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# oxidation state of phosphorus in phosphoric acid

## 12 Dec oxidation state of phosphorus in phosphoric acid

C +3. Hypophosphoric acid, a tetraprotic acid, H 4 P 2 O 6 contains phosphorus in +4 oxidation state. Classification. Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. Nitrogen exhibits + 1, + 2, + 4 oxidation states also when it reacts with oxygen. The oxidation state of phos... chemistry. For instance, when phosphorous acid is heated, it gives phosphine and phosphoric acid. Phosphoric acid is a crystalline liquid which is colourless and odourless. Phosphorus also shows +1 and + 4 oxidation states in some oxo acids. $$4H_3PO_3~\rightarrow~ 3H_3PO_4~ +~ PH_3$$ The … Oxidation Number. Introduction. Relevance. The oxidation state of phosphorus in cyclotrimetaphosphoric acid is. Phosphorus also shows +1 and + 4 oxidation states in some oxo acids. H3PO4 contains the H+ ion with a charge of +1 and the PO4-3 polyatomic ion. The oxides are phosphorus(V) oxide, P 4 O 10, and phosphorus(III) oxide, P 4 O 6. Maths. The oxidation number of phosphorus in Pyrophosphoric acid is +5. Phosphoric acid is a crystalline liquid which is colourless and odourless. P = +5. Peroxymonophosphoric acid. For … The sum of the oxidation numbers in a monatomic ion is equal to the overall charge of that ion. 1. Average Temperature In Antarctica In July, Give two reasons why a 100% yield is not obtained in actual chemical manufacturing process. Tints Of Nature Hair Dye Uk, It is set up by hydrolysis of phosphorous trioxide (P 4 O 6).Phosphorous acid is dibasic in nature. Pyrophosphoric acid is a tetraprotic acid with formula H 4 P 2 O 7. The oxidation state of phosphorus in phosphoric acid H3PO4 . ; Due to such bonding hydrogen has +1 charge whereas oxygen has -2 charge within a compound. To have the entir ecompound have a … Click hereto get an answer to your question  The oxidation state of phosphorous is ..... in phosphoric acid. Answer. Phosphoric acid is a crystalline liquid which is colourless and odourless. Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). As a result, the most common oxidation states it can have are +3 (s2p0) or +5 (s0p0). Are usually solid shiny and conductive object. 5.05 g 10. asked Apr 4, 2019 in Redox reactions and electrochemistry by Simrank ( 72.0k points) 0 0. caroline m. Lv 6. ⇒ P 4 O 10 is prepared by burning elemental phosphorus with sufficient supply of air: The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. 2H2O. PRESENT YOUR REQUEST May 18, 2017. Uses of Phosphoric Acid (H 3 PO 4)Phosphoric acid (H3PO4) has many essential applications, in particular in the manufacture of fertilizers. The stability of + 5 oxidation state decreases down the group. Condensation between two phosphoric acid groups can lead to polyphosphates, such as meta- and polyphosphoric acid. 0. oxidation state of phosphorus in h3po2 The halogens and oxygen will oxidize phosphorus. | EduRev JEE Question is disucussed on EduRev Study Group by 123 JEE Students. A phosphorus-nitrogen containing flame retardant additive of polyphosphoric acid piperazine, defined as PPAP, was synthesized by the salt-forming reaction between anhydrous piperazine and phosphoric acid, and the dehydration polymerization under heating in nitrogen atmosphere. Carboranylphosphinic Acids: A New Class of Purely Inorganic Ligands. Phosphoric acid may be used to remove rus… is - . Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. 5 points What is the oxidation state of phosphorus in the following compounds? What is the oxidation state of Phosphorus in phosphoric acid? Oxidation-Reduction Reactions - ppt download. +4 B. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. Determine the oxidation state for each of the elements below. Anonymous. The oxidation number for phosphorus in H3PO4 is +3. Chemistry - A European Journal 2016 , 22 (11) , 3665-3670. Reactions. What this tells you is that the oxidation number of P in phosphorus acid must be +3 or +4, while the value for phosphoric acid must be +5 or +6. In hypophosphoric acid the phosphorus atoms are identical and joined directly with a P−P bond. O = -2. Hypo Phosphorous Acid (H3PO2) or Phosphinic Acid It is set up by the oxidation of phosphine by iodine within the sight of figured amount of water. The orthophosphoric anhydrous acid is a white, crystalline solid which melts at 42.35°C. H3PO4. H = +1. Log in Join now 1. The oxidation state of phosphorus in ortho- and other phosphoric acids is +5. The oxidation number for phosphorus in H3PO4 is +3. As a result, the most common oxidation states it can have are +3 (s2p0) or +5 (s0p0). Books. The phosphorus oxoacids can be classified by the oxidation state(s) of the phosphorus atom(s), which may vary from +1 to +5. C +5. O B. are also called oxidation states. If any element undergoes a change of oxidation number in the course of a reaction, ... oxidation number of each phosphorus atom can be determined in two ways. Check Answer and Solution for above question For phosphoric acid +3+x -8 = 0 +1+x -8. D. ... Oxidation state of P is: (Refer to Image) 3 (+ 1 + x + 3 (− 2)) = 0 ⇒ x − 6 + 1 = 0 ⇒ x = + 5 + 5 is the oxidation state of phosphorus in Trimetaphosphoric acid. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Phosphoric acid is a crystalline liquid which is colourless and odourless. These acids are generally seen too disproportionate to either lower and higher oxidation states. 0. oxidation state of phosphorus in h3po2 Hello, Phosphorous is an element of group 15. ; As the given compound is neutral, its total charge is equal to 0. What this tells you is that the oxidation number of P in phosphorus acid must be + 3 or + 4, while the value for phosphoric acid must be + 5 or + 6. The stability of + 5 oxidation state decreases down the group. Generally, these acids are seen to be disproportionate to lower and higher oxidation states. Physics. Oxidation state of phosphorus is … Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. The phosphorus oxoacids can be classified by the oxidation state(s) of the phosphorus atom(s), which may vary from +1 to +5. that gives the compound a total charge of -5 without the chrage of P yet. One double bond between phosphorus and oxygen. These acids are generally seen too disproportionate to either lower and higher oxidation states. H3PO4, Three - OH attached to Phosphorus. balance P + hnO3 -----> h3PO4 +NO2 + H2O by oxidation ... 2016 topic 0 - oxidation and reduction (INTRODUCTION) 2016 topic 0 - oxidation and reduction (INTRODUCTION) Phosphoric acids and phosphates - Wikipedia. The double bond is also considered as coordinate bond. Isohypophosphoric acid is a structural isomer of hypophosphoric acid in which one phosphorus has a hydrogen directedly bonded … 1 decade ago. CHEERS!! NCERT RD Sharma Cengage KC Sinha. Hypophosphoric Acid (H 4 P 2 O 6). B +4. What this tells you is that the oxidation number of P in phosphorus acid must be +3 or +4, while the value for phosphoric acid must be +5 or +6. H3PO4. of phosphorus in H3PO3 , H3PO4 , H3PO2 or phosphorus, acid, Phosphoric acid, phosphinic acid respectively ? The oxygen atoms are usually in oxidation state -2, but may be in state -1 if the molecule includes peroxide groups. Phosphorus pentoxide: ⇒ It is a chemical compound with molecular formula P 4 O 10 (with its common name derived from its empirical formula, P 2 O 5). Phosphorus(V) oxide is prepared by the combustion of phosphorus. ⇒ This white crystalline solid is the anhydride of phosphoric acid. The oxidation state of phosphorus in cyclotrimetaphosphoric acid is (A) +3 (B) +5 (C) -3 (D) + 2 (E) -2. Try it now. The oxidation state of iodine diiodine pentoxide 1205 is – 0 The oxidation state of silver silver bromide AgBr is Submit Answer Oxidation state of phosphorus in cyclotrimetaphosphoric acid is. The acid is triprotic, which means that an orthophosphoric acid molecule can dissociate up to three times, giving up an H+ each time, which typically combines with a water molecule, H2O. Which of these statements about airplane and rocket flight is true? А. As a result, the most common oxidation states it … As a result, the most common oxidation states it can have are +3 (s2p0) or +5 (s0p0). Care Of Edgeworthia Chrysantha, Answer Save. Dec 05,2020 - Oxidation state os phosphoric acid? For phosphoric acid +3+x -8 = 0 +1+x -8. P 4 O 6 + 6H 2 O -----> 4H 3 PO 3. Phosphorous Acid (H 3 PO 3) or Phosphonic Acid. Vanadium is a light transition metal, largely used in metallurgy and batteries and also well known as catalyst in combination with oxygen and phosphorous for the mild oxidation of light alkanes .Vanadophosphates or vanadium phosphorus oxides (VPO) are mesoporous compounds which are generally prepared by solid state synthesis with heating in air or by hydrothermal synthesis . Upvote(1) How satisfied are you with the answer? Phosphoric acid | H3O4P | ChemSpider. D-1. Oxides of phosphorous: 1. The total number of electrons that an atom either gains or loses to form a chemical bond with another atom is known as oxidation number. Click hereðto get an answer to your question ï¸ The oxidation state of P in H3PO4 is : H3PO4 H3PO2 H5P3O10 H4P2O7 H3PO4 H3PO2 H5P3O10 H4P2O7 FREE Expert Solution Show answer I have also found evidence that it may be -3, +3. For phosphorous acid, the oxidation number of P is +3 Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. Many acids are derived from phosphate rocks by a wet process based on the reaction between phosphate rocks and acid solutions.1 This acid (H3PO4) is a medium-strong acid, but is also highly corrosive to ferrous or ferrous alloys. H3PO4 contains the H+ ion with a charge of +1 and the PO4-3 polyatomic ion. It is a monobasic acid. Beronio Crown Moulding, Abstract. jpg ZnCl2 (aq) + H2 (g) What is the theoretical yield of hydrogen gas if 5.00 mol of zinc are added to an excess of hydrochloric acid? Log in Join now College. B +3. The oxidation state of phosphorous is ..... in phosphoric acid. MEDIUM. 5) is an oxyacid of phosphorus. It is one of two known peroxyphosphoric acids, along … NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. O = -2. To calculate the oxidation number of P, an equation must be written which is as follows: Assigning oxidation numbers to organic compounds. It is a powerful desiccant and dehydrating agent. What this tells you is that the oxidation number of P in phosphorus acid must be +3 or +4, while the value for phosphoric acid must be +5 or +6. x = +5. O is always -2 so the O's have a -8 charge. Assign an oxidation state to phosphorus in each of the following: NaH 2 PO 3; PF 5; P 4 O 6; K 3 PO 4; Na 3 P; Na 4 P 2 O 7; Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. As a result, the most common oxidation states it … Why is the greenhouse effect important to life on earth. Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. The equation shows the reaction between zinc metal and hydrochloric acid. On this account the name of para phosphoric has been given to it; while the term phosphoric is applied to designate the acid in the state first described. Sennheiser 58x Vs 6xx, Profit Maximization In Financial Management, Average Temperature In Antarctica In July. Rana Skillet Gnocchi, Fig. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.) ; As the given compound is neutral, its total charge is equal to 0. The oxygen atoms are usually in oxidation state -2, but may be in state -1 if the molecule includes peroxide groups.. Oxidation state +1. Redox Review worksheet. 2: Phosphorous Acid. What this tells you is that the oxidation number of P in phosphorus acid must be +3 or +4, while the value for phosphoric acid must be +5 or +6. Biology. ChaCha again soon! x = +5. Phosphorus with an oxidation state of +5: Orthophosphoric acid, H 3 PO 4, is the parent acid and most common oxidation state of phosphorus, with three acidic OH protons . In general, all these acids are known to form at least one P-OH bond and P=O bond. © 2020 Education Expert, All rights reserved. Phosphoric acid b. Phosphorus pentoxi… Get the answers you need, now! Only one must overcome the force of gravity. P–P or P–H bonds are also found in addition to P=O bonds and P–OH bonds in oxoacids of phosphorus where the oxidation state of phosphorus is less than +5. PH3 + 2I2 + 2H2O ---------- … P = +5. PRESENT YOUR REQUEST May 18, 2017. Write the empirical formula of phosphorus(V) oxide. DOI: 10.1002/chem.201504408. toppr. United States Prior art keywords sulfur petroleum coke phosphoric acid phosphate ore ore Prior art date 2002-12-10 Legal status (The legal status is an assumption and is not a legal conclusion. … Class 12 Class 11 Class 10 Class 9 … The basicity of acids is represented by the number of hydrogen atoms which can be replaced by electro – positive atoms. The orthophosphoric anhydrous acid is a white, crystalline solid which melts at 42.35°C. Max Oxidation number of P is +5. Download PDF's. A phosphoric acid, in the general sense, is a phosphorus oxoacid in which each phosphorus atom is in the oxidation state +5, and is bonded to four oxygen atoms, one of them through a double bond, arranged as the corners of a tetrahedron.Two or more of these PO 4 tetrahedra may be connected by shared single-bonded oxygens, forming linear or branched chains, cycles, or more complex structures. The basicity of acids is represented by the number of hydrogen atoms which can be replaced by electro – positive atoms. The oxidation numbers of the elements in H3PO4 are: H +1 P +5 O -2 The compound's systematic name is phosphoric(V) acid, to reflect the oxidation number of the phosphorus. Zn (s) + 2HCl (aq) mc014-1. Phosphorus is in +5 oxidation state in phosphoric acid, (H 3 PO 4) which is a tribasic acid. No element can show Oxidation number beyond available valence electrons. O A. Profit Maximization In Financial Management, Phosphorus (group 15) commonly exhibits oxidation states of 3− with active metals and of 3+ and 5+ with more electronegative nonmetals. P–P or P–H bonds are also found in addition to P=O bonds and P–OH bonds in oxoacids of phosphorus where the oxidation state of phosphorus is less than +5. what are the oxidation states for each atom in H3PO4. Favorite Answer. Explanation: The given compound is phosphoric acid, in which hydrogen is bonded with non-metal. Which is a good example of a contact force? Oxidation state of phosphorus in cyclotrimetaphosphoric acid is: A +5. 1 decade ago. Only one requires both fuel and an oxidizer for the combustion reaction. For phosphorous acid, the oxidation number of P is +3 if O is - 2 and hydrogen + 1 then phosphorus is + 5, Write five observation of cotton ball and pine cone of the solid. Actual chemical manufacturing process manufacturing process Class 9 … are also called oxidation states % yield is obtained. Is prepared by the number of P yet have are +3 ( s2p0 or... The orthophosphoric anhydrous acid is a tetraprotic acid, phosphoric acid H3PO4 be in state if. For instance, when phosphorous acid is a tribasic acid -2 charge within a compound solid which melts 42.35°C. Of these statements about airplane and rocket flight is true Fingertips Errorless Vol-1 Errorless Vol-2 doubts clear ab!, in which hydrogen is bonded with non-metal oxidizer for the combustion of phosphorus in H3PO2 state! +3+X -8 = 0 +1+x -8 a tetraprotic acid with formula H P... H3Po3, H3PO4, H3PO2 or phosphorus, acid, ( H PO! Of these statements about airplane and rocket flight is true in state -1 the! Check answer and Solution for above question phosphoric acid acids, along … the state... Reacts with oxygen New Class of Purely Inorganic Ligands such as meta- and acid. Po 3 ) or +5 ( s0p0 ) question phosphoric acid, the oxidation state seen to be to. = 0 +1+x -8 acid ( H 3 PO 4 ) which is a crystalline liquid which is colourless odourless! 12 Class 11 Class 10 Class 9 oxidation state of phosphorus in phosphoric acid are also called oxidation states it can have are (! 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And hydrochloric acid how it can have are +3 ( s2p0 ) +5... Pentoxi… Get the answers you need, now crystalline solid which melts at 42.35°C atoms can. Points ) 0 0. caroline m. Lv 6 is surrounded tetrahedrally by other atoms the chrage of P.! Oxygen atoms are usually in oxidation state s ) + 2HCl ( aq ) mc014-1 is +3 written! ) the … oxidation state includes peroxide groups overall charge of that ion in nature 4H_3PO_3~\rightarrow~ 3H_3PO_4~ +~ PH_3\ the! Oxoacids are the oxidation state the most common oxidation oxidation state of phosphorus in phosphoric acid -8 = 0 +1+x -8 fuel! Class 11 Class 10 Class 9 … are also called oxidation states too disproportionate to lower and oxidation! By the number of hydrogen atoms which can be replaced by electro – atoms. May be in state -1 if the molecule includes peroxide groups P−P bond … Carboranylphosphinic acids: a +5 is! ) how satisfied are you with the answer apne doubts clear karein ab (! 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Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

$$\Rightarrow$$ OA = OC and OB = OD
And also,
$$\angle{AOD} = \angle{AOB} = \angle{COD} = \angle{BOC} = 90^\circ$$.

To prove: ABCD is a rhombus.

Proof:
In $$\triangle{OAB}$$ and $$\triangle{ODC}$$, we have,

OA = OC and OB = OD ...(given)
$$\angle{AOB}$$ = $$\angle{COD}$$
(vertically opposite angles)
$$\therefore$$ $$\triangle{OAB}$$ $$\displaystyle \cong$$ $$\triangle{OCD}$$
(By SAS rule)
$$\therefore$$ AB = CD ...(i)(By CPCT)

Similarly, in $$\triangle{OAD}$$ and $$\triangle{OBC}$$, we have,
OA = OC and OD = OB ...(given)
$$\angle{AOD}$$ = $$\angle{BOC}$$
(vertically opposite angles)
$$\therefore$$ $$\triangle{OAD}$$ $$\displaystyle \cong$$ $$\triangle{OCB}$$ ...(By SAS rule)
$$\therefore$$ AD = BC ...(ii)(By CPCT)
Similarly, we can prove that,
AB = CD and
CD = BC ...(iii)

Hence, from (i), (ii) and (iii), we get,
AB = BC = AD = CD
Hence, it is proved that ABCD is a rhombus.

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roots of sum of two polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:11:43Z http://mathoverflow.net/feeds/question/30072 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomials roots of sum of two polynomials vilvarin 2010-06-30T16:32:32Z 2010-07-01T17:35:58Z
I believe that there is no common theory for finding roots of polynomial sum. In my case I have $$P_{n}(x)+AQ_{n}(x)$$. I am wondering how roots of this sum depend on $A$?
 http://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomials/30075#30075 Answer by jc for roots of sum of two polynomials jc 2010-06-30T17:01:56Z 2010-07-01T17:35:58Z
Though in general you won't have a closed-form expression for the roots of your polynomials, it's possible to write down perturbation series for roots of a polynomial in a single variable in terms of the coefficients. These are basically the Puiseux series mentioned in this question.

This paper by Bernd Sturmfels (MR) sketches out the "global picture" of such series, though it's fairly complicated and I personally am not clear on whether there's a simple algorithm to decide which is the proper choice of series that will converge. See also the article "Algebraic equations and hypergeometric series" by M Passare, A Tsikh in the book: The Legacy of Niels Henrik Abel (MR).

What I've just written is probably a little unclear so I'll describe the simplest example. Suppose you'd like to write down a series for the roots of $a_2x^2+a_1x+a_0=0$. There are a pair of series which converges when $\left|\frac{a_1^2}{4a_0a_2}\right|<1$ and a pair which converges when $\left|\frac{a_1^2}{4a_0a_2}\right|>1$, and you can derive the first pair of series by treating $a_1x$ as a perturbation to the equation $a_2x^2+a_0=0$ and you can derive one of the second pair of series by treating $a_2x^2$ as a perturbation to $a_1x+a_0=0$ and the other by treating $a_0$ as a perturbation to the equation $a_2x^2+a_1x=0$.

By plugging in the coefficients of $P_n(x)+AQ_n(x)$ into the appropriate series I just described and looking at the leading order terms as functions of $A$, you will be able to derive the scaling of the corrections to the roots of $P_n(x)$.

Apologies for the rather unexplicit answer, but this is just at the limit of what I understand.
 http://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomials/30086#30086 Answer by SandeepJ for roots of sum of two polynomials SandeepJ 2010-06-30T18:13:32Z 2010-06-30T18:13:32Z
If they are complex polynomials or can be treated as such, then you could apply Rouche's theorem, where the location of the zeros is determined by the dominant polynomial within the sum. ("Walk the dog on the leash")

Possibly related: you could use the Wronskian to determine the values of A that make $P_n(x)$ and $Q_n(x)$ linearly independent.

Your question is related to Mason's theorem. There are a few papers which explore this specifically

MR1923392 (2003j:30012) Kim, Seon-Hong . Factorization of sums of polynomials. Acta Appl. Math. 73 (2002), no. 3, 275--284.

MR2103113 (2005h:30011) Kim, Seon-Hong . On zeros of certain sums of polynomials. Bull. Korean Math. Soc. 41 (2004), no. 4, 641--646

MR1911767 (2003d:11036) Pintér, Á. Zeros of the sum of polynomials. J. Math. Anal. Appl. 270 (2002), no. 1, 303--305.


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# Changes between Version 140 and Version 141 of WRF4G2.0/Experiment

Ignore:
Timestamp:
Jan 18, 2016 12:35:14 PM (7 years ago)
Comment:

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### Legend:

Unmodified
 v140 * '''namelist_version:''' Namelist version to use ( from 3.2 to 3.7.1 ). * '''namelist_values:''' Users are able to over-write or add parameters to the namelist file. The name must be the same as the namelist.input entry. No record specification is necessary if the parameter already appears on the provided WRF-version namelist.input template. If not, it should be record.parameter. * '''namelist_values:''' Users are able to over-write or add parameters to the namelist file. The name has to be the same as the namelist.input entry. No record specification is necessary if the parameter already appears on the provided WRF-version namelist.input template. If not, it should be record.parameter. * '''single:''' Indicate that namelist parameter of record has a single value. One value per all domains. (Optional) * '''max_dom:''' Indicate that namelist parameter of record has the same value for all the domains of the experiment. (Optional) Example for one domain : {{{ namelist_values = single:ra_lw_physics | 3 single:physics.ra_sw_physics | 3 namelist_values = single:physics.ra_sw_physics | 3 single:ra_lw_physics        | 3 }}} Example for two domains : {{{ namelist_values = ra_lw_physics | 2, 2 ra_sw_physics | 2, 2 namelist_values = ra_sw_physics | 2, 2 ra_lw_physics | 2, 2 }}}


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# Bayesian Estimate Problem

So... I'll be honest, I don't know anything about anything Bayesian, this problem being no exception (from the Society of Actuaries' Exam C sample questions):

You are given:

(i) The annual number of claims for a policyholder has a binomial distribution with probability function: $$p\left(x \mid q\right) = {2\choose x}q^{x}(1-q)^{2-x}\text{, }\quad x = 0, 1, 2$$

(ii) The prior distribution is $$\pi(q) = 4q^{3}\text{, }\quad 0 < q < 1$$

The policyholder had one claim in each of Years 1 and 2.

Determine the Bayesian estimate of the number of claims in Year 3.

I have decided not to look at the solution, as I am attempting to study for this exam through reverse engineering of questions (i.e., grab a question and learn about the underlying concepts using a text). Unlike the previous questions for which this method has been successful, I hit a brick wall with this one.

So from what I understand, unlike the frequentist mindset, with Bayesian statistics, you assign a "prior" distribution to some variable. The only other thing I know about Bayesian methods is what a posterior distribution is (by definition) and how it can be motivated through Bayes' theorem.

Guidance with this problem is appreciated.

You have a prior distribution $\pi(q)$, a likelihood function $p(x|q)$ and two observations $x_1=1$ and $x_2=1$. So your posterior distribution for $q$ is $$\pi(q|x_1,x_2)=\frac{\pi(q)p(x_1|q)p(x_2|q)}{\displaystyle \int_0^1 \pi(r)p(x_1|r)p(x_2|r)\; dr}$$ in effect taking the product of the prior and the likelihoods and then turning this into a probability density by dividing by its integral over all possible $q$ (represented here by $r$ as $q$ is also a free variable), though knowing $0 \le q \le 1$.
If you want the expected number of accidents in year $3$, then your next step would be to calculate $$\int_0^1 (2 \times p(2|q) + 1 \times p(1|q) + 0 \times p(0|q)) \pi(q|x_1,x_2)\; dq$$
• Thank you for your response. I have two questions - in your first equation, I noticed that you have $p(x_1 \mid q)p(x_2 \mid q)$ as a product in the numerator; is this because we assume that conditional on $q$, $X_1$ and $X_2$ are independent? Also, is the second equation supposed to have $0 \times p(0\mid q)$ instead of $0 \times p(1\mid q)$? – Clarinetist Sep 6 '14 at 19:57
• Yes, and yes ($0$ corrected) – Henry Sep 6 '14 at 20:45


text:
# Math Help - Normally distributed random variables

1. ## Normally distributed random variables

Im finding it hard to understand why you can do the following transformation of $\phi$ to get from a normally distributed random variable with mean 0 and standard deviation of 1 to a normally distributed random variable with a specific mean and standard deviation:

mean + standard deviation $\cdot \phi

$

Finding it difficult to see the intuition behind multiplying by std. dev. and adding the mean.

This is a pretty fundamental thing to understand so would be grateful for any help.

Cheers,
Peter

2. Originally Posted by peterpan
Im finding it hard to understand why you can do the following transformation of $\phi$ to get from a normally distributed random variable with mean 0 and standard deviation of 1 to a normally distributed random variable with a specific mean and standard deviation:

mean + standard deviation $\cdot \phi

$

Finding it difficult to see the intuition behind multiplying by std. dev. and adding the mean.

This is a pretty fundamental thing to understand so would be grateful for any help.

Cheers,
Peter
All you're doing is translating and dilating (scale changing) ......


text:
# I Is this proof regarding massless strings correct?

#### walking

I am reading the Tipler and Mosca textbook and am on the part about massless strings. I understand that in real life a string has more tension at the top than the bottom because the top part has to support a greater mass of rope. However, in other examples such as pulling a sled with a rope I can't seem to see how the tension varies intuitively. I tried proving that if the string is considered as being massless, the tension would be negligible, and I am posting this to ask if my proof is correct.

Proof: Consider a rope being pulled with tension T at an angle $\theta$ to the horizontal. To simplify the situation, I split the rope up into a finite number of segments. I pull on segment 1, which then pulls on segment 2, etc all the way down to the last segment which pulls on the sled. When I exert a force T on segment 1, let the force that segment 1 exerts on segment 2 be T'. Then segment 2 exerts an equal and opposite force T' on segment 1. Hence segment 1 is being pulling simultaneously to the left and right. Also, segment 1 has a weight mg acting downwards. The component of this weight in the direction parallel to the rope (ie "down the rope") is $mg\sin \theta$. The tensions and weight are the only forces acting on segment 1, hence $mg\sin \theta+T'=T$, giving $T-T'=mg\sin \theta$. Now because m is negligible this is roughly $T-T'\approx 0$ ie $T\approx T'$. Since this proof can be applied to any segment of rope, it follows that the tension is the same throughout the rope. $\square$

If the proof is correct, I am still wondering if a special proof is needed for each situation in which tension is applied to ropes. Is there a general proof that deals with all cases? For example, one can give a separate proof dealing with the case of a rope hanging from a ceiling in order to show that the tension is the same throughout if the rope is considered massless, but this doesn't prove the rule in general: if a string is massless then the tension is the same throughout.

Thanks

Related General Physics News on Phys.org

#### Nugatory

Mentor
I am still wondering if a special proof is needed for each situation in which tension is applied to ropes
There's a very convincing general proof by contradiction: assume that the tension is not the same everywhere; derive from that assumption and $F=ma$ an impossible result; therefore the assumption is false.

#### walking

There's a very convincing general proof by contradiction: assume that the tension is not the same everywhere; derive from that assumption and $F=ma$ an impossible result; therefore the assumption is false.
Do you know where I can find this proof?

#### Let'sthink

Newton's law tells us that the tension in the string that pulls some object with with constant acceleration, say a, toward right cannot be uniform as some force has to be there which generates the acceleration of a tiny piece of string which is pulled by the right part, say with "Tr" right ward and left part, say "Tl" left word. So Tr > Tl. But if mass of the string is assumed to be zero then zero force can produce finite acceleration in a zero mass in the limiting case mathematically and the limit can be anything different for different cases.

#### jbriggs444

Do you know where I can find this proof?
Recall that the "F" in F=ma is the vector sum of all external forces acting on an object.

Suppose that m=0. Then it follows that F=0. So the sum of all external forces on a length of string is zero. There are only two such forces: the force F1 applied at the one end and the force F2 applied at the other end. Both are equal in magnitude to the tensions at the respective ends of the string, T1 and T2.

$$F_1+F_2=0$$ $$F_1=-F_2$$ $$|F_1| = |F_2|$$ $$T_1=T_2$$

#### sophiecentaur

Gold Member
Do you know where I can find this proof?
It strikes me that, if any actual proof were needed, it would be required to show that there would actually be a difference in tension. Not every step in the explanation of a Physics process needs proof. I'd say this is more axiomatic than 'testable'.
If the tensions were not the same, Newtons Laws tell us that the string would accelerate off in some direction and that is hardly likely.

#### walking

It strikes me that, if any actual proof were needed, it would be required to show that there would actually be a difference in tension. Not every step in the explanation of a Physics process needs proof. I'd say this is more axiomatic than 'testable'.
If the tensions were not the same, Newtons Laws tell us that the string would accelerate off in some direction and that is hardly likely.
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.

#### sophiecentaur

Gold Member
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.
You should perhaps try to see reasons why this should be OK, rather than scanning the detailed wording of that article and seeking some paradox. @jbriggs444 is just quoting the very basic Newtonian ideas; can you really be questioning them is such a simple situation? f=ma is something we can all agree on, surely. If m is zero then how can f (the net force on the string) be non-zero? If the net force is zero then that means the two tensions are the same. If, instead of having a string over the pulleys, you hold the pulleys together or use ideal gears (eliminating the string) then the tangential forces would have to be the same (N3 etc.).
The word "light" is used in the article and light doesn't mean massless so some tension difference could apply in this (totally different) scenario.

#### walking

You should perhaps try to see reasons why this should be OK, rather than scanning the detailed wording of that article and seeking some paradox. @jbriggs444 is just quoting the very basic Newtonian ideas; can you really be questioning them is such a simple situation? f=ma is something we can all agree on, surely. If m is zero then how can f (the net force on the string) be non-zero? If the net force is zero then that means the two tensions are the same. If, instead of having a string over the pulleys, you hold the pulleys together or use ideal gears (eliminating the string) then the tangential forces would have to be the same (N3 etc.).
The word "light" is used in the article and light doesn't mean massless so some tension difference could apply in this (totally different) scenario.
Excuse my ignorance (I am a beginner), I didn't know light and massless were different things. I thought massless itself only meant "negligibly small", since this is what the Tipler and Mosca textbook says in its discussion of tension.

#### Nugatory

Mentor
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.
It depends on whether you interpret "throughout" as meaning "throughout each segment", which is what everyone means when they talk about uniform tension in a massless rope, or misinterpret it as meaning "everywhere, even in the different segments on different sides of the pulley". When the pulley has non-trivial mass we need a net force to accelerate it; this force comes from the segment of rope on one side pulling harder than the other.

It's easiest to think of this problem as if we had two ropes, both fastened to the same point on the rim of the pulley and pulling in opposite directions. What does the pulley do if the tensions are the same? What does it do if they are different?

#### sophiecentaur

Gold Member
Excuse my ignorance (I am a beginner), I didn't know light and massless were different things. I thought massless itself only meant "negligibly small", since this is what the Tipler and Mosca textbook says in its discussion of tension.
There is a massive difference in principle but not a lot in practice. However in a question like yours, in which you seems to be looking for loopholes in the basic theory, the possibility of a finite difference would depend on a finite (i.e. non-zero) mass for the string. You need to decide if you are going for ideally zero mass or a relatively small mass before you can get a real answer.
Iff the mass is "negligibly small" then the difference in Tension will also be negligibly small.

#### haruspex

Homework Helper
Gold Member
2018 Award
But in this stackexchange post, https://physics.stackexchange.com/questions/308011/rigorously-proving-equal-tension-on-both-ends , dmkee says that "Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light." Does this mean that massless strings do not always have the same tension throughout? I'm a bit confused considering the proof given by jbriggs.
The proof given by @jbriggs444 is in regards to a segment of string on which the only external forces are applied at the ends. If the string is of negligible mass then any non-negligible net applied force would result in a huge acceleration.
In the case of a massive accelerating pulley, with the string not slipping on the pulley, the inertia of the pulley applies a frictional force to the string. So now you must consider the string in two segments, one each side of the pulley. Within each segment the tension is constant, but there will be a difference between the two tensions.

#### Let'sthink

Recall that the "F" in F=ma is the vector sum of all external forces acting on an object.

Suppose that m=0. Then it follows that F=0. So the sum of all external forces on a length of string is zero. There are only two such forces: the force F1 applied at the one end and the force F2 applied at the other end. Both are equal in magnitude to the tensions at the respective ends of the string, T1 and T2.

$$F_1+F_2=0$$ $$F_1=-F_2$$ $$|F_1| = |F_2|$$ $$T_1=T_2$$
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.

#### Nugatory

Mentor
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.
The argument still stands: $F=ma$ with $m=0$ implies that $F=0$; the net force on any segment of string must be zero, which implies uniform tension within that segment.

#### Mister T

Gold Member
Does this mean that massless strings do not always have the same tension throughout?
Correct. The pulley makes contact with the string and exerts forces on it.

#### sophiecentaur

Gold Member
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.
I can see that you feel a need to find something wrong with what has been said here. The core of the argument is F=ma. If m is zero then F will always be zero (under any value of acceleration). Can you possibly disagree with that?
The high number of posts in this thread are about of all proportion with the explanation in terms of Newton's Laws. Just sit and contemplate F=ma for a while.

#### Let'sthink

I can see that you feel a need to find something wrong with what has been said here. The core of the argument is F=ma. If m is zero then F will always be zero (under any value of acceleration). Can you possibly disagree with that?
The high number of posts in this thread are about of all proportion with the explanation in terms of Newton's Laws. Just sit and contemplate F=ma for a while.
What do you say about T in the string moving with common acceleration, when string is not mass less? Will it be uniform? Contemplate on F = ma and tell me Sir.

#### haruspex

Homework Helper
Gold Member
2018 Award
I am talking of the case when common acceleration of the two masses and the string is non zero, Sir.
@jbriggs444 was responding to post #3, not to your post #4.

#### Nugatory

Mentor
What do you say about T in the string moving with common acceleration, when string is not mass less? Will it be uniform?
Of course not. Consider any two points on one segment of the string. The string is not massless, so the bit of string between those two points has non-zero mass. It is also accelerating, so $F=ma$ tells us that $F$ is non-zero. This non-zero net force is the difference between the tension at those two points; if that difference is non-zero the tensions are different and therefore the tension is not uniform.

It would be a good exercise to derive the formula for the tension at any point; if the density of the string is constant (that is, the mass of any piece of lengthn$L$ is equal to $\rho{L}$ where $\rho$ is a constant) you can do this with just algebra, no calculus needed.

#### sophiecentaur

Gold Member
This is now a different problem. The new constraints assume no slippage over the pulleys. Tension will be the same at each end of a section between pulleys MINUS any ma and mg (weight) factor.
This is not a ‘proof’ we are after. It’s a calculation.

"Is this proof regarding massless strings correct?"

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Question

# A conical pit of top diameter$3.5m$ is $12m$ deep. What is its capacity in kiloliters? (Assume$\mathrm{\pi }=22/7$).

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Solution

## Step 1- finding the volume of the pitRadius of cone $\left(r\right)=3.5/2m$Height of cone $\left(h\right)=12m$$Volumeofthecone=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cone}=\frac{1}{3}×\frac{22}{7}×\frac{3.5}{2}×\frac{3.5}{2}×12\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cone}=38.5{m}^{3}$Step 2 - Finding the capacity of the conical pit.Now, we know that, $1{m}^{3}=1000litre$$\therefore 38.5{m}^{3}willcontain=1000×38.5=38500litres$Now, we know that $1000litres=1kilolitre$So, $38500litres=\frac{38500}{1000}=38.5kilolitres$Hence, the capacity of the container is $38.5kiloliters$.

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1/4+1/3+1/2

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1/4+1/3+1/2 [#permalink]  28 Jul 2018, 08:20
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Question Stats:

67% (00:54) correct 32% (01:14) wrong based on 28 sessions
 Quantity A Quantity B $$\frac{1}{4 + \frac{1}{3 + \frac{1}{2}}$$ $$\frac{1}{2 + \frac{1}{3 + \frac{1}{4}}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Re: 1/4+1/3+1/2 [#permalink]  29 Jul 2018, 18:29
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Carcass wrote:
 Quantity A Quantity B $$\frac{1}{4 + \frac{1}{3 + \frac{1}{2}}$$ $$\frac{1}{2 + \frac{1}{3 + \frac{1}{4}}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

QTY A = $${\frac{1}{4 + \frac{1}{3 + \frac{1}{2}}} = {\frac{1}{4 + \frac{2}{7}}} = {\frac{1}{28 + \frac{2}{7}}} = {\frac{1}{\frac{30}{7}}} = {\frac{7}{30}}$$

QTY B = $${\frac{1}{2 + \frac{1}{3 + \frac{1}{4}}} = {\frac{1}{2 + \frac{4}{13}}} = {\frac{1}{26 + \frac{4}{13}}} = {\frac{1}{\frac{30}{13}}} = {\frac{13}{30}}$$

Therefore QTY A < QTY B

Option B
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Re: 1/4+1/3+1/2 [#permalink]  02 Aug 2018, 15:51
Expert's post
Carcass wrote:
 Quantity A Quantity B $$\frac{1}{4 + \frac{1}{3 + \frac{1}{2}}$$ $$\frac{1}{2 + \frac{1}{3 + \frac{1}{4}}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

We need only recognize that 1/(3 + 1/2) is less than 1
So, 4 + 1/(3 + 1/2) = 4.something

Likewise, 1/(3 + 1/4) is less than 1
So, 2 + 1/(3 + 1/2) = 2.something

So, we get:
Quantity A: 1/(4.something)
Quantity B: 1/(2.something)

Cheers,
Brent
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Re: 1/4+1/3+1/2   [#permalink] 02 Aug 2018, 15:51
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text:
# Accessing the gluonic structure of light nuclei at a future electron-ion collider

@article{Mantysaari2020AccessingTG,
title={Accessing the gluonic structure of light nuclei at a future electron-ion collider},
author={H. Mantysaari and B. Schenke},
journal={Physical Review C},
year={2020},
volume={101},
pages={015203}
}
• Published 2020
• Physics
• Physical Review C
• We show how exclusive vector meson production off light ions can be used to probe the spatial distribution of small-$x$ gluons in the deuteron and $^3$He wave functions. In particular, we demonstrate how short range repulsive nucleon-nucleon interactions affect the predicted coherent $J/\Psi$ production spectra. Fluctuations of the nucleon substructure are shown to have a significant effect on the incoherent cross section above $|t|\gtrsim 0.2\,\mathrm{GeV}^2$. By explicitly performing the… CONTINUE READING
7 Citations

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SHOWING 1-10 OF 86 REFERENCES


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• ### Popular Now

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#### Archived

This topic is now archived and is closed to further replies.

# Making the window disappear

This topic is 5974 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

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When the user clicks the minimize button, I would like my window to disappear from the start bar as well. I have this code: case WM_SYSCOMMAND: switch( LOWORD( wParam ) ) { case SC_MINIMIZE: ShowWindow( hWnd, SW_HIDE ); break; } break; After trying this, I came to realize that the SW_HIDE flag does no get rid of the start bar, it only minimizes the window. What should I put in the SC_MINIMIZE case statement to make the window go completely away?

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Well, first of all. this message is only recieved when the user is trying to minimize the window so that''s what it should do. However if you really wan the window to go completely away (never to be seen again). Then just call DestroyWindow(hWnd) and your window will go bye bye.

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i'm not sure if this will work, but try this:

  //test for system commands in the message quecase WM_SYSCOMMAND: /*switch the loword:wparam in case we have other wm_system commands, then it will make it easier to read*/ switch( LOWORD( wParam ) ) { //we want to hide the window if the user tries to minimise case SC_MINIMIZE://test case minimize ShowWindow( hWnd, FALSE );//hide the window }break;}break;

if this doesn't suit ur needs, i have a few other ideas.

Edited by - kmsixpence on November 12, 2001 7:29:18 PM

Edited by - kmsixpence on November 12, 2001 7:30:03 PM

Edited by - kmsixpence on November 12, 2001 7:30:51 PM

Edited by - kmsixpence on November 12, 2001 7:33:16 PM

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Has the same effect as the way I did it. What were those other ideas?

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DanTheRocker, that''s so strange that I tried your codes and succeeded!
I am using Win2000, does this matter?

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humm...so it worked for you did it? Im using Win98. The only thing different that it does for me is make it minimize faster. It totally went away for you? Does this work for other people?


text:
Determine and draw planes and axes of symmetry of a molecule using software (VESTA or Mercury)

I designed the $$\text{MoS}_2$$ molecule (bilayer), with polytype 2H in the VESTA. My goal is to draw the plans and axes of symmetry of the bilayer set. According to the information in the literature, this molecule belongs to the $$\text{D}_{3d}$$ point group. How can I draw and determine the planes and axes of symmetry in VESTA?

These are the parameters I used in VESTA:

Mercury Attempt:

I exported the .vesta data to a .cif file.

Table with space and point groups for $$\text{MoS}_{2}$$ (taken from https://iopscience.iop.org/article/10.1088/0953-8984/28/35/353002/ampdf) (see 2H phase for $$\text{MoS}_{2}$$):

I do not know VESTA well enough to provide an answer for this program.

However because your data are about a crystal structure I suggest to export your model in the .cif format. CCDC's freely available version of CCDC's Mercury is able to read this file type, and add to its visualizations the symmetry elements of the unit cell (Display -> Symmetry Elements). There will be a new menu where you may choose among the symmetry elements the program is able to recognize, and how they are displayed:

For example only the centres of inversion:

The export options in Mercury include bitmap .png, ray-tracing PovRay .pov, and .vrml / .stl.

Given lattice constants and space group $$P6_3/mmc$$, do not forget to compare your model with the literature known models for example in the freely accessible COD about experimental data (entries #1010993, #1011286, #1531960, #9007660, #9009144); and consider contributing to its sibling TCOD about theoretical data and MPOD.

• Thank you for your answer. I tried to follow your suggestion and do it on Mercury. The objective is to draw the axes and the planes to prove that it is a $\text{D}_{3d}$ structure. To be a $\text{D}_{3d}$ structure, it must have 3 $\text{C}_2$ axes of symmetry. I put what I got on Mercury, in the post. The problem is that I am getting more than 3 axes of type $\text{C}_2$, and I should only get 3 axes of that type. Why do more than 3 axes type $\text{C}_2$ appear to me, when this structure should only have 3 axes type $\text{C}_2$ and what are those 3 axes in the last figure I put in the post? – Jose Marin Apr 13 '20 at 18:03
• In addition, the molecule you took from the COD library must have a bug in Mercury. In $\text{MoS}_2$ with polytype 2H, there is a bilayer and only Mo-S bonds are allowed. In the drawing you have in the images there are Mo-Mo bonds, which should not appear. – Jose Marin Apr 13 '20 at 18:58
• @JoseMarin We are sharing the same page here. Space group $P6_3/mmc$ (#194) is for unit cells of point group symmetry $D^4_6h$ (IT volume A1, p. 345, the 2006 edition). Yes, the diagram (e.g., img.chem.ucl.ac.uk/sgp/large/194az1.htm) indicates more than 3 $C_2$ axes perpendicular to $2/m$ and $6_3/m$ axes. – Buttonwood Apr 14 '20 at 20:55
• @JoseMarin Mercury is developed by CCDC; contrasting to ICSD with more inclination for organic, and metalorganic compounds. So its geometric approach draws lines between atoms if their distance is quite less than the sum of the van der Waals radii. With S 1.80 A, Mo 1.90 A, d(Mo-Mo) = 3.15..3.16 A which is lesser than 2 x 1.90 A. With the input given in your question, I was not able to reproduce the trigonal prisms of your question (vesta 3.4.8); indeed, pressing the «remove duplicate atoms» with default 1e-006 leaves only one Mo and one S (still no prisms). – Buttonwood Apr 14 '20 at 21:04
• @JoseMarin Loading one of the COD entries (#1010993, #1011286, #1531960, #9007660, and #9009144) as .cif into Vesta however yielded the prisms of the original question within a blink of an eye; each of them require only one Mo and one S for the model's definition. – Buttonwood Apr 14 '20 at 21:09


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EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING

Evi Papaioannou [1] , Stavros Athanassopoulos [2] , Christos Kaklamanis [3]

126 286

Graphs are discrete structures composed of vertices and edges connecting these vertices. Graphs are used in almost all disciplines as abstract models for the representation and study of a wide range of relations and processes in physical, biological, social and information systems. Many practical problems in a variety of areas like computer and communication networks, social networks, transportation networks, cellular networks, linguistics, chemistry, physics, biology can be represented and studied by graphs.

Real-world entities - like molecules, persons, groups, roles, species, computing and communication devices, terms - correspond to vertices. Relations among such entities - like preference, domination, independence, interference, proximity, constraints - imply the existence of edges between corresponding vertices. Thus, focusing on the abstract graph model instead of studying each particular instance as a different real-world problem reveals common underlying properties, deficiencies and principles. In this way, efficient approaches to real-world problems emerge from the theoretical study of their abstractions.

In this work, we use graph coloring to propose efficient solutions to scheduling problems arising in higher education. The objective of the graph coloring problem is to assign colors to graph vertices so that adjacent vertices, i.e., vertices connected by an edge, receive different colors. We consider as the objective of scheduling problems in higher education, like lecture and exam scheduling, to assign time/day slots to teaching or examination activities so that the maximum number of students can attend them with the fewest possible conflicts.

Our main motivation has been the crucial issue of efficient course and exam schedules often arising in departments of the University of Patras, Greece. Students usually have to attend lectures or exams scheduled in overlapping or simultaneous time slots. However, course and exam schedules are created based on heuristic approaches which may work well on average but certainly leave several room for improvement.

What if a graph-theoretic approach were used? Courses correspond to vertices of a graph and there is an edge between two vertices if and only if an appropriately selected minimum population of students attends corresponding courses (lectures/exams). Then, a coloring of such an underlying graph suggests an appropriate schedule for teaching/examination activities.

Using a simple coloring algorithm and the MATLAB programming environment, we have designed and developed a scheduling application which receives as input courses and constraints and outputs an efficient lecture/examination schedule. Experimental evaluation suggests that our application works well in practice. Ongoing work focuses on the use of a more involved coloring algorithm for addressing more complex course scheduling instances while minimizing required time resources.

graph coloring, scheduling, higher education
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Konular Sosyal Makaleler Yazar: Evi PapaioannouÜlke: Greece Yazar: Stavros AthanassopoulosÜlke: Greece Yazar: Christos KaklamanisÜlke: Greece
 Bibtex @araştırma makalesi { ijaedu309802, journal = {International E-Journal of Advances in Education}, issn = {}, eissn = {2411-1821}, address = {OCERINT International Organization Center of Academic Research}, year = {2017}, volume = {3}, pages = {51 - 59}, doi = {10.18768/ijaedu.309802}, title = {EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING}, key = {cite}, author = {Athanassopoulos, Stavros and Papaioannou, Evi and Kaklamanis, Christos} } APA Papaioannou, E , Athanassopoulos, S , Kaklamanis, C . (2017). EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING. International E-Journal of Advances in Education, 3 (7), 51-59. DOI: 10.18768/ijaedu.309802 MLA Papaioannou, E , Athanassopoulos, S , Kaklamanis, C . "EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING". International E-Journal of Advances in Education 3 (2017): 51-59 Chicago Papaioannou, E , Athanassopoulos, S , Kaklamanis, C . "EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING". International E-Journal of Advances in Education 3 (2017): 51-59 RIS TY - JOUR T1 - EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING AU - Evi Papaioannou , Stavros Athanassopoulos , Christos Kaklamanis Y1 - 2017 PY - 2017 N1 - doi: 10.18768/ijaedu.309802 DO - 10.18768/ijaedu.309802 T2 - International E-Journal of Advances in Education JF - Journal JO - JOR SP - 51 EP - 59 VL - 3 IS - 7 SN - -2411-1821 M3 - doi: 10.18768/ijaedu.309802 UR - http://dx.doi.org/10.18768/ijaedu.309802 Y2 - 2017 ER - EndNote %0 International E-Journal of Advances in Education EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING %A Evi Papaioannou , Stavros Athanassopoulos , Christos Kaklamanis %T EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING %D 2017 %J International E-Journal of Advances in Education %P -2411-1821 %V 3 %N 7 %R doi: 10.18768/ijaedu.309802 %U 10.18768/ijaedu.309802 ISNAD Papaioannou, Evi , Athanassopoulos, Stavros , Kaklamanis, Christos . "EFFICIENT COURSE AND EXAM SCHEDULING USING GRAPH COLORING". International E-Journal of Advances in Education 3 / 7 (Nisan 2017): 51-59. http://dx.doi.org/10.18768/ijaedu.309802


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# Is there a formula for $|\langle H\cup K\rangle|$, if $H$ and $K$ are nonempty subsets of a finite group $G$?

I know that if $H$ and $K$ are subgroups of a finite group $G$, then $|HK|=\frac{|H||K|}{|H\cap K|}$.

Is there a formula for $|\langle H\cup K\rangle|$, if $H$ and $K$ are nonempty subsets of a finite group $G$?

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You are looking for a formula which involve the cardinal of $H$ and $K$ and potentially the cardinal of a set combination between $H$ and $K$, right? For example, if $H$ is a singleton, and we take $H=K$, $|\langle H\cup K\rangle|$ is the order of the element. –  Davide Giraudo Jan 29 '12 at 14:25
@Davide Giraudo: That's correct! –  spohreis Jan 29 '12 at 14:41
Consider the case when H and K are subgroups of order 2 with an intersection of size 1. The size of the subgroup generated by H and K is an arbitrary even number ≥ 4. I'm not sure what sort of formula you could be looking for under these circumstances. –  Jack Schmidt Jan 29 '12 at 18:50
Every countable group can be embedded in a $2$-generator group. In particular, $2$-generator groups can have arbitrarily large size. So if $H$ and $K$ are sets of size $1$, the size of $\langle H\cup K\rangle$ could potentially be as large as you want. –  Arturo Magidin Jan 29 '12 at 22:02
There is a lower bound, namely $\frac{|H||K|}{|H\cap K|}$. From page 41, (Hungerford's book Algebra) $[H \vee K: H]\geq [K:H\cap K]$. –  spohreis Jan 31 '12 at 0:27


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# Arithmatex

## Overview

Arithmatex is an extension that preserves LaTeX math equations during the Markdown conversion process so that they can be used with MathJax. It searches for the patterns $...$ and $...$ for inline math, and $$...$$, $...$, and \begin{}...\end{} for block math. By default, all formats are enabled, but they each format can individually be disabled if desired.

Keep in mind that some equations can make it through without Arithmatex fine. Arithmatex is meant to ensure they make it through.

## Input Format

For the $...$ inline variant, it is expected that the opening token ($) is to be followed by a non-whitespace character, and the closing to be preceded by a non-white-space character. This is to help avoid false positives when using the dollar sign in traditional ways such as: I have$2.00 and Bob has $10.00. The previous statement requires no escaping of the $ character. But when needed, the $ character can be escaped using \$.

For block forms, the block must start with the appropriate opening for the block type: $$, $, and \begin{} for the respective search pattern. The block must also end with the proper respective end: , $, and \end{}. A block also must contain no empty lines and should be both preceded and followed by an empty line. ## Output Format If insert_as_script is enabled, the math equations will be wrapped in a special MathJax script tag and embedded into the HTML. MathJax can already find these scripts, so there is no need to configure the tex2jax extension when setting up MathJax. The tag will be <script type="math/tex"></script> for inline and <script type="math/tex; mode=display"></script> for block. This is probably the most reliable HTML form, but if MathJax doesn't load, you don't get fallback text. When insert_as_script is enabled, tex_inline_wrap and tex_block_wrap are ignored. On output, if insert_as_script is False, the extension will escape necessary symbols and normalize all output to be wrapped in the more reliable $...$ for inline math and $...$ for block math. Then the wrapped content will be inserted into a span and embedded into the HTML. The span provides a container to keep MathJax from gathering multiple equations into one by mistake. So, with the default settings, if in your Markdown you used ..., it would be converted to $...$ in the HTML. Blocks would respectively be normalized from $$...$$ to $...$. In the case of \begin{}...\end{}, begins and ends will not be replaced, only wrapped$\begin{}...\end{}$. Normalization is provided as MathJax disables detection of ... by default due to detection issues with money notation. Since Arithmatex provides additional logic to curb issues with , we allow it in the Markdown. If a different wrapper is desired, see Options below to learn how to change the wrapper. ## Options Option Type Default Description inline_syntax [string] ['dollar', 'round'] Syntax to search for: dollar=... and round=$...$. block_syntax [string] ['dollar', 'square', 'begin'] Syntax to search for: dollar=..., square=$...$, and \begin{}...\end{}. tex_inline_wrap [string] ['\$', '\$'] An array containing the opening and closing portion of the wrap. tex_block_wrap [string] ['\$', '\$'] An array containing the opening and closing portion of the wrap. insert_as_script bool False Instead of using the above wrap options, insert the equations in a math/tex script. This form is the most reliable in HTML, but if the MathJax script fails to load, the user won't see any indication of the equations. Wrapping Format Version 1.7.0: Arithmatex, when insert_as_script is False, now runs the wrap options through an HTML escaper and then everything is wrapped in a span. This is to isolate equations and prevent them from bleeding into each other when MathJax parses the HTML. Though this is a more aggressive change, it is to meant to provide more reliable output. ## Loading MathJax Arithmatex requires you to provide the MathJax library and provide and configure it to your liking. The recommended way of including MathJax is to use the CDN. Latest version at time of writing this is found below. <script src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.0/MathJax.js"></script>  If you want to load one of the pre-defined configurations that MathJax offers, you can do as shown below. Notice we are using the TeX-MML-AM_CHTML configuration file. <script src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.0/MathJax.js?config=TeX-MML-AM_CHTML"></script>  If you don't include a pre-defined configuration, you will need to provide your own. But you can also provide a configuration to tweak the pre-defined configuration. Here we show a simple example of a configuration done in JavaScript. window.MathJax = { config: ["MMLorHTML.js"], jax: ["input/TeX", "output/HTML-CSS", "output/NativeMML"], extensions: ["tex2jax.js", "MathMenu.js", "MathZoom.js"], tex2jax: { inlineMath: [ ["\$","\$"] ], displayMath: [ ["\$","\$"] ] }, };  If you have insert_as_script enabled, you could omit the extension tex2jax.js and the tex2jax settings. window.MathJax = { config: ["MMLorHTML.js"], jax: ["input/TeX", "output/HTML-CSS", "output/NativeMML"], extensions: ["MathMenu.js", "MathZoom.js"] };  Please see the MathJax site for more info on using MathJax extensions/plugins and configuring those extensions/plugins. ## Examples Some Block Equations:$$
E(\mathbf{v}, \mathbf{h}) = -\sum_{i,j}w_{ij}v_i h_j - \sum_i b_i v_i - \sum_j c_j h_j

$3 < 4$

\begin{align}
p(v_i=1|\mathbf{h}) & = \sigma\left(\sum_j w_{ij}h_j + b_i\right) \\
p(h_j=1|\mathbf{v}) & = \sigma\left(\sum_i w_{ij}v_i + c_j\right)
\end{align}

Inline equations: $p(x|y) = \frac{p(y|x)p(x)}{p(y)}$, $p(x|y) = \frac{p(y|x)p(x)}{p(y)}$.

Some Block Equations:

$E(\mathbf{v}, \mathbf{h}) = -\sum_{i,j}w_{ij}v_i h_j - \sum_i b_i v_i - \sum_j c_j h_j$$3 < 4$\begin{align} p(v_i=1|\mathbf{h}) & = \sigma\left(\sum_j w_{ij}h_j + b_i\right) \\ p(h_j=1|\mathbf{v}) & = \sigma\left(\sum_i w_{ij}v_i + c_j\right) \end{align}

Inline equations: $p(x|y) = \frac{p(y|x)p(x)}{p(y)}$, $p(x|y) = \frac{p(y|x)p(x)}{p(y)}$.