

"The different settlement ceramic assemblages present dominant technical sub-groups, which are different in aspect from one site to another (tab. 3). ... The ceramic entity A is characterised, from a technical and techno-petrographic point of view, by a discontinuity between one or two dominant sub-groups (quantitatively major) and a few satellite sub-groups (quantitatively minor). Ceramic entity B, on the contrary, shows a petrographic continuity with the dominant sub-groups A."

Composites
"Both of the emission spectra present the characteristic emission bands originating from the transition 5 D 4 → 7 F J (J = 6,5,4,3), with the transition 5 D 4 → 7 F 5 green emission as the dominant group."

"Both of the emission spectra present the characteristic emission bands originating from the transition 5 D 4 → 7 F J (J=6,5,4,3), with the transition 5 D 4 → 7 F 5 green emission as the dominant group."

"Both of the emission spectra exhibit the characteristic emission of Tb3+ arising from the transition 5 D 4 → 7 F J (J = 6, 5, 4, 3), with the transition 5 D 4 → 7 F 5 green emission as the dominant group."

"Also, the characteristic emission bands originating from the transition 5 D 4 → 7 F J (J = 6, 5, 4, 3), with the transition 5 D 4 → 7 F 5 green emission as the dominant group are observed in the emission spectra of the pure Tb(III) complex (Tb-L) and the hybrid materials as shown".

"The ω-hydroxymonocarboxylates represented the dominant group with a contribution higher than one quarter of the total."

Crystals
Def. "[a] solid composed of an array of atoms or molecules possessing long-range order and arranged in a pattern which is periodic in three dimensions", from Wiktionary crystal, is called a crystal.

“The dominant group V source is arsenic, although antimony and phosphorous sources are not atypical.”

“We may assume, however, as has been suggested by Dunning (see the discussion in ref. 32) that these macrospirals are the result of a periodic perturbation due to an interaction of the spirals forming the dominant group of cooperating spirals.”

Diamonds
"For as-deposited diamond, C---H is the dominant group on the hydrogen-terminated surfaces, and it has been found that ---OH [23] and C=O groups [24] are in fact generated on oxidized diamond surfaces."

Dielectrics
"Our theory applies to parameters (to be identified below) for which the second group is dominant, allowing for the third group, labeled ψrem (mnemonic remainder) to be neglected in the calculation of the saddle-points." "The dominance requirement constrains the remaining (first and third) terms of ψrem to be significantly smaller than the terms of the second group along the contour of integration. Since the two terms of the dominant group have comparable magnitudes, it suffices to make the comparison with only the second term in the dominant group."

"Such work indicated that group III interdiffusion in InGaAs/InP is associated with a redshift, while a blueshift is associated with a dominant group V interdiffusion."

"Hence we refer to this dominantly amplified magnon group as the dominant group."

Fibers
"For example, As2 and P2 can sometimes be the dominant group V species in contrast to the equilibrium calculation, and, in addition, several kinds of internal compound between As and P are found."

"Nearly 90% of L, in the "as-grown" ribbons belongs to class B and class C, with class B being the predominant group (Fig. 3a)."

"Free fatty acids were the most dominant group of lipids in rice straw extract and the second most abundant class of lipophilic substances in rye straw extractives."

Glasses
Notation: let the symbol KREEP be an acronym for potassium (the chemical symbol is K), rare earth element (REE), and phosphorus (chemical symbol is P).

"Because KREEP glasses are the dominant group at the Fra Mauro sampling site and because the Fra Mauro formation is considered part of the Imbrian ejecta blanket, KREEP basalts appear to have been part of the lunar crust prior to formation of the Imbrium Basin." Bold added.

"This likely is the result of the fact that the six-membered ring group becomes the dominant structural group in the vicinity of x = 0.5, whereas the tetrahedral boron group is becoming the least prominent structural group in the glass."

"In the other series of glasses, Li, Na, and K, in particular, that can be formed out to x = 0.75, the IR spectra clearly show that this structural group becomes the dominant group in the glass at this composition [4,5,43]."

"Thus, it does not appear to be the dominant group in the glass network."

Martensites
“In Section 3, we use group theory to present the system description of the microstructure of 2H martensite, including the correspondence variant dominant group.”

“The dominant group always consisted of four habit-plane variants grouped round a particular {llO} pole of the austenite, except in the case of the crystal with compression axis close to [001] which developed only two habit planes.”

Metals
Regarding the straggling of alpha particles in metal foils, "the Bi'" and Am'" sources each have two closely spaced alpha-particle groups and the Pu'" source has three such groups, it was found that the resolution obtained with the apparatus enabled one to obtain the location of the dominant group in each source with great accuracy."

Metallurgy
"Then, metallurgical reactions occur between the fuel slug and the cladding. ... Lanthanide elements, a dominant group of ﬁssion products, will be less signiﬁcant because the solubility of lanthanide elements in the U—Pu—Zr alloys is limited (~O.6 wt% or less)(9) and the eutectic point in the binary system of any lanthanide element and Fe is higher than that in the Pu—Fe system".

Nanomaterials
“Mesopores were still the dominant group of pores for all derivatives, as confirmed by the good agreement between the values of V tot and V mp D.”

Piezoelectrics
Notation: let the symbol PZT stand for lead zirconate titanate (Pb[ZrxTi1-x]O3, 0 ≤ x ≤ 1).

“Among [the] variety of piezoelectric materials, PZTs have been the dominant group for more than 40 years for their excellent properties.”

Plastics
Plastics also refers to polymers.

Notation: let the symbol PE stand for polyethylene.

"Finally, there is a dominant group (PE) that is still growing on the market."

"The first class consists of polymers such as cellulose (cotton, wood, etc.) and protein (gelatine, wool, etc.), while the dominant group in the second class is rubber, both synthetic and natural."

Precious metals
"The alluvial sample shows multiple populations that reflect all the samples collected at the mine, a dominant group which corresponds to the mine shaking table gold, and a minor population with higher silver which is similar to that shown by the soil sample."

Sands
"The behavior of some sandy soils is conditioned by a dominant group of pores larger than 1.5 µ."

"The dominant group in which the median grain-size falls is indicated (M: mud; FS: fine sand; MS: medium sand) as well as the number of observations in that group (based on sedisurf@database, Gent University, Renard Centre of Marine Geology)."

"In the hypersthene dominant group the hornblende content does not exceed 20 vol.%."

Semiconductors
Notation: let the symbol CP indicate band structure critical points.

"The dominant group speed of a carrier can be estimated to occur at the energy, above or below the CP where is the reduced effective mass and is Planck's constant." "The largest of the resulting fragments from the break-up of the prior dominant group is a cluster of eight papers dealing almost exclusively with n-type semiconductors, with papers originating primarily from two research labs."

Degradation of "Id-Vd characteristics before and after stressing for (a) the charge-trapping-dominant (group Í) and (b) the interface-trap- generation dominant (group 2) devices. Note the similar degradation characteristics for electron and hoie trappings."

Stones
"Although calcitic marbles represent the dominant group in the studied rocks, dolomitic marbles are also present (Raspenava marble, Strážné marble, Bílá Voda marble, Bohdaneč marble, Český Šternberk marble)."

Superconductors
Notation: let the symbol dmit indicate the 2-thioxo-1,3-dithiole-4,5-dithiolate dianion, C3S52-.

"Various crystals, and consequently various conductivities, can be obtained by using different monocations in the 1:2 salts of Z[Ni(dmit)2]2, which form the most dominant group in the Z[M(dmit)2]n family." Bold added.

Notation: let the symbol ERDA stand for elastic recoil detection analysis.

"The dominant group in the lower half results from the first oxygen-bearing layer in the Si02(0.5 pm)/Al(0.5 pm)/Si02(0.5 pm)/Al(2.0 pm) target stack."

Notation: let the symbol TN (or TN) stand for the Néel transition temperature.

"Below TN, the description of the experimental data requires at least a twoband picture with a dominant group of electrons showing standard BCS-Eliashberg-type behavior with a clear single gap whereas a second group of electrons dominates κ(T,H)." Bold added.

Woods
"The mean ratio of crown width to sapling height (DrIh) was significantly greater for the suppressed saplings than for the apically dominant group (t = 2.78, df= 18, P < .02) (Table 5), and DrIh was negatively correlated with h for the combined data set (r = -0.76, N = 20, P < .001)."

Notation: let the symbol CWH stand for capillary water height.

"There was a significant difference in CWH for sapwood, between specimens from "wet dominant" group and specimens from both "dry suppressed" and "dry dominant" group during absorption for 7 days and 14-15 days."

"The analysis of the the obtained results concerning the selected wood physical and mechanical properties of both pine species (Tables from 2 to 5) showed that the examined wood properties were higher, generally speaking, in the black pine trees in comparison with the Scots pine, with the exception of the co-dominant group of trees in the first of the two pine tree species."

Hypotheses

 * 1) As each material is usually best suited for certain uses, each can be a dominant group within these uses.

title: Dominant group/Materials science


text:
Talk:Digital Logic 1

Changed "Verilog" to "HDLs"; Verilog only represents a portion of commonly used hardware descriptive languages. I am happy to contribute information on the other major language, VHDL.

title: Talk:Digital Logic 1


title: File:VHDL.Signal.1.A.20120223.pdf


text:
User talk:John7~enwikiversity

No promotional content, please
Please refrain from adding promotional content to Wikiversity or its sister projects. Thanks, Mathonius 04:20, 23 February 2012 (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 John7. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name John7~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 23:35, 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) 06:12, 19 April 2015 (UTC)

title: User talk:John7~enwikiversity


title: File:Math.2.B.Hilbert.20120223.pdf


title: File:Pea2.s12.sec45.e.djvu


text:
User talk:필리핀 홈스테이

No promotional content, please
Please refrain from adding promotional content or spam to Wikiversity and its sister projects. Thanks, Mathonius 17:38, 23 February 2012 (UTC)

title: User talk:필리핀 홈스테이


text:
User:9484syoon/enes100/personal work page

Write problem/project Goal
Design and create the standbeest

My First Task
Analysis the problem of old project and think about the walking mechanism

Summary of actual work over first weekend
How is what you did different than what you planned?

Week1 Narrative
Tell a detailed story describing what you did for your team over the weekend.

My Second Task
Record what you are planning on doing for the team during the second weekend.

Summary of actual work over second weekend
How is what you did different than what you planned?

Week2 Narrative
Tell a detailed story describing what you did for your team over the weekend.

My Third task
Record what you are planning on doing for the team during the third weekend.

Summary of actual work over third weekend
How is what you did different than what you planned?

Week3 Narrative
Tell a detailed story describing what you did for your team over the weekend.

My Fourth task
''Record what you are planning on doing for the team during the final weekend of project work. ''

Summary of actual work over fourth weekend
How is what you did different than what you planned?

Week4 Narrative
Tell a detailed story describing what you did for your team over the weekend.

Complete Team Page
Fill out the Team Form (should have already copied the form, created the team page, linked to it and started filling it out).


text:
User talk:Engsihn

How to upload pictures... wikimedia.org you want Wiki Media Commons http://commons.wikimedia.org/wiki/Main_Page

Left column under Participate>Upload

Follow the steps.

For formatting into wikiversity look at examples of how other people did it by clicking the picture in one tab to see its file name in wikimedia commons and opening the text in the edit page of wikiversity to see the text format. The format is...

~{no spaces between brackets} (hit shift forward slash [above enter] to put in the "|")

Afterwards you can go back to Wikimedia Commons to view your uploads with "my uploads" in the top right by your sign in name

title: User talk:Engsihn


text:
User talk:Mfashola/enes100

Max is working hard and completing his responsibilities. Has a good amount of work for this next week, but I am not worried about whether he will come to class empty handed or not on Tuesday. Greg Jaksec.

Hey, where you at? We're holding up but could use your help. If you need any help with anything just let me know. Greg Jaksec.

Not sure if you are still in this class or not. If you could let me know that would be cool. Greg Jaksec.

title: User talk:Mfashola/enes100


text:
Controversies in Science/Are humans causing global warming/a critique of Evidence for the impact of global warming on the long-term population dynamics of common birds

(Review Paper) Cited in

Points made
Temperature anomalies affect birds' reproduction rates
 * The temperature does have an effect on birds reproduction because of how many birds are able to survive during a different climate or temperature year round in a country like Canada with all four season.Bomor394 (talk) 02:29, 11 April 2012 (UTC)

Methods
Spring 2003 was an exceptionally warm spring for France, and a capture and recapture technique was used to monitor bird populations in several sites. They studied a total of 32 species of birds which were likely affected more by local factors than by any global effect.
 * This article is good at showing how abnormal temperatures affect different species of birds. The selection of birds seemed to have been studied carefully.Lfamc 02:52, 15 February 2012 (UTC)
 * This study should have had a little more clarification about the different temperatures that the birds were affected by.Bomor394 (talk) 02:34, 11 April 2012 (UTC)

Results
In general, as temperature anomalies increased, productivity of the birds increased. There was quite a bit of variation in the results, most likely due to drought conditions which had not been taken into account during the study.

 * Although this is an interesting study, it is not clear how humans are affecting global warming. And if humans are indeed causing global warming, wouldn't this be beneficial to these particular species of birds?Sjorg89 03:17, 15 February 2012 (UTC)


title: File:VHDL.Signal.1.A.20120224.pdf


text:
Volcanoes/Mount Vesuvius

Mt. Vesuvius is characterized as a "stratovolcano", or composite volcano, which means it is made up of pyroclastic material and lava. Vesuvius is the only volcano in the European mainland to have erupted within the last hundred years. Mount Vesuvius was known for its eruption in AD 79 and was named after the biologist Pliny. Since then, this volcano has erupted more than 50 times. Scientists have studied that this volcano is dormant. Its last eruption was in 1944, but is still capable of erupting. Mt. Vesuvius is located on the coast of the Bay of Naples about six miles from a short distance inland from the shore.

Formation
Mount Vesuvius is caused by the African plate sinking underneath the Eurasian plate which therefore created a convergent boundary in the Bay of Naples, Italy. Later, Mount Vesuvius reached the surface and had attached to Italy due to the deposits from the eruptions since 479 A.D. As the surface cooled with the mantle is still heated 4.6 million years ago, part of the African plate melted to magma, rises up to the surface as a result of creating a volcano.

Mt. Vesuvius is composed of a mix of layers of lava flows and volcanic ash. It lies inside Mount Somma, a summit caldera. Two tectonic plates, African and Eurasian, collided. The heat that was made melted the rocks. Because magma is less dense, it was pushed upward. It found a weak point at the Earth's surface and it broke through forming a volcano.

Appearance
Mount Vesuvius is a composite volcano because its eruptions vary often from pyroclastic material to lava flows. It can be explosive like its most famous eruption in August 24, 79 A.D. which caused ashfalls, volcanic impact, and mudflows onto Pompeii and Herculaneum, or nonexplosive like in May 1, 1855 which a lava flow invades Massa and San Sebastiano. Mount Vesuvius stands as high as 4190 feet. It's composed of a volcanic cone, named Gran Cono, inside a summit caldera, called Mount Somma.

This volcano has a humpbacked shape and a large cone surrounded by the caldera caused by the collapse of Mount Somma. The height of Vesuvius is 4,202 ft. The slopes of the mountain filled with lava flows. Many layers of lava, volcanic ash, and pumice make up the mountain. Vesuvius is at a convergent boundary where the African plate is being subducted meaning one tectonic plate is moving under another.

History and Eruptions
"For the last two million years, the region of central and southern Italy has been the scene of active volcanism, characterized by explosive eruptions. This region is known among geologists as the Roman Comagmatic Province and extends along the Tyrrhenian coast, from the Vulsini area in northern Latium to Vesuvius in the south. In north Latium the volcanic activity became largely extinct about one hundred thousand years ago, but the Romans have ever since enjoyed the beautiful volcanic landscape created by this activity, such as the scenic Lake Bolsena in the caldera of Vulsini volcano, and Lake Vico. Further south, however, in the Campanian volcanic region, activity has continued up to the present day, as recorded in legend, history, and in the volcanic stratigraphy of Vesuvius and the Phlegraean Fields."

3800 b2k
The image on the right shows the mold of one of a group of huts found partially buried, its interior, including pottery, perfectly preserved, dated to 3800 b2k.

The second image down on the right shows a human victim of the Avellino eruption.

"Lirer et al.(1973) studied the ash-and pumice-fall deposit from the Bronze Age Avellino eruption and established that fallout occurred primarily to the east and east-northeast of the volcano during this event."

3460 b2k
The image on the right shows fossil footprints in the surge ash deposit of the Avellino eruption.

79
This is 1921 b2k (1921 years before 2000).

The first eruption, in August 24, 79 A.D., is one of the most famous of all the eruptions. Greek and Roman scholars, like Strabo and Diodorus Siculus, knew Mt. Vesuvius even before it erupted. Before this historic eruption, toxic fumes covered Pompeii. The citizens of Pompeii had no idea what was happening. Ashfalls forced buildings to collapse, pumices crashed into property, and lahars wiped out its captives. Pliny the Elder, admiral of the Roman fleet, is one of Mt. Vesuvius's captives as well as many other residents in Pompeii and Herculaneum. Pliny the Younger, who is Pliny the Elder's nephew and an author, recorded this eruption in Cape Misenum, 20 km away from Mt. Vesuvius.

c. 172
This is c.1828 b2k.

In c. 172 A.D., Claudius Galenus, whose profession is a physician, stated that "the matter in it (Mt. Vesuvius) is still burning".

203
This is 1797 b2k.

40 km away, Dio Cassius, in 203 A.D. recorded an eruption in Capua.

472
This is 1528 b2k

More than 200 years later, Galenus's hypothesis is proven correct; two eruptions are reported in November 6, 472 and 512 (date unknown).

512
This is 1488 b2k.

Even though it exploded in almost 2,000 years, it is still active today.

1631
Before the eruption, people near the volcano felt some earthquakes. They weren't that scared because earthquakes were common in that town. The seismic activity was much stronger a few days before the eruption of Mt. Vesuvius. The livestock seemed restless the night before the eruption. On December 16, there were explosions, ash clouds, fissures at the base of the volcano sending down lava and toxic smoke. The next day, the summit has a huge explosion with lava flows and mud flows. This eruption caused about 4,000 people to die. They died because of the pouring, hot lava and ash.

1767
The drawing in the right image shows the eruption of Vesuvius in 1767.

1779
During the 1779 eruption of Vesuvius as depicted in the painting of the right, lightning was observed to strike at or near the column of eruption.

1793
An eruption in 1793 has been documented by the painting on the right that has Vesuvius in the background.

circa 1845
On the right is a painting by Enrico La Pira depicting the circa 1845 eruption of Vesuvius.

1872
The photograph on the right taken on 26 April 1872 has apparently been taken during the day time with the Sun at the back of the photographer. No lightning discharges have been captured in the exposure.

1888
This photograph shows a visit of Kaiser Wilhelm I while Vesuvius is undergoing a mild eruption in the background.

1895
The image shows the surroundings of Naples and the eruption of Mount Vesuvius, July 5 1895, as seen from the Observatory.

Vesuvius today
There is a Vesuvius National Park that was founded on June 5, 1995. This park was founded to preserve animals and plants. It holds about 612 plants and 227 animals. Also, it covers 135 square kilometers. Since, the volcano is dormant, the Italian government has created an evacuation plan to clear out the red zone 72 hrs. ahead of an eruption.

title: Volcanoes/Mount Vesuvius


text:
Survey research and design in psychology/Assessment/Lab reports/2

===Lab report 2===

Task
Describe and graph the correlational relationship between: (a) two variables, at least one of which has a nominal or ordinal level of measurement; and (b) two variables, neither of which have a nominal or ordinal level of measurement.

Marking criteria
In addition to the generic guidelines, this report should include:
 * 1) 5%. Title/Abstract: Provide a succint overview of the study.
 * 2) Sumarise the study's purpose, method, findings, and implications in less than 150 words.
 * 3) 10%. Introduction:  Clearly introduce the topic and the hypotheses:
 * 4) 5%. Background: Briefly introduce the theoretical concepts and related research, using relevant academic citations
 * 5) 5%. Research questions: Propose logically-derived research question(s) (which are addressed in the Results)
 * 6) 10%. Method: Clearly explain how the study was conducted
 * 7) 5%. Participants:
 * 8) Provide a solid overall picture and description of the sample
 * 9) Note: It may be appropriate to provide detailed description of specific aspects of the sample in subsequent lab reports; only a general overview is required for this lab report
 * 10) 5%. Procedure:
 * 11) Describe the research procedure, with sufficient detail to permit critique of the methodology and full replication.
 * 12) Include an electronic APA style citation to the survey administration guidelines (but do not provide these in an appendix)
 * 13) Describe your own survey administration procedures and experiences (e.g., where/when/how did you collect data, response rate, anomalies (i.e., what didn't go according to plan, what could have unwittingly affected the results))
 * 14) 50%. Results: Appropriate tests conducted and reported in APA style
 * 15) 25%. Correlation 1: Describe and present descriptive statistics and graph(s) for the the correlational relationship between at least two variables, at least one of which has a nominal or ordinal level of measurement.
 * 16) 25%. Correlation 2: Describe and present descriptive statistics and graph(s) for the the correlational relationship between at least two variables, neither of which have a nominal or ordinal level of measurement.
 * 17) 25%. Discussion: Explain results and implications:
 * 18) Explain what the results indicate about the research question(s).
 * 19) Consider the implications of what was found.
 * 20) Comment on the strengths and weaknesses of the Procedure. How could it be (practically) improved?

General feedback

 * 1) Word count
 * 2) Many reports were over the word count. However, word count penalty was applied leniently (only for word count > 1000 words from Introduction to Discussion) due to an old link about word count on the coversheet (to 2011 guidelines). The coversheet word count link has been corrected to the 2012 guidelines. Word count penalty will be applied more strictly for Lab reports 3, 4 and 5.
 * 3) There was considerable discussion about this on the Moodle discussion forum in order to clarify what had happened, including explaining why the sample HD report was over the word count but not penalised, and to address the concerns of those who had followed the word count rules and felt that it was unfair that others were not penalised (effectively others had been allowed greater word count to address the marking criteria). The convener apologised for the error and explained that the decision had been taken towards being lenient in applying the word count penalty due to the error in having links to the 2011 word count guidelines (which were more lenient) on the assignment coversheet rather than the 2012 word count guidelines.
 * 4) APA style
 * 5) Use of APA style for reporting statistical results often not used (e.g., r and p etc. in italics, leaving spaces before and after = (treat it as a word in a grammatical sentence)
 * 6) Title
 * 7) A title page was sometimes not included
 * 8) A running head was often not included (put it in the header)
 * 9) Abstract
 * 10) Often the results were not clearly summarised
 * 11) Sometimes correlational results were confused as indicating a causal relationship between variables
 * 12) Introduction
 * 13) Often too long
 * 14) Should be focused on succintly reporting on the theoretical and/or research justification for each hypothesis/research question
 * 15) Method
 * 16) Participants:
 * 17) Often a simplistic description of the sample was provided
 * 18) Procedure:
 * 19) Generally well done
 * 20) The sampling method was often not described; many incorrectly described it as random sampling
 * 21) Often the survey admin guidelines were not referenced
 * 22) Sometimes the researcher's own procedure was not described in enough detail
 * 23) Results
 * 24) Generally well done
 * 25) Sometimes the wrong correlational tests were used (e.g., Pearson product-moment correlation instead of a chi-square test of contingencies)
 * 26) Sometimes bivariate graphs were not presented, and often these were not edited into APA style
 * 27) Often the direction, strength and statistical significance of the relationship between the variables was not clearly explained
 * 28) Coefficient of determination was often not reported and discussed
 * 29) Testing normality: Shapiro-Wilks is an overly sensitive test with a large sample size. Hence in this unit it has been recommended that to use skewness and kurtosis and visual inspection of a histogram to determine whether normality assumptions are met.
 * 30) Discussion
 * 31) Generally well done
 * 32) Sometimes causality was assumed - be wary of interpreting correlations as causality
 * 33) Stronger discussions explained how methodological limitations may have affected the results in the current study; weaker discussions tended to more vague and general about study limitations
 * 34) Formatting
 * 35) Often multiple line spaces were used instead of page breaks


title: File:1css-Nhtml.jpg


text:
University of Florida/Egm4313/s12.teamboss/R4

Statement
4.1 from lecture notes [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec7c-1.djvu R4.1 Lect. 7c pgs. 19-22]

Given the general form of polynomial excitation.

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$$  \displaystyle y''+ay'+by=\sum_{j=0}^{n}d_{j}x^{j} $$     (1.1)
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 *
 * }

The particular solution that satisfies:

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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x^{j} $$     (1.2)
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 *
 * }

The first and second derivative of the particular solution that solves the original polynomial excitation equation.

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$$  \displaystyle y_{p}'(x)=\sum_{j=0}^{n-1}c_{j+1}(j+1) x^{j} $$     (1.3)
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 *
 * }

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$$  \displaystyle y_{p}''(x)=\sum_{j=0}^{n-2}c_{j+2}(j+2)(j+1) x^{j} $$     (1.4)
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 *
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The particular solutions are put into the polynomial excitation equation to give the general summation form:

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$$  \displaystyle \sum_{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum_{j=0}^{n}d_{j}x^{j} $$     (1.5)
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 *
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Obtain the equations associated with $$d_{1}\!$$, coefficients of $$x\!$$; $$d_{2}\!$$, coefficients of $$x^{2}\!$$; $$d_{n-2}\!$$, coefficients of $$x^{n-2}\!$$; $$d_{n-1}\!$$, coefficients of $$x^{n-1}\!$$; $$d_{n}\!$$, coefficients of $$x^{n}\!$$. Five total equations for coefficients.

Also set up the matrix $$\mathbf{A}\!$$ that satisfies $$\mathbf{A}\mathbf{c}=\mathbf{d}\!$$.

Solution
The given equation associated with $$d_{0}\!$$ taking $$j=0\!$$

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$$  \displaystyle 2c_{2}+ac_{1}+bc_{0}=d_{0} $$     (1.6) Taking $$j=1\!$$
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 *
 * }
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$$  \displaystyle [c_{1+2}(1+2)(1+1)+ac_{1+1}(1+1)+bc_{1}]x^{1}=d_{1}x^{^{1}} $$     (1.7)
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 *
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$$  \displaystyle [6c_{3}+2ac_{2}+bc_{1}]x=d_{1}x $$     (1.8)
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 *
 * }

The equation associated with $$d_{1}\!$$, coefficients of $$x\!$$
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$$  \displaystyle 6c_{3}+2ac_{2}+bc_{1}=d_{1} $$     (1.9) Taking $$j=2\!$$
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 *
 * }
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$$  \displaystyle [c_{2+2}(2+2)(2+1)+ac_{2+1}(2+1)+bc_{2}]x^{2}=d_{2}x^{^{2}} $$     (1.10)
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 *
 * }
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$$  \displaystyle [12c_{4}+3ac_{3}+bc_{2}]x^{2}=d_{2}x^{2} $$     (1.11)
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 *
 * }

The equation associated with $$d_{2}\!$$, coefficients of $$x^{2}\!$$
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$$  \displaystyle 12c_{4}+3ac_{3}+bc_{2}=d_{2} $$     (1.12) Taking $$j=n-2\!$$
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 *
 * }
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$$  \displaystyle [c_{(n-2)+2}(n-2+2)(n-2+1)+ac_{(n-2+1)}(n-2+1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2} $$     (1.13)
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 *
 * }
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$$  \displaystyle [c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2} $$     (1.14)
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 *
 * }

The equation associated with $$d_{n-2}\!$$, coefficients of $$x^{n-2}\!$$
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$$  \displaystyle c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}=d_{n-2} $$     (1.15) For $$n-1\!$$ (the summation term only goes to $$j=n-2\!$$)
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 *
 * }
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$$  \displaystyle ac_{n}nx^{n-1}+bc_{n-1}x^{n-1}=d_{n-1}x^{n-1} $$     (1.16)
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 *
 * }

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$$  \displaystyle ac_{n}n+bc_{(n-1)}]x^{n-1}=d_{n-1}x^{n-1} $$     (1.17)
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 *

 * }

The equation associated with $$d_{n-1}\!$$, coefficients of $$x^{n-1}\!$$
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$$  \displaystyle ac_{n}n+bc_{(n-1)}=d_{n-1} $$     (1.18) For $$n\!$$.
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 *

 * }
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$$  \displaystyle bc_{n}x^{n}=d_{n}x^{^{n}} $$     (1.19)
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 *

 * }

The equation associated with $$d_{n}\!$$, coefficients of $$x^{n}\!$$
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$$  \displaystyle bc_{n}=d_{n} $$     (1.20) Now set up the equation:
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$$  \displaystyle \mathbf{A}\mathbf{c}=\mathbf{d} $$     (1.21)
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 *

 * }

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$$  \displaystyle \begin{bmatrix} Equation for d_{0}\\ Equation for d_{1}\\ Equation for d_{2}\\ ...\\ Equation for d_{n-2}\\ Equation for d_{n-1}\\ Equation for d_{n} \end{bmatrix}\begin{bmatrix} c_{0}\\ c_{1}\\ c_{2}\\ ...\\ c_{n-2}\\ c_{n-1}\\ c_{n} \end{bmatrix}=\begin{bmatrix} d_{0}\\ d_{1}\\ d_{2}\\ ...\\ d_{n-2}\\ d_{n-1}\\ d_{n} \end{bmatrix} $$     (1.22)
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 *

 * }

Therefore the matrix $$\mathbf{A}\!$$ that satisfies the matrix equation is:
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$$  \displaystyle \mathbf{A}=\begin{bmatrix} b& a& 2&  &  &  & \\ & b&  2a&  6&  &  & \\ & &  b&  3a&  12&  & \\ ...& ...&  ...&  ...&  ...&  ...& ...\\  &  &  &  &  b&  a(n-1)& n(n-1)\\ & &  &  &  &  b& an\\ & &  &  &  &  & b \end{bmatrix} $$     (1.23)
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 *

 * }

Author
Solved and Typed By - Chris Stewart Egm4313.s12.team1.stewart (talk) -- 21:12, 11 March 2012 (UTC)

Reviewed By - --Egm4313.s12.team1.durrance (talk) 23:14, 11 March 2012 (UTC)

Statement
Consider the L2-ODE-CC with $$ \sin x \!$$ as excitation (see R4.2 Lect. 7c pgs. 26-27):

$$y''-3y'+2y = r(x) \!$$ $$r(x) = \sin x \!$$

and the initial conditions

$$y(0) = 1, y'(0) = 0 \!$$

1) Use the Taylor series for $$ \sin x \!$$ to reproduce the figure on p.7-24.

2) Let $$ y_{p,n}(x) \!$$ be the particular soln corresponding to the excitation $$ r_n(x) \!$$:

$$y''_{p,n}+ay_{p,n}'+by_{p,n}=r_n(x) \!$$

Let $$ r_n(x) \!$$be the truncated Taylor series of $$ \sin x \!$$:

$$ r_n(x):= \sum_{k=0}^{n}\frac{(-1)^kt^{2k+1}}{(2k+1)!}=t-\frac{t^3}{3!}+...+\frac{(-1)^nt^{2n+1}}{(2n+1)!} \!$$

Let $$ y_n(x)\!$$ be the overall soln for the L2-ODE-CC: $$ y''_n+ay'_n+by_n=r_n(x) \!$$ With the same initial conditions as stated above.

Find $$ y_n(x)\!$$ for n=3,5,9; plot these solns for x in the interval [0,4π].

3)Find the exact overall soln $$ y(x)\!$$, and plot it in the above figure to compare with $$ y_n(x)\!$$ for n=3,5,9.

Solution
To approximate the value of the excitation, the Taylor expansion must be found:
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$$  \displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(\hat x)}{n!}(x-\hat x)^n $$     (2.0) For the sine function, the Taylor series approximated to 13 places is:
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 *

 * }
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$$  \displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!} $$     (2.1)
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 *

 * }

Any lower approximation would include all of the terms above without any terms with a higher order than the desired order:

Plotting every order approximation of the Taylor series up to n=13 with the actual sine function produces the following: Figure 4.2-1 In order to find the overall solution for the L2-ODE-CC corresponding to the Taylor series expansion of the sine function, both the homogenous and particular solutions must be found. The homogenous equation can be found through this method:
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$$  \displaystyle y''-3y'+2y=0 $$     (2.2)
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 *

 * }
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$$  \displaystyle \lambda ^2-3\lambda+2=0 \rightarrow \lambda = 1,2 $$     (2.3)
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 *

 * }
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$$  \displaystyle y_{h,n}(x)=C_1e^x+C_2e^{2x} $$     (2.4)
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 *

 * }

Next, the excitation must be expanded to the desired n. The following shows the excitation expanded to n=3, 5, and 9:
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$$  \displaystyle r_3(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} $$     (2.5)
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$$  \displaystyle r_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!} $$     (2.6)
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$$  \displaystyle r_9(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}-\frac{x^{15}}{15!}+\frac{x^{17}}{17!}-\frac{x^{19}}{19!} $$     (2.7)
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 *

 * }

Next, the particular solution must be found. The particular solution will be of the form:
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$$  \displaystyle y_{p,n}=\sum_{i=0}^{n}k_ix^i $$     (2.8) Using the derivation discussed in R4.1, a matrix equation in the form Ak = d can be found, where k is the matrix containing the coefficients of the particular solution and d is the matrix containing the coefficients of each power of x in the expansion of the excitation. The general formula for A is:
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 *

 * }
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$$  \displaystyle \begin{bmatrix} b& a & 2 & &  & &\\ & b & 2a & 6 & & &\\ & & b & 3a & 12 & &\\ & &  & ... & ... & ... &\\  &  &  &  & b & a(n-1) & n(n-1)\\ & &  &  &  &b &an\\ & &  &  &  & &b \end{bmatrix} $$     (2.9) n=3 The matrix A is found as:
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 *

 * }
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$$  \displaystyle A = \begin{bmatrix} 2 & -3 & 2 & &  &  &  & \\  & 2 & -6 & 6 &  &  &  & \\  &  & 2 & -9 & 12 &  & &  \\  &  &  & 2 &  -12& 20 &  & \\  &  &  &  & 2 & -15 & 30 & \\  &  &  &  &  & 2 & -18 & 42\\  &  &  &  &  &  & 2 &-21 \\  &  &  &  &  &  &  & 2 \end{bmatrix} $$     (2.10) Therefore, the matrix equation is:
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$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & &  &  &  & \\  & 2 & -6 & 6 &  &  &  & \\  &  & 2 & -9 & 12 &  & &  \\  &  &  & 2 &  -12& 20 &  & \\  &  &  &  & 2 & -15 & 30 & \\  &  &  &  &  & 2 & -18 & 42\\  &  &  &  &  &  & 2 &-21 \\  &  &  &  &  &  &  & 2 \end{bmatrix} \begin{bmatrix} k_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7 \end{bmatrix}= \begin{bmatrix} 0\\ 1\\ 0\\ -\frac{1}{6}\\ 0\\ \frac{1}{120}\\ 0\\ -\frac{1}{5040} \end{bmatrix} $$     (2.11)
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$$  \displaystyle \begin{bmatrix} k_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7 \end{bmatrix}= \begin{bmatrix} -0.1992\\ -0.3984\\ -0.3984\\ -0.0990\\ -0.0078\\ -0.0031\\ -0.0010\\ -9.9206\times 10^{-5} \end{bmatrix} $$     (2.12) Therefore the particular solution is:
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$$  \displaystyle y_{p,3}(x)=-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (2.13) And the overall solution for n=3 is:
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$$  \displaystyle y_{3}(x)=C_1e^x+C_2e^{2x}-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (2.14) Using the initial conditions:
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$$  \displaystyle 1=C_1+C_2-0.1992 \& 0=C_1+2C_2-0.3984 $$     (2.15) Solving yields $$C_1=2 \& C_2=-0.8008 \!$$. Therefore the overall solution is:
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$$  \displaystyle y_{3}(x)=2e^x-0.8008e^{2x}-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (2.16) n=5 The matrix equation is: Therefore the particular solution is:
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$$  \displaystyle y_{p,5}(x)=-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$     (2.18) And the overall solution for n=5 is:
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$$  \displaystyle y_{5}(x)=C_1e^x+C_2e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$     (2.19) Using the initial conditions:
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 * }
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$$  \displaystyle 1=C_1+C_2-0.2 \& 0=C_1+2C_2-0.3999 $$     (2.20) Solving yields $$C_1=2.0001 \& C_2=-0.8001 \!$$. Therefore the overall solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle y_{5}(x)=2.0001e^x-0.8001e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2 $$     (2.21) n=9 The matrix equation is: Therefore the particular solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,9}(x)=-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (2.23) And the overall solution for n=9 is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{9}(x)=C_1e^x+C_2e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (2.24) Using the initial conditions:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 1=C_1+C_2-0.2 \& 0=C_1+2C_2-0.4 $$     (2.25) Solving yields $$C_1=2 \& C_2=-0.8 \!$$. Therefore the overall solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle y_{9}(x)=2e^x-0.8e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (2.26) The following plot shows the overall solution of the ODE at n=3,5,9 over the domain [0,4π]: Figure 4.2-2
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The near perfect overlap between all three graphs shows that the approximations converge quickly for very low values of n.

Next, the exact ODE will be calculated to find the accuracy of the above approximations. The homogeneous solution is the same as the homogeneous solution above, only with different coefficients. The particular solution will be in the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p(x)=K\cos x + M\sin x $$ (2.27)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_p'(x)=-K\sin x + M\cos x $$ (2.28)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_p''(x)=-K\cos x - M\sin x $$ (2.29) Plugging these values into the ODE:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -K\cos x - M\sin x + 3K\sin x - 3M\cos x + 2K\cos x + 2M\sin x =\sin x $$ (2.30) Separating sine and cosine terms yield two linear equations which can be used to solve for the unknown coefficients:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle K-3M=0 \& 3K+M=1 $$     (2.31) Solving these equations yield K = 0.3 and M = 0.1.
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_p(x)=0.3\cos x + 0.1\sin x $$ (2.32) Therefore the exact overall solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y(x)=C_1e^x + C_2e^{2x}+0.3\cos x + 0.1\sin x $$ (2.33)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y'(x)=C_1e^x + 2C_2e^{2x}-0.3\sin x + 0.1\cos x $$ (2.34) Using the initial conditions yield the equations:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 1=C_1 + C_2 + 0.3 \&  0=C_1 + 2C_2 + 0.1 $$     (2.35)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving these two equations yield $$C_1=1.5 \& C_2=-0.8 \!$$. Therefore the exact overall solution is:
 * {| style="width:100%" border="1"

$$  \displaystyle y(x)=1.5e^x - 0.8e^{2x}+0.3\cos x + 0.1\sin x $$ (2.36)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following figure shows the plot of the exact solution over the plot in Fig. 4.2-2: Figure 4.2-3 The overlap of the plots shows that the Taylor series approximation approach to the ODE is actually accurate to a very large degree with respect to the exact solution of the ODE.

Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 04:02, 14 March 2012 (UTC)

Reviewed By - Chris Stewart Egm4313.s12.team1.stewart (talk) 04:52, 14 March 2012 (UTC)

Statement
Consider the L2-ODE-CC (5) p7b-7 with log(1+x) as excitation:

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

y''-3y'+2y=r(x) $$     (3.0)
 * $$\displaystyle
 * $$\displaystyle
 *

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

r(x) = \log (1+x) $$     (3.1)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

And the initial conditions
 * {| style="width:100%" border="0"

y(-\frac{3}{4}) =1, y'(-\frac{3}{4}) =0 $$     (3.2)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

1. Develop log(1+x) in Taylor series, about $$\displaystyle \hat x =0$$ to reproduce the figure on p.7-25

2. Let $$\displaystyle r_{n}(x)$$ be the truncated Taylor series, with n terms--which is also the highest degree of the Taylor (power) series -- of lg(1+x).

Find $$\displaystyle y_{n}(x)$$, for n=4,7,11, such that:

with the same initial conditions (2).

Plot $$\displaystyle y_{n}(x)$$ for n = 4,7,11 for x in $$\displaystyle [-\frac{3}{4},3]$$.

3. Use the matlab command ode45 to integrate numerically (5) p.7b-7 with (1)-(2) p 7-28 o obtain the numerical soln for y(x). Plot y(x) in the same figure with $$\displaystyle y_{n}(x)$$

Solution
1. Developing the Taylor series First, we take the generic Taylor series formula:

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

f(x)=\sum_{n=0}^{\infty }\frac{f^{(n)} (\hat x)}{n!} (x- \hat x)^n $$     (3.3)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Then, log(1+x) is expanded into a power series:
 * {| style="width:100%" border="0"

log(1+x)= \frac{1}{1!} (x)^1 + \frac{-1}{2!} (x)^2 + \frac{1}{3!} (x)^3 + \frac{-1}{4!} (x)^4 ... $$     (3.4)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

2. Yn(x) at different n values First we create the characteristic equation in standard form:
 * {| style="width:100%" border="0"

$$\displaystyle{\lambda^2-3 \lambda +2=0}$$ (3.5) Then, by setting it equal to zero, we can find what $$ \lambda \!$$ equals:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle{(\lambda - 2)(\lambda -1) = 0}$$ (3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{\lambda = 2,\lambda =1}$$ (3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Given two, distinct, real roots, the homogeneous solution looks like this:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h(x)=C_1e^{2x} + C_2e^{x} $$     (3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution at n=4 is of the form:

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

$$  \displaystyle y_p (x)= A_4 x^4+ A_3 x^3 +A_2 x^2 + A_1 x + A_0 $$     (3.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

It's derivative would look like this:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p '(x)= 4A_4 x^3 + 3A_3 x^2 + 2 A_2 x + A_1 $$     (3.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And the second derivative to follow would then become:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p ''(x)= 12A_4 x^2 + 6A_3 x + 2A_2 $$     (3.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Based on the coefficients, the following system of equations exists:
 * {| style="width:100%" border="0"

$$  \displaystyle 2A_4 = \frac{-1}{24} $$     (3.12)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle - 12A_4 + 2A_3 = \frac{1}{6} $$     (3.13)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 12 A_4 - 9A_3 + 2A_2 = \frac{-1}{2} $$     (3.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6 A_3 - 6A_2 + 2A_1 = 1 $$     (3.15)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 2 A_2 - 3A_1 + 2A_0 = 0 $$     (3.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The results of this set of equations make the coefficients of A's:
 * {| style="width:100%" border="0"

$$  \displaystyle A_4 = \frac{-1}{48} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle A_3 =\frac{-1}{24} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle A_2 =\frac{-3}{16} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle A_1 =\frac{1}{16} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle A_0 =\frac{9}{32} $$ The resulting particular equation looks like this:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y_p (x)= \frac{-1}{48}x^4 + \frac{-1}{24}x^3 + \frac{-3}{16} x^2 + \frac{1}{16} x + \frac{9}{32} $$     (3.17) By adding the particular and homogeneous solutions, we get the complete solution:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{-1}{48}x^4 + \frac{-1}{24}x^3 + \frac{-3}{16} x^2 + \frac{1}{16} x + \frac{9}{32} + C_1e^{2x} + C_2e^{x} = y_{n=4}(x) $$     (3.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We consider the initial conditions by taking the first derivative of the complete solution:

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

$$  \displaystyle 4\frac{-1}{48}x^3 + 3\frac{-1}{24}x^2 + 2\frac{-3}{16} x + \frac{1}{16} + 2 C_1e^{2x} + C_2e^{x} = y' $$ (3.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
 * {| style="width:100%" border="0"

$$  \displaystyle 1= \frac{-1}{48}(-.75)^4 + \frac{-1}{24}(-.75)^3 + \frac{-3}{16} (-.75)^2 + \frac{1}{16} (-.75) + \frac{9}{32} + C_1e^{2(-.75)} + C_2e^{-.75} = y_{n=4}(-.75) $$     (3.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0= 4\frac{-1}{48}(-.75)^3 + 3\frac{-1}{24}(-.75)^2 + 2\frac{-3}{16} (-.75) + \frac{1}{16} + 2 C_1e^{2(-.75)} + C_2e^{(-.75)} = y'_{n=4}(-.75) $$     (3.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving the equations proves that $$C_1 = -4.8322, C_2 = 4.0745 \!$$: The resulting complete solution at n=4 with consideration for initial conditions then becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{-1}{48}x^4 + \frac{-1}{24}x^3 + \frac{-3}{16} x^2 + \frac{1}{16} x + \frac{9}{32} + -4.8322e^{2x} + 4.0745e^{x} = y_{n=4}(x) $$     (3.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution at n=7 is found using this matrix equation:

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

$$  \displaystyle {\begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 & 0 \\ 0 & 0 & 2 & -9 & 12 & 0 & 0 \\ 0 & 0 & 0 & 2 & -12 & 20 & 0 \\ 0 & 0 & 0 & 0 & 2 & -15 & 30 \\ 0 & 0 & 0 & 0 & 0 & 2 & -18\\ 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{bmatrix}} \cdot \begin{pmatrix} {A}_{0}\\ {A}_{1}\\ {A}_{2}\\ {A}_{3}\\ {A}_{4}\\ {A}_{5}\\ {A}_{6}
 * style="width:95%" |
 * style="width:95%" |

\end{pmatrix} =\begin{Bmatrix} 0\\ 1\\ -1/2\\ 1/6\\ -1/24\\ 1/120\\ -1/720

\end{Bmatrix} $$     (3.23)
 *

 * }

The values for A then look like this:
 * {| style="width:100%" border="0"

$$  \displaystyle A_0 =\frac{-21}{128} $$ $$  \displaystyle A_1 =\frac{-21}{64} $$
 * style="width:95%" |
 * style="width:95%" |

$$  \displaystyle A_2 =\frac{-21}{64} $$

$$  \displaystyle A_3 =\frac{-5}{96} $$

$$  \displaystyle A_4 =\frac{-5}{192} $$

$$  \displaystyle A_5 =\frac{-1}{480} $$

$$  \displaystyle A_6 =-6.944E-4 $$
 * }

The resulting particular equation looks like this:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p (x)= -6.944E-4x^7 + \frac{-1}{480}x^6 + \frac{-5}{192}x^5 + \frac{-5}{96}x^4 + \frac{-21}{64}x^3 + \frac{-21}{64} x^2 + \frac{-21}{128} x $$ (3.24) By adding the particular and homogeneous solutions, we get the complete solution:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -6.944E-4x^7 + \frac{-1}{480}x^6 + \frac{-5}{192}x^5 + \frac{-5}{96}x^4 + \frac{-21}{64}x^3 + \frac{-21}{64} x^2 + \frac{-21}{128} x + C_1e^{2x} + C_2e^{x} = y_{n=7}(x) $$     (3.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We consider the initial conditions by taking the first derivative of the complete solution:

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

$$  \displaystyle (7)-6.944E-4x^6 + (6)\frac{-1}{480}x^5 + (5)\frac{-5}{192}x^4 + (4)\frac{-5}{96}x^3 + (3)\frac{-21}{64}x^2 + (2)\frac{-21}{64} x + \frac{-21}{128} + 2 C_1e^{2x} + C_2e^{x} = y'_{n=7}(x) $$     (3.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
 * {| style="width:100%" border="0"

$$  \displaystyle 1= -6.944E-4(-.75)^7 + \frac{-1}{480}(-.75)^6 + \frac{-5}{192}(-.75)^5 + \frac{-5}{96}(-.75)^4 + \frac{-21}{64}(-.75)^3 + \frac{-21}{64} (-.75)^2 + \frac{-21}{128} (-.75) + C_1e^{2(-.75)} + C_2e^{(-.75)} = y_{n=7}(-.75) $$     (3.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=  (7)-6.944E-4(-.75)^6 + (6)\frac{-1}{480}(-.75)^5 + (5)\frac{-5}{192}(-.75)^4 + (4)\frac{-5}{96}(-.75)^3 + (3)\frac{-21}{64}(-.75)^2 + (2)\frac{-21}{64} (-.75) + \frac{-21}{128} + 2 C_1e^{2(-.75)} + C_2e^{(-.75)} = y'_{n=7}((-.75)) $$     (3.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving the equations proves that $$C_1 = -3.3921, C_2 = 3.5789 \!$$: The resulting complete solution at n=7 with consideration for initial conditions then becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle -6.944E-4x^7 + \frac{-1}{480}x^6 + \frac{-5}{192}x^5 + \frac{-5}{96}x^4 + \frac{-21}{64}x^3 + \frac{-21}{64} x^2 + \frac{-21}{128} x + (-3.3921)e^{2x} + (3.5789)e^{x} = y_{n=7}(x) $$     (3.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution at n=11 is found using this matrix equation:

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

(3.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The values for A then look like this:
 * {| style="width:100%" border="0"

$$  \displaystyle A_0 =\frac{2481}{2048} $$ $$  \displaystyle A_1 =\frac{36635}{3072} $$
 * style="width:95%" |
 * style="width:95%" |

$$  \displaystyle A_2 =\frac{5897}{1024} $$

$$  \displaystyle A_3 =\frac{8987}{4608} $$

$$  \displaystyle A_4 =\frac{1417}{3072} $$

$$  \displaystyle A_5 =\frac{415}{4608} $$

$$  \displaystyle A_6 =\frac{33}{2560} $$

$$  \displaystyle A_7 =.001434 $$

$$  \displaystyle A_8 =5.27E-5 $$

$$  \displaystyle A_9 =3.44E-6 $$

$$  \displaystyle A_{10} =1.37786E-7 $$
 * }

The resulting particular equation looks like this:

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

$$  \displaystyle y_p (x)= 1.37786E-7x^{11} + 3.44E-6x^{10} + 5.27E-5x^9 + .001434x^ 8 + \frac{33}{2560}x^7 + \frac{415}{4608}x^6 +  \frac{1417}{3072}x^5 +  \frac{8987}{4608} x^4 + \frac{5897}{1024} x^3 +  \frac{36635}{3072} x^2 + \frac{2481}{2048} x $$ (3.31) By adding the particular and homogeneous solutions, we get the complete solution:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 1.37786E-7x^{11} + 3.44E-6x^{10} + 5.27E-5x^9 + .001434x^ 8 + \frac{33}{2560}x^7 + \frac{415}{4608}x^6 +  \frac{1417}{3072}x^5 +  \frac{8987}{4608} x^4 + \frac{5897}{1024} x^3 +  \frac{36635}{3072} x^2 + \frac{2481}{2048} x  + C_1e^{2x} + C_2e^{x} = y_{n=11}(x) $$     (3.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We consider the initial conditions by taking the first derivative of the complete solution:

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

$$  \displaystyle (11)1.37786E-7x^{10} + (10)3.44E-6x^9 + (9)5.27E-5x^8 + (8).001434x^ 7 + (7)\frac{33}{2560}x^6 + (6)\frac{415}{4608}x^5 +  (5)\frac{1417}{3072}x^4 +  (4)\frac{8987}{4608} x^3 + (3)\frac{5897}{1024} x^2 +  (2)\frac{36635}{3072} x + \frac{2481}{2048}  + C_1e^{2x} + C_2e^{x} = y'_{n=11}(x) $$     (3.33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
 * {| style="width:100%" border="0"

$$  \displaystyle 1= 1.37786E-7(-.75)^{11} + 3.44E-6(-.75)^{10} + 5.27E-5(-.75)^9 + .001434(-.75)^ 8 + \frac{33}{2560}(-.75)^7 + \frac{415}{4608}(-.75)^6 +  \frac{1417}{3072}(-.75)^5 +  \frac{8987}{4608} (-.75)^4 + \frac{5897}{1024} (-.75)^3 +  \frac{36635}{3072} (-.75)^2 + \frac{2481}{2048} (-.75)  + C_1e^{2(-.75)} + C_2e^{(-.75)} = y_{n=11}(-.75) $$     (3.34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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$$  \displaystyle 0=  (11)1.37786E-7(-.75)^{10} + (10)3.44E-6(-.75)^9 +  (9)5.27E-5(-.75)^8 + (8).001434(-.75)^ 7 + (7)\frac{33}{2560}(-.75)^6 + (6)\frac{415}{4608}(-.75)^5 +  (5)\frac{1417}{3072}(-.75)^4 +  (4)\frac{8987}{4608} (-.75)^3 + (3)\frac{5897}{1024} (-.75)^2 +  (2)\frac{36635}{3072} (-.75) + \frac{2481}{2048}  + C_1e^{2(-.75)} + C_2e^{(-.75)} = y'_{n=11}(-.75) $$     (3.35)
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Solving the equations proves that $$C_1 = 56.1374, C_2 = -32.64 \!$$: The resulting complete solution at n=11 with consideration for initial conditions then becomes:
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$$  \displaystyle 1.37786E-7x^{11} + 3.44E-6x^{10} + 5.27E-5x^9 + .001434x^ 8 + \frac{33}{2560}x^7 + \frac{415}{4608}x^6 +  \frac{1417}{3072}x^5 +  \frac{8987}{4608} x^4 + \frac{5897}{1024} x^3 +  \frac{36635}{3072} x^2 + \frac{2481}{2048} x  + 56.1374e^{2x} + -32.64e^{x} = y_{n=11}(x) $$     (3.36)
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'''Plot of all n=4 (red), n=7 (blue), n=11 (green). '''

3. Plotting the actual y(x) against approximations

Author
Solved and Typed By - Egm4313.s12.team1.silvestri (talk) 23:17, 12 March 2012 (UTC) Reviewed By ---Egm4313.s12.team1.rosenberg (talk) 03:10, 13 March 2012 (UTC)

Problem R4.4
4.4 from lecture notes [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec7c-1.djvu R4.1 Lect. 7c pgs. 29-30]

Extend the accuracy of the solution beyond $$ \widehat{x} = 1\!$$.

Part 1 Statement
1. Back up away a little from the brink of non-convergence at $$ x = 1\!$$ for the Taylor series of $$ log(1+x)\!$$ about $$ \widehat{x} = 0\!$$, and consider the point $$ x_1 = 0.9\!$$.

Find the value of $$ y_n(x_1), y'_n(x_1)\!$$ that will serve as initial conditions for the next iteration to extend the domain of accuracy of the analytical solution. Find $$n\!$$ sufficiently high so that $$ y_n(x_1), y'_n(x_1)\!$$ do not differ from the numerical solution by more than $$ 10^{-5}\!$$.

Part 2 Statement
2. Develop $$ log(1+x)\!$$ in Taylor series about $$ \widehat{x} = 1\!$$ for $$ n = 4, 7, 11\!$$, and plot these truncated series vs the exact function.

What is now the domain of convergence? (by observation of your results.)

Part 2 Solution
First, we have the general form of a Taylor series:

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f(x)=\sum_{n=0}^{\infty }\frac{f^{(n)} (\hat x)}{n!} (x- \hat x)^n $$     (4.2.0)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Next, we set $$ f(x) = r(x) = log(1+x)\!$$ and take several derivatives of $$ f(x)\!$$:

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f(x)=log(1+x) \rightarrow f'(x) = \frac{1}{(x+1)}, f(x) = \frac{-1}{(x+1)^2}, f'(x) = \frac{2}{(x+1)^3}... $$
 * $$\displaystyle
 * $$\displaystyle
 * }

Using these derivatives, we can now develop $$ log(1+x)\!$$ in Taylor series about $$ \hat x = 1\!$$:

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r(x) = log(1+x) = f(x)=\sum_{n=0}^{\infty }\frac{f^{(n)} (1)}{n!} (x- 1)^n = \frac{1}{0!}(x-1)^0 + \frac{1}{1!}(x-1)^1 - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{6}{4!}(x-1)^4 + ... $$     (4.2.1)
 * $$\displaystyle
 * $$\displaystyle
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 * }

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r(x) = log(1+x) = f(x)=\sum_{n=0}^{\infty }\frac{(-1)^{n-1}f^{(n)}(x-1)^n}{n!} $$     (4.2.2)
 * $$\displaystyle
 * $$\displaystyle
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 * }

Now the Taylor series for $$ n = 4, 7, 11\!$$ can be developed:

For $$ n = 4\!$$:

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$$  \displaystyle f(x)=\sum_{n=0}^{\infty }\frac{(-1)^{3}f^{(4)}(x-1)^4}{4!} = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{6}(x-1)^3 - \frac{1}{24}(x-1)^4 $$     (4.2.3)
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For $$ n = 7\!$$:

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$$  \displaystyle f(x)=\sum_{n=0}^{\infty }\frac{(-1)^{6}f^{(7)}(x-1)^7}{7!} = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{6}(x-1)^3 - \frac{1}{24}(x-1)^4 + \frac{1}{120}(x-1)^5 - \frac{1}{720}(x-1)^6 + \frac{1}{5040}(x-1)^7 $$     (4.2.4)
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For $$ n = 11\!$$:

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$$  \displaystyle f(x)=\sum_{n=0}^{\infty }\frac{(-1)^{10}f^{(11)}(x-1)^{11}}{11!} = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{6}(x-1)^3 - \frac{1}{24}(x-1)^4 + \frac{1}{120}(x-1)^5 - \frac{1}{720}(x-1)^6 + \frac{1}{5040}(x-1)^7 - \frac{1}{40320}(x-1)^8 + \frac{1}{362880}(x-1)^9 - \frac{1}{3628800}(x-1)^{10} + \frac{1}{39916800}(x-1)^{11} $$     (4.2.5)
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Part 3 Statement
3. Find $$ y_n(x)\!$$, for $$ n = 4, 7, 11\!$$, such that: $$ y''_n + ay'_n + by_n = r_n(x)\!$$ for $$ x\!$$ in $$ [0.9,3]\!$$ with the initial conditions found i.e., $$ y_n(x_1), y'_n(x_1)\!$$.

Plot $$ y_n(x)\!$$ for $$ n = 4, 7, 11\!$$ for $$ x\!$$ in $$ [0.9,3]\!$$.

Part 3 Solution
Using the same characteristic in problem 4.3 (3.5), we see that the homogenous solution is the same (3.8):

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$$  \displaystyle y_h(x)=C_1e^{2x} + C_2e^{x} $$     (4.3.0)
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The particular solution at n=4 is of the form:

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$$  \displaystyle y_p (x)= A_4 (x-1)^4+ A_3 (x-1)^3 +A_2 (x-1)^2 + A_1 (x-1) + A_0 $$     (4.3.1)
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It's derivative would look like this:
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$$  \displaystyle y_p '(x)= 4A_4 (x-1)^3 + 3A_3 (x-1)^2 + 2 A_2 (x-1) + A_1 $$     (4.3.2)
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And the second derivative to follow would then become:
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$$  \displaystyle y_p ''(x)= 12A_4 (x-1)^2 + 6A_3 (x-1) + 2A_2 $$     (4.3.3)
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Based on the coefficients, the following system of equations exists:
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$$  \displaystyle 2A_4 = \frac{-1}{4} $$     (4.3.4)
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$$  \displaystyle - 12A_4 + 2A_3 = \frac{1}{3} $$     (4.3.5)
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$$  \displaystyle 12 A_4 - 9A_3 + 2A_2 = \frac{-1}{2} $$     (4.3.6)
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$$  \displaystyle 6 A_3 - 6A_2 + 2A_1 = 1 $$     (4.3.7)
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$$  \displaystyle 2 A_2 - 3A_1 + 2A_0 = 0 $$     (4.3.8)
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The results of this set of equations make the coefficients of A's:
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$$  \displaystyle A_4 = \frac{-1}{8} $$
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$$  \displaystyle A_3 =\frac{-7}{12} $$
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$$  \displaystyle A_2 =\frac{-17}{8} $$
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$$  \displaystyle A_1 =\frac{-33}{8} $$
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$$  \displaystyle A_0 =\frac{-65}{16} $$ The resulting particular equation looks like this:
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$$  \displaystyle y_p (x)= \frac{-1}{8}(x-1)^4 + \frac{-7}{12}(x-1)^3 + \frac{-17}{8}(x-1)^2 + \frac{-33}{8}(x-1) + \frac{-65}{16} $$     (4.3.9) By adding the particular and homogeneous solutions, we get the complete solution:
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$$  \displaystyle y_{n=4}(x) = C_1e^{2x} + C_2e^{x} + \frac{-1}{8}(x-1)^4 + \frac{-7}{12}(x-1)^3 + \frac{-17}{8}(x-1)^2 + \frac{-33}{8}(x-1) + \frac{-65}{16} $$     (4.3.10)
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The particular solution at n=7 is found using this matrix equation:

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$$  \displaystyle {\begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 & 0 \\ 0 & 0 & 2 & -9 & 12 & 0 & 0 \\ 0 & 0 & 0 & 2 & -12 & 20 & 0 \\ 0 & 0 & 0 & 0 & 2 & -15 & 30 \\ 0 & 0 & 0 & 0 & 0 & 2 & -18\\ 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{bmatrix}} \cdot \begin{pmatrix} {A}_{0}\\ {A}_{1}\\ {A}_{2}\\ {A}_{3}\\ {A}_{4}\\ {A}_{5}\\ {A}_{6}
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\end{pmatrix} =\begin{Bmatrix} 0\\ 1\\ -1/2\\ 1/3\\ -1/4\\ 1/5\\ -1/6

\end{Bmatrix} $$     (4.3.11)
 *

 * }

The values for A then look like this:
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$$  \displaystyle A_0 =\frac{-99}{2} $$
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$$  \displaystyle A_1 =-99 $$

$$  \displaystyle A_2 =-49 $$

$$  \displaystyle A_3 =\frac{-95}{6} $$

$$  \displaystyle A_4 =\frac{-15}{4} $$

$$  \displaystyle A_5 =\frac{-13}{20} $$

$$  \displaystyle A_6 =\frac{-1}{12} $$
 * }

The resulting particular equation looks like this:
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$$  \displaystyle y_p (x)= \frac{-1}{12}(x-1)^7 + \frac{-13}{20}(x-1)^6 + \frac{-15}{4}(x-1)^5 + \frac{-95}{6}(x-1)^4 + -49(x-1)^3 + -99(x-1)^2 + \frac{-199}{2}(x-1) $$     (4.3.12) By adding the particular and homogeneous solutions, we get the complete solution:
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$$  \displaystyle y_{n=7}(x) = C_1e^{2x} + C_2e^{x} + \frac{-1}{12}(x-1)^7 + \frac{-13}{20}(x-1)^6 + \frac{-15}{4}(x-1)^5 + \frac{-95}{6}(x-1)^4 + -49(x-1)^3 + -99(x-1)^2 + \frac{-199}{2}(x-1) $$     (4.3.13)
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The particular solution at n=11 is found using this matrix equation:

Part 4 Statement
4. Use the matlab command ode45 to integrate numerically $$ y'' - 3y' + 2y = r(x)\!$$ with $$ r(x) = log(1+x)\!$$ and the initial conditions $$ y_n(x_1), y'_n(x_1)\!$$ to obtain the numerical solution for $$ y(x)\!$$.

Plot $$ y(x)\!$$ in the same figure with $$ y_n(x)\!$$.

Author
Solved and Typed By - --Egm4313.s12.team1.wyattling (talk) 19:44, 14 March 2012 (UTC)

Reviewed By - Egm4313.s12.team1.essenwein (talk) 19:52, 14 March 2012 (UTC)