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Search: MSC category 22E ( Lie groups {For the topology of Lie groups and homogeneous spaces, see 57Sxx, 57Txx; for analysis thereon, see 43A80, 43A85, 43A90} )

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1. CJM Online first

Adler, Jeffrey D.; Lansky, Joshua M.
 Lifting Representations of Finite Reductive Groups I: Semisimple Conjugacy Classes Suppose that $\tilde{G}$ is a connected reductive group defined over a field $k$, and $\Gamma$ is a finite group acting via $k$-automorphisms of $\tilde{G}$ satisfying a certain quasi-semisimplicity condition. Then the identity component of the group of $\Gamma$-fixed points in $\tilde{G}$ is reductive. We axiomatize the main features of the relationship between this fixed-point group and the pair $(\tilde{G},\Gamma)$, and consider any group $G$ satisfying the axioms. If both $\tilde{G}$ and $G$ are $k$-quasisplit, then we can consider their duals $\tilde{G}^*$ and $G^*$. We show the existence of and give an explicit formula for a natural map from the set of semisimple stable conjugacy classes in $G^*(k)$ to the analogous set for $\tilde{G}^*(k)$. If $k$ is finite, then our groups are automatically quasisplit, and our result specializes to give a map of semisimple conjugacy classes. Since such classes parametrize packets of irreducible representations of $G(k)$ and $\tilde{G}(k)$, one obtains a mapping of such packets. Keywords:reductive group, lifting, conjugacy class, representation, Lusztig seriesCategories:20G15, 20G40, 20C33, 22E35

2. CJM 2014 (vol 66 pp. 993)

Beuzart-Plessis, Raphaël
 Expression d'un facteur epsilon de paire par une formule intÃ©grale Let $E/F$ be a quadratic extension of $p$-adic fields and let $d$, $m$ be nonnegative integers of distinct parities. Fix admissible irreducible tempered representations $\pi$ and $\sigma$ of $GL_d(E)$ and $GL_m(E)$ respectively. We assume that $\pi$ and $\sigma$ are conjugate-dual. That is to say $\pi\simeq \pi^{\vee,c}$ and $\sigma\simeq \sigma^{\vee,c}$ where $c$ is the non trivial $F$-automorphism of $E$. This implies, we can extend $\pi$ to an unitary representation $\tilde{\pi}$ of a nonconnected group $GL_d(E)\rtimes \{1,\theta\}$. Define $\tilde{\sigma}$ the same way. We state and prove an integral formula for $\epsilon(1/2,\pi\times \sigma,\psi_E)$ involving the characters of $\tilde{\pi}$ and $\tilde{\sigma}$. This formula is related to the local Gan-Gross-Prasad conjecture for unitary groups. Keywords:epsilon factor, twisted groupsCategories:22E50, 11F85

3. CJM Online first

Feigin, Evgeny; Finkelberg, Michael; Littelmann, Peter
 Symplectic Degenerate Flag Varieties A simple finite dimensional module $V_\lambda$ of a simple complex algebraic group $G$ is naturally endowed with a filtration induced by the PBW-filtration of $U(\mathrm{Lie}\, G)$. The associated graded space $V_\lambda^a$ is a module for the group $G^a$, which can be roughly described as a semi-direct product of a Borel subgroup of $G$ and a large commutative unipotent group $\mathbb{G}_a^M$. In analogy to the flag variety $\mathcal{F}_\lambda=G.[v_\lambda]\subset \mathbb{P}(V_\lambda)$, we call the closure $\overline{G^a.[v_\lambda]}\subset \mathbb{P}(V_\lambda^a)$ of the $G^a$-orbit through the highest weight line the degenerate flag variety $\mathcal{F}^a_\lambda$. In general this is a singular variety, but we conjecture that it has many nice properties similar to that of Schubert varieties. In this paper we consider the case of $G$ being the symplectic group. The symplectic case is important for the conjecture because it is the first known case where even for fundamental weights $\omega$ the varieties $\mathcal{F}^a_\omega$ differ from $\mathcal{F}_\omega$. We give an explicit construction of the varieties $Sp\mathcal{F}^a_\lambda$ and construct desingularizations, similar to the Bott-Samelson resolutions in the classical case. We prove that $Sp\mathcal{F}^a_\lambda$ are normal locally complete intersections with terminal and rational singularities. We also show that these varieties are Frobenius split. Using the above mentioned results, we prove an analogue of the Borel-Weil theorem and obtain a $q$-character formula for the characters of irreducible $Sp_{2n}$-modules via the Atiyah-Bott-Lefschetz fixed points formula. Keywords:Lie algebras, flag varieties, symplectic groups, representationsCategories:14M15, 22E46

4. CJM Online first

Henniart, Guy; Sécherre, Vincent
 Types et contragrÃ©dientes Soit $\mathrm{G}$ un groupe rÃ©ductif $p$-adique, et soit $\mathrm{R}$ un corps algÃ©briquement clos. Soit $\pi$ une reprÃ©sentation lisse de $\mathrm{G}$ dans un espace vectoriel $\mathrm{V}$ sur $\mathrm{R}$. Fixons un sous-groupe ouvert et compact $\mathrm{K}$ de $\mathrm{G}$ et une reprÃ©sentation lisse irrÃ©ductible $\tau$ de $\mathrm{K}$ dans un espace vectoriel $\mathrm{W}$ de dimension finie sur $\mathrm{R}$. Sur l'espace $\mathrm{Hom}_{\mathrm{K}(\mathrm{W},\mathrm{V})}$ agit l'algÃ¨bre d'entrelacement $\mathscr{H}(\mathrm{G},\mathrm{K},\mathrm{W})$. Nous examinons la compatibilitÃ© de ces constructions avec le passage aux reprÃ©sentations contragrÃ©dientes $\mathrm{V}^Äe$ et $\mathrm{W}^Äe$, et donnons en particulier des conditions sur $\mathrm{W}$ ou sur la caractÃ©ristique de $\mathrm{R}$ pour que le comportement soit semblable au cas des reprÃ©sentations complexes. Nous prenons un point de vue abstrait, n'utilisant que des propriÃ©tÃ©s gÃ©nÃ©rales de $\mathrm{G}$. Nous terminons par une application Ã  la thÃ©orie des types pour le groupe $\mathrm{GL}_n$ et ses formes intÃ©rieures sur un corps local non archimÃ©dien. Keywords:modular representations of p-adic reductive groups, types, contragredient, intertwiningCategory:22E50

5. CJM 2013 (vol 66 pp. 241)

Broussous, P.
 Transfert du pseudo-coefficient de Kottwitz et formules de caractÃ¨re pour la sÃ©rie discrÃ¨te de $\mathrm{GL}(N)$ sur un corps local Soit $G$ le groupe $\mathrm{GL}(N,F)$, oÃ¹ $F$ est un corps localement compact et non archimÃ©dien. En utilisant la thÃ©orie des types simples de Bushnell et Kutzko, ainsi qu'une idÃ©e originale d'Henniart, nous construisons des pseudo-coefficients explicites pour les reprÃ©sentations de la sÃ©rie discrÃ¨te de $G$. Comme application, nous en dÃ©duisons des formules inÃ©dites pour la valeur du charactÃ¨re d'Harish-Chandra de certaines telles reprÃ©sentations en certains Ã©lÃ©ments elliptiques rÃ©guliers. Keywords:reductive p-adic groups , discrete series, Harish-Chandra character, pseudo-coefficientCategory:22E50

6. CJM 2013 (vol 66 pp. 566)

Choiy, Kwangho
 Transfer of Plancherel Measures for Unitary Supercuspidal Representations between $p$-adic Inner Forms Let $F$ be a $p$-adic field of characteristic $0$, and let $M$ be an $F$-Levi subgroup of a connected reductive $F$-split group such that $\Pi_{i=1}^{r} SL_{n_i} \subseteq M \subseteq \Pi_{i=1}^{r} GL_{n_i}$ for positive integers $r$ and $n_i$. We prove that the Plancherel measure for any unitary supercuspidal representation of $M(F)$ is identically transferred under the local Jacquet-Langlands type correspondence between $M$ and its $F$-inner forms, assuming a working hypothesis that Plancherel measures are invariant on a certain set. This work extends the result of MuiÄ and Savin (2000) for Siegel Levi subgroups of the groups $SO_{4n}$ and $Sp_{4n}$ under the local Jacquet-Langlands correspondence. It can be applied to a simply connected simple $F$-group of type $E_6$ or $E_7$, and a connected reductive $F$-group of type $A_{n}$, $B_{n}$, $C_n$ or $D_n$. Keywords:Plancherel measure, inner form, local to global global argument, cuspidal automorphic representation, Jacquet-Langlands correspondenceCategories:22E50, 11F70, 22E55, 22E35

7. CJM 2013 (vol 66 pp. 354)

Kellerhals, Ruth; Kolpakov, Alexander
 The Minimal Growth Rate of Cocompact Coxeter Groups in Hyperbolic 3-space Due to work of W. Parry it is known that the growth rate of a hyperbolic Coxeter group acting cocompactly on ${\mathbb H^3}$ is a Salem number. This being the arithmetic situation, we prove that the simplex group (3,5,3) has smallest growth rate among all cocompact hyperbolic Coxeter groups, and that it is as such unique. Our approach provides a different proof for the analog situation in ${\mathbb H^2}$ where E. Hironaka identified Lehmer's number as the minimal growth rate among all cocompact planar hyperbolic Coxeter groups and showed that it is (uniquely) achieved by the Coxeter triangle group (3,7). Keywords:hyperbolic Coxeter group, growth rate, Salem numberCategories:20F55, 22E40, 51F15

8. CJM 2012 (vol 66 pp. 102)

Birth, Lidia; Glöckner, Helge
 Continuity of convolution of test functions on Lie groups For a Lie group $G$, we show that the map $C^\infty_c(G)\times C^\infty_c(G)\to C^\infty_c(G)$, $(\gamma,\eta)\mapsto \gamma*\eta$ taking a pair of test functions to their convolution is continuous if and only if $G$ is $\sigma$-compact. More generally, consider $r,s,t \in \mathbb{N}_0\cup\{\infty\}$ with $t\leq r+s$, locally convex spaces $E_1$, $E_2$ and a continuous bilinear map $b\colon E_1\times E_2\to F$ to a complete locally convex space $F$. Let $\beta\colon C^r_c(G,E_1)\times C^s_c(G,E_2)\to C^t_c(G,F)$, $(\gamma,\eta)\mapsto \gamma *_b\eta$ be the associated convolution map. The main result is a characterization of those $(G,r,s,t,b)$ for which $\beta$ is continuous. Convolution of compactly supported continuous functions on a locally compact group is also discussed, as well as convolution of compactly supported $L^1$-functions and convolution of compactly supported Radon measures. Keywords:Lie group, locally compact group, smooth function, compact support, test function, second countability, countable basis, sigma-compactness, convolution, continuity, seminorm, product estimatesCategories:22E30, 46F05, 22D15, 42A85, 43A10, 43A15, 46A03, 46A13, 46E25

9. CJM 2012 (vol 64 pp. 721)

Achab, Dehbia; Faraut, Jacques
 Analysis of the Brylinski-Kostant Model for Spherical Minimal Representations We revisit with another view point the construction by R. Brylinski and B. Kostant of minimal representations of simple Lie groups. We start from a pair $(V,Q)$, where $V$ is a complex vector space and $Q$ a homogeneous polynomial of degree 4 on $V$. The manifold $\Xi$ is an orbit of a covering of ${\rm Conf}(V,Q)$, the conformal group of the pair $(V,Q)$, in a finite dimensional representation space. By a generalized Kantor-Koecher-Tits construction we obtain a complex simple Lie algebra $\mathfrak g$, and furthermore a real form ${\mathfrak g}_{\mathbb R}$. The connected and simply connected Lie group $G_{\mathbb R}$ with ${\rm Lie}(G_{\mathbb R})={\mathfrak g}_{\mathbb R}$ acts unitarily on a Hilbert space of holomorphic functions defined on the manifold $\Xi$. Keywords:minimal representation, Kantor-Koecher-Tits construction, Jordan algebra, Bernstein identity, Meijer $G$-functionCategories:17C36, 22E46, 32M15, 33C80

10. CJM 2012 (vol 65 pp. 66)

Deng, Shaoqiang; Hu, Zhiguang
 On Flag Curvature of Homogeneous Randers Spaces In this paper we give an explicit formula for the flag curvature of homogeneous Randers spaces of Douglas type and apply this formula to obtain some interesting results. We first deduce an explicit formula for the flag curvature of an arbitrary left invariant Randers metric on a two-step nilpotent Lie group. Then we obtain a classification of negatively curved homogeneous Randers spaces of Douglas type. This results, in particular, in many examples of homogeneous non-Riemannian Finsler spaces with negative flag curvature. Finally, we prove a rigidity result that a homogeneous Randers space of Berwald type whose flag curvature is everywhere nonzero must be Riemannian. Keywords:homogeneous Randers manifolds, flag curvature, Douglas spaces, two-step nilpotent Lie groupsCategories:22E46, 53C30

11. CJM 2012 (vol 64 pp. 497)

Li, Wen-Wei

12. CJM 2011 (vol 64 pp. 669)

Pantano, Alessandra; Paul, Annegret; Salamanca-Riba, Susana A.
 The Genuine Omega-regular Unitary Dual of the Metaplectic Group We classify all genuine unitary representations of the metaplectic group whose infinitesimal character is real and at least as regular as that of the oscillator representation. In a previous paper we exhibited a certain family of representations satisfying these conditions, obtained by cohomological induction from the tensor product of a one-dimensional representation and an oscillator representation. Our main theorem asserts that this family exhausts the genuine omega-regular unitary dual of the metaplectic group. Keywords:Metaplectic group, oscillator representation, bottom layer map, cohomological induction, Parthasarathy's Dirac Operator Inequality, pseudospherical principal seriesCategory:22E46

13. CJM 2011 (vol 64 pp. 123)

Lee, Jae-Hyouk
 Gosset Polytopes in Picard Groups of del Pezzo Surfaces In this article, we study the correspondence between the geometry of del Pezzo surfaces $S_{r}$ and the geometry of the $r$-dimensional Gosset polytopes $(r-4)_{21}$. We construct Gosset polytopes $(r-4)_{21}$ in $\operatorname{Pic} S_{r}\otimes\mathbb{Q}$ whose vertices are lines, and we identify divisor classes in $\operatorname{Pic} S_{r}$ corresponding to $(a-1)$-simplexes ($a\leq r$), $(r-1)$-simplexes and $(r-1)$-crosspolytopes of the polytope $(r-4)_{21}$. Then we explain how these classes correspond to skew $a$-lines($a\leq r$), exceptional systems, and rulings, respectively. As an application, we work on the monoidal transform for lines to study the local geometry of the polytope $(r-4)_{21}$. And we show that the Gieser transformation and the Bertini transformation induce a symmetry of polytopes $3_{21}$ and $4_{21}$, respectively. Categories:51M20, 14J26, 22E99

14. CJM 2011 (vol 64 pp. 481)

Chamorro, Diego
 Some Functional Inequalities on Polynomial Volume Growth Lie Groups In this article we study some Sobolev-type inequalities on polynomial volume growth Lie groups. We show in particular that improved Sobolev inequalities can be extended to this general framework without the use of the Littlewood-Paley decomposition. Keywords:Sobolev inequalities, polynomial volume growth Lie groupsCategory:22E30

15. CJM 2011 (vol 64 pp. 409)

Rainer, Armin
 Lifting Quasianalytic Mappings over Invariants Let $\rho \colon G \to \operatorname{GL}(V)$ be a rational finite dimensional complex representation of a reductive linear algebraic group $G$, and let $\sigma_1,\dots,\sigma_n$ be a system of generators of the algebra of invariant polynomials $\mathbb C[V]^G$. We study the problem of lifting mappings $f\colon \mathbb R^q \supseteq U \to \sigma(V) \subseteq \mathbb C^n$ over the mapping of invariants $\sigma=(\sigma_1,\dots,\sigma_n) \colon V \to \sigma(V)$. Note that $\sigma(V)$ can be identified with the categorical quotient $V /\!\!/ G$ and its points correspond bijectively to the closed orbits in $V$. We prove that if $f$ belongs to a quasianalytic subclass $\mathcal C \subseteq C^\infty$ satisfying some mild closedness properties that guarantee resolution of singularities in $\mathcal C$, e.g., the real analytic class, then $f$ admits a lift of the same class $\mathcal C$ after desingularization by local blow-ups and local power substitutions. As a consequence we show that $f$ itself allows for a lift that belongs to $\operatorname{SBV}_{\operatorname{loc}}$, i.e., special functions of bounded variation. If $\rho$ is a real representation of a compact Lie group, we obtain stronger versions. Keywords:lifting over invariants, reductive group representation, quasianalytic mappings, desingularization, bounded variationCategories:14L24, 14L30, 20G20, 22E45

16. CJM 2011 (vol 63 pp. 1238)

Bump, Daniel; Nakasuji, Maki
 Casselman's Basis of Iwahori Vectors and the Bruhat Order W. Casselman defined a basis $f_u$ of Iwahori fixed vectors of a spherical representation $(\pi, V)$ of a split semisimple $p$-adic group $G$ over a nonarchimedean local field $F$ by the condition that it be dual to the intertwining operators, indexed by elements $u$ of the Weyl group $W$. On the other hand, there is a natural basis $\psi_u$, and one seeks to find the transition matrices between the two bases. Thus, let $f_u = \sum_v \tilde{m} (u, v) \psi_v$ and $\psi_u = \sum_v m (u, v) f_v$. Using the Iwahori-Hecke algebra we prove that if a combinatorial condition is satisfied, then $m (u, v) = \prod_{\alpha} \frac{1 - q^{- 1} \mathbf{z}^{\alpha}}{1 -\mathbf{z}^{\alpha}}$, where $\mathbf z$ are the Langlands parameters for the representation and $\alpha$ runs through the set $S (u, v)$ of positive coroots $\alpha \in \hat{\Phi}$ (the dual root system of $G$) such that $u \leqslant v r_{\alpha} < v$ with $r_{\alpha}$ the reflection corresponding to $\alpha$. The condition is conjecturally always satisfied if $G$ is simply-laced and the Kazhdan-Lusztig polynomial $P_{w_0 v, w_0 u} = 1$ with $w_0$ the long Weyl group element. There is a similar formula for $\tilde{m}$ conjecturally satisfied if $P_{u, v} = 1$. This leads to various combinatorial conjectures. Keywords:Iwahori fixed vector, Iwahori Hecke algebra, Bruhat order, intertwining integralsCategories:20C08, 20F55, 22E50

17. CJM 2011 (vol 63 pp. 1364)

Meinrenken, Eckhard
 The Cubic Dirac Operator for Infinite-Dimensonal Lie Algebras Let $\mathfrak{g}=\bigoplus_{i\in\mathbb{Z}} \mathfrak{g}_i$ be an infinite-dimensional graded Lie algebra, with $\dim\mathfrak{g}_i<\infty$, equipped with a non-degenerate symmetric bilinear form $B$ of degree $0$. The quantum Weil algebra $\widehat{\mathcal{W}}\mathfrak{g}$ is a completion of the tensor product of the enveloping and Clifford algebras of $\mathfrak{g}$. Provided that the Kac-Peterson class of $\mathfrak{g}$ vanishes, one can construct a cubic Dirac operator $\mathcal{D}\in\widehat{\mathcal{W}}(\mathfrak{g})$, whose square is a quadratic Casimir element. We show that this condition holds for symmetrizable Kac-Moody algebras. Extending Kostant's arguments, one obtains generalized Weyl-Kac character formulas for suitable equal rank'' Lie subalgebras of Kac-Moody algebras. These extend the formulas of G. Landweber for affine Lie algebras. Categories:22E65, 15A66

18. CJM 2011 (vol 63 pp. 1307)

Dimitrov, Ivan; Penkov, Ivan
 A Bott-Borel-Weil Theorem for Diagonal Ind-groups A diagonal ind-group is a direct limit of classical affine algebraic groups of growing rank under a class of inclusions that contains the inclusion $$SL(n)\to SL(2n), \quad M\mapsto \begin{pmatrix}M & 0 \\ 0 & M \end{pmatrix}$$ as a typical special case. If $G$ is a diagonal ind-group and $B\subset G$ is a Borel ind-subgroup, we consider the ind-variety $G/B$ and compute the cohomology $H^\ell(G/B,\mathcal{O}_{-\lambda})$ of any $G$-equivariant line bundle $\mathcal{O}_{-\lambda}$ on $G/B$. It has been known that, for a generic $\lambda$, all cohomology groups of $\mathcal{O}_{-\lambda}$ vanish, and that a non-generic equivariant line bundle $\mathcal{O}_{-\lambda}$ has at most one nonzero cohomology group. The new result of this paper is a precise description of when $H^j(G/B,\mathcal{O}_{-\lambda})$ is nonzero and the proof of the fact that, whenever nonzero, $H^j(G/B, \mathcal{O}_{-\lambda})$ is a $G$-module dual to a highest weight module. The main difficulty is in defining an appropriate analog $W_B$ of the Weyl group, so that the action of $W_B$ on weights of $G$ is compatible with the analog of the Demazure action" of the Weyl group on the cohomology of line bundles. The highest weight corresponding to $H^j(G/B, \mathcal{O}_{-\lambda})$ is then computed by a procedure similar to that in the classical Bott-Borel-Weil theorem. Categories:22E65, 20G05

19. CJM 2011 (vol 63 pp. 1083)

Kaletha, Tasho
 Decomposition of Splitting Invariants in Split Real Groups For a maximal torus in a quasi-split semi-simple simply-connected group over a local field of characteristic $0$, Langlands and Shelstad constructed a cohomological invariant called the splitting invariant, which is an important component of their endoscopic transfer factors. We study this invariant in the case of a split real group and prove a decomposition theorem which expresses this invariant for a general torus as a product of the corresponding invariants for simple tori. We also show how this reduction formula allows for the comparison of splitting invariants between different tori in the given real group. Keywords:endoscopy, real lie group, splitting invariant, transfer factorCategories:11F70, 22E47, 11S37, 11F72, 17B22

20. CJM 2011 (vol 63 pp. 1137)

Moy, Allen
 Distribution Algebras on p-adic Groups and Lie Algebras When $F$ is a $p$-adic field, and $G={\mathbb G}(F)$ is the group of $F$-rational points of a connected algebraic $F$-group, the complex vector space ${\mathcal H}(G)$ of compactly supported locally constant distributions on $G$ has a natural convolution product that makes it into a ${\mathbb C}$-algebra (without an identity) called the Hecke algebra. The Hecke algebra is a partial analogue for $p$-adic groups of the enveloping algebra of a Lie group. However, $\mathcal{H}(G)$ has drawbacks such as the lack of an identity element, and the process $G \mapsto \mathcal{H}(G)$ is not a functor. Bernstein introduced an enlargement $\mathcal{H}\,\hat{\,}(G)$ of $\mathcal{H}(G)$. The algebra $\mathcal{H}\,\hat{\,} (G)$ consists of the distributions that are left essentially compact. We show that the process $G \mapsto \mathcal{H}\,\hat{\,} (G)$ is a functor. If $\tau \colon G \rightarrow H$ is a morphism of $p$-adic groups, let $F(\tau) \colon \mathcal{H}\,\hat{\,} (G) \rightarrow \mathcal{H}\,\hat{\,} (H)$ be the morphism of $\mathbb{C}$-algebras. We identify the kernel of $F(\tau)$ in terms of $\textrm{Ker}(\tau)$. In the setting of $p$-adic Lie algebras, with $\mathfrak{g}$ a reductive Lie algebra, $\mathfrak{m}$ a Levi, and $\tau \colon \mathfrak{g} \to \mathfrak{m}$ the natural projection, we show that $F(\tau)$ maps $G$-invariant distributions on $\mathcal{G}$ to $N_G (\mathfrak{m})$-invariant distributions on $\mathfrak{m}$. Finally, we exhibit a natural family of $G$-invariant essentially compact distributions on $\mathfrak{g}$ associated with a $G$-invariant non-degenerate symmetric bilinear form on ${\mathfrak g}$ and in the case of $SL(2)$ show how certain members of the family can be moved to the group. Keywords:distribution algebra, p-adic groupCategories:22E50, 22E35

21. CJM 2011 (vol 63 pp. 1107)

Liu, Baiying
 Genericity of Representations of p-Adic $Sp_{2n}$ and Local Langlands Parameters Let $G$ be the $F$-rational points of the symplectic group $Sp_{2n}$, where $F$ is a non-Archimedean local field of characteristic $0$. Cogdell, Kim, Piatetski-Shapiro, and Shahidi constructed local Langlands functorial lifting from irreducible generic representations of $G$ to irreducible representations of $GL_{2n+1}(F)$. Jiang and Soudry constructed the descent map from irreducible supercuspidal representations of $GL_{2n+1}(F)$ to those of $G$, showing that the local Langlands functorial lifting from the irreducible supercuspidal generic representations is surjective. In this paper, based on above results, using the same descent method of studying $SO_{2n+1}$ as Jiang and Soudry, we will show the rest of local Langlands functorial lifting is also surjective, and for any local Langlands parameter $\phi \in \Phi(G)$, we construct a representation $\sigma$ such that $\phi$ and $\sigma$ have the same twisted local factors. As one application, we prove the $G$-case of a conjecture of Gross-Prasad and Rallis, that is, a local Langlands parameter $\phi \in \Phi(G)$ is generic, i.e., the representation attached to $\phi$ is generic, if and only if the adjoint $L$-function of $\phi$ is holomorphic at $s=1$. As another application, we prove for each Arthur parameter $\psi$, and the corresponding local Langlands parameter $\phi_{\psi}$, the representation attached to $\phi_{\psi}$ is generic if and only if $\phi_{\psi}$ is tempered. Keywords:generic representations, local Langlands parametersCategories:22E50, 11S37

22. CJM 2011 (vol 63 pp. 591)

Hanzer, Marcela; Muić, Goran
 Rank One Reducibility for Metaplectic Groups via Theta Correspondence We calculate reducibility for the representations of metaplectic groups induced from cuspidal representations of maximal parabolic subgroups via theta correspondence, in terms of the analogous representations of the odd orthogonal groups. We also describe the lifts of all relevant subquotients. Categories:22E50, 11F70

23. CJM 2011 (vol 63 pp. 327)

Jantzen, Chris
 Discrete Series for $p$-adic $SO(2n)$ and Restrictions of Representations of $O(2n)$ In this paper we give a classification of discrete series for $SO(2n,F)$, $F$ $p$-adic, similar to that of MÅglin--TadiÄ for the other classical groups. This is obtained by taking the MÅglin--TadiÄ classification for $O(2n,F)$ and studying how the representations restrict to $SO(2n,F)$. We then extend this to an analysis of how admissible representations of $O(2n,F)$ restrict. Category:22E50

24. CJM 2010 (vol 62 pp. 1340)

Mœglin, C.
 Holomorphie des opÃ©rateurs d'entrelacement normalisÃ©s Ã  l'aide des paramÃ¨tres d'Arthur In this paper we prove holomorphy for certain intertwining operators arising from the theory of Eisenstein series. To do that we need to normalize using the Langlands--Shahidi's normalization arising from the twisted endoscopy and the associated representations of the general linear group. Categories:22E50, 22E35

25. CJM 2010 (vol 62 pp. 1310)

Lee, Kyu-Hwan
 Iwahori--Hecke Algebras of $SL_2$ over $2$-Dimensional Local Fields In this paper we construct an analogue of Iwahori--Hecke algebras of $\operatorname{SL}_2$ over $2$-dimensional local fields. After considering coset decompositions of double cosets of a Iwahori subgroup, we define a convolution product on the space of certain functions on $\operatorname{SL}_2$, and prove that the product is well-defined, obtaining a Hecke algebra. Then we investigate the structure of the Hecke algebra. We determine the center of the Hecke algebra and consider Iwahori--Matsumoto type relations. Categories:22E50, 20G25
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## Thursday, February 20, 2014

### Opposed Checks in D&D are the Same as Coin Flips

This is a lesson in statistics and probability, as applied to the popular Dungeons and Dragons role playing game.  I'm having trouble lately with the LaTeX embedder: if you see a lot of dollar signs and slashes, then check your plugins and permissions on your browser and allow MathJax to work, so you can see the equations better.

At least since version 3.0, the Dungeons and Dragons rule book has featured a rule of opposed checks.  These are supposed to represent, using dice, the opposition of two separate skills: so, your ability to Hide versus the orc's ability to Spot; your ability to tie a rope versus the orc's ability to escape from bonds.  You roll your skill, the orc rolls his skill, you apply modifiers, and the higher outcome wins.

Even before this, rolling dice was a common way to set the difficulty of something in old versions and in other non-d20 games.  How hard is the door to force?  You didn't think of it, now you're on the spot, so you roll a die to figure out how hard it is.  Then you tell the PCs to beat that number on their own roll.  Makes sense.

Doing checks this way is, from a probabilistic point of view, about as good as flipping a coin. The probability of the PC winning is slightly more than 50-50.  In terms of DCs, an opposed check (before modifiers) is equivalent to a DC of 10.5.

I'll prove it.

When you roll a die, a number comes up.  A random number, hopefully.  If the dice has $n$ sides, then this number is between $1$ and $n$.  It is customary to denote a random number with a capital letter: in this case, I'm going to call $X$ the result of rolling the die; $X=1, 2, 3,\ldots, n$, depending on what we roll.

If we consider some number between $1$ and $n$, say 6, then the probability that $X=6$ is, as we all know, $1/n.$  It is common to write this as $\Pr(X=6) = \frac{1}{n}.$  And, of course, it isn't just for $X=6$ that this is the case, but for any number $x$ between $1$ and $n$.  More generally, for any such $x$, we write $\Pr(X=x) = \frac{1}{n}.$

In our case, we are going to roll the same die twice.  This gives us two random numbers (the results of the two dice), which we will call $X_1$ and $X_2.$  For concreteness, suppose this is an opposed strength check between the PC and an Orc.  We'll say that $X_1$ is the die we (the GM) roll for the Orc, and $X_2$ is the die that the PC rolls.  We want to know $\Pr(X_2 \geq X_1)$, that is, the probability that the second result is higher than the first or in game terms, the probability that the PC wins his contest against the Orc.

This could go a number of ways.  The GM might roll a 1, in which case the PC is guaranteed to win, or the GM might roll a $n$, in which case the PC has to also roll $n$ or lose, with other possibilities in between.  But we don't want to consider the probability of the PC winning given some particular roll from the GM, because that's trivial.  So what we want to do instead is consider all of these possibilities.

We look at
$$\Pr(X_2\geq X_1) = \sum_{x=1}^n \Pr(X_2 \geq x) \cdot \Pr(X_1=x),$$
which means that we consider the probability of the GM rolling some number $X_1=x$, then multiply by the probability of the PC winning given this roll, then consider this for all the possible $x$ the GM might roll and add these together.  That gives us the probability of the PC winning his roll, regardless of what the GM rolls.

Breaking this down, we find
$$\Pr(X_2\geq X_1) = \sum_{x=1}^n \sum_{y=x}^n \frac{1}{n} \frac{1}{n} = \frac{1}{n^2}\sum_{x=1}^n\sum_{y=x}^n 1 = \frac{1}{n^2}\sum_{x=1}^n (n-x+1).$$
Stopping for a second, for people less familiar with this stuff, the $\sum_{y=x}^n$ term means that we add up every value of $X_2$, starting at $x$, and ending at $n$.  Concretely, if we're rolling a d20, and the Orc's roll is $X_1=15$, then we add up contributions from $y=15,16,17,18,19,20,$; that's $6 = 20-15+1$ terms we consider.  More generally, it is $n-x+1$ terms, which is why we wrote $(n-x+1)$ there.  Moving on,
$$\Pr(X_2\geq X_1) = \frac{1}{n^2} \left(\sum_{x=1}^n n - \sum_{x=1}^n x + \sum_{x=1}^n 1\right) = \frac{1}{n^2}\left(n^2 - \sum_{x=1}^n x + n\right).$$
The term $\sum_{x=1}^n x$ means the sum of the first $n$ numbers.  So,
$$\sum_{x=1}^n = 1 + 2 + 3 + 4 + 5 + \cdots + n.$$
Those who've had Calculus will be familiar with this, but other people maybe no so much.  There's a beautiful formula due to Gauss, arguably the first person to discover it, whose proof is even more beautiful.  Consider the following image:

This shows a bunch of stacks of squares, increasing from 1 to 2 to 3, on up to $n$.  The area of these squares is $\sum_{x=1}^n x.$  Now consider a second one of exactly the same size: the two interlock, forming a rectangle:

The width of the rectangle is $n$ and the height is $n+1$, so its areas is $n(n+1)$.  But the area of the rectangle is equal to twice the area of the stacked squares!  Therefore,
$$\sum_{x=1}^n x = \frac{n(n+1)}{2}.$$

So then, carrying on with our equation, we now have
$$\Pr(X_2 \geq X_1) = \frac{1}{n^2}\left(n^2 - \frac{n(n+1)}{2} + n\right) = 1 - \frac{n+1}{2n} + \frac{1}{n} = \frac{2n - n - 1 + 2}{2n} = \frac{n+1}{2n},$$
which, as I said, is slightly better than 50% probability.

For a 6-sided die, it is $\frac{7}{12}.$  For a 20-sided die, it is $\frac{21}{40}$, which is a 52.5% chance of success, which corresponds to a DC of 10.5.  A flat DC 10 is a 55% chance of success.  So rolling an opposed roll for the Orc is the same as considering the Orc's passive check.

That is without modifiers.  To include modifiers, start at flat DC 10, and modify as
$$DC = 10 + (\text{Orc's mod}) - (\text{PC's mod}).$$
For an Orc with +3 and a PC with -1, the check would be at
$$DC = 10 + 3 - (-1) = 14.$$
This will be (mostly) statistically equivalent to a modified opposed check ($\pm$ a 2.5% sliver of probability)

I did this all for dice, which are what is relevant to RPG players.  For dice, the result is not quite $1/2$ because the rolls can only equal certain specific results (like 1, or 7) and a tie goes to the player, but in a general case, it is actually true that the probability of a second random number being larger than the first random number is exactly $1/2$: that is, $\Pr(X_2>X_1) = \frac{1}{2}.$  I'll prove it.

So, consider  $X_1, X_2$, which are still random numbers, but not necessarily from a die.  For instance, we might push blocks on ice, and $X_2$ and $X_1$ gives the distance the blocks travel before coming to rest.  Or throw darts at a wall and $X_2,X_1$ are the distances from a bullseye.  Or something.  It's also not necessarily that case that every possible value is equally likely.  For a fair die, every number has probability $1/n$ or coming up; for throwing darts at a bullseye, if we're any good, then we will be more likely to be near the bullseye.  Let $\Pr(X=x) = p(x)$, where $p(x)$ is just some function: give it a value $x$ and it gives you a probability $p$.  Here $p(x)$ is called the "probability distribution function".  For simplicity, we also consider $\Pr(X\leq x) = F(x)$, called the "cumulative density function".  This is the probability of $X$ being less than some value $x$; as we'll see, a separate symbol for this is really useful.

As before, we have
$$\Pr(X_2\geq X_1) = \int \Pr (X_2\geq x)\cdot\Pr(X_1=x)dx = \int (1-F(x))p(x) dx = \int p(x)dx - \int F(x)p(x)dx.$$
You may be wondering what the weird S is, the $\int$ thing.  That's an integral sign, and it basically just means "add up all the possible values of $x$."  It's different from the $\sum$ symbol in that $\sum$ considers only discrete values while $\int$ considers continuous spectra of values.  We have used here the fact that $1 = Pr(X\leq x) + Pr(X\geq x) = F(x) + \Pr(X\geq x)$ to express this in terms of $F$.

If we add up all the probabilities of things happening, we should get 100%; that is, $\int p(x)dx = 1.$  This makes sense; the probability that we roll a 1 or a 2 or a 3, etc, is 1.  So
$$\Pr(X_2\geq X_1) = 1 - \int F(x)p(x)dx.$$
To fully evaluate this, we can write it another way.  Think what happens if, instead of rolling for the orc first then making the PC roll higher than that, we have the PC roll, then roll for the orc and make sure the orc rolls lower.  It's the same thing in the end, but can be written as:
$$\Pr(X_2\geq X_1) = \int \Pr(X_2 = x) \Pr(X_1 \leq x) = \int p(x) F(x) dx.$$
Comparing these two,
$$\int p(x) F(x)dx = \Pr(X_2\geq X_1) = 1 - \int p(x) F(x)x,$$
which must mean $\Pr(X_2 \geq X_1) = \int p(x)F(x)dx = 0.5.$

So, the long and short of it is, if we have two random numbers that we produce in the same way, one after the other, and we want to know the probability that the second is larger than the first, then this is 50%.  In terms of D&D, this means that if you generate the DC for a skill check by rolling a die, then have the PC roll to beat that die, then you may as well flip a coin to accomplish the same thing.  This also means you can fix the DC of the opposed rolls at 10, and just add the Orc's bonuses and subtract the PC's bonus to increase the DC; it achieves the exact same thing.

If this result is unsatisfying to you, consider using a different system of opposed rolls that changes the dice used by each party, for slightly swingier results.

Note: this was originally written on 2/20/2014, but was updated on 5/7/2018 to make the wording more clear.

Anonymous said...

This is very interesting, I also never thought about the sum first n positive integers series like that. But I was wondering, if instead of the two rolls trying to beat each other, they are instead trying to succeed at beating a target number (akin to Basic roleplaying where to make a successful attack you roll under your attacking skill and to block a successful attack you can attempt to roll under your skill at blocking.) Is there a way to simplify it similar to this? Or is it just not really applicable.

Reece said...

Let me make sure I understand the question.

Your attacking skill is 19, so you need to roll 19 or under to hit? Or your block skill is 12, so you need to roll under 12 to block? Is that the gist?

If you are rolling a 20-sided die, then the probability of rolling less than $x$ is $p(x-1)$ (I hope the math encodes correctly there). That's based on the PC, and doesn't change.

But, if the DM didn't figure out the Orc's block skill beforehand, then maybe he'll roll for it. Let's call that roll $X_1$. Then the DM rolls again (call it $X_2$) to see if the Orc blocks, and has to get less than $X_1$. So we need to know $\Pr(X_2<X_1),$ which is the probability of the Orc blocking. This would work in an analogous way to the above.

I think the answer that you get is 19/40, which is just a little bit less than 1/2.

Reece said...

Apparently the math does encode!

Let me take away all of the fun of doing the sum yourself:

$$\sum_{x=1}^n \Pr(X_2x)\cdot\Pr(X_1=x) = \sum_{x=1}^n \left(\sum_{k=1}^{x-1} \frac{1}{n}\right)\cdot \frac{1}{n} = \frac{1}{n^2} \sum_{x=1}^n \sum_{k=1}^{x-1} 1 = \frac{1}{n^2} \sum_{x=1}^n (x-1) = \frac{1}{n^2} \frac{n(n-1)}{2} = \frac{1}{2} - \frac{1}{2n}.$$

For $n=20$, this becomes $\frac{1}{2}−\frac{1}{40}=\frac{19}{40}$.

Hope I did that right. It's late here. I'm so sorry if you received a dozen replies in your inbox, I had trouble getting the equation above to come out right and couldn't edit the comment.

Anonymous said...

Cool, thanks for that. You got the question right yeah. So if you are rolling to determine the difficulty of something, you might as well just use the median value of the die. Thats pretty interesting.

Reece said...

Probabilistically, they're the same. That doesn't mean that your players will respond the same way, though :P

format: plain


text:
## anonymous one year ago The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = -16t2 - 64t + 80, where t is the time elapsed that the ball is in the air. What is the instantaneous velocity of the ball when it hits the ground?

1. phi

First, find the time "t" when the ball hits the ground (i.e. when s(t)= 0) by solving for "t" in $-16t^2 -64t+80- 0$ Next, find the derivative of the height with respect to t, $$\frac{d s}{dt}$$ This is the instantaneous velocity of the ball as a function of t. Evaluate this expression at the time found from the first step.


text:
# how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them?

This is an interview question.

Given n red balls and m blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that.

My solution:

suppose we have c containers, distribute n/c read balls to each c. If c == 1, put all of them together, it is n/(m+n)

 If c == 2,  put 1 red in c1 and all left red and all blue ones in c2 in this way , we have:
1/2 + 1/2 *(n-1) /(m+n-1) > 1/2

If c == 3,   put a red in c1, put a red in c2, put left red and all blue in c3, we have:
1/3 + 1/3 + 1/3 * (n-2)/(m+n-2) > 2/3

If c == n,  put a red in each of p -1, and all left red and blue in pth container, we have:
( Sum of (1/n) from 1 to n-1 ) +  ( 1/n * 1/(m+1) )
(n-1)/n + 1/n * 1/(m+1) == 1 (almost)

As n is large, the (n-1)/n is very close to 1 so that we maximize the probability to get a red balls.

Any better ideas ?

-
off-topic. Best fit is probably math.stackexchange (NOT mathoverflow.net) –  Jason S Dec 26 '11 at 2:16
They are asking you to implement gerrymandering. Put one red ball in each container, then put all the remaining balls in a single container. –  Raymond Hettinger Dec 26 '11 at 2:17
I wonder what's wrong with : Put one red and one blue into first container, then into 2nd, then into nth, then restart or stop when you run out of any colored balls. –  ScarletAmaranth Dec 26 '11 at 2:17
Are there any constraints to how many balls can/need to go in every container? Can we just put all the blue balls in a single container? –  adi92 Dec 26 '11 at 2:18
I must be missing something. If I have n red and m blue balls, no matter how they are split, if I pick a random ball from a random container, the probability to get a red ball is the same n / n + m –  Mathias Dec 26 '11 at 20:40
show 1 more comment

## migrated from stackoverflow.comMay 3 '12 at 13:38

This question came from our site for professional and enthusiast programmers.


text:
## Found 21 Documents (Results 1–21)

MSC:  30D35

### The properties of powers of matrices over a $$p$$-division ring. (Chinese. English summary)Zbl 0962.15008

MSC:  15B33 15A03 18A10

### Existence of positive radial solutions of Dirichlet problems for elliptic equations $$\Delta u+g(|X|) f(u)=0$$ in annulus. (Chinese. English summary)Zbl 0961.35061

MSC:  35J65 34B15 35A05

### Adaptive output-feedback control of a class of nonlinear systems with bounded disturbances. (Chinese. English summary)Zbl 0966.93063

MSC:  93C40 93C10

### Exponential stability of Hopfield-type stochastic neural networks with variable delays. (Chinese. English summary)Zbl 0965.93098

Reviewer: Xu Daoyi (Chengdu)
MSC:  93E15 93C23 93D20

MSC:  60G10

### Strong consistency for kernel neighbor estimator of density function of spherical data. (Chinese. English summary)Zbl 0959.62035

MSC:  62G07 62G20 62H11

### A new algorithm for wavefront fitting using Zernike polynomial. (Chinese. English summary)Zbl 0960.65019

MSC:  65D10 65F25

### Asymptotic properties of a kind of sieve MLE. (Chinese. English summary)Zbl 0959.62024

MSC:  62F12 62G20

### Estimation of failure rate and its applications in the case of zero-failure data. (Chinese. English summary)Zbl 0959.62093

MSC:  62N05 62N02 62F15

### Initial value problems for nonlinear second-order impulsive integro-differential equations in Banach space. (Chinese. English summary)Zbl 0962.34045

MSC:  34G20 45J05

### Bifurcation analysis of one kind of reaction-diffusion equations. (Chinese. English summary)Zbl 0962.34030

MSC:  34C23 34B15

### Geometric singular perturbation theory for ordinary differential equations with multifrequency. (Chinese. English summary)Zbl 0962.34042

MSC:  34E15 34A26 34C29

MSC:  62F12

### Inverse problem of a hyperbolic integro-differential equation. (Chinese. English summary)Zbl 0961.35175

MSC:  35R30 45K05

MSC:  05C70

### Global asymptotic stability of periodic Lotka-Volterra systems with delays. (Chinese. English summary)Zbl 0962.34058

MSC:  34K20 92D25 34K13

### Attractor for Cauchy problem of dissiptive KdV type equation. (Chinese. English summary)Zbl 0961.35141

MSC:  35Q53 37L30 35Q55

all top 3


text:
# Project Euler Problem 436 – Unfair Wager

Spoiler Alert! This blog entry gives away the solution to problem 436 of Project Euler. Please don’t read any further if you have yet to attempt to solve the problem on your own. The information is intended for those who failed to solve the problem and are looking for hints or test data to help them track down bugs. It is posted not before the problem has a hundred solvers already.</br>
First I provide references of the major theoretical background that you probably need to know to solve the problem. The last section presents the solution from start to finish.

The post reflects my approach to the problem. Even though the final outcome was accepted by Project Euler doesn’t mean the information is correct or elegant. Algorithms won’t necessarily be the most efficient ones, but they are guaranteed to run within the time limit of one minute.

## References

This is a pen and paper problem that requires very little knowledge of maths. You mainly need to calculate conditional probabilities and use the distribution of the sum of uniform variables.

## Notation

I’ll use the following notation:

• $u_i$: The ith uniform number.
• $s_i$: The sum after i draws
• $n$: The number of draws to get the sum over 1
• $m$: The number of draws to get the sum over 2
• $p$: The first sum over 1, i.e. $p=s_n$
• $x$: The number that the first player gets.
• $y$: The number that the second player gets.

The capital letters denote the corresponding random variables and $f_X(x)$ denotes the probability density function of $X$ at $x$ .

## Solution

I use the following steps to solve the problem:

1. Sum after i picks
2. First Player – Distribution of x
3. Second Player – Distribution of y
4. Joint distribution of x and y
5. Probability of y bigger than x

### Sum after i picks

Let

The probability density function (pdf) for the sum after one step corresponds to the pdf of the continuous uniform distribution:

where $\mathbf{1}_{A}$ denotes the indicator function.

For each subsequent pick, the sum $S_n$ is obtained by adding a uniform random variable to $S_{n-1}$ . The uniform random variable is independent of the sum $S_{n-1}$ , and therefore the pdf of $S_{n}$ is obtained by convolution of $f_{S_{n-1}}$ and $f_U$ :

By induction we get
$% $

(We will only need the distribution below one).

### First Player – Distribution of x

Let $n$ denote the number of picks it takes to get the sum over 1. Consequently, $s_n$ is the first sum over 1, i.e., the sum where the second player takes over. The last summand of $s_n$ is the number $x^{(n)}$ that the first player gets if she draws $n$ numbers:

Because

and because $s_{n-1} \leq 1$ and $s_n > 1$ , we know that

so the joint probability density distribution of $x^{(n)}$ and $S_n$ is

The previous density applies to a given $n$ . Let $x$ be the number of the first player, and let $p$ be the first sum over 1. To get the unconditional probability $f_{P,X}$, we sum up $f_{X^{(n)},S_{n}}(x,s)$ (2) over all potential $n$ :

### Second Player – Distribution of y

The first player leaves the sum at $p$ an the second player needs to get the sum over 2. I prefer to shift the values to the point of origin: Let the second player start at 0 and let $y$ be the first value over $2-p$ . Also let $s_{i}’$ be the sum of the first i number that the second player picked. In other words, if the first player picked $n$ numbers that add up to $s_n$ , and the second player picks $m$ numbers that bring the total to $s_{n+m}$ , then $s_{m}’ = s_{n+m} - s_n$ .

We already know the distribution of $s’_i$ below $2-p$ from the first player from Equation 1:

To get to the distribution of $y$ we have to distinguish two fundamentally different cases.

#### Case 1: y is less or equal than 2-p

If $y$ is smaller than $2-p$ , then the second player needed at least two picks to get the sum over $2-p$ , and the penultimate sum is at greater than $2-p-y$ (since adding $y$ brings the sum over $2-p$ ) and at most $2-p$ (because if the sum would be bigger than that, it would be the last sum already).

Assume that the second player needs $m > 1$ picks and his last number is $y^{(m)}$ . Therefore, we have

Hence, the pdf of $y$ is:

#### Case 2: y is larger than 2-p

The second case is special, because this time the second player has a chance to get her number with the first pick. We can cover this case by defining

which means the sum of zero uniform random value is always 0 and the corresponding pdf is 1 at 0, and 0 for any other value.

Secondly, the lower bound of the integration needs to be adjusted: since $2-p-y$ is always smaller than zero, and the pdfs of $S_n$ are zero for negative values, we start integrating at 0.

The rest of the calculation is the same, except this time we include $m=1$ :
As before, let’s assume that the second player needs $m > 1$ picks and his last number is $y^{(m)}$ . Therefore, we have

Hence, the pdf of y is:

So all in all the pdf of y given P is:

### Joint distribution of x and y

The joint distribution of the picks of the first and second player is:

Since $L\leq 1 + x$ we have

The indicator is therefore always one if $y < 1-x$ and

If, on the other hand, $x + y \geq 0$ , then the indicator is only one if $L < 2 - y$ :

And we have:

### Probability of y bigger than x

Player two wins if $y$ greater than $x$ :

First, we tackle the case $x+y<1$ . Since also $y>x$ we have $0 \leq x < 1/2$ and therefore:

There are two possibilities for the case $x+y\geq 1$ :

1. $0 \leq x \leq 1/2$ and $1-x \leq y \leq 1$
2. $1/2 \leq x \leq 1$ and $x \leq y \leq 1$

Therefore,

So finally we have from Equations 3 and 4:


text:
# Families of birth-death processes with similar time-dependent behaviour

R.B. Lenin, P.R. Parthasarathy, Willem R.W. Scheinhardt, Erik A. van Doorn

Research output: Book/ReportReportProfessional

### Abstract

We consider birth-death processes taking values in ${\cal N} \equiv \{0,1,\ldots\}$, but allow the death rate in state $0$ to be positive, so that escape from ${\cal N}$ is possible. Two such processes with transition functions $\{p_{ij}(t)\}$ and $\{{\tilde p}_{ij}(t)\}$ are said to be {\it similar} if, for all $i,j \in {\cal N},$ there are constants $c_{ij}$ such that ${\tilde p}_{ij}(t) = c_{ij}p_{ij}(t)$ for all $t \geq 0.$ We determine conditions on the birth and death rates of a birth-death process for the process to be a member of a family of similar processes, and we identify the members of such a family.
Original language Undefined Enschede University of Twente, Department of Applied Mathematics 14 0169-2690 Published - 1999

### Publication series

Name Memorandum / Department of Applied Mathematics Department of Applied Mathematics, University of Twente 1486 0169-2690

• EWI-3306
• IR-65675
• METIS-141298
• MSC-60J80

### Cite this

Lenin, R. B., Parthasarathy, P. R., Scheinhardt, W. R. W., & van Doorn, E. A. (1999). Families of birth-death processes with similar time-dependent behaviour. (Memorandum / Department of Applied Mathematics; No. 1486). Enschede: University of Twente, Department of Applied Mathematics.
Lenin, R.B. ; Parthasarathy, P.R. ; Scheinhardt, Willem R.W. ; van Doorn, Erik A. / Families of birth-death processes with similar time-dependent behaviour. Enschede : University of Twente, Department of Applied Mathematics, 1999. 14 p. (Memorandum / Department of Applied Mathematics; 1486).
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author = "R.B. Lenin and P.R. Parthasarathy and Scheinhardt, {Willem R.W.} and {van Doorn}, {Erik A.}",
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Lenin, RB, Parthasarathy, PR, Scheinhardt, WRW & van Doorn, EA 1999, Families of birth-death processes with similar time-dependent behaviour. Memorandum / Department of Applied Mathematics, no. 1486, University of Twente, Department of Applied Mathematics, Enschede.

Families of birth-death processes with similar time-dependent behaviour. / Lenin, R.B.; Parthasarathy, P.R.; Scheinhardt, Willem R.W.; van Doorn, Erik A.

Enschede : University of Twente, Department of Applied Mathematics, 1999. 14 p. (Memorandum / Department of Applied Mathematics; No. 1486).

Research output: Book/ReportReportProfessional

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AB - We consider birth-death processes taking values in ${\cal N} \equiv \{0,1,\ldots\}$, but allow the death rate in state $0$ to be positive, so that escape from ${\cal N}$ is possible. Two such processes with transition functions $\{p_{ij}(t)\}$ and $\{{\tilde p}_{ij}(t)\}$ are said to be {\it similar} if, for all $i,j \in {\cal N},$ there are constants $c_{ij}$ such that ${\tilde p}_{ij}(t) = c_{ij}p_{ij}(t)$ for all $t \geq 0.$ We determine conditions on the birth and death rates of a birth-death process for the process to be a member of a family of similar processes, and we identify the members of such a family.

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Lenin RB, Parthasarathy PR, Scheinhardt WRW, van Doorn EA. Families of birth-death processes with similar time-dependent behaviour. Enschede: University of Twente, Department of Applied Mathematics, 1999. 14 p. (Memorandum / Department of Applied Mathematics; 1486).


text:
# Flood-Fill and scanLine algorithms are line-based floods but I want square based floods

I have a flood-fill algorithm (Flood-fill) to fill a 24x24 matrix as follows (matrix is 24x24 here but will be much bigger in production):

Main code:

var cspots,    // number of spots per group
gArr=[];   // global array which contains all group spots

var tArr = new Array(gArr.length);   // touch array for flood-fill
for(var spot in inArr) {
for (var tspot in tArr)  // initialise touch array
tArr[tspot]=0;
for(gspot in gArr) {     // find lowest open y*24+x ordinal
if (gArr[gspot][0] == 0)
break;
tArr[gspot]=1;
}
cspots = inArr[spot].GD;
userFill(gArr[gspot][1],gArr[gspot][2],inArr[spot].KY,tArr);
}

function userFill(x,y,elem,tArr) {
var gord, qt=0;
if (!cspots) return;
if ((x >= 0) && (x <= 23) && (y >= 0) && (y <= 23)) {
gord = y*24 + x;
if (gArr[gord][0] != 0 || tArr[gord])
return;
gArr[gord][0] = elem;
tArr[gord] = 1;
--cspots;
userFill(x+1,y,elem,tArr);
userFill(x-1,y,elem,tArr);
// before the y-change we need to see if there are any open spots on this line
for(gord=y*24; gord<=(y*24)+23; gord++) {
if (gArr[gord][0] == 0) {
qt=1;
break;
}
}
if (!qt) {
userFill(x,y+1,elem,tArr);
userFill(x,y-1,elem,tArr);
}
}
};

This is a standard flood-fill recursive algorithm (with an accompanying touch array to mark any touches) with the additional code that I check if all x-values are set to non-zero on each x-plane before changing the y-value. This produces a matrix like this:

The problem is that it doesn't look very good (imo) as most of the areas are strung-out along the x-plane. What I want is each different group area to be in the shape of a square as much as I can. Sort-of like this example (using letters to indicate the different group areas):

V V V W W W W X X X X X
V V Y W W W W X X X X Z
Y Y Y W W W W Z Z Z Z Z
Y Y   W W W W Z Z Z Z Z
... and so on

So I have changed the userFill to look at a boxX variable which is just the (sqrt of each area)+1, which hopefully I can use to limit each area to make a square-shape. And a preX variable to store the anchor point from each group area so I know how many spots have been added. Here's the new userFill:

Main code:

var tArr = new Array(gArr.length);
for(var spot in inArr) {
for (var tspot in tArr)  // initialise touch array
tArr[tspot]=0;
for(gspot in gArr) {     // find lowest open y*24+x ordinal
if (gArr[gspot][0] == 0)
break;
tArr[gspot]=1;
}
cspots = inArr[spot].GD;
boxX = Math.ceil(Math.sqrt(cspots));
preX = gArr[gspot][1];
userFill(gArr[gspot][1],gArr[gspot][2],inArr[spot].KY,tArr);
}

function userFill(x,y,elem,tArr) {
var gord, qt=0;
if (!cspots) return;
if ((x >= 0) && (x <= 23) && (y >= 0) && (y <= 23)) {
gord = y*24 + x;
if (gArr[gord][0] != 0 || tArr[gord])
return;
gArr[gord][0] = elem;
tArr[gord] = 1;
--cspots;
// before the x-change we need to see if we have done a boxX number of changes to maintain square-shape
if (Math.abs(x-preX) == boxX) {
userFill(preX,y+1,elem,tArr);
userFill(preX,y-1,elem,tArr);
return;
}
userFill(x+1,y,elem,tArr);
userFill(x-1,y,elem,tArr);
// before the y-change we need to see if there are any open spots on this line
for(gord=y*24; gord<=(y*24)+boxX; gord++) {
if (gArr[gord][0] == 0) {
qt=1;
break;
}
}
if (!qt) {
userFill(x,y+1,elem,tArr);
userFill(x,y-1,elem,tArr);
}
}
};

The only difference is that I check if boxX spots have been added and then call userFill recursively to change the y-plane.

Here's the output and it looks better as most areas are square-like but obviously it needs work (missing most of the spots, pale-blue group area is very oddly-shaped and not square-like at all), but I wonder if there is a better algorithm out there that changes a flood-fill from line-based to square based.

UPDATE:

I think I figured out an algorithm. You start with a single point at the lowest point in the matrix. You then add a surrounding square of points around that point followed by a larger square until cspots is exhausted. So:

0     (first spot in matrix (0,0))

1 1
0 1   (add 3 spots to make it a 2x2 square)

2 2 2
1 1 2
0 1 2   (add 5 spots to make it a 3x3 square)

and so on

This can be done even if a number of squares are drawn and you are building on top of other squares. For example (X and Y are preexisting squares):

  X
X X Y Y
X X Y Y

Place first spot of new group in the lowest point in matrix (0,2)

0 X
X X Y Y
X X Y Y

Try and build 1st level square (an 'X' will be blocking one of the 1-level spots)

1 1
0 X
X X Y Y
X X Y Y

Try to build 2nd level square

2 2 2
1 1 2
0 X 2
X X Y Y
X X Y Y

3rd level

3 3 3 3
2 2 2 3
1 1 2 3
0 X 2 3
X X Y Y
X X Y Y

4th level

4 4 4 4 4
3 3 3 3 4
2 2 2 3 4
1 1 2 3 4
0 X 2 3 4
X X Y Y
X X Y Y

.. and so on. I just need to find the implementation now.

UPDATE 2

I have created the breadth-first algorithm as suggested below and it works much better. Here's the code and the image produced (which is very square-like for each of the 10 groups).

function bfsFill(x,y,elem,tArr) {
var gord, i=0, pt, queue=[], cnt=0;
if (!cspots) return;
if (isOutOfBounds(x,y)) return;
queue.push([x,y]);
while(cspots>0 && queue.length>0) {
pt = queue.shift();
gord = pt[1]*24 + pt[0];
tArr[gord] = 1;
gArr[gord][0] = elem;
--cspots;
var rArr = neighbours(pt);
async.eachSeries(rArr, function(el, cb2) {
if (!isOutOfBounds(el[0],el[1])) {
gord = el[1]*24 + el[0];
if (tArr[gord] == 0 && gArr[gord][0] == 0) {
for(var qi in queue) {
if (queue[qi][0] == el[0] && queue[qi][1]==el[1]) {
cb2();
return;
}
}
queue.push(el);
}
}
cb2();
}, function(err) {
});
}
};

Image file produced:

• Hello there! Welcome to the CG stack! I am afraid your question is sort of hard to follow. Particularly, it doesn't quite make clear what you intend your result to be... Do you want your code to always draw a square of fixed width and height without changing what was already there? Do you want your code to draw a shape as similar as possible to a square using exactly cspots spots? Please clear up what you want to achieve so that we can help you out! – Sebastián Mestre Nov 22 '18 at 0:12
• @SebastiánMestre Yes I would like to draw a shape as similar to a square using exactly cspots spots for each group as cspots can contain any number. The total of all groups cspots value will equal (24*24) so as the drawing progresses, the areas will become less and less square-like but I would like to keep the semblance of a square. In this example, there are 10 groups of varying cspots values and they need to be all drawn within the 24*24 matrix as square-like as possible. – ImTalkingCode Nov 22 '18 at 0:22
• It's probably a good idea to add the explanation of cspots (or inArr) to the beginning of the question so that you don't have to read through the comments first. Nice problem by the way, it kind of reminds me of the knapsack problem. – user9485 Nov 22 '18 at 0:43
• @Chris. Sorry. I've updated it with a possible solution. – ImTalkingCode Nov 22 '18 at 1:15
• @ImTalkingCode What do you mean by cspots? – x-rw Nov 23 '18 at 8:03

## 1 Answer

Your code, as seen on your post at the time of posting this answer, looks sort of like this (the following was thrown together rather quickly and may or may not work) :

function fill ( x, y, touched, elem ) {
if ( count <= 0 ) return;
if ( isOutOfBounds(x,y) ) return;

const idx = y*24 + x;

if ( gArr[idx][0] != 0 || touched[idx] ) return;

touched[idx] = true;
gArr[idx][0] = elem;
count--;

fill(x+1, y, touched, elem);
fill(x-1, y, touched, elem);
fill(x, y+1, touched, elem);
fill(x, y-1, touched, elem);
}

the problem with this style of writing a flood fill is that it always takes the path to the right before considering the other 3 paths it can take. This leads the program to use up its spots by moving only to the right. Such a search is called a Depth-First Search (DFS)

Instead, it would be nice if we could move around our grid starting from a point and growing out instead of blindly following the first path we find. A well known algorithm that does such a thing exists and it is called Breadth-First Search (BFS). It goes roughly as follows (once again, this code probably wouldn't work, even if you implemented the missing bits):

function fill ( starting_point, count, elem ) {
if ( count <= 0 ) return;
if ( isOutOfBounds(starting_point) ) return;

let touched = Array(gArr.length).fill(false);
let queue = [starting_point];

while ( count > 0 && queue.length > 0 ) {
let p = queue.shift();

touched[p.idx] = true;
gArr[p.idx][0] = elem;

for ( let neighbor of neighbors(p) ) {
if ( !isOutOfBounds( neighbor ) ) {
if ( !touched[ neigbor.idx ] && gArr[ neigbor.idx ][0] == 0 ){
touched[ neighbor.idx ] = true;
queue.push(neighbor);
}
}
}
}
}

What this is does is that it puts points to be painted in a queue and then looks at them in the same order that they were put in.

Naturally, the first set of points that are added into the queue are the neighbors of the starting point of your flood fill, then the neighbors of those points, and so on. Effectively building a square of increasing size.

Hope this helps!

• Worked very well. I need to only push a neighbour on the queue only if it was not currently a queue entry. The problem was a point (x,y) may have been in the queue and a new neighbour search would find the same (x,y) and since the touch array had not been updated, 2 entries of the same point would be in the queue and this would be replicated many times so the queue got massive and performance really suffered. Also, the multiple entries of the same point would decrement cspots more than once for the same point causing only a section of each group to be drawn. – ImTalkingCode Nov 22 '18 at 5:18
• Ohh, i just fixed the code to take that into account. good catch! – Sebastián Mestre Nov 22 '18 at 9:42
• You may run into problems if at some point none of the remaining groups fit entirely in any of the remaining untouched components. I'm not sure if sorting the groups by size or choosing the neighbors in a particular order would solve this. – user9485 Nov 22 '18 at 11:11
• Chris, I need to do a lot of testing to determine if the algorithm creates any orphan spots/areas which does not allow additional group areas to be added. – ImTalkingCode Nov 22 '18 at 23:17
• @SebastiánMestre you can share jsfidle(code complete) please – x-rw Nov 23 '18 at 1:54


text:
# Microfiches

Time Limit: Java: 2000 ms / Others: 2000 ms

Memory Limit: Java: 65536 KB / Others: 65536 KB

## Description

Before computers were widely used, there was another fairly convenient way of archiving data, where librarians archived periodicals on microfiches. A microfiche is a rectangular card made of high-resolution magnet that can store the image of a newspaper page on an area as small as one square millimeter. One may view the newspaper from a microfiche by a device with strong magnification capability. In order to keep a large microfiche collection in perfect order and provide good search service to the users, Gholam, one ingenious librarian, invented a microfiche storage machine.

There are n microfiches in the collection, and each of them has a unique id number. There is a catalogue that allows one to easily find the id number of the microfiche desired.

Gholam proposes to put each microfiche in a special frame, as shown in the figure below. The upper left corner of the frame is clipped, to make sure that there is only one orientation possible for a framed microfiche in its stack (explained below).

Figure 1: A microfiche with id number 010011, mounted on a frame. The shaded black portion is the microfiche and the surrounding white part is the frame containing the microfiche.

The id number of a microfiche is encoded, in binary, on the frame as follows. Let g be the number of bits necessary to represent an id number, indexed from 0 to g - 1. The left side of the frame contains g information cells. A "1" is represented as a dent on the information cell, and a "0" is represented as a hole. The first (uppermost) information cell corresponds to the most significant bit of the id number; the last information cell corresponds to its least significant bit. See the above figure for an illustration.

The microfiche storage machine contains three stacks, stack 0, stack 1, and stack 2, capable of holding n microfiches each. If you look at these stacks from the top, each of them is shaped like a rectangle with the upper left corner clipped. Initially, stack 0 is used for storage of microfiches; stacks 1 and 2 are only used for performing the move operation described below. The machine used to perform more operations, but move is the only operation that is considered in this problem.

Figure 2: The microfiche with a hole in position 3 gets pulled at (A), and the one with a dent in position 3 stays in the stack (B)

move (i, k, j): This operation moves all the microfiches with a "0" in information cell k (0 k g - 1) from stack i to the top of stack j such that the original relative ordering of the pulled microfiches from stack i , and that of the ones originally in stack j are preserved. In practice, a rod is inserted through all the holes (encoded "0"s) and dents (encoded "1"s) of the information cells in position k. Then the machine pulls the rod to the left, and thus the microfiches that have a hole (a "0") at position k will be pulled out (e.g. the microfiche of Figure 2.A), but those which have a dent ("1") in position k (e.g. Fig 2.B) would be left in the stack.

Gholam wants to find a certain microfiche with a known id number. He achieves this goal by performing a sequence of move operations, such that after performing the operations, there will be one stack that holds the desired microfiche only. Your task is to write a program to find the minimum length of such a sequence.

## Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test data are two integers n (1 <= n <= 10000) and g (1 <= g <= 15) which are the number of microfiches and the number of bits, respectively. Following the first line, there are n lines each containing the binary representation of a g-bit integer, with no leading or trailing spaces. The desired microfiche is the first one appearing in the test case.

## Output

There should be one output line per test case containing the minimum number of move operations needed to find the desired microfiche.

## Sample Input

1
4 3
010
011
100
110

## Sample Output

2

None

## Source

Asia 2002, Tehran (Iran), Preliminary


text:
## Playful Math Carnival #95 via Life Through A Mathematician’s Eyes

Check out the new playful math education carnival at Life Through A Mathematician’s Eyes. Parabolas, fractions, numbers square puzzles, edible math manipulatives, and all sorts of mathy fun:

Excitement!! The MTaP Math Educational Blog Carnival is at its 95 issue and I am extremely excited to host it. Moreover, February is one of my favorite months so I am extra excited for the opportunity.
First thing, let us think a little about the interesting properties of the number 95…

## Review and Giveaway

There’s still time to enter the book giveaway at Our Home on the Range blog:

Here on the Range, I’m determined to establish an environment where math is not just numbers and answers. I firmly believe my children can learn all the math they want, when they’re ready, as long as they don’t convince themselves they can’t learn it, they don’t like it, or that it’s too hard. To reach this goal, math must be a regular part of our lives in a way that encourages conversation and exploration.

• Let’s Play Math could be the very introduction a young family needs as they contemplate the first few years of homeschooling. First Son’s early years may have been completely different if I had read this book when he was five.
• It could be a fantastic book for a family with a child that’s struggling (in homeschool or otherwise) with math. A few years ago, when First Son first showed signs of a potentially life-long hatred of all things numerical, reading this book may have helped me adapt the curriculum we were then using to meet his needs and enrich him. (We ended up switching and I’m happy with that, but I could have avoided quite a bit of angst.)
• This book would be perfect for a parent who has always struggled with inadequacies in math or for someone like me, who always did just fine in math but never understood the claims of math’s beauty or fascination. I find myself excited to explore some of the resources the author has gathered together for my own growth and new challenges.

—Kansas Mom

## Let’s Play Math Back Cover Blurbs

Advance review comments for Let’s Play Math: How Families Can Learn Math Together—and Enjoy It:

Want to help your kids with math? Don’t help with the homework. Get them to engage with math by doing things together — many of which don’t even look like math. Let’s Play Math is charming, intelligent, and practical; full of family fun and sound advice.

—Ian Stewart, author of Professor Stewart’s Casebook of Mathematical Mysteries

This book is well researched, well annotated, and includes loads of activities you can try with kids K-12 at home.

—Jennifer Bardsley, credentialed teacher and author of TeachingMyBabytoRead.com

This is the math helper I wish I’d had years ago.

—Anne White, author of Minds More Awake

A crash course in how to enjoy math with your children! Denise Gaskins uses her years of experience to show parents how to teach math with games, stories, puzzles, manipulatives, and living books. Full of useful advice and pedagogical insight, this book is a treasure trove for parents who want to help their children appreciate the beauty, history, and fun of math but don’t know where to start.

—Kate Snow, KatesHomeschoolMath.com and author of Preschool Math at Home

## How to Update Your Let’s Play Math Ebook

I love how seeing math as a playful game can help even the busiest parents enjoy learning alongside their children.

My math books grew from more than a quarter-century of playing math with children — my own and those of our friends — at my house, at the library, in the park, and in group workshops. The kids and I learned from each other as we shared the adventure of learning mathematics.

Now that the publication dust has settled and the typos and formatting glitches have been sorted out, I’ve updated all the Let’s Play Math ebook files to match my shiny new, greatly expanded paperback edition.

### Do You Need an Update?

If your version shows the family playing together on the cover, then you’re all set. The copyright notice should say “Ebook version 3.0.”

But if you have an earlier edition of Let’s Play Math, you will probably want to update to this new, revised edition. Be warned: this is a totally new file, so you’ll lose all your highlights and bookmarks. But the expanded version has SO many wonderful changes, believe me, it’s well worth the inconvenience.

The following instructions are for Amazon.com. If you bought your book somewhere else, check the book dealer’s webpage — if the new cover (above) is showing, then they have the updated file. Find the Customer Service section of their website and follow their procedure.

### Amazon Policy Change

I’ve asked Amazon to release the new ebook files to everyone who bought the earlier edition — as used to be their standard policy. But they’ve had too many complaints about people losing their bookmarks, so they no longer issue updates except by customer request.

The only way to get the updated file is to contact Customer Service and ask for it yourself.

Step by step instructions:

2. Under “What can we help you with?” click the option “Digital Services.”
3. Under “Tell us more about your issue,” click “Kindle eBooks” and then “Something else.”
4. Under “Enter short summary of issue,” copy and paste the following text:
I bought the ebook Let's Play Math (ASIN B0095POAX4). This book has recently been updated with new material. Would you please send the updated ebook files to my account? Thank you.
5. Choose how you want to be contacted. I always pick “Email,” but you can pick a more immediate option if you like.

And then wait for the support people to do their magic. You should get the new file within 24 hours.

What if you got a bad copy of a paperback book?

The machines that print POD (Print On Demand) books — like my math book or my daughter’s fantasy novels — are basically giant computer printers. As everyone knows, printers get glitches. And the humans who take a book off the machine and put it in a shipping box won’t necessarily notice the mistake.

If you ever get a messed-up copy of my math paperbacks (or any book by anyone, for that matter), you can always contact customer service to have it replaced.

Just follow the same steps as above, but choose the options that make sense for whatever complaint you have. Or if you didn’t buy from Amazon, then find the Customer Service section for whatever book dealer you used, and follow their procedure.

## Let’s Play Math Paperback Edition Now Available

Finally, the revised and much-expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together—and Enjoy It is finished and loaded up on Amazon worldwide. Other bookstores will follow as soon as they update their files.

My book shows you new ways to explore math as a family adventure:

• Introduce your kids to the “Aha!” factor, the thrill of solving a challenging puzzle.
• Help them build thinking skills with toys, games, and library books.
• Find out how to choose math manipulatives, or make your own.
• And learn how to tackle story problems with confidence.

True mathematical thinking involves the same creative reasoning that children use to solve puzzles. Let’s Play Math turns math into a learning adventure for the whole family. Your children will build a stronger foundation of understanding when you teach math as a game, playing with ideas.

This blog originally grew out of my long-out-of-print books for homeschoolers, and now it has come full circle. The new edition of my book ripened on the vine, expanding to include families of every schooling style, with useful tips and resources for classroom teachers, too.

Wouldn’t you like to share in the harvest?

### Detailed Review (and a Giveaway)

If you’d like to know more about Let’s Play Math — and have a chance to win a free copy — check out Kate Snow’s review post:

## Understanding Math: Conclusion

Click to read the earlier posts in this series: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?; Understanding Math, Part 3: Is There Really a Difference?; Understanding Math, Part 4: Area of a Rectangle; Understanding Math, Part 5: Multiplying Fractions; and Understanding Math, Part 6: Algebraic Multiplication.

Earlier in this blog post series, I gave you three middle-school math rules. But by exploring the concept of rectangular area as a model of multiplication, we discovered that in a way they were all the same.

The rules are not arbitrary, handed down from a mathematical Mount Olympus. They are three expressions of a single basic question: what does it mean to measure area?

There was only one rule, one foundational pattern that tied all these topics together in a mathematical web.

Many children want to learn math instrumentally, as a tool for getting answers. They prefer the simplicity of memorizing rules to the more difficult task of making sense of new ideas. Being young, they are by nature short-term thinkers. They beg, “Just tell me what to do.”

But if we want our children to truly understand mathematics, we need to resist such shortcuts. We must take time to explore mathematics as a world of ideas that connect and relate to each other in many ways. And we need to show children how to reason about these interconnected concepts, so they can use them to think their way through an ever-expanding variety of problems.

Our kids can only see the short term. If we adults hope to help them learn math, our primary challenge is to guard against viewing the mastery of facts and procedures as an end in itself. We must never fall into thinking that the point of studying something is just to get the right answers.

We understand this in other school subjects. Nobody imagines that the point of reading is to answer comprehension questions. We know that there is more to learning history than winning a game of Trivial Pursuit. But when it comes to math, too many parents (and far too many politicians) act as though the goal of our children’s education is to produce high scores on a standardized test.

### What If I Don’t Understand Math?

If you grew up (as I did) thinking of math as a tool, the instrumental approach may feel natural to you. The idea of math as a cohesive system may feel intimidating. How can we parents help our children learn math, if we never understood it this way ourselves?

Don’t panic. Changing our worldview is never easy, yet even parents who suffer from math anxiety can learn to enjoy math with their children. All it takes is a bit of self-discipline and the willingness to try.

You don’t have to know all the answers. In fact, many people have found the same thing that Christopher Danielson described in his blog post “Let the children play” — the more we adults tell about a topic, the less our children learn. With the best of intentions we provide information, but we unwittingly kill their curiosity.

• If you’re afraid of math, be careful to never let a discouraging word pass your lips. Try calling upon your acting skills to pretend that math is the most exciting topic in the world.
• Encourage your children to notice the math all around them.
• Search out opportunities to discuss numbers, shapes, symmetry, and patterns with your kids.
• Investigate, experiment, estimate, explore, measure — and talk about it all.

### The Science of Patterns

Patterns are so important that American mathematician Lynn Arthur Steen defined mathematics as the science of patterns.

As biology is the science of life and physics the science of energy and matter, so mathematics is the science of patterns. We live in an environment steeped in patterns — patterns of numbers and space, of science and art, of computation and imagination. Patterns permeate the learning of mathematics, beginning when children learn the rhythm of counting and continuing through times tables all the way to fractals and binomial coefficients.

— Lynn Arthur Steen, 1998
Reflections on Mathematical Patterns, Relationships, and Functions

If you are intimidated by numbers, you can look for patterns of shape and color. Pay attention to how they grow, and talk about what your children notice. For example, some patterns repeat exactly, while other patterns change as they go (small, smaller, smallest, or loud, louder, loudest).

Nature often forms fractal-like patterns: the puffy round-upon-roundness of cumulus clouds or broccoli, or the branch-upon-branchiness of a shrub or river delta. Children can learn to recognize these, not as a homework exercise but because they are interesting.

### Math the Mathematician’s Way

Here is the secret solution to the crisis of math education: we adults need to learn how to think like mathematicians.

For more on what it means to think about math the mathematician’s way, check out my Homeschooling with Math Anxiety blog post series:

As we cultivate these characteristics, we will help our children to recognize and learn true mathematics.

CREDITS: “Frabjous 01” photo (top) by Windell Oskay and “Back to School” photo (middle) by Phil Roeder via Flicker (CC BY 2.0). “I Can Solve Problems” poster by Nicole Ricca via Teachers Pay Teachers. This is the final post in my Understanding Math series, adapted from my book Let’s Play Math: How Families Can Learn Math Together—and Enjoy It, available at your favorite online book dealer.

## Understanding Math: Algebraic Multiplication

Click to read the earlier posts in this series: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?; Understanding Math, Part 3: Is There Really a Difference?; Understanding Math, Part 4: Area of a Rectangle; and Understanding Math, Part 5: Multiplying Fractions.

We’ve examined how our vision of mathematical success shapes our children’s learning. Do we think math is primarily a tool for solving problems? Or do we see math as a web of interrelated concepts?

Instrumental understanding views math as a tool. Relational understanding views math as an interconnected system of ideas. Our worldview influences the way we present math topics to our kids. And our children’s worldview determines what they remember.

In the past two posts, we looked at different ways to understand and teach rectangular area and fraction multiplication. But how about algebra? Many children (and adults) believe “math with letters” is a jumble of abstract nonsense, with too many formulas and rules that have to be memorized if you want to pass a test.

Which of the following sounds the most like your experience of school math? And which type of math are your children learning?

### Instrumental Understanding: FOIL

Every mathematical procedure we learn is an instrument or tool for solving a certain kind of problem. To understand math means to know which tool we are supposed to use for each type of problem and how to use that tool — how to categorize the problem, remember the formula, plug in the numbers, and do the calculation.

When you need to multiply algebra expressions, remember to FOIL: multiply the First terms in each parenthesis, and then the Outer, Inner, and Last pairs, and finally add all those answers together.

### Relational Understanding: The Area Model

Each mathematical concept is part of a web of interrelated ideas. To understand mathematics means to see at least some of this web and to use the connections we see to make sense of new ideas.

The concept of rectangular area has helped us understand fractions. Let’s extend it even farther. In the connected system of mathematics, almost any type of multiplication can be imagined as a rectangular area. We don’t even have to know the size of our rectangle. It could be anything, such as subdividing a plot of land or designing a section of crisscrossed colors on plaid fabric.

We can imagine a rectangle with each side made up of two unknown lengths. One side has some length a attached to another length b. The other side is x units long, with an extra amount y stuck to its end.

We don’t know which side is the “length” and which is the “width” because we don’t know which numbers the letters represent. But multiplication works in any order, so it doesn’t matter which side is longer. Using the rectangle model of multiplication, we can see that this whole shape represents the area $\left ( a+b \right )\left ( x+y \right )$ .

But since the sides are measured in pieces, we can also imagine cutting up the big rectangle. The large, original rectangle covers the same amount of area as the four smaller rectangular pieces added together, and thus we can show that $\left ( a+b \right )\left ( x+y \right )=ax+ay+bx+by$ .

With the FOIL formula mentioned earlier, our students may get a correct answer quickly, but it’s a dead end. FOIL doesn’t connect to any other math concepts, not even other forms of algebraic multiplication. But the rectangular area model will help our kids multiply more complicated algebraic expressions such as $\left ( a+b+c \right )\left ( w+x+y+z \right )$ .

Not only that, but the rectangle model gives students a tool for making sense of later topics such as polynomial division. And it is fundamental to understanding integral calculus.

To be continued. Next up, Understanding Math Part 7: The Conclusion…

CREDITS: “Math Workshop Portland” photo (top) by US Department of Education via Flicker (CC BY 2.0). This is the sixth post in my Understanding Math series, adapted from my book Let’s Play Math: How Families Can Learn Math Together—and Enjoy It, available at your favorite online book dealer.

## Math Teachers at Play #94 via mathematicsandcoding

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# Triple infinite summation of a 3D Fourier series for Madelung Potential

I'm trying to evaluate the equation below excluding the case when $$n_x=n_y=n_z=0$$. I know this equation converges everywhere except where x,y, and z are all multiples of $$2\pi$$. I've attempted breaking the summation into 3 parts which avoid the unwanted case. Unfortunately Mathematica wasn't able to solve that. I also tried defining a function to use in the place of the fraction with a value of 0 when $$n_x=n_y=n_z=0$$. That hasn't worked either. I've also been trying to break the problem down into smaller pieces and looking at special cases with little success. Any help at all would be appreciated. $$\sum_{n_x=0}^\infty \sum_{n_y=0}^\infty \sum_{n_z=0}^\infty \frac{\cos(n_x x)\cos(n_y y)\cos(n_z z)}{n_x^2+n_y^2+n_z^2}$$

This equation would give what I call the Madelung Potential. Essentially the potential produced by a positive point charge in a cubic lattice submerged in a sea of negative charge so that the total charge is 0.


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Quartic B-spline collocation method to the numerical solutions of the Burgers’ equation. (English) Zbl 1130.65103
Summary: The collocation method using quartic B-splines is described for the numerical solutions of the Burgers’ equation. The effect of the quartic B-splines in the collocation method is sought. The same method is applied to the time split Burgers’ equation. Numerical comparison of results of both algorithms and some other published numerical results are done by studying three standard problems.

##### MSC:
 65M70 Spectral, collocation and related methods (IVP of PDE) 35Q53 KdV-like (Korteweg-de Vries) equations
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##### References:
 [1] Bateman, H.: Some recent researches on the motion of the fluids. Monthly weather rev 43, 163-170 (1915) [2] Burgers, J. M.: A mathematical model illustrating the theory of turbulence. (1948) · Zbl 0030.04503 [3] Hopf, E.: The partial differential equation ut+UUx - $\nu$Uxx=0. Commun pure appl math 3, 201-230 (1950) · Zbl 0039.10403 [4] Cole, J. D.: On a quasi-linear parabolic equations occurring in aerodynamics. Quart appl math 9, 225-236 (1951) · Zbl 0043.09902 [5] Miller EL. Predictor -- corrector studies of Burgers’ model of turbulent flow. M.S. Thesis, University of Delaware, DE, Newark, 1966. [6] Rubin, S. G.; Graves, R. A.: Viscous flow solutions with a cubic spline approximation. Comput fluids 3, 1-36 (1975) · Zbl 0347.76020 [7] Prenter, P. M.: Splines and variational methods. (1975) · Zbl 0344.65044 [8] Rubin, S. G.; Khosla, P. K.: Higher-order numerical solutions using cubic splines. Aiaa j 14, 851-858 (1976) · Zbl 0344.65048 [9] Davies, A. M.: A numerical investigation of errors arising in applying the Galerkin method of the solution of nonlinear partial differential equations. Comput methods appl mech engrg 11, 341-350 (1977) · Zbl 0364.65093 [10] Davies, A. M.: Application of the Galerkin method to the solution of Burgers equation. Comput methods appl mech engrg 14, 305-321 (1978) · Zbl 0387.76014 [11] Jain, P. C.; Holla, D. N.: Numerical solutions of coupled Burgers equations. Int J non-linear mech 13, 213-222 (1978) · Zbl 0388.76049 [12] Jain, P. C.; Lohar, B. L.: Cubic spline technique for coupled nonlinear parabolic equations. Comput math appl 5, 179-185 (1979) · Zbl 0421.65063 [13] Varo&gbreve, E.; Lu; Finn, W. D. L.: Space-time finite elements incorporating characteristics for the Burgers equation. Int J numer methods engrg 16, 171-184 (1980) · Zbl 0449.76076 [14] Christie, I.; Griffiths, D. F.; Mitchell, A. R.; Sanz-Serna, J. M.: Product approximation for nonlinear problems in the finite element method. IMA J numer anal 1, 253-266 (1981) · Zbl 0469.65072 [15] Herbst, B. M.; Schoombie, S. W.; Mitchell, A. R.: A moving Petrov -- Galerkin method for transport equations. Int J numer methods engrg 18, 1321-1336 (1982) · Zbl 0485.65093 [16] Nguyen, H.; Reynen, J.: A space -- time finite element approach to Burgers equation. Numerical methods for non-linear problems 2, 718-728 (1982) [17] Caldwell, J.: Applications of cubic splines to the nonlinear Burgers equation. Numerical methods for nonlinear problems 3, 253-261 (1987) [18] Ali A.H.A., Gardner L.R.T., Gardner G.A. A Galerkin approach to the solution of Burgers’ equation. University of Wales, Bangor, Math. Preprint 90.04, 1990. [19] Kakuda, K.; Tosaka, N.: The generalized boundary element approach to Burgers equation. Int J numer methods engrg 29, 245-261 (1990) · Zbl 0712.76070 [20] Zhang, D. S.; Wei, G. W.; Kouri, D. J.; Hoffman, D. K.: Burgers equation with high Reynolds number. Phys fluids 9, 1853-1855 (1997) · Zbl 1185.76843 [21] Wei, G. W.; Zhang, D. S.; Kouri, D. J.; Hoffman, D. K.: A robust and reliable approach to nonlinear dynamical problems. Comput phys commun 111, 87-92 (1998) · Zbl 0935.65108 [22] Kutluay, S.; Bahadır, A. R.; Özdeş, A.: Numerical solution of one-dimensional Burgers equation: explicit and exact-explicit finite difference methods. J comput appl math 103, 251-261 (1999) · Zbl 0942.65094 [23] S. Kutluay , A. Esen , I. Da&gbreve: Numerical solutions of the Burgers equation by the least squares quadratic B-spline finite element method. J comput appl math 167, 21-33 (2004) · Zbl 1052.65094 [24] &idot; Da&gbreve, .; ; Irk, D.; Saka, B.: A numerical solution of the Burgers equation using cubic B-splines. Appl math comput 163, 199-211 (2005) · Zbl 1060.65652 [25] &idot; Dağ ; Saka, B.; Boz, A.: B-spline Galerkin methods for numerical solutions of the Burgers equation. Appl math comput 166, 506-522 (2005) · Zbl 1073.65099 [26] Ramadan, M. A.; El-Danaf, T. S.; Alaal, F.: A numerical solution of the Burgers equation using septic B-splines. Chaos, solitons & fractals 26, 795-804 (2005) · Zbl 1075.65127