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Posted on Categories:Network Analysis, 数据科学代写, 统计代写, 统计代考, 网络分析

# 统计代写|网络分析代写Network Analysis代考|SOC379 Handling Real-World Network Data Sets

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## 统计代写|网络分析代写Network Analysis代考|Handling Real-World Network Data Sets

In today’s modern world, large data sets are frequently created and applied for a variety of purposes, whether they are in the form of gigabytes of data contained in a single file or hundreds of files, each containing a small amount of data, with both of these approaches encountering difficulties and issues that must be addressed and resolved in order to provide the required performance. The massive size of data, which is typically unsuitable for processing and management by a single machine and cannot be contained in a single file.

Real-world networks are large in size and intricate in structure, which complicates data handling and management. As examples of this type of data, millions or even billions of newly created user accounts on popular social media platforms, such as Facebook, Instagram, and Twitter, among others, could provide a reasonable estimate of the size of this subject’s data, where the importance of these data sets are to analyze or improve their activity and troubleshoot problems.

These data would be useful if they could be analyzed to gain better information and come up with better ideas to improve the structure, such as when Facebook is interested in finding out about the number of new users who are active and those who are less active. This would be a working model that needs to be analyzed and run to produce the desired results.
A network consists of nodes or elements and the edges that connect them; in other words, the structural relationships between these nodes define and create the network’s structure. Indeed, each of these created user accounts will be represented by a node, and the connections between them will define the network and its behavior.

Real-world network data sets are created to simulate the structure of the real world, which we encounter on a daily basis. Although mathematical equations and network models are used to analyze and handle this type of data, the complexity of this system does not allow for precise numerical data; however, approximate results are useful when sampling a network of this size.

Alternatively, dealing with big amounts of data may result in out-ofmemory issues, the creation of needless blank spots on the screen, and sluggish functionality. The practice of performing long-term monitoring from permanent gauges and high-frequency readings in the field is now becoming more common, resulting in data sets that are substantially bigger than those necessary for pressure transient analysis [1].

## 统计代写|网络分析代写Network Analysis代考|Aspects of the Network

The term “connection” has been the subject of increasing public fascination in contemporary world society over the last decade, with the concept of a network at its heart. A pattern of interconnections between a collection of objects is a definition of this concept. To put it another way, a network is any collection of things in which certain pairs of these items are linked to another via links. Numerous different types of relationships or connections can be used to define links, depending on the context [2].

Assume that each email address is a system object; thus, a massive number of accounts exist in this network assumption. These accounts, on the other hand, are unable to provide information; they are analogous to a collection of unused dots in the environment. This inactive information can be generalized by adding links or, in the other words, connections to the dots that indicate the relationship between the objects. In this case, when an email is sent to one or more other email addresses, a network is created because we have objects and their relationships. More precisely, this system’s objects are referred to as nodes or elements of the network, whereas the lines are referred to as edges. In other words, when two or more nodes discover a relationship between them, edges are formed. Indeed, a network is formed when nodes and defined provide a description of the network.

The purpose of Figure $3.1$ is to demonstrate how a network in small size looks like, as well as to provide a better vision and understanding of a large-scale network. The small network depicted in Figure $3.1$ represents the friendship connection between 34 members of a karate club. Each member of this club is represented by a numerical circle, and the friendship that exists between them is demonstrated by the connection.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


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# Have \cleardoublepage working in oneside scrreprt

That seems quite elementary, but I didn't find the answer. How can I have the \cleardoublepage command work in a oneside environment? That seems kind of odd to ask for that, but I have to send an application file with a bunch of documents added to a .tex file via pdfpages. And I don't know whether it will be seen only on screen, printed on a one-sided printer or printed on a two-sided printer.

So I want the layout to be one-sided for the file not to look too odd on a computer screen or if printed onesidedly. But I also want new parts of the document to start on an odd page, so that my id card is not printed on the back of my research statement in case of two-sided printing.

Here is an example :

\documentclass[oneside,openright]{scrreprt}
\begin{document}
Foo1.
\newpage
Foo2.
\cleardoublepage
Foo3.
\cleardoublepage
Foo4
\end{document}

I would like "Foo4" to be on page 5, as it is in twoside mode.

-
Welcome to TeX.sx! –  hpesoj626 Mar 14 '13 at 12:06
Thanks for the welcome. –  Thomas Richard Mar 14 '13 at 16:59

You can define your own command:

\documentclass[oneside,openright]{scrreprt}

\newcommand\cleartooddpage{\clearpage
\ifodd\value{page}\else\null\clearpage\fi}

\begin{document}
Foo1.
\newpage
Foo2.
\cleartooddpage
Foo3.
\cleartooddpage
Foo4
\end{document}

Add \thispagestyle{empty} after \null if you don't want headers and footers on the empty page.

-
Works perfectly. Thanks ! –  Thomas Richard Mar 14 '13 at 16:59
BTW: KOMA-Script provides \ifthispageodd{<true>}{<false>} (which however needs two compilation runs) –  clemens Mar 14 '13 at 18:27
@cgnieder Testing \ifodd\value{page} after a \clearpage is safe. –  egreg Mar 14 '13 at 19:08
I know and I'm sure you wouldn't have suggested it if it wasn't. That's why I only made a “by the way” :) –  clemens Mar 14 '13 at 19:11

format: plain


text:
# How to add mulitple authors in document properties [closed]

I know how to add my User data to the document properties, but I don't know how to add another author. How can I do this?

I am using Libreoffice Writer 4.2.7.2 on Ubuntu 14.04 LTS.

edit retag reopen merge delete

### Closed for the following reason the question is answered, right answer was accepted by Alex Kemp close date 2016-03-06 14:02:20.571654

I also want to see the second author in the exported pdf.

( 2014-12-16 17:24:09 +0200 )edit

Sort by » oldest newest most voted

You could possibly use the Comment window under the Description tab. If you do so you can link this information into the document using Insert > Fields > Other > Doc information tab > Comments

I made this test in LibO 4.3.4.1 using an XP machine.

more

I can think of a number of ways to do this. The right for your depends on your needs/objectives.

Probably the "right way" to do this (with the standard writer document and no plugins), would be to: - create a "Custom Property" named "Authors" (and include yourself).

FileMenu, Properties, CustomPropertiesTab ... enter Properties Name = "Authors"; Type=Text, Value: Delimited List of Authors

You could then use this field as you wish, formatting would be possible but might take some thinking if your particular. The documents metadata would complete and accurate.

You note that your user data User data allows you to be named as Author by inserting the "Author" field. I think technically this is a leap in faith on the part of the designers, and for those lucky enough to have secretaries typing dictated documents could easily lead to inaccuracy. (Something like the solution here would be necessary to provide for accuracy, or a hack like having your secretary pretend to be you when he creates the document and then insert his name in the builtin custom property of "Typist."

(I believe that if you were using CMIS that the answer could be more elegant, but I think that's beyond the scope of your question as well as my knowledge - I've tried to look into that but I couldn't get a CMIS system running.)

more


text:
2017 Jan 19

# Amitsur Algebra: Yiftach Barnea, "Old and New Results on Subgroup Growth in Pro-p Groups."

12:00pm to 1:00pm

## Location:

Manchester Building, Room 209
Title: Old and New Results on Subgroup Growth in Pro-p Groups. Abstract: I will survey our current knowledge about subgroup growth in pro-p growth. In particular I will present new solutions to long standing open problems in the area: 1. What is the minimal subgroup growth of non-$p$-adic analytic pro-$p$ groups? (Joint work with Benjamin Klopsch and Jan-Christoph Schlage-Puchta.) 2. What are the subgroup growths of the Grigorchuk group and the Gupta-Sidki groups? (Joint work with Jan-Christoph Schlage-Puchta.)
2017 Jun 22

# Amitsur Algebra: Jan Dobrowolski

12:00pm to 1:00pm

## Location:

Manchester 209
Title: Inp-minimal ordered groups. Abstract. The main goal of the talk is to present the proof of the theorem stating that inp-minimal (left)-ordered groups are abelian. This generalizes a previous result of P. Simon for bi-ordered inp-minimal groups.
2016 Dec 29

# Amitsur Algebra: Igor Rivin, "Random integer matrices"

12:00pm to 1:00pm

## Location:

Manchester Building, Room 209
Title: Random integer matrices Abstract: I will discuss various models of random integer matrices, and their (occasionally surprising) properties. Some of the work discussed is joint with E. Fuchs.
2017 May 04

# Amitsur Algebra: Jasbir Chahal, " A tale of three elliptic curves"

12:00pm to 1:00pm

## Location:

Manchester 209
Title: A tale of three elliptic curves. Abstract: We will show how the arithmetic of three elliptic curves answers three old questions in the Euclidean geometry.
2016 Dec 08

# Amitsur Algebra: George Glauberman (Chicago)

12:00pm to 1:15pm

## Location:

Manchester Building, Room 209
Title: Fixed points of finite groups on modules Abstract: Suppose G is a finite group, p is a prime, S is a Sylow p-subgroup of G, and V is a G-module over a field of characteristic p. In some situations, an easy calculation shows that the fixed points of G on V are the same as the fixed points of the normalizer of S in G. Generalizations of this result have been obtained previously to study the structure of G for p odd. We plan to describe a new generalization for p = 2. (This is part of joint work with J. Lynd that removes the classification of finite simple groups
2017 Mar 02

# Amitsur Algebra: Lev Glebsky, " Almost Congruence Extension Property for subgroups of free groups"

12:00pm to 1:00pm

## Location:

Manchester Building, Room 209
Title: Almost Congruence Extension Property for subgroups of free groups. Abstract. The talk essentially based on: https://arxiv.org/abs/1606.02345 Let G be a group and H
2017 Feb 09

# Amitsur Algebra: Evan DeCorte, "The spectral method for geometric colouring problems"

12:00pm to 1:00pm

## Location:

Manchester Building, Room 209
Title: The spectral method for geometric colouring problems
2017 Jun 29

12:00pm to 1:00pm

## Location:

Manchester 209
Title: Stability patterns in representation theory and applications Abstract: Many natural sequences of objects come equipped with group actions, e.g. the symmetric group on n letters acting on a space X_n. This leads to fundamental instability of invariants, such as homology, arising from the representation theory of the sequence of groups. Representation stability is a new and increasingly important set of ideas that describe a sense in which such sequence of representations (of different groups) stabilizes.
2015 Nov 19

# Groups & dynamics: Lei Yang (HUJI) "Equidistribution of expanding translates of curves in homogeneous spaces and Diophantine approximation"

10:00am to 11:00am

## Location:

Ross 70
Title: Equidistribution of expanding translates of curves in homogeneous spaces and Diophantine approximation. Abstract: We consider an analytic curve $\varphi: I \rightarrow \mathbb{M}(n\times m, \mathbb{R}) \hookrightarrow \mathrm{SL}(n+m, \mathbb{R})$ and embed it into some homogeneous space $G/\Gamma$, and translate it via some diagonal flow
2015 Nov 12

# Groups & dynamics: Elon Lindenstrauss (HUJI), "Rigidity of higher rank diagonalizable actions in positive characteristic"

10:00am to 11:00am

## Location:

Ross 70
Title: Rigidity of higher rank diagonalizable actions in positive characteristic
2015 Dec 29

# Dynamics lunch: Tom Gilat (HUJI): "Measure rigidity for `dense' multiplicative semigroups (following Einsiedler and Fish)"

12:00pm to 1:00pm

## Location:

Manchester building, Hebrew University of Jerusalem, (Coffee lounge)
2015 Nov 24

# Dynamics & probability: Yaar Salomon (Stonybrook) "The Danzer problem and a solution to a related problem of Gowers"

2:00pm to 3:00pm

## Location:

Manchester building, Hebrew University of Jerusalem, (Room 209)
The Danzer problem and a solution to a related problem of Gowers Is there a point set Y in R^d, and C>0, such that every convex set of volume 1 contains at least one point of Y and at most C? This discrete geometry problem was posed by Gowers in 2000, and it is a special case of an open problem posed by Danzer in 1965. I will present two proofs that answers Gowers' question with a NO. The first approach is dynamical; we introduce a dynamical system and classify its minimal subsystems. This classification in particular yields the negative answer to Gowers'
2015 Dec 15

# Dynamics & probability: Omri Solan (TAU) - Divergent trajectories in SL_3(R)/SL_3(Z)

2:00pm to 4:30pm

## Location:

Manchester building, Hebrew University of Jerusalem, 209
Abstract:
2015 Nov 03

# Dynamics lunch: Or Landesberg (HUJI)

12:00pm to 1:00pm

## Location:

Manchester building, Hebrew University of Jerusalem, (Room 209)
Title: On the Mixing Property for Hyperbolic Systems [following a paper by Martine Babillot]
2015 Dec 02

# Dynamics & probability: Ron Rosenthal (ETHZ) "Local limit theorem for certain ballistic random walks in random environments"

2:00pm to 3:00pm

## Location:

Ross 70
Title: Local limit theorem for certain ballistic random walks in random environments Abstract: We study the model of random walks in random environments in dimension four and higher under Sznitman's ballisticity condition (T'). We prove a version of a local Central Limit Theorem for the model and also the existence of an equivalent measure which is invariant with respect to the point of view of the particle. This is a joint work with Noam Berger and Moran Cohen.


text:
Conductor-discriminant formula

In mathematics, the conductor-discriminant formula or Führerdiskriminantenproduktformel, introduced by Hasse (1926, 1930) for abelian extensions and by Artin (1931) for Galois extensions, is a formula calculating the relative discriminant of a finite Galois extension $L/K$ of local or global fields from the Artin conductors of the irreducible characters $\mathrm{Irr}(G)$ of the Galois group $G = G(L/K)$.

Statement

Let $L/K$ be a finite Galois extension of global fields with Galois group $G$. Then the discriminant equals

$\mathfrak{d}_{L/K} = \prod_{\chi \in \mathrm{Irr}(G)}\mathfrak{f}(\chi)^{\chi(1)},$

where $\mathfrak{f}(\chi)$ equals the global Artin conductor of $\chi$.[1]

Example

Let $L = \mathbf{Q}(\zeta_{p^n})/\mathbf{Q}$ be a cyclotomic extension of the rationals. The Galois group $G$ equals $(\mathbf{Z}/p^n)^\times$. Because $(p)$ is the only finite prime ramified, the global Artin conductor $\mathfrak{f}(\chi)$ equals the local one $\mathfrak{f}_{(p)}(\chi)$. Because $G$ is abelian, every non-trivial irreducible character $\chi$ is of degree $1 = \chi(1)$. Then, the local Artin conductor of $\chi$ equals the conductor of the $\mathfrak{p}$-adic completion of $L^\chi = L^{\mathrm{ker}(\chi)}/\mathbf{Q}$, i.e. $(p)^{n_p}$, where $n_p$ is the smallest natural number such that $U_{\mathbf{Q}_p}^{(n_p)} \subseteq N_{L^\chi_\mathfrak{p}/\mathbf{Q}_p}(U_{L^\chi_\mathfrak{p}})$. If $p > 2$, the Galois group $G(L_\mathfrak{p}/\mathbf{Q}_p) = G(L/\mathbf{Q}_p) = (\mathbf{Z}/p^n)^\times$ is cyclic of order $\varphi(p^n)$, and by local class field theory and using that $U_{\mathbf{Q}_p}/U^{(k)}_{\mathbf{Q}_p} = (\mathbf{Z}/p^k)^\times$ one sees easily that $\mathfrak{f}_{(p)}(\chi) = (p^{\varphi(p^n)(n - 1/(p-1))})$: the exponent is

$\sum_{i=0}^{n-1}(\varphi(p^n) - \varphi(p^i)) = n\varphi(p^n) - 1 - (p-1)\sum_{i=0}^{n-2}p^i = n\varphi(p^n) - p^{n-1}.$

Notes

1. ^ Neukirch 1999, VII.11.9.


text:
Question 1 1 pts The EPS/EBIT analysis for Equity Bank indicates the following EPS values: 100%...

Question:

Question 1 1 pts The EPS/EBIT analysis for Equity Bank indicates the following EPS values: 100% Debt: 0.0021; 0.049 100% Stock: 0.0279;0.056 50/50 Debt/Stock Combo: 0.0245;0.0523 The best financing alternative for Equity Bank is: No answer text provided. 100% Stock 50/50 Debt/Stock Combo. 100% Debt

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Claim: The mean pulse rate (in beals per Minule (a) and belayadult males equab68.6 bpmrandom sample159 adult males the mean pulse rale 68.3 bpm and the slandard davialion11.4 bpm. Complete partsExpress the original claim symbolic lorm.bpm (Type linteger decimal. Do nol round )Idently the null and allernalive hypotheses:bpmbpm (Type Integers decimals. Do not round:)
Claim: The mean pulse rate (in beals per Minule (a) and belay adult males equab 68.6 bpm random sample 159 adult males the mean pulse rale 68.3 bpm and the slandard davialion 11.4 bpm. Complete parts Express the original claim symbolic lorm. bpm (Type linteger decimal. Do nol round ) Idently the nul...
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text:
\def\X{\mathbb X} \def\imp{\rightarrow} Defining discrete mathematics is hard because defining mathematics is hard.

3 In other words, induction is a style of argument we use to convince ourselves and others that a mathematical statement is always true. \def\dbland{\bigwedge \!\!\bigwedge}

What should you do?

Assuming all 10 people at the party each shake hands with every other person (but not themselves, obviously) exactly once, how many handshakes take place?

\def\nrml{\triangleleft}

This thread is archived. In discrete math, we don't have derivatives, so we look at differences. Renowned for her lucid, accessible prose, Epp explains complex, abstract concepts with clarity and precision, helping students develop the ability to think abstractly as they study each topic. CSE 16 Applied Discrete Mathematics Summer 2020 (June 22 – August 14) Description: Introduction to applications of discrete mathematical systems. Discrete math could still ask about the range of a function, but the set would not be an interval.

\def\pow{\mathcal P} \def\rem{\mathcal R}

\newcommand{\isom}{\cong} Work on the 4th edition is done on the edition branch.. Compilation Instructions

\def\circleC{(0,-1) circle (1)} This set of numbers is NOT discrete. Warning: With great power, comes great responsibility. http://www.vumultan.com/Books/MTH001-Discrete-Mathematics-with-Application-by-Susanna-S-Epp.pdf. \def\rng{\mbox{range}} Consider the function which gives the number of children of each person reading this.


text:
### Prompt Cards

These two group activities use mathematical reasoning - one is numerical, one geometric.

### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Exploring Wild & Wonderful Number Patterns

EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.

# 4 Dom

##### Stage: 1, 2, 3 and 4 Challenge Level:

Well done to all of you who found a solution to this challenge. Chloe and Mia from Scotts Primary School both managed to solve this challenge whilst in competition with Mrs Briggs, the teaching assistant. They both finished really quickly apparently and were really pleased with themselves (quite rightly!). Their teacher, Miss Lilley, and the rest of the class were very proud of them!

Zareah from St Joseph's Convent gave us some good general advice:

You just need to even it out. Don't put all the larger numbers in one corner and all the smaller numbers in another corner. Put a small number with a large number and soon you'll get it out!

Milo and Darren from SVS used a similar method to Zareah. Angus from West Linton Primary used this strategy for solving the problem:

First I added up the dots ($5+1+6+4+2+6+4+3=31$) then divided it by $4$ to find the average dots on a domino.
There are one and a half dominoes in a side.
$31\div4 =8$ (roughly)
$8\times 1\frac{1}{2}= 12$
Then I fitted $12$ domino spots into each side.

What a good idea, Angus. This estimation really helped you! Here is Angus' picture of the solution:

Lisa, who didn't give her school, described a very convincing way of working out the solution:

This is the lowest side total possible using the domino with the highest total: $6+4+1=11$.
The sides can therefore not add up to less then $11$.
However, for the $5+1$ domino to be part of a side totalling $11$, there would need to be another $5$. Because there is not another $5$ the lowest possible total of the sides is $12$.

This is the highest total possible using the domino with the lowest total: $5+1+6=12$.
Combining this with the information we already have, we know the sides must add up to $12$.
It is now simply the case of working out the number needed to make $12$.

$6+4+2=12$ There is only one $2$, so this domino must go here.

$6+2+4=12$ The remaining domino must therefore be placed this way round.

$3+4+5=12$ If the $5+1$ domino was the other way round it would need to be rotated to make the final side add up.

Very well explained, Lisa. We can see that this is the same solution as Angus', just rotated. A pupil from Wootton Upper School had a slightly different method:

Part 1: How many dots on each side?
The arrangement shows that each side of the square is composed of the sum of both sides of a single domino plus one of the sides of another domino.
The sum of the dots on each single domino is as follows: ($5$,$1$) = $6$, ($6$,$4$) = $10$, ($4$,$3$) = $7$ and ($6$,$2$) = $8$.
Therefore, the smallest sum possible is $8$ i.e. $6$ (from $5$,$1$) + $2$ (from $6$,$2$) whereas the largest sum possible is $16$ i.e. $10$ (from $6$,$4$) + $6$ (from $5$,$1$).

She then conjectured that, since the four sides of the square must have equal numbers of dots, a compromise of the extreme values is needed so each side could total $12$ dots.

Part 2: Arrangement
The ($5$,$1$) domino needs a $6$ from one side of another domino since $6 + 6 = 12$
Similarly, ($6$,$4$) needs a $2$ as $10 + 2 = 12$
($4$,$3$) needs a $5$ as $7 + 5 = 12$
($6$,$2$) needs as $4$ as $8 + 4 = 12$

For $6 + 6$, the $6$ can be obtained from ($6$,$4$) or ($6$,$2$)
For $10 + 2$, the $2$ can only be obtained from ($6$,$2$)
For $7 + 5$, the $5$ can only be obtained from ($5$,$1$)
For $8 + 4$, the $4$ can be obtained from ($6$,$4$) or ($4$,$3$)

Looking back at the square, it should be acknowledged that the dominoes are connected from head to tail like a chain. This is what we need to do with our numbers - arrange them so they form a nice chain.

For the above reason, we use ($6$,$4$) to obtain the $6$, and ($4$,$3$) to obtain the $4$ so that each domino is linked up consistently and they all get their required number of dots to total $12$.

The result obtained should be: ($6$,$4$) - ($2$,$6$) - ($4$,$3$) - ($5$,$1$) - ($6$,$4$) etc.


text:
Here's a brilliant puzzle we stumbled across...

You spin a wheel and it randomly lands of $£1$, $£2$ or END. If you land on $£1$ or $£2$, you bank the money and spin the wheel again. You keep spinning until you land on END, at which point you win the money you have banked. On average, how much do you expect to win?

For an extra challenge...

What if the wheel now has values $£1$, $£2$, ..., $£\text{n}$ as well as END and and you play the same game? How much do you expect to win now?

Post solutions below!

If you would like to be sent a list of our interviews and puzzles each week, sign up to Black Swans here.


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# Area under the parabola is always 1. But why?

I have the following matlab function:

function y=par(x,s)
%parabolic density function with basis = 2s
y=0*x;
ind=-s<x & x< s;
y(-s<x & x< s)=(3/(4*(s^3)))*(s*s-x(ind).*x(ind));

This function implements

$$y_s(x) = \begin{cases} \frac{3}{4s^3}(s^2-x^2) & \text{if } -s<x<s, \\ 0 & \text{otherwise.} \end{cases}$$

The area under the parabola is always one. I tested it with the following matlab procedure:

s = 0.5:0.1:100.5;
for i=1:length(s)
end

q is always 1 here.

btw. quad is a matlab function to evaluate the integral. Here it calculates $q = \int_{-s}^s y_s(x)\,dx$.

Can somebody explain to me why it is always one ?

-
I'd appreciate if this were written in standard math format, withoug the matlab. I think your question is likely really mathematical, and has nothing to do with unravelling computer code. –  coffeemath Oct 15 '12 at 12:11
that's true. I need some time to convert it into math format. –  superbly Oct 15 '12 at 13:15
How much calculus do you know? –  Rahul Nov 9 '12 at 12:52

The integral of the function is $1$, which is the area under the parabola:

$\displaystyle\int_{-s}^s\frac{3}{4s^3}(s^2-x^2)dx= \frac{3}{4s^3}\left(\int_{-s}^{s}s^2dx-\int_{-s}^{s}x^2dx \right) =\frac{3}{4s^3}\left(2ss^2-2\int_{0}^{s}x^2dx \right)= \\ =\displaystyle\frac{3}{4s^3}\left(2s^3-2\frac{s^3}{3}\right)=1$

-

It may help to rewrite your integrand as

$$y_s(x) = \frac{3}{4s^3}(s^2 - x^2) = \frac 1s \frac 34 \left( 1 - \left( \frac xs \right)^2 \right).$$

Now, let's first consider the case $s = 1$, which simplifies the integrand to

$$y_1(x) = \frac 34 (1 - x^2),$$

which is to be integrated from $-1$ to $1$. From high school calculus, I'm sure you'll remember that

$$\int_0^1 x^2 \,dx = \frac 13$$

which implies that

$$\int_0^1 (1 - x^2) \,dx = 1 - \frac 13 = \frac 23$$

and thus, since $(-x)^2 = x^2$,

$$\int_{-1}^1 (1 - x^2) \,dx = 2 \cdot \frac 23 = \frac 43.$$

Multiply that by $\frac 34$, and you get $1$.

Now, what about $s \ne 1$? The rewritten integrand has a factor of $\frac 1s$ in front of it, so the height of the parabola is divided by a factor of $s$. On the other hand, the argument $x$ is also divided by $x$, which effectively multiplies the width of the parabola (and, not coincidentally, the range over which it is integrated) by $s$. Those two effects cancel out, leaving the area unchanged.

In fact, we can generalize this result: for any function $f$ integrable on $[a,b]$ and any non-zero constant $s$,

\begin{aligned} \int_a^b f(x) dx &= \frac 1s \int_{sa}^{sb} f\left(\frac xs\right) \,dx \\ &= \int_{sa}^{sb} \frac 1s f\left(\frac xs\right) \,dx . \end{aligned}

-


text:
# Any other ways to solve?

Joe is swimming in a river 50 metres away from the bank. He would like to get back to his parents who are 100m along the river bank from where he is. If he can swim at 3 m/sec and run at 7m/sec, how far along the bank should he land in order to get back as quickly as possible?

So far, I've figured out how to do this:

Let $$x$$ be the distance along the bank that he should land in order to get back as quickly as possible.

Since he's 50 m away from the bank, he has to swim a distance of $$\sqrt{x^2 + 50^2}$$ metres, and then run a distance of $$100-x$$ metres.

The total amount of time taken to do this is $$\frac{7}{100-x} + \frac{3}{\sqrt{x^2 + 50^2}}$$.

Now, I can tabulate everything on an Excel spreadsheet, and figure out the answer from there, which is: he should land roughly 23.7 metres along the bank from where he is. Is there a more elegant way of getting around this?

dextercioby
Homework Helper
I'm sorry,but the formula is wrong...U divided speed by distance to find time...It is wrong.It's the other way around.
The total time is
$$t(x)=\frac{100-x}{7}+\frac{\sqrt{2500+x^{2}}}{3}$$

Do you know calculus??

Daniel.

learningphysics
Homework Helper
recon said:
Joe is swimming in a river 50 metres away from the bank. He would like to get back to his parents who are 100m along the river bank from where he is. If he can swim at 3 m/sec and run at 7m/sec, how far along the bank should he land in order to get back as quickly as possible?

So far, I've figured out how to do this:

Let $$x$$ be the distance along the bank that he should land in order to get back as quickly as possible.

Since he's 50 m away from the bank, he has to swim a distance of $$\sqrt{x^2 + 50^2}$$ metres, and then run a distance of $$100-x$$ metres.

The total amount of time taken to do this is $$\frac{7}{100-x} + \frac{3}{\sqrt{x^2 + 50^2}}$$.

Now, I can tabulate everything on an Excel spreadsheet, and figure out the answer from there, which is: he should land roughly 23.7 metres along the bank from where he is. Is there a more elegant way of getting around this?

Do you know calculus? If you do, then the way I did it was to calculate the derivative of the time function, and solve for 0.

Thanks a lot, you two. I just thought that there might be a more elegant way of solving this problem ,rather than using brute force through calculus or tabulating. I'm not sure if I'm allowed to solve this problem through calculus.

Edit: Recon, basically graph it and find its vertex, lowest point. Good enough answer...

Last edited:
dextercioby
Homework Helper
Yapper said:
Recon, basically graph it and find its vertex. Same point that using calculus gives you

Calculus yields the exact answer to
$$x_{extrema}=\pm \frac{75}{\sqrt{10}}$$

U can check that the physically acceptable solution (with the plus) corresponds to a maximum,plugging the value in the second derivative.The answer should be <0.

Daniel.

Isnt the second derivative a constant?

dextercioby
Homework Helper
Nope,the second derivative is a pretty ugly (sic) function.

Differentiate 2 times the original functions and convince yourself.

Daniel.

dextercioby, to answer a question you asked earlier, I know only a little bit of calculus.

I tried calculus and came out with this nasty looking equation:

$$\frac{\frac{21}{49}^2 \times 2500}{1 - \frac{21}{49}^2} = x^2$$

Solving for x yields the correct answer. However, I would like to know how you managed to simplify x to $$\frac{75}{\sqrt{10}}$$

dextercioby
Homework Helper
The derivative of the function is
$$t'(x)=-\frac{1}{7}+\frac{x}{3\sqrt{2500+x^{2}}}$$

Equate it to zero and square the resulting equation.

Daniel.

Oh ok i see. I get [3sqrt(2500+x^2)-((x^2)/3sqrt(2500+x^2))]/[9(2500+x^2)] Is that right for the second derivative?

dextercioby
Homework Helper
It looks okay...It might have helped by tired eyes if u had used LaTex.

Daniel.

1/49=(x^2)/(22500+9x^2)

9(2500+x^2)/49 = x^2 do I just use the quadratic equation? That gets messy... how did you get the nice numbers?

How do i use latex?

dextercioby
Homework Helper
Yapper said:
1/49=(x^2)/(22500+9x^2)

9(2500+x^2)/49 = x^2 do I just use the quadratic equation? That gets messy... how did you get the nice numbers?

How do i use latex?

You're on the right track.
$$22500+9x^{2}=49x^{2}$$
Solve for "x" and chose the "+" solution

Daniel.

Oh duh... lol 40x^2 - 22500 ahaha

+/- sqrt(-4*40*(-22500)) / 80 woohoo... lets see....

2*2*150*sqrt(10) / 80

7.5sqrt(10) * sqrt(10)/sqrt(10) = 75/sqrt(10) yay!!! Is this how you did it? or is there an easier methoid, you got it fast.

dextercioby


text:
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text:
# 2.E: Inclusion-Exclusion (Exercises)

• Page ID
7239
•

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

## 2.1: The Inclusion-Exclusion Formula

Ex 2.1.1 List all 6 solutions to the restricted equation in example 2.1.1, and list the corresponding 6 submultisets.

Ex 2.1.2 Find the number of integer solutions to $$x_1+x_2+x_3+x_4=25$$, $$1\le x_1\le6$$, $$2\le x_2\le 8$$, $$0\le x_3\le8$$, $$5\le x_4\le9$$.

Ex 2.1.3 Find the number of submultisets of $$\{25\cdot a,25\cdot b,25\cdot c,25\cdot d\}$$ of size 80.

Ex 2.1.4 Recall that $$\stwo{n}{k}$$ is a Stirling number of the second kind (definition 1.8.1). Prove that for $$n\ge k\ge 0$$,  \stwo{n}{k}={1\over k!}\sum_{i=0}^k (-1)^{k-i}i^n{k\choose i}.  Do $$n=0$$ as a special case, then use inclusion-exclusion for the rest. You may assume, by convention, that $$0^0=1$$.

### 2.2: Forbidden Position Permutations

Ex 2.2.1 Prove that $\ds D_n=nD_{n-1}+(-1)^n$ when $n\ge2$.

Ex 2.2.2 Prove that $D_n$ is even if and only if $n$ is odd.

Ex 2.2.3 Provide the missing details for example 2.2.1. What is $\ds\lim_{n\to\infty} {Q_n\over n!}$?

Ex 2.2.4 Find the number of permutations of $1,2,\ldots,8$ that have no odd number in the correct position.

Ex 2.2.5 Find the number of permutations of $1,2,\ldots,8$ that have at least one odd number in the correct position.

Ex 2.2.6 How many permutations of $$[n]$$ have exactly $k$ numbers in their correct positions?

Ex 2.2.7 Give a combinatorial proof that $$n!=\sum_{k=0}^n {n\choose k}D_{n-k}.$$

Ex 2.2.8 A small merry-go-round has 8 seats occupied by 8 children. In how many ways can the children change places so that no child sits behind the same child as on the first ride? The seats do not matter, only the relative positions of the children.

Ex 2.2.9 On the way into a party everyone checks a coat and a bag at the door. On the way out, the attendant hands out coats and bags randomly. In how many ways can this be done if

(a) No one gets either their own coat or their own bag?

(b) One may get one's own coat, or bag, but not both.

Ex 2.2.10 Suppose $n$ people are seated in $m\ge n$ chairs in a room. At some point there is a break, and everyone leaves the room. When they return, in how many ways can they be seated so that no person occupies the same chair as before the break?


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# Why Privacy Needs Composition

We’re back! In our last post we discussed some of the subtle pitfalls of formulating the assumptions underlying average-case relaxations of differential privacy. This time we’re going to look at the composition property of differential privacy—that is, the fact that running two independent differentially private algorithms on your data and combining their outputs is still differentially private. This is a key property of differential privacy and is actually closely related to the worst-case nature of differential privacy.

Composition is really the crucial property that has made differential privacy successful. Data analysis doesn’t happen in a vacuum, and the greatest threat to privacy comes from combining multiple pieces of information. These pieces of information can come from a single source that releases detailed statistics, or they could come from separate sources. So it’s critical to understand how the composition of multiple pieces of information can affect privacy.

In this post we’ll give some examples to illustrate why we need composition, and why composition is challenging for average-case relaxations of differential privacy. Composition is what allows you to design sophisticated differentially private algorithms out of simple building blocks, and it’s what allows one organization to release differentially private statistics without having to understand the entire ecosystem of related information that has been or will be released. As we’ll see, the challenges of composing average-case privacy guarantees are also very closely related to the subtleties that arise in thinking about the adversary’s beliefs.

### Differencing Attacks

Let’s start with a simple example of composition that was alluded to in our last post.

You’ve just started a new job and signed up for the health insurance provided by your employer. Thus, your employer is able to obtain aggregated data from the insurance provider. In particular, your employer can ask “How many of our employees have submitted claims for condition X?” However, your employer should not be able to find out whether or not you have condition X. For concreteness, condition X could be a mental health condition, drug addiction, being pregnant, terminal cancer, or an expensive chronic illness. Each of these could result in some kind of employment discrimination.

The employer may find out that 417 employees have condition X. That’s OK; on its own, this number reveals very little about whether or not you have condition X, as long as your employer is uncertain about how many employees other than you have condition X. We can formalize this as some kind of average-case or Bayesian privacy guarantee. Thus the health-insurance company is comfortable releasing this number exactly. But, yesterday, before you started your job, it also seemed reasonable to allow your employer to ask the exact same question, and yesterday the answer was 416. Thus your employer concludes that you have condition X.

In this example, we see how two pieces of information—the count before you started and the count after you started—each of which seems innocuous on its own can be combined to reveal private information. This is a simple example of a differencing attack and composition is important in part because it prevents these attacks.

This example involves only two pieces of information. However, an attack could combine many pieces of information. For example, the counts could be broken down by sex, race/ethnicity, age, location, and tobacco use.1 Additional data may also be obtained from other sources, such as public records, social media, voluntary disclosures, healthcare providers, financial records, employment records, or even illicit sources. The possibilities for attacks grow rapidly as more information is made available. And an employer is only one example of a potential privacy adversary.

The point of this example is that it’s easy to argue that one piece of information is harmless to privacy by making plausible-looking assumptions about the adversary. But this intuition rapidly breaks down once you consider the bigger picture where there are many pieces of information that can complete the puzzle. That’s why we need rigorous methods for understanding privacy and its composition.

### Quantifying Composition

How does differential privacy prevent a differencing attack like the one we just discussed? The simplest way is to add a little bit of random noise to each answer. On the first day, instead of releasing the exact count 416, we could release a noisy count, say, 420. Then on the second day, instead of releasing the true count 417, we release another noisy count, say, 415. More precisely, it is common to add noise to counts drawn from a Laplace or Gaussian distribution. These figures are still close enough to the true values to be useful, but the difference of 1 is now obscured by the noise, so your privacy is protected.

Since the noise is unknown to any potential adversary, it introduces uncertainty that protects the contribution that an individual makes to the count. Taking the difference of two independent noisy counts results in something that is still noisy. However, we must be careful to quantify this privacy guarantee, particularly when it comes to composition.

So, how much noise do we need to add? Let’s go back to the example and suppose the insurance company provides noisy answers where the noise has mean zero and some fixed variance. Your employer could simply ask the same question again and again and each time receive a different noisy answer. Averaging these noisy answers will effectively reduce the variance of the added noise and allow the true answer to be discerned. That leaves us back where we started.

There are actually two kinds of composition to consider. There is non-adaptive composition, where the questions to be asked are pre-specified and thus independent of the data, and there is adaptive composition, where the questions may themselves depend on the results of prior access to the data. Adaptive composition arises in an interactive system where queries are submitted one-by-one and each answer is returned before the next query is submitted. So far, we have really only considered non-adaptive composition.

Any interactive system must take adaptive composition into account. A natural algorithm which asks adaptive questions is gradient descent for minimizing a function that is determined by private data (e.g., for logistic regression on medical records). At each step, the algorithm asks for a gradient of the function, which depends on the private data, at the current point. Then the point is updated according to the reported gradient and the process repeats. Since the updated point depends on the previous answer, the next gradient computation is adaptive.

The good news is that differential privacy can handle adaptive composition just fine. However, to handle adaptive composition, it’s really important that you have a worst-case privacy definition like differential privacy. As we will see below, average-case variants of differential privacy cannot handle adaptive composition. Intuitively, the problem is that whatever distributional assumption you might make about the data or query a priori is unlikely to hold when you condition on past interactions with the same data or related data.

Here’s a technical example that shows the difficulty of adaptive composition. Our data $$x \in \{-1,+1\}^n$$ is a vector of $$n$$ bits, one bit per person.4 Because we’re considering average-case differential privacy, we’ll model this vector as uniformly random. Consider the following slightly odd algorithm $$M_2(x,v)$$—it takes a vector $$v \in \{-1,+1\}^n$$ from the user, and, if the correlation $$\langle x, v \rangle / n$$ between $$v$$ and $$x$$ is smaller than $$\varepsilon/2$$, the query returns $$\emptyset$$, but, if the correlation between $$v$$ and $$x$$ is larger than $$\varepsilon/2$$, the query returns the dataset $$x$$. In isolation this algorithm satisfies an average-case version of differential privacy, because if $$n$$ is large enough and $$x$$ is uniformly random, then it’s very unlikely that the user can guess a vector $$v$$ that causes this algorithm to output anything other than $$\emptyset$$. This algorithm may seem contrived; it is a simple stand-in for any algorithm that behaves very well most of the time, but fails completely on some rare inputs.

Now, consider another, more familiar differentially private algorithm called randomized response [W65]. For those not familiar, this algorithm $$M_1(x)$$ outputs a vector $$y \in \{-1,+1\}^n$$, where $$y_i$$ is slightly more likely to be $$x_i$$ than $$-x_i$$. Specifically, we set $$y_i = x_i$$ with probability $$(1+\varepsilon)/2$$ and $$y_i = - x_i$$ otherwise. This satisfies $$\log(\frac{1+\varepsilon}{1-\varepsilon})$$-differential privacy or, roughly, $$2\varepsilon$$-differential privacy. The upshot is that we obtain a vector $$y$$ where the correlation between $$x$$ and $$y$$ is about $$\varepsilon$$, i.e. $$\langle x , y \rangle / n \approx \varepsilon$$.

OK, so $$M_1$$ and $$M_2$$ both satisfy strong average-case versions of differential privacy when the data is uniform, but what about their composition? Well, the bad news is that running $$y = M_1(x)$$ followed by $$M_2(x,y)$$ is going to return the dataset $$x$$ with probability approaching 100%! That’s because $$y$$ was designed precisely to be a vector with correlation about $$\varepsilon$$ with $$x$$, and this is exactly the key that gets $$M_2$$ to unlock the dataset.

What went wrong here is that, even if $$x$$ really is uniformly random, it’s very far from it when conditioned on the output $$y=M_1(x)$$. To analyze $$M_2(x,y)$$ we must look at the distribution of $$x$$ conditioned on $$y$$. This distribution is going to be messy and may as well be a worst-case distribution, which means we must leave the realm of average-case privacy.

### Conclusion

Composition guarantees that, as long as each part of your system is differentially private, then the overall system is too. It would be difficult to build sophisticated systems without this property. And it’s what allows one organization to release differentially private statistics without having to worry about what other information might be out there. In short, composition is what allows differential privacy to deal with the complexities of the real world.

It is unlikely that differential privacy would have taken off as a field of research without this composition property. Any proposal for an alternative approach to privacy-preserving data analysis should first be evaluated in terms of how it handles composition.

This post only scratches the surface. In particular, we haven’t talked about the quantitative aspects of composition; that’s where the fun really begins. We will leave you with some pointers to further reading on the topic:

• [DSSU17] This is a survey of attacks which explains quantitatively the relationship between noise, number of questions, and privacy risks.
• [KOV15] [MV15] [DR16] [BS16] [M17] [DRS19] On the positive side, these papers analyze how differential privacy composes, yielding sharp quantitative bounds.

1. A good rule of thumb is that, if the number of released values is much larger than the number of people, then a privacy attack is probably possible. This is analogous to the rule from algebra that, if the number of constraints (released values) is greater than the number of unknown variables (people’s data), then the unknowns can be worked out.

2. Exactly quantifying how much noise is needed as the number of questions grows leads to the concept of a “privacy budget.” That is, we must precisely quantify how differential privacy degrades under composition. This is a very deep topic and is something we hope to discuss in future posts.

3. The queries do not need to be random [DY08]. The queries simply need to be “sufficiently distinct”, which can be formulated precisely as being nearly orthogonal vectors. Random, or even pseudorandom queries (e.g., hash functions), will almost certainly satisfy this property. In general, it is fairly likely that a set of queries will have this property and allow a reconstruction attack; that is, it is hard to avoid this phenomenon.

4. This representation of the dataset as a vector of bits $$x \in \{-1,+1\}^n$$ is an abstraction. The entries in the dataset would actually be something like a set of pairs $$( u_i, x_i )$$ for $$i = 1, \cdots, n$$, where $$u_i$$ is various information that identifies the individual concerned (name, address, race, date of birth, etc.).

Posted by Thomas Steinke and Jonathan Ullman on August 16, 2020.
Categories: Definitions

[cite this]


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# Tag Info

0

Try the Yarrow suite of Algorithms by Bruce Schneier

1

The most secure CSPRNG would be one which produces a stream indistinguishable from random data without knowledge of the key, and which takes a key large enough that an exhaustive search of the key space is impossible. It turns out that there are many such CSPRNGs, so there's more than one answer: ChaCha and Salsa20 are 256-bit stream ciphers which can ...

9

Some time ago (more than 5 years, I think) in a local forum one guy gave a "cryptographic challenge" to the community. He gave an encrypted string and a small piece of code that produced it. The goal was to find what the encrypted string was. The encryption key was based on System.Random(), which is the spot where it could be attacked. At the time my work ...

45

You asked for the practical impact, so the answer is that for \$120 I could probably have your entire password database done by tomorrow. Here is your program, or something similar to it: using System; using System.Text; using System.Security.Cryptography; class Program { static void Main(string[] args) { byte[] pwd = new byte[128]; ... 7 The practical impact is that if the attacker can determine the time when the password was created, they can feed that time as a seed to System.Random, and will get exactly the same password. System.Random uses a time value with a precision of 1 second. Assuming password creation time is known down to a second (like a user registration date/time from a ... 6 To aid the intuition, please consider my new password generation scheme "seems trustworthy until people inspect details" or "stupid." The way that stupid works is you throw a die and pick a password. If you throw 1, your password is "dhjousacbjlfswsgolFHfhjQPC." If you throw 2 it is "vmcseykogcsKcsNTLhczdg." And so on. Those look like pretty good passwords ... 19 The official documentation for System.Random explicitly says it should not be used for generating passwords. It’s predictable, and seeded only from the system clock. This means System.Random has at most 20 bits of entropy to anyone who has a clock accurate to within a second. Indeed, try creating two new instances in quick succession on different threads; ... 27 From what you have described, it sounds like your system works as follows: Consult the system clock to find a 32-bit seed$s$. Use System.Random to generate a passphrase$p = G(s)$. (Here$G$is shorthand for whatever computation happens inside System.Random.) Hash the passphrase with PBKDF(2?) into output$x = H(p, \sigma)$, where$\sigma$is a salt known ... 3 If someone chooses to run your program with a seed that you can't predict but an adversary can, you won't be able to tell. There is no way around this: nothing about the software can tell you what an adversary does or doesn't know; you can only use cryptography to ensure that if an adversary doesn't know a key and the message, then they can't decrypt the ... 3 How often should I replace the key? The SUPERCOP implementation replaces the 32-byte key every 768 bytes. It's a latency/throughput tradeoff. Generating a longer batch of output raises the maximum latency of any query to the PRNG but also increases the throughput by using proportionally more CPU cycles for generating data and fewer for generating ... 2 I'd guess that you can simply split this into two problems: create 64 - n random bits, call this R shuffle n bits where m bits (at any location) are set to 1, call this P Finally you can simply perform R | P (presuming big endian notation). Shuffling lists of elements is an operation present in almost any language. If there is any inefficiency it would be ... 1 The problem for choosing$k$bits from$64$ultimately comes down to computing a uniformly random integer$r$with$0 \leq r < \frac{64!}{k!(64-k)!}$then decoding it to determine which bits. The$k!$in the denominator is annoying, but we can ignore it, because we can just allow our algorithm to have$k!\$ random numbers that map to the same output (...

Top 50 recent answers are included


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## Quiz

### Formulas question type

This discussion has been locked because a year has elapsed since the last post. Please start a new discussion topic.

Coding error detected, it must be fixed by a programmer: \$post->parent must be set.


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## Kyoto Journal of Mathematics

### The approximate pseudorandom walk accompanied by the pseudostochastic process corresponding to a higher-order heat-type equation

#### Abstract

As is well known, a standard random walk is approximate to the stochastic process corresponding to the heat equation. Lachal constructed the approximate pseudorandom walk which is accompanied by the pseudostochastic process corresponding to an even-order heat-type equation. We have two purposes for this article. The first is to construct the approximate pseudorandom walk which is accompanied by the pseudostochastic process corresponding to an odd-order heat-type equation. The other is to propose a construction method for the approximate pseudorandom walk which is accompanied by the pseudostochastic process corresponding to an even-order heat-type equation. This method is different from that of Lachal.

#### Article information

Source
Kyoto J. Math., Volume 57, Number 4 (2017), 693-716.

Dates
Revised: 4 January 2016
Accepted: 25 May 2016
First available in Project Euclid: 9 June 2017

https://projecteuclid.org/euclid.kjm/1496973625

Digital Object Identifier
doi:10.1215/21562261-2017-0012

Mathematical Reviews number (MathSciNet)
MR3725258

Zentralblatt MATH identifier
06825575

#### Citation

Nakajima, Tadashi; Sato, Sadao. The approximate pseudorandom walk accompanied by the pseudostochastic process corresponding to a higher-order heat-type equation. Kyoto J. Math. 57 (2017), no. 4, 693--716. doi:10.1215/21562261-2017-0012. https://projecteuclid.org/euclid.kjm/1496973625

#### References

• [1] L. Beghin, K. J. Hochberg, and E. Orsingher,Conditional maximal distributions of processes related to higher-order heat-type equations, Stochastic Process. Appl.85(2000), 209–223.
• [2] L. Beghin, E. Orsingher, and T. Ragozina,Joint distributions of the maximum and the process for higher-order diffusions, Stochastic Process. Appl.94(2001), 71–93.
• [3] K. J. Hochberg,A signed measure on path space related to Wiener measure, Ann. Probab.6(1978), 433–458.
• [4] A. Lachal,First hitting time and place, monopoles and multipoles for pseudo-processes driven by the equation $\frac{\partial}{\partial_{t}}=\pm\frac{\partial^{N}}{\partial x^{N}}$, Electron. J. Probab.12(2007), 300–353.
• [5] A. Lachal,From pseudorandom walk to pseudo-Brownian motion: First exit time from a one-sided or a two-sided interval, Int. J. Stoch. Anal.2014, no. 520136.
• [6] M. Motoo, personal communication, August 1994.
• [7] T. Nakajima and S. Sato,On the joint distribution of the first hitting time and the first hitting place to the space-time wedge domain of a biharmonic pseudo process, Tokyo J. Math.22(1999), 399–413.
• [8] T. Nakajima and S. Sato,An approach to the pseudoprocess driven by the equation $\frac{\partial}{\partial t}=-A\frac{\partial^{3}}{\partial x^{3}}$ by a random walk, Kyoto J. Math.54(2014), 507–528.
• [9] Y. Nikitin and E. Orsingher,On sojourn distributions of processes related to some higher-order heat-type equations, J. Theoret. Probab.13(2000), 997–1012.
• [10] K. Nishioka,Monopoles and dipoles in biharmonic pseudo-process, Proc. Japan Acad. Ser. A Math. Sci.72(1996), 47–50.
• [11] K. Nishioka,The first hitting time and place of a half-line by a biharmonic pseudo process, Japan. J. Math. (N.S.)23(1997), 235–280.
• [12] K. Nishioka,Boundary conditions for one-dimensional biharmonic pseudo process, Electron. J. Probab.6(2001), no. 13.
• [13] S. Sato,An approach to the biharmonic pseudo process by a random walk, J. Math Kyoto Univ.42(2002), 403–422.
• [14] Y. Shimoyama,On behaviors of the stochastic pseudo processes governed by third-order heat-type equations, Kobe J. Math.30(2013), 49–82.


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American Institute of Mathematical Sciences

October  2010, 26(4): 1141-1151. doi: 10.3934/dcds.2010.26.1141

Sufficient conditions for the regularity to the 3D Navier-Stokes equations

 1 Department of Mathematics, Florida International University, Miami, FL 33199, United States

Received  December 2008 Revised  February 2009 Published  December 2009

In this paper we consider the three-dimensional Navier-Stokes equations subject to periodic boundary conditions or in the whole space. We provide sufficient conditions, in terms of one direction derivative of the velocity field, namely, $u_z$, for the regularity of strong solutions to the three-dimensional Navier-Stokes equations.
Citation: Chongsheng Cao. Sufficient conditions for the regularity to the 3D Navier-Stokes equations. Discrete & Continuous Dynamical Systems - A, 2010, 26 (4) : 1141-1151. doi: 10.3934/dcds.2010.26.1141
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2019 Impact Factor: 1.338


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# Fourier Series - Asymmetric Square Wave

1. Oct 27, 2013

### DmytriE

Good morning everyone,

I am taking a signals and systems course where we are now studying the fourier series. I understand that this is for signals that are periodic. But I get hung up when determining the fourier coefficients. In the video by Alan Oppenheim, he derives the equation for the fourier series. Below is the analysis equation.

$a_{k} = \frac{1}{T_{0}} ∫ x(t)*e^{jk\omega_{0}t}$

He goes through an example using an asymmetric square wave with an amplitude of 1. I understand the bounds that he chooses $(-T_0/2, 0)$ and $(0, T_0/2)$.

This leads to a fourier coefficient equation of the following:
$a_{k} = \frac{1}{T_{0}} [∫ (-1)*e^{jk\omega_{0}t} + ∫ (1)*e^{jk\omega_{0}t}]$

To compute the general equation for $a_k$ should I treat the ±1 as a function and use u-substitution / integration by parts? If so, can someone at least show the first step or 2? I haven't done I.B.P. or U-substitution in some time. If I can treat the x(t) as a constant then how can I integrate to get the answer below?

General equation: $a_k = \frac{1}{jkπ} (1 - (-1^{k}))$

This example is from M.I.T. Open Courseware Alan V. Oppenheim Signals and Systems course Lecture 7 approximately 20 minutes into the video.

2. Oct 27, 2013

### AlephZero

You don't need to do all that. "1" and "-1" are just constants. If you are studying this, you should know $\int e^{at}dt$, and that formula works when $a$ is a complex number as well as when $a$ is real.

You should also know $e^{j\omega t} = \cos \omega t + j \sin \omega t$.

You also have an equation connecting $T_0$ and $\omega_0$, which is why they both disappeared in the final equation for $a_k$ (and that's also where the $\pi$ came from).

If you are confused by the $(-1)^k$ part, just work out the first few values of $a_1$, $a_2$, etc and see what happens.

3. Oct 27, 2013

### DmytriE

This is precisely what I am confused about. Is there a rule that applies to the $(-1)^k$ or do I have to input a couple numbers for $a_k$?