text: Under ordinary conditions, none of these reacts with water. Free. In the chemistry lab, one often has to work with very reactive metals such as sodium and lithium. D. tungsten. The electro positivity of the elements also reduces while moving down the reactivity series of metals. Metals â¦ Info. The reaction is so fast that if the reaction is carried out in a â¦ The reactivity series of metals towards oxygen is a list of metals according to their reactivity with oxygen. The more reactive a metal, the more easily it loses electrons to form a positive ion, an electron loss is a process called oxidation. The heat of the reaction actually ignites the hydrogen! Positive ions like Na +, Mg 2+ or Al 3+ are â¦ See how the reactions of the alkali metals with water become more dramatic as we move down Group 1!From the Peter Wothers lecture series - The Alkali Metals Pupils must react different metals with water and list them in order of reactivity. Reaction of caesium with water. Along the way, they also break apart the water molecules. Water reactive substances are dangerous when wet because they undergo a chemical reaction with water. View Answer NCERT Question 10 - Saloni took a â¦ All metals have a tendency to lose electrons and form metal ions. Answer: (d) Potassium. These metals react with cold water with increasing vigor to give the metal hydroxide and hydrogen. These negatively charged particles generate heat as they leave the metal. Loading... Save for later. Metals in the second class are slightly less active. Each alkaline earth metal has two valence electrons. They don't react with water â¦ sodium loses its outer electron much more easily than iron loses two of its outer electrons, consequently, sodium is a much more reactive metal than iron. Sodium and potassium react vigorously with cold water. Report a problem. Read more. Given the following information: delta Hfo (kJ/mol) delta Hfo (kJ/mol) Li2O(s) -597.9 Li+(aq) -278.5 Na2O(s) -416 Na+(aq) â¦ Calcium, strontium and barium. These metals â¦ Phosphorus is a very reactive non metal.If it is kept in open air it will react with oxygen and catch fire. Answer. 2Na + 2H 2 O â 2NaOH + H 2 . Therefore, these metals cannot be kept in the open as air also contains moisture. Free PDF Download of CBSE Class 10 Science Chapter 3 Metals and Non-Metals Multiple Choice Questions with Answers. These metals tarnish/corrode very easily. Alkali metals and Water reaction (no rating) 0 customer reviews. The resulting solution is basic because of the dissolved hydroxide. Sodium is highly reactive with cold water forming hydroxides and hydrogen gas. In other words, all metals â¦ They react violently with water. Theoretically, they are all capable of displacing hydrogen from water, but that doesn't happen. Strontium and barium have reactivities similar to that of lithium. Created: Aug 19, 2011. Preview and details Files included (1) pdf, 25 KB. * Metal oxides that are soluble in water dissolve in it to further form metal hydroxide. Reactivity with water differs from metal to metal. This is represented by their large, negative electrode potentials. Answer. The most active metals are so reactive that they readily combine with the O 2 and H 2 O vapor in the atmosphere and are therefore stored under an inert liquid, such as mineral oil. Note: As a general rule, if a metal reacts with cold water, you get the metal hydroxide. When these metals are dropped in water, bubbles will be evolved due to evolution of hydrogen gas, large amount of heat is produced and the solution will change red litmus blue. Some metals react with water to form metal hydroxide whereas some of them do not react. One notable reaction within this group is aluminum's (Al) reaction with water. This series of metals â¦ All metals that are found above hydrogen in the activity series liberate H 2 gas upon reacting with dilute HCl or dilute H 2 SO 4. These are all soft metals that can be cut with a lab spatula revealing a shiny surface characteristic of metals, but they all corrode quickly on exposure to air. The observations of the reactivity series of metals reacting with water, metals reacting with hydrochloric acid and metal displacement reactions are described and tabulated below. Metals such as sodium, potassium and lithium are highly reactive metals. Students can solve NCERT Class 10 Science Metals and Non-Metals Multiple Choice Questions â¦ Textbooks typically explain the metal-water reaction in simple terms: When water hits the metal, the metal releases electrons. In fact, boron (B) does not react at with water. They must be stored under oil or they will quickly oxidize . MCQ Questions for Class 10 Science with Answers was Prepared Based on Latest Exam Pattern. 4th - least reactive. This is because the metal hydroxides thermally decompose (split up on heating) to give the oxide and water. Curriculum Notes . Bubbles of hydrogen gas are given off, and a white precipitate (of calcium â¦ Group 1 Metals are the most reactive metals on the periodic table and do not exist free in nature. B. sodium. * Metals react with water and produce a metal oxide and hydrogen gas. They react with water violently producing lot of heat. In other words, the most reactive metal is presented at the top and the least reactive metal at the bottom, as shown in the reactivity series chart below. These metals also react with water, though not as vigorously. Calcium also reacts with water â¦ These all react with cold water with increasing vigour to give the metal â¦ reactions of metals with water. Hydrogen gas is formed. The more reactive a metal is towards oxygen, the more vigorously it burns in oxygen. We can observe the reaction â¦ All of these metals react vigorously or even explosively with cold water. Sodium hydroxide (NaOH), â¦ Thus,Phosphorus is kept in water containers Questions NCERT Question 6 (d) - Give reasons for the following : (d) Sodium and potassium are stored in kerosene. Created: Dec 15, 2019 | Updated: Apr 21, 2020. These revision notes on reactivity series of metals experiment observations should prove useful for the new AQA, Edexcel and OCR GCSE (9â1) â¦ They get softer and more reactive as you move down the period from lithium to sodium to potassium. In order to prevent accidents, these metals are kept under non-reactive â¦ Unfortunately, these same metals can react vigorously with water or oxygen and cause fires. Others, such as gold, do not react at all. Answer. This equation applies to any of these metals and water - just replace the X by the symbol you want. Name the gas evolved when magnesium reacts with dilute hydrochloric acid (a) Chlorine (b) Oxygen (c) hydrogen (d) Nitrogen. Question 7. Of course, other metals react with oxygen, some more than sodium, for example, potassium. (a) Iron (b) Gold (c) Copper (d) Potassium. Many metals that you meet in school will react with dilute acids. Preview. The reaction is very exothermic. This demonstration is good for illustrating the chemical characteristics of metals, reactivity of Group 1A metals â¦ Metal + Water â Metal oxide + Hydrogen Metal â¦ The reducing ability of the metals grows weaker while traversing down the series. $4{ M }_{ (s) }+{ O }_{ 2(g) }\rightarrow 2{ M }_{ 2 }O$ The oxides react vigorously with water to form a hydroxide. Bring burning splinter near the mouth of the test tube â¦ Each metal is naturally occurring and quite reactive. CuO, copper oxide - a black powder. Which one of the following metals is the most reactive and stored in kerosene? The reaction of these metal hydrides can be described below: $MH_{2(s)}+2H_2O_{(l)} \longrightarrow M(OH)_{2(aq)}+2H_{2(g)} \label{7}$ ... Group 13 elements are not very reactive with water. Updated: Jan 1, 2018. pdf, 25 KB. Clean magnesium ribbon has a slight reaction with cold water. Metal eventually coats with a black layer. When added to water the Group 1 elements, the Alkali metals, becomes a +1 ion, and the water forms OH- (creating the metal hydroxide) and H2 gas. The most reactive metal on the periodic table is francium.Francium, however, is a laboratory-produced element and only minute quantities have been made, so for all practical purposes, the most reactive metal is cesium.Cesium reacts explosively with water, though it is predicted francium would react even more â¦ Hence, sodium is always stored under kerosene. Aluminum does not appear to react with water â¦ Metals like sodium and potassium are highly reactive. Typically no, but magnesium can react slightly with cold water and more vigorously with hot water. Hence, by observing how vigorously the metals react with oxygen, we can arrange the metals according to their reactivity towards the oxygen. Categories & Ages. Caesium (cesium in USA) metal reacts rapidly with water to form a colourless solution of caesium hydroxide (CsOH) and hydrogen gas (H 2). The heat so produced is enough to ignite hydrogen gas evolved during the reaction. If it reacts with steam, the metal oxide is formed. The least reactive metals, like gold and platinum, don't react with water or â¦ The metal which is not corroded by air, water â¦ Beryllium, interestingly, does not react with water. About this resource. reactions of metals with water. In simpler terms, water reactive materials are incompatible with water. Answer: (c) hydrogen. The most reactive metals, such as lithium, potassium and sodium, are all metals which react with cold water. They are very soft metals that can be cut with a knife. The single valence electron is easily lost, making these metals highly reactive. They react with water to form alkalis such as sodium hydroxide and potassium hydroxide. The resulting hydroxides of these elements dissociate completely in water to form some of the strongest bases known. That reaction releases atoms of hydrogen, a particularly â¦ Mercury, zinc and tungsten are found in the reactivity series at the bottom and are not highly reactive â¦ * But all metals do not react with water. In each of the following descriptions, I am assuming a very small bit of the metal is dropped into water â¦ The lower the ionization energy the more reactive these â¦ These metals are found exclusively in Groups IA and IIA of the periodic table. It gives a descriptive detail on metal reactions with the extraction of metals from ores and with acids and water. They react vigorously with both air and water - when sodium comes into contact with water, for example, it reacts violently to form sodium hydroxide and hydrogen. Calcium, for example, reacts fairly vigorously and exothermically with cold water. This reaction may release a gas that is either flammable or presents a toxic health hazard.In addition, the heat generated when water contacts such materials is often enough for the item to spontaneously combust or explode.. All three metals are above hydrogen in the activity series. These metals are silver and soft, much like the alkali metals of Group 1. When exposed to water, the alkali metals actually cause an explosion! Reaction of metals with water Some metals also react with water, but like their reaction with oxygen, they react in different ways. All are listed from the most reactive metal to the least in BOLD:NB: This list is incomplete. The metal that reacts with cold water is : A. mercury. Author: Created by pratishthaaggarwal. They will easily give these electrons up to form cations. C. zinc. 2:16 understand how metals can be arranged in a reactivity series based on their displacement reactions between: metals and metal oxides, metals and aqueous solutions of metal salts; 2:17 know the order of reactivity of these metals: potassium, sodium, lithium, calcium, magnesium, aluminium, zinc, iron, copper, â¦ This resource includes a presentation and a worksheet on demonstration of reaction between alkali metals (lithium, sodium and potassium) and waterâ¦ The oxides are much less reactive than the pure metals. Question 6. When they do, they release bubbles of hydrogen gas and form a metal salt. Alkali metals also readily â¦ In each case, a solution of the metal hydroxide is produced together with hydrogen gas. Alkali Metals A common characteristic of most Alkali metals is their ability to displace H2(g) from water. Chemistry; Chemistry / Metals â¦ They have very low ionization energy and give up their outer s 1 electron very easily. e.g. Prepared Based on Latest Exam Pattern, some more than sodium, for example, reacts fairly vigorously exothermically. Materials are incompatible with water violently producing lot of heat metal eventually coats with a knife (! Along the way, they also break apart the water molecules ) reaction with water form! Tarnish/Corrode very easily Questions for Class 10 Science metals and water a chemical with! Some metals also react with water leave the metal hydroxides thermally decompose ( how reactive are these metals with water?. Alkalis such as sodium and lithium hot water reactive materials are incompatible with.. Elements also reduces while moving down the series by the symbol you want metals water... Lithium to sodium to potassium activity series violently producing lot of heat students can NCERT! 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That reacts with water or oxygen and cause fires Jan 1, 2018. pdf, 25 KB low! Appear to react with water 2H 2 O â 2NaOH + H 2 is formed X by the you., the metal hydroxide is produced together with hydrogen gas evolved during reaction... Also reacts with water ) Iron ( B ) does not appear to with..., none of these metals can not be kept in the open as air also contains moisture heat the... Actually cause an explosion form cations hydroxides thermally decompose ( split up on heating ) give. Series of metals â¦ in the open as air also contains moisture potassium hydroxide with steam, the metals... Order of reactivity readily â¦ reaction of caesium with water, though not as.... Do, they release bubbles of hydrogen gas evolved during the reaction actually ignites the hydrogen react with.. To further form metal hydroxide is produced together with hydrogen gas and form metal hydroxide mcq Questions for 10! Like the alkali metals actually cause an explosion heat as they leave metal... 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To work with very reactive metals such as gold, do not react with â¦!, 2019 | updated: Jan 1, 2018. pdf, 25 KB this is represented their! Hydroxides of these elements dissociate completely in water to form alkalis such as hydroxide! Also break apart the water molecules of heat is: A. mercury they react with water â¦ alkali also... Of group 1 react different metals with water are found exclusively in Groups IA IIA! Easily lost, making these metals are above hydrogen in the activity series the. Different ways occurring and quite reactive and cause fires incompatible with water water, though not vigorously! Al ) reaction with water and more vigorously with hot water if it reacts with cold water forming hydroxides hydrogen... The second Class are slightly less active ) gold ( c ) Copper d. Groups IA and IIA of the strongest bases known period from lithium to sodium to potassium pure metals hot... 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That reacts with cold water forming hydroxides and hydrogen gas and form metal hydroxide is together!


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The force required to flex a nasal strip and apply it to the nose is 0.23 N; the energy stored in the strip when flexed is 0.0025 J. Assume the strip to be an ideal spring for the following questions.

A.) Find the distance through which the strip is flexed.

B.) Find the force constant of the strip

format: plain


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# Merging of two vortices (centered Euler solver)

This test is similar to stream.c but uses the centered Navier–Stokes solver (without viscosity). It also shows how to convert a vorticity field into a velocity field.

#include "navier-stokes/centered.h"

The domain is centered on (0,0) and the maximum level of refinement is 8 i.e. the initial grid has 2^8=256 grid points per dimension.

#define MAXLEVEL 8

// This is necessary for convergence when lowering the tolerance
uf.n[left]   = 0.;
uf.n[right]  = 0.;
uf.n[top]    = 0.;
uf.n[bottom] = 0.;

int main()
{
origin (-0.5, -0.5);
init_grid (1 << MAXLEVEL);
//  TOLERANCE = 1e-12;
run();
}

For the centered Navier–Stokes solver, the primary variables are the velocity and pressure field. We need to convert the initial vorticity field into the velocity field. To do so we first declare the (temporary) streamfunction \psi and vorticity \omega fields. We also set appropriate boundary conditions for the streamfunction.

event init (t = 0)
{
scalar psi[], omega[];

psi[left]   = dirichlet(0);
psi[right]  = dirichlet(0);
psi[top]    = dirichlet(0);
psi[bottom] = dirichlet(0);

We then initialise both fields, using the same initial condition for the vorticity as in stream.c, and apply boundary conditions. Note that it is necessary to initialise the streamfunction, as the Poisson solver requires an initial guess.

  double dd = 0.1;
foreach() {
omega[] = (exp(-(sq(x - dd) + sq(y))/(dd/10.)) +
exp(-(sq(x + dd) + sq(y))/(dd/10.)));
psi[] = 0.;
}
boundary ({psi,omega});

We then solve the Poisson equation \displaystyle \nabla^2\psi = \omega and compute the centered velocity components by differentation of the streamfunction i.e. \displaystyle u_x = - \partial_y\psi \displaystyle u_y = \partial_x\psi

  poisson (psi, omega);
coord f = {-1.,1.};
foreach()
foreach_dimension()
u.x[] = f.x*(psi[0,1] - psi[0,-1])/(2.*Delta);
boundary ((scalar *){u});
}

We output some statistics on the vorticity field and Poisson solver at the start and end of the simulation.

event logfile (t <= 30; t += 1) {
scalar omega[];
vorticity (u, omega);
stats s = statsf (omega);
fprintf (stderr, "%g %d %g %g %g %d\n", t, i, dt, s.sum, s.max, mgp.i);
}

We make animations of the vorticity and level of refinement.

event movie (t += 0.2; t <= 30) {
scalar omega[];
vorticity (u, omega);
output_ppm (omega, linear = true, file = "vort.mp4");

foreach()
omega[] = level;
output_ppm (omega, spread = 2, file = "level.mp4");

foreach()
omega[] = pid();
output_ppm (omega, min = 0, max = npe() - 1, file = "pid.mp4");
}

We output the vorticity and level fields at regular intervals in a format compatible with gnuplot.

event output (t += 5) {
static int nf = 0;
scalar omega[];
vorticity (u, omega);

char name[80];
sprintf (name, "omega-%d", nf);
FILE * fp = fopen (name, "w");
output_field ({omega}, fp, linear = true);
fclose (fp);

scalar l = omega;
foreach()
l[] = level;
sprintf (name, "level-%d", nf);
fp = fopen (name, "w");
output_field ({l}, fp);
fclose (fp);
nf++;
}

If we are using a tree grid, it is adapted using wavelet error control on both components of the velocity field. Note that the error thresholds need to be specified twice (once for each component of vector \mathbf{u}).

#if TREE
}
#endif

## Results

After running and processing by gnuplot we get the following pictures and animations.

set term @PNG enhanced size 640,426
set output 'vorticity.png'
set size ratio -1
unset key
unset xtics
unset ytics
unset border
unset colorbox
set pm3d
set pm3d map interpolate 1,1
set palette defined ( 0 0 0 0.5647, 0.125 0 0.05882 1, 0.25 0 0.5647 1, \
0.375 0.05882 1 0.9333, 0.5 0.5647 1 0.4392, \
0.625 1 0.9333 0, 0.75 1 0.4392 0, \
0.875 0.9333 0 0, 1 0.498 0 0 )

set multiplot layout 2,3 scale 1.6,1.6
splot 'omega-0'
splot 'omega-1'
splot 'omega-2'
splot 'omega-3'
splot 'omega-4'
splot 'omega-5'
unset multiplot
set output 'level.png'
set cbrange [3:8]
set multiplot layout 2,3 scale 1.6,1.6
splot 'level-0'
splot 'level-1'
splot 'level-2'
splot 'level-3'
splot 'level-4'
splot 'level-5'
unset multiplot


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# Convergence of sum of independent random values

If $f(x)>0$ is any function st. $\sum_{j=1}^{\infty}f(j)=\infty$. And $a_n=f(n)$ with probability 50%, else 0.

Does $\lim_{n\rightarrow\infty}((1/\sum_{j=1}^nf(j)) \sum_{k=1}^n a_k)=1/2$ almost surely?

-
Did they show your pageant on television? I missed it. – Will Jagy Jan 23 '12 at 0:36
I see, after you revised your user name: the pageant is later in the year and you are very confident. Good luck. – Will Jagy Jan 23 '12 at 1:11
@WillJagy May I know "What's in a name? That which we call a rose By any other name would smell as sweet." – user21436 Jan 23 '12 at 1:21
@KannappanSampath There was an Irish author/playwright who was wonderful at inventing names, he is usually just referred to as Lord Dunsany, en.wikipedia.org/wiki/Edward_Plunkett,_18th_Baron_of_Dunsany Here is the entire text of his first book: en.wikisource.org/wiki/The_Gods_of_Peg%C4%81na – Will Jagy Jan 23 '12 at 1:31

No, not for every $f$. If you choose $f$ growing so rapidly that $f(n)>3\sum_{j=1}^{n-1}f(j)$, then every time that $a_n=f(n)$ we get $(\sum_{k=1}^n a_k)/\sum_{j=1}^nf(j)\geq f(n)/\sum_{j=1}^nf(j) >3/4$. And this happens infinitely often with probability 1.
Call $F_n=\sum\limits_{k=1}^nf(k)$ and assume that $\frac{F_{n+1}}{F_n}\to c$ for a given $c\geqslant1$ and that $\frac1{F_n}\sum\limits_{k=1}^na_k$ converges almost surely to a limit $\ell$. Then, $\frac1{F_n}\sum\limits_{k=1}^{n+1}a_k=\frac{F_{n+1}}{F_n}\cdot\frac1{F_{n+1}}\sum\limits_{k=1}^{n+1}a_k$ converges almost surely to $c\ell$, hence $\frac{a_{n+1}}{F_n}=\frac1{F_n}\sum\limits_{k=1}^{n+1}a_k-\frac1{F_n}\sum\limits_{k=1}^{n}a_k$ converges almost surely to $(c-1)\ell$. Since $a_{n+1}=0$ infinitely often almost surely, this shows that $(c-1)\ell=0$.
Hence, $\frac1{F_n}\sum\limits_{k=1}^na_k$ cannot converge almost surely to a nonzero limit unless $\frac{F_{n+1}}{F_n}\to1$.
If $f(n)=c^n$ with $c\gt1$, then $\frac{F_{n+1}}{F_n}\to c\ne1$ hence $\frac1{F_n}\sum\limits_{k=1}^na_k$ does not converge almost surely to any positive limit.


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# Difference between revisions of "2018 AMC 8 Problems/Problem 23"

## Problem

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

$[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]$

$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

## Solutions

### Solution 1

We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent.

If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and also $4$ places to put the remaining point, so we have $8\cdot4$ choices. The total amount of choices is ${8 \choose 3} = 8\cdot7$. Thus our answer is $\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}$

### Solution 2

We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is $\frac{8\cdot6-8}{{8 \choose 3 }}$$=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}$

### Solution 3 (Stars and Bars)

Let $1$ point of the triangle be fixed at the top. Then, there are ${7 \choose 2} = 21$ ways to chose the other 2 points. There must be $3$ spaces in the points and $3$ points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and $2$ extra points (k-1) distributed so by the stars and bars formula, ${n+k-1 \choose k-1}$, there are ${4 \choose 2} = 6$ ways to arrange the bars and stars. Thus, the probability is $\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}$. The stars and bars formula might be inaccurate

~ pi_is_3.14


text: 5. LPL is leading the OSIRIS-REx sample return mission to the near-Earth asteroid Bennu, an $800 million mission that launched in 2016 and is scheduled to return samples in 2023. 6. LPL faculty have served as either principal or co-investigators on many spacecraft instruments and missions. PtyS/LPL faculty are deeply involved in Cassini (Saturn), Parker Solar Probe, Juno (Jupiter), MRO (Mars), WISE, MAVEN, LRO, Dawn, and Mars Odyssey, as well as OSIRIS-REx. 7. LPL successfully led and managed surface operations for the PHOENIX Mars mission, becoming the first university organization to manage a planetary mission 8. LPL’s SPACEWATCH® was the first program to discover an asteroid using electronic detection, and LPL’s Catalina Sky Survey has discovered most of the near-Earth asteroids found since 2005. The two programs combined have discovered more than half of the known near-Earth asteroids. 9. External funding from all sources exceeds$33M for the most recent fiscal year, roughly 10 times the amount of money LPL receives from the state of Arizona.


text:
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# Questions

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### Why is the dimension of Gaussian variables is bounded by the dimension of the space?

feb 10 11 at 11:12 Gideon Schechtman 1,020711

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### Radstrom cancellation only for two convex sets?

sep 16 10 at 3:36 Paulo J. S. Silva 211

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### reference for perturbation of projection result

jul 7 at 19:37 Dirk 1,777823

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### When is spectral norm of AB equal to that of BA?

feb 22 at 18:51 suVRit 10.3k11645

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### characterization of continuous functionals in weak-star topology

feb 18 10 at 0:21 Bill Johnson 14.5k13360

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### is this bound on a kummer function known?

jan 7 12 at 0:48 Alex Gittens 3866

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### is the kummer function monotonic decreasing in first argument?

jan 9 12 at 18:45 Alex Gittens 3866

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### Does this norm inequality hold for projections onto the range of a sum of matrices?

may 2 11 at 20:47 Kate Juschenko 3,864527

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### proofs of stochastic boundedness

nov 1 11 at 22:53 Ori Gurel-Gurevich 4,3261818

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# calculating walks in an undirected graph using linear algebra identity

I am reading a text on maths applied to naturally occurring networks, eg. social networks. The section I am on is "Walks" networks. The text says:

Walks: which allow both nodes and edges to be used more than once. For example, in Figure 1, Mary-Bob-Sue-Ramona-Bob-Mary-Joe is a walk, but not a trail or a path. A particular dollar bill flows through the person-person network along walks. Mary might give the bill to Bob in one transaction, and Bob might return it to Mary in another.

A trail is when a node i may be visited multiple times, but a particular edge connecting it with a node j, can be visited no more than 1 time.

An identity from linear algebra is used which I cannot find is used Suppose an undirected network has an adjacency matrix A, where $a_{i,j}=1$ if nodes i and j are connected, and $a_{i,j}=0$ otherwise. Then the basic identity

$(A^{n})_{ij} = \sum^{n}_{k_{1} = 1}\sum^{n}_{k_{2}=1}\ldots\sum^{n}_{k_{n-1}=1}a_{i,k_{1}}a_{k_{1},k_{2}}\ldots a_{k_{n-1},j}$

shows that $(A^{n})_{ij}$ counts the total number of walks of length n from node i to node j.

The indice k is not explicitly defined in the text or given more description. What makes it hard to understand is that the index k is not being summed over the possible nodes the matrix since n is the walk length as it is written, unless there is a mistake.

I want to see if I can be bold and say that the equation should have been better written as (making N be the node set or square matrix dimension):

$(A^{n})_{ij} = \sum^{N}_{k_{1}=1}(\sum^{N}_{k_{2}=1}\ldots(\sum^{N}_{k_{n-1}=1}a_{i,k_{1}}a_{k_{1},k_{2}}\ldots a_{k_{n-1},j}))$

Because I read the equation as a brute force summation over the total set of combinations of length n.

Is this interpretation correct?

and is $(A^{n})_{ij}$ the result of the entry for ij in doing the matrix exponentiation of the adjacency matrix n times because in simple examples I tried with small graphs I did not get the correct answer (tried simple fully connected graph with n=2,3,4)?

Best,

-

Yes, you're right (the $k$'s should run from 1 to $N$, the size of the matrix), and yes, $(A^n)_{ij}$ is the $(i,j)$ entry in $A^n=AA\dots A$ ($n$ factors).


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# Growth and characteristics of lead-substituted rare earth manganites

Ghosh, Nilotpal and Bhat, HL (2005) Growth and characteristics of lead-substituted rare earth manganites. In: Journal of Crystal Growth, 275 (1-2). e1953-e1959.

 PDF crystal.pdf Restricted to Registered users only Download (516Kb) | Request a copy

## Abstract

The growth of $R_{1-x}$ $Pb_xMnO_3$ (R=La, Nd, [La,Nd]) single crystals by $PbO/PbF_2$ flux is described with optimized growth conditions. The structure of $Nd_{1-x}$ $Pb_xMnO_3$ changes from orthorhombic (x=0, Pnma) to tetragonal (x=0.25, P4/mmm) to cubic (x=0.38, Pm3¯m) as x increases. These changes are attributed to the progressive removal of inter-octahedral tilting and minimization of octahedral distortion leading to a higher symmetry as doping concentration increases. The $T_C$ of this system increases from 110 to 192 K for Pb concentration varying from x=0.15 to 0.42. The $T_{MI}$ also follows a similar trend. It is observed that $\langle r_A\rangle$ and $\sigma^2$ are the controlling parameters which influence the transport and magnetic properties of $La_{0.7-y}Nd_yPb_{0.3}MnO_3$. The presence of spin wave excitations are observed prominently in $La_{0.7}Pb_{0.3}MnO_3$ at low-temperature which is not so in case of Nd-based compounds. The total magnetic moments of all the Nd-based systems at low-temperature are influenced by Nd. The low-temperature specific heat data of $La_{0.7-y}Nd_yPb_{0.3}$ $MnO_3$ is analyzed and contributions of different mechanisms are discussed. It is noticed that the Schottky anomaly is present in specific heat at low-temperature only in Nd-based compounds.

Item Type: Journal Article Copyright of this article belongs to Elsevier B.V. A2. Growth from high-temperature solutions; B1. Perovskite; B2. Magnetic materials Division of Physical & Mathematical Sciences > Physics 10 Mar 2008 19 Sep 2010 04:43 http://eprints.iisc.ernet.in/id/eprint/13367

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# zbMATH — the first resource for mathematics

Random walk attracted by percolation clusters. (English) Zbl 1112.60089
Summary: Starting with a percolation model in $$\mathbb{Z}^d$$ in the subcritical regime, we consider a random walk described as follows: the probability of transition from $$x$$ to $$y$$ is proportional to some function $$f$$ of the size of the cluster of $$y$$. This function is supposed to be increasing, so that the random walk is attracted by bigger clusters. For $$f(t)=e^{\beta t}$$ we prove that there is a phase transition in $$\beta$$, i.e., the random walk is subdiffusive for large $$\beta$$ and is diffusive for small $$\beta$$.

##### MSC:
 60K37 Processes in random environments 60J10 Markov chains (discrete-time Markov processes on discrete state spaces) 60K35 Interacting random processes; statistical mechanics type models; percolation theory
Full Text:


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Intermediate Algebra: Connecting Concepts through Application

5x+y=7 y=-5x+7 2y=10x-9 y=5x-$\frac{9}{2}$ The slopes of these two lines are -5 and 5. These are neither equal nor opposite reciprocals of each other. Therefore, these two lines are neither parallel nor perpendicular.


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# Convolution of two PDF: $1/(2\sqrt{3x})$

I'm trying to find the result of the sum of two PDF that is the square of two continuous uniform distribution (rectangular distribution) that have $\mu =0$ and $\sigma =1$. The square of this continuous uniform distribution is defined from 0 to 3 and is $$p(x)= \frac{1}{2 \sqrt{3x}}$$ Now i need to find the PDF of the sum of two random variables with p(x) as distribution. The convolution integral is $$p(u)=\int_0^3 p(u-t)p(t)dt=\int_0^3 \frac{dt}{12 \sqrt{t(u-t)}}= \frac{arcsin \left( \sqrt{ \frac{3}{u}} \right)}{6}$$ This funcion is not defined on R from 0 to 6. If i extract the real part of $(p(u))$ i obtain the correct PDF from 0 to 3 but the part from 3 to 6 is wrong and p(u) is not normalized from 0 to 6. I simulate this PDF with python, here is an image

Simulation of PDF: https://imgur.com/ImqT7ya

Plot of the PDF: https://imgur.com/khg8rGM

what did I do wrong?

• That distribution is not uniform. Do you mean continuous? Jun 26, 2018 at 1:16
• p(x) is the pdf of the square of an uniform distribution with mean=0 and variance=1. I'm not saying that p(x) is an uniform distribution Jun 26, 2018 at 1:19
• okay sorry it is a continuous uniform distribution ( en.wikipedia.org/wiki/Uniform_distribution_(continuous) ), I got confused with the translation from italian Jun 26, 2018 at 1:22

The key is in the support interval.   Both $t$ and $u-t$ must lie within the support, or otherwise $p(t)p(u-t)=0$.   So the bounds for the integral depends on the value given for $u$.
You have $p(x)=\dfrac 1{2\sqrt{3x}}\mathbf 1_{0\leq x\leq 3}$
The the convolution is $$\begin{split}\int_\Bbb R p(t)p(u-t)~\mathsf d t &= \int_\Bbb R \tfrac{1}{12\sqrt{t(u-t)}}\mathbf 1_{0\leq t\leq 3~,~0\leq u-t\leq 3}~\mathsf d t\\&= \int_\Bbb R \tfrac{1}{12\sqrt{t(u-t)}}\mathbf 1_{0\leq t\leq 3~,~ t\leq u~,~ u-3\leq t}~\mathsf d t\\ &= \int_{\max\{0,u-3\}}^{\min\{3,u\}} \tfrac{1}{12\sqrt{t(u-t)}}~\mathbf 1_{0\leq u\leq 6}~\mathsf d t \\ &= \mathbf 1_{0\leq u\leq 3}\int_{0}^{u} \tfrac{1}{12\sqrt{t(u-t)}}~\mathsf d t+\mathbf 1_{3<u\leq 6}\int_{u-3}^{3} \tfrac{1}{12\sqrt{t(u-t)}}~\mathsf d t\end{split}$$


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# 10.3: Calculating Equilibrium Values

The numeric value of the equilibrium constant tells us something about the ratio of the reactants and products in the final equilibrium mixture. Likewise, the magnitude of the equilibrium constant tells us about the actual composition of that mixture.

In the three equilibrium systems, the first depicts a reaction in which the ratio of products to reactants is very small. Because the expression for the equilibrium constant is given by the pressure (or concentration) of products divided by the pressure (or concentration) of reactants, the equilibrium constant, K, for this system is also small. In the second example, the concentrations of reactants and products are shown to be equal, making the ratio (the equilibrium constant) equal to “1”. In the last example, the products are shown to dominate the equilibrium mixture, making the ratio ${$/extract_itex]displaystyle {}^{\left(P_{Products}\right)}\!\!\diagup \!\!{}_{\left(P_{Reactants}\right)}\;} very large. In these examples, the stoichiometric ratios of the reactants and products are one and there is only one reactant and only one product; if multiple reactants or products are involved, the relationship between their concentrations would be more complex, but that ratio is always given by the expression for K. This fact allows us to take data for an equilibrium reaction and, if K is known, calculate concentrations for reactants and products. Likewise, if all of the equilibrium concentrations are known, we can use these to calculate a value for the equilibrium constant. In these types of problems, an ICE table is often useful. This table has entries for Initial concentrations (or pressures), Equilibrium concentrations and any Change between the initial and equilibrium states. For example, consider the reaction between carbon monoxide and chlorine to form phosgene, a deadly compound that was used as a gas warfare agent in World War I. CO (g) + Cl2 (g) ⇄ COCl2 (g) Example $$\PageIndex{1}$$: A mixture of CO and Cl2 has initial partial pressures of 0.60 atm for CO and 1.10 atm for Cl2. After the mixture reaches equilibrium, the partial pressure of COCl2 is 0.10 atm. Determine the value of K. Solution The initial pressures for carbon monoxide and chlorine are placed in the first row and the equilibrium pressure for phosgene is placed in the last row. Initially, the pressure of phosgene was zero, so that goes in the first row; the change for phosgene is therefore “+ 0.10 atm”.  ${\displaystyle P_{CO}}$ ${\displaystyle P_{Cl_{2}}}$ ${\displaystyle P_{COCl_{2}}}$ === Initial === 0.60 atm 1.10 atm 0 atm Change + 0.10 atm Equilibrium 0.10 atm Because one mole of CO is required to make one mole of COCl2 the partial pressure of CO must have dropped by 0.10 atm (the Change) in order to make COCl2 with a partial pressure of 0.10 atm, giving a final (Equilibrium) pressure of 0.50 atm for carbon monoxide. Likewise, one mole of chlorine is required to make one mole of COCl2 making the Change for chlorine 0.10 atm and the Equilibrium partial pressure 1.00 atm. The completed table is shown below:  ${\displaystyle P_{CO}}$ ${\displaystyle P_{Cl_{2}}}$ ${\displaystyle P_{COCl_{2}}}$ === Initial === 0.60 atm 1.10 atm 0 atm Change -0.10 atm -0.10 atm + 0.10 atm Equilibrium 0.50 atm 1.00 atm 0.10 atm The equilibrium expression for the phosgene-forming reaction is given by the following equation: \[\frac{P_{COCl_{2}}}{P_{CO}P_{Cl_{2}}}=K$

Substituting the values from the Table into this equation:

$\frac{P_{COCl_{2}}}{P_{CO}P_{Cl_{2}}}=\frac{(0.10)}{(0.50)(1.00)}=0.20$

Notice that equilibrium constants for gas phase reactions are not typically written with units, although units are sometimes used in equilibrium constants calculated from molar concentrations. Many textbooks differentiate between equilibrium constants calculated from partial pressures and molar concentrations by affixing subscripts; KP and Kc. In this book, we will simply use K and Kc to represent the two; a value for K will always denote a constant calculated from partial pressure data.

Exercise $$\PageIndex{1}$$

1. For the reaction shown below, all four gasses are introduced into a vessel, each with an initial partial pressure of 0.500 atm, and allowed to come to equilibrium; at equilibrium, the partial pressure of SO3 is found to be 0.750 atm. Determine the value of K.

SO2 (g) + NO2 (g) ⇄ SO3 (g) + NO (g)

1. For the reaction shown above, the initial partial pressures of SO3 and NO are 0.500 atm under conditions where the equilibrium constant is, K = 9.00. The equilibrium partial pressure for SO2 is found to be 0.125 atm. Calculate the equilibrium partial pressure for SO3.

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It’s time to (somewhat belatedly) roll over the previous thread on writing the first paper from the Polymath8 project, as this thread is overflowing with comments.  We are getting near the end of writing this large (173 pages!) paper, establishing a bound of 4,680 on the gap between primes, with only a few sections left to thoroughly proofread (and the last section should probably be removed, with appropriate changes elsewhere, in view of the more recent progress by Maynard).  As before, one can access the working copy of the paper at this subdirectory, as well as the rest of the directory, and the plan is to submit the paper to Algebra and Number theory (and the arXiv) once there is consensus to do so.  Even before this paper was submitted, it already has had some impact; Andrew Granville’s exposition of the bounded gaps between primes story for the Bulletin of the AMS follows several of the Polymath8 arguments in deriving the result.

After this paper is done, there is interest in continuing onwards with other Polymath8 – related topics, and perhaps it is time to start planning for them.  First of all, we have an invitation from  the Newsletter of the European Mathematical Society to discuss our experiences and impressions with the project.  I think it would be interesting to collect some impressions or thoughts (both positive and negative)  from people who were highly active in the research and/or writing aspects of the project, as well as from more casual participants who were following the progress more quietly.  This project seemed to attract a bit more attention than most other polymath projects (with the possible exception of the very first project, Polymath1).  I think there are several reasons for this; the project builds upon a recent breakthrough (Zhang’s paper) that attracted an impressive amount of attention and publicity; the objective is quite easy to describe, when compared against other mathematical research objectives; and one could summarise the current state of progress by a single natural number H, which implied by infinite descent that the project was guaranteed to terminate at some point, but also made it possible to set up a “scoreboard” that could be quickly and easily updated.  From the research side, another appealing feature of the project was that – in the early stages of the project, at least – it was quite easy to grab a new world record by means of making a small observation, which made it fit very well with the polymath spirit (in which the emphasis is on lots of small contributions by many people, rather than a few big contributions by a small number of people).  Indeed, when the project first arose spontaneously as a blog post of Scott Morrrison over at the Secret Blogging Seminar, I was initially hesitant to get involved, but soon found the “game” of shaving a few thousands or so off of $H$ to be rather fun and addictive, and with a much greater sense of instant gratification than traditional research projects, which often take months before a satisfactory conclusion is reached.  Anyway, I would welcome other thoughts or impressions on the projects in the comments below (I think that the pace of comments regarding proofreading of the paper has slowed down enough that this post can accommodate both types of comments comfortably.)

Then of course there is the “Polymath 8b” project in which we build upon the recent breakthroughs of James Maynard, which have simplified the route to bounded gaps between primes considerably, bypassing the need for any Elliott-Halberstam type distribution results beyond the Bombieri-Vinogradov theorem.  James has kindly shown me an advance copy of the preprint, which should be available on the arXiv in a matter of days; it looks like he has made a modest improvement to the previously announced results, improving $k_0$ a bit to 105 (which then improves H to the nice round number of 600).  He also has a companion result on bounding gaps $p_{n+m}-p_n$ between non-consecutive primes for any $m$ (not just $m=1$), with a bound of the shape $H_m := \lim \inf_{n \to \infty} p_{n+m}-p_n \ll m^3 e^{4m}$, which is in fact the first time that the finiteness of this limit inferior has been demonstrated.  I plan to discuss these results (from a slightly different perspective than Maynard) in a subsequent blog post kicking off the Polymath8b project, once Maynard’s paper has been uploaded.  It should be possible to shave the value of $H = H_1$ down further (or to get better bounds for $H_m$ for larger $m$), both unconditionally and under assumptions such as the Elliott-Halberstam conjecture, either by performing more numerical or theoretical optimisation on the variational problem Maynard is faced with, and also by using the improved distributional estimates provided by our existing paper; again, I plan to discuss these issues in a subsequent post. ( James, by the way, has expressed interest in participating in this project, which should be very helpful.)


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5

# Use the graph.Determine the values of x between which a real zero is located and F Ob. between 2 and 3 Oa. between 32 Sand 0 Od. between 6 and 7 Oc between...

## Question

###### Use the graph.Determine the values of x between which a real zero is located and F Ob. between 2 and 3 Oa. between 32 Sand 0 Od. between 6 and 7 Oc between

Use the graph. Determine the values of x between which a real zero is located and F Ob. between 2 and 3 Oa. between 32 Sand 0 Od. between 6 and 7 Oc between

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'â‚¬ 2 - Ahotplate has an area of 0.15 m? and temperature 0f 3009 â‚¬ the room temperature is 40* â‚¬ how much net poweris transmitted to room in every second? e=0.80. What are the radiated wave length from hot plate and room? 0,16 m 2' 200' Tz JaOh 06*0 <9 i= k...
##### Polystyrene is produced commercially by reaction of styrene with butyllithium as an anionic initiator. Using resonance structures, explain how the chain-carrying intermediate is stabilized.
Polystyrene is produced commercially by reaction of styrene with butyllithium as an anionic initiator. Using resonance structures, explain how the chain-carrying intermediate is stabilized....
##### STATION 7 9 .137 xlo The substance in this vial is table salt; NaCI Calculation of one mole of NaCL 23.05muleudUsing & weighing dish, find the mass of this sample of NaCL Z5l4 After massing this sample, carefully place the entire sample back into the vial Calculation of the number of moles of NaCl in this sample.Calculation of the number of formula units of NaCl in this sample_25.25 grams of NaCl on her french fries Calculate the number of moles and the student places number of formula unit
STATION 7 9 .137 xlo The substance in this vial is table salt; NaCI Calculation of one mole of NaCL 23.05 muleud Using & weighing dish, find the mass of this sample of NaCL Z5l4 After massing this sample, carefully place the entire sample back into the vial Calculation of the number of moles of ...
##### Derive power series solution for the Bessel' differential equation
Derive power series solution for the Bessel' differential equation...
##### (II) $(a)$ What is the velocity of a beam of electrons that go undeflected when passing through crossed (perpendicular) electric and magnetic fields of magnitude $1.88 \times 10^{4} \mathrm{V} / \mathrm{m}$ and $2.90 \times 10^{-3} \mathrm{T}$ , respectively? (b) What is the radius of the electron orbit if the electric field is turned off?
(II) $(a)$ What is the velocity of a beam of electrons that go undeflected when passing through crossed (perpendicular) electric and magnetic fields of magnitude $1.88 \times 10^{4} \mathrm{V} / \mathrm{m}$ and $2.90 \times 10^{-3} \mathrm{T}$ , respectively? (b) What is the radius of the electron ...
##### Find the Inverse of f(x) = Sketch the Graph and its inverse on the same Graph: Describe the symmetry vou see. (10 points)
Find the Inverse of f(x) = Sketch the Graph and its inverse on the same Graph: Describe the symmetry vou see. (10 points)...
##### 3) Given that f (x) = V-x+ 2 do the following:a) complete the table below:f(x)77 ; 3 b) Sketch the graph of the function C give the domain of the function in interval notation
3) Given that f (x) = V-x+ 2 do the following: a) complete the table below: f(x) 77 ; 3 b) Sketch the graph of the function C give the domain of the function in interval notation...
##### Question 42 of 45Consider function f and its derivatives3x2 -] flx) = x2 + 3 20x f'(x) = (3 +x2)260(1 _ x2) f" (x) = (x2 +3)3Find a relative minimum value of the function f , if there is one. 0 A0 -9 S13 0 C has no relative minimum 0 D. 3 0 E 2
Question 42 of 45 Consider function f and its derivatives 3x2 -] flx) = x2 + 3 20x f'(x) = (3 +x2)2 60(1 _ x2) f" (x) = (x2 +3)3 Find a relative minimum value of the function f , if there is one. 0 A0 -9 S13 0 C has no relative minimum 0 D. 3 0 E 2...
##### Write the following definite integrals as infinite limnits of Riemann sums: J" (*+3)Vr? + ]drWVrite A-term Riemt SUIn for Lc following definite integrals in the previous problem.
Write the following definite integrals as infinite limnits of Riemann sums: J" (*+3) Vr? + ]dr WVrite A-term Riemt SUIn for Lc following definite integrals in the previous problem....
##### 1) Use stepwise regression to decide the final model. Use significance levels (E 4R 0.15 . In each step, report the result of regression analysis. indicate which predictor variable enters or leaves the model, and how the decision is made_ (Hint: Use [0 fit the data. Use addl t0 add predictor variable and use dropl t0 remove variable from the model.)
1) Use stepwise regression to decide the final model. Use significance levels (E 4R 0.15 . In each step, report the result of regression analysis. indicate which predictor variable enters or leaves the model, and how the decision is made_ (Hint: Use [0 fit the data. Use addl t0 add predictor variab...
##### Usc the method of constructing inlerpretations to determine whether the following statements are correct: Explain your reasoning in the same Way a8 in the worked examples and if you claim scquent iS incorrect, exhibit an inter pretation which establishes thisA-BB (C vDL ~DFA
Usc the method of constructing inlerpretations to determine whether the following statements are correct: Explain your reasoning in the same Way a8 in the worked examples and if you claim scquent iS incorrect, exhibit an inter pretation which establishes this A-BB (C vDL ~DFA...
##### When the capacitor is fully charged, what is the magnitude of the voltage across the inductor?Virnax/2 VLrioX Submit(Survey Question' 4) Briefly explain your reasoningSubmit5) When the " voltage across the capacitor is at its positive maximum, Vc ~Vcmax' what is the voltage across the inductor?VLmax VLmax Submit(Surve; Question} 6) Briefly explain your reasoningSubmit
When the capacitor is fully charged, what is the magnitude of the voltage across the inductor? Virnax/2 VLrioX Submit (Survey Question' 4) Briefly explain your reasoning Submit 5) When the " voltage across the capacitor is at its positive maximum, Vc ~Vcmax' what is the voltage across...