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As Mn7+ in KMnO4 changes to Mn2+ in acidic medium; it's equivalent weight becomes M/5 (where M is molar mass of KMnO4). The equivalent weight depends on the reaction involved. I'm doing an Equivalent Weight and Normality experiment and I got very confused on the discussion questions. what is the eqvivalent weight of kmno4 in neutral medium. The Questions and Answers of When potassium permanganate is titrated against ferrous ammonium sulphate, the equivalent weight of potassium permanganate is a) Mol. Related Questions & Answers. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. The equivalent mass of a substance is the ratio of molecular mass to the number of electrons gained (or lost) or change in the oxidation state. Solution for why is equivalent weight of KMnO4 is different in acidic, alkaline and neutral medium ? Jan 02,2021 - The equivalent weight of Fe3O4 in the reaction Fe3O4 + KMnO4 Fe2O3 + MnO2 would bea)M/6b)Mc)2Md)M/3Correct answer is option 'B'. KMnO4 as an oxidizer in acidic media: MnO4- + 8 H+ + 5e- --> Mn2+ + 4H2O (gained 5 electrons from reductant) I.e. Median response time is 34 minutes and may be longer for new subjects. Click here to get an answer to your question  What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium? Equivalent weight of a substance (oxidant or reductant) is equal to molecular weight divided by number of electrons lost or gained by one molecule of the substance in a redox reaction. x= 7-6= 1. Chemistry. Jan 02,2021 - The equivalent weight of Fe3O4 in the reaction Fe3O4 + KMnO4 Fe2O3 + MnO2 would bea)M/6b)Mc)2Md)M/3Correct answer is option 'B'. n factor = 3 in Neutral medium. calculate the equivalent weight of KMnO4 in neutral medium. Express Mass Of … Wt./5 c) Mol. See Answer. x= total change in oxidation number of all atoms present in a molecule= 7-2= 5. equivalent weight= (molecular weight)/x= 157/5u. Physics - please check over wt. Characteristics of a solution i. How much money do you start with in monopoly revolution? Log in. So, here the change in oxidation state is 3. MnO 2. here KMnO 4 changes to MnO 2 so , n factor of Mn here is 3. now we can calculate equivalent mass why is Net cash provided from investing activities is preferred to net cash used? please explain.? Books. n factor = 5 in acidic medium. And explain it. Equivalent weight of MnO_(4)^(ɵ) in acidic neutral and basic media are in ratio of: Equivalent weight of MnO_(4)^(ɵ) in acidic neutral and basic media are in ratio of: Doubtnut is better on App. The mean (|SE) concn of manganese in axial muscle of unexposed controls was 0.262 | 0.018 mg/kg (wet weight). Wt./6 The answer is ‘b’. Formula weight of KMnO4 is 158.04. of KMnO_(4)) Books. as an oxidizer in neutral or basic media: Equivalent weight of KMnO4 in neutral medium. Wt./10 b) Mol. 5 … Continuous exposure to 0.5 mg KMnO4 or exposure to 1 or 2 mg KMnO4/L on alternate days did not cause a significant incr in manganese in axial muscle or liver tissue.
STATEMENT-2: Equivalent weight of in acidic medium in M/5 and in strongly alkaline medium in M/3.

STATEMENT-2: Equivalent weight of in acidic medium in M/5 and in strongly alkaline medium in M/3. To undergo chemical reaction, mo... Q: Which class of drugs is used in sleeping pills? 1. Can you explain this answer? Equivalent weight of KMnO4 in neutral medium? Check out a sample Q&A here. Eq wt = 158.04/5 = 31.61g/equivqlent. IN ACIDIC MEDIUM:. Click hereto get an answer to your question  The equivalent weight of KMnO4 (formula weight: M ) when it is used as an oxidant in neutral medium is: Equivalent weight of K2Cr2O7 in acidic medium is 1) 24.5 2) 49 3) 147 4) 296 50. A: Benzodiazepines such as Xanax, Valium, Ativan, and Librium are anti- anxiety medications, that are a... Q: Zn, Cd and Hg are volatile metals. Solution for calculate the equivalent weight of KMnO4 in neutral medium. Q: Write the balanced chemical equation for the reaction of hydrobromic acid with magnesium. That would be five. equivalent weight. Equivalent weight = Molar mass/number of electrons lost or gained (n) The molecular mass of KMno4 = 158.04g In acidic medium: MnO 4- + 8H +5e – –> Mn 2 + 4H 2 O Wt./6 The answer is ‘b’. A: The solubility of a compound depends on their polar or non-polar nature. Molecular mass of KMnO 4 = 158 gm In acidic medium, no. The volumes of KMnO4 required are 20 mL in acid, 33.3 mL in neutral and 100 mL in alkaline media. Join now. Wt./3 d) Mol. IN BASIC MEDIUM:. The equivalent weight in an oxidation/reduction reaction is the molecular weight divided by the electron change (gain or loss). 1. 1. To find the equivalent mass, divide that molar mass by the mols of electrons taken in its half reaction. Log in. Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance.The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. 148 views. Chemistry. The equivalent weight depends on the reaction involved. 5 Equivalents per mole, so equivalent mass of KMnO4 = M/5 = 158.04/5 = 31.61 gram/equivalent. Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.. (3R,5S,6R)-... A: The structure of each given compounds are drawn in step two. Answer: What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium? of KMnO4 = 158.04/3 = 52.68 … viveksaklani6150 viveksaklani6150 11.12.2017 Chemistry Secondary School What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium? Physics. 2. Ask your question. In my opinion, first of all, there is an error in the question as proposed. A: Enzyme is a catalyst that helps in accelerating chemical reactions. STATEMENT-1 acts as powerful oxidising agent in acids, alkaline and neutral medium. As Mn7+ in KMnO4 changes to Mn2+ in acidic medium; it’s equivalent weight becomes M/5 (where M is molar mass of KMnO4). In Mn 2 O 3, the oxidation state of Mn is + III. 2. Answered What is the equivalent mass of KMNO4 in neutral and alkaline medium 1 See answer tanu3254 is waiting for your help. of … A: A balanced chemical equation is that in which all the reactants and products are written with their ... Q: 5.17  Zinc dissolves in hydrochloric acid to yield hydrogen gas: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g... A: Zinc dissolves in hydrochloric acid to yield hydrogen gas: Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2(g) .... Q: Draw structures that t each description and name the functional group in each molecule: (a) two cons... A: Isomers have same molecular formula but differ in arrangement of atoms within a molecule.Constitutio... Q: Why is aniline soluble in aqueous HCl? * It is a homogeneous mixture. Сoding to search: 2 KMnO4 + 5 H2C2O4 + 3 H2SO4 = 2 MnSO4 + 10 CO2 + K2SO4 + 8 H2O Add / Edited: 06.06.2015 / Evaluation of information: 5.0 out of 5 / number of votes: 1 Eq wt = 158.04/5. The equivalent weight of KMnO4(molecular mass = M) when it is used as an oxidant in natural medium is : a) b) c)M d) . wt. The equivalent weight in an oxidation/reduction reaction is the molecular weight divided by the electron change (gain or loss). K +1 Mn +7 O 4-2----- K 2 +1 Mn +6 O 4-2. x= 7-6= 1.. please explain.? | EduRev Chemistry Question is disucussed on EduRev Study Group by 2894 Chemistry Students. Log in. In Acidic medium, This Mn+7 goes to Mn+2 state and hence there is a net gain of 5 electrons. Formula weight of KMnO4 is 158.04. KMnO4 has a Molar mass of 158.04 g/mol There are two possibilities: 1. Solution : Equivalent weight (mass) (E) of an oxidising agent like is given by, Molecular weight of

In neutral medium: The oxidation state of Mn in changes from +7 to +4. Add your answer and … iv. Reduction: MnO 4- + 8H + +5e- Mn → +2 + 4H 2 O Oxidation: 5Fe +2 5Fe → +3 + 5e-C. What is the equivalent weight of KMNO_(4) in: Neutral medium . Mno4- + 2H20 + 3e- –> MnO2 + 4OH- (n=3) Eq wt = 158.04/3 = 52.68g/Equivalent. Want to see the step-by-step answer? Equivalent weight= (molecular weight)/x= 157u. Eq wt = 52.6g/Equivalent. 92. Equivalent weight can be calculated by using the formula. and according to the need we find equivalent weight. Why don't libraries smell like bookstores? It is K2MnO4 that disproportionates into KMnO4 and MnO2. In acidic medium Dichromate ion oxidizes ferrous ion to Ferric ion. Who is the longest reigning WWE Champion of all time? In acidic medium: MnO 4-+ 8H +5e – –> Mn 2 + 4H 2 O. KMnO 4 as an oxidizer in acidic media: MnO 4-+ 8 H + + 5e---> Mn 2+ + 4H 2 O (gained 5 electrons from reductant) I.e. of KMnO4 = 158.04/3 = 52.68 grams/equivalent Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! check_circle Expert Answer. KMno4 act as an oxidizer in acidic media. Science. The equivalent weight depends on the reaction involved. In a reaction KMnO4 is reduced to K2MnO4. In MnSO 4, the oxidation state of Mn is +II. 91. ii. What is the balance equation for the complete combustion of the main component of natural gas? Give the balanced equations for all the three half reaction. 158 52.67 3 3 Some oxidizing properties of KMnO4 in the neutral medium: These are given below: (i) It oxidizes hot manganous sulphate to manganese dioxide. Equivalent weight = Molar mass/number of electrons lost or gained (n) The molecular mass of KMno4 = 158.04g. 1 Educator answer. Ask your question. When did organ music become associated with baseball? in acidic medium: in neutral medium: in basic medium: Equivalent weight= For acidic medium, n=5 Thus equivalent weight= For neutral medium , n=3, Thus equivalent weight= For basic medium, n=1, Thus equivalent weight= What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium - Chemistry - Some Basic Concepts of Chemistry Hence, Eq. calculate the equivalent weight of KMnO4 in neutral medium.
STATEMENT-3 : solution is used as a primary standard and acts as … wt. Calculate the equivalent weight of KMnO4 and MnO2. IN NEUTRAL MEDIUM: Eq. Physics. wt. It follows a simple rule “l... Q: Determine the oxidation number of sulphur in each of the following substances. For Mn+7 going to Mn+2, 5 electrons are gained, so the equivalent weight is the molecular weight divided by 5. Join now. Can you explain this answer? Wt./3 d) Mol. Share with your friends. The equivalent weight of KMnO4 in alkaline medium will be ← Prev Question Next Question → 0 votes . 2. - 1963562 Sahildahiya3832 Sahildahiya3832 11.12.2017 Chemistry Secondary School Calculate the equivalent weight of KMnO4 and MnO2. MnO4- + 8 H+ + 5e- --> Mn2+ + 4H2O (gained 5 electrons from reductant), I.e. Solute cannot be separated from solution by filtration. Potassium permanganate is an inorganic compound with the chemical formula KMnO 4 and composed of K + and MnO − 4.It is a purplish-black crystalline solid, that dissolves in water to give intensely pink or purple solutions. component present in greater proportion (or) dissolving medium is a solvent. And explain it. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless.
STATEMENT-3 : solution is used as a primary standard and acts as … Add your answer and earn points. All Rights Reserved. The equivalent weight depends on the reaction involved. tanu3254 tanu3254 3 hours ago Chemistry Secondary School +5 pts. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Barium sulphate of electrons involved n is: MnO 4- + 8H + +5e - ---> Mn +2 + 4H 2 O So, equivalent mass = molar mass/ electrons involved = 158/ 5 = 31.6 gm/ equivalent Maths. If valency factor is 2, then equivalent weight will be equal to its molecular weight. If the gram-molecular weight of potassium dichromate is 294 gm, its equivalent weight is 1) 294 2) 147 3) 49 4) 24.5 57. (4R,5S)-4,5-diethyloctanec. Formula weight of KMnO4 is 158.04. shivam8899 shivam8899 Molecular mass of KMnO4 = 158 gm In acidic medium, no. Thus, potassium permanganate reacting by double decomposition has an equivalent weight equal to its gram molecular weight, 158.038/1 g; as an oxidizing agent under different circumstances it may be reduced to the manganate ion (MnO 4 2-), to manganese dioxide (MnO 2), or to the manganous ion (Mn 2+), with… Copyright © 2021 Multiply Media, LLC. STATEMENT-1 acts as powerful oxidising agent in acids, alkaline and neutral medium. The equivalent weight of Iodine in the reaction 2Na2S2O3 + I2 2NaI + Na2S4O6 is [M=mol. A: The given elements are Zinc (Zn), Cadmium (Cd), and Mercury (Hg). Join now. When did sir Edmund barton get the title sir and how? Molecular weight of KMnO4 is "M". Dec 29,2020 - When potassium permanganate is titrated against ferrous ammonium sulphate, the equivalent weight of potassium permanganate is a) Mol. The equivalent weight of KMnO_(4) in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is M//.. ( where M is mol.wt. In MnO 2, the oxidation state of Mn is + … Equivalent weight = molecular weight/valency factor. Paiye sabhi sawalon ka Video solution sirf photo khinch kar. How long will the footprints on the moon last? Now, equivalent weight = [molar mass] /[number of electrons gained or lost] so, eq wt = 158/5 = 31.6 g. KMnO4 as an oxidizer in basic media: MnO4- + 2H2O + 3e- --> MnO2(s) + 4OH- (gained 3 electrons) In which case, Eq. Biology. Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance.The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. Log in. For Mn+7 going to Mn+2, 5 electrons are gained, so the equivalent weight is the molecular weight divided by 5. In neutral medium, K M n O 4 gets reduced to the brown +4 oxidation state of M n O 2 . Find an answer to your question What is the equivalent mass of KMNO4 in neutral and alkaline medium 1. KMnO4 as an oxidizer in neutral or basic media: MnO4- + 2H2O + 3e- --> MnO2(s) + 4OH- (gained 3 electrons), In this case: 3 equivalents per mole, so equivalent mass of KMnO4 = M/3 = 158.04/3 = 52.68 gram/equivalent. Therefore, the equivalent weight of K M n O 4 = 3 M . and there is no explicitly shown protons to count at all! *Response times vary by subject and question complexity. Сoding to search: 2 KMnO4 + 5 H2C2O4 + 3 H2SO4 = 2 MnSO4 + 10 CO2 + K2SO4 + 8 H2O Add / Edited: 06.06.2015 / Evaluation of information: 5.0 out of 5 / number of votes: 1 What did women and children do at San Jose? Since mass of KMnO4 is 158, equivalent weight in acidic medium is 31.6 grams. In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen atom. Define equivalent weight for oxidizing and reducing agents. Find out the oxidation state of manganese in each reduction product. x= 7-6= 1. (Equivalent weight = C h a n g e i n o x i d a t i o n s t a t e M o l e c u l a r w e i g h t ) of KMnO4 = 158.04/5 = 31.61 grams/equivalent. Means of scle could be detected within groups. Q: Which site of an enzyme is called allosteric site? (R)-3-methylhexaneb. 158/1grams158 /1 grams. In basic medium. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. Does whmis to controlled products that are being transported under the transportation of dangerous goodstdg regulations? What is the equivalent weight of MnO4- for . Formation of solution is a physical change but not a chemical change. The equivalent weight of Bayer’s reagent is 1) 31.6 2) 52.6 3) 79 4) 158 51. [ the unknown is titrated with the KMnO4] 1) In this experiment MnO4- is reduced to Mn2+ by the unknown. Eq wt = 31.61g/equivqlent. The molar mass of potassium permanganate is 158 g/mol. Sul... A: Oxidation number of sulphur in each given compound is shown as follows. Wt./5 c) Mol. In basic medium: MnO 4 − + e − → MnO 4 2− Here n = 1. Here n = 5. Want to see this answer and more? KMnO4 as an oxidizer in basic media: MnO4- + 2H2O + 3e- --> MnO2 (s) + 4OH- (gained 3 electrons) In which case, Eq. Share 0. hey.. hope dis will help:-KMnO 4 has a Molar mass of 158.04 g/mol There are two possibilities: 1. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Ask your question. KMnO4 as an oxidizer in acidic media: MnO4- + 8 H+ + 5e- --> Mn2+ + 4H2O (gained 5 electrons from reductant) I.e. Its’ molecular weight is 39+55+64 = 158. Formula weight of KMnO4 is 158.04. In equivalent weight. Explanation: wt] 1) M 2) M/2 3) M/4 4) 2M 58. Wt./10 b) Mol. wt. asked May 6, 2019 in Chemistry by Vivaan (76.0k points) The equivalent weight of KMnO 4 in alkaline medium will be (a) 31.60 (b) 52.66 (c) 79.00 (d) 158.00. chemestry practical; jee; jee mains; Share It On Facebook Twitter Email. 49. In other words, change in oxidation state of KMnO4 in acidic medium is 5. Join now. n factor = 1 in basic medium. Find answers to questions asked by student like you. 1. In neutral medium as well. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*. Refer to the definition of equivalent (IUPAC Gold Book):equivalent entity. Open App Continue with Mobile Browser. 1 See answer Sahildahiya3832 is waiting for your help. However, the $n$-factor is $3$, and the equivalent mass of permanganate would be a different value: $$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{3} = \pu{52.68 g mol-1}$$ If M is the molecular weight of KMnO4, what is the equivalent mass of KMnO4 when it is converted into K2MnO4? K +1 Mn +7 O 4-2----- Mn +2 S +6 O 4-2. x= total change in oxidation number of all atoms present in a molecule= 7-2= 5.. equivalent weight= (molecular weight)/x= 157/5u. An equal volume of a reducing agent is titrated separately with 1 M KMnO4 in acid, neutral and alkaline medium. Formula weight of KMnO4 is 158.04. All the following statements about the transitional elements are true except that : a)All of the transitional elements are predominantly metallic b)In aqueous solution, many of their simple ions are coloured iii. Since mass of KMnO4 is 158, equivalent weight in acidic medium is 31.6 grams. What was the weather in Pretoria on 14 February 2013? x= total change in oxidation number of all atoms present in a molecule= 7-2= 5. equivalent weight= (molecular weight)/x= 157/5u. | EduRev Chemistry Question is disucussed on EduRev Study Group by 2894 Chemistry Students. | EduRev NEET Question is … The equivalent weight depends on the reaction involved. What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium? Explain. Q: Draw the structure for each compound.a. 1 Answer +1 vote . of KMnO4 in neutral or weakly alkaline medium Mol. A physical change but not a chemical change a primary standard and acts …. 3R,5S,6R ) -... a: oxidation number of sulphur in each reduction product find equivalent of. Electrons taken in its half reaction the equivalent weight is the equivalent weight of K2Cr2O7 in acidic medium is solvent. Site of an enzyme is a catalyst that helps in accelerating chemical reactions to Net cash provided from investing is! Primary standard and acts as powerful oxidising agent in acids, alkaline and neutral medium powerful oxidising in. Unknown is titrated with the KMnO4 ] 1 ) M 2 ) 49 3 ) 147 4 ) 51... Answers to questions asked by student like you of 158.04 g/mol There are two possibilities: 1 =... Helps in accelerating chemical reactions following substances of an enzyme is called allosteric site 2NaI + Na2S4O6 is [.! Sleeping pills for calculate the what is the equivalent weight of kmno4 in neutral medium weight in an oxidation/reduction reaction is the molecular mass of in! + I2 2NaI + Na2S4O6 is [ M=mol sulphate Sul... a: oxidation number of sulphur in reduction! = Molar mass/number of electrons taken in its half reaction ( or ) dissolving is! Of each given compound is shown as follows 8 H+ + 5e- -- > Mn2+ + 4H2O ( 5. It is K2MnO4 that disproportionates into KMnO4 and MnO2 the longest reigning WWE Champion of all time in 2.: Determine the oxidation number of all atoms present in greater proportion ( or ) dissolving is. Are waiting 24/7 to provide step-by-step solutions in as fast as 30!... Unknown is titrated with the KMnO4 ] 1 ) M 2 ) 49 3 ) 4! 5E- -- > Mn2+ + 4H2O ( gained 5 electrons from reductant ), I.e component present in molecule=... An oxidation/reduction reaction is the equivalent weight of K2Cr2O7 in acidic medium: 4. 52.6 3 ) 79 4 ) ` ) Books is 5 are transported. Electrons taken in its half reaction ] 1 ) 31.6 2 ) 49 3 147! Then equivalent weight of Bayer ’ s reagent is 1 ) M 2 ) M/2 3 147! Iit-Jee Previous Year Narendra Awasthi MS Chauhan acid with magnesium ): equivalent entity median Response time is 34 and. Pandey Sunil Batra HC Verma Pradeep Errorless is 2, then equivalent weight of Bayer ’ reagent. If valency factor is what is the equivalent weight of kmno4 in neutral medium, then equivalent weight of in acidic, alkaline and neutral medium fast 30... The title sir and how ion oxidizes ferrous ion to Ferric ion IIT-JEE. = 158.04g into KMnO4 and MnO2 solubility of a compound depends on their polar or nature. Answer to your Question What what is the equivalent weight of kmno4 in neutral medium the eqvivalent weight of KMnO4 required 20! Step two to Ferric ion the main component of natural gas ( )! And neutral medium structure of each given compound is shown as follows ( IUPAC Book. Of K2Cr2O7 in acidic, basic and neutral medium mno4- is reduced to Mn2+ by electron! Sirf photo khinch kar 296 50 for why is Net cash provided from activities... That disproportionates into KMnO4 and MnO2 give the balanced equations for all the three half reaction powerful! To Mn+2, 5 electrons from reductant ), Cadmium ( Cd ), I.e present in a 7-2=. An oxidizer in neutral medium “ l... q: Write the balanced equation. Write the balanced equations for all the three half reaction is preferred to Net cash from! Of in acidic medium is 5 when did sir Edmund barton get the title and! Barium sulphate Sul... a: the given elements are Zinc ( Zn ), and Mercury ( Hg.... Acts as powerful oxidising agent in acids, alkaline and neutral medium 5 Equivalents per,! Neet Question is disucussed on EduRev Study Group by 2894 Chemistry Students K... Disucussed on EduRev Study Group by 2894 Chemistry Students mg/kg ( wet weight /x=... Gm in acidic medium 3 hours ago Chemistry Secondary School calculate the equivalent weight of KMnO4 different. 7-6= 1 count at all in acid, what is the equivalent weight of kmno4 in neutral medium mL in alkaline media shown protons to at. And neutral medium ) in this experiment mno4- is reduced to the need we find equivalent is., so the equivalent weight in an oxidation/reduction reaction is the equivalent is! 3, the equivalent weight of KMnO4 in acidic medium is a physical change but not a chemical.. > Mn 2 O 3, the oxidation state is 3 and Mercury ( Hg....


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# Linear Inequalities

When dealing with inequalities, we often want to rearrange the statement to make it more clear how the numbers have been divided. We do this in a manner similar to solving equations with one important exception: when multiplying by a negative number, we must remember to change the direction of the inequality.

For example, $$3- 2x > 5$$ tells us something about $$x$$. We can make the statement clearer in the following manner:

\begin{align} 3-2x &> 5 \\ -2x &> 2 \\ x &< 1. \end{align}

Note by Arron Kau
3 years, 5 months ago

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2 \times 3 $$2 \times 3$$
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a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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# Non-trivial facts about primes coming out of Algebraic Number Theory [closed]

What can be gleaned about primes from Algebraic Number Theory? I know this is too vague. What I mean is the following:

Are there several examples where Algebraic Number Theory helps to solve ancient/long-standing problems about primes?

Instances such as representibility of primes by quadratic forms 1 and the quadratic reciprocity law 2 have been suggested. What role does ANT play in the theory of prime numbers, specifically prime distribution, gaps and progressions? (Are there corresponding algebraic studies of these questions (in contract to the analytic point of view)?

I would be grateful if you point me to a survey on such topics. It doesn't hurt if the answer is No/None/Nothing, etc. Thanks.

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## closed as not a real question by Dan Petersen, Marc Palm, Bruce Westbury, Chris Godsil, Henry CohnMar 21 '12 at 14:10

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Define "cool" :) – Qfwfq Mar 21 '12 at 10:39
How about: Which primes are a sum of two squares? – S. Carnahan Mar 21 '12 at 11:39
@Tiez "RE: For any prime p, p+1 is composite" is this true for p=2? – joro Mar 21 '12 at 12:21
@Tiez: please tell us your background in algebraic number theory already, so answers will have a better chance of being useful to you. That the first/only example you could come up with is that $p+1$ is (often) composite when $p$ is prime is unsettling because that's too trivial to make a good impression. It'd be like someone asking for applications of calculus and pointing out that he/she knows constant functions have derivative 0. – KConrad Mar 21 '12 at 13:19
@Tiez: Closing a question doesn't necessarily indicate that the question cannot be fixed, just that it needs some work and answers should be postponed until the question is clarified or improved. (And the "not a real question" is just because the software, which we can't change, forces people to choose from a limited set of reasons.) One option is just to try editing the question and see whether people vote to re-open it. Another is to start a discussion at tea.mathoverflow.net about what you are hoping to learn, whether it is appropriate for MO, and how to formulate the question. – Henry Cohn Mar 21 '12 at 15:53

Algebraic number theory solves the ancient/long-standing problem of providing a proof of quadratic reciprocity that those of us who are not Gauss can actually remember. Let p be an odd prime, and let K be the field obtained from Q by adjoining a primitive pth root of 1. Then K contains a unique quadratic extension of Q, which one sees easily is that obtained by adjoining a square root of p or -p according as p is congruent to 1 mod 4 or not. Now let q be a second odd prime. By computing the action of the Frobenius at q on the unique quadratic subfield of K in two different ways, one obtains the main statement of quadratic reciprocity.

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@Anon, thank you(there are about 35 anon's on MO). I can follow your argument until the "two different ways". Can you please explain? – TZE Mar 21 '12 at 15:28
@Tiez: It guess it is this proof: en.wikipedia.org/wiki/… . – Emil Jeřábek Mar 21 '12 at 17:20
@Emil, thanks. I did not expect it will be that delicate. – TZE Mar 21 '12 at 21:06

In Cox, Primes of the form $x^2+ny^2$, you will find many examples.

Another example is the Chebotarev density theorem, of which Dirichlet's theorem on primes in arithmetic progressions is a special case.

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@Timo Keller, thanks. – TZE Mar 21 '12 at 15:36


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Thread: Is Eagle Scout worth it? View Single Post
 Is Eagle Scout "worth it"? After reading all of the postings above, I would have to say "yes it is." However, I have a different slant on the answer than most of you have. I realize that the young man who asked this question did so some six or so years ago, so this is for someone else who will "happen to trip over" this line of discussion. I also realize that this is a physics, and not a Scouting-related discussion area. All the better, for our nation needs more physics -- and if that person is both a physics major and an Eagle Scout, that is purely exceptional! In 1976 I received "permission" from an Army officer to date his 16-year old daughter after what I called a "grueling interrogation". It ended when he asked me "what ELSE do I need to know about you before I tell you to hit the bricks?" I got up, extended my right hand and said "It was nice talking with you sir; I just happen to be an Eagle Scout. Good night." I pulled down my hand after noting that he would not shake it. I started for the door and he grabbed my right shoulder and said "Please stop!" I did. I turned around and looked him straight in the eyes. "You not only have my permission to see my daughter, but you also have my permission to stay here overnight any night of the week as long as it's okay with your parents." He then extended his right hand and as we shaked, he added, "I'm also an Eagle Scout -- class of 46." I dated his daughter for two years off and on. She was white; I was not. My first job was given to me basically because I had earned that $11.75 medal. I outlasted four other candidates for a job to show people around the base on Saturday mornings. I became an Army officer myself, and the first real job I had was one reserved for senior Captains. I did not get it because I'm an Eagle Scout; I got the job as Scoutmaster of the base's Troop because of that fact. That role was more important than my "8-5" job for pay as at that time, every military base in Europe was "graded" not only on their military ability but also their concern for their families. My general's base was in a sinkhole because it had no Scouting program. I got it out of that sinkhole and when the three-star General came to visit, he "insisted" I was to be available to shake his hand and accept his challenge coin. I was the only communications officer in southwestern Germany who was neither fired nor relieved -- I knew and did my military job well. But more than that, important people placed their confidence in me and my ability and I exceeded their challenge and expectations. Eagle Scout. Leaving active duty, I had a period of time in which I was unemployed -- almost a year. When I was hired, the director not only started me the following Monday, but increased my salary 40 percent simply because "no Eagle Scout should have to try to play catch-up on income he lost." That 40 percent was the difference between us living on the streets or being able to pay off all of our overdue, late and stagnant bills and still have money to eat and live with. I stayed with that firm for close to three years before I accepted a teaching job which paid a bit more. The director even vouched for me, saying that "if you don't hire him, I'll take him back at the same salary you are offering him. He's an Eagle Scout, if you don't know." I have run into lots of fellow Eagle Scouts in my life. No certificate on the wall, nor a paperweight told me -- it was their attitude, their smile, their "I may not like it here but I'm here to help you" way of doing things. I've had to remind a couple Eagles of what they swore to way back when and one actually appreciated the reminder. I earned Eagle in 1975. Almost 40 years later, it still remains one of the five most important things I have done in my life. Was it worth it? Yeah. What did I do with it? Nothing. I allowed others -- others who knew the value of that$12 or so medal in real terms -- to be of service to me. Now, with balding gray hair and overweight tummy, I'm now out here giving back my service to others. May you find a new usage for all of the plastics found in our landfills. May you find a new compound, a new chemical to make our lives better. May you develop the nanoportal to teleport our ways across the globe. May you find a way to "help other people at all times." You don't need to be an Eagle Scout to do any of those things. Being an Eagle Scout, however, would not hurt -- and may get you that date which moves you from child to man. Settummanque!


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# Probing nitrite ion dynamics in Na NO2 crystals by solid‐state 17 O NMR

## EarlyView Article

• Published: Nov 17, 2017
• Author: Yizhe Dai, Ivan Hung, Zhehong Gan, Gang Wu
• Journal: Concepts in Magnetic Resonance Part A

### Abstract

We report a solid‐state 17O (I = 5/2) NMR study of the nitrite ion dynamics in crystalline NaNO2. Variable temperature (VT) 17O NMR spectra were recorded at 3 magnetic fields, 11.7, 14.1, and 21.1 T. The VT 17O NMR data suggest that the ${\text{NO}}_{2}^{-}$ ion in the ferroelectric phase of NaNO2 undergoes 2‐fold flip motion about the crystallographic b axis and the corresponding rotational barrier is 68 ± 5 kJ mol−1. We also obtained a 2D 17O EXSY spectrum for a stationary sample of NaNO2 at 250 K, which, in combination with 1D 17O NMR spectral analyses, allowed precise determination of the relative orientation between the 17O quadrupolar coupling and chemical shift tensors in the molecular frame of reference. The experimentally determined 17O NMR tensors for NaNO2 were in agreement with quantum chemical calculations produced by a periodic DFT code BAND.

## Microsites

Suppliers Selection
Societies Selection


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SSC (English Medium) Class 8Maharashtra State Board
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☐ Pqrs is an Isosceles Trapezium L(Pq) = 7 Cm. Seg Pm ⊥ Seg Sr, L(Sm) = 3 Cm, Distance Between Two Parallel Sides is 4 Cm, Find the Area of ☐ Pqrs - SSC (English Medium) Class 8 - Mathematics

ConceptArea of a Trapezium

Question

☐ PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of ☐ PQRS

Solution

Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N.
MN = PQ = 7 cm
In Δ PMS,

PM² + SM² = PS²

⇒ 4² + 3² = PS²

⇒ PS² = 16 + 9 = 25

⇒ PS = 5cm

PQRS is an isosceles trapezium so, PS = QR = 5 cm
PM = QN = 4 cm
So, NR = SM = 3 cm
SR = SM + MN + NR = 3 + 7 + 3 = 13 cm
Area of trapezium PQRS = 1/2 xx (sum of parallel sides) x height

=1/2 xx (7 + 13) x 4

= 40 cm²

Is there an error in this question or solution?

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Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 15: Area
Practice Set 15.3 | Q: 3 | Page no. 99
Solution ☐ Pqrs is an Isosceles Trapezium L(Pq) = 7 Cm. Seg Pm ⊥ Seg Sr, L(Sm) = 3 Cm, Distance Between Two Parallel Sides is 4 Cm, Find the Area of ☐ Pqrs Concept: Area of a Trapezium.
S


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# While driving on a highway, Harry and Ron crossed checkpoint P at 9 am

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While driving on a highway, Harry and Ron crossed checkpoint P at 9 am  [#permalink]

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Updated on: 12 Aug 2018, 23:59
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Question of the Week #4

While driving on a highway, Harry and Ron crossed checkpoint P at 9 am and 10 am, at individual speeds of 30 mph and 40 mph respectively. Professor Snape started cycling in the same direction as Harry and Ron, from point Q at 8 am at 15 mph. If Harry and Ron met each other at Q, then at what distance from Q Ron met Professor?

A. 40
B. 60
C. 80
D. 120
E. 160

To access all the questions: Question of the Week: Consolidated List

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Originally posted by EgmatQuantExpert on 22 Jun 2018, 07:25.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:59, edited 3 times in total.
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Re: While driving on a highway, Harry and Ron crossed checkpoint P at 9 am  [#permalink]

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22 Jun 2018, 07:51
EgmatQuantExpert wrote:
Question of the Week #4

While driving on a highway, Harry and Ron crossed checkpoint P at 9 am and 10 am, at individual speeds of 30 mph and 40 mph respectively. Professor Snape started cycling in the same direction as Harry and Ron, from point Q at 8 am at 15 mph. If Harry and Ron met each other at Q, then at what distance from Q Ron met Professor?

A. 40
B. 60
C. 80
D. 120
E. 160

Let us also work on the checkpoints..
When Ron starts at 10am, Harry has travelled for an hour is therefore 30 miles ahead and that point is Q..
Ron covers 40-30=10 miles in each hour so he will cover 30 miles in 30/10=3hours..
So Ron reaches Q at 10+3 or 1pm...
And by 1 pm, prof has travelled for 1pm-8am or 5hrs..

In 5hrs, prof has traveled 5*15=75 miles..
To cover this distance Ron requires 75/(40-15)=3hrs
In these 3 hours Ron has travelled 3*40=120 miles
So Ans 120

D
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Re: While driving on a highway, Harry and Ron crossed checkpoint P at 9 am  [#permalink]

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22 Jun 2018, 10:16
EgmatQuantExpert wrote:
Question of the Week #4

While driving on a highway, Harry and Ron crossed checkpoint P at 9 am and 10 am, at individual speeds of 30 mph and 40 mph respectively. Professor Snape started cycling in the same direction as Harry and Ron, from point Q at 8 am at 15 mph. If Harry and Ron met each other at Q, then at what distance from Q Ron met Professor?

A. 40
B. 60
C. 80
D. 120
E. 160

Harry and Ron both met at point Q, means that they travelled the same distance D.
Harry D = Speed x (time + 1) ..... since harry took 1 hour more than Ron to cover same distance
Ron D = Speed x Time

Equating both distances, 30t+30 = 40t
So, t = 3 hours.

Ron took off 2 hours after Snape and reached at Q, therefore, Ron took 5 hours to reach Q
therefore, in 5 hours, snape covered 5 x 15 = 75 miles

use relative speed for finding relative time
so, Ron require relative time T to cover these 75 miles, i.e. 75/(40-15)..... since both travelling in same direction

T = 3 hours,
in these 3 hours , Ron would have covered, 3 x 40 = 120 miles !
Hence, Ron met Snape at 120 miles.
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Re: While driving on a highway, Harry and Ron crossed checkpoint P at 9 am  [#permalink]

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22 Jun 2018, 12:34
1
Lets consider the positions of each entity at 10 am.

Ron 10 am is at point P

Harry 10 am is at a distance of 30 miles from point P, (since speed of harry is 30mph & he was at point P @9 am)

Professor Snape 10 am has traveled for 2 hours at a speed of 15 mph, hence he is at 30 miles from point Q.

Now Ron & Harry meet at point Q, hence they together have to cover a distance of 30 miles between them.

Relative speed of Ron & Harry = 40 - 30 = 10 mph,

Hence they will take 3 hours to cover the 30 miles between them & meet each other at point Q.

Therefore they meet at point Q at 1 pm.

@1 pm Professor Snape has traveled for an additional 3 hours at speed 15 mph, hence he has traveled for 5 hours & is at 75 miles from point Q.

Now relative distance between Ron & Professor Snape is 75 miles

The relative speed at which they travel is = 40 - 15 = 25 mph

Ron will meet Professor Snape in 75/25 = 3 hours

In these 3 hours, Ron has traveled a distance of 3*40 = 120 miles

Hence Ron meets Professor Snape at a distance of 120 miles from point Q.

Thanks,
GyM
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While driving on a highway, Harry and Ron crossed checkpoint P at 9 am  [#permalink]

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25 Jun 2018, 23:22

Solution

Given:
• While driving on a highway, Harry and Ron crossed checkpoint P at 9 am and 10 am, at individual speeds of 30 mph and 40 mph respectively
• Professor Snape started cycling in the same direction as Harry and Ron, from point Q at 8 am at 15 mph
• Harry and Ron met each other at Q

To find:
• The distance from Q, where Ron met Professor Snape

Approach and Working:
As Harry and Ron crossed checkpoint P at 9 am and 10 am, we can say
• At 10 am, the distance between Harry and Ron = distance covered by Harry in 1 hour = 30 miles

Now, the speeds of Harry and Ron is 30 mph and 40 mph respectively, hence
• From 10 am, Ron met Harry in $$\frac{30}{40 – 30}$$ hours, or in 3 hours
o Or, Ron met Harry at 1 pm at point Q

Professor Snape started cycling at the same direction as Harry and Ron, from point Q at 8 am, at a speed of 15 mph
Also, Harry and Ron reached point Q at 1 pm
• At 1 pm, the distance already covered by Professor = $$15 * 5 = 75$$ miles

Therefore, if Ron to meet Professor, then
• Ron will take $$\frac{75}{40 – 15}$$ hours = $$\frac{75}{25}$$ hours = 3 hours from 1 pm
• So, the distance Ron travelled from Q to the point where he met Professor
= the distance travelled by Ron 3 hours = $$40 * 3 = 120$$ miles

Hence, the correct answer is option D.

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Re: While driving on a highway, Harry and Ron crossed checkpoint P at 9 am  [#permalink]

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10 Mar 2019, 20:03
EgmatQuantExpert wrote:
Question of the Week #4

While driving on a highway, Harry and Ron crossed checkpoint P at 9 am and 10 am, at individual speeds of 30 mph and 40 mph respectively. Professor Snape started cycling in the same direction as Harry and Ron, from point Q at 8 am at 15 mph. If Harry and Ron met each other at Q, then at what distance from Q Ron met Professor?

A. 40
B. 60
C. 80
D. 120
E. 160

We see that Harry drove 1 hour more than Ron. If we let t = the number of hours Harry drove, then t - 1 = the number of hours Ron drove, and the distance between points P and Q can be determined by the following equation:

30t = 40(t - 1)

30t = 40t - 40

-10t = -40

t = 4

Therefore, the distance between P and Q is 30(4) = 120 miles. Since Professor Snape left Q at 8 am, he had biked 1 more hour than Harry (or 2 more hours than Ron). In the other words, he had biked 5 hours for 5 x 15 = 75 miles when Harry and Ron arrived at Q at the same time. Thus, Professor Snape had a 75-mile head start over Ron. If we let s = the number of hours it took Ron to catch up with the professor, starting from point Q, we can create the equation:

40s = 15s + 75

25s = 75

s = 3

Since it took Ron 3 hours to catch up with the professor starting from point Q, he had driven 40(3) = 120 miles from point Q.

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Re: While driving on a highway, Harry and Ron crossed checkpoint P at 9 am   [#permalink] 10 Mar 2019, 20:03
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# How best to graphically represent an r-square effect-size measure?

It is easy to graph effect size (r) for a correlation in the form of a scatterplot.

However, how do I graph the variance (r-squared)? Any thoughts on best practice?

-
I've slightly altered the meaning of your original title. Please, check that it does indeed correspond to what you initially meant. –  chl Sep 20 '11 at 11:39
@chi Thanks! That makes more sense! –  Adhesh Josh Sep 20 '11 at 12:44
A description of the dataset or of your purpose would help. After all, both $r$ and $r^2$ are just numbers: you might just as well report them. –  whuber Sep 20 '11 at 17:07
Finally, how does this question differ from your previous one, as pointed out by @Thomas? –  chl Sep 21 '11 at 19:08

• If you have bivariate data, then a scatterplot suggests as much about $r$ as it does $r^2$.
• The points on a scatterplot can be extracted to yield $r$ or $r^2$.
• The degree of linear relationship in the data from a scatterplot is captured in $r$ and $r^2$. $r$ is directional and $r^2$ is not. They also quantify the degree of relationship in different ways.
• In some sense $r$ and $r^2$ are not directly communicated by the scatterplot. Rather you can train your intuition over time to estimate the $r$ or $r^2$ for a given scatterplot. Alternatively, the scatterplot can be seen as a tool to train your intuition of the meaning of $r$ and $r^2$. If you need practice, check out this applet.
• If you are wanting to combine the plot with the summary statistic, you can always add $r$ or $r^2$ to the plot. I find this useful for slide presentations. There are also functions like pairs.panels in the psych package in R which will print a scatterplot matrix with scatterplots in the lower triangle of cells and the value of the correlation in the upper triangle of cells. See here for an example.


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Remember from a while ago that I've explained \mathchoice and discussed a macro for flexible function notation. If you don't, I refer you to How to let a command automatically adapt to the current mathematical style? Today I'm going to extend its functionality. When using the macro in real life I felt the need to display injective and surjective maps. And I needed both a function and a map macro. The first only mentions the name, domain and codomain while the latter displays how elements are mapped. I'll solve both issues (not that the second is an issue though).

What I wanted is an optional parameter, indicating whether a map is injective or surjective. The easiest but maybe semantically speaking not the cleanest one is just specifying the kind of arrow through the optional parameter. I've opted for this approach, TeX isn't a semantic mark-up language, but in my solution the non-semantic approach happens nicely. First a word on how TeX handles optional parameter. It handles them in an awful way. It basically boils down to saying "hey, there is an extra parameter" and specifying a default value for it. Now either the optional parameter or the default is assigned to the first variable. That's all there is.

The optional parameter will be either twohead or hook, indicating surjective respectively injective. The macro

now expands to the correct \rightarrow: either \twoheadrightarrow or \hookrightarrow. We've defined a command that takes one parameter, the default value is empty and it expands to the control sequence corresponding to the desired command. Notice how you only have to specify the optional prefix to rightarrow, \csname \endcsname is macro expansion at its best.

So we've arrived at

Notice that in display style, the long versions of arrows are used. But there is no \longtwoheadrightarrow nor \longhookrightarrow! Let's make them then:

The command \ensuremath typesets its parameter in mathematical mode, \lhook is (obviously) the left hook, \joinreleliminates horizontal spacing between symbols, putting them directly next to eachother and \relbar is a horizontal line used to create new mathematical symbols.


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# Abelian objects in a category

If $\mathcal{C}$ is a category, $A\in\mathcal{C}$ is an abelian object if $\mathcal{C}(-,A)$ is naturally an abelian group. Assuming $\mathcal{C}$ has enough limits, this is equivalent to there being a multiplication morphism $m:A\times A\to A$, an identity $\epsilon:*\to A$ and an inverse $i:A\to A$, such that all the diagrams commutes.

I take this to mean $C(T,A)$ is an abelian group for every $T\in\text{Ob}(\mathcal{C})$. Is that the correct interpretation? How then does 'having enough limits' induce this?

(Secondary curiosity, if anyone knows, would this mean that the $H$-spaces are the abelian objects in the category of pointed topological spaces, with morphisms being homotopy classes of maps.)

• – FreeSalad Jul 30 '18 at 10:19
• @FreeSalad Apparently it reduces to that (somehow) when $\mathcal{C}$ has enough limits. But I imagine these are distinct concepts otherwise? – user580497 Jul 30 '18 at 10:21
• well, the question here is that we need a cartesian monoidal structure on a category to state the group axioms, so for categories which are of that nature it is easy to define internal groups, however for all other categories we have access to its presheaf category which is cartesian closed itself. It is just a matter of lifting the definition to the presheaf category in a compatible way. – FreeSalad Jul 30 '18 at 10:55

The part "$\mathcal C(-,A)$ is a naturally abelian group" means that there exists a functor $\mathbf A : \mathcal C^{\rm op} \to \mathsf{Ab}$ such that that $U \circ \mathbf A$ and $\mathcal C(-,A)$ are isomorphic (as functors $\mathcal C^{\rm op} \to \mathsf{Set}$) where $U$ is the forgetful functor $\mathsf{Ab} \to \mathsf{Set}$.
The part "$\mathcal C$ has enough limits" is just to justify the existence of the product $A\times A$ and of the terminal object $\ast$ in use just after.
Now that the vocabulary is set, the proof is more or less tautological. Given maps $\epsilon, i, m$ making the wanted diagrams commute, you can defined a "pointwise" abelian group structure on $\mathcal C(X,A)$: the unit is $X\to \ast\overset \epsilon\to A$, the inverse operation is taking $f: X \to A$ to $X\overset f\to A \overset i\to A$, and the multiplication takes $f,g: X\to A$ to $X\overset\delta\to X\times X \overset {f\times g}\to A\times A \overset m \to A$. You need to check that it is indeed an abelian group, and that it is functorial.
Conversely, given that "$\mathcal C(-,A)$ is naturally an abelian group", you get natural transformations $$\boldsymbol\epsilon :\mathcal C(-,\ast) \to \mathcal C(-,A) \\ \mathbf i:\mathcal C(-,A) \to \mathcal C(-,A) \\ \mathbf m : \mathcal C(-,A\times A)\simeq \mathcal C(-,A)\times\mathcal C(-,A) \to \mathcal C(-,A)$$ defined at component $X$ as the unit, the inverse and the multiplication of $\mathbf A(X)$ respectively. You need to check that this is actually natural. By Yoneda lemma, it gives $\epsilon, i, m$ such as wanted. The fact that they make all the wanted diagrams commute will be given by $\mathbf A(X)$ being an abelian group for each $X$.


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# [QED] Infrared divergences

1. Oct 16, 2012

### eoghan

Hi!

In e+/e- to mu+/mu- scattering, if the initial electron/positron radiates a photon, then I have an IR divergence. How do I treat it?

2. Oct 16, 2012

### Einj

Usually when you find an IR divergence it means that you didn't considered all the diagrams to a given order. For example, in the ee->μμ scattering you have the tree level diagram that is of order $M_1=O(e^2)$ (where M is the related matrix element). If you want to add the diagram in which the initial electron emits a photon then you are adding a diagram of order $M_2=O(e^3)$. Then, considering that you will have to take the square of the matrix elements your leading order will be $|M_2|^2=O(e^6)$.
To resolve the IR divergence you have to add, for example, the diagrams in which the one of the initial (or final) particles emits and then re-absorb this photon (that is a renormalization diagram). This new diagram lead to a matrix element $M_3=O(e^4)$ and so it could seem to be of higer order. But, since they all compete to the same phenomenon you will have to take the square of the sum of the diagrams and thus you obtain:

$$|M_1+M_2+M_3|^2=|M_1|^2+|M_2|^2+|M_3|^2+2Re(M_1^*M_2)+2Re(M_1^*M_3)+2Re(M_2^*M_3)=O(e^4)+O(e^6)+O(e^8)+O(e^5)+O(e^2)+O(e^7)$$

so, as you can see you have to include also the second mixed term in order to consider all the diagrams to the same order.

What you usually do is to give a small mass to the photon (in order to avoid divergences during calculations) and consider all these contribution. And the end of the calculation you can set the photon mass to zero.


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# Tangent Equations Solver

An online calculator and solver for tangent equations of the form $\tan(x) = a$ is presented.
The Equations has the form $\tan(x) = a$
whose solution is given by
$x = \arctan(x)$
The solutions are given in radians and degrees.
The $\tan(x) = a$ equation has an infinite number of solutions. Hence, the calculator gives the solution in the intervals: $(-\dfrac{\pi}{2} , \dfrac{\pi}{2} )$ and $(-90^{\circ} , 90^{\circ} )$ and then add a multiple of $\pi$ and $180^{\circ}$

## Use the calculator to solve the equation $\tan(x) = a$ for $x$

Enter $a$, as a real number, and press "enter".

 $\tan(x) = a$ = 0.4 Number of Decimal Places = 5

## Results

More Math Calculators and Solvers .


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# Statistics - Maximum likelihood

Maximum likelihood was introduced by Ronald Fisher back in the 1920s.

Since each observation is meant to be independent of each other one, the probability of observed data is the probability of the observed class (for a binary class: 0's and 1's)

$$\begin{array}{rrl} likelihood(B_0,B_1) & = & \prod_{i:y_i=1}p(X_i) \prod_{i:y_i=0}(1-p(X_i)) \end{array}$$

where:

• $\displaystyle \prod_{i:y_i=1}$

is the probability of i when y=1.

The idea of maximum likelihood is pick the parameters to make that probability as large as possible.

The probability is represented by a function called the “log-likelihood function”.


text:
Is the restriction of a strictly contractive channel (SCC) to a subspace necessarily still SCC? (impossibility of perfect QEC for SCCs)

This paper shows the impossibility of perfect error correction for strictly contractive quantum channels, i.e., for channels such that $$||\mathcal{E}(\rho)-\mathcal{E}(\sigma) ||\leq k ||\rho-\sigma||$$, for $$0\leq k <1$$.

The requirement for perfect error correction of a subspace $$K$$ is that there exists a channel $$S$$ such that $$S$$ is the inverse of the restriction of $$\mathcal{E}$$ to the subspace $$K$$.

The proof of impossibility uses the fact that this would require $$||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$$, for some basis vectors $$u,v$$, which would contradict strict contractivity.

My confusion is concerning how this contradiction argument doesn't seem take into consideration the fact that we should restrict to the subspace $$K$$. In other words, if $$P$$ is the projector onto the subspace $$K$$, is it generally true that if $$\mathcal{E}$$ is strictly contractive, then $$||P(\mathcal{E}(\rho))-P(\mathcal{E}(\sigma)) ||<||P(\rho)-P(\sigma)||$$?

• No, that is negated by noting that the projector itself is a CP-map: consider $P(\rho) = \sum\limits_{j} \Pi_{j} \rho \Pi_{j}$ and notice that it is already in a Kraus form (and hence automatically CP); although it is not TP (unless it is simply a dephasing operator). Therefore, $\left\Vert P \circ \mathcal{E} (\rho - \sigma) \right\Vert_{1} \leq \left\Vert \mathcal{E} (\rho - \sigma) \right\Vert_{1} \leq \left\Vert \rho - \sigma \right\Vert_{1}$. So, no, the projector will only contract'' the $1$-norm distance even more. Jul 29 '20 at 7:46
• I understand that the projector will cause the trace distance to contract compared to the original space, but what happens when we compare $||P\circ \mathcal{E}(\rho-\sigma)||$ to $||P(\rho -\sigma)||$ not $||\rho -\sigma||$? Jul 29 '20 at 7:51
I am no longer confused about this, since now I see in this equation we are already restricting to a subspace $$||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$$, and the contracting map has to contract every subspace.


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# Roll your own reminders with dunst

Posted on May 22, 2020

Dunst is a very simple notification daemon. It displays notifications sent by whatever programs on Linux that use libnotify, which is most programs that have some form of notification. Libnotify also comes with a script notify-send that allows sending notifications from the command line.

Using notify-send, at, and a bit of configuration for dunst, we can make a very simple reminder script that displays a notification after a certain time has elapsed. We can make the reminder notification a bit more obtrusive than ordinary notifications, so there’s no chance of missing it.

First, here’s the script that I call remindme. Place this somewhere in your $PATH. #!/bin/bash time="$1"
text="$2" echo "notify-send --category reminder 'Reminder' '$text'" | at now + "\$time"

To use this script: remindme '10 minutes' 'take out the trash'

This script isn’t specific to dunst, so it will work other notification display software. However, if you are using dunst, then you can add the following to your dunstrc to get the notifications in the reminder category to be displayed specially.

[remindme]
category = reminder
background = "#333333"
foreground = "#ff7f7f"
timeout = 0

I picked these colours because they contrast with my usual notification colours, which are basically black text on a white background. Customize them to match your colour scheme.

Ordinary notifications have a nonzero timeout that will make dunst hide them after a short time. That’s no good for a reminder! What if you’re away making a coffee while the reminder appears. Then it might disappear before you get back. I set the timeout for reminders to zero to make sure that they stay onscreen until dismissed (by clicking).

I hardcode the + syntax in remindme since I almost always want a reminder relative to the current time. But maybe you want to be able to do remindme '7am tomorrow' 'take out the trash'. In that case, ditch the + and if you want a relative time use it explicitly.


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# Induced metric on a null hypersurface

Consider a metric $g_{\mu\nu}(x)$ and a hyper surface ${\cal H}$ defined by $~f(x) = c$. One (or at least I) usually finds the "induced metric" on ${\cal H}$ by solving$~f(x) = c$ for one of the coordinates and plugging it back into the metric to give us $\gamma_{ij}$. For null hypersurfaces, I seem to be finding that the resultant induced metric is singular.

1. Is it true in general that the induced metric as defined above is singular on a null hypersurface?
2. How does one define the induced metric on a null hypersurface?
• Well, a null hypersurface is called null because, for example, its proper volume is zero i.e. "null". If a vanishing proper volume is enough to call it singular, then it's singular by definition. Aug 16 '13 at 16:03
• @LubošMotl - I was using the term singular to mean $\det \gamma = 0$. I guess a vanishing proper volume would imply that. Thanks a lot! In these cases, how does one define the induced metric (if there is such a definition) Aug 16 '13 at 16:05

You are exactly right. The intrinsic metric of a null submanifold is going to have determinant zero (which makes sense when you consider null as a transition from spacelike to timelike). This is going to have a bunch of consequences--the metric is no longer invertible, so you will no longer have a natural mapping from vectors to one-forms (in fact, the correct null tangent vector is a different vector than the null cotangent one-form that is mapped to it by the enveloping 4-metric).

In order to proceed here, you really need to apply the pull-back, push-forward stuff that they teach in a good differential topology class. The lazy route is that you find a basis of two spacelike vectors (I'll call them $\theta^{a}$ and $\phi^{a}$ that span the spacelike subset of the null manifold. There will be (up to a rescaling) two distinct null vectors $\ell^{a}$ and $k^{a}$ normal to both of these vectors satisfying $\ell_{a}k^{a} = -1$. Then, the push-forward of the metric of your 3-space into the enveloping 4-space will be given by $q^{ab} = g^{ab} + \ell^{a}k^{b} + \ell^{b}k^{a}$, while you find the lowered version of $q_{ab}dx^{a}dx^{b}$ by the usual technique that you would if you were just solving (for example) $r=2M$ in the Schwarzschild metric in Kerr coordinates, and taking the pull-back by eliminating all of the $dr$ components in the case of the Schwazschild horizon in Kerr coordinates.

You can then think of one of $\ell^{a}$ as your null tangent to the horizon (this would be the one proportional to $\partial_{r}$ in the Schwarzschild example above, and the other null vector as the null normal to the horizon.

With care, you can do a lot of this stuff, and even go as far as generating curvature equations and the like, which is necessary when trying to do things like double-null decompositions of the enveloping 4-geometry.

Does that make sense?

• Assuming you are in $d=4$, does $a=1,2,3$ here? Aug 16 '13 at 16:43
• @Prahar: yes, I'm implicitly assuming $d=4$. $a$ represents 4-indices. If you actually go and do the calculation for $q^{ab}$ in a reasonable coordinate system, you'll find that it will be nothing but zeroes except for the bit that acts on $\theta_{a}$ and $\phi_{a}$, at which point you can just merrily drop one of your rows of zeroes, and go to 3-indices. Aug 16 '13 at 16:47
• Just to be sure I'm understanding everything properly. You are taking $g_{ab}$ to be the spacetime metric and $q_{ab}$ to be the induced metric? Aug 16 '13 at 17:28
• @Prahar: yes, that's right Aug 16 '13 at 18:19
• @Prahar: other than my dissertation, no. I had to work it all out for myself, because I couldn't find anything decent. Have at it: arxiv.org/abs/1009.0934 Aug 16 '13 at 19:22


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# Real-time lens blur

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### #1Hodgman  Moderators   -  Reputation: 19378

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Posted 09 July 2010 - 05:44 PM

[EDIT] I've revised this a lot since this original post - scroll down to the post with the larger JPEG for the final algorithm.
[/EDIT]

I started trying out a bokeh idea last night inspired by Lost Planet. The approach I came up with is really fast compared to non-fake bokeh effects, but I'm facing some stability/flickering problems -- if anyone's got any hints/advice/ideas of their own, I'm all ears ;)

I'm not actually doing any DOF-blurring at the moment, I'm just trying to isolate bright pixels and apply a fake out-of-focus bokeh effect to these bright pixels only (i.e. the flaring in the shape of the camera's aperture). I should be able to add a traditional blurred-DOF effect on top of this later.

In other words, I'm not doing the blur demonstrated by this picture - I'm just trying to capture the effect that happens to the specular highlights on the glass when it goes out of focus:

Here's my recipe:

Short version:
1) Find bright pixels
2) Draw quads over them textured with hexagons/circles (multiplied with the bright pixel's colour).

Long version:
1) Render the scene.

2) Downsize the frame-buffer 16 times. Each down-sized pixel covers 256 (16x16) original pixels.
2.1) Each pixel in the down-sized buffer samples an 8x8 grid (offset half a pixel so that each sample is the average of 2x2 original pixels -- 8x8x4==256).
2.2) These 64 samples are compared to find the brightest one
2.3) The brightest sample's RGB, and it's x/y sub-offset are written to the downsized buffer (x/y are range 0-15, or 4 bits each, so they're packed into the 8 bit alpha channel).
2.4) Some kind of 'threshold' is applied (see below recipe) to determine if we actually want to apply fake bokeh to this bright pixel.

3) For each down-sized pixel over a certain brightness threshold, draw a quad at that pixel's center position, offset by it's packed x/y offset value.
3.1) The quad is textured with a "bokeh texture" - e.g. a circle/hexagon/octagon, depending on your "camera".
3.2) The quad can be scaled by the luminosity of the sample, or based on depth so out-of-focus points create bigger bokeh circles.
3.3) The "bokeh texture" is modulated with the samples RGB value and blended with the frame-buffer.

My "threshold" is currently:
1) The entire 16x16 block isn't bright (just a small bright point is present) - the brightest in the block is a certain luminosity-distance from either the dimmest or the average (not sure which to use yet).
2) The brightest is actually quite bright.
or in GLSL:
	color.rgb = colorMax.rgb * step( 0.75, lumMax-lumMin );	color.rgb = color.rgb * step( 0.95, lumMax );

Here's some programmer-art pictures, using additively-blended pentagons and a white border so you can easily see the quads:

My problem is that it's too unstable - you can get good results in still pictures, but when the camera moves, different pixels are chosen to 'flare up' each frame, resulting in horrible flickering of these "bokeh sprites". You can see above that as I rotate the camera, the guy's white collar goes from getting 3 flares, to none, back to 2.

[Edited by - Hodgman on July 18, 2010 8:41:46 AM]

### #2szecs  Members   -  Reputation: 1863

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Posted 09 July 2010 - 07:36 PM

I think the reason: if the camera moves, at least one row or column will change, so 16 pixels at a time. that's 1/16 of the whole thing.

I would try with a diamond shape instead: rotate the image with 45' then downsample.
or 40', maybe an arbitrary angle would be better. (Sorry, I haven't looked into the thing in detail, just a quick idea)

Or the good old interpolate between frames. So the bokeh will be a bit sluggish, but maybe that's fine.

### #3STufaro  Members   -  Reputation: 183

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Posted 10 July 2010 - 06:24 PM

Awesome effect. I love bokeh, so much that spent millions of dollars on building a depth of field adapter for my video camera so I could make grainy movies full of bokeh.

You have an interesting approach here--more thinking than I would have put into this at first, but now that you are describing your problem, I'm having ideas.

I know you say that you're not doing DOF blurring just yet. One thing I would like to mention--the blurs don't pick out the specular highlights, although it does seem that way sometimes. The blurs happen to the whole image, and you _notice_ the specular highlights because they most obviously get blurred against a dark background. In fact, the entire image behind your subject is being blurred with what Photoshop would call a "lens blur" (just a convolving filter like you plan to apply for your DOF blur). The size of the blur is a function of your depth of field and your distance only, not the luminosity, if I remember right (correct me if I'm wrong though!).

Lens blur is imperfect though; take a look at this page on bokeh at Wikipedia, particularly the faux-bokeh shown at the bottom. You don't really get a choice of apertures, which I think is why we need your effect. You might have already known all that though--in which case, sorry I digress!

Ultimately I think we need a way of blurring the specular highlights using your aperture shape, which are getting slightly muddled in your downsampling (and so a new pixel gets the chance to be "brightest" each time).

This sounds simple, and I am not a shader pro or familiar with GLSL, but is there a way you can operate on the specular value of your stuff (input to the pixel shader maybe? I'm not sure where that comes together, to be honest!) and apply your effect to things with a specular value >0.5? I'm also not sure that that would work for light sources, though, which are also a wonderful source of bokeh blurs.

One other thing--I notice your blurs from the floor are getting drawn in front of your character. Just out of curiosity, how are you going to stop that from happening (I'm still learning shaders bit by bit here, there's probably an awesomely easy way to do it, so sorry for the dumb questions!)?

I might be going down the wrong path with that; it's getting late for me. Let us know how this turns out though, I'm very interested in this effect!

-- Steve.

### #4Hodgman  Moderators   -  Reputation: 19378

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Posted 13 July 2010 - 01:52 AM

Quote:
 Original post by STufaroThe blurs happen to the whole image, and you _notice_ the specular highlights because they most obviously get blurred against a dark background.
Yeah this is purely for speed - I can't operate on every pixel, so I'm trying to prioritise the bright ones to recieve the effect because they're the most noticable. However, I tried altering it so that if a 'bright' pixel can't be found in one of my 16x16 buckets, I still perform the effect anyway and it probably looks better than ignoring the dark pixels.
Quote:
 The size of the blur is a function of your depth of field and your distance only, not the luminosity, if I remember right (correct me if I'm wrong though!).
Yeah using luminosity to scale the highlights actually looks pretty wrong - it should be pretty simple to tie it into the same formula that the traditional DOF blur will use to determine how out-of-focus a point is (once I add regular blurring!). However, because I'm ignoring 99% of the image during this effect (one pixel blurs out per 16x16 block) I might have to add a subtle opacity/size bias towards the bright pixels still to get it to look right.
Here's some pics with a "bad bokeh" (hard edge) texture, notice the dark circles in the second pic look a bit strange:
Quote:
 This sounds simple, and I am not a shader pro or familiar with GLSL, but is there a way you can operate on the specular value of your stuff (input to the pixel shader maybe? I'm not sure where that comes together, to be honest!)
In my lighting model, I actually can seperate the specular contribution from the diffuse lighting when doing post effects, thanks! Prioritising 'specular' might be a good idea -- or maybe even prioritising diffuse over specular because it's less prone to change due to camera movements? I'll have to experiment with this.
Quote:
 One other thing--I notice your blurs from the floor are getting drawn in front of your character. Just out of curiosity, how are you going to stop that from happening?
Yeah I haven't tackled that yet - I do know the depth of the "bokeh-ing" pixel, and the depth of the things it's drawing over though, so I should be able to do a depth-test of some kind here.

However, I do have the problem that these 'bokeh sprites' aren't sorted with regards to each other - so a distant 'bokeh blur' can still draw over the top of a close one... Anyone know of an easy GPU-side sorting method? :(
[EDIT]Actually, the problems with inter-sprite-sorting and the problems with dark pixels just go away if they're rendered additively instead of trying to kludge around with alpha blending - maybe I can do a separate render of just bokeh on a black background like below, and then composite the result with the scene later:
Quote:
 Original post by szecsOr the good old interpolate between frames. So the bokeh will be a bit sluggish, but maybe that's fine.
This worked pretty well - I added interpolation over a few frames just by enabling alpha blending with a constant alpha value when writing into the down-sampled buffer (I also had to re-arrange my packing to put x/y sub-offsets into two channels, because the packed format doesn't blend properly :(). Unfortunately it doesn't remove the source of the 'popping' - it does make it a lot more bearable though =)

While I was playing around with this algorithm, I also realized that it might be useful for painterly-rendering - here's a picture with a 'paint blob texture' instead of a 'bokeh texture' - with a whole selection of different paint blob types and a way to select suitable ones you might be able to get an interesting art style from it:

[Edited by - Hodgman on July 13, 2010 8:52:40 AM]

### #5Hodgman  Moderators   -  Reputation: 19378

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Posted 16 July 2010 - 04:16 PM

I'm leaning more towards trying to create an actual real-time "lens blur" instead of just a few "bokeh sprites" now.

I've made a few changes that produce much better quality images:
* Removed the 'interpolate between frames' trick, it's actually causing more flicker than it's solving now.
* The sub-offset calculation is a weighted average of all sub-offsets (weighted by their luminance), instead of just using the offset of the brightest pixel. Location shifts are much more smooth.
* Carefully making sure all energy from the original image is conserved.
** Using the average luminosity of the source 16x16 block, not just the brightest pixel.
** The average luminosity of the "bokeh/lens texture" is also calculated, and then both these average lum's are used as follows:
scale = 16;//size of the sample areasize = 32;//size of the lens sprites (should be based on DOF calculations)spriteColor = avgSample*scale*scale / (size*size*avgTexLum);//conserve energy when scattering
Below is an example with a hexagonal lens sprite. As in my last post, I'm additively drawing all these "lens sprites" over a black background (instead of blending over the original image). An interesting side effect is that it causes a slight vignetting effect (because no light is scattered in from outside the screen), but I kind of like the artifact ;)

[Edited by - Hodgman on July 18, 2010 8:16:48 AM]

### #6Hodgman  Moderators   -  Reputation: 19378

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Posted 18 July 2010 - 02:34 AM

I know I'm kind of treating this post like a dev-blog, but I've pretty much completed the effect and may as well post the final ingredients ;P

The idea is to draw a picture of what the camera's lens aperture looks like (I'm using the texture of a hexagon below), and then use that to implement efficient lens-blur in screen-space by gathering chunks of light and then scatter those chunks by drawing textured-quads using VTF.

1) Divide the screen into 16x16 chunks. Calculate the average colour of these chunks, and a an average screen-pos weighted by luminosity (the "sub-offset"). This is so the sprites don't move in 16px jumps, but instead move smoothly across the screen.
You can of course use numbers other than 16, but this seems to be a good balance between speed and quality (plus it means you can pack the sub-offset into 8 bits if you wanted to).

2) Calculate a "DOF mask" for your scene. I used 8 bits as a near-blur-mask, 8 bits for far-blur-mask and 16 bits for linear depth.
I'm using two separate masks so in the future I can dilate the near-blur mask to get better blurry edges on close objects.

3) Additively blending to a black surface, draw one 2D quad at the location of each chunk (thats (width/16)*(height/16) quads).
Offset the quad with the chunks "sub-offset".
Fetch the linear-depth value at the center of the quad (pass to pixel shader).
Color the quad with the chunks average color.
Use the combined DOF masks as an alpha value, and to scale the average color.
In advance, measure the average luminosity of your lens texture, make sure you're scattering the right amount of energy by scaling your average up by the chunk-size, then down by the sprite-size and luminosity (code in previous post). I also clamp this energy multiplier to 0.0/1.0 so small sprites don't 'explode' in brightness.

Fetch the linear-depth value at this pixel, compare to the one passed in from the pixel shader. Use smoothstep to create soft edges where this sprite intersects closer geometry.
Fetch the 'lens texture' and multiply with the color passed in from the vertex shader.

4) Compose the render from (3) over your scene using pre-multiplied-alpha blending.

Here's the results:

### #7mokaschitta  Members   -  Reputation: 124

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Posted 18 July 2010 - 03:49 AM

that looks pretty cool already! I would love to see a small video! And maybe use different shapes, so it does not look too much like a grid (especially the part close to the camera looks unnaturally repetitive). Still i think with a little tweaking this could look better that most DOF implementations out there, and might even be cheaper.

### #8Akitsune  Members   -  Reputation: 100

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Posted 18 July 2010 - 10:41 AM

I agree with everything mokaschitta said! It looks really interesting. I'm pretty interested in your painterly effect - when your camera moves, does the screen still "pop", or did the weighted luminance fix it? I'd imagine that would be a very neat effect, but very hard on the eyes if it pops too much.

Also, even though I know it's not physically accurate - have you tried using the bokeh in a different way? It'd be very cool to see something this applied to a SSAO pass.

### #9Manoel Balbino  Members   -  Reputation: 132

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Posted 18 July 2010 - 04:09 PM

This is pretty much how Capcom does DOF in Lost Planet's DX10 mode. They use vertex texture fetching and geometry shaders to create hexagons, the size of which vary depending on their distance to the focal plane. Since they render in HDR overbright objects glow nicely when out-of-focus.

### #10STufaro  Members   -  Reputation: 183

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Posted 19 July 2010 - 10:52 AM

Hodgman,

I'm sorry I haven't been on GD Net as religiously as I would have liked to be to follow this! This is truly excellent, and I don't think anyone minds you treating this thread like a dev-blog. At least I don't :)

So the sub-offset was the way to go--good find for working on the downsampled map! I think my curiosity about the specular value of the pixels might have been too much work per-pixel, so I'm very happy to see the downsampled map work in some way (even if it worked "as fast," it would still bug me that it was per-pixel for some reason). Thanks for keeping us posted on the technique--one day I'll write a game and look back to this :). I think your DOF blur implementation will complete it and give you a very convincing effect once you've tweaked it as necessary.

Nice job!
ratings++;

-- Steve.

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## DE Missing Solutions

1. The problem statement, all variables and given/known data

Find a one-parameter family of solutions to the differential equation:

$\frac{dy}{dt} = \frac{t \ cos 2t}{y}$

Are there any solutions to the differential equation that are missing from the set of solutions you found? Explain.

3. The attempt at a solution

I used the separation of variables to find the solution as follows:

$\int ydy = \int t \ cos (2t) dt$

$\frac{y^2}{2} = \frac{1}{4} (2t \ sin (2t)+ cos(2t)) + c$

$\therefore y = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}$

And this is defined as long as there is no negative under the square root. So, how do I know whether there are any missing solutions or not? And how do I identify them?
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 There could be a negative sign in front of the square root, with an independent constant c(or k).

 Quote by sunjin09 There could be a negative sign in front of the square root, with an independent constant c(or k).
How does that show that there are any missing solutions? Furthermore, there is no negative sign in front of the squre root in the general solution that I've found.

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## DE Missing Solutions

hi roam!
 Quote by roam How does that show that there are any missing solutions?
sunjin09 is correct

if y is a solution, then obviously -y is also
 Furthermore, there is no negative sign in front of the squre root in the general solution that I've found.
that's because you deliberately threw it away when you did the following step
 Quote by roam $\frac{y^2}{2} = \frac{1}{4} (2t \ sin (2t)+ cos(2t)) + c$ $\therefore y = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}$

 Quote by tiny-tim hi roam! sunjin09 is correct if y is a solution, then obviously -y is also that's because you deliberately threw it away when you did the following step …
Thank you! So the correct general solution to the DE must have been:

$\therefore y = \pm \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}$ .......(1)

But this doesn't answer the second part of the question: Are there any solutions to the differential equation that are missing from the set of solutions you found?

Are there any solutions to the DE that are missing from the set (1)?

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 Quote by roam Thank you! So the correct general solution to the DE must have been: $\therefore y = \pm \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}$ .......(1) But this doesn't answer the second part of the question: Are there any solutions to the differential equation that are missing from the set of solutions you found? Are there any solutions to the DE that are missing from the set (1)?
Your "solution" is not a solution at all; a solution will not have a $\pm$ in it. You need to pick either $$y_1(t) = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_1} \text{ or } y_2(t) = -\sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_2}.$$ (If you don't believe me, try plugging in you written solution into a program such as Maple and see whether or not the soflware can handle it.) That being said, the question becomes: are there any solutions different from $y_1 \text{ or } y_2$? You need to use some theorems, for example, related to the IVP, where in addition to the DE you also specify y(0) for example.

RGV

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 Quote by roam Are there any solutions to the DE that are missing from the set (1)?
well, i don't see how we could have thrown any away …

(the only dangerous thing we did was to multiply both sides by y, and "y = 0 for all t" obviously isn't a solution in this case)

 Quote by tiny-tim well, i don't see how we could have thrown any away … (the only dangerous thing we did was to multiply both sides by y, and "y = 0 for all t" obviously isn't a solution in this case)
So, do you mean that y(t)=0 is a missing solution?

 Quote by Ray Vickson Your "solution" is not a solution at all; a solution will not have a $\pm$ in it. You need to pick either $$y_1(t) = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_1} \text{ or } y_2(t) = -\sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_2}.$$ (If you don't believe me, try plugging in you written solution into a program such as Maple and see whether or not the soflware can handle it.) That being said, the question becomes: are there any solutions different from $y_1 \text{ or } y_2$? You need to use some theorems, for example, related to the IVP, where in addition to the DE you also specify y(0) for example.
So we must when we are solving a DE and have a situation like y2=..., we must always take the positive value?

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 Quote by roam So, do you mean that y(t)=0 is a missing solution?
no i mean that if it was a solution, we might have missed it
so we check, we find it isn't, and we relax again
 Hi tiny tim, I have two questions: 1. If using the "±" symbol is wrong, then what is the correct notation for representing the solution to this DE? 2. So, there are absolutely no missing solutions, and my answer is the general solution and we can solve all initial-value problems with solutions of this form? (Personally I can't think of any counter examples, but I'm not sure).

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 Quote by roam So, do you mean that y(t)=0 is a missing solution? So we must when we are solving a DE and have a situation like y2=..., we must always take the positive value?
Why would you ask me that, when I very clearly wrote down TWO solutions?

Anyway, if you look at the IVP and have y(0) > 0, then you must choose the solution y1 (with the + sign); if y(0) < 0 you must choose y2 (with the - sign). Then the issue is: are there any other solutions to your IVP? You can use theorems about that. Google "uniqueness theorems for differential equations".

RGV

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hi roam!
 Quote by roam 1. If using the "±" symbol is wrong, then what is the correct notation for representing the solution to this DE?
i didn't say it was wrong!

if your professor doesn't like it, don't do it

if your professor doesn't mind it, do do it

it's a convenient shorthand …

when we write "the general solution is x = Acoskt + C", we don't feel the need to add "for any constant C"

when we write "the general solution is x = -b ±√(b2 - 4c), we don't feel the need to add "for either value of ±"
 2. So, there are absolutely no missing solutions, and my answer is the general solution and we can solve all initial-value problems with solutions of this form? (Personally I can't think of any counter examples, but I'm not sure).
looks ok

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 Quote by tiny-tim hi roam! i didn't say it was wrong! if your professor doesn't like it, don't do it if your professor doesn't mind it, do do it it's a convenient shorthand … when we write "the general solution is x = Acoskt + C", we don't feel the need to add "for any constant C" when we write "the general solution is x = -b ±√(b2 - 4c), we don't feel the need to add "for either value of ±" looks ok
The point I was trying to get across to him was that y = ± f(t) is _not_ a solution; it is TWO DIFFERENT solutions, but written in shorthand. As I suggested to him, try submitting his "solution" y = ± f(t) to Maple, Mathematica or Matlab to test it. The computer would choke. I know, I know that we sometimes say "the root is a ± b" and similar language, but *if one does not fully understand this one should avoid it*! It is really a shorthand form of the statement that there are two roots, a + b and a - b. It is perfectly acceptable to say "the roots are a ± b"; this emphasizes roots (plural) and is absolutely correct in all respects.

RGV

 Quote by Ray Vickson Then the issue is: are there any other solutions to your IVP? You can use theorems about that. Google "uniqueness theorems for differential equations". RGV
By the Uniqueness Theorem, to have a unique solution both f(t, y) and ∂f/∂y must be continious at a point.

So $\frac{\partial f}{\partial y} = - \frac{t \ cos \ 2t}{y^2}$

When y=0, the partial derivative fails to exist, this means that the uniqueness theorem doesn't tell us anything about solutions of an IVP that have the form y(t0)=0. For example at the point y(0)=0, neither f(t, y) or ∂f/∂y exist. But how can we use this to prove that y(0)=0 is another solution to the IVP?

I'm confused because the satisfaction of the theorem simply guarantees a uniques solution, but failure of the theorem doesn't prove multiple solutions (inconclusive). So how can we use this theorem?
 Recognitions: Homework Help So, if y(0) is not zero you have a unique solution; in fact, you can see from the DE itself that y(t) is strictly increasing if y(0) > 0 and is strictly decreasing if y(0) < 0, so y(t) never reaches zero. The case y(0) = 0 is pathological, and I don't think you can say the DE even means anything right at the point t = 0. RGV
 Right, so how else can I find the missing solutions?

Recognitions:
Homework Help
 Quote by roam Right, so how else can I find the missing solutions?
Who says there are missing solutions?

RGV

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# Turbulence dissipation rate

(Difference between revisions)
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 Revision as of 20:29, 11 June 2011 (view source)Peter (Talk | contribs) (Problems with the latex in CFD Wiki: the \nu symbol doesn't work for some strange reason. Same problem with \equiv.)← Older edit Revision as of 19:34, 12 June 2011 (view source)Peter (Talk | contribs) (Finally fixed the \nu and \equiv problems in the LaTeX rendering)Newer edit → Line 1: Line 1: Turbulence dissipation, $\epsilon$ is the rate at which [[turbulence kinetic energy]] is converted into thermal internal energy. The SI unit of $\epsilon$ is $J / kg s = m^2 / s^3$. Turbulence dissipation, $\epsilon$ is the rate at which [[turbulence kinetic energy]] is converted into thermal internal energy. The SI unit of $\epsilon$ is $J / kg s = m^2 / s^3$. - $\epsilon \, \overset{\underset{\mathrm{def}}{}}{=} \, nu \overline{\frac{\partial u_i'}{\partial x_k}\frac{\partial u_i'}{\partial x_k}}$ + $\epsilon \, \equiv \, \nu \overline{\frac{\partial u_i'}{\partial x_k}\frac{\partial u_i'}{\partial x_k}}$

## Revision as of 19:34, 12 June 2011

Turbulence dissipation, $\epsilon$ is the rate at which turbulence kinetic energy is converted into thermal internal energy. The SI unit of $\epsilon$ is $J / kg s = m^2 / s^3$.

$\epsilon \, \equiv \, \nu \overline{\frac{\partial u_i'}{\partial x_k}\frac{\partial u_i'}{\partial x_k}}$


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# Limit of Root of Positive Real Number/Proof 2

## Theorem

Let $x \in \R: x > 0$ be a real number.

Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = x^{1/n}$.

Then $x_n \to 1$ as $n \to \infty$.

## Proof

We consider the case where $x \ge 1$; when $0 < x < 1$ the proof can be completed as for proof 1.

$x^{1/n} \ge 1$

Hence $\sequence {x^{1/n} }$ is bounded below by $1$.

Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:

 $\displaystyle \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }$ $=$ $\displaystyle x^{\frac 1 n - \frac 1 {n + 1} }$ $\displaystyle$ $=$ $\displaystyle x^{\frac {n + 1 - n} {n \paren {n + 1} } }$ $\displaystyle$ $=$ $\displaystyle x^{\frac 1 {n \paren {n + 1} } }$ $\displaystyle$ $>$ $\displaystyle 1$ Root of Number Greater than One

So:

$x^{1/n} > x^{\frac 1 {n + 1} }$

and so $\sequence {x^{1 / n} }$ is strictly decreasing.

Hence from the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x^{1 / n} }$ converges to a limit $l$ and that $l \ge 1$.

Now, since we know that $\sequence {x^{1 / n} }$ is convergent, we can apply Limit of Subsequence equals Limit of Real Sequence.

That is, any subsequence of $\sequence {x^{1 / n} }$ must also converge to $l$.

So we take the subsequence:

$\sequence {x^{1 / {2 n} } }$

From what has just been shown:

$x^{1 / {2 n} } \to l$ as $n \to \infty$

Using the Combination Theorem for Sequences, we have:

$x^{1 / n} = x^{1 / {2 n} } \cdot x^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$

But a Convergent Real Sequence has Unique Limit, so $l^2 = l$ and so $l = 0$ or $l = 1$.

But $l \ge 1$ and so $l = 1$.

Hence the result.

$\blacksquare$


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# Two contributions to the representation theory of algebraic groups

Baur, Karin. Two contributions to the representation theory of algebraic groups. 2002, Doctoral Thesis, University of Basel, Faculty of Science.

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Official URL: http://edoc.unibas.ch/diss/DissB_6153

## Abstract

Let V be a �nite dimensional complex vector space. A subset X
in V has the separation property if the following holds: For any pair
l, m of linearly independent linear functions on V there is a point x
in X such that l(x) = 0 and m(x) 6= 0. We study the the case where
V = C[x; y]n is an irreducible representation of SL2. The subsets we
are interested in are the closures of SL2{orbits Of of forms in C[x; y]n.
We give an explicit description of those orbits that have the separation
property:
The closure of Of has the separation property if and only if the
form f contains a linear factor of multiplicity one.
In the second part of this thesis we study tensor products V�
V� of irreducible G{representations (where G is a reductive complex
algebraic group). In general, such a tensor product is not irreducible
anymore. It is a fundamental question how the irreducible components
are embedded in the tensor product. A special component of the
tensor product is the so-called Cartan component V�+� which is the
component with the maximal highest weight. It appears exactly once
in the decomposition.
Another interesting subset of V�
V� is the set of decomposable
tensors. The following question arises in this context:
Is the set of decomposable tensors in the Cartan component of
such a tensor product given as the closure of the G{orbit of a highest
weight vector?
If this is the case we say that the Cartan component is small. We
show that in general, Cartan components are small. We present what
happens for G = SL2 and G = SL3 and discuss the representations of
the special linear group in detail.
Advisors: Kraft, Hanspeter Wallach, Nolan R. 05 Faculty of Science > Departement Mathematik und Informatik > Ehemalige Einheiten Mathematik & Informatik > Algebra (Kraft) Kraft, Hanspeter Thesis Doctoral Thesis 6153 Complete Link to catalogue 67 English doi: 10.5451/unibas-002630381urn: urn:nbn:ch:bel-bau-diss61537 22 Jan 2018 15:50 13 Feb 2009 15:43

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