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## Found 21,490 Documents (Results 1–100)

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### The 3-path-connectivity of the $$k$$-ary $$n$$-cube. (English)Zbl 07589675

MSC:  05Cxx 68Mxx 68Rxx
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### Universal quantum semigroupoids. (English)Zbl 07584342

MSC:  16S10 16T99 16W50
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### Closure operators associated with relations. (English)Zbl 07568615

MSC:  54A05 54H30 08A02
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MSC:  70E55
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### Adapting paths against zero-determinant strategies in repeated prisoner’s dilemma games. (English)Zbl 07593000

MSC:  91A20 91A05
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### Hamiltonian and long paths in bipartite graphs with connectivity. (English)Zbl 07592409

MSC:  05C45 05C12 05C40
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### Fifth-order A-WENO schemes based on the path-conservative central-upwind method. (English)Zbl 07592126

MSC:  65Mxx 35Lxx 76Mxx
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### An asymptotic radius of convergence for the Loewner equation and simulation of $$SLE_{\kappa}$$ traces via splitting. (English)Zbl 07591717

MSC:  60Hxx 60Jxx 30Cxx
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### Realization utility with path-dependent reference points. (English)Zbl 07591710

MSC:  91G80 35Q91 49J40
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### Unidirectional Littelmann paths for crystals of type $$A$$ and rank 2. (English)Zbl 07590583

MSC:  17B10 17B35
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### Limit laws for two distance-based indices in random recursive tree models. (English)Zbl 07589927

MSC:  05C05 60F05
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### Strength-based concurrent shape and fiber path optimization of continuous fiber composites. (English)Zbl 07589848

MSC:  74-XX 90-XX
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### Concurrent optimization of topological configuration and continuous fiber path for composite structures – a unified level set approach. (English)Zbl 07589827

MSC:  74-XX 90-XX
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### The overfullness of graphs with small minimum degree and large maximum degree. (English)Zbl 07589623

MSC:  05C07 05C15
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### Path connectedness of Volterra type integral operators on Bergman and Dirichlet type spaces. (English)Zbl 07589100

MSC:  30B40 30D10
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MSC:  90Cxx
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MSC:  90Cxx
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### Monotone path-connectedness of strict suns. (English)Zbl 07587447

MSC:  41Axx 54Cxx 52Axx
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MSC:  83-XX
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### Leavitt path algebras for power graphs of finite groups. (English)Zbl 07584874

MSC:  16S88 05C25
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### Lie solvable Leavitt path algebras. (English)Zbl 07584710

MSC:  16S88 17B30
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### A translation of an analogue of Wiener space with its applications on their product spaces. (English)Zbl 07584431

MSC:  28C20 46G12 46T12
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### Efficiency analysis of the forward error correction at the transport protocol level. (English. Russian original)Zbl 07583570

Autom. Remote Control 83, No. 7, 1059-1077 (2022); translation from Avtom. Telemekh. 2022, No. 7, 59-81 (2022).
MSC:  93C55 93E03
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### Two optimal path planning problems for a moving object in the case of degeneration of necessary extremum conditions. (English. Russian original)Zbl 07583567

Autom. Remote Control 83, No. 7, 1011-1035 (2022); translation from Avtom. Telemekh. 2022, No. 7, 3-32 (2022).
MSC:  93C20 93C10
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### Safe interval path planning and flatness-based control for navigation of a mobile robot among static and dynamic obstacles. (English. Russian original)Zbl 07583560

Autom. Remote Control 83, No. 6, 903-918 (2022); translation from Avtom. Telemekh. 2022, No. 6, 96-117 (2022).
MSC:  93C85 93B45
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MSC:  74-XX
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### An algorithm for updating betweenness centrality scores of all vertices in a graph upon deletion of a single edge. (English)Zbl 07583048

MSC:  05C38 05C12 05C35
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MSC:  05C50
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MSC:  82-XX
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### Alexandroff homology and path homology. (English. Russian original)Zbl 07581879

Math. Notes 112, No. 1, 159-162 (2022); translation from Mat. Zametki 112, No. 1, 148-152 (2022).
MSC:  55N35 57M15 05C20
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### Gas path parameter prediction of aero-engine based on an autoregressive discrete convolution sum process neural network. (English)Zbl 07581684

MSC:  68-XX 82-XX
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### On size multipartite Ramsey numbers of large paths versus wheel on five vertices. (English)Zbl 07581635

MSC:  05C55 05D10
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### Algorithm to check the existence of $$H$$ for a given $$G$$ such that $$A(G)A(H)$$ is graphical. (English)Zbl 07581625

MSC:  05C85 05C50
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### On $$H$$-irregularity strength of cycles and diamond graphs. (English)Zbl 07581623

MSC:  05C70 05C78
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### Decomposition of hypercube graphs into paths and cycles of length four. (English)Zbl 07579612

MSC:  05C70 05C38 05C12
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MSC:  74-XX
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### An efficient routing heuristic for a drone-assisted delivery problem. (English)Zbl 07578501

MSC:  90-XX 91-XX
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### Fast and optimal sequence-to-graph alignment guided by seeds. (English)Zbl 07577831

Pe’er, Itsik (ed.), Research in computational molecular biology. 26th annual international conference, RECOMB 2022, San Diego, CA, USA, May 22–25, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13278, 306-325 (2022).
MSC:  92D20 05C90
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### A shift Gray code for fixed-content Łukasiewicz words. (English)Zbl 07577713

Bazgan, Cristina (ed.) et al., Combinatorial algorithms. 33rd international workshop, IWOCA 2022, Trier, Germany, June 7–9, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13270, 383-397 (2022).
MSC:  68Rxx 68Wxx
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### Chevalley formula for anti-dominant minuscule fundamental weights in the equivariant quantum $$K$$-group of partial flag manifolds. (English)Zbl 07577510

MSC:  17Bxx 20Fxx 05Exx
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### Connectedness in structures on the real numbers: o-minimality and undecidability. (English)Zbl 07576906

MSC:  03C64 03D35 54D99
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### A sufficient condition for mobile sampling in terms of surface density. (English)Zbl 07576441

MSC:  68-XX 62-XX
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### A near-optimal algorithm for shortest paths among curved obstacles in the plane. (English)Zbl 07575637

MSC:  68Q25 68Q20 68P05
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### Stress-driven nonlinear behavior of curved nanobeams. (English)Zbl 07575006

MSC:  74-XX 76-XX
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### Gallai-Ramsey numbers for rainbow $$P_5$$ and monochromatic fans or wheels. (English)Zbl 07574776

MSC:  05C55 05D10 05C15
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### Extended path partition conjecture for semicomplete and acyclic compositions. (English)Zbl 07574749

MSC:  05C70 05C38 05C20
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### Comparison theorem for path dependent SDEs driven by $$G$$-Brownian motion. (English)Zbl 07574577

MSC:  60J65 47G20 60G52
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### Path independence of the additive functionals for stochastic differential equations driven by $$G$$-Lévy processes. (English)Zbl 07574061

MSC:  60H10 60G51
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### Decorated Dyck paths, polyominoes, and the delta conjecture. (English)Zbl 07573750

Memoirs of the American Mathematical Society 1370. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-7157-6/pbk; 978-1-4704-7170-5/ebook). iii, 119 p. (2022).
MSC:  05-02 05E05 05B50
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### Decomposing $$2k$$-regular graphs into paths of length $$k$$. (English)Zbl 07572745

MSC:  05C70 05C38 05C12
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### The two-stripe symmetric circulant TSP is in P. (English)Zbl 07572298

Aardal, Karen (ed.) et al., Integer programming and combinatorial optimization. 23rd international conference, IPCO 2022, Eindhoven, The Netherlands, June 27–29, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13265, 319-332 (2022).
MSC:  90C27
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### Faster goal-oriented shortest path search for bulk and incremental detailed routing. (English)Zbl 07572276

Aardal, Karen (ed.) et al., Integer programming and combinatorial optimization. 23rd international conference, IPCO 2022, Eindhoven, The Netherlands, June 27–29, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13265, 15-28 (2022).
MSC:  90C35 90C27 90C10
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### Phase space Feynman path integrals of parabolic type on the torus as analysis on path space. (English)Zbl 07571553

MSC:  81S40 35S05 35K25
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MSC:  93-XX
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### The median of Sierpiński graphs. (English)Zbl 07567749

MSC:  05C12 05C35 05C38
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### The microlocal irregularity of Gaussian noise. (English)Zbl 07566092

MSC:  60G15 60G60
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### Coexistence of two kinds of superfluidity at finite temperatures in optical lattices. (English)Zbl 07565269

MSC:  82-XX 81-XX
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MSC:  62-XX
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### Heavy traffic limits for queues with non-stationary path-dependent arrival processes. (English)Zbl 1491.60159

MSC:  60K25 60F17 90B22
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### Group inverses of adjacency matrices of cycles, wheels and brooms. (English)Zbl 07562939

MSC:  05C50 15A09
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### Decomposing 10-regular graphs into paths of length 5. (English)Zbl 07562657

MSC:  05C38 05C70
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MSC:  90-XX
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MSC:  90-XX
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MSC:  60H10
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### Construction of reduced chemical mechanisms orientated toward specific applications: a case study of primary reference fuel. (English)Zbl 07561985

MSC:  80Axx 76-XX
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### On paths with three blocks $$P(k,1,l)$$. (English)Zbl 07560305

MSC:  05C38 05C45
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### Primal-dual path-following methods and the trust-region updating strategy for linear programming with noisy data. (English)Zbl 07560295

MSC:  65L20 65K05 65L05
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### Strongly graded Leavitt path algebras. (English)Zbl 07559651

MSC:  16S88 16W50
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### Distance constrained vehicle routing problem to minimize the total cost: algorithms and complexity. (English)Zbl 07558393

MSC:  90C35 90C27 90B06
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MSC:  90C31
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MSC:  90C35
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MSC:  90C27
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### Stochastic control of a class of dynamical systems via path limits. (English)Zbl 07557898

MSC:  60F10 90C40 93E03
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### 3-D path planning system for autonomous vehicle considering the rollover and path length. (English)Zbl 07557780

MSC:  93C85 93C42
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### The vectorial kernel method for walks with longer steps. (English)Zbl 07557724

MSC:  05A15 05A16
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### The crossing numbers of join products of paths with three graphs of order five. (English)Zbl 07557566

MSC:  05C76 05C10 05C38
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### KMS states and continuous orbit equivalence for ultragraph shift spaces with sinks. (English)Zbl 07556781

MSC:  16S88 37B10 46L55
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### Invariant ideals in Leavitt path algebras. (English)Zbl 07556776

MSC:  16D25 16S88
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# Chapter 2 - Atoms, Molecules, and Ions - Exercises: 2.73e

$HgBr_2$

#### Work Step by Step

Mercury can have a variety of different charges. However, the textbook specifies that it is Mercury (II). The (II) indicates that this Mercury ion has a charge of 2+. Bromine is located one group to the left of the noble gases, meaning that it has a charge of 1-. In order for the charges of the ions to cancel each other out, there must be two Bromine ions for every one Mercury ion. Because metals come before non-metals in chemical formulas, the chemical formula for Mercury (II) Bromide is $HgBr_2$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

format: plain


text:
Browse Questions

# What is the decreasing order of strength of the bases. $OH^{-}.NH_2^{-}.HC \equiv C^{-}$ and $CH_3 -CH_2^{-}$?

$(a) \;CH_3CH_2^{-} > NH^{-} _2 > HC \equiv C^{-} > OH^{-} \\ (b)\;HC \equiv C^{-} > CH_3CH_2^{-} > NH_2^{-} > OH^{-} \\ (c)\;OH^{-} > NH_2^{-} > HC \equiv C^{-} > CH_3CH_2^{-} \\ (d)\;NH_2 ^{-} > HC \equiv C^{-} > OH^{-} > CH_3CH_2^{-}$

The strength of their conjugate acids are in the order :
$H_2O > HC \equiv CH > NH_3 > CH_3 -CH_3$
Stronger the acid weaker is its conjugate base.
Therefore correct order of strength their conjugate bases is
$CH_3 - CH_2^{-} > NH_2^{-} > HC \equiv C^{-} > OH^{-}$
Hence a is the correct answer.


text:
Ping-Pong Lemma

Question. Let G be a group and $a, b \in G$.  When is $\langle a, b \rangle \cong F_2$?

Ping-Pong Lemma: Let G be a group acting on a set X and $a, b \in G$. Assume:

1. a and b have infinite orders.
2. There exist $X_1, X_2 \subseteq X$ such that $X_2 \nsubseteq X_1$ and $a^m X_2\subseteq X_1$, $b^m X_1 \subseteq X_2$ for all $m \neq 0$.

Then $\langle a, b \rangle \cong F_2$.

Proof. Consider $\varphi : F_2 = \langle x, y \rangle \twoheadrightarrow \langle a,b \rangle \leq G$ such that $\varphi (x) = a$ and $\varphi (y) = b$.  Choose a reduced word w for a nontrivial element in $F_2$, i.e., either $w = x^{m_1} y^{n_1} x^{m_2} \cdots$ or $w = y^{m_1} x^{n_1} y^{m_2} \cdots$ and $m_i, n_i \neq 0$.

Case 1: $w = x^{m_1} y^{n_1} \cdots x^{m_k}$.  Then $\varphi (w) X_2 = a^{m_1} b^{n_1} \cdots a^{m_k} X_2$ and $a^{m_k} X_2 \subseteq X_1$, so $\varphi (w) X_2 \subseteq a^{m_1} b^{n_1} \cdots b^{n_{k-1}} X_1$ and $b^{n_{k-1}} X_1 \subseteq X_2$, and so on until $\varphi (w) X_2 \subseteq a^{m_1} X_2 \subseteq X_1$, so $\varphi (w) X_2 \neq X_2$.  Therefore $\varphi (w) \neq 1$.

Case 2: $w = y^{m_1} x^{n_1} \cdots y^{m_k}$.  Then, by Case 1, $\varphi (xwx^{-1}) \neq 1$, so $\varphi (x) \varphi (w) \varphi (x^{-1}) \neq 1$.  Therefore $\varphi (w) \neq 1$.

Case 3: $w = x^{m_1} y^{n_1} \cdots y^{n_k}$.  (similar to above)

Case 4: $w = y^{m_1} x^{n_1} \cdots x^{n_k}$.  (similar to above)

Free Groups Are Linear

Theorem 3. $F_2$ is linear.

Proof. $\mathrm{SL}_2\mathbb{R}$ acts on $\mathbb{R}^2$ by linear transformations.  Let $a = \left(\begin{array}{cc} 1&2 \\ 0&1 \\ \end{array}\right)$ and $b = \left(\begin{array}{cc} 1&0 \\ 2&1 \\ \end{array}\right)$.  Then $a^m = \left(\begin{array}{cc} 1&2m \\ 0&1 \\ \end{array}\right)$ and $b^m = \left(\begin{array}{cc} 1&0 \\ 2m&1 \\ \end{array}\right)$. Let $X_1 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| > |y|\right\}$ and $X_2 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| < |y|\right\}$.  Then $a^m X_2 \subseteq X_1, b^m X_1 \subseteq X_2$ for all $m \neq 0$, so $\langle a, b \rangle \cong F_2$ by the Ping-Pong Lemma.

Corollary 1. Finitely generated free groups are linear, hence residually finite.

Proof. The case of $\mathbb{Z}$ is obvious; otherwise, this follows from $F_n \hookrightarrow F_2$ as proved in Exercise 1.

Separability

Definition. Let G be a group.  The profinite topology on G is the coarsest topology such that every homomorphism from G to a finite group (equipped with the discrete topology) is continuous.

Definition. A subgroup H is separable in G if H is closed in the profinite topology of G.

Exercise 3. Let G be a group. $X \subseteq G$ is separable if and only if for all $g \in G \smallsetminus X$, there exists a homomorphism to a  finite group $q : G \rightarrow Q$ such that $q(g) \not \in q(X)$.  Note that if X = {1}, this is equivalent to: G is RF if and only if {1} is separable.

Hint. For the “if” direction, let $g \in G$ and consider $q^{-1} \circ q(g)$.  For the other direction, use the definition of subbase and that $p : G \rightarrow P, q: G \rightarrow Q \Rightarrow \text{ker}(p \times q) = \text{ker}(p) \cap \text{ker}(q)$.

Definition. Let G be a group.

1. G is Extended RF (ERF) if any subgroup of G is separable.
2. G is Locally ERF (LERF, subgroup separable) if any finitely generated subgroup is separable.

Lemma 3. Let G be a group. A subgroup H of G is separable if and only if for all $g \in G \smallsetminus H$, there exists a finite-index subgroup $K \leq G$ such that $H \leq K \leq G$ and $g \not \in K$.

Proof. In the “only if” direction, by the previous exercise, for all $g \in G \smallsetminus H$, there exists a homomorphism to a  finite group $q : G \rightarrow Q$ such that $q(g) \not \in q(H)$. Then $g \not \in q^{-1} \circ q(H) = K$.  Conversely, let $g \in G \smallsetminus H$. By hypothesis, there exists a finite-index subgroup $K \leq G$ such that $H \leq K \leq G$ and $g \not \in K$.  Let $L = \cap_{h \in G} K^h$. Note that this is a finite number of intersections ($|G/N_G (K)|$, to be precise). There exists a finite quotient $q : G \rightarrow G/L$.  Then  $q^{-1} \circ q(H) = H \cdot L$.  Therefore, $g \not \in q^{-1} \circ q(H)$, i.e., $q(g) \not \in q(H)$, and the lemma follows by the previous exercise.

Scott’s Criterion (1978). Let X be a Hausdorff topological space and $G = \pi_1 (X,x)$.  Let $X' \rightarrow X$ be a covering and $H = \pi_1 (X',x')$. Then H is separable in G if and only if for any compact $\Delta \subseteq X'$, there exists and intermediate finite-sheeted cover $\hat{X} \rightarrow X$ such that $X' \rightarrow \hat{X}$ embeds $\Delta$ into $\hat{X}$.

Exercise 4. Let $H \leq G$ be a separable subgroup.

1. If $G' \leq G$, then $G' \cap H$ is separable in G’.
2. If $G \leq G'$ has finite index, then $H$ is separable in G’.


text:
##### Site index · List index
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From tillrohrmann <...@git.apache.org>
Date Thu, 07 May 2015 09:48:13 GMT
Github user tillrohrmann commented on a diff in the pull request:

--- Diff: docs/libs/ml/optimization.md ---
@@ -0,0 +1,222 @@
+---
+mathjax: include
+title: "ML - Optimization"
+displayTitle: ML - Optimization
+---
+
+
+{:toc}
+
+
+## Mathematical Formulation
+
+The optimization framework in Flink is a developer-oriented package that can be used
to solve
+[optimization](https://en.wikipedia.org/wiki/Mathematical_optimization)
+problems common in Machine Learning (ML) tasks. In the supervised learning context, this
usually
+involves finding a model, as defined by a set of parameters $w$, that minimize a function
$f(\wv)$
+given a set of $(\x, y)$ examples,
+where $\x$ is a feature vector and $y$ is a real number, which can represent either a
real value in
+the regression case, or a class label in the classification case. In supervised learning,
the
+function to be minimized is usually of the form:
+
+$$+$$+ f(\wv) := + \frac1n \sum_{i=1}^n L(\wv;\x_i,y_i) + + \lambda\, R(\wv) + \label{eq:objectiveFunc} + \ . +$$ +$$
+
+where $L$ is the loss function and $R(\wv)$ the regularization penalty. We use $L$ to
measure how
+well the model fits the observed data, and we use $R$ in order to impose a complexity
cost to the
+model, with $\lambda > 0$ being the regularization parameter.
+
+### Loss Functions
+
+In supervised learning, we use loss functions in order to measure the model fit, by
+penalizing errors in the predictions $p$ made by the model compared to the true $y$ for
each
+example. Different loss function can be used for regression (e.g. Squared Loss) and classification
+(e.g. Hinge Loss).
+
+Some common loss functions are:
+
+* Squared Loss: $\frac{1}{2} (\wv^T \x - y)^2, \quad y \in \R$
+* Hinge Loss: $\max (0, 1 - y ~ \wv^T \x), \quad y \in \{-1, +1\}$
+* Logistic Loss: $\log(1+\exp( -y ~ \wv^T \x)), \quad y \in \{-1, +1\}$
+
+Currently, only the Squared Loss function is implemented in Flink.
+
+### Regularization Types
+
+[Regularization](https://en.wikipedia.org/wiki/Regularization_(mathematics)) in machine
learning is
+imposes penalties to the estimated models, in order to reduce overfitting. The most common
penalties
+are the $L_1$ and $L_2$ penalties, defined as:
+
+* $L_1$: $R(\wv) = \|\wv\|_1$
+* $L_2$: $R(\wv) = \frac{1}{2}\|\wv\|_2^2$
+
+The $L_2$ penalty penalizes large weights, favoring solutions with more small weights
rather than
+few large ones.
+The $L_1$ penalty can be used to drive a number of the solution coefficients to 0, thereby
+producing sparse solutions.
+The optimization framework in Flink supports the $L_1$ and $L_2$ penalties, as well as
no
+regularization. The
+regularization parameter $\lambda$ in $\eqref{objectiveFunc}$ determines the amount of

+regularization applied to the model,
+and is usually determined through model cross-validation.
+
+
+In order to find a (local) minimum of a function, Gradient Descent methods take steps
in the
+direction opposite to the gradient of the function $\eqref{objectiveFunc}$ taken with
+respect to the current parameters (weights).
+In order to compute the exact gradient we need to perform one pass through all the points
in
+a dataset, making the process computationally expensive.
+An alternative is Stochastic Gradient Descent (SGD) where at each iteration we sample
one point
+from the complete dataset and update the parameters for each point, in an online manner.
+
+In mini-batch SGD we instead sample random subsets of the dataset, and compute the gradient
+over each batch. At each iteration of the algorithm we update the weights once, based
on
+the average of the gradients computed from each mini-batch.
+
+An important parameter is the learning rate $\eta$, or step size, which is currently
determined as
+$\eta = \frac{\eta_0}{\sqrt{j}}$, where $\eta_0$ is the initial step size and $j$ is
the iteration
+number. The setting of the initial step size can significantly affect the performance
of the
+algorithm. For some practical tips on tuning SGD see Leon Botou's
+
+The current implementation of SGD  uses the whole partition, making it
+effectively a batch gradient descent. Once a sampling operator has been introduced in
+mini-batch SGD will be performed.
+
+
+### Parameters
+
+  The stochastic gradient descent implementation can be controlled by the following parameters:
+
+

+

+        Parameter
+        Description
+

+

+

+        Loss Function
+
+

+            The class of the loss function to be used. (Default value:
+            SquaredLoss, used for regression tasks)
+

+
+

+

+        RegularizationType
+
+

+            The type of regularization penalty to apply. (Default value:
+            NoRegularization)
+

+
+

+

+        RegularizationParameter
+
+

+            The amount of regularization to apply. (Default value:0)
+

+
+

+

+        Iterations
+
+

+            The maximum number of iterations. (Default value: 10)
+

+
+

+

+        Stepsize
+
+

+            Initial step size for the gradient descent method.
+            This value controls how far the gradient descent method moves in the opposite
+            (Default value: 0.1)
+

+
+

+

+

--- End diff --

I think the PredictionFunction parameter is missing.

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text:
#### Vol. 10, No. 6, 2017

 Recent Issues
 The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print) Author Index To Appear Other MSP Journals
Bergman kernel and hyperconvexity index

### Bo-Yong Chen

Vol. 10 (2017), No. 6, 1429–1454
##### Abstract

Let $\Omega \subset {ℂ}^{n}$ be a bounded domain with the hyperconvexity index $\alpha \left(\Omega \right)>0$. Let $\varrho$ be the relative extremal function of a fixed closed ball in $\Omega$, and set $\mu :=|\varrho |{\left(1+|log|\varrho ||\right)}^{-1}$ and $\nu :=|\varrho |{\left(1+|log|\varrho ||\right)}^{n}$. We obtain the following estimates for the Bergman kernel. (1) For every $0<\alpha <\alpha \left(\Omega \right)$ and $2\le p<2+2\alpha \left(\Omega \right)∕\left(2n-\alpha \left(\Omega \right)\right)$, there exists a constant $C>0$ such that ${\int }_{\Omega }|{K}_{\Omega }\left(\phantom{\rule{0.3em}{0ex}}\cdot \phantom{\rule{0.3em}{0ex}},w\right)∕\sqrt{{K}_{\Omega }\left(w\right)}{|}^{p}\le C|\mu \left(w\right){|}^{-\left(p-2\right)n∕\alpha }$ for all $w\in \Omega$. (2) For every $0, there exists a constant $C>0$ such that $|{K}_{\Omega }\left(z,w\right){|}^{2}∕\left({K}_{\Omega }\left(z\right){K}_{\Omega }\left(w\right)\right)\le C{\left(min\left\{\nu \left(z\right)∕\mu \left(w\right),\nu \left(w\right)∕\mu \left(z\right)\right\}\right)}^{r}$ for all $z,w\in \Omega$. Various applications of these estimates are given.

 Dedicated to Professor John Erik Fornaess on the occasion of his 70th birthday
##### Keywords
Bergman kernel, hyperconvexity index
Primary: 32A25
Secondary: 32U35


text:
Transforming Categorical to Numerical variable

I have a categorical variable with 4 levels ('8 c', '6 c','NAN','Others') and I want to convert it to numerical form. an Obvious way is to simply remove the 'c' part from the first two categories and replace NAN with 0. However, I was wondering about the 'Others' level?

What could be the best way to transform this level? Please note that the variable represents the number of cylinders for a given car.

• Firstly, what is the meaning of (others) in this column?
– user119783
Jul 16 '21 at 9:19
• it wasn't specified in the dataset description but I guess it means any other value than 8 c or 6 c
– user120794
Jul 16 '21 at 9:47
• Could you please tell me what are the possible max number of cylinders you may have other than 6 and 8? Can this column have 0 number of cylinder, does that make sense? I think I need to understand more the dataset to give you a better solution. If the dataset is publicly available, you can share it with us.
– user119783
Jul 16 '21 at 10:27
• I'm trying to analyse this dataset: kaggle.com/austinreese/craigslist-carstrucks-data
– user120794
Jul 16 '21 at 11:39

I spent some time exploring this dataset:

There are some findings I want to share it with you:

• Number of samples is 426880 samples.
• Number of categories in the cylinder column are: 3,4,5,6,7,8,10,12,others,Blank cell. You could take a look at the cylinder list at the beginning. here
• There are no 7, 9, and 11 cylinders. Then, 'others' could contain more likely either 1 or 2 cylinders. click here
• Number of samples contain the 'others' value in the cylinder column is 1298 samples which is equivalent to 0.3% of the total number of samples.

You can handle missing value and 'others' as following:

1. Others: due to the very small amount of samples containing 'others', you could remove all these samples. Alternatively, you could replace them with 1 or 2 cylinders where 2 is the most popular.

2. Blank cells: from my naive knowledge of cars, if the cars are from the same manufacturer and the same model, the same type of fuel, they have probably the same number of cylinders. Then, you can replace the blank values with the number of cylinders from the other cars which have the number of cylinders. Please, see the example below:

Another example:

Another example:

• If you find the car model is missing. Then, I recommend removing these samples (they are 2673).

I would suggest in this case 2 steps as part of your data preparation:

• substitute 'NAN' for 'Others', since both labels are giving you no info and can be considered as unknown values
• once you have finally 3 labels ('8 c', '6 c','Others'), apply one hot encoding, since you only have 3 possible categories (which prevents your dataset from being too sparse) and at the same time you do not assume that the unknown values are 0 cylinders.
• But he wants to convert them to ordinal data(6<8) not one-hot code.
– user119783
Jul 16 '21 at 10:42
• I thought at the beginning like you with dealing with nan and others together. But then I thought that nan values can contain 6, 8 or any value but others contains any value but not 6 and 8 because this is the meaning of others.
– user119783
Jul 16 '21 at 10:46
• Ok to your second point, nice appreciation. About the first point, I think she/he should not force ordinal encoding, since does not know whether the 'other' values are higher or lower than 6, 8... Jul 16 '21 at 16:13

It really depends what your variable refers to, and which kind of model you want to use.

A few things you can do :

• OneHotEncoding : will create binary variables for each possibility for your variable : in your case, it'll create 4 variables '8 c', '6 c','NAN','Others', that take 1 or 0. This way, each possible variable output is now a binary variable, independant from others. Example : 'Var'='8 c' becomes '8 c' = 1, '6 c' = 0, 'NAN' = 0, 'Others' = 0

• Manual coding with order : You can yourself transform '8 c' as 1, '6 c' as 2, 'NAN' as 3 and 'Others' as 4 for example, but this will mean for your model than '8 c' and '6 c' are closer than '8 c' and 'Others' for example. That's a method you can use for ordinal values (when V1 > V2 > ... )

• Target Encoding : Often used when you have too much possible values, and can't afford to OneHot (which creates as much variables as potential values). Target Encoding will code your values as numbers, refering to their potential link to the target variable (y in your modelisation). I don't recommend it in your case since you only have 4 potential values.

• You can't replace nan with 0 because nan means no information. There are some techniques to deal with this.
– user119783
Jul 16 '21 at 10:52
• can you please share with me the name of one of these techniques?
– user120794
Jul 16 '21 at 11:40
• @Hamzawi : it doens't replace NaN to 0, it creates as much BINARY variables as potential values, and when Value is NaN, it assigns '1' to NaN value and 0 to others. Jul 16 '21 at 12:00
• @maryamBlushi OneHotEncoding and TargetEncoding are the names of the techniques Jul 16 '21 at 12:00
• @MaryamBlushi. If you mean techniques to replace Nan values in your dataset. There is a lot you could take a look at here: analyticsindiamag.com/…
– user119783
Jul 16 '21 at 13:37


text:
## Advances in Operator Theory

### Almost Hadamard matrices with complex entries

#### Abstract

We discuss an extension of the almost Hadamard matrix formalism, to the case of complex matrices. Quite surprisingly, the situation here is very different from the one in the real case, and our conjectural conclusion is that there should be no such matrices, besides the usual Hadamard ones. We verify this conjecture in a number of situations, and notably for most of the known examples of real almost Hadamard matrices, and for some of their complex extensions. We discuss as well some potential applications of our conjecture, to the general study of complex Hadamard matrices.

#### Article information

Source
Adv. Oper. Theory, Volume 3, Number 1 (2018), 137-177.

Dates
Received: 9 February 2017
Accepted: 12 May 2017
First available in Project Euclid: 5 December 2017

Permanent link to this document
https://projecteuclid.org/euclid.aot/1512497957

Digital Object Identifier
doi:10.22034/aot.1702-1114

Mathematical Reviews number (MathSciNet)
MR3730344

Zentralblatt MATH identifier
1373.15047

#### Citation

Banica, Teodor; Nechita, Ion. Almost Hadamard matrices with complex entries. Adv. Oper. Theory 3 (2018), no. 1, 137--177. doi:10.22034/aot.1702-1114. https://projecteuclid.org/euclid.aot/1512497957

#### References

• T. Banica, B. Collins and J.-M. Schlenker, On orthogonal matrices maximizing the 1-norm, Indiana Univ. Math. J. 59 (2010), 839–856.
• T. Banica and I. Nechita, Almost Hadamard matrices: the case of arbitrary exponents, Discrete Appl. Math. 161 (2013), 2367–2379.
• T. Banica, I. Nechita and J.-M. Schlenker, Submatrices of Hadamard matrices: complementation results, Electron. J. Linear Algebra 27 (2014), 197–212.
• T. Banica, I. Nechita and K. Życzkowski, Almost Hadamard matrices: general theory and examples, Open Syst. Inf. Dyn. 19 (2012), 1–26.
• N. Barros e Sá and I. Bengtsson, Families of complex Hadamard matrices, Linear Algebra Appl. 438 (2013), 2929–2957.
• K. Beauchamp and R. Nicoara, Orthogonal maximal abelian $*$-subalgebras of the $6\times6$ matrices, Linear Algebra Appl. 428 (2008), 1833–1853.
• G. Björck, Functions of modulus $1$ on ${\rm Z}_n$ whose Fourier transforms have constant modulus, and cyclic $n$-roots, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 315 (1990), 131–140.
• A. T. Butson, Generalized Hadamard matrices, Proc. Amer. Math. Soc. 13 (1962), 894–898.
• C. J. Colbourn and J. H. Dinitz, Handbook of combinatorial designs, Second edition. Discrete Mathematics and its Applications (Boca Raton). Chapman & Hall/CRC, Boca Raton, FL, 2007.
• U. Haagerup, Orthogonal maximal abelian $*$-subalgebras of the $n\times n$ matrices and cyclic $n$-roots, in “Operator algebras and quantum field theory”, International Press (1997), 296–323.
• U. Haagerup, Cyclic $p$-roots of prime lengths $p$ and related complex Hadamard matrices, arxiv: 0803.2629.
• J. Hadamard, Résolution d'une question relative aux déterminants, Bull. Sci. Math. 2 (1893), 240–246.
• K.J. Horadam, Hadamard matrices and their applications, Princeton University Press, Princeton, NJ, 2007.
• M. Idel and M. M. Wolf, Sinkhorn normal form for unitary matrices, Linear Algebra Appl. 471 (2015), 76–84.
• B.R. Karlsson, Three-parameter complex Hadamard matrices of order 6, Linear Algebra Appl. 434 (2011), 247–258.
• H. Kharaghani and B. Tayfeh-Rezaie, A Hadamard matrix of order 428, J. Combin. Des. 13 (2005), 435–440.
• W. de Launey and D. A. Levin, A Fourier-analytic approach to counting partial Hadamard matrices, Cryptogr. Commun. 2 (2010), 307–334.
• K. H. Leung and B. Schmidt, New restrictions on possible orders of circulant Hadamard matrices, Des. Codes Cryptogr. 64 (2012), 143–151.
• H. J. Ryser, Combinatorial mathematics, The Carus Mathematical Monographs, No. 14 Published by The Mathematical Association of America; distributed by John Wiley and Sons, Inc., New York 1963.
• D. R. Stinson, Combinatorial designs: constructions and analysis, With a foreword by Charles J. Colbourn. Springer-Verlag, New York, 2004.
• J. J. Sylvester, Thoughts on inverse orthogonal matrices, simultaneous sign-successions, and tesselated pavements in two or more colours, with applications to Newton's rule, ornamental tile-work, and the theory of numbers, Phil. Mag. 34 (1867), 461–475.
• F. Szöllősi, Exotic complex Hadamard matrices and their equivalence, Cryptogr. Commun. 2 (2010), 187–198.
• W. Tadej and K. Życzkowski, A concise guide to complex Hadamard matrices, Open Syst. Inf. Dyn. 13 (2006), 133–177.
• W. Tadej and K. Życzkowski, Defect of a unitary matrix, Linear Algebra Appl. 429 (2008), 447–481.


text:
If causality reversal is an argument against velocities exceeding the velocity of light, can the same argument be made for sound?

Reversal of causality is stated as an argument against any v>c, like in this thought experiment.

But if we were a species that used echolocation as our primary sense, and the ball was travelling faster than the speed of sound, then wouldn't the train passengers (in this case too) experience the reverse order of events ? They would then conclude that nothing can travel faster than sound. What am I getting wrong here ?

(I know that there are other ways to show that things can't accelerate to the speed of light.. but I am just asking with reference to the logic of the above thought experiment)

"They would then conclude that nothing can travel faster than sound."

Someone using echolocation to continuously observe objects moving uniformly along straight trajectories would find that some objects remain always "visible", on both the inbound and the outbound legs (at relative subsonic velocities), while others "disappear" after passing by (supersonic velocities). But eventually they would also find that at least some of these "disappearing" supersonic objects may become "visible" again if the observer ran fast enough after them, for instance by riding a train. And then they would necessarily discover that if this works on the ground, it also works on the train, etc.

So an echolocating species would be bound to understand that velocities faster than sound are entirely possible.

Regarding reversal of causality:

An observer probing an incoming supersonic object through echolocation would indeed receive return signals in backwards order, in an apparent reversal of causality. Then would loose sight of the target on the outbound leg. The sounds emitted by supersonic objects would likewise be received in backwards order, on both legs. But the correct causal sequence may still be observed by recording position vs. time locally, at any given points along the trajectory.

Or the echolocators may discover better means of observation, like e.m. waves. At which point they would also discover the speed of light limit and relativity ;D

In any case, the echolocation "reversal of causality" would remain "apparent", as opposed to an actual reversal where events occur in backwards order even when observed through point-local recordings.

A couple of observations on the train scenario itself:

I see two tacit assumptions here:

1. That if the ball is moving faster than the speed of sound relative to the ground, then it is also moving faster than the speed of sound relative to the train. Obviously not necessarily. A ball moving faster than sound relative to the ground may still move slower than sound relative to the train. On the other hand a light pulse is always traveling at the speed of light, in any frame.

2. Possibly that the echolocation signals used by passengers on the train go at the speed of sound relative to the ground, not relative to the carrier medium (air) in the car. Completely untrue as far as sound is concerned.

This latter idea reminds me of an anecdote about the first trains designed to go at the "unheard of velocity" of ... 30mph: there was a public outcry against them out of concern that air in the cars would "naturally" swoosh out and leave the passengers in great peril of asphyxiation. The point: the air in the train car remains at rest with respect to the car, but moves at the velocity of the car relative to the ground. Any echolocation signals propagating forward within the car travel faster-than-sound relative to the ground. No such thing happens with light, there is no carrier medium for light that moves along with the train car.

The species in question that used echolocation would experience the sequence of events in the wrong order but their subjective experience is not the objective truth of the causality of the events. It is just their psychological interpretation of the environment which wasn't meant to deal with this ultra rare scenario.

The species that used echolocation would conclude that objects CAN move faster than sound because they don't obey causality (assuming they could differentiate that the order is indeed the wrong way round from the type of sound). Whereas it is impossible to witness the scenario in thought experiment so we conclude the opposite that nothing does travel faster than light.

First of all the fact that there is a limit to the speed that anything can move is a consequence of the homogeneity and isotropy of space-time. This limit is the speed of any massless particle. The point I am trying to make is that the limit to speed is a property of space-time. Now can similar arguments be made for sound? No. First of all what is sound? Sound is ultimately is the oscillation of matter pressure and displacement $\textit{through}$ a medium like air or water. That medium is contained in space-time.This means sound waves have nothing to do with the structure of space-time itself and hence nothing to with causality.

• My question is regarding the above thought experiment.. we can use sound instead of light (with an echolocating observer), and the above will "prove" that causality breaks beyond the speed of light.. – Daud Dec 20 '16 at 5:23
• @Daud This is why I hate these thought experiments. They totally miss the point. The structure of special relativity has nothing to with light (although that is how Einstein found it). This makes mute the point, "What about if I did things with sound?" My point is that causality has to do with the structure of space-time and that is something you are not investigating if you use sound. You can investigate it if you use light because photons are massless and move at the maximum speed. – Amara Dec 20 '16 at 13:37


text:
# How to find perpendicular vector to another vector?

How do I find a vector perpendicular to a vector like this: $$3\mathbf{i}+4\mathbf{j}-2\mathbf{k}?$$ Could anyone explain this to me, please?

I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components...

When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks.

-
Choose two coordinates, switch them, add a minus sign, and complete with zeroes. For example: choosing i and j might yield 4i-3j, choosing i and k might yield 2i+3k, and choosing j and k might yield 2j+4k. – Did Apr 26 '12 at 19:09
@Didier thanks for letting me know but as you told,we have got 3 solutions. 4i-3j,2i+3k,2j+4k its not single vector.I need a vector something like ai+bj+ck which is perpendicular to other vector.sorry but I Just started to learn vectors. – niko Apr 26 '12 at 19:15
$2j+4k=0i+2j+4k$. – David Mitra Apr 26 '12 at 19:19
Pick any vector not colinear to your vector and take their cross product. – N. S. Apr 26 '12 at 20:26
Not to de-rail the thread, but does anyone know why this particular question has over 15k views? – Jesse Madnick Feb 27 at 7:51

There exist infinite vector in 3 dimension that are perpendicular to a fixed one. They should only satisfy the following formula: $$(3\mathbf{i}+4\mathbf{j}-2\mathbf{k}) \cdot v=0$$

For finding all of them, just choose 2 perpendicular vectors, like $v_1=(4\mathbf{i}-3\mathbf{j})$ and $v_2=(2\mathbf{i}+3\mathbf{k})$ and any linear combination of them is also perpendicular to the original vector: $$v=((4a+2b)\mathbf{i}-3a\mathbf{j}+3b\mathbf{k}) \hspace{10 mm} a,b \in \mathbb{R}$$

-
 What is this kind of notation called? I have never seen a vector being defined like $(3i + 4j - 2k)$. The notation I've seen so far would be $\left(\begin{array}{c}3\\4\\2\end{array}\right)$, therefore I do not really understand your answer. :( – Niklas R Oct 30 '12 at 22:20 There are many possible notation, I choose to use the same notation of the question, but other choice are good as well. $i$,$j$,$k$ refers to vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, so it is basically the same thing after you do vector-scalar multiplication. – carlop Dec 14 '12 at 13:49

You just need to find any vector $v \neq 0$ such that $v \cdot (3\mathbf{i}+4\mathbf{j}-2\mathbf{k}) = 0$.

There is no unique solution, any one will do. To save typing, let $p = 3\mathbf{i}+4\mathbf{j}-2\mathbf{k}$.

Pick a vector $x$, that is not on the line through the origin and $p$. Take $x = 3\mathbf{i}$, for example.

Construct a vector perpendicular to $p$ in the following way: Find a value of $t$ so that $(v+t p) \cdot p = 0$. Then the vector $(v+t p)$ will be perpendicular to $v$.

In my example, $(v+t p) = (3 + 3 t)\mathbf{i}+4 t \mathbf{j}-2t\mathbf{k}$, and $(v+t p) \cdot p = 9 + 29 t$. By choosing $t=-\frac{9}{29}$, the vector $v+t p$ is now perpendicular to $p$.

-

A suggested solution without a branch could be: Construct an array of 2 vector elements in the following way:

arr[0] = (c,c,-a-b) arr[1] = (-b-c, a,a)
int selectIndex = ((c != 0) && (-a != b)) // this is not a branch
perpendicularVector = arr[selectIndex]

If (c, c, -a-b) is zero, selectIndex is 1 and the other vector will be selected.

-

A related problem is to construct an algorithm that finds a non-zero perpendicular vector without branching. If the input vector is N = (a,b,c), then you could always choose T = (c,c,-a-b) but T will be zero if N=(-1,1,0). You could always check to see if T is zero, and then choose T = (-b-c,a,a) if it is, but this requires a test and branch. I can't see how to do this without the test and branch.

-

The dot product of two perpendicular vectors are always 0 so if you (ai+bj+ck)*(di+ej+fk)=0 you can solve for the different variables. If you have one vector than the infinite amount of perpendicular vectors will form a plane that is perpendicular to the original vector. If you know one or two of the coordinates of the desired perpendicular line than you can find the corresponding vector(s) on that plane.

-

Remember: There exist infinite vector in 3 dimension that are perpendicular to a fixed one. Now, Let v ≠0 be the vector whose is xi+yj+zk. So , v is perpendicular to the vector 3i+4j-2k. Therefore, v.(3i+4j−2k)=0. (xi+yj+zk). (3i+4j−2k)=0 3x+4y-2z=0 ……………….(1) where i.i =j.j=k.k=1.

Now, there are three unknown variable such x, y and z in (1). You can choose any two variable whatever you like. Let y=2 and z=1,
then x=-2 from (1),

One of the vector is (-2i+2j+k). Similarly, you can choose one of two variables from (1) , then find the third variable. So, you can find infinite perpendicular vectors to the vector 3i+4j-2k.

-


text:
#### Volume 16, issue 3 (2016)

 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
L-space surgery and twisting operation

### Kimihiko Motegi

Algebraic & Geometric Topology 16 (2016) 1727–1772
##### Abstract

A knot in the $3$–sphere is called an L-space knot if it admits a nontrivial Dehn surgery yielding an L-space, ie a rational homology $3$–sphere with the smallest possible Heegaard Floer homology. Given a knot $K$, take an unknotted circle $c$ and twist $K$ $n$ times along $c$ to obtain a twist family $\left\{{K}_{n}\right\}$. We give a sufficient condition for $\left\{{K}_{n}\right\}$ to contain infinitely many L-space knots. As an application we show that for each torus knot and each hyperbolic Berge knot $K$, we can take $c$ so that the twist family $\left\{{K}_{n}\right\}$ contains infinitely many hyperbolic L-space knots. We also demonstrate that there is a twist family of hyperbolic L-space knots each member of which has tunnel number greater than one.

##### Keywords
L-space surgery, L-space knot, twisting, seiferter, tunnel number
##### Mathematical Subject Classification 2010
Primary: 57M25, 57M27
Secondary: 57N10


text:
A fourth-order compact time-splitting method for the Dirac equation with time-dependent potentials
In this paper, we present an approach to deal with the dynamics of the Dirac equation with time-dependent electromagnetic potentials using the fourth-order compact time-splitting method ($S_\text{4c}$). To this purpose, the time-ordering technique for time-dependent Hamiltonians is introduced, so that the influence of the time-dependence could be limited to certain steps which are easy to treat. Actually, in the case of the Dirac equation, it turns out that only those steps involving potentials need to be amended, and the scheme remains efficient, accurate, as well as easy to implement. Numerical examples in 1D and 2D are given to validate the scheme.


text:
# QAuxiliaryModularEtaQuotientPackage¶

QAuxiliaryModularEtaQuotientPackage helps to do computations with eta quotients that are modular functions.

aLigozat: (PositiveInteger , Integer , Integer ) -> Integer
aLigozat(m, c, delta) computes an entry of the matrix of Ligozat corresponding to the index (c, delta) counted in divisors.
coprimePart: (Integer , Integer ) -> Integer
coPrimePart(n, c) splits n into u*v such that gcd(u,c)=1 and returns u. We assume that all input numbers are positive.
cusps2OfGamma0: PositiveInteger -> List Fraction Integer
Do not use that function! It is just another implementation of cuspsOfGamma0.
cuspsOfGamma0: PositiveInteger -> List Fraction Integer
According to cite[Lemma~2.45]{Radu:PhD:2010}, there are $eulerphi(gcd(N/c, c))$ different cusps $frac{a}{c}$ of $Gamma_0(N)$ that correspond to a divisor $c$ of $N$. cusps(m) returns all the cusps for Gamma_0(m) sorted first by denominator and then (among the cusps with the same denominator) by the numerator of the cusp.
cuspToMatrix: (PositiveInteger , Fraction Integer ) -> Matrix Integer
cuspToMatrix(m, a/c) returns a matrix gamma=[[a,b],[c,d]] such that splitMatrix(gamma, m) returns two matrices [m1, m2] with m1*m2 = [[m*a, c], [m*b, d]] and m1 being in SL_2(Z) and m2 being a triangular matrix with 0 <= m2(1,2) < m/c.
dedekindPsi: PositiveInteger -> PositiveInteger
dedekindPsi(n) computes the value of the Dedekind psi function. https://en.wikipedia.org/wiki/Dedekind_psi_function
doubleCosetRepresentatives: PositiveInteger -> List Matrix Integer
doubleCosetRepresentatives(nn) returns a list of double coset representatives of $Gamma_0(nn) backslash Gamma_0(1) / SL_2(setZ)_infty$ where $SL_2(setZ)_infty$ are matrices of the form [[1,h],[0,1]] with h being an integer.
genusOfGamma0: PositiveInteger -> NonNegativeInteger
genusOfGamma0(level) returns the genus of the congruence subgroup Gamma_0(level). It corresponds to the series https://oeis.org/A001617 .
indexOfGamma0: PositiveInteger -> PositiveInteger
indexOfGamma0(n) computes the index of Gamma_0(n) in SL2Z. indexOfGamma0(n) is equal to dedekindPsi(n).
jacobiLowerStar: (Integer , Integer ) -> Integer
See, for example, Definition 2.26 in the PhD thesis of Silviu Radu. http://www3.risc.jku.at/publications/download/risc_5453/main.pdf
jacobiUpperStar: (Integer , Integer ) -> Integer
See, for example, Definition 2.26 in the PhD thesis of Silviu Radu. http://www3.risc.jku.at/publications/download/risc_5453/main.pdf
matrixAtkinLehner: (PositiveInteger , PositiveInteger ) -> Matrix Integer
matrixAtkinLehner(m, t) with t a Hall divisor (https://en.wikipedia.org/wiki/Hall_subgroup) of m returns the 2x2 matrix with entries [[t, -a], [m, t*b]] such that n=m/t and t*b + n*a = 1, see cite{Kohnen:WeierstrassPointsAtInfinity:2004}.
matrixLigozat: PositiveInteger -> Matrix Integer
matrixLigozat(m) returns the (n,n) matrix with entries a_m(di,dj) given by Notation 3.2.6 in cite{Ligozat:CourbesModulaires:1975} where di and dj run over all positive divisors of m. This matrix is A_N as described after Theorem 5.2 in cite{Hemmecke+Radu:EtaRelations:2018}.
minimalSquareRootOfUnity: PositiveInteger -> PositiveInteger
minimalSquareRootOfUnity(p) returns an integer n such that in Z(x) (where x is a primitive n-th root of unity) there exists an element y such that y^2=p. p must be a radical integer, i.e. should not contain the square of a prime.
nu2: PositiveInteger -> Integer
cite[p.~25]{Shimura:ArithmeticTheory:1994}
nu3: PositiveInteger -> Integer
cite[p.~25]{Shimura:ArithmeticTheory:1994}
numberOfCuspsOfGamma0: PositiveInteger -> PositiveInteger
numberOfCuspsOfGamma0(m) returns the number of cusps of level m.
numberOfGaps: (PositiveInteger , List PositiveInteger ) -> NonNegativeInteger
numberOfGaps(n, l) computes reduce(+, [floor(x/n) for x in l], 0).
primePower: (PositiveInteger , PositiveInteger ) -> NonNegativeInteger
primePower(d, p) returns e such that gcd(d/p^e, p)=1, i.e. the highest power e such that p^e is a factor of d.
rationalSquareRoot: Fraction Integer -> Record(rat: Fraction Integer , rootOf: PositiveInteger )
if z = rationalSquareRoot(x) then x = (z.rat)^2 * z.root
rightCosetRepresentatives: (PositiveInteger , PositiveInteger ) -> List Matrix Integer
rightCosetRepresentatives(nn, mm) returns a list of right coset representatives of $Gamma_0(mm*nn) backslash Gamma_0(mm)$.
rStarConditionMatrix: PositiveInteger -> Matrix Integer
rStarConditionMatrix(nn) returns a matrix mat such that if n=#divisors(nn) and v is such that mat*v=0 then v(1..n) is in R^*(nn), in other words, the nullspace of mat “generates” R^*(nn).
rStarConditions: (PositiveInteger , List Integer ) -> Integer
rStarConditions returns 0 if all conditions are fulfilled. Otherwise it returns a positive number in the range 1 to 4 that corresponds to the condition that is not met.
rStarNonNegativeMatrix: (PositiveInteger , List Integer ) -> Matrix Integer
rStarNonNegativeMatrix(nn, idivs) returns a matrix mat which is a vertical concatenation of rStarConditionMatrix(nn) and (a part of) matrixLigozat(nn) whose rowindices correspond to idivs. It computes matrix B_N equation (8) in cite{Hemmecke+Radu:EtaRelations:2018} (DOI:10.1016/j.jsc.2018.10.001) joined with the respective rows of matrix A_N from the same article.
splitMatrix: (Integer , Integer , Integer , Integer ) -> Record(red: Matrix Integer , triang: Matrix Integer )
splitMatrix(a, b, c, d) returns a record rec that contains two matrices rec.red=m1 and rec.triang=m2 such that mat=m1*m2 according to cide[Lemma~4.8]{Radu:PhD:2010}, but in a variation as described in qetafun.spad, i.e. such that 0<=m2(1,2)<det(mat)/g where g = gcd(a, c) and mat=[[a,b],[c,d]]. Input condition: a*d-b*c>0.
splitMatrix: (Matrix Integer , Integer ) -> Record(red: Matrix Integer , triang: Matrix Integer )
splitMatrix(mat, delta) computes splitMatrix(mat, delta, 1, 0). Input condition: determinant(mat)>0.
splitMatrix: (Matrix Integer , Integer , PositiveInteger , NonNegativeInteger ) -> Record(red: Matrix Integer , triang: Matrix Integer )
splitMatrix(mat, delta, m, lambda) computes splitMatrix(a, b, c, d) where the arguments are the matrix entries of matrix [[delta, delta*lambda], [0, m]]*mat. Input condition: determinant(mat)>0.
upsilonExponent: (Integer , Integer , Integer , Integer ) -> IntegerMod 24
upsilonExponent(a,b,c,d) returns z in {0..23} such that exp(pi*i*z/12) is a multiplier in the transformation of an eta function by the matrix [[a,b],[c,d]] from SL_2(Z). A definition can be found in cite[Lemma~2.27]{Radu:PhD:2010}.
upsilonExponent: Matrix Integer -> IntegerMod 24
upsilonExponent(m) for a matrix m from SL_2(Z) returns upsilonExponent(m(1,1), m(1,2), m(2,1), m(2,2)).
width: (PositiveInteger , Integer ) -> Integer
width(m, c) returns the width of the cusp a/c for Gamma_0(m) and some a. The width is independent of a and given by m/gcd(c^2,m). cite[Lemma~2.37]{Radu:PhD:2010}


text:
# Find the Mean (Arithmetic) 7.1 , 2.1

7.1 , 2.1
The mean of a set of numbers is the sum divided by the number of terms.
7.1+2.12


text:
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text:
• Revision id: 476205
• Created:
• Creator: kscarfone
• Is current revision? Yes
• Comment Updated tags

{{DomRef}}

## Summary

Adopts a node from an external document. The node and its subtree is removed from the document it's in (if any), and its `ownerDocument` is changed to the current document. The node can then be inserted into the current document.

Supported since Gecko 1.9 (Firefox 3)

## Syntax

```node = document.adoptNode(externalNode);
```
• `node` is the adopted node that can be used in the current document. The new node's `parentNode` is `null`, since it has not yet been inserted into the document tree.
• `externalNode` is the node from another document to be adopted.

{{todo}}

## Notes

In general the `adoptNode` call may fail due to the source node coming from a different implementation, however this should not be a problem with browser implementations.

{{WRONG_DOCUMENT_ERR_note}}

## Revision Source

```

{{DomRef}}

Summary

Adopts a node from an external document. The node and its subtree is removed from the document it's in (if any), and its ownerDocument is changed to the current document. The node can then be inserted into the current document.

Supported since Gecko 1.9 (Firefox 3)

Syntax

node is the adopted node that can be used in the current document. The new node's parentNode is null, since it has not yet been inserted into the document tree.

externalNode is the node from another document to be adopted.

Example

{{todo}}

Notes

In general the adoptNode call may fail due to the source node coming from a different implementation, however this should not be a problem with browser implementations.

{{WRONG_DOCUMENT_ERR_note}}

Specification


text:
# Impossible Integral

Calculus Level pending

Taking the integral from zero to infinity can be tedious. To the nearest integer number find this integral for lim s to infinity x^2(x-s)^xdx

×


text:
# Lesson 7Striving for IndependencePractice Understanding

## Set

### 5.

A group of students was observed to determine whether or not they wear glasses and whether or not they wear a hooded sweatshirt. The data is shown in the table. Use the data to respond to the problems.

Glasses

No Glasses

Total

Hoodie

No Hoodie

Total

#### a.

How many students were observed?

#### f.

For this sample, are wearing glasses and wearing a hoodie independent events? Why or why not?

### 6.

The principal at a school checked student class schedules to see if they were enrolled in world languages or music classes. She compiled the data in the table below. Use the data to find the probabilities and answer the following problems.

Music

No Music

Total

Language

No Language

Total

#### a.

How many total outcomes are possible?

#### f.

Is scheduling of music and world language classes a set of independent events? Why or why not?

### 7.

Shorts

No Shorts

Total

Hat

No Hat

Total

#### a.

How many total outcomes are possible?

#### f.

Are wearing shorts and wearing a hat independent events? Why or why not?

### 8.

People at an amusement park were surveyed to see if they rode the roller coaster and if they had purchased a souvenir. The data is shown in the table below. Use the data to find the probabilities and respond to the following problems.

Souvenir

No Souvenir

Total

Ride Coasters

Not Ride

Total

#### a.

How many total outcomes are possible?

#### f.

Are purchasing a souvenir and riding roller coasters independent events? Why or why not?

## Go

Data gathered on the shopping patterns during the months of April and May of high school students from Peanut Village revealed the following. of students purchased a new pair of shorts (call this event ), of students purchased a new pair of sunglasses (call this event ), and of students purchased both a pair of shorts and a pair of sunglasses.

### 9.

Find the probability that a student purchased a pair of sunglasses given that you know they purchased a pair of shorts.

### 10.

Find the probability that a student purchased a pair of shorts or purchased a new pair of sunglasses.

### 11.

Given that you know a student has purchased at least one of the items, what is the probability that they purchased only one of the items?

### 12.

Are the two events, and , independent of one another? Why or why not?

The table provided displays data collected from individuals concerning whether or not to extend the length of the school year. Use the table to respond to problems 13–15.

For

Against

No Opinion

Total

Youth (5 to 19)

Seniors (55+)

Total

### 13.

Given the condition that a person is an adult, what is the probability that they are in favor of extending the school year?

### 14.

Given the condition that a person is against extending the school year, what is the probability they are a senior?

### 15.

What is the probability that a person has no opinion given that they are a youth?


text:
# Johnson and Shot Noise. First Draft

Abstract

Two sources of noise, Johnson noise and shot noise, are investigated in this experiment. The Johnson noise which is the voltage fluctuations across a resistor that arose from the random motion of electrons. It was measured across different resistances and at different bandwidths at room temperature, resulting in a calculation of the Boltzmann constant of $$1.46 \cdot 10^{-23} \textrm{ m}^2 \textrm{ kg} \textrm{ s}^{-2} \textrm{ K}^{-1} \pm2.5 \cdot 10^{-21} \textrm{ m}^2 \textrm{ kg} \textrm{ s}^{-2} \textrm{ K}^{-1}$$ and $$1.46 \cdot 10^{-23} \textrm{ m}^2 \textrm{ kg} \textrm{ s}^{-2} \textrm{ K}^{-1} \pm2.6 \cdot 10^{-21} \textrm{ m}^2 \textrm{ kg} \textrm{ s}^{-2} \textrm{ K}^{-1}$$. The shot noise occurs due to the quantization of charge, and was measured by varying current in the system, with which we calculated the electron charge of $$1.64 \cdot 10^{-19} \pm 7.0 \cdot 10^{ -22}$$. They agree quite well with the accepted values of $$1.38064852 \cdot 10^{-23} \textrm{ m}^2 \textrm{ kg} \textrm{ s}^{-2} \textrm{ K}^{-1}$$, and $$1.64 \cdot 10^{-19} \textrm{C}$$ for the Boltzmann constant and electron charge respectively. Errors are discussed.

Please compare your description of Johnson Noise and Shot Noise below in your Aims and Introduction Sections with the short description I provided you in the getting started instructions for this lab (reprinted below):

By measuring Johnson noise — the fluctuating voltage that arises due to the random walk motion of electrons in a resistor (even when current isn’t flowing through it) —- as a function of temperature and frequency “bandwidth”, we can determine Boltzmann’s constant $$k_B$$, the fundamental constant connecting temperature $$T$$ to energy $$k_B T$$.

By measuring Shot noise — the fluctuations in a measured current of electrons due to quantization of charge — we can determine the fundamental unit of charge $$e$$ (the charge on an electron).

Who would have thought you could discover fundamental physics from simple “noise?” That’s pretty neat!

Notice how this version tries to provide a description of

1. what Johnson noise and Shot noise are

2. what we need to measure (and as a function of what variables)

3. and what we aim to determine from the measurement (for example, values for fundamental constants of physics $$k_B$$ and $$e$$)

.

Do you think your classmates in PHY350 would understand these three points from your paper, as written? If not, you need to rewrite your paper in a way that they can.

Do you understand these 3 points yourself? If not, your responsibility is to review the physics — I recommend reviewing the introductions to sections 1 and 3 of your teachspin manual! — and make sure you do! Then, rewrite the paper to make it clear you understand what you were doing (and why).


text:
# Show $V=\ker T\oplus Sp\left \{ w \right \}$

Let $V$ be a finite-dimensional vector space and let $T: V\rightarrow V$ be such that $\dim \text{Im} T=1$.

Show that there exists some $w\in V$ such that $V=\ker T\oplus \text{Sp}\left \{ w \right \}$

According to inuition, I'm pretty sure $0\neq w\in \text{Im} T$, so $\text{Sp}\left \{ w \right \}=\text{Im} T$. We're supposed to take a basis of $\ker T$- $\left \{ v_1,\cdots ,v_{n-1} \right \}$ (where $\dim V=n$). However I don't know where to make progress....

• Since the dimension of the image is one, there exists some $w$ such that Im$T$ =$\text{span}(w)$. Then all you need to do is show that any vector can be expressed as the sum of two vectors, one from the kernel and one from the image. Indeed $Tw \neq 0$ as you suggest. – John Martin Sep 2 '16 at 14:09
• How do you know $T(w)\neq 0$? – Theorem Sep 2 '16 at 14:14
• Second tentative.... Let $u\not =0$ in $Im(T)$, and $w$ such that $T(w)=u$ (hence $w\not =0$). For $x\in V$, we have $T(x)=bu$. Write $x=x-bw+bw$ – Kelenner Sep 2 '16 at 14:33
Let $w$ be any vector such that $T(w)\neq 0$. If $v\in \ker T\cap span\{w\}$ then $v=cw$ and $T(v)=0$ so $cT(w)=0$ thus $c=0$.
On the other hand if $v\in V$ is any vector then there is a $c$ such that $T(v)=cT(w)$ because the image is one dimensional. Now it is easy to see $v-cw\in \ker T$.


text:
## Le Monde puzzle [#1112]

Posted in Books, Kids, R with tags , , , on October 3, 2019 by xi'an

Another low-key arithmetic problem as Le Monde current mathematical puzzle:

Find the 16 integers x¹,x²,x³,x⁴,y¹,y²,y³,y⁴,z¹,z²,z³,z⁴,w¹,w²,w³,w⁴ such that the groups x¹,y¹,z¹,w¹, &tc., are made of distinct positive integers, the sum of the x’s is 24, of the y’s 20, of the z’s 19 and of the w’s 17. Furthermore, x¹<max(y¹,z¹,w¹), x²<max(y²,z²,w²), x³<max(y³,z³,w³)=z³, and x⁴<max(y⁴,z⁴,w⁴), while x¹<x².

There are thus 4 x 3 unknowns, all bounded by either 20-3=17 or 19-3=16 for x³,y³,z³. It is then a case for brute force resolution since drawing all quadruplets by rmultinom until all conditions are satisfied

```valid=function(x,y,z,w){
(z[3]>max(y[3],w[3]))&(x[1]<x[2])&(sum(x)==24)&
(sum(y)==20)&(sum(z)==19)&(sum(w)==17)}
```

returns quickly several solutions. Meaning I misread the question and missed the constraint that the four values at each step were the same up to a permutation, decreasing the number of unknowns to four, a,b,c,d (ordered). And then three because the sum of the four is 20, average of the four sums. It seems to me that the first sum of x’s being 24 and involving only a,b, and c implies that 4c is larger than 24, ie c>6, hence d>7, while a>0 and b>1, leaving only two degrees of freedom for choosing the four values, meaning that only

1 2 7 10
1 2 8 9
1 3 7 9
1 4 7 8
2 3 7 8

are possible. Sampling at random across these possible choices and allocating the numbers at random to x,y,z, and w leads rather quickly to the solution

```     [,1] [,2] [,3] [,4]
[1,]    3    7    7    7
[2,]    2    8    2    8
[3,]    7    2    8    2
[4,]    8    3    3    3
```

## poor statistics

Posted in Books, pictures, R, Statistics, Travel, Wines with tags , , , , , , , , , , , , on September 24, 2019 by xi'an

I came over the weekend across this graph and the associated news that the county of Saint-Nazaire, on the southern border of Brittany, had a significantly higher rate of cancers than the Loire countries. The complete study written by Solenne Delacour, Anne Cowppli-Bony, amd Florence Molinié, is quite cautious about the reasons for this higher rate, even using a Bayesian Poisson-Gamma smoothing (and the R package empbaysmooth), and citing the 1991 paper by Besag, York and Mollié, but the local and national medias are quick to blame the local industries for the difference. The graph above is particularly bad in that it accumulates mortality causes that are not mutually exclusive or independent. For instance, the much higher mortality rate due to alcohol is obviously responsible for higher rates of most other entries. And indicates a sociological pattern that may or may not be due to the type of job in the area, but differs from the more rural other parts of the Loire countries. (Which, like Brittany, are already significantly above (50%) the national reference for alcohol related health issues.), and may not be strongly connected to exposition to chemicals. For instance, the rates of pulmonary cancers are mostly comparable to the national average, if higher than the rest of the Loire countries and connect with a high smoking propensity. Lymphomas are not significantly different from the regional reference. The only type of cancer that can be directly attributed to working conditions are the mesothelioma, mostly caused by asbestos exposure, which was used in ship building, a specialty of the area. Among the many possible reasons for the higher mortality of the county, the study mentions a lower exposure to medical testings (connected with the sociological composition of the area). Which would indicate the most effective policies for lowering these higher cancer and mortality rates.

## Le Monde puzzle [#1111]

Posted in R, Statistics with tags , , , , , , on September 18, 2019 by xi'an

Another low-key arithmetic problem as Le Monde current mathematical puzzle:

Notice that there are 10 numbers less than, and prime with 11, 100 less than and prime with 101, 1000 less than, and prime with 1111? What is the smallest integer N such that the cardinal of the set of M<N prime with N is 10⁴? What is the smallest multiple of 1111 using only different digits? And what is the largest?

As it is indeed a case for brute force resolution as in the following R code

```library(numbers)
homanycoprim=function(n){
many=1
for (i in 2:(n-1)) many=many+coprime(i,n)
return(many)}

smallest=function(){
n=1e4
many=homanycoprim(n)
while (many!=1e4) many=homanycoprim(n<-n+1)
return(n)}
```

which returns n=10291 as the smallest value of N.  For the other two questions, the usual integer2digit function is necessary

```smalldiff=function(){
n=1111;mul=1
while (mul<1e6) {
x=as.numeric(strsplit(as.character(n*mul), "")[[1]])
while (sum(duplicated(x))!=0){
mul=mul+1
x=as.numeric(strsplit(as.character(n*mul), "")[[1]])}
print(n*mul);mul=mul+1}}
```

leading to 241,087 as the smallest and 9,875,612,340 as the largest (with 10 digits).

## Le Monde puzzle [#1110]

Posted in Books, Kids, R, Travel with tags , , , , , , , , , on September 16, 2019 by xi'an

A low-key sorting problem as Le Monde current mathematical puzzle:

If the numbers from 1 to 67 are randomly permuted and if the sorting algorithm consists in picking a number i with a position higher than its rank i and moving it at the correct i-th position, what is the maximal number of steps to sort this set of integers when the selected integer is chosen optimaly?

As the question does not clearly state what happens to the number j that stood in the i-th position, I made the assumption that the entire sequence from position i to position n is moved one position upwards (rather than having i and j exchanged). In which case my intuition was that moving the smallest moveable number was optimal, resulting in the following R code

```sorite<-function(permu){ n=length(permu) p=0 while(max(abs(permu-(1:n)))>0){
j=min(permu[permu<(1:n)])
p=p+1
permu=unique(c(permu[(1:n)<j],j,permu[j:n]))}
return(p)}
```

which takes at most n-1 steps to reorder the sequence. I however checked this insertion sort was indeed the case through a recursive function

```resorite<-function(permu){ n=length(permu);p=0 while(max(abs(permu-(1:n)))>0){
j=cand=permu[permu<(1:n)]
if (length(cand)==1){
p=p+1
permu=unique(c(permu[(1:n)<j],j,permu[j:n]))
}else{
sol=n^3
for (i in cand){
qermu=unique(c(permu[(1:n)<i],i,permu[i:n]))
rol=resorite(qermu)
if (rol<sol)sol=rol}
p=p+1+sol;break()}}
return(p)}
```

which did confirm my intuition.

## Le Monde puzzle [#1109]

Posted in Kids, R with tags , , , , , on September 4, 2019 by xi'an

A digital problem as Le Monde current mathematical puzzle:

Noble numbers are such that they only involve different digits and are multiple of all their digits. What is the largest noble number?

Hmmmm…. Brute force? Since the maximal number of digits is 10, one may as well try:

```k=soz=9
for (t in 1:1e3){
sol=1
while (sol<10^(k-1)){
u=sample(0:8,k);i=digit2int(u)
if (max(i%%u[u>0])==0) soz=max(soz,sol<-i)}}
```

which returns 9875643120… (I made the conscious choice to exclude zero from the dividers. Which was not a choice spelled out in the original question.)

## Gibbs sampling with incompatible conditionals

Posted in Books, Kids, R, Statistics with tags , , , , , , on July 23, 2019 by xi'an

An interesting question (with no clear motivation) on X validated wondering why a Gibbs sampler produces NAs… Interesting because multi-layered:

1. The attached R code indeed produces NAs because it calls the Negative Binomial Neg(x¹,p) random generator with a zero success parameter, x¹=0, which automatically returns NAs. This can be escaped by returning a one (1) instead.
2. The Gibbs sampler is based on a Bin(x²,p) conditional for X¹ and a Neg(x¹,p) conditional for X². When using the most standard version of the Negative Binomial random variate as the number of failures, hence supported on 0,1,2…. these two conditionals are incompatible, i.e., there cannot be a joint distribution behind that returns these as conditionals, which makes the limiting behaviour of the Markov chain harder to study. It however seems to converge to a distribution close to zero, which is not contradictory with the incompatibility property: the stationary joint distribution simply does not enjoy the conditionals used by the Gibbs sampler as its conditionals.
3. When using the less standard version of the Negative Binomial random variate understood as a number of attempts for the conditional on X², the two conditionals are compatible and correspond to a joint measure proportional to $x_1^{-1} {x_1 \choose x_2} p^{x_2} (1-p)^{x_1-x_2}$, however this pmf does not sum up to a finite quantity (as in the original Gibbs for Kids example!), hence the resulting Markov chain is at best null recurrent, which seems to be the case for p different from ½. This is unclear to me for p=½.

## a non-riddle

Posted in Books, Kids, R with tags , , on July 12, 2019 by xi'an

Unless I missed a point in the last riddle from the Riddler, there is very little to say about it:

Given N ocre balls, N aquamarine balls, and two urns, what is the optimal way to allocate the balls to the urns towards drawing an ocre ball with no urn being empty?

Both my reasoning and a two line exploration code led to having one urn with only one ocre ball (and no acquamarine ball) and all the other balls in the second urn.

```odz<-function(n,m,t) 2*m/n+(t-2*m)/(t-n)
probz=matrix(0,trunc(N/2)-1,N-1)
for (n in 1:(N-1))
for (m in 1:(trunc(N/2)-1))
probz[m,n]=odz(n,m,N)
```


text:
# Solved Question Papers - IIT JEE/ChemSol1996

## Q1 - Objective Questions

### Ideal Gas Mixtures

For a student who missed the tricky part of this question, it is going to be very confusing, because he would be left wondering if there is anything that has to be taken upto liquid helium temperature to be liquefied. The point here is that ideal gases are assumed to have zero intermolecular forces. No matter what the conditions are, they just do not have any IMF whatsoever to be liquefied! Check back the assumptions for ideal gas.

### Balmer Series

The Balmer series corresponds to the energy level transition from some orbit to n=2. The shortest wavelength would be the one with the most energetic transition - naturally, n  ∞

${\displaystyle v=R_{H}\left({\frac {1}{2^{2}}}-{\frac {1}{\infty ^{2}}}\right)={\frac {R_{H}}{4}}}$

If we were to find the wavelength, using 1240/E (simplified equation of E = hc/λ) would’ve been advisable. 27419.25 how do you this find cm-1

#### A little warning

The point here is, people mess up in calculation. They might use the latter equation and then take its reciprocal - this at a time when they just require plugging in the value of Rhydberg's constant from the question paper!

This is the point of such easy questions, people who just browse the theory and equations before an examination under an illusion that the numerical problems would require only value substitution will slog here! The following applies to the students studying Science at SCHOOL in a laid back environment.

Personally, I have met many many people who take the Phy/Chem textbook and start reading it, skipping even the easiest in-text problems! Remember, we skip those questions because we do not want to think about them, because it is painful to think and hinder your pace when you want to READ pages in a hurry. These are those students who have a hilarious criterion of what they understand and what they don’t! If they have heard some terms from a chapter and have an idea of what the topics are, they feel they're ready for the chapter in the examination. If they have never bothered to read it, they say they are yet to study it! In the end, each and every examination they answer is disastrous.

Now, isn't it funny? And believe me; a lot of such students are people who have scored 90+ percents in their SSC, signifying that they are quite capable of studying HSC. You make the choice. Why ever do you want to read a Physics/Chemistry chapter when you do not even want to think about it? And why do you not want to think about it? Is it because you do not want to know that you did not understand what you read just five minutes ago?

My personal advice is very simple – give a text one reading at the most before you move onto the exercises. The most productive part of your studying would be searching the textbook for details when you need them in exercises!

### Rate Constant

Just recall the Arrhenius equation. K = A(e)^(-Ea/RT) When T tends towards infinity, the exponential term will tend towards 1, giving the value of the rate constant at infinite temperature = A.

### Ratio of RMS velocities

Simple application of RMS = ${\displaystyle {\sqrt {\frac {3RT}{M}}}}$

Taking the ratio, The required answer is ${\displaystyle {\sqrt {\frac {M_{\left(oxygen\right)}T_{\left(hydrogen\right)}}{T_{\left(oxygen\right)}M_{\left(hydrogen\right)}}}}=1}$

#### Note

By the way, most people, including relatively better placed students; make a little mistake of substituting molar masses in gm/mol in the equation of RMS speed. Whether or not you convert it into Kg/Mol, it will not matter here (because its a ratio that we calculated) but will introduce a BIG error (if at all you can call it an error and not a mistake!) in the calculation of the absolute value.

### Orbital Angular Momentum

This is a trivial piece of information. The orbital angular momentum is ${\displaystyle {\frac {{\sqrt {\left(l\right)}}\left(l+1\right)h}{2\pi }}}$.

(nh/2π) is the angular momentum.

At this point of time, you might want to revise all the quantum numbers and their significance, including names etc. Such important but trivial info is very valuable for some rank deciding marks at the JEE. For example, checkout the Matrix Match question of JEE 2008 on quantum numbers. Not many people get it right because of lack of revision.

#### Note to editors

Write a note distinguishing the two quantities and supply convincing information, maybe from wikipedia.

The editor can be anyone - you, some professor or even me when I am free.

## Question 3 - On Organic Chemistry

### (i) Mechanism of dehydration of alcohol

For those who have a general degree of awareness in Organic Chemistry, the answer is protonation of alcohol, followed by generation of a 2 degree carbocation. This intermediate immediately changes to the more stable three degree carbocation by a methyl shift. This carbocation is the one which eliminates the beta Proton to give the alkene.

The reaction given is the generation of alkene in the presence of acid from an alcohol. More precisely, it is the dehydration of that alcohol by the acid.

The mechanism is very simple,

• First the alcohol is protonated and H2O leaves.
• Then, since the carbocation so formed has no nucleophile to react with, it eliminates a β proton to give the alkene. Now, the product that we would expect is 2,2dimethylbut-1-ene.

The produced product is not that. Infact, the actual product that is obtained has a different carbon skeleton than the alcohol! This is because the formed carbocation is 2 degree. This particular carbon skeleton permits the formation of a more stable carbocation (3 degree) through a methyl shift. This is the major point.

• β deprotonation occurs on the rearranged carbocation and we get the alkene.

#### Checkpoints

We made a little point that deprotonation occurs because the carbocation has no nucleophile to react with? But, isn't the lone pair on oxygen in some other alcohol molecule a nucleophile? Will it not react with the carbocation? You give us the answer. [Hint: Check the mechanisms of dehydration of alcohol, there are two possible products under different reaction conditions.]

Also, some people are of the opinion that a two degree carbocation does not form at all! that is to say, it rearranges as soon as it forms to give a three degree carbocation. Now, the logic conveyed is a very obvious one and cannot be explained in more simple words. The point is, can you visualize whatever that was said? Think of a transition state mechanism as in your SN2 reactions.

### Order of acidity

The order of acidity is decided qualitatively by the stability of the conjugate base. But the sad thing is, many many students do not even know what a conjugate acid/base pair is. Being unable to solve Ionic equilibrium questions is one thing and not knowing the basics is another!

The conjugate bases of the substituted phenols will all have a negative charge on the whole molecule. We say it is on the whole molecule and not on the oxygen atom because the anion is a resonance hybrid of various canonical structures with the negative charge on different positions. The resonance hybrid with maximum dispersion of the negative charge will be the most stable.

Warning: - The following explanation provides the most basic reasoning that sprouts from the actual discussion on resonance hybrid. It will be beneficial for those who study their textbooks.

• Phenoxide - draw the resonance structures and see, the negative charge is dispersed on the two ortho positions, the para position and the oxygen atom.
• Phenoxide with methyl on para position - Draw the resonance structure with negative charge on para position. The negative charge is on the carboon bonded to the methyl group, which is an electron donating group. This canonical structure is destabilized compared to the unsubstituted structure. So, its contribution is less towards the resonance hybrid, meaning that the resonance hybrid has less charge dispersion. 2<1 (acidity)
• m-nitrophenoxide - See below
• p-nitrophenoxide - In the resonance structure with negative charge para to oxygen, the nitro group withdraws the electron density (how? That's another resonance structure! -m effect), thus stabilizing that particular resonance structure. Contrast this with (2) and you'll realize that (4)>(1)>(2)
• m-nitrophenoxide - There is no resonance structure with the negative charge on the meta position, so NO2 cannot withdraw electron density through -m effect. But then, -I effect due to its electronegativity still operates, and stabilizes it more than the phenoxide ion, although weakly.

### Cannizarro reaction

Factual - it is the hydride ion transfer. Note carefully that a hydride ion is transferred and not a proton!

NCERT behaved very strangely and eliminated the mechanisms of Cannizarro and Aldol Condensation reactions and much more from it's course. Remember that Organic Chemistry is the only subject whose syllabus for AIEEE/JEE demands greater detail than the CBSE syllabi. Infact, studying organic chem on the surface has made it dull and rote-based for the normal student!

The mechanism of the cannizarro reaction is: -

• Nucleophillic addition of the hydroxide ion on the carbonyl carbon.
• The anion becomes an -ate anion by eliminating a hydride ion and the acidic H.
• The hydride ion acts as a nucleophile for addition over some other carbonyl carbon, creating an alcohol.

### Hoffmann isocyanide test

Well, that says it. This is a direct question! ...Direct question even by board exam standards! However, since they demand a structure, you need to provide the location of bonds et-al. This is the only hurdle, but is not a big challenge if you have drawn the resonance structures of the cyanide ion some times.

## Question 5 - On Ions

### Cell potential

Now THIS, is a very very interesting problem! Simple to the end but very intriguing for the untrained mind!

There are quite a few problems relating the solubility of a salt with some cell potential. Out of experience I can say that those problems do not have a good reputation! So, what happens when you see a similar one in an examination like the JEE? Loss of confidence, what else! Keep cool! Tough questions at the JEE are all about common sense. Let's see why.

We have been given the standard cell potential of the Cu2+/Cu redox couple and we are to find its cell potential at pH 14. We keep one fact in mind that cell potentials depend upon the concentration of reactants as given in the Nernst equation. But then, what has a 1 Molar concentration of hydroxide ions got to do with the concentration of cupric ions? EVERYTHING! Cupric hydroxide is a precipitate and the Q value of its precipitation cannot exceed its kSP. In other words, cupric and hydroxide ions are in equilibrium with the precipitate of cupric hydroxide.

At equilibrium, from the kSP of cupric hydroxide, we get the concentration of cupric ions at pH14 to be 10(-19) M. Plug this value in the Nernst equation and get your answer. Some calculation is required, so you are not home yet, but still, you can celebrate the gain of 3 marks which not everybody would have been able to fetch in examination conditions!

The only point to be kept in mind is that the number of electrons involved in this redox half cell is 2, so n=2 in the Nernst equation. The rest is just calculation.

#### A little doubt

Now this is a doubt on my part. Any student or teacher who has the answer is very much welcome to answer it here!

The question has asked us to find the cell potential at pH 14. Now we used the pH 14 data to come to the conclusion that the concentration of cupric ions is 10(-19). If we give it a thought, the concentration can be anything less than this particular value then why did we carry on solving the problem with that particular value? Is it some assumption regarding cell potential that requires saturated solution? Or is it something much more trivial?

### On Salt hydrolysis

This one is a direct question again. Make sure that you get it right anytime you do it!

Let h be the degree of hydrolysis and C be the concentration. Set up a reaction table,

• The equilibrium constant would then be ${\displaystyle {\frac {Ch^{2}}{1-h}}}$

This is equal to the Kb of the cyanide ions, that is 10(-4.7). Remember that the sodium ions are spectator ions in the hydrolysis. Oh sorry, do I have to mention that cyanide ions will be in equilibrium with HCN, creating hydroxide ions? (I also apologize for any arrogant text! That just happens to be my native writing style!)

Look at the equation carefully. The denominator cannot be grater than one, so if the fraction as of the order 1/100000, it is only because degree of hydrolysis of cyanide ions itself is very small. That is to say, h<<1, or 1-h is approx 1.

We carry on with Ch2 = Kb

${\displaystyle h={\sqrt {\frac {K_{b}}{C}}}ch={\sqrt {C}}K_{b}}$

The expression Ch is nothing but the concentration of hydroxide ions in water.

Just take the logarithms of the equation. We get $\displaystyle –pOH=\frac{\log C}{2}\frac{\log K_b}{2}$

Giving due respect to the signs of pKb and logKb, we get pOH = 2.5, pH = 11.5

#### Let us laugh a little

The general trend in such questions is to specify the pKa of the conjugate acid, hydrocyanic acid in this case. Most people are then trained to carrying on the question through Ka values – using Kb = Kw/Ka.

So when I came across an anonymous (and published) solution of this question by a JEE tutor, I was not surprised that he had elongated this simple problem by four more steps in giving the relations between pKa, pKb and pKw.

${\displaystyle h={\sqrt {\frac {K_{w}}{CK_{a}}}}}$

Then after the logarithm steps, he replaced pKa wih 14 – pKb (and using about three lines to explain this). It would take very subtle observation to realize he was spinning round in circles.

#### Disclaimer

I mention it clearly that the above section was not written with immodest intentions or under a superiority complex. It is just a modest way to make people realize that even JEE specific classes do not guarantee quality education. Very few are competent enough and even fewer stick to the basics while nurturing open thought. BCPL, Kota is one of the best in this regard. Please keep in mind that this is just a personal opinion!

## Question 6 - Chemical Bonding

### Alkali Earth Metal Carbonates

Stability of these carbonates increases as the size of the AE Metal increases. So, IV < II < III < I. Many sources serves this statement as a fact, however I could not explain this on the basis of Fajan’s rule or otherwise. You are welcome to edit it here on wikibooks if you can provide a legitimate explanation.

### Structure of molecules

• NF3 - Same as NH3. It is a tetrahedron with one vertex occupied by a lone pair so that the overall molecule appears pyramidal.
• Nitrate ion – The central atom is nitrogen in sp2 hybrid state. One oxygen is bonded through one σ and one π bond; the second one with a co-ordinate bond and the last one has a negative charge on it and is bonded by one σ bond.
• BF3 - Of course sp2 hybrid with trigonal planar structure!
• Hydronium ion - Three bond pairs and a lone pair, makes it the same as Ammonia. That is, sp3 hybrid with pyramidal shape. Actually, water has two lone pairs. But when a hydronium ion is formed from it, one lone pair of oxygen is shared with the 1S orbital of a proton making it the third bond pair.
• Hydrazoic acid – Might give you the illusion of ammonia like structure – but however can one think of a sp3 hybrid hydrogen? Consider the structure below. One might want to refer the structure of the tri-iodide ion at the moment as well!

#### Note

Note carefully that the molecules all look similar to the untrained student! And it is not an exaggeration to say that even sincere 12th students will not be able to tell the difference between these similar molecules! Hybridization’s theory is one thing and practicing it over a lot of molecules is another! I mention this here solely to emphasize that practice is the key! Most people make the mistake of reading the same text again and again. Why don’t read it once and then solve all the problems at the end of the chapter? Alternately, why not grab some question papers and search for questions from the particular topic? I devote a lot of space in this book just to emphasize that practice allows a student to take decisions at a much better speed! DECISION MAKING allows a student to solve questions at a better speed and also to avoid conflicting thoughts while attempting questions!

### Calcium Carbide

On hydrolysis, CaC2 gives ethyne. This is enough to tell that the anionic part of calcium carbide is the same as ethyne minus two protons. The answer is then B

### Cesium Tribromide

Alkali metals having a trivalent cation is an unheard statement for me. Tribromide ion is a possibility because we know I3- ion does exist. We just discussed its structure above. B

# = Q7 - Organic Chemistry

## Q8 - Inorganic Chemistry

### Dipole moments

Benzene is a non polar molecule, because the vector sum of all of the bond dipoles in the molecule is zero. When one of the hydrogen is replaced by a methyl group, the resultant dipole moment is equivalent to a single C-H dipole moment in a methane molecule. Use your existing knowledge of vectors to come to this conclusion.

In p-dichlorobenzene all bond moments add up to zero. In o-dichlorobenzene, only two bond moments of antiparallel C-H bonds cancels. The resultant is a little less than vector sum of two C-Cl bond dipoles at 60o. [Why? Don’t forget the C-H bonds antiparallel to the C-Cl bonds!] Similarly, in m-dichlorobenzene the resultant dipole moment is a little less than two C-Cl bond dipoles placed at 120o.

So, Toluene>Benzene. In other words, toluene>p-dichlorobenene. o-dichlorobenzene>m-dichlorobenzene. Amongst the meta isomer and toluene, the meta isomer has a higher dipole moment. [Why?]

B

#### Notes

Regarding ortho/para

## Q9 – Organic Chem

### 3-ethyl-2-pentene

HBr in the presence of peroxide.
Br2/H2O
OMDM

## Q10 – Objectives again

### Electrolysis

Obtaining sodium out of the electrolysis of any aqueous sodium salt is out of the question? Why would the Hall Heroultz process be important if it was so easy to obtain sodium? S2O82- is the reduction product of sulphate ions – so we cant obtain it at the anode. A is the answer then!

${\displaystyle r_{\left(H_{2}\right)}={\frac {VolumeX}{Timetaken}}={\frac {X}{5sec}}}$
$\displaystyle {\frac {\frac {X}{T_{H}_{2}}}{\frac {X}{T_{O}_{2}}}}={\sqrt {\frac {M_{O}_{2}}{M_{H}_{2}}}}$


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Browse Questions

# Which of the following facts about the complex $[Cr(NH_3)_6]Cl_3$ is wrong

$\begin{array}{1 1}(a)\;\text{The complex is an outer orbital complex}\\(b)\;\text{The complex gives white precipitate with silver nitrate solution}\\(c)\;\text{The complex involves }d^2sp^3\;\text{hybridisation and is octahedral in shape}\\(d)\;\text{The complex is paramagnetic}\end{array}$

The complex is an outer orbital complex is wrong.
Hence (a) is the correct answer.


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# Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise: 48

$12\frac{15}{16} \text{ft}$

#### Work Step by Step

Add the given lengths to find the total length of the welded piece:. Total length $=3\frac{1}{4} + 2\frac{3}{8} + 3\frac{1}{8} + 4\frac{3}{16} \\=(3+2+3+4) + (\frac{1}{4}+\frac{3}{8}+\frac{1}{8} + \frac{3}{16}) \\=12+ (\frac{1}{4}+\frac{4}{8}+\frac{3}{16})$ Make the fractions similar using their LCD of $16$ to obtain: Total length $\\=12+(\frac{4}{16}+\frac{8}{16}+\frac{3}{16}) \\=12+\frac{15}{16} \\=12\frac{15}{16} \text{ft}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# What is an algebraic equation?

Jul 18, 2018

An equation is a statement that two expressions are equal, so we have:

$\text{expression = expression}$

#### Explanation:

An expression is a maths sentence made up of terms which can have numbers and or variables as factors.

$3 x , \text{ "3x+5," } 2 {x}^{2} - 5 x + 3$ are all examples of expressions.

An equation is a statement that two expressions are equal, so we have:

$\text{expression = expression}$

An equation is solved, which means to find the value(s) of the variable(s) which makes the equation true,

$3 x - 5 = 22 \text{ } \rightarrow x = 9$

${x}^{2} + 4 x - 45 = 0 \text{ } x = - 9 \mathmr{and} x = 5$

An equation which contains two or more variables does not have a unique solution, but rather many, many solutions,

$x + y = 10$ has infinitely many solutions.