text:
# American Institute of Mathematical Sciences

January  1996, 2(1): 141-149. doi: 10.3934/dcds.1996.2.141

## On continuous dependence under approximation for groundwater flow models with distributed and pointwise observations

 1 CRSC & Department of Mathematics, North Carolina State University, Raleigh, NC 27695-8205, United States, United States

Received  July 1995 Published  October 1995

We present in this paper some results on continuous dependence for parameters in a groundwater flow model. These results are crucial for theoretical and computational aspects of least squares estimation of parameters. As is typically the case in field studies, the form of the data is pointwise observation of hydraulic head and hydraulic conductivity at a discrete collection of observation well sites. We prove continuous dependence results for the solution of the groundwater flow equation, with respect to conductivity and boundary values, under certain types of numerical approximation.
Citation: B.G. Fitzpatrick, M.A. Jeffris. On continuous dependence under approximation for groundwater flow models with distributed and pointwise observations. Discrete & Continuous Dynamical Systems - A, 1996, 2 (1) : 141-149. doi: 10.3934/dcds.1996.2.141
 [1] Michiel Bertsch, Carlo Nitsch. Groundwater flow in a fissurised porous stratum. Networks & Heterogeneous Media, 2010, 5 (4) : 765-782. doi: 10.3934/nhm.2010.5.765 [2] Juan A. Calzada, Rafael Obaya, Ana M. Sanz. Continuous separation for monotone skew-product semiflows: From theoretical to numerical results. Discrete & Continuous Dynamical Systems - B, 2015, 20 (3) : 915-944. doi: 10.3934/dcdsb.2015.20.915 [3] Dorothee Knees, Andreas Schröder. Computational aspects of quasi-static crack propagation. Discrete & Continuous Dynamical Systems - S, 2013, 6 (1) : 63-99. doi: 10.3934/dcdss.2013.6.63 [4] Nicoletta Del Buono, Cinzia Elia, Roberto Garrappa, Alessandro Pugliese. Preface: "Structural Dynamical Systems: Computational aspects". Discrete & Continuous Dynamical Systems - B, 2018, 23 (7) : i-i. doi: 10.3934/dcdsb.201807i [5] Louis Caccetta, Ian Loosen, Volker Rehbock. Computational aspects of the optimal transit path problem. Journal of Industrial & Management Optimization, 2008, 4 (1) : 95-105. doi: 10.3934/jimo.2008.4.95 [6] Nur Aidya Hanum Aizam, Louis Caccetta. Computational models for timetabling problem. Numerical Algebra, Control & Optimization, 2014, 4 (3) : 269-285. doi: 10.3934/naco.2014.4.269 [7] Simone Göttlich, Oliver Kolb, Sebastian Kühn. Optimization for a special class of traffic flow models: Combinatorial and continuous approaches. Networks & Heterogeneous Media, 2014, 9 (2) : 315-334. doi: 10.3934/nhm.2014.9.315 [8] Rolf Bronstering. Some computational aspects of approximate inertial manifolds and finite differences. Discrete & Continuous Dynamical Systems - A, 1996, 2 (4) : 417-454. doi: 10.3934/dcds.1996.2.417 [9] Yalin Zhang, Guoliang Shi. Continuous dependence of the transmission eigenvalues in one dimension. Inverse Problems & Imaging, 2015, 9 (1) : 273-287. doi: 10.3934/ipi.2015.9.273 [10] Jiří Benedikt. Continuous dependence of eigenvalues of $p$-biharmonic problems on $p$. Communications on Pure & Applied Analysis, 2013, 12 (3) : 1469-1486. doi: 10.3934/cpaa.2013.12.1469 [11] Giuseppe Maria Coclite, Angelo Favini, Gisèle Ruiz Goldstein, Jerome A. Goldstein, Silvia Romanelli. Continuous dependence in hyperbolic problems with Wentzell boundary conditions. Communications on Pure & Applied Analysis, 2014, 13 (1) : 419-433. doi: 10.3934/cpaa.2014.13.419 [12] Christophe Chalons. Theoretical and numerical aspects of the interfacial coupling: The scalar Riemann problem and an application to multiphase flows. Networks & Heterogeneous Media, 2010, 5 (3) : 507-524. doi: 10.3934/nhm.2010.5.507 [13] Vladimir Korotkov, Vladimir Emelichev, Yury Nikulin. Multicriteria investment problem with Savage's risk criteria: Theoretical aspects of stability and case study. Journal of Industrial & Management Optimization, 2020, 16 (3) : 1297-1310. doi: 10.3934/jimo.2019003 [14] Juvencio Alberto Betancourt-Mar, Víctor Alfonso Méndez-Guerrero, Carlos Hernández-Rodríguez, José Manuel Nieto-Villar. Theoretical models for chronotherapy: Periodic perturbations in hyperchaos. Mathematical Biosciences & Engineering, 2010, 7 (3) : 553-560. doi: 10.3934/mbe.2010.7.553 [15] Pierre Aime Feulefack, Jean Daniel Djida, Atangana Abdon. A new model of groundwater flow within an unconfined aquifer: Application of Caputo-Fabrizio fractional derivative. Discrete & Continuous Dynamical Systems - B, 2019, 24 (7) : 3227-3247. doi: 10.3934/dcdsb.2018317 [16] Rafał Kamocki, Marek Majewski. On the continuous dependence of solutions to a fractional Dirichlet problem. The case of saddle points. Discrete & Continuous Dynamical Systems - B, 2014, 19 (8) : 2557-2568. doi: 10.3934/dcdsb.2014.19.2557 [17] Pavel Krejčí, Thomas Roche. Lipschitz continuous data dependence of sweeping processes in BV spaces. Discrete & Continuous Dynamical Systems - B, 2011, 15 (3) : 637-650. doi: 10.3934/dcdsb.2011.15.637 [18] Ramon Quintanilla. Structural stability and continuous dependence of solutions of thermoelasticity of type III. Discrete & Continuous Dynamical Systems - B, 2001, 1 (4) : 463-470. doi: 10.3934/dcdsb.2001.1.463 [19] P.E. Kloeden, Pedro Marín-Rubio. Equi-Attraction and the continuous dependence of attractors on time delays. Discrete & Continuous Dynamical Systems - B, 2008, 9 (3&4, May) : 581-593. doi: 10.3934/dcdsb.2008.9.581 [20] Paola Goatin, Philippe G. LeFloch. $L^1$ continuous dependence for the Euler equations of compressible fluids dynamics. Communications on Pure & Applied Analysis, 2003, 2 (1) : 107-137. doi: 10.3934/cpaa.2003.2.107

2019 Impact Factor: 1.338


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### Round-Optimal Secure Multiparty Computation with Honest Majority

Prabhanjan Ananth, Arka Rai Choudhuri, Aarushi Goel, and Abhishek Jain

##### Abstract

We study the exact round complexity of secure multiparty computation (MPC) in the honest majority setting. We construct several round-optimal $n$-party protocols, tolerating any $t<\frac{n}{2}$ corruptions. - Security with abort: We give the first construction of two round MPC for general functions that achieves security with abort against malicious adversaries in the plain model. The security of our protocol only relies on one-way functions. - Guaranteed output delivery: We also construct protocols that achieve security with guaranteed output delivery: (i) Against fail-stop adversaries, we construct two round MPC either in the (bare) public-key infrastructure model with no additional assumptions, or in the plain model assuming two-round semi-honest oblivious transfer. In three rounds, however, we can achieve security assuming only one-way functions. (ii) Against malicious adversaries, we construct three round MPC in the plain model, assuming public-key encryption and Zaps. Previously, such protocols were only known based on specific learning assumptions and required the use of common reference strings. All of our results are obtained via general compilers that may be of independent interest.

Note: Added citation to concurrent work

Available format(s)
Publication info
Keywords
Secure Computation
Contact author(s)
prabhanjan @ csail mit edu
achoud @ cs jhu edu
agoel10 @ jhu edu
abhishek @ cs jhu edu
History
2018-09-10: last of 5 revisions
See all versions
Short URL
https://ia.cr/2018/572

CC BY

BibTeX

@misc{cryptoeprint:2018/572,
author = {Prabhanjan Ananth and Arka Rai Choudhuri and Aarushi Goel and Abhishek Jain},
title = {Round-Optimal Secure Multiparty Computation with Honest Majority},
howpublished = {Cryptology ePrint Archive, Paper 2018/572},
year = {2018},
note = {\url{https://eprint.iacr.org/2018/572}},
url = {https://eprint.iacr.org/2018/572}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.

format: plain


text:
# Render any Literate Haskell to PDF, HTML, or similar

How to I convert a Literate Haskell file to something easier on the eyes? Here's what happens when I try to do the obvious thing:

$latex Check.lhs This is pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) entering extended mode (./Check.lhs LaTeX2e <2009/09/24> Babel <v3.8l> and hyphenation patterns for english, usenglishmax, dumylang, noh yphenation, loaded. ! LaTeX Error: Environment code undefined. See the LaTeX manual or LaTeX Companion for explanation. Type H <return> for immediate help. ... l.7 \begin{code}  I'm not familiar with the usual commands for processing tex files. What's a simple, no-nonsense way I can use to convert just about any .lhs file to e.g. PDF or HTML? - ## 2 Answers In this space you can use lhs2TeX, which has quite a few options but can give quite nice results on formatting. It has pretty extensive documentation, which you can find on the homepage. An example of usage (directly from the manual) would be \documentclass{article} %include polycode.fmt \begin{document} This is the famous ‘‘Hello world’’ example, written in Haskell: \begin{code} main :: IO () main = putStrLn "Hello, world!" \end{code} \end{document}  The \documentclass{article}, \begin{document} and \end{document} are all TeX commands to indicate what kind of document you want, and where it starts/ends. The %include polycode.fmt is actually a comment in TeX, but will indicate to lhs2TeX what format to use when processing the .lhs file. You can compile the example with $ lhs2TeX -o HelloWorld.tex HelloWorld.lhs
\$ pdflatex HelloWorld.tex

For HTML you can also try Pandoc, which can give you all sorts of output formats (pdf, html, rtf, etc).

-

You want to preprocess your literate files with lhs2tex.

-


text:
Chapter 13
Chapter 13
Chapter 13
Chapter 14
Chapter 14

### 100

Divide 28 ÷ 3. How many are left over?
REMAINDER is 1 because 28 ÷ 3 = 9 R1

### 100

Solve 63 ÷ 8. DON'T FORGET TO CHECK!!
63 ÷ 8 = 7 R7

### 100

68 students are separated into equal groups with 3 students left over. Tell how many groups could have been formed and how many students were in each group.
5 groups with 13 students each OR 13 groups with 5 students each

### 100

Solve 260 ÷ 2. DON'T FORGET TO CHECK!!
260 ÷ 2 = 130. WATCH OUT FOR THAT ZERO!!!

### 100

Solve 350 ÷ 5 using a basic fact and a pattern.
350 ÷ 5 = 70

### 200

Divide 48 ÷ 7. How many are left over?
REMAINDER is 6 because 48 ÷ 7 = 6 R6

### 200

Together, Megan and Sandra spent $15.50 for 2 t-shirts. Megan's t-shirt cost$3.50 less than Sandra's. How much did each girl pay for her t-shirt?
Megan paid $6 and Sandra paid$9.50.

### 200

Solve 62 ÷ 9. DON'T FORGET TO CHECK!!
62 ÷ 9 = 6 R8

### 200

Solve 495 ÷ 6. DON'T FORGET TO CHECK!!
495 ÷ 6 = 82 R3

### 200

Solve 365 ÷ 8. DON'T FORGET TO CHECK!!
365 ÷ 8 = 45 R5

### 300

Solve 82 ÷ 6. DON'T FORGET TO CHECK!!
82 ÷ 6 = 13 R4

### 300

Solve 34 ÷ 8. DON'T FORGET TO CHECK!!
34 ÷ 8 = 4 R2

### 300

Divide 38 ÷ 5. How many are left over?
REMAINDER is 3 because 38 ÷ 5 = 7 R3

### 300

Solve 392 ÷ 9. DON'T FORGET TO CHECK!!
392 ÷ 9 = 43 R5

### 300

Solve 48,000 ÷ 8 using a basic fact and a pattern.
48,000 ÷ 8 = 6,000

### 400

There are 26 students in the fourth grade. The students are split into equal groups of 5 students. How many students are left over?
1 student is left over because 26 ÷ 5 = 5 R1

### 400

At day camp, 44 campers are split into 6 equal group. How many campers are left over?
2 campers are left over because 44 ÷ 6 = 7 R2

### 400

Solve 73 ÷ 4. DON'T FORGET TO CHECK!!
73 ÷ 4 = 18 R1

### 400

Solve 406 ÷ 3. DON'T FORGET TO CHECK!!
406 ÷ 3 = 135 R1

### 400

Solve 742 ÷ 7. DON'T FORGET TO CHECK!!
742 ÷ 7 = 106

### 500

Solve 85 ÷ 5. DON'T FORGET TO CHECK!!
85 ÷ 5 = 17

### 500

35 cupcakes were divided evenly onto plates with 5 left over. Tell how many plates and how many cookies were on each plate.
5 plates and 6 cookies on each OR 6 plates and 5 cookies on each

### 500

Together, Billy and Chris spent $14.25 for 2 sandwiches. Billy's sandwich cost$4.25 less than Chris's. How much did each boy pay for his sandwich?
Billy paid $5 and Chris paid$9.25.

### 500

Solve 439 ÷ 4. DON'T FORGET TO CHECK!!
439 ÷ 4 = 109 R 3

### 500

Solve 2,400 ÷ 6 using a basic fact and a pattern.
2,400 ÷ 6 = 400


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# Prominence of Pairing in Inclusive ($p$ ,2$p$) and ($p$, $pn$) Cross Sections from Neutron-Rich Nuclei

* Corresponding author
Abstract : Fifty-five inclusive single nucleon-removal cross sections from medium mass neutron-rich nuclei impinging on a hydrogen target at $\sim$250 MeV=nucleon are measured at the RIKEN Radioactive Isotope Beam Factory. Systematically higher cross sections are found for proton removal from nuclei with an even number of protons as compared to odd-proton number projectiles for a given neutron separation energy. Neutron removal cross sections display no even-odd splitting, contrary to nuclear cascade model predictions. Both effects are understood through simple considerations of neutron separation energies and bound state level densities originating in pairing correlations in the daughter nuclei. These conclusions are supported by comparison with semimicroscopic model predictions, highlighting the enhanced role of lowlying level densities in nucleon-removal cross sections from loosely bound nuclei
Document type :
Journal articles
Domain :

Cited literature [79 references]

https://hal-cea.archives-ouvertes.fr/cea-02122388
Contributor : Anna Corsi <>
Submitted on : Tuesday, May 7, 2019 - 12:09:51 PM
Last modification on : Friday, September 18, 2020 - 1:44:06 PM
Long-term archiving on: : Tuesday, October 1, 2019 - 7:24:02 PM

### File

PhysRevLett.122.162503.pdf
Publisher files allowed on an open archive

### Citation

N. Paul, A. Obertelli, C Bertulani, A. Corsi, P. Doornenbal, et al.. Prominence of Pairing in Inclusive ($p$ ,2$p$) and ($p$, $pn$) Cross Sections from Neutron-Rich Nuclei. Physical Review Letters, American Physical Society, 2019, 122, pp.162503. ⟨10.1103/PhysRevLett.122.162503⟩. ⟨cea-02122388⟩

Record views


text:
Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Find a vector of magnitude 6 ,

Question:

Find a vector of magnitude 6 , which is perpendicular to both the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $4 \hat{i}-\hat{j}+3 \hat{k}$

Solution:


text:
Ch.9: Sampling Distributions: MeanWorksheetSee all chapters

# Small Sample Confidence Interval for a Population Mean

See all sections

Concept #1: Determining the Small Sample Confidence Interval for a Population Mean

Concept #2: Determining the Small Sample Confidence Interval for a Population Mean: Intro

Practice: Books get more and more expensive every semester. 25 randomly selected students in your school spent, on average $500 with a standard deviation of$50. Construct a 98% confidence interval for the true spending on books.

Practice: People always purchase gifts on Black Friday. The average spending on this day is $1,000 with a standard deviation of$256. These estimates are based on a sample of 16 randomly selected Americans. Construct a 95% confidence interval for the true spending of Americans on Black Friday.

Practice: College students are said to binge drink more than any other population of young adults. You randomly select 9 people in your school and find that the average number of drinks consumed each week is 10 with a standard deviation of 5. Construct a 90% confidence interval for the true average number of drinks college students at your school consumer each week.


text:
# how to compile tex file with -shell-escape in AucTeX using C-c C-c on ubuntu

I want to compile my tex file with -shell-escape option in Emacs + AucTex using C-c C-c key binding.

It gave me an error to run with -shell-escape. I know I can do it through terminal by latex -shell-escape la.tex but I want AucTeX to do it.

I even added a local varible % TeX-command-extra-options: -shell-escape but it gave the same error

You have to put it into a Local Variables block.

\documentclass{article}
\begin{document}
\end{document}

%%% Local Variables:
%%% TeX-command-extra-options: "-shell-escape"
%%% End:

When I open the file Emacs warns me:

The local variables list in test.tex
contains values that may not be safe (*).

Do you want to apply it?  You can type
y  -- to apply the local variables list.
n  -- to ignore the local variables list.
!  -- to apply the local variables list, and permanently mark these
values (*) as safe (in the future, they will be set automatically.)

* TeX-command-extra-options : "-shell-escape"

When I answer y and use the C-c C-c shortcut, I see in the log

Running LaTeX' on test' with lualatex --jobname=test  -file-line-error -shell-escape  -interaction=nonstopmode "\input" test.tex''

Your output might differ slightly because I have changed some of Emacs' and AucTeX's default settings.

• well. when I added the variable, the variable block entered automatically, but the warning didn't open. So I did M-x revert-buffer and warning came and every thing went smoothly. So thank you. – ذوالفقار عبد الحليم Dec 27 '18 at 17:18


text:
# Key Concepts

Variables: measures that can take on different sizes.

Data: recorded values of variables.

Time series data: a set of measurements made sequentially at different points in time.

High (low) frequency data series have short (long) intervals between observations.

Cross-section data: values for different variables recorded at a point in time.

Longitudinal data follow the same units of observation through time.

Index number: value for a variable, or an average of a set of variables, expressed relative to a given base value, as seen in Equation 2.2:

$$\text{Value of index} = \displaystyle\frac{\text{Absolute value in year}\,t} {\text{Absolute value in base year}} \times \:100.$$

$$\textbf{Percentage change}= \displaystyle\frac{\text{change in values}} {\text{original value}} \times \:100.$$

Consumer price index: the average price level for consumer goods and services.

Inflation rate: the annual percentage increase in the consumer price index.

Deflation rate: the annual percentage decrease in the consumer price index.

Real price index: a nominal price index divided by the consumer price index, scaled by 100.

Nominal price index: the current dollar price of a good or service.

Nominal earnings: earnings measured in current dollars.

Real earnings: earnings measure in constant dollars to adjust for changes in the general price level.

Scatter diagram plots pairs of values simultaneously observed for two variables.

Econometrics is the science of examining and quantifying relationships between economic variables.

Regression line represents the average relationship between two variables in a scatter diagram.

Positive economics studies objective or scientific explanations of how the economy functions.

Normative economics offers recommendations that incorporate value judgments.

Economic equity is concerned with the distribution of well-being among members of the economy.


text:
# How do you Evaluate $\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1}dx$ Without Using Residue Calculus

So I'm having trouble computing this integral:

$$\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1}dx$$

After seeing some other answers on this website, I've realized that many people are using residues to get an answer. However, I know nothing about complex analysis, so I wanted to know if there is a better way of doing it without it.

I started by finding the indefinite integral which is:

$$-\dfrac{\sinh\left(1\right)\left(\operatorname{Si}\left(x+\mathrm{i}\right)+\operatorname{Si}\left(x-\mathrm{i}\right)\right)-\mathrm{i}\cosh\left(1\right)\left(\operatorname{Ci}\left(x+\mathrm{i}\right)-\operatorname{Ci}\left(x-\mathrm{i}\right)\right)}{2}$$

but finding what this function comes out to when finding the limit of it towards $\infty$ and $-\infty$ I get $-\pi\sinh(1)$

Now, I could be very wrong with both the indefinite integral and the value. Am I doing something wrong? Am I forgetting to take into account something I should be? Thanks in advanced!

• Your antiderivative seems fine. And if we denote this by $I(x)$ with aid of Mathematica, it indeed yields the correct value as we have $$\lim_{x\to\infty}I(x) = -\frac{\pi}{2}\sinh 1, \qquad \lim_{x\to-\infty} I(x) = \frac{\pi}{2}\sinh 1 - \pi \cosh 1.$$ So my bet is that you forgot to take the branch cut of $\operatorname{Ci}$ into consideration – Sangchul Lee Nov 5 '17 at 1:01
• @SangchulLee Huh, that's odd, do you know how to come about with the answer for the one with x approaching negative infinity? Also, I'm pretty sure I took account of the branch cut of Ci – Tom Himler Nov 5 '17 at 1:09
• In fact the best explanation for this comes from complex analysis. The extra factor appears when the contour crosses the singularity of $\frac{\cos z}{z}$ at $z = 0$. I am currently working to find an explanation that does not use complex analysis. – Sangchul Lee Nov 5 '17 at 1:16
• Here is one explanation: for $x \in \mathbb{R}$ we have $$\operatorname{Ci}(x\pm i) = \operatorname{Ci}(-x\mp i) - \operatorname{Ci}(\mp i) + \operatorname{Ci}(\pm i).$$ So if you know that $\operatorname{Ci}(x\pm i) \to 0$ as $x\to\infty$, it follows that $\operatorname{Ci}(x\pm i) \to \pm (\operatorname{Ci}(i) - \operatorname{Ci}(-i))$ as $x\to-\infty$. Now the answer follows from $$\operatorname{Ci}(i) - \operatorname{Ci}(-i) = i\pi.$$ – Sangchul Lee Nov 5 '17 at 1:48
• @SangchulLee That's interesting, because actually replacing $i\pi$ for the definite integral calculation, it still doesn't seem to give $\frac{\pi}{e}$ Also want to preface that I'm still probably doing it wrong. – Tom Himler Nov 5 '17 at 2:00

Here's a much nicer way using the standard differentiation under the integral sign technique. Let's first consider the general case$$I(a)=\int\limits_{-\infty}^{\infty}\frac {\cos ax}{1+x^2}\, dx$$Through integration by parts on $u=1/(1+x^2)$ and $dv=\cos ax\, dx$, we have$$a\cdot I(a)=2\int\limits_{-\infty}^{\infty}\frac {x\sin ax}{(1+x^2)^2}\, dx$$Differentating with respect to $a$, we obtain$$a\cdot I'(a)+I(a)=2I(a)-2\int\limits_{-\infty}^{\infty}\frac {\cos ax}{(1+x^2)^2}\, dx$$Combining like terms, we realize that the right-hand side is similar to what we obtained before. So differentiating (again), we finally see that$$a\cdot I''(a)=-2\int\limits_{-\infty}^{\infty}\frac {x\sin ax}{(1+x^2)^2}\, dx=a\cdot I(a)$$Solving the simple differential equation that follows, the general solution is $I(a)=C_1e^{a}+C_2e^{-a}$. When we set $a=0$, we see that $C_1=0$ and similarly, $C_2=\pi$ for $a\to\infty$. Therefore, the general solution is$$\int\limits_{-\infty}^{\infty}\frac {\cos ax}{1+x^2}\, dx=\color{blue}{\frac {\pi}{e^{|a|}}}$$

• A way of using the Feynman's Trick would really clarify actually if you could do that. – Tom Himler Nov 5 '17 at 1:16
• @TomHimler That's the problem. I'm not sure what's supposed to be the correct substitution. I have a gut-feeling that it can be integrated using Feynman's trick, probably because I've seen a variant somewhere where differentiation can be used, but I'm not sure. I will update the answer as I make progress. – Crescendo Nov 5 '17 at 1:17
• I would really appreciate it if you can. In the meantime, I'll try to see if I can do it as well. – Tom Himler Nov 5 '17 at 1:18
• @TomHimler I think I've got it. Although it does involve a little bit of differential equations, namely solving$$I''(a)-I(a)=0$$You can do that, right? – Crescendo Nov 5 '17 at 1:27
• Yes, thank you, I really appreciate the effort you've made to help me. – Tom Himler Nov 5 '17 at 1:33

Do you know the Fourier inversion theorem ? $$\frac{2}{1+x^2} = \int_{-\infty}^\infty e^{-|t|}e^{i x t}dt$$ where $\frac{d}{dt} e^{-|t|} \in L^1$ thus $$\int_{-\infty}^\infty \frac{2 e^{-i xu}}{1+x^2}dx = \lim_{A \to \infty} \int_{-A}^A e^{-ixu} (\int_{-\infty}^\infty e^{-|t|}e^{i x t}dt)dx=\lim_{A \to \infty}\int_{-\infty}^\infty e^{-|t|} A\frac{\sin(A(t-u))}{A(t-u)}dt$$ which converges to $2\pi e^{-|u|}$ by integration by parts

Step 1. Let $s > 0$ and $\beta \in \mathbb{C}$. Then using the gaussian integral $\int_{-\infty}^{\infty} e^{-sx^2} \, dx = \sqrt{\pi/s}$, we have

\begin{align*} \int_{-\infty}^{\infty} e^{-s(x-\beta)^2} \, dx &= \int_{-\infty}^{\infty} \bigg( e^{-sx^2} + \int_{0}^{1} \overbrace{ 2s\beta(x-\beta t) e^{-s(x-\beta t)^2} }^{= \frac{\partial}{\partial t} e^{-s(x-\beta t)^2}} \, dt \bigg) \, dx \\ &= \sqrt{\frac{\pi}{s}} -\beta \int_{0}^{1} \int_{-\infty}^{\infty} (-2s)(x-\beta t) e^{-s(x-\beta t)^2} \, dx dt \quad {\small(\because\text{Fubini})}\\ &= \sqrt{\frac{\pi}{s}} -\beta \int_{0}^{1} \left[ e^{-s(x-\beta t)^2} \right]_{x=-\infty}^{x=\infty} \, dt = \bbox[border:1px dashed green,6px]{ \sqrt{\frac{\pi}{s}} }. \end{align*}

Then plugging $\beta=\pm\mathrm{i}/2s$, we check that

$$\int_{-\infty}^{\infty} e^{-sx^2}\cos(x) \, dx = \sqrt{\frac{\pi}{s}} e^{-1/4s}.$$

Step 2. Using the previous step,

\begin{align*} I := \int_{-\infty}^{\infty} \frac{\cos x}{x^2+1} \, dx &= \int_{-\infty}^{\infty} \cos x \left( \int_{0}^{\infty} e^{-(x^2+1)s} \, ds \right) \, dx \\ &= \int_{0}^{\infty} \left( \int_{-\infty}^{\infty} e^{-sx^2}\cos(x) \, dx \right) e^{-s} \, ds \quad {\small(\because\text{Fubini})}\\ &= \int_{0}^{\infty} \sqrt{\frac{\pi}{s}} e^{-\left( s + \frac{1}{4s}\right)} \, ds \\ &= \int_{0}^{\infty} \sqrt{2\pi} e^{-\frac{1}{2}\left( t^2 + \frac{1}{t^2}\right)} \, dt. \quad {\small(2s=t^2)} \end{align*}

Finally here is a very slick way of computing the last integral. Applying the substitution $t\mapsto 1/t$ shows that

$$\int_{0}^{\infty} e^{-\frac{1}{2}\left( t^2 + \frac{1}{t^2}\right)} \, dt = \int_{0}^{\infty} \frac{1}{t^2} e^{-\frac{1}{2}\left( t^2 + \frac{1}{t^2}\right)} \, dt.$$

So averaging,

$$I = \sqrt{\frac{\pi}{2}} \int_{0}^{\infty} \left(1 + \frac{1}{t^2}\right) e^{-\frac{1}{2}\left( t - \frac{1}{t}\right)^2 - 1} \, dt.$$

Finally, applying the substitution $u = t - \frac{1}{t}$ proves

$$I = \sqrt{\frac{\pi}{2}} \int_{-\infty}^{\infty} e^{-\frac{u^2}{2} - 1} \, du = \frac{\pi}{e}.$$

• I think you’ve forgotten to pull $e^{-1}$ out of the last integral – adfriedman Nov 5 '17 at 1:44
• @adfriedman, Oh yes, I made a silly typo by writing $2$ instead of $e$. – Sangchul Lee Nov 5 '17 at 1:49

Let $f(a)$ be given by the convergent improper integral

$$f(a)=\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx \tag1$$

Since the integral $\int_0^\infty \frac{x\sin(ax)}{x^2+1}\,dx$ is uniformly convergent for $|a|\ge \delta>0$, we may differentiate under the integral in $(1)$ for $|a|>\delta>0$ to obtain

\begin{align} f'(a)&=-\int_0^\infty \frac{x\sin(ax)}{x^2+1}\,dx\\\\ &=-\int_0^\infty \frac{(x^2+1-1)\sin(ax)}{x(x^2+1)}\,dx\\\\ &=-\int_0^\infty \frac{\sin(ax)}{x}\,dx+\int_0^\infty \frac{\sin(ax)}{x(x^2+1)}\,dx\\\\ &=-\frac{\pi}{2}+\int_0^\infty \frac{\sin(ax)}{x(x^2+1)}\,dx\tag2 \end{align}

Again, since the integral $\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx$ converges uniformly for all $a$, we may differentiate under the integral in $(2)$ to obtain

$$f''(a)=\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx=f(a)\tag 3$$

Solving the second-order ODE in $(3)$ reveals

$$f(a)=C_1 e^{a}+C_2 e^{-a}$$

Using $f(0)=\pi/2$ and $f'(0)=-\pi/2$, we find that $C_1=0$ and $C_2=\frac{\pi}{2}$ and hence $f(a)=\frac{\pi e^{-a}}{2}$. Setting $a=1$ and exploiting even symmetry yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1}\,dx=\frac{\pi}{e}}$$

as expected!


text:
Mirror operated in collaboration with local support

# K-Theory and Homology

## Authors and titles for recent submissions

[ total of 8 entries: 1-8 ]
[ showing up to 25 entries per page: fewer | more ]

### Wed, 20 May 2020

[1]
Title: Gorenstein objects in the n-Trivial extensions of abelian categories
Subjects: K-Theory and Homology (math.KT); Category Theory (math.CT)

### Tue, 19 May 2020

[2]
Title: The stable Adams operations on Hermitian K-theory
Subjects: K-Theory and Homology (math.KT); Algebraic Geometry (math.AG)
[3]
Title: On $K(1)$-local $\mathrm{TR}$
Authors: Akhil Mathew
Subjects: K-Theory and Homology (math.KT); Algebraic Topology (math.AT)
[4]  arXiv:2005.07996 (cross-list from math.FA) [pdf, ps, other]
Title: The Index of Some Mixed Order Dirac-Type Operators and Generalised Dirichlet-Neumann Tensor Fields
Comments: Key Words: Dirac Operator, Picard's Extended Maxwell System, Fredholm Index, Cohomology, Hilbert Complex, Elasticity Complex, Biharmonic Complex, Harmonic Dirichlet and Neumann Tensors Achnowledgements: We cordially thank Walter Zulehner, JKU Linz, for many inspiring and fruitful discussions and kind help related to the Poincare maps necessary for the construction of the Neumann fields
Subjects: Functional Analysis (math.FA); Mathematical Physics (math-ph); Analysis of PDEs (math.AP); Geometric Topology (math.GT); K-Theory and Homology (math.KT)

### Fri, 15 May 2020

[5]
Title: $η$-periodic motivic stable homotopy theory over fields
Comments: 60 pages v2: compare Adams operations with those of Fasel--Haution, add corollary about cellularity, mention some further prior work, add missing flatness hypothesis in Lemma 4.8 and expand on its proof, fix some typos
Subjects: K-Theory and Homology (math.KT); Algebraic Geometry (math.AG); Algebraic Topology (math.AT)

### Thu, 14 May 2020

[6]  arXiv:2005.06141 (cross-list from math.OA) [pdf, ps, other]
Title: A Cuntz-Pimsner Model for the $C^*$-algebra of a Graph of Groups
Subjects: Operator Algebras (math.OA); K-Theory and Homology (math.KT)
[7]  arXiv:2005.06119 (cross-list from math.DG) [pdf, ps, other]
Title: An index theorem for higher orbital integrals
Title: $E_\infty$-cells and general linear groups of infinite fields


text:
# Using texi2dvi inside R to generate a pdf as a temporary file

I need to generate a pdf from a tex file, as a temporary file. However I'm facing an error. I believe it's due to the fact that the function tools::texi2dvi is just a call for the system command with the same name. Since the Rtmp directory just allow R to create files, I can't create the pdf file there.

There's what I'm trying to the, and the error I get:

> tf <- tempfile()
> brew("template.brew", tf)
> tools::texi2dvi(tf)

Error in tools::texi2dvi(tf) :
Running 'texi2dvi' on '/tmp/Rtmp9n4JFU/filea5811e6a32a' failed.
Messages:
/usr/bin/texi2dvi: etex exited with bad status, quitting.

I believe one workaround would be create a new temporary file using tempfile() and ask tools::texi2dvi to write the output on it. However I don't know if the system is allowed to edit a file. And also, I couldn't find a way to set tools::texi2dvi output file.

Does anyone know a workaround?

Thanks!

-
Try using setwd() to change the working directory before calling tex2dvi. –  Andrie Feb 9 '12 at 12:22


text:
## Polar Form of Complex Numbers

### Learning Objectives

By the end of this section, you will be able to:

• Plot complex numbers in the complex plane.
• Find the absolute value of a complex number.
• Write complex numbers in polar form.
• Convert a complex number from polar to rectangular form.
• Find products of complex numbers in polar form.
• Find quotients of complex numbers in polar form.
• Find powers of complex numbers in polar form.
• Find roots of complex numbers in polar form.

“God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.

We first encountered complex numbers in Precalculus I. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem.

## Plotting Complex Numbers in the Complex Plane

Plotting a complex number $a+bi$ is similar to plotting a real number, except that the horizontal axis represents the real part of the number, $a$, and the vertical axis represents the imaginary part of the number, $bi$.

### How To: Given a complex number $a+bi$, plot it in the complex plane.

1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis.
2. Plot the point in the complex plane by moving $a$ units in the horizontal direction and $b$ units in the vertical direction.

### Example 1: Plotting a Complex Number in the Complex Plane

Plot the complex number $2 - 3i$ in the complex plane.

### Solution

From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction.

Figure 1

### Try It 1

Plot the point $1+5i$ in the complex plane.

Solution

## Finding the Absolute Value of a Complex Number

The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or $|z|$. It measures the distance from the origin to a point in the plane. For example, the graph of $z=2+4i$, in Figure 2, shows $|z|$.

Figure 2

### A General Note: Absolute Value of a Complex Number

Given $z=x+yi$, a complex number, the absolute value of $z$ is defined as

$|z|=\sqrt{{x}^{2}+{y}^{2}}$

It is the distance from the origin to the point $\left(x,y\right)$.

Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, $\left(0,\text{ }0\right)$.

### Example 2: Finding the Absolute Value of a Complex Number with a Radical

Find the absolute value of $z=\sqrt{5}-i$.

### Solution

Using the formula, we have

$\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(\sqrt{5}\right)}^{2}+{\left(-1\right)}^{2}}\hfill \\ |z|=\sqrt{5+1}\hfill \\ |z|=\sqrt{6}\hfill \end{array}$

Figure 3

### Try It 2

Find the absolute value of the complex number $z=12 - 5i$.

Solution

### Example 3: Finding the Absolute Value of a Complex Number

Given $z=3 - 4i$, find $|z|$.

### Solution

Using the formula, we have

$\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(3\right)}^{2}+{\left(-4\right)}^{2}}\hfill \\ |z|=\sqrt{9+16}\hfill \\ \begin{array}{l}|z|=\sqrt{25}\\ |z|=5\end{array}\hfill \end{array}$

The absolute value $z$ is 5.

Figure 4

### Try It 3

Given $z=1 - 7i$, find $|z|$.

Solution

## Writing Complex Numbers in Polar Form

The polar form of a complex number expresses a number in terms of an angle $\theta$ and its distance from the origin $r$. Given a complex number in rectangular form expressed as $z=x+yi$, we use the same conversion formulas as we do to write the number in trigonometric form:

$\begin{array}{l}x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}$

We review these relationships in Figure 5.

Figure 5

We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point $\left(x,y\right)$. The modulus, then, is the same as $r$, the radius in polar form. We use $\theta$ to indicate the angle of direction (just as with polar coordinates). Substituting, we have

$\begin{array}{l}z=x+yi\hfill \\ z=r\cos \theta +\left(r\sin \theta \right)i\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}$

### A General Note: Polar Form of a Complex Number

Writing a complex number in polar form involves the following conversion formulas:

$\begin{array}{l}\hfill \\ x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}$

Making a direct substitution, we have

$\begin{array}{l}z=x+yi\hfill \\ z=\left(r\cos \theta \right)+i\left(r\sin \theta \right)\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}$

where $r$ is the modulus and $\theta$ is the argument. We often use the abbreviation $r\text{cis}\theta$ to represent $r\left(\cos \theta +i\sin \theta \right)$.

### Example 4: Expressing a Complex Number Using Polar Coordinates

Express the complex number $4i$ using polar coordinates.

### Solution

On the complex plane, the number $z=4i$ is the same as $z=0+4i$. Writing it in polar form, we have to calculate $r$ first.

$\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{0}^{2}+{4}^{2}}\hfill \\ r=\sqrt{16}\hfill \\ r=4\hfill \end{array}$

Next, we look at $x$. If $x=r\cos \theta$, and $x=0$, then $\theta =\frac{\pi }{2}$. In polar coordinates, the complex number $z=0+4i$ can be written as $z=4\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)$ or $4\text{cis}\left(\frac{\pi }{2}\right)$.

Figure 6

### Try It 4

Express $z=3i$ as $r\text{cis}\theta$ in polar form.

Solution

### Example 5: Finding the Polar Form of a Complex Number

Find the polar form of $-4+4i$.

### Solution

First, find the value of $r$.

$\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(-4\right)}^{2}+\left({4}^{2}\right)}\hfill \\ r=\sqrt{32}\hfill \\ r=4\sqrt{2}\hfill \end{array}$

Find the angle $\theta$ using the formula:

$\begin{array}{l}\cos \theta =\frac{x}{r}\hfill \\ \cos \theta =\frac{-4}{4\sqrt{2}}\hfill \\ \cos \theta =-\frac{1}{\sqrt{2}}\hfill \\ \theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}\hfill \end{array}$

Thus, the solution is $4\sqrt{2}\text{cis}\left(\frac{3\pi }{4}\right)$.

### Try It 5

Write $z=\sqrt{3}+i$ in polar form.

Solution

## Converting a Complex Number from Polar to Rectangular Form

Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given $z=r\left(\cos \theta +i\sin \theta \right)$, first evaluate the trigonometric functions $\cos \theta$ and $\sin \theta$. Then, multiply through by $r$.

### Example 6: Converting from Polar to Rectangular Form

Convert the polar form of the given complex number to rectangular form:

$z=12\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)$

### Solution

We begin by evaluating the trigonometric expressions.

$\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\\\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$

After substitution, the complex number is

$z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)$

We apply the distributive property:

$\begin{array}{l}z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\hfill \\ \text{ }=\left(12\right)\frac{\sqrt{3}}{2}+\left(12\right)\frac{1}{2}i\hfill \\ \text{ }=6\sqrt{3}+6i\hfill \end{array}$

The rectangular form of the given point in complex form is $6\sqrt{3}+6i$.

### Example 7: Finding the Rectangular Form of a Complex Number

Find the rectangular form of the complex number given $r=13$ and $\tan \theta =\frac{5}{12}$.

### Solution

If $\tan \theta =\frac{5}{12}$, and $\tan \theta =\frac{y}{x}$, we first determine $r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{.}$ We then find $\cos \theta =\frac{x}{r}$ and $\sin \theta =\frac{y}{r}$.

$\begin{array}{l}z=13\left(\cos \theta +i\sin \theta \right)\hfill \\ =13\left(\frac{12}{13}+\frac{5}{13}i\right)\hfill \\ =12+5i\hfill \end{array}$

The rectangular form of the given number in complex form is $12+5i$.

### Try It 6

Convert the complex number to rectangular form:

$z=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$

## Finding Products and Quotients of Complex Numbers in Polar Form

Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.

### A General Note: Products of Complex Numbers in Polar Form

If ${z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right)$ and ${z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right)$, then the product of these numbers is given as:

$\begin{array}{l}\hfill \\ \begin{array}{l}{z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\cos \left({\theta }_{1}+{\theta }_{2}\right)+i\sin \left({\theta }_{1}+{\theta }_{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}={r}_{1}{r}_{2}\text{cis}\left({\theta }_{1}+{\theta }_{2}\right)\hfill \end{array}\hfill \end{array}$

Notice that the product calls for multiplying the moduli and adding the angles.

### Example 8: Finding the Product of Two Complex Numbers in Polar Form

Find the product of ${z}_{1}{z}_{2}$, given ${z}_{1}=4\left(\cos \left(80^\circ \right)+i\sin \left(80^\circ \right)\right)$ and ${z}_{2}=2\left(\cos \left(145^\circ \right)+i\sin \left(145^\circ \right)\right)$.

### Solution

Follow the formula

$\begin{array}{l}{z}_{1}{z}_{2}=4\cdot 2\left[\cos \left(80^\circ +145^\circ \right)+i\sin \left(80^\circ +145^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(225^\circ \right)+i\sin \left(225^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\hfill \end{array}$

## Finding Quotients of Complex Numbers in Polar Form

The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.

### A General Note: Quotients of Complex Numbers in Polar Form

If ${z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right)$ and ${z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right)$, then the quotient of these numbers is

$\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\cos \left({\theta }_{1}-{\theta }_{2}\right)+i\sin \left({\theta }_{1}-{\theta }_{2}\right)\right],{z}_{2}\ne 0\\ \frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\text{cis}\left({\theta }_{1}-{\theta }_{2}\right),{z}_{2}\ne 0\end{array}$

Notice that the moduli are divided, and the angles are subtracted.

### How To: Given two complex numbers in polar form, find the quotient.

1. Divide $\frac{{r}_{1}}{{r}_{2}}$.
2. Find ${\theta }_{1}-{\theta }_{2}$.
3. Substitute the results into the formula: $z=r\left(\cos \theta +i\sin \theta \right)$. Replace $r$ with $\frac{{r}_{1}}{{r}_{2}}$, and replace $\theta$ with ${\theta }_{1}-{\theta }_{2}$.
4. Calculate the new trigonometric expressions and multiply through by $r$.

### Example 9: Finding the Quotient of Two Complex Numbers

Find the quotient of ${z}_{1}=2\left(\cos \left(213^\circ \right)+i\sin \left(213^\circ \right)\right)$ and ${z}_{2}=4\left(\cos \left(33^\circ \right)+i\sin \left(33^\circ \right)\right)$.

### Solution

Using the formula, we have

$\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\cos \left(213^\circ -33^\circ \right)+i\sin \left(213^\circ -33^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\cos \left(180^\circ \right)+i\sin \left(180^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[-1+0i\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\hfill \end{array}$

### Try It 7

Find the product and the quotient of ${z}_{1}=2\sqrt{3}\left(\cos \left(150^\circ \right)+i\sin \left(150^\circ \right)\right)$ and ${z}_{2}=2\left(\cos \left(30^\circ \right)+i\sin \left(30^\circ \right)\right)$.

Solution

## Finding Powers and Roots of Complex Numbers in Polar Form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer $n,{z}^{n}$ is found by raising the modulus to the $n\text{th}$ power and multiplying the argument by $n$. It is the standard method used in modern mathematics.

### A General Note: De Moivre’s Theorem

If $z=r\left(\cos \theta +i\sin \theta \right)$ is a complex number, then

$\begin{array}{l}{z}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\\ {z}^{n}={r}^{n}\text{cis}\left(n\theta \right)\end{array}$

where $n$ is a positive integer.

### Example 10: Evaluating an Expression Using De Moivre’s Theorem

Evaluate the expression ${\left(1+i\right)}^{5}$ using De Moivre’s Theorem.

### Solution

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write $\left(1+i\right)$ in polar form. Let us find $r$.

$\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}}\hfill \\ r=\sqrt{2}\hfill \end{array}$

Then we find $\theta$. Using the formula $\tan \theta =\frac{y}{x}$ gives

$\begin{array}{l}\tan \theta =\frac{1}{1}\hfill \\ \tan \theta =1\hfill \\ \theta =\frac{\pi }{4}\hfill \end{array}$

Use De Moivre’s Theorem to evaluate the expression.

$\begin{array}{l}{\left(a+bi\right)}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\hfill \\ {\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\cos \left(5\cdot \frac{\pi }{4}\right)+i\sin \left(5\cdot \frac{\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=-4 - 4i\hfill \end{array}$

## Finding Roots of Complex Numbers in Polar Form

To find the nth root of a complex number in polar form, we use the $n\text{th}$ Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding $n\text{th}$ roots of complex numbers in polar form.

### A General Note: The nth Root Theorem

To find the $n\text{th}$ root of a complex number in polar form, use the formula given as

${z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left[\cos \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)+i\sin \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)\right]$

where $k=0,1,2,3,…,n - 1$. We add $\frac{2k\pi }{n}$ to $\frac{\theta }{n}$ in order to obtain the periodic roots.

### Example 11: Finding the nth Root of a Complex Number

Evaluate the cube roots of $z=8\left(\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right)$.

### Solution

We have

$\begin{array}{l}{z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left[\cos \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)\right]\hfill \\ {z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)\right]\hfill \end{array}$

There will be three roots: $k=0,1,2$. When $k=0$, we have

${z}^{\frac{1}{3}}=2\left(\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\right)$

When $k=1$, we have

$\begin{array}{l}{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)\right]\begin{array}{cccc}& & & \end{array}\text{ Add }\frac{2\left(1\right)\pi }{3}\text{ to each angle.}\hfill \\ {z}^{\frac{1}{3}}=2\left(\cos \left(\frac{8\pi }{9}\right)+i\sin \left(\frac{8\pi }{9}\right)\right)\hfill \end{array}$

When $k=2$, we have

$\begin{array}{ll}{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)\right]\begin{array}{cccc}& & & \end{array}\hfill & \text{Add }\frac{2\left(2\right)\pi }{3}\text{ to each angle.}\hfill \\ {z}^{\frac{1}{3}}=2\left(\cos \left(\frac{14\pi }{9}\right)+i\sin \left(\frac{14\pi }{9}\right)\right)\hfill & \hfill \end{array}$

Remember to find the common denominator to simplify fractions in situations like this one. For $k=1$, the angle simplification is

$\begin{array}{l}\frac{\frac{2\pi }{3}}{3}+\frac{2\left(1\right)\pi }{3}=\frac{2\pi }{3}\left(\frac{1}{3}\right)+\frac{2\left(1\right)\pi }{3}\left(\frac{3}{3}\right)\hfill \\ =\frac{2\pi }{9}+\frac{6\pi }{9}\hfill \\ =\frac{8\pi }{9}\hfill \end{array}$

### Try It 8

Find the four fourth roots of $16\left(\cos \left(120^\circ \right)+i\sin \left(120^\circ \right)\right)$.

Solution

## Key Concepts

• Complex numbers in the form $a+bi$ are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the x-axis as the real axis and the y-axis as the imaginary axis.
• The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: $|z|=\sqrt{{a}^{2}+{b}^{2}}$.
• To write complex numbers in polar form, we use the formulas $x=r\cos \theta ,y=r\sin \theta$, and $r=\sqrt{{x}^{2}+{y}^{2}}$. Then, $z=r\left(\cos \theta +i\sin \theta \right)$.
• To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by $r$.
• To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property.
• To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles.
• To find the power of a complex number ${z}^{n}$, raise $r$ to the power $n$, and multiply $\theta$ by $n$.
• Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent.

## Glossary

argument
the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis
De Moivre’s Theorem
formula used to find the $n\text{th}$ power or nth roots of a complex number; states that, for a positive integer $n,{z}^{n}$ is found by raising the modulus to the $n\text{th}$ power and multiplying the angles by $n$
modulus
the absolute value of a complex number, or the distance from the origin to the point $\left(x,y\right)$; also called the amplitude
polar form of a complex number
a complex number expressed in terms of an angle $\theta$ and its distance from the origin $r$; can be found by using conversion formulas $x=r\cos \theta ,y=r\sin \theta$, and $r=\sqrt{{x}^{2}+{y}^{2}}$

## Section Exercises

1. A complex number is $a+bi$. Explain each part.

2. What does the absolute value of a complex number represent?

3. How is a complex number converted to polar form?

4. How do we find the product of two complex numbers?

5. What is De Moivre’s Theorem and what is it used for?

For the following exercises, find the absolute value of the given complex number.

6. $5+3i$

7. $−7+i$

8. $−3−3i$

9. $\sqrt{2}−6i$

10. $2i$

11. $2.2−3.1i$

For the following exercises, write the complex number in polar form.

12. $2+2i$

13. $8−4i$

14. $−\frac{1}{2}−\frac{1}{2}i$

15. $\sqrt{3}+i$

16. $3i$

For the following exercises, convert the complex number from polar to rectangular form.

17. $z=7\text{cis}\left(\frac{\pi}{6}\right)$

18. $z=2\text{cis}\left(\frac{\pi}{3}\right)$

19. $z=4\text{cis}\left(\frac{7\pi}{6}\right)$

20. $z=7\text{cis}\left(25^{\circ}\right)$

21. $z=3\text{cis}\left(240^{\circ}\right)$

22. $z=\sqrt{2}\text{cis}\left(100^{\circ}\right)$

For the following exercises, find z1z2 in polar form.

23. $z_{1}=2\sqrt{3}\text{cis}\left(116^{\circ}\right)\text{; }\left(118^{\circ}\right)$

24. $z_{1}=\sqrt{2}\text{cis}\left(205^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right)$

25. $z_{1}=3\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right)$

26. $z_{1}=3\text{cis}\left(\frac{5\pi}{4}\right)\text{; }z_{2}=5\text{cis}\left(\frac{\pi}{6}\right)$

27. $z_{1}=\sqrt{5}\text{cis}\left(\frac{5\pi}{8}\right)\text{; }z_{2}=\sqrt{15}\text{cis}\left(\frac{\pi}{12}\right)$

28. $z_{1}=4\text{cis}\left(\frac{\pi}{2}\right)\text{; }z_{2}=2\text{cis}\left(\frac{\pi}{4}\right)$

For the following exercises, find $\frac{z_{1}}{z_{2}}$ in polar form.

29. $z_{1}=21\text{cis}\left(135^{\circ}\right)\text{; }z_{2}=3\text{cis}\left(65^{\circ}\right)$

30. $z_{1}=\sqrt{2}\text{cis}\left(90^{\circ}\right)\text{; }z_{2}=2\text{cis}\left(60^{\circ}\right)$

31. $z_{1}=15\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=3\text{cis}\left(40^{\circ}\right)$

32. $z_{1}=6\text{cis}\left(\frac{\pi}{3}\right)\text{; }z_{2}=2\text{cis}\left(\frac{\pi}{4}\right)$

33. $z_{1}=5\sqrt{2}\text{cis}\left(\pi\right)\text{; }z_{2}=\sqrt{2}\text{cis}\left(\frac{2\pi}{3}\right)$

34. $z_{1}=2\text{cis}\left(\frac{3\pi}{5}\right)\text{; }z_{2}=3\text{cis}\left(\frac{\pi}{4}\right)$

For the following exercises, find the powers of each complex number in polar form.

35. Find $z^{3}$ when $z=5\text{cis}\left(45^{\circ}\right)$.

36. Find $z^{4}$ when $z=2\text{cis}\left(70^{\circ}\right)$.

37. Find $z^{2}$ when $z=3\text{cis}\left(120^{\circ}\right)$.

38. Find $z^{2}$ when $z=4\text{cis}\left(\frac{\pi}{4}\right)$

39. Find $z^{4}$ when $z=\text{cis}\left(\frac{3\pi}{16}\right)$.

40. Find $z^{3}$ when $z=3\text{cis}\left(\frac{5\pi}{3}\right)$.

For the following exercises, evaluate each root.

41. Evaluate the cube root of z when $z=27\text{cis}\left(240^{\circ}\right)$.

42. Evaluate the square root of z when $z=16\text{cis}\left(100^{\circ}\right)$.

43. Evaluate the cube root of z when $z=32\text{cis}\left(\frac{2\pi}{3}\right)$.

44. Evaluate the square root of z when $z=32\text{cis}\left(\pi\right)$.

45. Evaluate the cube root of z when $z=8\text{cis}\left(\frac{7\pi}{4}\right)$.

For the following exercises, plot the complex number in the complex plane.

46. $2+4i$

47. $−3−3i$

48. $5−4i$

49. $−1−5i$

50. $3+2i$

51. $2i$

52. $−4$

53. $6−2i$

54. $−2+i$

55. $1−4i$

For the following exercises, find all answers rounded to the nearest hundredth.

56. Use the rectangular to polar feature on the graphing calculator to change $5+5i$ to polar form.

57. Use the rectangular to polar feature on the graphing calculator to change $3−2i$

58. Use the rectangular to polar feature on the graphing calculator to change $−3−8i$

59. Use the polar to rectangular feature on the graphing calculator to change $4\text{cis}\left(120^{\circ}\right)$ to rectangular form.

60. Use the polar to rectangular feature on the graphing calculator to change $2\text{cis}\left(45^{\circ}\right)$ to rectangular form.

61. Use the polar to rectangular feature on the graphing calculator to change $5\text{cis}\left(210^{\circ}\right)$ to rectangular form.


text:
Heavy-traffic approximations for linear networks operating under$\alpha$-fair bandwidth-sharing policies

P.M.D. Lieshout, S.C. Borst, M.R.H. Mandjes

Research output: Chapter in Book/Report/Conference proceedingConference contributionAcademicpeer-review

Abstract

We consider the flow-level performance of a linear network supporting elastic traffic, where the service capacity is shared among the various classes of users according to a weighted alpha-fair policy. Assuming Poisson arrivals and exponentially distributed service requirements for each class, the dynamics of the user population may be described by a Markov process. While valuable stability results have been established for the family of alpha-fair policies, the distribution of the number of active users has remained intractable in all but a few special cases. In order to gain further insight in the flow-level performance in more general scenarios, we develop approximations for the mean number of users based on the assumption that one or two of the nodes experience heavy-traffic conditions.In case of just a single 'bottleneck' node, we exploit the fact that this node approximately behaves as a two-class Discriminatory Processor-Sharing model. In the case that there are two nodes critically loaded, we rely on the observation that the joint workload process at these nodes is asymptotically independent of the fairness coefficient alpha, provided all classes have equal weights. In particular, the distribution of the joint workload process is roughly equal to that for an unweighted Proportional Fair policy, which is exactly known. In both cases, the numbers of users at non-bottleneck nodes can be approximated by that in an M/M/1 queue with reduced service capacity. Extensive numerical experiments indicate that the resulting approximations tend to be reasonably accurate across a wide range of parameters, even at relatively moderate load values. The approximations for the mean number of users also provide useful estimates for the mean transfer delays and user throughputs.
Original language English Proceedings First International Conference on Performance Evaluation Methodologies and Tools (ValueTools, Pisa, Italy, October 11-13, 2006) L. Lenzini, R. Cruz New York NY, USA Association for Computing Machinery, Inc 1-59593-504-5 https://doi.org/10.1145/1190095.1190154 Published - 2006

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text:
# [metapost] misprint in mpman

Sergey Metelev metelev.sv at gmail.com
Fri Mar 14 18:15:06 CET 2008

I don't know where to report an misprint in mpman.pdf so I decide to
post it here. I look at the last version of mpman.tex also, at the
http://tug.org/metapost/src/manual/ and it is still there.

It is a line 976 of mpman.tex source, where one can read:
&&\hbox{\verb|x3+y3=x6+y6=1.1in;|}

and later, in the line 981:
z3=(.3in,.6in); z6=(-.3in,1.2in);

but (for example) -.3+1.2 not equal to 1.1 (I even tried it in Metapost :)))

--
С уважением,
Sergey                          mailto:metelev.sv at gmail.com


text:
# 2 Epsilon-Delta Definition

## Limits of Functions

We can easily understand the limits of functions if we clearly understand the limits of sequences. The intuitive picture of the limit of a sequence is that

the value of $a_n$ infinitely approaches a constant value $\alpha$ as the index $n$ increases.

As was explained in the previous chapter, this picture is precisely characterized by the following definition:

For any $\varepsilon > 0$, there exists a natural number $N(\varepsilon)$ such that for all $n \geq N(\varepsilon)$, we have $| \, a_n \, - \, \alpha \, | < \varepsilon$.

On the other hand, the intuitive picture of the limit of a function is that

the value of $f(x)$ infinitely approaches a constant value $\alpha$  as the value of $x$ with $x \neq a$ approaches $a$.

If we regard the variable $x$ as corresponding to the index $n$, we find that this picture of the limit of a function is essentially the same as that of a sequence.

The difference between these pictures is in the phrase “as the value of $x$ with $x \neq a$ approaches $a$.”

In the case of sequences, “as the index $n$ increases” is expressed by the phrase “for all $n \geq N(\varepsilon)$” since the value of $N(\varepsilon)$ is generally large.

In contrast, in the case of functions, “as the value of $x$ with $x \neq a$ approaches $a$” is expressed by:
$$0 < | \, x \, - \, a \, | < \delta(\varepsilon),$$
where $\delta$ denotes the Greek letter called “delta.”

Notice that $x$ is close to $a$ if $\delta(\varepsilon)$ is small. In fact, recalling that $| \, x \, - \, a \, |$ means the distance between $x$ and $a$, the inequality $| \, x \, - \, a \, | < \delta(\varepsilon)$ implies that $x$ is in a ball (interval) with radius $\delta(\varepsilon)$ and center $a$. In addition, we have $x \neq a$ because of $0 < | \, x \, - \, a \, |$.

Thus, we arrive at the following definition of the limit of a function:

For any $\varepsilon > 0$, there exists a positive real number $\delta(\varepsilon)$ such that for all $x$ satisfying $0 < | \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, \alpha \, | < \varepsilon$.

When $f(x)$ satisfies this condition, we say that $f(x)$ converges to $\alpha$ as $x \to a$, which is denoted by $\displaystyle\lim_{ x \to a} f(x) = \alpha$. Moreover, $\alpha$ is called the limit value of $f(x)$ as $x \to a$. This is the definition of the limit of a function by the epsilon-delta argument. The key point of this definition is that $\delta$ is a function of $\varepsilon$.

The existence of the function $\delta(\varepsilon)$ guarantees the convergence of a function in the same way as that of a sequence.

I understand that $\delta$ is a function of $\varepsilon$, and that the existence of $\delta(\varepsilon)$ guarantees the convergence of a function. However, I wonder whether the value of $\delta(\varepsilon)$ decreases as $\varepsilon$ decreases. In the case of sequences, the value of $N(\varepsilon)$ increases as $\varepsilon$ decreases.

In general, the value of $\delta(\varepsilon)$ decreases as $\varepsilon$ decreases. I think that the following figure is helpful for understanding this property intuitively. Suppose that the value of $\varepsilon$ in Figure 1 is smaller than that in Figure 2. Then, the value of $\delta(\varepsilon)$ in Figure 1 is smaller than that in Figure 2.

In order to understand the definition of the limit of a function, we will show $\displaystyle\lim_{x \to 2} f(x) = 4$, where $f(x) = x^2$. That is,

for any $\varepsilon > 0$, there exists a positive real number $\delta(\varepsilon)$ such that for all $x$ satisfying $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, 4 \, | < \varepsilon$.

Note the last inequality $| \, f(x) \, - \, 4 \, | < \varepsilon$ in the above condition. Since
$$| \, f(x) \, - \, 4 \, | = | \, x^2 \, - \, 4 \, | = | \, x \, - \, 2 \, | | \, x \, + \, 2 \, |,$$
we seek the range of $x$ where $| \, x \, - \, 2 \, | | \, x \, + \, 2 \, |< \varepsilon$ holds. We consider that the value of $| x + 2 |$ can be approximated by $4$ because we take $x$ in a small neighborhood of $2$ for examining the limit of $f(x)$ as $x \to 2$. Therefore, we assume that $$| x + 2 | < 5$$ holds, where we add $1$ to $4$ with a margin. Then, for all $x$ satisfying $| x \, - 2 | < \displaystyle\frac{\varepsilon}{5}$, we have $$|f(x) \, - \, 4| = |x+2| |x-2| < 5 | x-2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon$$
The above argument holds under the assumption $| x + 2 | < 5$. This inequality is equivalent to $-5 < x + 2 < 5$; i.e., $-7 < x < 3$. Hence, we can obtain $| x + 2 | < 5$ as long as we restrict the range of $x$ to satisfy $1 < x < 3$; i.e., $| x \, - 2 | < 1$. Thus, if $$| x \, - 2 | < 1 \ \ \ {\rm and} \ \ \ | x \, - 2 | < \displaystyle\frac{\varepsilon}{5},$$ then, noting $|x +2 | < 5$ by $|x -2 | <1$, we have $$|f(x) \, - \, 4| = |x+2| |x-2| < 5 | x-2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$ Now, we define $\delta(\varepsilon)$ as follows:
$$\delta(\varepsilon) = \min( \frac{\varepsilon}{5}, \ 1 ),$$
where $\min(x, y)$ denotes the minimum value of $x$ and $y$. We note that $\delta(\varepsilon) \leq \displaystyle\frac{\varepsilon}{5}$ and $\delta(\varepsilon) \leq 1$ hold.

Let $x$ satisfy $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$. Then, noting that $\delta(\varepsilon) \leq 1$, we have $| x \, - 2 | < 1$, which leads to $| x + 2 | < 5$. Therefore, we have
$$|f(x) \, - \, 4| = |x + 2| |x - 2| < 5 | x -2| < 5 \, \delta(\varepsilon) \leq 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$
Thus, for any $\varepsilon > 0$, there exists a real positive number $\delta(\varepsilon) = \min( \displaystyle\frac{\varepsilon}{5}, 1 )$ such that for all $x$ satisfying $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, 4 \, | < \varepsilon$.

Exercise 2  Let $f(x) = \sqrt{x}$. Show that $\displaystyle\lim_{ x \to 1 }f(x) = 1$.

## Continuity of Functions

When the graph of a function is neither cut off nor has a hole, we say that the function is continuous. To be precise, we say that a function $f(x)$ is continuous at $x = a$ if

(A) the value of $f(x)$ is defined at $x = a$,

and

(B) the limit value $\displaystyle\lim_{x \to a} f(x)$ exists, and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds.

For example, the function presented in Figure 1 is continuous at $x = a$ because the limit value $\displaystyle\lim_{x \to a} f(x)$ exists and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds; the graph of the function is neither cut off nor has a hole at $x = a$.

In contrast, the functions presented in Figures 2 and 3 are not continuous at $x = a$. In the case of Figure 2, the limit value $\displaystyle\lim_{x \to a} f(x)$ does not exist; the graph of the function is cut off at $x = a$. In the case of Figure 3, $\displaystyle\lim_{x \to a} f(x)$ exists, but $\displaystyle\lim_{x \to a} f(x) \neq f(a)$ does not hold; the graph of the function has a hole at $x = a$.

By using the definition of the limit of a function, the above condition B that the limit value $\displaystyle\lim_{x \to a} f(x)$ exists and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds is given by:

For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $0 < | \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

However, noting that the value of $f(x)$ is defined at $x = a$, the last inequality in this statement $| \, f(x) \, - \, f(a) \, | < \varepsilon$ holds at $x = a$. In fact, we have $| \, f(a) \, - \, f(a) \, | = 0 < \varepsilon$. Therefore, the above statement can be slightly modified as follows:

For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

This is exactly the condition that $f(x)$ is continuous at $x = a$ when the value of $f(x)$ is defined at $x = a$.

In order to understand the definition of the continuity of a function, we will show that $f(x) = x^2$ is continuous at $x = a$. Since it is clear that the value of $f(x)$ is defined for all $x$, our goal is to show the following:

For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

As was explained before, noting the last inequality in the above assertion $| \, f(x) \, - \, f(a) \, | < \varepsilon$, we seek the range of $x$ where this inequality holds. Since
$$| \, f(x) \, - \, f(a) \, | = | \, x^2 \, - \, a^2 \, | = | \, x \, - \, a \,| | \, x \, + \, a \,|,$$
we must seek the range of $x$ where $|x \, - \, a| | x + a | < \varepsilon$ holds.

When $x$ is close to $a$, we can consider that $| x + a |$ is approximated by $2|a|$. Therefore, noting that $x$ is close to $a$, we assume that $$| x + a | < 2|a| + 1$$
holds, where we add $1$ to $2|a|$ with a margin. Then, for all $x$ satisfying $| x \, - \, a | < \displaystyle\frac{\varepsilon}{2|a| + 1}$, we have
$$| \, f(x) \, - \, f(a) \, | = | \, x \, - \, a \,| | \, x \, + \, a \, | < \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1 ) = \varepsilon.$$
This result is obtained under the assumption $| x + a | < 2|a| + 1$. However, we have not yet specified the range of $x$ where this inequality holds.

Suppose $| x \, - \, a | < \rho$, where $\rho$ will be specified later. Then, we have $|x| < |a| + \rho$ because $x$ is in a ball (interval) with radius $\rho$ and center $a$. Therefore, we have $$\begin{array}{l} | x + a | \leq |x| + |a| \\[2ex] \ \ \ \ < |a| + \rho + |a| \\[2ex] \ \ \ \ = 2|a| + \rho \end{array}$$ Hence, when $\rho \leq 1$, we have
$$| x + a | < 2|a| + 1.$$ Consequently, noting the case that $\rho = 1$, we see that $| x \, - \, a | < 1$ implies $| x + a | < 2|a| + 1$.

Thus, if $$| x \, - a | < 1 \ \ \ {\rm and} \ \ \ | x \, - a | < \displaystyle\frac{\varepsilon}{2|a| + 1},$$ then, noting $| x + a | < 2|a| + 1$ by $| x - a | <1$, we have $$| \, f(x) \, - \, f(a) \, | = | \, x \, - \, a \,| | \, x \, + \, a \, | < \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1 ) = \varepsilon.$$

Now, we define
$$\delta(\varepsilon) = \min\left( \frac{\varepsilon}{2|a| + 1}, \ 1 \right).$$
Notice that $| x \, - \, a | < \delta(\varepsilon)$ implies $| x \, - \, a | < 1$ and  $| x \, - \, a | < \displaystyle\frac{\varepsilon}{2|a|+1}$ because $\delta(\varepsilon) \leq 1$ and $\delta(\varepsilon) \leq \displaystyle\frac{\varepsilon}{2|a|+1}$.

Therefore, for all $x$ satisfying $| x \, - \, a | < \delta(\varepsilon)$, we have $| x + a | < 2|a| + 1$ by $| x \, - \, a | < 1$, which leads to $$\begin{array}{l} | f(x) \, - \, f(a) | = | x \, - \, a| | x \, + \, a | \\[2ex] \ \ \ \ < | x \, - \, a| (2|a| + 1) \\[2ex] \ \ \ \ < \delta(\varepsilon) (2|a|+ 1) \\[2ex] \ \ \ \ = \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1) = \varepsilon \end{array}$$ Hence, we see that $f(x) = x^2$ is continuous at $x = a$.

From the definition of the continuity of a function, I understand that $\delta$ is a function of $\varepsilon$ because $\delta$ depends on $\varepsilon$. However, in the above example, we define $\delta = \min\left( \displaystyle\frac{\varepsilon}{2|a| + 1}, \ 1 \right)$. Therefore, I think that $\delta$ can be regarded as a function of $a$ as well as of $\varepsilon$.

That’s a very good point. You are completely right.

It is usual that a function is defined on a set on the real line ${\bf R}$. This set is called the “domain.” In many cases, the domain of a function is given by an interval on ${\bf R}$. For example, when the domain of a function $f(x)$ is the interval $I = [a, \, b]$, the interval $I$ is given by the set
$$[a, \, b] = \{ \, x \, | \, a \leq x \leq b \, \}.$$
In this case, the value of the function $f(x)$ is defined for $x$ satisfying $a \leq x \leq b$. We can define a function on an interval that does not include its endpoints, such as
$$(a, \, b) = \{ \, x \, | \, a < x < b \, \}.$$
Moreover, the real line denoted by ${\bf R}$ can be regarded as an interval, and is often expressed by ${\bf R } = (-\infty, \, \infty)$.

Suppose that the domain of a function $f(x)$ is an interval $I$. Then, we say that $f(x)$ is continuous on $I$ if for any $a \in I$, $f(x)$ is continuous at $x = a$. That is, the definition that $f(x)$ is continuous on $I$ is as follows:

For any $a \in I$ and for any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon ; a) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon ; a)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

It should be noted that $\delta$ is a function of $\varepsilon$ and $a$ in this definition. For example, the function $f(x) = x^2$ is continuous on ${\bf R}$. In fact, as was explained above, for any $a \in {\bf R}$ and for any $\varepsilon > 0$, there exists a real positive number $\delta(\varepsilon ; a)$ defined by
$$\delta(\varepsilon ; a) = \min\left( \displaystyle\frac{\varepsilon}{2|a| + 1}, \ 1 \right)$$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon ; a)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.

I understand that when a function $f (x)$ is continuous on an interval $I$, $\delta$ is a function of $a \in I$ and $\varepsilon > 0$. However, I wonder whether there exists a function such that $\delta$ is a function of only $\varepsilon$?

Indeed. When $\delta$ is a function of only $\varepsilon$, we say that $f(x)$ is uniformly continuous on $I$. As seen in the following exercise, there are many uniformly continuous functions. The uniform continuity of functions often plays an important role in advanced mathematical analysis.

Exercise 3　Show that $f(x) = x^2 -x$ is continuous at any $a \in {\bf R}$.

Exercise 4   Show that $f(x) = x^2$ is uniformly continuous on an interval $[a, \, b]$, where $a$ and $b$ are finite real numbers.

## Concluding Remarks

We come to the end of our explanation for the basis of the epsilon-delta argument. The idea of the epsilon-delta argument is to formulate our naive intuition for the limits of functions as the existence problem of a function $\delta = \delta(\varepsilon)$. Although this idea is quite simple, the epsilon-delta argument requires some technical training in using logic symbols, estimating inequalities, negating statements, and so on. We will be happy if this website can help calculus beginners to understand the basic underlying ideas and to overcome the technical barriers to using the epsilon-delta argument.

Can you explain the underlying basic idea of the epsilon-delta definition of the limits of functions (the continuity of functions) through the parable of an airplane approaching an airport in the previous chapter? I think that it is a good exercise to confirm your understanding for the epsilon-delta definition of limits of functions.

epsilon-delta
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text:
, 26.02.2020 16:10, jack626259

# A letter is chosen at random from the word probability find the probability that it is a vowel

### Other questions on the subject: Math

Math, 20.08.2019 01:00, yachna18
Teh least interger by which 3696 be multiplied to make it a perfect cube is ?
Probability of selecting 2 balls such that none of them is blue. here are total 7 balls. 5 balls are not blue. select 2 balls from those 5 balls (these are the favourable cases) select 2 balls from those 7 balls (these are total cases) p = 5c2/7c2 p= 10/21


text:
Validity of duplicateCorrelation after ComBat in this scenario
1
0
Entering edit mode
@vizio12341337-22616
Last seen 18 months ago

Hi,

I did an analysis a few years ago using ComBat adjusted microarray values fed into limma and was wondering if it's acceptable (Maybe not ideal, but informative). We've been able to validate many of these genes since. I've since read that when using limma with a ComBat modified matrix, the batch covariate should still be used in the subsequent limma design. I did not do that here however, duplicateCorrelation was used on SampleID which is nested within Batch. Does this possibly serve a similar purpose without also putting Batch in design on top of duplicateCorrelation of SampleID? My code follows:

> SampleTable
Sample SampleID  Batch Treatment
1   A_Tx1        A Batch1       Tx1
5   B_Tx1        B Batch1       Tx1
9   C_Tx1        C Batch2       Tx1
13  D_Tx1        D Batch2       Tx1
2   A_Tx2        A Batch1       Tx2
6   B_Tx2        B Batch1       Tx2
10  C_Tx2        C Batch2       Tx2
14  D_Tx2        D Batch2       Tx2
3   A_Tx3        A Batch1       Tx3
7   B_Tx3        B Batch1       Tx3
11  C_Tx3        C Batch2       Tx3
15  D_Tx3        D Batch2       Tx3
4   A_Tx4        A Batch1       Tx4
8   B_Tx4        B Batch1       Tx4
12  C_Tx4        C Batch2       Tx4
16  D_Tx4        D Batch2       Tx4

modcombat = model.matrix(~1, data=SampleTable) ### Treatment could be included here alternatively
combat_edata = ComBat(dat=edata, batch=SampleTable$Batch, mod=modcombat, par.prior=TRUE, prior.plots=FALSE) design <- model.matrix(~0+SampleTable$Treatment)
corfit <- duplicateCorrelation(combat_edata,design,block=SampleTable$SampleID) fitV <- lmFit(combat_edata, design, block=SampleTable$SampleID, cor=corfit\$consensus.correlation)
fitV <- eBayes(fitV)
fitV2 <- eBayes(contrasts.fit(fitV, contrasts = c(1,0,-1,0)))
sig <- topTable(fitV2, coef = 1 , number = 10000000, p.value = 0.05)

Running it with Batch included in the design in addition to duplicateCorrelation on SampleID produces similar gene lists with the provided code.

Venn Diagram: https://ibb.co/fDLQ7MP

Is this a non-incorrect approach? Thanks!

limma ComBat SVA • 180 views
1
Entering edit mode
@gordon-smyth
Last seen 59 minutes ago
WEHI, Melbourne, Australia

You already know that inputing ComBat adjusted values into limma as if they were unadjusted is not correct. For your experiment, the effect would be relatively small. Given your code, limma will think that the residual df is 12 when it really should be closer to 11. That will tend to make all the estimated standard errors about 4% smaller than they should be and the t-statistics about 4% larger than they should be. You will therefore get a bit too much DE.

The use of duplicateCorrelation ameliorates the problem somewhat but, no, it does not make your model equivalent to including Batch in the design matrix.

I don't quite understand what the ultimate problem is though. You say that get a similar gene list even if you do include Batch in the design, so why not do that?

As another alternative, I would consider defining:

design <- mode.matrix(~SampleID + Treatment)

and doing away with both ComBat and duplicateCorrelation.

0
Entering edit mode

Hi Dr. Smyth,

Thanks so much for the response. This is what I expected, but wanted to double check. Luckily, the genes that are sensitive to adding Batch to the design are not ones we're particularly interested in for this study. Thanks again!


text:
# Getting 5 or 7 and returning the opposite? [closed]

I get 5 or 7 and if i get 5 i need to return 7 if i get 7 i need to return 5.

i need to do this in 1 mathematical formula.

I have those:

12 - x

35 / x

There are more solutions ?

Example in code:

public static int Transform(int x)
{
return (12-x);
//return (35/x);
}

-

## closed as not a real question by Aryabhata, Guess who it is., Zev Chonoles, Henry, Ross MillikanMay 11 '11 at 17:18

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

There are infinitely many more... –  Aryabhata May 11 '11 at 15:51
Example: $\sqrt{74 - x^2}$. This can be easily generalized. –  Aryabhata May 11 '11 at 15:53
Why are you unsatisfied with the simple ones you already have? –  Guess who it is. May 11 '11 at 15:57
@Danpe, this isn't a puzzle site. –  quanta May 11 '11 at 16:02
If $h$ is an invertible function we can define $f(x) = h^{-1}(h(5) + h(7) - h(x))$. This $f$ has the property that you want. This covers the two methods you have (with $h(x)=x$ and $h(x)=\log(x)$, respectively, and quite a few more. –  Thomas Andrews May 11 '11 at 16:08

A way to verify this answer: $2 = 5\ \text{XOR}\ 7$. (So basically return x ^ (5 ^ 7);) –  Aryabhata May 11 '11 at 16:25


text:
# To find: The equivalent polar equation for the given rectangular-coordinate equation.Given:\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
${x}^{2}+{y}^{2}+8x=0$

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Concept used: Conversion formula for coordinate systems are given as
a)
From polar to rectangular:
$x=r\mathrm{cos}\theta$
$y=r\mathrm{sin}\theta$
b)
From rectangular to polar:
$r=±\sqrt{{x}^{2}+{y}^{2}}$
$\mathrm{cos}\theta =\frac{x}{r};\mathrm{sin}\theta =\frac{y}{r};\mathrm{tan}\theta =\frac{x}{y}$
Calculation:
Converting into equivalent polar equation
${x}^{2}+{y}^{2}+8x=0$
Put $\left\{x\right\}=\left\{r\right\}\mathrm{cos}\left\{\theta \right\},\left\{y\right\}=\left\{r\right\}\mathrm{sin}\left\{\theta \right\},$
$x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta$
$⇒{\left(r\mathrm{cos}\theta \right)}^{2}+{\left(r\mathrm{sin}\theta \right)}^{2}+8r\mathrm{cos}\theta =0$
$⇒{r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta +8r\mathrm{cos}\theta =0$
$⇒{r}^{2}\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)+8r\mathrm{cos}\theta =0\left\{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1\right\}$
$⇒{r}^{2}+8r\mathrm{cos}\theta =0$
$⇒r+8\mathrm{cos}\theta =0$
Thus, desired equivalent polar equation would be $r+8\mathrm{cos}\theta =0$


text:
#### Vol. 4, No. 1, 2006

 Recent Issues Volume 18 Volume 16 Volume 15 Volume 14 Volume 13 Volume 12 Volume 11 Volume 10 Volume 9 Volume 8 Volume 6+7 Volume 5 Volume 4 Volume 3 Volume 2 Volume 1
 The Journal About the Journal Editorial Board Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN (electronic): 2640-7345 ISSN (print): 2640-7337 Author Index To Appear Other MSP Journals
Large caps with free pairs in dimensions five and six

### Jeffrey Bryan Farr and Petr Lisonek

Vol. 4 (2006), No. 1, 69–88
##### Abstract

A cap in PG$\left(N,q\right)$ is said to have a free pair of points if any plane containing that pair contains at most one other point from the cap. In an earlier paper we determined the largest size of caps with free pairs for $N=3$ and $4$. In this paper we use product constructions to prove similar results in dimensions $5$ and $6$ that are asymptotically as large as possible. If $q>2$ is even, we determine exactly the largest size of a cap in PG$\left(5,q\right)$ with a free pair. In PG$\left(5,3\right)$ we give constructions of a maximal size $42$-cap having a free pair and of the complete $48$-cap that contains it. Additionally, we give some sporadic examples in higher dimensions.

##### Keywords
cap, free pair, Galois space
Primary: 51E22