text:
# Euler Phi Function is Multiplicative

## Theorem

$m \perp n \implies \phi \left({m n}\right) = \phi \left({m}\right) \phi \left({n}\right)$

where $m, n \in \Z_{>0}$.

## Proof

Let $R = \left\{{r_1, r_2, \ldots, r_{\phi \left({m}\right)}}\right\}$ and $S = \left\{{s_1, s_2, \ldots, s_{\phi \left({n}\right)}}\right\}$ be the reduced residue systems for the respective moduli $m$ and $n$.

We are to show that the set of $\phi \left({m}\right) \phi \left({n}\right)$ integers:

$T = \left\{{n r + m s: r \in R, s \in S}\right\}$

is a reduced residue system for modulus $m n$.

We need to establish the following:

We prove each in turn:

• Suppose $p$ is a prime divisor of $\gcd \left\{{n r + m s, m n}\right\}$ where $r \in R, s \in S$.

As $p$ divides $m n$ but $m \perp n$, $p$ either divides $m$ or $n$ but not both, from Divisors of Product of Coprime Integers.

Suppose WLOG that $p \mathop \backslash m$.

Then as $p \mathop \backslash n r + m s$, we have $p \mathop \backslash n r$ and hence $p \mathop \backslash r$.

But then $p \mathop \backslash \gcd \left\{{m, r}\right\} = 1$ which is a contradiction.

Similarly if $p \mathop \backslash n$.

So there is no such prime and hence $n r + m s \perp m n$.

• Suppose $n r + m s = n r' + m s' \left({\bmod\, m n}\right)$, where $r, r' \in R, s, s' \in S$.

Then:

$n \left({r - r'}\right) + m \left({s - s'}\right) = k \left({m n}\right)$ for some $k \in \Z$.

As $m$ divides two of these terms it must divide the third, so $m \mathop \backslash n \left({r - r'}\right)$.

Now $m \perp n$ so by Euclid's Lemma $m \mathop \backslash \left({r - r'}\right)$, or $r \equiv r' \left({\bmod\, m}\right)$.

But $r$ and $r'$ are part of the same reduced residue system modulo $m$, so $r = r'$.

Similarly for $n$: we get $s = s'$.

Hence distinct elements of $T$ can not be congruent modulo $m n$.

• Let $k \in \Z: k \perp m n$.

Since $m \perp n$, from Set of Integer Combinations equals Set of Multiples of GCD we can write $k = n r' + m s'$ for some $r', s' \in \Z$.

Suppose there exists some prime number $p$ such that $p \mathop \backslash m$ and $p \mathop \backslash r'$.

Such a prime would be a common divisor of both $k$ and $m n$, contradicting $k \perp m n$.

Hence $r' \perp m$ and so is congruent modulo $m$ to one of these integers in $R$.

By the same argument, $s' \perp n$ and so is congruent modulo $n$ to one of these integers in $S$.

Writing $r' = r + a m, \, s' = s + b n$ we have:

$k = n r' + m s' = n r + m s + m n \left({a + b}\right) \equiv n r + m s \left({\bmod\, m n}\right)$.

Hence the result.

$\blacksquare$


text:
Lemma 70.10.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. For $U$ affine and étale over $X$ the set $\mathcal{S}_ X(U)$ is the set of nonzerodivisors in $\mathcal{O}_ X(U)$.

Proof. Follows from Lemma 70.7.5. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

format: plain


text:
# BUS 670 Employment Law

This document of BUS 670 Employment Law comprises:

Respond to chapter 51, Problem 3 on Title VII:Dianne Rawlinson, a female applicant who was rejected for employment as a prison guard in the Alabama prison system, challenged certain state rules restricting her employment prospects under Title VII. They were (1) requirements that all prison employees be at least 5 feet 2 inches tall and weigh at least 120 pounds, and (2) a rule expressly prohibiting women from assuming close-contact prison guard positions in maximum-security prisons (most of which were all male). What method of proving a Title VII case should Rawlinson use in attacking the height and weight requirements? Does she need to use one of these methods to attack the second rule? What argument should the state use if Rawlinson establishes that the height and weight requirements have an adverse impact? What Title VII defense might the state have for the second rule? With regard to the second rule, assume that at this time Alabama

• Proftech
1 orders completed
$14.29 ANSWER Tutor has posted answer for$14.29. See answer's preview

*** 670 Employment ***


text:
## needspace.sty?

Is there something like needspace.sty for SWP?  (I tried dropping it into the .tex file's directory, added it to packages, and ran SWP, but then SWP just waited forever.)

regards,

/iaw

### When you say SW waited

When you say SW waited forever, this implies that you selected to compile for PDF.  The console output from pdflatex is directed to a TrueTeX Previewer window, but this window is most often behind other windows.  If there is a LaTeX error, you need to interact with the console to somehow allow the LaTeX compile to complete.  The console window also shows the error.  It's difficult to guess what happened without knowing the error.

The CTAN directory for needspace contains a .ins and .tex file (plus a README and the documentation in a PDF).  You will need to run LaTeX on the .ins file (use the TrueTeX formatter, entering in the .ins file name and ignoring the .tex filter that shows only .tex files in the open dialog).  You should then have needspace.sty that needs to be in some directory below TCITeX\TeX (to be consistent with other CTAN packages, use TCITeX\TeX\LaTeX\contrib\needspace).

After completing the above, you can add the package to the document and then use the macros defined by the package in your document.

If you continue to have difficulty, post the error messages you are seeing.

### You need to put it in the

You need to put it in the latex tree : try

c:\swp55\tcitex\tex\latex\contrib\needspace

(where you create the needspace folder in the existing contrib one).


text:
# Refraction seismology - travel time for wave

I am taking an introductory class in seismology, but have some difficulties understanding the logic behind the formula used to calculate the time it takes for a refracted wave to return to the surface from which it was sent out originally.

So, say we have a seismic wave sent out from a horizontal layer. We also assume that the interfaces below are horizontal, and parallel to the top layer. If the wave hits the first interface at the so-called critical angle, it will be reflected back to the surface with the same angle at which it hit the interface. Now, the formula given in the book states that in order to calculate travel time, we use:

$$t = \frac{2 h_1 cos(i_c)}{v_1}$$

where $h_1$ is the vertical height of the layer, $i_c$ is the critical angle, and $v_1$ is the velocity of the wave as it travels through the layer.

As mentioned, I don't quite see the logic of this formula. I know that time = distance/velocity. Surely the shortest possible distance to travel back and forth between two horizontal layers, separated with distance $h_1$ is simply $2 h_1$. So by that logic, the shortest possible time a wave could possibly use, would be:

$$t = \frac{2 h_1}{v_1}$$

But in the given formula we multiply the numerator with $cos(i_c)$, which would yield an even shorter value for distance. And I just can not see how this makes sense at all. Now, if we were to use the cosine-term in the denominator, then the formula would make perfect sense to me. But why isn't it so? Any help here would be greatly appreciated!

-
I think, and I remark that I think (I need to review my optics notes) that you can derive that formula via Snell's law and imposing minimum condition, so that the sine in Snell gives you a cosine in the derivative –  Jorge Jan 24 '13 at 0:37
Thanks. I would really, really be grateful if you could provide me with the derivation. –  user12277 Jan 24 '13 at 8:19


text:
## “On Nonlinear PDEs and their Applications”

March 1-2, 2019

Department of Mathematics
2515 Speedway
Robert Lee Moore Hall (RLM) 8.136
The University of Texas at Austin
Austin, TX 78712

During the conference, we will honor the 70th birthday of Dr. Luis A. Caffarelli, who has made fundamental contributions to the field of partial differential equations and their applications.

Invited Lecturers

##### Ioannis Athanasopoulos, Crete University, HeraklionAri Arapostathis, UT AustinHenri Berestycki, EHESS ParisRafael de la Llave, Georgia Inst. of TechnologyYanyan Li, Rutgers UniversityFanghua Lin, Courant Institute, NYUFrancesco Maggi, UT AustinManolis Milakis, University of CyprusStefania Patrizi, UT AustinJean Michel Roquejoffre, U. Paul Sabatier, Toulouse FranceYannick Sire, John Hopkins UniversityTakis Souganidis, The University of ChicagoAlexis Vasseur, UT Austin

Program Schedule

Friday, March 1st

09:15am – 10:00am — Fanghua Lin
“A few Problems Related to Periodic Homogenization”

10:05am – 10:50am — Francesco Maggi
“Plateau’s problem as a singular capillarity problem”

11:00am – 11:20am — Coffee break

11:20am – 12:05pm — Ioannis Athanasopoulos
“A Fractional Heat Obstacle Problem”

12:10pm – 12:55pm — Stefania Patrizi
“Dislocations dynamics: from microscopic models to macroscopic crystal plasticity”

01:00pm – 02:30pm  — Lunch break

02:30pm – 03:15pm — Manolis Milakis
“Regularity for fully nonlinear Parabolic equations with oblique boundary data”

03:20pm – 04:05pm — Alexis Vasseur
“De Giorgi holder regularity theory applied to kinetic-type equations”

04:05pm – 04:25pm  — Coffee break

04:25pm – 05:10pm — Jean-Michel Roquejoffre
“Large time behaviour in multi-dimensional Fisher-KPP type equations”

05:15pm – 06:00pm — Yannick Sire
“Extremal metrics and isoperimetry on riemannian manifolds”

06:45pm — Dinner

****************************

Saturday, March 2nd

09:15am – 10:00am — Takis Souganidis

10:05am – 10:50am — Ari Arapostathis
“A lower bound on the rate of convergence for a class of Lévy driven SDEs arising in the study of stochastic networks”

10:55am – 11:40am  — Yanyan Li
“Existence and uniqueness of Green’s function to a nonlinear Yamabe problem”

11:45am – 12:05pmCoffee break

12:05pm – 12:50pm — Rafael de la Llave
“Some examples in the global theory of minimizing geodesics and minimal surfaces”

12:55pm  – 01:40pm — Henri Berestycki
Predators-prey model with competition : the emergence of territoriality

01:45pm – Closing remarks (Irene M. Gamba)

Abstracts

Ari Arapostathis
“A lower bound on the rate of convergence for a class of Lévy driven SDEs arising in the study of stochastic networks”

We first describe how non-local operators arise in the study of stochastic queueing networks. This happens when the arrivals are heavy-tailed, or the system suffers from asymptotically negligible service interruptions. In these models, the Itô equations in the so called “diffusion” approximation are driven by either (1) a Brownian motion and a pure-jump Lévy process, or (2) an anisotropic Levy process with independent one-dimensional symmetric alpha-stable components, or (3) an anisotropic Lévy process and a pure-jump Lévy process. We identify conditions on the parameters in the drift, the Lévy measure and/or covariance function which result in subexponential and/or exponential ergodicity, and we show that these conditions are sharp. In addition, we show that for the queueing models described above with no abandonment, the rate of convergence to the stationary distribution in total variation is polynomial, and we provide a sharp quantitative characterization of this rate via matching upper and lower bounds. From a PDE perspective this rate of convergence describes the large-time behavior of the solution of the associated parabolic equation to that of the elliptic. Deriving an upper bound for the rate of convergence is standard. It is the lower bound has some novelty.  This talk is based on joint work with Hassan Hmedi, Guodong Pang and Nikola Sandric.

Ioannis Athanasopoulos
“A Fractional Heat Obstacle Problem”
In the class of the non-dynamic Fractional Obstacle Problems of parabolic type, it is shown existence, uniqueness, apriori bounds of the solution and optimal regularity of the space derivatives of the solution. Furthermore, at free boundary points of positive parabolic density, it is proven that the time derivative of the solution is H\”{o}lder continuous. At such free boundary points, space-time regularity of the corresponding free boundary is obtained for any fraction $s\in(0,1)$

Henri Berestycki
Predators-prey model with competition : the emergence of territoriality”
In this talk, I report on joint work with Alessandro Zilio in which we propose a new model for predators like wolves that can divide into several hostile packs. We aim at showing how territories and packs are formed among certain predators. This model, an extension of the original Lotka-Volterra system, rests on basic principles of predator – prey interaction and competition.

We analyze stationary states and various asymptotic behaviors of this system, especially when the competition parameter between packs becomes unbounded. To analyze the limiting state, we mainly rely on methods developed by L. Caffarelli and F-H. Lin. One particular novel feature of our approach is to treat the number of components (packs) as a variable. The questions we address are to understand the conditions under which predators segregate into packs, how many packs there can be, whether there is an advantage to have such hostile packs, and to discuss the different territory configurations that can arise. We can then compare the implications of this model with various observations.

Rafael de la Llave
“Some examples in the global theory of minimizing geodesics and minimal surfaces”
In a famous paper on the theory of flat chains. H. Federer introduces an example  that he attributes to F. Almgren.  The example is really an example of  metric that has the property that none of the miniming periodic orbits is really a class A minimizer. In that respect, it is more dramatic than the better known example of Hedlund in which only three mininizing periodic orbits are class-A minimizing.  There are many related examples.

Yanyan Li
“Existence and uniqueness of Green’s function to a nonlinear Yamabe problem”
For a given finite subset S of a compact Riemannian manifold (M; g) whose Schouten curvature tensor belongs to a given cone, we establish a necessary and sufficient condition for the existence and uniqueness of a conformal metric on M∖S
M
S
such that each point of S corresponds to an asymptotically flat end and that the Schouten tensor of the new conformal metric belongs to the boundary of the given cone.  This is a joint work with Luc Nguyen.

Fang-Hua Lin
“A few Problems Related to Periodic Homogenization”
A solution of the heat equation is called an ancient solution if it is
defined for all time. One can classify completely all positive ancient solutions or ancient solutions of polynomial growth solutions for the standard heat equation. Generalizations of such results for heat equations on certain classes of complete Riemannian manifolds and uniformly parabolic equations are possible. It is particularly interesting to understand the case that equations have periodic coefficients. We will discuss a few questions with regarding more explicit representations. There are also related questions concerning Nodal and Critical points of solutions of elliptic (parabolic) equations with periodic coefficients. Some recent progress and open problems will be presented.

Francesco Maggi
“Plateau’s problem as a singular capillarity problem”
We introduce a length scale in Plateau’s problem by modeling soap films as liquid with small volume rather than as surfaces, and study the relaxed problem and its relation to minimal surfaces. This is based on joint works with Antonello Scardicchio (at ICTP Trieste), Darren King and Salvatore Stuvard (at UT Austin).

Manolis Milakis
“Regularity for fully nonlinear Parabolic equations with oblique boundary data”
I will give a brief overview of our recent results regarding boundary regularity, in parabolic H\”older spaces, for viscosity solutions of fully nonlinear parabolic equations with oblique boundary conditions. This is a joint work with Georgiana Chatzigeorgiou.

Stefania Patrizi
“Dislocations dynamics: from microscopic models to macroscopic crystal plasticity”
We present some recent results obtained  for equations of evolutionary type, run by fractional and possibly anisotropic fractional operators. The models considered arise natural in crystallography, in which the solution of the equation has the physical meaning of the atom dislocation inside the crystal structure. Since different scales come into play in such description, different models have been adopted in order to deal with phenomena at atomic, microscopic, mesoscopic and macroscopic scale. We show that, looking at the asymptotic states of the solutions of equations modeling the dynamics of dislocations at a given scale, one can deduce the model for the motion of dislocations at a larger scale. The results presented  are joint works with R. Monneau and E. Valdinoci.

Jean-Michel Roquejoffre
“Large time behaviour in multi-dimensional Fisher-KPP type equations”
The Fisher-KPP (Kolmogorov, Petrovskii, Piskunov) equation in variable is one of the simplest looking reaction-diffusion equations. In one space dimension the level sets of the solution starting from a Heaviside initial datum will converge, in some Galilean reference frame, and up to an additional nontrivial logarithmic time delay, to constants. This result was proved by Bramson in the early 80’s, using elaborate probabilistic arguments. This result was extended by Gärtner in the following way: in several space dimensions the solution starting from a compactly supported initial datum will, still in a Galilean reference frame, and up to a logarithmic time delay that is different from the 1D one, will be bounded from above and below. The goal of this talk is to further this result: in every direction, the level sets stabilise to constant positions that, in general, will depend on the direction. Joint work with L. Rossi and V. Roussier.

Yannick Sire
“Extremal metrics and isoperimetry on riemannian manifolds”
Several years ago, N. Nadirashvili and myself started a program devoted to the study of critical metrics for eigenvalues of the Laplace-Beltrami operator on surfaces. This led to several existence and regularity results. In this talk, I will describe the main features of the results and how to extend them to several contexts or other type of operators such as Schrodinger operators.  I will emphasize in particular on some open problems and some new directions in the higher dimensional case. This type of problem is at the interface of PDEs, real and complex geometry and also, surprisingly, algebraic geometry.

Alexis Vasseur
“De Giorgi holder regularity theory applied to kinetic-type equations”
In this talk, we will present recent results of holder regularity for solutions to kinetic equations. It includes a result of partial regularity for the homogenous Landau equation (in collaboration with Francois Golse and Cyril Imbert). We will also present a result of holder regularity for a class of hypo-elliptic nonlocal equation (result of Logan Stokols)

The Organizing Committee

Irene Gamba, ICES, UT Austin
Yannick Sire, Johns Hopkins University


text:
# Continued Fractions

• A finite continued fraction is an expression of the form

• Finite continued fractions with integer terms represent rational numbers.
• Every rational number can be expressed as a finite continued fraction using the Euclidean Algorithm, but the expression is not unique.
• The convergent of is (the value of) the continued fraction .
• The convergents oscillate about the value of the continued fraction.
• The convergents of a continued fraction may be computed from the continued fraction using a recursive algorithm. The algorithm produces rational numbers in lowest terms.

Definition. Let , ... be real numbers, with , ..., positive. A finite continued fraction is an expression of the form

To make the writing easier, I'll denote the continued fraction above by . In most cases, the 's will be integers.

Example.

In short form, .

A little bit of thought should convince you that you can express any rational number as a finite continued fraction in this way. In fact, the expansion corresponds to the steps in the Euclidean algorithm. First,

Rewrite these equations as

You can get the continued fraction I found above by substituting the third equation into the second, and then substituting the result into the first.

Since this is just the Euclidean algorithm, I can use the first two columns of the Extended Euclidean Algorithm table to get the numbers in the continued fraction expansion:

Notice that the successive quotients 2, 1, 3, and 4 are the numbers in the continued fraction expansion.

Example. Find the finite continued fraction expansion for .

Lemma. Every finite continued fraction with integer terms represents a rational number.

Proof. If , then is rational.

Inductively, suppose that a finite continued fraction with "levels" is a rational number. I want to show that

is rational.

By induction,

is rational.

So

is the sum of two rational numbers, which is rational as well.

Example. The continued fraction expansion of a rational number is not unique. For example,

And in general,

I want to talk about infinite continued fractions --- things that look like

In preparation for this, I'll look at the effect of truncating a continued fraction.

Definition. The convergent of the continued fraction is

Note that for , .

Example.

And as well.

The next result gives an algorithm for computing the convergents of a continued fraction. It's important for theoretical reasons, too --- I'll need it for several of the proofs that follow. For the theorem, I won't assume that the 's are integers, since I will need the general result later on.

Theorem. Let , , ..., be positive real numbers. Let

Then the k-th convergent of is .

Proof. First, note that

by regarding the last two terms as a single term.

Note also that , ..., and , ..., are the same for these two fractions, since they only differ in the k-th term.

Now I'll start the proof --- it will go by induction on k. For ,

And for ,

Suppose , and assume that result holds through the k-th convergent. Then

Now this is the k-th convergent of a continued fraction, so by induction this is , where and refer to (as opposed to ). But what are the and for this fraction? They're given inductively by

Now , , , are the same for and , as I noted at the start. On the other hand, the k-th term of is . So

(The next to the last equality also follows by induction.) This shows that the result holds for , so the induction step is complete.

Example.

There is a pattern to the computation of the p's and q's which makes things pretty easy. To get the next p, for instance, multiply the current a by the last p and add the next-to-the-last p.

Notice that the convergents oscillate, and that the fractions which give the convergents are always in lowest terms.

Example.

Again, notice that the convergents oscillate, and that the fractions for the convergents are always in lowest terms.

I'll prove that the convergent fractions are in lowest terms first.

Theorem. Let , , ..., be positive real numbers. Let

Then

Corollary. Let , , ..., be positive integers. Let

For , is in lowest terms.

Proof. implies that .

Proof of the Theorem. I'll induct on k. For ,

Take , and assume the result holds for k. Then

This proves the result for , so the general result is true by induction.

Example. I'll show later that (the golden ratio) has the infinite continued fraction expansion . Here are the first ten convergents:

In fact, . In this case, you can see formally that should be . Let

The roots are . Since the fraction is positive, take the positive root to obtain .

Contact information


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# A doctor begins a skin cancer patient on a new treatment

Author Message
Current Student
Joined: 29 Jan 2005
Posts: 5218
Followers: 26

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A doctor begins a skin cancer patient on a new treatment [#permalink]

### Show Tags

23 Nov 2005, 06:34
00:00

Difficulty:

(N/A)

Question Stats:

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### HideShow timer Statistics

A doctor begins a skin cancer patient on a new treatment that has met with great success in other patients. The new treatment involves a new drug and is less painful and invasive than some of the other available options and is more advanced and specialized. However, after she begins the new treatment, the patents health begins to decline.

Which of the following, if true, would give the best partial explanation of the discrepancy below?

(A) While this drug does kill these cancer cells, this is just one application for which it is commonly prescribed.
(B) The drug is not particularly discerning between cancer and non-cancer cells.
(C) The patient began other treatments right after she started taking the drug.
(D) Right after beginning the drug, the patient recovered from a case of pneumonia.
(E) Of all the cancer drugs on the market, the prescribed drug has the highest success rate.
SVP
Joined: 28 May 2005
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Location: Dhaka
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### Show Tags

23 Nov 2005, 06:42
I am torn between C and D.

but on the exam day I will choose C.
_________________

hey ya......

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Posts: 1482
Location: Germany
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### Show Tags

23 Nov 2005, 07:48

E) is out
B) is out b/c the arg says the drug is spezialized. moreover it states that the drug is successful with other patients. out
D) out b/c recover means that he gets better
A) too general

C) is left ! why ? b/c it may habe negative side effects with other drugs
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Current Student
Joined: 29 Jan 2005
Posts: 5218
Followers: 26

Kudos [?]: 404 [0], given: 0

### Show Tags

23 Nov 2005, 21:14
OA is B

OE:

Step 1: ID Question Type and Task
The question asks you to explain the discrepancy which indicates that this is a Paradox Question.

Task: Find the seemingly contradictory issue and pose it in the form of a question.

Step 2: Read the Argument and Extract Necessary Information:
The necessary information is the paradox, identify the paradox and pose it to yourself as a question.

Paradox: Why would the patients health decline after starting a drug that is successful in other patients and by most standards better than other drugs?

Step 3: Formulate an Answer to the Question
Formulate a resolution to the paradox (if possible).

Remember, the best answer provides a connection between the two sides of the paradox without changing or discounting either side.

(A) Keep it. This might explain why what was expected didnt occur.

(B) Keep. This presents a partial explanation for the ineffectiveness of the treatment, by suggesting that it caused as many problems as it cured.

(C) Keep it. This might explain the discrepancy.

(D) Doesnt explain. It is irrelevant that the patient just recovered from pneumonia.

(E) Doesn`t explain. This further confirms the paradox since the success should be expected in this patient.

Step 5: Identify differences in remaining choices

(A) This choice is partially irrelevant since the other treatments for which it is used are not relevant to our issue here.

(C) This choice requies the assumption that the treatments are in conflict, but does not directly point to negative consequences of other treatments.

(B) This provides the best explanation, though it does not address why the treatment was successful in other patients it does clarify why it was successful in this patient
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24 Nov 2005, 21:41
A - Cannot be inferred.
B - Correct.
C - Maybe correct.
D - Out of Scope
E - Out of Scope.

Between B & C I vote for B.
24 Nov 2005, 21:41
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text:
## The Poisson formula for groups with hyperbolic properties.(English)Zbl 0984.60088

Given a Markov operator $$P$$ on a Lebesgue measure space $$(X,m)$$, there exists a space $$\Gamma$$ (called the Poisson boundary of $$P$$) equipped with a family of probability measures $$\nu_x$$, $$x\in X$$, such that the Poisson formula establishes an isometry between the space of $$P$$-harmonic functions (that is, functions $$f$$ satisfying $$Pf=f$$) from the space $$L^{\infty}(X,m)$$ to the space $$L^{\infty}(\Gamma)$$. The Poisson boundary is defined as the space of ergodic components of the time shift $$T$$ in the space of sample paths of the Markov chain of $$X$$ associated with the operator $$P$$, the measures $$\nu_x$$ being the images of the measures in the path space corresponding to starting the Markov chain at points $$x\in X$$. In the case where the space $$X$$ is endowed with additional (geometric, algebraic, etc.) structures, the operator $$P$$ is supposed to have some properties which make it compatible with this structure, and it is natural to ask for a description of the Poisson boundary (which in principle is a measure-theoretic object) in terms of this structure. This paper addresses the question of identifying the Poisson boundary for the random walk operator determined by a probability measure $$\mu$$ on a countable group $$G$$. The author develops new methods and obtains new results on the subject, generalizing previous results of himself and of other authors. The methods consist in describing the Poisson boundary of certain groups, via entropy estimates for conditional random walks. The boundaries are identified then as natural topological boundaries. The classes of groups concerned are those of word hyperbolic groups and discontinuous groups of isometries on Gromov hyperbolic spaces, groups with infinitely many ends, cocompact lattices in Cartan-Hadamard manifolds and discrete subgroups of semi-simple Lie groups.

### MSC:

 60J50 Boundary theory for Markov processes 28D20 Entropy and other invariants 53C22 Geodesics in global differential geometry 31C05 Harmonic, subharmonic, superharmonic functions on other spaces 20F67 Hyperbolic groups and nonpositively curved groups
Full Text:


text:
1

# Why is my number script not working properly?

Edited 5 days ago

Whenever I playtest the game with 2 players, only their number gets shown, but not anyone else's.

For example: 2 players are in the game, they both have BillboardGuis on top of them. Both of those players can see "Your number is: THIS NUMBER" on their character, but when they look at each other's billboards, it just says "Your number is: LOADING...".

Here's the image link to explain it even further: http://pasteboard.co/nmpCqo9GrRD4.png

Here are the scripts I used:

The 1st script:

local billboard = script.BillboardGui

local NumText = billboard:Clone()
end)
end)

The 2nd script:

local billboardamount = game.Players.LocalPlayer.Character.Head:WaitForChild("BillboardGui").TextLabel

local RandNum = Instance.new("NumberValue", game.Players.LocalPlayer)
RandNum.Value = math.random

while wait(0.1) do
billboardamount.Text = "Your number is: "..
end

0

I think issue is that second script is in localscript so it will only show to player, what you should do is you can make remote event and fire it, then put script in serverscriptstorage and onserverevent change text, like this.

LocalScript

local billboardamount = game.Players.LocalPlayer.Character.Head:WaitForChild("BillboardGui").TextLabel

local RandNum = Instance.new("NumberValue", game.Players.LocalPlayer)
RandNum.Value = math.random

while wait(1) do
game.ReplicatedStorage.RemoteEvent:FireServer(RandNum)
end

ServerScript

game.ReplicatedStorage.RemoteEvent.OnServerEvent:Connect(function(player,randnum)


text:
# pyAgrum.causal documentation¶

Causality in pyAgrum mainly consists in the ability to build a causal model, i.e. a (observational) Bayesian network and a set of latent variables and their relation with observation variables and in the abilidy to compute using do-calculus the causal impact in such a model.

Causality is a set of pure python3 scripts based on pyAgrum’s tools.

Tutorial

Reference


text:
## Precalculus (6th Edition) Blitzer

$25^o$
RECALL: An angle $\theta$, where $-360^o \lt \theta \lt 0$, is coterminal with: $\theta+ 360^o$ Thus, theusing the formula above, given angle is coterminal with: $=-335^o + 360^o \\=25^o$ Thus, $-335^o$ is coterminal with $25^o$. $25^o$ is in Quadrant I so $-335^o$ is also in Quadrant I. RECALL: The following are the means on how to find the reference angle of an angle $\theta$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $180^o-\theta$ (3) Quadrant III: $\theta - 180^o$ (4) Quadrant IV: $360^o-\theta$ Use formula (1) above to obtain: reference angle of $-335^o$ = reference angle of $25^o$, which is $25^o$.


text:
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# Create a Fashionable Dress Shoe in Illustrator

In this tutorial, you'll learn how to create a dress shoe with the help of gradient meshes. We will also explore how to create custom vector textures.

## Tutorial Assets

You can find this tutorial's reference photo, as well as other fashionable dress shoe reference photos, on Thinkstock.

## Step 1

Start by opening up your image in Photoshop. Use the Marquee Tools to extract one shoe and save it as a separate image. I would normally recommend the Pen Tool; however, there's no need to be precise, as we are just using this image as a template.

## Step 2

Copy and Paste the new image into Illustrator (I've flipped it for preference). In the Layers Panel, click the Flyout menu and select "Template."

## Step 3

Next, create a new layer called "Colors" (This is where we will store the initial colors sampled from the shoe).

## Step 4

With the "Colors" layer active, use the Eyedropper Tool (I) to sample a dominant color from the shoe and switch to the Blob Brush (Shift + B) to paint a swatch on your artboard (notice that the color is the original color from the shoe and not the screened color as it appears on the layer).

## Step 5

Continue to select the major colors from the shoe. It may help to number the colors to keep track of which colors go where.

## Step 6

Now we can add these colors to our swatches in groups, just select a group of colors at a time and click on the new Color Croup icon in to the swatches panel. Do this for all color groupings.

Here, all the color groups have been added. I also made sure that the colors were in orders as I have labeled on the "Colors" layer.

## Step 7

Before we start working on building the shoe, we should plan out how we will approach the gradient maps. I've quickly drawn a grid that I think best follows the contours of the shoe. As you'll see by the red boxes, we will have four separate gradient maps that we need to create: One for the front of the shoe, two for the middle (due to the complexity of the shape) and one for the back.

## Step 8

To start applying gradient maps, create a new layer called "Front" and use the Rectangle Marquee Tool to draw a shape over the front part of the shoe.

## Step 9

Go to Object > Create Gradient Mesh. Set the values as shown and Click OK. We want to start with a basic shape first. We will add additional control points later.

## Step 10

Position four corners of the mesh in to the logical 'corners' of the shoe. It may help to use Outline Mode (Command + Y). Use the handles to manipulate the mesh to follow the outline of the shoe.

## Step 11

Select the Gradient Mesh Tool and click on the edges of the Gradient Mesh to add additional rows and columns. As you add new mesh lines, adjust the handles so they follow the general direction of the grid we sketched out in Step 7.

## Step 12

Now, we can start to color the mesh. We do this by assigning a color to a point on the mesh. Use a combination of the Direct Selection Tool and the Lasso Tool to select the points on your mesh, and use the Swatches Panel to add a color to the point(s).

Use the color guide we created in Steps 3-4 to help the process of assigning colors. You may need to manually adjust some points so that the colors match the reference image. To do this, simply use the Direct Selection Tool to move and use the handles to ease in and out.

Your colored mesh should resemble the following:

## Step 13

Repeat Steps 8-12 with the back part of the shoe.

## Step 14

The middle of the shoe is a little different. We have two gradient meshes we need to create (refer back to the image we sketched of the gradients):

Using previous techniques create a simple gradient mesh for both sections. The trick is to make sure both meshes have the same amount of rows and that they are relatively in line with each other.

## Step 15

We need to connect both gradient meshes. First, make sure Snap to Point is on (View > Snap to Point). Next, Using the Direct Selection Tool, snap each point of the shared edge to its corresponding neighbor.

## Step 16

To fix the gaps in between each point, you'll need to match up the handles of both points. To do this, it helps to have Smart Guides turned on.

Select both points, and use the Direct Selection Tool to move the handles of both corresponding points to the same location. Smart Guides will not only snap to the handle, but it will also let you know when you've reached the handle.

## Step 17

Go ahead and add color to both meshes (make sure to add the same color to the points that overlap).

## Step 18

To add Texture to the shoe, go back into Photoshop and create a new document 1000 px by 1000 px. Add Noise by going to Filter > Noise > Add Noise. Use the settings shown below.

## Step 19

Next, go to Filter > Blur > Gaussian Blur with the following settings.

## Step 20

Go to Image > Adjustments > Posterize and set the levels to 5.

Your image should now look like this:

Adjust the Levels (Command + L) to increase the contrast.

## Step 21

Copy and paste this image into a new layer in Illustrator and use Illustrator's Live Trace to bring up the Live Trace settings.

For this texture, we will keep the default settings. Click Expand to covert it to a vector and drag the newly created texture into the Symbols Panel. A property dialogue box will open, you may want to rename the symbol, but any other changes are unnecessary, just click OK to complete.

## Step 22

We can now start applying this texture to each part of the shoe. To do this, select the front section of the shoe and go to Object > Path > Offset Path. Set this to 0. The result will be 2 new shapes, the original gradient, and the offset path. We will be working with the offset path.

## Step 23

Drag the Texture symbol from the Symbols Panel onto your document and position it so it covers the entire front of the shoe. Since order matters when working with clipping paths, send the texture to the back of the layer stack (Object > Arrange > Send to Back).

Select both the texture and the offset path, then go to Object > Clipping Mask > Make. You now have texture in the shape of your shoe, but it is in the back of your layer stack. Just go to Object > Arrange > Bring to Front to fix this.

## Step 24

In the Transparency panel, set the Opacity of the texture to 3% and set the Blend Mode to Hard Light.

## Step 25

Repeat steps 22-24 to add texture to the remaining sections of the shoe (Note: The middle has two shapes; these should be combined into one shape using the Unite option in the Pathfinder Panel before applying the texture). The result should look like this:

## Step 26

The next step is to add the seams on the shoe. On a new layer called "Seams," use the Rectangle Tool to create a thin rectangle. Next, use the Direct Selection Tool to move each corner into position. Lastly, convert these corner points to end points, then use the Direct Selection Tool to manipulate the shape to follow the seam on the reference image.

## Step 27

Select the Mesh Tool and start to add mesh lines as shown.

## Step 28

Using the reference image, add color to the new mesh.

## Step 29

Repeat these steps for the remaining seams.

## Step 30

Following the same techniques that we just learned, go ahead and create the shoe's tongue on a new layer. Do the same for the inside of the shoe (the inside does not require a texture).

## Step 31

To create the shoe's sole, start by creating a rectangle. Then, use the Mesh Tool (U) to add mesh lines as shown.

## Step 32

By sticking to the basic rectangle shape, we can add a gradient mesh with much more control. Go ahead and use the additional points to help fine-tune the contours of the shoe's sole.

## Step 33

Use the same techniques to build the shoe's heel.

## Step 34

Use the reference photo to add more mesh lines around the treads. It may help to view in Outline Mode (Command + Y).

## Step 35

Once you're done with the treads, go ahead and add some more definition to the gradient mesh. Use the reference image to add color.

## Step 36

To add texture to the sole, select each shape and go to Effect > Artistic > Film Grain.

Use the following settings:

Your image should resemble the following:

## Step 37

The sole still seems flat. Since the shoe will be resting on the ground, the sole should naturally appear darker closer to the ground plane. To achieve this effect, first make a copy (Command + C) of both shapes that make up the sole, and paste them in front (Command + F).

Next, change the color of the gradient mesh to all white. Now, use the Mesh Tool to add a horizontal mesh line halfway up the one shape (this does not need to be done to the heel shape). You can now add black to the bottom points of the sole.

Change the two new meshes to a mode of Multiply with a Transparency of 55%.

## Step 38

Now, we'll continue to build the rest of the sole using the techniques from Steps 31 and 32. Add another rectangle and use the Mesh Tool (U) to add lines. Form this shape and add colors to the mesh.

## Step 39

Follow the same techniques to add another shape to act as the seam for the shoe leather and rubber sole. This time, add the Film Grain Effect (like we did in Step 36).

## Step 40

To finish the shoe, we will add some stitching. Do this by drawing a path next to each seam and giving it a dark brown color with a dotted line stroke.

## Step 41

Do the same for the inside of the shoe.

## Step 42

Lastly, add a gradient mesh underneath the shoe for the shoe's shadow.

## You're Done!

Your final image should resemble the following:

One subscription.


text:
# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2016 | June | Q#3

Question

a.   Simplify

Giving your answer in the form  , where a is an integer.

b.    Hence, or otherwise, simplify

Giving your answer in the form , where b and c is are integers and .

Solution

a.

We are given;

Since ;

Since ;

b.

We are given;

We have found in (a) that;

Therefore;

If we need a rational number in the denominator of a fraction, we need to follow procedure of  “denominator rationalization” as given below.

ü If the denominator is of the form  then multiply both numerator and denominator by .

ü If the denominator is of the form  then multiply both numerator and denominator by  .

ü If the denominator is of the form  then multiply both numerator and denominator by  .

Since ;


text:
# Abstract Nonsense

## Second Orthogonality Relation For Irreducible Characters

Point of post: In this post we discuss the second orthogonality relation for the irreducible characters. In particular, we prove that $\displaystyle \sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(g)\overline{\chi^{(\alpha)}(h)}=c(g,h)\#\left(\bold{C}_G(g)\right)$ where $c(g,h)$ takes the value one if $g$ and $h$ are conjugate and zero otherwise and $\bold{C}_G(g)$ is the centralizer of $g$ in $G$.

Motivation

In the past we’ve seen that if we ‘fix $\alpha,\beta\in\widehat{G}$ and let $g$ ‘vary’ over $G$ (in the form of the sum) that there is an interesting orthogonality relation. Namely, this is just the orthogonality relation $\displaystyle \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha)}(g)\overline{\chi^{(\beta)}(g)}=\delta_{\alpha,\beta}$. In this post we explore what happens if the ‘fixing’ and ‘varying’ are reversed. In other words, we compute $\displaystyle \sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(g)\overline{\chi^{(\alpha)}(h)}$. This shall serve as an interesting tool inall that comes. We call this the second orthogonality relation for the irreducible characters.

Second Orthogonality Relation for the Irreducible Characters

Theorem (Second Orthogonality Relation for the Irreducible Characters): For a finite group $G$ and $g\in G$ denote the centralizer of $k$ by $\bold{C}_G(g)$. Then, for $g,h\in G$ let $c(g,h)$ take the value one if $g$ and $h$ are conjugate  and zero otherwise. Then,

$\displaystyle \sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(g)\overline{\chi^{(\alpha)}(h)}=c(g,h)\#\left(\bold{C}_G(g)\right)$

Proof: Let $\mathcal{C}_1,\cdots,\mathcal{C}_k$ be the number of conjugacy classes in $G$ and let $g_1,\cdots,g_k$ be a set of representatives such that $g_j\in \mathcal{C}_j$ for $j=1,\cdots,k$. Recall then that $\#\left(\widehat{G}\right)=k$ and so the elements of $G$ can be labeled $\alpha_1,\cdots,\alpha_k$. We then consider the $k\times k$ matrix $M=[M_{i,j}]$ where $M_{i,j}=\chi^{(\alpha_i)}(g_j)$. We then let $D$ denote the matrix $\text{diag }\left(\#\left(\mathcal{C}_1\right),\cdots,\#\left(\mathcal{C}_k\right)\right)$. Recall though from the first orthogonality relation that

$\displaystyle |G|\delta_{i,j}=\sum_{g\in G}\chi^{(\alpha_i)}(g)\overline{\chi^{(\alpha_j)}(g)}$

But, recalling that the irreducible characters are class functions and thus constant on conjugacy classes it easily follows that this may be rewritten

$\displaystyle |G|\delta_{i,j}=\sum_{r=1}^{k}\#\left(\mathcal{C}_k\right)\chi^{(\alpha_i)}(g_r)\overline{\chi^{(\alpha_j)}(g_r)}$

This then implies that

$\displaystyle |G|I_k=MD M^{\ast}$

But, recall the fact that for square matrices a left inverse is necessarily a right inverse, and thus the above implies that

$|G|I=DM^\ast M$

But, the general $(i,j)^{\text{th}}$ entry of this right hand matrix is

$\displaystyle \sum_{r=1}^{k}\#\left(\mathcal{C}_i\right)\overline{\chi^{(\alpha_k)}(g_i)}\chi^{(\alpha_k)}(g_j)$

Dividing both sides and recalling by the orbit stabilizer theorem that $\displaystyle \#\left(\bold{C}_G(g_i)\right)=\frac{|G|}{\#\left(\mathcal{C}_i\right)}$ we see that

$\displaystyle \#\left(\bold{C}_G(g_i)\right)=\sum_{r=1}^{k}\chi^{(\alpha_k)}(g_j)\overline{\chi^{(\alpha_k)}(g_i)}$

And recalling all of our notation and taking the conjugate of both sides gives the desired result. $\blacksquare$

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

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March 7, 2011 -

## 5 Comments »

1. […] by the second orthogonality relation one has […]

Pingback by Representation Theory: Relation Between the Kernels of Characters and Normal Subgroups « Abstract Nonsense | March 7, 2011 | Reply

2. […] if then by proposition #1 (since and so ) we have that . So, using the second orthogonality relation we see […]

Pingback by Representation Theory: Burnside’s Theorem « Abstract Nonsense | March 11, 2011 | Reply

3. […] Proof: This is just a restatement of the second orthongonality relation for irreducible characters. […]

Pingback by Representation Theory: Character Tables « Abstract Nonsense | March 21, 2011 | Reply

4. […] using the second orthogonality relation and denoting the conjugacy class of by we may rewrite this […]

Pingback by Representation Theory: The Square Root Function and its Relation to Irreducible Characters « Abstract Nonsense | March 24, 2011 | Reply

5. […] where is any element of . Recall though that . Thus, it follows from the second orthogonality relation that […]

Pingback by Representation Theory: The Connection Between the Square Root Function and the Number of Self-Conjugate Irreps (Cont.) « Abstract Nonsense | March 25, 2011 | Reply


text:
# Average of a function pow

For hours I have tried to find the average of the next function :

p*r^t

Where p and r are constants and t is the variable. I want to know the average between t=0 and t=[value].

Can you help me?

The application is the consumption of food of a population which grow exponentially (p = members of the population at t=0 and r is the factor of reproduction). So I want to know, between two hours, what quantity of food was consumed.

Regards.

Edit for more details :

by exemple : If my population grow at 1% by hour and i've 10.000 members at 0h, at 24h my population = 12679 members (10.000 * 1.01^24). So at 0h, my population consume 100 food/h and at 24h, it consume 126.79 food/h. So i want to calculate, if my stock is 100.000 food at 0h00, the new quantity of food at 24h.

The algorythme to determinate the moment where food stocks are empty (without mathematic determination) is :

const populationStart = 10;
// Population grow at 1% each hour
const grow = 1.01;
const baseFood = 100_000;
// Each member of population eat 1 food / h
const consumeFood = 1;

let currentPopulation = populationStart;
let currentFood = baseFood;
function nextPopulation() {
return Math.log(Math.round(currentPopulation + 1) / currentPopulation) / Math.log(grow);
}

let totalTime = 0;
while (currentFood > 0) {
// Calculate moment where next baby born
const next = nextPopulation();
// Consume food quantity of current stable population quantity
currentFood -= currentPopulation * consumeFood * next;
// Grow population
currentPopulation *= Math.pow(grow, next);
totalTime += next;
// Show result
console.log('New pop', currentPopulation, 'food stocks :', currentFood, 'time:', totalTime);
}

With this program, my output is :

New pop 11 food stocks : 99904.21405960187 time: 9.578594039813165
New pop 12 food stocks : 99808.02377598119 time: 18.32316527805758
New pop 13 food stocks : 99711.49306697701 time: 26.36739102840552
New pop 14 food stocks : 99614.6718002056 time: 33.81518078005248
New pop 15 food stocks : 99517.59963094103 time: 40.748907156094006
New pop 16 food stocks : 99420.30861304936 time: 47.23497501553837
[...]
New pop 1006.0000000000001 food stocks : 133.70547554179498 time: 463.4169783608241
New pop 1007.0000000000001 food stocks : 33.25622154409804 time: 463.5168285139033
New pop 1007.9999999999999 food stocks : -67.19308199061292 time: 463.61657956011453

So between 463.51h and 463.61h, i've between 1006 and 1007 members and food stocks are empty. If want a mathematical formule to find "463.xxh" without a programmatic loop, and a formule to calculate "after 47.23h : i've consumed 579.69 food"

The average between what and what of what? Do you want an average time, or an average amplitude? You did not specify whether you want a probability density function or not, and this is a statistical site, not a mathematical site, so the answer is in terms of statistics.

Edit: After the OP explained the question well enough to specify what was actually being asked, it turns out that the question seems to be an algebra question, not a stats question.

$$P \,\left(1 + \frac{r}{n}\right)^{\,n\;t}\;\;,$$ where $$P$$ is the initial principal balance,

$$r$$ is the interest rate,

$$n$$ is the number of times interest is compounded per time period, and

$$t$$ is the number of time periods.

Since the population is eating food, we organize the principal to be the amount of food consumed in any time period to be $$P$$. Then, the total amount is merely the the sum of the increasing principal for each period, which we can write as $$\Sigma_{i=1}^n P \,\left(1 + \frac{r}{n}\right)^{\,n\;t}\;\;,$$

Now, if instead we desire the integral solution (as in continuous compounding), we take the limit as $$n$$ goes to infinity: $$\underset{n\to \infty }{\text{lim}}P \left(\frac{r}{n}+1\right)^{n\, \tau }=P\, e^{r\, \tau }$$ The integral from $$\tau$$ goes from $$0$$ to $$t$$ is then $$\int_0^t P e^{r \tau } \, d\tau=\frac{P \left(e^{r t}-1\right)}{r}$$

(Note the use of tau, $$\tau$$, as a dummy variable. This is a formal requirement for definite integration.) This seems to be the answer you sought. It is a statement equivalent to the total amount of food eaten by a "continuously" growing population up to time $$t$$. In other words, $$F(t)=P\frac{e^{r t}-1}{r},$$

where $$r$$ is the population growth rate assumed equal to the food eating growth rate per unit time period, $$t$$ is in time periods, $$F(t)$$ is the total amount of food eaten from time is zero to $$t$$, and $$P$$ is the rate of food consumption of an initial population per time period, for example tons of food eaten per year for 10,000 people, if the rate is in years, where we do not need to know exactly what the population is, just the rate at which a fixed number of people eat food. Note, you did not ask for what the population is at any time, but it's growth rate is assumed to be identical to $$r$$. However, if you want that number then it is assumed to be proportional to the rate at which food is eaten, which latter is $$P e^{r\,t}$$, as above.

Put some numbers in and check it for yourself. It is, in any case, a math problem and seems to have nothing to do with stats per se.

Old answer, under the assumption that the question was statistical follows. Unfortunately, that does not seem to be the case: The indefinite integral is $$\int p r^t \, dt=\frac{p r^t}{\ln (r)}$$

The definite integral between $$x=0$$ and $$x=t$$ is

$$\int_0^t p r^x \, dx=\frac{p r^t}{\ln (r)}\;\;,$$

where $$x$$ is a dummy variable for time, and $$t$$ is the elapsed time. As a probability density function from $$p r^t$$, we write the function $$f(t)=-r^t \ln(r)\text{ for }r<1\;\;,$$

because by definition $$\int_0^{\infty } -r^t \ln (r) \, dt=1,$$ for $$r<1$$. Notice that the variable $$p$$ has disappeared. It is not needed for a density, but if you wish to scale the density to have a particular amplitude (for example for curve fitting or for scaling to match some other amplitude), then multiply the density by $$p$$. The cumulative density function is $$F(t)=1-r^t\text{ for } r<1\;\;.$$

Now finally, the mean value of the independent variable of a density function is $$\overline{f(t)}=\int_0^\infty t f(t) dt$$, thus (assuming $$t$$ is time) is the mean residence time (MRT), equal to $$\text{MRT}_{f(t)}=r\;\;.$$ At this point, you must specify which average you want: an average functional height or an "averaged" time. There is generally not an incomplete residence time from $$x=0,t$$, so there is some question as to what that would mean. Why do you want what you asked for, and what were you asking?

• My goal is to calculate the value of food consumed between two times: If my population is 10.000 at 0h00, how many food is consumed until 10h if one member of my population eat 1 food by 24h. But i need to take in account that population has growed between 0h00 and 10h by pr^t. So i need to know the average of population between 0h and 10h to simplify the calcul to "foodByPeople x averagePopulation x time" – Chklang Nov 23 '20 at 8:23
• Actually, if the sample size changes in time, then the meaning of each event changes as well, in time. For example, if at $t=0$ there are 10000 in the sample, but the first event happens at 0h01, when there are 10001 in the sample, then the probability of that event is 1/10001. So I wouldn't use an average value, I would use probabilities. – Carl Nov 23 '20 at 18:33
• I don't understand. If my population grow at 1% by hour and i've 10.000 members at 0h, at 24h my population = 12679 members (10.000 * 1.01^24). So at 0h, my population consume 100 food/h and at 24h, it consume 126.79 food/h. So i want to calculate, if my stock is 100.000 food at 0h00, the new quantity of food at 24h. It's not a probability, it's a calcul for a litle game. – Chklang Nov 24 '20 at 10:06
• I've edited my question to add an exemple and an algorythme to calculate it (and it's this algorythme that i want to replace by a "cute mathematical formula") – Chklang Nov 24 '20 at 11:15
• The question you are asking makes assumptions. Which assumptions do you want to make? You specify population growth as a power function. This leads to fractions of a person eating, But real people come in whole numbers, 1,2,3,4....n and not real numbers like $\sqrt{2}$ people. The growth rate may be an average but in practice the number of people born is somewhat random, it may average to be approximately a power function, but in real life it is somewhat more random than that. Do you want to assume fractional people and exact growth, or do you want to know a range of outcomes that are likely? – Carl Nov 25 '20 at 2:02


text:
# Fourier Transform: Understanding change of basis property with ideas from linear algebra

The notion of Fourier transform was always a little bit mysterious to me and recently I was introduced to functional analysis. I am a beginner in this field but still I am almost seeing that the Fourier transform can be viewed as a change of basis in a space of functions. I read the following article here which tries to build an intuition:

Now, I can see that the Fourier and Inverse Fourier tranforms are projecting and projecting back a function $f(x)$ onto and from the basis of complex exponentials, $e^{i2\pi sx}$, respectively: $$F(s) = \int_{-\infty}^{\infty}f(x)e^{-i2\pi sx}dx$$

$$f(x) = \int_{-\infty}^{\infty}F(s)e^{i2\pi sx}ds$$

1)If I understand correctly, this operation is akin to regular linear algebra change of basis operations $a=Mb$ and $b=M^{-1}a$. Roughly, in this case $M$ is a matrix of uncountable many rows and columns where each row is $e^{-i2\pi sx}$, a function of $x$, and similarly $M^{-1}$ has rows as $e^{i2\pi sx}$, functions of $s$. Is this interpretation correct?

2)I don't have the exact rigour for this but intuitively think, if 1) is a correct interpretation then we should obtain from the "infinite dimensional" matrix multiplication $MM^{-1}$ something which resembles an infinite dimensional identity matrix. To test that, I built the inner product where $s$ is held fixed and equal in both terms from two objects $M$ and $M^{-1}$, which should correspond to a "diagonal" element of $MM^{-1}$: $\int_{-\infty}^{\infty}e^{i2\pi sx}e^{-i2\pi sx}dx = \int_{-\infty}^{\infty}e^{0}dx=\infty$. So this is not $1$ as expected from an identity matrix. What is the reason of that?

If you have an orthonormal basis $\{ e_{k} \}_{k=1}^{N}$ on a finite-dimensional space, such as what you would obtain with Gram-Schmidt, then every vector $x$ is expressed as $$x = \sum_{k} (x,e_{k})e_{k}.$$ This extends to $L^{2}[0,2\pi]$ using $e_{k} =\frac{1}{\sqrt{2\pi}}e^{ikx}$: $$f = \sum_{k}(f,e_{k})e_{k}$$ The importance of this basis is that it consists of eigenvectors of $Lf=\frac{1}{i}\frac{d}{dx}$ because $Le_{k}=ke_{k}$. So this basis diagonalizes the differentiation operator. Finally, the same thing holds in a continuous sense on $L^{2}(\mathbb{R})$ with \begin{align} f & = \int_{k} (f,e_{k})e_{k} \\ & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-ikt}dt\right)e^{ikx}dk \end{align} This is a generalization rather than a precise extension because $e_{k}=\frac{1}{\sqrt{2\pi}}e^{ikx}$ is not--strictly speaking--an eigenfunction of $L=\frac{1}{i}\frac{d}{dx}$ because $e_{k} \notin L^{2}(\mathbb{R})$ due to the fact that the function is not square integrable. However, for every $\delta > 0$, the following is square integrable $$e_{k,\delta}=\frac{1}{\sqrt{2\pi}}\int_{k-\delta}^{k+\delta}e_{k}(x)dk,$$ and, in the norm of $L^{2}$, it becomes closer and closer to an eigenvector with eigenvalue $k$ as $\delta\downarrow 0$: $$\|Le_{k,\delta}-ke_{k,\delta}\| < \delta\|e_{k,\delta}\|.$$ So the Fourier transform is the coefficient function and the expansion of $f$ looks very much like a "continuous" (i.e., integral) expansion of $f$ in approximate eigenfunctions of the differentiation operator. (The $e_{k,\delta}$ are even mutually orthogonal if the intervals $(k-\delta,k+\delta)$ do not overlap.)

As a final note, to make this generalization more precise, $$\|x\|^{2} = \sum_{k}|(x,e_k)|^{2}$$ also holds for the continuous orthogonal expansion: $$\|f\|^{2} = \int_{k} |(x,e_{k})|^{2}dk.$$ This is how Parseval saw it, who is the person after whom Parseval's identity is named: $$\int_{-\infty}^{\infty}|f(x)|^{2}dx = \int_{-\infty}^{\infty}|\hat{f}(k)|^{2}dk.$$

• Thanks for the answer. Sorry if this is too obvious but to make it clearer: The general form of the Fourier transform: $\int_{-\infty}^{\infty}f(t)e^{-ikt}dt$ is actually not defined, since $e^{ikt}$ does not have a finite integral and not in $L^2(\mathbb{R})$ space. What is actually going on is that we are defining $e_{k,\delta}$ as basis vectors like $\sum_{k} f(t) e_{k,\delta}$ and then letting $\delta \to 0$. Is this right? May 29 '15 at 14:18
• @UfukCanBiçici : The function $e^{ikt}$ is not square integrable on $\mathbb{R}$, even though wave packets of such things are: $\int_{k-\delta}^{k+\delta}e^{ikx}dk \in L^{2}$, and these are approximate eigenfunctions with eigenvalue $k$. For $f \in L^{2}$, you may view the Fourier transform as a limiting coefficient function for expansions as you mention. Essentially, $\hat{f}$ becomes the coefficient of $e^{ikx}$ in the continuous expansion $f=\int_{-\infty}^{\infty}\hat{f}(k)\frac{ikx}{\sqrt{2\pi}}dk$ May 29 '15 at 14:26
• Then roughly speaking, can we say that this is why my rather intuitive matrix multiplication $MM^{-1}$ does not result in something meaningful, like an identity matrix; since I am enforcing $L^2$ dot product on functions which are not square integrable to begin with? May 29 '15 at 14:37
• @UfukCanBiçici : As a final note, it is worth mentioning that how you state it is consistent with how the notion of "continuous spectrum" arose. Discrete spectrum is where you have a discrete set of eigenvalues, and the continuous case is thought of as a limit of the discrete, transitioning to the continuous. May 29 '15 at 14:37
• @UfukCanBiçici : Concerning $M$, that's right. You have a limiting case. The generalization comes through a Unitary operator $U$, i.e., one which is invertible and norm preserving, $\|Uf\|=\|f\|$. In this context, the Fourier transform turns application of $\frac{1}{i}\frac{d}{dt}$ into multiplication, which is really an application of a "continuous" diagonal matrix. You have $\frac{1}{i}\frac{d}{dt} f = U^{-1}MUf$ where $U$ is the Fourier transform and $(Mf)(\lambda)=\lambda f(\lambda)$ is multiplication. That's the Mathematical generalization through a unitary change of basis. May 29 '15 at 14:43

You can imagine derivative as a matrix such as this, as the corresponding values of each element approach zero (limit definition):

Or integration as like this(riemann sum):

Moreover, fourier transform already has a matrix representation for discrete case

https://en.wikipedia.org/wiki/DFT_matrix

You need extend this matrix to infinity and shrink the corresponding interval to zero (riemann sum) to get the continuous transform.

$$\int_{-\infty}^{\infty}e^{0}dx=\infty$$

Think of this as summing up all the ones in the diagonal. The diagonal components have the value 1, they correspond to an infinitesimal interval (dx) and when you sum infinitely many of them you get infinity.

Alternatively think about what would happen if you multiplied the derivative matrix and the integral matrix. You would get the identity matrix. By integrating you would be summing the ones on the diagonal.

• Where do we use the matrix $A$ in our case exactly? And is obtaining an infinite dimensional identity matrix with diagonal elements at infinity expected then, in this case? May 29 '15 at 13:44
• You don't use infinite matrices. Matrices are used by computers because it is the only thing they can do well. May 29 '15 at 13:50
• I mean, from an intuitive point of view, not for using in a practical application. My goal is to get an understanding of the Fourier Transform with analogies to finite dimensional matrix-vector operations commonly used in linear algebra. May 29 '15 at 13:52
• We don't use the matrix A. We would use the matrix W in the wiki article. Derivative and integral matrices were given as an example of continuous linear transformations being turned into matrices (combined with the related limit definition). May 29 '15 at 13:54


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`Journal of HistologyVolume 2013 (2013), Article ID 827089, 8 pageshttp://dx.doi.org/10.1155/2013/827089`
Research Article

## Tumor Expression of the Carcinoembryonic Antigen Correlates with High Mitotic Activity and Cell Pleomorphism Index in Lung Carcinoma

1Laboratory of Specific Recognition and Biological Activity, Department of Quality Control, Center of Molecular Immunology, 216 Street and 15 Avenue Atabey, Playa, P.O. Box 16040, 11600 Havana, Cuba
2Department of Pathology, Manuel Fajardo General Hospital, Zapata and D Street Vedado, Plaza de la Revolución, 10400 Havana, Cuba
3Division of Clinical Trials, Center of Molecular Immunology, 216 Street and 15 Avenue Atabey, Playa, P.O. Box 16040, 11600 Havana, Cuba

Received 25 March 2013; Accepted 17 May 2013

Copyright © 2013 Rancés Blanco et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

At present, some research efforts are focusing on the evaluation of a variety of tumor associated antigens (TAAs) for a better understanding of tumor biology and genetics of lung tumors. For this reason, we evaluated the tissue expression of carcinoembryonic antigen (CEA) and ior C2 (a cell surface O-linked glycoprotein carbohydrate chain TAA) in lung carcinomas, as well as its correlation with a variety of clinicopathological features. The tissue expression of CEA was evidenced in 22/43 (51.16%) lung carcinomas and it was correlated with mitotic activity, cell pleomorphism indexes, and age of patients. The expression of ior C2 was observed in 15/43 (34.88%) tumors but no correlation with the clinicopathological features mentioned above was obtained. No correlation between both CEA and ior C2 antigens expression and the overall survival (OS) of non-small-cell lung cancer patients was also observed. However, CEA-negative patients displayed higher OS rates as compared with positive ones (69.74 versus 58.26 months). Our results seem to be in agreement with the role of CEA expression in tumor cell proliferation, inhibition of cell polarizations and tissue architecture distortion. The significance of ior C2 antigen in these malignancies and it potential use in diagnosis, prognosis, and/or immunotherapy must be reevaluated.

#### 1. Introduction

Lung tumors are one of the leading causes of cancer-related mortality around the world [1]. There are two main variants of the disease, non-small cell lung cancer (NSCLC) and small-cell lung cancer (SCLC). However, NSCLC represents more than 80% of all lung carcinomas [2]. In patients with NSCLC, some genetic and regulatory abnormalities have been considered responsible for the tumor survival advantage. The alterations in gene expression that occur in cells during the malignant transformation usually conduce to the aberrant expression of antigens whether existing or not in normal cells. In this way, some research efforts are focusing on the evaluation of a variety of tumor associated antigens (TAAs) for a better understanding of tumor biology and genetics of lung tumors [3].

Carcinoembryonic antigen (CEA) is a glycoprotein expressed during embryonic and fetal development. It is frequently expressed on the apical surface of gastrointestinal epithelium, although it can also be found in other human epithelium, including lung tissues [4, 5]. Serum expression of CEA has been considered a sensitive and valuable tumor marker for diagnosis, prognosis, and therapy monitoring in lung cancer [6, 7]. Nevertheless, up to now, the evaluation of CEA expression has been mainly restricted to the serum analysis [69] and the function of tumor CEA expression has remained unelucidated [10].

On the other hand, the ior C2 is an O-linked-tumor-specific glycoprotein with very limited presence in most normal tissues and overexpressed in the surface of some human malignancies. This TAA was demonstrated in these malignancies using the ior C5 Mab, a highly specific IgG1 against ior C2 antigen, by means of immunohistochemical methods [11, 12]. Additionally, immunoscintigraphy with 99mTc-labeled Mab ior C5 has been considered a useful procedure for the diagnosis and followup of the patients with colorectal tumors, its metastasis, and recurrences [1317].

In the present work, we evaluated the tissue expression of CEA and ior C2 antigens in lung carcinoma, as well as its correlation with a variety of clinicopathological features such as mitotic activity index, cell pleomorphism, and overall survival of patients.

#### 2. Materials and Methods

##### 2.1. Monoclonal Antibodies

We used the ior cea1 [18] and ior C5 [12] Mabs, both obtained and produced by the Center of Molecular Immunology (Havana, Cuba). These Mabs are two murine IgG1 highly specific against the carcinoembryonic and a cell surface O-linked glycoprotein carbohydrate chain (ior C2) antigens, respectively.

##### 2.2. Tissue Specimens and Previous Processing

A number of 43 routinely processed, formalin-fixed, and paraffin-embedded archival samples with diagnosis of lung cancer were obtained from the pathology department of the National Institute of Oncology and Radiobiology. The samples were taken after obtaining the approval consent by the institutional ethical committee. Five micrometer serial sections from each block were obtained in a micrometer (Leitz 1512, Germany) and mounted on plus slides (Dako S2024, Carpinteria, CA, USA). All sections were attached to the slide by heating in a 70°C oven for 1 h. Afterward, the slides were kept at room temperature until they were used.

##### 2.3. Immunohistochemical Staining

The slides were dewaxed in xylene and rehydrated in decreasing ethanol series as usual and endogenous peroxidase activity was blocked with 0.03% hydrogen peroxide in methanol for 30 minutes at room temperature. Afterward all sections were washed in distilled water for 10 minutes, then were rinsed with TBS (Tris/saline buffer solution) for 5 minutes, were placed in a humid chamber, and incubated with the primary mouse ior cea1 and ior C5 Mabs for 1 h at room temperature. Negative controls were performed by substituting primary antibody for TBS and sections of colonic adenocarcinoma of known positivity for these antigens were taken as positive control. After two rinses in TBS, the slides were incubated with a rabbit anti-mouse biotinylated secondary antibody (Dako E0354, Carpinteria, CA, USA) and ABComplex/HRP (Dako E0355, Carpinteria, CA, USA) both for 30 minutes at room temperature dilution 1 : 100. Between incubations, slides were washed with TBS for 10 minutes. Afterward, enzymatic activity was visualized with DAB substrate chromogenic solution (Dako K3465, Carpinteria, CA, USA) and the tissues were counterstained with Mayer’s hematoxylin (Dako S2020, Carpinteria, CA, USA). The samples were dehydrated and mounted with a synthetic medium.

##### 2.4. Immunohistochemical Evaluation

The presence of brown staining (DAB reaction product) on both cell membrane and cytoplasm of malignant cells was considered as positive reactivity for ior cea1 and ior C5 Mabs. Sections only stained with Mayer’s hematoxylin (blue color) were taken as negative.

All evaluations were performed using an Olympus BX51 brightfield microscope with magnification of 200x and 400x (10x ocular with 20x and 40x objectives, resp.). The intensity of reaction of each tumor tissues was qualitatively estimated and expressed as follows: negative (−), weak (+), moderate (++), and intense (+++). A combination of these patterns was used to express intermediate levels of immunostaining. The most representative regions of each section were selected and the percentage of tumor cells stained with Mabs in them was estimated using the 10x objective lens (100x magnification). It was classified as 0 (negative to less than 5%), 1 (6%–50%), and 2 (more than 50%). All microscopic analyses were performed by two different observers.

##### 2.5. Pathological Features Evaluation

The evaluation of some pathological features was performed for each tumor tissues using the hematoxylin and eosin (HE) staining. Morphologic parameters such as histopathological classification, grade of differentiation, and degree of cellular pleomorphism as well as mitotic and necrosis indexes were evaluated for an expert pathologist (Charles E. Rengifo) as previously described in [19]. Briefly, the degree of cell pleomorphism was expressed by the evaluation of some cytomorphologic characteristics such as cell and nuclear size, cellular shape, chromatin pattern, nucleoli, and amount of cytoplasm and was scored as follows: 0 (no evident cell pleomorphism) 1 (low), 2 (moderate), and 3 (high) cell pleomorphism.

For the mitotic activity index (MAI), the most cellular area of each tumor section (containing the subjectively highest number of mitoses) was selected. Then, only unequivocal mitotic figures were counted in 10 high-power fields with 400x magnification (10x ocular, 40x objective) of each sample. Finally, the MAI was calculated by dividing mitotic cells out of total cells counted and expressed as previously described for cell pleomorphism.

In the case of the degree of tumor necrosis (necrosis index), a low-power field with 100x magnification (10x ocular, 10x objective) of each section was used. It was scored subjectively as follows: 0 (no necrosis), 1 (less than 50% of necrosis areas per field), and 2 (more than 50% of necrosis areas per field).

##### 2.6. Statistical Analysis

GraphPad Prism 5 software (2007 GraphPad Software Inc., La Jolla, CA, USA) was used for data analysis. The correlation between the reactivity of both ior cea1 and ior C5 Mabs with clinical and pathological features was assessed by Spearman ranks correlation coefficients. Survival distribution was estimated by the Kaplan-Meier method. Survival comparison was performed by two-sided logrank tests. For all tests, was considered statistically significant.

#### 3. Results

##### 3.1. Patient Description and Pathological Features

Tables 1 and 2 showed a summary of patient characteristics and some pathological features. The gender ratio was close to 2 : 1 in favor of males. The median of patient age at presentation was 55.58 years (ranged from 23 to 76 years). Median overall survival of patients was 65.55 months (ranged from 3 to 93.0 months) and 35.02 months for NSCLC and SCLC, respectively.

Table 1: Patients characteristics.
Table 2: Tumor characteristics.
##### 3.2. High Tumor Expression of CEA Was Evidenced in Lung Tumors

The tissue expression of CEA was evidenced in 22/43 (51.16%) lung carcinomas (Table 3). In 15/22 (68.18%) of tumors, a weak or focal reaction located in both cell clusters and its secretion with ior cea1 Mab was detected, while in 6/22 (27.27%) of samples a moderate-to-intense staining with this Mab was observed. In the latter, the pattern of staining of ior cea1 Mab was finely granular and was highly expressed in both the plasmatic membrane and the cytoplasm of malignant cells (Figure 1).

Table 3: Immunohistochemical expression of CEA.
Figure 1: Microphotographs of lung squamous cell carcinomas sections. (a) and (c) represent hematoxylin and eosin staining. Vote malignant cells exhibiting a moderate mitotic activity index and high degree of cell pleomorphism, respectively. (b) and (d) represent tissue detection of CEA antigen. Note the moderate (b) and intense (d) reactivity of ior cea1 Mab. The reactivity of this Mab was mainly located in the membrane and also in the cytoplasm of malignant cells. White bar = 100 μm.
##### 3.3. CEA Expression Correlates with Age of Patient, High Mitotic Activity Index, and Cell Pleomorphism in Lung Carcinoma

A significant correlation was detected between the tissue expression of CEA and both MAI (, ; Spearman test) and pleomorphism index (, ; Spearman test). In addition, the expression of CEA correlates with the age of patients (, ; Spearman test). No correlation between CEA expression and the rest of the clinicopathological features was observed.

##### 3.4. The Expression of Ior C2 Antigen Was Limited in Lung Tumors

The expression of ior C2 antigen was observed in 15/43 (34.88%) lung tumors and was mainly localized in cell clusters and its secretion (score 1). According to the histopathological classification, 4/14 (28.57%) squamous cell carcinoma, 5/14 (25.71%) adenocarcinoma, and 3/9 (33.33%) large cell carcinoma as well as 2/4 (50.00%) of other minor types represented were recognized by the ior C5 Mab (Figure 2). The reactivity of ior C5 Mab was also evidenced in 2/2 of SCLC.

Figure 2: Hematoxylin and eosin staining of lung adenocarcinoma (a) and mucoepidermoid carcinoma (c) sections. Note: malignant cells exhibiting a moderate degree of cell pleomorphism and high mitotic activity index, respectively. (b) and (d) represent tissue detection of C2 antigen. Note the moderate reactivity of ior C5 Mab mainly located in cell secretion and cell clusters, respectively. White bar = 100 μm.

No correlation between the expression of ior C2 antigen and the clinicopathological features was obtained.

##### 3.5. Relationship between Antigens Expression and Overall Survival of Patients

No correlation between both CEA and ior C2 antigens overexpression and the overall survival of NSCLC patients was observed ( and ; logrank test, resp.) (Figures 3 and 4). However, CEA-negative patients displayed higher overall survival rates as compared with positive ones (69.74 versus 58.26 months).

Figure 3: Kaplan-Meier estimate of overall survival among NSCLC patients showing different levels of tissue CEA expression (; logrank test).
Figure 4: Kaplan-Meier estimate of overall survival among NSCLC patients showing different levels of tissue expression of ior C2 antigen (; logrank test).

#### 4. Discussion

Despite the recent advances in lung cancer therapy, this disease still has a very poor prognosis, representing one of the leading causes of cancer-related mortality worldwide [1]. For that reason, the identification of prognostic factors, including tumor markers, clinicopathological indicators and genetic alterations, is of the utmost importance for patients with lung cancer [3].

The overexpression of CEA has been reported in 30%–70% of patients with NSCLC and was most frequently observed in patients with adenocarcinoma and advanced stage carcinoma [20]. In this study, we showed the tissue reactivity of ior cea1 Mab, a murine IgG1 highly specific raised against carcinoembryonic antigen, in about the half of lung cancer sections evaluated. The pattern of staining with this Mab was finely granular and was localized in both, cell membrane and cytoplasm of malignant epithelial cells. A similar immunohistochemical result using a different anti-CEA Mab was previously described [10]. Moreover, the in vivo tissue reactivity of ior cea1 Mab labelled with 99mTc was demonstrated in both primary and metastatic colorectal carcinoma by the radioimmunoscintigrafic technique [16, 17, 21].

Several studies have shown that high CEA levels are a potential marker of poor prognosis in NSCLC regardless of treatment [6, 20]. Meanwhile, the determination of which amount of tissue CEA expression in lung tumors has clinical and/or biological relevance still remains unclear [10]. We obtained no differences between CEA tissue expression and overall survival of patients. Although, patients with high CEA expression displayed lower overall survival rates as compared with negative ones. In this way, the potential prognostic value of ior cea1 Mab reactivity in NSCLC should be evaluated using a higher number of cases.

It is known that CEA is a cell surface adhesion protein related with the cell-to-cell adhesion [22], blocking cell differentiation effects and cellular transformation [23]. Additionally, the antiapoptotic effects of CEA expression in cancer cells have been demonstrated [24].

The overexpression of CEA has been considered to be directly related to the tumor aggressive behavior [8, 25]. Here, we found that high levels of CEA expression correlate with an increase in tumor cell proliferation measured by means of mitotic activity index. In previous studies, the expression of CEA in the proliferative phase of the human colonocytes cell cycle has been reported to be limited [26]. Nevertheless, induced constitutive expression of CEA at all phases of the growth cycle, as in most of colorectal carcinomas [27], has been related to the appearance of dramatic tumorigenic effects [26].

Previously, it was demonstrated that CEACAM6, a closely related family member of CEA, is able to induce lung cells proliferation, probably by interfering with the contact-inhibiting signal, leading to undifferentiated anchorage-independent cell growth [28]. In the present work, we obtained a correlation between CEA expression and the degree of cell pleomorphism measured by the evaluation of some cytomorphologic characteristics. Deregulated overexpression of CEA in human cells that remain potential division and have not yet differentiated can decrease the cellular differentiation, block cellular polarization, and distort normal cellular architecture both in vitro and in vivo [26].

In previous studies, little differences in the biological behavior of NSCLC in older and younger individuals have been reported [29]. In this way, molecular phenotypic studies have been suggested to investigate this controversial topic [30]. Here, we found that younger patients exhibited high level of tissue CEA expression as compared with older patients. These results are in agreement with the suggestion that younger NSCLC patients generally display a more accelerated growth and progress than that in older patients [31]. However, NSCLC had been considered a similar entity in older and younger individuals, since the same clinicopathological prognostic factors were valid in both age groups [29, 30].

Our results seem to be in agreement with the important role of tissue CEA expression in lung tumor biology, although no correlation of CEA expression and some pathological features such as tumor size, tumor state, and recurrence was evidenced.

On the other hand, the ior C5 Mab is a murine IgG1 that recognizes the TAA ior C2. This antigen is a cell surface O-linked glycoprotein carbohydrate chain very limited in most normal tissues and heterogeneously expressed in a variety of human malignancies such as ovary, breast, uterus, and lung tumors, although it was mostly evidenced in colorectal tumors [12, 32]. In primary and metastatic colorectal carcinomas as well as in ovarian adenocarcinomas, these results were confirmed by immunoscintigraphy technique with ior C5 Mab labeled with 99mTc-labeled [1317, 33].

In lung tumors, the expression of ior C2 antigen was only restricted to adenocarcinomas [12, 32]. In the present work, the tissue reactivity of ior C5 Mab was detected in about the 35% of lung carcinomas, not taking into account the histopathological classification. Curiously, the staining with this Mab was mostly evidenced in SCLC. The pattern of recognition was finely granular and was located in both epithelial cells and its secretion, similar to that described in [12, 32].

In spite of the expression of the ior C2 antigen in tumors that has been demonstrated, to our knowledge the clinical and/or biological significance in lung cancer still remains unclear. Unfortunately, in this study we obtained no correlation between ior C2 antigen expression and the overall survival of patients neither with the rest of the clinicopathological features assessed. In this way, a higher series of lung tumors must be evaluated in order to explore the role of ior C2 antigen in these malignancies, as well as its potential use in diagnosis, prognosis, and/or immunotherapy.

#### 5. Conclusions

In summary, we reported the tissue expression of CEA antigen by means of ior cea1 Mab, in both SCLC and NSCLC. The correlation of CEA expression with age of patient, high mitotic activity index, and the degree of cell pleomorphism was also demonstrated. Our results seem to be in agreement with the role of CEA expression in tumor cell proliferation as well as in the inhibition of cell polarization and the distortion of tissue architecture. Additionally, we showed the immunohistochemical expression of ior C2 antigen, a cell surface O-linked glycoprotein carbohydrate chain, in the same system. However, no correlation between ior C2 antigen expression and the overall survival of patients neither with the rest of the clinicopathological features was obtained. In this way, the significance of ior C2 antigen for lung tumors biological behavior still remains unclear.

#### Conflict of Interests

The authors report no conflict of interests.

#### Authors’ Contribution

Rancés Blanco and Charles E. Rengifo contributed equally to this work.

#### Acknowledgments

The authors are grateful to Idania Suárez López-Gavilán B.Sc., for the editorial assistance. Financial support was provided by the Center of Molecular Immunology.

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