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# American Institute of Mathematical Sciences

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December  2014, 19(10): 3057-3085. doi: 10.3934/dcdsb.2014.19.3057

## Inside dynamics of solutions of integro-differential equations

 1 INRA, UR 546 Biostatistique et Processus Spatiaux (BioSP), F-84914 Avignon, France, France, France, France

Received  July 2013 Revised  January 2014 Published  October 2014

In this paper, we investigate the inside dynamics of the positive solutions of integro-differential equations \begin{equation*} \partial_t u(t,x)= (J\star u)(t,x) -u(t,x) + f(u(t,x)), \ t>0 \hbox{ and } x\in\mathbb{R}, \end{equation*} with both thin-tailed and fat-tailed dispersal kernels $J$ and a monostable reaction term $f.$ The notion of inside dynamics has been introduced to characterize traveling waves of some reaction-diffusion equations [23]. Assuming that the solution is made of several fractions $\upsilon^i\ge 0$ ($i\in I \subset \mathbb{N}$), its inside dynamics is given by the spatio-temporal evolution of $\upsilon^i$. According to this dynamics, the traveling waves can be classified in two categories: pushed and pulled waves. For thin-tailed kernels, we observe no qualitative differences between the traveling waves of the above integro-differential equations and the traveling waves of the classical reaction-diffusion equations. In particular, in the KPP case ($f(u)\leq f'(0)u$ for all $u\in(0,1)$) we prove that all the traveling waves are pulled. On the other hand for fat-tailed kernels, the integro-differential equations do not admit any traveling waves. Therefore, to analyse the inside dynamics of a solution in this case, we introduce new notions of pulled and pushed solutions. Within this new framework, we provide analytical and numerical results showing that the solutions of integro-differential equations involving a fat-tailed dispersal kernel are pushed. Our results have applications in population genetics. They show that the existence of long distance dispersal events during a colonization tend to preserve the genetic diversity.
Citation: Olivier Bonnefon, Jérôme Coville, Jimmy Garnier, Lionel Roques. Inside dynamics of solutions of integro-differential equations. Discrete & Continuous Dynamical Systems - B, 2014, 19 (10) : 3057-3085. doi: 10.3934/dcdsb.2014.19.3057
##### References:

show all references

##### References:
 [1] Liang Zhang, Bingtuan Li. Traveling wave solutions in an integro-differential competition model. Discrete & Continuous Dynamical Systems - B, 2012, 17 (1) : 417-428. doi: 10.3934/dcdsb.2012.17.417 [2] Narcisa Apreutesei, Nikolai Bessonov, Vitaly Volpert, Vitali Vougalter. Spatial structures and generalized travelling waves for an integro-differential equation. Discrete & Continuous Dynamical Systems - B, 2010, 13 (3) : 537-557. doi: 10.3934/dcdsb.2010.13.537 [3] Samir K. Bhowmik, Dugald B. Duncan, Michael Grinfeld, Gabriel J. Lord. Finite to infinite steady state solutions, bifurcations of an integro-differential equation. Discrete & Continuous Dynamical Systems - B, 2011, 16 (1) : 57-71. doi: 10.3934/dcdsb.2011.16.57 [4] Karel Hasík, Jana Kopfová, Petra Nábělková, Sergei Trofimchuk. On pushed wavefronts of monostable equation with unimodal delayed reaction. Discrete & Continuous Dynamical Systems, 2021, 41 (12) : 5979-6000. doi: 10.3934/dcds.2021103 [5] Narcisa Apreutesei, Arnaud Ducrot, Vitaly Volpert. Travelling waves for integro-differential equations in population dynamics. Discrete & Continuous Dynamical Systems - B, 2009, 11 (3) : 541-561. doi: 10.3934/dcdsb.2009.11.541 [6] Giuseppe Maria Coclite, Mario Michele Coclite. Positive solutions of an integro-differential equation in all space with singular nonlinear term. Discrete & Continuous Dynamical Systems, 2008, 22 (4) : 885-907. doi: 10.3934/dcds.2008.22.885 [7] Elena Trofimchuk, Manuel Pinto, Sergei Trofimchuk. Pushed traveling fronts in monostable equations with monotone delayed reaction. Discrete & Continuous Dynamical Systems, 2013, 33 (5) : 2169-2187. doi: 10.3934/dcds.2013.33.2169 [8] Rui Huang, Ming Mei, Yong Wang. Planar traveling waves for nonlocal dispersion equation with monostable nonlinearity. Discrete & Continuous Dynamical Systems, 2012, 32 (10) : 3621-3649. doi: 10.3934/dcds.2012.32.3621 [9] Walter Allegretto, John R. Cannon, Yanping Lin. A parabolic integro-differential equation arising from thermoelastic contact. Discrete & Continuous Dynamical Systems, 1997, 3 (2) : 217-234. doi: 10.3934/dcds.1997.3.217 [10] Shihchung Chiang. Numerical optimal unbounded control with a singular integro-differential equation as a constraint. Conference Publications, 2013, 2013 (special) : 129-137. doi: 10.3934/proc.2013.2013.129 [11] Frederic Abergel, Remi Tachet. A nonlinear partial integro-differential equation from mathematical finance. Discrete & Continuous Dynamical Systems, 2010, 27 (3) : 907-917. doi: 10.3934/dcds.2010.27.907 [12] Yi Cao, Jianhua Wu, Lihe Wang. Fundamental solutions of a class of homogeneous integro-differential elliptic equations. Discrete & Continuous Dynamical Systems, 2019, 39 (3) : 1237-1256. doi: 10.3934/dcds.2019053 [13] Yubo Chen, Wan Zhuang. The extreme solutions of PBVP for integro-differential equations with caratheodory functions. Conference Publications, 1998, 1998 (Special) : 160-166. doi: 10.3934/proc.1998.1998.160 [14] Tianling Jin, Jingang Xiong. Schauder estimates for solutions of linear parabolic integro-differential equations. Discrete & Continuous Dynamical Systems, 2015, 35 (12) : 5977-5998. doi: 10.3934/dcds.2015.35.5977 [15] Patricio Felmer, Ying Wang. Qualitative properties of positive solutions for mixed integro-differential equations. Discrete & Continuous Dynamical Systems, 2019, 39 (1) : 369-393. doi: 10.3934/dcds.2019015 [16] Michel Chipot, Senoussi Guesmia. On a class of integro-differential problems. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1249-1262. doi: 10.3934/cpaa.2010.9.1249 [17] Elena Bonetti, Elisabetta Rocca. Global existence and long-time behaviour for a singular integro-differential phase-field system. Communications on Pure & Applied Analysis, 2007, 6 (2) : 367-387. doi: 10.3934/cpaa.2007.6.367 [18] Wan-Tong Li, Wen-Bing Xu, Li Zhang. Traveling waves and entire solutions for an epidemic model with asymmetric dispersal. Discrete & Continuous Dynamical Systems, 2017, 37 (5) : 2483-2512. doi: 10.3934/dcds.2017107 [19] Shi-Liang Wu, Tong-Chang Niu, Cheng-Hsiung Hsu. Global asymptotic stability of pushed traveling fronts for monostable delayed reaction-diffusion equations. Discrete & Continuous Dynamical Systems, 2017, 37 (6) : 3467-3486. doi: 10.3934/dcds.2017147 [20] Nestor Guillen, Russell W. Schwab. Neumann homogenization via integro-differential operators. Discrete & Continuous Dynamical Systems, 2016, 36 (7) : 3677-3703. doi: 10.3934/dcds.2016.36.3677

2020 Impact Factor: 1.327

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# How can I add another plot to the \addplot command in PGFPlots?

My goal is to add another plot to the \addplot command and this small example hopefully shows what I'm trying to do:

\documentclass{standalone}
\usepackage{pgfplotstable}
\begin{document}

procs, none
16, 6.079748e+03
32, 3.624823e+03
64, 1.685646e+03
128, 9.180763e+02
256, 5.977239e+02
512, 4.410280e+02
}\scalability

\begin{tikzpicture}
\begin{loglogaxis}[
ylabel={Time (sec)},
xtick=data,
xticklabels from table={\scalability}{procs},
table/x=procs,
set scaling/.code={
\edef\xcol{\pgfkeysvalueof{/pgfplots/table/x}}
\pgfplotstablegetelem{0}{\xcol}\of{\scalability}
\edef\proc{\pgfplotsretval}
\pgfplotstablegetelem{0}{\pgfkeysvalueof{/pgfplots/y}}\of{\scalability}
\edef\offset{\pgfplotsretval}
\pgfplotsset{table/y expr=\offset * \proc / \thisrowno{0}}
\pgfplotsextra{
\draw (1,1) -- (256,256);              % this is drawn
\addplot table[y=none] {\scalability}; % this is not
}
},
scaling/.style={
forget plot,
set scaling,
},
forget plot style={opacity=0.4},
]

\end{loglogaxis}
\end{tikzpicture}

\end{document}

What I'd like to see is:

but tinkering with the .code functionality (and also execute at end plot) nothing seems to be able to call another \addplot only lower level TikZ commands (\draw, \path, etc.):

Is there any way to get another \addplot appended?

-
I think this is going to be quite hard. It would probably be easier to define a new command that calls both of the plots. Do you have a particular reason for not wanting that? –  Jake Jun 19 '12 at 21:31
Well, for starters, the execute and .code keys seemed like the elegant way to do this. Also, if defining a \newcommand, it seemed quite hacky to pass optional key-value pairs down to pgfplots; i.e. \mynewplotcommand +[optional styles to append]; \mynewplotcommand [optional styles to override] If passing that down to pgfplots is easier than it seems, then it would indeed work. –  Sean Farley Jun 20 '12 at 0:02

Your approach does not work because \addplot is designed to be placed within the axis -- and \pgfplotsextra is designed to NOT process things inside of an axis (but to postpone their processing).

If you want to use \addplot, you have to place it into the axis.

But from what I understand from your application, I believe that you can simply place your expression evaluation stuff directly in front of the plot in question:

\documentclass{standalone}
\usepackage{pgfplotstable}
\begin{document}

procs, none
16, 6.079748e+03
32, 3.624823e+03
64, 1.685646e+03
128, 9.180763e+02
256, 5.977239e+02
512, 4.410280e+02
}\scalability

\begin{tikzpicture}
\begin{loglogaxis}[
ylabel={Time (sec)},
xtick=data,
xticklabels from table={\scalability}{procs},
table/x=procs,
table/y=none,
forget plot style={opacity=0.4},
]

\pgfplotstablegetelem{0}{\pgfkeysvalueof{/pgfplots/table/x}}\of{\scalability}
\edef\proc{\pgfplotsretval}
\pgfplotstablegetelem{0}{\pgfkeysvalueof{/pgfplots/table/y}}\of{\scalability}
\edef\offset{\pgfplotsretval}

\addplot +[forget plot] table[y expr=\offset * \proc / \thisrowno{0}] {\scalability};


text:
# Math Help - Help with Circles please

1. ## Help with Circles please

The question states:
Maraiah is making a graphic design on a computer using a coordinate grid. She began with a circle that has a center (0,0). Then, she drew a congruent cirlce with the equation x^2+y^2+12x+4y+5=0.

B)what translation did Maraiah make to draw the second circle?

What i did was, since the second equation is congruent to the fist one, i replaced the x and y with 0. However, the final statement results to be false. 5=0. And using (x-h)^2+(y-k)^2=r^2, i figured the radius to be 0.... i have no clue on how to this problem. Please guide me! Thanks guys.

2. Hello, jjkins!

Maraiah is making a graphic design on a computer using a coordinate grid.
She began with a circle that has a center (0,0).
Then, she drew a congruent cirlce with the equation $x^2+y^2+12x+4y+5\:=\:0$

Complete the square . . .

We have: . $(x^2 + 12x \qquad) + (y^2 + 4y \qquad) \;=\;-5$

. . . . . . . . $(x^2 + 12x + {\color{blue}36}) + (y^2 + 4y + {\color{red}4}) \;=\;-5 + {\color{blue}36} + {\color{red}4}$

. . . . . . . . . . . $(x+6)^2\quad+\quad(y+2)^2\quad \;= \quad\;35$

Radius of both circles: . $\sqrt{35}$

B) What translation did Maraiah make to draw the second circle?
The center of the second circle is: . $(\text{-}6,\,\text{-}2)$

She moved the original circle: 6 units to the left and 2 units down.


text:
Following up from my previous post, here’s the next problem:

The public key is $(n, e) = (221, 11)$. What is the RSA Decryption of $7$?

• What is RSA? wikipedia: RSA (Rivest–Shamir–Adleman) is one of the first public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem, the encryption key is public and distinct from the decryption key which is kept secret (private). In RSA, this asymmetry is based on the practical difficulty of factoring the product of two large prime numbers, the “factoring problem”.

• How does RSA work?
Let’s look at an example from wikipedia:

1. Choose two distinct prime numbers, such as
$p = 61$ and $q = 53$
2. Compute $n = pq$ giving
$n = 61 \times 53 = 3233$
3. Compute the Carmichael’s totient function of the product as
$\lambda (n)=\operatorname {lcm} (p-1,q-1)$ giving
$\lambda (3233)=\operatorname {lcm} (60,52)=780$
4. Choose any number $1 \lt e \lt 780$ that is coprime to $780$. Choosing a prime number for $e$ leaves us only to check that $e$ is not a divisor of $780$. Let $e = 17$.
5. Compute $d$, the modular multiplicative inverse of $e \bmod \lambda(n)$ yielding, $d = 413$
Worked example for the modular multiplicative inverse:
$d\times e \equiv 1{\bmod {\lambda }}(n)$
$413\times 17 \equiv 1{\bmod {7}}80$

The public key is $(n = 3233, e = 17)$. For a padded plaintext message m, the encryption function is $c(m)=m^{17} \bmod 3233$

The private key is $(n = 3233, d = 413)$. For an encrypted ciphertext c, the decryption function is $m(c)=c^{413} \bmod 3233$

For this problem we have $n$, $e$, and $c$. We have to find $d$ and calculate $m$ using the above decryption function.
$n = 221 = 13 \times 17$. Carmichael’s totient function, $\lambda (221)=\operatorname {lcm} (12,16)=48$. We already have $e = 11$.
$d$ is the modular multiplicative inverse of $e \bmod \lambda(n)$ i.e.,
$d \times e \equiv 1 \bmod \lambda (n)$ or $d \times 11 \equiv 1 \bmod 48$.

We can write a simple program to calculate d:

d = 0
while (11 * d) % 48 != 1: d+=1
print(d)

35

Now we can find $m$ as: $m = c ^ d \bmod n = 7 ^{35} \bmod 221 = 54$

Thus, 54 is our final answer.

• What is Carmichael’s totient function?
wikipedia: The Carmichael function associates to every positive integer $n$ a positive integer $\lambda (n)$, defined as the smallest positive integer $m$ such that
$$a^m \equiv 1 \bmod n$$
for every integer $a$ between $1$ and $n$ that is coprime to $n$.

wikipedia: The exponent of the multiplicative group of integers mod n (previous post), that is, the least common multiple of the orders in the cyclic groups, is given by the Carmichael function $\lambda (n)$. In other words, $\lambda (n)$ is the smallest number such that for each $a$ coprime to $n$, $a^{\lambda(n)} \equiv 1 \pmod n$ holds.

• How can we find the modular multiplicative inverse using pen and paper? By making use of the Extended Euclidean Algorithm. Here’s a youtube video that walks you through the procedure and here’s a Khan Academy article that explains it well. Adding these here for myself for future reference.

Now let’s try another one:

The public key is $(n, e) = (437, 13)$. What is the RSA Digital Signature of $17$?

• What is RSA Digital Signature?
This article explained it very well. From the article, here is the “textbook” definition of RSA signing compared with the Encryption and Decryption functions.

• $Enc(m; e) = R(m,e)$
• $Dec(c; d) = R(c,d)$
• $Sign(m; d) = R(m,d)$
• $Ver(m; s; e) = R(s,e) == m$

We are basically applying the sender’s private key on the message they are sending to generate a signature. That means, for this problem, we have to find $m^d \bmod n$.

We can follow the procedure from the previous problem now. $d$, or the private key, is the modular multiplicative inverse of $e \bmod \lambda (n)$. Here, $\lambda (n)$ is the Carmichael totient function, calculated as $\lambda (n) = lcm (p-1, q-1)$, where $p$ and $q$ are the prime factors of $N$.

Through trial and error, we find that the prime factors of $n = 437$ are $19$ and $23$. Of course this would be a very difficult problem if the prime numbers were sufficiently large; that’s the whole idea behind RSA encryption. Anyways, now that we have $p$ and $q$, we can find $\lambda (437) = lcm (18, 22) = 198$. Now, using the same algorithm I used for the previous problem,

d = 0;
while (d * 13) % 198 != 1: d+=1
print(d)

61

Therefore, the solution to this problem is $17^{61}\bmod 437 = 195$.

Until next time!


text:
# Dividing Polynomials Calculator

The calculator will divide two polynomials (quadratic, binomial, trinomial, etc.), with steps shown.

First polynomial:

Second polynomial:

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

## Solution

Your input: divide $x^{3} + 7 x^{2} + 1$ by $x - 1$

Perform polynomial long division or synthetic division (use the polynomial long division calculator or the synthetic division calculator to see the steps).

$\frac{x^{3} + 7 x^{2} + 1}{x - 1}=x^{2} + 8 x + 8+\frac{9}{x - 1}$

Answer: $\frac{x^{3} + 7 x^{2} + 1}{x - 1}=x^{2} + 8 x + 8+\frac{9}{x - 1}$


text:
Differences

This shows you the differences between two versions of the page.

 cs-312:hw7.5 [2014/12/31 15:57]ringger created cs-312:hw7.5 [2014/12/31 16:25] (current)ringger 2014/12/31 16:25 ringger 2014/12/31 15:57 ringger created 2014/12/31 16:25 ringger 2014/12/31 15:57 ringger created Line 14: Line 14: - ''What does a good two-column proof look like? '' Statements are found in the left column and corresponding justifications for each statement are found on the right.  Each statement and reason should be numbered for ease of reference.  Each statement must be a mathematical statement that has a truth value.  For example, an equation (e.g., <math>2x+5 = 1) is either true or not, whereas an expression (e.g., <math>2x+5) does not have truth value.  Each step must be justified by a reason in the right column.  If you cannot provide a clear and convincing reason for a particular statement in your proof, then you need to give the matter further thought and you may consider adding additional steps to improve the argument. + ''What does a good two-column proof look like? '' Statements are found in the left column and corresponding justifications for each statement are found on the right.  Each statement and reason should be numbered for ease of reference.  Each statement must be a mathematical statement that has a truth value.  For example, an equation (e.g., $2x+5 = 1$) is either true or not, whereas an expression (e.g., $2x+5$) does not have truth value.  Each step must be justified by a reason in the right column.  If you cannot provide a clear and convincing reason for a particular statement in your proof, then you need to give the matter further thought and you may consider adding additional steps to improve the argument.


text:
# math

Level pending

a chemist has 20 Liters of 60% acid solution . how many Liters of 90% solution must be added to make a mixture of 75%

×


text:
# Poissonness plot

#### noetsi

##### Fortran must die
This is essentially the equivalent of a QQ plot for Poisson distribution, which can not use QQ plots since the distribution is of discrete data. Michael Friendly tells you how to use these and gives SAS code on 51 on in his book "Visualizing Categorical Data"

Sample code

Title ' Poissoness Plot: Horskicks';
Data horskicks;
input deaths corpsyrs;
label deaths = 'number of Deaths'
corpsyrs =' Number of Corps years':
datalines;

0 109
1 65
2 22
3 3
4 1
;

%poisplot (data=horskick, count=Deaths, freq=corpsyrs);

Obviously you need to run the macro %poisplot first per the link. Hoaglin created the poissonnees plot. Tukey helped create similar ones for logarithmic, geometric and binomila distributions. That dude got around...

Last edited:

#### hlsmith

##### Less is more. Stay pure. Stay poor.
We need the code for all of the distributions! Keep posting.

#### Dason

Is there a question or are you just sharing? You can do this very easily in R as well using the distplot function in the vcd package.

Code:
x <- 0:4
kicks <- c(109, 65, 22, 3, 1)
dat <- rep(x, kicks)

library(vcd)
distplot(dat, "poisson")

N <- sum(kicks) # total number of observations
y <- log((factorial(x) * kicks)/sum(kicks))

o <- lm(y ~ x) # We get the slope/intercept given in plot
plot(x, y)
abline(o)

#### Dason

We need the code for all of the distributions! Keep posting.
noetsi didn't even share the code for the poisson

#### noetsi

##### Fortran must die
The input code above inputs a actual poisson distribution - one based on a famous analysis. The macro will generate everything else. Note you will get lots of error messages when you run this, at least in EG, there are issues with the macro and more recent SAS code I believe. But it generates the same results the author finds in his book so the warnings don't seem to matter.

I ran a test to see if this works. First I ran a simulation from a known poisson distribution.

%Let N = 500;
Data Poisson (keep=x);
call streaminit(4321);
lambda = 4;
do i = 1 to &N;
X= rand('Poisson',lambda);
output;
End;
Run;

Then I tested this with the method shown above (test is the data set that has the counts and frequency in it, this is obviously an easier way to do this then input them manually).

title 'Poissoness Plot: Death by Horsekicks';
data horsekick1;
SET test1;
label count = 'Number of Deaths'
freq ='number of corps years' ;

%poisplot (data=horsekick1, count=count, freq=freq);

This showed correctly the distribution was poisson. Note the leveled poisson plot suggested to me, incorrectly, that the distribution was not poisson. I am not sure I know how to run this.

Then I generated a binomial distribution.

%let N =500;
Data binomial (keep=x);
Call streaminit(4321);
p = 1/2;
Do i=1 to &N;
x = RAND('Binomial',p,10);
output;
End;
Run;

I tested this with the code I used above again.

title 'Poissoness Plot: Death by Horsekicks';
data horsekick3;
SET test3;
label count = 'Number of Deaths'
freq ='number of corps years' ;

%poisplot (data=horsekick3, count=count, freq=freq);

The result is a curve around the diagnonal which the author states to be an indication the distribution is not poisson. Which of course is correct.

I have not got to the other plots - this stuff is not easy for me Distplot is the macro apparently for these plots similar to what Dason noted in R.

This is useful by the author.

http://www.datavis.ca/courses/grcat/grcfoils1_4.pdf

Last edited:

#### noetsi

##### Fortran must die
Is there a question or are you just sharing? You can do this very easily in R as well using the distplot function in the vcd package.

Code:
x <- 0:4
kicks <- c(109, 65, 22, 3, 1)
dat <- rep(x, kicks)

library(vcd)
distplot(dat, "poisson")

N <- sum(kicks) # total number of observations
y <- log((factorial(x) * kicks)/sum(kicks))

o <- lm(y ~ x) # We get the slope/intercept given in plot
plot(x, y)
abline(o)
Just sharing for a change. Does R created leveled Poissoness plots? These are the one element I have trouble with in SAS - although I may be doing them wrong.

#### noetsi

##### Fortran must die
Whatever you do don't run the macros word gskip etc attached to this macro. It totally messes up poisplot and there appears to be no way to fix this even opening up different EG or SAS and starting from scratch

Sorry I forgot to post the link to the macro...

http://euclid.psych.yorku.ca/SCS/vcd/poisplot.html

This is the link to the macro distplot used for the other distribution types.

http://www.datavis.ca/books/vcd/distplot.html

"The DISTPLOT macro constructs plots of a discrete distribution designed to diagnose whether the data follows one of the standard distributions: the Poisson, Binomial, Negative Binomial, Geometric, or Log Series, specified by the DIST= parameter. The usual (PLOT=DIST) plot is constructed so that the points lie along a straight line when the data follows that distribution. An influence plot (PLOT=INFL) shows the influence of each observation on the choice of the distribution parameter(s)."

Last edited:

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Will play around with it next week. Out of the office with a sick kid. Currently watching the first season of Manhattan. Takes two episodes to get into it but now I am on episode 11.

#### noetsi

##### Fortran must die

When I run distplot I run into problems.

This is the error message.

WARNING: Apparent invocation of macro WORDS not resolved.
WARNING: Apparent invocation of macro WORDS not resolved.
ERROR: The macro DISTPLOT will stop executing.

The warnings may be meaningless, I get the samewhen using the poissonness macro and the results are fine. According to the link by Friendly the only things you have to include are the following:

%distplot(data=qtest6,count=count,freq=freq,dist=binomial);

but that generated the error message above. Unfortunately the way the macro is formatted makes it very hard to read and find the problem {which I might not understand anyhow}

#### Dason

This is the error message.

WARNING: Apparent invocation of macro WORDS not resolved.
WARNING: Apparent invocation of macro WORDS not resolved.
ERROR: The macro DISTPLOT will stop executing.

The warnings may be meaningless, I get the samewhen using the poissonness macro and the results are fine.
You get the same errors? Are you sure they're the same for the poisplot macro as is above?

#### noetsi

##### Fortran must die
Actually it said that it could not find three macros including word. aftr I ran that it still can't find LABEL and GSKIP. I will look for those and run them. Thanks for assistance dason. Friendly does not suggest you need to run those macros in the book or his web site.

#### noetsi

##### Fortran must die
I ran all three macros word label and gskip. It runs now, but no graph is produced. I get the following error message.

ERROR: No valid observations are found.
24 order=(. to . by 1 )
_
22
200
ERROR 22-322: Syntax error, expecting one of the following: a quoted string, a numeric constant, a datetime constant.

ERROR 200-322: The symbol is not recognized and will be ignored.

WARNING: AXIS statement 1 not found. The default axis description will be used.
WARNING: Graph has missing data for the required role ( y * count ), no graph will be produced.
WARNING: No output was generated due to all missing data.

The problem for me is that the macro is so long figuring out where the error is is very difficult.

Maybe HLSMITH can figure out what is wrong, or dason I do think that you have to run all three macros word label and gskip although that does not fully correct the problem. Looking at the data produced it appears that the confidence intervals are not being generated.

Last edited:

#### noetsi

##### Fortran must die
What I found really anoying about this is that it was working fine. Then I ran the macros noted above to fix something and I could never get the basic macro to work again. Even when I started a brand new session of SAS.

#### noetsi

##### Fortran must die
hlsmith I have actually run that successfully. But it is a lot of work to get that going, much less than Friendly. And it does not apply to many distributions that friendly addressed.


text:
# Need the Integral of

1. May 14, 2004

### Swiss Army

Could someone show me how the integral of f(v) equals 1, where f(v) is

(4/(Sqrt Pi)) x (m/2kT)^(3/2) x (v^2) x (e^(-mv^2/2kT))

I got through the integral and it didn't equal 1, but it's supposed to, so could someone help me out.

2. May 14, 2004

### Hurkyl

Staff Emeritus
Could you show us what you have done so far?

And what is the region of integration?

3. May 14, 2004

### Swiss Army

Ooops, sorry. The integration is from zero to infinity. I'll type in what I've done so far if someone can tell me how to get the integral and various other mathematical characters in the post, it's much easier that way. But so far my methods have included substitution, integration by parts, and taking the limit of the function.

4. May 15, 2004

### Zurtex

Look here: https://www.physicsforums.com/misc/howtolatex.pdf

Or go to this thread and click on various examples and the code will pop up in a small window so you see how it is done: https://www.physicsforums.com/showthread.php?t=8997

5. May 15, 2004

### Parth Dave

you should be able to get the indefinate integral using integration by parts. use v^2 = du and e^(-mv^2/2kt) = v. Than in the new integral you create, you can substitute e^u and it should work out fine (in other words, do integration by parts and than use substitution). Whenever you have a parametre for an integral as an infinite, you have to take the limit of the integral as it approaches that infinite. For example, in this case you would have to than show that (lim x--> infinite of F(x)) - F(0) = 1 (thats what you do when you have infinite parameters for integration). Try it out and see what you get.

6. May 18, 2004

### Swiss Army

hmmm...

Yeah, those were the methods that I used to integrate it, integration by parts, substitution, and taking the limit. I talked to my Calculus 3 teacher yesterday about it, and she said it was something like a complex integral where you can't just solve it using those methods...or something like that.

7. May 18, 2004

### Parth Dave

when you do integration by parts you will obtain a new integral that you have to solve. This [unless im mistaken] can be solved using substitution. Altho you may have to do integration by parts again (i suspect you would)

Last edited: May 18, 2004
8. May 18, 2004

### NSX

Yeah .. it seems that way.

I tried to use IBP & substitution.

I assume k, m, T, & $$\pi$$ are constants.

Then I get:

$$\frac{4m^\frac{3}{2}}{\sqrt{\pi}(2kT)^\frac{3}{2}e^\frac{m}{2kT}}\int_{0}^{\infty}\frac{v^2}{e^v^2}dv$$

I'm not sure how to do in an integral the likes of $\int\frac{v^2}{e^v^2}$, or $\intv^2e^-v^2$ for that matter.

With substitution of $u=v^2$, I get:

$$\frac{1}{2}\int\frac{u^\frac{1}{2}}{e^u}du$$

Last edited: May 20, 2004
9. May 18, 2004

### franznietzsche

Maybe i'm just more tired than i tihnk, but with that substitution i believe you should get:

$$\frac{1}{2}\int\frac{u}{e^u}du$$

10. May 18, 2004

You are tired.

11. May 18, 2004

### Ebolamonk3y

Huh? Makes no sense..

12. May 19, 2004

### franznietzsche

oh...yeah...i didn't change the varible of integration, i left it $$dv$$ even though i typed $$du$$. So yeah, disregard that post then.

edit: actually he never put the $$dv$$ up there, so i'm assuming that was my error.

13. May 19, 2004

### Ebolamonk3y

yay! problem solved..

14. May 19, 2004

### franznietzsche

Just the roblem of my mistakes when sleep-deprived. Oh well...i suppose i could go actually try to solve it now...here i go

15. May 19, 2004

### Ebolamonk3y

Dun try now... get sleep and then try. :)

16. May 19, 2004

### franznietzsche

actually i think it can't be solved analytically...integration by parts fails, given that $$\frac{1}{e^x^2}$$ cannot be integrated. So...i can't seem to come up with any standard single variable calculus techniques that would help...so yeah, sleep first, boggle later.

17. May 19, 2004

### Swiss Army

The thing is, the integral of the whole expression is supposed to divide out to 1. The integral is supposed to end up being the inverse of the constant you pull out when you integrate, giving you a 1/1 or 1 answer. This suggests that standard integration techniques won't work, because the when you integrate, you have to end up with the Sqrt( Pi ) / 4 in the answer to cancel out the 4/Sqrt ( Pi ) in the constant.

18. May 19, 2004

### NSX

Mathematica?
Matlab?

haha

But like you said in your post, if you require concepts from Calc III, it's beyond what I can do.

19. May 19, 2004

### Tom Mattson

Staff Emeritus
This integral can be done the same way that the integral of exp(-x2) can be done. First, write the integral of x2exp(-x2) from zero to infinity. Then write the integral of y2exp(-y2) from zero to infinity (they're both exactly the same as your integral). Now multiply the integrands together double integrate over x and y. When you convert to polar coordinates, you will get an integral that can be done by parts. Just don't forget to take the square root at the end.

Note:

x2y2=r4sin2(θ)cos2(θ)
x2+y2=r2
dx dy=rdr dθ

20. Oct 21, 2007

### Morto

Hello!

Sorry to hash up this very old thread, but I need help with this exact problem and I don't understand the solution given.

I'm trying to find the integral of x2exp(-x2 / a2) from -infinity to infinity.

I've done what's recommended here, but I end up with a double integral of 1/4 sin2(2θ) r5 exp[-r2] drdθ from zero to infinity and zero to infinity. How do I solve this?


text:
Publication
Title
Interferon-$\alpha$ in acute myeloid leukemia : an old drug revisited
Author
Abstract
 Interferon-α (IFN-α), a type I IFN, is a well-known antitumoral agent. The investigation of its clinical properties in acute myeloid leukemia (AML) has been prompted by its pleiotropic antiproliferative and immune effects. So far, integration of IFN-α in the therapeutic arsenal against AML has been modest in view of the divergent results of clinical trials. Recent insights into the key pharmacokinetic determinants of the clinical efficacy of IFN along with advances in its pharmaceutical formulation, have sparked renewed interest in its use. This paper reviews the possible applicability of IFN-α in the treatment of AML and provides a rational basis to re-explore its efficacy in clinical trials.
Language
English
Source (journal)
Leukemia. - Baltimore, Md, 1987, currens
Publication
Baltimore, Md : 2011
ISSN
0887-6924 [print]
1476-5551 [online]
Volume/pages
25:5(2011), p. 739-748
ISI
000290471100001
Full text (Publisher's DOI)
Full text (publisher's version - intranet only)
UAntwerpen
 Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address


text:
# image processing – Color regions defined by many lines

A kludgey implementation that struggles at some of the “points” of the zones:

1. Define a function that takes an equation defining the boundaries and turns it into an inequality satisfied if a point is on the same side of the boundary as the origin is.

``````PerpBisIneq(pt1_, pt2_) := Block({eq = PerpBis(pt1, pt2)},
If(eq === True, True,
If((First(eq) /. {x -> 0, y -> 0}) > 0, Greater @@ eq,
Less @@ eq))
)
``````
2. Define a function that returns a list of inequalities as functions of the point `{x,y}`.

``````ineqs({x_, y_}) =
Flatten(Table(PerpBisIneq({0, 0}, {i, j}), {i, -2, 2}, {j, -2, 2}))
``````
3. Create a `RegionPlot` defined by the region of the plan in which precisely $$n-1$$ elements of `ineqs` are `False`.

``````brillouinZone(n_) := RegionPlot(
Count(ineqs({x, y}), False) == n - 1, {x, -5, 5}, {y, -5, 5},
PlotStyle -> ColorData(109)(n), PlotPoints -> 100)
``````
4. Plot the Brillouin zones.

``````Show(
Table(ContourPlot(Evaluate@PerpBis({0, 0}, {i, j}), {x, -5, 5}, {y, -5, 5}), {i, -2, 2}, {j, -2, 2}),
Table(brillouinZone(n), {n, 1, 5})
)
``````

A few notes on how this could be refined:

• The equation for `i = j = 0` always evaluates as `True`, requiring the first layer of case-checking in `PerpBisIneq`. The second layer of case-checking is not necessary if `gamma` is always negative, but I’m not sure if this is the case.

• The setting for `PlotPoints` in the definition of `BrillouinZone` is necessary to get Mathematica to return a “nice” plot. Lower settings cause it to miss some zones entirely.

• There is probably a nicer way of getting Mathematica to logically reduce the condition “precisely $$n$$ of these inequalities are `False`” than the way that I’ve done it, in which Mathematica is (I think) effectively sampling the plane point-wise. I suspect there’s a way to do it with `Reduce` or something similar, but I’m not immediately sure what it is.

• The color scheme can be changed by substituting another number for `109` in the definition of `BrillouinZone`.

EDIT: I was able to use `Reduce` to find a set of logical predicates for the nth Brillouin zone. But it’s ugly as sin and slow as molasses, so I can’t say I recommend it. It might, however, be helpful if you ever need to rigorously define a Brillouin zone (for, say, some kind of integration), so I’m including it for reference.

``````Brillouinpred({x_, y_}, n_) :=
Reduce(Or @@ Apply(And, Subsets(ineqs({x,y}),{Length(ineqs({x, y})) - (n + 1)}), {1}) &&
! (Or @@ Apply(And, Subsets(ineqs({x,y}), {Length(ineqs({x, y})) - n}), {1})))
``````

This returns a logical condition on `x` and `y` which is `True` if `{x,y}` is in the nth Brillouin zone and `False` otherwise. This is done by noting that if we have a set of $$N$$ predicates $$P = {P_1, P_2, …, P_N}$$, then the statement that “no more than $$m$$ of the predicates are true” is equivalent to
$$mathcal{P}_m = bigvee_{Q subset P,, |Q| = N-m} left( bigwedge_{P_i in Q} P_i right).$$
The outer “or” runs over all subsets of cardinality $$N-m$$, and the inner “and” runs over the elements of each such subset. One can then construct the predicate “exactly $$m$$ predicates are true” by demanding that $$mathcal{P}_m wedge neg mathcal{P}_{m-1}$$. The code above explicitly constructs this statement and then applies `Reduce` to it for simplification purposes. However, note that this creates a set of $${N choose m}$$ logical predicates, each containing $$N-m$$ subpredicates, as an intermediate step; so Mathematica takes some time to solve this.


text:
# How do you find abs( 2 + 3i )?

$| 2 + 3 i | = \sqrt{13}$
$| a + b i | = \sqrt{{a}^{2} + {b}^{2}}$
Hence, $| 2 + 3 i | = \sqrt{{2}^{2} + {3}^{2}} = \sqrt{4 + 9} = \sqrt{13}$


text:
# Monic Irreducible Polynomials over Finite Field

Let $F=\mathbb{F}_{q}$ be a finite field (so $q=p^k$ for some prime $p$ and positive integer $k$), and let $\varphi(d)$ denote the number of monic irreducible polynomials of degree $d$ in $F[X]$. I'm supposed to show that $\displaystyle{\sum_{d \mid n} d \varphi(d) = q^n}$.

I see there are previous questions about this topic and even a paper, but all (save one) seem to employ the use of the Möbius function and Möbius inversion - both topics I have not covered yet in class. There is also this answer, but it appears to hinge upon the extension having prime degree. Is there some way to show this without explicitly coming up with a formula for the number of irreducible monic polynomials of a given degree in $F[X]$?

Any help would be greatly appreciated.

-
Consider elements $\alpha\in F$ and their minimal polynomials $m_\alpha(x)$ over the prime field. What degrees are possible for $m_\alpha$? Why do they all appear as $m_\alpha$ for some $\alpha$? How many elements of $F$ share the same minimal polynomial? Derive an equation! –  Jyrki Lahtonen Mar 19 '13 at 4:54
Do you know that $\mathbb F(q^r)\subset\mathbb F(q^s)$ if and only if $r|s$? –  Lubin Mar 19 '13 at 4:54
As an aside, Mobius inversion is employed after that formula is already attained (to find $\varphi$ explicitly) - it is not used to obtain it. –  anon Mar 19 '13 at 4:56
@Lubin, Do you mean $\mathbb{F}_{q^r} \subset \mathbb{F}_{q^s}$ iff $r \mid s$? If so, then yes, I do know that. –  JoeDub Mar 19 '13 at 4:57
Joe, $m_\alpha$ is not the minimal polynomial over $F$, but over the prime field of $p$ elements. Nevertheless, they do not share roots, but have as many roots as is indicated by their degree. –  Jyrki Lahtonen Mar 19 '13 at 5:04

The splitting field of $X^{q^n}-X$ is ${\bf F}_{q^n}$. Every irreducible $\pi(X)$ of degree $d$ splits in ${\bf F}_{q^n}$ and every element of ${\bf F}_{q^n}$ is a root of $X^{q^n}-X$, and thus $\pi(X)\mid(X^{q^n}-X)$. Furthermore $X^{q^n}-X$ has no repeated roots so each irreducible $\pi(X)$ of degree $d\mid n$ must appear in its factorization precisely once. Therefore we have the conclusion

$$X^{q^n}-X=\prod_{d\mid n}\prod_{\deg\pi=d}\pi(X).$$

Taking degrees yields $\displaystyle q^n=\sum_{d\mid n}d\varphi(d)$ (and from here Möbius inversion yields $\varphi(d)$).

-
Somehow it wasn't clicking that I was examining polynomial degrees. When you put it that way, it's quite clear to me now what is happening. –  JoeDub Mar 19 '13 at 5:14


text:
# OWL 2 Profiles Features List

## OWL 2 EL

Supported class restrictions:

– existential quantification to a class expression or a data range

– existential quantification to an individual or a literal

– self-restriction

– enumerations involving a single individual or a single literal

– intersection of classes and data ranges

Supported axioms, restricted to allowed set of class expressions:

– class inclusion, equivalence, disjointness

– object property inclusion (w. or w.o. property chains), and data property inclusion

– property equivalence

– transitive object properties

– reflexive object properties

– domain and range restrictions

– assertions

– functional data properties

– keys

NOT supported in OWL 2 EL (with respect to OWL 2 DL):

– universal quantification to a class expression or a data range

– cardinality restrictions

– disjunction

– class negation

– enumerations involving more than one individual

– disjoint properties

– irreflexive, symmetric, and asymmetric object properties

– inverse object properties, functional and inverse-functional object properties

## OWL 2 QL

The supported axioms in OWL 2 QL take into account what one can use on the left-hand side of the inclusion operator ($$\sqsubseteq$$, SubClassOf) and what can be asserted on the right-hand side:

• Subclass expressions restrictions:

– a class

– existential quantification (ObjectSomeValuesFrom) where the class is limited to owl:Thing

– existential quantification to a data range (DataSomeValuesFrom)

• Super expressions restrictions:

– a class

– intersection (ObjectIntersectionOf)

– negation (ObjectComplementOf)

– existential quantification to a class (ObjectSomeValuesFrom)

– existential quantification to a data range (DataSomeValuesFrom)

Supported Axioms in OWL 2 QL:

– Restrictions on class expressions, object and data properties occurring in functionality assertions cannot be specialized

– subclass axioms

– class expression equivalence (involving subClassExpression), disjointness

– inverse object properties

– property inclusion (not involving property chains and SubDataPropertyOf)

– property equivalence

– property domain and range

– disjoint properties

– symmetric, reflexive, irreflexive, asymmetric properties

– assertions other than individual equality assertions and negative property assertions (DifferentIndividuals, ClassAssertion, ObjectPropertyAssertion, and DataPropertyAssertion)

NOT supported in OWL 2 QL (with respect to OWL 2 DL):

– existential quantification to a class expression or a data range in the subclass position

– self-restriction

– existential quantification to an individual or a literal

– enumeration of individuals and literals

– universal quantification to a class expression or a data range

– cardinality restrictions

– disjunction

– property inclusions involving property chains

– functional and inverse-functional properties

– transitive properties

– keys

– individual equality assertions and negative property assertions

## OWL 2 RL

OWL 2 RL Supported in OWL 2 RL:

– More restrictions on class expressions (see table 2 of [MGH+09]; e.g., no SomeValuesFrom on the right-hand side of a subclass axiom)

– All axioms in OWL 2 RL are constrained in a way that is compliant with the restrictions in Table 2.

– Thus, OWL 2 RL supports all axioms of OWL 2 apart from disjoint unions of classes and reflexive object property axioms.

A quick one-liner of the difference is: No $$\forall$$ and $$\neg$$ on the left-hand side, and $$\exists$$ and $$\sqcup$$ on right-hand side of \(\sqsubseteq\.


text:
# Measuring in the computational basis in the single qubit gate QFT implementation

I've come across this paper about a single-qubit-gate-only QFT implementation. In the paper it is claimed that measuring a qubit after applying the Hadamard gate (it isn't called Hadamard gate in the paper but its description matches that of the Hadamard gate) is no different from measuring before the gate.

Now what I assumed is that the measurement before the gate would have been made in the $$\{|0⟩,|1⟩\}$$ basis and the measurement after the gate would have been made in the $$\{|+⟩,|-⟩\}$$ basis.

However, I came across these conflicting sources about the term "Computational Basis" regarding measurement:

So my questions are:

1. Does saying "measure in the computational basis" means that in whatever basis you qubits are in you measure them according to that basis to get classical bits or does it always refer to measuring in the {|0⟩,|1⟩} basis?
2. If it is the latter then how does the claim hold when we know that $$H|0\rangle=\frac{1}{\sqrt 2}(|0⟩+|1⟩)$$ and $$H|1\rangle=\frac{1}{\sqrt 2}(|0⟩-|1⟩)$$ so the probabilities of measuring an outcome after H should be $$\frac{1}{2}$$ regardless of the original state of the qubit?

Measuring in the 'Computational Basis' always means to measure in the $$\{|0\rangle,|1\rangle\}^n$$ basis for $$n$$ qubits.
As for your second question, I feel that you misinterpreted the author. It is certainly true that on applying the $$H$$ gate on $$|0\rangle$$ or $$|1\rangle$$ you would get the states $$|0\rangle$$ and $$|1\rangle$$ with equal probabilities. But the underlying idea is this: let $$q$$ be a qubit in the circuit given in the paper. Let $$m$$ many gates are applied on $$q$$. For some $$k\le m$$, see that if only the first $$k$$ gates actually transform the state of $$q$$ (by some rotation) and if for the rest of the $$m-k$$ gates $$q$$ just acts as a control qubit, then the last $$m-k$$ gates cannot change the state of $$q$$. So, if the probability of obtaining $$|0\rangle$$ on measuring the qubit $$q$$ as $$|0\rangle$$ was $$p$$ before applying the last $$m-k$$ gates, then even after applying the $$m-k$$ gates, the probability of obtaining $$|0\rangle$$ on measuring $$q$$ remains $$p$$. The author then uses this fact and uses the measured outcomes to control the other gates. An example might help understand. Take, for instance, the following three qubit circuit:
In this circuit, $$q_0$$ gets rotated in the first gate and just acts as a control qubit for the next two gates. Now notice that if you were to measure $$q_0$$ just after the first gate, you would get $$|0\rangle$$ and $$|1\rangle$$ with equal probabilities. But if you measure $$q_0$$ after all the gates have been applied, the probabilities of obtaining $$|0\rangle$$ and $$|1\rangle$$ still remain the same.
Next, notice that the output of this circuit is $$|\psi\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$$. So you get the states $$|000\rangle$$ and $$|111\rangle$$ with equal probabilities on measuring all the qubits of the circuit after all the gates have been applied. Now, on the other hand let us first measure the qubit $$q_0$$ just after $$H$$, then apply an $$X$$ gate to $$q_1$$ and $$q_2$$ only if the measured outcome of $$q_0$$ is 1 and then measure the qubits $$q_1$$ and $$q_2$$. Now, if the measurement of $$q_0$$ yields $$|0\rangle$$ then the other two gates have no effect on $$q_1$$ and $$q_2$$. So after the final measurement, the complete measured state would be $$|000\rangle$$. Similarly, if the measurement of $$q_0$$ yields $$|1\rangle$$ then the other two gates flips the qubit state in $$q_1$$ and $$q_2$$. So after the final measurement, the complete measured state would be $$|111\rangle$$. Now, if you perform this for some $$d$$ times, then for an expected $$d/2$$ times you obtain $$|0\rangle$$ during the first measurement and so obtain $$|000\rangle$$ as the final measurement and for an expected $$d/2$$ times you obtain $$|1\rangle$$ during the first measurement and so obtain $$|111\rangle$$ as the final measurement. Now, see that in both the cases, you obtain the same probability distribution. But in the first case we use two two-qubit gates but in the second case, we use no two-qubit gates. So irrespective of whether you measure a qubit after all the gates in a circuit or whether you measure a qubit just after all the rotations on it have been applied, the output probability distribution turns out to be the same. This is the idea that the author tries to convey and use in constructing the single-qubit-gate-only QFT implementation in the paper.


text:
(a) Evaluate A in the scalar equality (2 + 4)4 =A.

(b) Evaluate A, B, and C in the vector equality2.00 +4.00 = A +B + C .
 A = B = C =

Explain how you arrive at your answers to convince a student whothinks that you cannot solve a single equation for three differentunknowns.

(c) The functional equality or identity given below is true for allvalues of the variables x and t, which aremeasured in meters and in seconds, respectively.
A + B cos (Cx + Dt +E) = (2.00 mm) cos (4x + 4t + 5)
Evaluate the constants.
 A = mm B = mm C = rad/m D = rad/s E = rad

Explain how you arrive at your answers.