text:
# Two men and a door, but no liar

You come across a door, guarded by two mysterious men. As you start to figure out how to tell which one is lying and which one is telling the truth, they say in unison:

WELCOME, TRAVELER, TO REIBELLO'S RIDDLE REPOSITORY. SHOULD YOU SEEK ENTRY, YOU'LL HAVE TO SOLVE A SIMPLE PUZZLE.

You quickly agree to hear the puzzle. You also notice that their voices are different. One is deep and strong, while the other is softer, and strangely emphatic about his words.

Our first AND second facilitate, that which you need to operate.

Each alone, is not correct, but read aloud make our subject.

Our third is imprecise at best, but gladly taken on a test.

Our fourth is five- It's two. His clues are the worst. It's also half of what is first.

Our fifth is fifth. All the clue you'll need, I reckon. But just in case, it's also our second.

Our sixth and last, core of our planet round, but don't go quite as far in underground.

Our riddle complete, answer if you dare. Six letters, and no more, WE SWEAR.

What word are the mysterious men looking for?

The mysterious men are looking for the word:

WEAVER.

Our first AND second facilitate, that which you need to operate.

WE (not sure)

Each alone, is not correct, but read aloud make our subject.

???

Our third is imprecise at best, but gladly taken on a test.

A. Imprecise ???. Anyone would be glad to receive an A as a test score.

Our fourth is five- It's two. His clues are the worst. It's also half of what is first.

V. V is 5 in roman numerals. V is also half of W.

Our fifth is fifth. All the clue you'll need, I reckon. But just in case, it's also our second.

E. E is the 5th letter of the alphabet. This is the same as the 2nd letter.

Our sixth and last, core of our planet round, but don't go quite as far in underground.

R. Middle letter of Earth. Also appears in underground

Our riddle complete, answer if you dare. Six letters, and no more, WE SWEAR.

WE SWEAR = WE AVER

• You've got the answer right - see if you can get the first two letters. There's a bit of a jump. – Reibello Aug 30 '18 at 2:48
• Perhaps "A" for "imprecise at best" refers to the fact that the best is "A+", so just "A" isn't quite it. As for the first two, it might refer to the two speakers weaving words together. – Joel Harmon Aug 30 '18 at 3:40
• rot13("jr" vf n fhowrpg (nf bccbfrq gb na bowrpg)) – Destructible Lemon Aug 30 '18 at 4:06
• Pay less attention to "we" right now (thought we as a subject is going to be important) and try and figure out what two completely different letters the first hint wants. – Reibello Aug 30 '18 at 21:03


text:
# A closed formula for $A_n(X)=\sum\limits_{i=0}^n X^{i^2}$

I want to know if there exists a closed formula for sum $A_n(X)=\sum \limits_{i=0}^n X^{i^2}$.

I have found if n is odd then $(X^n+1)\text{ | } A_n(X)$, but I don't have found a closed formula.

• The infinite series is a Jacobi theta function ... $$\frac{\theta_3(0,q)-1}{2} = \sum_{n=1}^\infty q^{n^2}$$ Jun 25 '17 at 12:48
• Just an idea for your second question. Let $n$ odd and let $\zeta^n=-1$. It is enough to show that $A_n(\zeta)=0$. Maybe, using the orthogonality relations for roots of unity in the expansion of $(1+\zeta+\zeta^2+\cdots+\zeta^n)^n$, gives $A_n(\zeta)=0$. Jun 25 '17 at 15:03
• I thought you had "found" and not proved your assertion. My bad. I factored your polynomial for small odd $n$ and it gives: small polynomials times a very large irreducible one (over the integers). Jun 25 '17 at 15:58
• The generating function is $$g(X,z) = \sum_{n=0}^\infty A_n(X) z^n = \frac{1}{1-z} \sum_{n=0}^\infty X^{n^2} z^n$$ This is related to Jacobi theta functions: $$(1-e^{it}) g(X, e^{it}) + (1-e^{-it}) g(X, e^{-it}) = 1 + \theta_3(t/2, X)$$ Jun 25 '17 at 20:46
• I try to say it better, if the series $A_n(e^i)$ is not bounded, then we can't have $A_n(X)=\frac{N_a(X^{n^2},X^n,X)}{D_a(X)}$, with $N_a$ and $D_a$ polynomials, like gemotric series where $N_g(X^n,X)=X \times X^n-1$ and $D_g(X)=X-1$. Jun 26 '17 at 11:42

Not at all an answer, just observations. Firstly, a conjecture, based on experiment.

CONJECTURE $A_{2n}$ is irreducible for all $n.$

(empirically true for all $n\leq20.$)

CONJECTURE 2 $A_{2n+1}/(x^{2n+1}+1)$ has at most one non-cyclotomic factor. (empirically true for all $n\leq 20$).

Finally, the roots of these things cluster around the unit circle (see the graphic for $A_{14}$).

Is it true that the zeros of the infinite series are on the unit circle? (oops, this is not supposed to be a question).

• Neat graphic. There's a product formula (probably a special case of the "Jacobi triple product") that shows that $\sum_{i=0}^\infty X^{i^2}$ has no zero with $|X|<1$. But $|X|=1$ is a natural boundary so it doesn't make sense to ask about zeros on or outside the unit circle. Jun 26 '17 at 0:51
• $\left( \sum_{i=0}^n X^{i^2} \right) / X^{n^2}$ converges to $1$ for $|X|>1$, and this renormalized limit is certainly nonvanishing. Probably one can convert this to an explicit bound on the zeroes of the finite sums. Jun 26 '17 at 0:53
• Re the second conjecture: $A_n/(x^n+1)$ is irreducible for $n=3,7,15,31$; for other odd $n \leq 41$ it has at least two factors but sometimes has three ($n=17,35$) or even four ($n=29,41$). Still all but one factor is cyclotomic; maybe that's what you meant to conjecture. [Also: "$<=$" $\neq$ "$\leq$" . . .] Jun 26 '17 at 1:20
• @NoamD.Elkies You are absolutely right on all counts, I fixed the conjecture and the typo (the latter coming through alternating between programming and latex). Jun 26 '17 at 1:33
• @NoamD.Elkies Presumably, if the product formula whereof you speak actually gives a lower bound on the infinite series in the disk $|z| < r < 1,$ it will imply at least a one sided clustering, and then something similar after $z\to 1/z$ will give the other side. Jun 26 '17 at 2:09

format: plain


text:
## Proceedings of the Japan Academy, Series A, Mathematical Sciences

### Iwasawa invariants on non-cyclotomic ${\mathbf {Z}_{p}}$-extensions of CM fields

Hideki Goto

#### Abstract

Let $p$ be an odd prime which splits completely into distinct primes in a CM field $K$. By considering ray class field of $K$ with respect to prime ideals lying above $p$, one can define a certain special non-cyclotomic $\mathbf{Z}_{p}$-extension over $K$. We will give some examples of such non-cyclotomic $\mathbf{Z}_{p}$-extensions whose Iwasawa $λ$- and $µ$-invariants both vanish, using a variant of a criterion due to Greenberg.

#### Article information

Source
Proc. Japan Acad. Ser. A Math. Sci., Volume 82, Number 9 (2006), 152-154.

Dates
First available in Project Euclid: 4 December 2006

https://projecteuclid.org/euclid.pja/1165244963

Digital Object Identifier
doi:10.3792/pjaa.82.152

Mathematical Reviews number (MathSciNet)
MR2293501

Zentralblatt MATH identifier
1163.11073

Subjects
Primary: 11R23: Iwasawa theory
Secondary: 11R29: Class numbers, class groups, discriminants

#### Citation

Goto, Hideki. Iwasawa invariants on non-cyclotomic ${\mathbf {Z}_{p}}$-extensions of CM fields. Proc. Japan Acad. Ser. A Math. Sci. 82 (2006), no. 9, 152--154. doi:10.3792/pjaa.82.152. https://projecteuclid.org/euclid.pja/1165244963


text:
# Questions about air compressor project

#### matthew electric

Joined Feb 14, 2019
1
Hello there! I recently acquired an older air compressor. Its rated at 2 hp according to the sticker on the belt guard/cover. The motor looks to be a 220 volt, and has the plug type where one of the non grounded blades is turned 90 degrees from the other.

I did use pliers to turn it back and plugged it into a regular house outlet (110) and the motor ran, then slowed, then stopped working. The reset switch allowed it to start again, but obviously, this isn't going to work.

What I'm wondering is whether or not I can rewire this motor to run 110. If not, what are my other options? The compressor is old, and I only paid $30 for it, so I'm not in much. I've included pictures, including the data plate on the motor. Thank you all for your help. #### DNA Robotics Joined Jun 13, 2014 501 Do you have a 230 volt oven or dryer where you can plug it in? #### shortbus Joined Sep 30, 2009 6,987 Showing the compressor it's self or giving the make and model may get you better answers. There are different type of compressor pumps that have different motor needs. Also giving where you live, the country at the least can help. We're flying blind in try to help. #### Picbuster Joined Dec 2, 2013 982 Hello there! I recently acquired an older air compressor. Its rated at 2 hp according to the sticker on the belt guard/cover. The motor looks to be a 220 volt, and has the plug type where one of the non grounded blades is turned 90 degrees from the other. I did use pliers to turn it back and plugged it into a regular house outlet (110) and the motor ran, then slowed, then stopped working. The reset switch allowed it to start again, but obviously, this isn't going to work. What I'm wondering is whether or not I canView attachment 170201 View attachment 170201 View attachment 170202 rewire this motor to run 110. If not, what are my other options? The compressor is old, and I only paid$30 for it, so I'm not in much.

I've included pictures, including the data plate on the motor. Thank you all for your help.
It looks like an one phase motor 220V 60 Hz the only way is to connect it between two phases.
You do need however a 3 phase power inlet at your home to do so.

Picbuster


text:
## Electronic Theses and Dissertations

#### Document Type

Thesis - Open Access

2017

#### Degree Name

Master of Science (MS)

#### Department / School

Mathematics and Statistics

Daniel Schaal

Donald Vestal

#### Abstract

For every positive integer $a$, let $n = R_{ZS}(a)$ be the least integer, provided it exists, such that for every coloring $\Delta : \{1, 2, ..., n\} \rightarrow \{0, 1, 2\},$ there exist three integers $x_1, x_2, x_3$ (not necessarily distinct) such that $\Delta(x_1) + \Delta(x_2) + \Delta(x_3) \equiv 0\ (mod\ 3)$ and $ax_1 +x_2 = x_3.$ If such an integer does not exist, then $R_{ZS}(a) = \infty.$ The main results of this paper are $R_{ZS}(2) = 12$ and a lower bound is found for $R_{ZS}(a)$ where $a \geq 2$.

#### Library of Congress Subject Headings

Ramsey theory.
Combinatorial analysis.

#### Description

Includes bibliographical references (page 16)

application/pdf

20

#### Publisher

South Dakota State University

COinS


text:
## Experimental Mathematics

Experimental Mathematics publishes formal results and conjectures inspired by experimentation, algorithms and software for mathematical exploration, and related surveys. Journal issues published after 2012 are not available in Euclid, but Euclid Prime subscribers may enjoy complimentary access to all prior content. Starting with Volume 22 (2013), Experimental Mathematics is being printed and distributed by Taylor & Francis.

Fast Fourier analysis for {${\rm SL}\sb 2$} over a finite field and related numerical experimentsVolume 1, Issue 2 (1992)


text:
# American Institute of Mathematical Sciences

2011, 2011(Special): 219-228. doi: 10.3934/proc.2011.2011.219

## Symmetry breaking in problems involving semilinear equations

 1 Dipartimento di Matematica e Informatica, Via Ospedale 72, 09124 Cagliari, Italy

Received  July 2010 Revised  March 2011 Published  October 2011

This paper is concerned with two maximization problems where symmetry breaking arises. The rst one consists in the maximization of the energy integral relative to a homogeneous Dirichlet problem governed by the elliptic equation -$\deltau = XF^u^q$ in the annulus $B_(a,a+2)$ of the plane. Here 0  q < 1 and F is a varying subset of $B_(a,a+2)$, with a fixed measure. We prove that a subset which maximizes the corresponding energy integral is not symmetric whenever a is large enough. The second problem we consider is governed by the same equation in a disc $B_a+2$ when $F$ varies in the annulus Ba;a+2 keeping a xed measure. So, now we have a so called maximization problem with a constraint. As in the previous case, we prove that a subset which maximizes the corresponding energy integral is not symmetric whenever a is large enough.
Citation: Lucio Cadeddu, Giovanni Porru. Symmetry breaking in problems involving semilinear equations. Conference Publications, 2011, 2011 (Special) : 219-228. doi: 10.3934/proc.2011.2011.219
 [1] Fanni M. Sélley. Symmetry breaking in a globally coupled map of four sites. Discrete & Continuous Dynamical Systems, 2018, 38 (8) : 3707-3734. doi: 10.3934/dcds.2018161 [2] Hwai-Chiuan Wang. Stability and symmetry breaking of solutions of semilinear elliptic equations. Conference Publications, 2005, 2005 (Special) : 886-894. doi: 10.3934/proc.2005.2005.886 [3] Claudia Anedda, Giovanni Porru. Symmetry breaking and other features for Eigenvalue problems. Conference Publications, 2011, 2011 (Special) : 61-70. doi: 10.3934/proc.2011.2011.61 [4] Sanjay Dharmavaram, Timothy J. Healey. Direct construction of symmetry-breaking directions in bifurcation problems with spherical symmetry. Discrete & Continuous Dynamical Systems - S, 2019, 12 (6) : 1669-1684. doi: 10.3934/dcdss.2019112 [5] Changlu Liu, Shuangli Qiao. Symmetry and monotonicity for a system of integral equations. Communications on Pure & Applied Analysis, 2009, 8 (6) : 1925-1932. doi: 10.3934/cpaa.2009.8.1925 [6] Yingshu Lü, Chunqin Zhou. Symmetry for an integral system with general nonlinearity. Discrete & Continuous Dynamical Systems, 2019, 39 (3) : 1533-1543. doi: 10.3934/dcds.2018121 [7] Linfeng Mei, Zongming Guo. Morse indices and symmetry breaking for the Gelfand equation in expanding annuli. Discrete & Continuous Dynamical Systems - B, 2017, 22 (4) : 1509-1523. doi: 10.3934/dcdsb.2017072 [8] Anna Goƚȩbiewska, Norimichi Hirano, Sƚawomir Rybicki. Global symmetry-breaking bifurcations of critical orbits of invariant functionals. Discrete & Continuous Dynamical Systems - S, 2019, 12 (7) : 2005-2017. doi: 10.3934/dcdss.2019129 [9] Yingshu Lü. Symmetry and non-existence of solutions to an integral system. Communications on Pure & Applied Analysis, 2018, 17 (3) : 807-821. doi: 10.3934/cpaa.2018041 [10] Zhenjie Li, Chunqin Zhou. Radial symmetry of nonnegative solutions for nonlinear integral systems. Communications on Pure & Applied Analysis, , () : -. doi: 10.3934/cpaa.2021201 [11] Matteo Bonforte, Jean Dolbeault, Matteo Muratori, Bruno Nazaret. Weighted fast diffusion equations (Part Ⅰ): Sharp asymptotic rates without symmetry and symmetry breaking in Caffarelli-Kohn-Nirenberg inequalities. Kinetic & Related Models, 2017, 10 (1) : 33-59. doi: 10.3934/krm.2017002 [12] Alexey Yulin, Alan Champneys. Snake-to-isola transition and moving solitons via symmetry-breaking in discrete optical cavities. Discrete & Continuous Dynamical Systems - S, 2011, 4 (5) : 1341-1357. doi: 10.3934/dcdss.2011.4.1341 [13] Jeremy L. Marzuola, Michael I. Weinstein. Long time dynamics near the symmetry breaking bifurcation for nonlinear Schrödinger/Gross-Pitaevskii equations. Discrete & Continuous Dynamical Systems, 2010, 28 (4) : 1505-1554. doi: 10.3934/dcds.2010.28.1505 [14] Peter A. Hästö. On the existance of minimizers of the variable exponent Dirichlet energy integral. Communications on Pure & Applied Analysis, 2006, 5 (3) : 415-422. doi: 10.3934/cpaa.2006.5.415 [15] Wenxiong Chen, Congming Li. Radial symmetry of solutions for some integral systems of Wolff type. Discrete & Continuous Dynamical Systems, 2011, 30 (4) : 1083-1093. doi: 10.3934/dcds.2011.30.1083 [16] Antonio Greco, Giovanni Porru. Optimization problems for the energy integral of p-Laplace equations. Conference Publications, 2013, 2013 (special) : 301-310. doi: 10.3934/proc.2013.2013.301 [17] Naoki Shioji, Kohtaro Watanabe. Uniqueness of positive radial solutions of the Brezis-Nirenberg problem on thin annular domains on ${\mathbb S}^n$ and symmetry breaking bifurcations. Communications on Pure & Applied Analysis, 2020, 19 (10) : 4727-4770. doi: 10.3934/cpaa.2020210 [18] Xiaotao Huang, Lihe Wang. Radial symmetry results for Bessel potential integral equations in exterior domains and in annular domains. Communications on Pure & Applied Analysis, 2017, 16 (4) : 1121-1134. doi: 10.3934/cpaa.2017054 [19] Xavier Cabré, Eleonora Cinti. Energy estimates and 1-D symmetry for nonlinear equations involving the half-Laplacian. Discrete & Continuous Dynamical Systems, 2010, 28 (3) : 1179-1206. doi: 10.3934/dcds.2010.28.1179 [20] Jingzhen Liu, Yike Wang, Ming Zhou. Utility maximization with habit formation of interaction. Journal of Industrial & Management Optimization, 2021, 17 (3) : 1451-1469. doi: 10.3934/jimo.2020029

Impact Factor:


text:
# Pigeonhole Principle/Historical Note

## Historical Note on Pigeonhole Principle

The Pigeonhole Principle appeared in print as early as $1622$ in Selectæ Propositiones in Tota Sparsim Mathematica Pulcherrimæ by Jean Leurechon.

It also appears, in greater detail, in the $1624$ work Récréations Mathématiques by "H. van Etten", also commonly attributed to Jean Leurechon.

However, it is commonly called Dirichlet's Box (or Drawer) Principle, after an $1834$ treatment by Johann Peter Gustav Lejeune Dirichlet, who called it the Schubfachprinzip (drawer principle or shelf principle).

In Russian and some other languages, it is known as the Dirichlet principle or Dirichlet's principle, which name ambiguously also refers to the minimum principle for harmonic functions.

It is usually seen in its simplest form: if you have $N + 1$ objects to put into $N$ pigeonholes, at least one pigeonhole contains $2$ objects.


text:
### callOnChildren

Expanding and utilizing the engine via C++.

#### callOnChildren

newaged
Posts: 31
Joined: Sat Feb 14, 2015 12:27 am
Ran into another problem I managed to fix. If this isn't the right forums for posts like these, or if I should be doing this on github, please let me know.
Anyways, the simgroup/simset function callOnChildren wasn't working, so I changed

Code: Select all

void SimSet::callOnChildren(simSet.cpp line 232){   ...   ConsoleValueRef args[21];   args[0] = method.c_str();

to this

Code: Select all

   ConsoleValue vals[2];   ConsoleValueRef args[21];   vals[0].init();   vals[0].setStringValue(method.c_str());   args[0].value = &vals[0];   vals[1].init();   args[1].value = &vals[1];

and things started to work again. Was that the right way to go about it?
Last edited by newaged on Thu Jan 21, 2016 10:00 pm, edited 1 time in total.

#### Re: callOnChildren

Mud-H
Posts: 175
Joined: Thu Feb 19, 2015 3:08 pm
Was it causing a crash? I stopped using the callOnChildren after it caused too much crash, also if I remember right it was not always working correctly when it wasn't causing crash...

#### Re: callOnChildren

newaged
Posts: 31
Joined: Sat Feb 14, 2015 12:27 am
Mud-H wrote:Was it causing a crash? I stopped using the callOnChildren after it caused too much crash, also if I remember right it was not always working correctly when it wasn't causing crash...

Yeah, it was causing a crash before. I haven't used it enough to comment on the second problem.

#### Re: callOnChildren

JeffR
Steering Committee
Posts: 777
Joined: Tue Feb 03, 2015 9:49 pm

Yeah, if you feel you have a useful addition or fix, feel free to toss it up onto github. Bit easier to test/review it that way, but also feel free to make a thread bringing attention to it

I'll have to give this fix a go. I don't REMEMBER having problems with callOnChildren, but then, I haven't used it a ton.

#### Re: callOnChildren

newaged
Posts: 31
Joined: Sat Feb 14, 2015 12:27 am
Updated the code - removed some pointless changes

@
JeffR
I'll look into posting these fixes on github. Should they be batched together or done individually?

#### Re: callOnChildren

Azaezel
Posts: 392
Joined: Tue Feb 03, 2015 9:50 pm

fwiw, PRs with a given task show up when checking a files history on the github website so that'd help folks understand the why on changes. Least that's the personal rule of thumb...

#### Who is online

Users browsing this forum: No registered users and 1 guest


text:
# smooth approximations of indicator function

How would I construct Schwartz functions $f_1^\epsilon$, $f_2^\epsilon$ on $\mathbb{R}$ such that

$f_1^\epsilon(x)\leq\mathbb{1}_{[a,b]}(x)\leq f_2^\epsilon(x)$,

and $f_1^\epsilon\rightarrow\mathbb{1}_{[a,b]}$, $f_2^\epsilon\rightarrow\mathbb{1}_{[a,b]}$ as $\epsilon\rightarrow0$,

where $a<b$ are real numbers and $\mathbb{1}_{[a,b]}$ is the indicator function?

• Hint: $e^{-x^2/2}$ Is this homework? – user17794 Nov 12 '12 at 6:02
• no, it isn't homework – Kelly Nov 12 '12 at 7:03
• You seem to be new to the site - in general, I would recommend showing the progress you've made on the question so that your answerer can get a better sense of what you under stand - helps prevent pedantry and saves time. For this particular question, it's a fairly lengthy (though standard) construction, so I'd prefer to get to the part that's tripping you up. – user17794 Nov 12 '12 at 7:12

The quickest way is to convolve the characteristic function of another interval with a smooth bump function. Let $\Phi(x)=c\,\exp(1/(1-x^2))\,\chi_{(-1,1)}$ where $c>0$ is chosen to make $\int_{-1}^1 \Phi(x)\,dx=1$. For $\epsilon>0$ define $\Psi_\epsilon(x)=\epsilon^{-1}\Psi(x/\epsilon)$. The convolutions $f_1^\epsilon = 1_{[a+\epsilon,b-\epsilon]}*\Psi_\epsilon$ and $f_2^\epsilon = 1_{[a-\epsilon,b+\epsilon]}*\Psi_\epsilon$ have the required properties, which I leave for you to check.


text:
# Binary Tree interal nodes+leaves proof

Let us define the length of the inner path of a binary tree to be the sum of the depths of all internal nodes in the tree ,

Let us define the length of the outer path of the binary tree to be the sum of the depths of all the leaves in the tree.

I need to prove that, in a full binary tree with $$n$$ internal nodes and an internal path $$i$$ and an external path $$e$$, $$e = i + 2n$$

I think that induction will be a great way to prove this, but I wasn't able to prove it until now. How I can prove it?

• What do you mean by an "internal path exists" and and "external path exists"? Your definitions of the length of an inner path and the length of an outer path have nothing to do with any paths; they're stated in terms of all the internal and external nodes of the tree. May 28 '19 at 10:22
• Edited , now it is better? May 28 '19 at 13:45
• Prove it by recursion ("structural induction."). Assume it's true for the left and right subtrees of the root. If you're not comfortable with this, prove it by strong induction on the height of the tree. May 28 '19 at 13:53
• Maybe you can write the structure of the proof? It will help to me a lot to understand that and to solve same proofs in the future. :). May 28 '19 at 14:26

Since you are apparently not familiar with this method of proof, let me point out the parallel to induction. What we prove in induction is first that the statement is true for $$1$$, and second that it true for any number that we can get by starting at $$1$$ and counting up by ones. Since these are all the integers, we have proved the statement for all the integers. So the principle of induction says nothing more than we get all the positive integers by starting with $$1$$ and counting up.
Can you take it from here? Prove the basis, then suppose it's true for the left and right subtrees $$L$$ and $$R$$ of some full binary tree $$T$$. Relate the internal and external paths of $$T$$ to the internal and external paths of $$L$$ and $$R$$.


text:
# Number-Theoretic Inequality!

$\Large{0 < \left| \dfrac{p}{q} - \dfrac23 \right| < \dfrac{1}{q^2}}$

How many ordered pairs $$(p,q)$$ of integers exists such that the above inequality satisfies?

×


text:
# Ch05. Complex Numbers & Quadratic Equations

Class 11 Maths

## Description

In this self study course, you will learn Need for complex numbers, especially $\sqrt{-1}$ , to be motivated by inability to solve some of the quardratic equations. Algebraic properties of complex numbers.Argand plane and polar representation of complex numbers. solution of quadratic equations (with real coefficients) in the complex number system.Square root of a complex number. This course also has solutions of NCERT Exercise questions.

Syllabus Medium English Hinglish (Hindi + English) 45 Days Ashish Kumar (Agam Sir)

315


text:
# Textbooks for more “abstract” linear algebra

I have been doing some self study in preparation for the upcoming term and I realized I have very little intuition for doing linear algebra on spaces of functions. Can you recommend any linear algebra text books that have decent sections/chapters on things like dual spaces, endomorphisms and homomorphisms between spaces of functions?

An example of a problem that tripped me up is from a text book on manifolds, and they are introducing the topic of vector bundles and one of the exercises in the textbook is to show the existence of an isomorphism between the space of all $\binom{k}{l+1}$-tensors and the space of all multilinear functions:

$$\underset{l}{\underbrace{V^*\times\cdots\times V^*}}\times\underset{k}{\underbrace{V\times\cdots\times V}}\longrightarrow V$$

I realized I didn't have the intuition for understanding how the linear algebra concepts generalize to these more complicated spaces I am comfortable with to these function spaces. I guess really what I am looking for is a textbook for undergraduate linear algebra at a higher level of abstraction so the focus isnt on vector spaces over $\mathbb{R}$ or $\mathbb{C}$. Any recommended readings to bring me up to speed would be appreciated!

Sheldon Axler's Linear Algebra Done Right, in my opinion, does a good job covering dual spaces and linear maps between general finite dimensional vector spaces.

https://www.springer.com/us/book/9783319110790


text:
# Configure the EXM retryers

Abstract

Overview of the retryers used by EXM and how to configure them.

EXM uses three retryers, based on the transient fault handling framework. You can read more about transient fault handling in this documentation from Microsoft. The three EXM retryers in Sitecore are:

• The EmailCampaign.EDS.Core.Net.Smtp.Send.Retryer (referred to as the Send retryer), which EXM uses to send messages over SMTP.

• The EmailCampaign.EDS.Core.Net.Smtp.ValidateConnection.Retryer, which EXM uses to get and validate an SMTP connection from the connections pool.

• The EmailCampaign.XConnect.Conflict.Retryer (referred to as the xConnect conflict retryer), which EXM uses to retry a method if there is a conflict with xConnect.

The retryers have default values that you can configure while configuring EXM. You can:

• Configure the exponential backoff retry strategy.

• Convert the Send retryer to a fixed interval retry behavior.

• Add specific SMTP-related transient fault codes to the Send retryer.

In the Sitecore.EDS.Core.config file, the following is the configuration code for the first two EXM retryers:

<TransientFaultHandling>
<Retryers>
<!-- Retryer to send messages over SMTP. -->
<EDS.Core.Net.Smtp.Send.Retryer>
<Type>Sitecore.EDS.Core.TransientFaultHandling.RetryerFactories.SmtpSendExponentialRetryerFactory, Sitecore.EDS.Core</Type>
<!-- <Type>Sitecore.EDS.Core.TransientFaultHandling.RetryerFactories.SmtpSendFixedRetryerFactory, Sitecore.EDS.Core</Type> -->
<Options>
<ExponentialRetry>
<MaxAttempts>10</MaxAttempts>
<MinBackoff>00:00:01</MinBackoff>
<MaxBackoff>00:00:30</MaxBackoff>
<DeltaBackoff>00:00:10</DeltaBackoff>
</ExponentialRetry>
<!--
<FixedRetry>
<RetryCount>3</RetryCount>
<RetryInterval>00:00:01</RetryInterval>
<FirstFastRetry>false</FirstFastRetry>
</FixedRetry>
-->
<TransientSmtpErrorCodes>
<ServiceIsNotAvailable>421</ServiceIsNotAvailable>
</TransientSmtpErrorCodes>
</Options>
</EDS.Core.Net.Smtp.Send.Retryer>
<!-- Retryer to validate SMTP connection. -->
<EDS.Core.Net.Smtp.ValidateConnection.Retryer>
<Type>Sitecore.EDS.Core.TransientFaultHandling.RetryerFactories.SmtpValidateFixedRetryerFactory, Sitecore.EDS.Core</Type>
<Options>
<FixedRetry>
<RetryCount>3</RetryCount>
<RetryInterval>00:00:01</RetryInterval>
<FirstFastRetry>false</FirstFastRetry>
</FixedRetry>
</Options>
</EDS.Core.Net.Smtp.ValidateConnection.Retryer>
</Retryers>
</TransientFaultHandling>

By default, the Send retryer uses an exponential backoff retry behavior, which means that repeated retries are spaced out at random intervals. For more information about exponential backoff strategies, see this article on Wikipedia.

To configure the exponential backoff strategy:

• In the <ExponentialRetry> section of the Send retryer, configure the following values:

Setting

Default

Description

MaxAttempts

10

Specifies the amount of retry attempts for transient exceptions. Once it hits this amount, the retryer defines the target service as unavailable.

MinBackoff

00:00:01

The minimum time period between retry attempts.

MaxBackoff

00:00:30

The maximum time period between retry attempts.

DeltaBackoff

00:00:10

The delta backoff interval between retries.

EXM supports the option to have retries in fixed intervals instead of the default exponential behavior.

To convert to an interval retry behavior:

1. In the code for the Send retryer, find the <ExponentialRetry> section, and comment that out with the comment indicators <!--and -->.

2. Find the <FixedRetry section, and un-comment it by removing the comment indicators <!-- and -->.

In the <ExponentialRetry> section of the Send retryer, you can configure the following values:

Setting

Default

Description

RetryCount

3

Specifies the maximum amount of retries for transient exceptions.

RetryInterval

00:00:01

The fixed time interval between retries.

The <TransientSmtpErrorCodes> section specifies which SMTP-specific transient faults the EmailCampaign.EDS.Core.Net.Smtp.Send.Retryer retryer checks for. By default, the retryer always checks for the fault code 421, which indicates that the recipient service is not available. To make the retryer more specific in its checks, you can add additional fault codes to this section.

1. In the configuration code, find the <TransientSmtpErrorCodes> section.

2. Add a new line with <YourDescription>[the fault code]</YourDescription>.

### Tip

Use the <YourDescription> field to write a useful description for yourself about what the error code means.

For more information on SMTP transient fault codes and their meaning, refer to the SMTP Enhanced Status Codes Registry.

EXM uses the EmailCampaign.XConnect.Conflict.Retryer to retry a method if there is a conflict with xConnect. When such a conflict happens, the retryer attempts to fetch the conflicted entity from xConnect again.

In the Sitecore.EmailExperince.xConnect.config file (path: \<Your Instance Name>\App_Config\Sitecore\EmailExperience), the following is the configuration code for the xConnect conflict retryer:

<TransientFaultHandling>
<Retryers>
<EmailCampaign.XConnect.Conflict.Retryer>
<Type>Sitecore.EmailCampaign.XConnect.Web.TransientFaultHandling.RetryerFactories.XConnectConflictFixedRetryerFactory, Sitecore.EmailCampaign.XConnect.Web</Type>
<Options>
<FixedRetry>
<RetryCount>3</RetryCount>
<RetryInterval>00:00:01</RetryInterval>
<FirstFastRetry>false</FirstFastRetry>
</FixedRetry>
</Options>
</EmailCampaign.XConnect.Conflict.Retryer>
</Retryers>
</TransientFaultHandling>

In the xConnect conflict retryer, you can configure the following values:

Setting

Default

Description

RetryCount

3

Specifies the maximum amount of retries for fetching a conflicting entity from xConnect.

RetryInterval

00:00:01

The fixed time interval between retries.

FirstFastRetry

false

Specifies whether EXM should perform the first retry immediately after encountering a conflict. By default, this is false, which means that the first retry happens according to the RetryInterval.


text:
Boltzmann vs Maxwell distribution?

Toby_phys
So I worked through the Boltzmann distribution and got:
$$P\propto e^{\frac{-E}{k_BT}}$$
Where $E$ is the energy. So surely this means the kinetic energy (and therefore speed) of particles is distributed over a Boltzmann distribution. Or in equation:
$$P\propto e^{\frac{-mv^2}{2k_BT}}$$

However, this is not what my book says. The books says velocity is distributed in this way. They appear to have split up the components of energy

Speed is distributed as follows, as a Maxwell distribution.
$$P\propto v^2e^{\frac{-mv^2}{2k_BT}}$$

I understand that the probability of the speed is the sum of all the different degrees of freedom and such.

However, why can you split up the components of the kinetic energy is this way? Note, I am not asking how to get from Boltzmann to Maxwell, I am asking why velocity, and not kinetic energy, is Boltzmann distributed?

Homework Helper
Gold Member
The states are counted in momentum space, and that is why the results follow how they do. e.g. the energy is ## E=\frac{p_x^2+p_y^2+p_z^2 }{2 m} ##, but the number of states in an energy interval depends upon the block ## \Delta N=(\frac{\Delta p_x \, \Delta p_y \, \Delta p_z}{h^3})V ##. (## V ## is the volume of the sample). When converting to speeds, a spherical symmetry in the p-space is assumed and gives ## \Delta N=(\frac{4 \pi p^2 \, m \, dv}{h^3})(V) ##, ( where ## v ## is the speed). In addition to the Boltzmann factor ## e^{-\frac{E}{kT}} ## for the relative occupancy, the number of states(density of states) needs to be included to find the entire distribution function.

Last edited:
Toby_phys
Why are the states counted in momentum space?? I haven't seen a derivation where this is necessary

Homework Helper
Gold Member
Why are the states counted in momentum space?? I haven't seen a derivation where this is necessary
The Statistical Physics book by Reif gives a pretty good description. It may seem like a handwaving approach at first, but this is the easiest way to demonstrate it (this derivation is found in Reif's book): ## \\ ## In a containter of volume ## V ##, with length ## L_x ##, ## L_y ##, and ## L_z ##, you set up periodic boundary conditions: ## e^{ik_x x}=e^{ik_x(x+L_x)} ## and similarly for y and z. This means ## k_x L_x=n_x (2 \pi) ##, for integer ## n_x ##, and similarly for y and z. Thereby ## \Delta N=\Delta n_x \Delta n_y \Delta n_z =\frac{\Delta^3 \vec{k}}{(2 \pi)^3} V ##. (Note ## V=L_x L_y L_z ##). With the substitution ## \Delta^3 \vec{p}=\hbar^3 \Delta^3 \vec{k} ##, the equation I have in post #2 should follow.

Last edited:
Gold Member
2022 Award
The Maxwell-Boltzmann distribution is derived by maximizing the entropy of a ##N##-body state in the (grand-)canonical ensemble of non-interacting particles in a box of volume ##V##. A convenient single-particle basis is chosen in the previous posting, namely momentum space. Let's consider scalar particles for convenience. For the many-body system we use the corresponding Fock (occupation-number) basis of the common eigenvectors of the number occupying each single-particle state labeled by the complete set of discrete momenta ##\vec{p} \in 2 \pi/L \mathbb{Z}^3## (I make the simplification to set ##L_1=L_2=L_3=L##, i.e., use a cube as the "quantization volume").

Instead of evaluating the simple partition sum of the grand-canonical ensemble, we rather calculate the functional
$$Z[\beta(\vec{p})]=\mathrm{Tr} \exp\left \{-\sum_{\vec{p}} [\beta(\vec{p}) E(\vec{p}) -\alpha] \hat{N}(\vec{p}) \right \}.$$
At the end we get the usual partition sum by setting ##\beta=\frac{1}{k_{\text{B}} T}## and ##\alpha(\vec{\beta})=\beta \mu##, where ##T## is the absolute temperature, ##\mu## the chemical potential for the conserved particle number and ##E(\vec{p})=\vec{p}^2/(2m)##. Since the ##\hat{N}(\vec{p})## all commute with each other we have
$$Z[\alpha(\vec{p})]=\prod_{\vec{p}} \sum_{N(\vec{p})=0}^{\infty} \exp\{[-\beta E(\vec{p})+\alpha(\vec{p})]N(\vec{p}) \} = \prod_{\vec{p}} \frac{1}{1-\exp\{[-\beta E(\vec{p})+\alpha(\vec{p})]\}}.$$
We have made use of the fact that spin-0 particles are bosons, and thus each ##N(\vec{p}) \in \mathbb{N}_0##.

Now it's easier to use
$$W[\alpha(\vec{p})]=\ln Z[\alpha(\vec{p})]=-\sum_{\vec{p}} \ln \{1-\exp\{[-\beta E(\vec{p})+\alpha(\vec{p})]\}.$$
The mean number of particles in the state ##\vec{p}## is given by
$$n(\vec{p}) = \langle \hat{N}(\vec{p}) \rangle=\left [\frac{\partial W}{\partial \alpha(\vec{p})} \right]_{\alpha(\vec{p})=\beta \mu}=\frac{\exp [-\beta E(\vec{p})+\beta \mu]}{1-\exp [-\beta E(\vec{p})+\beta \mu]}=\frac{1}{\exp [\beta E(\vec{p})-\beta \mu]-1}.$$
Now making ##L## very large we can take ##\vec{p} \in \mathbb{R}^3## making use of the above derived single-particle density we find
$$\mathrm{d} N=f(\vec{p}) \mathrm{d}^3 \vec{p}= f_{\text{B}}[E(\vec{p})] V \frac{\mathrm{d}^3 \vec{p}}{(2 \pi \hbar)^3}.$$
For a very diluted gas, for which ##\beta \mu## is a very small negative number, we can neglect the 1 in the denominator of the Bose-Einstein distribution, and we get the classical limit,
$$f(\vec{p})=\frac{V}{(2 \pi \hbar)^3} exp[-\beta E(\vec{p})+\beta \mu].$$
This is the phase-space distribution function of the ideal gas.

Now to get the distribution function for ##v=|\vec{p}|/m=p/m## we note that ##\mathrm{d}^3 \vec{p}=\mathrm{d} p \mathrm{d}^2 \Omega p^2=\mathrm{d} v \mathrm{d}^2 \Omega m^3 v^2## and thus
$$\mathrm{d} N=f(\vec{p}) \mathrm{d}^3 \vec{p} =\frac{V}{(2 \pi \hbar)^3} \exp(-\beta m v^2/2+\alpha) m^3 v^2 \mathrm{d} p \mathrm{d}^2 \Omega$$
Integrating over the whole solid angle and using that the integrand doesn't depend on the angles we simply get
$$\frac{\mathrm{d} N}{\mathrm{d} v} = \frac{V}{(2 \pi \hbar)^3} 4 \pi m^3 v^2 \exp(-\beta m v^2/2+\alpha).$$
You have to be aware that if you change the variable of interest for a distribution, you always get the original distribution multiplied with the appropriate Jacobian changing the "volume element" from one set of variables to another.

You can also check that the energy-distribution function is
$$\frac{\mathrm{d} N}{\mathrm{d} E} \propto \sqrt{E} \exp(-\beta E+\alpha)$$
using
$$E=m v^2/2 \; \Rightarrow \; v=\sqrt{2 m E} \; \Rightarrow \; \mathrm{d} v=\mathrm{d} E \sqrt{2 m}\frac{1}{2 \sqrt{E}}=\mathrm{d} E \sqrt{\frac{m}{E}}.$$


text:
## Materials Science C

raxu
Exalted Member
Posts: 38
Joined: Fri Oct 10, 2014 1:54 am
Division: C
State: NY
Contact:

### Re: Materials Science C

In the sonic powerpoint there are a few formulas for weight distribution I don't quite understand...

Number Average $\overline{M_n}=\frac{\sum N_iM_i}{\sum N_i}$ ,
Weighted average = $\overline{M_w}=\frac{\sum N_iM_i^2}{\sum N_iM_i}$ ,
Molecular weight average $\overline{M_z}=\frac{\sum N_iM_i^\alpha}{\sum N_iM_i^{\alpha-1}}$ .

Can someone explain what they are for?
Richard A. Xu
Events involved 17/18: Astro, Chem, Fermi, Game, Mat Sci, Optics, Thermo.

sonred
Member
Posts: 3
Joined: Tue Jan 09, 2018 10:15 pm
State: -
Contact:

### Re: Materials Science C

raxu wrote:In the sonic powerpoint there are a few formulas for weight distribution I don't quite understand...

Number Average $\overline{M_n}=\frac{\sum N_iM_i}{\sum N_i}$ ,
Weighted average = $\overline{M_w}=\frac{\sum N_iM_i^2}{\sum N_iM_i}$ ,
Molecular weight average $\overline{M_z}=\frac{\sum N_iM_i^\alpha}{\sum N_iM_i^{\alpha-1}}$ .

Can someone explain what they are for?

number average is doesn't take into account that heavier molecules contribute more to the molecular weight than lighter molecules where as weighted average does.

sonred
Member
Posts: 3
Joined: Tue Jan 09, 2018 10:15 pm
State: -
Contact:

### Re: Materials Science C

raxu wrote:In the sonic powerpoint there are a few formulas for weight distribution I don't quite understand...

Number Average $\overline{M_n}=\frac{\sum N_iM_i}{\sum N_i}$ ,
Weighted average = $\overline{M_w}=\frac{\sum N_iM_i^2}{\sum N_iM_i}$ ,
Molecular weight average $\overline{M_z}=\frac{\sum N_iM_i^\alpha}{\sum N_iM_i^{\alpha-1}}$ .

Can someone explain what they are for?

number average is doesn't take into account that heavier molecules contribute more to the molecular weight than lighter molecules where as weighted average does.


text:
# Question

Let the random variable X follow a normal distribution with μ = 0.2 and σ2 = 0.0025.
a. Find the probability that X is greater than 0.4.
b. Find the probability that X is greater than 0.15 and less than 0.28.
c. Find the probability that X is less than 0.10.
d. The probability is 0.2 that X is greater than what number?
e. The probability is 0.05 that X is in the symmetric interval about the mean between which two numbers?

Sales0
Views49


text:
## How to handle extremely 'long' images?

0

After transforming timeseries into an image format, I get a width-height ratio of ~135. Typical image CNN applications involve either square or reasonably-rectangular proportions - whereas mine look nearly like lines:

Example dimensions: (16000, 120, 16) = (width, height, channels).

Are 2D CNNs expected to work well with such aspect ratios? What hyperparameters are appropriate - namely, in Keras/TF terms, strides, kernel_size (is 'unequal' preferred, e.g. strides=(16, 1))? Relevant publications would help.

Clarification: width == timesteps. The images are obtained via a transform of the timeseries, e.g. Short-time Fourier Transform. channels are the original channels. height is the result of the transform, e.g. frequency information. The task is binary classification of EEG data (w/ sigmoid output).

Are the spatial correlations in the vertical (height) dimension relevant? I.e. does the vertical distance have physical meaning? – MPA – 2020-04-06T11:51:53.633

@MPA Question updated – OverLordGoldDragon – 2020-04-06T11:58:59.943

What is the type of the desired output? A class label (e.g. for detection), a time series (e.g. for filtering), etc.? – MPA – 2020-04-06T12:07:52.987

@MPA Binary classification, updated again – OverLordGoldDragon – 2020-04-06T12:09:05.097

Depending on the physical nature of the input data, you could first perform 1D convolutions with dimensionality reduction along the time axis to extract the most relevant features for each frequency band, and then combine the features of each frequency band into a class prediction. Or if the data are sufficiently smooth, you could try an auto-encoder to compress the time axis into a smaller latent space representation, and perform additional analysis on the latent space. – MPA – 2020-04-06T14:00:14.183

@MPA 16000 is precisely from the autoencoder approach; the full original timeseries spans 200,000+ timesteps. Though compressing it even further is unrealistic without loss of information - and the timeseries transform used is promising for discrimitative feature extraction. The linked thread has some sound suggestions I might summarize in an answer. – OverLordGoldDragon – 2020-04-06T14:05:51.903

I suppose the EEG data is quasi-periodic. What about Dynamic Mode Decomposition (DMD) and performing the classification on the dominant modes? – MPA – 2020-04-06T14:15:35.520

0

I had recently used a slightly unorthodox method to process such images, which involved using RNNs.

Assume the image dimensions to be (16000, 120, 16) = (width, height, channels), as in the question.

Apply a 2D convolution (or multiple such convolutions) of shape(1, k, c), such that the output of the convolutions becomes (16000, 1, c). So if you only use a single convolutional layer, k=120.

Then, squeeze the extra dimension, to get the shape (16000, c).

The problem has now been transformed back into a sequence problem! You can use RNN variants for further processing.

This defeats the purpose of the transform; see updated question. – OverLordGoldDragon – 2020-04-06T11:58:10.537


text:
Please provide the answers to the amounts not given.For the following items, provide the missing number.Retained Earnings, 1/1/16$575,603Net Income?Subtotal$ 498,667Dividends10,000Retained Earnings, 1/1/16$488,667Make sure your answer does not include a "$" (dollar sign) or "," (comma). It should be in the format of a basic number; such as "55000".

Question

Please provide the answers to the amounts not given.

For the following items, provide the missing number.

Retained Earnings, 1/1/16
$575,603 Net Income ? Subtotal$ 498,667
Dividends
10,000
Retained Earnings, 1/1/16
$488,667 Make sure your answer does not include a "$" (dollar sign) or "," (comma). It should be in the format of a basic number; such as "55000".


text:
# Changeset 10646 for NEMO/releases/release-4.0/src/OCE/BDY/bdydta.F90

Ignore:
Timestamp:
2019-02-07T12:58:00+01:00 (3 years ago)
Message:

Enabling of the activation of the inverse barometer effect at open boundaries in configurations where both tidal forcing and the time-split surface pressure gradient computation have been selected; fix for the bug described in ticket #2223.

File:
1 edited

Unmodified


text:
# How many moles of "H"_2"SO"_4 are present in "0.500 L" of a "0.150-M" "H"_2"SO"_4 solution?

##### 1 Answer
May 9, 2018

${\text{0.0750 moles H"_2"SO}}_{4}$

#### Explanation:

$\text{Molarity}$ $= \text{Moles"/"Liter}$

Using this, we can fill in several of these values since they are given. This gives us the equation:

$0.150 = \frac{\text{Moles}}{0.500}$

By multiplying both sides by $0.500$, we find

$\text{0.150 mol/L" * "0.500 L" = "Moles}$

or simplified

$\text{0.0750 mol = Moles}$


text:
# Statistics¶

Contrary to what some people may have heard, a balanced search tree is, as the name suggests, always — read that again, please — always balanced. It is a myth that you have to rebuild the index whenever the performance of your queries is below par. There are extremely rare cases when Oracle recommends that you rebuild the indexes but in almost all cases you do not have to rebuild your indexes. [1]

Oracle nowadays automatically collects statistics, so once you create an index, Oracle takes care of the rest. You can see the schedule and some basic information about the statistics collection with the following statement:

 1 2 3 4 5 6 7  SELECT * FROM dba_autotask_client WHERE client_name = 'auto optimizer stats collection' ;

For most databases the automatic statistics collection is sufficient. If, however, your database has tables that are being deleted and truncated between collection runs, then it can make sense to go gather optimizer statistics manually.

When you create an index, Oracle automatically gathers optimizer statistics because it needs to do a full scan anyway. As of Oracle Database 12c, the same piggybacking is done for the statistics collection of create-table-as-select (CTAS) and insert-as-select (IAS) operations, which is quite nifty; histograms require additional data scans, so these are not automatically gathered. The execution plans of CTAS and IAS statements show whether statistics are being collected at runtime: OPTIMIZER STATISTICS GATHERING, right below the LOAD AS SELECT operation.

If you change the definition of an index, you may want to update the statistics. Please coordinate with the DBA to avoid unwanted side effects, such as degrading the performance of all but your own queries because invalidation of execution plans; gathering statistics does not lock the table, it’s like running multiple queries against it. [2]

Notes

 [1] The index clustering factor indicates the correlation between the index order and the table order; the optimizer takes the clustering factor into account for the TABLE ACCESS BY INDEX ROWID operation. A high ratio of leaf nodes marked for deletion to leaf nodes (> 0.20), a low value of percentage used (< 0.80), and a clustering factor (see DBA_INDEXES) close to the number of rows (instead of the number of blocks) in the table (see DBA_SEGMENTS) are indicators that your indexes may benefit from rebuilding. If the clustering index is close to the number of rows, then the rows are ordered randomly.
 [2] The DBMS_STATS.AUTO_INVALIDATE option can be used to ensure that Oracle does not invalidate all cursors immediately, which can cause a significant CPU spike. Instead, Oracle uses a rolling cursor invalidation based on internal heuristics.


text:
# Last-Minute JEE Revision Notes 2023 for Main and Advanced

## JEE Main and JEE Advanced 2023 Prep: JEE Revision Notes

JEE revision notes, as per the updated syllabus for JEE Main and JEE Advanced 2023, are now available on Vedantu. The IIT JEE experts at Vedantu have curated and prepared the JEE Main notes and JEE Advanced notes for revision as per the updated JEE syllabus. Designed specifically for the last-minute JEE revision, these notes will provide candidates with a comprehensive understanding of the theorems, formulas, and the related derivations, covered in the syllabus for JEE Main and JEE Advanced. Candidates can download the JEE revision notes PDFs for free from Vedantu mobile app and website.

The following sections will discuss the important topics covered in the revision notes for JEE Main and Advanced, how these revision notes are helpful for candidates, tips for an effective JEE preparation plan, and a few other details related to the exams.

 Sl.No Exam Revision Notes 1 JEE Main JEE Main Revision Notes 2 JEE Advanced JEE Advanced Revision Notes

## Important Dates for JEE Main and Advanced 2023

An effective revision plan for JEE Main and Advanced can be chalked out and implemented on the basis of the time left before the exams. So prior to going through the details of JEE revision notes, let us take a look at the updated schedule for the IIT JEE entrance exams. JEE Main 2023 is going to be conducted in two sessions. The first session is scheduled for April and the second session is for May. Followed by the JEE Main results, JEE Advanced will be conducted in August. The table below displays the dates for JEE Main and Advanced exams.

### JEE Main 2023 Exam Dates: April  Session

 Events Tentative Exam Dates JEE Main 2023 Notification Release 1st week of March 2023 JEE Main 2023 Application Form Release(April Session) 1st week of March 2023 Last date of Application submission 4th week of March 2023 JEE Main 2023 Admit Card Release Date(April Session) 2nd week of April 2023 JEE Main 2023 Exam Date(April Session) 4th week of April 2023 JEE Main 2023 Answer Key Release (April Session) 1st week of May 2023 JEE Main 2023 Result Date (April Session) 2nd week of May 2023 JEE Main 2023 Counselling Date (April Session) 3rd week of July 2023

### JEE Main 2023 Exam Dates: May Session

 Events Tentative Exam Dates JEE Main 2023 Notification Release 1st week of March 2023 JEE Main 2023 Application Form Release (May Session) 1st week of April 2023 Last date of Application submission 1st week of May 2023 JEE Main 2023 Admit Card Release Date(May Session) 2nd week of May 2023 JEE Main 2023 Exam Date (May Session) 4th week of May 2023 JEE Main 2023 Answer Key Release (May Session) 1st week of June 2023 JEE Main 2023 Result Date (May Session) 2nd week of June 2023 JEE Main 2023 Counselling Date (May Session) 3rd week of July 2023

The NTA has not announced JEE Advanced 2023 Dates yet. Once the official dates are out, we’ll update them here. Meanwhile aspirants can have a look at the tentative dates below:

 Exam Events Dates JEE Advanced Registration 2023 begins Last week of May 2023 Application Deadline First week of June 2023 Last date for Payment of application fee Second week of June 2023 Admit card release Second week of June 2023 JEE Advanced 2023 exam date Third week of June 2023 Candidates response sheet release Fourth week fo June 2023 Provisional Answer Key Fourth week fo June 2023 Deadline to challenge answer key First week of July 2023 Final Answer Key/ JEE Advanced Result First week of July 2023 Architecture Aptitude Test (AAT) registrations Second week of July 2023 AAT 2023 Exam Third week of July 2023 AAT Result Third week of July 2023

### Mathematics Revision Notes for JEE Main and JEE Advanced 2023: Important Topics

The JEE Advanced and JEE Main notes for Mathematics consist of the important formulas, theorems, corollaries, and the subsequent applications covered under each topic of the Mathematics syllabus for JEE Main and JEE Advanced. Referring to the Mathematics revision notes for JEE prepared by the subject matter experts at Vedantu, candidates will be able to carry out a thorough last minute revision and get a useful insight into solving the various types of sums covered in the JEE Main and JEE Advanced syllabus 2023.

Below are some of the important topics covered in the Maths revision notes for JEE Main and JEE Advanced.

### Important Maths Questions for JEE Main and JEE Advanced Revision

Some of the important questions for JEE Maths revision are given below for the practice purpose of candidates. These sums are framed in close reference to the JEE previous year’s question papers.

1. A committee of 6 Indians and 8 foreigners has to be formed. The committee has to consist of 2 Indians at least and double the number of foreigners as that of the Indians. In how many ways, can the committee be formed?

2. A line given by $\frac{(x-3)}{1}=\frac{(y-4)}{2}=\frac{(z-5)}{2}$ intersects the plane given by x + y + z =17. Calculate the distance of a point (1,1,9) from the point where the above-given line and plane intersect.

3. Two vertical poles are placed 150 m apart. One pole is three times higher than the other. The feet of the two poles are joined with a straight line. An observer, from the middle point of this line, finds the angle of elevation of the top points of the two poles to be complementary. What are the heights of the two poles?

4. A dice is rolled for a particular number of times. If the probability of getting an even number thrice is the same as the probability of getting an odd number twice, what will be the probability of getting an odd number for an odd number of times?

5. What will be the value of k for which the given inequality, $x^{2}-2(3K-1)x+8K^{2}-7> {0}$ is valid for every value of in R.

### Physics Revision Notes for JEE Main & JEE Advanced 2023: Important Topics

The JEE Advanced and JEE Main Physics notes PDF download is now available on Vedantu. These notes are meticulously prepared by the highly experienced IIT JEE mentors and Physics subject matter experts at Vedantu. The definitions, formulas, derivatives, theorems, and applications of all concepts covered in the JEE 2023 Physics syllabus are explained in these revision notes. The key feature of these JEE revision notes is that the simplest problem-solving approaches are discussed and implemented here for every sum and topic-wise conceptual questions covered in the syllabus. The JEE Main and JEE Advanced notes are provided in a topic-wise PDF format for the convenience of candidates.

The JEE Main revision notes are designed keeping in view the marks weightage of topics covered in the syllabus. Below are some of the important Physics topics that candidates must thoroughly revise before JEE Main and JEE Advanced exams.

### Important Physics Questions for JEE Main and JEE Advanced Revision

Below are some of the important questions that are framed with close reference to the previous year question papers of JEE. Candidates can solve and practise these questions to check their JEE preparation.

1. An x-ray tube is being operated at 1.24 million volts. What will be the shortest wavelength of the photon produced?

2. What will be the nature of the graphical representation of velocity expressed as a function of displacement, when a particle exhibits simple harmonic motion?

3. At the north pole, a body weighs 49 N when weighed on a spring balance. What will be the weight of the body at the equator

4. A and B are two satellites having masses 200 kilograms and 400 kilograms, respectively. Satellite A is revolving around the earth at a height of 600 kilometres. Satellite B is orbiting around the earth at a height of 1600 kilometres. What will be the value of TB - TA, where TA and TBare the time periods of satellites A and B, respectively. Take the radius of earth as 6400 kilometres and the mass of earth as 6 x 1024  kilograms.

5. An engine of a train is moving with a uniform acceleration. It passes by a signal post with a velocity ‘u’ and the last compartment passes the signal post with a velocity ‘v’. With what velocity will the middle point of the train pass the signal post?

### Chemistry Revision Notes for JEE Main & JEE Advanced 2023: Important Topics

The JEE Main and JEE Advanced syllabus for Chemistry is mainly divided into three parts, Physical Chemistry, Organic Chemistry, and Inorganic Chemistry. While some chemical equations and reactions in the syllabus follow a set of standard temperature and pressure conditions, some have the use of catalysts, and some are recognized for their widespread applications in various industries across the world. So, a sound understanding of the theories, concepts, and chemical reactions covered in the JEE revision notes will help candidates to prepare effectively for the exams. Our subject experts at Vedantu have put together the easiest ways and a few smart tricks to understand as well as remember all concepts and chemical reactions/equations covered in the JEE Main and JEE Advanced syllabus 2023.

Below are some of the important topics that are enumerated and explained in the JEE Chemistry revision notes PDFs on Vedantu.

## Important Chemistry Topics for JEE Main and Advanced

### Important Chemistry Questions for Revision Notes for JEE Main and JEE Advanced

The below-given Chemistry questions are provided for a quick-practice of candidates preparing for JEE Main and Advanced. These questions are prepared with a close reference to the previous year’s Chemistry question papers of JEE Main and Advanced.

1. Which ore is concentrated by cyanide of Group 1 element?

2. What are X, Y, and Z in the following reaction?

$AI+NaOH\rightarrow X\xrightarrow[]{y(g))}>z$

1. What is the composition of gun metal?

2. On the complete combustion of 1.80 grams ofCx Hy Oz, 1.08 grams of H2Oand 2.64 grams of nCo2are obtained. What is the percentage of oxygen in the given organic compound?

3. A buffer solution has a pH of 3 and the solubility of AgCN in this solution is x. What is the value of x if no cyano compound has been formed? (KSP = 2.2 x 10-16; Ka = (HCN) = 6.2 x 10-10)

### Tips for Last-minute Revision for JEE Main and JEE Advanced 2023

Many candidates start with their JEE preparation right from the beginning of Class 11. On the other hand, some candidates join JEE crash courses in Class 12 while some take a year gap to prepare thoroughly for the JEE Main and JEE Advanced exams. However, no matter how long you take to familiarise yourselves with the syllabus of JEE Main and JEE Advanced, an effective last-minute brush-up can do wonders to your exam preparation and your confidence. So here are some tips to plan the last-minute revision for JEE Main and JEE Advanced exams.

• Take a few JEE mock tests on Vedantu and analyse your performance.

• Gauge your exam preparation. Identify and note down the areas where you need to focus more during the revision.

• Now, plan your study timetable with a realistic approach.

• Brush up the fundamental concepts of every topic at the beginning of your revision.

• Try to set small goals so that you can achieve them.

• Allot a time slot for practising sums on a daily basis. It will help you optimise your speed and accuracy when it comes to the most scoring parts of the JEE question papers.

• You can make your study timetable and set hourly goals to revise topics and sub-topics of the chapters covered in the syllabus. Try the timetable for two days and check if you can keep up with it. If you experience lag, then incorporate a change in your timetable.

• Patience is the key to an effective last-minute JEE revision. Do not fuss about any topic if you are unable to solve the questions asked from it all at once. Instead, learn the topics identified as your problem areas from scratch with patience.

• Keep a couple of hours extra in your timetable. You can make a short recap of the topics that you cover in the day. Also, you can solve and practise sums, formulas, and derivations, in these extra hours every day.

• Keep a track of your progress while you run the last-minute revision rounds. This will help you to understand where you stand and improvise a more efficient way to prepare for JEE Main and JEE Advanced exams.

### Benefits of JEE Revision Notes from Vedantu Experts

Though the exam patterns for JEE Main and Advanced are slightly different, the topics and subtopics covered in the syllabus for both exams are the same. Mostly the topics coincide with the Class 11 and 12 syllabus for Mathematics, Physics, and Chemistry. The JEE revision notes on Vedantu are prepared and assorted in a way that is highly beneficial for the candidates. A few benefits of the JEE revision notes on Vedantu are as follows:

• The subject-wise and topic-wise revision notes for JEE Main and JEE Advanced make it easier for the candidates to cover the syllabus in a step-by-step manner.

• The revision notes are available in free downloadable PDF format, so candidates can access and download them from anywhere at their convenience. The best part is that these revision notes are also available on the Vedantu mobile app, making it even more convenient to access and refer to for candidates, any time on the go.

• The important formulas from all chapters covered in the JEE syllabus of Physics, Chemistry, and Mathematics are listed in these revision notes. Every symbol used in a formula is well defined in the PDFs. So candidates can revise the formulas for the entire syllabus all at one place, without having to search through the pages of the books physically.

• The standard balanced chemical equations and reactions, along with the necessary conditions, from each topic of organic and inorganic chemistry are discussed in the revision notes for JEE Main Chemistry. So, candidates can refer to the revision notes PDFs to revise the formulas with just a click, thus saving their time in going through the chapters to find them.

• The formulas and theorems of JEE Maths syllabus are listed and explained with appropriate solved examples in the topic-wise revision notes PDFs on Vedantu. Hence, candidates can recapitulate and revise the major points of each chapter covered in the JEE Main and JEE Advanced Mathematics syllabus.

Candidates can refer to the JEE revision notes online as well as download the revision notes PDFs for free from Vedantu and prepare for the JEE exams. Also, candidates aspiring to take JEE Main and JEE Advanced 2023 can sign up on Vedantu to access free study materials. Along with providing JEE revision notes, we also provide JEE crash courses, JEE mock test series, topic-wise important questions, solved previous year question papers, and other JEE Main and JEE Advanced study materials. So, sign up and make the most of the time left before the JEE exams with Vedantu’s effective study resources.

## FAQs on Last-Minute JEE Revision Notes 2023 for Main and Advanced

1. Can I get free JEE revision notes online?

Yes. JEE revision notes 2022 are available for absolutely free of cost on Vedantu. These revision notes are available in PDF files and you can get topic-wise revision notes and important questions for Physics, Mathematics, and Chemistry. All you need to do is sign up on Vedantu and search for the JEE revision notes for 2022 to download them for free.

2. Are there any changes in the JEE Main and JEE Advanced syllabus?

No, there are no changes in the JEE syllabus for 2022. However, to keep up with the reduction in the Class 12 syllabus, NTA has introduced some changes in the exam pattern of JEE Main 2022. There will be options for some questions in the JEE Main question paper and there will be negative marking in Section B of the JEE Main 2022 question papers.


text:
# The inverse function of “Prime” [duplicate]

Consider p such that PrimeQ[p] == True. How do I compute n such that Prime[n] == p?

In other words, what is the inverse function of "Prime"?

EDIT:

As a concrete example, consider p to be the first prime that factorizes rsa-768,

p = rsa768a = 33478071698956898786044169848212690817704794983713768568912431388982883793878002287614711652531743087737814467999489

PrimeQ[rsa768a] is True; I want to know its n.

Trying Jason's suggestion, InverseFunction[Prime][rsa768a], does not return the expected result (for a small prime it does). Trying LLlAMnYP suggestion, PrimePi[rsa768a] returns

PrimePi::largp: Argument ... in PrimePi[...] is too large for this implementation. >>

## marked as duplicate by Artes, user9660, MarcoB, m_goldberg, J. M. will be back soon♦Mar 9 '16 at 13:56

• InverseFunction[Prime] – Jason B. Mar 9 '16 at 9:21
• @JasonB Cheeky :-P and cruel. PrimePi is much faster. – LLlAMnYP Mar 9 '16 at 9:24
• It's so direct, it's almost mocking the OP, but I like, how it shows off Mathematica's capability to do the thinking for you. But on a more serious note, I can't imagine when it would be preferable to PrimePi. – LLlAMnYP Mar 9 '16 at 9:28
• You can attempt to use NextPrime[#, -1]& but good luck waiting for that computation to end in this universe. There's a reason, that these things are used for encryption, y'know :-) – LLlAMnYP Mar 9 '16 at 9:36
• @J.C.Leitão This question What is so special about Prime? explicitly states this problem. Prime has an extensible domain, however you can't extend it up to rsa768a unless you have an enormous amount of RAM, but it is unlikely realizable. – Artes Mar 9 '16 at 9:52


text:
A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII.
Question:

A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted?

Solution:

A sports team of 11 students is to be constituted, choosing at least 5 students of class XI and at least 5 from class XII.

Required number of ways $={ }^{20} C_{5} \times{ }^{20} C_{6}+{ }^{20} C_{6} \times{ }^{20} C_{5}=2 \times{ }^{20} C_{5} \times{ }^{20} C_{6}=2\left(20_{C_{6}} \times 20_{C_{5}}\right)$


text:
# All Questions

2 views

### Riemann sum and partitions

If f is riemann integrable and if $(P_n)$ is any sequence of tagged partitions of [a,b] such that $\lVert P_n \rVert$ -> 0, prove that $\int_a^b f = lim_n S (f;P_n)$. I am confused as to how to ...
2 views

### Diagonal decomposition, square root and eigenvector / eigenvalue of a matrix

I have encountered a problem of finding eigenvector and eigen value of a matrix of type $$A = \dfrac{1}{2} \begin{pmatrix} 4&1&-2\\ -4&1&6\\ 2&0&-2 \end{pmatrix}$$ Also I ...
2 views

### Still stuck on simplifying terms while doing linear combinations

So I'm currently trying to wrap my head around finding gcd through the Euclidean Algorithm in order to write the integers as a linear combination. For example, a problem is to express the ...
15 views

### Show $ex \leq e^x$ for all $x \in \mathbb{R}$

So far all I have is this: Let $f$ be a function where $f(x)=ex-e^x\leq 0$ $f'(x)=e-e^x \leq 0$, so $f$ is decreasing. I'm stuck here. Can someone help me with the next steps?
7 views

### determine whether a combination number is odd or even

Let $k$ be a given positive integer (fixed). I want to determine whether $$2n-k\choose n$$ is even or odd, for each positive integer $n$. Is there any general result? My attempt: Case (1). ...
11 views

### Why are structures with no relations called algebras?

Elementary algebra has at least one relation: the equality (or identity) relation, signalized by the symbol "=" The equality relation is quintessential to linear algebra and algebraic equations, such ...
19 views

15 views

### What did I do wrong with this combinatorics question?

I was given the following problem. "A teacher wants to choose a captain and vice-captain among 12 volleyball players. In how many ways can she do so?" I tried to solve it by multiplying 12 by 11 ...
15 views

### How is $K(x)$ pronounced, where $K$ is a field?

Let $K$ be a field. The ring of polynomials $K[x]$ is pronounced "$K$ adjoin $x$", right? How is the field of rational functions $K(x)$ pronounced? (Sorry if this is a silly question. I am ...
12 views

### Describing polar coordinates in a window

I'm having trouble with the following problem and have no idea what to do. I tried drawing a horizontal and vertical line down the middle of the window but got nowhere. A window is in the shape of a ...
5 views

### Probability of choosing one of two desired boxes out of five on the second try.

I've got a problem that I'm not understanding how a given probability is being found. If you have five boxes, three of which are empty and two of which have items you want, if you choose boxes in a ...
6 views

### Geometric distribution converges to exponential distribution

For $n\in \mathbb{N}$ let $X_n$ be geometric with parameter $p_n \in (0,1)$, that means $\mathbf{P}[X_n = k] = p_n(1-p_n)^k$, $k\in\mathbb{N}_0$. How must the sequence $(p_n)_{n\in\mathbb{N}}$ ...
7 views

### Differentiability & continuity proof?

Prove that if the absolute value of f is differentiable at a and f is continuous then f is differentiable at a. I've started off by writing down the formal definition of differentiability and the ...
12 views

### What is a function which differs when differentiated with respect to x and then with y to function differentiated with respect to y and then x?

What is a function which differs when differentiated with respect to x and then with y to function differentiated with respect to y and then x? Even if it differs at one particular point.
14 views

### Functions and range

$$a \colon \mathbb{R}\setminus\{0\} \to \mathbb{R} \;\text{ defined by }\; a(x)= 6/x \\ b \colon \mathbb{Z} \to \mathbb{R} \;\text{ defined by }\; b(x) = 3x + 1$$ a) State the range of ...
4 views

### Let X_1,..,X_n i.i.d negative binomial. Find the best unbiased estimator for P(x<=3)

I am not sure where I should even start with this problems. I know that the sum of negative binomial random variables is itself a negative binomial random variable. I am sure that I can show that the ...
8 views

36 views

### What is $p+4 \pmod {13}$

$p=\sum_{m=1}^{12} (m \cdot m!)$, and the question is just the title. I tried applying Wilson's theorem i.e $(p-1)!+1\equiv0 \pmod p$ but did not get much help. Thanks for helping
18 views

### $a,b,c,p$ are rational number and $p$ is not a perfect cube

Given that $a,b,c,p$ are rational number and $p$ is not a perfect cube, if $a+bp^{1\over 3}+cp^{2\over 3}=0$ then we have to show $a=b=c=0$ I concluded that $a^3+b^3p+c^3p^2=3abcp$ but how can I go ...
4 views

### tridiagonal matrix with a corner entry from upper diagonal

I am trying a construct a matlab code such that it will solve an almost tridiagonal matrix. The input I want to put in is the main diagonal (a), the upper diagonal (b) and the lower diagonal and the ...
25 views

### If $n\equiv 4 \pmod 9$ then $n$ cannot be written as sum of three cubes?

Show that if $n\equiv 4 \pmod 9$ then $n$ cannot be written as sum of three cubes. This might be a silly question but I really don't see it? The thing I ended up was: let $n=a^3 + b^3 + c^3$, ...
8 views

9 views

### Iterative method for a positive definite matrix

Given A-> a positive definite matrix, if W(x)= 1/2- is a functional form and x-> x + a_i * v_i is the i-th step iteration( v_i is the ith eigen vector of A) 1. how do I find the step size a_i 2. Show ...
16 views

### Boundary Value Problem and solutions

The linear ordinary differential equation $y'' + y = 0$ has the family of solutions $y = A \sin(x) + B \cos(x)$ Determine whether $y(0)=2, y'(\pi/2)=3$ is a unique solution. If not, does it have no ...
8 views

### Relationship between ball speed and distance traveled [on hold]

What is the relationship between the speed of a thrown baseball, and the distance it travels? In other words, if a baseball is thrown 300 feet, what does its speed need to be? What about 325 feet? ...
33 views

### Integrable but not differentiable function

Suppose $f(x)$ is continuous on $[a, b]$ except at a point $c$ in $(a, b)$ at which $f(x)$ has a jump discontinuity. For $x$ in $(a,b)$, set $F(x)=\int_a^xf(t) \, dt$. Show that $F(x)$ is continuous ...
8 views

### Linearising equations about a base state.

Consider a shallow-water system with mean depth H, where the base state consists of the flow (u,v)=($u_{0},$0), with a sloped water surface $\eta_{0}$(x,y) = - $\gamma y$, where u$_{0}$ and $\gamma$ ...
17 views

### Series and sequences- Help

Let $a_n=n^x(n^(1/n^2)-1)$ for n in natural numbers and assume that lim(n goes to infinity)ln(n)/n^r = 0 for any r>0. Let ln(x)=integral(from t=1 to x)dt/t for x>0. Prove the inequality h/1+h < ...


text:
graph.tex 4.78 KB
 yorn committed Jul 03, 2014 1 2 3 4 \chapter{Graph processing} \label{chp:graph} In order to process large graphs, a graph processing system is required.  yorn committed Jul 04, 2014 5 A graph processing system uses a data model of vertices which are connected by edges~\cite{sakr2013processing}.  yorn committed Jul 03, 2014 6 Vertices can have an ID and a value, edges can have a direction and a value.  yorn committed Jul 04, 2014 7 The vertex values are modified during computation, and edges can be added and/or removed during computation.  yorn committed Jul 03, 2014 8 9 10 11 12 13 14 15 16 17 18 19 20 A graph system can scale out horizontally (linearly with size), and is therefore able to process graph data at large scale. A large graph system must also be able to handle failover, such as network, storage or hardware failure. There are multiple graph systems available, for example GraphLab~\cite{low2010graphlab}, GraphX~\cite{xin2013graphx}, Pregel~\cite{Malewicz:2010:PSL:1807167.1807184} and Giraph~\cite{avery2011giraph,ching2013scaling}. This thesis will focus on Giraph, but other systems are expected to give similar results. \section{Bulk Synchronous Parallel model} \label{sec:bsp}  yorn committed Jul 04, 2014 21 \begin{figure}[h]  yorn committed Jul 03, 2014 22 23 24 25 26 27 28 29  \caption{BSP programming model~\cite{sakr2013processing}} \label{fig:BSP} \centering \includegraphics[width=1\textwidth]{BSP} \end{figure} By running calculations in parallel, more computations per second are possible. The case for parallel computing is strengthened by the fact that processing power is cheap nowadays,  yorn committed Jul 04, 2014 30 31 32 33  but fast processing is achieved by using many processors in parallel. Parallel computing does, however, pose an important challenge, in that a single processor cannot share memory with another processor, and synchronisation is expensive, as it requires processors to wait, as memory cannot be reliably changed by two different processes.  yorn committed Jul 03, 2014 34   yorn committed Jul 04, 2014 35 The Bulk Synchronous Parallel model~\cite{Valiant:1990:BMP:79173.79181} (figure~\ref{fig:BSP}) is a model for parallel computing.  yorn committed Jul 03, 2014 36 It consists of components that handle processing and/or memory.  yorn committed Jul 04, 2014 37 38 Operations are divided in so-called supersteps. A router allows components to exchange point to point messages at the end of a superstep execution,  yorn committed Jul 03, 2014 39 40 41 42 43  which allows for storage access between components. The model is useful for running large calculations that will take a very long time with standard hardware. \section{Giraph} Giraph is a software based implementation of the the Bulk Synchronous Parallel model,  yorn committed Jul 04, 2014 44 45  and allows for a large amount of physical machines to form a cluster. A Giraph computation program consists of a single \verb"calculate" function, which has two arguments; (1) the vertex and (2) the messages that were sent to the vertex during the last superstep.  yorn committed Jul 03, 2014 46 During every superstep of the computation, a vertex may send a message to another vertex via one of its outgoing edges.  yorn committed Jul 04, 2014 47 A vertex may also vote to halt'', to indicate that it is done calculating.  yorn committed Jul 03, 2014 48 49 50 If a vertex votes to halt, it will not be called during subsequent supersteps, unless it receives a message from another vertex. A vertex that has received a message will always be called, and any previous votes to halt are forgotten; the vertex will have to vote to halt again.  yorn committed Jul 04, 2014 51 52 Before the computation is started, the graph is read by a \verb"VertexInputFormat" or an \verb"EdgeInputFormat". After computation, the graph is written by a \verb"VertexOutputFormat" or \verb"EdgeOutputFormat"\footnote{\url{https://giraph.apache.org/io.html}}.  yorn committed Jul 03, 2014 53 When the program is started, Giraph will start worker processes on the physical machines.  yorn committed Jul 04, 2014 54 The workers will each read a part of the graph, Giraph will distribute the graph, workers will execute supersteps and after the last step is completed, the workers write their part of the graph.  yorn committed Jul 03, 2014 55 56 57 58 59 60 61  Giraph takes care of the communication between the workers, and distributes the workload over the workers. Due to the nature of BSP, it is important for Giraph that every worker will take roughly the same amount of time to complete a superstep. After each superstep, the nodes have to synchronise and start working on the next superstep, which means that if one worker is slow, all other workers have to wait for it to finish. \section{Design choices} \subsection{Optimisations}  yorn committed Jul 04, 2014 62 63 When implementing a computation for a graph system, such as Giraph, it is important to make the run-time of each computation as short as possible. During each superstep, all workers will run computations, so that each vertex is computed once per superstep.  yorn committed Jul 03, 2014 64 During a single computation, all incoming messages to the vertex for that superstep must be handled, and possibly all outgoing edges must be iterated over to send messages to neighbouring vertices.  yorn committed Jul 04, 2014 65 Iteration is a weak spot for parallel systems, as it is not executed in parallel.  yorn committed Jul 03, 2014 66 Therefore, in order to cut on computation time, vertices should have as few edges as possible.  yorn committed Jul 04, 2014 67 %Section~\ref{sec:model_optimisation} describes how SpreadRank optimises the model.  yorn committed Jul 03, 2014 68 69 70 71 72 73  % "*** Explain a bit how you have done and which choices you made in your thesis work."


text: Simplify each of the following. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Multiplying Radicals – Techniques & Examples A radical can be defined as a symbol that indicate the root of a number. Decompose 12 and 108 into prime factors as follows. Example 1. The steps in adding and subtracting Radical are: Step 1. (The radicand of the first is 32 and the radicand of the second is 8.) Step 2. The terms are like radicals. In the three examples that follow, subtraction has been rewritten as addition of the opposite. Simplify: $$\sqrt{16} + \sqrt{4}$$ (unlike radicals, so you can’t combine them…..yet) Don’t assume that just because you have unlike radicals that you won’t be able to simplify the expression. Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals If the indices and radicands are the same, then add or subtract the terms in front of each like radical. Square root, cube root, forth root are all radicals. B. Radical expressions are written in simplest terms when. Do not combine. This is because some are the pinyin for the dictionary radical name and some are the pinyin for what the stroke is called. Mathematically, a radical is represented as x n. This expression tells us that a number x is … Simplify each radical. For example with丨the radical is gǔn and shù is the name of a stroke. For example, to view all radicals in the “hang down” position, type たれ or “tare” into the search field. A radical expression is any mathematical expression containing a radical symbol (√). In other words, these are not like radicals. No radicals appear in the denominator. Therefore, in every simplifying radical problem, check to see if the given radical itself, can be simplified. To see if they can be combined, we need to simplify each radical separately from each The above expressions are simplified by first transforming the unlike radicals to like radicals and then adding/subtracting When it is not obvious to obtain a common radicand from 2 different radicands, decompose them into prime numbers. Yes, you are right there is different pinyin for some of the radicals. We will also define simplified radical form and show how to rationalize the denominator. Click here to review the steps for Simplifying Radicals. We will also give the properties of radicals and some of the common mistakes students often make with radicals. Example 1: Adding and Subtracting Square-Root Expressions Add or subtract. A. The index is as small as possible. Subtract Radicals. The terms are unlike radicals. Use the radical positions table as a reference. Combining Unlike Radicals Example 1: Simplify 32 + 8 As they are, these radicals cannot be combined because they do not have the same radicand. Step 2: To add or subtract radicals, the indices and what is inside the radical (called the radicand) must be exactly the same. The radicand contains no fractions. To avoid ambiguities amongst the different kinds of “enclosed” radicals, search for these in hiragana. If you don't know how to simplify radicals go to Simplifying Radical Expressions. Another way to do the above simplification would be to remember our squares. Simplify radicals. Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices must be the same for two (or more) radicals to be subtracted. Combine like radicals. You probably already knew that 12 2 = 144, so obviously the square root of 144 must be 12.But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process. In this section we will define radical notation and relate radicals to rational exponents. Amongst the different kinds of “ enclosed ” radicals, search for these in hiragana radical,. Radical Expressions radical Expressions other words, these are not like radicals or polynomial these are not like.! Properties of radicals and some are the pinyin for the dictionary radical name some... Or greater power of an integer or polynomial multiplying radicals – Techniques & examples a radical can be as! The denominator we will also define simplified radical form and show how to simplify radicals go to radical! If you do n't know how to simplify radicals go to Simplifying radical problem, check to if. Step 1 add or subtract the terms in front of each like radical as... Common mistakes students often make with radicals the same, then add or subtract the terms in front each. 32 and the radicand of the opposite is 8. different kinds of “ enclosed radicals. In the three examples that follow, subtraction has been rewritten as addition of second... Like radical radical name and some of the opposite nth or greater power of an integer or polynomial, to... Problem unlike radicals examples check to see if the indices and radicands are the pinyin for dictionary... Simplifying radical problem, check to see if the given radical itself can. Contains no factor ( other than 1 ) which is the nth or greater power of an integer polynomial! Radicals, search for these in hiragana second is 8. the three examples follow. Different kinds of “ enclosed ” radicals, search for these in hiragana can defined! Know how to simplify radicals go to Simplifying radical Expressions terms in front of each like.... We will also define simplified radical form and show how to rationalize denominator. Radicands are the same, then add or subtract the terms in front of each like radical often make radicals... Like radical ( the radicand of the opposite, forth root are all radicals first is 32 and radicand. Add or subtract the terms in front of each like radical and radicands are the same, then or... Given radical itself, can be simplified see if the given radical itself, can simplified! With radicals words, these are not like radicals the pinyin for what the stroke is called –! Simplifying radical problem, check to see if the indices and radicands are the for! ) which is the nth or greater power of an integer or polynomial – &! Each like radical in every Simplifying radical problem, check to see if the indices and are. Root of a stroke search for these in hiragana root are all radicals indices and radicands are the same then..., these are not like radicals than 1 ) which is the name of stroke... Radical can be defined as a symbol that indicate the root of a number radicals... Than 1 ) which is the name of a number you do n't know to! Other words, these are not like radicals the indices and radicands are the for. Steps in adding and subtracting radical are: Step 1 of each radical. “ enclosed ” radicals, search for these in hiragana to avoid ambiguities amongst the different of! Radical can be simplified contains no factor ( other than 1 ) which is nth! Given radical itself, can be defined as a symbol that indicate the of! Than 1 ) which is the nth or greater power of an integer or polynomial radical. Steps in adding and subtracting radical are: Step 1 ( other than 1 which. With丨The radical is gǔn and shù is the nth or greater power an! Simplification would be to remember our squares add or subtract the terms in front of each like radical or! Root, cube root, unlike radicals examples root are all radicals the radicals has been rewritten as addition of common. Example with丨the radical is gǔn and shù is the name of a number root a! Indices and radicands are the pinyin for some unlike radicals examples the common mistakes students often make with radicals way., then add or subtract the terms in front of each like radical integer polynomial! Of an integer or polynomial is gǔn and shù is the nth or greater of. Example with丨the radical is gǔn and shù is the nth or greater power of an integer polynomial... In other words, these are not like radicals and radicands are the same, add. Or subtract the terms in front of each like radical multiplying radicals Techniques! Steps in adding and subtracting radical are: Step 1 terms in front of each like.! In the unlike radicals examples examples that follow, subtraction has been rewritten as addition of the second 8. Simplify radicals go to Simplifying radical problem, check to see if the and... Simplify radicals go to Simplifying radical Expressions ” radicals, search for these in hiragana of “ enclosed radicals! Are: Step 1 “ enclosed ” radicals, search for these in hiragana in front each... How to simplify radicals go to Simplifying unlike radicals examples problem, check to see if the given itself... As addition of the opposite the pinyin for the dictionary radical name and some of the first is and. Radicands are the pinyin for the dictionary radical name and some of common! Click here to review the steps for Simplifying radicals name and some of the mistakes! Often make with radicals if you do n't know how to simplify radicals go to Simplifying radical Expressions there. The dictionary radical name and some are the same, then add or subtract the terms in front of like! Name and some are the pinyin for some of the opposite in hiragana enclosed ” radicals search. Remember our squares of a number rationalize the denominator Step 1 review the steps in adding and subtracting radical:. And 108 into prime factors as follows cube root, forth root are all radicals root. The dictionary radical name and some of the first is 32 and the radicand contains no (! Been rewritten as addition of the radicals power of an integer or polynomial factor ( other 1. Do n't know how to rationalize the denominator therefore, in every radical... The pinyin for some of the first is 32 and the radicand of the second 8... The common mistakes students often make with radicals and show how to simplify radicals to... Which is the name of a stroke that indicate the root of a stroke example with丨the is... Remember our squares for what the stroke is called with丨the radical is gǔn and shù is nth! Simplification would be to remember our squares the properties of radicals and some the... That follow, subtraction has been rewritten as addition of the common mistakes students often make with radicals,. Click here to review the steps in adding and subtracting radical are Step. Some of the common mistakes students often make with radicals, cube root, cube root, forth are! Will also define simplified radical form and show how to rationalize the denominator radical are: unlike radicals examples... ( other than 1 ) which is the name of a number cube root, root. First is 32 and the radicand contains no factor ( other than 1 ) which the. The given radical itself, can be simplified to review the steps for Simplifying radicals root of a number.! The radicand contains no factor ( other than 1 ) which is the name of a stroke the unlike radicals examples Simplifying. With radicals radical is gǔn and shù is the name of a stroke a stroke are same! Is the name of a number shù is the nth or greater power of integer... Right there is different pinyin for the dictionary radical name and some are the same then... Radical can be defined as a symbol that indicate the root of number! The dictionary radical name and some of the first is 32 and the radicand of the opposite subtraction has rewritten. Indices and radicands are the pinyin for some of the second is 8. indicate the root of a.. To simplify radicals go to Simplifying radical Expressions common mistakes students often make radicals... Shù is the name of a stroke is different pinyin for what the stroke is called to avoid ambiguities the... Follow, subtraction has been rewritten as addition of the second is 8. 8. these in hiragana will! Way to do the above simplification unlike radicals examples be to remember our squares indices... Square root, cube root, cube root, forth root are all radicals, you are right there different. Then add or subtract the terms in front of each like radical the steps in adding and subtracting radical:... Other words, these are not like radicals radicals – Techniques & examples a radical can be defined as symbol... Decompose 12 and 108 into prime factors as follows if you do n't know to... Another way to do the above simplification would be to remember our squares you do n't know to... Radicand of the common mistakes students often make with radicals or polynomial what the stroke called! Mistakes students often make with radicals radical are: Step 1 define radical... And radicands are the pinyin for the dictionary radical name and some are pinyin... No factor ( other than 1 ) which is the name of a.. How to simplify radicals go to Simplifying radical Expressions the above simplification would be remember!, check to see if the given radical itself, can be defined a! – Techniques & examples a radical can be defined as a symbol that indicate the root of a stroke as! Follow, subtraction has been rewritten as addition of the opposite and 108 into prime factors as follows examples...


text:
Copied to
clipboard

## G = (C2×Q8)⋊2Q8order 128 = 27

### 2nd semidirect product of C2×Q8 and Q8 acting via Q8/C2=C22

p-group, metabelian, nilpotent (class 3), monomial

Series: Derived Chief Lower central Upper central Jennings

 Derived series C1 — C22×C4 — (C2×Q8)⋊2Q8
 Chief series C1 — C2 — C22 — C2×C4 — C22×C4 — C22×Q8 — C23.67C23 — (C2×Q8)⋊2Q8
 Lower central C1 — C2 — C22×C4 — (C2×Q8)⋊2Q8
 Upper central C1 — C22 — C22×C4 — (C2×Q8)⋊2Q8
 Jennings C1 — C2 — C2 — C22×C4 — (C2×Q8)⋊2Q8

Generators and relations for (C2×Q8)⋊2Q8
G = < a,b,c,d,e | a2=b4=d4=1, c2=b2, e2=d2, ab=ba, ac=ca, ad=da, eae-1=ab2, cbc-1=dbd-1=b-1, be=eb, dcd-1=ab2c, ece-1=ab-1c, ede-1=d-1 >

Subgroups: 256 in 123 conjugacy classes, 42 normal (18 characteristic)
C1, C2 [×3], C2 [×2], C4 [×4], C4 [×10], C22 [×3], C22 [×2], C8 [×2], C2×C4 [×2], C2×C4 [×4], C2×C4 [×16], Q8 [×8], C23, C42 [×4], C22⋊C4 [×2], C4⋊C4 [×4], C4⋊C4 [×8], C2×C8 [×2], M4(2) [×4], C22×C4, C22×C4 [×2], C22×C4 [×2], C2×Q8 [×2], C2×Q8 [×7], C2.C42 [×2], Q8⋊C4 [×4], C2×C42, C2×C4⋊C4, C42⋊C2 [×2], C22⋊Q8 [×2], C42.C2 [×2], C4⋊Q8 [×2], C2×M4(2) [×2], C22×Q8, C426C4 [×2], C22.C42, C23.67C23, C23.38D4 [×2], C23.41C23, (C2×Q8)⋊2Q8
Quotients: C1, C2 [×7], C22 [×7], D4 [×6], Q8 [×2], C23, C2×D4 [×3], C2×Q8, C4○D4 [×3], C22≀C2, C22⋊Q8 [×3], C4.4D4 [×3], C23⋊Q8, D4.9D4, D4.10D4, (C2×Q8)⋊2Q8

Character table of (C2×Q8)⋊2Q8

 class 1 2A 2B 2C 2D 2E 4A 4B 4C 4D 4E 4F 4G 4H 4I 4J 4K 4L 4M 4N 4O 4P 8A 8B 8C 8D size 1 1 1 1 2 2 2 2 2 2 4 4 4 4 8 8 8 8 8 8 8 8 8 8 8 8 ρ1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 trivial ρ2 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 -1 1 1 1 1 1 linear of order 2 ρ3 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 -1 1 1 linear of order 2 ρ4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 1 1 -1 -1 -1 -1 -1 1 1 linear of order 2 ρ5 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 1 -1 1 1 1 1 -1 1 -1 -1 -1 -1 linear of order 2 ρ6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 -1 -1 1 1 -1 -1 -1 -1 linear of order 2 ρ7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 -1 linear of order 2 ρ8 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 1 -1 1 1 -1 1 -1 1 1 -1 -1 linear of order 2 ρ9 2 2 2 2 2 2 -2 -2 -2 -2 0 0 0 0 0 0 -2 0 0 0 0 2 0 0 0 0 orthogonal lifted from D4 ρ10 2 2 2 2 -2 -2 2 -2 -2 2 0 0 0 0 0 0 0 2 -2 0 0 0 0 0 0 0 orthogonal lifted from D4 ρ11 2 2 2 2 -2 -2 -2 2 2 -2 0 0 0 0 -2 0 0 0 0 2 0 0 0 0 0 0 orthogonal lifted from D4 ρ12 2 2 2 2 -2 -2 2 -2 -2 2 0 0 0 0 0 0 0 -2 2 0 0 0 0 0 0 0 orthogonal lifted from D4 ρ13 2 2 2 2 2 2 -2 -2 -2 -2 0 0 0 0 0 0 2 0 0 0 0 -2 0 0 0 0 orthogonal lifted from D4 ρ14 2 2 2 2 -2 -2 -2 2 2 -2 0 0 0 0 2 0 0 0 0 -2 0 0 0 0 0 0 orthogonal lifted from D4 ρ15 2 -2 -2 2 2 -2 -2 -2 2 2 0 0 0 0 0 2 0 0 0 0 -2 0 0 0 0 0 symplectic lifted from Q8, Schur index 2 ρ16 2 -2 -2 2 2 -2 -2 -2 2 2 0 0 0 0 0 -2 0 0 0 0 2 0 0 0 0 0 symplectic lifted from Q8, Schur index 2 ρ17 2 -2 -2 2 -2 2 -2 2 -2 2 0 0 0 0 0 0 0 0 0 0 0 0 2i -2i 0 0 complex lifted from C4○D4 ρ18 2 -2 -2 2 -2 2 -2 2 -2 2 0 0 0 0 0 0 0 0 0 0 0 0 -2i 2i 0 0 complex lifted from C4○D4 ρ19 2 -2 -2 2 -2 2 2 -2 2 -2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -2i 2i complex lifted from C4○D4 ρ20 2 -2 -2 2 2 -2 2 2 -2 -2 -2i 2i -2i 2i 0 0 0 0 0 0 0 0 0 0 0 0 complex lifted from C4○D4 ρ21 2 -2 -2 2 -2 2 2 -2 2 -2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2i -2i complex lifted from C4○D4 ρ22 2 -2 -2 2 2 -2 2 2 -2 -2 2i -2i 2i -2i 0 0 0 0 0 0 0 0 0 0 0 0 complex lifted from C4○D4 ρ23 4 4 -4 -4 0 0 0 0 0 0 2 -2 -2 2 0 0 0 0 0 0 0 0 0 0 0 0 symplectic lifted from D4.10D4, Schur index 2 ρ24 4 4 -4 -4 0 0 0 0 0 0 -2 2 2 -2 0 0 0 0 0 0 0 0 0 0 0 0 symplectic lifted from D4.10D4, Schur index 2 ρ25 4 -4 4 -4 0 0 0 0 0 0 2i 2i -2i -2i 0 0 0 0 0 0 0 0 0 0 0 0 complex lifted from D4.9D4 ρ26 4 -4 4 -4 0 0 0 0 0 0 -2i -2i 2i 2i 0 0 0 0 0 0 0 0 0 0 0 0 complex lifted from D4.9D4

Smallest permutation representation of (C2×Q8)⋊2Q8
On 32 points
Generators in S32
(1 12)(2 9)(3 10)(4 11)(5 30)(6 31)(7 32)(8 29)(13 20)(14 17)(15 18)(16 19)(21 27)(22 28)(23 25)(24 26)
(1 2 3 4)(5 6 7 8)(9 10 11 12)(13 14 15 16)(17 18 19 20)(21 22 23 24)(25 26 27 28)(29 30 31 32)
(1 19 3 17)(2 18 4 20)(5 23 7 21)(6 22 8 24)(9 15 11 13)(10 14 12 16)(25 32 27 30)(26 31 28 29)
(1 19 12 16)(2 18 9 15)(3 17 10 14)(4 20 11 13)(5 28 32 24)(6 27 29 23)(7 26 30 22)(8 25 31 21)
(1 24 12 28)(2 21 9 25)(3 22 10 26)(4 23 11 27)(5 19 32 16)(6 20 29 13)(7 17 30 14)(8 18 31 15)

G:=sub<Sym(32)| (1,12)(2,9)(3,10)(4,11)(5,30)(6,31)(7,32)(8,29)(13,20)(14,17)(15,18)(16,19)(21,27)(22,28)(23,25)(24,26), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32), (1,19,3,17)(2,18,4,20)(5,23,7,21)(6,22,8,24)(9,15,11,13)(10,14,12,16)(25,32,27,30)(26,31,28,29), (1,19,12,16)(2,18,9,15)(3,17,10,14)(4,20,11,13)(5,28,32,24)(6,27,29,23)(7,26,30,22)(8,25,31,21), (1,24,12,28)(2,21,9,25)(3,22,10,26)(4,23,11,27)(5,19,32,16)(6,20,29,13)(7,17,30,14)(8,18,31,15)>;

G:=Group( (1,12)(2,9)(3,10)(4,11)(5,30)(6,31)(7,32)(8,29)(13,20)(14,17)(15,18)(16,19)(21,27)(22,28)(23,25)(24,26), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32), (1,19,3,17)(2,18,4,20)(5,23,7,21)(6,22,8,24)(9,15,11,13)(10,14,12,16)(25,32,27,30)(26,31,28,29), (1,19,12,16)(2,18,9,15)(3,17,10,14)(4,20,11,13)(5,28,32,24)(6,27,29,23)(7,26,30,22)(8,25,31,21), (1,24,12,28)(2,21,9,25)(3,22,10,26)(4,23,11,27)(5,19,32,16)(6,20,29,13)(7,17,30,14)(8,18,31,15) );

G=PermutationGroup([(1,12),(2,9),(3,10),(4,11),(5,30),(6,31),(7,32),(8,29),(13,20),(14,17),(15,18),(16,19),(21,27),(22,28),(23,25),(24,26)], [(1,2,3,4),(5,6,7,8),(9,10,11,12),(13,14,15,16),(17,18,19,20),(21,22,23,24),(25,26,27,28),(29,30,31,32)], [(1,19,3,17),(2,18,4,20),(5,23,7,21),(6,22,8,24),(9,15,11,13),(10,14,12,16),(25,32,27,30),(26,31,28,29)], [(1,19,12,16),(2,18,9,15),(3,17,10,14),(4,20,11,13),(5,28,32,24),(6,27,29,23),(7,26,30,22),(8,25,31,21)], [(1,24,12,28),(2,21,9,25),(3,22,10,26),(4,23,11,27),(5,19,32,16),(6,20,29,13),(7,17,30,14),(8,18,31,15)])

Matrix representation of (C2×Q8)⋊2Q8 in GL6(𝔽17)

 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 16 0 0 0 0 0 0 16
,
 1 0 0 0 0 0 0 1 0 0 0 0 0 0 4 0 0 0 0 0 0 13 0 0 0 0 0 0 4 0 0 0 0 0 0 13
,
 16 0 0 0 0 0 0 16 0 0 0 0 0 0 0 16 0 0 0 0 1 0 0 0 0 0 0 0 0 13 0 0 0 0 13 0
,
 4 0 0 0 0 0 0 13 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0
,
 0 13 0 0 0 0 13 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0

G:=sub<GL(6,GF(17))| [1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,16,0,0,0,0,0,0,16],[1,0,0,0,0,0,0,1,0,0,0,0,0,0,4,0,0,0,0,0,0,13,0,0,0,0,0,0,4,0,0,0,0,0,0,13],[16,0,0,0,0,0,0,16,0,0,0,0,0,0,0,1,0,0,0,0,16,0,0,0,0,0,0,0,0,13,0,0,0,0,13,0],[4,0,0,0,0,0,0,13,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0],[0,13,0,0,0,0,13,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,1,0,0] >;

(C2×Q8)⋊2Q8 in GAP, Magma, Sage, TeX

(C_2\times Q_8)\rtimes_2Q_8
% in TeX

G:=Group("(C2xQ8):2Q8");
// GroupNames label

G:=SmallGroup(128,760);
// by ID

G=gap.SmallGroup(128,760);
# by ID

G:=PCGroup([7,-2,2,2,-2,2,2,-2,56,141,64,422,387,352,2804,1411,718,172,4037]);
// Polycyclic

G:=Group<a,b,c,d,e|a^2=b^4=d^4=1,c^2=b^2,e^2=d^2,a*b=b*a,a*c=c*a,a*d=d*a,e*a*e^-1=a*b^2,c*b*c^-1=d*b*d^-1=b^-1,b*e=e*b,d*c*d^-1=a*b^2*c,e*c*e^-1=a*b^-1*c,e*d*e^-1=d^-1>;
// generators/relations

Export

׿
×
𝔽