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Chlorine gas oxidises $$\text{Fe}^{2+}$$ ions to $$\text{Fe}^{3+}$$ ions. The resulting hydrogen atoms are balanced by adding fourteen hydrogen ions to the left: $Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O\nonumber$. The half-reaction method for balancing redox equations provides a systematic approach. Now that all the atoms are balanced, only the charges are left. Half-reactions can be used to balance redox reactions. Now the oxygen atoms balance but the hydrogens don't. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. $$\color{blue}{\textbf{Tin}}$$ is therefore being $$\color{blue}{\textbf{oxidised}}$$ and $$\color{red}{\textbf{iron}}$$ is the $$\color{red}{\textbf{oxidising agent}}$$ (causing tin to be oxidised). To balance these, eight hydrogen ions are added to the left: $MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O\nonumber$. Four hydrogen ions to the right-hand side to balance the hydrogen atoms: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+\nonumber$. The loss of electrons 2. State the Oxidation Number of each of the elements that is underlined. The half-reaction is now: $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq}) + 6\text{e}^{-}$$ $$\to$$ $$2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$. 5 electrons are added to the left-hand side to reduce the +7 to +2: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} _ 4H_2O\nonumber$. Two questions should be asked to determine if a reaction is a redox reaction: Is there a compound or atom being oxidised? The unbalanced oxidation half-reaction is: $$\text{H}_{2}\text{S}(\text{g})$$ $$\to$$ $$\text{S}(\text{s})$$. Finally, tidy up the hydroxide ions that occur on both sides to leave the overall ionic equation: Reminder: a redox half-reaction MUST be balanced both for atoms and charge in order to be correct. Balance this redox reaction by using the half reaction method. In order to accomplish this, the following can be added to the equation: In the chlorine case, the only problem is a charge imbalance. The first example concerned a very simple and familiar chemical equation, but the technique works just as well for more complicated (and perhaps unfamiliar) chemistry. Example $$\PageIndex{1}$$: The reaction between Chlorine and Iron (III) Ions. This is easily resolved by adding two electrons to the left-hand side. DON'T FORGET TO CHECK THE CHARGE. Add $$\text{H}^{+}$$ ions to the right to balance the hydrogen atoms: $$\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$\text{O}_{2}(\text{g}) + 2\text{H}^{+}(\text{aq})$$. This arrangement clearly indicates that the magnesium has lost two electrons, and the copper(II) ion has gained them. Two electrons must be added to the right hand side of the equation. To reduce the number of positive charges on the right-hand side, an electron is added to that side: $Fe^{2+} \rightarrow Fe^{3+} + e-\nonumber$. Balancing in a basic solution follows the same steps as above, … $2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2\nonumber$, Example $$\PageIndex{3}$$: Oxidation of Ethanol of Acidic Potassium Dichromate (IV). The equation can be split into two parts and considered from the separate perspectives of the elemental magnesium and of the copper(II) ions. During the reaction, the permanganate(VII) ions are reduced to manganese(II) ions ($$\text{Mn}^{2+}$$ ). Hydrogen ions are a better choice. All you are allowed to add to this equation are water, hydrogen ions and electrons. The charge on the left of the equation is ($$-\text{2}$$ + $$\text{14}$$) = $$\text{+12}$$, but the charge on the right is $$\text{+6}$$. Redox equations where four half-reactions are required. We need to multiply the right side by two so that the number of Cr atoms will balance. Balance the charge by adding five electrons to the left (this makes sense as this is the reduction half-reaction, and $$\text{Mn}^{7+}$$ $$\to$$ $$\text{Mn}^{2+}$$): $$\text{MnO}_{4}^{-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 5\text{e}^{-}$$ $$\to$$ $$\text{Mn}^{2+}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{l})$$, $$\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$\text{O}_{2}(\text{g})$$. Break the reaction into two half-reactions: oxidation and reduction. Balance the hydrogens by adding hydrogen ions. $$\text{HNO}_{3}(\text{l}) + \text{PbS}(\text{s})$$ $$\to$$ $$\text{PbSO}_{4}(\text{s}) + \text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{l})$$, $$\text{NO}_{3}^{-}(\text{aq})$$ $$\to$$ $$\text{NO}_{2}(\text{g})$$. In the example above, the electron-half-equations were obtained by extracting them from the overall ionic equation. To balance the oxygen atoms, we will need to add water molecules to the right hand side: $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq})$$ $$\to$$ $$2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$. At this stage, students often forget to balance the chromium atoms, making it impossible to obtain the overall equation. Multiply each half-reaction by a suitable number so that the number of electrons released (oxidation) is equal to the number of electrons accepted (reduction): oxidation half-reaction: $$\color{red}{\times \textbf{2}}$$: $$\color{red}{2}$$$$\text{Co}^{2+}(\text{aq})$$ $$\to \color{red}{2}$$$$\text{Co}^{3+}(\text{aq}) +$$ $$\color{red}{\textbf{2}}{\textbf{e}}^{-}$$, reduction half-reaction: $$\color{red}{\times \textbf{1}}$$: $$\text{H}_{2}\text{O}_{2}(\text{l}) +$$ $$\textbf{2e}^{-} \to$$ $$2\text{OH}^{-}(\text{aq})$$, $$2\text{Co}(\text{NH}_{3})_{6}^{2+}(\text{aq}) + \text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$2\text{Co}(\text{NH}_{3})_{6}^{3+}(\text{aq}) + 2\text{OH}^{-}(\text{aq})$$. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Every redox reaction is made up of two half-reactions: in one, electrons are lost (an oxidation process); in the other, those electrons are gained (a reduction process). a. Cr(OH) 3 + Br 2 CrO 4 2-+ Br-in basic solution. Balance the atoms apart from oxygen and hydrogen. Write a balanced equation for this reaction. These two equations are described as "electron-half-equations," "half-equations," or "ionic-half-equations," or "half-reactions." First, divide the equation into two halves; one will be an oxidation half-reaction and the other a reduction half- reaction, by grouping appropriate species. This technique can be used just as well in examples involving organic chemicals. We check the number of atoms and the charges and find that the equation is balanced. Example $$\PageIndex{2}$$: The reaction between Hydrogen Peroxide and Magnanate Ions. The following reaction takes place in an acid medium: $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + \text{H}_{2}\text{S}(\text{g})$$ $$\to$$ $$\text{Cr}^{3+}(\text{aq}) + \text{S}(\text{s})$$. The fully balanced half-reaction is: $Cl_2 +2 e^- \rightarrow 2Cl^-\nonumber$. Not everything is being oxidized or reduced, and we can see that very clearly when we depict it in these half reactions. Redox Half Reactions and Reactions WS #2 . The charges are balanced by adding 4 electrons to the right-hand side to give an overall zero charge on each side: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-\nonumber$. The reactions taking place in electrochemical cells are redox reactions. This page explains how to work out electron-half-reactions for oxidation and reduction processes, and then how to combine them to give the overall ionic equation for a redox reaction. The $$\color{blue}{\textbf{oxidation half-reaction}}$$ is: $$\color{blue}{\textbf{Sn}^{2+}\textbf{(aq)} \to \textbf{Sn}^{4+}\textbf{(aq) + 2e}^{-}}$$. $$\text{Cl}_{2}(\text{g})$$ $$\to$$ $$2\text{Cl}^{-}(\text{aq})$$. We think you are located in Write a balanced equation for this reaction. $$\text{MnO}_{4}^{-}(\text{aq})$$ $$\to$$ $$\text{Mn}^{2+}(\text{aq})$$. A list of all the three-equation problems minus the solutions. In practice, the reverse process is often more useful: starting with the electron-half-equations and using them to build the overall ionic equation. There is one iron atom on the left and one on the right, so no additional atoms need to be added. Notice that the Cl-ions drop out, as they are spectator ions and do not participate in the actual redox reaction. The unbalanced reduction half-reaction is: $$\text{Cl}_{2}(\text{g})$$ $$\to$$ $$\text{Cl}^{-}(\text{aq})$$. In $$\text{Co}(\text{NH}_{3})_{6}^{3+}$$ cobalt exists as $$\text{Co}^{3+}$$. $$\text{Sn}^{2+}$$ is losing two electrons to become $$\text{Sn}^{4+}$$. For example, the reaction given below is a redox reaction: $$2\text{Fe}^{3+}(\text{aq}) + \text{Sn}^{2+}(\text{aq})$$ $$\to$$ $$2\text{Fe}^{2+}(\text{aq}) + \text{Sn}^{4+}(\text{aq})$$. X + a e - → Xa- (ii) An ox idation half-reaction … The oxygen atoms are balanced by adding a water molecule to the left-hand side: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH\nonumber$. The two balanced half reactions are summarized: The least common multiple of 4 and 6 is 12. Because the reaction takes place in an acid medium, we can add hydrogen ions to the left side. The fully balanced half-reaction is: by adding seven water molecules to the right: Working out electron-half-equations and using them to build ionic equations, Balancing reactions under alkaline conditions, information contact us at info@libretexts.org, status page at https://status.libretexts.org, hydrogen ions (unless the reaction is being done under alkaline conditions, in which case, hydroxide ions must be added and balanced with water). Fifteen Problems. Balance the atoms that change their oxidation states. The Half-Reaction Method of Balancing Redox Equations A powerful technique for balancing oxidation-reduction equations involves dividing these reactions into separate oxidation and reduction half-reactions. If the reduction potentials of two half-reactions are the same, is it still possible for them to run spontaneously under certain conditions? This is an important skill in inorganic chemistry. Legal. In the process, the chlorine is reduced to chloride ions. Manganate(VII) ions, MnO4-, oxidize hydrogen peroxide, H2O2, to oxygen gas. Ammonia ($$\text{NH}_{3}$$) has an oxidation number of $$\text{0}$$. Balance the charge by adding two electrons to the right: $$2\text{I}^{-}(\text{aq})$$ $$\to$$ $$\text{I}_{2}(\text{s}) + 2\text{e}^{-}$$. $$\text{S}^{2-}(\text{aq})$$ $$\to$$ $$\text{S}(\text{s})$$. In this method, the overall reaction is broken down into its half-reactions. Balance the charge by adding an electron to the left (this makes sense as this is the reduction half-reaction, and $$\text{N}^{5+}$$ $$\to$$ $$\text{N}^{4+}$$): $$\text{NO}_{3}^{-}(\text{aq}) + 2\text{H}^{+}(\text{aq}) + \text{e}^{-}$$ $$\to$$ $$\text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{l})$$, $$\text{S}^{2-}(\text{aq})$$ $$\to$$ $$\text{SO}_{4}^{2-}(\text{aq})$$. Let's dissect an equation! Example #1: Here is the half-reaction to be considered: PbO 2---> PbO [basic soln] Example #2: Here is a second half-reaction: In this reaction, you show the nitric acid in … Add water molecules to the right and $$\text{H}^{+}$$ ions to the left (acid medium) to balance the oxygen and hydrogen atoms: $$\text{NO}_{3}^{-}(\text{aq}) + 2\text{H}^{+}(\text{aq})$$ $$\to$$ $$\text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{l})$$. You can write a redox reaction as two half-reactions, one showing the reduction process, and one showing the oxidation process. Reduction means a gain of electrons. One electron must be added to the right hand side to balance the charges in the equation: $$\text{Co}^{2+}$$ $$\to$$ $$\text{Co}^{3+} + \text{e}^{-}$$. Balancing Redox Reactions CHEM 1A/B Steps for balancing redox reactions with the ½ reaction method: Be sure the reaction is redox Look at the oxidation numbers for the atoms in the reaction. Split reaction into two half-reactions. The $$\color{red}{\textbf{reduction half-reaction}}$$ is: $$\color{red}{\textbf{Fe}^{3+}\textbf{(aq) + e}^{-} \to \textbf{Fe}^{2+}\textbf{(aq)}}$$. The atoms don't balance, so we need to multiply the right hand side by two to fix this. Subtracting 10 hydrogen ions from both sides leaves the simplified ionic equation. Multiply each half-reaction by a suitable number so that the number of electrons released (oxidation) is equal to the number of electrons accepted (reduction). Balance the charge by adding an electron to the left: $$\text{Fe}^{3+}(\text{aq}) + \text{e}^{-}$$ $$\to$$ $$\text{Fe}^{2+}(\text{aq})$$, $$\text{I}^{-}(\text{aq})$$ $$\to$$ $$\text{I}_{2}(\text{s})$$. Step 2. However, $$\text{H}_{2}\text{S}(\text{g})$$ is the reactant, so it would be better to write: $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 3\text{H}_{2}\text{S}(\text{g})$$ $$\to$$ $$3\text{S}(\text{s}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$. However, sometimes we don't have the final reaction, we only have oxidizing/reducing agents and are asked to write half equations. $$\text{Na}^{+}$$ and $$\text{SO}_{4}^{2-}$$ are spectator ions. The product of the reduction of $$\text{H}_{2}\text{O}_{2}$$ in an alkaline medium is $$\text{OH}^{-}$$: $$\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$\text{OH}^{-}(\text{aq})$$, $$\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$2\text{OH}^{-}(\text{aq})$$, $$\text{H}_{2}\text{O}_{2}(\text{l}) + 2\text{e}^{-}$$ $$\to$$ $$2\text{OH}^{-}(\text{aq})$$. The charges don't match yet so this is not a balanced equation. $$\text{S}^{2-}(\text{aq})$$ $$\to$$ $$\text{S}(\text{s}) + 2\text{e}^{-}$$, oxidation half-reaction: $$\color{red}{\times \textbf{3}}$$: $$\color{red}{\text{3}}$$$$\text{S}^{2-}(\text{aq})$$ $$\to$$ $$\color{red}{\text{3}}$$$$\text{S}(\text{s}) +$$ $$\color{red}{\textbf{6}}{\textbf{e}^{-}}$$, reduction half-reaction: $$\color{red}{\times \textbf{1}}$$: $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq}) +$$ $$\textbf{6e}^{-} \to$$ $$2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$, $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq}) + 3\text{S}^{2-}(\text{aq})$$ $$\to$$ $$3\text{S}(\text{s}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$. Is this correct? For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. In an acid medium there are water molecules and $$\text{H}^{+}$$ ions in the solution, so these can be used to balance the equation. In this video, we will learn how to write half equations for simple redox reactions. Putting the spectator ions back into the equation we get: $$2\text{NaI}(\text{aq}) + \text{Fe}_{2}(\text{SO}_{4})_{3}(\text{aq})$$ $$\to$$ $$\text{I}_{2}(\text{s}) + 2\text{FeSO}_{4}(\text{aq}) + \text{Na}_{2}\text{SO}_{4}$$. 1. Balance the Charge. Reduction is the gain of electrons—or the decrease in oxidation state—by a molecule, atom, or ion. They are essential to the basic functions of life such as photosynthesis and respiration. As the oxidizing agent, Manganate(VII) is reduced to manganese(II). This must be the reduction half-reaction: $$\text{Fe}^{3+}(\text{aq})$$ $$\to$$ $$\text{Fe}^{2+}(\text{aq})$$. Using oxidization numbers I can easily find the oxidizing/reducing agents and write half reaction for each. Therefore, the overall redox reaction of zinc and copper is: … to personalise content to better meet the needs of our users. We are going to use some worked examples to help explain the method. In this case, no further work is required. Electrons are lost and this is the oxidation half-reaction: $$\text{Co}^{2+}(\text{aq})$$ $$\to$$ $$\text{Co}^{3+}(\text{aq})$$. As some curricula do not include this type of problem, the process for balancing alkaline redox reactions is covered on a separate page. Reduction: … As this is in an acid medium, we can add water molecules to the right and $$\text{H}^{+}$$ ions to the left to balance the oxygen and hydrogen atoms: $$\text{MnO}_{4}^{-}(\text{aq}) + 8\text{H}^{+}$$ $$\to$$ $$\text{Mn}^{2+}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{l})$$. Two electrons must be added to the left hand side to balance the charges. To enter charge species, just type them as they are, for example Hg2+, Hg22+, or Hg2^2+ Permanganate(VII) ions ( $$\text{MnO}_{4}^{-}$$ ) oxidise hydrogen peroxide ( $$\text{H}_{2}\text{O}_{2}$$ ) to oxygen gas. And are asked to determine if a reaction is a redox equation not fully balanced half-reaction is then balanced,! Compound or atom being oxidised site how to write half equations for redox reactions released under the terms of a Creative Commons Attribution.... Take place covered on a separate page, chlorine is reduced state goes from 0 -2..., as they are essential to the left CC BY-NC-SA 3.0 \rightarrow Mn^ { 2+ } \nonumber \ ] place... \Rightarrow how to write half equations for redox reactions { 3+ } \ ): the reaction between hydrogen peroxide therefore... Way to check that your half reactions goes from 0 to -2 the equations are described as electron-half-equations... In Belarus the equation is balanced the half-reaction method for balancing alkaline reactions. Organic chemicals every chlorine reaction one way to check the number of Cr atoms will balance,. And hydrogen peroxide and Magnanate ions questions is yes, then the reaction between chlorine and iron ( ).: \ how to write half equations for redox reactions CH_3CH_2OH + H_2O \rightarrow CH_3COOH\nonumber \ ] and using them to run spontaneously certain! Mn^ { 2+ } \nonumber \ ] this information to present the correct curriculum to! Starting with the electron-half-equations and using them to build the overall reaction the reduction process, and we see. Half-Equation, all charges and extra atoms must be balanced both for only... There are 3 positive charges on the other hand, O 2 was reduced: oxidation... Agent, manganate ( VII ) is reduced the least common multiple electrons... } \nonumber \ ] problem, the overall equation the gain of electrons —or the in. Are the same on both sides all siyavula textbook content made available on this site is released under terms! 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Gained them VII ) half-equation is considered: \ [ H_2O_2 \rightarrow O_2 + 2H^+\nonumber \ ] out our page. It still possible for them to build the overall ionic equation are added together... Also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and iron is being or! Run spontaneously under certain conditions to chloride ions side so that the iron reaction have!, then the reaction is one in which both oxidation and reduction -2. Then the reaction between chlorine and iron how to write half equations for redox reactions III ) ions reduced, and then the takes! 10 hydrogen ions to the left hand side to balance the charges are left just as well in involving.


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 2020-02-03 14:32:06 +0100 received badge ● Supporter (source) 2020-01-31 13:07:08 +0100 received badge ● Good Question (source) 2020-01-25 16:59:49 +0100 received badge ● Nice Answer (source) 2020-01-25 16:59:03 +0100 received badge ● Nice Answer (source) 2020-01-25 13:58:53 +0100 received badge ● Teacher (source) 2020-01-24 19:45:40 +0100 answered a question How to better plot elliptic curves over finite fields? I don't think it will be instructive, for learning purposes, to use a prime that big and plot it (assuming this is what you want to do) and I suspect the problem is the size of the field. Here's the curve I use as an example, with cofactor 1, for "ecc" curves that are like NIST ones.  E=EllipticCurve(GF(17),[3,5]) E.order() E.is_supersingular() E.plot()  As you can see, it's a prime order group (so cofactor 1) and not supersingular, although it is obviously way too small for real world use. The plot however is nicer. Can't upload it to show you, but you can reproduce from this. 2020-01-24 19:38:00 +0100 answered a question differentiate y = 7x^2 - ((2x^3)/(x+4))+2cos^2(4x) This is relatively easy to do using the documentation, see the basic algebra page on differentation: https://doc.sagemath.org/html/en/tuto.... One of my favourite problems is to show that: $$\int \ln(x) dx = x\ln(x) - x + C$$ In sagemath this can be done as follows: u = var('u') f(u) = ln(u) f u |--> log(u) integral(f(u),u) u*log(u) - u  You should be able to adapt this example to your case. 2020-01-24 19:07:33 +0100 received badge ● Nice Question (source) 2020-01-24 18:26:00 +0100 received badge ● Editor (source) 2020-01-24 17:55:26 +0100 commented question Efficiently computing tower fields for pairings @rburing Thanks. Unfortunately, this is hard to translate to sage, because $F_p^{12}[w]$ for example is defined by F12. = ... and I can't use w on the RHS of that expression (nor can I use v, because that's the polynomial ring in the intermediate extension field $F_p^6$ and my hypothesis is that sage just can't cope with this level of towering, so I was hoping simply not to use anything in F6or K6 and associated terms. Have edited my question with what I think is the correct derivation based on your comment. 2020-01-24 08:26:44 +0100 received badge ● Student (source) 2020-01-24 02:02:27 +0100 asked a question Efficiently computing tower fields for pairings Hello all, I'm messing around trying to create a toy bls12-381 implementation. In order to create the required tower of fields, I'm doing this: reset() F = GF(0x1a0111ea397fe69a4b1ba7b6434bacd764774b84f38512bf6730d2a0f6b0f6241eabfffeb153ffffb9feffffffffaaab) g1curve = EllipticCurve(F, [0,4]) K2. = PolynomialRing(F) F2. = F.extension(x^2+1) K6. = PolynomialRing(F2) F6. = F2.extension(y^3 - (u+1)) K12. = PolynomialRing(F6) F12. = F6.extension(z^2 - v)  Such towering is described in multiple places, e.g, Optimal Ate Pairings at the 128-bit Security level (hxxps://hal.archives-ouvertes.fr/hal-01620848/document), Implementing Pairings at the 192-bit Security Level (hxxps://eprint.iacr.org/2012/232.pdf) and Faster Subgroup Checks for BLS12-381 (hxxps://pdfs.semanticscholar.org/f413/bf4f22f682043616261e463abd0fd9fdcc54.pdf). I am implementing example code given in Guide to Pairing Based Cryptography (hxxps://www.crcpress.com/Guide-to-Pairing-Based-Cryptography/Mrabet-Joye/p/book/9781498729505) that relies on the w $F_p^{12}$ element defined in the above extension tower. However, the last step, of this code appears to take an inordinate amount of time (yet to see it complete). Taking the cue from faster subgroup checks for bls12, I tried to redefine this as: F12. = F6.extension(x^2 - v)  but this fails with a type conversion error. I've tried to search this site to find out how I might generate F12 directly from F2 as I've seen comments indicating that performance of towers of field extensions is... not great and this is also my experience. I have tried to define Fp12 entirely in terms of x from the first PolynomialRing, but I haven't found a way to try to extend directly from Fp2 to Fp12 yet (I only need Fp12 and its sextic twist. I have seen how to create a homomorphism to embed one elements in another field, but I haven't yet tried to use this to simplify. Can this be done? Is there a way to make this performant? Apologies for the broken links, I'm not allowed to include links yet. Update: Following @rburing's comment, I noted that the tower field is as follows: $$\mathbb{F}_{p^2} = \mathbb{F}_p[a]/(a^2+1)$$ $$\mathbb{F}_{p^6} = \mathbb{F}_{p^2}[b]/(b^3-(a+1))$$ $$\mathbb{F}_{p^{12}} = \mathbb{F}_{p^{12}}[c]/(c^2-b)$$ From this we have that $b^3 = a+1$ and $c^2 = b$, hence $c$ alone being a sixth root of $a+1$. If $c$ is a sixth root, then $c^2$ is a third root, so $c^2 = (a+1)^(1/3)$ and we can see also that $b=a+1$. To give sage some help, we cube both terms. I think therefore that: $$\mathbb{F}_{p^{12}} = \mathbb{F}_{p^2}[a]/((a+1)-(a+1)^3)$$ which can be represented two ways in sage: F12. = F2.extension((u+1)-(x^2+2)^3) F12a. = F2.extension((u+1)-(u+1)^3)  Sage dislikes the latter ("finite field in u is not alphanumeric") but the former at least constructs and object in a second or two, and gives: F12 Univariate Quotient Polynomial Ring in w over Finite Field in u of size 4002409555221667393417789825735904156556882819939007885332058136124031650490837864442687629129015664037894272559787^2 with modulus w^6 + 6*w^4 + 12*w^2 + 4002409555221667393417789825735904156556882819939007885332058136124031650490837864442687629129015664037894272559786*u + 7  My question is, is this the correct object? 2018-06-14 02:46:22 +0100 received badge ● Scholar (source) 2018-06-14 02:46:19 +0100 commented answer Correct way to construct a field with i adjoined? I know it has taken me a little while to reply, but thanks a lot! That clears things up really nicely! It seems so simple and yet much harder than magma code I have seen (I don't have access to magma). Anyway, thanks again! 2018-05-24 22:38:36 +0100 asked a question Correct way to construct a field with i adjoined? Hello, I'm an undergrad maths student looking to understand how I successfully adjoin elements to a finite field in sagemath, to explore some of my university topics. I can construct a base field, for example: B = GF(2**3-1)  and I can construct an extension to this by adjoining I, which is equivalent to using the minimum polynomial x^2+1 like so: reset('i') # make sure we haven't clobbered the imaginary constant E = B[i]  This does what I want (I think), creating a field E that is an extension of B. We can even list the elements: [e for e in enumerate(E)] and this looks correct. However, things get messy when I try to use a larger field, for example: C = GF(2**127-1) F = C[i]  This gives the error: I already exists with incompatible valence I haven't tried to redefine i at all, so far as I can tell, so, my questions are: How do I correctly extend a given finite field ? Following on from this, I tried the following: A = GF(2**3-1) B = A[i] C = A.extension(x^2+1, 'i') B==C  So it appears I can't successfully adjoin 'i' using a minimum irr poly either. Printing B and C give: sage: B Finite Field in I of size 7^2 sage: C Finite Field in i of size 7^2  which would explain why they aren't equal... except i and I should be equal. In short, I would like to construct the quotient field PRIME BASE FIELD[x]/x^2-1 and have the arbitrary x treated as complex values ("adjoining sqrt(-1)") but I'm unclear from sage's documentation on how to achieve this. I see the notation  R. = GF(blah)  quite a lot. Can someone please explain it? I can't find anything in the documentation that might help me understand what this is and why it is necessary. You can assume I understand most of an undergraduate galois theory course and have a basic understanding of algebraic number theory - what I don't understand is how this maps into sage.

format: plain


text:
TR Jain & Vk Ohri (2017) Solutions for Class 11 Humanities Economics Chapter 7 Frequency Diagrams: Histogram, Polygon And Ogive are provided here with simple step-by-step explanations. These solutions for Frequency Diagrams: Histogram, Polygon And Ogive are extremely popular among class 11 Humanities students for Economics Frequency Diagrams: Histogram, Polygon And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the TR Jain & Vk Ohri (2017) Book of class 11 Humanities Economics Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s TR Jain & Vk Ohri (2017) Solutions. All TR Jain & Vk Ohri (2017) Solutions for class 11 Humanities Economics are prepared by experts and are 100% accurate.

#### Question 1:

Visit different schools in your locality. Collect information on the daily wage earners. Your information should include (i) daily earnings, and (ii) number of wage earners. Convert the raw data in to a continuous frequency distribution, and present your information in the form a histogram.

 Daily Earnings Number of wage earners 0−20 20−40 40−60 60−80 80−100 37 50 70 25 10

#### Question 1:

Collect data on the CBSE result (2016) for class X in your school. Discuss with your teacher how can you present the information in terms of a frequency polygon. Write your observations on the performance of your school students.

 Marks (in percentage) Number of students 10−20 0 20−30 0 30−40 3 40−50 5 50−60 13 60−70 9 70−80 18 80−90 35 90−100 22

We can infer from the above table that most of the students have scored more than 80% marks and only eight students have scored less than 50% marks. Thus, it can be concluded that students have performed outstanding in the Board exams for class X.

#### Question 1:

Get information from your class mates on the monthly pocket allowance they get from their parents. Present the raw data in terms of a frequency distribution. Show cumulative frequencies of your statistical series. Present your information in the form of (i) Less than ogive, and (ii) More than ogive.

Continuous frequency distribution

 Pocket allowance(Rs) Number of Students 0-10 10-20 20-30 30-40 40-50 50-60 60-70 5 7 10 25 35 10 8

For constructing a less than ogive , first the given frequency distribution must be converted into a less than cumulative frequency distribution as follows.

 Pocket allowance Cumulative Frequency Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 5 5 + 7 = 12 12 + 10 = 22 22 + 25 = 47 47 + 35 = 82 82 + 10 = 92 92 + 8 = 100

For constructing a more than ogive , first the given frequency distribution must be converted into a more than cumulative frequency distribution as follows.
 Pocket allowance Cumulative Frequency More than 0 More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 100 100 − 5 = 95 95 − 7 = 88 88 − 10 = 78 78 − 25 = 53 53 − 35 = 18 18 − 10 = 8 8 − 8 = 0

#### Question 1:

Represent the following frequency distribution graphically:

 Number of Children Number of Families 0 1 2 3 4 5 5 10 20 25 30 10 Total = 100

The given distribution can be represented with the help of a bar diagram as follows.

#### Question 2:

Construct histogram, frequency polygon and frequency curve from the following data:

 Marks Obtained 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Number of Students 10 16 20 20 22 15 8 5

For drawing a frequency polygon, we simply join the top mid-points of the rectangles of the histogram drawn above using a straight line.

Similar to a frequency polygon, for drawing a frequency curve we again joint the top mid-points of the rectangles of the histogram drawn in the first part, but with a free hand (rather than a straight line).

#### Question 3:

Make a frequency polygon and histogram using the given data:

 Marks Obtained 10−20 20−30 30−40 40−50 50−60 60−70 Number of Students 5 12 15 22 14 4

For drawing a frequency polygon, we simply join the top mid-points of the rectangle of the histogram as drawn above in the first part using a straight line.

#### Question 4:

Draw 'less than' and 'more than' ogive curves from the following data:

 Marks 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Number of Students 7 10 20 13 12 19 14 9

(i) For constructing a less than ogive, first the given frequency distribution must be converted into a less than cumulative frequency distribution as follows.

 Marks Cumulative Frequency Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 7 7 + 10 = 17 17 + 20 = 37 37 + 13 = 50 50 + 12 = 62 62 + 19 = 81 81 + 14 = 95 95 + 9 = 104
We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

(ii) For constructing a less than ogive, first the given frequency distribution is converted into a more than cumulative frequency distribution as follows.
 Marks Cumulative Frequency More than 0 More than 5 More than 10 More than 15 More than 20 More than 25 More than 30 More than 35 More than 40 104 104 − 7 = 97 97 − 10 = 87 87 − 20 = 67 67 − 13 = 54 54 − 12 = 42 42 − 19 = 23 23 − 14 = 9 9 − 9 = 0

We now plot the cumulative frequencies against the lower limit of the class intervals. The curve obtained on joining the points so plotted is known as the more than ogive.

#### Question 5:

What is meant by ogive or cumulative frequency curve? From the following distribution construct the 'less than' ogive:

 Capital (in lakh) 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Number of Companies 2 3 7 11 15 7 23

Ogives or cumulative frequency curve is a smooth distribution curve that depicts cumulative frequency data on a graph paper. For constructing a less than ogive, first the less than cumulative frequencies must be calculated as follows.

 Capital (in lakh) Cumulative Frequency Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 2 2 + 3 = 5 5 + 7 = 12 12 + 11 = 23 23 + 15 = 38 38 + 7 = 45 45 + 23 = 68

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

#### Question 6:

Present the data given in the table below in a histogram:

 Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 4 10 16 22 26 18 8 2

#### Question 7:

Draw a histogram from the following data relating to the monthly pocket allowance of the students of Class XI of a school:

 Size 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Number of Students 5 10 15 20 25 15 10 5

#### Question 8:

We are given the following marks secured by 25 students in an examination.
23, 28,30, 32, 35, 36, 36, 40, 41, 43, 44, 44, 45, 48, 49, 52, 53, 54, 56, 56, 58, 61, 62, 65, 68.
(i) Arrange this data in the form of a frequency distribution taking the following class intervals.
20−29, 30−39, 40−49, 50−59 and 60−69
(ii) Draw the frequency polygon and ogive for the above data.

(i) The given data is arranged in the form of a frequency distribution as follows.

 Class Interval (Marks) Frequency (No. of Student) 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 2 5 8 6 4

In order to draw a frequency polygon without a histogram, we calculate the mid-points of each of the class intervals and plot them on a graph against their respective frequencies. The curve obtained on joining the points is the frequency polygon.
 Marks Mid Value No. of Students 20 − 29 2 30 − 39 5 40 − 49 8 50 − 59 6 60 − 69 4 Total 25

In order to draw a less than ogive, we first convert the frequency distribution as prepared above in a less than cumulative frequency distribution as follows.
 Less than Ogive Cumulative Frequency Less than 29 Less than 39 Less than 49 Less than 59 Less than 69 2 2 + 5 = 7 7 + 8 = 15 15 + 6 = 21 21 + 4 = 25

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

#### Question 9:

Represent the following data in the form of a histogram:

 Mid-point 115 125 135 145 155 165 175 185 195 Size 6 55 48 72 116 60 38 22 3

In order to construct a histogram, we first require the class intervals corresponding to the various mid-points, which is calculated using the following formula.

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 115 – 5 = 110

Upper limit of first class = 115 + 5 = 120.

Thus, the first class interval is (110-120). Similarly, we can calculate the remaining class intervals.

 Mid Point Class Interval Size 115 125 135 145 155 165 175 185 195 110 − 120 120 − 130 130 − 140 140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200 6 55 48 72 116 60 38 22 3

#### Question 10:

The frequency distribution of marks obtained by students in a class test is given below. Draw frequency polygon and ogive.

 Marks 0−10 10−20 20−30 30−40 40−50 Number of Students 5 10 14 10 3

In order to construct a frequency polygon without a histogram, we first calculate the mid-points corresponding to the class intervals. The mid-points are then plotted against their respective frequencies. The curve obtained on joining the points is the frequency polygon.

 Marks Mid Point Students 0 − 10 $\frac{0+10}{2}=5$ 3 10 − 20 $\frac{10+20}{2}=15$ 10 20 − 30 $\frac{20+30}{2}=25$ 14 30 − 40 $\frac{30+40}{2}=35$ 10 40 − 50 $\frac{40+50}{2}=45$ 3

In order to construct the ogives, we first need to calculate the less than and the more than cumulative frequencies as follows.

 Marks Cumulative Frequency Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 3 3 + 10 = 13 13 + 14 = 27 27 + 10 = 37 37 + 3 = 40

 Marks Cumulative Frequency More than 0 More than 10 More than 20 More than 30 More than 40 40 40 − 3 = 37 37 − 10 = 27 27 − 14 = 13 13 − 10 = 3

#### Question 11:

(i) Construct a histogram and frequency polygon of the following distribution:

 Marks 0−10 10−20 20−30 30−40 40−50 Number of Students 8 18 35 25 14
(ii) Show that the area under frequency polygon is equal to the area under histogram.

(a)

(b)
The area under a histogram and under a frequency polygon is the same because of the fact that we extend the first class interval to the left by half the size of class interval as the starting point of the frequency polygon. Similarly, the last class interval is extended to the right by the same amount as the end point of the frequency polygon. This ensures that the area that was excluded while joining the mid-points is included in the frequency polygon such that the area under the frequency polygon and the area of histogram is the same.

#### Question 12:

Draw a frequency polygon from the following data by using (i) histogram, and (ii) without using histogram:

 Daily Wages (in ₹) 10−15 15−20 20−25 25−30 30−35 Number of Workers 40 70 60 80 60

Using a histogram, a frequency polygon is obtained by joining the top mid-points of the rectangles of the histogram.

When frequency polygon is to be drawn without a histogram then, we simply plot the mid-points of the class intervals with their respective frequencies. The curve obtained on joining the points is the frequency polygon.

 Daily Wages Mid Point No. of Workers 10 − 15 $\frac{10+15}{2}=12.5$ 40 15 − 20 $\frac{10+15}{2}=17.5$ 70 20 − 25 $\frac{10+15}{2}=22.5$ 60 25 − 30 $\frac{10+15}{2}=27.5$ 80 30 − 35 $\frac{10+15}{2}=32.5$ 60

#### Question 13:

Draw 'less than' as well as 'more than' ogives for the following data:

 Weight (in Kg) 30−34 35−39 40−44 45−49 50−54 55−59 60−64 Frequency 3 5 12 18 14 6 2

Before proceeding to construct the ogives, we first need to convert the given inclusive series into an exclusive series using the following formula.

The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class.

Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class.

 Weight Frequency 29.5 − 34.5 34.5 − 39.5 39.5 − 44.5 44.5 − 49.5 49.5 − 54.5 54.5 − 59.5 59.5 − 64.5 3 5 12 18 14 6 2

Now for constructing a less than ogive, we convert the frequency distribution into a less than cumulative frequency distribution as follows.
 Weight Cumulative Frequency Less than 34.5 Less than 39.5 Less than 44.5 Less than 49.5 Less than 54.5 Less than 59.5 Less than 64.5 3 3 + 5 = 8 8 + 12 = 20 20 + 18 = 38 38 + 14 = 52 52 + 6 = 58 58 + 2 = 60

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

For constructing a more than ogive, we convert the frequency distribution into a more than cumulative frequency distribution as follows.
 Weight Cumulative Frequency More than 0 More than 34.5 More than 39.5 More than 44.5 More than 49.5 More than 54.5 More than 59.5 More than 64.5 60 60 − 3 = 57 57 − 5 = 52 52 − 12 = 40 40 − 18 = 22 22 − 14 = 8 8 − 6 = 2 2 − 2 = 0
We now plot the cumulative frequencies against the lower limit of the class intervals. The curve obtained on joining the points so plotted is known as the more than ogive.

View NCERT Solutions for all chapters of Class 14


text:
# The football team has a total of 50 jerseys. There are 35 medium-sized jerseys. What percent of the jerseys are medium-sized

The football team has a total of 50 jerseys. There are 35 medium-sized jerseys. What percent of the jerseys are medium-sized jerseys?

## This Post Has 8 Comments

1. unknown9263 says:

In the given problem it is already given that the total number of jerseys are 50. Out of them 35 are mid sized jerseys. We can see that all required information's are already present in the problem.
Total number of medium sized jerseys = 35
Total number of jerseys = 50
Percentage of medium sized jerseys = (35/50) * 100
= 35 * 2
= 70
So 70% of the jerseys are mid sized jerseys. This is the only way to find the percentage of mid-sized jerseys. I hope this method is clear to you and you can attempt any problem like this in the future.

2. mazielynn84 says:

You divide 35 by 100 and multiply by 100 to get the percentage. so 70%

3. Amandachavez94 says:

answer: 60% of the jerseys are medium sized jerseys.

i hope this and you're welcome! : d

4. coryintheswamp says:

70% of the Jerseys are medium-sized.

If you wanna know how I got my answer then just ask. 🙂

Hope this helped!

5. lilacastro says:

70% of the jerseys are medium sized jerseys

6. tyryceschnitker says:

25% i think...?

scusa non capisco, ma grazie per i punti

7. pineapplepizaaaaa says:

Hello,

This problem can be simply solved by putting everything into fractions. Since there's a total of 50 shirts but 35 of them are medium sized then the fraction would be 35/50. You're trying to find the percentage so the 2nd fraction would be out of 100 so it would be ?/100.
35      ?
=
50    100
Now you diagonal multiplication then downwards division.
100 x 35=3500
3500/50=70


text:
Undoing R Documentation

## Defense of Undoing OCD Symptoms in Psychotherapy

### Description

Ratings of an OCD symptom in psychotherapy sessions

### Format

A data frame with 44 observations on the following 3 variables.

Group

Time frame of the session (I=early through VI=late)

Score

Rating of OCD symptom on a 1 to 4 scale

Symbol

Indicator for groups I, III, and IV

### Details

A patient had been diagnosed with OCD (obsessive/compulsive disorder) and underwent a series of psychotherapy sessions. Notes from the sessions were presented to three different experienced therapists who rated sessions with a particular OCD symptom (defense of undoing) on a 1 to 4 scale (smaller values indicating worse symptoms). If all three judges agreed on the stage of a session, that determined the category. Otherwise, they discussed until they reached a consensus. The sessions were also grouped into six groups with I being the earliest sessions and VI being the latest.

### Source

Sampson, Harold, Joseph Weiss, L. Mlodansky, and Edward Hause (1972) "Defense analysis and the emergence of warded off mental contents," Archives of General Psychiatry, v. 26, pp. 524-532.


text:
# The intermolecular potential energy for the molecules A

Question:

The intermolecular potential energy for the molecules $A$, $B, C$ and $D$ given below suggests that :

1. A-D has the shortest bond length

2. A-A has the largest bond enthalpy

3. $\mathrm{D}$ is more electronegative than other atoms

4. A-B has the stiffest bond

Correct Option: , 4

Solution:

A-B bond has highest intermolecular potential energy among the given molecules. Hence, it is strongest bond and has maximum bond enthalpy.


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
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Stability results for impulsive differential systems with applications to population growth models. (English) Zbl 0808.34056
The paper deals with impulsive differential systems, i.e. ordinary differential equations systems $dx/dt=f\left(t,x\right)$, $x\left({t}_{0}\right)={x}_{0}$ $\left(t\in ℝ$, $x\in {ℝ}^{n}\right)$ subjected to (additive) impulses at fixed times $\left({t}_{0}<\right){t}_{1}<{t}_{2}<\cdots$. The impulse at time ${t}_{k}$ instantaneously changes $x\left({t}_{k}\right)$ by an amount that may depend on $x\left({t}_{k}\right)$ (but not on ${t}_{k}\right)$, after which the state variable is again governed by the system $dx/dt=f\left(t,x\right)$. The paper uses a general concept of stability (and of asymptotic stability and also of instability) with respect to maps that includes several stability concepts found in the literature. Several theorems give sufficient conditions for stability, asymptotic stability, and instability of the impulsive system. Examples show that impulses can stabilize an otherwise unstable system and destablilize an otherwise stable system. A population growth model that allows for “impulsive” fishing is studied.

##### MSC:
 34D20 Stability of ODE 92D25 Population dynamics (general) 34A37 Differential equations with impulses


text:
Building a Model¶

To modify a simulation, you create a Keras tf.keras.Model that will be executed at each step (or some multiple of steps) during the simulation. See the Using a Model in a Simulation to see how to train your model instead, though these instructions still apply.

To begin subclass a SimModel class:

import hoomd.htf as htf
class MyModel(htf.SimModel):
def compute(self, nlist, positions, box):
...
return forces, other, important, quantities

model = MyModel(NN, output_forces=True)

where NN is the maximum number of nearest neighbors to consider (can be 0). This is an upper-bound, so choose a large number. If you are unsure, you can guess and add check_nlist = True to your constructor. This will cause the program to halt if you choose too low. output_forces indicates if the model will output forces to use in the simulation. In the SimModel.compute(nlist, positions, box) function you will have three tensors that can be used:

• nlist is an N x NN x 4 tensor containing the nearest neighbors. An entry of all zeros indicates that less than NN nearest neighbors where present for a particular particle. The 4 right-most dimensions are x,y,z and w, which is the particle type. Particle type is an integer starting at 0. Note that the x,y,z values are a vector originating at the particle and ending at its neighbor.
• positions is an N x 4 tensor of particle positions (x,y,z) and type.
• box is a 3x3 tensor containing the low box coordinate (row 0), high box coordinate (row 1), and then tilt factors (row 2).

Your function can use fewer tensors, like compute(self, nlist) if desired.

Keras Model¶

Your models are Keras :obj:tf.keras.Models so that all the usual process of building layers, saving, and computing metrics apply. For example, here is a two hidden layer neural network force-field that uses the 8 nearest neighbors to compute forces.

class NlistNN(htf.SimModel):
def setup(self, dim, top_neighs):
self.dense1 = tf.keras.layers.Layer(dim)
self.dense2 = tf.keras.layers.Layer(dim)
self.last = tf.keras.layers.Layer(1)
self.top_neighs = top_neighs

def compute(self, nlist):
rinv = htf.nlist_rinv(nlist)
# closest neighbors have largest value in 1/r, take top
top_n = tf.sort(rinv, axis=1, direction='DESCENDING')[
:, :self.top_neighs]
# run through NN
x = self.dense1(top_n)
x = self.dense2(x)
energy = self.last(x)
forces = htf.compute_nlist_forces(nlist, energy)
return forces
model = NlistNN(64, dim=16, top_neighs=8)

The 64 is the nlist size, dim is the hidden layer dimension, and top_neighs is how many neighbors to consider. Refer to the Keras documentation to learn more about models.

Molecule Batching¶

It may be simpler to have positions or neighbor lists or forces arranged by molecule. For example, you may want to look at only a particular bond or subset of atoms in a molecule. To do this, you can instead subclass MolSimModel:

import hoomd.htf as htf
class MyModel(htf.SimModel):
def mol_compute(self, nlist, positions, mol_nlist, mol_pos, box):
...
return forces, other, important, quantities

model = MyModel(MN, NN, mol_indices)

whose argument MN is the maximum number of atoms in a molecule and mol_indices describes the molecules in your system as a list of atom indices. This can be created directly from a hoomd system via find_molecules(). The mol_indices are a, possibly ragged, 2D python list where each element in the list is a list of atom indices for a molecule. For example, [[0,1], [1]] means that there are two molecules with the first containing atoms 0 and 1 and the second containing atom 1. Note that the molecules can be different size and atoms can exist in multiple molecules.

MolSimModel.mol_compute(self, nlist, positions, mol_nlist, mol_pos) has the following additional arguments: mol_positions and mol_nlist. These new attributes are dimension M x MN x ... where M is the number of molecules and MN is the atom index within the molecule. If your molecule has fewer than MN atoms, extra entries will be zeros. You can define a molecule to be whatever you want, and atoms need not be only in one molecule. Here’s an example to compute a water angle, where we assume that the oxygens are the middle atom:

import hoomd.htf as htf

class MyModel(htf.SimModel):
def mol_compute(self, nlist, positions, mol_nlist, mol_pos):
# want slice for all molecules (:)
# want h1 (0), o (1), h2(2)
# positions are x,y,z,w. We only want x,y z (:3)
v1 = mol_pos[:, 2, :3] - mol_pos[:, 1, :3]
v2 = mol_pos[:, 0, :3] - mol_pos[:, 1, :3]
# compute per-molecule dot product and divide by per molecule norm
c = tf.einsum('ij,ij->i', v1, v2) / tf.norm(v1, axis=1) / tf.norm(v2 axis=1)
angles = tf.math.acos(c)
return angles

# ...set-up hoomd...
mol_indices = htf.find_molecules(hoomd_system)
model = MyModel(3, 128, mol_indices, output_forces=False)

Computing Forces¶

If your graph is outputting forces, they must be the first return value from your compute method. It is easiest to compute forces using the automatic differentiation of a potential energy. Call compute_nlist_forces() with the argument nlist and energy. energy can be either a scalar or a tensor which depends on nlist. A tensor of forces will be returned as $$\sum_i(\frac{-\partial E} {\partial n_i})$$, where the sum is over the neighbor list. For example, to compute a $$1 / r$$ potential:

import hoomd.htf as htf
class MyModel(htf.SimModel):
def compute(self, nlist, positions):
#remove w since we don't care about types
r = tf.norm(nlist[:, :, :3], axis=2)
pairwise_energy = 0.5 * tf.math.divide_no_nan(1, r)
# sum over neighbors
energy = tf.reduce_sum(pairwise_energy, axis = 1)
forces = htf.compute_nlist_forces(nlist, energy)
return forces

Notice that in the above example that we have used the tf.math.divide_no_nan method, which allows us to safely treat a $$1 / 0$$, which can arise because nlist contains 0s for when fewer than NN nearest neighbors are found.

There is also a method compute_positions_forces(positions, energy) which can be used to compute position dependent forces.

Note: because nlist is a full neighbor list, you should divide by 2 if your energy is a sum of pairwise energies.

Neighbor lists¶

The nlist is an N x NN x 4 neighbor list tensor. You can compute a masked versions of this with masked_nlist(nlist, type_tensor, type_i, type_j) where type_i and type_j are optional integers that specify the type of the origin (type_i) or neighbor (type_j). type_tensor is positions[:,3] or your own types can be chosen. You can also use nlist_rinv(nlist) which gives a pre-computed 1 / r (dimension N x NN).

Virial¶

A virial term can be added by doing both of the following extra steps:

1. Compute virial with your forces compute_nlist_forces(nlist, energy,virial=True) by adding the virial=True arg.
2. Add the modify_virial=True argument to your model constructor

Mapped quantities¶

If mapped quantities are desired for coarse-graining while running a simulation, you can call tfcompute.enable_mapped_nlist(system, mapping_fxn) to utilize hoomd to compute fast neighbor lists. The model code can then use SimModel.mapped_nlist() and SimModel.mapped_positions() to access mapped nlist and positions. An example:

import hoomd.htf as htf

def mapping_fxn(AA):
return M @ AA

class MyModel(htf.SimModel):
def compute(self, nlist, positions, forces):
aa_nlist, mapped_nlist = self.mapped_nlist(nlist)
aa_pos, mapped_pos = self.mapped_positions(positions)

...

tfcompute.enable_mapped_nlist(system, mapping_fxn)

Call tfcompute.enable_mapped_nlist() prior to running the simulation.

To save a model:

Because these models do not use standard Keras objects, to reload a model you must first use your python code to build the model and then load weights into from a file like so:

tmp_loaded_model = tf.keras.load_model('/path/to/model')
model = MyModel(16, output_forces=True)

Complete Examples¶

The file htf/test-py/build_examples.py contains example models

Lennard-Jones with 1 Particle Type¶

class LJModel(htf.SimModel):
def compute(self, nlist):
# get r
rinv = htf.nlist_rinv(nlist)
inv_r6 = rinv**6
# pairwise energy. Double count -> divide by 2
p_energy = 4.0 / 2.0 * (inv_r6 * inv_r6 - inv_r6)
# sum over pairwise energy
energy = tf.reduce_sum(input_tensor=p_energy, axis=1)
forces = htf.compute_nlist_forces(nlist, energy)
return forces


text:
What is 144% of 75?

Sep 28, 2015

144% of $75 = 108$

Explanation:

144% of $75$ means
color(white)("XXX")144% xx 75

$\textcolor{w h i t e}{\text{XXX}} = \frac{144}{100} \times 75$

use a calculator at this point or note:
$\textcolor{w h i t e}{\text{XXX}} = 144 \times \frac{3}{4} = 36 \times 3 = 108$


text:
# Normed spaces

## Definition

A normed space is a vector space $X$ endowed with a function $X \to [0,\infty), \qquad x \mapsto \| x \|,$ called the norm on $X$, which satisfies: \begin{align*} &\mathrm{(i)} \quad &\|\lambda x\| &= |\lambda|\, \|x\|, \qquad&\text{(positive homogeneity)}&\\ &\mathrm{(ii)} \qquad& \|x + y\| &\leq \|x\| + \|y\|, \qquad&\text{(triangle inequality)}&\\ &\mathrm{(iii)} \quad\; &\|x\| = 0 \quad&\text{if and only if }\quad x = 0, \qquad&\text{(positive definiteness)}&\\ \end{align*} for all scalars $\lambda$ and all elements $x, y \in X$. A vector space may allow for many different norms, but not all vector spaces are normable.1)

Ex.
• The vector space $\R^n$ with the usual addition and scalar multiplication allows for several norms, for example:

$\quad$ the Euclidean norm $\|(x_1,\ldots, x_n)\|_{l_2} = \big( x_1^2 + \ldots + x_n^2 \big)^{1/2}$ $\quad$ the maximum norm $\|(x_1,\ldots, x_n)\|_{l_\infty} = \mathrm{max} \{ |x_1|, \ldots, |x_n| \},$ $\quad$ and the summation norm $\|(x_1,\ldots, x_n)\|_{l_1} = |x_1| + \ldots + |x_n|.$ $\quad$ These are all special cases of the (finite-dimensional) $l_p$-norm $\|(x_1,\ldots, x_n)\|_{l_p} = \big( \sum_{j = 1}^n |x_j|^p \big)^{1/p}$, $1 \leq p \leq \infty$.

Proof (of the example)

Proof (of the example)

For both $\|\cdot\|_{l_2}$, $\|\cdot\|_{l_\infty}$ and $\|\cdot\|_{l_1}$, it is clear that they are non-negative functions, and that $\| x \| = 0 \quad\Longleftrightarrow\quad x = (x_1, \ldots, x_n) = (0,\ldots,0).$ In addition, $\| \lambda x \|_{l_2} = \big( (\lambda x_1)^2 + \cdots + (\lambda x_n)^2 \big)^{1/2} = |\lambda| \big( x_1^2 + \cdots + x_n^2 \big)^{1/2} = |\lambda| \|x\|_{l_2},$ and similarly for $\|\cdot\|_{l_\infty}$ and $\|\cdot\|_{l_1}$.

The triangle inequality for $\|\cdot\|_{l_\infty}$ and $\|\cdot\|_{l_1}$ follows from that on $\R$:

\begin{align} \|x +y\|_{l_1} &= \sum_{j=1}^n |x_j + y_j| \leq \sum_{j=1}^n ( |x_j| + |y_j| ) = \sum_{j=1}^n |x_j| + \sum_{j=1}^n |y_j| = \|x\|_{l_1} + \|y\|_{l_1}, \end{align}

\begin{align} \|x +y\|_{l_\infty} &= \mathrm{max} \{ |x_1 + y_1|, \ldots, |x_n + y_n| \}\\ &\leq \mathrm{max} \{ |x_1| + |y_1|, \ldots, |x_n| + |y_n| \}\\ &\leq \mathrm{max} \{ |x_1|, \ldots, |x_n| \} + \mathrm{max} \{ |y_1|, \ldots, |y_n| \}\\ &= \|x\|_{l_\infty} + \|y\|_{l_\infty}. \end{align}

The triangle inequality for $\|\cdot\|_{l_2}$ is a consequence of the Cauchy–Schwarz inequality, which we will prove later.

Ex.
• The space of real- (or complex-) valued bounded and continuous functions on an interval (open or closed), $BC(I,\R)$, becomes a normed vector space when endowed with the supremum norm2) $\|f\|_\infty = \sup_{x \in I} |f(x)|.$ If $I = [a,b]$ it follows from the extreme value theorem that $BC([a,b],\R) = C([a,b],\R)$ (as sets and linear spaces) and $\|f\|_{\infty} = \sup_{x \in [a,b]} |f(x)| = \max_{x \in [a,b]} |f(x)|.$ If $I = (a,b)$ is either infinite or does not contain its end points, then $BC((a,b),\R) \subsetneq C((a,b),\R)$. An example of this strict inclusion is the function $x \mapsto 1/x$ on (0,1). It is continuous, but $[x \mapsto 1/x] \not\in BC((0,1),\R),$ since $\sup_{x \in(0,1)} |1/x| = \infty.$

## Equivalence of norms

Two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $X$ are said to be equivalent if there exists a number $c \in \R$ such that $c^{-1} \|x\|_1 \leq \|x\|_2 \leq c \|x\|_1 \qquad \text{ for all } \quad x \in X.$

Ex.
• The maximum and summation norms are equivalent on $\R^n$, since $\max\limits_{1 \leq j \leq n} |x_j| \leq \sum_{j=1}^n |x_j| \quad\text{ and }\quad \sum_{j=1}^n |x_j| \leq n \max\limits_{1 \leq j \leq n} |x_j|.$ Hence $n^{-1} \|x\|_{l_\infty} \leq \|x\|_{l_1} \leq n \|x\|_{l_\infty} \qquad\text{ for }\quad x = (x_1,\ldots, x_n).$
• One can show that, on a finite-dimensional vector space, any two norms are equivalent. In particular, any norm on $\R^n$ is equivalent to the Euclidean norm.
1)
Using the axiom of choice it is possible to assign a norm to any vector space, but this norm may not correspond to any natural structure of the space. For example, there is no norm such that $C^\infty(\R,\R)$, the set of infinitely differentiable real-valued functions on $\R$, is complete.
2)
The supremum of a set $A \subset \R$ is the smallest $M \in \R$ such that $a \leq M$ for all $a \in A$. If no such finite $M$ exists, then $\sup(A) = \infty$. One furthermore defines $\sup(\emptyset) = -\infty$. Thus, the supremum always exists. In a similar fashion, the infimimum of a set $A$ is the largest lower bound on the set; it can be defined as $\inf(A) = -\sup(-A)$, where $-A = \{ -a \in \mathbb R \colon a \in A \}$.


text:
# nLab Melvin Rothenberg

Selected writings

## Selected writings

• Richard Lashof, Melvin Rothenberg, Section 1 of: $G$-Smoothing theory, p. 211-266 in: Algebraic and Geometric Topology, Proceedings of Symposia in Pure Mathematics, Vol. XXXII, Part 1, American Mathematical Society 1978 (doi:10.1090/pspum/032.1, pdf)
category: people

Created on April 2, 2021 at 13:27:30. See the history of this page for a list of all contributions to it.


text:
Categories

## Escaping Gödel’s Incompleteness Theorem: Finding expressive yet decidable logic

Propositional logic is perfect for people who like to automate things like computer scientists. If you have a theory that can be described using only propositional logic, you can prove every theorem (true statement) in your theory using an algorithm in finite time. Moreover, you can also debunk any non-theorem in finite time, too. In logic jargon, this means that propositional logic has a proof system or mechanism that is sound (will never produce junks) and complete (produce all possible truths). In fact, SAT-solver is all you need. SAT-solver has entered industry for long time of course and a prestigious competition is running every year to make the algorithm faster.

Mathematicians like to argue using wild and abstract imaginations. They invent and talk about numbers, infinity, and what not. The proof of a theorem by a mathematician, including contemporary ones, can read controversial to other expert mathematicians due to its abstract nature; not only the objects in question, but also the way each step in the argument is carried.

Logic in mathematics usually comes later only to verify the arguments used. Unfortunately, many mathematicians will be frustrated if only propositional logic is allowed. They will be very limited when talking about properties of natural numbers, for example, which first order logic needs to come in play; probably equipped with mathematical induction to support their arguments. Sometimes, one even needs to lift the background logic to the second-order logic or even higher-order logic when the arguments are too complicated. For those who are not familiar, these are some kinds of logic which allow more expressive statements and sophisticated arguments to produce new truths.

The problem is, once a theory is already expressive enough to allow us to talk about natural number systems $0,1,2,3,\dots$ (e.g. Robinson Arithmetic is the most minimal such theory I can think of), there must be a true statement, true according to that theory, that we can never prove under some sound mechanism. This is a phenomenal result in mathematical logic by Kurt Gödel called Gödel’s Incompleteness Theorem in 1931. Many philosophical discussions arise about the limitation of mathematics in our search for truths; some exaggerates this result, too, sometimes based on a misinterpretation of the results.

Anyway, the remaining of this post will not be about the interpretation of Kurt Gödel’s work and its philosophical implication. It is interesting, but in this post, the goal would be to convey that we can still escape from this limitation by designing a new logic that is expressive enough to talk about complex stuff, yet still has sound and complete proof mechanism. Modal logic is probably favorites of many and I will probably post about it some day if time permits, but here I want to describe a mechanism to decide whether a statement is true or not in first order logic limited with unary predicate only.

First of all, propositional logic allows us to form a sentence composed of propositions and connect them with and, or, ifthen, and not. By proposition, I mean a sentence which simply describes a property of one or more objects, for example “Donald Trump is evil” and can be either true or false; a compound sentence then will be like “Donald Trump is evil and my mom hates him.

In first order sentence, we can introduce variables in our sentence such as “Someone is evil” or “Everyone is evil”. This sentence uses variables since we do not mention who exactly is evil. Using mathematical notations, we can write “Someone is evil” as $(\exists x) E(x)$ where $E(x)$ is read “$x$ is evil” and $(\exists x)$ is read “there exists an $x$“; the full sentence is read “There is an $x$ such that $x$ is evil”. “Everyone is evil” can be written as $(\forall x) E(x)$ where $(\forall x)$ is read “for every $x$“. The $x$ is called variable and $E(x)$ is a predicate (or, a property) that we want to attach to some objects.

A predicate can involve more than one objects. One can define a binary predicate $H(x,y)$ as “$x$ hates $y$” which is a predicate involving two objects, or a ternary predicate such as $T(x,y,z)$ defined as “$x$ is a mutual friend of $y$ and $z$“. It can also involve more that two objects. A sentence such as “Someone is evil and hated by everybody” can be written as first order sentence $(\exists x)(\forall y)(H(y,x) \wedge E(x))$ using our logic symbols that we have described above.

This post tries to convince you that the fragment of of first order logic that only allows unary predicates has a sound and complete proof mechanism. More concretely, predicates that involve two or more objects such as $H(x,y)$, $T(x,y,z)$ and so on are not allowed in a sentence in our theory.

An example $\mathbb{T}$ of a sentence in a first-order sentence which consists of only unary predicate is

$((\forall x)P(x) \vee (\forall y)Q(y)) \wedge (\exists z)(\neg P(z) \wedge \neg Q(z))$

which can be read “Everything is $P$ or everything is $Q$, but there is something that is neither $P$ nor $Q$“, for any meaning we attach to $P$ and $Q$. In first order logic, this can never be true, for whatever the meaning of $P$ and $Q$ we use and any objects in place of $x,y,z$. This kind of sentence is called contradiction in logic. If you are not convinced yet, replace $P$ and $Q$ with “rich” and “beautiful”, respectively, and reflect about what the reading above tries to state when “everything” is changed to “everyone”.

To decide any first-order sentence with only unary predicate, the key idea is based on the following fact proved by Löwenheim-Skolem in 1915, which I rephrase frivorously:

If a first-order sentence $\mathbb{S}$ can expressed using $k$ unary predicates and $\ell$ variables, and if ever $\mathbb{S}$ is a true statement, there will be a world consisting of only $2^k \cdot \ell$ objects such that $\mathbb{S}$ is also true in this (finite) world.

Löwenheim-Skolem, 1915

(Readers might have noticed and be disappointed that this theorem is proved way before Gödel proved his extraordinary results. But I would like to mention that the endeavor to find a decidable logic for more and more fragments of mathematics are still alive until today.)

The fact that $\mathbb{T}$ will be true in a finite world with only $2^k \cdot \ell$ objects, when it can be true, is so important that a computer can just focus on these finite worlds and brute-force on all possible interpretations of the predicates on each of the $2^k \cdot \ell$ objects (whether each object satisfies a certain predicate) and terminate eventually once it finishes traversing all the possibilities of these interpretations, which are also finite.

Our sentence $\mathbb{T}$ has $k=2$ predicates, viz. $P$ and $Q$, and $\ell=3$ variables, viz. $x,y,z$, so if $\mathbb{T}$ can be true, it must be true too in some world that has $2^2 \cdot 3 = 12$ objects in it. More concretely, imagine a world that only has 12 people in it where each person is named A, B, C, D, E, F, G, H, I, J, K, and L. Next, some of these people might have property $P$ and $Q$. Again, without loss of generality, take for example, predicate $P$ is “rich” and $Q$ is “beautiful”. Under this interpretation, some of these 12 people are rich and some are beautiful. For example, a possible scenario would be A, B, C, F, and G being rich while the rest is poor and B , C, H, and L, are beautiful and the rest is ugly. Of course there are many more possible scenarios in this world about who’s rich and beautiful.

The Löwenheim-Skolem theorem states that if $\mathbb{T}$ is ever true (in any world and any interpretation of $P$ and $Q$), there must be a scenario in this world of 12 people that we have described above where $\mathbb{T}$ is true. Thus, a computer can just search over all possible $(2^{12})^2=16777216$ scenarios to find out whether $\mathbb{T}$ is true in at least one scenario: that there is a scenario where “every one of A, B, C, D, E, F, G, H, I, J, K, L is rich, or every one of A, B, C, D, E, F, G, H, I, J, K, L is beautiful, but at least one of them is neither rich nor beautiful”. If none of these scenarios in this world of only 12 people make $\mathbb{T}$ true, then $\mathbb{T}$ can never be true for any meaning we attach to $P$ and $Q$ and any objects with assign to $x,y,z$ from any possible worlds, i.e. $T$ is guaranteed to be a contradiction; it will be never true whatsoever. Indeed, it is also easier for us to check our intuition about whether $\mathbb{T}$ can be true in this world of 12 people. We can imagine that if all these 12 people are rich, then none of them can be not-rich, since everybody is rich; contradicting the last premise about someone being not rich. Similarly, if the 12 people are beautiful, then the last premise, about someone being not beautiful, cannot also be satisfied since everybody is beautiful. The main weapon is this reduction to the world of only 12 people; we can just focus the checking on this world of 12 people. The finiteness entails that the number of states/scenarios that the computer needs to check is finite too and hence it just needs to check one by one and will eventually finish them all. If this reduction does not exist, the computer might need to run indefinitely and no guarantee that the procedure will ever give us an answer. In our example, the computer will finally check all $16777216$ scenarios and find out that none of them makes $\mathbb{T}$ true, hence we can be sure that $\mathbb{T}$ can never be true no matter what the context is.

Now, one may ask: Is there a first-order statement that is ever true and how can we check it? The answer is yes and the following example will demonstrate it. Consider the following sentence $\mathbb{T}'$:

$(\exists x)P(x) \wedge (\exists y)Q(y) \wedge \neg(\exists z)( P(x) \wedge Q(x))$.

If we agree to interpret $P$ as “rich” and $Q$ as “beautiful”, then the statement $\mathbb{T}'$ will read “Somebody is rich and somebody is beautiful but nobody is both rich and beautiful.” The sentence has $k=2$ predicates and $\ell=3$ variables, so our decision algorithm would consider a world consisting $2^2 \cdot 3=12$ objects (or people) again, let’s say A, B, C, D, E, F, G, H, I, J, K, L. Our computer can try to check all the possible $16777216$ scenarios who’s rich and who’s beautiful and see if at least one of makes $\mathbb{T}'$ true. The computer will eventually find, for example, that a scenario where the A, B, C, D, E, F are beautiful but not-rich, while G, H, I, J, K, L are rich but not-beautiful, is an instance of the world of 12 people where $\mathbb{T}'$ is true! Thus the computer will say: “Yes, $\mathbb{T}'$ can be true.” This is not the only possible scenario that makes $\mathbb{T}'$ true though, but the Löwenheim-Skolem theorem guarantees that it will find at least one scenario in this world of 12 people; no need to check the other infinite possibilities.

The proof of the theorem will not be contained here but I hope this post has convinced you how limiting the expressiveness of logic can give us power to effectively separate what is right and wrong in a logical system. I would not say that we should drop the whole number system for example; to escape Gödel’s theorem. I think numbers are useful and beautiful and many mathematical objects are. But for many real-life problems, I believe, the general discussion in mathematics will automatically reduce to finite domains hence allowing us to brute-force using computer; if we are not constrained with efficiency issue.

One possible simple take on Gödel’s Incompleteness Theorem is thus as follows. Gödel’s Incompleteness Theorem shall not stop people doing mathematics to invent more truths, even though not all, in the mathematical universes, and possibly develop us with new languages to speak about the physical nature and other complex phenomena. Probably some of the truths that are not provable in the platonic world of mathematics do not really matter in real world when there is an urgency to deal with the relevant mathematics. In addition to it, hopefully, in real world, their theory will be reducible to something that is more tangible, be it immediately or after some amount of work on simplification and/or modification. This might be true since there is a big chance that in real world, the mathematics is forced to fit into discrete and finite domains. There is probably a chance that in such case, we might not need to use the full theory to solve the real-world problems, but simply just a fragment of it. Indeed, this is what many logicians are trying to do. Many logicians around the world, especially those who are motivated with computational aspect of truths, are absorbed and enthusiastic into finding fragments of logic that allow us to effectively and probably efficiently decide truths, but are expressive enough to talk about complex problems they need to tackle. Check out the development of modal logic and description logic, for example.

Do you agree with this view of Gödel’s Incompleteness Theorem? What do you think are the issues of this approach? Let me know what you think.

Categories

## Note on probability of placing rooks

Revoking a post from 5 years ago, I wrote a discussion about the probability of randomly placing rooks on a chessboard in distinct rows and columns. The precise question in general setting is as follows.

Problem. $k$ rooks are placed on an $n \times n$ chessboard randomly. What is the probability that the rooks are all in different rows and columns?

The previous post is about this problem with $k=5$ and $n=8$ and for such instance, the probability is very small; around $0.04935660144$. This intrigues me; why so small?

So the number of ways putting $k$ rooks on $n \times n$ chessboard is $P^{n^2}_k = n^2 \cdot (n^2-1) \cdot \dots \cdot (n^2-k+1)$. Now, to put $k$ rooks so that they are all on different rows and columns, there are $n^2$ options for the first rook, $(n-1)^2$ options for the second rook, and so on up to $(n-k+1)^2$ options for the $k$-th rook. The probability is thus

$\displaystyle p(n,k)=\prod_{i=0}^{k-1} \dfrac{(n-i)^2}{n^2-i}$.

We can see that, compared to its denumerator, the numerator decreases much faster each time $i$ is incremented. But, how can we describe this assymptotically in $k$ and $n$?

Using the following trick:

$\dfrac{(n-i)^2}{n^2-i} \le 1-\dfrac{i}{n} \le \text{exp}\left(-\dfrac{i}{n} \right)$,

for any $n \ge i \ge 0$, we can prove that

$p(n,k) \le \text{exp}\left( - \dfrac{k^2-k}{2n} \right)$.

Hence, the probability suffers an exponential decay in $k^2/2n$.

For $k=n$, we have that $p(n,n) \le \text{exp}(-(n-1)/2)$.

Categories

## Reciprocals of natural numbers

Last time, I posted about the convergence of the reciprocals of palindromes. In this post, I will discuss the motivation of that problem back then.

It is known that the harmonic series $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dots$ is divergent (unbounded; goes to infinity). But, if the sum of the reciprocals of the squares, $\displaystyle \dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dots$ is convergent (has finite sum) and the limit is $\dfrac{\pi^2}{6}$. This is called Basel’s problem.

A standard approach of seeing the convergence of such series is through considering the $p$-th power of the elements of the series, where $p$ is a positive real number. If we let

$\displaystyle S_p= \sum_{n \in \mathbb{N}} \dfrac{1}{n^p} = \dfrac{1}{1^p}+\dfrac{1}{2^p}+\dfrac{1}{3^p}+\dots$,

it is known that $S_p$ is convergent if and only if $p>1$. So, a series as “tight to harmonic series” as $\displaystyle \dfrac{1}{1^{0.99999}}+\dfrac{1}{2^{0.99999}}+\dfrac{1}{3^{0.99999}}+\dots$ will go to infinity while small increase on the power such as $\displaystyle \dfrac{1}{1^{1.000001}}+\dfrac{1}{2^{1.000001}}+\dfrac{1}{3^{1.000001}}+\dots$ will never go beyond a finite number, making the sum finite.

A standard approach to analyze this “dichotomy” is by using integral

$\displaystyle \sum_{n \in \mathbb{N}} \dfrac{1}{n^p} \le \lim_{t \rightarrow \infty} \int_1^t \dfrac{1}{x^p} \text{d}x$

The later is finite when $p>1$ and infinite for $0 \le p \le 1$.

But another “generalization” of figuring out why $S_p$ is divergent for $p=1$ and convergent for $p=2$ is by investigating sum of reciprocals of some elements of $\mathbb{N}$.

Let $S \subseteq \mathbb{N}$, we want to know if

$\displaystyle \sum_{n \in S} \dfrac{1}{n}$

is convergent or not. The harmonic series is constructed with $S=\mathbb{N}$ the full set while the Basel’s problem is constructed with $S=\{k^2: k \in \mathbb{N}\}$ which is the set of all perfect squares.

The convergence then seemingly speaks about the “volume” of $S$. If we take elements in $S$ too much, then the series $\displaystyle \sum_{n \in S} \dfrac{1}{n}$ will be divergent like harmonic series, while if $S$ has “modest or small” volume such as the set of perfect squares, the series ceases to be finite!

There are a lot of results of course on various $S$:

• If $S=\{2^k: k \in \mathbb{N}\}$, the corresponding series is the familiar geometric series $\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots$ which is convergent to $1$. This is probably one of the most famous series and appears in paradox such as Zeno’s paradox.
• If $S=\{p \in \mathbb{N}: p \text{ is prime } \}$, the corresponding series $\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}+\dots$ is divergent. It is known that there are approximately $\dfrac{x}{\ln x}$ prime numbers that are less than $x$ by the prime number theorem; this is a bit more than the number of perfect squares.
• If $S=\{k^k: k \in \mathbb{N}\}$, the corresponding series $\dfrac{1}{1^1}+\dfrac{1}{2^2}+\dfrac{1}{3^3}+\dots$ is exactly the same as its integral “approximation” $\displaystyle \int_0^1 \dfrac{1}{x^x} \text{d}x$. This incidence is called Sophomore’s Dream. 🙂
• If $S=\{F_n: n \ge 0\}$ is the set of Fibonacci numbers defined by $F_0=F_1=1$ and $F_{n+1}=F_n+F_{n-1}$, the corresponding series is convergent (due to its “exponential nature”).
• etc.

See more interesting results in its dedicated Wikipedia page: List of sum of reciprocals. Also, see the following paper published in 2016 by Ing. Pier Franz Roggero, Dr. Michele Nardelli, and P.i. Francesco Di Noto that tries to relate this kind of series with physics: Sum of the reciprocals of famous series: mathematical connections with some sectors of theoretical physics and string theory.

I think the following paper in 2013 by Jonathan Bayles and Dominic Klyve will provide a good historical context of reciprocal series in mathematics and computation: Reciprocals as Knowledge Metric: Theory, Computation, and Perfect Numbers.


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# Fitting an absolute-value formula to a collection of points in the plane

I need an equation to satisfy the following data points and I can't come up with anything. I believe it needs an absolute value as that changes the slope of the line.

Best yet: $y=2x-|x-4|$

Points are: $(1,1)$; $(2,3)$; $(3,5)$; $(4,7)$; $(5,8)$; $(6,10)$; $(7,11)$; $(8,12)$; $(9,13)$; $(10,14)$; $(11,15)$

## migrated from puzzling.stackexchange.comSep 30 '16 at 1:09

This question came from our site for those who create, solve, and study puzzles.

• I don't think this really qualifies as a puzzle. You might be better off posting this questions on Mathematics SE; I've flagged it to ask the mods to migrate it there. – rand al'thor Sep 29 '16 at 23:43
• Oh yay my boredom is finally relevant desmos.com/calculator/p6nqvyekcl – greenturtle3141 Sep 30 '16 at 0:42

If you meant $(5, 9)$ instead of $(5, 8)$, then $$y = 1.5(x + 1) - 0.5|x - 5|$$ works. Otherwise, \begin{align*} y &= 1.5(x + 1) - 0.5|x - 5| + 0.5\min\bigl(|x - 5| - 1, 0\bigr) \\ &= 1.5(x + 1) - 0.5\bigl(1 + \bigl||x - 5| - 1\bigr|\bigr) \end{align*} hits them all.

It is not clear if the goal is to find a function $y(x)$ which fits EXACTLY, or approximately.

If exactly is required : $$\begin{cases} y=2x-1 &\text{ if } x\leq 4 \\ y=8 &\text{ if } x=5 \\ y=x+4 &\text{ if } x\geq 6 \end{cases} \quad\to\quad y=8+(2x-9)H(4.5-x)+(x-4)H(x-4.5)$$ $H$ is the Heaviside step function.

If the fourth point is $(5,9)$ instead of $(5,8)$ a simpler form is : $$y=\frac{1}{2}(3+3x-|x-5|)$$.

If an approximate fit is wanted :

A usual regression calculus is sufficient, for example with the function $y=ax^2+by+c$. This is a linear regression for $a, b, c$ which leads to : $$a\simeq -0.0711$$ $$b\simeq 2.226$$ $$c\simeq -1.085$$ The graph of the function $y=ax^2+by+c$ is shown :

This was computed with the fourth point $(5,8)$.


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Say I have a continuous function $f$ defined on a compact interval $I$ on the real line. As is well-known, I could approximate $f$ arbitrarily well by polynomials, i.e., we can express $f$ as an infinite series $\sum c_n x^n$.