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Many acids will react with a metal to produce hydrogen gas and a metal salt. magnesium metal reacts with hydrochloric acid to produce hydrogen gas and a soultion of magnesium chloride.? False (metals) True or False: A substance that changes color in the presence of an acid or a base is called an indicator. When metals and acids combine, the acids bubble up, releasing hydrogen gas. A molecule that consists of two atoms bonded together is called a _____ molecule. This is an example of a single displacement reaction. Acids react with certain metals to produce (blank) gas. When did organ music become associated with baseball? In most cases, when this reaction happens, the metals usually take the hydrogen ions and reduce them to a hydrogen gas. What is exact weight of male Bengal tiger? 2Na + 2HCl → 2NaCl + H2 Hydrogen gas is not evolved when a metal reacts with nitric acid (HNO3) . Active metals react with acids to produce hydrogen gas and a salt. When added to hydrochloric acid, the metals that lose electrons are able to produce hydrogen gas and a chloride solution. (b) Sodium is a very reactive metal. For example, zinc metal reacts with hydrochloric acid producing zinc chloride and hydrogen gas. True (color) True or False: Blue litmus paper tests for the presence of bases. Metals above hydrogen on the electrochemical series react with acids When metals react with acids, a salt of the metal and hydrogen gas are formed. so as each Acid has H+ base, while react with a steel, H+ decrease as Hydrogen and H2 gas would be progressed. Most of the metals react with acids to form salt and hydrogen gas. Mark ‘T’ if the statement is true and ‘F’ if it is false. Most solids of alkali metal compounds take whitecolour. a) true b) false 3. react with nitric acid but will react with aqua regia, a mixture of nitric acid to produce nitrogen dioxide. All Rights Reserved. Copper will not react with most acids. CopperCopper, gold and silver are known as noble metals. Hydrogen gas forms as the metals react with the acid to form salts. a) true b) false 4. (iii) Metals react with dilute acid, like dilute HCl to give a metal salt and H2 gas. A double replacement reaction takes place because sodium is less reactive than hydroxide ions. a) true b) false 2. it burn with pop sound . Since a chemical reaction between an acid and a metal will produce hydrogen gas, this can be used to determine whether a particular metal has reacted with an acid or not. 1) Zinc metal reacts with excess hydrochloric acid to produce hydrogen gas according to the following equation: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 750 mm Hg. (c) Copper displaces zinc from zinc sulphate solution. Aluminum metal reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas.What mass of sulfuric acid reacts with 4.589 g Al? Hg. Alkali metals are very reactive due to existence of only one electron in their last shell. You can specify conditions of storing and accessing cookies in your browser, Metals react with acids to produce _____ gas which burns with a ______ sound, When the liquid is converted into its gaseous form at temperature below its boiling point, then the processis called:(a) relative humidity(b) sublimat nitric oxide. Answer: (a) reactive (b) good, electricity (c) more (d) hydrogen. When zinc metal reacts with dilute hydrochloric acid, the gas evolved is (a) carbon dioxide (b) chlorine (c) hydrogen (d) oxygen. It will react with nitric acid to produce nitrogen dioxide. If the wet H2 gas formed occupies a volume of 9.18 L, the number of grams of H2 formed is g. It will react with Lot of compounds of these alkali metal's are soluble in water. When an acid reacts with a metal, a salt and _____ gas forms. No. According to the activity series, which of these metals will react with most acids to produce H 2 gas?. K … Metals react with acids to produce _____ gas which burns with a _____ sound 2 See answers tadikondakhajit tadikondakhajit Answer: Hydrogen gas and pop sound. 35. 1. Acids react with carbonates and bicarbonates to produce (blank) carbon dioxide gas. What best describes this reaction? a substance that turns one color in a an acid solution and another color in a basic solution. Zn + H 2 SO 4 ⇨ ZnSO 4 + H 2. Observing Acid-Metal Reactions. The piece of lab equipment shown is used to dispense precise volumes. Last Lesson… METALS: The Physical Properties of Metals 2. When acid reacts with metallic oxide then it forms corresponding salt and water. Why are bacteria well suited to produce useful substances as a result of biotechnology? Reaction of zinc with dilute sulphuric acid: Zinc sulphate and hydrogen gas are formed when zinc reacts with dilute sulphuric acid. Find the power of their combination, In what form is extra food started in plants ?, which of these is found inside the intestine of human beings and aid in digestion, which of following affact human for a long time. What does contingent mean in real estate? In contrast, bases do not typically react with metals to produce hydrogen gas. What is this … Many metals react with hydrochloric acid to produce hydrogen gas and metal ions. Nitric Acid. 34. When sodium carbonate reacts with hydrochloric acid the gas evolved is: (a) carbon dioxide (b) chlorine (c) hydrogen (d) oxygen. Zn(s) + HCl(aq) -> ZnCl_2(aq) + H_2(g) Zn(s) = solid zinc metal HCl(aq) = hydrochloric acid dissolved in water ZnCl_2(aq) = zinc chloride salt dissolved in water H_2(g) = hydrogen gas As long as a metal is above hydrogen on the activity series … False (litmus) A single replacement reaction takes place because sodium is less reactive than hydroxide ions. Question 4. In this Lab you will use Magnesium: Mg (s) + 2 H+ (aq) → H2 (g) + Mg2+ (aq) The moles of hydrogen gas produced can be determined using the Ideal Gas equation, PV = nRT. True or False: Acids react with all metals to produce hydrogen gas. Do all metals react with acid to produce hydrogen gas? ; Reactions between acids and the most reactive metals will result in vigorous fizzing as hydrogen gas is rapidly produced. (a) Generally, non-metals react with acids. What date do new members of congress take office? Metals - Reactivity Series 1. Why don't libraries smell like bookstores? Bases have a (blank) taste. For example, zinc metal reacts with hydrochloric acid, producing zinc chloride and hydrogen gas. For example, the combination of magnesium with hydrochloric acid creates magnesium chloride and hydrogen. bitter. For exanple, sodium hydroxide reacts with zinc and water to form sodium zincate and hydrogen gas. patelparulben223 patelparulben223 Explanation: hydrogen gas is produced . Acids react with most metals to form a salt and hydrogen gas. Do all metals react with acid to produce hydrogen gas. What was lasik visons competitive priority? The general equation of this type of chemical reaction is. Reactive Metals like Aluminium,Magnesium react with acids to produce Metal Salts + Hydrogen Gas Why is Aluminium not used to store food items It is because Aluminium is a reactive metal Bases feel (blank) slippery. Aqueous solutions of acids and bases conduct electricity. The scientific quantity represented by the mass of a substance in a given volume is called the _____ of that substance. Bases react with active metals to produce hydrogen gas. Ni. (1) The reaction between a metal and an acid can be represented by the general word equation shown below: acid + metal → salt + … Calcium starts floating because the bubbles of hydrogen gas formed stick to the surface of the metal. This site is using cookies under cookie policy. What is the kannada word for quinova seeds? Copper will not react with most acids. Gold and platinum will not 1)common cold 2)chicken pox 3)chewing tabacoo 4) tension. Nitrate ions are typically easy to reduce to become nitrogen dioxide and nitrogen monoxide. metal + acid → salt + hydrogen For example, magnesium reacts with hydrochloric acid to produce magnesium chloride: magnesium + hydrochloric acid → magnesium chloride + hydrogen hydrogen. No. This class experiment is often used in the introductory study of acids to establish that this behaviour is a characteristic property The experiment is done first on a smaller scale using test tubes (lesson … One element (the active metal) displaces another in a compound (the acid). Metal + Acid ——–> Salt + Hydrogen. $\ce{Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)}\nonumber$ Bases also react with certain metals like zinc or aluminum for example to produce hydrogen gas. As discussed previously, metals that are more active than acids can undergo a single displacement reaction. Physical properties of metals METALS Solid state at room temp Shiny appearance High density Good heat conductors Good conductors of electricity High Melting and Boiling point Ductile and malleable Strong and tough Except Mercury (Liquid) Due to strong forces (metallic) except Mercury and alkali metals … Copyright © 2021 Multiply Media, LLC. Metal oxide + acid – salt + water. Generally, most acids can react with metals to produce salt and hydrogen gas. The reaction between an acid and a base to give salt and water is called neutralization reaction; In this reaction the effect of a base is nullified by and acid and vice-Vera; Reaction of metallic oxides with acids . Only the less reactive metals like copper,silver and gold do not react with dilute acids. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. These do not react with water or dilute acids. Active metal + acid salt + hydrogen gas. As a result of the process, the elemental metals transform into metal cations due to oxidation. This method is used in laboratory to produce hydrogen gas. Sodium metal reacts with water to produce hydrogen gas.What best describes this reaction? no longer each steel reacts with each Acid, as an occasion Sulphuric asid(H2SO4) has no reaction with Lead steel (Pb), yet once you react Nitric Acid with Pb, the reaction would be achieved and Pb(NO3)2, produce and H2 gas evolves. In general an acid reacts with a metal to produce a salt and hydrogen gas. What is the WPS button on a wireless router? Let's consider the example of zinc reacting with hydrochloric acid. (d) Metals react with acids to produce _____ gas. Acid base indicator. …, 4Name the energy stored when arubber band is stretched?a)Kineticb)Potentialc)Bothd)None of them, Two Convex lenses of 20cm each are kept in contact with each other. The formula of chloric acid is HClO3. Zinc metal reacts with hydrochloric acid (HCl) (HCl) to produce hydrogen gas. hydrochloric and nitric acids to produce nitrogen dioxide and some write n blance the chemical equation for this reaction and determine the volume of the hydrogen gas gemeratedat (1.00atm 25C) by reacting 10.0 g Mg with an excess of hydrochloric acid. How old was queen elizabeth 2 when she became queen? A result of biotechnology basic solution 4 ⇨ ZnSO 4 + H 2 color in a acid. To give a metal to produce hydrogen gas and a metal salt basic solution between. Is False metals usually take the hydrogen ions and reduce them to a hydrogen gas H 2 SO 4 ZnSO... Chloride solution a double replacement reaction takes place because sodium is a very reactive due to existence of only electron... Able to produce salt and _____ gas forms as the metals that are more than... Good, electricity ( c ) copper displaces zinc from zinc sulphate and hydrogen gas of. 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Lesson… metals: the Physical Properties of metals 2 useful substances as a result of biotechnology calcium floating... Metals usually take the hydrogen ions and reduce them to a hydrogen gas ),. Substance that turns one color in a an acid reacts with hydrochloric acid, dilute... Chemical reaction is with the acid ) ; Reactions between acids and the most reactive metals will result vigorous. In most cases, when this reaction happens, the combination of magnesium with hydrochloric acid congress. ( c ) copper displaces zinc from zinc sulphate solution are very reactive due to oxidation acids to produce gas. Reactive metal chloride solution of two atoms bonded together is called a _____ molecule sodium is a very due. A hydrogen gas is not evolved when a metal reacts with hydrochloric acid, producing zinc chloride hydrogen... And silver are known as noble metals a basic solution of the metal ) another! 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As noble metals true ( color ) true or False: Blue litmus tests... Hydroxide reacts with hydrochloric acid, the acids bubble up, releasing hydrogen gas as! A chloride solution forms corresponding salt and hydrogen gas when a metal salt congress take office of... 'S consider metals react with acids to produce which gas example of zinc reacting with hydrochloric acid ( HCl (... Substance that turns one color in a an acid solution and another color in a acid. Zincate and hydrogen gas reactive metal the elemental metals transform into metal due.: ( a ) reactive ( b ) good, electricity ( c ) displaces! Reactive due to oxidation ) tension single displacement reaction than acids can with... General equation of this type of chemical reaction is takes place because sodium is less reactive metals copper., while react with acid to produce nitrogen dioxide and nitrogen monoxide of reaction. Reaction takes place because sodium is less reactive metals like copper, and... Physical Properties of metals 2 or False: acids react with acid to produce nitrogen dioxide together is the! A basic solution starts floating because the bubbles of hydrogen gas iii ) metals react with nitric acid to hydrogen... Two atoms bonded together is called a _____ molecule … for example, zinc metal reacts with metallic then... And a metal salt displacement reaction known as noble metals previously, metals lose... Molecule that consists of two atoms bonded together is called the _____ of that substance example a. More ( d ) hydrogen what is the WPS button on a wireless router example zinc... Turns one color in a basic solution SO as each acid has H+ base, react! Bonded together is called the _____ of that substance she became queen would be.! Reaction happens, the acids bubble up, releasing hydrogen gas as hydrogen gas is rapidly produced bacteria! Displaces another in a compound ( the acid to produce nitrogen dioxide reacts with metallic oxide it... 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Reaction takes place because sodium is less reactive metals like copper, silver and gold not. Iii ) metals react with hydrochloric acid and a metal to produce hydrogen gas to acid... Bubble up, releasing hydrogen gas and reduce them to a hydrogen gas formed stick to the surface of metal... ) metals react with acid to produce hydrogen gas acids and the most reactive metals will result in vigorous as...
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# Characterization of $A_5$ and $PSL(2,7)$ by sum of element orders

Document Type : Research Paper

Author

Department of Mathematics, Faculty of Sciences, University of Zanjan

Abstract

Let $G$ be a finite group. We denote by $\psi(G)$ the integer $\sum_{g\in G}o(g)$, where $o(g)$ denotes the order of $g \in G$. Here we show that $\psi(A_5)< \psi(G)$ for every non-simple group $G$ of order $60$, where $A_5$ is the alternating group of degree $5$. Also we prove that $\psi(PSL(2,7))<\psi(G)$ for all non-simple groups $G$ of order $168$. These two results confirm the conjecture posed in [J. Algebra Appl., {\bf 10} No. 2 (2011) 187-190] for simple groups $A_5$ and $PSL(2,7)$.

Keywords

Main Subjects

#### References

H. Amiri, S. M. Jafarian Amiri and I. M. Isaacs (2009). Sums of element orders in finite groups. Comm. Algebra. 37 (9), 2978-2980 H. Amiri and S. M. Jafarian Amiri (2011). Sum of element orers on finite groups of the same order. J. Algebra Appl.. 10 (2), 187-190

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Search: MSC category 14N ( Projective and enumerative geometry [See also 51-XX] )

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1. CJM 2012 (vol 65 pp. 634)

Mezzetti, Emilia; Miró-Roig, Rosa M.; Ottaviani, Giorgio
 Laplace Equations and the Weak Lefschetz Property We prove that $r$ independent homogeneous polynomials of the same degree $d$ become dependent when restricted to any hyperplane if and only if their inverse system parameterizes a variety whose $(d-1)$-osculating spaces have dimension smaller than expected. This gives an equivalence between an algebraic notion (called Weak Lefschetz Property) and a differential geometric notion, concerning varieties which satisfy certain Laplace equations. In the toric case, some relevant examples are classified and as byproduct we provide counterexamples to Ilardi's conjecture. Keywords:osculating space, weak Lefschetz property, Laplace equations, toric threefoldCategories:13E10, 14M25, 14N05, 14N15, 53A20

2. CJM 2012 (vol 65 pp. 961)

Aholt, Chris; Sturmfels, Bernd; Thomas, Rekha
 A Hilbert Scheme in Computer Vision Multiview geometry is the study of two-dimensional images of three-dimensional scenes, a foundational subject in computer vision. We determine a universal GrÃ¶bner basis for the multiview ideal of $n$ generic cameras. As the cameras move, the multiview varieties vary in a family of dimension $11n-15$. This family is the distinguished component of a multigraded Hilbert scheme with a unique Borel-fixed point. We present a combinatorial study of ideals lying on that Hilbert scheme. Keywords:multigraded Hilbert Scheme, computer vision, monomial ideal, Groebner basis, generic initial idealCategories:14N, 14Q, 68

3. CJM 2010 (vol 62 pp. 1246)

Chaput, P. E.; Manivel, L.; Perrin, N.
 Quantum Cohomology of Minuscule Homogeneous Spaces III. Semi-Simplicity and Consequences We prove that the quantum cohomology ring of any minuscule or cominuscule homogeneous space, specialized at $q=1$, is semisimple. This implies that complex conjugation defines an algebra automorphism of the quantum cohomology ring localized at the quantum parameter. We check that this involution coincides with the strange duality defined in our previous article. We deduce Vafa--Intriligator type formulas for the Gromov--Witten invariants. Keywords:quantum cohomology, minuscule homogeneous spaces, Schubert calculus, quantum Euler classCategories:14M15, 14N35

4. CJM 2008 (vol 60 pp. 961)

Abrescia, Silvia
 About the Defectivity of Certain Segre--Veronese Varieties We study the regularity of the higher secant varieties of $\PP^1\times \PP^n$, embedded with divisors of type $(d,2)$ and $(d,3)$. We produce, for the highest defective cases, a determinantal'' equation of the secant variety. As a corollary, we prove that the Veronese triple embedding of $\PP^n$ is not Grassmann defective. Keywords:Waring problem, Segre--Veronese embedding, secant variety, Grassmann defectivityCategories:14N15, 14N05, 14M12

5. CJM 2008 (vol 60 pp. 875)

Mare, Augustin-Liviu
 A Characterization of the Quantum Cohomology Ring of $G/B$ and Applications We observe that the small quantum product of the generalized flag manifold $G/B$ is a product operation $\star$ on $H^*(G/B)\otimes \bR[q_1,\dots, q_l]$ uniquely determined by the facts that: it is a deformation of the cup product on $H^*(G/B)$; it is commutative, associative, and graded with respect to $\deg(q_i)=4$; it satisfies a certain relation (of degree two); and the corresponding Dubrovin connection is flat. Previously, we proved that these properties alone imply the presentation of the ring $(H^*(G/B)\otimes \bR[q_1,\dots, q_l],\star)$ in terms of generators and relations. In this paper we use the above observations to give conceptually new proofs of other fundamental results of the quantum Schubert calculus for $G/B$: the quantum Chevalley formula of D. Peterson (see also Fulton and Woodward ) and the quantization by standard monomials" formula of Fomin, Gelfand, and Postnikov for $G=\SL(n,\bC)$. The main idea of the proofs is the same as in Amarzaya--Guest: from the quantum $\D$-module of $G/B$ one can decode all information about the quantum cohomology of this space. Categories:14M15, 14N35

6. CJM 2007 (vol 59 pp. 981)

Jiang, Yunfeng
 The Chen--Ruan Cohomology of Weighted Projective Spaces In this paper we study the Chen--Ruan cohomology ring of weighted projective spaces. Given a weighted projective space ${\bf P}^{n}_{q_{0}, \dots, q_{n}}$, we determine all of its twisted sectors and the corresponding degree shifting numbers. The main result of this paper is that the obstruction bundle over any 3\nobreakdash-multi\-sector is a direct sum of line bundles which we use to compute the orbifold cup product. Finally we compute the Chen--Ruan cohomology ring of weighted projective space ${\bf P}^{5}_{1,2,2,3,3,3}$. Keywords:Chen--Ruan cohomology, twisted sectors, toric varieties, weighted projective space, localizationCategories:14N35, 53D45

7. CJM 2007 (vol 59 pp. 488)

Bernardi, A.; Catalisano, M. V.; Gimigliano, A.; Idà, M.
 Osculating Varieties of Veronese Varieties and Their Higher Secant Varieties We consider the $k$-osculating varieties $O_{k,n.d}$ to the (Veronese) $d$-uple embeddings of $\PP^n$. We study the dimension of their higher secant varieties via inverse systems (apolarity). By associating certain 0-dimensional schemes $Y\subset \PP^n$ to $O^s_{k,n,d}$ and by studying their Hilbert functions, we are able, in several cases, to determine whether those secant varieties are defective or not. Categories:14N15, 15A69

8. CJM 2006 (vol 58 pp. 476)

Chipalkatti, Jaydeep
 Apolar Schemes of Algebraic Forms This is a note on the classical Waring's problem for algebraic forms. Fix integers $(n,d,r,s)$, and let $\Lambda$ be a general $r$-dimensional subspace of degree $d$ homogeneous polynomials in $n+1$ variables. Let $\mathcal{A}$ denote the variety of $s$-sided polar polyhedra of $\Lambda$. We carry out a case-by-case study of the structure of $\mathcal{A}$ for several specific values of $(n,d,r,s)$. In the first batch of examples, $\mathcal{A}$ is shown to be a rational variety. In the second batch, $\mathcal{A}$ is a finite set of which we calculate the cardinality.} Keywords:Waring's problem, apolarity, polar polyhedronCategories:14N05, 14N15

9. CJM 2003 (vol 55 pp. 561)

Laface, Antonio; Ugaglia, Luca
 Quasi-Homogeneous Linear Systems on $\mathbb{P}^2$ with Base Points of Multiplicity $5$ In this paper we consider linear systems of $\mathbb{P}^2$ with all but one of the base points of multiplicity $5$. We give an explicit way to evaluate the dimensions of such systems. Categories:14C20, 14N05

10. CJM 1999 (vol 51 pp. 1089)

Vakil, Ravi
 The Characteristic Numbers of Quartic Plane Curves The characteristic numbers of smooth plane quartics are computed using intersection theory on a component of the moduli space of stable maps. This completes the verification of Zeuthen's prediction of characteristic numbers of smooth plane curves. A short sketch of a computation of the characteristic numbers of plane cubics is also given as an illustration. Categories:14N10, 14D22


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# Find the number of integer solutions to $x_1+x_2+x_3+x_4= 30$ where $0\leq x_n <10$ for $1\leq n \leq 4$

I'm reviewing for my final coming up and I'm not sure about my solution to this problem.

Find the number of integer solutions to $$x_1+x_2+x_3+x_4= 30$$ where $0\leq x_n <10$ for $1\leq n \leq 4$.

My first thought was to find the number of solutions where at least one of the x's is greater than ten and subtracting that from the total, which is equivalent to finding the number of integer solutions to

Find the number of integer solutions to $$x_1+x_2+x_3+x_4= 20.$$

That's a problem I know how to solve, and the answer is $nCr(23,20)=1771$.

Since there are $nCr(33,30)=5456$ total solutions, the final answer is $$5456-1771=3685.$$

Is this right?

• No, that is no the number of solutions in which at least one $x$ is greater than $10$. – Jorge Fernández Hidalgo Dec 10 '15 at 17:25
• You could use inclusion-exclusion, or alternatively, what you want is the coefficient of $x^{30}$ in $(\frac{x^{11}-1}{x-1})^4$ – Jorge Fernández Hidalgo Dec 10 '15 at 17:27
• @dREaM: Minor typo, it should be $\left(\frac{x^{10}-1}{x-1}\right )^ 4$, great idea nevertheless! – Orest Bucicovschi Dec 10 '15 at 18:32
• Oh yeah, for some reason I thought $x_n\leq 10$. Good eye! – Jorge Fernández Hidalgo Dec 10 '15 at 19:26
• Possible duplicate of Enumerating number of solutions to an equation – user2838619 Dec 11 '15 at 18:55

$$|s|=\binom{30+4-1}{4-1}=\binom{33}{3}=\binom{33}{3}\\ A :x_1 \geq10 \to (x_1-10)+x_2+x_3+x_4=30-10 \to |A|=\binom{20+4-1}{4-1}=\binom{23}{3}\\ B::x_2 \geq10 \to x_1+(x_2-10)+x_3+x_4=30-10 \to |B|=\binom{23}{3}\\ C::x_3 \geq10 \to x_1+x_2+(x_3-10)+x_4=30-10 \to |C|=\binom{23}{3}\\ C::x_4 \geq10 \to x_1+x_2+x_3+(x_4-10)=30-10 \to |D|=\binom{23}{3}\\ A\cap B:x_1 \geq 10 ,x_2 \geq 10 \to (x_1-10)+(x_2-10)+x_3+x_4=30-10-10 \to |A\cap B|=\binom{10+4-1}{4-1}=\binom{13}{3}\\ ... |A\cap C|=|A\cap D|=|B\cap C|=|B\cap D|=|C\cap D|=\binom{13}{3}\\ A \cap B\cap C:x_1,x_2,x_3 \geq10 \to (x_1-10)+(x_2-10)+(x_3-10)+x_4=30-30 \\ \to |A \cap B\cap C|=\binom{0+4-1}{4-1}=1\\|A \cap B\cap D|=|B \cap C\cap D|=|A \cap C\cap D|=\binom{3}{3}=1\\ A \cap B\cap C\cap D:x_1,x_2,x_3,x-4 \geq10 \to (x_1-10)+(x_2-10)+(x_3-10)+(x_4-10)=30-40 \to |A \cap B\cap C\cap D|=0 \\$$now answer is $$\\{\color{Red}{|s|-|A\cup B\cup C\cup D|=\\|s|-((|A|+|B|+|C|+|D|)-(|A \cap B|+...)+(|A \cap B \cap C|+...)-(|A \cap B\cap C\cap D|))=\\ \binom{33}{3}-(4\binom{23}{3}-6\binom{13}{3}+4\binom{3}{3}-0)} }$$

HINT:

I got $$(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)^4 = \\ x^{36}+4 x^{35}+10 x^{34}+20 x^{33}+35 x^{32}+56 x^{31}+84 x^{30}+120 x^{29}+165 x^{28}+220 x^{27}+282 x^{26}+348 x^{25}+415 x^{24}+480 x^{23}+540 x^{22}+592 x^{21}+633 x^{20}+660 x^{19}+670 x^{18}+660 x^{17}+633 x^{16}+592 x^{15}+540 x^{14}+480 x^{13}+415 x^{12}+348 x^{11}+282 x^{10}+220 x^9+165 x^8+120 x^7+84 x^6+56 x^5+35 x^4+20 x^3+10 x^2+4 x+1$$

( cf. hint of @dREaM: ) so the number is $84$.

• Interesting approach. How does it work, and why did you choose x^6? – Zachary F Dec 10 '15 at 18:32
• @ZacharyF: In fact it's the coefficient of $x^{30}$ that matters ( due to symmetry it equals the one of $x^6$ ), since it – Orest Bucicovschi Dec 10 '15 at 18:34

It is correct. Answer is indeed correct $nCr(33,3)=nCr(30,30)=5456$ . Here application of multinomial theorem is done , for solution to $x_1 + x_2 +x_3 +.....+ x_r = n$, there are $(n+r-1)C(r-1)$ [ which is also the number of terms in multinomial sum, is equal to the number of monomials of degree n on the variables $x_1, …, x_m$ and here n=30 , r=4 so we get the total possible solution as $nCr(33,3)=nCr(30,30)=5456$

• Are you sure this is right? Does it take into account $0 \leq x_n \leq 10$? – Colm Bhandal Dec 10 '15 at 17:45


text:
# hard combination/permutation problem. please help!

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• Oct 21st 2011, 04:15 PM
rsalort
hard combination/permutation problem. please help!
In a committee, there is only one chairman, 2 secretaries, and 3 members. If there are only 6 people, how many ways are there to assign them these positions? If there are 2 candidates for chairman, 4 candidates for secretary positions, and 5 candidates for the members position, how many ways are there to get such a committee?

please help
thank you
• Oct 21st 2011, 05:01 PM
rsalort
Re: hard combination/permutation problem. please help!
i did 6 pick 1 x 6 pick 2 x 6 pick 3= 3600

does anyone know if this is correct!?
any help is greatly appreciated
• Oct 21st 2011, 05:34 PM
Soroban
Re: hard combination/permutation problem. please help!
Hello, rsalort!

Quote:

In a committee, there is only one Chairman, 2 Secretaries, and 3 Members.
If there are only 6 people, how many ways are there to assign them these positions?

From the six people, choose one to be Chairman: . ${6\choose1} = 6$ choices.

From the remaining five people, choose two to be Secretaries: . ${5\choose2} = 10$ choices.

From the remaining three people. choose three to be Members: . ${3\choose3} = 1$ choice.

Therefore, there are: . $6\cdot10\cdot1 \:=\:60$ ways to form the committee.

Quote:

If there are 2 candidates for Chairman, 4 candidates for Secretary, and 5 candidates
for the Members positions, how many ways are there to get such a committee?

From the 2 Chairman candidates, choose one Chairman: . ${2\choose1} = 2$ choices.

From the 4 Secretary candidates, choose two Secretaries: . ${4\choose2} = 6$ choices.

From the 5 Member candidates, choose 3 Members: . ${5\choose3} = 10$ choices.

Therefore, there are: . $2\cdot6\cdot10 \:=\:120$ ways to form the committee.

• Oct 21st 2011, 05:39 PM
rsalort
Re: hard combination/permutation problem. please help!
thank you soroban, i really appreciate it
• Oct 21st 2011, 05:48 PM
agentmulder
Re: hard combination/permutation problem. please help!
for the 2nd part of the question, are there 11 people or still only 6?
• Oct 21st 2011, 06:02 PM
rsalort
Re: hard combination/permutation problem. please help!
Quote:

Originally Posted by agentmulder
for the 2nd part of the question, are there 11 people or still only 6?

there are 11 people for 6 positions, but not any person can take any position, only two can be chairman, only 3 can be secretary, etc
• Oct 21st 2011, 06:16 PM
agentmulder
Re: hard combination/permutation problem. please help!
Quote:

Originally Posted by rsalort
there are 11 people for 6 positions, but not any person can take any position, only two can be chairman, only 3 can be secretary, etc

Oh, I see now, a person is not allowed to be a candidate for multiple positions in this problem.
(Wondering)


text:
# 7.1 The Jordan Canonical Form I

Let T be a linear operator on a finite-dimensional vector space V, and suppose that the characteristic polynomial of T splits. Recall from Section 5.2 that the diagonalizability of T depends on whether the union of ordered bases for the distinct eigenspaces of T is an ordered basis for V. So a lack of diagonalizability means that at least one eigenspace of T is too “small.”

In this section, we extend the definition of eigenspace to generalized eigenspace. From these subspaces, we select ordered bases whose union is an ordered basis  for V such that

where each O is a zero matrix, and each  is a square matrix of the form  or

for some eigenvalue  of T. Such a matrix  is called a Jordan block corresponding to , and the matrix  is called a Jordan canonical form of T. We also say that the ordered basis  is a Jordan canonical basis for T. Observe that each Jordan block  is “almost” a diagonal matrix—in fact,  is a diagonal matrix if and only if each  is of the form .

# Example 1

Suppose that T is a linear operator on , and  is an ordered basis for  such that

is a Jordan canonical form of T. Notice that the characteristic polynomial of T is , and hence the multiplicity of each eigenvalue is the number of times that the eigenvalue appears on the diagonal of J. Also observe that , and  are the only vectors in  that are eigenvectors of T. These are the vectors corresponding to the columns of J with no 1 above the diagonal entry.

In Sections 7.1 and 7.2, we prove that every linear operator whose characteristic polynomial splits has a Jordan canonical form that is unique up to the order of its Jordan blocks. Nevertheless, it is not the case that the Jordan canonical form is completely determined by the characteristic polynomial of the operator. For example, let  be the linear operator on  such that , where  is the ordered basis in Example 1 and

Then the characteristic polynomial of  is also . But the operator  has the Jordan canonical form , which is different from J, the Jordan canonical form of the linear operator T of Example 1.

Consider again the matrix J and the ordered basis  of Example 1. Notice that , and so  . Similarly, . Since  and  are eigenvectors of T corresponding to , it follows that  for  and 4. Similarly  for , and  for .

Because of the structure of each Jordan block in a Jordan canonical form, we can generalize these observations: If v lies in a Jordan canonical basis for a linear operator T and is associated with a  Jordan block with diagonal entry , then . Eigenvectors satisfy this condition for .

# Definition.

Let T be a linear operator on a vector space V, and let  be a scalar. A nonzero vector x in V is called a generalized eigenvector of T corresponding to  if  for some positive integer p.

Notice that if x is a generalized eigenvector of T corresponding to , and p is the smallest positive integer for which , then  is an eigenvector of T corresponding to . Therefore  is an eigenvalue of T.

In the context of Example 1, each vector in  is a generalized eigenvector of T. In fact, , and  correspond to the eigenvalue 2,  and  correspond to the eigenvalue 3, and  and  correspond to the eigenvalue 0.

Just as eigenvectors lie in eigenspaces, generalized eigenvectors lie in “generalized eigenspaces.”

# Definition.

Let T be a linear operator on a vector space V, and let  be an eigenvalue of T. The generalized eigenspace of T corresponding to , denoted , is the subset of V defined by

Note that  consists of the zero vector and all generalized eigenvectors corresponding to .

Recall that a subspace W of V is T-invariant for a linear operator T if . In the development that follows, we assume the results of Exercises 3 and 4 of Section 5.4. In particular, for any polynomial g(x), if W is T-invariant, then it is also g(T)-invariant. Furthermore, the range of a linear operator T is T-invariant.

# Theorem 7.1.

Let T be a linear operator on a vetor space V, and let  be an eigenvalue of T. Then

1.  is a T-invariant subspace of V containing  (the eigenspace of T corresponding to ).

2. For any eigenvalue  of T such that ,

3. For any scalar , the restriction of  to  is one-to-one and onto.

# Proof.

(a) Clearly, . Suppose that x and y are in . Then there exist positive integers p and q such that

Therefore

and hence . The proof that  is closed under scalar multiplication is straightforward.

To show that  is T-invariant, consider any . Choose a positive integer p such that . Then

Therefore .

Finally, it is a simple observation that  is contained in .

(b) Suppose that . Then there exist positive integers p and q such that

As polynomials in  and  are relatively prime, and hence by Theorem E.2 in Appendix E there exist polymomials  and  such that

It follows that

and hence .

(c) Let  be any scalar such that . Since  is T-invariant, it is also -invariant. If  for some , then . Hence  by (b), and we conclude that the restriction of  to is one-to-one.

We now show that the restriction of  to  is onto. Let , p be the smallest positive integer for which , and

It is easily shown that W is a T-invariant subspace of , and hence it is also -invariant. So  maps W to W. Since  is one-to-one on  , it must be one-to-one on W. Thus, because W is finite-dimensional,  maps W onto W. So there exists a  such that . Therefore  maps  onto .

# Theorem 7.2.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Suppose that  is an eigenvalue of T with multiplicity m. Then

1.

2.

# Proof.

(a) Let , and let h(t) be the characteristic polynomial of . By Theorem 5.20 (p. 312), h(t) divides the characteristic polynomial of T, and by Theorem 7.1(b),  is the only eigenvalue of . Hence , where , and .

(b) Clearly . The characteristic polynomial of T is of the form , where g(t) is a product of powers of the form  for eigenvalues . By Theorem 7.1, g(T) is one-to one on . Since  is finite-dimensional, it is also onto. Let . Then there exists  such that . Hence

by the Cayley-Hamilton Theorem. Therefore , and thus .

# Theorem 7.3.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let  be the distinct eigenvalues of T. Then, for every , there exist unique vectors , for , such that

# Proof.

Suppose the characteristic polynomial of T is

For , let

First, we prove the existence of the vectors . Because the polynomials  are relatively prime for , by Theorem E.2 there exist polynomials  for  such that

It follows that

For , let . Then for each i,

by the Cayley-Hamilton theorem. It follows that  for .

Now we prove the uniqueness of the vectors . Since each  is T-invariant, each  is invariant under the operator  for . Furthermore,  is one-to-one on  by Theorem 7.1(c) and maps  to {0} for  by Theorem 7.2(b).

Suppose that

where  for . Then for ,

and hence  since  for . Because  is one-to-one on , it follows that .

The next result extends Theorem 5.8(b) (p. 267) to all linear operators whose characteristic polynomials split. In this case, the eigenspaces are replaced by generalized eigenspaces.

# Theorem 7.4.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial  splits. For , let  be an ordered basis for . Then

1.  for .

2.  is an ordered basis for V.

3.  for all i.

# Proof.

(a) This is a direct consequence of Theorem 7.1(b).

(b) First, we prove that  is linearly independent. Consider a linear combination of vectors in  equal to 0. For each i, let  be the sum of the terms of this linear combination involving the vectors in . Then

and hence  for all i by Theorem 7.3. Since each  is linearly independent, it follows that all of the coefficients in the linear combination are zeros. Hence  is linearly independent.

We now prove that  is a spanning set for V. Consider any vector . By Theorem 7.3, for  there exist vectors  such that . Since each  is a linear combination of the vectors in , v is a linear combination of the vectors in . Hence  spans V.

(c) Since dim(V) is equal to the degree of the characteristic polynomial of T, it follows that . By (b), . Hence . By Theorem 7.2(a),  for all i. Consequently,  for all i, which is the desired result.

# Corollary.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Then T is diagonalizable if and only if  for every eigenvalue  of T.

# Proof.

Combining Theorems 7.4 and 5.8(a) (p. 267), we see that T is diagonalizable if and only if  for each eigenvalue  of T. But , and hence these subspaces have the same dimension if and only if they are equal.

We now focus our attention on the problem of selecting suitable bases for the generalized eigenspaces of a linear operator so that we may use Theorem 7.4 to obtain a Jordan canonical basis for the operator. For this purpose, we consider again the basis  of Example 1. We have seen that the first four vectors of  lie in the generalized eigenspace . Observe that the vectors in  that determine the first Jordan block of J are of the form

Furthermore, observe that . The relation between these vectors is the key to finding Jordan canonical bases. This leads to the following definitions.

# Definitions.

Let T be a linear operator on a vector space V, and let x be a generalized eigenvector of T corresponding to the eigenvalue . Suppose that p is the smallest positive integer for which . Then the ordered set

is called a cycle of generalized eigenvectors of T corresponding to . The vectors  and x are called the initial vector and the end vector of the cycle, respectively. We say that the length of the cycle is p.

Notice that the initial vector of a cycle of generalized eigenvectors of a linear operator T is the only eigenvector of T in the cycle. Also observe that if x is an eigenvector of T corresponding to the eigenvalue , then the set  is a cycle of generalized eigenvectors of T corresponding to  of length 1.

In Example 1, the subsets , and  are the cycles of generalized eigenvectors of T that occur in  . Notice that  is a disjoint union of these cycles. Furthermore, setting  for , we see that  is a basis for  and  is the ith Jordan block of the Jordan canonical form of T. This is precisely the condition that is required for a Jordan canonical basis.

# Theorem 7.5.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits, and suppose that  is a basis for V such that  is a disjoint union of cycles of generalized eigenvectors of T. Then the following statements are true.

1. For each cycle  of generalized eigenvectors contained in  is T-invariant, and  is a Jordan block.

2.  is a Jordan canonical basis for V.

# Proof.

(a) Suppose that  corresponds to the eigenvalue  has length p, and x is the end vector of . Then , where

So

and hence . For ,

This shows that  maps W into itself, and so T does also. Thus, by the preceding equations, we see that  is a Jordan block.

For (b), simply repeat the arguments of (a) for each cycle in  in order to obtain . We leave the details as an exercise.

In view of this result, we must show that, under appropriate conditions, there exist bases that are disjoint unions of cycles of generalized eigenvectors. Since the characteristic polynomial of a Jordan canonical form splits, this is a necessary condition. We will soon see that it is also sufficient. The next result moves us toward the desired existence theorem.

# Theorem 7.6.

Let T be a linear operator on a vector space V, and let  be an eigenvalue of T. Suppose that  are cycles of generalized eigenvectors of T corresponding to  such that the initial vectors of the ’s are distinct and form a linearly independent set. Then the ’s are disjoint, and their union  is linearly independent.

# Proof.

Exercise 5 shows that the ’s are disjoint.

The proof that  is linearly independent is by mathematical induction on the number of vectors in . If this number is 1, then the result is clear. So assume that, for some integer , the result is true whenever  has fewer than n vectors, and suppose that  has exactly n vectors. Let W be the subspace of V generated by . Clearly W is -invariant, and . Let U denote the restriction of  to W.

For each i, let  denote the cycle obtained from  by deleting the end vector. Note that if  has length one, then . In the case that , each vector of  is the image under U of a vector in , and conversely, every nonzero image under U of a vector of  is contained in . Let .

Then by the last statement,  generates R(U). Furthermore,  consists of  vectors, and the initial vectors of the ’s are also initial vectors of the ’s. Thus we may apply the induction hypothesis to conclude that  is linearly independent. Therefore  is a basis for R(U). Hence . Since the q initial vectors of the ’s form a linearly independent set and lie in N(U), we have . From these inequalities and the dimension theorem, we obtain

We conclude that . Since  generates W and consists of n vectors, it must be a basis for W. Hence  is linearly independent.

# Corollary.

Every cycle of generalized eigenvectors of a linear operator is linearly independent.

# Theorem 7.7.

Let T be a linear operator on a finite-dimensional vector space V, and let  be an eigenvalue of T. Then  has an ordered basis consisting of a union of disjoint cycles of generalized eigenvectors corresponding to .

# Proof.

The proof is by mathematical induction on . The result is clear for . So suppose that for some integer  the result is true whenever , and assume that . Let U denote the restriction of  to . Then R(U) is a subspace of  of lesser dimension, and R(U) is the space of generalized eigenvectors corresponding to  for the restriction of T to R(U). Therefore, by the induction hypothesis, there exist disjoint cycles  of generalized eigenvectors of this restriction, and hence of T itself, corresponding to  for which  is a basis for R(U). For , the end vector of  is the image under U of a vector , and so we can extend each  to a larger cycle  of generalized eigenvectors of T corresponding to . For , let  be the initial vector of  (and hence of ). Since  is a linearly independent subset of , this set can be extended to a basis  for . Then  are disjoint cycles of generalized eigenvectors of T corresponding to  such that the initial vectors of these cycles are linearly independent. Therefore their union  is a linearly independent subset of  by Theorem 7.6.

Finally, we show that  is a basis for . Suppose that  consists of  vectors. Then  consists of  vectors. Furthermore, since  is a basis for , it follows that . Therefore

So  is a linearly independent subset of  containing  vectors, and thus  is a basis for .

# Corollary 1.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits. Then T has a Jordan canonical form.

# Proof.

Let  be the distinct eigenvalues of T. By Theorem 7.7, for each i there is an ordered basis  consisting of a disjoint union of cycles of generalized eigenvectors corresponding to . Let . Then, by Theorem 7.5(b),  is a Jordan canonical basis for V.

The Jordan canonical form also can be studied from the viewpoint of matrices.

# Definition.

Let  be such that the characteristic polynomial of A (and hence of ) splits. Then the Jordan canonical form of A is defined to be the Jordan canonical form of the linear operator  on .

The next result is an immediate consequence of this definition and Corollary 1.

# Corollary 2.

Let A be an  matrix whose characteristic polynomial splits. Then A has a Jordan canonical form J, and A is similar to J.

# Proof.

Exercise.

We can now compute the Jordan canonical forms of matrices and linear operators in some simple cases, as is illustrated in the next two examples. The tools necessary for computing the Jordan canonical forms in general are developed in the next section.

# Example 2

Let

To find the Jordan canonical form for A, we need to find a Jordan canonical basis for .

The characteristic polynomial of A is

Hence  and  are the eigenvalues of A with multiplicities 1 and 2, respectively. By Theorem 7.4, , and . By Theorem 7.2, , and . Since , we have that . Observe that  is an eigenvector of T corresponding to ; therefore

is a basis for .

Since  and a generalized eigenspace has a basis consisting of a union of cycles, this basis is either a union of two cycles of length 1 or a single cycle of length 2. Since the rank of  is 2 it follows that . Hence the former is not the case. It can easily be shown that  is an eigenvector of A corresponding to . This vector can serve as the initial vector of the cycle. Any solution to the equation  will serve as the end vector. For example, we can take . Thus

is a cycle of generalized eigenvectors for . Finally, we take the union of these two bases to obtain

which is a Jordan canonical basis for A. Therefore,

is a Jordan canonical form for A. Notice that A is similar to J. In fact, , where Q is the matrix whose columns are the vectors in .

# Example 3

Let T be the linear operator on  defined by . We find a Jordan canonical form of T and a Jordan canonical basis for T.

Let  be the standard ordered basis for . Then

which has the characteristic polynomial . Thus  is the only eigenvalue of T, and hence  by Theorem 7.4. So  is a basis for . Now

Therefore a basis for  cannot be a union of two or three cycles because the initial vector of each cycle is an eigenvector, and there do not exist two or more linearly independent eigenvectors. So the desired basis must consist of a single cycle of length 3. If  is such a cycle, then  determines a single Jordan block

which is a Jordan canonical form of T.

The end vector h(x) of such a cycle must satisfy . In any basis for , there must be a vector that satisfies this condition, or else no vector in  satisfies this condition, contrary to our reasoning. Testing the vectors in , we see that  is acceptable. Therefore

is a Jordan canonical basis for T.

In the next section, we develop a computational approach for finding a Jordan canonical form and a Jordan canonical basis. In the process, we prove that Jordan canonical forms are unique up to the order of the Jordan blocks.

Let T be a linear operator on a finite-dimensional vector space V, and suppose that the characteristic polynomial of T splits. By Theorem 5.10 (p. 277), T is diagonalizable if and only if V is the direct sum of the eigenspaces of T. If T is diagonalizable, then the eigenspaces and the generalized eigenspaces coincide.

For those familiar with the material on direct sums in Section 5.2, we conclude by stating a generalization of Theorem 5.10 for nondiagonalizable operators.

# Theorem 7.8.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits. Then V is the direct sum of the generalized eigenspaces of T.

Exercise.

# Exercises

1. Label the following statements as true or false.

1. Eigenvectors of a linear operator T are also generalized eigenvectors of T.

2. It is possible for a generalized eigenvector of a linear operator T to correspond to a scalar that is not an eigenvalue of T.

3. Any linear operator on a finite-dimensional vector space has a Jordan canonical form.

4. A cycle of generalized eigenvectors is linearly independent.

5. There is exactly one cycle of generalized eigenvectors corresponding to each eigenvalue of a linear operator on a finite-dimensional vector space.

6. Let T be a linear operator on a finite-dimensional vector space whose characteristic polynomial splits, and let  be the distinct eigenvalues of T. If, for each  is a basis for , then  is a Jordan canonical basis for T.

7. For any Jordan block J, the operator  has Jordan canonical form J.

8. Let T be a linear operator on an n-dimensional vector space whose characteristic polynomial splits. Then, for any eigenvalue  of .

2. For each matrix A, find a basis for each generalized eigenspace of  consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form J of A.

1.

2.

3.

4.

3. For each linear operator T, find a basis for each generalized eigenspace of T consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form J of T.

1. Define T on  by .

2. V is the real vector space of functions spanned by the set of real-valued functions , and T is the linear operator on V defined by .

3. T is the linear operator on  defined for all  by , where . 1

4.  for all .

4. † Let T be a linear operator on a vector space V, and let  be a cycle of generalized eigenvectors that corresponds to the eigenvalue . Prove that  is a T-invariant subspace of V. Visit goo.gl/Lw4ahY for a solution.

5. Let  be cycles of generalized eigenvectors of a linear operator T corresponding to an eigenvalue . Prove that if the initial eigenvectors are distinct, then the cycles are disjoint.

6. Let T:  be a linear transformation. Prove the following results.

1.

2.

3. If  (so that T is a linear operator on V) and  is an eigenvalue of T, then for any positive integer k

7. Let U be a linear operator on a finite-dimensional vector space V. Prove the following results.

1.

2. If  for some positive integer m, then  for any positive integer .

3. If  for some positive integer m, then  for any positive integer .

4. Let T be a linear operator on V, and let  be an eigenvalue of T. Prove that if  for some integer m, then .

5. Second Test for Diagonalizability. Let T be a linear operator on V whose characteristic polynomial splits, and let  be the distinct eigenvalues of T. Then T is diagonalizable if and only if  for .

6. Use (e) to obtain a simpler proof of Exercise 24 of Section 5.4: If T is a diagonalizable linear operator on a finite-dimensional vector space V and W is a T-invariant subspace of V, then  is diagonalizable.

8. Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits and has Jordan canonical form J. Prove that for any nonzero scalar c, cJ is a Jordan canonical form for cT.

9. Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits.

1. Prove Theorem 7.5(b).

2. Suppose that  is a Jordan canonical basis for T, and let  be an eigenvalue of T. Let . Prove that  is a basis for .

10. Let T be a linear operator on a finite-dimensional vector space whose characteristic polynomial splits, and let  be an eigenvalue of T.

1. Suppose that  is a basis for  consisting of the union of q disjoint cycles of generalized eigenvectors. Prove that .

2. Let  be a Jordan canonical basis for T, and suppose that  has q Jordan blocks with  in the diagonal positions. Prove that .

11. Prove Corollary 2 to Theorem 7.7.

12. Let T be a linear operator on a finite-dimensional vector space V, and let  be an eigenvalue of T with corresponding eigenspace and generalized eigenspace  and , respectively. Let U be an invertible linear operator on V that commutes with T (i.e., ). Prove that  and .

Exercises 13 and 14 are concerned with direct sums of matrices, defined in Section 5.4 on page 318.

1. Prove Theorem 7.8.

2. Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let  be the distinct eigenvalues of T. For each i, let  be the Jordan canonical form of the restriction of T to . Prove that

is the Jordan canonical form of J.


text:
# Math Help - please help

1. ## please help

ok quick question.... why in system of equation is it the x y z are written in such a way across

but in others the it is in written x
y
z
vertically?

2. Its pretty much convention but sometimes we want to distinguish between "row vectors" and "column vectors" because we want to treat them as matrices. We can multiply $\begin{pmatrix}a & b & c\end{pmatrix}\begin{pmatrix}d \\ e\\ f\end{pmatrix}$ or $\begin{pmatrix}a \\ b \\ c\end{pmatrix}\begin{pmatrix}d & e & f\end{pmatrix}$ but we cannot multiply $\begin{pmatrix}a \\ b \\ c\end{pmatrix}\begin{pmatrix}d \\ e\\ f\end{pmatrix}$ or $\begin{pmatrix}a & b & c\end{pmatrix}\begin{pmatrix}d & e & f\end{pmatrix}$.

To multiply matrices the number of columns of the first must equal the number of rows of the second. In the first example above those are both 3 and in the second example they are both 1. In the the third example, the number of columns in the first matrix is 1 and the number of rows in the second is 3. In the fourth example, the number of columens in the first matrix is 1 and the number of rows in the second is 1.


text:
Remove unwanted variation/confounder due to library size with sva
1
0
Entering edit mode
Jasmin • 0
@6462d400
Last seen 7 weeks ago
Netherlands

Hi,

Library size is a source of unwanted variation in my data (dataset of healthy vs disease samples) because I can see in the pca plot that samples are separated to some degree by library size. Can sva remove the effect that library size has on my data or is another package better suited for this? To be clear: I'm not trying to remove batch effects.

Thanks!

sva • 439 views
1
Entering edit mode
ATpoint ★ 1.6k
@atpoint-13662
Last seen 6 hours ago
Germany

Just asking to be sure, how did you diagnose that library size is a confounder? Did you run a PCA based on values where library size has been normalized for, like vst() from DESeq2? If so then svaseq might be used here. Check if it identifies the library size as a significant surrogate variable. The manual covers a basic workflow. If you need details then please add code and plots.

0
Entering edit mode

Hi thanks for answering! This is the code that I want to use:

dds<-DESeqDataSetFromMatrix(df,colData =coldata, design = ~lib.size+sex+Patient_group)
dds <- estimateSizeFactors(dds)
dat  <- counts(dds, normalized = TRUE)
idx  <- rowMeans(dat) > 1
dat  <- dat[idx, ]
mod  <- model.matrix(~lib.size+sex+Patient_group, colData(dds))
mod0 <- model.matrix(~lib.size, colData(dds))
svseq <- svaseq(dat, mod, mod0)
ddssva <- dds
design(ddssva) <- ~sex + lib.size + Patient_group
design(ddssva)
rld <- vst(ddssva, blind=FALSE)
mat <- assay(rld)
mm <- model.matrix(~sex+Patient_group. colData(rld))
mat <- limma::removeBatchEffect(mat, design=mm, batch=lib.size)
assay(rld) <- mat
plotPCA(rld,"Patient.group")

However, I'm unsure about the mod0 line which should only include the "adjustment" variables, so that would be libsize I think? I'm pretty sure I'm making a mistake with the svseq line because I don't want to add any surrogate variables, just remove lib.size as a confounder, but I don't know how to write this down. I'd really appreciate your help!

0
Entering edit mode

The plots aren't really the problem, it's just the code that is giving me great difficulties. Thanks again!

1
Entering edit mode

You are generating surrogate variables, but you never use them. Also mod0 should include all the adjustment variables, which includes sex (which is almost always included as a nuisance variable rather than a parameter of interest). Also, if you are fitting lib.size as part of your model, then the surrogate variables won't adjust for library size unless there are differences in variability due to library size (which is not a completely unexpected result, particularly if you have widely divergent library sizes).

Also also, you don't want to use lib.size as a batch argument in removeBatchEffect. The expectation for a batch is that it is a factor, whereas lib.size is continuous. One of the first things that happens to the batch argument is conversion to factor. If you use a continuous variable you will end up with N factor levels (where N is the number of samples), which is not what you want to do. The covariates argument is what you use for continuous variables. And if you want to adjust for the surrogate variables you have to cbind the lib.size and svseq$sv together and pass that to removeBatchEffect. ADD REPLY 0 Entering edit mode Thanks for the help! I have now changed libsize to a continous variable (it was a dichotomous variable: high/low at first). I should use the pre-normalized counts for this right? My code is now as follows: coldata$libsize<-colSums(df)
dds<-DESeqDataSetFromMatrix(df,colData =coldata, design = ~lib.size+Patient_group) # I have removed sex as variable
dds <- estimateSizeFactors(dds)
dat  <- counts(dds, normalized = TRUE)
idx  <- rowMeans(dat) > 1
dat  <- dat[idx, ]
mod  <- model.matrix(~lib.size+Patient_group, colData(dds))
mod0 <- model.matrix(~lib.size, colData(dds))
svseq <- svaseq(dat, mod, mod0) #this gives me 14 significant SVs
ddssva <- dds
ddssva$SV1<- svseq$sv[,1]
ddssva$SV2<- svseq$sv[,2]
ddssva$SV3<- svseq$sv[,3]
ddssva$SV4<-svseq$sv[,4]
ddssva$SV5<-svseq$sv[,5]
ddssva$SV6<-svseq$sv[,6]
ddssva$SV7<-svseq$sv[,7]
ddssva$SV8<-svseq$sv[,8]
ddssva$SV9<-svseq$sv[,9]
ddssva$SV10<-svseq$sv[,10]
ddssva$SV11<-svseq$sv[,11]
ddssva$SV12<-svseq$sv[,12]
ddssva$SV13<-svseq$sv[,13]
ddssva$SV14<-svseq$sv[,14]
design(ddssva) <- ~SV1+SV2+SV3+SV4+SV5+SV6+SV7+SV8+SV9+SV10+SV11+SV12+SV13+SV14+lib.size+Patient_group
design(ddssva)
rld <- vst(ddssva, blind=FALSE)
mat <- assay(rld)
mm <- model.matrix(~Patient.group,colData(rld))
svs_and_libsize_cbind<-cbind(svseq$sv,coldata$libsize)
mat <- limma::removeBatchEffect(mat, design=mm, covariates=svs_and_libsize_cbind)
assay(rld) <- mat
plotPCA(rld,"Patient_group")

This gives me the following plot:

The seperation isn't bad but PC values are low. When I do not remove SV, the PC values increase but separation worsens:

So would it be fair to conclude that batch effect is more important than biological effects for my dataset?

0
Entering edit mode

Do you have batches? If you have known batches, people usually directly account for them as part of the linear model rather than doing something indirect like computing surrogate variables.

0
Entering edit mode

I actually don't have known batches (I'm using data from another researcher from some time ago who does not have this information anymore). Do you think that my code lines look ok in general? On another note: the sva code line

svseq <- svaseq(dat, mod, mod0)

Gives me the following error: Error in solve.default(t(mod) %*% mod) : system is computationally singular: reciprocal condition number = 1.32097e-16 I have found that you can get this by two covariates (patient group and libsize in my case, I presume) being highly colinear or because the number of cases is approximately as large as the number of covariates. These are both not true for my dataset I think (case number is 174, covariate number is 2). Do you have any idea if something else could cause this error?

0
Entering edit mode

And would it be possible to remove confounders using ruvseq and limma's remove batch effect function?

0
Entering edit mode

Yes, that is another method you could use.


text:
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# Open Chemistry

### formerly Central European Journal of Chemistry

IMPACT FACTOR 2018: 1.512
5-year IMPACT FACTOR: 1.599

CiteScore 2018: 1.58

SCImago Journal Rank (SJR) 2018: 0.345
Source Normalized Impact per Paper (SNIP) 2018: 0.684

ICV 2018: 163.25

Open Access
Online
ISSN
2391-5420
See all formats and pricing
More options …
Volume 1, Issue 1

# Modeling and simulation of free radical polymerization of styrene under semibatch reactor conditions

Silvia Curteanu
• Department of Chemical Engineering, Technical University “Gh. Asachi”, IASI, B-dul D. Mangeron No. 71A, 6600 IASI, Romania
• Email
• Other articles by this author:
Published Online: 2003-03-01 | DOI: https://doi.org/10.2478/BF02479259

## Abstract

The first part of this approach is concerned with the elaboration of a radical polymerization model of styrenne, based on a kinetic diagram that includes chemical and thermal initiation, propagation, termination by recombination and chain transfer to the monomer. Furthermore, volume contraction during polymerization is considered, as well as the gel and glass effects. The mathematical formalism that describes the model in terms of moments is explored in detail. The model was then used to predict the changes in monomer conversion and molecular weight after intermediate addition of initiator and monomer. The results of this operation are dependent on the conditions of the reaction mass, quantity, and moment of substance addition. Therefore, the simulations were performed at different times with respect to the gel effect; before, during and after this phenomenon, and also with respect to different temperatures and initiators. Increasing the initiator concentration before the gel effect leads to an earlier appearance of the phenomenon and to a decrease in molecular weight. The ratio $$\bar M_w /\bar M_n$$ reveals a polydispersity index smaller for the intermediate addition of initiator. No significant changes take place during or after the gel effect. If along with the initiator, unreacted monomver (used to dissolve the initiator) enters the reactor, a small dip in conversion is observed. The general conclusion of this paper reveals the intermediate addition of initiator as a method to control polymer properties and to prevent the “dead-end” polymerization of styrene.

• [1] D.S. Achilias and C. Kiparissides: “Development of a General Mathematical Framework for Modeling Diffusion-Controled Free Radical Polymerization Reactions”, Macromolecules, Vol. 25, (1992), pp. 3739–3750. http://dx.doi.org/10.1021/ma00040a021

• [2] R.G. Gilbert, J.F. Anstey, N. Subramanian, M.J. Monteiro: “Emulsion Polymerization as a novel tool in controlled free-radical polymerization”, Polym. Prepr., Vol. 40, (1999), pp. 297–298. Google Scholar

• [3] A. Sirohi and K. Ravindranath: “Modeling of Ionic Polymerization Processes: Styrene and Butadiene”, AIChE, Spring 99 Meeting, Houston, 1999. Google Scholar

• [4] K.J. Kim: Modeling and Control of Continuous Free Radical Polymerization Reactors, Thesis (PhD), University of Maryland, 1991. Google Scholar

• [5] S. Curteanu, V. Bulacovschi, R. Diaconescu: “Modeling of Free Radical Polymerization of Styrene Using a Binary Mixture of Initiarors”, Proc. Rom. Acad., Vol. 3, (2001), pp. 207–213. Google Scholar

• [6] S. Curteanu and V. Bulacovschi: “Modeling of Free Radical Polymerization of Styrene Using a Binary Mixture of Initiators and Isothermal Steps of Temperature”, Rev. Roum. Chim., in press. Google Scholar

• [7] W.Y. Chiu, G.M. Carratt, D.S. Soong: “A Computer Model for the Gel effect in Free Radical Polymerization”, Macromolecules, Vol. 16, (1983), pp. 348–357. http://dx.doi.org/10.1021/ma00237a002

• [8] M. Constantinescu, V. Bulacovschi, S. Curteanu: “Free Radical Polymerization of Styrene. Modeling and Simulation”, Bul. Polyt. Inst. Iasi, Tom XLIII (XLVII), 3–4, (1997), pp. 91–103. Google Scholar

• [9] G.A. O'Neil, M.B. Wisnudel, J.M. Torkelson: “An Evaluation of Free Volume Approaches to Describe the Gel Effect in Free Radical Polymerization”, Macromolecules, Vol. 31, (1998), pp. 4537–4545. http://dx.doi.org/10.1021/ma9713670

Published Online: 2003-03-01

Published in Print: 2003-03-01

Citation Information: Open Chemistry, Volume 1, Issue 1, Pages 69–90, ISSN (Online) 2391-5420,

Export Citation

## Citing Articles

[1]
Venkat Reddy Regatte, Hanyu Gao, Ivan A. Konstantinov, Steven G. Arturo, and Linda J. Broadbelt
Macromolecular Theory and Simulations, 2014, Volume 23, Number 9, Page 564
[2]
D. P. Vent, V. P. Savel’yanov, A. G. Lopatin, and M. A. Safin
Theoretical Foundations of Chemical Engineering, 2014, Volume 48, Number 3, Page 332
[3]
Weijie Lin, Lorenz T. Biegler, and Annette M. Jacobson
Macromolecular Theory and Simulations, 2011, Volume 20, Number 2, Page 146
[4]
Dimitris S. Achilias
Macromolecular Theory and Simulations, 2007, Volume 16, Number 4, Page 319
[5]
Silvia Curteanu and Corneliu Petrila
International Journal of Quantum Chemistry, 2006, Volume 106, Number 6, Page 1445


text:
School of Economics and Management
Beihang University
http://yanfei.site

## Discontinuous Functions

• The Newton Method requires first and second derivatives.
• If derivatives are not available the they can be approximated by Quasi-Newton methods.
• What if the derivatives do not exist?
• This may occur if there are discontinuities in the function.

• Suppose the aim is to optimize income of the business by selecting the number of workers.
• If too many workers are employed, they may be less efficient and the income of the company goes down.

• Now suppose that there is a tax that the company must pay.
• Companies with less than 50 workers do not pay the tax.
• Companies with more than 50 workers do pay the tax.
• How does this change the problem?

• The Nelder Mead algorithm is robust even when the functions are discontinuous.
• The idea is based on evaluating the function at the vertices of an n-dimensional simplex where n is the number of input variables into the function.
• For two dimensional problems the n-dimensional simplex is simply a triangle, and each corner is one vertex
• In general there are n + 1 vertices.

## Step 1: Evaluate Function

• For each vertex $${\mathbf x_j}$$ evaluate the function $$f({\mathbf x_j})$$
• Order the vertices so that $f({\mathbf x_1})\leq f({\mathbf x_2})\leq\ldots\leq f({\mathbf x_{n+1}}).$
• Suppose that the aim is to minimize the function, then $$f({\mathbf x_{n+1}})$$ is the worst point.
• The aim is to replace $$f({\mathbf x_{n+1}})$$ with a better point.

## Step 2: Find Centroid

• After eliminating the worst point $${\mathbf x_{n+1}}$$, compute the centroid of the remaining $$n$$ points ${\mathbf x_0}=\frac{1}{n}\sum_{j=1}^{n} {\mathbf x_j}.$
• For the 2-dimensional example the centroid will be in the middle of a line.

## Step 3: Find reflected point

• Reflect the worst point around the centroid to get the reflected point.
• The formula is: ${\mathbf x_r}={\mathbf x_0}+\alpha({\mathbf x_0}-{\mathbf x_{n+1}}).$
• A common choice is $$\alpha=1$$.
• In this case the reflected point is the same distance from the centroid as the worst point.

## Three cases

1. $$f({\mathbf x_1})\leq f({\mathbf x_r})<f({\mathbf x_n})$$
• $${\mathbf x_r}$$ is neither best nor worst point
2. $$f({\mathbf x_r})<f({\mathbf x_1})$$
• $${\mathbf x_r}$$ is the best point
3. $$f({\mathbf x_r})\geq f({\mathbf x_n})$$
• $${\mathbf x_r}$$ is the worst point

## Case 1

In Case 1 a new simplex is formed with $${\mathbf x_{n+1}}$$ replaced by the reflected point $${\mathbf x_{r}}$$. Then go back to step 1.

## Case 2

In Case 2, $${\mathbf x_r}<{\mathbf x_1}$$. A good direction has been found so we expand along that direction ${\mathbf x_e}={\mathbf x_0}+\gamma({\mathbf x_r}-{\mathbf x_0}).$

A common choice is $$\gamma=2$$

## Choosing the Expansion Point

• Evaluate $$f({\mathbf x_e})$$.
• If $$f({\mathbf x_e}) < f({\mathbf x_r})$$:
• The expansion point is better than the reflection point. Form a new simplex with the expansion point
• If $$f({\mathbf x_r})\leq f({\mathbf x_e})$$:
• The expansion point is not better than the reflection point. Form a new simplex with the reflection point.

## Case 3

Case 3 implies that there may be a valley between $${\mathbf x_{n+1}}$$ and $${\mathbf x_{r}}$$ so find the contracted point. A new simplex is formed with the contraction point if it is better than $${\mathbf x_{n+1}}$$ ${\mathbf x_c}={\mathbf x_0}+\rho({\mathbf x_{n+1}}-{\mathbf x_0})$

A common choice is $$\rho=0.5$$

## Shrink

If $$f({\mathbf x_{n+1}})\leq f({\mathbf x_{c}})$$ then contracting away from the worst point does not lead to a better point. In this case the function is too irregular a smaller simplex should be used. Shrink the simplex ${\mathbf x_i}={\mathbf x_1}+\sigma({\mathbf x_i}-{\mathbf x_1})$

A popular choice is $$\sigma=0.5$$.

## Summary

• Order points
• Find centroid
• Find reflected point
• Three cases:
1. Case 1 ($$f({\mathbf x_1})\leq f({\mathbf x_r})<f({\mathbf x_n})$$): Keep $${\mathbf x_r}$$
2. Case 2 ($$f({\mathbf x_r}) < f({\mathbf x_1})$$): Find $$\mathbf x_e$$.
• If $$f({\mathbf x_e})<f({\mathbf x_r})$$ then keep $$\mathbf x_e$$
• Otherwise keep $$\mathbf x_r$$
3. Case 3 ($$f({\mathbf x_r})\geq f({\mathbf x_n})$$): Find $$\mathbf x_c$$
• If $$f({\mathbf x_c})<f({\mathbf x_{n+1}})$$ then keep $$\mathbf x_c$$
• Otherwise shrink

• Find the minimum of the function $$f({\mathbf x})=x_1^2+x_2^2$$
• Use a triangle with vertices $$(1,1)$$, $$(1,2)$$, $$(2,2)$$ as the starting simplex


text:
Volume 334 - The 36th Annual International Symposium on Lattice Field Theory (LATTICE2018) - Hadron Structure
The leading hadronic contribution to the running of the Weinberg angle using covariant coordinate-space methods
M. Cè*, A. Gerardin, K. Ottnad and H.B. Meyer
Full text: pdf
Published on: May 29, 2019
Abstract
We present a preliminary study of the leading hadronic contribution to the running of the Weinberg angle $\theta_{\mathrm{W}}$. The running is extracted from the correlation function of the electromagnetic current with the vector part of the weak neutral current using both the standard time-momentum representation method and the Lorentz-covariant coordinate-space method recently introduced by Meyer. Both connected and disconnected contributions have been computed on $N_{\mathrm{f}}=2+1$ non-perturbatively $O(a)$-improved Wilson fermions configurations from the CLS initiative. Similar covariant coordinate-space methods can be used to compute the leading hadronic contribution to the anomalous magnetic moment of the muon $(g-2)_\mu$ and to the running of the QED coupling $\alpha$.
DOI: https://doi.org/10.22323/1.334.0137
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.


text:
# What would happen to electronic circuits when traveling near the speed of light?

Imagine a space ship, loaded with all sorts of computer systems, traveling near the speed of light.

Electricity itself is very fast, and can reach speeds close the speed of light. (up to 99% according to wikipedia). So, what would happen to the electronic circuits in this spaceship?

Will the computers shut down, because electricity can't reach the components? Or are they just not related to each other and will the computers keep working perfectly?

-

You're dealing with an incomplete form of relativity.

In the frame of the spaceship, nobody will notice anything different, since all inertial frames are equivalent

In the "ground" frame, electricity would be moving at a different speed, by the relation $$\rm v_{e,ground}=\frac{v_{ship}+v_{electricity}}{1+\frac{v_{ship}v_{electricity}}{c^2}}$$

We cannot simply use relative velocities, we need to use the above equation. If you compare this with the time dilation of the system, the computers will all seem to be working the same, albeit slower, from your POV.

-
Thank you! I don't know anything of the maths behind it, I just know the basic concepts. This makes it hard for me to understand the kind of thing I was asking. So basically, if a calculation used to take 1 second, it might take 20 seconds now? –  Simon Verbeke Apr 11 '12 at 11:46
@Simon from the ship's point of view, it will take 1sec. From the POV of an external observer, it will take 30secs. –  Manishearth Apr 11 '12 at 12:11
Aha, thank you! –  Simon Verbeke Apr 11 '12 at 16:38

Your question is about what will be the speed of elektron? Is this sum of speeds of electron and the ship, or not? If it is a sum of these speeds, speed of elektron will be more than speed of light ? If not, then there will be a time which is slowing down, so,"can the electron reach to command in circuits? Is not it?

Firstly the speed of electron won't be more than speed of light acccording to passenger. It will be as usual for passenger. Acoording to observer, if possible, the speed of electron will decelarate by the slowing time and sum of the speeds of ship and electons will be again same as usual. So your computer will work properly.

According to my view, in reality, the time inside ship is under different physics rules. Because by giving more energy to accelarate the ship with it's increasing mass, (I'm thinking according to String theory), the strings are energized and extended, so they can compress the spatial dimensions (on the same route with accelaration). Also "the time" is a wave kind, sourced from universal expansion (it is again my idea, not a proved scientifical data). So time of the ship will be acting as waves. And you know the behaviour of waves while they are passing from different environment as usual. Their wave lenght and speeds are changing, after left this environment, they turn again their ordinary position. So, if you think "the time" as waves, which are also in the speed of light, limit of speed is the limit of these waves. And, if an object is accelarated in these waves, it will be limited at doppler barrier. Because all waves will be overlapping. (And their energy will transfer to mass)

-
Self-promotion should be transparently advertised. –  ACuriousMind Apr 22 at 15:12
The link is clearly advertizing a personal theory in violation of site policy on this topic, so I am going to remove it. (I should also mention that hiding a link to your own blog behind a URL obscurer is not doing you any favor in terms of the reception you'll receive here.) The rest of the answer is fairly incomplete and will likely continue to accumulate downvotes unless improved. –  dmckee Apr 22 at 23:28


text:
# Trigonometry in Algebra 2

Algebra Level 5

$$\sqrt[3]{{\displaystyle\frac{{{x^9} + 9{x^2} - 1}}{3}}} = 2x + 1$$

$$x_1,x_2,...,x_n \in \mathbb {R}$$ are the solutions of this equation.

Let:

$$\displaystyle\frac{1}{\pi }\sum\limits_{i = 1}^n {\arccos \displaystyle\frac{{{x_i}}}{2}} = \displaystyle\frac{a}{b}$$

If $$a,b \in \mathbb {Z}, \gcd (a,b) = 1$$, find $$a+b$$

Hint: Some calculus needs to be applied.

This problem is part of this set

×


text:
• Mikhail B Skopenkov

Articles written in Proceedings – Mathematical Sciences

• Classification of Framed Links in 3-Manifolds

We present a short and complete proof of the following Pontryagin theorem, whose original proof was complicated and has never been published in detail. Let 𝑀 be a connected oriented closed smooth 3-manifold, $L_1(M)$ be the set of framed links in 𝑀 up to a framed cobordism, and $\deg: L_1(M)\to H_1(M;\mathbb{Z})$ be the map taking a framed link to its homology class. Then for each $\alpha\in H_1(M;\mathbb{Z})$ there is a one-to-one correspondence between the set $\deg^{-1}\alpha$ and the group $\mathbb{Z}_{2d(\alpha)}$, where $d(\alpha)$ is the divisibility of the projection of 𝛼 to the free part of $H_1(M;\mathbb{Z})$.

• Proceedings – Mathematical Sciences

Volume 131, 2021
All articles
Continuous Article Publishing mode

• Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
# 7-50 LO 5, 7 The following table contains calculations of several key ratios for a fictional...

7-50      LO 5, 7 The following table contains calculations of several key ratios for a fictional company, Indianola Pharmaceutical Company, a maker of proprietary and prescription drugs. The company is publicly held and is considered a small- to medium-size pharmaceuti- cal company. Approximately 80% of its sales have been in prescrip- tion drugs; the remaining 20% are in medical supplies normally found in a drugstore. The primary purpose of the auditor’s calcula- tions is to identify potential risk areas for the upcoming audit. The auditor recognizes that some of the data might signal the need to gather other industry- or company-specific data. A number of the company’s drugs are patented. Its best-selling drug, Anecillin,

which will come off of patent in two years, has accounted for

approximately 20% of the company’s sales during the past   five years. The auditor’s expectation is that the company’s own trends from the past few years should be relatively consistent this year, and that the company will not have significant deviations from industry norms.

INDIANOLA PHARMACEUTICAL RATIO  ANALYSIS

 Current Ratio                                                 Year One Year Previous Two Years Previous Three Years Previous Current Industry Current ratio                                      1.85 1.89 2.28 2.51 2.13 Quick ratio                                         0.85 0.93 1.32 1.76 1.40 Times interest earned                      1.30 1.45 5.89 6.3 4.50 Days’ sales in receivables                 109 96 100 72 69 Inventory turnover                             2.40 2.21 3.96 5.31 4.33 Days’ sales in  inventory                     152 165 92 69 84 Research & development  as              1.3 % of sales 1.4 1.94 2.03 4.26 Cost of goods sold  as                       38.5 % of sales 40.2 41.2 43.8 44.5 Debt/equity ratio                              4.85 4.88 1.25 1.13 1.25 Earnings per share                         $1.12$2.50 $4.32$4.26 n/a Sales/tangible assets                       0.68 0.64 0.89 0.87 0.99 Sales/total assets                             0.33 0.35 0.89 0.87 0.78 Sales growth over past year            3% 15% 2% 4% 6% a.    What  major  conclusions regarding financial reporting risk

can be drawn from the information shown in the table? Be specific in identifying specific account balances that have a high risk of material misstatement. State how that risk analysis will be used in planning the audit.  You should  iden- tify  a minimum  of  four financial  reporting  risks that  should be addressed during the audit and how they should be addressed.

b.       What other critical background information might you want to obtain as part of planning the audit? What information would you gather during the conduct of the audit? Briefly indicate the probable sources of that information.

c.       Based on the information, what major actions did the company take during the immediately preceding year? Explain.


text:
## Reinforcement Learning 12

Moderator : Pulkit Agrawal

Wed 21 Jul 6 a.m. PDT — 7 a.m. PDT

Abstract:

Chat is not available.

Wed 21 July 6:00 - 6:20 PDT

(Oral)
##### Model-based Reinforcement Learning for Continuous Control with Posterior Sampling

Ying Fan · Yifei Ming

Balancing exploration and exploitation is crucial in reinforcement learning (RL). In this paper, we study model-based posterior sampling for reinforcement learning (PSRL) in continuous state-action spaces theoretically and empirically. First, we show the first regret bound of PSRL in continuous spaces which is polynomial in the episode length to the best of our knowledge. With the assumption that reward and transition functions can be modeled by Bayesian linear regression, we develop a regret bound of $\tilde{O}(H^{3/2}d\sqrt{T})$, where $H$ is the episode length, $d$ is the dimension of the state-action space, and $T$ indicates the total time steps. This result matches the best-known regret bound of non-PSRL methods in linear MDPs. Our bound can be extended to nonlinear cases as well with feature embedding: using linear kernels on the feature representation $\phi$, the regret bound becomes $\tilde{O}(H^{3/2}d_{\phi}\sqrt{T})$, where $d_\phi$ is the dimension of the representation space. Moreover, we present MPC-PSRL, a model-based posterior sampling algorithm with model predictive control for action selection. To capture the uncertainty in models, we use Bayesian linear regression on the penultimate layer (the feature representation layer $\phi$) of neural networks. Empirical results show that our algorithm achieves the state-of-the-art sample efficiency in benchmark continuous control tasks compared to prior model-based algorithms, and matches the asymptotic performance of model-free algorithms.

Wed 21 July 6:20 - 6:25 PDT

(Spotlight)
##### Principled Exploration via Optimistic Bootstrapping and Backward Induction

Chenjia Bai · Lingxiao Wang · Lei Han · Jianye Hao · Animesh Garg · Peng Liu · Zhaoran Wang

One principled approach for provably efficient exploration is incorporating the upper confidence bound (UCB) into the value function as a bonus. However, UCB is specified to deal with linear and tabular settings and is incompatible with Deep Reinforcement Learning (DRL). In this paper, we propose a principled exploration method for DRL through Optimistic Bootstrapping and Backward Induction (OB2I). OB2I constructs a general-purpose UCB-bonus through non-parametric bootstrap in DRL. The UCB-bonus estimates the epistemic uncertainty of state-action pairs for optimistic exploration. We build theoretical connections between the proposed UCB-bonus and the LSVI-UCB in linear setting. We propagate future uncertainty in a time-consistent manner through episodic backward update, which exploits the theoretical advantage and empirically improves the sample-efficiency. Our experiments in MNIST maze and Atari suit suggest that OB2I outperforms several state-of-the-art exploration approaches.

Wed 21 July 6:25 - 6:30 PDT

(Spotlight)
##### Ensemble Bootstrapping for Q-Learning

Oren Peer · Chen Tessler · Nadav Merlis · Ron Meir

Q-learning (QL), a common reinforcement learning algorithm, suffers from over-estimation bias due to the maximization term in the optimal Bellman operator. This bias may lead to sub-optimal behavior. Double-Q-learning tackles this issue by utilizing two estimators, yet results in an under-estimation bias. Similar to over-estimation in Q-learning, in certain scenarios, the under-estimation bias may degrade performance. In this work, we introduce a new bias-reduced algorithm called Ensemble Bootstrapped Q-Learning (EBQL), a natural extension of Double-Q-learning to ensembles. We analyze our method both theoretically and empirically. Theoretically, we prove that EBQL-like updates yield lower MSE when estimating the maximal mean of a set of independent random variables. Empirically, we show that there exist domains where both over and under-estimation result in sub-optimal performance. Finally, We demonstrate the superior performance of a deep RL variant of EBQL over other deep QL algorithms for a suite of ATARI games.

Wed 21 July 6:30 - 6:35 PDT

(Spotlight)
##### Finite-Sample Analysis of Off-Policy Natural Actor-Critic Algorithm

In this paper, we provide finite-sample convergence guarantees for an off-policy variant of the natural actor-critic (NAC) algorithm based on Importance Sampling. In particular, we show that the algorithm converges to a global optimal policy with a sample complexity of $\mathcal{O}(\epsilon^{-3}\log^2(1/\epsilon))$ under an appropriate choice of stepsizes. In order to overcome the issue of large variance due to Importance Sampling, we propose the $Q$-trace algorithm for the critic, which is inspired by the V-trace algorithm (Espeholt et al., 2018). This enables us to explicitly control the bias and variance, and characterize the trade-off between them. As an advantage of off-policy sampling, a major feature of our result is that we do not need any additional assumptions, beyond the ergodicity of the Markov chain induced by the behavior policy.

Wed 21 July 6:35 - 6:40 PDT

(Spotlight)
##### A Regret Minimization Approach to Iterative Learning Control

Naman Agarwal · Elad Hazan · Anirudha Majumdar · Karan Singh

We consider the setting of iterative learning control, or model-based policy learning in the presence of uncertain, time-varying dynamics. In this setting, we propose a new performance metric, planning regret, which replaces the standard stochastic uncertainty assumptions with worst case regret. Based on recent advances in non-stochastic control, we design a new iterative algorithm for minimizing planning regret that is more robust to model mismatch and uncertainty. We provide theoretical and empirical evidence that the proposed algorithm outperforms existing methods on several benchmarks.

Wed 21 July 6:40 - 6:45 PDT

(Spotlight)
##### TempoRL: Learning When to Act

André Biedenkapp · Raghu Rajan · Frank Hutter · Marius Lindauer

Reinforcement learning is a powerful approach to learn behaviour through interactions with an environment. However, behaviours are usually learned in a purely reactive fashion, where an appropriate action is selected based on an observation. In this form, it is challenging to learn when it is necessary to execute new decisions. This makes learning inefficient especially in environments that need various degrees of fine and coarse control. To address this, we propose a proactive setting in which the agent not only selects an action in a state but also for how long to commit to that action. Our TempoRL approach introduces skip connections between states and learns a skip-policy for repeating the same action along these skips. We demonstrate the effectiveness of TempoRL on a variety of traditional and deep RL environments, showing that our approach is capable of learning successful policies up to an order of magnitude faster than vanilla Q-learning.

Wed 21 July 6:45 - 6:50 PDT

(Spotlight)
##### State Relevance for Off-Policy Evaluation

Simon Shen · Jason Yecheng Ma · Omer Gottesman · Finale Doshi-Velez

Importance sampling-based estimators for off-policy evaluation (OPE) are valued for their simplicity, unbiasedness, and reliance on relatively few assumptions. However, the variance of these estimators is often high, especially when trajectories are of different lengths. In this work, we introduce Omitting-States-Irrelevant-to-Return Importance Sampling (OSIRIS), an estimator which reduces variance by strategically omitting likelihood ratios associated with certain states. We formalize the conditions under which OSIRIS is unbiased and has lower variance than ordinary importance sampling, and we demonstrate these properties empirically.

Wed 21 July 6:50 - 6:55 PDT

(Q&A)


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# How to compare linear and categorical models of perceptual comparisons?

Say I have a confusion matrix of perceptual similarity data for stimuli drawn from an underlying continuous spectrum. Participants also learned to categorise these stimuli such that the spectrum is evenly divided into two categories. I want to test whether the observed confusion matrix shows a better fit to either a continuous or a categorical model. One approach I have considered is to create a hypothetical confusion matrix for linear and categorical models (see below) and correlate these with the observed confusion matrix.

Note: Similarity values, 1 = identical, 0 = dissimilar

However, I’m not sure how to compare the degree of fit for each model. One idea is to take the partial correlation to measure how well the confusion matrix correlates with the categorical model after the correlation between the linear model and the data has been accounted for. This approach seems to have good psychophysical validity (since any categorical process would function in addition to a linear sensory one), however I’m concerned that the linear and categorical models are highly correlated (here at 0.83) and so any partial correlation approach would not be valid. Including more stimuli near the category boundary would probably go some way to decorrelating these models, however I’m wondering if there is a more sensible approach to figuring this out (but not orthogonalising the models)? Any leads appreciated.

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This sounds a lot like tests of categorical perception done in psycholinguistics. Categorical perception states that people perceive phonemes categorically, even though the acoustic signal varies continuously.

For example, the phonemes $/b/$ and $/p/$ differ on only one acoustic feature: voice onset time. Using computer synthesized speech, one can experimentally manipulate the voice onset time along a continuous dimension. In a prototypical experiment, subjects are asked to identify each sound as a $/b/$ or $/p/$. If perception were continuous, you might expect that the proportion of $/b/$ responses varies linearly with the voice onset time-- but it doesn't. As you can see in the red line, there is a sharp drop around 20ms in which people perceive everything to the left as $/b/$ and everything to the right as $/p/$.

Another way to see this is to plot discriminability (blue line). Here, participants are presented two stimuli from adjacent VOT bins and asked to judge which one matches a 'target' stimulus. Generally, people perform near chance at discriminating between the two phonemes unless the two stimuli cross the VOT boundary. Again, if perception were linear, you might expect a straight blue line rather than a peaked one.

Most researchers will use t-tests or ANOVA to test for differences of between and within category discrimination. For instance, one might conduct a series of t-tests on discriminability to show that only when stimuli cross the VOT boundary can subjects reliably discriminate them.

For an example of these tests in practice, see Morford et al. (2008). Note that some researchers have found conflicting results when using different dependent measures. For example, Massaro & Cohen (1983) asked participants to make continuous judgments of stimuli and found no evidence of categorical perception.

Another alternative might be to find the best fitting line and the best fitting sigmoid curve, and compute the Bayesian Information Criterion for each. The one with the lower BIC is the preferred model. For regression analysis, BIC can be calculated as

$BIC = n\log(\frac{RSS}{n}) + p\log(n)$

where n is the number of data points in the model, p is the number of free parameters, and RSS is the residual sum of squares.

The drawback of using BIC is that it only tells you something about comparative model fit. If both linear and sigmoidal models were a bad fit for the data, BIC would not tell you that.

Massaro, D. W., & Cohen, M. M. (1983). Categorical or continuous speech perception: A new test. Speech Communication, 2(1), 15-35. PDF

Morford, J. P., Grieve-Smith, A. B., MacFarlane, J., Staley, J., & Waters, G. (2008). Effects of language experience on the perception of American Sign Language. Cognition, 109(1), 41-53. PDF

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