text:
User talk:Gpdimonderose

 Hello Gpdimonderose, 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! --Mathonius (talk) 17:33, 6 April 2012 (UTC)

Your editing experiments
You can continue to participate at Wikiversity and keep other community members from reverting or removing your edits as vandalism by conducting your editing experiments in the sandbox, and in your own user space when you login or create an account. You can contact me or the Wikiversity community with any questions you may have. Thank you. --Mathonius (talk) 17:35, 6 April 2012 (UTC)

title: User talk:Gpdimonderose


text:
User:Blibrestez55

 Collection:Burundi_Librestez_Portal_Interdisciplinary_Research_Studies-Wikiversity) by Duchess B. Breinstein is licensed under a Creative Commons Attribution 3.0 Unported License. Based on a work at http://en.wikiversity.org. Permissions beyond the scope of this license may be available at http://creativecommons.org/licenses/by-sa/3.0/.

Burundi_Librestez Business Statistics....

Category:References

http://en.wikiversity.org/w/index.php?title=User:Blibrestez55&action=edit# – — … ° ≈ ± − × ÷ ← → · § Author:Duchess B. Breinstein 108.197.53.95 15:31, 28 October 2012 (UTC)

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

title: User:Blibrestez55


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User talk:Blibrestez55

Burundi_Librestez Research Journals

Hi, welcome to Wikiversity!

I noticed you had added a number of categories to your profile. Note that using Category:Category adds you to that category rather than simply showing your interest in the category. By using the format :Category:Category, you can have a link to the category without appearing in the category yourself.

Category:Research journals

– — … ° ≈ ± − × ÷ ← → · §Blibrestez55 (discuss • contribs) 22:50, 19 February 2013 (UTC)

Blibrestez55 (discuss • contribs) 22:46, 19 February 2013 (UTC)

Author:Duchess B. Breinstein 108.197.53.95 18:07, 28 October 2012 (UTC)

See http://meta.wikimedia.org/wiki/Help:Wikitext_examples for more information.

CuriousOne (talk) 15:56, 28 October 2012 (UTC)

Category Display
Hi! I just wanted to let you know that I adjusted your category membership to category display instead. Your user page was showing up as a an available Wikiversity course. Please let me know if you have any questions. -- Dave Braunschweig (discuss • contribs) 16:17, 20 July 2013 (UTC)

Book -> Collection
We don't have a Book namespace here at Wikiversity. The corresponding namespace is Collection. Your collection was moved to Collection:Burundi Librestez Intelligences. -- Dave Braunschweig (discuss • contribs) 04:04, 24 August 2014 (UTC)

title: User talk:Blibrestez55


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User:Egm4313.s12.team12/R6

The list for the level 1 and level 2 problems are here. All of the level 1 problems have to be done. Just to note: 6.2 has two components, two problems from the book and then a problem that he lists in the Section 9 notes. The two problems in the book are just finding if the functions listed are even or odd, so it is not complicated.

Level 2
Note: These problems are only extra credit and are submitted separately from the regular report.

title: User:Egm4313.s12.team12/R6


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User talk:Cliff

 Hello Cliff, 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! Jtneill - Talk - c 23:01, 6 April 2012 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

User rename
Thanks for your prompt. FYI, user name done. Sincerely, James. -- Jtneill - Talk - c 23:01, 6 April 2012 (UTC)

title: User talk:Cliff


text:
User:Violettetakesmanhattan

♥♥ Violette Non Inferieur ♥♥
Student, New Zealand.

Interests
History, Synthetic Biology, Science, Mathematics.

title: User:Violettetakesmanhattan


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User talk:Violettetakesmanhattan

Hello world! --Violettetakesmanhattan (talk) 23:05, 6 April 2012 (UTC)

title: User talk:Violettetakesmanhattan


text:
Genetics/12p13.31

Human chromosome region 12p13.31 region contains six genes: CD69, CLEC2D, CLECL1, KLRB1, LOC374443, OVOS. Deletions of this region of the human chromosome have been linked to multiple myeloma.

title: Genetics/12p13.31


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Genetics/CD69

CD69 (Cluster of Differentiation 69) is a human transmembrane C-Type lectin protein encoded by the CD69 gene. It is one of six genes located in human chromosome region 12p13.31. CD69 is expressed on activated T-cells and B-cells. Its induction stimulates TGF-β1 and TGF-β2 synthesis and down regulates immune response.

title: Genetics/CD69


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User talk:Prof. Anderson Guimarães

 Hello Prof. Anderson Guimarães, 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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 * Take a guided tour and learn to edit.
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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! -- Ruy Pugliesi ◥ 00:39, 7 April 2012 (UTC)

Pêcheux
You can continue to participate at Wikiversity and keep other community members from reverting or removing your edits as vandalism by conducting your editing experiments in the sandbox, and in your own user space when you login or create an account. You can contact me or the Wikiversity community with any questions you may have. Thank you. -- Ruy Pugliesi ◥ 00:39, 7 April 2012 (UTC)

title: User talk:Prof. Anderson Guimarães


text:
Lectin

Lectins are sugar-binding proteins (not to be confused with glycoproteins, which are proteins containing sugar chains or residues) that are highly specific for their sugar moieties. They play a role in biological recognition phenomena involving cells and proteins. For example, some viruses use lectins to attach themselves to the cells of the host organism during infection. Lectins may be disabled by specific mono- and oligosaccharides, which bind to them and prevent their attachment to cell membranes.

Etymology
The name "lectin" is derived from the Latin word legere, meaning, among other things, "to select".(lek'tin)

History
Although they were first discovered more than 100 years ago in plants, they are now known to be present throughout nature. It is generally believed that the earliest description of a lectin was given by Peter Hermann Stillmark in his doctoral thesis presented in 1888 to the University of Dorpat. Stillmark isolated ricin, an extremely toxic hemagglutinin, from seeds of the castor plant (Ricinus communis) The first lectin to be purified on a large scale and available on a commercial basis was concanavalin A, which is now the most-used lectin for characterization and purification of sugar-containing molecules and cellular structures. The legume lectins are probably the most well-studied lectins.

Biological functions
Most lectins are non-enzymatic in action and non-immune in origin. Lectins occur ubiquitously in nature. They may bind to a soluble carbohydrate or to a carbohydrate moiety that is a part of a glycoprotein or glycolipid. They typically agglutinate certain animal cells and/or precipitate glycoconjugates.

Functions in animals
Lectins serve many different biological functions in animals, from the regulation of cell adhesion to glycoprotein synthesis and the control of protein levels in the blood. They may also bind soluble extracellular and intercellular glycoproteins. Some lectins are found on the surface of mammalian liver cells that specifically recognize galactose residues. It is believed that these cell-surface receptors are responsible for the removal of certain glycoproteins from the circulatory system. Another lectin is a receptor that recognizes hydrolytic enzymes containing mannose-6-phosphate, and targets these proteins for delivery to the lysosomes. I-cell disease is one type of defect in this particular system. Lectins are also known to play important roles in the immune system by recognizing carbohydrates that are found exclusively on pathogens, or that are inaccessible on host cells. Examples are the lectin complement activation pathway and mannose-binding lectin.

Functions in plants
The function of lectins in plants (legume lectin) is still uncertain. Once thought to be necessary for rhizobia binding, this proposed function was ruled out through lectin-knockout transgene studies.

The large concentration of lectins in plant seeds decreases with growth, and suggests a role in plant germination and perhaps in the seed's survival itself. The binding of glycoproteins on the surface of parasitic cells is also believed to be a function. Several plant lectins have been found to recognise non-carbohydrate ligands that are primarily hydrophobic in nature, including adenine, auxins, cytokinin, and indole acetic acid, as well as water-soluble porphyrins. It has been suggested that these interactions may be physiologically relevant, since some of these molecules function as phytohormones. are another major family of protein ANCs, which are specific sugar-binding proteins exhibiting reversible carbohydrate-binding activities. Lectins are similar to antibodies in their ability to agglutinate red blood cells; however, lectins are not the product of immune system. The toxicity of lectins has been identified by consumption of food with high content of lectins, which can lead to diarrhea, nausea, bloating, vomiting, even death (as from ricin). Many legume seeds have been proven to contain high lectin activity, termed as hemagglutinating activity. Soybean is the most important grain legume crop, the seeds of which contain high activity of soybean lectins (soybean agglutinin or SBA). SBA is able to disrupt small intestinal metabolism and damage small intestinal villi via the ability of lectins to bind with brush border surfaces in the distal part of small intestine. Heat processing can reduce the toxicity of lectins, but low temperature or insufficient cooking may not completely eliminate their toxicity, as some plant lectins are resistant to heat. (It is believed that undercooking red kidney beans increases toxicity.) In addition, lectins can result in irritation and over-secretion of mucus in the intestines, causing impaired absorptive capacity of the intestinal wall.

Use in medicine and medical research
Purified lectins are important in a clinical setting because they are used for blood typing. Some of the glycolipids and glycoproteins on an individual's red blood cells can be identified by lectins. In neuroscience, the anterograde labeling method is used to trace the path of efferent axons with PHA-L, a lectin from the kidney bean.
 * A lectin from Dolichos biflorus is used to identify cells that belong to the A1 blood group.
 * A lectin from Ulex europaeus is used to identify the H blood group antigen.
 * A lectin from Vicia graminea is used to identify the N blood group antigen.
 * A lectin from " Coconut milk" is used to identify Theros antigen.
 * A lectin from " Dorex is used to identify R antigen.

A lectin (BanLec) from bananas inhibits HIV-1 in vitro

Use in studying carbohydrate recognition by proteins
Lectins from legume plants, such as PHA or concanavalin A, have been widely used as model systems to understand the molecular basis of how proteins recognize carbohydrates, because they are relatively easy to obtain and have a wide variety of sugar specificities. The many crystal structures of legume lectins have led to a detailed insight of the atomic interactions between carbohydrates and proteins.

Use as a biochemical tool
Concanavalin A and other commercially available lectins have been widely used in affinity chromatography for purifying glyco proteins.

In general, proteins may be characterized with respect to glycoforms and carbohydrate structure by means of affinity chromatography, blotting, affinity electrophoresis and affinity immunoelectrophoreis with lectins as well as in microarrays as in evanescent-field fluorescence-assisted lectin microarray.

Use in biochemical warfare
One example of the powerful biological attributes of lectins is the biochemical warfare agent ricin. The protein ricin is isolated from seeds of the castor oil plant and comprises two protein domains. Abrin from the jequirity pea is similar:
 * One domain is a lectin that binds cell surface galactosyl residues and enables the protein to enter cells
 * The second domain is an N-glycosidase that cleaves nucleobases from ribosomal RNA, resulting in inhibition of protein synthesis and cell death.

Digestion and immune distress
Foods with high concentrations of lectins, such as beans, cereal grains, seeds, and nuts, may be harmful if consumed in excess in uncooked or improperly cooked form. Adverse effects may include nutritional deficiencies, and immune (allergic) reactions. Possibly, most effects of lectins are due to gastrointestinal distress through interaction of the lectins with the gut epithelial cells. A recent in vitro study has suggested that the mechanism of lectin damage may occur by interfering with the repair of already-damaged epithelial cells.

For various reasons foods have been genetically modified with lectins, such as potatoes containing the GNA gene from the snowdrop plant. In 1998 Árpád Pusztai said in an interview on a World in Action programme that his group had observed damage to the intestines and immune systems of rats fed the genetically modified potatoes. He also said "If I had the choice I would certainly not eat it", and that "I find it's very unfair to use our fellow citizens as guinea pigs". These remarks started the Pusztai affair.

Lectin and Leptin Resistance
Lectin may cause leptin resistance, affecting its functions (signal have high levels of leptin and several effects gathering to protect from lipid overload), as indicated by studies on effects of single nucleotide polymorphisms on the function of leptin and the leptin receptor. Such leptin resistance may translate into diseases, notably it could be responsible for obesity in humans who have high levels of leptin.

title: Lectin


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User talk:Hr.hanafi

 Hello Hr.hanafi, 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! -- Ruy Pugliesi ◥ 00:50, 7 April 2012 (UTC)

You worked on the "Introduction to Computers" course.
I noticed the course, "Introduction to Computers", featured in the Wikiversity. I am so glad that someone has begun to encourage others to study this important subject. As advertised on the "Featured Articles" page, you have collected a good variety of pictures to illustrate many topics.

I really enjoyed reading through the course. :-) However, I have two concerns: (1) I suspect that you are not a native speaker of American English. This is not necessarily a problem. The Wiki will benefit greatly by addressing a truly international audience. (2) I would like to see a better discussion of the important distinctions between "data" and "information".

I don't know yet if I can be of any help. I'm a new contributor to the Wikiversity myself. Ray Calvin Baker (talk) 23:07, 18 May 2012 (UTC)

Dear Ray Calvin Baker

Receiving your message doubled my enthusiasm and interests in doing my job of putting information on Wikipedia. I express my heartful thanks to you for your comments. You are right I am not a native English speaker, even so very interested in putting information on the net. My full name is Hamid Reza Hanafi from Iran. I am 44 years old and a high school teacher. I have studied Electronic engineering getting a bachelor a couple of years ago. I teach computer science in high school. My big concern is to write in English, so it would be a very great favor of yours if you could at all help me in correcting the English texts I put on the net now and then.By the way, a friend of mine has helped me with writing the present comment.Regards

Hr.hanafi 03:23, 22 May 2012 (UTC)

Distinction between "Data" and "Information"?

http://commons.wikimedia.org/wiki/File:Data_and_information.png

http://en.wikiversity.org/wiki/Data_Codeing_and_Information_Decodeing

Hr.hanafi (talk) 09:16, 23 May 2012 (UTC)

Copyright problems on photos
Hello. There is a problem with one or more files you have uploaded. Every file you upload must have a clear statement of the copyright position. If you took the photos yourself, add PD-self if you are placing it unreservedly in the public domain. If you wish to retain minimal rights, add Cc-by-sa-3.0. If you did not take the photo, please add an explanation of why it can be hosted here. I am afraid that if you do not do this, the file wil be deleted. Let me know if you have any problems.--Collingwood (talk) 20:04, 6 July 2012 (UTC)

Looking for help
Hi there, I saw your activity on some computer science topics here on wikiversity. I have been trying to fix up the main computer science department page (http://en.wikiversity.org/wiki/Computer_Science) But nobody is there. The pages (including electrical engineering) need a lot of help - if you have some free time I would love some coworkers! Azotochtli (talk) 13:28, 8 August 2012 (UTC)

IC3
Hamid, thanks for your comment regarding Introduction to Computers. I've started to work on a similar project. I'm not sure if you've heard of IC3. It's a minimum level computer certification adopted as an international standard. It is comprised of three tests, as indicated in the article.

I teach at a community college (2 year college) in the US. We have high school students in our area who take a computer class in high school and then take the IC3 certification exam. If they pass the exam, we will award proficiency credit at our college for the introduction to computers course (CIS 100).

What I would like to do is develop a set of open materials that anyone can use to study for the IC3 certification exams. It provides a good combination of open education and industry certification to validate that learning. If you're interested, please join me.

Also, if you put any computer content together and want me to take a look and edit for grammar, just let me know. I regularly grade assignments from students for whom English is a second language, so I'm used to the sentence structure.

-- Dave Braunschweig (discuss • contribs) 02:30, 5 February 2013 (UTC)

Dear Dave Braunschweig

My full name is Hamid Reza Hanafi from Iran and teach computer science in high school.I am very interested in putting computer content on the internet for example:data. I am not a native English speaker then my big concern is to write in English.I will glad help me.Regards--Hr.hanafi (discuss • contribs) 05:22, 5 February 2013 (UTC)

 * Hamid, I've added a navigation sidebar to most of the IC3 pages. I left Computing Fundamentals alone for now because I know you like the round boxes, and because the Introduction to Computers course was highly rated based on its design.  Please take a look at some of the other IC3 pages and let me know what you think.  Is there a better way we can have both round boxes and full course navigation?  To see where I've used this successfully in the past, take a look at Internet protocol analysis.  Thanks!  -- Dave Braunschweig (discuss • contribs) 01:30, 7 February 2013 (UTC)

Mr Dave Braunschweig, It's very good.I agree. Thanks! --Hr.hanafi (discuss • contribs) 09:12, 7 February 2013 (UTC)

Hi
Hi Hamid, Just to introduce myself, I am based in London and I am currently doing an e-learning course in e-learning! I am interested in collaborative working on developing the IC3 pages. I have already had some interactions with Dave Braunschweig. I think we will all work well together. Leutha (discuss • contribs) 14:27, 8 February 2013 (UTC)

Data vs. Information
I noticed your edit at IC3/Computer Hardware on data vs. information. While I agree that data is probably the better term here, the objectives and skills come directly from the IC3 guidelines at http://www.certiport.com/portal/common/documentlibrary/IC3_GS4_Computing_Fundamentals.pdf and their word is information. I think in this case it's better for us to use the certifying body's terminology, as that is what students are likely to see on the exam.

By the way, my wife has been working with students using the IC3 materials, and they have been successfully completing the certification and are finding the materials helpful.

Dave Braunschweig (discuss • contribs) 00:37, 19 November 2013 (UTC)

Files Missing Information
Thanks for uploading files to Wikiversity. All files must have source and license information to stay at Wikiversity. The following files are missing and/or License tags, and will be deleted if the missing information is not added. See Uploading files for more information.

 * File:MR900440379.JPG
 * File:MR900440424.JPG

MaintenanceBot (discuss • contribs) 15:25, 16 April 2016 (UTC)

title: User talk:Hr.hanafi


text:
Gene transcriptions/TATA binding proteins/Associated factors

When there is no TATA box nucleotide sequence in the gene core promoter region of the DNA next to a gene, say A1BG of the human genome, a TATA binding protein associated factor (TAF) will bind sequence specifically and force the TATA box binding protein to bind non-sequence specifically to the DNA in the core promoter.

Notations
Notation: let the symbol TAF stand for TATA binding protein associated factor. Notation: let the symbol TBP stand for TATA binding protein.

Genetics
Genetics involves the expression, transmission, and variation of inherited characteristics.

Def. a "branch of biology that deals with the transmission and variation of inherited characteristics, in particular chromosomes and DNA" is called genetics.

Gene transcriptions
DNA is a double helix of interlinked nucleotides surrounded by an epigenome. On the basis of biochemical signals, an enzyme, specifically a ribonucleic acid (RNA) polymerase, is chemically bonded to one of the strands (the template strand) of this double helix. The polymerase, once phosphorylated, begins to catalyze the formation of RNA using the template strand. Although the catalysis may have more than one beginning nucleotide (a start site) and more than one ending nucleotide (a stop site) along the DNA, each nucleotide sequence catalyzed that ultimately produces approximately the same RNA is part of a gene. The catalysis of each RNA representation from the template DNA is a transcription, specifically a gene transcription. The overall process is also referred to as gene transcription.

Theoretical TBP associated factors
Here's a theoretical definition:

Def. any factor associating with the TATA binding protein (TBP) is called a TBP associated factor (TAF).

TATA boxes
A TATA box is a common type of core promoter sequence in eukaryotes which is a short DNA sequence.

The TATA box (also called Goldberg-Hogness box) is a DNA sequence (cis-regulatory element) found in the promoter region of genes in archaea and eukaryotes; approximately 24% of human genes contain a TATA box within the core promoter.

The TATA box is a binding site of either general transcription factors or histones.

In the direction of transcription along the DNA strand, the TATA box has the core DNA sequence [3'-TATAAA-5'] or a variant, which is usually followed by three or more adenine [(A)] bases, specifically [3'-TATAAA(A)AAA-5' on the template strand].

"[M]ost of the diversity within metazoan core promoters appears to involve the variable occurrence of consensus or near-consensus TATA, Inr, and DPE elements."

The TATA box can be an AT-rich sequence "located at a fixed distance upstream of the transcription start site".

TATA binding proteins
The TATA-binding protein (TBP) is a general transcription factor that binds specifically to a DNA sequence called the TATA box. This DNA sequence is found about 30 base pairs upstream of the transcription start site in some eukaryotic gene promoters. TBP, along with a variety of TBP-associated factors, make up the TFIID, a general transcription factor that in turn makes up part of the RNA polymerase II preinitiation complex. As one of the few proteins in the preinitiation complex that binds DNA in a sequence-specific manner, it helps position RNA polymerase II over the transcription start site of the gene.

Initiator elements
Notation: let the symbol Inr denote an initiator element.

Notation: let the symbol +1 designate the nucleotide that is the transcription start site (TSS).

Most human genes lack a TATA box and use an Inr or downstream promoter element instead. As in other metazoans, for genes lacking a TATA box, the Inr is functionally analogous, with a base pair (bp) consensus 5'-YYA+1NWYY-3', to direct transcription initiation. On the template strand (used as a template for RNA synthesis), the consensus sequence is 3'-YYA+1NWYY-5'.

The Inr is the only element in metazoan protein-encoding genes known to be a functional analog of the TATA box, in that it is sufficient for directing accurate transcription initiation in genes that lack TATA boxes. An Inr for mammalian RNA polymerase II can be defined as a DNA sequence element that overlaps a TSS and is sufficient for
 * 1) determining the start site location in a promoter that lacks a TATA box and
 * 2) enhancing the strength of a promoter that contains a TATA box.

"Although any isolated TAF may not exhibit sequence-specific interactions at the Inr element in the absence of a TATA-box, a combination of TAFs may bind sequence specifically to the Inr element regardless of the TATA-box and/or DPE (Chalkley and Verrijzer, 1999)." Bold added.

TAF1
GeneID: 6872, TAF1 "binds to core promoter sequences encompassing the transcription start site. It also binds to activators and other transcriptional regulators, and these interactions affect the rate of transcription initiation." TAF1 "is part of a complex transcriptional unit (TAF1/DYT3)".

"Yeast TAF1 can be divided into four regions including a putative histone acetyltransferase domain and TBP, TAF, and promoter binding domains."

"TAF1 [has] been systematically dissected into ... functional domains: an N-terminal TBP-binding domain termed TAND, a TAF-TAF interaction domain, a putative histone acetyltransferase (HAT) domain, ... a promoter recognition domain [(PB1), and] a ... domain that interacts with TAF7". The promoter recognition domain is approximately at one end of the gene for TAF1.

On human chromosome X (number 23, NC_000023), specifically NC_000023.10, TAF1 is located 3'-70586113[-70685855]-5', 99,742 nt. The 3'-UTR begins at 70586114.

TAF1 isoform 1
Isoform 1 (variant 1) "represents the longer transcript and encodes the longer isoform".

TAF1 isoform 2
Isoform 2 (variant 2) "uses an alternate in-frame splice site, compared to variant 1, resulting in a shorter isoform (2) that lacks an internal 21 aa segment, compared to isoform 1."

TAF1/DYT3
TAF1/DYT3 is "a complex transcript system that is composed of at least 43 exons. Thirty-eight exons code for [TAF1] ... Five downstream exons (d1-d5) ... can either form transcripts with TAF1 exons or be transcribed independently." "Transcripts including exons d1, d3, d4, d5, plus TAF1 exons make up transcript "variant 1." Major "variant 2" is composed of various TAF1 exons plus exons d3 and d4. Alternately, d exons can generate transcripts independent of TAF1 exons 1-38. Exons d2, d3, and d4 make up "variant 3" and exons d3 and d4 constitute "variant 4."" The "additional five exons are located 3' to exon 38 ... ("downstream" exons 1-5)".

TAF1/TAF2
"[A] [TAF1]-[TAF2] complex selects sequences that match the Initiator (Inr) consensus."

TAF1A
GeneID: 9015, TAF1A, TATA box binding protein (TBP)-associated factor for RNA polymerase I. It has two isoforms.

TAF1B
GeneID: 9014, TAF1B, TATA box binding protein (TBP)-associated factor for RNA polymerase I.

TAF1C
GeneID: 9013, TAF1C, TATA box binding protein (TBP)-associated factor for RNA polymerase I. It has two isoforms.

TAF1L
GeneID: 138474, TAF1L "is expressed in male germ cells, and the product has been shown to function interchangeably with the TAF1 product."

TAF2
GeneID: 6873, TAF2 "is stably associated with the TFIID complex. It contributes to interactions at and downstream of the transcription initiation site, interactions that help determine transcription complex response to activators."

TAF3
GeneID: 83860, TAF3 is part of the "set of TBP-associated factors (TAFs) [which] contribute to promoter recognition and selectivity and act as antiapoptotic factors"

TAF4
GeneID: 6874, TAF4 "has been shown to potentiate transcriptional activation by retinoic acid, thyroid hormone and vitamin D3 receptors. In addition, [it] interacts with the transcription factor CREB, which has a glutamine-rich activation domain, and binds to other proteins containing glutamine-rich regions."

TAF4B
GeneID: 6875, TAF4B is "a cell type-specific TAF that may be responsible for mediating transcription by a subset of activators in B cells."

TAF5
GeneID: 6877, TAF5 is "an integral subunit of TFIID associated with all transcriptionally competent forms of that complex. [It] interacts strongly with two TFIID subunits that show similarity to histones H3 and H4, and it may participate in forming a nucleosome-like core in the TFIID complex."

TAF6
GeneID: 6878, TAF6 "binds weakly to TBP but strongly to TAF1". It has four isoforms.

TAF7
GeneID: 6879, TAF7 "interacts with the largest TFIID subunit, as well as multiple transcription activators. [It] is required for transcription by promoters targeted by RNA polymerase II."

TAF7L
GeneID: 54457, TAF7L "could be a spermatogenesis-specific component of the DNA-binding general transcription factor complex TFIID." It has two isoforms.

TAF8
GeneID: 129685, TAF8 "contains an H4-like histone fold domain, and interacts with several subunits of TFIID including TBP and the histone-fold protein TAF10."

TAF9
GeneID: 6880, TAF9 "binds to the basal transcription factor GTF2B as well as to several transcriptional activators such as p53 and VP16." It has four isoforms.

TAF10
GeneID: 6881, TAF10 "is associated with a subset of TFIID complexes. Studies with human and mammalian cells have shown that this subunit is required for transcriptional activation by the estrogen receptor, for progression through the cell cycle, and may also be required for certain cellular differentiation programs."

TAF11
GeneID: 6882, TAF11 "is present in all TFIID complexes and interacts with TBP. This subunit also interacts with another small subunit, TAF13, to form a heterodimer with a structure similar to the histone core structure."

TAF12
GeneID: 6883, "TAF12 interacts directly with TBP as well as with TAF2I [(TAF11)]."

TAF13
GeneID: 6884, TAF13 "interacts with TBP and with two other small subunits of TFIID, TAF10 and TAF11."

TAF15
GeneID: 8148, TAF15 is in "a subunit of TFIID present in a subset of TFIID complexes. Translocations involving chromosome 17 and chromosome 9, where the gene for the nuclear receptor CSMF is located, result in a gene fusion product that is an RNA binding protein associated with a subset of extraskeletal myxoid chondrosarcomas."

TAF15 has two isoforms.

General transcription factor II Ds
Before the start of transcription, the transcription factor II D (TFIID) complex, binds to the core promoter of the gene.

Hypotheses

 * 1) TAFs are not involved in the transcription of A1BG.


text:
Genetics/CLEC2D

C-type lectin domain family 2 member D is a protein that in humans is encoded by the CLEC2D gene. It is called C-type lectin-like because it has lost the Ca2+-dependent carbohydrate recognition domain present in most conventional C-type lectins while maintaining partial structural similarities. It is one of six genes located in human chromosome region 12p13.31. Several CLEC2D gene transcripts have been described. They are expressed primarily in hematopoietic cells. Lectin-like transcript 1 (LLT1) is expressed on the surface of natural killer cells and some T-cells. Its ligand is CD161.

title: Genetics/CLEC2D


text:
Genetics/CD161

Killer cell lectin-like receptor subfamily B, member 1, also known as KLRB1, NKR-P1A or CD161 (cluster of differentiation 161), is a human gene.

NKR-P1A, the receptor encoded by the KLRB1 gene, recognizes Lectin Like Transcript-1 (LLT1) as a functional ligand.

title: Genetics/CD161


text:
Talk:Tricolour Voting Game

Metrics

 * Metrics, April, 2012
 * Metrics, May, 2012
 * Metrics, June, 2012
 * Metrics, July, 2012

Game Use
Fabian, for the uninitiated player it might be helpful to have a clearer indication at the very beginning of the game's purpose and what it is attempting to simulate. Let's talk about this.

title: Talk:Tricolour Voting Game


text:
Category:Ludic Science Club

Pages relating to Ludic Science Club

title: Category:Ludic Science Club


text:
Data Coding and Information Decoding

Main Course Launch Page - Data Size and Speeds Exit Supplmental Material

Supplemental Course Navigation

Unique Code
Data consists of numbers, text, sounds, images, videos, etc. In order to define them in the real world, we are using numbers (0-9), letters (A-Z) and symbols (@,[,\, etc) or a combination thereof. For example:

 * 2012 (number)
 * Wikiversity (letters)
 * 3<5 (combination of numbers and symbols)
 * X=(Y*Z)+W-U (combination of alphabet and symbols)
 * N123 (combination of letters and numbers)

Computers, however, don't understand the definitions for numbers (0-9), letters (A-Z) and symbols (@,[,\, etc), so in order to process those pieces of information a unique code must be assigned to each of them. The unique code for numbers, letters, and symbols is a binary numeral. Data2 codeing.png

In practical use, each "0" and "1" in binary number uses a Physical attribute that is electrical or magnetic. The smallest part of that Physical attribute is named "Bit", the existence of the Physical attribute is "1", and the absence of the Physical attribute is "0"; 8 bits grouped up are called 1 Byte (so 1 byte=8 bits).

The binary numeral system, or base-2 number system, represents numeric values using two symbols - 0 and 1.

Examples:

 * numerical value " 0 " is 0. The unique code for 0 is →  00000000
 * numerical value " 1 " is 1. The unique code for 1 is →  00000001
 * numerical value " 2 " is 10. The unique code for 2 is →  00000010
 * numerical value " 3 " is 11. The unique code for 3 is →  00000011
 * numerical value " 4 " is 100. The unique code for 4 is →  00000100
 * numerical value " 5 " is 101. The unique code for 5 is →  00000101
 * numerical value " 6 " is 110. The unique code for 6 is →  00000110
 * numerical value " 7 " is 111. The unique code for 7 is →  00000111
 * numerical value " 8 " is 1000. The unique code for 8 is →  00001000
 * numerical value " 9 " is 1001. The unique code for 9 is →  00001001

Alphabet (A,...Z):

 * The unique code for " A " is →  01000001
 * The unique code for " B " is →  01000010
 * The unique code for " C " is →  01000011
 * The unique code for " D " is →  01000100
 * The unique code for " E " is →  01000101
 * The unique code for " F " is →  01000110
 * The unique code for " G " is →  01000111
 * The unique code for " H " is →  01001000
 * The unique code for " I " is →  01001001
 * The unique code for " J " is →  01001010
 * The unique code for " K " is →  01001011
 * The unique code for " L " is →  01001100
 * The unique code for " M " is →  01001101
 * The unique code for " N " is →  01001110
 * The unique code for " O " is →  01001111
 * The unique code for " P " is →  01010000
 * The unique code for " Q " is →  01010001
 * The unique code for " R " is →  01010010
 * The unique code for " S " is →  01010011
 * The unique code for " T " is →  01010100
 * The unique code for " U " is →  01010101
 * The unique code for " V " is →  01010110
 * The unique code for " W " is →  01010111
 * The unique code for " X " is →  01011000
 * The unique code for " Y " is →  01011001
 * The unique code for " Z " is →  01011010

for symbols :

 * The unique code for " @ " is →  01000000
 * The unique code for " [ " is →  01011011
 * The unique code for " \ " is →  01011100

Distinction between "Data" and "Information"

Supplemental Course Navigation

title: Data Coding and Information Decoding


text:
User:Happyhouseinc

test page happyhouseinc

title: User:Happyhouseinc


title: File:Pea2.s12.sec48.k.djvu


text:
Template:PD Symptomatic Relief Pages

Therapy > Symptomatic Relief

Sub Pages:
 * Bradykinesia
 * Dyskinesia
 * Tremor

title: Template:PD Symptomatic Relief Pages


text:
University of Florida/Egm4313/s12.team11.R6

Intermediate Engineering Analysis Section 7566 Team 11 Due date: April 11, 2012.

R6.1
Solved by Daniel Suh

Problem Statement

 * Find the fundamental period of $$\cos (n\omega x) \!$$ and $$\sin (n\omega x) \!$$
 * Show that these functions also have period $$p\!$$.
 * Show that the constants $$a_{0}\!$$ is also a periodic function with period $$p\!$$.

Solution

 * Find the fundamental period, and show that these functions have a period of $$p\!$$.

$$p=2L=2\frac{\pi}{\omega}\!$$

$$L = \frac{\pi}{\omega}\!$$

$$f(x+np) = f(x)\!$$

$$f(x) = cos (n\omega x)\!$$

$$f(x+n\frac{2\pi}{n\omega }) = cos (n\omega (x+\frac{2\pi}{n\omega }))\!$$

$$f(x+\frac{2\pi}{\omega}) = cos (n\omega x+2\pi)\!$$

$$f(x+2L) = cos (n\omega x+2\pi)\!$$

$$p=2L\!$$ is the fundamental period for $$cos (n\omega x)\!$$.

$$f(x+np) = f(x)\!$$

$$f(x) = sin (n\omega x)\!$$

$$f(x+n\frac{2\pi}{n\omega }) = sin (n\omega (x+\frac{2\pi}{n\omega }))\!$$

$$f(x+\frac{2\pi}{\omega }) = sin (n\omega x+2\pi)\!$$

$$f(x+2L) = sin (n\omega x+2\pi)\!$$

$$p=2L\!$$ is the fundamental period for $$sin (n\omega x)\!$$.

 * Show that the constants $$a_{0}\!$$ is also a periodic function with period $$p\!$$.

From Fourier Series, $$a_{0}=\frac{1}{2L}\int_{-L}^{L}f(x)dx\!$$

$$f(x) = \cos (n\omega x)\!$$

$$a_{0}=\frac{\omega}{2\pi}\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}cos (n\omega x)dx\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{sin (n\omega x)}{n\omega}|_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}]\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{2sin (n\pi)}{n\omega}]\!$$

$$a_{0}=0\!$$

$$f(x) = \sin (n\omega x)\!$$

$$a_{0}=\frac{\omega}{2\pi}\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}sin (n\omega x)dx\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{cos (n\omega x)}{n\omega}|_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}]\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{-cos{n\pi}+cos{n\pi}}{n\omega}]\!$$

$$a_{0}=0\!$$

Problem Statement
Solve the problems 11 and 12 from Kreyszig p.491

Solution
First, for problem 11: $$f(x)=x^2\!$$ from $$<-1< x< 1>\!$$ where $$p=2\!$$ We know that $$p=2=2L\!$$ so that $$L=1\!$$

We check is the function is even, odd or neither. Since $$f(x)\!$$ satisfies the condition $$f(x)=f(-x)\!$$ we know that the function is even. Given that the function is even and that $$b_n=0\!$$ for even functions, we exclude the $$b_n\!$$ term from the Fourier Series. We find $$ a_0 \!$$: $$a_0=\frac{1}{2L}\int_{-L}^{L}f(x)dx\!$$ Plugging in, $$a_0=\frac{1}{2}\int_{-1}^{1}x^2dx=\frac{1}{2}\left [ \frac{1}{3}x^3 \right ]_{-1}^{1}=\frac{1}{2}\left [ \frac{1}{3}+\frac{1}{3} \right ]=\frac{1}{3} \!$$ We find $$a_n\!$$ $$a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx\!$$ $$a_n=1\int_{-1}^{1}x^2cos(n\pi x)dx\!$$ We can use the equivalent expression: $$a_n=2\int_{0}^{1}x^2cos(n\pi x)dx\!$$ Now we evaluate the integral by using integration by parts. Below are the steps: Integration by parts uses $$\int fdg=fg-\int gdf\!$$ $$\begin{matrix} f=x^2 & dg=cos(\pi nx)dx\\ df=2xdx & g=\frac{sin(\pi nx)}{\pi n} \end{matrix}\!$$ Then, we have: $$\left [ \frac{x^2sin(\pi nx)}{\pi n} \right ]_{0}^{1}-\frac{2}{\pi n}\int_{0}^{1}xsin(\pi nx)dx\!$$ Repeating for the second term: $$\begin{matrix} f=x & dg=sin(\pi nx)\\ df=dx & g=\frac{-cos(\pi nx)}{\pi n} \end{matrix}\!$$ Then, we have: $$\left [ \frac{-xcos(\pi nx)}{\pi n} \right ]_{0}^{1}-\frac{1}{\pi n}\int_{0}^{1}cos(\pi nx)dx\!$$ Bringing all the terms together: $$2\left [ \left [ \frac{x^2sin(\pi nx)}{\pi n} \right ]_{0}^{1}-\frac{2}{\pi n}\left [  \frac{-xcos(\pi nx)}{\pi n} \right ]_{0}^{1}-\frac{1}{\pi n}\int_{0}^{1}cos(\pi nx)dx\right ]\!$$ Then, $$2\left [ \left [ \frac{x^2sin(\pi nx)}{\pi n} \right ]_{0}^{1}-\frac{2}{\pi n}\left [  \frac{-xcos(\pi nx)}{\pi n} \right ]_{0}^{1}-\frac{1}{\pi n}\left [ \frac{sin(\pi nx)}{\pi n} \right ]_{0}^{1} \right ]\!$$ Evaluating at the boundaries and simplifying: $$2\left [ \left [0-0\right ]+\frac{2}{\pi^2 n^2}\left [cos(\pi n) \right ]-\frac{1}{\pi n}\left [0-0 \right ]\right ]\!$$ Further, $$\frac{4}{\pi^2 n^2}cos(\pi n)\!$$ We can simplify one step further by acknowledging that $$cos(\pi n)=(-1)^n\!$$. $$ \therefore a_n=\frac{4}{\pi^2 n^2}(-1)^n\!$$ Finally, $$f(x)=a_0+\sum_{n=1}^{\infty}a_ncos(\frac{\pi n}{L})x\!$$ The Fourier series is: $$f(x)=\frac{1}{3}+\sum_{n=1}^{\infty}\frac{4}{\pi^2 n^2}(-1)^ncos(\pi n)x\!$$

Now problem 12: $$f(x)=1-\frac{x^2}{4}\!$$ from $$<-2< x< 2>\!$$ where $$p=2\!$$ We know that $$p=4=2L\!$$ so that $$L=2\!$$

We check is the function is even, odd or neither. Since $$f(x)\!$$ satisfies the condition $$f(x)=f(-x)\!$$ we know that the function is even. Given that the function is even and that $$b_n=0\!$$ for even functions, we exclude the $$b_n\!$$ term from the Fourier Series. We find $$ a_0 \!$$: $$a_0=\frac{1}{2L}\int_{-L}^{L}f(x)dx\!$$ Plugging in, $$a_0=\frac{1}{4}\int_{-2}^{2}1-\frac{x^2}{4}dx=\frac{1}{4}\left [ x-\frac{1}{12}x^3 \right ]_{-2}^{2}=\frac{1}{4}\left [ \frac{24-8}{12}-\frac{24+8}{12} \right ]=\frac{2}{3} \!$$ We find $$a_n\!$$ $$a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx\!$$ $$a_n=\frac{1}{2}\int_{-2}^{2}(1-\frac{x^2}{4}cos(n\pi x)dx\!$$ Now we evaluate the integral by using integration by parts. This process is similar to the one shown in the solution for problem 11 above and therefore will be performed in Wolfram Alpha. The following command was given in Wolfram Alpha: integral (1-(x^2/4))*cos(pi*x) from -2 to 2. Wolfram Alpha yields the following $$\frac{2}{\pi^2 n^2}\!$$

Finally, $$f(x)=a_0+\sum_{n=1}^{\infty}a_ncos(\frac{\pi n}{L})x\!$$ The Fourier series is: $$f(x)=\frac{2}{3}+\sum_{n=1}^{\infty}\frac{-2}{\pi^2 n^2}cos(\frac{\pi n}{2})x\!$$

Part B
Solved by Francisco Arrieta

Problem Statement
Find the Fourier series expansion for $$ f(x) \!$$ on p. 9.8 as follows:

Part 1
Develop the Fourier series expansion of $$ f(\bar x) \!$$

Plot $$ f(\bar x) \!$$ and the truncated Fourier series $$ f_n(\bar x) \!$$

$$ f_n(\bar x):=\bar a_0 +\sum_{k=1}^{n}[\bar a_k\cos k \omega \bar x+\bar b_k\sin k \omega \bar x] \!$$

for n=0,1,2,4,8. Observe the values of $$ f_n(\bar x) \!$$ at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of $$ f(x) \!$$

Solution

 * $$ p=2L=4 \!$$

 * $$ L=2 \!$$

 * $$ \omega =\frac{2\pi}{p} \!$$

$$ \bar a_0=\frac{1}{2L}\int_{-L}^{L}f(\bar x)d\bar x \!$$

 * $$ \bar a_0=\frac{1}{4}\int_{-2}^{2}f(\bar x)d\bar x \!$$

 * $$ \bar a_0=\frac{A}{2} \!$$

$$ \bar a_k=\frac{1}{L}\int_{-L}^{L}f(\bar x)\cos k\omega \bar xd\bar x \!$$

 * $$ \bar a_k=\frac{1}{2}\int_{-2}^{2}f(\bar x)\cos \frac{k\pi \bar x}{2}d\bar x \!$$

 * $$ \bar a_k=\frac{2A}{k\pi}\sin \frac{k\pi}{2} \!$$

Note:

If k is even $$ \bar a_k=0 \!$$

If k=1, 5, 9... $$ \bar a_k=\frac{2A}{k\pi} \!$$

If k=3, 7, 11... $$ \bar a_k=\frac{-2A}{k\pi} \!$$

$$ \bar b_k=\frac{1}{L}\int_{-L}^{L}f(\bar x)\sin k\omega \bar xd\bar x \!$$

Note:

$$ \bar b_k=0 \!$$ for n=1, 2...

Then Fourier series becomes a Fourier cosine series:

$$ f_k(\bar x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\cos \frac{k\pi}{2}\bar x)\!$$

For n=0

$$ f_k(\bar x)=\frac{A}{2}\!$$

For n=1

$$ f_k(\bar x)=\frac{A}{2}+(\frac{2A}{\pi}\cos \frac{\pi}{2}\bar x)\!$$

For n=5

$$ f_k(\bar x)=\frac{A}{2}+\frac{2A}{\pi}(\cos \frac{\pi}{2}\bar x-\frac{1}{3}\cos \frac{3\pi}{2}\bar x+\frac{1}{5}\cos \frac{5\pi}{2}\bar x)\!$$

After transforming the variable of $$ f(\bar x) \!$$

$$ f_k(x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\cos \frac{k\pi}{2}(x-1.25))\!$$

Part 2
Do the same as above, but using $$ f(\tilde x) \!$$ to obtain the Fourier expansion of $$ f(x) \!$$ ; compare to the result obtained above

Solution

 * $$ p=2L=4 \!$$

 * $$ L=2 \!$$

 * $$ \omega =\frac{2\pi}{p} \!$$

$$ \tilde a_0=\frac{1}{2L}\int_{0}^{2L}f(\tilde x)d\tilde x \!$$

 * $$ \tilde a_0=\frac{1}{4}\int_{0}^{4}f(\tilde x)d\tilde x \!$$

 * $$ \tilde a_0=\frac{A}{2} \!$$

$$ \tilde a_k=\frac{1}{L}\int_{0}^{2L}f(\tilde x)\cos k \omega \tilde x d\tilde x \!$$

 * $$ \tilde a_k=\frac{1}{2}\int_{0}^{4}f(\tilde x)\cos k \omega \tilde x d\tilde x \!$$

 * $$ \tilde a_k=\frac{A}{k\pi}\sin \pi k \!$$

Note:

$$ \bar b_k=0 \!$$ for n=1, 2..

$$ \tilde b_k=\frac{1}{L}\int_{0}^{2L}f(\tilde x)\sin k \omega \tilde x d\tilde x \!$$

 * $$ \tilde b_k=\frac{1}{2}\int_{0}^{4}f(\tilde x)\sin k \frac{\pi}{2} \tilde x d\tilde x \!$$

 * $$ \tilde b_k=\frac{-A}{\pi k}(\cos \pi k -1) \!$$

Note:

If k is even $$ \bar a_k=0 \!$$

If k=1, 3, 5... $$ \bar a_k=\frac{2A}{k\pi} \!$$

Then Fourier series becomes a Fourier sine series:

$$ f_k(\tilde x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\sin \frac{k\pi}{2}\tilde x)\!$$

For n=0:

$$ f_k(\tilde x)=\frac{A}{2}\!$$

For n=1:

$$ f_k(\tilde x)=\frac{A}{2}+(\frac{2A}{\pi}\sin \frac{\pi}{2}\tilde x) \!$$

For n=5:

$$ f_k(\tilde x)=\frac{A}{2}+\frac{2A}{\pi}(\sin \frac{\pi}{2}\tilde x+\frac{1}{3}\sin \frac{3\pi}{2}\tilde x+\frac{1}{5}\sin \frac{5\pi}{2}\tilde x) \!$$

After transforming the variable of $$ f(\bar x) \!$$

$$ f_k(x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\sin \frac{k\pi}{2}(x-.25))\!$$

Problem Statement
Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.

Solution
15.

$$f(x)=\left\{\begin{matrix} -x-\pi, -\pi<x<-\pi/2\\ x, -\pi/2<x<\pi/2\\ \pi-x, \pi/2<x<\pi \end{matrix}\right.$$

Since $$f(-x)=-f(x)$$, the function is odd.

$$a_0=(1/\pi)*[-1 \int_{-\pi}^{-\pi/2}(x+\pi)dx+ \int_{-\pi/2}^{\pi/2}(x)dx+ \int_{\pi/2}^{\pi}(\pi-x)dx]$$ After simplification, $$a_0=0$$. Using Equation (5) on page 490, the exapansion becomes; $$f(x)=8k/\pi^2[sin(\pi*x/L)-sin(3\pi*x/L)/9...)$$ With $$L=\pi,k=\pi/2$$ this becomes; $$f(x)=4/\pi(sinx-sin(3x)/9+sin(5x)/25-sin(6x)/36....)$$ for n is "odd". This means when n is even, $$f(x) =0$$

17. $$f(x)=1-|x|, -1\leq x \leq1$$

$$f(x)=\left\{\begin{matrix} 1+x-\pi, -1<x<0\\ 1-x, 0<x<1\\ \end{matrix}\right.$$

$$a_0=1/2[\int_{-1}^{0}(1+x)dx + \int_{0}^{1}(1-x)dx]$$ $$a_0=1/2[1-1/2+1-1/2]$$ $$a_0=1/2$$

$$a_n=1/1[\int_{-1}^{0}(1+x)cos(n\pi*x/1)dx+ \int_{0}^{1}(1-x)cos(n\pi*x/1)dx]$$ Simplifying results in; $$a_n=2[1/(n^2\pi^2)_(-1)^n/(n^2\pi^2)]$$ So, $$ a_n=4/(n^2\pi^2)$$ when n is odd. $$ b_n=1/1[\int_{-1}^{0}(1+x)sin(n\pi*x/1)dx+ \int_{0}^{1}(1-x)sin(n\pi*x/1)dx]$$ Simplifying results in; $$b_n=0$$ $$f(x)=1/2+ \sum_{n=0}^{\infty }4/(n^2\pi^2)*cos(n\pi*x)$$

Since $$a_n=0$$ for even n, $$f(x)$$ becomes $$f(x)=1/2$$ for even n.

R6.4
solved by Luca Imponenti

Problem Statement
Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:
 * $$y''-3y'+2y=f(x)\!$$

and the initial conditions
 * $$y(0)=1 \, \ y'(0)=0\!$$

1. Find $$y_n(x)\!$$ such that:
 * $$y_n''+ay_n'+by_n=r_n(x)\!$$

with the same initial conditions as above.

Plot $$y_n(x)\!$$ for $$n=2,4,8\!$$ for x in $$[0,10]\!$$

2. Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.
 * Level 1: $$n=0,1\!$$

Fourier Series
One period of the window function p9.8 is described as follows

 * $$f(x) =

\begin{cases} 0 & \ \ -1.75 < x <l 0.25 \\ A & \ \ 0.25< x < 2.25 \\ \end{cases} $$

From the above intervals one can see that the period, $$p=4\!$$ and therefore $$L=2\!$$ Applying the Euler formulas from $$-1.75$$ to  $$2.25\!$$ the Fourier coefficients are computed:

$$a_0=\frac{1}{2L}\int_{-1.75}^{2.25}f(x)\ dx\!$$

$$a_0=\frac{1}{4}(\int_{-1.75}^{0.25}0\ dx+\int_{0.25}^{2.25}A\ dx)\!$$

$$a_0=+\frac{1}{4}(0+\int_{0.25}^{2.25}A\ dx)\!$$

$$a_0=\frac{1}{4}(2.25A-0.25A)\!$$

$$a_0=\frac{A}{2}\!$$

The integral from $$-1.75\!$$ to  $$0.25\!$$ can be omitted from this point on since it is always zero.

$$a_n=\frac{1}{L}\int_{0.25}^{2.25}f(x)cos(\frac{n\pi x}{L})\ dx\!$$

$$a_n=\frac{1}{2}\int_{0.25}^{2.25}Acos(\frac{n\pi x}{2})\ dx\!$$

$$a_n=\frac{2A}{2n\pi}(sin(\frac{2.25n\pi}{2})-sin(\frac{0.25n\pi}{2})\!$$

$$a_n=\frac{A}{n\pi}(sin(\frac{9n\pi}{8})-sin(\frac{n\pi}{8})\!$$

and

$$b_n=\frac{1}{L}\int_{0.25}^{2.25}f(x)sin(\frac{n\pi x}{L})\ dx\!$$

$$b_n=\frac{1}{2}\int_{0.25}^{2.25}Asin(\frac{n\pi x}{2})\ dx\!$$

$$b_n=\frac{2A}{2n\pi}(cos(\frac{2.25n\pi}{2})-cos(\frac{0.25n\pi}{2})\!$$

$$b_n=\frac{A}{n\pi}(Acos(\frac{n\pi}{8})-Acos(\frac{9n\pi}{8})\!$$

The coefficients give the Fourier series:

$$f(x)=a_0+\sum_{n=1}^{\infty}[a_ncos(\frac{n\pi x}{L})+b_nsin(\frac{n\pi x}{L})]\!$$

$$f(x)=\frac{A}{2}+\sum_{n=1}^{\infty}[\frac{A}{n\pi}(sin(\frac{9n\pi}{8})-sin(\frac{n\pi}{8}))cos(\frac{n\pi x}{2})\!$$
 * $$+\frac{A}{n\pi}(cos(\frac{n\pi}{8})-cos(\frac{9n\pi}{8}))sin(\frac{n\pi x}{2})]\!$$

Homogeneous Solution
Considering the homogeneous case of our ODE:

$$y''-3y'+2y=0\!$$

The characteristic equation is

$$\lambda^2-3\lambda+2=0\!$$

$$(\lambda-1)(\lambda-2)=0\!$$

$$\lambda_1=1, \lambda_1=2\!$$

Therefore our homogeneous solution is of the form

$$y_h=c_1e^x+c_2e^{2x}\!$$

Particular Solution
Considering the case with f(x) as excitation

$$y''-3y'+2y=\frac{A}{2}+\sum_{k=1}^{n}\frac{A}{k\pi}[(sin(\frac{9k\pi}{8})-sin(\frac{k\pi}{8}))cos(\frac{k\pi x}{2})\!$$
 * $$+(cos(\frac{k\pi}{8})-cos(\frac{9k\pi}{8}))sin(\frac{k\pi x}{2})]\!$$

The solution will be of the form

$$y_n=A_0+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})\!$$

Taking the derivatives

$$y_n'=-A_n\frac{\pi}{2}\sum_{k=2}^{n}ksin(\frac{k\pi x}{2})+B_n\frac{\pi}{2}\sum_{k=2}^{n}kcos(\frac{k\pi x}{2})\!$$

$$y_n''=-A_n\frac{\pi^2}{4}\sum_{k=3}^{n}k^2cos(\frac{k\pi x}{2})-B_n\frac{\pi^2}{4}\sum_{k=3}^{n}k^2sin(\frac{k\pi x}{2})\!$$

Plugging these back into the ODE:

$$-\frac{\pi^2}{4}\sum_{k=3}^{n}A_kk^2cos(\frac{k\pi x}{2})-\frac{\pi^2}{4}\sum_{k=3}^{n}B_kk^2sin(\frac{k\pi x}{2})-3[-\frac{\pi}{2}\sum_{k=2}^{n}A_kksin(\frac{k\pi x}{2})\!$$

$$+\frac{\pi}{2}\sum_{k=2}^{n}B_kkcos(\frac{k\pi x}{2})]+2[A_0+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})]\!$$

$$=\frac{A}{2}+\sum_{k=1}^{n}\frac{A}{k\pi}[(sin(\frac{9k\pi}{8})-sin(\frac{k\pi}{8}))cos(\frac{k\pi x}{2})+(cos(\frac{k\pi}{8})-cos(\frac{9k\pi}{8}))sin(\frac{k\pi x}{2})]\!$$

Setting the two constants equal

$$2A_0=\frac{A}{2}\!$$

$$A_0=\frac{A}{4}\!$$

This is valid for all values of n. Since the coefficients of the excitation $$sin(\frac{9k\pi}{8})-sin(\frac{k\pi}{8})\!$$ and  $$cos(\frac{k\pi}{8})-cos(\frac{9k\pi}{8})\!$$ are zero for all even n, then the coefficients $$A_k\!$$  and  $$B_k\!$$ will also be zero, so we must only find these coefficients for odd n's. Now carrying out the sum to $$n=1\!$$ and comparing like terms yields the following sets of equations. Written in matrix form:
 * $$ \begin{bmatrix}2 & 0\\

\\0 & 2\end{bmatrix}* \begin{bmatrix} A_1 \\ B_1\end{bmatrix}= \begin{bmatrix} \ \frac{A}{\pi}(sin(\frac{9\pi}{8})-sin(\frac{\pi}{8})) \\   \ \frac{A}{\pi}(cos(\frac{\pi}{8})-cos(\frac{9\pi}{8}))\end{bmatrix}\!$$

Assuming $$A=1\!$$ this matrix can be solved to obtain

$$A_1=-0.1218 \, \ B_1=0.2941\!$$

For the remaining coefficients to be solved all sums will be used so a more general equation may be written:
 * $$ \begin{bmatrix}

2-\frac{(\pi k)^2}{4} & \frac{-3\pi k}{2}\\ \frac{-3\pi k}{2} & 2-\frac{(\pi k)^2}{4}\end{bmatrix}* \begin{bmatrix} A_k \\ B_k\end{bmatrix}= \begin{bmatrix} \ \frac{A}{\pi k}(sin(\frac{9\pi k}{8})-sin(\frac{\pi k}{8})) \\ \ \frac{A}{\pi k}(cos(\frac{\pi k}{8})-cos(\frac{9\pi}{8}))\end{bmatrix}\!$$

Results of these calculations are shown below:

$$A=\begin{bmatrix} A_1 \\ A_2 \\ .\\ .\\ A_7 \\ A_8\end{bmatrix}= \begin{bmatrix} -0.1218 \\ 0 \\ 0.0084 \\ 0 \\ 0.0014 \\ 0 \\ 0.0001 \\ 0\end{bmatrix}  , \ B=\begin{bmatrix} B_1 \\ B_2 \\ .\\ .\\ B_7 \\ B_8\end{bmatrix}= \begin{bmatrix} 0.2941 \\ 0 \\ 0.0019 \\ 0 \\ 0.0014 \\ 0 \\ 0.0007 \\ 0\end{bmatrix}\!$$

The solution to the particular case can be written for all n (assuming A=1):

$$y_n=\frac{1}{4}+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})\!$$

General Solution
The general solution is

$$y=y_h+y_p\!$$

where

$$y_h=c_1e^x+c_2e^{2x}\!$$

Different coefficients $$c_1 \, \ c_2\!$$ will be calculated for each n. These coefficients are easily solved for by applying the given initial conditions. Below are the calculations for n=2.

$$y=c_1e^x+c_2e^{2x}+\frac{1}{4}+\sum_{k=1}^{2}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{2}B_ksin(\frac{k\pi x}{2})\!$$

Applying the first initial condition $$y(0)=1\!$$

$$y(0)=c_1+c_2+A_1+A_2=1\!$$

Taking the derivative

$$y'=c_1e^x+2c_2e^{2x}-\sum_{k=2}^{2}\frac{k\pi}{2}A_ksin(\frac{k\pi x}{2})+\sum_{k=2}^{2}\frac{k\pi}{2}B_kcos(\frac{k\pi x}{2})\!$$

Applying the second initial condition $$y'(0)=0\!$$

$$y'(0)=c_1+2c_2+\frac{2\pi}{2}B_2=0\!$$

Solving the two equations for two unknowns yields:

$$c_1=2.2436 \, \ c_2=-1.1218\!$$

So the general solution for n=2 is:

$$y=2.2436e^x-1.1218e^{2x}+\frac{1}{4}+\sum_{k=1}^{2}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{2}B_ksin(\frac{k\pi x}{2})\!$$

Below is a plot showing the general solutions for n=2,4,8:

Matlab Plots
Using ode45 the following graph was generated for n=0:

and for n=1

R6.5
Solved by: Gonzalo Perez

Part R4.2, p.7c-26
For each value of n=3,5,9, re-display the expressions for the 3 functions $$ y_{p,n} (x), y_{h,n} (x), y_n (x) \! $$, and plot these 3 functions separately over the interval $$ [0,20 \pi] \! $$.

R4.2 p.7-26: Consider the L2-ODE-CC (5) p.7b-7 with $$ sin x \! $$ as excitation:

$$ y''-3y'+2y=r(x) \! $$              (5)p.7-7

$$ r(x) = sin x \! $$                     (1)p.9-15

and the initial conditions:

$$ y(0) = 1, y'(0) = 0. \! $$                      (3b)p.3-7

Exact solution: $$ y(x) = y_h (x) + y_p (x) \! $$                   (2)p.9-15

Re-display the expressions for $$ y_p (x), y_h (x), y(x) \! $$.

Superpose each of the above plot with that of the exact solution.

Solution
The graphs are to be separately graphed as follows.

For n = 3, the code that will generate the graph looks like this:

For n = 6:

For n = 9:

Part R4.3, p.7c-28
Understand and run the TA's code to produce a similar plot, but over a larger interval $$ [0,10] \! $$. Do zoom-in plots about the points $$ x=-0.5,0,+0.5 \! $$ and comment on the accuracy of different approximations.

R4.3 p.7-28: Consider the L2-ODE-CC (5) p.7b-7 with $$ log(1+x) \! $$ as excitation:

$$ y''-3y'+2y=r(x) \! $$     (5)p.7-7

$$ r(x) = log(1+x) \! $$     (1)p.7-28

and the initial conditions:

$$ y(-\frac{3}{4})=1, y'(-\frac{3}{4})=0 \! $$    (2)p.7-28

Solution
For n=4 (from x = 0 to x = 10):

For n=4 (zoom-in plots around x = -0.5):

For n=4 (zoom-in plots around x = 0):

For n=4 (zoom-in plots around x = +0.5):

For n=7 (from x = 0 to x = 10):

For n=11 (from x = 0 to x = 10):

Where the black line represents n=11 and the blue line represents log(1+x).

The problem asks for n=4, 7, and 11. The TA had up to n=16, but the problem does not ask for this. Manipulating the code and graphs to only account for these n values, n=16 has been disregarded.

Combined plots for n=7 and n=11 (zoom-in plots around x = -0.5):

Combined plots for n=7 and n=11 (zoom-in plots around x = 0):

Combined plots for n=7 and n=11 (zoom-in plots around x = 0.5):

Part R4.4, p.7c-29
Understand and run the TA's code to produce a similar plot, but over a larger interval $$ [0.9,10] \! $$, and for n=4,7. Do zoom-in plots about $$ x=1,1.5,2,2.5 \! $$ and comments on the accuracy of the approximations.

R4.4 p.7-29: Extend the accuracy of the solution beyond $$ \hat{x}=1 \! $$.

Solution
The general code that can be used to graph the plots of n=4,7,11 is:

With the above code, we can add the following code to graph n=4.

For n = 4 (from x = 0.9 to x = 10):

For n = 7 (from x = 0.9 to x = 10):

For n=11 (from x = 0.9 to x = 10):

Repeating the same common part of the code, let's graph the other plots around x = 1, 1.5, 2, 2.5:

For n = 4, 7, 11 at x = 1:

For n = 4, 7, 11 at x = 1.5:

For n = 4, 7, 11 at x = 2:

For n = 4, 7, 11 at x = 2.5:

For the last part, the first (and unchanged TA's code) is the following:

The second part includes the change that had to be made in order to solve this problem:

Problem Statement
Given: For the following differential equation: $$ y_{p}''+4y_{p}'+3y_{p} = 2 e^{2x} cos(3x) \!$$ With a particular solution of the form: $$ y_{p} = x e^{-2x}(Mcos(3x) + Nsin(3x)) \!$$ Verify that this solution has a final expression of $$ y_{p} = 6 e^{-2x} (Ncos(3x) - Msin(3x)) \!$$ as follows: 1) Simplify the term $$ y_{p}'' \!$$ 2) Simplify the 2nd term $$4y_{p}'\!$$ and combine with the simplified first term 3) Finally add the 3rd term $$ 13y_{p}\!$$ 4) Find the final expression for $$ y_{p}(x) \!$$

Solution
1) To find the derivative and simplify $$ y_{p} \!$$ we will use www.wolframalpha.com . The "Simplify" command will be exploited as well as the notation for taking a derivative in mathematica. The following was entered into www.wolframalpha.com: Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]] The answer that was yielded is as follows: $$ y_{p} = -6Me^{-2x}sin(3x) + 12Me^{-2x}x sin(3x) - 4Me^{-2x}cos(3x) - 5Me^{-2x} x cos(3x)\!$$ $$ - 4Ne^{-2x}sin(3x) -5Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) - 12Ne^{-2x} x cos(3x) \!$$  2) In the same fashion, $$4y_{p}'\!$$ was evaluated and simplified using www.wolframalpha.com The following was entered into www.wolframalpha.com: simplify[4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]]] The answer that was yielded is as follows: $$ 4y_{p}' = -12Me^{-2x} x sin(3x) + 4M e^{-2x}cos(3x) - 8Me^{-2x} x cos(3x) + 4Ne^{-2x}sin(3x) - 8Ne^{-2x} x sin(3x) + 12Ne^{-2x} x cos(3x) \!$$ This term can be added to the previous, and then simplified yet again using www.wolframalpha.com The following was entered into www.wolframalpha.com Simplify[D[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x],x]+ 4*[D[x*exp(-2x)*(Mcos(3x) + Nsin(3x)),x]] The answer that was yielded is as follows: $$ y_{p}'' + 4y_{p}' = -6Me^{-2x}sin(3x) - 13Me^{-2x} x cos(3x) -13Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) \!$$ 3) The final term $$ 13y_{p} \!$$ is now added to this result, as follows: $$ y_{p} + 4y_{p}' + 13y_{p} = -6Me^{-2x}sin(3x) - 13Me^{-2x} x cos(3x) -13Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) + 13Me^{-2x} x cos(3x) +13Ne^{-2x} x sin(3x) \!$$  $$ y_{p} +  4y_{p}' + 13y_{p} = -6Me^{-2x}sin(3x) + (13-13)Me^{-2x} x cos(3x) + (13-13)Ne^{-2x} x sin(3x) + 6Ne^{-2x}cos(3x) \!$$  $$ y_{p} +  4y_{p}' + 13y_{p} = -6Me^{-2x}sin(3x) + 6Ne^{-2x}cos(3x) \!$$ $$ y_{p} +  4y_{p}' + 13y_{p} = 6e^{-2x}(Ncos(3x)-Msin(3x) ) \!$$    4) As verified above, the final term is as follows: $$ y_{p}(x) = 6e^{-2x}(Ncos(3x)-Msin(3x) ) \!$$ --Egm4313.s12.team11.sheider (talk) 22:30, 10 April 2012 (UTC)

R6.7
Solved by Solved by Daniel Suh

Problem Statement
Find the separated Ordinary Differential Equations for the heat equation.

$$\frac{\partial u}{\partial t}=\kappa \frac{\partial^2 u}{\partial x^2} \cdots (1)\!$$

$$u(x,t) = F(x)\cdot G(t)\!$$

Solution
$$\frac{\partial u}{\partial t} = F(x)\cdot \dot{G}(t) \cdots (2)\!$$

$$\frac{\partial^2 u}{\partial x^2} = F''(x)\cdot G(t) \cdots (3)\!$$

Combine $$(2)\!$$ and $$(3)\!$$ into $$(1)\!$$

$$F(x)\cdot \dot{G}(t) = \kappa F''(x)\cdot G(t)\!$$

$$\frac{\dot{G}(t)}{\kappa G(t)} = \frac{F''(x)}{F(x)} = c\!$$ (constant)

Separate

$$F''(x)-cF(x) = 0\!$$ $$G'(t)-\kappa cG(t) = 0\!$$


text:
User:Egm4313.s12.team11.arrieta/Report 6

Problem R6.2
Solved by Francisco Arrieta

Problem Statement
Find the Fourier series expansion for $$ f(x) \!$$ on p. 9.8 as follows:

Part 1
Develop the Fourier series expansion of $$ f(\bar x) \!$$

Plot $$ f(\bar x) \!$$ and the truncated Fourier series $$ f_n(\bar x) \!$$

$$ f_n(\bar x):=\bar a_0 +\sum_{k=1}^{n}[\bar a_k\cos k \omega \bar x+\bar b_k\sin k \omega \bar x] \!$$

for n=0,1,2,4,8. Observe the values of $$ f_n(\bar x) \!$$ at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of $$ f(x) \!$$

Solution

 * $$ p=2L=4 \!$$

 * $$ L=2 \!$$

 * $$ \omega =\frac{2\pi}{p} \!$$

$$ \bar a_0=\frac{1}{2L}\int_{-L}^{L}f(\bar x)d\bar x \!$$

 * $$ \bar a_0=\frac{1}{4}\int_{-2}^{2}f(\bar x)d\bar x \!$$

 * $$ \bar a_0=\frac{A}{2} \!$$

$$ \bar a_k=\frac{1}{L}\int_{-L}^{L}f(\bar x)\cos k\omega \bar xd\bar x \!$$

 * $$ \bar a_k=\frac{1}{2}\int_{-2}^{2}f(\bar x)\cos \frac{k\pi \bar x}{2}d\bar x \!$$

 * $$ \bar a_k=\frac{2A}{k\pi}\sin \frac{k\pi}{2} \!$$

Note:

If k is even $$ \bar a_k=0 \!$$

If k=1, 5, 9... $$ \bar a_k=\frac{2A}{k\pi} \!$$

If k=3, 7, 11... $$ \bar a_k=\frac{-2A}{k\pi} \!$$

$$ \bar b_k=\frac{1}{L}\int_{-L}^{L}f(\bar x)\sin k\omega \bar xd\bar x \!$$

Note:

$$ \bar b_k=0 \!$$ for n=1, 2...

Then Fourier series becomes a Fourier cosine series:

$$ f_k(\bar x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\cos \frac{k\pi}{2}\bar x)\!$$

For n=0

$$ f_k(\bar x)=\frac{A}{2}\!$$

For n=1

$$ f_k(\bar x)=\frac{A}{2}+(\frac{2A}{\pi}\cos \frac{\pi}{2}\bar x)\!$$

For n=5

$$ f_k(\bar x)=\frac{A}{2}+\frac{2A}{\pi}(\cos \frac{\pi}{2}\bar x-\frac{1}{3}\cos \frac{3\pi}{2}\bar x+\frac{1}{5}\cos \frac{5\pi}{2}\bar x)\!$$

After transforming the variable of $$ f(\bar x) \!$$

$$ f_k(x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\cos \frac{k\pi}{2}(x-1.25))\!$$

Part 2
Do the same as above, but using $$ f(\tilde x) \!$$ to obtain the Fourier expansion of $$ f(x) \!$$ ; compare to the result obtained above

Solution

 * $$ p=2L=4 \!$$

 * $$ L=2 \!$$

 * $$ \omega =\frac{2\pi}{p} \!$$

$$ \tilde a_0=\frac{1}{2L}\int_{0}^{2L}f(\tilde x)d\tilde x \!$$

 * $$ \tilde a_0=\frac{1}{4}\int_{0}^{4}f(\tilde x)d\tilde x \!$$

 * $$ \tilde a_0=\frac{A}{2} \!$$

$$ \tilde a_k=\frac{1}{L}\int_{0}^{2L}f(\tilde x)\cos k \omega \tilde x d\tilde x \!$$

 * $$ \tilde a_k=\frac{1}{2}\int_{0}^{4}f(\tilde x)\cos k \omega \tilde x d\tilde x \!$$

 * $$ \tilde a_k=\frac{A}{k\pi}\sin \pi k \!$$

Note:

$$ \bar b_k=0 \!$$ for n=1, 2..

$$ \tilde b_k=\frac{1}{L}\int_{0}^{2L}f(\tilde x)\sin k \omega \tilde x d\tilde x \!$$

 * $$ \tilde b_k=\frac{1}{2}\int_{0}^{4}f(\tilde x)\sin k \frac{\pi}{2} \tilde x d\tilde x \!$$

 * $$ \tilde b_k=\frac{-A}{\pi k}(\cos \pi k -1) \!$$

Note:

If k is even $$ \bar a_k=0 \!$$

If k=1, 3, 5... $$ \bar a_k=\frac{2A}{k\pi} \!$$

Then Fourier series becomes a Fourier sine series:

$$ f_k(\tilde x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\sin \frac{k\pi}{2}\tilde x)\!$$

For n=0:

$$ f_k(\tilde x)=\frac{A}{2}\!$$

For n=1:

$$ f_k(\tilde x)=\frac{A}{2}+(\frac{2A}{\pi}\sin \frac{\pi}{2}\tilde x) \!$$

For n=5:

$$ f_k(\tilde x)=\frac{A}{2}+\frac{2A}{\pi}(\sin \frac{\pi}{2}\tilde x+\frac{1}{3}\sin \frac{3\pi}{2}\tilde x+\frac{1}{5}\sin \frac{5\pi}{2}\tilde x) \!$$

After transforming the variable of $$ f(\bar x) \!$$

$$ f_k(x)=\frac{A}{2}+\sum_{k=1}^{n}(\frac{2A}{k\pi}\sin \frac{k\pi}{2}(x-.25))\!$$


text:
Talk:US States/West Virginia

This article should be moved to http://en.wikiversity.org/wiki/West_Virginia

A mess.
This project is a mess. Needs lots of cleanup. referring to WV. CezauroBlanca (discuss • contribs) 15:33, 27 November 2019 (UTC)

 * It looks like it could all be deleted with redirects to Wikipedia. It's not adding value beyond what is already available there. -- Dave Braunschweig (discuss • contribs) 16:57, 27 November 2019 (UTC)

title: Talk:US States/West Virginia


text:
Talk:Introduction to Category Theory/Equivalence of categories

Categories are defined in lesson 4. This section seems to be in the wrong spot.


text:
User:Thehelpfulone

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title: User:Thehelpfulone


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User talk:Thehelpfulone

Hi! Welcome to my user talk page, please see my Meta-Wiki user page for more information about me. If you'd like to get in contact with me, please contact me on my Meta-Wiki talk page or by e-mail. The Helpful One

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