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# Examples of Mathematical Sequences

Now, you’ll get to see sequences where you find the next number in the sequence by multiplying the previous number by another number, or subtracting a number from the previous number.

Example 1

The sequence

 $1,\phantom{\rule{0.33em}{0ex}}2,\phantom{\rule{0.33em}{0ex}}4,\phantom{\rule{0.33em}{0ex}}8,\phantom{\rule{0.33em}{0ex}}16,\dots$

has the property that you multiply the previous number by $2$ to find the next number:

$1\cdot 2=2$, the first number of the sequence multiplied by $2$ is the second number of the sequence.

$2\cdot 2=4$, the second number of the sequence multiplied by $2$ is the third number of the sequence.

$4\cdot 2=8$, the third number of the sequence multiplied by $2$ is the fourth number of the sequence.

$8\cdot 2=16$, the fourth number of the sequence multiplied by $2$ is the fifth number of the sequence.

$16\cdot 2=32$, the fifth number of the sequence multiplied by $2$ is the sixth number of the sequence.

Example 2

The sequence

 $40,\phantom{\rule{0.33em}{0ex}}35,\phantom{\rule{0.33em}{0ex}}30,\phantom{\rule{0.33em}{0ex}}25,\phantom{\rule{0.33em}{0ex}}20,\dots$

has the property that $5$ is subtracted from the previous number in the sequence to get the next number in the sequence:

$40-5=35$, the first number of the sequence minus $5$ is the second number of the sequence.

$35-5=30$, the second number of the sequence minus $5$ is the third number of the sequence.

$30-5=25$, the third number of the sequence minus $5$ is the fourth number of the sequence.

$25-5=20$, the fourth number of the sequence minus $5$ is the fifth number of the sequence.

$20-5=15$, the fifth number of the sequence minus $5$ is the sixth number of the sequence.

Are there sequences where you find the next number by dividing by a specific number?

Yes, there are. An example of this is the sequence

 $192,\phantom{\rule{0.33em}{0ex}}96,\phantom{\rule{0.33em}{0ex}}48,\phantom{\rule{0.33em}{0ex}}24,\phantom{\rule{0.33em}{0ex}}12,\phantom{\rule{0.33em}{0ex}}6,\phantom{\rule{0.33em}{0ex}}3,\dots$

The pattern here is that each number in the sequence is the previous number divided by $2$. $192:2=96$, the first number of the sequence divided by $2$ is the second number of the sequence.


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Proof for Sine of 75 degrees in Trigonometric method

The exact value of sine of angle seventy five degrees in fraction form is equal to the quotient of square root of six plus square root of two by four.

$\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{6}+\sqrt{2}}{4}$

The sine of angle five pi by twelve radian can be proved exactly in fraction form in trigonometric mathematics. So, let us learn how to derive the sine seventy five degrees in trigonometry.

Angle sum identity of sine function

The angle $75^\circ$ can be written as the sum of two known angles $30^\circ$ and $45^\circ$. Hence, the sine of angle $75$ degrees can be written as the sine of sum of them.

$\sin{(75^\circ)}$ $\,=\,$ $\sin{(45^\circ+30^\circ)}$

The sine of sum of the angles $45^\circ$ and $30^\circ$ can be expanded by the sine angle sum identity.

$\sin{(x+y)}$ $\,=\,$ $\sin{x}\cos{y}$ $+$ $\cos{x}\sin{y}$

Let $x \,=\, 45^\circ$ and $y \,=\, 30^\circ$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\sin{(45^\circ+30^\circ)}$ $\,=\,$ $\sin{(45^\circ)}\cos{(30^\circ)}$ $+$ $\cos{(45^\circ)}\sin{(30^\circ)}$

Substitute values of sine and cosine functions

According to the trigonometry, the exact values of sin 30 degrees, sin 45 degrees, cos 30 degrees and cos 45 degrees are already known to us.

$(1).\,\,\,$ $\sin{(45^\circ)}$ $\,=\,$ $\dfrac{1}{\sqrt{2}}$

$(2).\,\,\,$ $\cos{(30^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}}{2}$

$(3).\,\,\,$ $\cos{(45^\circ)}$ $\,=\,$ $\dfrac{1}{\sqrt{2}}$

$(4).\,\,\,$ $\sin{(30^\circ)}$ $\,=\,$ $\dfrac{1}{2}$

Now, we substitute them in the expansion of the sine of sum of angles $30$ and $45$ degrees expression.

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\sin{(45^\circ+30^\circ)}$ $\,=\,$ $\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2}$ $+$ $\dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2}$ $+$ $\dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}$

Evaluate the sine 75 degrees by simplification

The sine of angle $75$ degrees is expressed as a mathematical expression in numerical form. In order to calculate it, we have to simplify the right hand expression of the equation.

$\sin{(75^\circ)}$ $\,=\,$ $\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2}$ $+$ $\dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}$

Each term in the right hand side expression of the equation represents the multiplication of two fractions. So, let’s evaluate the product of the fractions for each term by the multiplication of the fractions.

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{1 \times \sqrt{3}}{\sqrt{2} \times 2}$ $+$ $\dfrac{1 \times 1}{\sqrt{2} \times 2}$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{1 \times \sqrt{3}}{2 \times \sqrt{2}}$ $+$ $\dfrac{1 \times 1}{2 \times \sqrt{2}}$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}}{2\sqrt{2}}$ $+$ $\dfrac{1}{2\sqrt{2}}$

The right hand side expression expresses an addition of the two like fractions. So, we can find their sum by the addition of the like fractions.

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$

It can be further simplified by eliminating the square root of $2$ from the denominator of the right hand side expression of the equation.

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ $\times$ $1$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ $\times$ $\dfrac{\sqrt{2}}{\sqrt{2}}$

Now, multiply the fractions for simplifying the mathematical expression of sine of $5\pi$ by $12$.

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{(\sqrt{3}+1) \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}}$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{2} \times (\sqrt{3}+1)}{2 \times \sqrt{2} \times \sqrt{2}}$

In the denominator, the factors can be expressed in exponential notation.

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{2} \times (\sqrt{3}+1)}{2 \times (\sqrt{2})^2}$

In the numerator, the multiplication of square root of $2$ can be distributed to the sum of the square root of $3$ and one by the distributive property of multiplication over addition.

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{2} \times \sqrt{3}+ \sqrt{2} \times 1}{2 \times 2}$

$\implies$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{2 \times 3}+\sqrt{2}}{4}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(75^\circ)}$ $\,=\,$ $\dfrac{\sqrt{6}+\sqrt{2}}{4}$

Therefore, the exact value for the sine of angle seventy five degrees in fraction form is equal to the quotient of square root of six plus square root of two by four.

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# R wavelets package inverse dwt does not always recover original signal

I recently worked with R's "wavelets" package. I tried the very simple dwt and inverse dwt as follows:

yi = idwt(dwt(y));
plot(y,yi);

where y is a vector. Sometimes y=yi, however, other times, yi does not equal y, they are very different, so that the scatterplot is completely random. This also seems to happen randomly, even with default setting, or following some suggestions, use the option boundary="reflection" . Why is that? The limited documentation says idwt "should" recover the signal. This looks very wrong to me.

Does anyone know a "working" package of wavelet analysis for R?


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In equation ___, the relationship between the velocity and the shear stress was assumed to be linear. Not all the materials obey this relationship. There is a large class of materials which shows a non-linear relationship with velocity for any shear stress. This class of materials can be approximated by a single polynomial term that is $$a = bx^n$$. From the physical point of view, the coefficient depends on the velocity gradient. This relationship is referred to as power relationship and it can be written as $\tau = \overbrace{K (\frac{dU}{dx})^{n-1}}^{viscosity} \left(\frac{dU}{dx}\right)\tag{16}$ The new coefficients $$(n, K)$$ in equation ___ are constant. When $$n = 1$$ equation represent Newtonian fluid and $$K$$ becomes the familiar $$\mu$$. The viscosity coefficient is always positive. When $$n$$ is above one, the liquid is dilettante. When $$n$$ is below one, the fluid is pseudoplastic. The liquids which satisfy equation ___ are referred to as purely viscous fluids. Many fluids satisfy the above equation. Fluids that show increase in the viscosity (with increase of the shear) referred to as thixotropic and those that show decrease are called rheopectic fluids (see Figure 1.5).
Materials which behave up to a certain shear stress as a solid and above it as a liquid are referred as Bingham liquids. In the simple case, the liquid side'' is like Newtonian fluid for large shear stress. The general relationship for simple Bingham flow is $\tau_{xy} = -\mu \cdot \pm \tau_{0} \quad if \lvert\tau_{yx}\rvert > \tau_{0}\tag{17}$ $\frac{dU_{x}}{dy} = 0 \quad if\lvert\tau_{yx}\rvert < \tau_{0}\tag{18}$ There are materials that simple Bingham model does not provide adequate explanation and a more sophisticate model is required. The Newtonian part of the model has to be replaced by power liquid. For example, according to Ferraris at el concrete behaves as shown in Figure 1.7. However, for most practical purposes, this kind of figures isn't used in regular engineering practice.


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## Calculus: Early Transcendentals 8th Edition

$\Sigma^{\infty}_{n=1} \frac{1}{1+(\frac{2}{3})^{n}}$ $\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty}\frac{1}{1+(\frac{2}{3})^{n}}$ $=\frac{1}{1+0}$ $=1 \ne 0$ Since $\lim\limits_{n \to \infty}a_{n} \ne 0$, the series is divergent.


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# Damn Damping

Level pending

The free damping equation is 2y"+3y'+8y=0.Find the free damping factor in the form $\frac{a}{b}$ where a and b are co-prime positive integers.Find a+b.

×


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An irregular quadrilateral has lengths $1, 4, 7$ and $8$ units. What is the maximum area of the quadrilateral possible?

To find the area of an irregular quadrilateral, five values are needed; four sides along with an angle or the size of at least one diagonal. However, I am not able to figure out how to get one angle or the size of one diagonal?

• As you point out, figuring out that value is the problem. Give the unknown angle or diagonal a name, say $x$. Write the area as a function of $x$. Then use calculus to find the maximum. (There may be a geometric theorem that solves this problem, but I don't know one.) – Ethan Bolker May 23 '18 at 12:34

Given any quadrilateral $Q$ with side $a,b,c,d$. Let $s = \frac12(a+b+c+d)$ be the semi-perimeter and $2\theta$ be the sum of two opposite angles of $Q$. The area of $Q$ is given by the Bretschneider's formula

$$\verb/Area/ = \sqrt{(s-a)(s-b)(s-b)(s-c) - abcd\cos^2\theta}$$

It is clear this area is maximized when $\cos\theta = 0$, i.e. when $Q$ is a cyclic quadrilateral. In that case, the formula reduce to Brahmagupta's formula:

$$\verb/Area/ = \sqrt{(s-a)(s-b)(s-b)(s-c)}$$

For the problem at hand, $(a,b,c,d) = (1,4,7,8) \implies s = 10$. The desired maximum area is $$\sqrt{(10-1)(10-4)(10-7)(10-8)} = 18$$

• Thank You @achille hui ... Great help! ... This is the correct solution ... However, the answer is 18..! :-) – Math Tise May 24 '18 at 6:41
• Thank You again @achille ... :-) – Math Tise May 24 '18 at 6:41
• @MathTise Hmm... looks like I plug some wrong numbers into the correct formula. $\sqrt{(10-1)(10-4)(10-7)(10-8)}$ do equal to $18$. – achille hui May 24 '18 at 7:22

Let the diagonal be $d$. By Heron's formula

$$2A=\sqrt{(1+4+d)(1+4-d)(1-4+d)(-1+4+d)}+\sqrt{(7+8+d)(7+8-d)(7-8+d)(-7+8+d)}\\ =\sqrt{(5^2-d^2)(d^2-3^2)}+\sqrt{(15^2-d^2)(d^2-1^2)}.$$

You can optimize for $d^2$.

This leads to a polynomial equation as

$$\left(\sqrt{p(d_2)}+\sqrt{q(d_2)}\right)'=0$$ when

$$p'^2(d_2)\,q(d_2)=p(d_2)\,q'^2(d_2).$$

As the polynomials in the two members have the same leading coefficient, the equation is cubic in $d_2$.


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Question

# In the equation, rate$={\mathrm{Z}}_{\mathrm{AB}}{\mathrm{e}}^{\frac{-\mathrm{Ea}}{\mathrm{RT}}}$what does ${\mathrm{Z}}_{\mathrm{AB}}$ means?

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Solution

## Collision Theory:For a chemical reaction to begin, the reacting molecules must approach and collide. The bonds between the interacting molecules are severed as a result of these collisions, and the product is created with new bonds.To explain the reaction rates in a gaseous state, Arrhenius presented a simple collision theory. Kinetic molecular theory or reaction rate collision theory is the name given to this theory.Arrhenius equation is given by rate$={\mathrm{Ae}}^{\frac{-\mathrm{Ea}}{\mathrm{RT}}}$ where A is a pre-exponential factor or orientation factor which tells the orientation of the molecule.For reaction $\mathrm{A}+\mathrm{B}\to \mathrm{AB}$, the Pre exponential factor is also given by A$={\mathrm{PZ}}_{\mathrm{AB}}{\mathrm{e}}^{\frac{-\mathrm{Ea}}{\mathrm{RT}}}$ where P represents the probability factor, ZAB represents collision frequency, Ea represents activation energy, R represents gas constant, and T represents temperature. Therefore, the collision frequency of the reactants A and B are represented by ZAB.

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Translator Disclaimer
VOL. 1 | 2018 Chapter 4. Points algébriques de degré donné sur la courbe de Picard
Moussa FALL, Oumar SALL

Editor(s) Hamet SEYDI, Gane Samb LO, Aboubakary DIAKHABY

## Abstract

We give a parameterization of the algebraic points of given degree over $\mathbb{Q}$ on the Picard curve given by $y^{3}=x^{4}-1$. This result extends a previous result of Klassen and Schaefer(1996) on the set of algebraic points of degree at most $3$ over $\mathbb{Q}$ in the same curve.

## Information

Published: 1 January 2018
First available in Project Euclid: 26 September 2019

Digital Object Identifier: 10.16929/sbs/2018.100-01-04

Subjects:
Primary: 11D68, 12F05, 14H50, 14hH40


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# A dig-your-own potato farm charges $1.27 per pound of potatoes plus$0.20 per pound to clean the potatoes. Kevin digs 30 pounds of potatoes. The farm cleans 8 pounds for him. How much does Kevin pay in all?

###### Question:

A dig-your-own potato farm charges $1.27 per pound of potatoes plus$0.20 per pound to clean the potatoes. Kevin digs 30 pounds of potatoes. The farm cleans 8 pounds for him. How much does Kevin pay in all?

### How would fiscal discipline by the government help the economy?

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The 2016 annual report for Mega Mills disclosed that 1 billion shares of common stock have been authorized. At the end of 2015, 800 million shares had been issued and the number of shares in treasury stock was 109 million. During 2016, the only common share transactions were that 17 million common s...

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The Peloponnesian War a. broke out when Greece was united under Pericles. b. was the result of the plague. c. began because older Spartans wanted to return to fighting. d. began with quarrels over trade....

### Does 2x+7=3x+1 explain

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### DUE TODAY PLEASE HELP.....WILL GIVE BRANIST!!!!!!! Instructions: Create a comic strip detailing the adventure of your character as the character is exposed to thermal energy, causing it to undergo phase changes from a solid, to a liquid, to a gas. Place drawings inside the boxes and written content on the lines below each box. Your presentation must include the following: • title and introduction of your character, including what substance it is made of • source of thermal energy your character

DUE TODAY PLEASE HELP.....WILL GIVE BRANIST!!!!!!! Instructions: Create a comic strip detailing the adventure of your character as the character is exposed to thermal energy, causing it to undergo phase changes from a solid, to a liquid, to a gas. Place drawings inside the boxes and written content ...

### In what way is Montresor an unreliable narrator? How does the use of an unreliable narrator influence readers’ views of Montresor’s actions? Cite evidence from the text to support your response.

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-- 0.009910--


text:
• Create Account

### #Actualsliders_alpha

Posted 05 May 2012 - 12:14 PM

the thing is, the world is generated as the player explore (minecraft clone).
if the player only goes toward the east this one file containing everything will be filled with 75% of 0.

Someone else told me that actually using a database such as MySQL could be usefull.
1 column for the coordinates (primary key), 1 column for the data.
Since I only open a session at the beginning of the game I won't lose time like when I need to acces 900 separated text files.

In addition to that I can even uses indexes for equality seach (on the coordinates) to speed those database access.

....
calculating the coordinate is not time consuming since the player position on the sector grid is always being tracked.
a sector contains more information but I'm only storing the data, since the coordinate are stored inside, storing the file with it's coordinate name is also quick.
public class Sector{
byte[][][] data;
int x,y;
int DL;

//some methods
}

I also noticed something weird today, using System.currentTimeMillis()
opening a file = 0ms
writing = 65ms
closing = 0 ms
same thing for reading, shouldn't opening the file be time consuming?

### #2sliders_alpha

Posted 05 May 2012 - 12:10 PM

the thing is, the world is generated as the player explore (minecraft clone).
if the player only goes toward the east this one file containing everything will be filled with 75% of 0.

Someone else told me that actually using a database such as MySQL could be usefull.
1 column for the coordinates (primary key), 1 column for the data.
Since I only open a session at the beginning of the game I won't lose time like when I need to acces 900 separated text files.

In addition to that I can even uses indexes for equality seach (on the coordinates) to speed those database access.

....
calculating the coordinate is not time consuming since the player position on the sector grid is always being tracked.
a sector contains more information but I'm only storing the data, since the coordinate are stored inside, storing the file with it's coordinate name is also quick.
public class Sector{
byte[][][] data;
int x,y;
int DL;

//some methods
}

### #1sliders_alpha

Posted 05 May 2012 - 12:10 PM

the thing is, the world is generated as the player explore (minecraft clone).
if the player only goes toward the east this one file containing everything will be filled with 75% of 0.

Someone else told me that actually using a database such as MySQL could be usefull.
1 column for the coordinates (primary key), 1 column for the data.
Since I only open a session at the beginning of the game I won't lose time like when I need to acces 900 separated text files.

In addition to that I can even uses indexes for equality seach (on the coordinates) to speed those database access.

....
calculating the coordinate is not time consuming since the player position on the sector grid is always being tracked.
a sector contains more information but I'm only storing the data, since the coordinate are stored inside, storing the file with it's coordinate name is also quick.
public class Sector{
byte[][][] data;
int x,y;
int DL;
}

PARTNERS


text:
# Tag Info

6

I think the Investopedia article does not use clear terminology which leads to confusion. In Economics, utility can be split across two categories as ordinal and cardinal utility. In the case of ordinal utility assigning numeric values is meaningless. For example, if A is proffered to B, assigning 200 utils to A and 100 utils to B is meaningless because that ...

4

Utility functions as ordinarily used are not a measure of well-being comparable among people, but a representation of preferences. Moreover, preferences could principally be elicited from choice experiments. A utility function assigns real numbers to alternatives so that one alternative is preferred to the other if and only if it is assigned a higher number. ...

4

$u = \max(x, y)$ represents the preferences over two substitute goods that cannot be consumed together. For example - tea and coffee. In the event that the consumer gets x quantity of tea and y quantity of coffee, consumer choose to consume only one of the them depending on the quantity. He always choose the one that is offered in larger quantity and throws ...

3

Your thinking is correct that, in some ways, $x_1, x_2$ are substitute goods. We define substitute goods which have the following property: $$\left.\frac{\partial x_i}{\partial p_j}\right|_{u=\bar u}>0$$ The case of $U(x_1,x_2)=\max\{x_1,x_2\}$ is that of a boundary solution as the indifference curves are now concave to the origin. So equilibrium solution ...

2

This is trivially not true. Consider simple example of utility: $$u(x) = x^{1/2}$$ Expected utility $E(u(x)) = E[x^{1/2}]$ Inverse utility is $u^{-1} \implies x = u^2$ clearly generally $E(u) \neq u^{-1}$.

2

There are no simple methods for estimating cardinal utility (ordinal utility would be a different matter - you could just observe few of your choices). This is not because cardinal utility would necessary be unmeasurable. Although this is not completely settled question (see Moscati (2018) Measuring Utility: From the Marginal Revolution to Behavioral ...

2

Yes. In general not. Let's say the individual has initial wealth $W$ and the gamble $g$ has payouts $0$ and $G$, each with probability $1/2$. As you say, the certainty equivalent $C$ of the gamble is the amount $C$ with $$u(W+C)=(u(W)+u(W+G))/2.$$ Now the same individual would be willing to pay at most $P$ to enter the gamble, where u(W)=(u(W-P)+u(W+G-P))/...

1

If you assume that agents' utility functions over pairs $(x,t)$ of consumption bundles $x$ and monetary transfers $t$ are quasilinear in money, $u_i(x,t)=v_i(x)+t$, then $v_i(x)$ measures $i$'s WTP for $x$. In a utilitarian framework the sum of utilities is then a social welfare function satisfying Arrow's axioms (apart from universal domain of course) and ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# The IAC Stripe82 Legacy Survey: Improved Sky-rectified Images

Román, J.; Trujillo, I.
Referencia bibliográfica

Research Notes of the American Astronomical Society, Volume 2, Issue 3, article id. 144, 0 pp. (2018).

Fecha de publicación:
8
2018
Descripción
The IAC Stripe 82 Legacy Survey is a new co-addition of the SDSS Stripe 82 data (Abazajian et al. 2009), especially reduced to preserve the faintest surface brightness features of this data set. The survey maps a 2.5 degree wide stripe along the Celestial Equator in the Southern Galactic Cap (-50$^{\circ}$ $<$ R.A. $<$ 60$^{\circ}$, -1.25$^{\circ}$ $<$ Dec.$<$ 1.25$^{\circ}$) with a total of 275 square degrees in all the five SDSS filters (u,g,r,i,z). The new reduction includes an additional and deeper band (rdeep), which is a combination of g, r and i bands. The average seeing of the Stripe82 dataset is around 1 arcsec. The mean surface brightness limits are $\mu_{lim}$[3$\sigma$,10$\times$10 arcsec$^{2}$] = 27.9, 29.1, 28.6, 28.1 and 26.7 mag arcsec$^{-2}$ for the u, g, r, i and z bands respectively. The significant depth of the data and the emphasis on preserving the characteristics of the background (sky + diffuse light) through a non-aggressive sky subtraction strategy make this dataset suitable for the study of the low surface brightness Universe. In this research note we present a new data release with improved sky-rectified images. This new data-set is published on the survey webpage and is publicly available for the community.


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# A container with a volume of 15 L contains a gas with a temperature of 290^o K. If the temperature of the gas changes to 350 ^o K without any change in pressure, what must the container's new volume be?

$\approx 18.1 L$.

#### Explanation:

According to ideal gas law,

$P \cdot V = n \cdot R \cdot T$

Since,
$P , n , R$ are constant throughout the process,

$\frac{V}{T} = \frac{n \cdot R}{P} = c o n s \tan t$

Therefore,

$V \propto T$

${V}_{\text{initial"/T_"initial"=V_"final"/T_"final}}$

So,
${V}_{\text{final"=(V_"initial"*T_"final")/T_"initial}} = \frac{15 \cdot 350}{290} \approx 18.1 L$

The answer is $18.1 L$

May 15, 2016

${V}_{2} = 18.1 L$

#### Explanation:

$P = \text{constant}$
${V}_{1} = 15 L$
${T}_{1} = {290}^{o} K$

${T}_{2} = {350}^{o} K$
V_2=?

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

$\frac{15}{290} = {V}_{2} / 350$

${V}_{2} = \frac{15 \cdot 350}{290}$

${V}_{2} = 18.1 L$


text:
Lessons:16Length:1.5 hours
• Overview
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# 3.3 Update Query Basics

Uh oh! You've identified a problem with your data. Or maybe you just need to change some fields. Either way, it's important to understand how to make changes to your data once it is in your tables. Let's go over the basics of the UPDATE query.

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• Can. J. Math. (IF 0.881) Pub Date : 2020-11-04
Man-Chun Lee; Luen-Fai Tam

In this work, we obtain a local maximum principle along the Ricci flow $g(t)$ under the condition that $\mathrm {Ric}(g(t))\le {\alpha } t^{-1}$ for $t>0$ for some constant ${\alpha }>0$ . As an application, we will prove that under this condition, various kinds of curvatures will still be nonnegative for $t>0$ , provided they are non-negative initially. These extend the corresponding known results

更新日期：2020-11-25
• Can. J. Math. (IF 0.881) Pub Date : 2020-10-12
Christopher Manon; Jihyeon Jessie Yang

We construct a family of compactifications of the affine cone of the Grassmannian variety of $2$ -planes. We show that both the tropical variety of the Plücker ideal and familiar valuations associated to the construction of Newton–Okounkov bodies for the Grassmannian variety can be recovered from these compactifications. In this way, we unite various perspectives for constructing toric degenerations

更新日期：2020-11-21
• Can. J. Math. (IF 0.881) Pub Date : 2020-10-28
John Cullinan; Jeffrey Yelton

Let A be a two-dimensional abelian variety defined over a number field K. Fix a prime number $\ell$ and suppose $\#A({\mathbf {F}_{\mathfrak {p}}}) \equiv 0 \pmod {\ell ^2}$ for a set of primes ${\mathfrak {p}} \subset {\mathcal {O}_{K}}$ of density 1. When $\ell =2$ Serre has shown that there does not necessarily exist a K-isogenous $A'$ such that $\#A'(K)_{{tor}} \equiv 0 \pmod {4}$ . We extend

更新日期：2020-11-17
• Can. J. Math. (IF 0.881) Pub Date : 2020-09-17
Ágnes Backhausz; Balázs Szegedy

We present a new approach to graph limit theory that unifies and generalizes the two most well-developed directions, namely dense graph limits (even the more general $L^p$ limits) and Benjamini–Schramm limits (even in the stronger local-global setting). We illustrate by examples that this new framework provides a rich limit theory with natural limit objects for graphs of intermediate density. Moreover

更新日期：2020-11-16
• Can. J. Math. (IF 0.881) Pub Date : 2020-10-12
Jiawei Lin; Greg Martin

Let $a_1$ , $a_2$ , and $a_3$ be distinct reduced residues modulo q satisfying the congruences $a_1^2 \equiv a_2^2 \equiv a_3^2 \ (\mathrm{mod}\ q)$ . We conditionally derive an asymptotic formula, with an error term that has a power savings in q, for the logarithmic density of the set of real numbers x for which $\pi (x;q,a_1)> \pi (x;q,a_2) > \pi (x;q,a_3)$ . The relationship among the $a_i$ allows

更新日期：2020-11-13
• Can. J. Math. (IF 0.881) Pub Date : 2020-09-02
Niels Lubbes

We classify real two-dimensional orbits of conformal subgroups such that the orbits contain two circular arcs through a point. Such surfaces must be toric and admit a Möbius automorphism group of dimension at least two. Our theorem generalizes the classical classification of Dupin cyclides.

更新日期：2020-09-02
• Can. J. Math. (IF 0.881) Pub Date : 2020-09-02
Maria Fărcăşeanu; Mihai Mihăilescu; Denisa Stancu-Dumitru

The asymptotic behavior of solutions to a family of Dirichlet boundary value problems, involving differential operators in divergence form, on a domain equipped with a Finsler metric is investigated. Solutions are shown to converge uniformly to the distance function to the boundary of the domain, which takes into account the Finsler norm involved in the equation. This implies that a well-known result

更新日期：2020-09-02
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-24
Gregory Arone; Markus Szymik

Suppose that $N_1$ and $N_2$ are closed smooth manifolds of dimension n that are homeomorphic. We prove that the spaces of smooth knots, $\operatorname {\mathrm {Emb}}(\mathrm {S}^1, N_1)$ and $\operatorname {\mathrm {Emb}}(\mathrm {S}^1, N_2),$ have the same homotopy $(2n-7)$ -type. In the four-dimensional case, this means that the spaces of smooth knots in homeomorphic $4$ -manifolds have sets

更新日期：2020-08-24
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-20
Daniel López Neumann

We construct quantum invariants of balanced sutured 3-manifolds with a ${\text {Spin}^c}$ structure out of an involutive (possibly nonunimodular) Hopf superalgebra H. If H is the Borel subalgebra of ${U_q(\mathfrak {gl}(1|1))}$ , we show that our invariant is computed via Fox calculus, and it is a normalization of Reidemeister torsion. The invariant is defined via a modification of a construction of

更新日期：2020-08-20
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-20
Gwénaël Massuyeau; Delphine Moussard

We prove a “splicing formula” for the LMO invariant, which is the universal finite-type invariant of rational homology three-spheres. Specifically, if a rational homology three-sphere M is obtained by gluing the exteriors of two framed knots $K_1 \subset M_1$ and $K_2\subset M_2$ in rational homology three-spheres, our formula expresses the LMO invariant of M in terms of the Kontsevich–LMO invariants

更新日期：2020-08-20
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-20
Weichao Guo; Huoxiong Wu; Dongyong Yang

A new characterization of $\text {CMO}(\mathbb R^n)$ is established replying upon local mean oscillations. Some characterizations of iterated compact commutators on weighted Lebesgue spaces are given, which are new even in the unweighted setting for the first order commutators.

更新日期：2020-08-20
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-11
Zakhar Kabluchko; Daniel Temesvari; Christoph Thäle

A new approach to prove weak convergence of random polytopes on the space of compact convex sets is presented. This is used to show that the profile of the rescaled Schläfli random cone of a random conical tessellation, generated by n independent and uniformly distributed random linear hyperplanes in $\mathbb {R}^{d+1}$ , weakly converges to the typical cell of a stationary and isotropic Poisson hyperplane

更新日期：2020-08-11
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-10
Zoltán Füredi; Tao Jiang; Alexandr Kostochka; Dhruv Mubayi; Jacques Verstraëte

An ordered hypergraph is a hypergraph whose vertex set is linearly ordered, and a convex geometric hypergraph is a hypergraph whose vertex set is cyclically ordered. Extremal problems for ordered and convex geometric graphs have a rich history with applications to a variety of problems in combinatorial geometry. In this paper, we consider analogous extremal problems for uniform hypergraphs, and determine

更新日期：2020-08-10
• Can. J. Math. (IF 0.881) Pub Date : 2020-08-07
Daniel Gonçalves; Benjamin Steinberg

Given an action ${\varphi }$ of inverse semigroup S on a ring A (with domain of ${\varphi }(s)$ denoted by $D_{s^*}$ ), we show that if the ideals $D_e$ , with e an idempotent, are unital, then the skew inverse semigroup ring $A\rtimes S$ can be realized as the convolution algebra of an ample groupoid with coefficients in a sheaf of (unital) rings. Conversely, we show that the convolution algebra of

更新日期：2020-08-07
• Can. J. Math. (IF 0.881) Pub Date : 2020-07-30
Janne Heittokangas; Hui Yu; Mohamed Amine Zemirni

A classical theorem of Frei states that if $A_p$ is the last transcendental function in the sequence $A_0,\ldots ,A_{n-1}$ of entire functions, then each solution base of the differential equation $f^{(n)}+A_{n-1}f^{(n-1)}+\cdots +A_{1}f'+A_{0}f=0$ contains at least $n-p$ entire functions of infinite order. Here, the transcendental coefficient $A_p$ dominates the growth of the polynomial coefficients

更新日期：2020-07-30
• Can. J. Math. (IF 0.881) Pub Date : 2020-07-17
Andrew Graham; Daniel R. Gulotta; Yujie Xu

Let f and g be two cuspidal modular forms and let ${\mathcal {F}}$ be a Coleman family passing through f, defined over an open affinoid subdomain V of weight space $\mathcal {W}$ . Using ideas of Pottharst, under certain hypotheses on f and g, we construct a coherent sheaf over $V \times \mathcal {W}$ that interpolates the Bloch–Kato Selmer group of the Rankin–Selberg convolution of two modular forms

更新日期：2020-07-27
• Can. J. Math. (IF 0.881) Pub Date : 2020-07-27
Omer Angel; Alexander E. Holroyd; Tom Hutchcroft; Avi Levy

We show that the Mallows measure on permutations of $1,\dots ,n$ arises as the law of the unique Gale–Shapley stable matching of the random bipartite graph with vertex set conditioned to be perfect, where preferences arise from the natural total ordering of the vertices of each gender but are restricted to the (random) edges of the graph. We extend this correspondence to infinite intervals, for which

更新日期：2020-07-27
• Can. J. Math. (IF 0.881) Pub Date : 2020-07-14

The aim of this paper is to study circular units in the compositum K of t cyclic extensions of ${\mathbb {Q}}$ ( $t\ge 2$ ) of the same odd prime degree $\ell$ . If these fields are pairwise arithmetically orthogonal and the number s of primes ramifying in $K/{\mathbb {Q}}$ is larger than $t,$ then a nontrivial root $\varepsilon$ of the top generator $\eta$ of the group of circular units of K is

更新日期：2020-07-14
• Can. J. Math. (IF 0.881) Pub Date : 2020-07-01
Junxian Li

In this paper, we are interested in obtaining large values of Dirichlet L-functions evaluated at zeros of a class of L-functions, that is, \begin{align*}\max_{\substack{F(\rho)=0\\ T\leq \Im \rho \leq 2T}}L(\rho,\chi), \end{align*} where $\chi$ is a primitive Dirichlet character and F belongs to a class of L-functions. The class we consider includes L-functions associated with automorphic representations

更新日期：2020-07-01
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-29
Pierre Magal; Ousmane Seydi

In this paper, we extend to the non-Hille–Yosida case a variation of constants formula for a nonautonomous and nonhomogeneous Cauchy problems first obtained by Gühring and Räbiger. By using this variation of constants formula, we derive a necessary and sufficient condition for the existence of an exponential dichotomy for the evolution family generated by the associated nonautonomous homogeneous problem

更新日期：2020-06-29
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-29
Matthew Daws; Bence Horváth

We study ring-theoretic (in)finiteness properties—such as Dedekind-finiteness and proper infiniteness—of ultraproducts (and more generally, reduced products) of Banach algebras. While we characterise when an ultraproduct has these ring-theoretic properties in terms of its underlying sequence of algebras, we find that, contrary to the $C^*$ -algebraic setting, it is not true in general that an ultraproduct

更新日期：2020-06-29
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-29
Sören Kleine

We study the growth of p-primary Selmer groups of abelian varieties with good ordinary reduction at p in ${{Z}}_p$ -extensions of a fixed number field K. Proving that in many situations the knowledge of the Selmer groups in a sufficiently large number of finite layers of a ${{Z}}_p$ -extension over K suffices for bounding the over-all growth, we relate the Iwasawa invariants of Selmer groups in different

更新日期：2020-06-29
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-24
Remkes Kooistra; Alan Thompson

We present a systematic study of threefolds fibred by K3 surfaces that are mirror to sextic double planes. There are many parallels between this theory and the theory of elliptic surfaces. We show that the geometry of such threefolds is controlled by a pair of invariants, called the generalized functional and generalized homological invariants, and we derive an explicit birational model for them, which

更新日期：2020-06-24
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-16
Hong-Quan Li; Peter Sjögren

Let $v \ne 0$ be a vector in ${\mathbb {R}}^n$ . Consider the Laplacian on ${\mathbb {R}}^n$ with drift $\Delta _{v} = \Delta + 2v\cdot \nabla$ and the measure $d\mu (x) = e^{2 \langle v, x \rangle } dx$ , with respect to which $\Delta _{v}$ is self-adjoint. This measure has exponential growth with respect to the Euclidean distance. We study weak type $(1, 1)$ and other sharp endpoint estimates for

更新日期：2020-06-16
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-11
Hiroyuki Osaka; Tamotsu Teruya

We correct an error in the statement in a proposition and a theorem in Jiang–Su absorption for inclusions of unital C*-algebras. Canad. J. Math. 70(2018), 400–425. This error was found by Dr. M. Ali Asadi-Vasfi and communicated to the authors by N. Christopher Phillips of the University of Oregon, who also suggested the outline for the following correct proofs.

更新日期：2020-06-11
• Can. J. Math. (IF 0.881) Pub Date : 2020-06-08
Michael Willis

We define a limiting ${\mathfrak {sl}_N}$ Khovanov–Rozansky homology for semi-infinite positive multicolored braids. For a large class of such braids, we show that this limiting homology categorifies a highest-weight projector in the tensor product of fundamental representations determined by the coloring of the braid. This effectively completes the extension of Cautis’ similar result for infinite

更新日期：2020-06-08
• Can. J. Math. (IF 0.881) Pub Date : 2020-05-26
Alejandro Poveda; Assaf Rinot; Dima Sinapova

We introduce a class of notions of forcing which we call $\Sigma$ -Prikry, and show that many of the known Prikry-type notions of forcing that centers around singular cardinals of countable cofinality are $\Sigma$ -Prikry. We show that given a $\Sigma$ -Prikry poset $\mathbb P$ and a name for a non-reflecting stationary set T, there exists a corresponding $\Sigma$ -Prikry poset that projects to

更新日期：2020-05-26
• Can. J. Math. (IF 0.881) Pub Date : 2020-05-21
S.K. Roushon

We observe an inductive structure in a large class of Artin groups of finite real, complex and affine types and exploit this information to deduce the Farrell–Jones isomorphism conjecture for these groups.

更新日期：2020-05-21
• Can. J. Math. (IF 0.881) Pub Date : 2020-05-21
Jiajie Hua; Qingyun Wang

Let $\Theta =(\theta _{j,k})_{3\times 3}$ be a nondegenerate real skew-symmetric $3\times 3$ matrix, where $\theta _{j,k}\in [0,1).$ For any $\varepsilon>0$ , we prove that there exists $\delta>0$ satisfying the following: if $v_1,v_2,v_3$ are three unitaries in any unital simple separable $C^*$ -algebra A with tracial rank at most one, such that \begin{align*}\|v_kv_j-e^{2\pi i \theta_{j,k}}v_jv_k\|<\delta 更新日期：2020-05-21 • Can. J. Math. (IF 0.881) Pub Date : 2020-05-20 Timothée Marquis Let A be a symmetrisable generalised Cartan matrix, and let\mathfrak {g}(A)$be the corresponding Kac–Moody algebra. In this paper, we address the following fundamental question on the structure of$\mathfrak {g}(A)$: given two homogeneous elements$x,y\in \mathfrak {g}(A)$, when is their bracket$[x,y]$a nonzero element? As an application of our results, we give a description of the solvable 更新日期：2020-05-20 • Can. J. Math. (IF 0.881) Pub Date : 2020-05-19 Vinoth Nandakumar; Gufang Zhao Bernstein, Frenkel, and Khovanov have constructed a categorification of tensor products of the standard representation of$\mathfrak {sl}_2$, where they use singular blocks of category$\mathcal {O}$for$\mathfrak {sl}_n$and translation functors. Here we construct a positive characteristic analogue using blocks of representations of$\mathfrak {s}\mathfrak {l}_n$over a field$\mathbf {k}$of characteristic 更新日期：2020-05-19 • Can. J. Math. (IF 0.881) Pub Date : 2020-05-15 Bertrand Lemaire; Jean-Loup Waldspurger Soient$F$un corps global, et$G$un groupe réductif connexe défini sur$F$. On prouve que si deux données endoscopiques de$G$sont équivalentes en presque toute place de$F$, alors elles sont équivalentes. Le résultat est encore vrai pour l’endoscopie (ordinaire) avec caractère. On donne aussi, pour$F$global ou local et$G$quasi-simple simplement connexe, une description des données endoscopiques 更新日期：2020-05-15 • Can. J. Math. (IF 0.881) Pub Date : 2020-05-14 Fei Hu; Sichen Li Let X be a normal projective variety of dimension n and G an abelian group of automorphisms such that all elements of$G\setminus \{\operatorname {id}\}$are of positive entropy. Dinh and Sibony showed that G is actually free abelian of rank$\le n - 1$. The maximal rank case has been well understood by De-Qi Zhang. We aim to characterize the pair$(X, G)$such that$\operatorname {rank} G = n - 2$更新日期：2020-05-14 • Can. J. Math. (IF 0.881) Pub Date : 2020-04-24 Mengyuan Zhang We study bundles on projective spaces that have vanishing lower cohomologies using their short minimal free resolutions. We partition the moduli$\mathcal{M}$according to the Hilbert function H and classify all possible Hilbert functions H of such bundles. For each H, we describe a stratification of$\mathcal{M}_H$by quotients of rational varieties. We show that the closed strata form a graded lattice 更新日期：2020-04-24 • Can. J. Math. (IF 0.881) Pub Date : 2019-08-27 Naveed S. Bari; Eugenie Hunsicker We answer Mark Kac’s famous question, “Can one hear the shape of a drum?” in the positive for orbifolds that are 3-dimensional and 4-dimensional lens spaces; we thus complete the answer to this question for orbifold lens spaces in all dimensions. We also show that the coefficients of the asymptotic expansion of the trace of the heat kernel are not sufficient to determine the above results. 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-10-31 Ehud Moshe Baruch; Soma Purkait We study genuine local Hecke algebras of the Iwahori type of the double cover of$\operatorname{SL}_{2}(\mathbb{Q}_{p})$and translate the generators and relations to classical operators on the space$S_{k+1/2}(\unicode[STIX]{x1D6E4}_{0}(4M))$,$M$odd and square-free. In [9] Manickam, Ramakrishnan, and Vasudevan defined the new space of$S_{k+1/2}(\unicode[STIX]{x1D6E4}_{0}(4M))$that maps Hecke 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-05-07 Ugo Bindini We consider a multimarginal transport problem with repulsive cost, where the marginals are all equal to a fixed probability$\unicode[STIX]{x1D70C}\in {\mathcal{P}}(\mathbb{R}^{d})$. We prove that, if the concentration of$\unicode[STIX]{x1D70C}$is less than$1/N$, then the problem has a solution of finite cost. The result is sharp, in the sense that there exists$\unicode[STIX]{x1D70C}$with concentration 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-06-13 Jakub Byszewski; Jakub Konieczny We conjecture that bounded generalised polynomial functions cannot be generated by finite automata, except for the trivial case when they are ultimately periodic. 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-08-30 Peng Gao; Liangyi Zhao In this paper we prove some one-level density results for the low-lying zeros of families of quadratic and quartic Hecke$L$-functions of the Gaussian field. As corollaries, we deduce that at least 94.27% and 5%, respectively, of the members of the quadratic family and the quartic family do not vanish at the central point. 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-07-16 Shaoxiong Hou; Deping Ye This paper provides a functional analogue of the recently initiated dual Orlicz–Brunn–Minkowski theory for star bodies. We first propose the Orlicz addition of measures, and establish the dual functional Orlicz–Brunn–Minkowski inequality. Based on a family of linear Orlicz additions of two measures, we provide an interpretation for the famous$f$-divergence. Jensen’s inequality for integrals is also 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-05-07 Pınar Kılıçer; Elisa Lorenzo García; Marco Streng We give a bound on the primes dividing the denominators of invariants of Picard curves of genus 3 with complex multiplication. Unlike earlier bounds in genus 2 and 3, our bound is based, not on bad reduction of curves, but on a very explicit type of good reduction. This approach simultaneously yields a simplification of the proof and much sharper bounds. In fact, unlike all previous bounds for genus 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-09-03 Robert Laterveer; Charles Vial This note is about certain locally complete families of Calabi–Yau varieties constructed by Cynk and Hulek, and certain varieties constructed by Schreieder. We prove that the cycle class map on the Chow ring of powers of these varieties admits a section, and that these varieties admit a multiplicative self-dual Chow–Künneth decomposition. As a consequence of both results, we prove that the subring 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2019-05-21 Eran Nevo; Guillermo Pineda-Villavicencio; Julien Ugon; David Yost We study$n$-vertex$d$-dimensional polytopes with at most one nonsimplex facet with, say,$d+s$vertices, called almost simplicial polytopes. We provide tight lower and upper bound theorems for these polytopes as functions of$d,n$, and$s$, thus generalizing the classical Lower Bound Theorem by Barnette and the Upper Bound Theorem by McMullen, which treat the case where$s=0$. We characterize 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2020-03-31 None 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2020-03-31 None 更新日期：2020-04-20 • Can. J. Math. (IF 0.881) Pub Date : 2020-04-14 Claudia Anedda; Fabrizio Cuccu; Silvia Frassu Let$\Omega \subset \mathbb {R}^N$,$N\geq 2$, be an open bounded connected set. We consider the fractional weighted eigenvalue problem$(-\Delta )^s u =\lambda \rho u$in$\Omega $with homogeneous Dirichlet boundary condition, where$(-\Delta )^s$,$s\in (0,1)$, is the fractional Laplacian operator,$\lambda \in \mathbb {R}$and$ \rho \in L^\infty (\Omega )$. We study weak* continuity, convexity 更新日期：2020-04-14 • Can. J. Math. (IF 0.881) Pub Date : 2020-04-14 Wilmar Bolaños; Guillermo Mantilla-Soler In the mid 80’s Conner and Perlis showed that for cyclic number fields of prime degree p the isometry class of integral trace is completely determined by the discriminant. Here we generalize their result to tame cyclic number fields of arbitrary degree. Furthermore, for such fields, we give an explicit description of a Gram matrix of the integral trace in terms of the discriminant of the field. 更新日期：2020-04-14 • Can. J. Math. (IF 0.881) Pub Date : 2020-04-12 Tsuyoshi Kajiwara; Yasuo Watatani We introduce a dimension group for a self-similar map as the$\mathrm {K}_0$-group of the core of the C*-algebra associated with the self-similar map together with the canonical endomorphism. The key step for the computation is an explicit description of the core as the inductive limit using their matrix representations over the coefficient algebra, which can be described explicitly by the singularity 更新日期：2020-04-12 • Can. J. Math. (IF 0.881) Pub Date : 2020-04-02 Joel Brewster Lewis; Alejandro H. Morales We enumerate factorizations of a Coxeter element in a well-generated complex reflection group into arbitrary factors, keeping track of the fixed space dimension of each factor. In the infinite families of generalized permutations, our approach is fully combinatorial. It gives results analogous to those of Jackson in the symmetric group and can be refined to encode a notion of cycle type. As one application 更新日期：2020-04-02 • Can. J. Math. (IF 0.881) Pub Date : 2020-03-30 Seidon Alsaody We study exceptional Jordan algebras and related exceptional group schemes over commutative rings from a geometric point of view, using appropriate torsors to parametrize and explain classical and new constructions, and proving that over rings, they give rise to nonisomorphic structures. We begin by showing that isotopes of Albert algebras are obtained as twists by a certain$\mathrm F_4$-torsor with 更新日期：2020-03-30 • Can. J. Math. (IF 0.881) Pub Date : 2020-03-30 Abhishek Kumar Shukla; Ben Williams We construct a scheme$B(r; {\mathbb {A}}^n)$such that a map$X \to B(r; {\mathbb {A}}^n)$corresponds to a degree-n étale algebra on X equipped with r generating global sections. We then show that when$n=2$, i.e., in the quadratic étale case, the singular cohomology of$B(r; {\mathbb {A}}^n)({\mathbb {R}})$can be used to reconstruct a famous example of S. Chase and to extend its application to 更新日期：2020-03-30 • Can. J. Math. (IF 0.881) Pub Date : 2020-03-11 Ernst Kuwert; Tobias Lamm We study immersed surfaces in $${\mathbb R}^3$$ that are critical points of the Willmore functional under boundary constraints. The two cases considered are when the surface meets a plane orthogonally along the boundary and when the boundary is contained in a line. In both cases we derive weak forms of the resulting free boundary conditions and prove regularity by reflection. 更新日期：2020-03-11 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-27 Alex Iosevich; Doowon Koh; Sujin Lee; Thang Pham; Chun-Yen Shen In this paper, we completely solve the$L^{2}\to L^{r}$extension conjecture for the zero radius sphere over finite fields. We also obtain the sharp$L^{p}\to L^{4}$extension estimate for non-zero radii spheres over finite fields, which improves the previous result of the first and second authors significantly. 更新日期：2020-02-27 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-24 Stephane Geudens; Marco Zambon We study coisotropic submanifolds of b-symplectic manifolds. We prove that b-coisotropic submanifolds (those transverse to the degeneracy locus) determine the b-symplectic structure in a neighborhood, and provide a normal form theorem. This extends Gotay’s theorem in symplectic geometry. Further, we introduce strong b-coisotropic submanifolds and show that their coisotropic quotient, which locally 更新日期：2020-02-24 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-24 Ziba Fazelpour; Alireza Nasr-Isfahani A ring$\unicode[STIX]{x1D6EC}$is called right Köthe if every right$\unicode[STIX]{x1D6EC}$-module is a direct sum of cyclic modules. In this paper, we give a characterization of basic hereditary right Köthe rings in terms of their Coxeter valued quivers. We also give a characterization of basic right Köthe rings with radical square zero. Therefore, we give a solution to Köthe’s problem in these 更新日期：2020-02-24 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-24 Łukasz Kosiński; Włodzimierz Zwonek Motivated by works on extension sets in standard domains, we introduce a notion of the Carathéodory set that seems better suited for the methods used in proofs of results on characterization of extension sets. A special stress is put on a class of two-dimensional submanifolds in the tridisc that not only turns out to be Carathéodory but also provides examples of two-dimensional domains for which the 更新日期：2020-02-24 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-24 Peter Crooks; Maarten van Pruijssen This work is concerned with Bielawski’s hyperkähler slices in the cotangent bundles of homogeneous affine varieties. One can associate such a slice with the data of a complex semisimple Lie group$G$, a reductive subgroup$H\subseteq G$, and a Slodowy slice$S\subseteq \mathfrak{g}:=\text{Lie}(G)$, defining it to be the hyperkähler quotient of$T^{\ast }(G/H)\times (G\times S)$by a maximal compact 更新日期：2020-02-24 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-12 B. Monson; Egon Schulte In the classical setting, a convex polytope is said to be semiregular if its facets are regular and its symmetry group is transitive on vertices. This paper continues our study of alternating semiregular abstract polytopes, which have abstract regular facets, still with combinatorial automorphism group transitive on vertices and with two kinds of regular facets occurring in an alternating fashion. 更新日期：2020-02-12 • Can. J. Math. (IF 0.881) Pub Date : 2020-02-12 Zhang Chao; José L. Torrea In this paper we analyze the convergence of the following type of series $$\begin{eqnarray}T_{N}^{{\mathcal{L}}}f(x)=\mathop{\sum }_{j=N_{1}}^{N_{2}}v_{j}\big(e^{-a_{j+1}{\mathcal{L}}}f(x)-e^{-a_{j}{\mathcal{L}}}f(x)\big),\quad x\in \mathbb{R}^{n},\end{eqnarray}$$ where${\{e^{-t{\mathcal{L}}}\}}_{t>0}$is the heat semigroup of the operator${\mathcal{L}}=-\unicode[STIX]{x1D6E5}+V$with$\unicode[STIX]{x1D6E5}\$

更新日期：2020-02-12
• Can. J. Math. (IF 0.881) Pub Date : 2020-02-06
Kenji Sakugawa

The generalized Soulé character was introduced by H. Nakamura and Z. Wojtkowiak and is a generalization of Soulé’s cyclotomic character. In this paper, we prove that certain linear sums of generalized Soulé characters essentially coincide with the image of generalized Beilinson elements in K-groups under Soulé’s higher regulator maps. This result generalizes Huber–Wildeshaus’ theorem, which is a cyclotomic

更新日期：2020-02-06
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Two Tailors, a and B Earn Rs 15 and Rs 20 per Day Respectively. a Can Stitch 6 Shirts and 4 Pants While B Can Stitch 10 Shirts and 4 Pants per Day. - Mathematics

Sum

Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants  while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at a minimum labour cost?

Solution

Let tailor A work for x days and tailor B work for days.
In one day, A can stitch 6 shirts and 4 pants whereas can stitch 10 shirts and 4 pants
Thus, in  x days A can stitch 6x shirts and 4y pants whereas B can stich 10y shirts and 4y pants.
pants.
It is given that the minimum requirement of the shirts and pants are respectively 60 and 32.
Thus,

$6x + 10y \geq 60$
$4x + 4y \geq 32$

Further it is given that A and B earn Rs 15 and Rs 20 per day respectively.
Thus, A earn Rs 15x and B earn Rs 20y .
Let Z denotes the total cost

$\therefore Z = 15x + 20y$
Days cannot be negative.

∴ $x, y \geq 0$
Min Z =  $15x + 20y$ subject to

$6x + 10y \geq 60$

$4x + 4y \geq 32$

$x, y \geq 0$
First we will convert inequations into equations as follows :
6x + 10y = 60, 4x + 4y = 32, x = 0 and y = 0

Region represented by 6x + 10y ≥ 60:
The line 6x + 10y = 60 meets the coordinate axes at A1(10, 0) and B1(0, 6) respectively. By joining these points we obtain the line6x + 10y = 60. Clearly (0,0) does not satisfies the 6x + 10y = 60. So,the region which does not contains the origin represents the solution set of the inequation 6x + 10y ≥ 60.

Region represented by 4x + 4y ≥ 32:
The line 4x + 4y =32 meets the coordinate axes at C1(8, 0) and D1(0, 8) respectively. By joining these points we obtain the line 4x + 4y = 32.Clearly (0,0) does not satisfies the inequation 4x + 4y ≥ 32. So,the region which does not contains the origin represents the solution set of the inequation 4x + 4y ≥ 32.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and ≥ 0.
The feasible region determined by the system of constraints 6x + 10y ≥ 60,4x + 4y ≥ 32, x ≥ 0, and y ≥ 0 are as follows.
Thus, the mathematical formulation of the given linear programming problem is
The corner points are D1(0, 8), E1(5, 3) and  A1(10, 0).
The values of Z at these corner points are as follows

 Corner point Z = 15x + 20y D1 160 E1 135 A1 150

The minimum value of Z is 135 which is attained at E1(5, 3).
Thus, for minimum labour cost, A should work for 5 days and B should work for 3 days.
Concept: Graphical Method of Solving Linear Programming Problems
Is there an error in this question or solution?

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RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.4 | Q 3 | Page 50


text:
## anonymous one year ago evaluate the integral

1. anonymous

$\int\limits_{D}^{}\frac{ \left( x^3+x^2+x+1 \right) }{ x^4 }~dx$ if D is the lower quarter circle thats centered at zero and joining $$\sf \frac{-1-i}{\sqrt{2}}$$ and $$\sf \frac{1-i}{\sqrt{2}}$$ in the counterclockwise direction (meaning positive sense)

2. anonymous

@dan815


text:
Electromagnetic radiation depends on structural details of the source system of electric charge and electric current. Direct analysis can be intractable if the structure is unknown or complicated. Multipole analysis offers a way to separate the radiation into moments of increasing complexity. Since the electromagnetic field depends more heavily on lower-order moments than on higher-order moments, the electromagnetic field can be approximated without knowing the structure in detail.

### Linearity of moments

Since Maxwell's equations are linear, the electric field and magnetic field depend linearly on source distributions. Linearity allows the fields from various multipole moments to be calculated independently and added together to give the total field of the system. This is the well-known principle of superposition.

### Origin dependence of multipole moments

Multipole moments are calculated with respect to a fixed expansion point which is taken to be the origin of a given coordinate system. Translating the origin changes the multipole moments of the system with the exception of the first non-vanishing moment.[4][5] For example, the monopole moment of charge is simply the total charge in the system. Changing the origin will never change this moment. If the monopole moment is zero then the dipole moment of the system will be translation invariant. If both the monopole and dipole moments are zero then the quadrupole moment is translation invariant, and so forth. Because higher-order moments depend on the position of the origin, they cannot be regarded as invariant properties of the system.

### Field dependence on distance

The field from a multipole moment depends on both the distance from the origin and the angular orientation of the evaluation point with respect to the coordinate system.[4] In particular, the radial dependence of the electromagnetic field from a stationary ${\displaystyle 2^{\ell }}$-pole scales as ${\displaystyle 1/r^{\ell +2}}$.[2] That is, the electric field from the electric monopole moment scales as inverse distance squared. Likewise, the electric dipole moment creates a field that scales as inverse distance cubed, and so on. As distance increases, the contribution of high-order moments becomes much smaller than the contribution from low-order moments, so high-order moments can be ignored to simplify calculations.

The radial dependence of radiation waves is different from static fields because these waves carry energy away from the system. Since energy must be conserved, simple geometric analysis shows that the energy density of spherical radiation, radius ${\displaystyle r}$, must scale as ${\displaystyle 1/r^{2}}$. As a spherical wave expands, the fixed energy of the wave must spread out over an expanding sphere of surface area ${\displaystyle 4\pi r^{2}}$. Accordingly, every time-dependent multipole moment must contribute radiant energy density that scales as ${\displaystyle 1/r^{2}}$, regardless of the order of the moment. Hence, high-order moments cannot be discarded as easily as in static case. Even so, the multipole coefficients of a system generally diminish with increasing order, usually as ${\displaystyle 1/(2\ell +1)!!}$, so radiation fields can still be approximated by truncating high-order moments.[5]

## Time-dependent electromagnetic fields

### Sources

Time-dependent source distributions can be expressed using Fourier analysis. This allows separate frequencies to be analyzed independently. Charge density is given by

${\displaystyle \rho (\mathbf {x} ,t)=\int _{-\infty }^{\infty }d\omega \,{\hat {\rho }}(\mathbf {x} ,\omega )e^{-i\omega t}}$

and current density by

${\displaystyle \mathbf {J} (\mathbf {x} ,t)=\int _{-\infty }^{\infty }d\omega \,{\hat {\mathbf {J} }}(\mathbf {x} ,\omega )e^{-i\omega t}}$.[6]

For convenience, only a single angular frequency ω is considered from this point forward; thus

${\displaystyle \rho (\mathbf {x} ,t)=\rho (\mathbf {x} )e^{-i\omega t}}$
${\displaystyle \mathbf {J} (\mathbf {x} ,t)=\mathbf {J} (\mathbf {x} )e^{-i\omega t}}$

The superposition principle may be applied to generalize results for multiple frequencies.[5] Vector quantities appear in bold. The standard convention of taking the real part of complex quantities to represent physical quantities is used.

The intrinsic angular momentum of elementary particles (see Spin (physics)) may also affect electromagnetic radiation from some source materials. To account for these effects, the intrinsic magnetization of the system ${\displaystyle \mathbf {M} (\mathbf {x} ,t)}$ would have to be taken into account. For simplicity however, these effects will be deferred to the discussion of generalized multipole radiation.

### Potentials

The source distributions can be integrated to yield the time-dependent electric potential and magnetic potential φ and A respectively. Formulas are expressed in the Lorenz Gauge in SI units.[5][6]

${\displaystyle \phi (\mathbf {x} ,t)={\frac {1}{4\pi \epsilon _{0}}}\int d^{3}\mathbf {x'} \int dt'{\frac {\rho (\mathbf {x'} ,t')}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}\delta \left(t'-(t-{\frac {\|\mathbf {x} -\mathbf {x'} \|_{2}}{c}})\right)}$
${\displaystyle \mathbf {A} (\mathbf {x} ,t)={\frac {\mu _{0}}{4\pi }}\int d^{3}\mathbf {x'} \int dt'{\frac {\mathbf {J} (\mathbf {x'} ,t')}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}\delta \left(t'-(t-{\frac {\|\mathbf {x} -\mathbf {x'} \|_{2}}{c}})\right)}$

In these formulas c is the speed of light in vacuum, ${\displaystyle \delta }$ is the Dirac delta function, and ${\displaystyle \|\mathbf {x} -\mathbf {x'} \|_{2}}$ is the Euclidean distance from the source point x to the evaluation point x. Integrating the time-dependent source distributions above yields

${\displaystyle \phi (\mathbf {x} ,t)={\frac {1}{4\pi \epsilon _{0}}}e^{-i\omega t}\int d^{3}\mathbf {x'} \rho (\mathbf {x'} ){\frac {e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}}$
${\displaystyle \mathbf {A} (\mathbf {x} ,t)={\frac {\mu _{0}}{4\pi }}e^{-i\omega t}\int d^{3}\mathbf {x'} \mathbf {J} (\mathbf {x'} ){\frac {e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}}$

where k=ω/c. These formulas provide the basis for analyzing multipole radiation.

### Multipole expansion in near field

The near field is the region around a source where the electromagnetic field can be evaluated quasi-statically. If target distance from the multipole origin ${\displaystyle r=\|\mathbf {x} \|_{2}}$ is much smaller than the radiation wavelength ${\displaystyle \lambda =2\pi /k}$, then ${\displaystyle kr\ll 1}$. As a result, the exponential can be approximated in this region as:

${\displaystyle e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}=1+O(kr)}$

See Taylor expansion. By using this approximation, the remaining x dependence is the same as it is for a static system, the same analysis applies.[4][5] Essentially, the potentials can be evaluated in the near field at a given instant by simply taking a snapshot of the system and treating it as though it were static - hence it is called quasi-static.[5] See near and far field and multipole expansion. In particular, the inverse distance ${\displaystyle 1/\|\mathbf {x} -\mathbf {x'} \|_{2}}$ is expanded using spherical harmonics which are integrated separately to obtain spherical multipole coefficients.

### Multipole expansion in far field: Multipole radiation

At large distances from a high frequency source, ${\displaystyle \lambda \ll r}$, the following approximations hold:

${\displaystyle {\frac {1}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}={\frac {1}{r}}+O(1/r^{2})}$
${\displaystyle e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}=e^{ik(r-\mathbf {n} \cdot \mathbf {x'} +O(1/r))}=e^{ikr-ik\mathbf {n} \cdot \mathbf {x'} }(1+O(1/r))}$

Since only the first-order term in ${\displaystyle 1/r}$ is significant at large distances, the expansions combine to give

${\displaystyle {\frac {e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}={\frac {e^{ikr}}{r}}(1-ik(\mathbf {n} \cdot \mathbf {x'} )+{\frac {(-ik)^{2}}{2}}(\mathbf {n} \cdot \mathbf {x'} )^{2}+...)+O(1/r^{2})}$

Each power of ${\displaystyle \mathbf {n} \cdot \mathbf {x'} }$ corresponds to a different multipole moment. The first few moments are evaluated directly below.

The zeroth order term, ${\displaystyle {\frac {e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}\rightarrow {\frac {e^{ikr}}{r}}}$, applied to the scalar potential gives

${\displaystyle \phi _{\text{Electric monopole}}(\mathbf {x} ,t)={\frac {1}{4\pi \epsilon _{0}}}{\frac {e^{ikr-i\omega t}}{r}}\int d^{3}\mathbf {x'} \rho (\mathbf {x'} )={\frac {e^{ikr-i\omega t}}{4\pi \epsilon _{0}r}}q}$

where the total charge ${\displaystyle q=\int d^{3}\mathbf {x'} \rho (\mathbf {x'} )}$ is the electric monopole moment oscillating at frequency ω. Conservation of charge requires q=0 since

${\displaystyle q(t)=\int d^{3}\mathbf {x'} \rho (\mathbf {x'} ,t)=\int d^{3}\mathbf {x'} \rho (\mathbf {x'} )e^{-i\omega t}=qe^{-i\omega t}}$.

If the system is closed then the total charge cannot fluctuate which means the oscillation amplitude q must be zero. Hence, ${\displaystyle \phi _{\text{Electric monopole}}(\mathbf {x} ,t)=0}$. The corresponding fields and radiant power must also be zero.[5]

### Electric dipole potential

Electric dipole radiation can be derived by applying the zeroth-order term to the vector potential.[5]

${\displaystyle \mathbf {A} _{\text{Electric dipole}}(\mathbf {x} ,t)={\frac {\mu _{0}}{4\pi }}{\frac {e^{ikr-i\omega t}}{r}}\int d^{3}\mathbf {x'} \mathbf {J} (\mathbf {x'} )}$

Integration by parts yields[7]

${\displaystyle \int d^{3}\mathbf {x'} \mathbf {J} (\mathbf {x'} )=-\int d^{3}\mathbf {x'} \mathbf {x'} (\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x'} ))}$.

and the charge continuity equation shows

${\displaystyle {\frac {\partial \rho (\mathbf {x} ,t)}{\partial t}}+\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x} ,t)=\left(-i\omega \rho (\mathbf {x} )+\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x} )\right)e^{-i\omega t}=0}$.

It follows that

${\displaystyle \mathbf {A} _{\text{Electric dipole}}(\mathbf {x} ,t)={\frac {-i\omega \mu _{0}}{4\pi }}{\frac {e^{ikr-i\omega t}}{r}}\int d^{3}\mathbf {x'} \mathbf {x'} \rho (\mathbf {x'} )}$

Similar results can be obtained by applying the first-order term, ${\displaystyle {\frac {e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}\rightarrow {\frac {e^{ikr}}{r}}(-ik)(\mathbf {n} \cdot \mathbf {x'} )}$ to the scalar potential. The amplitude of the electric dipole moment of the system is ${\displaystyle \mathbf {p} =\int d^{3}\mathbf {x'} \mathbf {x'} \rho (\mathbf {x'} )}$, which allows the potentials to be expressed as

${\displaystyle \phi _{\text{Electric dipole}}(\mathbf {x} ,t)={\frac {-ik}{4\pi \epsilon _{0}}}{\frac {e^{ikr-i\omega t}}{r}}\mathbf {n} \cdot \mathbf {p} }$
${\displaystyle \mathbf {A} _{\text{Electric dipole}}(\mathbf {x} ,t)={\frac {-i\omega \mu _{0}}{4\pi }}{\frac {e^{ikr-i\omega t}}{r}}\mathbf {p} }$

### Electric dipole fields

Once the time-dependent potentials are understood, the time-dependent electric field and magnetic field can be calculated in the usual way. Namely,

${\displaystyle \mathbf {E} (\mathbf {x} ,t)=-\mathbf {\nabla } \phi (\mathbf {x} ,t)-{\frac {\partial \mathbf {A} (\mathbf {x} ,t)}{\partial t}}}$
${\displaystyle \mathbf {B} (\mathbf {x} ,t)=\mathbf {\nabla } \times \mathbf {A} (\mathbf {x} ,t)}$,

or, in a source-free region of space, the relationship between the magnetic field and the electric field can be used to obtain

${\displaystyle \mathbf {H} (\mathbf {x} ,t)={\frac {1}{\mu _{0}}}\mathbf {\nabla } \times \mathbf {A} (\mathbf {x} ,t)}$
${\displaystyle \mathbf {E} (\mathbf {x} ,t)={\frac {iZ_{0}}{k}}\mathbf {\nabla } \times \mathbf {H} (\mathbf {x} ,t)}$

where ${\displaystyle Z_{0}={\sqrt {\mu _{0}/\epsilon _{0}}}}$ is the impedance of free space. The electric and magnetic fields that correspond to the potentials above are

${\displaystyle \mathbf {H} _{\text{Electric dipole}}(\mathbf {x} ,t)={\frac {ck^{2}}{4\pi }}(\mathbf {n} \times \mathbf {p} ){\frac {e^{ikr-i\omega t}}{r}}}$
${\displaystyle \mathbf {E} _{\text{Electric dipole}}(\mathbf {x} ,t)=Z_{0}(\mathbf {H} _{\text{Electric dipole}}\times \mathbf {n} )}$

which is consistent with spherical radiation waves.[5]

### Pure electric dipole power

The power density, energy per unit area per unit time, is expressed by the Poynting vector ${\displaystyle \mathbf {S} =\mathbf {E} \times \mathbf {H} }$. It follows that the time averaged power density per unit solid angle is given by

${\displaystyle {\frac {dP(\mathbf {x} )}{d\Omega }}={\frac {r^{2}}{2}}\Re (\mathbf {n} \cdot \mathbf {E} \times \mathbf {H} )}$.

The dot product with ${\displaystyle \mathbf {n} }$ extracts the emission magnitude and the factor of 1/2 comes from averaging over time. As explained above, the ${\displaystyle r^{2}}$ cancels the radial dependence of radiation energy density. Application to a pure electric dipole gives

${\displaystyle {\frac {dP_{\text{Electric dipole}}(\mathbf {x} )}{d\Omega }}={\frac {c^{2}Z_{0}}{32\pi ^{2}}}k^{4}\|\mathbf {p} \|_{2}^{2}\sin ^{2}\theta }$

where θ is measured with respect to ${\displaystyle \mathbf {p} }$.[5] Integration over a sphere yields the total power radiated:

${\displaystyle P_{\text{Electric dipole}}={\frac {c^{2}Z_{0}}{12\pi }}k^{4}\|\mathbf {p} \|_{2}^{2}}$

### Magnetic dipole potential

The first-order term, ${\displaystyle {\frac {e^{ik\|\mathbf {x} -\mathbf {x'} \|_{2}}}{\|\mathbf {x} -\mathbf {x'} \|_{2}}}\rightarrow {\frac {e^{ikr}}{r}}(-ik)(\mathbf {n} \cdot \mathbf {x'} )}$, applied to the vector potential gives magnetic dipole radiation and electric quadrupole radiation.[5]

${\displaystyle \mathbf {A} _{\text{Magnetic dipole / Electric quadrupole}}(\mathbf {x} ,t)={\frac {\mu _{0}}{4\pi }}{\frac {e^{ikr-i\omega t}}{r}}(-ik)\int d^{3}\mathbf {x'} (\mathbf {n} \cdot \mathbf {x'} )\mathbf {J} (\mathbf {x'} )}$

The integrand can be separated into symmetric and anti-symmetric parts in n and x

${\displaystyle (\mathbf {n} \cdot \mathbf {x'} )\mathbf {J} (\mathbf {x'} )={\frac {1}{2}}\left((\mathbf {n} \cdot \mathbf {x'} )\mathbf {J} (\mathbf {x'} )+(\mathbf {n} \cdot \mathbf {J} (\mathbf {x'} ))\mathbf {x'} \right)+{\frac {1}{2}}(\mathbf {x'} \times \mathbf {J} (\mathbf {x'} ))\times \mathbf {n} }$

The second term contains the effective magnetization due to the current ${\displaystyle \mathbf {M} _{\text{effective}}(\mathbf {x'} )=1/2(\mathbf {x'} \times \mathbf {J} (\mathbf {x'} ))}$ and integration gives the magnetic dipole moment.

${\displaystyle \int d^{3}\mathbf {x'} \mathbf {M} _{\text{effective}}(\mathbf {x'} )=\mathbf {m} }$
${\displaystyle \mathbf {A} _{\text{Magnetic dipole}}(\mathbf {x} ,t)={\frac {-ik\mu _{0}}{4\pi }}{\frac {e^{ikr-i\omega t}}{r}}\mathbf {m} \times \mathbf {n} }$

Notice that ${\displaystyle \mathbf {A} _{\text{Magnetic dipole}}}$ has a similar form to ${\displaystyle \mathbf {H} _{\text{Electric dipole}}}$. That means the magnetic field from a magnetic dipole behaves similarly to the electric field from an electric dipole. Likewise, the electric field from a magnetic dipole behaves like the magnetic field from an electric dipole. Taking the transformations

${\displaystyle \mathbf {E} _{\text{Electric dipole}}\rightarrow Z_{0}\mathbf {H} _{\text{Magnetic dipole}}}$
${\displaystyle \mathbf {H} _{\text{Electric dipole}}\rightarrow {\frac {-1}{Z_{0}}}\mathbf {E} _{\text{Magnetic dipole}}}$
${\displaystyle \mathbf {p} \rightarrow \mathbf {m} /c}$

on previous results yields magnetic dipole results.[5]

### Magnetic dipole fields

${\displaystyle \mathbf {E} _{\text{Magnetic dipole}}(\mathbf {x} ,t)={\frac {-k^{2}Z_{0}}{4\pi }}(\mathbf {n} \times \mathbf {m} ){\frac {e^{ikr-i\omega t}}{r}}}$
${\displaystyle \mathbf {H} _{\text{Magnetic dipole}}(\mathbf {x} ,t)={\frac {-1}{Z_{0}}}(\mathbf {E} _{\text{Magnetic dipole}}\times \mathbf {n} )}$[5]

### Pure magnetic dipole power

The average power radiated per unit solid angle by a magnetic dipole is

${\displaystyle {\frac {dP_{\text{Magnetic dipole}}(\mathbf {x} )}{d\Omega }}={\frac {Z_{0}}{32\pi ^{2}}}k^{4}\|\mathbf {m} \|_{2}^{2}\sin ^{2}\theta }$

where θ is measured with respect to the magnetic dipole ${\displaystyle \mathbf {m} }$. The total power radiated is:

${\displaystyle P_{\text{Magnetic dipole}}={\frac {Z_{0}}{12\pi }}k^{4}\|\mathbf {m} \|_{2}^{2}}$[5]

The symmetric portion of the integrand from the previous section can be resolved by applying integration by parts and the charge continuity equation as was done for electric dipole radiation.

${\displaystyle {\frac {1}{2}}\int d^{3}\mathbf {x} \left((\mathbf {n} \cdot \mathbf {x'} )\mathbf {J} (\mathbf {x'} )+(\mathbf {n} \cdot \mathbf {J} (\mathbf {x'} ))\mathbf {x'} \right)={\frac {-i\omega }{2}}\int d^{3}\mathbf {x'} \mathbf {x'} (\mathbf {n} \cdot \mathbf {x'} )\rho (\mathbf {x'} )}$
${\displaystyle \mathbf {A} _{\text{Electric quadrupole}}(\mathbf {x} ,t)={\frac {-k\omega \mu _{0}}{8\pi }}{\frac {e^{ikr-i\omega t}}{r}}\int d^{3}\mathbf {x'} \mathbf {x'} (\mathbf {n} \cdot \mathbf {x'} )\rho (\mathbf {x'} )}$

This corresponds to the traceless electric quadrupole moment tensor ${\displaystyle Q_{\alpha \beta }=\int d^{3}\mathbf {x'} (3x'_{\alpha }x'_{\beta }-\|\mathbf {x'} \|_{2}^{2}\delta _{\alpha \beta })\rho (\mathbf {x'} )}$. Contracting the second index with the normal vector ${\displaystyle [Q(\mathbf {n} )]_{\alpha }=\sum _{\beta }Q_{\alpha \beta }n_{\beta }}$ allows the vector potential to be expressed as

${\displaystyle \mathbf {A} _{\text{Electric quadrupole}}(\mathbf {x} ,t)={\frac {-k\omega \mu _{0}}{8\pi }}{\frac {e^{ikr-i\omega t}}{r}}{\frac {1}{3}}\mathbf {Q(n)} }$[5]

The resulting magnetic and electric fields are:

${\displaystyle \mathbf {H} _{\text{Electric quadrupole}}(\mathbf {x} ,t)={\frac {-ick^{3}}{24\pi }}{\frac {e^{ikr-i\omega t}}{r}}\mathbf {n} \times \mathbf {Q(n)} }$
${\displaystyle \mathbf {E} _{\text{Electric quadrupole}}(\mathbf {x} ,t)=Z_{0}(\mathbf {H} _{\text{Electric quadrupole}}\times \mathbf {n} )}$[5]

The average power radiated per unit solid angle by an electric quadrupole is

${\displaystyle {\frac {dP_{\text{Electric quadrupole}}(\mathbf {x} )}{d\Omega }}={\frac {c^{2}Z_{0}}{1152\pi ^{2}}}k^{6}\|(\mathbf {n} \times \mathbf {Q(n)} )\times \mathbf {n} \|_{2}^{2}}$

where θ is measured with respect to the magnetic dipole ${\displaystyle \mathbf {m} }$. The total power radiated is:

${\displaystyle P_{\text{Electric quadrupole}}={\frac {c^{2}Z_{0}}{1440\pi }}k^{6}\sum _{\alpha ,\beta }Q_{\alpha \beta }^{2}}$[5]

As the multipole moment of a source distribution increases, the direct calculations employed so far become too cumbersome to continue. Analysis of higher moments requires more general theoretical machinery. Just as before, a single source frequency ${\displaystyle \omega }$ is considered. Hence the charge, current, and intrinsic magnetization densities are given by

${\displaystyle \rho (\mathbf {x} ,t)=\rho (\mathbf {x} )e^{-i\omega t}}$
${\displaystyle \mathbf {J} (\mathbf {x} ,t)=\mathbf {J} (\mathbf {x} )e^{-i\omega t}}$
${\displaystyle \mathbf {M} (\mathbf {x} ,t)=\mathbf {M} (\mathbf {x} )e^{-i\omega t}}$

respectively. The resulting electric and magnetic fields share the same time-dependence as the sources.

${\displaystyle \mathbf {E} (\mathbf {x} ,t)=\mathbf {E} (\mathbf {x} )e^{-i\omega t}}$
${\displaystyle \mathbf {H} (\mathbf {x} ,t)=\mathbf {H} (\mathbf {x} )e^{-i\omega t}}$

Using these definitions and the continuity equation allows Maxwell's equations to be written as

${\displaystyle \mathbf {\nabla } \cdot \mathbf {E} (\mathbf {x} )=-{\frac {iZ_{0}}{k}}\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x} )}$
${\displaystyle \mathbf {\nabla } \cdot \mathbf {H} (\mathbf {x} )=-\mathbf {\nabla } \cdot \mathbf {M} (\mathbf {x} )}$
${\displaystyle \mathbf {\nabla } \times \mathbf {E} (\mathbf {x} )=ikZ_{0}\left(\mathbf {H} (\mathbf {x} )+\mathbf {M} (\mathbf {x} )\right)}$
${\displaystyle \mathbf {\nabla } \times \mathbf {H} (\mathbf {x} )=-{\frac {ik}{Z_{0}}}\mathbf {E} (\mathbf {x} )+\mathbf {J} (\mathbf {x} )}$

These equations can be combined by taking the curl of the last equations and applying the identity ${\displaystyle \mathbf {\nabla } \times (\mathbf {\nabla } \times \mathbf {V} )=\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {V} )-\mathbf {\nabla } ^{2}\mathbf {V} }$. This gives the vector forms of the non-homogeneous Helmholtz equation.

${\displaystyle (\nabla ^{2}+k^{2})\mathbf {E} (\mathbf {x} )=-\left[ikZ_{0}\mathbf {J} (\mathbf {x} )+ikZ_{0}\mathbf {\nabla } \times \mathbf {M} (\mathbf {x} )+{\frac {iZ_{0}}{k}}\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x} ))\right]}$
${\displaystyle (\nabla ^{2}+k^{2})\mathbf {H} (\mathbf {x} )=-\left[k^{2}\mathbf {M} (\mathbf {x} )+\mathbf {\nabla } \times \mathbf {J} (\mathbf {x} )+\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {M} (\mathbf {x} ))\right]}$

### Solutions of the wave equation

The homogeneous wave equations that describes electromagnetic radiation with frequency ${\displaystyle \omega }$ in a source-free region have the form.

${\displaystyle (\mathbf {\nabla } ^{2}+k^{2})\mathbf {\Psi } (\mathbf {x} )=0}$

The wave function ${\displaystyle \mathbf {\Psi } (\mathbf {x} )}$ can be expressed as a sum of vector spherical harmonics

${\displaystyle \mathbf {\Psi } (\mathbf {x} )=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }f_{\ell m}(kr)\mathbf {X} _{\ell m}(\theta ,\phi )}$
${\displaystyle f_{\ell m}(kr)=A_{\ell m}^{(1)}h_{\ell }^{(1)}(kr)+A_{\ell m}^{(2)}h_{\ell }^{(2)}(kr)}$

Where ${\displaystyle \mathbf {X} _{\ell m}(\theta ,\phi )=\mathbf {L} Y_{\ell m}(\theta ,\phi )/{\sqrt {\ell (\ell +1)}}}$ are the normalized vector spherical harmonics and ${\displaystyle h_{\ell }^{(1)}}$ and ${\displaystyle h_{\ell }^{(2)}}$ are spherical Hankel functions. See spherical Bessel functions. The differential operator ${\displaystyle \mathbf {L} =-i\mathbf {x} \times \mathbf {\nabla } }$ is the angular momentum operator with the property ${\displaystyle L^{2}Y_{\ell m}=\ell (\ell +1)Y_{\ell m}}$. The coefficients ${\displaystyle A_{\ell m}^{(1)}}$ and ${\displaystyle A_{\ell m}^{(2)}}$ correspond to expanding and contracting waves respectively. So ${\displaystyle A_{\ell m}^{(2)}=0}$ for radiation. To determine the other coefficients, the Green's function for the wave equation is applied. If the source equation is

${\displaystyle (\mathbf {\nabla } ^{2}+k^{2})\mathbf {\Psi } (\mathbf {x} )=-\mathbf {V} (\mathbf {x} )}$

then the solution is:

${\displaystyle \Psi _{\alpha }(\mathbf {x} )=\sum _{\beta }\int d^{3}\mathbf {x'} G_{\alpha \beta }(\mathbf {x} ,\mathbf {x'} )V_{\beta }(\mathbf {x'} )}$

The Green function can be expressed in vector spherical harmonics.

${\displaystyle G_{\alpha \beta }(\mathbf {x} ,\mathbf {x'} )=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }ikh_{\ell }^{(1)}(kr)j_{\ell }(kr')X_{\ell m\alpha }(\theta ,\phi )X_{\ell m\beta }^{*}(\theta ',\phi ')}$

Note that ${\displaystyle \mathbf {X} _{\ell m}^{*}=Y_{\ell m}^{*}\mathbf {L} /{\sqrt {\ell (\ell +1)}}}$ is a differential operator that acts on the source function ${\displaystyle \mathbf {V} }$. Thus, the solution to the wave equation is:

${\displaystyle \mathbf {\Psi } (\mathbf {x} )=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }{\frac {ik}{\sqrt {\ell (\ell +1)}}}h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi )\int d^{3}\mathbf {x'} j_{\ell }(kr')Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {L'} \cdot \mathbf {V} (\mathbf {x'} )}$

### Electric multipole fields

Applying the above solution to the electric multipole wave equation

${\displaystyle (\nabla ^{2}+k^{2})\mathbf {H} (\mathbf {x} )=-\left[k^{2}\mathbf {M} (\mathbf {x} )+\mathbf {\nabla } \times \mathbf {J} (\mathbf {x} )+\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {M} (\mathbf {x} ))\right]}$

gives the solution for the magnetic field:[5]

${\displaystyle \mathbf {H} ^{(E)}(\mathbf {x} )=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }a_{\ell m}^{(E)}h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi )}$
${\displaystyle a_{\ell m}^{(E)}={\frac {ik}{\sqrt {\ell (\ell +1)}}}\int d^{3}\mathbf {x'} j_{\ell }(kr')Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {L'} \cdot \left[k^{2}\mathbf {M} (\mathbf {x'} )+\mathbf {\nabla '} \times \mathbf {J} (\mathbf {x'} )+\mathbf {\nabla '} (\mathbf {\nabla '} \cdot \mathbf {M} (\mathbf {x'} ))\right]}$

The electric field is:

${\displaystyle \mathbf {E} ^{(E)}(\mathbf {x} )={\frac {iZ_{0}}{k}}\mathbf {\nabla } \times \mathbf {H} ^{(E)}(\mathbf {x} )}$

The formula can be simplified by applying the identities

${\displaystyle \mathbf {L} \cdot \mathbf {V} (\mathbf {x} )=i\mathbf {\nabla } \cdot (\mathbf {x} \times \mathbf {V} (\mathbf {x} ))}$
${\displaystyle \mathbf {L} \cdot (\mathbf {\nabla } \times \mathbf {V} (\mathbf {x} ))=i\nabla ^{2}(\mathbf {x} \cdot \mathbf {V} (\mathbf {x} ))-{\frac {i\partial }{r\partial r}}(r^{2}\mathbf {\nabla } \cdot \mathbf {V} (\mathbf {x} ))}$
${\displaystyle \mathbf {L} \cdot \mathbf {\nabla } s(\mathbf {x} )=0}$

to the integrand, which results in[5]

${\displaystyle a_{\ell m}^{(E)}={\frac {-ik^{2}}{\sqrt {\ell (\ell +1)}}}\int d^{3}\mathbf {x'} j_{\ell }(kr')Y_{\ell m}^{*}(\theta ',\phi ')\left[-ik\mathbf {\nabla } \cdot (\mathbf {x'} \times \mathbf {M} (\mathbf {x'} ))-{\frac {i}{k}}\nabla ^{2}(\mathbf {x'} \cdot \mathbf {J} (\mathbf {x'} ))-{\frac {c\partial }{r'\partial r'}}(r'^{2}\rho (\mathbf {x'} ))\right]}$

Green's theorem and integration by parts manipulates the formula into

${\displaystyle a_{\ell m}^{(E)}={\frac {-ik^{2}}{\sqrt {\ell (\ell +1)}}}\int d^{3}\mathbf {x'} j_{\ell }(kr')Y_{\ell m}^{*}(\theta ',\phi ')\left[-ik\mathbf {\nabla } \cdot (\mathbf {x'} \times \mathbf {M} (\mathbf {x'} ))+ik\mathbf {x'} \cdot \mathbf {J} (\mathbf {x'} )\right]+cY_{\ell m}^{*}(\theta ',\phi ')\rho (\mathbf {x'} ){\frac {\partial }{\partial r'}}(r'j_{\ell }(kr'))}$

The spherical bessel function ${\displaystyle j_{\ell }(kr')}$ can also be simplified by assuming that the radiation length scale is much larger than the source length scale, which is true for most antennas.

${\displaystyle j_{\ell }(kr')={\frac {(kr')^{\ell }}{(2\ell +1)!!}}+O((kr')^{\ell +2})}$

Retaining only the lowest order terms results in the simplified form for the electric multipole coefficients:[5]

${\displaystyle a_{\ell m}^{(E)}={\frac {-ick^{\ell +2}}{(2\ell +1)!!}}\left({\frac {\ell +1}{\ell }}\right)^{1/2}[Q_{\ell m}+Q_{\ell m}']}$
${\displaystyle Q_{\ell m}=\int d^{3}\mathbf {x'} r'^{\ell }Y_{\ell m}^{*}(\theta ',\phi ')\rho (\mathbf {x'} )}$
${\displaystyle Q_{\ell m}'=-{\frac {ik}{c(\ell +1)}}\int d^{3}\mathbf {x'} r'^{\ell }Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {\nabla } \cdot (\mathbf {x'} \times \mathbf {M} (\mathbf {x'} ))}$

${\displaystyle Q_{\ell m}}$ is the same as the electric multipole moment in the static case if it were applied to the static charge distribution ${\displaystyle \rho (\mathbf {x} )}$ whereas ${\displaystyle Q_{\ell m}'}$ corresponds to an induced electric multipole moment from the intrinsic magnetization of the source material.

### Magnetic multipole fields

Applying the above solution to the magnetic multipole wave equation

${\displaystyle (\nabla ^{2}+k^{2})\mathbf {E} (\mathbf {x} )=-\left[ikZ_{0}\mathbf {J} (\mathbf {x} )+ikZ_{0}\mathbf {\nabla } \times \mathbf {M} (\mathbf {x} ))+{\frac {iZ_{0}}{k}}\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x} ))\right]}$

gives the solution for the electric field:[5]

${\displaystyle \mathbf {E} ^{(M)}(\mathbf {x} )=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }a_{\ell m}^{(M)}h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi )}$
${\displaystyle a_{\ell m}^{(M)}={\frac {ik}{\sqrt {\ell (\ell +1)}}}\int d^{3}\mathbf {x'} j_{\ell }(kr')Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {L'} \cdot \left[ikZ_{0}\mathbf {J} (\mathbf {x} )+ikZ_{0}\mathbf {\nabla } \times \mathbf {M} (\mathbf {x} ))+{\frac {iZ_{0}}{k}}\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {J} (\mathbf {x} ))\right]}$

The magnetic field is:

${\displaystyle \mathbf {H} ^{(M)}(\mathbf {x} )=-{\frac {i}{kZ_{0}}}\mathbf {\nabla } \times \mathbf {E} ^{(M)}(\mathbf {x} )}$

As before, the forumula simplifies to:

${\displaystyle a_{\ell m}^{(M)}={\frac {-ik^{2}}{\sqrt {\ell (\ell +1)}}}\int d^{3}\mathbf {x'} j_{\ell }(kr')Y_{\ell m}^{*}(\theta ',\phi ')\left[\mathbf {\nabla } \cdot (\mathbf {x'} \times \mathbf {J} (\mathbf {x'} ))-k^{2}\mathbf {x'} \cdot \mathbf {M} (\mathbf {x'} )\right]+Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {\nabla } \cdot \mathbf {M} (\mathbf {x'} ){\frac {\partial }{\partial r'}}(r'j_{\ell }(kr'))}$

Retaining only the lowest order terms results in the simplified form for the magnetic multipole coefficients:[5]

${\displaystyle a_{\ell m}^{(M)}={\frac {-ik^{\ell +2}}{(2\ell +1)!!}}\left({\frac {\ell +1}{\ell }}\right)^{1/2}[M_{\ell m}+M_{\ell m}']}$
${\displaystyle M_{\ell m}={\frac {1}{\ell +1}}\int d^{3}\mathbf {x'} r'^{\ell }Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {\nabla } \cdot (\mathbf {x'} \times \mathbf {J} (\mathbf {x'} ))}$
${\displaystyle M_{\ell m}'=\int d^{3}\mathbf {x'} r'^{\ell }Y_{\ell m}^{*}(\theta ',\phi ')\mathbf {\nabla } \cdot \mathbf {M} (\mathbf {x'} )}$

${\displaystyle M_{\ell m}}$ is the magnetic multipole moment from the effective magnetization ${\displaystyle \mathbf {x} \times \mathbf {J} (\mathbf {x} )/2}$ while ${\displaystyle M_{\ell m}'}$ corresponds to the intrinsic magnetization ${\displaystyle \mathbf {M} (\mathbf {x} )}$.

### General solution

The electric and magnetic multipole fields combine to give the total fields:[5]

${\displaystyle \mathbf {E} (\mathbf {x} ,t)=\Re \left(\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }\left[a_{\ell m}^{(M)}h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi )+{\frac {iZ_{0}}{k}}a_{\ell m}^{(E)}\mathbf {\nabla } \times (h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi ))\right]e^{-i\omega t}\right)}$
${\displaystyle \mathbf {H} (\mathbf {x} ,t)=\Re \left(\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }\left[a_{\ell m}^{(E)}h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi )-{\frac {i}{kZ_{0}}}a_{\ell m}^{(M)}\mathbf {\nabla } \times (h_{\ell }^{(1)}(kr)\mathbf {X} _{\ell m}(\theta ,\phi ))\right]e^{-i\omega t}\right)}$

Note that the radial function ${\displaystyle h_{\ell }^{(1)}(kr)}$ can be simplified in the far field limit ${\displaystyle 1/r\ll 1}$.

${\displaystyle h_{\ell }^{(1)}(kr)=(-i)^{\ell +1}{\frac {e^{ikr}}{kr}}+O(1/r^{2})}$


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# Bibliography

 Van Allen Probes Bibliography is from August 2012 through September 2021 Notice: Clicking on the DOI link will open a new window with the original bibliographic entry from the publisher. Clicking on a single author will show all publications by the selected author. Clicking on a single keyword, will show all publications by the selected keyword.

Electron Acceleration in the Heart of the Van Allen Radiation Belts

 Author Reeves, G.; Spence, H.; Henderson, M.; Morley, S.; Friedel, R.; Funsten, H.; Baker, D.; Kanekal, S.; Blake, J.; Fennell, J.; Claudepierre, S.; Thorne, R.; Turner, D.; Kletzing, C.; Kurth, W.; Larsen, B.; Niehof, J.; Keywords Van Allen Probes Abstract The Van Allen radiation belts contain ultrarelativistic electrons trapped in Earth\textquoterights magnetic field. Since their discovery in 1958, a fundamental unanswered question has been how electrons can be accelerated to such high energies. Two classes of processes have been proposed: transport and acceleration of electrons from a source population located outside the radiation belts (radial acceleration) or acceleration of lower-energy electrons to relativistic energies in situ in the heart of the radiation belts (local acceleration). We report measurements from NASA\textquoterights Van Allen Radiation Belt Storm Probes that clearly distinguish between the two types of acceleration. The observed radial profiles of phase space density are characteristic of local acceleration in the heart of the radiation belts and are inconsistent with a predominantly radial acceleration process. Year of Publication 2013 Journal Science Volume 341 Number of Pages 991-994 Section Date Published 07/2013 ISBN URL http://www.sciencemag.org/cgi/doi/10.1126/science.1237743 DOI 10.1126/science.1237743


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# Prove the following inequality??

Prove that $$(1+k)^{p}<\frac{(1-k^{p+1})}{(1-k)},$$ Here k,p $\in$ (0,1). I have proved this ineqality for p=1/2. The proof is as follows. \begin{eqnarray} (1+k)^{0.5}<\frac{(1-k^{1.5})}{(1-k)} \\ (1+k)^{0.5}<\frac{(1-\sqrt{k})(k+1+\sqrt{k})}{(1-\sqrt{k})(1+\sqrt{k})}\\ (1+k)^{0.5}(1+\sqrt{k})<(k+1+\sqrt{k})\\ \text{Geometric Mean}<\text{Arithmatic Mean} \end{eqnarray}, However I am not able to show this inequality for p $\in$ (0,1) with k lying between 0 to 1. I have checked the inequality to hold numerically. Please have a look at it.

• have you heard about Bernoulli's inequality? – Vasya Apr 12 '18 at 17:32
• Thanks for your comment. I saw the inequality right now, don't know if it could be used here somehow. Do you think it would be helpful – Rahul Sharma Apr 12 '18 at 17:48

## 1 Answer

You can indeed show it using Bernouilli's inequality. It states that for $x > -1$ and $0 < r < 1$, $(1 + x)^r < 1 + rx$. The given inequality can be rewritten as $$(1 - k)(1 + k)^p < 1 - k^{p + 1}.$$ Applying Bernouilli's inequality to the LHS, we get $$(1 - k)(1 + k)^p < (1 - k)(1 + pk) = 1 - k - pk^2 + pk.$$ If we can show that this is strictly less than $1 - k^{p + 1}$ or, equivalently, that $$k^p < 1 + pk - p,$$ we are done. This follows however from $k^p = (1 + (k - 1))^p < 1 + p(k - 1)$ by Bernouilli's inequality.