Thank you! View all posts by Whit Ford, am very thankful 2 the information above.it is very helpful to me. The notation that follows a capital Pi describes only the term that is to be multiplied. A “series” is the sum of the first N terms of a sequence. Chapters & Colonies . Two of these hybrids from each C atom overlap with H 1s orbitals, while the third overlaps with an sp 2 hybrid on the other C atom. (mathematics) Braid group algebra. and therefore I will always be checking this site. A description of the double bond is the sigma-pi model shown in Figure 1. Search for "sigma symbol" in these categories . These symbols can be copy and pasted anywhere you want. All that matters in this case is the difference between the starting and ending term numbers… that will determine how many twos we are being asked to add, one two for each term number. Yes No. (Physics, scattering) Cross_section_(physics). i=2: (14/12)(8/12)(6/12) The pi bond is the "second" bond of the double bonds between the carbon atoms and is shown as an elongated green lobe that extends both above and below the plane of the molecule. Differentiating this would turn the right side into the reciprocal of the original sum times its derivative = a mess. The Sigma symbol, , is a capital letter in the Greek alphabet. So, my opinion would be: sure! I will modify my response shortly. The Pi symbol, , is a capital letter in the Greek alphabet call “Pi”, and corresponds to “P” in our alphabet. then there is no need for a notation to represent repeated exponentiation, since exponents that are products already represent repeated exponentiation. Pi (capitale Π, minuscule π ou parfois ϖ ; en grec πι) est la 16 e lettre de l'alphabet grec, précédée par omicron et suivie par rhô. Thanks , Very useful post. Therefore, additional notations are not needed to describe repeated subtraction or division… Which is quite convenient. Ways to make pi symbol, HTML unicode entities and more. Character Palette allows you to view and use all characters and symbols, including pi, available in all fonts (some examples of fonts are "Arial", "Times New Roman", "Webdings") installed on your computer. Post was not sent - check your email addresses! Choose from 500 different sets of phi sigma rho flashcards on Quizlet. The convention is to increase it, just like with Sigma and Pi notation, but they also support decreasing indeces. I have a student asking whether there is a symbol for exponentiation of a sequence? You can put it in Facebook, Youtube or Instagram. it would appear as though the quantity in parentheses is becoming increasingly negative (a sum of growing negative numbers), and therefore the value probably goes to negative infinity. Sigma Symbol, The Eighteenth Letter in Greek Alphabet – ©Niakris6 at ShutterStock The sigma sign has been used in mathematics, chemistry, and other fields of study for a very long time. Equation for Xn in terms of P1,P2,……Pn. Sigma Pi Fraternity, International has chartered over 230 chapters with 123 currently active plus 6 additional colonies in the United States and Canada and is headquartered in Lebanon, Tennessee. In my mind, this rounds up each time the value is divided by (1-r). Summation notation does not provide an easy way that I can think of to do what you describe. I like it … And I hope it will help other students too to acheive their goals …. Sigma Symbol in Greek Alphabet. sigma and pi? Those are only the first numbers of pi, there's a symbol for pi precisely because you can't precisely specify pi with any decimal number, only to a certain degree of accuracy. Excellent blog and makes maths symbols and operations simple to understand. for i = 0 to 32. please help improve this site's prominence in search results by including a link to this site in appropriate places elsewhere on the internet - perhaps in a response to a math question, or in a comment on a math blog, etc. The “square root of negative one” is not i ; it is represented by i . Acest articol despre o literă grecească este deocamdată un ciot. A shielding constant. Interesting question! Configure your keyboard layout in Windows so that you can type all additional symbols you want as easy as any other text. 117. If the index limit above the Pi symbol is a variable, as in the example you gave: I am not aware of such notation, and furthermore, I am not aware of situations where such notation would be needed. If I have interpreted the expression you show correctly, it is neither an arithmetic nor a geometric sequence. To facilitate this, a variable is usually listed below the Sigma with an equal sign between it and the starting term number. Quick Help to Type the Pi symbol. Is there a way to rewrite the following expression using both sigma and pi? Sigma notation provides a compact way to represent many sums, and is used extensively when working with Arithmetic or Geometric Series. Sir,If I have equation like this : Σ "sigma" Ce symbole mathématiques permet d'exprimer plus simplement certaine somme et donc certaines expressions Exemples : Exemple 1 : la somme de 1 à 100 peut s'écrire : Exemple 2 : la somme des nombres impairs de 1 à 13 peut s'écrire : Un nombre pair est de la forme 2j où j est un entier naturel , 2j + 1 est donc l'écriture d'un nombre impair. Do you know of situations that require repeating exponentiation to model them? It has been represented by … Since its inception, the fraternity has initiated more than 100,000 men and has 5,300 undergraduate members. It’s basically a for loop in scripting, makes so much sense. I want to do that to signify that the matrices do not commute. If you can illustrate it please. However, your expression leaves me uncertain as to whether you are analyzing the situation correctly or not. But if you are trying to give a general answer, you should show each term individually so that the person reading your answer can see any pattern that is developing, and understand how to fill in the “…” used to represent all the terms that are not shown. In Gematria it has the value of 200. 1 Translingual. Raccourci clavier pour Pi π. Alt + 227. You are correct – this can be represented using a combination of Sigma and Pi notation: In the above notation, i is the index variable for the Sum, and provides the starting number for each product. can there be a bigger value at the base and smaller value at top of the PI operator? Good suggestion – it makes more sense to keep the expression the same as the previous example! When you don't have time in the morning to piece together the perfect outfit, a t-shirt and jeans will always do the trick. Copy and paste pi symbol or look below to find out how to type pi symbol on keyboard. hello sir, Sigma Σ is one of the most popular mathematic signs which means a summation of something. So for instance, if I wanted to round the above to the nearest whole after each division (or multiplication) step I think I could write: ⌈⌈⌈I÷(1-r)⌉÷(1-r)⌉÷(1-r)⌉…. In this tutorial, you’ll learn 4 different ways you can insert the sigma symbol in Word. Our men strive for excellence by living our core values to promote fellowship, develop character and leadership, advance heightened moral awareness, enable academic achievement, and inspire service. You can add the Greek language on any iOS devices. 2) The values grow in magnitude linearly by 2 each time. Comment taper tous les symboles et signes sur votre clavier. The post was tremendously helpful. of 14. Sigma notation is most useful when the “term number” can be used in some way to calculate each term. See more ideas about sigma pi, sigma, delta. This would be easy to do in a computer program, but not so much using summation notation. Sigma Pi Letter T-Shirt comes with Sigma Pi in 4-inch Greek Twill letters sewn on the front. What I am wondering about is this. thank you for the amazing and very helpful post. multiply until n reaches 143 (i.e. Subtraction can be rewritten as the addition of a negative. Putting the three thoughts above together, I get: using Sigma or Pi notation, or possibly both. Make your mail stand out from the crowd with our wonderful selection of designs made with love, just for you! The difficulty you describe is that you wish to specify what happens to the result of that product, and capital Pi notation does not provide any means to do that. Xi Sigma Pi was originally founded at the University of Washington on November 24, 1908, and is now comprised of over 40 chapters across the USA and Canada. For repeated exponentiation I would assume that form rather than (x^a)^b. Achetez AZSTEEL Sigma Pi T-Shirt 3.14 Symbol Math Science | for Men Women Comfortable Fit Wearable Anywhere, White and Black in Sizes S-3xl livraison gratuite retours gratuits selon … Raccourci clavier pour Tau τ. Alt + 231. You can input pi symbol using it. I’ll challenge him to find a need for it and maybe he can create his own notation. Hello, I am trying to utilize the Pi notation to represent a repeating multiplication, but one that rounds up to the nearest whole after each time there is a multiplication(or division). A factor of (2n) will produce such numbers, but when n=1 this will have a value of 2, not 3… so I need to add 1 to each value: . To type the Pi (π) Symbol anywhere (like in Word or Excel), press Option + Z shortcut for Mac.If you are on Windows, simply press down the Alt key and type 227 using the numeric keypad on the right side of your keyboard.For Microsoft Word, just type 03C0 and then press Alt + X to get the symbol. La manipulation de sommes, via le symbole (sigma), repose sur un petit nombre de règles. Sigma Pi Fraternity was founded in 1897 at Vincennes University, in Vincennes, Indiana. Notation is a convention, a commonly shared interpretation of some symbols. Indeed a very lucid exposition of Sigma and Pi notations! I suppose a problem could be posed this way if you are being asked to come up with an expression for such a product that does not involve Pi notation: is there some closed form expression involving “n” that represents this product? For Phi Sigma Pi, Steven has served as National Secretary/Treasurer, VP of Membership Development and National President. Pi Symbol in Greek Alphabet. Before I continue please forgive my mathematical illiteracy, I am taking an amateur interest in this. You earned the right to be called a Phi Sigma Pi by getting it done in the classroom. Este utilizată în matematic ă pentru a desemna o sumă de elemente. Good point, however, x^a^b is not the same as x^ab. We work hard daily to ensure that you have a diverse and complete selection of Sigma Pi T-Shirts to choose from. 1,375 sigma symbol stock photos, vectors, and illustrations are available royalty-free. The alternative form of sigma (ς) must be used in word-final position. Election is a lifelong membership and includes a once-year complimentary membership in the Society of Physics Students (SPS). If I wanted to represent something being rounded up I think I could use ceiling function brackets: ⌈⌉. does work? In Javascript you should write like a = "this \u2669 symbol" if you want to include a special symbol in a string. (-1)^n will change sign every time “n” grows by one, but when n=1 it is negative – which is the wrong sign for the first time. Change ), You are commenting using your Facebook account. The Sigma Pi fellowships strive for excellence and moral and character development by living by their e… 1.1.1 Usage notes; 1.2 See also; 2 Greek. GREEK CAPITAL LETTER PI ← Ο [U+039F] Greek and Coptic: Ρ → [U+03A1] Contents. Jan 6, 2017 - Explore Something Greek's board "Delta Sigma Pi", followed by 11296 people on Pinterest. Note that “two cubed” and “sigma pi” are puns — they depend upon the sound of what they are. Tabs Dropdowns Accordions Side Navigation Top Navigation Modal Boxes Progress Bars Parallax Login Form HTML Includes Google Maps Range Sliders Tooltips Slideshow Filter List Sort List. Sigma Pi has chosen for its mascot, the owl, a ubiquitous symbol of wisdom that well-suits its members devotion to the alma mater and to high academic achievement, but the values of the Fraternity are not for college days alone. The letter is pronounced as "p". The table below is a Gem in The Rough. Basically this is where k = n. It’s important to emphasize that. And all of them can produce pi text symbol. So maybe we do still need something? I simply cannot figure out how to represent that using big Pi Π. Once that has been evaluated, you can evaluate the next sigma to the left. Annexe E Liste des symboles mathématiques usuels (LATEX) Vous trouverez ci-dessous la liste des commandes LATEX permettant de produire les symboles mathématiques les plus courants. Symbolscopyandpaste.com consists of combination of discrete math symbols and latex math symbols like root symbol, sigma symbol, square root symbol, pi symbol text, under root symbol, infinity text symbol, sigma text symbol, square root of pie, square root of infinity, pi emoji, sigma sign and much more. Užití ve vědě. An online LaTeX editor that's easy to use. This plane contains the six atoms and all of the sigma bonds. Using Pi notation in the exponent achieves the desired purpose. Misconception: many students in the Pacific may have this worng notion that a sigma . But, perhaps I do not understand the situation you seek to describe. If you're using it often you can add a shortcut in Settings ➜ General ➜ Keyboard ➜ Text Replacement. If the sum of a bunch of terms in known as a “summation of a series”, then what is the product of a bunch of terms known as in mathematics? The Stefan–Boltzmann constant. HTML CSS JavaScript Python SQL … The Steven A. DiGuiseppe Excellence in Administration Award was created in his honor. Then, when you need to type it, hold down … πℼ Pi symbol sign Find out how to type Pi sign π directly from your keyboard. LIKE US. Try these curated collections. It can also help you lookup Unicode codes for entering symbols with keyboard. My answer to that would be: I probably would not use Sigma Notation to write such a simple expression. Sizing example: 3 … Moreover, π is a transcendental number – a number that is not the root of any nonzero polynomial having rational coefficients. GeoGebra Applets However, I have never worked with infinite products. It has been represented by the Greek letter "π" since the mid-18th century. An upper bound would be provided by an infinite geometric sequence, but I am uncertain what might best provide a lower bound. The Sigma and Pi expression I used to answer the previous question did not have a value specified for “N”, so any value given for the expression will have to be in terms of “N”… as your question is. So like E(x+n) for n=1 to 3 would produce (x+1)^(x+2)^(x+3)… Or maybe((x+1)^(x+2))^(x+3). Reading this post it seems like this would be easy to use the big Pi Π notation. I was finding how to use Sigma notation, and finally found such a good one. expression is So, the sum of the first two terms would indeed be 30. This site is valid XHTML 1.0 Strict, CSS | Get Buzz via RSS or Atom | Colophon | LegalXHTML 1.0 Strict, CSS | Get Buzz via RSS or Atom | Colophon | Legal It is a weaker type of covalent bond as compared to sigma bonds (σ bonds). ( Log Out /  The Pi symbol, , is a capital letter in the Greek alphabet call “Pi”, and corresponds to “P” in our alphabet. No new vocabulary is needed. and if you leave the final index as “n” becomes: Is there some closed form expression that represents this product? For example, X1 means we have One term say P3 and rest two are (1-P) and summation of such product terms for 3 values(P1,P2 and P3). Π π Pi: Ε ε Epsilon: Ρ ρ Ro: Ζ ... Ͱ ͱ Heta: Ϸ ϸ Șo: Ϻ ϻ San: diacritice: Litera Σ (sigma mare) este o literă din alfabetul grec (corespondent pentru litera S în limba română). n=112, n=113 etc. Gargi, Loops in programming languages can be written to decrease the index each time just as easily as they can increase it. As n grows, the constant power of 2 in the expression will dominate the initial results a lot more, but the infinite number of subtractions from it will eventually catch up to its value, no matter how large it is. I have modified the post. https://mythologian.net/greek-alphabet-greek-letters-symbols-and-meanings Our Executive Office is located in Nashville, Tennessee. I still cannot think of either an application for such an expression or a notation for it. 3) There are seven terms, so n will need a starting value of 1, and an ending value of 7. One other thought… if It is used in the same way as the Sigma symbol described above, except that succeeding terms are multiplied instead of added: Sigma (summation) and Pi (product) notation are used in mathematics to indicate repeated addition or multiplication. The coefficients will be “32 Choose i”, or. Since So, if n=3, then It came from the Phoenician letter pē, which meant mouth. X2=(1-P1)P2.P3+(1-P2)P2P3+(1-P3)P2P3 The student in question is actually only 11 years old and somehow I don’t think that he will accept the “not needed” reason! If it ends with, or continues beyond tan(np/2n), which will always be undefined, then my first impression is that there would be no limit to the product. Sigma Pi Phi (ΣΠΦ) is the first successful and oldest Black Greek-lettered organization.It has always been non-collegiate, designed for professionals at mid-career or older. In Modern Greek, this sound is voiced to /z/ before /m/, /n/, /v/, /ð/ or /ɣ/. So Sigma notation describes repeated subtraction when its argument is a negative quantity. Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the molecule. Pi sign is one of the most popular mathematic constants and means a ratio of a circle perimeter to its diameter. If j went from one to three each time, the expression on the right would have to be (i + j – 1). Atomic Term Symbol (Explained in detail) | All types of problems asked in CSIR - Duration: 23:52. However, this particular example can be “simplified” by collecting like terms to become Section: Internet Tutorial: Greek Letters Fabulous Code Chart for Greek Letters & Symbols … With more than 140 collegiate chapters on campuses throughout the U.S., Phi Sigma Pi is one of the largest and most active fraternities out there. Thanks for your clear explanations. If I wanted to take, let’s say “I”, and multiply “I” by a repeating multiple, let’s say “1/(1-r)”. 1) your expansion of the problem using square brackets Go to Settings, General, Keyboard, and Keyboard again. How can I type a pi symbol on an iPad? There are several in the posting… Ooops – I just realized you were asking about my reply to the comment. Sigma Pi Sigma exists to honor outstanding scholarship in physics, to encourage interest in physics among students at all levels, to promote an attitude of service, and to provide a fellowship of persons who have excelled in physics. Sorry, your blog cannot share posts by email. i=1: (14/12)(8/12) Name Unicode Glyph Unicode Name Description Aliases; alefsym: 02135: ALEF SYMBOL : Alpha: 00391: GREEK CAPITAL LETTER ALPHA : alpha: 003B1: GREEK SMALL LETTER ALPHA How would I derive the polynomial for the following expression: n = 112 which would raise the question: “why write it using Sigma Notation when you could just as easily write ?”. A more typical use of Sigma notation will include an integer below the Sigma (the “starting term number”), and an integer above the Sigma (the “ending term number”). To make use of them you will need a “closed form” expression (one that allows you to describe each term’s value using the term number) that describes all terms in the sum or product (just as you often do when working with sequences and series). Your answer options suggest that there is some expansion of a a logarithm that results in an infinite product of tangent functions, however I am not familiar with that. Thank you, this was very helpful. ), I’m interested in simplifying the polynomial to 32 terms and determine the exponents of y, Using Pi notation, I interpret your question to be, Using a binomial expansion, the terms will be Symbol . Math teacher, substitute teacher, and tutor (along with other avocations) GREEK PI SYMBOL: ϗ : 983: 03D7 : GREEK KAI SYMBOL ... GREEK CAPITAL REVERSED DOTTED LUNATE SIGMA SYMBOL Previous Next COLOR PICKER. Pi π is an irrational number, which means that it cannot be expressed exactly as a ratio of two integers; consequently, its decimal representation never ends and never repeats. These Sigma Pi fraternity stickers measure Approx 2" tall Greek Letter Sticker. I was just practicing the question wanted to know can 30….. n(n+1)(n+2) be the ans to the above sigma and product equation given by you. A “product” is the result of multiplying two or more “factors”. 2.1 Letter; 2.2 See also; Translingual Symbol Π. English Wikipedia has an article on: Capital pi notation. This site uses Akismet to reduce spam. In Gematria it has the value of 200. As is the case with most of the other alphabets, even English for that matter, two different symbols are used to represent sigma in uppercase and lowercase respectively. Furthermore is there a way of simplifying the notation and finding a result that is a function of n? But what if the Pi notation is not in closed form, such as. . I do not follow your thinking though when you say you wish to use a descending index value to indicate that matrices do not commute… I would not perceive a descending index value, or an ascending one, to indicate anything about the commutative property’s applicability to the resulting expression. (mathematics) Sum of divisors. How can u write this using summation notation: 3 -5+7 -9+11-13+15? You'll have to copy paste pi symbol π if you're on iOS or iPad OS. All rights to text and images hosted on this blog site are reserved by the author. what is the relation between the two when they are logarithmic differentiated? You can make frequently used technical non-fancy symbols like "√ ∑ π ∞ ∆ ™ © æ £ ¢" and åccénted letters on Mac using [Option] key. You press Alt and, while holding it, type a code on Num Pad while it's turned on. So you should wear lettered apparel that gets it done too. For example, if n=1, then the expression would be: Certificates. and if n=4, then So will provide the correct sign for the nth term. It's a text symbol. There actually are 3 different ways to type symbols on Linux with a keyboard. How should I proceed if I want to get it for n instead of 3. 3d sigma sigma logo sigma icon sigma sigma vector sigma letter 6 steps flow greece graffiti omega letter greek lettering. ( Log Out /  There is no emoji for pi at the moment. Division can be rewritten as multiplication by the reciprocal. Một bộ sưu tập các biểu tượng toán học, chẳng hạn như: Pi, Infinity, Sum, Sigma, Square Root, Integral ... các dấu hiệu & ký hiệu. I have a question, please. The letter is pronounced as "p". Welcome them to your fraternity house properly with Sigma Pi Signs, Flags and Banners. Pi symbol π is important for calculation of circular and spherical figures and stands for value of 3.141592653589793238.. . Actually n SHOULD be squared in his reply since he’s saying that that’s the LAST TERM in the product. Character map allows you to view and use all characters and symbols available in all fonts (some examples of fonts are "Arial", "Times New Roman", "Webdings") installed on your computer. Question. How to find the derivative of the pi notation. ( Log Out /  hope to receive your reply as soon as possible. Sigma Pi (ΣΠ) is an international social collegiate fraternity founded in 1897 at Vincennes University. • In both Ancient and Modern Greek, the sigma represents the voiceless alveolar fricative /s/. We have an awesome collection of mathematical symbols for you. Please, read a guide if you're running a laptop. However, while it is common to see uppercase and lowercase sigma signs in science textbooks and the likes, what many people are completely unaware of is the fact that the sigma sign has a third form as well. The symbol of Phi Beta Sigma Fraternity Incorporation is the White Dove What is the name of our publication? what can be the correct answer this equation? In such cases, just as in the example that resulted in a bunch of twos above, the term being added never changes: The “starting term number” need not be 1. It corresponds to “S” in our alphabet, and is used in mathematics to describe “summation”, the addition or sum of a bunch of terms (think of the starting sound of the word “sum”: Sssigma = Sssum). i.e. Had always skipped these symbols in technical papers and today is the first time i get to understand what they mean. Pi notation provides a compact way to represent many products. The Fraternity currently has over 285,000 total initiated members. Hi Mr. Ford, See more ideas about Phi sigma pi, Sigma pi, New mexico state university. Does Multiplication operator always increment? Change ). Cet article a pour objet de les énumérer et d’en donner des exemples d’utilisation, sans aucune prétention à l’originalité. There is often a pattern to them, a formula that can be used to determine the value of the next term in the sequence. For example, $\prod_{i = 3}^6 (x + 2^i) = (x + 2^3) (x + 2^4) (x + 2^5) (x + 2^6)$ In general, we have $\boxed{\prod_{i = a}^b f(i) = f(a) \cdot f(a+1) \cdot f(a+2) \cdot … \cdot f(b)}$ Properties of Pi Notation. So, even if it is not commonly used in a particular way, there is no strong reason I can think of why you couldn’t use it that way (if necessary, including a note or example describing how you intend the notation to be interpreted). You can keep the one you order from Greek Gear in the basement or closet of your chapter house for years of use. Series definitions almost always rely on summation notation. Raccourci clavier pour Theta Θ. Alt + 233 Thanks…. then there are an indeterminate number of factors in the product until such time as “n” is specified. Minususcule : Alt + 229. Let’s list the first few terms of this sequence individually to get a sense of how this series behaves: If sigma is for summation, and pi is for multiplication, are there any notations for division and subtraction? In this case only two of the p orbitals on each C atom are involved in the formation of hybrids. Learn phi sigma rho with free interactive flashcards. It came from the Phoenician letter pē, which meant mouth. The symbol for uppercase sigma is Σ Chỉ cần nhấp vào một dấu hiệu hoặc biểu tượng để sao chép nó vào bảng tạm và dán nó ở bất cứ đâu. Thanks! Understanding math topics without memorization. Which “k” are you referring to? I've compiled a list of shortcuts in my article and explained how to open keyboard viewer. This vector image was created with a text editor. If you need to differentiate a sum, I would not expect logarithmic differentiation to be very useful, as the laws of logarithms do not allow us to do anything with something like Pi Bond are formed by the overlapping of atomic orbital’s sidewise. If I were to see an upper index value that is smaller than the lower one, my first assumption would be that I would need to decrease the index by 1 for each iteration – which seems to be what you intend. HOW TO. Ooops – didn’t think of CharMap allows you to view and use all characters and symbols available in all fonts (some examples of fonts are "Arial", "Times New Roman", "Webdings") installed on your computer. Phi Sigma Pi Necklace - Phi Sigma Pi Gift - Big Gift - Little Gift - Gift For Big - Gift For Little - Wholesale Sorority - Sorority Gift FelicityAndBliss. The pi symbol will be shown on the virtual keyboard. The Sigma symbol can be used all by itself to represent a generic sum… the general idea of a sum, of an unspecified number of unspecified terms: But this is not something that can be evaluated to produce a specific answer, as we have not been told how many terms to include in the sum, nor have we been told how to determine the value of each term. Is represented by the Greek letter Sticker, perhaps I do not understand the situation you seek to describe Greek! The Greek letter  π '' since the mid-18th century please, read a guide if you 're iOS. Simply can not Figure out how to find a need for summation notation: 3 -9+11-13+15. Convention is to be a leader for others closet of your Chapter house for years use... … and I hope it will help other students too to acheive their goals … and!, which meant mouth will be negative vào bảng tạm và dán nó ở bất cứ đâu would that! Usually have no need for it and maybe he can create his own notation at New Mexico University... Π. English Wikipedia has an article on: capital Pi notation provides compact... Can be rewritten as multiplication by the Greek alphabet assume that form rather than x^a... Can I type a code on Num Pad while it 's turned on sense to keep the expression same..., such as a member of the p orbitals on each C atom are involved in the Formation of.... Using a backwards sigma symbol list of HTML and JavaScript entities for Pi at the moment we grant this... \U2669 symbol '' if you could provide an example of what you commenting! Also be considered as math symbols can also be used in word-final position deocamdată un ciot sum... Sp Sp2 Sp3, Organic Chemistry, bonding - Duration: 36:31 Arithmetic... Please reply in my mind, this sound is voiced to /z/ before /m/ /n/. I ”, or possibly both should wear lettered apparel that gets it done in product. X^A ) ^b, version control, hundreds of LaTeX templates, and found... The exponent achieves the desired purpose initiated members in terms of a circle to! Este deocamdată un ciot Alt + 228 an ending value of 1, and finally found such good. Matematic ă pentru a desemna o sumă de elemente a single “ term.... Vào bảng tạm và dán nó ở bất cứ đâu sigma and Pi is an social! Do with his investigation into combination formulae… he ’ s important to emphasize that s important emphasize... To my list atom for a while scholarship, leadership, and Pi notations (... “ series ” is the relation between the two when they are I was finding how to interpret notation! Finding how to type sum symbol with keyboard using this technique that has been evaluated, are... ) 1 ( b ) 2-log2 ( C ) 3 -log4 total of six sigma bonds also decreasing... What is the sigma-pi model shown in Figure 1 sigma Rho flashcards Quizlet... Six sigma bonds are formed by the Greek alphabet standard deviation symbol are used interchangeably for this character sigma pi symbol sur. = 112 expression is multiply until n reaches 143 ( i.e sequence ” is not ;! Appeared at that time ) …… } ^ ( 1/n ) on Desktops and Laptops! To open keyboard viewer it done in the molecule symbol '' in these categories and subtraction JavaScript should. Involves a limit and a power outside of any Pi notation this vector image was created with a text.! ) ^b different techniques depending on your system quite convenient of the on. And receive notifications of New posts by email and most Laptops running MS Windows more! If the Pi symbol in Greek alphabet constants and means a summation of something that you can evaluate next... Colorado State University was established in 1943 I derive the polynomial for the nth term a for loop in,! I think I could use ceiling function brackets: ⌈⌉ are there any for! 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# Number of lists in which an element is repeated consecutively exactly twice

I have an integer list that is n long and each value can be ranging from 1 .. n.

I need a formula that tells me how many of all possible lists for a given n, that have one or more consecutive sequences of a length of exactly 2 of the same number and no other consecutive sequences that are longer than 2.

For example for n=5:

These two should count:

{ 1, 1, 5, 3, 3 }
{ 2, 3, 2, 5, 5 }

Where as these should not:

{ 1, 1, 1, 2, 2 }
{ 1, 3, 2, 5, 4 }

I've been attempting to do this by looking at the possible sequences using the following formula where x = n-1:

(n) x n n = x * n^3
x (n) x n = x^2 * n^2
n x (n) x = x^2 * n^2
n n x (n) = x * n^3

And sum these four up.

However, these also needs to take overlaps between the four into account. This is where I could use a bit of help..? What would the formulas be for excluding the overlaps?

An alternative approach would also be welcome.

Going trough all sequences counting manually is not an option - I need this to work for for n larger than what makes that approach computationally feasible.

If it helps anyone, I've written a little program that runs trough all the sequences and counts have the following results:

n = 2
L[2]: 2
L[1]: 2

n = 3
L[3]: 3
L[2]: 12
L[1]: 12

n = 4
L[4]: 4
L[3]: 24
L[2]: 120
L[1]: 108

n = 5
L[5]: 5
L[4]: 40
L[3]: 280
L[2]: 1520
L[1]: 1280

Where n = 5, L[2]: 1520 is the result I've asked for a formula to in the above question.

• You are using the word permutation to mean something different from what the rest of us mean when we use that word. – Gerry Myerson Sep 29 '13 at 12:59
• I rephrased it a bit - I hope it makes better sense. – Kasper Middelboe Petersen Sep 29 '13 at 13:10
• Are ${ 2, 3, 2, 5, 5 }$ and ${ 3, 2, 2, 5, 5 }$ different sequences that both count? – miracle173 Sep 29 '13 at 13:37
• @miracle173 yes – Kasper Middelboe Petersen Sep 29 '13 at 13:40
• Does $1, 2, 3, 1$ count? Ie, can the doubled elements "wrap around"? – Jack M Sep 29 '13 at 13:58

How many sequences of length $q$ of numbers from $\{1,...,p\}$ are there such that consecutive elements are always different? For the first element of such a sequence we can selcet one of the $p$ different values of $\{1,...,p\}$. For the following $q-1$ positions we can select always all values from $\{1,...,p\}$ except the value of its predecessor in the sequence. These are $p-1$ possible values. So we have

$$p(p-1)^{q-1}$$

different sequences.

$\Omega_n$ is the number of sequences of length $n$ with elements from $\{1,\ldots,n\}$ such that no three consecutive elements have the same value but at least one pair of consecutive elements have the same value.

$\Omega_{n,k}$ is the number of sequences of length $n$ with elements from $\{1,\ldots,n\}$ such that no three consecutive elements have the same value but exactly $k$ pairs of consecutive elements have the same value.

For arbitrary $n$ there are $k \le \frac{n}{2}$ different possible values for the number of consecutive element pairs that have equal values.

We have

$$|\Omega_{n}|=\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }|\Omega_{n,k}|$$

Now we select a $k$. Choose a sequence of $n-k$ values from $\{1,\ldots,n\}$ such that two consecutive values always differ. There are $n (n-1)^{n-k-1}$ such sequences. For such a sequence we select $k$ of its elements (there are $\binom{n-k}{k}$ such possibilities) and insert an element with the same value after each selected value. So $$|\Omega_{n,k}|=\binom{n-k}{k}n(n-1)^{n-k-1}$$ and $$|\Omega_{n}|=\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }\binom{n-k}{k}n(n-1)^{n-k-1}$$

• For $n=3$ there are $12$ sequences.

• For $n=4$ there are t $120$ sequences.

• For $n=5$ there are $1520$ sequences.

The answer to the original question is deleted

• @KasperMiddelboePetersen The formula would be $n\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }\frac{(n-k)(n-k-1)\ldots(n-2k+1)}{k!}(n-1)^{n-k-1}$. If you are doing more combinatorics problems, it would be a worth-while investment of time to study binomial coefficients, just a bit. – Marc van Leeuwen Sep 29 '13 at 20:27
• @MarcvanLeeuwen thanks - I'd just figured it out and deleted the comment 30 seconds before you answered :) It will be a challenge to calculate this for a large value of n though. – Kasper Middelboe Petersen Sep 29 '13 at 20:32
• I think the only challenge is computing with large integers. For $n=69$, I get 4676495092345476699420593108779018542679373291685221536304769813013106968507850431624264673416798415845073262231682605593395200, no sweat. – Marc van Leeuwen Sep 29 '13 at 20:43

Here is a different solution that may interest you. Introduce three sequences $a_{n,k}$, $b_{n,k}$ and $c_{n,k}$ that count the number of strings over $\Sigma^k$ where $|\Sigma|=n,$ that end in a digit that is not repeated, a digit that is repeated twice and a digit that is repeated at least three times. In fact we take these sequences to be generating functions in two variables, where the variable $v$ counts occurrences of subsequences of length exactly two and $w$ counts occurrences of subsequences of length at least three.

This gives the following set of recurrences: $$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v b_{n,k-1} + (n-1) c_{n,k-1},$$ $$b_{n,k} = a_{n,k-1} \quad\text{and}\quad c_{n,k} = w b_{n,k-1} + c_{n,k-1}.$$

With these settings the generating function of all elements of $\Sigma^k$ classified according to the number of length 2 and length more than two subsequences is given by $$a_{n,k} + v b_{n,k} + w c_{n,k}.$$

Observe that $a_{n,1} = 1$ and $b_{n,1} = c_{n,1} = 0.$ Furthermore the last recurrence implies that $$c_{n,k} = w \sum_{q=1}^{k-1} b_{n,q} = w \sum_{q=1}^{k-2} a_{n,q}.$$

Taken together this yields a recurrence for $a_{n,k}:$ $$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v a_{n,k-2} + (n-1) w \sum_{q=1}^{k-3} a_{n,q}.$$

Introduce the trivariate generating function $$G_n(z) = \sum_{k\ge 1} a_{n,k} z^k.$$

Multiply the recurrence by $z^k$ and sum for $k\ge 4:$ $$\sum_{k\ge 4} a_{n,k} z^k = (n-1) z \sum_{k\ge 4} a_{n,k-1} z^{k-1} + (n-1) vz^2 \sum_{k\ge 4} a_{n,k-2} z^{k-2} \\+ \sum_{k\ge 4} (n-1) w z^k [z^{k-3}] \frac{1}{1-z} G_n(z).$$ Now $a_{n,1} = n, \; a_{n,2} = n(n-1)$ and $$a_{n,3} = n(n-1)^2 + n(n-1)v.$$ The equation derived from the recurrence now becomes $$G_n(z) - (n(n-1)^2 + n(n-1)v)z^3 - n(n-1)z^2 -nz \\= (n-1)z (G_n(z) - n(n-1)z^2 -nz) + (n-1)vz^2 (G_n(z) - nz) \\ + (n-1)w z^3 \sum_{k\ge 4} z^{k-3} [z^{k-3}] \frac{1}{1-z} G_n(z).$$ The sum term simplifies to $$(n-1)w z^3 \frac{1}{1-z} G_n(z).$$ We may now solve for $G_n(z),$ getting $$G_n(z) = -{\frac {nz \left( -1+z \right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{ z}^{2}-v{z}^{3}-w{z}^{3}n+w{z}^{3}+1}}.$$ Now the generating function for $b_{n,k}$ is $$z G_n(z) \quad\text{and the one for}\; c_{n,k} \; \text{is}\quad \frac{wz^2}{1-z} G_n(z).$$ It follows that the generating function $H_n(z)$ of $a_{n,k} + v b_{n,k} + w c_{n,k}$ is $$\left(1 + vz + \frac{w z^2}{1-z}\right) G_n(z)$$ or equivalently $$H_n(z) = -{\frac {zn \left( -1+z-vz+v{z}^{2}-w{z}^{2} \right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{z}^{2}-v{z}^{3}-w{ z}^{3}n+w{z}^{3}+1}}.$$ Now we do not permit sequences of length at least three, so we take $$[w^0] H_n(z) = -{\frac { \left( vz+1 \right) nz}{v{z}^{2}n-v{z}^{2}+nz-z-1}}.$$ In fact there must be at least one two-sequence, so we first calculate $$[w^0] H_n(z) - [v^0] [w^0] H_n(z) = {\frac {v{z}^{2}n}{ \left( nz-z-1 \right) \left( v{z}^{2}n-v{z}^{2}+nz-z-1 \right) }}$$ and put $v=1$, finally obtaining the generating function $$M_n(z) = {\frac {n{z}^{2}}{ \left( nz-z-1 \right) \left( n{z}^{2}-{z}^{2}+nz-z-1 \right) }}.$$ The partial fraction decomposition of $M_n(z)$ is given by $$-{\frac {n}{ \left( n-1 \right) \left( n{z}^{2}-{z}^{2}+nz-z-1 \right) }}+{ \frac {n}{ \left( n-1 \right) \left( nz-z-1 \right) }}.$$

Now the singularities are at $$\rho_0 = \frac{1}{n-1} \quad\text{and}\quad \rho_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{n^2+2n-3}}{2(n-1)}.$$ Expanding $M_n(z)$ into series we obtain the following closed form result for the number of admissible strings of length $k$ and an alphabet of $n$ symbols: $$[z^k] M_n(z) = n\frac{\rho_0^2 \rho_1 \rho_2}{(\rho_0-\rho_1)(\rho_0-\rho_2)} \rho_0^{-k}\\ + n\frac{\rho_0\rho_1^2 \rho_2}{(\rho_1-\rho_0)(\rho_1-\rho_2)} \rho_1^{-k} + n\frac{\rho_0\rho_1 \rho_2^2}{(\rho_2-\rho_0)(\rho_2-\rho_1)} \rho_2^{-k}.$$ The reader is asked to verify that indeed $$[z^n] M_n(z) = \sum_{k=1}^{\lfloor n/2 \rfloor} {n-k\choose k} n (n - 1)^{n-k-1}.$$ The above formula for $[z^k] M_n(z)$ is quite powerful. It gives the exact value of the number of admissible strings with any alphabet of size $n$ and of length $k$. For example, when there are $n=2$ symbols, the sequence starting at length $k=1$ is: $$0, 2, 4, 8, 14, 24, 40, 66, 108, 176.$$ With $n=7$ we get the following sequence: $$0, 7, 84, 798, 6804, 54684, 423360, 3194856, 23668848, 172939536.$$

The generating function $H_n(z)$ encapsulates the complete distribution of the $n^k$ strings classified according to the number of two-sequences (counted by $v$) and sequences of length at least three (counted by $w$).

Here is an example: $$[z^4] H_3(z) = 6\,{v}^{2}+36\,v+15\,w+24$$ for a total of $3^4=81$ terms (strings of length $4$ over $3$ symbols). Another example is: $$[z^6] H_5(z) = 80\,{v}^{3}+1920\,{v}^{2}+520\,vw+20\,{w}^{2}+6400\,v+1565\,w+5120$$ for a total of $5^6=15625$ terms (strings of length $6$ over $5$ symbols). It is a useful combinatorics excercise to verify some of these values with pen and paper. For example, a string of length four over three symbols containing a sequence of length at least three can consist of four equal symbols, giving a contribution of three, or a length three sequence at the front for a contribution of three times two or a length three sequence at the end, again for three times two, for a total of $3+6+6 = 15$ which is indeed the value from the generating function.


# Assigning values to permutations

$N$ objects can be arranged in $N!$ different orders. For example, $10$ playing cards can be stacked $10! = 3,628,800$ different ways. Is there a way to assign a numerical value to each permutation so that every integer from $1$ to $N!$ corresponds to exactly one permutation? Is there a way to derive the permutation from the corresponding numerical value?

• Well, there are many ways: just ennumerate arbitrarily the permutations from $1$ to $\,N!\,$ Mar 12 '13 at 3:05
• This is discussed in detail in Higher-Order Perl, section 4.3.1, pp. 128–135, which is available online. It is also discussed in exhaustive detail in volum IV of Knuth's The Art of Computer programming.
– MJD
Mar 12 '13 at 3:41
• Could you not number each card and then name permutations according to the order? Mar 12 '13 at 4:59

I don't know if there is a standard way of doing this but you could do something like this:

Lets take 4 playing cards b/c it is more managable. Order them some way, say label them a,b,c,d and put them in the order (a,b,c,d). Then

$$1:(a,b,c,d)\\ 2:(a,b,d,c)\\ 3:(a,c,b,d)\\ 4:(a,c,d,b)\\ 5:(a,d,b,c)\\ 6:(a,d,c,b)$$ So fixing "a" as the first entry gives 6 possible permutations, repeating this for b,c,d in the first entry will give you the other 18, for a total of 24. The method is to first fix an ordering and then permute only the last 2, then once thats done permute the last 3 and so on.

• No help so far. Mar 12 '13 at 15:47
• Im not sure what your asking then, you asked if there was a way to assign a number to each permutation. The answer is yes, and here is one way of doing it systematically.
– mv3
Mar 12 '13 at 16:03
• And if you know the original ordering of the objects, (a,b,c,d) then you can reconstruct the permutation from the value
– mv3
Mar 12 '13 at 16:08

Yes. The easiest way is to order them lexicographically. So for $\{0,1,2,3,4\}$ there are $120$ permutations, from $01234$ to $43210$. It is easiest if our permutation numbers run from $0$ to $119$. Of these $4!=24$ have each number first, so if you want permuation $n$, the first number is $a_0=\lfloor \frac n{24} \rfloor$. Then of those, there are $3!=6$ that have each of the remaining numbers first. To find it, compute $a_1=\lfloor \frac {n-24a_0}6 \rfloor$, then increment by $1$ if $a_0 \le a_1$ because you want the $a_1$st of what is left. Now recurse.

• I'm looking for a function. Suppose ten cards are in the order Ace-2-3-4-5-6-7-8-9-10 when arranged in permutation #1. What would permutation #1,000,000 look like? Suppose the cards are in the order 5-9-2-4-10-Ace-3-8-6-7. Which permutation # (an integer between 1 and 3,628,800) would that be? Maybe what I'm asking just isn't possible, but it would be useful to me if it can be done. Mar 12 '13 at 16:09
• @LeroyNimka: It is quite possible. For permutation 1,000,000 the first card is $\lfloor\frac {999,999}{9!}\rfloor+1=3$ where the -1 and +1 come because you are counting from 1. The ones starting with Ace and 2 use up $2\cdot 9!=725760$ so now we want permutation 274240 of the rest. $\lfloor\frac {274239}{8!}\rfloor+1=7$ so we want the seventh card, the $8$. That has used up $6 \cdot 8!=241920$, so we need the $32320$ of the rest, the seventh one, the 9. So we have started with 389. This can be written in a few tens of lines with a loop. Mar 12 '13 at 16:27
• @LeroyNimka: for the other direction, the 5 leads off starting with permutation $4 \cdot 9!+1=1451521. Having the 9 next adds$7 \cdot 8!=282240. The 2 next adds $7!=5040$, so we are up to 1738801. Again, keep going, and again a few tens of lines. Mar 12 '13 at 16:30

Notice that for one and the same letter kept, out of $$n$$, in one position, you have $$(n-1)!$$ possible choices. And that is all that you need.

$$4231, n=4$$

Before we reached $$4$$ we had $$(4-1)(n-1)!$$ permutations. With $$4$$ fixed we had $$(2-1)(n-2)!$$ options. With $$42$$ fixed we had $$(3-1-1)(n-3)!$$ options as $$2$$ is preceding $$3$$ and is to the left of it. With $$423$$ fixed we had $$(1-1)(n-4)!$$ options. It is $$18$$th permutation.

Formula is then:

$$P(a_1a_2...a_m)=\sum_{k=1}^{m}(\alpha_r(a_k)-1)(m-k)!$$

where $$\alpha_r(a_k)$$ is $$1$$-based alphabetical position of symbol $$a_k$$ reduced by the number of symbols that appear earlier in the alphabet and appear to the left of it.

Inverse is more or less obvious. Essentially you write N, the position, in factorial positional system, and read the permutation from the result.

$$18=3110_!$$

We start

$$1234$$ $$3110$$

Pick the one at position $$3$$, remove the position handled and the element ($$4$$).

$$4$$ $$123$$ $$110$$

Pick the next one at position $$1$$, remove the position handled and the element ($$2$$).

$$42$$ $$13$$ $$10$$

Pick the next one at position $$1$$, remove the position handled and the element ($$3$$).

$$423$$ $$1$$ $$0$$

Finally take the last element to have $$4231$$ as expected.

(You can read the order directly. Start from the left and use order in the alphabet, if you find the same position twice, just increment it. Therefore

$$3110_!$$ is $$3, 31, 311=312, 3120 \to 4231$$

$$1210_!$$ would be $$1,12,121=122=123,1230 \to 2341$$)


# What is the probability that this person is female?

There is a person behind a curtain - I do not know whether the person is female or male.

I know the person has long hair, and that 90% of all people with long hair are female

I know the person has a rare blood type AX3, and that 80% of all people with this blood type are female.

What is the probability the person is female?

NOTE: this original formulation has been expanded with two further assumptions: 1. Blood type and hair length are independent 2. The ratio male:female in the population at large is 50:50

(The specific scenario here is not so relevant - rather, I have an urgent project that requires I get my mind around the correct approach for answering this. My gut feel is that it's a question of simple probability, with a simple definitive answer, rather than something with multiple debatable answers according to different statistical theories.)

-
There are not multiple theories of probability, but it is notoriously true that people have difficulties thinking correctly about probabilities. (Augustus DeMorgan, a good mathematician, gave up the study of probability due to its difficulties.) Don't look at debates: look for appeals to principles of probability (such as the Kolmogorov axioms). Don't let this be resolved democratically: your question is attracting many ill-conceived answers which, even if some of them happen to agree, are merely collectively wrong. @Michael C gives good guidance; my reply tries to show you why he's right. –  whuber Jun 22 '12 at 3:46
@Whuber, if independence is assumed, would you agree that 0.97297 is the correct answer? (I believe that the answer might be anywhere between 0% and 100% without this assumption - your diagrams show this nicely). –  ProbablyWrong Jun 22 '12 at 4:53
Independence of what, precisely? Are you suggesting that female and male hairstyles are the same? As you say in your question, this particular scenario involving gender/hair/blood type may not be relevant: that tells me you seek to understand how to solve problems like this in general. To do that you will need to know which assumptions imply which conclusions. Thus you need to focus very carefully on the assumptions you are willing to make and determine exactly how much they allow you to conclude. –  whuber Jun 22 '12 at 13:07
The kind of independence to explore concerns the combination of all three characteristics. E.g., if AX3 is a marker for a syndrome that includes baldness in females (but not in males), then any long-haired person with AX3 is necessarily male, making the probability of being female 0%, not 97.3%. I hope this makes it obvious that anybody producing a definite answer to this question must be making additional assumptions, even if they do not explicitly acknowledge them. The truly useful answers, IMHO, would be those that show directly how different assumptions lead to different results. –  whuber Jun 22 '12 at 14:31
You're missing the probability that a female doesn't have long hair. That's a critical measure. –  Daniel R Hicks Jul 3 '12 at 19:54

Many people find it helpful to think in terms of a "population," subgroups within it, and proportions (rather than probabilities). This lends itself to visual reasoning.

I will explain the figures in detail, but the intention is that a quick comparison of the two figures should immediately and convincingly indicate how and why no specific answer to the question can be given. A slightly longer examination will suggest what additional information would be useful for determining an answer or at least obtaining bounds on the answers.

### Legend

Cross-hatching: female / Solid background: male.

Top: long-haired / Bottom: short-haired.

Right (and colored): AX3 / Left (uncolored): non-AX3.

### Data

Top cross-hatching is 90% of the top rectangle ("90% of all people with long hair are female").

Total cross-hatching in the right colored rectangle is 80% of that rectangle ("80% of all people with this blood type are female.")

### Explanation

This diagram shows schematically how the population (of all females and non-females under consideration) can simultaneously be partitioned into females/non-females, AX3/non-AX3, and long haired/non-long haired ("short"). It uses area, at least approximately, to represent proportions (there's some exaggeration to make the picture clearer).

It is evident that these three binary classifications create eight possible groups. Each group appears here.

The information given states that the upper cross-hatched rectangle (long-haired females) comprises 90% of the upper rectangle (all long-haired people). It also states that the combined cross-hatched parts of the colored rectangles (long-haired females with AX3 and short-haired females with AX3) comprise 80% of the colored region at the right (all people with AX3). We are told that someone lies in the upper right corner (arrow): long-haired people with AX3. What proportion of this rectangle is cross-hatched (female)?

I have also (implicitly) assumed that blood type and hair length are independent: the proportion of the upper rectangle (long hair) that is colored (AX3) equals the proportion of the lower rectangle (short hair) that is colored (AX3). That's what independence means. It is a fair and natural assumption to make when addressing such questions like this, but of course it needs to be stated.

The position of the upper cross-hatched rectangle (long-haired females)is unknown. We can imagine sliding the top cross-hatched rectangle side-to-side and sliding the bottom cross-hatched rectangle side-to-side and possibly changing its width. If we do this so that 80% of the colored rectangle remains cross-hatched, such an alteration will change none of the stated information, yet it can alter the proportion of females in the upper right rectangle. Evidently the proportion could be anywhere between 0% and 100% and still be consistent with the information given, as in this image:

One strength of this method is it establishes the existence of multiple answers to the question. One could translate all this algebraically and, by means of stipulating probabilities, offer specific situations as possible examples, but then the question would arise whether such examples are really consistent with the data. For instance, if someone were to suggest that perhaps 50% of long-haired people are AX3, at the outset it is not evident that this is even possible given all the information available. These (Venn) diagrams of the population and its subgroups make such things clear.

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Whuber, assuming that blood type and hair length are independent, then surely the portion of long haired women with type AX3 should be the same as the portion of short haired women with AX3? I.e. you don't have flexibility to shift rectangles in the way you propose... If we assume also that men and women are 50:50 in the whole population, doesn't that give us enough info to solve this question with a single indisputable answer? –  ProbablyWrong Jun 21 '12 at 6:40
@whuber +1 very nice. –  Michael Chernick Jun 21 '12 at 10:42
ProbablyWrong, take a close look at the question in your comment: because it deals with women, it is making an additional assumption about independence conditional on gender. The assumption of (unconditional) independence of hair and blood type does not mention gender at all, so to understand what it means, erase the cross-hatching from the figures. This, I hope, indicates why we have the flexibility to situate the cross-hatching wherever we like within the upper and lower rectangles. –  whuber Jun 21 '12 at 14:46
@whuber, I like this. However, I have 2 questions / clarifications: 1. the figures seem to assume population proportions for long vs short hair (about 6:4) & ~AX3 vs AX3 (about 85:15), but this is not mentioned in the original question nor discussed in your explanations of the figures. I suspect the pop proportions are not relevant. Am I right / could you clarify that in the explanations? 2. I think this situation is ultimately working w/ the same phenomenon as Simpson's Paradox, only framed differently (coming at the issue from the other direction, as it were). Is that a fair assessment? –  gung Jun 27 '12 at 16:25
@gung, thank you for making those clarifications. The figures of course must represent some proportions in order to work at all, but any proportions not specifically pinned down in the problem statement are free to vary. (I did construct the figure so that about 50% of the population appears female, anticipating a later edit in which this was assumed.) The idea of applying this graphical representation to understanding Simpson's Paradox is intriguing; I think it has merit. –  whuber Jun 27 '12 at 16:34

This is a question of conditional probability. You know that the person has long hair and blood type Ax3 . Let$$\ \ \ \ \ A =\{\text{'The person has long hair'}\}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \{\text{'The person has blood type Ax3'}\} \\ C =\{\text{'The person is female'}\}.$$
So you seek $P(C|A\ \text{and}\ B)$. You know that $P(C|A)=0.9$ and $P(C|B)=0.8$.
Is that enough to calculate $P(C|A\ \text{and}\ B)$? Suppose $P(A\ \text{and}\ B\ \text{and}\ C)=0.7$. Then $$P(C|A\ \text{and}\ B)=P(A\ \text{and}\ B\ \text{and}\ C)/ P(A\ \text{and}\ B)=0.7/P(A\ \text{and}\ B).$$ Suppose $P(A\ \text{and}\ B)=0.8$. Then, by the above, $P(C|A\ \text{and}\ B)=0.875$. On the other hand if $P(A\ \text{and}\ B)=0.9$ we would then have $P(C|A\ \text{and}\ B)$=0.78.

Now both are possible when $P(C|A)=0.9$ and $P(C|B)=0.8$. So we can't tell for sure what $P(C|A\ \text{and}\ B)$ is.

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Hi Michael, If I read you correctly, you're saying the question as posed can't be answered, is that right? Or to put it another way, you'd need more information to answer this question? 1. Let's assume that the rare blood type in my original question doesn't have any impact on a person's desire or ability to grow their hair long. Can the question now be answered? 2. Would you agree that the answer must be GREATER than 0.9? (Because you have a second piece of independent information - blood type - that reinforces the hypothesis that the person is a female) –  ProbablyWrong Jun 21 '12 at 3:30
If $P(A\text{ and }B)$ is independant, then $P(A\text{ and }B)=P(A)P(B)$ and you'll need to specify what fraction of persons have long hair, i.e., $P(A)$ and what fraction of persons have blood type Ax3, i.e., $P(B)$. Also, you can't say that the answer must be greather than 0.9, which is equivalent to stating that $P(C|A\text{ and }B)>0.9$ (I really don't see why). –  Néstor Jun 21 '12 at 7:39
Thank you Nestor. –  Michael Chernick Jun 21 '12 at 10:00
@ProbablyWrong. Yes the problem as initially stated has insufficient information for a unique answer. –  Michael Chernick Jun 21 '12 at 10:04
Why does $$P(C|A\ \text{and}\ B)=P(A\ \text{and}\ B\ \text{and}\ C)\times P(A\ \text{and}\ B)??$$ I thought that $$P(C|A\cap B)=\frac{P(C \cap (A \cap B))}{P(A \cap B)}=\frac{P(A\cap B\cap C)}{P(A\cap B)}$$ using the definition of conditional probability. –  Dilip Sarwate Jun 21 '12 at 11:13

Fascinating discussion ! I am wondering if we specified P(A) and P(B) as well whether the ranges of P(C| A,B) will not be much narrower than the full interval [0,1], simply because of the many constraints we have.

Sticking to the notation introduced above:

A = the event that the person has long hair

B = the event that the person has blood type AX3

C = the event that person is female

P(C|A) = 0.9

P(C|B) = 0.8

P(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)

it does not seem possible to assume that events A and B are conditionally independent given C ! That leads directly to a contradiction: if $P(A \wedge B | C) = P(A| C) \cdot P(B| C) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)}$

then

$P(C| A \wedge B ) = P(A \wedge B | C) \cdot \left( \frac{P(C)}{P(A \wedge B)} \right) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)} \cdot \left( \frac{P(C)}{P(A \wedge B)} \right)$

If we now assume that A and B are independent as well: $P(A \wedge B) = P(A) P(B)$ most terms cancel and we end up with

$P(C| A \wedge B ) = \frac{P(C| A) \cdot P(C| B)}{P(C)} = \frac{0.9 \cdot 0.8}{0.5} > 1$

Following up on whuber's wonderful geometric representation of the problem: While it is true that generally speaking $P(C | A \wedge B)$ can assume any value in the interval $[0,1]$ the geometric constraints do narrow the range of possible values significantly for values of $P(A)$ and $P(B)$ that are not "too small". (Though we can also upper bound the marginals: $P(A)$ and $P(B)$)

Let us compute the {\bf smallest possible value} for $P(C | A \wedge B)$ under the following geometric constraints:

1. The fraction of the upper area (A TRUE) covered by the upper rectangle must be equal to $P(C|A)=0.9$

2. The sum of the areas of the two rectangles must be equal to $P(C)=0.5$

3. The sum of the fraction of the areas of the two colored rectangles (i.e. their overlap with event B) must be equal to $P(C|B)=0.8$

4. (trivial) The upper rectangle cannot be moved beyond the left boundary and should not be moved beyond its minimum overlap to the left.

5. (trivial) The lower rectangle cannot be moved beyond the right boundary and should not be moved beyond its maximum overlap to the right.

These constraints limit how freely we can slide the hashed rectangles and in turn generate lower bounds for $P(C | A \wedge B)$. The figure below (created with this R script ) shows two examples

Running through a range of possible values for P(A) and P(B) (R script) generates this graph

In conclusion, we can lower bound the conditional probability P(c|A,B) for given P(A), P(B)

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Markus, the first paragraph belongs as a separate question rather than within an answer. The subsequent material looks like a good observation but it is hard to follow without being told what $A, B,$ and $C$ represent. Please bear in mind that different users will see the answers in different sequences, according to their preferences and when the answers were last edited, so each answer has to be readable independently of the others (although of course you can link to other answers). –  whuber Jul 3 '12 at 14:34
@whuber: thanks for the useful comment ! I hope the new edits make it more readable and clear. –  Markus Loecher Jul 3 '12 at 18:36
@whuber and others: I had hoped to reignite the discussion but the thread seems to have gone inactive ? No more comments by anyone ? –  Markus Loecher Jul 8 '12 at 20:36

Make the hypotheses is that the person behind a curtain is a woman.

We area given 2 pieces of evidence, namely:

Evidence 1: We know the person has long hair (and we're told that 90% of all people with long hair are female)

Evidence 2: We know the person has a rare blood type AX3 (and we're told that 80% of all people with this blood type are female)

Given just Evidence 1, we can state that the person behind a curtain has a 0.9 probability value of being a woman (assuming 50:50 split between men and women).

Regarding the question posed earlier in the thread, namely "Would you agree that the answer must be GREATER than 0.9?", without doing any Math, I would say intuitively, the answer must be "yes" (it is GREATER than 0.9). The logic is that Evidence 2 is supporting evidence (again, assuming a 50:50 split for the number of men and women in the world). If we were told that 50% of all people with AX3 type blood were female, then Evidence 2 would be neutral and have no bearing. But since we're told that 80% of all people with this blood type are female, Evidence 2 is supporting evidence and logically should push the final probability of a woman above 0.9.

To calculate a specific probability, we can apply Bayes' rule for Evidence 1 and then use Bayesian updating to apply Evidence 2 to the new hypothesis.

Suppose:

A = the event that the person has long hair

B = the event that the person has blood type AX3

C = the event that person is female (assume 50%)

Applying Bayes rule to Evidence 1:

P(C|A) = (P(A|C) * P(C)) / P(A)

In this case, again if we assume 50:50 split between men and women:

P(A) = (0.5 * 0.9) + (0.5 * 0.1) = 0.5

So, P(C|A) = (0.9 * 0.5) / 0.5 = 0.9 (Not surprising, but it would be different if we didn't have 50:50 split between men and women)

Using Bayesian updating to apply Evidence 2 and plugging in 0.9 as the new prior probability, we have:

P(C|A AND B) = (P(B|C) * 0.9) / P(E)

Here, P(E) is the probability of Evidence 2, given the hypotheses that the person already has a 90% chance of being female.

P(E) = (0.9 * 0.8) + (0.1 * 0.2) [this is law of total probability: (P(woman)*P(AX3|woman) + P(man)*P(AX3|man)] So, P(E) = 0.74

So, P(C|A AND B) = (0.8 * 0.9) / 0.74 = 0.97297

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There are a few statements in your answer that do not make sense to me. (1) P(C|A)=0.9 by assumption. Nowhere was it said that P(C)=0.9. We assumed P(C)=0.5. (2) How did you get the result for P(E)? P(woman)=P(man)=0.5 by assumption where you write P(woman)=0.9. –  Michael Chernick Jun 21 '12 at 11:39
The value of P(C) is assumed at 0.5, which is what I've used. The value for P(E) is the probability of Evidence 2 after applying Evidence 1 (which leads to a new hypotheses that the probability that the person is female is 0.9). P(E) = (probability that the person is a woman (given Evience 1) * probability the the person has AX3 if a woman) + (probability that the person is a man (given Evience 1) * probability the the person has AX3 if a man) = (0.9 * 0.8) + (0.1 * 0.2) = 0.74 –  RandomAnswer Jun 21 '12 at 14:42
Your definition of probability of E is a bit confusing and the terms you are using to calculate it look different from what you wrote before. It really doesn't matter though. The answer is apparently correct based on Huu's nicely presented answer. –  Michael Chernick Jun 21 '12 at 14:57
@Michael Except it appears Huu made mistakes. –  whuber Jun 21 '12 at 15:12
This answer is simply wrong. There may be other errors, but this one is glaring. You state a definitive answer for P("Has Long Hair") (your P(A)), and then use that to give your final definitive answer. There simply isn't enough information to determine this, even assuming P(F) = 0.5. Your line to calculate P(A) seems to come from nowhere. Here is the correct formula using Bayes theroem: P(A) = P(A|F)P(F)/P(F|A) from which, using your stated assumptions, get to P(A) = P(A|F)*5/9. However we still don't know P(A|F), which could be anything. –  Bogdanovist Jun 22 '12 at 4:04
show 1 more comment

I believe now that, if we assume a ratio of men and women in the population at large, then there is a single indisputable answer.

A = the event that the person has long hair

B = the event that the person has blood type AX3

C = the event that person is female

P(C|A) = 0.9

P(C|B) = 0.8

P(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)

Then P(C|A and B) = [P(C|A) x P(C|B) / P(C)] / [[P(C|A) x P(C|B) / P(C)] + [[1-P(C|A)] x [1-P(C|B)] / [1-P(C)]]]

in this case, P(C|A and B) = 0.972973

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P[C|A and B)= P(A and B and C)/P(A and B)=P(A and B and C)/ [P(A|B) P(B)]. How did you get your formula? –  Michael Chernick Jun 21 '12 at 10:12
There is probably a way to add conditions so that you get a unique answer. –  Michael Chernick Jun 21 '12 at 10:19
To add by independence of A and B the formula simplifies to P(A and B and C}/[P(A) P(B)]=P(B and C|A)/P(B). –  Michael Chernick Jun 21 '12 at 10:22
The intent of my question was really for you to justify the formula. I don't understand how it would be derived. –  Michael Chernick Jun 21 '12 at 11:24
No, the answer that supposedly used Bayes Rule is incorrect. I'm not sure why you are confused, MC's formula above is correct and cannot be used to get any result, that's what his and Whuber's answers to the question explained! –  Bogdanovist Jun 22 '12 at 4:08

Note: In order to get a definitive answer, the below answers assume that the probability of a person, a long-haired man, and a long-haired women having AX3 are approximately the same. If more accuracy is desired, this should be verified.

You start out with the knowledge that the person has long hair, so at this point the odds are:

90:10

Note: The ratio of males to females in the general population does not matter to us once we find out the person has long hair. For example, if there were 1 female in a hundred in the general population, a randomly-selected long-haired person would still be a female 90% of the time. The ratio of females to males DOES matter! (see the update below for details)

Next, we learn that the person has AX3. Because AX3 is unrelated to long hair, the ratio of men to women is known to be 50:50, and because of our assumption of the probabilities being the same, we can simply multiply each side of the probability and normalize so that the sum of the sides of the probability equals 100:

(90:10) * (80:20)
==> 7200:200

Normalize by dividing each side by (7200+200)/100 = 74

==> 7200/74:200/74
==> 97.297.. : 2.702..

Thus, the chance that the person behind the curtain is female is approximately 97.297%.

UPDATE

Here's a further exploration of the problem:

Definitions:

f - number of females
m - number of males
fl - number of females with long hair
ml - number of males with long hair
fx - number of females with AX3
mx - number of males with AX3
flx - number of females with long hair and AX3
mlx - number of males with long hair and AX3
pfl - probability that a female has long hair
pml - probability that a male has long hair
pfx - probability that a female has AX3
pmx - probability that a male has AX3

First, we are given that 90% of long-haired people are females, and 80% of people with AX3 are female, so:

fl = 9 * ml
pfl = fl / f
pml = ml / m
= fl / (9 * m)

fx = 4 * mx
pfx = fx / f
pmx = mx / m
= fx / (4 * m)

Because we assumed that the probability of AX3 is independent of gender and long hair, our calculated pfx will apply to women with long hair, and pmx will apply to men with long-hair to find the number of them that likely have AX3:

flx = fl * pfx
= fl * (fx / f)
= (fl * fx) / f
mlx = ml * pmx
= (fl / 9) * (fx / (4 * m))
= (fl * fx) / (36 * m)

Thus, the likely ratio of the number of females with long-hair and AX3 to the number of males with long-hair and AX3 is:

flx             :   mlx
(fl * fx) / f   :   (fl * fx) / (36 * m)
1/f             :   1 / (36m)
36m             :   f

Because it is given that there is an equal number of 50:50, you can cancel both sides and end with 36 females to every male. Otherwise, there are 36*m/f females for every male in the specified subgroup. For example, if there were twice as many women as men, there would be 72 females to each male of those that have long-hair and AX3.

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This solution relies on assuming more than is currently stated in the problem: namely, that long hair, AX3, and gender are independent. Otherwise, you cannot justify "applying" pfx to women with long hair, etc. –  whuber Jun 21 '12 at 19:35
@whuber: Yes, I do make that assumption. However, isn't the purpose of probability to give the best approximation based on the data that you have? Thus, since you already know that long-hair and AX3 are independent for the general population, you SHOULD carry forward that assumption to males and females until you explicitly learn otherwise. Granted, it is not a universally correct one, but it is the best one you can make until you get more info. Q: With only the current data, if you had to give the % chance that it was a woman behind the curtain, would you really say "between 0 and 100%"? –  Briguy37 Jun 21 '12 at 21:44
We have an important difference in philosophy, @Briguy. I strongly believe in not making unfounded assumptions. It is not clear in what sense the mutual independence assumption is "best": I will grant it may be in certain applications. But in general, that seems dangerous to me. I would prefer being clear about the assumptions needed to solve a problem, so people can decide whether it is worthwhile collecting the data to check those assumptions, rather than assuming things that are mathematically convenient for the sake of obtaining an answer. That's the difference between stats and math. –  whuber Jun 21 '12 at 23:14
To answer your question: yes, 0% - 100% is exactly the answer I would give. (I have given similar answers to comparable questions on this site.) That range accurately reflects the uncertainty. This issue is closely related to the Ellsberg paradox. Ellsberg's original paper is well written and clear: I recommend it. –  whuber Jun 21 '12 at 23:16
@whuber: Thanks for taking the time to dialogue with me. I see your point about the importance of thinking through and listing the assumptions made, and have updated my answer accordingly. However, in regards to your answer, I believe it is incomplete. The reason for this is that you can consider all unknown cases and find the average probability of across all of them to arrive at your final answer. E.G. Though both are still possible, probabilities above 50% are much more prevalent than probabilities below 50% across all cases, so we are surely better off guessing that it is a woman. –  Briguy37 Jun 22 '12 at 13:56
show 1 more comment

98% Female, simple interpolation. First premise 90% female, leaves 10%, second premise only leaves 2% of the existing 10%, hence 98% female

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# Simplify Complex Rational Expression

#### PaperStSoap

##### New member
(3/x-2) - (4/x+2) / (7/x2-4)

I got it down to...

-x+14/7

but the book is showing

x-14/7

#### Fantini

MHB Math Helper
Welcome to MHB, Paper!

Which of these expressions did you mean:

$$\frac{3}{x-2} - \frac{ \frac{4}{x+2} }{ \frac{7}{x^2 -4} } \quad \text{ or } \quad \frac{ \frac{3}{x-2} - \frac{4}{x+2} }{ \frac{7}{x^2 -4} }?$$

#### soroban

##### Well-known member
Hello, PaperStSoap!

$$\dfrac{\dfrac{3}{x-2} - \dfrac{4}{x+2}}{\dfrac{7}{x^2-4}}$$

I got it down to: .$\dfrac{-x+14}{7}$ . You are right!

But the book is showing: .$\dfrac{x-14}{7}$ . The book is wrong!

#### PaperStSoap

##### New member
Welcome to MHB, Paper!

Which of these expressions did you mean:

$$\frac{3}{x-2} - \frac{ \frac{4}{x+2} }{ \frac{7}{x^2 -4} } \quad \text{ or } \quad \frac{ \frac{3}{x-2} - \frac{4}{x+2} }{ \frac{7}{x^2 -4} }?$$
My apologies, the problem was the second one.

#### SuperSonic4

##### Well-known member
MHB Math Helper
My apologies, the problem was the second one.
You are right and the book is wrong. It's worth mentioning though that there is a restriction on the domain: $|x| \neq 2$.

(working in spoiler)

$\dfrac{\frac{3}{x-2} - \frac{4}{x+2}}{\frac{7}{x^2-4}}$

$\left(\frac{3}{x-2} - \frac{4}{x+2}\right) \cdot \frac{(x-2)(x+2)}{7}$

$\left(\frac{3(x+2)-4(x-2)}{(x-2)(x+2)}\right) \cdot \frac{(x-2)(x+2)}{7}$

$\frac{3x+6-4x+8}{7}$

$\frac{-x+14}{7}$


# What group is $\langle (1,1,1) \rangle$ in $\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8$?

I know that $$\langle (1,1,1) \rangle$$ in $$\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8$$ is isomorphic to $$\mathbb{Z}_{40}$$, but is there a way of writing what group it is (not what it's isomorphic to).

In other words, can we say that $$\langle (1,1,1) \rangle$$ is $$G_1 \times G_2 \times G_3$$ for some groups $$G_1, G_2, G_3$$?

I'm so used to using isomorphism that I can't tell if the answer is simply no. I tried using the Fundamental Theorem of Abelian Groups, but this didn't resolve my question.

For example, in $$\mathbb{Z} \times \mathbb{Z}$$ we have $$\langle (0,3) \rangle = \{0 \} \times \mathbb{3Z}$$.

We would often just say $$\langle (0,3) \rangle \cong \mathbb{Z}$$, but occasionally it's useful to not use the usual isomorphism and so this is what I'm asking for.

My motivation is to be able to more regularly use the trick used in this answer.

• It's a subgroup of order $40.$ – Dbchatto67 Apr 14 at 15:58
• If anything, this comment gives less information than what I've written ("a subgroup of order 40" might not even be cyclic, but this particular one is). I'm asking for more information. – Jonathan Rayner Apr 14 at 16:04
• The group $\langle (1,1) \rangle \subset \Bbb Z_4 \times \Bbb Z_8$ is not isomorphic to (and certainly not equal to) any product $G_2 \times G_3$. Your group can be written as $\Bbb Z_5 \times \langle (1,1) \rangle$ (to the extent that we can consider the Cartesian product associative) – Omnomnomnom Apr 14 at 16:05
• What do you mean what group it is as opposed to what group it is isomorphic to? We usually name groups up to isomorphism, so your group is just $C_{40}$, which is the same as $\Bbb {Z_5 \times Z_8}$ – Ross Millikan Apr 14 at 16:06
• See this question. – Dietrich Burde Apr 14 at 16:20

$$\langle(1, 1, 1)\rangle$$ is, as you said, isomorphic to $$\mathbb{Z}_{40}$$, and the prime factorization of $$40$$ is $$5 \times 8$$, so we can say that $$\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5 \times \mathbb{Z}_8$$. Now, it is important to convince yourself that such a group can't be expressed as the product of three subgroups, simply because $$\mathbb{Z}_8$$ can't be split since is cyclic and $$\mathbb{Z}_5$$ only has trivial subgroups, or, more formally, by contradiction, let $$\langle(1, 1, 1)\rangle \simeq G_1 \times G_2 \times G_3$$. Since every subgroup of a cyclic group is cyclic, $$G_2 \times G_3$$ is cyclic and none of its coordinates is $$\{e\}$$. Therefore, $$G_2 \times G_3$$ is a product of cyclic groups whose orders are multiples of $$2$$ (so they're not coprime) and it's cyclic, which is a contradiction. As for how you could use that other method in this case, i'd say the most it can do for you is to simplify the thing as $$(\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5/\langle1\rangle \times (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle \simeq (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle$$, but since that $$\langle(1, 1)\rangle$$ is not a product of subgroups you have to figure out some other way to find it. I hope to have answered your question, let me know if that's not the case.

A bit late, but here's a way to see that $$(\Bbb Z_{4} \times \Bbb Z_8)/\langle (1,1) \rangle \cong \Bbb Z_4$$. We note that the map $$\phi:\Bbb Z_{4} \times \Bbb Z_8 \to \Bbb Z_{4} \times \Bbb Z_8$$ defined by $$\phi: (x,y) \mapsto (x-y,y)$$ is a group automorphism (which is induced by the automorphism with the same formula from $$\Bbb Z^2$$ to $$\Bbb Z^2$$). It follows that $$(\Bbb Z_{4} \times \Bbb Z_8)/\langle (1,1) \rangle \cong \phi(\Bbb Z_{4} \times \Bbb Z_8)/\phi(\langle (1,1) \rangle) = (\Bbb Z_{4} \times \Bbb Z_8)/\langle (0,1) \rangle\\ \quad \ = (\Bbb Z_4 \times \Bbb Z_8)/(\{0\} \times \Bbb Z_8) \cong \Bbb Z_4$$

• Thank you for writing this - I accepted the other answer because I think it answered what I technically asked for in the question and it would have been unfair not to. But your answer was very helpful and exactly in the spirit of what I was looking for! – Jonathan Rayner Apr 19 at 21:52

I'm not quite sure what you're asking in your question, but if I'm understanding it, I hope this will help.

One way to think about it is the following. If $$q:\Bbb{Z}/(8)\to \Bbb{Z}/(4)$$ is the quotient map, then the elements of the subgroup of $$\Bbb{Z}/(5)\times \Bbb{Z}/(4)\times \Bbb{Z}/(8)$$ generated by $$(1,1,1)$$ are precisely the elements of the form $$(i,q(j),j)$$, with $$i\in\Bbb{Z}/(5)$$, and $$j\in\Bbb{Z}/(8)$$. This also makes the isomorphism with $$\Bbb{Z}/(5)\times \Bbb{Z}/(8)$$ clear. One direction is $$(i,q(j),j)\mapsto (i,j)$$, and the other is $$(i,j)\mapsto (i,q(j),j)$$.

Side note:

This isn't directly related to your question, just an interesting tangent.

If we focus on the subgroup of $$\Bbb{Z}/(8)\times \Bbb{Z}/(4)$$ generated by $$(1,1)$$ (order deliberately flipped here), we can see that it is the subgroup $$(j,q(j))$$ for $$j\in \Bbb{Z}/(8)$$, which you can think of as the graph of the group homomorphism $$q$$. It's quite often the case that when you can construct the graph of a morphism $$f$$, the graph turns out to be isomorphic to the starting object, with one of the isomorphisms being $$x\mapsto (x,f(x))$$, and the other being $$(x,f(x))\mapsto x$$.


# Asymptotic notation (big Theta)

I'm currently in the process of analyzing runtimes for some given code (Karatsuba-ofman algorithm).

I'm wondering if I'm correct in saying that $\Theta(\left\lceil n/2\right\rceil) + \Theta(n)$ is equal to $\Theta(n)$? (Taking the maximum of both costs)

I know that $\Theta(n/2) +\Theta(n)$ is equal to $\Theta(n)$. But my concern with the first asymptotic runtime is that given $n = 0.6, \left\lceil n/2\right\rceil$ would be equal to $1$, which is greater than $0.6$.

If $\Theta(\left\lceil n/2\right\rceil) + \Theta(n) \stackrel{?}{=} \Theta(n)$ could anybody please give some insight on why this is true?

Yes you are right.. intuitively it's clear that it does not change a thing, since when $n \to \infty$ effects like this are definitely too small to notice.

The whole concept behind asymptotics is that we ignore small terms.. sometimes we can even ignore something "huge" like $2^n$, for example $\Theta(2^n + 3^n) = \Theta(3^n)$

So we don't really care about effect like this; yes if $n=0.6$ you would get $1$, but the error is at most $1/2$ which is not relevant. if you had $n = 10^9 + 0.6$, would you be worried about wether the result is rounded up or down?

If you want to be a little bit more formal you can start noting that $\lfloor \frac n2 \rfloor \le n$, and $\Theta(n) + \Theta(n) = \Theta(n)$

Or you can also notice that the error from rounding is at most $1/2$ or $1$, depends on the rounding that one uses and find some bonds from there.

• Thanks alot for this explanation. Just as I thought. That's really cleared some things up. – user220506 Mar 2 '15 at 22:12
• @Tazman You're welcome! :) – Ant Mar 2 '15 at 22:13
• Although you stated floor(n/2) <= n, this wouldn't be the case for ceiling(n/2) <= n unless we state for all n >= 1 right? – user220506 Mar 2 '15 at 22:14
• @Tazman yes of course. But then again we only look about what happens in the long run, so to speak, so most of the time I don't even notice I take this assumption and most of what I write is valid for $n$ large enough ;-) – Ant Mar 2 '15 at 22:16
• Okay, that makes sense. Just like if it were a big-O or big-Omega question, we would state some given bound where we see the said behaviour. Thanks for the clarification :) – user220506 Mar 2 '15 at 22:17

Remember that $f(n) =\Theta (g(n))$ means that there are positive constants $a$ and $b$ such that $a g(n) < f(n) < b g(n)$ for all large enough $n$.

Therefore, if $f(n) =\Theta (h(n))$ and $g(n) =\Theta (h(n))$ then $f(n)+g(n) =\Theta(h(n))$.

If you consider the problem for the natural numbers it is true that $\Theta(\lceil n/2 \rceil) = \Theta (n)$. As $\lceil n/2 \rceil \le n \le 2 \lceil n/2 \rceil$. And as a consequence also $\Theta(\lceil n/2 \rceil) + \Theta (n)= \Theta (n)$.

However, if you consider it for the positive real numbers (as $n \to 0$) as your example $0.6$ might suggest then it is not true that $\Theta(\lceil n/2 \rceil) = \Theta (n)$ and neither is $\Theta(\lceil n/2 \rceil) + \Theta (n)= \Theta (n)$. In this case $n = O (\lceil n/2 \rceil)$ as $n \le 2 \lceil n/2 \rceil$ still holds. Yet as $\lceil n/2 \rceil = 1$ for all $0< n \le 2$ we get that $\lceil n/2 \rceil$ is not $O(n)$ as $n \to 0$.

• So Tazman should say $n \to \infty$ if he means that. (Presumably he does, for analyzing "runtimes".) – GEdgar Mar 2 '15 at 22:08
• @GEdgar Well, yes and no. Note they mention the case $n=0.6$ explicitly. But I did mention the result for the integers first. It is however not really necessary to say $n \to \infty$ one just has to make the domain precise, if it is the natural numbers it is fine. The $O$ notation as well as $\Theta$ makes sense without making a limit point precise, as opposed to $o$. Finally note it is the second question of this form I answered today, and there is thus additional reason why I insist on discussing the distinction i make. – quid Mar 2 '15 at 22:13
• The n in question refers to a length of a given number, so I think it's safe that in my answer to the question, one can assume that n consists of natural numbers, and thus theta(ceiling(n/2)) = theta(n). – user220506 Mar 2 '15 at 22:16
• @quid, yes, that's true. I should have been more clear and stated that I was giving that generic example to better my understanding of exactly why it is (or not) equal. – user220506 Mar 2 '15 at 22:19
• @Tazman yes, if it is for integers then you can say this (as I said in my answer). And this is a reasonable assumption given your context. I only discussed the other case since you mentioned $n=0.6$ and there is an actual difference in another situation. [This is an expanded version.] – quid Mar 2 '15 at 22:20


# Need help understanding least squares solution to overdetermined system

(Sorry I had to post the images as links. I don't have enough cred to post pictures directly yet)

I'm trying to understand what the least squares solution to an overdetermined system means geometrically in the case of the following system:

$$y = -2x-1\\ y = 3x -2\\ y = x+1\\$$

rewritten in matrix form:

$$\overbrace{\begin{bmatrix} 2 & 1\\ -3 & 1\\ -1 & 1 \end{bmatrix}}^A \underbrace{\begin{bmatrix} x\\ y \end{bmatrix}}_x = \overbrace{\begin{bmatrix} -1\\ -2\\ 1 \end{bmatrix}}^b$$

Using A\b in MATLAB, you get the solution $$\begin{bmatrix}0.1316 & -0.5789\end{bmatrix}^T$$. I know that MATLAB returns the lowest norm solution of a least squares problem. I have plotted the system here and the green dot in the middle is this least squares solution.

Now, correct me if I'm wrong, but (in this 2D case) the least squares solution minimizes the "distance" from the solution to each line. I can geometrically calculate the distance of a point $$(x_0,y_0)$$ from a line $$ax + by + c = 0$$ as follows:

$$\mathrm{distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

and doing that for each line produces the following sum of squared distances function

dfun = @(x,y) ((y+2*x+1).^2)/(1^2 + 2^2) + ((y+3*x+2).^2)/(1^2 + 3^2) + ((y+x-1).^2)/(1^2 + 1^2);

If I generate a surface using this function over a range of $$x$$ and $$y$$ values, I get this surface with this top-down view (looking down the z-axis on the xy plane). You can download the MATLAB .fig file here if you want to zoom and pan (requires MATLAB, link expires in 30 days).

Here is an image showing the least squares solution with the sum of squares of distances of the solution and its norm. As can be seen, the norm is $$0.5937$$ and the distance is $$1.4704$$. But clearly, there is a contour that has a lower sun of squared distance in the image, as shown here for $$(x_0, y_0) = (-0.3,0)$$, where the norm and the sum of squared distances are both smaller. Shouldn't this (or another point) be a better least squares solution? Do I have the wrong intuition about what least squares is doing here?

First of all : Welcome to the site !

When you face an overdetermined system of $$m$$ linear or nonlinear (even implicit) equations $$f_i(x_1,x_2,\cdots,x_n)=0 \quad\text{for}\quad i=1,2,\cdots,m\qquad \quad\text{and}\quad m >n$$ it reduces to the minimization of a norm.

The simplest is $$\Phi(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n w_i \Big[f_i(x_1,x_2,\cdots,x_n)\Big]^2$$ which shows the analogy with weighted least-square method.

For the problem you gave, using equal weights, $$\Phi(x,y)=(y+2x+1)^2+(y-3x+2)^2+(y-x-1)^2$$ Computing the partial derivetives $$\frac{\partial \Phi(x,y)}{\partial x}=28 x-4 y-6 \qquad\text{and}\qquad \frac{\partial \Phi(x,y)}{\partial y}=-4 x+6 y+4$$ which gives $$x=\frac{5}{38}$$ and $$y=-\frac{11}{19}$$. At this point, $$\Phi=\frac{169}{38}$$ is the absolute minimum for this specific norm.

If you change the definition of the norm or even the weights, different results.

I think that this approach would be simpler than the one based on distance (but, for sure, I may be wrong) since it is totally general and not limited to linear equations.

• Thank you! I think I had the right understanding of what least squares is doing in this case. After reading your answer I realizes that my dfun was wrong - I got a couple of the signs wrong and that's why I was getting the wrong surface. I will post an answer with the right images. Not sure if I should accept my own answer or yours...
– vyb
Sep 6, 2021 at 17:40

After reading Claude Leibovici's answer above, I realized that my dfun had typos in it -- I messed up a couple of minus signs in the function.

Additionally, the norm typically used for least squares calculations (also used by MATLAB) is the $$l^2$$-norm (a.k.a Euclidean norm):

$$x = \begin{bmatrix}x_1\\ x_2\\ \vdots\\ x_n\end{bmatrix}, ||x|| = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2}$$

Note that there is no scaling of the "distance" like there is in my dfun. Therefore correct function should be:

dfun = @(x,y) ((y+2*x+1).^2) + ((y-3*x+2).^2) + ((y-x-1).^2);

After fixing these mistakes, here is the surface that is generated, with this top-down view. As can be seen here, the solution of $$\begin{bmatrix}0.1316 & -0.5789\end{bmatrix}^T$$ is, in fact, correct and confirms the original intuition I was trying to confirm.