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# How to extract the diagonal from a bivariate generating function

Let $F(s,t)= \sum_{i,j} f(i,j) s^i t^j$, which is a bivariate generating funcion of the number $f(i,j)$ for some enumeration problem. Sometimes we know about $F(s,t)$, but what we really need is the number $f(i,i)$ with the generating function $G(x)= \sum_i f(i,i) x^i$, called the diagonal of $F$.

The question is how to obtain the diagonal $G(x)$ from $F(s,t)$? Furthermore, how to get the asymtotic formula for $f(i,i)$ from $F(s,t)$?

One way to do this is shown as follows: $G(x)$ is the constant term of $F(s,\frac{x}{s})$ regarded as a Laurent series in $s$ whose coefficients are power series in x. And we can use Cauchy's integral theorem and Residue Theorem to compute.

This method works when $F(s,t)$ is rational. But when $F(s,t)$ is more complicated, it seems not workable. An example is $F(s,t)=\frac{4 s t}{\sqrt{1-4 s^2}\sqrt{1-4 t^2}(\sqrt{1-4 s^2}\sqrt{1-4 t^2}-4 s t)}$.

• Stanley's Enumerative Combinatorics II discusses this issue in Chapter 6, Section 3. Sep 19, 2010 at 17:21
• An authority on the asymptotics of $f(i,j)$ for rational $F(s,t)$ is Robin Pemantle. See math.upenn.edu/~pemantle/papers/Papers.html. I don't know what to do in general about nonrational $F(s,t)$. Sep 20, 2010 at 1:45
• @Richard, thanks for the reference. I found that the diagonal of a algebraic bivariate generating function is not necessarily algebraic. Sep 28, 2010 at 5:34
• An important example is given by $F(s,t)$ of the form $\frac{\phi(s)}{1+t\psi(s)}$, where the diagonal can be obtained with Lagrange-Buermann formula. This case can be recognized by equality $\left(\frac{\partial}{\partial t}\right)^2 \frac1{F(s,t)} = 0$. Nov 20, 2019 at 21:22


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# Electric Flux is zero

When is the Electric Flux zero through a closed surface? I know that if number of field lines entering is equal to leaving then it is true. Which I think is always true for a closed surface. So can there be any situation in which an isolated surface has 0 flux?

• Gauss's law states that the flux through a closed surface is proportional to the charge contained within the surface... – lemon Apr 23 '16 at 13:25
• So What ? Will that answer my question – Kumar Sambhav Apr 23 '16 at 13:37
• Yes... To put it even more simply: the flux through a closed surface is equal to the charge contained within it (in the right units). So under what circumstance will the flux be zero? – lemon Apr 23 '16 at 14:02

Electric flux through a closed surface $\rm S$ is $$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.$$
From $(1)\,,$ it can be concluded that when $Q_\textrm{int}= 0\,$ the electric flux is zero.

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# For non-zero integers a, b, c and d, is ab/cd positive?

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For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]

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20 Dec 2006, 03:25
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For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0

(2) abcd = -4
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Aug 2013, 03:09, edited 2 times in total.
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20 Dec 2006, 06:09
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Agree with D.

S1: One of ad or bc has to be negative for them to add to zero.

S2: One or three of abcd has to be negative for their product to be negative.
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Re: DS_For non-zero integers a, b, c and d... [#permalink]

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22 Dec 2006, 15:48
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mm007 wrote:
For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0

(2) abcd = -4

should be D.

i. one of the integer is -ve so ab/cd is -ve. sufficient
ii. either one or three of the integers is/are -ve, so ab/cd is again -ve. sufficient.
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Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]

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10 Oct 2013, 14:57
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mm007 wrote:
For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0

(2) abcd = -4

So this question is basically testing negatives and positives (Remember >0).

First Statement

ad = -bc. Now we could rearrange this to be a/c = -b/d. Now replacing in the original statement we would have (-b/d)(b/d) . Since this is basically the same fraction but with different signs then the result HAS to be negative. Therefore this statement is Sufficient

Second Statement

abcd = -4. Now here, we see that the result is -ve. So actually, we can either have 1 negative or 3 negatives. But either choice will give us ab/cd <0. Because the only thing we need is to have an odd number of negative signed numbers. I suggest to try it with different combinations and see it for yourself

Hope it helps
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21 Dec 2006, 08:12
Getting B ...

if ABCD <0 (either one or three of them is negetive)

then AB/CD < 0

B is SUFF
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21 Dec 2006, 16:29
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Given : a,b,c and d are all non-zero ints.

(1) ad + bc = 0
---------------------
ad = - bc --> one of abcd has a different sign that the others
ex: all +ve and one is -ve or all -ve and one is +ve
So, is ab/cd > 0 ? NO

statement 1 is sufficient

(2) abcd = -4
------------------
abcd = -ve # --> one abcd also has a different sign than the others
So, is ab/cd > 0 ? NO

statement 2 is sufficient

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22 Dec 2006, 15:08
From 1 we get the expression as -(b^2/d^2) hence negative

from 2 we get (-a^2b^2)/4

Both are negative

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24 Dec 2006, 17:51
a,b,c,d are all non zero.

Now we have:

multiply it by bd we get (since bd non zero):

ab(d^2) + (b^2)cd=0
Therefore: ab/cd = -(b^2)/(d^2)
or ab/cd <0 --------------------------- sufficient

abcd = -4
or (ab/cd)* (cd)^2 = -4
or ab/cd = -4/(cd)^2
or ab/cd <0 ...................................sufficient

D !

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For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]

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03 Aug 2013, 09:26
For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0
(2) abcd = -4

I dont agree with the OA.

IMO B.

Statement 1 : ad = -bc
1) If all integers a,b,c & d are -ve/+ve, then statement 1 holds true and ab/bc is positive.
2) If any one integer is -ve & other 3 integers are positive, statement 1 holds true and ab/bc is negative.

Let me know where my thinking is wrong?

Last edited by Zarrolou on 03 Aug 2013, 09:30, edited 1 time in total.
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Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]

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03 Aug 2013, 09:34
mohitvarshney wrote:
For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0
(2) abcd = -4

I dont agree with the OA.

IMO B.

Statement 1 : ad = -bc
1) If all integers a,b,c & d are -ve/+ve, then statement 1 holds true and ab/bc is positive.
2) If any one integer is -ve & other 3 integers are positive, statement 1 holds true and ab/bc is negative.

Let me know where my thinking is wrong?

If $$a,b,c,d$$ are all positive or negative (1) does not hold true,as
$$positive+positive>0$$ and $$negative + negative < 0$$ ( and not equal 0).

Hope it's clear.
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Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]

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From statement 1, we know that $$ad+bc=0$$
$$=>$$ $$ad=-bc$$
$$=>$$ $$-\frac{a}{b}=\frac{c}{d}$$
$$=>$$ $$-\frac{a*b}{b*b}=\frac{c*d}{d*d}$$
$$=>$$ $$-\frac{ab}{b^2}=\frac{cd}{d^2}$$
$$=>$$ $$\frac{ab}{cd}=-\frac{b^2}{d^2}$$
Thus from 1, we come to know that ab/cd is negative since $$b^2$$ and $$d^2$$ are positive

From statement 2, $$abcd=-4$$
Now the only way this can happen is if one of the term is negative. And if one of the term is negative, then ab/cd has to be negative.

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Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]

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10 Nov 2015, 19:49
Q: a, b, c, d are non-zero integers. Is ab/cd positive?

St1: ad + cb = 0, ad = -cb. One of the four elements is of an opposite sign, therefore either one or three elements are negative and ab/cd = -ve. Sufficient.

St2: abcd = -4, ab = -4/cd, ab/cd = -4/cd^2, therefore ab/cd is negative.
Re: For non-zero integers a, b, c and d, is ab/cd positive?   [#permalink] 10 Nov 2015, 19:49
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1 If y = O.abcd, where a, b, c, and d each represent a nonzero 2 01 May 2013, 12:06
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# Multiplying Expressions with Exponents

Stephen Sacchetti, Jennifer Beddoe
• Author
Stephen Sacchetti

Stephen graduated from Haverford College with a B.S. in Mathematics in 2011. For the past ten years, he has been teaching high school math and coaching teachers on best practices. In 2015, Stephen earned an M.S. Ed from the University of Pennsylvania where he currently works as an adjunct professor.

• Instructor
Jennifer Beddoe

Jennifer has an MS in Chemistry and a BS in Biological Sciences.

Learn about multiplying exponents with different bases, as well as multiplying exponents with the same base. Explore multiplying exponents rules, how to multiply exponents with different bases, and what happens when you multiply exponents. Updated: 11/08/2021

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## How Do You Multiply Exponents?

This lesson will focus on operations on numbers that have exponents. In particular, the main goal of this lesson will be to answer the question of when and how do you multiply exponents. After establishing the when and how there will be an exploration of multiplying exponents rules that outline how to combine exponents in a variety of situations.

### What are Exponents?

The four basic operations of arithmetic are addition, subtraction, multiplication, and division. Recall that multiplication is simply a shorter way to write repeated addition. For example, {eq}3 + 3 + 3 + 3 + 3 + 3 = 6 \times 3 {/eq}. Similarly, an exponent is an operation that is a shorter way to write repeated multiplication. For example, {eq}3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 {/eq}. In this case, 6 is called the exponent and 3 is called the base. At times, the word power is also used to describe an exponent. For instance, the expression in Fig. 1 might be called "two to the power of five."

More generally, any exponential expression can be written as {eq}b^a {/eq} where a is the exponent and b is the base.

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## Multiplying Exponents with the Same Base: Do You Add Exponents When Multiplying?

What do you do when multiplying exponents? Since exponents mean repeated multiplication, a natural question of what happens when you multiply exponents might arise. In this section, the idea of multiplying two bases raised to exponents will be explored. For example, consider the expression {eq}2^5 \times 2^3 {/eq}. It may be tempting here to assume that the 2's multiply together or that the 5 and 3 multiply together, but in fact, neither of these is correct. Instead, lean on the definition of an exponent to rewrite this expression before combining any terms:

{eq}2^5 \times 2^3 = (2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2) {/eq}

Noticing that there are eight total 2's here, this expression can be rewritten as {eq}2^8 {/eq}. This idea can be generalized to show that you add exponents when multiplying like bases: {eq}b^x \times b^y = b^{x + y} {/eq}

## Multiplying Exponents With Different Bases

But what happens when multiplying exponents with different bases? This cannot be done by adding the exponents as in the previous section because it is unclear what to do with the base. This section will show how to multiply exponents with different bases using two different methods of simplification.

### Exponential Multiplication: Simplification Method

The first method is the simplification method. Essentially, this method is used to write out the exponential expression as a long string of multiplication and then combine all of the terms. Consider the example: {eq}3^4 \times 4^3 {/eq}. Writing this out gives: {eq}3 \times 3 \times 3 \times 3 \times 4 \times 4 \times 4 {/eq} and none of the 3's can be combined with the 4's other than by multiplying out this product. Doing so gives an answer of 5184.

### Exponential Multiplication: Grouping Method

The second method of multiplying exponential expressions with different bases is grouping. The grouping method essentially says to rearrange the terms so that the multiplication can be done a bit more easily. Recall the example above of {eq}3^4 \times 4^3 {/eq} Notice that there is one more 3 than there are 4's. So we can combine the same number of 3's and 4's and have one 3 left over. For instance:

{eq}3 \times 3 \times 3 \times 3 \times 4 \times 4 \times 4 = 3 \times 4 \times 3 \times 4 \times 3 \times 4 \times 3 = 12 \times 12 \times 12 \times 3 = 12^3 \times 3 {/eq}

So it is possible to rewrite the expression using exponents: {eq}3^4 \times 4^3 = 3^3 \times 4^3 \times 3 = 12^3 \times 3 {/eq}.

More generally, it can be said that {eq}a^x \times b^x = (ab)^x {/eq} or, in words, multiplying different bases raised to the same power means that the bases can be multiplied together first and then raised to that power.

Notice, though, that this grouping method requires having the same exponent on each base. For instance, to group {eq}x^5 \times y^7 {/eq}, first rewrite it as {eq}x^5 \times y^5 \times y^2 {/eq} and then group the terms with the same exponents: {eq}(xy)^5 \times y^2 {/eq}

## Multiplying Exponents Rules: All Exponent Rules

There are seven rules to remember for multiplying bases with exponents. Each multiplying exponent rule will be shown below with an example to help illustrate it. Some of these rules do not involve multiplication but will be useful in solving more complicated problems involving multiplying exponents.

### Rule 1: Product of Powers

If multiplying exponential expressions with the same base, then add the exponents together: {eq}2^4 \times 2^7 = 2^{11} {/eq}

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Jianyue Zhang-thesis-4.pdf (45.9 MB)

# Effect of Ultrasonic Shot Peening on Mechanical Properties and Corrosion Resistance of MG Alloy Sheet

thesis
posted on 10.06.2019, 19:34
Magnesium alloys are regarded as the most promising structure materials in transportation and aerospace fields because of their low densities and high specific strengths. However, the unsatisfactory mechanical performance and corrosion resistance restrict their applications. Grain refinement is an effective way to improve the mechanical properties and widen the applications. Among which, ultrasonic shot peening shows a great potential in producing refined grains or even nanocrystalline. A nanocrystalline forms at the surface after ultrasonic shot peening treatment. The formed nanocrystalline has been proved to dramatically affect the mechanical properties, such as hardness, mechanical stress, wear resistance and fatigue life.

In this dissertation, the microstructure evolution of AZ31 Mg alloy after the ultrasonic shot peening as well as its effect on the mechanical properties are investigated. The grain size, the twin structure, the surface roughness and the residual stress distribution after ultrasonic shot peening are characterized. A gradient nanostructure is achieved through ultrasonic shot peening and the thickness of this gradient nanostructure increases with prolonging the treated time. The grains at the top surface after 5 min treatment is refined to 45 nm and further refined to 42 nm for 10 min and 37nm for 15 min treatment from the XRD result. A lamellar nanocrystaline is below the top surface and a lot of tensile twins are found at the heavily deformed grains below the nanocrystalline layer. Below the twinned layer, a residual stress is distributed as deep as 400 $\mu$m in the matrix. A rough surface is obtained and the surface roughness of the 5 min treated sample was 5.934 $\mu$m, increased to 6.161 $\mu$m for10 min and 6.236 $\mu$m for 15 min. The nanocrystalline leads to the improvement of the microhardness, from 65 HV of the as-received to 123 HV, 127 HV and 145 HV for 5 min, 10 min and 15 min treatment, respectively. The tensile stress and compression stress are also improved remarkably. The yield stress is increased from 127.7 MPa of as-received to 198 MPa of 10 min treated sample and the compression stress is improved from 73 MPa to 100 MPa. The improved yield stress is attributed to the grain refinement and the work hardening of the nanocrystalline.

The wear resistance of AZ31 Mg alloy is improved greatly after ultrasonic shot peening process. The coefficient of friction and the wear rate of the ultrasonic shot peening treated sample are both lower than that of as-received. The width of the wear track of ultrasonic shot peening treated is also narrower than that of as-received, and the worn surface has a lower surface roughness. In as-received samples, abrasion and oxidation dominate the wear mechanism at low sliding speed and low applied load. The increase of sliding speed or applied load resulted in the delamination. Severe wear such as thermal softening happens with the further increase of load value or sliding speed. In ultrasonic shot peening treated samples, oxidation, the abrasion and delamination are also existing while no severe wear is found. The improved wear resistance of the ultrasonic shot peened sample is due to the improved hardness and a higher activity of oxidation during wear process. The nanocrystalline on the top surface leads to the the transition boundary between the mild wear and severe wear to a higher sliding speed and higher applied load.

The corrosion resistance of AZ31 Mg alloy before and after ultrasonic shot peening is tested in 3.5 $\%$ NaCl solution. The corrosion resistance after ultrasonic shot peening is reduced greatly because of the Fe particles at the top surface, which was exfoliated from the shot during the treating process. After a 40 $\mu$m thick polishing, Fe particles are removed totally and the corrosion resistance is improved, compared with that of as-received. The anodic current density of the nanocrystallized surface after polishing is reduced because of the compression residual stress and a rapid formation of protective layer. Meanwhile, the grain boundary acts as a physical barrier for corrosion and reduces the corrosion rate.

The bending behavior of AZ31 Mg alloy before and after ultrasonic shot peening is studied by a V-bending test. The ultrasonic shot peening treated sample has a similar bending performance even though the ductility has been reduced after ultrasonic shot peening. A single side ultrasonic shot peening (either at the inner side or the outer side) changed the bending behavior because of its asymmetric structure. The ultrasonic shot peening at the inner side for 5 min improves the bendability and longer treated, such as 10 min and 15 min degenerates the bendability to as-received. The improved bendability of 5 min treated sample is due to the drawing back of the neutral layer. The ultrasonic shot peening at the outer side for 5 min also improves the bending performance and a longer treatment of 15 min further enhanced the bendability. The improved bendability after outer side treatment is due to the high yield stress of nanocrystalline at the convex, resulting in the smaller strain here.

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# Solving a system of equations with a specific shape of the solutions

I have a system of four equations, which are:

n=1;
U[r_] := (1 + r^2);
eq1 = uu'[r] + 1/r*uu[r] + I*n/r*vv[r] + k*ww[r] == 0;
eq2 = uu''[r] +
1/r*uu'[r] - ((n^2 + 1)/r^2 - k^2 + R*(k*U[r] - I*ω))*
uu[r] - I*2*n*vv[r]/r^2 - R*pp'[r] == 0;
eq3 = vv''[r] +
1/r*vv'[r] - ((n^2 + 1)/r^2 - k^2 + R*(k*U[r] - I*ω))*
vv[r] + I*2*n*uu[r]/r^2 - I*n*R*pp[r]/r == 0;
eq4 = ww''[r] +
1/r*ww'[r] - ((n^2 )/r^2 - k^2 + R*(k*U[r] - I*ω))*ww[r] -
R*uu[r]*U'[r] - R*k*pp[r] == 0;

The solution of the system are the functions {uu[r],vv[r],ww[r],pp[r]}. In this case, I want these functions has the shape:

uu[r_] := r^2*(S11 + S21*r^2 + S31*r^4);
vv[r_] := (S12 + S22*r^2 + S32*r^4);
ww[r_] := r*(S13 + S23*r^2 + S33*r^4);
pp[r_] := r*(S14 + S24*r^2 + S34*r^4);

So what I want to know is the value of the terms {S11,S21,S31,S12,S22,S32,S13,S23,S33,S14,S24,S34} which are all functions of {r,R,k,ω}

I tried to use Solve but doesn't works.

• How the dependent variables and S's are connected? – zhk May 5 at 3:17


text:
Absolute Colimits

In category theory a concept is called absolute if it is preserved by all functors. Identity arrows and composition are absolute by the definition of functor. Less trivially, isomorphisms are absolute. In general, anything that is described by a diagram commuting is absolute as that diagram will be preserved by any functor. This is generally the case, but if I tell you something is an absolute epimorphism, it’s not clear what diagram is causing that; the notion of epimorphism itself doesn’t reduce to the commutativity of a particular diagram.

Non-Examples

To start, we can show that certain colimits cannot be absolute, at least for $\Set$-enriched category theory. In particular, initial objects and coproducts are never absolute. Using the trick above, this is easily proven.

$\Hom(0,0)\cong 1 \not\cong 0$

$\Set(\Hom(X+Y,Z),1)\cong 1 \not\cong 2\cong\Set(\Hom(X,Z),1)+\Set(\Hom(Y,Z),1)$

Absolute Epimorphisms

What do absolute epimorphisms look like? We know that there are absolute epimorphisms because a split epimorphism is defined by a certain diagram commuting. Are there other absolute epimorphisms? To find out, we apply our trick.

Let $r:X\onto Y$ be our epimorphism. The we have the surjection $\Hom(Y,r):\Hom(Y,X)\onto\Hom(Y,Y)$ but this means that for every arrow $f:Y\to Y$, there’s an arrow $s:Y\to X$ such that $f = r \circ s$. As you can no doubt guess, we want to choose $f=id_Y$, and we then have that $r$ is a split epimorphism. Therefore split epimorphisms are the only examples of absolute epimorphisms.

Split Coequalizers

Now let’s consider the coequalizer case. Let $f,g:X\to Y$ and $e:Y\onto C$ be their coequalizer which we’ll assume is absolute. Before we pull out our trick, we can immediately use the previous result to show that $e$ has a section, i.e. an arrow $s : C\rightarrowtail Y$ such that $id_C=e\circ s$. Moving on we use the trick to get the diagram: $\Hom(Y,X)\rightrightarrows\Hom(Y,Y)\onto\Hom(Y,C)$

Next, we use the explicit construction of the coequalizer in $\Set$ which $\Hom(Y,C)$ is supposed to be canonically isomorphic to. That is, the coequalizer of $\Hom(Y,f)$ and $\Hom(Y,g)$ is $\Hom(Y,Y)$ quotiented by the equivalence relation generated by the relation which identifies $h,k:Y\to Y$ when $\exists j:Y\to X.h = f\circ j \land k = g\circ j$. Let $[h]$ represent the equivalence class of $h$. The claim that $\Hom(Y,C)$ is (with $\Hom(Y,e)$) a coequalizer of the above arrows implies that $e\circ h = \bar e([h])$ and $[h]=\bar e^{-1}(e\circ h)$ with $\bar e$ and $\bar e^{-1}$ forming an isomorphism. Of course our next move is to choose $h=id_Y$ giving $e=\bar e([id_Y])$. However, $e=e\circ s\circ e = \bar e([s\circ e])$ so we get $[id_Y]=[s\circ e]$ because $\bar e$ is invertible.

If we call the earlier relation $\sim$ and write $\sim^*$ for its reflexive, symmetric, transitive closure, then $[id_Y] = \{h:Y\to Y\mid id_Y\sim^* h\}$. Therefore $id_Y \sim^* s\circ e$. Now let’s make a simplifying assumption and assume further that $id_Y \sim s\circ e$, i.e. that $id_Y$ is directly related to $s\circ e$ by $\sim$. By definition of $\sim$ this means there is a $t : Y\to X$ such that $id_Y = f\circ t$ and $s\circ e = g\circ t$. Given $f$, $g$, and $e$ such that $e\circ f = e\circ g$ and equipped equipped with a $s : C\to Y$ and $t : Y\to X$ satisfying the previous two equations along with $q\circ s = id_C$ is called a split coequalizer. This data is specified diagrammatically and so is preserved by all functors, thus split coequalizers are absolute. All that we need to show is that this data is enough, on its own, to show that $e$ is a coequalizer.

Given any $q : Y\to Z$ such that $q\circ f = q\circ g$, we need to show that there exists a unique arrow $C\to Z$ which $q$ factors through. The obvious candidate is $q\circ s$ leading to us needing to verify that $q=q\circ s\circ e$. We calculate as follows: \begin{align} q \circ s \circ e & = q \circ g \circ t \\ & = q \circ f \circ t \\ & = q \end{align} Uniqueness then quickly follows since if $q = k\circ e$ then $q\circ s = k\circ e \circ s = k$. $\square$

There’s actually an even simpler example where $s\circ e = id_Y$ which corresponds to the trivial case where $f=g$.

Absolute Coequalizers

Split coequalizers show that (non-trivial) absolute coequalizers can exist, but they don’t exhaust all the possibilities. The obvious cause of this is the simplifying assumption we made where we said $id_Y\sim s\circ e$ rather than $id_Y\sim^* s\circ e$. In the general case, the equivalence will be witnessed by a sequence of arrows $t_i : Y\to X$ such that we have either $s\circ e = g\circ t_0$ or $s \circ e = f\circ t_0$, then $f\circ t_0 = g\circ t_1$ or $g\circ t_0 = f\circ t_1$ respectively, and so on until we get to $f\circ t_n = id_Y$ or $g\circ t_n = id_Y$. As a diagram, this is a fan of diamonds of the form $f\circ t_i = g\circ t_{i+1}$ or $g\circ t_i = f\circ t_{i+1}$ with a diamond with side $s\circ e$ on one end of the fan and a triangle with $id_Y$ on the other. All this data is diagrammatic so it is preserved by all functors making the coequalizer absolute. That it is a coequalizer uses the same proof as for split coequalizers except that we have a series of equalities to show that $q\circ s \circ e = q$, namely all those pasted together diamonds. There is no conceptual difficulty here; it’s just awkward to notate.

Absolute Colimits

The absolute coequalizer case captures the spirit of the general case, but you can see a description here. I’m not going to work through it, but you could as an exercise. Less tediously, you could work through absolute pushouts. If $P$ is the pushout of $Y \leftarrow X \to Z$, then the functors to consider are $\Hom(P,-)$ and $\Hom(Y,-)+\Hom(Z,-)$. For each, the pushout in $\Set$ can be turned into a coequalizer of a coproduct. For the first functor, as before, this gives us an inverse image of $id_P$ which will either be an arrow $P\to Y$ or an arrow $P\to Z$ which will play the role of $s$. The other functor produces a coequalizer of $\Hom(Y,Y)+\Hom(Z,Y)+\Hom(Y,Z)+\Hom(Z,Z)$. The generating relation of the equivalence relation states that there exists either an arrow $Y\to X$ or an arrow $Z\to X$ such that the appropriate equation holds. The story plays out much as before except that we have a sequence of arrows from two different hom-sets.


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# Sample Post: The Usage of Paper Towels

A friend of mine recently told me about a TED video on how to dry your hands with a single paper towel. The speaker, Joe Smith, presents some compelling numbers. He states that 13 billion (that’s $$13 \times 10^9$$) pounds of paper towels are used in the United States every year. He further suggests that if we restrict ourselves to using 1 sheet of paper towels every time we dry our hands, we could save 571,320,000 pounds of paper towels.

Certainly this figure sounds impressive! 500+ million pounds of paper towels is hard for me to even imagine. The problem is that whenever we consider the entire population of the United States, we will always get a large number. For example, if everyone drank 7 cups of water a day instead of the recommended 8, we could save $\frac{ 1 \text{ cup of water }}{\text{ 1 person, 1 day }} \times \frac{3 \times 10^8 \text{ people }}{ \text{ the US}} \times \frac{365 \text{ days }}{1 \text{ year }} \approx 9 \times 10^{10} \text{ cups per year in the US }$ or 90 trillion cups of water a year!  I’m not writing this to argue that we should drink less water. I’m simply saying that we can make impressive numbers by using large scales, like the population of the US, and long periods of time, like a year.

The figures in the TED talk are the same way. They are giving us a figure that happens in a large scale over a long period of time. So how can we determine if this is something we should actually focus on? Assuming we believe Smith’s oddly precise figure of 571,320,000 pounds of paper towels, let’s calculate the percentage saving on overall paper towel consumption.

$\frac{5.71 \times 10^8}{13 \times 10^9} \times 100\% \approx 4\%$

While 4% is not a lot, I think it’s big enough to warrant my attention (others may disagree on this). I still have two concerns, though:

We can, however, address the second problem. According to the EPA, the US uses 69 million tons of paper and paper board products each year. Smith’s method of drying, according to his numbers, we’d save $$5.71 \times 10^8$$ pounds of paper towels. Let’s convert that to tons: $5.71 \times 10^8 \text{ pounds of paper towels } \times \frac{ 1 \text{ ton }}{ 2,000 \text{ pound }} \approx 2.86 \times 10^5 \text{ tons of paper towels }$ or 286,000 tons. Compared with 69 million tons, that seems very small to me. Let’s find out the percent of paper saved from this method compared with the total amount of paper consumed: $\frac{2\times10^5}{6.9 \times 10^7} \approx 0.4\%$ The percentage is  in fact very small.


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Convergence for log 2

1. The series for $\log (1+x)$ is convergent(by estimating radius of convergence) only with $|x| <1$. Then how is it still true for $x = 1$?

2. How is it still apparently true when $x$ is a roots of unity other than $-1$? Such as in the answers to this question: Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$

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Perhaps you might add the series $\log \left( 1+x\right) =\displaystyle\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}x^{n}}{n}$, if that is what you mean. –  Américo Tavares Nov 24 '10 at 22:49
Knowing the radius of convergence doesn't, by itself, tell you anything about convergence on the boundary; you have to test for that separately. –  Qiaochu Yuan Nov 24 '10 at 23:03

Abel's theorem is the key to why you can say that the limit as you approach the boundary comes out to the same as evaluating the series at the boundary. Sivaram's answer addresses why the series converges at those boundary points.

Explicitly, Abel's theorem implies that

$$\lim_{r\nearrow 1}\log(1+zr)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}z^n$$

for each complex number $z$ with $|z|=1$ such that the series converges. By continuity of the logarithm, this will be $\log(1+z)$ in such cases.

-


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# A-level Physics (Advancing Physics)/Bosons/Worked Solutions

1. A stationary light source emits single photons at regular intervals. Draw a Feynman diagram to represent this.

2. Write two equations (including a W+ boson) which describe positron emission.

${\displaystyle p\to n+W^{+}}$

${\displaystyle W^{+}\to \nu +e^{+}}$

3. What is the charge on a W- boson?

-1.6 x 10−19C, since it has to balance the charge on a proton.


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# quantile

0th

Percentile

##### Sample Quantiles

The generic function quantile produces sample quantiles corresponding to the given probabilities. The smallest observation corresponds to a probability of 0 and the largest to a probability of 1.

Keywords
univar
##### Usage
quantile(x, ...)## S3 method for class 'default':
quantile(x, probs = seq(0, 1, 0.25), na.rm = FALSE,
names = TRUE, type = 7, ...)
##### Arguments
x
numeric vector whose sample quantiles are wanted, or an object of a class for which a method has been defined (see also details). NA and NaN values are not allowed in numeric vectors unless na.rm is TRUE.
probs
numeric vector of probabilities with values in $[0,1]$. (Values up to 2e-14 outside that range are accepted and moved to the nearby endpoint.)
na.rm
logical; if true, any NA and NaN's are removed from x before the quantiles are computed.
names
logical; if true, the result has a names attribute. Set to FALSE for speedup with many probs.
type
an integer between 1 and 9 selecting one of the nine quantile algorithms detailed below to be used.
...
further arguments passed to or from other methods.
##### Details

A vector of length length(probs) is returned; if names = TRUE, it has a names attribute.

NA and NaN values in probs are propagated to the result.

The default method works with classed objects sufficiently like numeric vectors that sort and (not needed by types 1 and 3) addition of elements and multiplication by a number work correctly. Note that as this is in a namespace, the copy of sort in base will be used, not some S4 generic of that name. Also note that that is no check on the correctly, and so e.g.quantile can be applied to complex vectors which (apart from ties) will be ordered on their real parts.

There is a method for the date-time classes (see "POSIXt"). Types 1 and 3 can be used for class "Date" and for ordered factors.

##### References

Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.

Hyndman, R. J. and Fan, Y. (1996) Sample quantiles in statistical packages, American Statistician 50, 361--365.

##### See Also

ecdf for empirical distributions of which quantile is an inverse; boxplot.stats and fivenum for computing other versions of quartiles, etc.

##### Aliases
• quantile
• quantile.default
##### Examples
library(stats) quantile(x <- rnorm(1001)) # Extremes & Quartiles by default quantile(x, probs = c(0.1, 0.5, 1, 2, 5, 10, 50, NA)/100) ### Compare different types p <- c(0.1, 0.5, 1, 2, 5, 10, 50)/100 res <- matrix(as.numeric(NA), 9, 7) for(type in 1:9) res[type, ] <- y <- quantile(x, p, type = type) dimnames(res) <- list(1:9, names(y)) round(res, 3)
Documentation reproduced from package stats, version 3.3, License: Part of R 3.3

### Community examples

Looks like there are no examples yet.


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# Abelian characteristic subgroup

This article describes a property that arises as the conjunction of a subgroup property: characteristic subgroup with a group property (itself viewed as a subgroup property): Abelian group
View a complete list of such conjunctions

## Definition

A subgroup $H$ of a group $G$ is termed an abelian characteristic subgroup if $H$ is abelian as a group (i.e., $H$ is an abelian subgroup of $G$ and also, $H$ is a characteristic subgroup of $G$, i.e., $H$ is invariant under all automorphisms of $G$.

## Examples

Here are some examples of subgroups in basic/important groups satisfying the property:

Group partSubgroup partQuotient part
A3 in S3Symmetric group:S3Cyclic group:Z3Cyclic group:Z2

Here are some examples of subgroups in relatively less basic/important groups satisfying the property:

Group partSubgroup partQuotient part
Center of dihedral group:D8Dihedral group:D8Cyclic group:Z2Klein four-group
Center of quaternion groupQuaternion groupCyclic group:Z2Klein four-group
Center of special linear group:SL(2,3)Special linear group:SL(2,3)Cyclic group:Z2Alternating group:A4
Center of special linear group:SL(2,5)Special linear group:SL(2,5)Cyclic group:Z2Alternating group:A5
Cyclic maximal subgroup of dihedral group:D8Dihedral group:D8Cyclic group:Z4Cyclic group:Z2
First agemo subgroup of direct product of Z4 and Z2Direct product of Z4 and Z2Cyclic group:Z2Klein four-group
First omega subgroup of direct product of Z4 and Z2Direct product of Z4 and Z2Klein four-groupCyclic group:Z2
Klein four-subgroup of alternating group:A4Alternating group:A4Klein four-groupCyclic group:Z3
Normal Klein four-subgroup of symmetric group:S4Symmetric group:S4Klein four-groupSymmetric group:S3

Here are some examples of subgroups in even more complicated/less basic groups satisfying the property:

Group partSubgroup partQuotient part
Center of M16M16Cyclic group:Z4Klein four-group
Center of dihedral group:D16Dihedral group:D16Cyclic group:Z2Dihedral group:D8
Center of direct product of D8 and Z2Direct product of D8 and Z2Klein four-groupKlein four-group
Center of semidihedral group:SD16Semidihedral group:SD16Cyclic group:Z2Dihedral group:D8
Cyclic maximal subgroup of dihedral group:D16Dihedral group:D16Cyclic group:Z8Cyclic group:Z2
Cyclic maximal subgroup of semidihedral group:SD16Semidihedral group:SD16Cyclic group:Z8Cyclic group:Z2
Derived subgroup of M16M16Cyclic group:Z2Direct product of Z4 and Z2
Derived subgroup of dihedral group:D16Dihedral group:D16Cyclic group:Z4Klein four-group
Direct product of Z4 and Z2 in M16M16Direct product of Z4 and Z2Cyclic group:Z2
Klein four-subgroup of M16M16Klein four-groupCyclic group:Z4
Non-central Z4 in M16M16Cyclic group:Z4Cyclic group:Z4

## Relation with other properties

### Stronger properties

Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
cyclic characteristic subgroup cyclic group and a characteristic subgroup of the whole group |FULL LIST, MORE INFO
characteristic central subgroup central subgroup (i.e., contained in the center) and a characteristic subgroup of the whole group |FULL LIST, MORE INFO
abelian critical subgroup |FULL LIST, MORE INFO
characteristic subgroup of center characteristic subgroup of the center Characteristic central subgroup|FULL LIST, MORE INFO
characteristic subgroup of abelian group the whole group is an abelian group and the subgroup is characteristic Characteristic central subgroup, Characteristic subgroup of center|FULL LIST, MORE INFO
maximal among abelian characteristic subgroups |FULL LIST, MORE INFO
abelian fully invariant subgroup abelian and a fully invariant subgroup -- invariant under all endomorphisms |FULL LIST, MORE INFO

### Weaker properties

Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
abelian normal subgroup abelian and a normal subgroup -- invariant under all inner automorphisms follows from characteristic implies normal follows from normal not implies characteristic in the collection of all groups satisfying a nontrivial finite direct product-closed group property |FULL LIST, MORE INFO
abelian subnormal subgroup abelian and a subnormal subgroup |FULL LIST, MORE INFO
class two characteristic subgroup characteristic subgroup that is also a group of nilpotency class two |FULL LIST, MORE INFO
nilpotent characteristic subgroup characteristic subgroup that is also a nilpotent group follows from abelian implies nilpotent follows from nilpotent not implies abelian Class two characteristic subgroup|FULL LIST, MORE INFO
solvable characteristic subgroup characteristic subgroup that is also a solvable group (via nilpotent) (via nilpotent) Nilpotent characteristic subgroup|FULL LIST, MORE INFO


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# desoldering stations

#### genec

Jul 26, 2015
13
hello all! i'm in need a desderoing station or gun. i'm not sure which is better..help? it needs to be under $200. also wanted to know if there is a station that has the hot air+desoldering gun both to one single station. #### (*steve*) ##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator Jan 21, 2010 25,505 For surface mount, the reflow tool can be used to solder or remove devices. For through-hole work, a vacuum tool is probably the best. I'm pretty sure I paid well under$US200 for mine. If you take a look at Dave Jones' review of a desoldering pump on eevblog.com, mine is similar.

There are other options, including those having the pump in the handpiece. They're cheaper, but I've not had any experience with them.

In order to get good performance I had to modify mine. I replaced a spring in the solder collection chamber with some material cut from a wire solder sponge. It traps more solder and allows greater airflow.

It's also a tool you really need to learn to use. It's *really* easy to do a very bad job with them -- in this case "bad" means damaging the PCB.

#### genec

Jul 26, 2015
13
well i wish i could get dave's pump, but where he got it from in australia they only sell locally, so thats a bust. i think i'm going to get just a station with a pump/vacuum type deal....so i get the station pump deal do i stil need the hot air wand thingy.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
well i wish i could get dave's pump, but where he got it from in australia they only sell locally, so thats a bust. i think i'm going to get just a station with a pump/vacuum type deal....so i get the station pump deal do i stil need the hot air wand thingy.

You will need the hot air want thingy if you're planning to do any reflow soldering or remove surface mount components.

You can do it with a normal soldering iron (at least for the larger stuff) and for the really small stuff you really need an oven as the rework tool can easily blow components off the board :-(

#### genec

Jul 26, 2015
13
well i'm working on psu from a computer and i'm horrible at the wick and solder iron.

J
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