text:
# Prove that pseudorandom generator is a one way function

Suppose the following PRG $G : \{0,1\}^n \rightarrow \{0,1\}^{n +l}$, I want to prove that $G$ is one way function (and not building one), for:

1. $l = \omega (\log n)$
2. $l = 1$

For $l = \log n$, suppose that $G$ isn't OWF, so there's an inverter $A$ which succeeds w.p. $\geq 1/p(n)$, trying to build a distinguisher $B$ from inverter $A$ to $G$, I can do as follows:

$B(y)$ assigns 1 to each input successfully inverted. So $\Pr[B(y) = 1 : y = f(x) = G(x)] \geq 1/p(n)$, and for random input $\Pr[B(U_{n + \log n})] \geq 2^{n}/2^{\log n + n}$. All in all, the distinguisher succeeds w.p.

$$|\Pr[B(y) = 1 : y = f(x) = G(x)] - \Pr[B(U_{n + \log n})]| \geq 1/n - 1/p(n) \geq 1/2n$$

But for $l=1$, I get a bound of $1/4$ ($1/2 -1/p(n) \geq 1/4$)... It doesn't feel correct, but does (?) follow the definition of distinguisher for PRGs.

What am I missing?

• From the question, it's completely unclear what you are actually doing, and then where the problem is. But maybe this question and answer is helpful to you. However, the properties of PRFs, PRGs and OWFs are to some degree related but not the same. For a correct proof the details matter. – tylo Nov 27 '17 at 14:29
• @tylo I updated the question, could you look again? – Napoleon Nov 27 '17 at 15:17
• What exactly is your question? Why 1/4 is sufficient advantage for a PRF distinguisher? Your proof itself - though missing a lot of details - would appear to be on the right track. – Maeher Nov 28 '17 at 1:12
• I want to prove that $G$ is OWF. 1/4 is enough, because its PRG should be computional indistinguishble for any negligible function, and 1/4 is greater for any negligeble function – Napoleon Nov 28 '17 at 7:48

$$l = \log n: \Pr[B(Un+\log n)]≥2^n/2^{n+\log n} \cdot 1/p(n)$$ Since even if the input of $B$ is from the image of $G$, the probability to invert it is $1/p(n)$. So you need to correct this case and it will help you with $l = 1$.
• Firat of all, shouldn't your bound be $\leq$ (cause otherwise you can't bound $|\Pr[B(G(U_n)) = 1] - \Pr[B(U_{n + \log n})]|$)? Second, if I follow your way, for $l = \log n$ we get bound of $1/2p(n)\log n$, and for $l = 1$ we've get a bound of $\geq 2^n/2^{n+1}/p(n) = 1/2p(n)$. Am I correct? And finally, I stiil don't understand what's wrong with my previous way. – Napoleon Nov 29 '17 at 14:36
• $1/p(n) - 2^{n}/(2^{n+1}*p(n)) = 1/p(n)$ so G isn't a PRG – sosolo Nov 29 '17 at 15:09


text:
## Lookahead: A Far-sighted Alternative of Magnitude-based Pruning

Sep 25, 2019 (edited Mar 11, 2020)ICLR 2020 Conference Blind SubmissionReaders: Everyone
• Original Pdf: pdf
• TL;DR: We study a multi-layer generalization of the magnitude-based pruning.
• Abstract: Magnitude-based pruning is one of the simplest methods for pruning neural networks. Despite its simplicity, magnitude-based pruning and its variants demonstrated remarkable performances for pruning modern architectures. Based on the observation that magnitude-based pruning indeed minimizes the Frobenius distortion of a linear operator corresponding to a single layer, we develop a simple pruning method, coined lookahead pruning, by extending the single layer optimization to a multi-layer optimization. Our experimental results demonstrate that the proposed method consistently outperforms magnitude-based pruning on various networks, including VGG and ResNet, particularly in the high-sparsity regime. See https://github.com/alinlab/lookahead_pruning for codes.
• Keywords: network magnitude-based pruning

format: plain


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## anonymous one year ago does anyone know how to do this

• This Question is Closed
1. anonymous

2. misty1212

HI!!

3. misty1212

|dw:1443299552389:dw|

4. misty1212

that rectangle has area $$4\times 11=44$$

5. misty1212

|dw:1443299625759:dw|

6. misty1212

the other one has area $$3\times =12$$ add the two up

7. anonymous

thank you so much i was really confused on how to do it

8. anonymous

9. anonymous

so with this one you just have to add or what do you do

10. anonymous

do you add the 8 + 5 + 6

11. misty1212

same as before, make two rectangles out of it

12. misty1212

or do $8\times 6-3\times 5$ big one minus missing piece

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# Six machines, each working at the same constant rate

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Six machines, each working at the same constant rate [#permalink]

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Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the Job in 8 days?

A. 2
B. 3
C. 4
D. 6
E. 8
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Re: Six machines, each working at the same constant rate [#permalink]

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26 Apr 2010, 22:50
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zz0vlb wrote:
Six machines, each working at the same constant rate, together can complete a certain job in 12 days.
How many additional machines, each working at the same constant rate, will be needed to complete the Job in 8 days?
a.2
b.3
c.4
d.6
e.8
[Reveal] Spoiler:
3

Source: GMAT Prep

let each machine work at rate of x days
1/x+1/x+1/x+1/x+1/x+1/x = 1/12
6/x = 1/12
x = 72
now y is number of machines needed to complete job in 8 days so,
y/72 = 1/8
y = 9
required = y-x = 3 hence b.

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Re: Six machines, each working at the same constant rate [#permalink]

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27 Apr 2010, 06:02
3
KUDOS
zz0vlb wrote:
Six machines, each working at the same constant rate, together can complete a certain job in 12 days.
How many additional machines, each working at the same constant rate, will be needed to complete the Job in 8 days?

a.2
b.3
c.4
d.6
e.8

[Reveal] Spoiler:
3

Source: GMAT Prep

Another solution which is faster is Since each machine works at a constant rate. The time needs to bought down from 12 to 8. So the new time is 2/3 of the original time. Thus to achieve this we need the rate to be 3/2 of original.

So 3/2* 6 = 9

So we need 9-6 = 3 more machines.
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27 Apr 2010, 09:49
Please explain the quick and logical approach
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Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?

A. 2
B. 3
C. 4
D. 6
E. 8

Let $$x$$ be the time needed for 1 machine to complete the job, so rate of one machine is $$\frac{1}{x}$$ (rate is the reciprocal of time) --> rate of 6 machines would be $$\frac{6}{x}$$.

As $$job=time*rate$$ --> $$1=12*\frac{6}{x}$$ --> $$x=72$$ days needed for 1 machine to complete the job.

To complete the job in 8 days $$\frac{72}{8}=9$$ machines are needed.

Difference: 9-6=3.

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27 Apr 2010, 14:39
Another solution which is faster is Since each machine works at a constant rate. The time needs to bought down from 12 to 8. So the new time is 2/3 of the original time. Thus to achieve this we need the rate to be 3/2 of original.

So 3/2* 6 = 9

So we need 9-6 = 3 more machines.
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Re: Six machines, each working at the same constant rate [#permalink]

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19 Jun 2010, 19:52
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Actually, this was more simple than I thought. When I first saw this, I started to panic thinking it's a rate question. But then I took a step back, and it was the simplest of calculations.

So here's the deal:

12 days -> 6 machines: 8 days -> ?

8 days means you definitely need more machines than the 6 and in such case the equation looks like-

12 * 6 = 8 * x

Therefore, x = 72 / 8 = 9

So 9-6=3 more machines

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16 Jun 2011, 13:56
6*12/1= (6+X)8

X=3

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18 Jun 2011, 18:56
Lets look at this way :As the rate of one machine is constant i e 1/(12*6)=1/72 .Even after addition of extra machines the rate will remain the same .Hence,if machines are x now ,the rate becomes 1/8x which is equal to 1/72.
X comes out to be 9 .
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6 machines require 12 days = 72 machine days
how many machines to do the job in 8 days = $$\frac{72 machine-days}{8 days} = 9 machines$$

so, 3 more machines.

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05 Aug 2011, 09:02
6 machines - 1 unit of work - 12 days

=> 1 machine = 1/72 of work per day

=> rate of each machine = 1/72

equating work done by 6 machines and 6+x machines

1 = (1/72)(6+x)(8)

=> x = 3

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05 Aug 2011, 13:17
6 machines 12 days...
therefore (6*12)/8=9number of machines required to complete the job in 8 days
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30 Sep 2011, 20:44
Hi, can someone tell me if this is really a level 700 question? I felt like this was so much easier than some of the other questions posted but its labelled as level 700.

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30 Sep 2011, 23:22
musicbelle wrote:
Hi, can someone tell me if this is really a level 700 question? I felt like this was so much easier than some of the other questions posted but its labelled as level 700.

Yes its definitely not a 700 level question but then rating of a question on a forum is as per the poster. A 700 level question for me can look like a 500 for you and vice-versa. I would rather concentrate on the take aways from each question

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06 Jan 2012, 06:28
Nice question. Nice solutions by many. Thanks all.
Easy one. 3 more machines are required.
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06 Jan 2012, 20:53
This is effectively an inverse proportionality problem.

6M --> 12 days --> 1 work
X --> 8 days --> 1 work

X x 8 = 6 x 12
X = 9

Therefore, additional machines = 9-6 = 3
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Re: Six machines, each working at the same constant rate [#permalink]

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04 Mar 2013, 00:09
zz0vlb wrote:
Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the Job in 8 days?

A. 2
B. 3
C. 4
D. 6
E. 8

6 * rate * 12 = # of Machine * rate * 8.

# of machine =( 6 *12) / 8 = 9.

9 - 6 = 3 additional machines required
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A one-proportion $z$-interval was constructed. A researcher wants to reduce the margin of error to one-fifth of its original margin of error.

What would she have to do to the sample size?

A

This cannot be answered without the sample size.

B

Divide the sample size by $5$.

C

Divide the sample size by $25$.

D

Multiply the sample size by $5$.

E

Multiply the sample size by $25$.

Select an assignment template


text:
### Arsenal Collection: Half Time

from nrich.maths.org/8821

When Arsenal played Aston Villa in the 2010/11 Premier League, the final score was $4 - 2$.

What could the half time score have been?
Can you find all the possible half time scores?
How will you make sure you don't miss any out?

Later in the season Arsenal played Tottenham Hotspur.  The final score was a draw, $3 - 3$.

Can you find all the possible half time scores for this match?


text:
# How do you divide ( -3x^3 - 2x^2+13x+4 )/(3-x )?

May 3, 2017

Using the Long Division method ->

#### Explanation:

First step I would do is remove the negative sign from the fraction.
$- \frac{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}{-} \left(x - 3\right)$ $= \frac{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}{x - 3}$
Proceed with Long Division.

..........................$\left(+ 3 {x}^{2} + 11 x + 20\right)$
$\textcolor{w h i t e}{\frac{x - 3}{\textcolor{b l a c k}{x - 3}} \frac{3 {x}^{3} + 2 {x}^{2} - {x}^{2} - 7 x - 7}{\textcolor{b l a c k}{\text{)} \overline{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}}}}$
.............$- \left(3 {x}^{3} - 9 {x}^{2}\right)$
..................------------------------------------
.................$\left(0 {x}^{3} + 11 {x}^{2} - 13 x\right)$
..........................$- \left(11 {x}^{2} - 33 x\right)$
..................------------------------------------
..........................$\left(- 0 {X}^{2} + 20 x - 4\right)$
...........................................$- \left(20 x - 60\right)$
.................--------------------------------------
................................................$\left(0 x + 56\right)$

Note: Ignore the full stops, I'm new at this :/

You may want to copy it on paper to make it clearer.

So, to simplify the fraction from the above:

$- \frac{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}{-} \left(x - 3\right)$ = $3 {x}^{2} + 11 x + 20 + \left(\frac{56}{x - 3}\right)$

May 3, 2017

This really is long division. It just looks different!

$3 {x}^{2} + 11 x + 20 + \frac{56}{x - 3}$

#### Explanation:

For convenience I have written $3 - x$ as $- x + 3$

$\text{ } - 3 {x}^{3} - 2 {x}^{2} + 13 x + 4$
$\textcolor{m a \ge n t a}{3 {x}^{2}} \left(- x + 3\right) \to \underline{- 3 {x}^{3} + 9 {x}^{2}} \leftarrow \text{ Subtract}$
$\text{ } 0 - 11 {x}^{2} + 13 x + 4$
$\textcolor{m a \ge n t a}{11 x} \left(- x + 3\right) \to \text{ "ul(-11x^2+33x )larr" Subtract}$
$\text{ "0" } - 20 x + 4$
$\textcolor{m a \ge n t a}{20} \left(- x + 3\right) \to \text{ "ul(-20x+60) larr" Subtract}$
" "color(magenta)(0-56 larr "Remainder")

$\frac{- 3 {x}^{3} - 2 {x}^{2} + 13 x + 4}{3 - x} = \textcolor{m a \ge n t a}{3 {x}^{2} + 11 x + 20 - \frac{56}{3 - x}}$

To match Ching's answer consider $- \frac{56}{3 - x}$

Not that $3 - x$ is the same as $- \left(+ x - 3\right)$

So we have $\text{ } - \left(\frac{56}{- \left(x - 3\right)}\right) \to + \frac{56}{x - 3}$

Giving: $\text{ } 3 {x}^{2} + 11 x + 20 + \frac{56}{x - 3}$


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### php - about laravel dompdf

I am creating a process to convert Excel to PDF with Laravel.

I'm trying to create it using dompdf.

I am referring to the following site, but I can't do it well in the command section.
laravel-dompdf PDF in Japanese environment with version 7.0 environment Create

``# php load_font.php ipag \$path_to_font_directory/ipag.ttf``

Where
"Fatal error: require_once (): Failed opening required'vendor/autoload.php'(include_path ='C: \ xampp \ php \ PEAR') in C: \ xampp \ htdocs \ sample \ vendor \ dompdf \ dompdf \ load_font.php on line 3 "

I get an error with

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I did not understand even after checking, so I will ask you a question.


text:
Solutions to the Monge-Ampere Equation with Polyhedral and Y-shaped Singularities

Surfaces which The Monge-Ampere equation $\mathrm{det}(D^2u) = 1$ arises in prescribed curvature problems and in optimal transport. An interesting feature of the equation is that it admits singular solutions. We will discuss new examples of convex functions on $\mathbb{R}^n$ that solve the Monge-Ampere equation away from finitely many points but contain polyhedral and $Y$-shaped singular structures. Along the way, we will discuss

The Stickiness Property of Nonlocal Minimal Surfaces

Surfaces that minimize a nonlocal perimeter functional exhibit quite different behaviors than the ones minimizing the classical perimeter. Among these peculiar features, an interesting property, which is also in contrast with the pattern produced by the solutions of linear equations, is given by the capacity, and the strong tendency, of adhering at the boundary. We

Generic Regularity in the Fractional Obstacle Problem

The fractional obstacle problem in $\mathbb R^n$ with obstacle $\varphi\in C^\infty(\mathbb{R}^n)$ can be written as $\min\{(-\Delta)^s u , u-\varphi\} = 0,\quad\textrm{in }\quad\mathbb{R}^n.$ The set $\{u = \varphi\} \subset \mathbb{R}^n$ is called the contact set, and its boundary is the free boundary, an unknown of the problem. The free boundary for the fractional obstacle problem can be divided between two subsets: regular points

Well-posedness of Non-Linear Partial Differential Equations in Weighted Sobolev Spaces

The well-posedness (existence, uniqueness, and stability) of the solutions for several nonlinear dispersive PDEs with initial data in Sobolev spaces has been intensively analyzed. However, Sobolev functions do not necessarily decay fast at infinity, which is in many applications an unrealistic assumption. In this talk, we discuss examples of existence and uniqueness results in weighted

Maiden Edition of the Aims-Cameroon Mathematics and its Applications Meeting (ACMAM) Holds in January 2022

AIMS Cameroon Research Center will organize the maiden edition of its annual scientific event named AIMS-Cameroon Mathematics and its Applications Meeting (ACMAM) from the 12th to the 14th of January, 2022. Lectures take place at the AIMS Cameroon Research Center in Limbe South West Region. The AIMS-Cameroon Mathematics and its Applications Meeting (ACMAM) is an

Regularity for C^{1,\alpha} interface transmission problems

Transmission problems originally arose in elasticity theory, and they are nowadays of great interest due to their many applications in different areas in science. Typically, in these problems, there is a fixed interface where solutions may change abruptly, and the primary focus is to study their behavior across this surface. In this talk, we will

Regularity Estimate for Nonlocal Space-Time Master Equations

We analyze regularity estimates for solutions to nonlocal space-time equations driven by fractional powers of parabolic operators in divergence form. These equations are fundamental in semipermeable membrane problems, biological invasion models and they also appear as generalized Master equations. We develop a parabolic method of semigroups that allows us to prove a local extension problem

Instability of Finite Time Blow-up For Incompressible Euler

In this talk, we will discuss the interaction between the stability, and the propagation of regularity, for solutions to the incompressible 3D Euler equation. It is still unknown whether a solution with smooth initial data can develop a singularity in finite time. We will describe how, in such a scenario, the solution becomes unstable as

Boundedness Properties for Hermite Pseudo-Multipliers

Fourier multipliers and pseudo-differential operators are defined by means of the Fourier transform and play an important role in the study of partial differential equations. In the same spirit, Hermite pseudo-multipliers are associated to Hermite expansions and they represent the counterparts to pseudo-differential operators in the Hermite setting. After some preliminaries, we will present results

Sharp Quantitative Faber-Krahn inequalities and the Alt-Caffarelli-Fredman Monotonicity Formula

The Colloquium will be focused on this research work by Mark Allen of Brigham Young University in the USA. He establishes the new sharp quantitative estimates for Faber-Krahn inequalities on Euclidean space, the round sphere and hyperbolic space and then applies these inequalities in order to establish a quantitative form of the Alt-Caffarelli-Friedman monotonicity formula.


text:
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# Ozone is depleted in the stratosphere by chlorine from $\mathrm{CF}_{3} \mathrm{Cl}$ according to this set of equations:$$\begin{array}{c}{\mathrm{CF}_{3} \mathrm{Cl}+\mathrm{UV} \text { light } \longrightarrow \mathrm{CF}_{3}+\mathrm{Cl}} \\ {\mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{ClO}+\mathrm{O}_{2}} \\ {\mathrm{O}_{3}+\mathrm{UV} \operatorname{light} \longrightarrow \mathrm{O}_{2}+\mathrm{O}} \\ {\mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2}}\end{array}$$What total volume of ozone at a pressure of 25.0 $\mathrm{mm} \mathrm{Hg}$ and a tem-perature of 225 $\mathrm{K}$ is destroyed when all of the chlorine from 15.0 $\mathrm{g}$ of$\mathrm{CF}_{3} \mathrm{Cl}$ goes through ten cycles of the above reactions?

## $1.61 \times 10^{3} \mathrm{L} \mathrm{O}_{3}$

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##### Stephanie C.

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##### Allea C.

University of Maryland - University College

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### Video Transcript

So this question were given, um, a series of reactions that causes those in depletion by chlorofluorocarbons. Um, and were given that we have 15 grams of CF three CEO i things going through 10 cycles of this reaction. So we're doing this first thing we're gonna do is we're gonna want to calculate how many moles of that chlorofluorocarbon we have, starting with just 15 grams of, uh see if three c l and I'm gonna convert two moles, so I'm gonna do that is using the more massive this which is one of 4.46 grams per mole, and then go to convert to malls which, if I look at that, gives me 0.14 malls. Now, if you look at the first direction, you'll see that there's a oneto one molar concentration of ah cf three clt, Jesse L. So I can say this is Mel's off. Corinne. All right, so now we're gonna look at this, um, reaction. And the first step of this reaction is that chlorine combines with ozone to create ah cielo and oxygen gas. Um, and then that same Cielo product is later used in the reaction. So If you look at this, every one mole of chlorine leads to two moles of ozone depleted. All right, eso What we can do is weaken, and also we're going through 10 cycles. So 10 cycles of this reaction means 20 moles of 03 depleted. Now, that's assuming that we're using one mole of chlorine, though, and that's not sure. So what we're gonna do is 20 times 200.14 That's going to give us the moles of over three that we deplete, which is going to be 2.8 on that We could do this is you could Alm you could double the 0.14 number, give us point to a and then times that by 10 to get the same two point number. And that is, um, moles of 03 are So now we can just use the ideal gas law where I can say the equals and r t overpay. I consult from there so I can save E is equal to 2.8 malls. Ah, and then our since I'm using. Since the problem gives everything in millimeters of mercury, where is gonna continue to use that? It's like 62.36 uh, millimeters of Mercury Times leaders over mole times. Calvin, the problem gives us a temperature of, um, 225 Children and a pressure of 25.0 millimeters of mercury. All right. So I won't go through and check our units. Ah, we have moles that cancels their and their Kelvin cancels their and their militarism record cancels there, and they're leaving us with only leaders. Ah, and that works very, very well, because the question is asking us how many leaders of ozone depleting or it's the volume of those, Um, all right, so if you do this and you run through this calculation, you under with a very large number of 1571.5 leaders of both three

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#### Topics

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##### Stephanie C.

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##### Allea C.

University of Maryland - University College

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text:
Volume 47, Issue 2
Regularity of Positive Solutions for an Integral System on Heisenberg Group

J. Math. Study, 47 (2014), pp. 208-220.

Published online: 2014-06

Cited by

Export citation
• Abstract

In this paper, we are concerned with the properties of positive solutions of the following nonlinear integral systems on the Heisenberg group $\mathbb{H}^n$, $$\left\{\begin{array}{ll} u(x)=\int_{\mathbb{H}^n}\frac{v^{q}(y)w^{r}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ v(x)=\int_{\mathbb{H}^n}\frac{u^{p}(y)w^{r}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ w(x)=\int_{\mathbb{H}^n}\frac{u^{p}(y)v^{q}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ \end{array}\right.$$
for $x\in \mathbb{H}^n$, where $0<\alpha<Q=2n+2$, $n\geq3$, $\beta\geq0$, $\alpha+\beta<Q$, and $p,q,r > 1$ satisfying $\frac{1}{p+1}$+ $\frac{1}{q+1} + \frac{1}{r+1} = \frac{Q+α+β}{Q}.$ We show that positive solution triples $(u,v,w)\in L^{p+1}(\mathbb{H}^n)\times L^{q+1}(\mathbb{H}^n)\times L^{r+1}(\mathbb{H}^n)$ are bounded and they converge to zero when $|x|→∞.$

• Keywords

Ground state solutions, Heisenberg group, nonlinear integral system.

45E10, 45G05

xiaowei19901207@126.com (Weiyang Chen)

littleli_chen@163.com (Xiaoli Chen)

• BibTex
• RIS
• TXT
@Article{JMS-47-208, author = {Weiyang and Chen and xiaowei19901207@126.com and 13257 and Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China and Weiyang Chen and Xiaoli and Chen and littleli_chen@163.com and 13258 and Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China and Xiaoli Chen}, title = {Regularity of Positive Solutions for an Integral System on Heisenberg Group}, journal = {Journal of Mathematical Study}, year = {2014}, volume = {47}, number = {2}, pages = {208--220}, abstract = {

In this paper, we are concerned with the properties of positive solutions of the following nonlinear integral systems on the Heisenberg group $\mathbb{H}^n$, $$\left\{\begin{array}{ll} u(x)=\int_{\mathbb{H}^n}\frac{v^{q}(y)w^{r}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ v(x)=\int_{\mathbb{H}^n}\frac{u^{p}(y)w^{r}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ w(x)=\int_{\mathbb{H}^n}\frac{u^{p}(y)v^{q}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ \end{array}\right.$$
for $x\in \mathbb{H}^n$, where $0<\alpha<Q=2n+2$, $n\geq3$, $\beta\geq0$, $\alpha+\beta<Q$, and $p,q,r > 1$ satisfying $\frac{1}{p+1}$+ $\frac{1}{q+1} + \frac{1}{r+1} = \frac{Q+α+β}{Q}.$ We show that positive solution triples $(u,v,w)\in L^{p+1}(\mathbb{H}^n)\times L^{q+1}(\mathbb{H}^n)\times L^{r+1}(\mathbb{H}^n)$ are bounded and they converge to zero when $|x|→∞.$

}, issn = {2617-8702}, doi = {https://doi.org/10.4208/jms.v47n2.14.05}, url = {http://global-sci.org/intro/article_detail/jms/9955.html} }
TY - JOUR T1 - Regularity of Positive Solutions for an Integral System on Heisenberg Group AU - Chen , Weiyang AU - Chen , Xiaoli JO - Journal of Mathematical Study VL - 2 SP - 208 EP - 220 PY - 2014 DA - 2014/06 SN - 47 DO - http://doi.org/10.4208/jms.v47n2.14.05 UR - https://global-sci.org/intro/article_detail/jms/9955.html KW - Ground state solutions, Heisenberg group, nonlinear integral system. AB -

In this paper, we are concerned with the properties of positive solutions of the following nonlinear integral systems on the Heisenberg group $\mathbb{H}^n$, $$\left\{\begin{array}{ll} u(x)=\int_{\mathbb{H}^n}\frac{v^{q}(y)w^{r}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ v(x)=\int_{\mathbb{H}^n}\frac{u^{p}(y)w^{r}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ w(x)=\int_{\mathbb{H}^n}\frac{u^{p}(y)v^{q}(y)}{|x^{-1}y|^\alpha|y|^\beta}\,dy,\\ \end{array}\right.$$
for $x\in \mathbb{H}^n$, where $0<\alpha<Q=2n+2$, $n\geq3$, $\beta\geq0$, $\alpha+\beta<Q$, and $p,q,r > 1$ satisfying $\frac{1}{p+1}$+ $\frac{1}{q+1} + \frac{1}{r+1} = \frac{Q+α+β}{Q}.$ We show that positive solution triples $(u,v,w)\in L^{p+1}(\mathbb{H}^n)\times L^{q+1}(\mathbb{H}^n)\times L^{r+1}(\mathbb{H}^n)$ are bounded and they converge to zero when $|x|→∞.$

Weiyang Chen & Xiaoli Chen. (2019). Regularity of Positive Solutions for an Integral System on Heisenberg Group. Journal of Mathematical Study. 47 (2). 208-220. doi:10.4208/jms.v47n2.14.05
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# Ubuntu – Not design having the softlink (symlink) and hardlink in the same “link”

I haven't used hardlinks for a long time and never really needed them until I was asked in an interview. I read their difference from symlinks here: What is the difference between a hard link and a symbolic link?

Is there any particular reason why the design is not having both of the capabilities of the symlink and the capabilities of the hardlink in the same link file?

You want to point to a file. Ok so you start with a hardlink functionality to cover situations where the filename is changed or the file is moved. If hardlink is not valid because it refers outside the filesystem or fails for some other reason have a fallback, the filepath of that file to refer to, in other words have a symlink.

Because what the user of an operating system wants by the end of the day is just have a link to a file.

Is there anything that could prevent the above design solution for links?

In general I don't think it's a good idea to hide fundamentally different things from the user. For most scenarios, soft links are fine. In my experience, hard links are mainly useful for backups. For example dirvish makes use of them.


text:
# What Are Closed Timelike Curves?

1. Aug 25, 2008

### Champion

i was going over some relativity search online and came across what is called a closed timelike curve and that i actually allows time travel, am i right?

2. Aug 25, 2008

### Fredrik

Staff Emeritus
A "curve" is a function $c:A\rightarrow M$ where A is some interval of real numbers and M is the spacetime manifold. "Timelike" means that if we call the tangent vector v and the metric tensor g, then g(v,v)<0 at all points on the curve. (If we define the metric with a +--- signature instead of -+++, it's ">" instead of "<"). "Closed" means that c(t+T)=c(t) for some real numbers t and T such that both t and t+T are in A.

A timelike curve is the type of curve that can represent that path of a massive object through spacetime. If it's closed, the object will meet a younger version of itself at some point.

Edit: I should probably add that the condition g(v,v)<0 can also be written

$$g_{\mu\nu}v^\mu v^\nu<0$$

and that in a local inertial frame, this is just

$$-(v^0)^2+(v^1)^2+(v^2)^2+(v^3)^2<0$$

Last edited: Aug 25, 2008
3. Aug 25, 2008

### Champion

i was thinking it over,about the constant energy.ctc's would violate 2nd law of therodynamics,also i looked up that GR allows ctc's but QM does not.


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What Is Loan To Value (LTV) And How Does It Affect The Size Of My Loan?

While this video simplifies things to help you remember, the loan to value ratio is the amount of money you borrow compared with the price or appraised value of the home you are purchasing. Each loan has a specific LTV limit. For example: With a 75% LTV loan on a home priced at $100,000 you could borrow up to$75,000 (75% of $100,000) and would have to pay$25000 as a down payment. The LTV ratio reflects the amount of equity borrowers have in their homes. The higher the LTV the less cash homebuyers are required to pay out of their own funds. So, to protect lenders against potential loss in case of default, higher LTV loans (80% or more) usually require mortgage insurance policies.
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# Breadth First Search to find connected primes

I've been working through the British Informatic Olympiad's 2016 Paper, and I've been struggling with some of the larger cases for question 3, specifically the larger cases on the mark scheme (the 3 before last specifically).

• Two prime numbers are connected if the difference between them is 2n for some whole number n ≥ 0.
• A path is a sequence of (at least two) prime numbers, without repetition, where adjacent numbers in the sequence are connected. If the first number in the sequence is p and the last number is q then we say the path is between p and q.
• The length of a path is the total number of prime numbers used. There may be multiple paths between two prime numbers; the lengths of these paths may be different.

Write a program to determine the length of the shortest path between two primes. Your program should accept three integers in order: l (4 ≤ l ≤ 224) indicating the highest value you are allowed to use, followed by the primes p then q (2 ≤ p < q < l). You will only be given input where there is a path between p and q using values below l. You should output the length of the shortest path.

Inputs I was struggling with:

614700 3643 90149 | outputs 18
987654 3643 90149 | outputs 16
1000000 2 968137  | outputs 18

How can I improve its performance?

import math

primes = None
upper_limit = 0
powers_of_two = []

class Node:
def __init__(self, value, parent):
self.value = value
self.parent = parent
for i in powers_of_two:
if (self.value + i < upper_limit
and self.parent != self.value + i
and self.value + i in primes):
yield Node(i+self.value, self)
if (self.value - i >= 2
and self.parent != self.value - i
and self.value - i in primes):
yield Node(self.value - i, self)

def sieve(n):
numbers = list(range(0, n))
for prime in numbers:
if prime < 2:
continue
elif prime > n ** 0.5:
break
for i in range(prime ** 2, n, prime):
numbers[i] = 0
return [x for x in numbers if x > 1]

def bfs(f, to):
f = Node(f, None)
q = [f]
visited = [f.value]
while len(q) != 0:
if q[0].value == to:
return q[0]
else:
#print(q[0].value, x.value)
if x.value not in visited:
visited.append(x.value)
q.append(x)
del q[0]
while True:
inp = [int(x) for x in input().split(' ')]
upper_limit = inp[0]
m = int(math.ceil(math.log2(inp[0]))) if math.log2(inp[0]) % 1 != 0 else int(math.log2(inp[0]))
for i in range(0, m):
powers_of_two.append(2**i)
primes = sieve(inp[0])
x = bfs(inp[1], inp[2])
counter = 0
while x is not None:
print(x.value)
x = x.parent
counter += 1

print("--------------------")
print(counter)

There are 3 major performance problems you currently have:

• the primes is currently a list and every time you do something in primes - this is O(n) average-time complexity (where n is the length of primes) - define primes as a set instead
• the visited is a list, same O(n) lookups - define it as a set instead
• the del q[0] is also a O(n) operation - removing an element from the beginning of a list requires rearranging all following elements - use a collections.deque in place of a list for O(1) deletes from both ends (.popleft() for the left-end deletion)

Revised version (works for me in an instant for all the test inputs):

import math
from collections import deque

class Node:
__slots__ = ['value', 'parent']

def __init__(self, value, parent):
self.value = value
self.parent = parent

for i in powers_of_two:
if self.value + i < upper_limit and self.parent != self.value + i and self.value + i in primes:
yield Node(i + self.value, self)
if self.value - i >= 2 and self.parent != self.value - i and self.value - i in primes:
yield Node(self.value - i, self)

def sieve(n):
numbers = list(range(0, n))
for prime in numbers:
if prime < 2:
continue
elif prime > n ** 0.5:
break

for i in range(prime ** 2, n, prime):
numbers[i] = 0

return {x for x in numbers if x > 1}

def bfs(f, to, primes):
f = Node(f, None)

q = deque([f])
visited = set([f.value])
while q:
if q[0].value == to:
return q[0]
else:
# print(q[0].value, x.value)
if x.value not in visited:
q.append(x)
q.popleft()

if __name__ == '__main__':
upper_limit, p, q = [int(x) for x in input().split(' ')]

m = int(math.ceil(math.log2(upper_limit))) if math.log2(upper_limit) % 1 != 0 else int(math.log2(upper_limit))
powers_of_two = [2 ** i for i in range(m)]

primes = sieve(upper_limit)

x = bfs(p, q, primes)

counter = 0
while x is not None:
print(x.value)
x = x.parent
counter += 1

print("--------------------")
print(counter)

You can also see some other changes:

• removed the while True loop (from what I understand you just need a single run)
• using if __name__ == '__main__':
• avoiding using globals and passing the primes set to the bfs() and get_adjacent() as an argument
• using __slots__ for the Node class
• used list unpacking for reading the input numbers
• used upper_limit, p and q variables to increase readability (p and q variable names are used in the problem description)
• According to this, average search time for a list is O(n) and average for a set is O(1) (with worse case being O(n)). – NonlinearFruit Mar 13 '17 at 12:59
• @Nonlinear The worst case very, very rarely comes up in practice, though, as long as you are (a) not willfully trying to make it happen (in which case it's easy) and (b) following all the expectations about what your __hash_code__ returns. – Nic Hartley Mar 13 '17 at 14:52
• This was really useful, also for my general knowledge of python :) I didn't know about __slots__ and also naively assumed that python lists were internally sets. Thank you! – It'sNotALie. Mar 13 '17 at 17:45

Just to add to @alecxe's answer, the sieve function can be re-written as:

def sieve(n):
numbers = range(2, n ** 0.5)
for prime in numbers:
for i in range(prime ** 2, n, prime):
numbers[i] = 0
return {x for x in numbers if x > 1}

But you should be careful with that ** 0.5, because floating point numbers may be rounded at some point.

EDIT: to fix the code above I started thinking about it and I didn't really like it, so here's another version which returns a set of primes:

def sieve(n):
primes = [False, False] + [True] * (n - 2)
for (prime, isprime) in enumerate(primes):
if isprime:
for i in range(prime*prime, n, prime):
primes[i] = False
return {prime for prime, val in enumerate(primes) if val}

• This code doesn't work for me, complains about firstly n ** 0.5 not being integral (I cast to int() for a fix), then numbers[i] not being assignable to (list()), and then numbers[i] being out of range (not sure where to fix it. Always good to see more Eratosthenes sieves though :) – It'sNotALie. Mar 13 '17 at 17:48


text:
# Proof By Mathematical Induction - Example Help

• Oct 16th 2007, 02:01 AM
name101
Proof By Mathematical Induction - Example Help
hello just started school 2 days ago. and we were told to copy down some examples and there was one were i think there is a mistake but not sure.

Use mathematical induction to prove
$3^n >= 2n + 5$ for all integers

i understand all the steps untill it does this.

$3^(k+1) >= 2(2k+5)$
$3^(k+1) >= 6+15$
$3^(k+1) >= 2k+7$ <This line i do not understand.
Therefore
$3^(k+1) >= 2(k+1) + 5$ <This line i do not understand.

could some one prove this is true or false..
• Oct 16th 2007, 04:37 AM
topsquark
Quote:

Originally Posted by name101
hello just started school 2 days ago. and we were told to copy down some examples and there was one were i think there is a mistake but not sure.

Use mathematical induction to prove
$3^n >= 2n + 5$ for all integers

i understand all the steps untill it does this.

$3^(k+1) >= 2(2k+5)$
$3^(k+1) >= 6+15$
$3^(k+1) >= 2k+7$ <This line i do not understand.
Therefore
$3^(k+1) >= 2(k+1) + 5$ <This line i do not understand.

could some one prove this is true or false..

There are two lines I don't understand here, and they aren't the ones you noted. I'm kinda lost on the first two.

Let me run through this one.

I presume you've already done the base case so I won't redo that.

We are assuming that, for some value of k:
$3^k \geq 2k + 5$

We need to show that this is true for the k + 1 case. So consider
$3^{k + 1} = 3 \cdot 3^k$

But we know that
$3^k \geq 2k + 5$

So
$3^{k + 1} = 3 \cdot 3^k \geq 3(2k + 5)$<-- (1)

Thus
$3^{k + 1} \geq 6k + 15$ <-- (2)

We also know that
$6k + 15 > 2(k + 1) + 5$ <-- This step is easy to show. I leave it to you.

Thus
$3^{k + 1} \geq 2(k + 1) + 5$

The lines I have labeled as (1) and (2) are what I think your lines 1 and 2 are meant to have been.

-Dan
• Oct 16th 2007, 06:23 AM
Soroban
Hello, name101!

Quote:

Use mathematical induction to prove: . $3^n \:\geq \:2n + 5$ for all integers
This is not true for all integers.
Could it have said: .for all $n \geq 2$ ?

Here's the way I would do the proof.
. . I suspect that "they" had the same idea.

Verify the base case. . $S(2)\!:\;\;3^2 \,\geq\,2\!\cdot2 + 5$ . . . True!

Assume $S(k)\!:\;\;3^k \:\geq\:2k + 5$

Multiply both side by 3:. $3\cdot3^k \:\geq \:3(2k+5)$

. . and we have: . $3^{k+1}\:\geq\:6k + 15$

Now we note that: . $6k + 15 \;=\;2k + 4k + 2 + 5 + 8 \;=\;2k + 2 + 5 + 4k + 8$

. . . . . . . . . . . . $=\:2(k+1) + 5 + (4k+8) \;>\;2(k+1) + 5$ .
. . . since $k$ is positive

Therefore, we have: . $3^{k+1} \:\geq \:2(k+1) + 5$
. . and the inductive proof is complete.

• Oct 16th 2007, 09:08 PM
name101
Quote:

Originally Posted by Soroban
[size=3]This is not true for all integers.
Could it have said: .for all $n \geq 2$ ?

Ahhh yes I should of mentioned that fact.

Quote:

Originally Posted by Soroban
Therefore, we have: . $3^{k+1} \:\geq \:2(k+1) + 5$
. . and the inductive proof is complete.

Ahh thank you both. there were a couple of steps that were missing in the book.(I dont think it was a very good example to be learning off) but as i said.

Thankyou
and ill be back with more questions :D

~Regards
Name101
• Oct 17th 2007, 03:52 AM
topsquark
Quote:

Originally Posted by name101
(I dont think it was a very good example to be learning off)

I think it is an excellent example to be learning from, given the nature of the substitutions that need to be made to solve the problem. That and the fact that there are many ways to go about doing it.

-Dan
• Oct 19th 2007, 01:44 AM
name101
Quote:

Originally Posted by topsquark
I think it is an excellent example to be learning from, given the nature of the substitutions that need to be made to solve the problem. That and the fact that there are many ways to go about doing it.

-Dan

I have been doing some more questions on the topic. and I revoke that statement.


text:
Find the direction cosines of a line which makes equal angles with the coordinate axes.

Question:

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution:

Let the direction cosines of the line make an angle α with each of the coordinate axes.

∴ l = cos αm = cos αn = cos α

$l^{2}+m^{2}+n^{2}=1$

$\Rightarrow \cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1$

$\Rightarrow 3 \cos ^{2} \alpha=1$

$\Rightarrow \cos ^{2} \alpha=\frac{1}{3}$

$\Rightarrow \cos \alpha=\pm \frac{1}{\sqrt{3}}$

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are $\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}$, and $\pm \frac{1}{\sqrt{3}}$.


text:
## 7.1 Sampling bowl activity

### 7.1.1 What proportion of this bowl’s balls are red?

Take a look at the bowl in Figure 7.1. It has a certain number of red and a certain number of white balls all of equal size. Furthermore, it appears the bowl has been mixed beforehand, as there does not seem to be any coherent pattern to the spatial distribution of the red and white balls.

Let’s now ask ourselves, what proportion of this bowl’s balls are red?

One way to answer this question would be to perform an exhaustive count: remove each ball individually, count the number of red balls and the number of white balls, and divide the number of red balls by the total number of balls. However, this would be a long and tedious process.

### 7.1.2 Using the shovel once

Instead of performing an exhaustive count, let’s insert a shovel into the bowl as seen in Figure 7.2. Using the shovel, let’s remove $$5 \cdot 10 = 50$$ balls, as seen in Figure 7.3.

Observe that 17 of the balls are red and thus 0.34 = 34% of the shovel’s balls are red. We can view the proportion of balls that are red in this shovel as a guess of the proportion of balls that are red in the entire bowl. While not as exact as doing an exhaustive count of all the balls in the bowl, our guess of 34% took much less time and energy to make.

However, say, we started this activity over from the beginning. In other words, we replace the 50 balls back into the bowl and start over. Would we remove exactly 17 red balls again? In other words, would our guess at the proportion of the bowl’s balls that are red be exactly 34% again? Maybe?

What if we repeated this activity several times following the process shown in Figure 7.4? Would we obtain exactly 17 red balls each time? In other words, would our guess at the proportion of the bowl’s balls that are red be exactly 34% every time? Surely not. Let’s repeat this exercise several times with the help of 33 groups of friends to understand how the value differs with repetition.

### 7.1.3 Using the shovel 33 times

Each of our 33 groups of friends will do the following:

• Use the shovel to remove 50 balls each.
• Count the number of red balls and thus compute the proportion of the 50 balls that are red.
• Return the balls into the bowl.
• Mix the contents of the bowl a little to not let a previous group’s results influence the next group’s.

Each of our 33 groups of friends make note of their proportion of red balls from their sample collected. Each group then marks their proportion of their 50 balls that were red in the appropriate bin in a hand-drawn histogram as seen in Figure 7.5.

Recall from Section 2.5 that histograms allow us to visualize the distribution of a numerical variable. In particular, where the center of the values falls and how the values vary. A partially completed histogram of the first 10 out of 33 groups of friends’ results can be seen in Figure 7.6.

Observe the following in the histogram in Figure 7.6:

• At the low end, one group removed 50 balls from the bowl with proportion red between 0.20 and 0.25.
• At the high end, another group removed 50 balls from the bowl with proportion between 0.45 and 0.5 red.
• However, the most frequently occurring proportions were between 0.30 and 0.35 red, right in the middle of the distribution.
• The shape of this distribution is somewhat bell-shaped.

Let’s construct this same hand-drawn histogram in R using your data visualization skills that you honed in Chapter 2. We saved our 33 groups of friends’ results in the tactile_prop_red data frame included in the moderndive package. Run the following to display the first 10 of 33 rows:

tactile_prop_red
# A tibble: 33 x 4
group            replicate red_balls prop_red
<chr>                <int>     <int>    <dbl>
1 Ilyas, Yohan             1        21     0.42
2 Morgan, Terrance         2        17     0.34
3 Martin, Thomas           3        21     0.42
4 Clark, Frank             4        21     0.42
5 Riddhi, Karina           5        18     0.36
6 Andrew, Tyler            6        19     0.38
7 Julia                    7        19     0.38
8 Rachel, Lauren           8        11     0.22
9 Daniel, Caroline         9        15     0.3
10 Josh, Maeve             10        17     0.34
# … with 23 more rows

Observe for each group that we have their names, the number of red_balls they obtained, and the corresponding proportion out of 50 balls that were red named prop_red. We also have a replicate variable enumerating each of the 33 groups. We chose this name because each row can be viewed as one instance of a replicated (in other words repeated) activity: using the shovel to remove 50 balls and computing the proportion of those balls that are red.

Let’s visualize the distribution of these 33 proportions using geom_histogram() with binwidth = 0.05 in Figure 7.7. This is a computerized and complete version of the partially completed hand-drawn histogram you saw in Figure 7.6. Note that setting boundary = 0.4 indicates that we want a binning scheme such that one of the bins’ boundary is at 0.4. This helps us to more closely align this histogram with the hand-drawn histogram in Figure 7.6.

ggplot(tactile_prop_red, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Proportion of 50 balls that were red",
title = "Distribution of 33 proportions red")

### 7.1.4 What did we just do?

What we just demonstrated in this activity is the statistical concept of sampling. We would like to know the proportion of the bowl’s balls that are red. Because the bowl has a large number of balls, performing an exhaustive count of the red and white balls would be time-consuming. We thus extracted a sample of 50 balls using the shovel to make an estimate. Using this sample of 50 balls, we estimated the proportion of the bowl’s balls that are red to be 34%.

Moreover, because we mixed the balls before each use of the shovel, the samples were randomly drawn. Because each sample was drawn at random, the samples were different from each other. Because the samples were different from each other, we obtained the different proportions red observed in Figure 7.7. This is known as the concept of sampling variation.

The purpose of this sampling activity was to develop an understanding of two key concepts relating to sampling:

1. Understanding the effect of sampling variation.
2. Understanding the effect of sample size on sampling variation.

In Section 7.2, we’ll mimic the hands-on sampling activity we just performed on a computer. This will allow us not only to repeat the sampling exercise much more than 33 times, but it will also allow us to use shovels with different numbers of slots than just 50.

Afterwards, we’ll present you with definitions, terminology, and notation related to sampling in Section 7.3. As in many disciplines, such necessary background knowledge may seem inaccessible and even confusing at first. However, as with many difficult topics, if you truly understand the underlying concepts and practice, practice, practice, you’ll be able to master them.

To tie the contents of this chapter to the real world, we’ll present an example of one of the most recognizable uses of sampling: polls. In Section 7.4 we’ll look at a particular case study: a 2013 poll on then U.S. President Barack Obama’s popularity among young Americans, conducted by Kennedy School’s Institute of Politics at Harvard University. To close this chapter, we’ll generalize the “sampling from a bowl” exercise to other sampling scenarios and present a theoretical result known as the Central Limit Theorem.

Learning check

(LC7.1) Why was it important to mix the bowl before we sampled the balls?

(LC7.2) Why is it that our 33 groups of friends did not all have the same numbers of balls that were red out of 50, and hence different proportions red?


text:
# symmetric tensor

Let $V$ be a vector space over a field. Let $S_{n}$ be the symmetric group on $\{1,\ldots,n\}$. An order-n tensor (http://planetmath.org/TensorProduct) $A\in V^{\otimes n}$ is said to be if $P(\sigma)A=A$ for all $\sigma\in S_{n}$, where $P(\sigma)$ is the permutation operator associated to $\sigma$. The set of symmetric tensors in $V^{\otimes n}$ is denoted by $S^{n}(V)$.

Title symmetric tensor SymmetricTensor 2013-03-22 16:15:41 2013-03-22 16:15:41 Mathprof (13753) Mathprof (13753) 5 Mathprof (13753) Definition msc 15A03


text:
# Why is the formal solution to a linear differential equation of exponential form?

So $x(t) = e^{ct}$ solves $dx/dt = cx$. This is clear enough from differentiation rules... But I fail to grasp, in some sense which I can't quite put my finger on, why it is so. Why can the solution to any linear differential equation be given as something related to the exponential function, keeping in mind the latter's shape? What is the connection here?

I found the introductory lines at this section of Wikipedia to be of some help: I can see how the exponential function provides a sort of trick for this summing of derivatives to work out. But I don't see the deeper connection, if any.

Additional questions, related to the link provided:

1. If the "exponential function is one of the few functions that keep its shape after differentiation" then what are the others, and why not use those?
2. I don't get the step from the first to the second equation. If we set $y=e^{zx}$, then how does $y^{(n)}$ equal $z^ne^{zx}$ rather than $e^{zxn}$? Or are those the same in some way I don't see?

Thanks!

Revised upon reading OPs comments Your specific questions: 1. There is no other function like that. Only the exponential satisfies $f'=f$, non trivially (of course $f=0$ also works). 2. You are confusing exponents with repeated differentiation notation. If $f=e^{zx}$ where z is a constant, then $f'=ze^{zx}, f''=z(ze^{zx})=z^2e^{zx}$ each diffentiation brings down an additional z from the exponenet and multiplies it by the existing function. That is why you get that.

After reading your comments, I think I have a better handle on where your confusion lies.

First, note that the exponential is a general solution only for linear differential equations with constant coefficients. If you have variable coefficients, then all bets are off, as you can get Bessel functions and other non-exponential solutions.

With that said, we are focusing on cases like the following: $af''+bf'+cf=0$

Now, what is this equation telling us? It is saying the left hand side (LHS) must be zero, so the weighted sum of the function and its first two derivatives must equal 0 everywhere $f$ is defined. Therefore, the derivatives of $f$ must be expressible in terms of linear combinations of each other so that you can cancel out the functions for all values of x.

You can see that by using the above equation to solve for $f$ or one of its two derivatives in terms of the other two:

$f=-\frac{1}{c}(af''+bf')$

$f'=-\frac{1}{b}(af''+cf)$ and

$f''=-\frac{1}{a}(bf'+cf)$

As you can see, each derivative is equal to a linear combination of the other two, hence all must be expressible as linear combinations of some set of functions. If that were not the case, then there wold be some values of x where the above equation takes a non-zero value. As an analogy, think of the following very simple equation:

$x^2 + 3x + af(x) = 0$, where we want to solve for $f$. By simple algebra (one step!) we know $f$ must be expressed in terms of $x^2$ and $x$, otherwise there will be at least one x where we can get a non-zero value (e.g., if $f(x) = x^3 + x^2 + x$) there is no value for $a$ that will satisfy the equation for all x.

Likewise, as we saw above, linear differential equations with constant coefficients will equal 0 for all x only if the function and its derivatives can be expressed as linear combinations of a finite number of functions, which will allow us to select coefficient values so that all the deriviatives cancel out.

This can be extended to the case where there are "repeated roots" in the characteristic equation. Here, we have N roots and N-r functions. Hence, we are missing some functions in our "function space", which is similar to a vector space, except the unit vectors $\overrightarrow{\mathbf{i}},\overrightarrow{\mathbf{j}},\overrightarrow{\mathbf{k}}$ etc are replaced by abstract "directions" represented by functions, so that an abstract "vector" in this space is $\overrightarrow{\mathbf{f}}=\langle c_1,c_2,c_3...c_n\rangle\cdot\langle f_1,f_2,f_3...f_n\rangle$. Therefore, we have a situation where r of the "vectors" are collinear, just like with matrix algebra, and so the space is underdetermined, just like when you have N equations and N+M unknowns. You need more equations to get a unique solution. That is why we use the functions $x^ne^{kx}$, for n={1...M} where M is the degree of multiplicity for a particular root. This function has the unique property that its derviatives will be expressible in terms of lower values of n, and hence will be expressible in this expanded "basis" of functions.

Overall, that is the key property of the exponential in linear differential equations with constant coefficients - derivatives of the exponential, or x times the exponential, produce functions that do not stray outside the set of functions forming the basis for the abstract space of solutions. Other functions will not do this, which means that the higher derivatives will produce functions that are not present in the actual proposed solution, and will therefore not be able to be cancelled for all x. As an example, take $f(x) = \sum c_i\frac{1}{x^n}$ No matter what you choose for the $c_i$ the derivatives will produce functions that are not in $f$, and hence there is no set of constant coefficients that will allows the derivates to cancel out to zero for all x.

Hope that was a bit clearer than before.

• Thanks a lot. First of all for the difference between $y^n$ and $y^{(n)}$! As to the main question: your answer helps, but I'm still confused. I will ask a few questions in separate comments. Apologies if my phrasing/notation is muddled or invalid... – Diploria Nov 20 '13 at 21:03
• (1) Eigenfunctions are new to me, so trying to go by analogy from eigenvectors. If we are asked for $A\vec{v}$ but A-multiplication is cumbersome, but we know $A\vec{u}=\lambda\vec{u}$, we express $\vec{v}$ as $\sum a_i\vec{u}_i$; then $\sum \lambda_i{u}a_i\vec{u}_i$ is $A\vec{v}$ (right?). Similarly (??), if we are asked $D[f(x)]+...+D^n[f(x)]$, we can use $D^{(i)}[e^{cx}]=c^ie^{cx}$... but then: express $f(x)$ as $\sum a_i c^i e^{cx}$ or $\sum a_i e^{cx}$? If the latter, then are all eigenfunctions the same (unlike with $\vec{u}$)? What are the analogies to $\vec{u}$ and $\lambda$ exactly? – Diploria Nov 20 '13 at 21:03
• (2) What happens if there are coefficients as in e.g. $4D^2[f(x)]+3D[f(x)]$? (How) do these get into the expression in terms of $e^x$? – Diploria Nov 20 '13 at 21:04
• (3) It's one thing to find the derivative of some function $f$ by expressing it in terms of its eigenfunctions (here exponential ones). But in my original question, we start with a differential equation which we solve for $f(x)$ which turns out to be expressable in terms of exponential functions: the 'bare' function which is expressed in terms of exponentials, not its derivative or a linear combination of derivatives. What is the connection then? If we know we can express a derivative (or sum of them) as a (sum) of exponentials, does it follow we can do the same for the original function? – Diploria Nov 20 '13 at 21:06
• (4) Is there any insight to be had from the shape of the exponential function? (This is more out of curiosity... I'd rather understand all of the above first.) – Diploria Nov 20 '13 at 21:07


text:
# Light bulb resistance

Electricity and Magnetism Level 2

Two light bulbs P and Q are identical in all respects, except that P's filament is thicker than Q's. If the same potential difference is applied to each, then $$\text{_________}.$$

×


text:
OGS
[tag] permeability_model

A hydraulic permeability model for the fracture. It describes the permeability calculation based on the current mechanical aperture, the initial aperture, and permeability model specific parameters.


text:
Array

Some parameters require an array of integers, floating point numbers, or a mixture of both.

For example, in
GRAVITY( "simple gravity" ) {
gravity = { 0, -9.81, 0 }
}
the gravity parameter requires a three-component vector of floating point numbers. This parameter requires exactly three numbers separated by commas. An array may be two dimensional. For example, in
PERIODIC_BOUNDARY_CONDITION( "a PBC" ) {
rotation_axis = { 0, 0, 0 ; 0, 0, 1 }
}
rotation_axis is defined by two points in space. The axis of rotation is in the z-direction, passing through the points (0,0,0) and (0,0,1). The rows of a two dimensional array are separated by semicolons. In the above example, rotation_axis is a 2x3 array. The array may also have mixed integer and floating point values. For example,
COORDINATE {
coordinates = { 1, 0.0, 0.0, 0.0 ;
2, -1.5, -1.5, -1.5 ;
... }
}
defines the nodal coordinates of the problem. Here the first column is the node number and columns two, three, and four define the x, y, and z components of the coordinates. The number of columns must be at least four and the number of rows defines the number of nodes in the system. The values of an array may be read from an ASCII file:
COORDINATE {
}
1 0.0 0.0 0.0
...


text:
> > > Hexagon Formula

# Hexagon Formula

A polygon is a two-dimensional closed shape. It is having many sides. Each side will be a straight line segment. In geometry, a hexagon is also a type of polygon with 6 sides. If the hexagon is regular, then the lengths of all the sides will be the same as well as all the angles will be equal. Clearly sides of a regular hexagon are congruent. This article will discuss the hexagon and hexagon formula to find out the measurements of it. Also, some suitable examples will help students to understand it. Let us learn the concept!

## What is Regular Hexagon?

A regular hexagon is a kind of polygon with 6 equal sides. Its properties are:

• It is having six sides and six angles.
• Lengths of all the sides are equal.
• Measurements of all the angles are equal.
• The total number of diagonals in it is 9.
• The sum of all interior angles will be 720 degrees. Also, each interior angle will be 120 degrees.
• The sum of all exterior angles will be 360 degrees. Also, each exterior angle will be 60 degrees.
• The area of a hexagon will be the region occupied inside its boundary.
• The perimeter of it will be the total length of all 6 sides of it.
• The shape has nine diagonals, means the lines between the interior angles.

Source: en.wikipedia.org

## Formulas for Hexagon

### 1] Area of a Hexagon

To find out the area of a hexagon, we have to divide it into small six isosceles triangles. Then, we will find the area of one of the triangles and finally, we will multiply by 6 to find the total area of the polygon.

$$A = \frac{3\sqrt{3}}{ 2} \times a^{2}$$

### 2] Permiter of a Hexagon

Perimeter will be the sum of all 6 side lengths. The formula for the perimeter of a hexagon is given as:

$$P = 6 \times a$$

### 3] Diagonal of a Hexagon

There is no standard formula to find out the diagonals of irregular hexagons. In regular hexagons, there are nine diagonals in six equilateral triangles. So, it is easy to determine the length of each diagonal line. If one side of the hexagon is given then it is easy to calculate the all diagonal. It is having diagonals of two different lengths.

Short diagonal : $$d_{1} = \sqrt{3} \times a$$

Long Diagonal: $$d_{2} = 2 \times a$$

 a Length of one side A Area of hexagon P Perimeter of Hexagon $$d_{1}$$ Short diagonal $$d_{2}$$ Long diagonal

## Solved Examples for Hexagon Formula

Q.1: Calculate the area and perimeter of a given regular hexagon whose side is 4 cm.

Solution: Given parameters are,

Side of the hexagon, a = 4 cm

Thus area of Hexagon will be,

$$A = \frac{3\sqrt{3}}{ 2} \times a^{2}\\$$

$$=\frac{3\sqrt{3}}{ 2} \times 4^{2}\\$$

$$= 24\sqrt{3} sqcm.$$

The perimeter of the hexagon will be,

$$P = 6 \times a = 6 \times 4$$

= 24 cm

So, the area of the hexagon will be $$24\sqrt{3} sqcm.$$, and perimeter will be 24

Q.2: Perimeter of a regular hexagonal board is 48 cm. Compute the area of this board.

Solution: Given parameters are:

Perimeter, P = 48 cm

Also, formula for perimeter is,

$$P = 6 \times a$$

Thus side,$$a = \frac {P}{6} = \frac {48}{6}$$

a = 8 cm

Now, Area of an Hexagon formula is:

$$A = \frac{3\sqrt{3}}{ 2} \times a^{2}\\$$

$$= \frac{3\sqrt{3}}{ 2} \times 8^{2}\\$$

$$= 96 \sqrt{3} sqcm$$

So, area of hexagon will be  $$96 \sqrt{3} \;sq\;cm$$

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## Illinois Journal of Mathematics

### A "nice" map colour theorem

K. S. Sarkaria

#### Abstract

A closed orientable triangulated surface is "nice" if its vertices can be assigned 4 colours in such a way that all 4 colours are used in the closed star of each edge. The 4-colouring can be interpreted as a simplicial map from the surface to the 4-vertex 2-sphere. If the surface has genus $(n-1)^2$, then the degree of this map is at least $n^2$. Conversely we show that, if $n$ is not divisible by 2 and 3, then there are "nice" surfaces of genus $(n-1)^2$ for which the degree of the above map is exactly $n^2$. Complex analytically "nice" surfaces can be viewed as minimally triangulated meromorphic functions of a Riemann surface.

#### Article information

Source
Illinois J. Math., Volume 46, Number 4 (2002), 1111-1123.

Dates
First available in Project Euclid: 13 November 2009

https://projecteuclid.org/euclid.ijm/1258138469

Digital Object Identifier
doi:10.1215/ijm/1258138469

Mathematical Reviews number (MathSciNet)
MR1988253

Zentralblatt MATH identifier
1030.57037

#### Citation

Sarkaria, K. S. A "nice" map colour theorem. Illinois J. Math. 46 (2002), no. 4, 1111--1123. doi:10.1215/ijm/1258138469. https://projecteuclid.org/euclid.ijm/1258138469


text:
# zbMATH — the first resource for mathematics

## Magurn, Bruce A.

Compute Distance To:
 Author ID: magurn.bruce-a Published as: Magurn, B. A.; Magurn, Bruce; Magurn, Bruce A.
 Documents Indexed: 23 Publications since 1978, including 4 Books
all top 5

#### Co-Authors

 14 single-authored 5 Vaserstein, Leonid N. 2 Jajodia, Sushil 2 Oliver, Bob 1 Dennis, Roger Keith 1 Lafont, Jean-François 1 Laubenbacher, Reinhard C. 1 Ortiz, Ivonne Johanna 1 Van der Kallen, Wilberd
all top 5

#### Serials

 4 Journal of Pure and Applied Algebra 3 Journal of Algebra 2 Communications in Algebra 2 Canadian Journal of Mathematics 2 Encyclopedia of Mathematics and Its Applications 1 Mathematical Proceedings of the Cambridge Philosophical Society 1 Inventiones Mathematicae 1 Journal of the London Mathematical Society. Second Series 1 Journal für die Reine und Angewandte Mathematik 1 Pacific Journal of Mathematics 1 Linear Algebra and its Applications
all top 5

#### Fields

 15 Associative rings and algebras (16-XX) 9 Category theory, homological algebra (18-XX) 9 $K$-theory (19-XX) 8 Group theory and generalizations (20-XX) 6 Number theory (11-XX) 5 Manifolds and cell complexes (57-XX) 2 Commutative algebra (13-XX) 2 Algebraic topology (55-XX) 1 General mathematics (00-XX) 1 Algebraic geometry (14-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Functional analysis (46-XX) 1 Geometry (51-XX)