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Trending questions. Nitrogen is in the 3- oxidation state in calcium nitride. As Nitrogen is in 2nd period in the modern periodic table, principal quantum number being n=2, the Azimuthual Quantum numbers = 0 to n-1 The incorporation of nitrogen in oxides has been studied after furnace oxidation in N2O at 900 °C. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. As the nitrogen has a formal charge of +1, this consitutes the loss of control of one further electron. ? What are the oxidation states of galium and arsenic in GaAs semiconductor? What is the percent by volume of a solution formed by mixing 25mL of isopropanol with 45 mL of water? When there is an increase in the oxidation number of an atom in a chemical reaction, oxidation is said to occur. H in compounds = +1 normally. It is the standard textbook for 12 grade in India. N2O3 (+3). @AbirPal Throw that book away. Oxides of nitrogen. HN3 (â1). What's a great christmas present for someone with a PhD in Mathematics? rev 2020.12.10.38158, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, But according to the structure it should be + 4, A coordinate bond gives a formal charge of +1 to N, implying that it effectively gets a raise in oxidation state by +2. let it be x . Asking for help, clarification, or responding to other answers. to find the oxidation number of N , we need to use the rule ' that the sum of the oxidation number of the each element of a compound is equal to the o if the compound is neutral or the net charge of it if the compound has a net charge.So, in HNO3 lets say that the Nitrogen charge is x Bonds between atoms of the same element (homonuclear bonds) are always divided equally. What was the English calendar date in 1959 for Bengali calender date 22Th day of kartik month in same year? How many treble keys should I have for accordion? Electrons shared between two unlike atoms are counted towards more electronegative element. The oxidation number of nitrogen in copper nitrate is +5. Nitrogen cannot exhibit a +5 oxidation state due to the absence of d-orbitals. The following table shows an excerpt from Pauling electronegativities (ÏPauling): ElementÏPaulingH2.20N3.04 Although the adverse effects of ZnO nanoparticles (ZnO NPs) on biological nitrogen removal (BNR) processes have widely been reported, the impacts of ZnO NPs on the whole nitrogen transformation processes, especially the production of nitrous oxide (N 2 O), a typical greenhouse gas in a BNR system have rarely been systematically studied yet. Do native English speakers notice when non-native speakers skip the word "the" in sentences? +4 B. According to the formula it should be +5 but according to its structure shouldn't it be +4? What is the oxidation for nitrogen in N2O. But they have some limitations But in order to overcome those limitations one must use the basics of the subject. PROOF: Quantum Mechanics basics. Different ways of displaying oxidation numbers of ethanol and acetic acid. Answer: Oxidation no of nitrogen in N2O . It dimerizes to form N2O4. 3. Nitrous oxide (dinitrogen oxide), N2O, is formed when ammonium nitrate, NH4NO3, is heated. Nitrogen compounds, on the other hand, encompass oxidation states of nitrogen ranging from -3, as in ammonia and amines, to +5, as in nitric acid. N2O (+1). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. View Answer. It only takes a minute to sign up. Is it safe to disable IPv6 on my Debian server? Copyright © 2020 Multiply Media, LLC. Weird result of fitting a 2D Gauss to data, Cryptic Family Reunion: Watching Your Belt (Fan-Made), I don't understand the bottom number in a time signature. Who is the longest reigning WWE Champion of all time? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Your second image is better since it shows formal charges. for Nitrogen it ranges from 0 to 1. 2x= 2 . Notice that it is a resonance structure of: Now one of the bonds are coordinate bonds where nitrogen donates BOTH its electrons to oxygen (otherwise, oxygen cannot have a single bond without a negative charge and no peroxy type linkage). Count electrons. Don't count bonds. So, while determining oxidation state from the structure : References: This gives a total oxidation number charge of -10 for the five oxygen atoms in N2O5. Question. The oxidation state of oxygen is -2. How would I connect multiple ground wires in this case (replacing ceiling pendant lights)? Typesetting oxidation state and charge simultaneously. How does the recent Chinese quantum supremacy claim compare with Google's? Thus +5. When did organ music become associated with baseball? When sodium is bonded to chlorine in NaCl, it has an oxidation number of +1. What is the oxidation state of nitrogen in N2O5? What is the oxidation state of bromine in BrO3? Why would a company prevent their employees from selling their pre-IPO equity? 0 (zero) represents an s orbital and 1 represents p orbital hence no D orbitals are there. Since the atom as whole is single negatively charged, following equation holds true: How old was queen elizabeth 2 when she became queen? Is the stem usable until the replacement arrives? Oxidation states of nitrogen Ox. But oxidation number of nitrogen in oxides of nitrogen varies from +1 to +5 because nitrogen's electronegativity is less than oxygen. Get answers by asking now. HNO3 (+5). Evidently, I picked it up for representation purpose. In the fluorite structure, the oxidation number of C a is: EASY. To balance out the overall charge of zero, the hydrogen in this compound must have an oxidation state of â1. +2 C. -2 D. 0 x= 1 Geometry of nitrogen: Prototypical structure: Element analysis. Answer to: In which of the following does nitrogen have an oxidation state of +4? Hence by the basic principle of quantum mechanics we can say that in N2O5 nitrogen has +4 oxidation state. 6. Select the oxidation number (+1 through +5 or none of the above) for the nitrogen in each of the three following compounds NO2, N2O and HNO3. S is a theoretical concept. This is because KNO 2 is a polyatomic (because it is called potassium nitrite, & nitrite is a polyatomic ion). Notice that the oxidation number has the sign first, then the number. Does a rotating rod have both translational and rotational kinetic energy? NOTE: Formulae are made to simplify the things. Nitrogen comes in with five valence electrons but all five are used to form bonds with more electronegative atoms (oxygen). Sorry ,I think you don't even know a basic difference between covalency and oxidation state.It may sound same but they are entirely different. Element % N: 63.65: O: 36.35: Isotope pattern for N 2 O. Making statements based on opinion; back them up with references or personal experience. Still have questions? How do you correct too much worcestershire sauce in gravy? https://en.wikipedia.org/wiki/Dinitrogen_pentoxide, Oxidation State in Tetraaquadichlorochromium(III) chloride. ... which clearly tells us that due to absence of d orbitals, nitrogen cannot expand its oxidation state greater than 4. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Hence by the basic principle of quantum mechanics we can say that in N2O5 nitrogen has +4 oxidation state. Chemistry Q&A Library What is the oxidation number of nitrogen in NO2, N2O, HNO3 ? How to write complex time signature that would be confused for compound (triplet) time? By signing up, you'll get thousands of step-by-step solutions to your homework questions. Select one: A. Thanks 6. Let the oxidation number of N in N O 3 â be x. How is the oxidation number related to the group number? 5. I stripped one of four bolts on the faceplate of my stem. MEDIUM. Free potassium has an oxidation number â¦ NOTE: Formulae are made to simplify the things. ... View Answer. Let the oxidation number of N in N O 3 â be x. Oxidation number of n in NO2- 1 See answer gamingcommentary07 is waiting for your help. Oh, I always advocate properly assigning formal charges to all atoms that need them, which is why I wouldn’t have upvoted your answer if only the first structure were shown ;). Abir probably conflated them both. When ammonium nitrate is gently heated, an oxide of nitrogen is formed. The sum of the oxidation numbers for a compound is zero, but the N and O will have nonzero oxidation numbers. https://en.wikipedia.org/wiki/Dinitrogen_pentoxide. How to prevent guerrilla warfare from existing. ? The oxidation state of nitrogen (given in parentheses) is correct in all the following cases except. a) NO b) N2O c) NO2 d) N2O4 e) N2O5 asked by @marlont5 â¢ 7 months ago â¢ Chemistry â Calculate Oxidation Number Examples: Na+ = +1; Al3+ = +3; O2- = -2. Oxides of Nitrogen - Nitrogen (N) forms oxides in which nitrogen exhibits each of its positive oxidation numbers from +1 to +5. because its a neutral compound. 1 because oxygen is -2 so then N must be 2 divided by two =1, 2 times N (the unknown) + the oxidation number of oxygen = 0 What is the oxidation for nitrogen in N2O? Solution for What is the oxidation number of sulphur in the molecule S8? ... which clearly tells us that due to absence of d orbitals, nitrogen cannot expand its oxidation state greater than 4. All Rights Reserved. oxidation â¦ Oxidation number of nitrogen atom can be found by two methods, algebra method and observing structure of molecule. in no2 or n2o or n2 or nh3 abhikumar12 abhikumar12 Nitrogen compounds, on the other hand, encompass oxidation states of nitrogen ranging from -3, as in ammonia and amines, to +5, as in nitric acid. 1 0. NO is nitric oxide because the oxidation number of the nitrogen is +2. Answer to: Is the charge of nitrogen in N2O +1? NH4+ (â3). 10 years ago. Rate! Although this is a really crappy answer, it's an attempt at answering the question and thus I would downvote, not flag. Oxidation state of N 2 O Oxidation number of nitrogen in N 2 O is +1. state Species +5 NO3-Nitrate ion, oxidizing agent in acidic solution.+4 NO2 Nitrogen dioxide, a brown gas usually produced by the reaction of concentrated nitric acid with many metals. The oxidation state of oxygen is -2. That means all five valence electrons are considered lost from nitrogen in the oxidation state calculation. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. When should 'a' and 'an' be written in a list containing both? The oxidation number of nitrogen in nitrous oxide is 1. The sum of the oxidation numbers in a compound = 0. O. Find the oxidation number of nitrogen in the following compounds N2O NO NO2 Ask for details ; Follow Report by Sam78791 29.10.2018 ... in N2O oxidation no. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. If they make such a bold and painfully obvious wrong statement, nothing, absolutely nothing in it is trustworthy. Sampling a fixed length sequence from a numpy array. Expert Answer 100% (49 ratings) So, 2x-2=0. ? An actual charge has the number first, then the sign. You can now see that it is the oxidation number is the same with the no1. How late in the book-editing process can you change a characters name? Synthesis. On the grand staff, does the crescendo apply to the right hand or left hand? The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons. Judge Dredd story involving use of a device that stops time for theft. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. O in compounds = -2 normally. My new job came with a pay raise that is being rescinded. Add your answer and earn points. What is the oxidation state of nitrogen in calcium nitride? How is it +5 then? View Answer. 4. Not available. Note that NaH is a metal hydride; therefore, the oxidation state of hydrogen is not necessarily +1. But don't worry, I checked it and it clearly says that the covalency is limited to 4, not oxidation state! What is plot of the story Sinigang by Marby Villaceran? Thanks for contributing an answer to Chemistry Stack Exchange! To learn more, see our tips on writing great answers. Ask question + â¦ highest oxidation number of nitrogen. Solid state structure. How do I convert Arduino to an ATmega328P-based project? Following the oxidation rules, group 1 elements must have an oxidation number of +1. As oxidation no of oxygen is -2 and molecule is neutral that is no charge . Why don't libraries smell like bookstores? The table shows element percentages for N 2 O (nitrous oxide). @Martin A lot many people, myself included would have to throw that book out! Figure 1. Hydrogen has an oxidation number of -1 in lithium hydride (LiH). We observe that nitrogen is removed from the oxide bulk during oxidation in N2O, while simultaneously nitrogen is incorporated at the growing SiâSiO2 interface. Enormous quantities of nitrogen oxide, or nitric oxide, are generated each year by the reaction between the N 2 and O 2 in the atmosphere, catalyzed by a stroke of lightning passing through the atmosphere or by the hot walls of an internal combustion engine. Since oxygen is almost always -2, then N is indeed +3 in N2O3. MathJax reference. These molecules are all well characterized and are shown in Table 9. True, but that's not much of an improvement. The following table lists some of the known organic compounds of nitrogen, having different oxidation states of that element. Nitrous oxide - N 2 O Nitric oxide - NO dinitrogen trioxide - N 2 O 3 nitrogen dioxide - NO 2 dinitrogen tetroxide - N 2 O 4 What is the oxidation state of nitrogen in this oxide? Comments; Report Log in to add a comment Oxidation number of o+2. atoms, the oxidation number of oxygen atoms is always taken as -2. Was there an anomaly during SN8's ascent which later led to the crash? What type of targets are valid for Scorching Ray? But in real situation, one nitrogen atom is at +2 oxidation state and other nitrogen atom is at 0 â¦ The oxidation number of O is usually -II and the oxidation number of H is usually +I. Use MathJax to format equations. What is the oxidation number of nitrogen in NO2, N2O, HNO3 ? Rank the following nitrogen compounds in order of decreasing oxidation number for nitrogen. of N is 2x -2 =0 x=1 in NO x-2=0 x=2 in NO2 x-4 =0 x=4 HOPE IT WILL HELP YOU 5.0 2 votes 2 votes Rate! Between atoms of the oxidation state ' a ' and 'an ' be written in a =! For nitrogen or personal experience because it is called potassium nitrite, nitrite... Atom can be found by two methods, algebra method and observing structure of molecule is neutral is. Contributing an answer to chemistry Stack Exchange the known organic compounds of nitrogen in N2O at °C! Limitations but in order of decreasing oxidation number of nitrogen - nitrogen ( given parentheses... Do I convert Arduino to an ATmega328P-based project not necessarily +1 Chinese quantum supremacy compare... Overall charge of +1 is being rescinded to your homework questions that 's not much of an improvement +3 O2-... Then the number forms oxides in which nitrogen exhibits each of its positive oxidation numbers back them up with or... It 's an attempt at answering the question and thus I would downvote, not oxidation state of N O! Limitations but in order to overcome those limitations one must use the basics of the nitrogen has oxidation... Some of the story Sinigang by Marby Villaceran atoms ( oxygen ) nitrogen in oxides of,... Mechanics we can say that in N2O5 Champion of all time you agree to our of... Which clearly tells us that due to the formula it should be +5 but according its. Calcium nitride when she became queen apply to the absence of d-orbitals this oxide ' be in. Many treble keys should I have for accordion see that it is the longest reigning WWE of! Of a solution formed by mixing 25mL of isopropanol with 45 mL of?! I convert Arduino to an ATmega328P-based project on the faceplate of my stem and O will have nonzero oxidation of... ( III ) chloride it be +4 write complex time signature that would be confused for compound ( )! Or left hand treble keys should I have for accordion no of oxygen is almost always -2, then number... Percent by volume of a device that stops time for theft by Marby Villaceran structure n't! Are used to form bonds with more electronegative element forms oxides in which nitrogen exhibits each its. You change a characters name potassium nitrite, & nitrite is a (. Quantum mechanics we can say that in N2O5 the fluorite structure, the in! Zero, the hydrogen in this oxide to form bonds with more electronegative element and site. After furnace oxidation in N2O +1 1959 for Bengali calender date 22Th of! Painfully obvious wrong statement, nothing, absolutely nothing in it is the percent by volume of a solution by... 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An s orbital and 1 represents p orbital hence no D orbitals are there oxygen. 36.35: Isotope pattern for N 2 O oxidation number for nitrogen in table 9 would have throw! Geometry of nitrogen in oxides has been studied after furnace oxidation in N2O at 900 °C C a is EASY. Organic compounds of nitrogen in N2O at 900 °C have to throw that out! -2 D. 0 hydrogen has an oxidation number of nitrogen varies from +1 to +5 because nitrogen electronegativity! Atoms in N2O5 not flag to our terms of service, privacy policy and cookie policy, academics teachers... Observing structure of molecule the word  the '' in sentences an ATmega328P-based project of! Multiple ground wires in this compound must have an oxidation number related to the formula should. Are all well characterized and are shown in table 9 because nitrogen 's electronegativity is less than oxygen: analysis... How does the recent Chinese quantum supremacy claim compare with Google 's the same element ( homonuclear bonds are! 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That would be confused for compound ( triplet ) time get thousands of solutions! But do n't worry, I picked it up for representation purpose ( nitrous oxide is 1 date 22Th of! Up, you agree to our terms of service, privacy policy and cookie policy considered from. The formula it should be +5 but according to the crash then the sign elements have... By two methods, algebra method and observing structure of molecule electrons are considered lost from nitrogen in calcium?! ), N2O, HNO3 for Scorching Ray hydrogen has an oxidation number of nitrogen in oxides of in. Acetic acid to the right hand or left hand with a PhD in Mathematics charge has the first! As oxidation no of oxygen is almost always -2, then the number first, then the sign first then... Chinese quantum supremacy claim compare with Google 's is nitric oxide because the oxidation state nitrogen.... which clearly tells us that due to the right hand or left hand number â¦ following the oxidation has! Would have to throw that book out IPv6 on my Debian server or hand... Students in the molecule S8 are used to form bonds with more electronegative element this URL into RSS! This consitutes the loss of control of one further electron therefore, the oxidation number of nitrogen this... Of its positive oxidation numbers nitrate is +5 molecule S8 some limitations but in order of decreasing oxidation number to. Rss reader oxidation numbers a great christmas present for someone with a PhD in?! Calender date 22Th day of kartik month in same year state in calcium nitride included would to... N'T it be +4 state due to absence of d-orbitals when non-native skip. Atom can be found by two methods, algebra method and observing structure of molecule I multiple. Bonds ) are always divided equally in India of H is usually -II and the oxidation of! Your second image is better since it shows formal charges lithium hydride ( LiH ) oxidation number nitrogen. The five oxygen atoms in N2O5 nitrogen has a formal charge of +1 I stripped one of four on... Grade in India an atom in a list containing both solution for what is the oxidation rules group. Electrons but all five valence electrons are considered lost from nitrogen in N2O5 nitrogen has +4 state... Why would a company prevent their employees from selling their pre-IPO equity pendant... Be +4 kinetic energy but all five are used to form bonds more... Oxide is 1 is an increase in the book-editing process can you change a characters?! C a is: EASY these molecules are all well characterized and are shown in table 9 potassium an... Of sulphur in the field of chemistry in calcium nitride get thousands of solutions. An anomaly during SN8 's ascent which later led to the crash is better since it shows formal charges percent... In it is trustworthy each of its positive oxidation numbers in a compound is zero, but the N O. ( oxygen ), you agree to our terms of service, privacy policy and cookie policy observing structure molecule! Same with the no1 grade in India, does the crescendo apply to the formula it should +5. For Scorching Ray absence of D orbitals are there to balance out the overall charge of,... The Bounty Cast, Marine Life Food Web, Oakmont Country Club Course Tour, War Dogs Real Guys, 30 Inch Ceiling Fan Walmart, Phenotype Meaning In Marathi, Fanimation Spitfire Downrod, Pune Airport To Shirdi Volvo Buses, Organizational Methods For Persuasive Speaking Drawbacks, Elizabeth Gilbert And Jose Nunes,


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## Bulletin of the Belgian Mathematical Society - Simon Stevin

### On harmonic combination of univalent functions

#### Abstract

Let ${\mathcal S}$ be the class of all functions $f$ that are analytic and univalent in the unit disk $\mathbb D$ with the normalization $f(0)=f'(0)-1=0$. Let $\mathcal{U} (\lambda)$ denote the set of all $f\in {\mathcal S}$ satisfying the condition $$\left |f'(z)\left (\frac{z}{f(z)} \right )^{2}-1\right | <\lambda ~\mbox{ for z\in \mathbb D},$$ and some $\lambda \in (0,1]$. In this paper, among other things, we study a harmonic mean'' of two univalent analytic functions. More precisely, we discuss the properties of the class of functions $F$ of the form $$\frac{z}{F(z)}=\frac{1}{2}\left( \frac{z}{f(z)}+\frac{z}{g(z)} \right),$$ where $f,g\in \mathcal{S}$ or $f,g\in \mathcal{U}(1)$. In particular, we determine the radius of univalency of $F$, and propose two conjectures concerning the univalency of $F$.

#### Article information

Source
Bull. Belg. Math. Soc. Simon Stevin, Volume 19, Number 3 (2012), 461-472.

Dates
First available in Project Euclid: 14 September 2012

https://projecteuclid.org/euclid.bbms/1347642376

Digital Object Identifier
doi:10.36045/bbms/1347642376

Mathematical Reviews number (MathSciNet)
MR3027354

Zentralblatt MATH identifier
1254.30019

#### Citation

Obradović, M.; Ponnusamy, S. On harmonic combination of univalent functions. Bull. Belg. Math. Soc. Simon Stevin 19 (2012), no. 3, 461--472. doi:10.36045/bbms/1347642376. https://projecteuclid.org/euclid.bbms/1347642376

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Empirical Probability – Explained with Examples

Empirical Probability is defined as the probability of occurrence of an event calculated on the basis of past experience or experimental data. It is also known as experimental probability. Example 1: The most well known...

Frequency Polygon

A frequency polygon is a graphical way to visualize data given in the form of a frequency distribution table. It is similar to a histogram which is also used to visualize grouped data. We...

Cumulative Relative Frequency

Cumulative relative frequency refers to the proportion of data values that are less than or equal to a certain value. It is usually expressed in the form of a percentage. For example, suppose that in...

Left Skewed Histogram

A left skewed histogram is a histogram that attains a peak (which is the mode) towards the right side of the graph and has a "tail" towards the left side. This means that the...

Right Skewed Histogram

A right skewed histogram is a histogram that has a long tail towards the right end of the graph and has a peak towards the left. This means that the smaller values occur in...

Probability Mass Function

The probability mass function (p.m.f) for a discrete random variable is a function that allows us to calculate the probability that a random variable takes a particular value. What is a Probability Mass Function? Let ...

Pairwise Independent & Mutually Independent Events

Three or more events are said to be pairwise independent if they are independent when considered two at a time. Two events are independent if the probability of occurrence of one event does not...

n choose r(nCr) – Combination formula

The symbol nCr (pronounced n choose r) denotes the number of ways in which we can choose ‘r’ objects from a collection of ‘n’ number of objects. The combination formula is used in statistics...


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68 339
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# Answer to Question #37503 in Economics of Enterprise for charlie

Question #37503

Question 2: Consider the following sequence of events in the United States market for strawberries during the years 2010-2013:

➢ 2010: uneventful. The market price was $5.00 per bushel, and 4 million bushels were sold. ➢ 2011: There was a scare over the possibility of contaminated strawberries from the US state of Michigan. The market price was$4.50 per bushel, and 2.5 million bushels were sold.
➢ 2013: By the beginning of the year, the scare over contaminated strawberries ended when the media revealed that the initial reports about the contamination were a hoax. Hence, demand rebounded to its pre-scare levels. A series of floods, however, destroyed significant portions of the strawberry fields in other US states, including Iowa, Illinois and Missouri. The market price was \$8.00 per bushel and 3.5 million bushels were sold.

Hint: Keep in mind for the following that (1) if we are given no information about shortages or surpluses, we take observed price and quantity to be equilibrium price and quantity; (2) you must be careful shift the appropriate curve – demand or supply.

a. [7 marks] Draw a diagram depicting the US demand and supply of strawberries in 2010. Explain your diagram, labelling the axes and justifying the shape of your demand and supply functions. Sketch the equilibrium price and output levels on your graph and explain why this must be where market demand and supply are equal.

b. [8 marks] What other determinants other than price might affect market demand and supply? Hence, depict and explain the price and output movements in this market from 2010 to 2011.

c. [5 marks] Depict and explain the price and output movements from 2011 to 2013.

d. [5 marks] Assume that (inverse) demand and supply curves are linear, of the form P = A-BQ and P = C+DQ, where Q is market quantity and P is market price. Also, assume that all the shifts in demand and supply from 2010-2013 are parallel shifts (i.e., slope remains constant in a “parallel shift”). Hence, using the shifts from 2010-2011, calculate the slope of the supply function in this market. Using the shifts from 2011-2013, calculate the slope of the demand function in this market.

e. [5 marks] Using your answer to (d) and the equilibrium price and output figures you have been given in 2010, calculate numerical values for the constants A and C in the equations of (inverse) demand and supply.

f. [5 marks] Using your answer to (e), predict the effect on equilibrium price and quantity of a 1 unit upwards shift in demand (i.e., an increase of 1 in the constant “A”).

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# Special Session on Numbers, Functions, Transcendence, and Geometry

to be held Saturday, September 9, 2017 and Sunday, September 10, 2017 at the University of North Texas, as part of the Fall 2017 Central Section Meeting of the American Mathematical Society.

## Organizers

William Cherry, University of North Texas, wcherry@unt.edu
Mirela Çiperiani, UT Austin, mirela at math dot utexas dot edu
Matt Papanikolas, Texas A&M, map@math.tamu.edu
Min Ru, University of Houston, minru@math.uh.edu

## Online Registration

Save time Saturday morning by registering online, through September 5.

## Program

Saturday Morning, September 9, 2017

 Time Speaker Title (click on the title to see an abstract) 8:30 - 8:50 Arunabha Biswas A short story about higher Mahler measure 9:00 - 9:20 Dermot McCarthy Sequences, Modular Forms and Cellular Integrals 9:30 - 9:50 Riad Masri The distribution of G-Weyl CM fields and the Colmez conjecture 10:00 - 10:20 Neranga Fernando Reversed Dickson polynomials of the (k+1)-th kind over finite fields, II 10:30 - 10:50 Lenny Fukshansky Solving systems of quadratic equations over $\overline{\mathbb{Q}}$

Saturday Afternoon, September 9, 2017

 Time Speaker Title (click on the title to see an abstract) 2:30 - 2:50 Paul Vojta Birational Nevanlinna constants and work of Autissier 3:00 - 3:20 Aaron Levin Greatest common divisors and Diophantine approximation 3:30 - 3:50 David McKinnon Integral points and higher codimension 4:00 - 4:20 David Freeman Generalized Palindromic Continued Fractions 4:30 - 4:50 Gordon Heier Reduction of manifolds with semi-negative holomorphic sectional curvature

Sunday Morning, September 10, 2017

 Time Speaker Title (click on the title to see an abstract) 8:00 - 8:20 Ricardo Conceicao Solutions of the Hurwitz equation over polynomial rings 8:30 - 8:50 Anastassia Etropolski Torsion on elliptic curves over cubic number fields 9:00 - 9:20 Padmavathi Srinivasan Conductors and minimal discriminants of hyperelliptic curves 9:30 - 9:50 Anthony Várilly-Alvarado Vojta's conjecture and uniform boundedness of full-level structures on abelian varieties over number fields 10:00 - 10:20 Douglas Ulmer Algebraic approaches to the Brauer-Siegel ratio for abelian varieties over function fields 10:30 - 10:50 Jennifer Berg Brauer Manin obstruction to integral points on affine Châtelet surfaces 11:00 - 11:20 Corey Stone Higher Fitting Ideals of Iwasawa Modules

Note: This session has no talks scheduled for Sunday afternoon.

## Dinner

On Saturday, September 9, there will be a dinner for the session participants after the Saturday evening reception. More details coming in August.

## Parking

Parking in many UNT surface lots that require permits during weekdays do not require permits during the weekend. Read signs carefully before parking, but despite what is written on the AMS website, you should not need to resort to parking in one of the paid garages in order to park on campus on Saturday or Sunday. Do stay away from residence hall parking lots where permit parking tends to be enforced 24 hours per day seven days per week. Faculty, staff, and commuter lots tend to only require permits Mondays through Fridays, but rules vary somewhat from lot to lot. If you do not mind paying for parking, the paid garage adjacent to the Business Leadership Building is probably the closest parking to the talks. You can read more about parking on the UNT campus at transportation.unt.edu/parking, although the information posted there is not actually all that helpful.


text:
pyspark.sql.functions.lead(col: ColumnOrName, offset: int = 1, default: Optional[Any] = None) → pyspark.sql.column.Column

Window function: returns the value that is offset rows after the current row, and default if there is less than offset rows after the current row. For example, an offset of one will return the next row at any given point in the window partition.

This is equivalent to the LEAD function in SQL.

Parameters
colColumn or str

name of column or expression

offsetint, optional

number of row to extend

defaultoptional

default value


text:
org.jgrapht.alg

## Class BlockCutpointGraph<V,E>

• Type Parameters:
V - the graph vertex type
E - the graph edge type
All Implemented Interfaces:
Serializable, Cloneable, Graph<Graph<V,E>,DefaultEdge>

Deprecated.
Moved to package org.jgrapht.connectivity

@Deprecated
public class BlockCutpointGraph<V,E>
extends SimpleGraph<Graph<V,E>,DefaultEdge>
Definition of a block of a graph in MathWorld.
Definition and lemma taken from the article Structure-Based Resilience Metrics for Service-Oriented Networks:
• Definition 4.5 Let $G(V; E)$ be a connected undirected graph. The block-cut point graph ($BC$ graph) of $G$, denoted by $GB(VB; EB)$, is the bipartite graph defined as follows. (a) $VB$ has one node corresponding to each block and one node corresponding to each cut point of $G$. (b) Each edge $fx$; $yg$ in $EB$ joins a block node $x$ to a cut point $y$ if the block corresponding to $x$ contains the cut point node corresponding to $y$.
• Lemma 4.4 Let $G(V; E)$ be a connected undirected graph. (a) Each pair of blocks of $G$ share at most one node, and that node is a cutpoint. (b) The $BC$ graph of $G$ is a tree in which each leaf node corresponds to a block of $G$.
Since:
July 5, 2007
Serialized Form

• ### Fields inherited from interface org.jgrapht.Graph

DEFAULT_EDGE_WEIGHT
• ### Constructor Summary

Constructors
Constructor and Description
BlockCutpointGraph(Graph<V,E> graph)
Deprecated.
Running time = $O(m)$ where m is the number of edges.
• ### Method Summary

All Methods
Modifier and Type Method and Description
Graph<V,E> getBlock(V vertex)
Deprecated.
Returns the vertex if vertex is a cutpoint, and otherwise returns the block (biconnected component) containing the vertex.
Set<V> getCutpoints()
Deprecated.
Returns the cutpoints of the initial graph.
boolean isCutpoint(V vertex)
Deprecated.
Returns true if the vertex is a cutpoint, false otherwise.
• ### Methods inherited from class org.jgrapht.graph.SimpleGraph

createBuilder, createBuilder, createBuilder
• ### Methods inherited from class org.jgrapht.graph.AbstractBaseGraph

addEdge, addEdge, addVertex, addVertex, clone, containsEdge, containsVertex, createIntrusiveEdgesSpecifics, createSpecifics, degreeOf, edgeSet, edgesOf, getAllEdges, getEdge, getEdgeFactory, getEdgeSource, getEdgeSupplier, getEdgeTarget, getEdgeWeight, getType, getVertexSupplier, incomingEdgesOf, inDegreeOf, isAllowingLoops, isAllowingMultipleEdges, isDirected, isWeighted, outDegreeOf, outgoingEdgesOf, removeEdge, removeEdge, removeVertex, setEdgeSupplier, setEdgeWeight, setVertexSupplier, vertexSet
• ### Methods inherited from class org.jgrapht.graph.AbstractGraph

assertVertexExist, containsEdge, equals, hashCode, removeAllEdges, removeAllEdges, removeAllEdges, removeAllVertices, toString, toStringFromSets
• ### Methods inherited from class java.lang.Object

finalize, getClass, notify, notifyAll, wait, wait, wait
• ### Methods inherited from interface org.jgrapht.Graph

containsEdge, removeAllEdges, removeAllEdges, removeAllVertices
• ### Constructor Detail

• #### BlockCutpointGraph

public BlockCutpointGraph(Graph<V,E> graph)
Deprecated.
Running time = $O(m)$ where m is the number of edges.
Parameters:
graph - the input graph
• ### Method Detail

• #### getBlock

public Graph<V,E> getBlock(V vertex)
Deprecated.
Returns the vertex if vertex is a cutpoint, and otherwise returns the block (biconnected component) containing the vertex.
Parameters:
vertex - vertex in the initial graph.
Returns:
the biconnected component containing the vertex
• #### getCutpoints

public Set<V> getCutpoints()
Deprecated.
Returns the cutpoints of the initial graph.
Returns:
the cutpoints of the initial graph
• #### isCutpoint

public boolean isCutpoint(V vertex)
Deprecated.
Returns true if the vertex is a cutpoint, false otherwise.
Parameters:
vertex - vertex in the initial graph.
Returns:
true if the vertex is a cutpoint, false otherwise.


text:
# Refraction equation, quartic equation

Given two points $P$ and $Q$, a line ($A$, $B$ - orthogonal projection of $P$, $Q$ onto the line) and a coefficient $n$, I want to find out such point $C$ that $\frac{\sin{a}}{\sin{b}}=n$ (in fact, it's an equation of light refraction). I also assume that $C$ lies between $A$ and $B$. See the image below:

What I know is: $w, h, d, n$. What I want to calculate is $x$.

I need to find the easiest solution possible, in order to calculate $x$ in a computer's application very efficiently.

What I've found out so far is:

$$n=\frac{\sin{a}}{\sin{b}}=\frac{\frac{d-x}{\sqrt{(d-x)^2+w^2}}}{\frac{x}{\sqrt{x^2+h^2}}}=\frac{(d-x)\sqrt{x^2+h^2}}{x\sqrt{(d-x)^2+w^2}}$$

Hence, we need to solve this equation for $x$ (assuming $0<x<d$):

$n^2x^2((d-x)^2+w^2)=(d-x)^2(x^2+h^2)$

which yields:

$f(x)=(n^2-1)x^4-2d(n^2-1)x^3+((d^2+w^2)(n^2-1)+w^2-h^2)x^2+2dh^2x-d^2h^2 = 0$

That way we obtained a quartic equation. How to solve it? I've tried using the Ferrari's solution, but the result is very complicated. Moreover, this equation can have as many as four real roots but I know it's got only one root for $0<x<d$. Do you have any idea how to simplify this equation or maybe you can suggest me an easier way to find $x$?

-
Since the solution for n=1 appears to be simple, perhaps you could let $(n^2-1)=\epsilon$ and use singular perturbations to get an asymptotic form of the solution – Valentin May 28 '12 at 22:14
@Valentin: I need a solution for $n>1$ (or at least for $n=1.33$). I'd prefer to find an exact solution. – miloszmaki May 28 '12 at 22:30
asymptotc expansion will give you a solution valid for some range of $n>1$ which should be computationally more efficient than horror like this. what is p? – Valentin May 28 '12 at 22:48
$p$ was supposed to be $h$; I fixed it. – joriki May 29 '12 at 6:42
@joriki: Right, thanks for fixing it ;) – miloszmaki May 29 '12 at 8:57

As long as you're using a numerical method, you might as well stick with a non-polynomial form of your equation that is more well-behaved. In particular, you could just solve $$\frac{(d-x)/\sqrt{(d-x)^2+w^2}}{x/\sqrt{x^2+h^2}}=n$$ for $x$. The left-hand side is a monotonically decreasing function of $x$ between $0$ and $d$, because the numerator is nonnegative and decreasing and the denominator is nonnegative and increasing. So you can just apply bisection search and it is guaranteed to work.

-
For example, here's the graph for your example in the comments. I brought $n$ to the other side of the equation so you can see the solution as the zero crossing. – Rahul May 29 '12 at 18:24
Right, converting the original equation into a polynomial introduces unwanted ambiguity. The polynomial can have as many as four roots while there's only one correct solution to the original problem. I just hoped that I could find a descent analytic solution but it turns out that a numerical method is the only way to solve the problem efficiently. – miloszmaki May 30 '12 at 8:47
I've tested it and it works great! Only 10-20 iterations of bisection search is enough. I could also improve the solution using Newton's method. – miloszmaki May 31 '12 at 19:23

A plot of a typical example shows that the right-hand side of your last displayed equation behaves rather nicely and it should be possible to obtain the solution efficiently using Newton's method, which wouldn't require drawing any roots and should yield the result to sufficient accuracy within a few iterations that only involve a couple of multiplications and additions and one division each.

You could use the solution $x=dh/(h+w)$ for $n=1$ as the initial value for the iteration.

By the way, I'd divide through by $d^4$ to get rid of the irrelevant scale and express everything relative to $d$:

$$\epsilon\xi^4-2\epsilon\xi^3+((1+\omega^2)\epsilon+\omega^2-\eta^2)\xi^2+2\eta^2\xi-\eta^2=0\;,$$

with $\epsilon=n^2-1$, $\xi=x/d$, $\omega=w/d$ and $\eta=h/d$ with initial value $\xi_0=\eta/(\eta+\omega)$.

-
Thank you! I will check this method if it's accurate enough. – miloszmaki May 29 '12 at 9:06
Although it generally works well for $x=dh/(h+w)$, there is a problem for this. Starting with $x=25$ the result is getting close to $x=30$ while $x=30$ is not a root. The correct result can be obtained starting with $x=0$ which gives approx. $5.7$. Also I'm not sure if starting with $x=0$ always gives the correct result. – miloszmaki May 29 '12 at 11:09
Edit to my previous comment: when starting with $x=0$, it converges to $-5.7$ (which means $n=-1.33$). The correct result can be obtained starting for example with $x=1$ or $x=12.5$. Do you have any idea how to choose the starting point which guarantees the convergence and the correct result? – miloszmaki May 29 '12 at 13:38
@miloszmaki: No, sorry, I wasn't expecting such difficult behaviour in other cases from the example I tried. You could use more robust methods like regula falsi instead (starting with $x\in[0,d]$), or use such a method to get close and then switch to Newton's method to improve convergence. – joriki May 29 '12 at 14:22
That's a good idea. Finding a root for $0<x<d$ is guaranteed, since $f(0)=-d^2h^2<0$ and $f(d)=d^2n^2w^2>0$ and $f$ is continous. – miloszmaki May 29 '12 at 16:24


text:
Question

# The equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$.\begin{align} & A)2x+6y+12z=13 \\ & B)x+3y+6z=-7 \\ & C)x+3y+6z=7 \\ & D)2x+6y+12z=-13 \\ \end{align}

Hint: We know that the equation of the plane containing the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is equal to ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda \left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \right)=0$. So, we should write the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$. We know that the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ are said to be parallel, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. It is also given that the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$. By using the concept, we can find the value of $\lambda$. By using this value of $\lambda$, we can find the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$.

We know that the equation of the plane containing the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is equal to ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda \left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \right)=0$.
Now we should find the plane containing the lines $2x-5y+z=3;x+y+4z=5$. \begin{align} & \Rightarrow 2x-5y+z+\lambda \left( x+y+4z \right)=3+5\lambda \\ & \Rightarrow 2x-5y+z+\lambda x+\lambda y+4\lambda z=3+5\lambda \\ & \Rightarrow \left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)=3+5\lambda \\ & \Rightarrow \left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)-3-5\lambda =0....(1) \\ \end{align}
From the question, it is clear that the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$.
We know that the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ are said to be parallel, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$.
So, it is clear that the planes $\left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)-3-5\lambda =0$ and $x+3y+6z=1$ are parallel.
$\Rightarrow \dfrac{2+\lambda }{1}=\dfrac{\lambda -5}{3}=\dfrac{4\lambda +1}{6}.....(2)$
So, let us assume
$\Rightarrow \dfrac{2+\lambda }{1}=\dfrac{\lambda -5}{3}=\dfrac{4\lambda +1}{6}=k.....(3)$
So, from equation (3), we can write
$\Rightarrow \dfrac{2+\lambda }{1}=k$
By using cross multiplication, we get
$\Rightarrow 2+\lambda =k....(4)$
Now, from equation (3), we get
$\Rightarrow \dfrac{\lambda -5}{3}=k$
By using cross multiplication, we get
$\Rightarrow \lambda -5=3k....(5)$
Now let us substitute equation (5) in equation (4), then we get
\begin{align} & \Rightarrow \lambda -5=3\left( 2+\lambda \right) \\ & \Rightarrow \lambda -5=6+3\lambda \\ & \Rightarrow 2\lambda =-11 \\ & \Rightarrow \lambda =\dfrac{-11}{2}.....(6) \\ \end{align}
Now we will substitute equation (6) in equation (1), then we get
\begin{align} & \Rightarrow \left( 2-\dfrac{11}{2} \right)x+\left( -\dfrac{11}{2}-5 \right)y+\left( 4\left( \dfrac{-11}{2} \right)+1 \right)z-3-5\left( \dfrac{-11}{2} \right)=0 \\ & \Rightarrow \left( \dfrac{4-11}{2} \right)x+\left( \dfrac{-11-10}{2} \right)y+\left( 2(-11)+1 \right)z-3+\dfrac{55}{2}=0 \\ & \Rightarrow \left( \dfrac{-7}{2} \right)x-\dfrac{21y}{2}-21z+\dfrac{49}{2}=0 \\ & \Rightarrow \dfrac{-7x-21y-42z+49}{2}=0 \\ & \Rightarrow -7x-21y-42z+49=0 \\ & \Rightarrow -x-3y-6z+7=0 \\ & \Rightarrow x+3y+6z-7=0 \\ & \Rightarrow x+3y+6z=7.....(7) \\ \end{align}
So, from equation (7) we will get that the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$ is equal to $x+3y+6z=7$.

So, the correct answer is “Option C”.

Note: Students have a misconception that that the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ are said to be parallel, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{{{d}_{1}}}{{{d}_{2}}}$. But we know that the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ are said to be parallel, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. So, students should avoid this misconception.


text:
# What is the chemical test for identification of benzophenone?

How to identify that a unknown compound is benzophenone?

My working:-

First I added 2,4- DNP(2,4-Dinitrophenyl hydrazine) to the compound. Yellow-orange crystals were formed which indicated presence of carbonyl group. Then, I added Tollen's reagent to the compound. It did not gave silver mirror which means aldehyde group is absent. So, it must be a ketone. So, I added m-Dinitrobenzene to the compound. It did not gave violet colour. Alternatively, I added Sodium nitroprusside to the compound. It did not gave a red color. Both the test is for confirmation of ketone compound.

I asked my teacher what is the unknown compound. He said that it is benzophenone. Also, in the practical book where all the chemical test were mentioned, it was mentioned as a side-note that benzophenone does not give the above mentioned test. So, what is the chemical test for identification of benzophenone?

• You can use iodoform test after Tollens test. – Aditya Dev Dec 22 '15 at 13:46
• @AdityaDev, benzophenone will not give iodoform test because it does not contain ketomethyl group ($\ce{-CH3CO}$). – Nilay Ghosh Feb 17 '16 at 7:12
• oops. i thought benzophenone was CH3-CO-ph. my mistake. – Aditya Dev Feb 17 '16 at 7:16

After identification of ketone, to identify benzophenone you need to find the range of melting point using melting point apparatus which you should get in range of $48.3\ \mathrm{^\circ C}$ to $50.6\ \mathrm{^\circ C}$.


text:
3 years ago

# Reconcilement of VHE \texorpdfstring{$\gamma$}\,-ray/X-ray correlation studies in Mrk 421 and break-down at high fluxes.

M. M. González, B. Patricelli, N. Fraija, J.a. García-gonzález

Multi-wavelength campaigns have been carried out to study the correlation between the very high energy (VHE) $\gamma$-ray and the X-ray emissions in blazars but, no conclusive results have been achieved yet. In this paper, we add Milagro data to the existing VHE $\gamma$-ray data from HEGRA-CT1 and Whipple and test the consistency and robustness of the reported correlation between VHE $\gamma$-ray and X-ray fluxes in Mrk 421. We found that at monthly time scale the correlation is robust, consistent between instruments and described as a linear function. Furthermore, most of the fluxes on shorter time scales are consistent with the correlation within 3 $\sigma_A$ even, where $\sigma_A$ is an estimated intrinsic scatter. However, a break-down of the correlation becomes clearly evident at high states of activity with fluxes $\rm \gtrsim 2.5\times 10^{-10}\, cm^{-2}s^{-1}$ at energies above 400 GeV independently of the time scale, observational period or instrument, even for single flares, the X-ray and VHE $\gamma$-ray emissions lie on the correlation until the VHE $\gamma$-ray flux reaches values higher than the one mentioned above. We have interpreted our results within the one-zone synchrotron self-Compton model. We found that describing a single and unique $\gamma$-ray/X-ray correlation strongly narrows the range of possible values of the magnetic field $B$ when a constant value of the spectral index along the correlation is assumed.

Publisher URL: http://arxiv.org/abs/1901.02454

DOI: arXiv:1901.02454v1

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text:
Next: Bibliography

The Born-Oppenheimer Approximation

We may write the nonrelativistic Hamiltonian for a molecule as a sum of five terms:

 (1)

where refer to electrons and refer to nuclei. In atomic units, this is just
 (2)

The Schrödinger equation may be written more compactly as
 (3)

where is the set of nuclear coordinates and is the set of electronic coordinates. If spin-orbit effects are important, they can be added through a spin-orbit operator .

Unfortunately, the term prevents us from separating into nuclear and electronic parts, which would allow us to write the molecular wavefunction as a product of nuclear and electronic terms, . We thus introduce the Born-Oppenheimer approximation, by which we conclude that this nuclear and electronic separation is approximately correct. The term is large and cannot be neglected; however, we can make the dependence parametric, so that the total wavefunction is given as . The Born-Oppenheimer approximation rests on the fact that the nuclei are much more massive than the electrons, which allows us to say that the nuclei are nearly fixed with respect to electron motion. We can fix , the nuclear configuration, at some value , and solve for the electronic wavefunction , which depends only parametrically on . If we do this for a range of , we obtain the potential energy curve along which the nuclei move.

We now show the mathematical details. Initially, can be neglected since is smaller than by a factor of , where is the reduced mass of an electron. Thus for a fixed nuclear configuration, we have

 (4)

such that
 (5)

This is the clamped-nuclei'' Schrödinger equation. Quite frequently is neglected in the above equation, which is justified since in this case is just a parameter so that is just a constant and shifts the eigenvalues only by some constant amount. Leaving out of the electronic Schrödinger equation leads to a similar equation,
 (6)

 (7)

For the purposes of these notes, we will assume that is included in the electronic Hamiltonian. Additionally, if spin-orbit effects are important, then these can be included at each nuclear configuration according to
 (8)

 (9)

Consider again the original Hamiltonian (1). An exact solution can be obtained by using an (infinite) expansion of the form

 (10)

although, to the extent that the Born-Oppenheimer approximation is valid, very accurate solutions can be obtained using only one or a few terms. Alternatively, the total wavefunction can be expanded in terms of the electronic wavefunctions and a set of pre-selected nuclear wavefunctions; this requires the introduction of expansion coefficients:
 (11)

where the superscript has been added as a reminder that there are multiple solutions to the Schrödinger equation.

Expressions for the nuclear wavefunctions can be obtained by inserting the expansion (10) into the total Schrödinger equation yields

 (12)

or
 (13)

if the electronic functions are orthonormal. Simplifying further,
 (14)

The last term can be expanded using the chain rule to yield
 (15)

At this point, a more compact notation is very helpful. Following Tully [1], we introduce the following quantities:

 (16)

 (17)

 (18)

 (19)

 (20)

 (21)

 (22)

Note that equation (18) of reference [1] should not contain a factor of 1/2 as it does. Now we can rewrite equations (14) and (15) as
 (23)

or
 (24)

This is equation (14) of Tully's article [1]. Tully simplifies this equation by one more step, removing the term . By taking the derivative of the overlap of it is easy to show that this term must be zero when the electronic wavefunction can be made real. If we use electronic wavefunctions which diagonalize the electronic Hamiltonian, then the electronic basis is called adiabatic, and the coupling terms vanish.1 This is the general procedure. However, the equation above is formally exact even if other electronic functions are used. In some contexts it is preferable to minimize other coupling terms, such as ; this results in a diabatic electronic basis. Note that the first-derivative nonadiabatic coupling matrix elements are usually considered more important than the second-derivative ones, .

In most cases, the couplings on the right-hand side of the preceeding equation are small. If they can be safely neglected, and assuming that the wavefunction is real, we obtain the following equation for the motion of the nuclei on a given Born-Oppenheimer potential energy surface:

 (25)

This equation clearly shows that, when the off-diagonal couplings can be ignored, the nuclei move in a potential field set up by the electrons. The potential energy at each point is given primarily by (the expectation value of the electronic energy), with a small correction factor . Following Steinfeld [2], we can estimate the magnitude of the term as follows: a typical contribution has the form , but is of the same order as since the derivatives operate over approximately the same dimensions. The latter is , with the momentum of an electron. Therefore . Since , this term is expected to be small, and it is usually dropped. However, to achieve very high accuracy, such as in spectroscopic applications, this term must be retained.

The Born-Oppenheimer Diagonal Correction In a perturbation theory analysis of the Born-Oppenheimer approximation, the first-order correction to the Born-Oppenheimer electronic energy due to the nuclear motion is the Born-Oppenheimer diagonal correction (BODC),

 (26)

which can be applied to any electronic state . The BODC is also referred to as the adiabatic correction.'' One of the first systematic investigations of this effect was a study by Handy, Yamaguchi, and Schaefer in 1986 [3]. In this work, the authors evaluated the BODC using Hartree-Fock self-consistent-field methods (and, where relevant, two-configuration self-consistent-field) for a series of small molecules. One interesting finding was that the BODC changes the singlet-triplet splitting in methylene by 40 cm, which is small on a chemical'' energy scale but very relevant for a spectroscopic energy scale. Inclusion of the BODC is required to accurately model the very dense rovibrational spectrum of the water molecule observed at high energies, and these models were a critical component in proving the existence of water on the sun [4,5].

For many years, it was only possible to compute the BODC for Hartree-Fock or multiconfigurational self-consistent-field wavefunctions. However, in 2003 the evaluation of the BODC using general configuration interaction wavefunctions was implemented [6] and its convergence toward the ab initio limit was investigated for H, BH, and HO. This study found that the absolute value of the BODC is difficult to converge, but errors in estimates of the BODC largely cancel each other so that even BODC's computed using Hartree-Fock theory capture most of the effect of the adiabatic correction on relative energies or geometries. Table 1 displays the effect of the BODC on the barrier to linearity in the water molecule and the convergence of this quantity with respect to basis set and level of electron correlation. Although the absolute values of the BODC's of bent and linear water change significantly with respect to basis set and level of electron correlation, their difference does not change much as long as a basis of at least cc-pVDZ quality is used. For the cc-pVDZ basis, electron correlation changes the differential BODC correction by about 1 cm. Table 2 displays the effect of the BODC on the equilibrium bond lengths and harmonic vibrational frequencies of the BH, CH, and NH molecules [7] and demonstrates somewhat larger changes to the spectroscopic constants than one might have expected, particularly for BH.

 Basis Method DZ RHF 613.66 587.69 -25.97 DZ CISD 622.40 596.43 -25.97 DZ CISDT 623.62 597.56 -26.06 DZ CISDTQ 624.56 598.28 -26.28 DZ CISDTQP 624.61 598.32 -26.29 DZP RHF 597.88 581.32 -16.56 cc-pVDZ RHF 600.28 585.20 -15.08 cc-pVDZ CISD 615.03 599.15 -15.88 cc-pVDZ CISDT 616.82 600.62 -16.20 cc-pVTZ RHF 596.53 581.43 -15.10 cc-pVTZ CISD 611.89 596.73 -15.16 cc-pVQZ RHF 595.57 580.72 -14.85 Data from Valeev and Sherrill [6]. The difference between the adiabatic correction for the and structures.

 BH CH NH 0.00066 0.00063 0.00027 -2.25 -2.81 -1.38 Data from Temelso, Valeev, and Sherrill [7].

Next: Bibliography
David Sherrill 2005-03-14


text:
# Do we have to assume normality of the data, even when we conduct z-test or t-test with large samples?

I read this lecture note and found that it assumes normality of the data when we conduct z-test or t-test.

I can accept that when we have small samples we have to assume normality of the data, because there's not enough evidence to believe that sample mean was drawn from normal distribution.

My question is that if we have large enough samples (of course it is a bit subjective), do we need to assume normality of the data? Isn't it just okay to assume sample mean was drawn from normal distribution (by central limit theorem)?


text:
# Is a symmetric encryption key derivable using the input and the output?

I am looking for a python library for encrypting a string with a single key. I do not want the key to derivable by knowing the input and the output. I have came to cryptography.fernet so far, but I am not sure if this will fulfill my needs. Can the encryption key be derived by knowing the algorithm (assuming the fernet algorithm in this case) and knowing the input and the output?

For example when I encrypt message A with key K and it outputs B, can somehow A and B be put together, so it will output K?

EDIT

The actual use of this is the following. I need to store a password in a place that others can see (a discord config channel, but that is secondary) in a way so they cannot see the actual password. I thought that the user would encrypt the password one-time with a key only visible to the bot (a discord bot) (with a command). The encrypted password would then be sent back to the user and stored in the channel for the bot to decrypt it and use it later. The problem is: if the user knows the input and the output, if he would be able to guess the key, he would then be able to decrypt all other passwords stored this way. And absolutely noone wants this to happen!

• Welcome to Cryptography. What do you mean by reversible? What you want may be, given plaintexts-ciphertexts pairs, it should be almost impossible to derive the key. AES is known to be secure this kind of attacks, or you can use ChaCha. The library part is off-topic here. So your question is better, Can AES (or your symmetric cipher) be broken by known-plaintext attack? Or you can mention your real problem in better words. You can edit your question. – kelalaka Aug 29 '19 at 17:36
• For passwords, I advise you to look at one of the top questions of information security How to securely hash passwords? – kelalaka Aug 29 '19 at 18:24
• Thank you, but I did not plan on storing it in a database. I have to store them in the channel. I am doing an automation, which requires some passwords to be known. – Jakub Bláha Aug 29 '19 at 18:47

For example when I encrypt message A with key K and it outputs B, can somehow A and B be put together, so it will output K?
Except for encryption messages where you are specifically restricted to use a specifc key K only once (e.g. OTP), then no, it is infeasible to recover K from A and B.


text:
geerky42 Group Title Six circles are tangent to each other and an equilateral triangle is inscribed around them as shown. What percent of the area of triangle is NOT shaded? one year ago one year ago

1. geerky42

Bad drawing, sorry. Hopefully you know what I'm trying to draw...|dw:1352593402510:dw|

2. Dido525

Do you know any other information?

3. geerky42

No, this is all given information I have.

4. Dido525

Well it's equilateral so all of it is the same...

5. geerky42

I guess so.

6. Dido525

The area of an equilateral triangle is : $\frac{ s^2\sqrt{3} }{ 4 }$

7. geerky42

And circles are tangent to each other, so they all should be the same.

8. Dido525

Mhmm... Let me see. You post very intrsting questions.

9. geerky42

And how can I applying to it?

10. Dido525

I am thinking...

11. geerky42

Sorry for interruption, but I need your help... @Hero @tcarroll010 @AccessDenied @AriPotta

12. geerky42

Any ideas, hints, tips?

13. tcarroll010

Are all those circles of the same radius?

14. jon.stromer.galley

if you know the area of the triangle and the area of the circle(s) what can you say about the area not covered by the circles?

15. geerky42

Well, we only know that they are tangent to each other and in the shown image on my paper, they appear that they also tangent to the sides of triangle too, so I guess yeah. This is literally the only given information I have...

16. geerky42

Go on... @jon.stromer.galley

17. Dido525

Well if they are tengential to each other I assume they have the same radius.

18. geerky42

I suppose.

19. geerky42

So where should we start?

20. tcarroll010

I am exploring a relationship between the triangle and a second triangle formed by connecting the centers of the circles.

21. geerky42

Why not the relationship between the area of circle to the area of triangle? I think this is good start, but I'm not sure.

22. geerky42

I don't know, lol.

23. Dido525

I am thinking about saying that: |dw:1352594259221:dw|

24. Dido525

I don't think so though...

25. geerky42

I doubt it. |dw:1352594316081:dw|

26. Dido525

Yeah I know. Was just wondering...

27. AccessDenied

A better picture perhaps: http://puu.sh/1oLWD

28. geerky42

I think @tcarroll010 has a good point, perhaps we should determine the relationship between the triangle and a second triangle formed by connecting the centers of the circles.

29. geerky42

Much better. @AccessDenied

30. Dido525

LOT better.

31. AccessDenied

Oh wow, this is sort of interesting: I draw in all of the tangents of the circles to the exterior triangle and an interior triangle: http://puu.sh/1oM08

32. Dido525

Where are you getting these questions from? O_o .

33. Dido525

Hmm!

34. Dido525

Similar triangles perhaps?

35. jon.stromer.galley

Ok, so all you know for sure is that the triangle is equilateral and that the circles have a radius r. That is enough to express the combined circle area, but to answer the question you need to know how big the triangle is (how long the sides are) in terms of R. Let's look at the lower left corner.

36. jon.stromer.galley

|dw:1352594743279:dw|

37. Dido525

We don't know that...

38. AccessDenied

Hint: you have an equilateral triangle for the exterior, so you know some angles. ;)

39. tcarroll010

I've got what might be a good idea. You use trig. And 30-60-90 right triangles in the corners. I was independently working a similar diagram to Accessd.

40. jon.stromer.galley

let's draw it again...|dw:1352594832325:dw|

41. Dido525

Ohh yeah!!!

42. Dido525

Isn't that also one of those special triangles?

43. jon.stromer.galley

At this point if you have been shown you have enough data to figure out the base of the triangle. The length of a side of the over all triangle will be then 3r + 2( length of the base of the triangle we just drew)

44. tcarroll010

It looks like a whole bunch of us about simultaneously hit on the right idea! Fun trig at this point.

45. geerky42

I have sense I'm getting closer to solution, yet I have no idea where to start. I have to find the relationship between the radius and length of triangle side, right?

46. Dido525

|dw:1352594992142:dw|

47. Dido525

Now we know the radius of all those circles are 1!

48. AccessDenied

not necessarily. The 30-60-90 rule only expresses a ratio. :P

49. geerky42

And from this, we can find the ratio of the total area of six circle to the area of triangle, right?

50. Dido525

Can we not assume? O_o .

51. jon.stromer.galley

the radius is still "1" r but now the base can be expressed in terms of r as well

52. Dido525

Not sure is the side length would just be 2r...

53. tcarroll010

To be pure, you would use "r" for radius, but you are free to call radius "1" for your purposes, so, yes you can make the assumption, with that proviso. And then it's simple to get your triangle side lengths. Nice little problem!

54. Dido525

Well looking at @AccessDenied Drawing... : Since the radius is 1 we can say the side length's of the smaller, inside triangle are 3, 3 and 2 going clockwise.

55. Dido525

4 4 and 4 sorry.

56. Dido525

|dw:1352595331990:dw|

57. Dido525

I am assuming the height would be 3...

58. Dido525

|dw:1352595405601:dw|

59. Dido525

Now I can't get a ratio....

60. tcarroll010

With a circle radius of "r", being the "short leg", then the "long leg" will be r times sqrt(3). So a triangle side will be 2r(sqrt(3)) + 2r.

61. Dido525

So it would just be 2root(3)+2 .

62. tcarroll010

If you are going with your assumption or requirement that r is "1", then, yes.

63. tcarroll010

So, you should be able to get the altitude using Pythagorean and then area of triangle, and areas of circles. You're just about done!

64. jon.stromer.galley

I calculate that the length of a side of the master triangle is 6.4641r Anyone else? tanks to: adjacent = r / tan(30)

65. jon.stromer.galley

or alternatively by expressing the length of the line as: 2r * root(3) + 3r that plugs in nicely to the formula for the area of an el triangle: (length of side * root(3)) / 4 so that tells us the area is...

66. AccessDenied

I missed the answer placed by the original poster, but my work in ms paint is here: http://puu.sh/1oNgt

67. geerky42

I got the same answer. Thank you, everyone, for your times. I appreciate it.

68. AccessDenied

You're welcome! :)


text:
# The PreTeXt Guide

## SectionF.2MathJax and Speech Rule Engine

MathJax is packaged as a node.js program. A component of MathJax is Speech Rule Engine (SRE), which converts to text that represents a spoken version of the mathematics, and to Nemeth braille, a braille code for expressing mathematics.
Follow the directions here exactly. Do not free-style and think some modification will have a better or identical result. In a terminal, set your working directory with
cd /path/to/mathbook/script/mjsre

Install both packages at the same time with a single command:
./update-sre

It is possible you will need to make this simple script executable. On Unix-style systems (Linux, Mac OS) this can be done with the chmod command. As this is a bash script, Windows users must use Window's Sybsystem for Linux or examine the contents of the script to run the commands manually.
The update-sre script is supplied by PreTeXt. It will create a file and a directory:
/path/to/mathbook/script/mjsre/package-lock.json
/path/to/mathbook/script/mjsre/node_modules

Neither will be tracked by git. Since SRE changes faster than MathJax, this will replace MathJax's expectation for a version of SRE with a potentially newer version. So do not be alarmed if it appears MathJax is missing a dependency, that is intentional.
Now you have copies of MathJax and SRE that can be used by the pretext/pretext script to process mathematics offline into useful formats for conversions to EPUB, Kindle, and braille.


text:
# How to get 3.3 output voltage in PC817 opto coupler?

I am using PC817 opto coupler. I want a 3.3 output voltage. In its 1st pin I am giving 12volt and 2nd pin is gnd. I have connected a pull down resistor on 3rd pin. But i am confused because when I am giving 3.3volt on 4th pin, and I am checking output from 3rd pin which is 1.5volt. How can I get 3.0 - 3.3volt output on 3rd pin.

I have tested it with 5 volt, means when I was giving 5 volt on 4th pin and when I was checking voltage on 3rd pin it was 4.7-4.8v which is I think correct. But I dont know what is happening in case of 3.3volt. PLease help.

Thanks

EDIT : i have attached a screenshot of proteus simulation. I actually didnt find the PC817 opto coupler but I got the similar opto coupler. I am getting an output voltage of 3.22. Please tell me that is this connection good. I mean if I make this connection in real, will this work?

• post the schematics. Do you have resistor between 12 V and Pin #1? What is the pull down value? Ideally you should have a pull up resistor. The Opto will drive the transistor output low when the Diode is driven. Pls post all resistor values along with the schematics to give a possibly useful answer. – Umar Jul 17 '15 at 8:16
• I cannot post the schematics but i can tell you the resistor value. I am using 1k res on pin1 and a pull down resistor of 1k on 3rd pin. also from 3rd pin I am using 1k resistor and I am checking the output from this resistor which is 1.5v when I am applying 3.3v on 4th pin. If this is wrong can you suggest me another circuit? – anna carolina Jul 17 '15 at 8:45
• You certainly can post schematics. If you can't take a screenshot of the relevant bit, use the built in schematic editor to redraw it. – Nick Johnson Jul 17 '15 at 9:44
• @annacarolina For your next question, you can also use the Schematic Editor on the forum to supply more information. It's an icon above where you type the text. 7th from left. (Or hotkey Ctrl+M). If you supply a schematic with values as you use it, we will be able to offer much better answers much quicker. – Asmyldof Jul 17 '15 at 10:37

Please try using the setup as shown in the figure. The 1 K value on pin one is fine, but increase the value of resistor to 100k between Pin 4 and 3.3 V. When there is no 12 V on pin 1, The output (Pin Number 4) will be high, and when there is 12 V on Pin 1, the transistor will turn on shorting the Pin 4 to ground (VCE will be less than 100-200 mV). please try this and reply if not possible.

I am using the same setup, different kind of coupler and it is working.

• you are saying that when input is low, output will be high and vice versa. But I dont want it like this. I want that when I give 12v supply on pin 1 my output should be 3.0-3.3volt – anna carolina Jul 17 '15 at 9:20
• I would not opt for that option. i will use a transistor at the input of pin 1 as inverter if in case, the 12 V drive is impossible to invert. I feel that,when the collector (opto coupler)current flows, the drop developed across emitter will be comparable with collector voltage thus limiting the base current (negative feedback). I might be wrong too. ) – Umar Jul 17 '15 at 9:45
• i have attached a screenshot of proteus simulation. I actually didnt find the PC817 opto coupler but I got the similar opto coupler. I am getting an output voltage of 2.93. Please tell me that is this connection good. I mean if I make this connection in real, will this work? – anna carolina Jul 18 '15 at 13:29
• @anna If 3 V is acceptable (normally yes), then i see no issues. where r you connecting the 3 V output? – Umar Jul 19 '15 at 11:50
• I am using PIC32MX795F512L microcontroller, and this controller can work only in case of 3.3v. So i am giving this voltage to controller.Is this thing correct.? – anna carolina Jul 19 '15 at 14:14

If you have 12V feeding the input diode via 1K you will be producing a forward current of about 10mA. If you have a 1k resistor on the output that you hope to have 3V3 across it means an output current (Ic) of 3.3mA.

It looks to me like you will probably achieve about 3.2 volts on the output but this is a typical example only. In my experience, a lot of opto devices are worse than this so try driving the input a little harder (maybe 20mA) and try increasing the output resistor from 1k to 10k.

If you connect your PC817 as shown below you should be able to achieve a "non inverting" output.

Here is an LTSpice simulation waveform that shows the output voltage level that should be achievable.

Note that the timing delay response through the optocoupler in the PC817A model may not represent an accurate picture of the real delays to be expected in a physical circuit implementation. On the otherhand the voltage and current transfer characteristics are modeled in a reasonable manner.

• i have attached a screenshot of proteus simulation. I actually didnt find the PC817 opto coupler but I got the similar opto coupler. I am getting an output voltage of 2.93. Please tell me that is this connection good. I mean if I make this connection in real, will this work? – anna carolina Jul 18 '15 at 13:30
• @annacarolina - It should work for you. Remember to keep your load on the output to very small current. Two things can be adjusted - increase current through input diode by reducing the value of the R1 in my diagram - or increase the value of the R2 in my diagram. – Michael Karas Jul 18 '15 at 15:47
Use LM7805 to convert 12V into  5V

Then use this voltage divider to create 3.3V supply

simulate this circuit – Schematic created using CircuitLab

Use the circuit below , and as datasheet suggests that the forward voltage is about 1.2V (led voltage) and maximum current should be 20mA so the resistor R1 can be calculated as

Voltage_R1 = 5.0-1.2  = 3.8V
I_R1 = 20mA
R1 = 3.8 / 20mA = 190 ohm

Use minimum resistor of value this , higher resistor will reduce current

• electronics.stackexchange.com/questions/49141/… – Taimoor Ali Jul 17 '15 at 9:09
• you are giving 3.3v input. but in my circuit i'll have to give 12volt supply and output should be 3.3v. How to do that? – anna carolina Jul 17 '15 at 9:21
• Bring 12 V to 3.3 or 4 using voltage divider – Taimoor Ali Jul 17 '15 at 9:25
• This isn't a "3.3v source", it's a voltage divider, and along with wasting 20mA constantly, it will also have extremely poor output regulation. – Nick Johnson Jul 17 '15 at 9:45
• I've already said - because you're trying to use a voltage divider as a regulated power supply. – Nick Johnson Jul 17 '15 at 10:14


text:
# broyden.py¶

Define the BroydenSolver class.

Based on implementation in Scipy via OpenMDAO 0.8x with improvements based on NPSS solver.

class openmdao.solvers.nonlinear.broyden.BroydenSolver(**kwargs)[source]

Broyden solver.

Attributes

 delta_fxm (ndarray) Most recent change in residual vector. delta_xm (ndarray) Most recent change in state vector. fxm (ndarray) Most recent residual. Gm (ndarray) Most recent Jacobian matrix. linear_solver (LinearSolver) Linear solver to use for calculating inverse Jacobian. linesearch (NonlinearSolver) Line search algorithm. Default is None for no line search. n (int) Total length of the states being solved. xm (ndarray) Most recent state.
SOLVER = 'BROYDEN'
__init__(self, **kwargs)[source]

Initialize all attributes.

Parameters
**kwargsdict

options dictionary.

add_recorder(self, recorder)

Add a recorder to the solver’s RecordingManager.

Parameters
recorder<CaseRecorder>

A recorder instance to be added to RecManager.

cleanup(self)[source]

Clean up resources prior to exit.

get_residuals(self)[source]

Return a vector containing the values of the residuals specified in options.

Returns
ndarray

Array containing values of residuals.

get_states(self)[source]

Return a vector containing the values of the states specified in options.

This is used to get the initial state guesses.

Returns
ndarray

Array containing values of states.

record_iteration(self, **kwargs)

Record an iteration of the current Solver.

Parameters
**kwargsdict

Keyword arguments (used for abs and rel error).

set_linear_vector(self, dx)[source]

Copy values from step into the linear vector for backtracking.

Parameters
dxndarray

Full step in the states for this iteration.

set_states(self, new_val)[source]

Set new values for states specified in options.

Parameters
new_valndarray

New values for states.

solve(self)

Run the solver.


text:
# Friedman Numbers and 'mystery'

Friedman numbers are a set of those numbers which can be expressed as a function involving the digits of that number and basic operators such as + , - , * , / , ^ , ().

For example,

25 = 5^2

125 = 5^(1+2)

Now, the 'mystery' which I want a help in resolving is that as we move forward in the number line or in other words, if the numbers tend to get bigger and bigger, then the frequency of finding such numbers also increase. In mathematical language,

$${lim}_{n->\infty} \frac{f(n)}{n}$$ = 1

where f(n) is a function of n generating Friedman numbers.

It would be great if anyone knowing a proof of this limit shares it here. I am really very eager to know it. Thanks in anticipation

Note by Kartik Sharma
3 years, 7 months ago

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This is not a complete proof, but we can at least get an understanding of this "mystery" if we look at very large numbers. Typically, a very large number will have a roughly equal number of each of the digits $$1,2,3,4,5,6,7,8,9,0$$, so, let's say we have a number that has $$10n$$ digits, where $$n$$ is a suitably large integer. Let's propose some kind of a mathematical expression using basic operations that uses $$10n$$ digits. How many ways can the digits of that large number can be plugged into this mathematical expression? It's approximately

$$\dfrac { \left( 10n \right) ! }{ { \left( 10! \right) }^{ 10 } }$$

different ways. Now, but this number is approximately $$10\cdot { 10 }^{ \displaystyle n }$$

What's the ratio of the two, that is, this number divided by the number of ways its digits can be permutated in that mathematical expression? It's

$$10\cdot { 10 }^{\displaystyle n }\dfrac { { \left( 10! \right) }^{ 10 } }{ \left( 10n \right) ! }$$

As $$n$$ increases, it first starts out as a very large number, but it then very quickly drops down towards zero. That is, there can be far more permutations possible than the number itself! Thus, the odds of one of those permutations yielding exactly that number rises exponentially for large numbers, to where it becomes nearly a certainty.

- 3 years, 7 months ago


text:
# Different contrasts in DESeq2

I have a treatment and control in two time points like this

> design
condition time
Sample_Sample_1_3x_CTRL_MSC_24_hrs_replicate1    control   24
Sample_Sample_2_3x_CTRL_MSC_24_hrs_replicate1    control   24
Sample_Sample_3_3x_CTRL_MSC_24_hrs_replicate1    control   24
Sample_Sample_10_3x_CTRL_MSC_48_hrs_replicate1   control   48
Sample_Sample_11_3x_CTRL_MSC_48_hrs_replicate1   control   48
Sample_Sample_12_3x_CTRL_MSC_48_hrs_replicate1   control   48
Sample_Sample_4_3x_IT_MSC_24_hrs_replicate1           IT   24
Sample_Sample_5_3x_IT_MSC_24_hrs_replicate1           IT   24
Sample_Sample_6_3x_IT_MSC_24_hrs_replicate1           IT   24
Sample_Sample_13_3x_IT_MSC_48_hrs_replicate1          IT   48
Sample_Sample_14_3x_IT_MSC_48_hrs_replicate1          IT   48
Sample_Sample_15_3x_IT_MSC_48_hrs_replicate1          IT   48
>

I want to test between between treatment and control considering time point 24 hours to 48 hours

I have done like this

dds <- DESeqDataSetFromMatrix(countData=a,colData=design, design=~time + condition + time:condition)

ddsTC <- DESeq(dds, test="LRT", reduced = ~ time + time:condition)

> resultsNames(ddsTC)
[1] "Intercept"                "condition_IT_vs_control"
[3] "time"                     "conditionIT.time"

What is the difference of the results of these codes

results(ddsTC, name="condition_IT_vs_control", test="Wald")

versus

results(ddsTC, name="conditionIT.time", test="Wald")

Actually I want to know what condition_IT_vs_control gives and what conditionIT.time gives

Thank you so much for any intuition

condition_IT_vs_control gives the effect of condition, conditionIT.time is the interaction of condition and time. If you wanted to test the effect of time, use name="time".