TITLE: I dont understand the motion of this particle
QUESTION [0 upvotes]: I have found the velocity to be equal to $\dfrac{1}{\sqrt{2}} - \cos t$ and acceleration to be equal to $\sin t$. I am not sure how to find the remaining questions.

REPLY [0 votes]: Your answers to (a) are correct. $f'(x)=\frac{\sqrt{2}}{2}-cos(t)$ and $f''(x)=sin(t)$.
Now, (b) is asking what is the total distance this particle travels over the first $\pi$ seconds. This is not to be confused with the concept of displacement. The question is not asking how far the particle is from it's starting point. Instead, it's asking the entire distance the particle has covered, even if it's back-tracked. An example, you walk to your car which is $10m$ away from you, and then you walk back to your starting point. Your displacement would be $0$, but the distance you've covered would be $20m$. Now, for your problem, we look at the velocity function of this particle which happens to be:
$$v(t)=f'(x)=\frac{\sqrt{2}}{2}-\cos(t)$$
So we set this function equal to zero because we want to see where our critical points are
$$0=\frac{\sqrt{2}}{2}-\cos(t)\Rightarrow \cos(t)=\frac{\sqrt{2}}{2}$$
We're only interested in the first $\pi$ seconds so our interval is $[0,\pi]$. What values of $t$ make this statement true on the interval $[0,\pi]$? We will call this value $j$, so we say $t=j$. With this information, we look back at our position function. Our velocity function can vary in behavior. The velocity could be positive for $1$s and then be negative for $0.4$s before returning to a positive value. This is where backtracking can occur and is lost in the position function. We know when the velocity is $0$ (i.e. a critical point) we are guaranteed the time prior that, the velocity is either increasing, decreasing, or constant. So we can evaluate our position function at $t=j$
$$f(j)=\frac{\sqrt{2}}{2}(j)-\sin(j)$$
This is our distance covered from $t=0$ to $t=j$. At this point our velocity might change and we might go backwards-- so we retrace our steps, but still cover some distance. We want to know this distance. Since our velocity function had only one critical point on our interval, we are interested in the position covered by the remainder time which is given by $t=\pi-j$. Thus, we evaluate our position function at this time.
$$f(\pi-j)=\frac{\sqrt{2}}{2}(\pi-j)-\sin(\pi-j)$$
We add these values to find our total distance covered. $x_{tot}=f(j)+f(\pi-j)$.
(c) When is velocity increasing? That is to say, when is it's derivative positive? You found the acceleration function already, so how do we find when it's increasing?
(d) Simply wants to know how many times the sign of our velocity changed over the critical points. From the interval before one critical point to the interval after, and so on. How many times did this sign change? You need to test values on the intervals. We found one critical point so let's use that (THOUGH THERE MAY BE MORE SINCE OUR INTERVAL IS NOW $[0,2\pi]$). What is the sign of the velocity from $[0,j]$, well we pick a random number in this interval and evaluate it the velocity function. That will give us a positive or negative number. That sign, positive or negative is the direction of the velocity. So with all critical points on $[0,2\pi]$ we have to look at the interval between endpoints to critical points, to critical points, to endpoints. (i.e. on intervals like $[0,j], [j,g], [g,2\pi]$, say g is another critical point). We have to see how many times the sign of the velocity function evaluated changes.
I understand this may seem long, but being thorough is important for your understanding. It's not a simple answer to the questions you've asked. It's extremely easy to get confused, especially since this is an applied problem to physics and not just a pure math example.


TITLE: Prove that the set of integers modulo m is precisely {[0], [1],...,[m-1]}
QUESTION [0 upvotes]: I need to prove that each of the congruence classes listed above are different, and that any other congruence class [a](for any integer a) has to be equal to one of the congruence classes listed.
I know in order to prove the second part I must show that a is a subset of a congruence class and that congruence class is a subset of a to show they're equal but I'm not sure how to go about the first part.

REPLY [0 votes]: The congruence class of an integer is the congruence class of its remainder when divided by $m$. Two remainders can't be congruent, since they're their own remainders. What is the list of the possibles remainders?


TITLE: I need help with an assignment question please for numerical methods
QUESTION [0 upvotes]: Consider the function $f (x) = xe^x - 2,$ we want to study the properties of $f (x)$ so that we can apply numerical methods to solve the equation $f (x) = 0$.
Which option is false ?

the function, $f (x)$ is well defined and continuous for all $x$ in the interval $(0,2)$
the function, $f (x)$ has no discontinuity and no singularities
the function, $f '(x)$ is well defined and continuous for all $x$ in the interval $(0,2)$
the function, $f '(x)$ has no discontinuity and no singularities
all of the above

I do not understand the different options given. How do I know if the function is well defined?

REPLY [1 votes]: Hint 1: Here is a plot of $f(x)$ and $f'(x)$ over the range (do they look continuous over the range, see any singularities over $\mathbb{R}$). Can you make an analytical argument over the range $(0,2)$ and $\mathbb{R}$, respectively, regarding continuity and singularities? Are both continuous? Do either of them have singularities?

Hint 2: Here is a plot of $f(x) = 0$.

Hint 3: From hint 2, did you try solving $f(x) = 0$ analytically? Can you?


TITLE: Finding the convex hull of m x n matrices
QUESTION [1 upvotes]: Given a set of $\mathbb{R}^{m \times n}$ matrices, I would like to find the matrices forming the vertices of their convex hull.
Would this be the same problem as finding the convex hull of  a set of vectors in $\mathbb{R}^{m + n}$? ie: Could I simply reshape the matrices into vectors and proceed with known algorithms to find the convex hull, and then convert back into $\mathbb{R}^{m \times n}$?
PS: Please be kind. I am a engineer with limited experience in this field. If you could point me to any references that could be helpful, I's appreciate it.
Thanks
Edit:

My set of $\mathbb{R}^{m \times n}$ matrices is finite and each matrix is discrete.
Some additional clarification: I am building this set up one matrix at a time. Each time I have a new matrix, I want to test if the new matrix is contained within the convex hull formed by the matrices already in the set. If it is, then I can ignore it. If it is not in the convex hull, then I add it to the set and extend the convex hull to include the new matrix.

REPLY [1 votes]: Since $\mathbb R^{m\times n}$ is isomorphic as a vector space to $\mathbb R^{mn}$ (note that that‘s not a plus) via the obvious map (taking all entries), you can indeed do the computation in $\mathbb R^{mn}$.


TITLE: Empty spectrum of an operator
QUESTION [0 upvotes]: Recently, I was studying the spectrum of an operator $T$ on a normed
linear space $X$. In order to ensure that the spectrum is non-empty, one
needs to assume that $X$ is a complex Banach space. But the books don't give
sufficient justification. I would like to have an example of a real
incomplete normed linear space as well as an example of a complex
incomplete normed linear space $X$ and a bounded operator $T:X \rightarrow X$ such
that the spectrum of $T$ is empty. In order to avoid confusion, I would like
to define the spectrum: A scalar $\lambda$ is called a spectral value of a
bounded operator $T$ on a normed linear space $X$ if $T - \lambda I$ is not
invertible. '$I$' is the identity operator on $X$ and the set of all spectral
values of $T$ is called the spectrum of $T$.

REPLY [2 votes]: I think the example in Previous question in mathoverflow works both in the real and in the complex case.


TITLE: Do linearly independent vector fields with vanishing Lie bracket always have integral manifolds which are level sets?
QUESTION [3 upvotes]: Suppose I have a smooth distribution on an open subset of $M = R^n_{++}$ consisting of $n-1$ vector fields $X^i$. I know three things about this distribution:

At each point $p \in M$ the set $(X^1_p,\ldots,X^{n-1}_p)$ is linearly independent and so spans an $(n-1)$-dimensional subspace of $R^n$.

For each pair $i,j, \; 1 \leq i < j \leq n-1, \; [X^i,X^j] = 0$ (so the system is in involution).

At each point of $M$ the normal to the subspace spanned by this system is non-negative, with the last coordinate strictly positive.

By 2, Frobenius tells us this system is integrable with its integral manifolds forming a foliation of $M$. I understand this also means that there are local coordinates $x^i$ such that $X^i = \partial/\partial x^i$. I have two questions:

In this particular case does there always exist a single function $f: M \rightarrow R$ (smooth or not) such that the integral manifolds (hypersurfaces) are level sets of $f$; i.e. each leaf $L$ of the foliation is of the form $f^{-1}(x)$ for some $x \in R$?

Is each leaf $L$ of the foliation globally the graph of some function $h:D \rightarrow R$ (i.e. $L = (x_1,\ldots,x_{n-1},h(x_1,\ldots,x_{n-1}))$ where $D$ is some domain in $R^{n-1}$?

My sense is that the first statement is not true but (because of the third assumption) that the second is. As to 1., it has been suggested to me that if I looked at the distribution from the perspective of the normal field (i.e. 1-form approach) that my assumption that the Lie bracket vanishes is equivalent to the "well-known" integrability conditions insuring there exists a function f with the required property, but I'm not sure this is right.
I would appreciate any insight into this (surely) simple issue that I'm just not seeing.
Edit: In case it helps I can give a concrete example of the vector fields I'm working with when $n=3$. $X^1 = (-m(x_1,x_2,x_3),1, 0), \; X^2 = (-n(x_1,x_2,x_3),0,1)$. Here the smooth functions $m$ and $n$ are strictly positive everywhere on $M$ and bounded away from zero and $\infty$ on every compact subset of $M$. I also simply impose the condition $[X^1,X^2] = mn_1 - nm_1 -n_2 +m_3 = 0$. Note that the normal field is $N_{(x_1,x_2,x_3)} = (1,m(x_1,x_2,x_3),n(x_1,x_2,x_3)) \gg 0$.
Further clarification: Assumptions on m are, for each fixed $x_2,x_3$ that $\lim_{x \rightarrow \infty}m(x,x_2,x_3) = 0, \lim_{x \rightarrow 0}m(x,x_2,x_3) = \infty$. A similar assumption holds for $n$; i.e. for each fixed $x_1,x_3$ $\lim_{x \rightarrow \infty}n(x_1,x,x_3) = 0, \lim_{x \rightarrow 0}n(x_1,x,x_3) = \infty$.

REPLY [0 votes]: I will try the case $n=2$. Consider a field of directions on $(0,\infty)^2$, with the field of normals
$n(x,y) = (n_1(x,y), n_2(x,y))\ $, $n_1\ge 0$, $n_2>0$.
The leafs of the foliations will be the graphs of functions $y = \phi(x)$ solutions of the differential equation
$$a_1 dx + a_2 dy = 0$$
or
$$\frac{dy}{dx} = -\frac{a_1(x,y)}{a_2(x,y)}$$
(in the case $n=2$, we get an ordinary differential equation, so there are no integrability conditions)
The graphs of different solutions will be the leaves of the foliation.
Note that to get a solution we fix $x_0$, $y_0>0$ and consider the solution that satisfies furthermore $\phi(x_0) = y_0$.  Now, there are in fact $\infty^1$, and not $\infty^2$ such solutions, since for all pairs $(x^*, y^*)$ on the graph of one, we get the same one.
In other words, the family of solutions ( of leaves if you want) depends on $1$ parameter ( something that they tell you in ODE class while hand-waving). However, globally labelling such solutions could be tricky.
To label the solutions( leaves), find a transversal curve $t\mapsto (\psi_1(t), \psi_2(t))$ that intersects each leaf in exactly one point. This has to work globally. (Foliations are  tricky...). Now I think we could consider the transversal $t \mapsto (t, t)$. In this way we'll get a function $f\colon (0, \infty)^2$ with the leaves the level curves. The description of the function is this:
for every $(x_0, y_0)$, the solution of the Cauchy problem
$$\frac{dy}{dx} = - \frac{n_1(x,y)}{n_2(x,y)}, \  y(x_0) = x_0 $$
has a unique fixed point $t = t(x_0, y_0)$. We take $f(x_0, y_0) = t$.
Notes:

We see that in this case every level set is a leaf so connected. This is a nice situation, where leaves can be labeled "linearly". However, you can start with a submersion $f$, but whose level sets are not connected. Now, look only at the distribution of hyperplanes, and so at the foliation. All is good, integrable, but trying to write each leaf separately as a level set would be a problem.

The case of general $n$ : the Frobenius integrability condition ensure that the solution of the Cauchy-Frobenius problem

$$\frac{\partial y}{\partial x_i} = - \frac{a_i(x', y)}{a_n(x',y)}$$
exists locally. Extend it all the way and get the graphs as the leaves of the foliations. Again, we have to label them. Each of the leaves will intersect the diagonal $(t, t, \ldots, t)$ in exactly one point. Now the function $f(x_{01}, \ldots, x_{0n-1}, x_{0n})$ is as follows: the solution $y_{(x'_0, x_{0,n})}$ of
$$\frac{\partial y}{\partial x_i} = -\frac{a_i(x',y)}{a_n(x',y)} \ \ ,  y(x'_0) = x_{0n}$$ has a unique "fixed point" $(t,t\ldots, t)$. Define $f((x'_0, x_{0n})=t$.


TITLE: Calculate the curvilinear integral $\int \limits_{(-1,2)}^{(1,-2)} (3x^2-2xy+y^2)dx + (2xy-x^2-3y^2)dy$
QUESTION [0 upvotes]: $$I = \int \limits_{(-1,2)}^{(1,-2)} (3x^2-2xy+y^2)dx + (2xy-x^2-3y^2)dy$$
I tried to use fact that $\int \limits_{Г_{AB}} Pdx + Qdy =  u(B) -u(A)$
Am I right that $u(x,y)  = \int \limits_{-1}^{x} (3x^2-2xy+y^2)dx + \int \limits_{2}^{y} (-2y-1-3y^2)dy + C$
I got  $u(x,y) = 1 + x^3 + y - x^2 y + y^2 + x y^2 + 14 - y - y^2 - y^3 + C$
So $I = u(1,-2) - u(-1, 2) = 30$
Is this method correct in this case?

REPLY [0 votes]: Note that the given line is $y = -2x$ for $x$ from $-1$ to $1$.
Then substitute, noting $\mathrm{d}y = -2 \mathrm{d}x$
$$I = \int_{-1}^1 3x^2 + 4x^2 + 4x^2 + \left[ -4x^2 - x^2 - 12x^2 \right] \cdot -2 \; \mathrm{d}x = \int_{-1}^1 45x^2 \mathrm{d}x = \left[15x^3\right]_{-1}^1 = 30$$


TITLE: Number coloring : same color sequence
QUESTION [6 upvotes]: The natural numbers are colored with two colors. Prove that there exists a sequence of natural numbers $ k_1 < k_2 < ... < k_n < ...$ with the property that the sequence $2k_1 < k_1 + k_2 < 2k_2 < k_2 + k_3 < 2k_3 < ...$ is of the same color.
Please suggest.

REPLY [2 votes]: If the two colors are red and blue, we begin by making two preliminary attempts at picking sequences like this:

Pick an arbitrary value $r_1$ such that $2r_1$ is red, and keep trying to pick numbers to continue it: for as long as you can pick a number $r_{i+1}$ bigger than the previous $r_i$ such that $r_i + r_{i+1}$ and $2r_{i+1}$ are both red, do so.
Pick an arbitrary value $b_1$ such that $2b_1$ is blue, and keep trying to pick numbers to continue it: for as long as you can pick a number $b_{i+1}$ bigger than the previous $b_i$ such that $b_i + b_{i+1}$ and $2b_{i+1}$ are both blue, do so.

It's possible that we might be lucky, and one (or both!) of these sequences can be continued forever. If not, we eventually end up with sequences $r_1 < r_2 < \dots < r_n$ and $b_1 < b_2 < \dots < b_m$ that cannot be extended further. This means:

For all $x > r_n$, at least one of $\{x + r_n, 2x\}$ must be blue.
For all $x > b_m$, at least one of $\{x + b_m, 2x\}$ must be red.

We assume, without loss of generality, that $b_m > r_n$, because either $b_m > r_n$ or $b_m < r_n$ must hold. (We can't have $b_m = r_n$, because $2b_m$ is blue and $2r_n$ is red.)
Choose some number $z > r_n + b_m$ that is even and colored red. (We'll later see how to deal with the case that all sufficiently large even numbers are blue.) Letting $x = z - r_n$, we know the following:

One of $\{x + r_n, 2x\}$ must be blue. But $x + r_n = z$ is red, so $2x$ is blue.
One of $\{x + b_m, 2x\}$ must be red. But $2x$ is blue, so $x + b_m$ is red.

Since $x + b_m = (z - r_n) + b_m = z + (b_m - r_n)$, we've shown that whenever $z > r_n + b_m$ is red, so is $z + (b_m - r_n)$. But by repeating this argument, we conclude that every term of the infinite arithmetic progression $$z, z + (b_m - r_n), z + 2(b_m - r_n), z + 3(b_m - r_n), \dots$$ is red.
Now we are ready to choose the sequence we're looking for. Let $k_i = \frac z2 + i(b_m - r_n)$. Then we have $k_1 < k_2 < k_3 + \cdots$, and all numbers of the form $2k_i$ or $k_i + k_j$ are red, because they're elements of this infinite arithmetic progression.
(Earlier, we set the "all sufficiently large even numbers are blue" case aside for later. We can now come back to it: if all even numbers $\ge N$ are blue, let $k_i = \frac{N}{2} + 2i$. Then all numbers of the form $2k_i$ or $k_i + k_j$ are even and greater than $N$, so they're all blue.)


TITLE: Understanding Theorem $5.1$ in Massey: $H^q(M;\mathbb{Z})$ and $\mathrm{Hom}(H_q(M;\mathbb{Z}),\mathbb{Z}$) are not isomorphic in general
QUESTION [2 upvotes]: Theorem $5.1$ in Massey p$.366$ is the following:
Let $M$ be a compact oriented $n-$manifold and $F$ a field. Then the bilinear form $$H^q(M;F) \otimes H^{n-q}(M;F) \longmapsto F$$ defined by $$u \otimes v \longmapsto \langle u  \smile v, \mu\rangle$$
is nonsingular.
I do understand the proof but I don't understand the observation made right after it which is the following:
It would be nice to have an analogue of Theorem $5.1$ for the case of cohomology with integer coefficients, rather than coefficient in a field. Since the groups $H^q(M;\mathbb{Z})$ and Hom($H_q(M;\mathbb{Z}),\mathbb{Z}$) are not isomorphic in general some modifications are needed.
I don't understand the latter, shouldn't be the isomorphism a consequence of the fact of universal coefficient Theorem for cohomology since Ext($A,B$) = 0 if either $A$ or $B$ is free? This seems the case taking $G = \mathbb{Z}$.
Any help reference or solution in order to understand the problem would be appreciated.

REPLY [1 votes]: No, this is not at all the case that $\operatorname{Ext}_R^1(A,B)=0$ when $B$ is a free $R$-module. It is the case when $A$ is projective (in particular when it's free), or when $B$ is injective. But when $R=\mathbb{Z}$, being injective means being divisible, so it is completely different from being free.
In short, there is no reason that $\operatorname{Ext}_{\mathbb{Z}}^1(A,\mathbb{Z})=0$ unless $A$ has some special property. In fact, $\operatorname{Ext}_{\mathbb{Z}}^1(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})\simeq \mathbb{Z}/n\mathbb{Z}$ for example.


TITLE: Palindromes on Keypad and divisibility by $111$
QUESTION [1 upvotes]: The integers 1 through 9 are arranged as follows on a rectangular keypad:
$\begin{array}{c c c}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{array}$
Consider the 6-digit palindromes formed by going back and forth across a diagonal, row, or column of the grid. (ex: 123321, 159951, 357753, 456654, ...).
Why are all these numbers divisible by 111? In other words, what property of these numbers makes them divisible by 111? I've tried casework on whether the number is formed on a row, diagonal, or column, but haven't gotten anywhere.
In addition, can we generalize the result (keypads with different combinations of numbers that still have the above property)?

REPLY [0 votes]: Let's see  $abccba$.  $k=  c - b = b - a = \pm3,2,$ or $4$.
So $abccba = 111111*b \pm (100001 - 1100)k = 111111*b \pm (98901)k = 111(1001*b + 891k)$.
So ... that's that.
By the way.  11 also divides it so 1221 divides them.  This will be true of any six digit palindrome where to the three unique digits average to the second digit.


TITLE: How to find convergence of conditionally convergent series obtained while calculating the electrostatic potential energy of a NaCl crystal?
QUESTION [1 upvotes]: I was reading Electricity and Magnetism by E M Purcell and there in the first chapter there is an attempt to estimate the electrostatic potential energy of the crystal lattice of a NaCl crystal.
During the calculation we obtain an infinite series which is conditionally convergent. Now, this reminded me of the Riemann Theorem regarding conditionally convergent sequences which says that a rearrangement of such a sequence can be made  to diverge or converge to any real number.
Now, we know that the order in which we are calculating the potential energy should not matter. But here, there is crude way described by the author to calculate the energy, and I'm unable to understand the exact reason behind the method employed.

It is clear, incidentally, that this series does not converge absolutely; if we were so foolish as to try to sum all the positive terms first, that sum would diverge.To evaluate such a sum, we should arrange it so that as we proceed
  outward, including ever more distant ions, we include them in groups
  that represent nearly neutral shells of material. Then if the sum is broken
  off, the more remote ions that have been neglected will be such an
  even mixture of positive and negative charges that we can be confident
  their contribution would have been small. This is a crude way to describe
  what is actually a somewhat more delicate computational problem. The
  numerical evaluation of such a series is easily accomplished with a computer.

I suspect that the method employed is justified in a sense that the assumption that it is an infinite series is wrong and hence the series will converge to a unique finite number. But, I'm not very sure and hence asking this question to know about others' views regarding this.
Thanks in advance.

REPLY [1 votes]: Now, this reminded me of the Riemann Theorem regarding conditionally convergent sequences which says that a rearrangement of such a sequence can be made to diverge or converge to any real number.

You are right. Just rearranging the order of summation on can get every real number or $ \pm \infty$ or even no limit at all. However it is false that the order in which we are calculating the potential energy should not matter. That is true for finite sums. Series are a different story. However series are unavoidable if one needs to say something about thermodynamic limit, i.e. if we need the energy per particle of the infinite system.
Although rarely cited, the real reason behind the conditional convergence of the Coulomb series is that for Coulomb point-like charges the boundary effects matter, they depend on the shape and charge of the surface  and do not vanish relatively to the bulk contribution with increasing the size of the sample, as it happens for short-range potentials. Therefore, there is a clear physical origin of the conditional convergence of the sum. The  only way to get a unique result is to decide which, among all possible boundary effects has to be chosen. It turns out that the choice of finite size cells which make vanishing the highest multipoles over the whole summation procedure  is a wise choice because it allows to get a unique result and energy per particle of different structures which are comparable without need of taking into account the pesence of surface terms.
Only after fixing the physical ambiguity of the problem, mathematics re-enter the game with special techniques (like Ewald summation or others) to speed up the convergence of a series made well behaving on the basis of a physical requirement.
A couple of relevant references are:

Roy, S. K. (1954). On the evaluation of certain lattice series. Canadian Journal of Physics, 32(8), 509-514.
Hajj, F. Y. (1979). A simple method for the potential in ionic crystals. The Journal of Chemical Physics, 70(9), 4369-4375.


TITLE: $A=\frac{6}{\frac{6}{\frac{6}{6-\cdots}-5}-5}-5 :A=\text{?}$
QUESTION [1 upvotes]: $$A=\cfrac 6 {\cfrac 6 {\cfrac 6 {6-\cdots}-5}-5}-5 :A=\text{?}$$
my try :

But I want to recurrence Sequence method!!
$$a_{n+1}=\frac{6}{a_n}-5: a_1=\text{?}$$
$$a_2=\frac 6 {a_1}-5,$$
$$a_3=\frac{6}{a_2}-5$$
???
thank you very much !

REPLY [2 votes]: $A=\dfrac{6}{A}-5\\A^2=6-5A\\A^2+5A-6=0\\(A-1)(A+6)=0\\A=1\text{ or }-6$


TITLE: On the probability of no. of radiating particles
QUESTION [0 upvotes]: Consider an experiment that consists of counting the number of $\beta $ particles
given of in one second time interval by one gram of a radio active material.
It is known that, on the average, 3.2 $\beta$-particles are given off .
What is
a good approximation to the probability that no more than 2 such $\beta$-
particles appear?
I don't know where to start with this problem. How does one calculate the probability that between $x$ particles are given  off ?  (I hope this makes sense )
Please help . thanks in advance

REPLY [0 votes]: Because of the nature of radioactive decays, in which the probability for an atom to decay in a short time interval is constant, and that there are an enormous number of atoms to decay (1 gram), the random variable that represents the total number of atoms that decay in a fixed time interval follows the Poisson distribution. Use that distribution to work out the probability that 0, 1, or 2 decays occur in one second, given that the expectation value is 3.2.


TITLE: Longitudinal magnification confusion in concave mirror
QUESTION [0 upvotes]: If you move a rod lying on the principal axis of any concave mirror, when the near end of the rod is at a distance equal to half it's own length from the mirror, then the instantaneous velocity of image is four times the object.
Can this be proved by calculation?
My try at doing this was by taking assumption that the rod has velocity say $V$ and the image then will have velocity $4V$. Let the rod length be $L$. Now i tried to prove that for the above relation of velocities the center of the rod will be at distance L from the concave mirror.
I differentiated the mirror equation$$
\frac{1}{v}+\frac{1}{u}~=~\frac{1}{f}
$$for two ends of the rod and replaced $u'$ with $V$ and $v'$ with $4V.$

REPLY [1 votes]: You have the right approach: indeed differentiating in that way the Gaussian mirror equation tells you how far axially the image of an on-axis point will move in response to that point's motion. I get, from $v^{-1} + u^{-1} = f^{-1}$:
$$-\frac{\dot{v}}{v^2}-\frac{\dot{u}}{u^2}=0$$
which gives you $\frac{\dot{v}}{\dot{u}} = -\frac{v^2}{u^2}$, which, given that the transverse magnification is $\frac{v}{u}$, is a statement of the well known result that the longitudinal magnification is the transverse magnification squared, since $\frac{\dot{v}}{\dot{u}}$ is the ratio of the relative longitudinal motions for small motions.


TITLE: solve the ode $-y''y+1=(y')^2$
QUESTION [2 upvotes]: solve the ode $$-y''y+1=(y')^2$$
I tried to use:   $$p=y$$  and      $$pp'=y''$$
then got $$p'y=\frac{1}{p}-p$$ and did not know what to do next..

REPLY [0 votes]: At this point you have a separable first order ODE (note that $p=p(y)$) that can be solved by separation of variables
$$
\int\frac{p\,dp}{1-p^2}=\frac{dy}{y}\implies 1-p^2=\frac{C}{y^2}\\\text{ or }y^2-(yy')^2=C
$$
so that with $u=y^2$
$$
4u-u'^2=4C
$$
which is autonomous and thus separable for both signs in the square root.


text:
\begin{document}
\title{On totally periodic $\omega$-limit sets}

\author{ Habib Marzougui and Issam Naghmouchi}

\address{ Habib Marzougui, University of Carthage, Faculty
of Sciences of Bizerte, Department of Mathematics,
Jarzouna, 7021, Tunisia; Issam Naghmouchi, University of Carthage, Faculty
of Sciences of Bizerte, Department of Mathematics,
Jarzouna, 7021, Tunisia.}
 \email{hmarzoug@ictp.it and habib.marzougui@fsb.rnu.tn; issam.nagh@gmail.com and issam.naghmouchi@fsb.rnu.tn}
 \thanks{This work is supported by the research laboratory: syst\`emes dynamiques et combinatoire: 99UR15-15}

\subjclass[2000]{37B45, 37E99}

\keywords{periodic, w-limit set, totally periodic, graph, dendrite, completely regular, hereditary locally connected}
\begin{abstract}
An $\omega$-limit set of a continuous self-mapping of a compact metric space $X$ is said to be totally periodic if all of
its points are periodic. We say that $X$ has the $\omega$-FTP property provided that for each continuous self-mapping $f$ of $X$,
every totally periodic $\omega$-limit set is finite. Firstly, we show that connected components of every totally periodic $\omega$-limit
set are
finite. Secondly, we show in one hand, that a zero-dimensional compact metric space has the $\omega$-FTP property, and in the other hand,
for the wide class of one-dimensional continua, we prove that
a hereditary locally connected $X$ has the
$\omega$-FTP property if and only if $X$ is completely regular. This holds in particular for $X$ being
 a local dendrite with discrete set of branch points, and in particular, for a graph. For higher dimension, we show that
any compact metric space $X$ containing a free topological $n$-ball $(n\geq 2)$ does not admit the $\omega$-FTP property. This holds
in particular, for any topological compact manifold of dimension greater than $1$.

\end{abstract}
\maketitle

\section{\bf Introduction}

 Let  $(X,d)$ be a compact metric space and $f$ be a self continuous map of $X$. Let $\mathbb{Z}_{+}$ and $\mathbb{N}$ be the
sets of non-negative integers and positive integers respectively.
Denote by $f^{n}$ the $n$-th iterate of $f$; that is, $f^{0} = \textrm{id}_{X}$: the identity and $f^{n} = f\circ f^{n-1}$ if
$n\in\mathbb{N}$. For any $x\in X$ the subset $O_{f}(x) = \{f^{n}(x):
\ n\in\mathbb{Z}_{+}\}$ is called the $f$-orbit of $x$. A point
$x\in X$ is called periodic of prime period $n\in\mathbb{N}$ if
$f^{n}(x)=x$ and $f^{i}(x)\neq x$ for $1\leq i\leq n-1$, the orbit
of such point is called periodic orbit. We denote by P($f)$ the set
of periodic points and by Fix$(f)$ the set of fixed points of $f$. A
subset $A\subset X$ is called \textit{$f$-invariant} if $f(A)\subset
A$, it is strongly $f$-invariant if $f(A)=A$. A $p$-tuple $(A_{0}, \dots, A_{p-1})$ of subsets of $X$ is called a \textit{periodic cycle} if
$f(A_{0})=A_{1}, f(A_{1})=A_{2}, \dots, f(A_{p-1})=A_{0}$. For a subset $A$ of
$X$, denote by $\overline{A}$ the closure of $A$. We define the $\omega$-limit set of a point $x\in X$ to be the set
$\omega_{f}(x) = \{y\in X: \exists\ n_{i}\in \mathbb{N},
n_{i}\to +\infty, \underset{i\to +\infty}\lim d(f^{n_{i}}(x), y)
= 0\}$.
\medskip

 The topological characterization of $\omega$-limit sets of $f$ is known only in few spaces, such as zero-dimensional spaces, this was found in
\cite{sp}, Theorem 13), graphs (which includes the intervals, circles), this was given by \cite{hr} and more generally,
for hereditally locally connected continuum (see below for the definition),
this was given by \cite{sp}. For higher-dimensional spaces, only
partial results are known; among them $\omega$-limit sets in the square \cite{ag}, \cite{jl}, \cite{ba}.
Apart from these results, the topological characterization of $\omega$-limit sets is unknown. In this note, we focus on the class of $\omega$-limit sets that are composed of periodic points.
We say that $\omega_{f}(x)$ is \textit{totally periodic} if \ $\omega_{f}(x)\subset P(f)$.
The topological characterization of such sets is not yet been examined.
The question whether a totally periodic $\omega$-limit set is finite, is addressed.
\medskip

\begin{defn}\label{def2}
 A compact metric space $X$ has the $\omega$-FTP property provided that for each continuous self-mapping $f$ of $X$, every totally periodic
$\omega$-limit set is finite (i.e. a periodic orbit).
 \end{defn}
\medskip

 The first result show that every totally periodic $\omega$-limit set has finitely many connected components.
\medskip

 \begin{thm}\label{thm1} Let $X$ be a compact metric space, $f$ be a continuous self-mapping of $X$ and $x\in X$. If $\omega_{f}(x)$ is totally periodic then
$\omega_{f}(x)$ has finitely many connected components that form a periodic cycle.
\end{thm}
\medskip

\begin{cor}\label{n3}
Let $X$ be a compact metric space, $f$ be a self continuous map of $X$. Let $n\geq 1$ be an integer and $x\in X$. If
$\omega_{f}(x)\subset \mathrm{Fix}(f^n)$ then the number of connected components of $\omega_{f}(x)$ divides $n$.
\end{cor}
\medskip

Recall that a continuum is a compact connected metric space. We give a complete answer of the question: whether a continuum has the $\omega$-FTP property, in the wide class of one-dimensional continua : the  hereditary locally connected continua.
\medskip

A space is called \textit{degenerate} provided it has only one point; otherwise it is called \textit{non-degenerate}. A continuum $X$ is said to be:
\medskip

 \textbullet \ hereditarily locally connected if every subcontinuum of $X$ is locally connected;

\textbullet \ completely regular if every non-degenerate subcontinuum of $X$ has non-empty interior.

\textbullet  \ dendrite if $X$ is locally connected and contains no simple closed curve;

\textbullet  \ local dendrite if every point of $X$ has a neighborhood which is a dendrite,

\textbullet  \ graph if $X$ can be written as the union of finitely many arcs any two of which are either disjoint or intersect only in
one or both of their end points.
\smallskip

Recall that if $X$ is a local dendrite, a point $x\in X$ is called a \textit{branch point} (\textit{resp.} an \textit{end point}) if for some neighborhood $U$ of $x$, $U$ is a dendrite and $U\backslash \{x\}$ has more than two connected components (\textit{resp.} $U\backslash \{x\}$ is connected).
Denote by $E(X)$ and $B(X)$ the set of all end points and branch points of $X$, respectively.
It is known that every completely regular continuum is
hereditarily locally connected (\cite{ku}, Theorem 50.IV.1). Note that any graph as well as any dendrite is hereditarily locally connected.
Thus every subcontinuum of a graph (resp. a dendrite) is a graph (resp. a dendrite).
Any graph as well as any local dendrite with branch points discrete
is completely regular. More information about hereditarily locally connected continua can be found in \cite{Nadler}.
Our second result can be stated as follows.
\medskip

 \begin{thm}\label{thm2} Let $X$ be hereditally locally connected continuum. Then $X$ has the $\omega$-FTP property if and only if $X$
is completely regular.
\end{thm}
\medskip

In particular:
\medskip

\begin{cor}\label{c2}
If $X$ is a local dendrite with $B(X)$ discrete then $X$ has the $\omega$-FTP property. In particular, this holds whenever $X$ being either:

\textbullet \ a graph,

or

\textbullet \ a dendrite with $B(X)$ discrete (in particular, a dendrite with $E(X)$ closed).
\end{cor}
\medskip

 \textbf{Remark 1}.  If $X$ is a dendrite and $f$ is a monotone continuous
map of $X$ into itself, then any totally periodic $\omega$-limit set is finite. This follows from (\cite{Nagh1}, Theorem C).
 \bigskip

 \textbf{Remark 2}. In \cite{Li}, Li introduced the definition of $\omega$-scrambled set for a continuous map $f$ of a compact metric space $X$ into itself
 as follows:
A subset $S$ of $X$ is called $\omega$-scrambled for $f$ if for any $x, \ y\in S$ with $x\neq y$:
\begin{itemize}
 \item[(i)] $\omega_{f}(x) \backslash \omega_{f}(y)$ is uncountable,

\item[(ii)]  $\omega_{f}(x)\cap \omega_{f}(y)$ is nonempty,

\item[(iii)] $\omega_{f}(x) \backslash P(f)$ is nonempty.
\end{itemize}
\medskip

Note that condition (iii) is equivalent to say that $\omega_{f}(x)$ is not totally periodic.
The definition of $\omega$-scrambled set is reduced only to conditions (i) and (ii)
when the space $X$ is either a completely regular continuum or a zero-dimensional compact space; this results from
Theorem \ref{thm2} and Corollary \ref{c1}.

\medskip

In higher dimension, let $n\geq 2$ be an integer and let $B_n$ denote the unit $n$-ball given by $B_n = \{x\in\mathbb{R}^n: \ \|x\|\leq 1\}$,
where $ \| \ \|$ in the Euclidean norm on $\mathbb{R}^n$, $S^{n-1}=\partial B_{n}$ its boundary and $\textrm{int}(B_{n})=B_n\backslash \partial B_{n}$ its interior. A topological $n$-ball is a topological
space homeomorphic to $B_n$.
Let $X$ be a topological space. A topological $n$-ball $B\subset X$ is called \textit{free} in $X$ if $h(\textrm{int}(B_{n}))$
is open in $X$ where $h$ is any homeomorphism from $B_n$ onto $B$.
Our third main result is the following.
\medskip

\begin{thm}\label{t4} Let $X$ be a compact metric space containing a free topological $n$-ball $(n\geq 2)$. Then there exists a homeomorphism of $X$ into itself having an infinite
$\omega$-limit set consisting of fixed points. In particular, $X$ does not admit the $\omega$-FTP property.
\end{thm}
\medskip

\begin{cor}\label{c18}
Every topological compact manifold of dimension $\geq 2$ does not admit the $\omega$-FTP property.
\end{cor}
\medskip

However, the following holds:
\medskip

\begin{prop}\label{p3} Let $X_{1},\dots, X_{n}$ are completely regular continua and let $f$ be the self map of
$X_{1}\times \dots \times X_{n}$ defined by
$f((x_{1}, \dots, x_{n})) = (f_{1}(x_{1}), \dots,f_{n}(x_{n}))$ where $f_{1}, \dots f_{n}$ are continuous self-mapping of $X_{i}$.
Then every totally periodic $\omega$-limit set of $f$ is finite.
 \end{prop}
\medskip

The proposition \ref{p3} holds, for example, whenever $X_{i}=[0,1]$ or $X_{i}= S^{1}$.
\medskip

This paper is organized as follows. In Section 2 we prove Theorem \ref{thm1} and
Corollary \ref{n3}. Section 3 is devoted to the one dimensional case, in particular we prove Theorem \ref{thm2}. Finally,
in Section 4, we study the higher dimension by proving Theorem \ref{t4} and Proposition \ref{p3}.
\medskip

\section{\bf The connected component of a totally periodic  $\omega$-limit set}
\medskip

Let $X$ be a compact metric space, $f$ be a self continuous map of $X$ and $x\in X$.
We recall that the $\omega$-limit sets possess the following basic properties:
\medskip

\begin{prop}[\cite{lsB}, Lemma 2, Chapter IV]\label{p1}
The set $\omega_{f}(x)$ is a non-empty, closed and strongly
invariant set.
\end{prop}
\medskip

\begin{prop}[\cite{lsB}, Lemma 4, Chapter IV]
An $\omega$-limit set $\omega_{f}(x)$ is finite if and only if it is
the orbit of some periodic point.
\end{prop}
\medskip

\begin{lem}[\cite{lsB}, Lemma 3, p. $71$] \label{n1}  Set
$L=\omega_{f}(x)$ and $F$ be any non-empty
 proper closed subset of $L$. Then $F \cap
 \overline{f(L\backslash F)} \neq \emptyset$.
\end{lem}
\medskip

\begin{lem}[\cite{K}, Proposition 3.5, p. $107$] \label{n2}
If $X$ is a compact metric space and totally disconnected then the clopen sets form a base for its topology.
\end{lem}
\medskip

\begin{lem} \label{prop1}
Assume that $\omega_{f}(x)$ is totally periodic. If $\omega_{f}(x)$ is totally disconnected then it is finite.
\end{lem}
\medskip

\begin{proof}
 Write $L = \omega_{f}(x)$. By Proposition \ref{p1}, we have $f(L)=L$. We claim that
the restriction map $f_{| L}:L \rightarrow L$ is an
homeomorphism:
 Let $x,y\in L$ with period $p$ and $q$ respectively.
If $f(x)=f(y)$ then $f^{pq}(x) = f^{pq}(y)$ hence $x = y$ then
$f_{|L}$ is bijective. As $L$ is compact then $f_{|L}$ is a homeomorphism.
By hypothesis, $L\subset P(f)$, so $ L =  \bigcup\limits_{n=1}^{+\infty}
\textrm{Fix}(f^{n})\cap L$. Hence $\textrm{Fix}(f^{p})
\cap L$ has non-empty interior in $L$ for some $p\geq 1$. By Lemma
\ref{n2}, there exists a non-empty clopen subset $U$ of $L$
such that $U \subset \textrm{Fix}(f^{p})$. Denote by $F=U \cup f(U)
\cup \dots \cup f^{p-1}(U)$. Then $F$ is clopen in $L$ and we have $f(F)= F$.

 Suppose that $L$ is infinite. One can choose $U$ so that $F\neq L$: indeed, if $U=\{y\}$, for some $y\in U$ then $F= \{y,\dots, f^{p-1}(y)\}$
and $F\neq L$. If  $U\neq \{y\}$ for some $y\in U$, we choose a non-empty clopen subset $U^{\prime}\subset U\backslash\{y\}$ and so
 $F:=U^{\prime} \cup f(U^{\prime})
\cup \dots \cup f^{p-1}(U^{\prime})\varsubsetneq U \cup f(U)
\cup \dots \cup f^{p-1}(U)$, hence  $F\neq L$.

Now by Lemma \ref{n1}, we have
$\overline{f(L\backslash F)} \cap F=f(L\backslash F) \cap F
\neq \emptyset $. Hence, there is a point $z\in L\backslash F$ such that $f(z)\in F$. As $z\in P(f)$ and  $f(F)= F$, thus $z\in F$, a contradiction. We conclude that $L$ is finite.
\end{proof}
\medskip

 As a consequence, we have the following result.

\begin{cor}\label{c1}
If $X$ is a compact metric space and totally disconnected, then $X$ has the $\omega$-FTP property. In particular, Cantor space has the $\omega$-FTP property.
\end{cor}
\medskip

 Now let $X$ be a metric compact space, $f$ be a self continuous map of $X$ and assume that $L=\omega_{f}(x)$ is totally periodic.
Denote by $Y=\overline{O_{f}(x)}=O_{f}(x) \cup L$. We have $f(Y)\subset Y$. For every $z\in Y$, $[z]$ means the connected component in $Y$ that
 contains $z$. Let $\mathcal{C}$ be the family of connected components of $Y$ which form a decomposition of $Y$.
  By collapsing these components to points, we get the quotient space $Y/\mathcal{C}$ which is a compact metric
  space and totally disconnected (cf. \cite{bdr}).
 We denote by $\pi: Y\longrightarrow Y/\mathcal{C}; y\longmapsto [y]$ and
 $\displaystyle{\tilde{f}:Y/\mathcal{C}\rightarrow
  Y/\mathcal{C}}$ the induced map defined by: $\forall \ [y]\in Y/\mathcal{C}, \quad \tilde{f}([y])=[f(y)]$ i.e.
$\tilde{f}\circ \pi = \pi \circ f$. It is plain that $\tilde{f}$ is well defined and continuous.
 \medskip

\begin{lem}\label{l33}
The following assertions hold:

 \begin{enumerate}

 \item for any $n\in \mathbb{Z_+}$,  $[f^{n}(x)] = \{f^{n}(x)\}$ and $[f^{n}(x)]$ is an isolated point in $Y/\mathcal{C}$.

\item If $y\in L$ then $[y]\subset L$.
 \item for any $y\in Y$, $f([y]) = [f(y)]$ and $\forall n\in \mathbb{Z_+}, f^{n}([y])=[f^{n}(y)]$.
 \item for any $z\in L$, $[z]$ is a periodic point for $\tilde{f}$.

\item If $L$ is infinite then $\{[y], y\in L\} = \omega_{\tilde{f}}([x])$.
 \end{enumerate}
\end{lem}
\medskip

\begin{proof} (1) If $f^{n}(x)\in L$ then $f^{n}(x)\in P(f)$ and so $Y = O_{f}(x)$ is finite. Therefore $f^{n}(x)$ is isolated in $Y$.
If $f^{n}(x)\notin L$, there is $\varepsilon_{1}>0$ such that
$B(f^{n}(x),\varepsilon_{1}) \cap L = \emptyset
 $, where $B(x_{0}, r)$ denotes the open ball of radius $r>0$ and center $x_{0}\in X$. In the other hand,
 there is $\varepsilon_{2} >0$ such that $B(f^{n}(x),\varepsilon_{2}) \cap \mathcal{O}_{f}(x)=\{f^{n}(x)\}$
(otherwise $f^{n}(x) \in L$, a contradiction). So $B(f^{n}(x),\varepsilon) \cap Y=\{f^{n}(x)\} $ where
$\varepsilon= \min(\varepsilon_{1},\varepsilon_{2})$. Thus $f^{n}(x)$ is isolated in $Y$ and hence $
 [f^{n}(x)]=\{f^{n}(x)\}$. As
$\pi^{-1}(\{[f^{n}(x)]\})=[f^{n}(x)] = \{f^{n}(x)\}$ which is open in $Y$, it follows that $[f^{n}(x)]$
 is isolated in $Y/\mathcal{C}$.
\\\\
(2) Let $y\in L$. If for some $n\in\mathbb{Z_+}$, $f^n(x)\in [y]$ then by assertion (1), $[y]=[f^n(x)]=\{f^n(x)\}$.
So $y=f^n(x)$ and hence $[y]=\{y\}\subset L$. If now $O_f(x)\cap [y]=\emptyset$ then it is clear that $[y]\subset L$.
\\\\
(3) By assertion (1), if $y\in O_f(x)$ then it is clear that $f([y])=[f(y)]$. Now let $y\in L$. Since $[y]$ is the connected component of $Y$ containing $y$, $f$ is continuous and  $f(Y)\subset Y$, so $f([y])\subset Y$ is
connected and contains $f(y)$, therefore, $f([y])\subset [f(y)]$ and so $f^{i}([y])\subset [f^{i}(y)]$, for every $i\in
\mathbb{N}$.
 Conversely, suppose that $f([y]) \varsubsetneq [f(y)]$.
By assertion (2) and as $f(L)= L$,
$[f(y)]\subset L$ and there exists $z\in L \backslash [y]$ such that $f(z)\in [f(y)]$. Let $p, \ k \geq 1$
such that $f^{p}(y)= y$ and $f^{k}(z)= z$. Then we get $f^{kp}(z)=z \in [f^{kp}(y)] = [y]$, a contradiction.
\\\\
(4) If $z\in L$ with period $k\geq 1$ then $\tilde{f}^{k}([z])= [f^{k}(z)]=[z]$ and so $[z]$ is a periodic point of $\tilde{f}$.
\\\\
(5) Since $L= \omega_{f}(x) = \underset{n\in \mathbb{N}}\cap\overline{\{f^{k}(x): k\geq n\}}$, we have $\pi(L)= \{[y],  y \in L\} \subset \omega_{\tilde{f}}([x])$. Conversely, let
$[y]\in \omega_{\tilde{f}}([x])$. If $y=f^{k}(x)$ for some $k\in
 \mathbb{Z}_+$, then $\omega_{\tilde{f}}([x])= \omega_{\tilde{f}}([y])$. Now by ($1$), $[y]$ is isolated in
$Y/\mathcal{C}$ and  $\{f^k(x)\}=[y]$. Hence $[y]= \tilde{f}^{k+p}([x])$
for infinitely many $p$. Therefore $[y]$ is $\tilde{f}$-periodic and so $f^k(x)$ is $f$-periodic point. Thus $L$ is finite, a contradiction.
We conclude that $y\in L$ and therefore $\omega_{\tilde{f}}([x])=
\{[y], \ y \in L\}$.
\end{proof}
\bigskip

\textit{Proof of Theorem \ref{thm1}}.  If $L$ is finite, the proof is clear. Assume that  $L$ is infinite. As $Y/\mathcal{C}$ is compact and totally disconnected metric space, and  $\omega_{f}(x) \subset P(f)$, then
 $\omega_{\tilde{f}}([x])$ is totally disconnected and by Lemma \ref{l33}, (4) and (5), $\omega_{\tilde{f}}([x]) \subset P(\tilde{f})$.  Therefore by Lemma \ref{prop1}, $\omega_{\tilde{f}}([x])$ is
finite i.e. $\omega_{\tilde{f}}([x])= \{[z], \tilde{f}([z]),\dots,\tilde{f}^{k-1}([z])\}$ (for some $z \in Y$ and $k\geq 1$). As
$\tilde{f}^{i}([z])= f^{i}([z])$ (Lemma \ref{l33}, (3)) we conclude that $L$ has a
 finite number of connected components: $[z],f([z]),\dots, f^{k-1}([z])$ that form a periodic cycle for $f$.
The proof is complete. \qed
\bigskip

\smallskip

\textit{Proof of Corollary \ref{n3}}.
As $\omega_f(x)=\omega_{f^n}(x)\cup\omega_{f^n}(f(x))\dots\cup \omega_{f^n}(f^{n-1}(x))$ then for each $0\leq i\leq n-1$,
$\omega_{f^n}(f^i(x))\subset \mathrm{Fix}(f^n)$. By Theorem \ref{thm1},
$\omega_{f^n}(f^i(x))$ has finitely many connected components $C_{1}, \dots, C_{p}$, so
each $C_{k}$, ($1\leq k\leq p$) is open in $\omega_{f^n}(f^i(x))$. Suppose that $\omega_{f^n}(f^i(x))$ is not connected, then $p\geq 2$ and by applying
 Lemma \ref{n1}, there exists $z\in \omega_{f^n}(f^i(x))\backslash C_{1}$ such that $f^{n}(z)\in C_{1}$, a contradiction since  $f^{n}(z)=z$.
We conclude that for each $1\leq i\leq n$, $\omega_{f^n}(f^i(x))$ is connected and therefore the number $l$ of connected components of
$\omega_f(x)$ is at most $n$. As the connected components of
$\omega_f(x)$ form a periodic cycle for $f$ then $l$ divides $n$. \qed
\bigskip

\section{\bf The $\omega$-FTP property in one-dimension}

\textbf{3.1. The example.} We are going to construct a continuous map $F_{0}$ on a dendrite $D_{0}$ admitting an $\omega$-limit set
 which is an arc formed only by fixed points.

\textbf{(a)} \ First, we define the space $D_{0}$ as a subset of the plane in the following way:

Let $n\in \mathbb{N}$. Denote by $S_{n} = \{\frac{i}{2^n}: \ 1\leq i\leq 2^n-1 \}$. For $n=1$, we let $\Omega_1=\{\frac{1}{2}\}$
and for each $n>1$, we let $$\Omega_{n} = S_{n} \backslash \cup_{i=1}^{n-1}\Omega_{i}.$$
For example, $\Omega_2 = \{\frac{1}{4}, \frac{3}{4}\}$, $\Omega_3 = \{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\}$.
For each $n\in\mathbb{N}$, the cardinal of $\Omega_n$ is $2^{n-1}$.
We define now a dense sequence $(a_n)_{n\in\mathbb{N}}$ in $[0,1]$ as follow:

 For $n=1$, we let $a_1=\frac{1}{2}$. For each $n\in\mathbb{N}$, set

\textbullet \ $\Omega_{2n}= \{a_i: 2^{2n-1}\leq i\leq 2^{2n}-1\}$ given in this order: \ $a_{2^{2n-1}}<\dots< a_{2^{2n}-1}$.

\textbullet \ $\Omega_{2n+1}=\{a_i: 2^{2n}\leq i\leq 2^{2n+1}-1\}$  given in this order: \ $a_{2^{2n}}>\dots> a_{2^{2n+1}-1}$.

For example, $a_{2}= \frac{1}{4}$,  $a_{3}= \frac{3}{4}$, $a_{4}= \frac{7}{8}$, $a_{5}= \frac{5}{8}$, $a_{6}= \frac{3}{8}$, $a_{7}= \frac{1}{8}$.

For any $n, k\in\mathbb{N}$ such that $a_{k}\in \Omega_n$, we let $I_k = \{a_k\}\times [0,\frac{1}{n}]$. The dendrite $D_{0}$ is then given by:
 $$D_{0}:=([0,1]\times \{0\})\cup \bigcup_{k\in\mathbb{N}}I_k.$$

\medskip

\textbf{(b)} \ Second we construct the map $F_{0}$ as follows: For any $k\in\mathbb{N}$ such that $a_{k}\in \Omega_n$, write
$I_{k} = \{a_{k}\}\times [0,\frac{1}{n}] = [A_k, B_k]$,
where $A_k = (a_k,0)$,  $B_k = (a_k,\frac{1}{n})$. We let $C_k = (a_k,\frac{2}{n})$. Then the map $F_{0}$ is defined inductively as follows:
 \begin{itemize}
 \item $[0,1]\times\{0\}= \textrm{Fix}(F_{0})$.\\
\item $(F_{0})_{| [A_k,C_k]}: [A_k,C_k]\to [A_k,A_{k+1}]$ be a homeomorphism that fixes the point $A_k$ and maps linearly
$[A_k,C_k]$ to $[A_k,A_{k+1}]$,\\
\item $(F_{0})_{\mid [C_k,B_k]}:[C_k,B_k]\to [A_{k+1},B_{k+1}]$ be a homeomorphism that maps linearly $[C_k,B_k]$ to $[A_{k+1},B_{k+1}]$
and sends $C_k$ to $A_{k+1}$ and $B_k$ to $B_{k+1}$.
 \end{itemize}

The map $F_{0}$ is continuous on $D_{0}$: Let $k\in\mathbb{N}$ and $M\in I_k\setminus A_k$. Then there is a sub-arc of $I_k$
containing $M$, which is open in $D_{0}$. Hence $F_{0}$ is continuous at $M$ since $F_{0}$ is a homeomorphism on $I_k$. Let
 $M=(x,0)\in [0,1]\times \{0\}$ and $(M_n)$ be a sequence of points in $D_0$ that converges to $M$. If $M_n$ lies eventually in
 $[0,1]\times \{0\}$ then $ F_0(M_n)= M_{n}$ and so $\underset{n\to +\infty}\lim F_0(M_n)=M$. If $M_n$ lies eventually in
 $I_{k(n)}\setminus A_{k(n)}$ then we distinguish three subcases:\\
   \begin{itemize}
     \item Case 1. $M_n$ lies finitely many times in each $I_{k(n)}\setminus A_{k(n)}$. In this case,
$\underset{n\to +\infty}\lim k(n)=+\infty$. Since $\underset{n\to +\infty}\lim \textrm{diam}(I_{k(n)})=0$, we obtain
 $\underset{n\to +\infty}\lim A_{k(n)}=M$. We also have $\underset{n\to +\infty}\lim \textrm{diam} ([A_{k(n)},A_{k(n)+1}])=0$ and for
eventually $n$, $F_0(M_n)\in [A_{k(n)},A_{k(n)+1}]\cup I_{k(n)+1}$. Therefore $\underset{n\to +\infty}\lim F_0(M_n)=M$.

\item Case 2. $M_n$ lies in a finite union of $I_k$ (i.e. $k(n)$ is bounded). In this case,
 then there exists $k_0\in\mathbb{N}$ such that $M_n$ lies eventually in $I_{k_0}$ and $A_{k_0}=M$. By the continuity of the restricted map
$(F_{0})_{\mid I_{k_0}}$, we get $\underset{n\to +\infty} \lim F_0(M_n)=M$.

     \item Case 3. There exists $k_0\in\mathbb{N}$ such that $M_n$ lies infinitely many times in $I_{k_0}$ and
$M_n\notin I_{k_0}$ for infinitely many times. In this case, the sequence $(M_n)_{n}$ has two subsequences $(M_{\alpha(n)})_{n}$ and
$(M_{\beta(n)})_{n}$ such that  $\mathbb{N}=\{\alpha(n): \ n\in\mathbb{N}\}\cup \{\beta(n): \ n\in\mathbb{N}\}$ where $(M_\alpha(n))$ satisfies
case 1 and  $(M_{\beta(n)})$ satisfies the case 2. So $\underset{n\to +\infty}\lim F_0(M_n)=M$.
   \end{itemize}

In result, $F_0$ is continuous on $[0,1]\times \{0\}$ and so $F_0$ is continuous on the whole dendrite $D_0$. The $F_{0}$-orbit of
the point $B_1$ is then $O_{F_{0}}(B_{1})= \{B_k: \ k\in\mathbb{N}\}$. Hence
$\omega_{F_{0}}(B_1) = [0,1]\times\{0\} = \textrm{Fix}(F_{0})$.
\bigskip
\bigskip

\bigskip
\bigskip
\bigskip
\bigskip
\bigskip
\bigskip
\textbf{3.2. Proof of Theorem \ref{thm2} }
\medskip

\textbullet \ Assume that $X$ is completely regular and let $L=\omega_{f}(x)\subset P(f)$. We claim that
$L$ is totally disconnected. Indeed, otherwise $L$ contains
a non-degenerate connected component $C$ necessarily with
non-empty interior in $X$. In particular $L$ is infinite and there exists $n\in\mathbb{N}$ such that $f^n(x)\in L$.
Thus $f^n(x)$ is a periodic point. As
$\omega_{f}(x) = \omega_{f}(f^n(x))$, then $L$ is finite, a
contradiction. Now by Lemma \ref{prop1}, $L$ itself is
finite.
\\\\
 \textbullet \ Conversely, suppose that $X$ is not completely regular.
\medskip

\textbf{(a)} First we prove that $X$ contains a homeomorphic copy of the dendrite $D_{0}$ (see the Example 3.1 above). Indeed, there is a non-degenerate locally connected
subcontinuum $Y$ of empty interior. As $Y$ itself is arcwise connected, it contains an arc $[a,b]$. Take $c\in (a,b)$ and $\varepsilon>0$.
As $X$ is locally arcwise connected, there is an arcwise connected neighborhood $U$ of $c$ in $X$ with diameter less than $\varepsilon$
 (see \cite{Nadler}, Theorem 8.25, p.131). As $Y$ has empty interior in $X$, there exists
$d\in U\setminus [a,b]$ and so an arc, say $[c,d]$ in $U$, joining $c$ and $d$. Let $r\in [c,d[$ be such that the sub-arc
$[c,r]$ of $[c,d]$ intersects $[a,b]$ only in the point $r$. Hence we have proved that for each $\eps>0$ and $c\in (a,b)$, there is $r\in [a,b]$ with $d(r,c)<\eps$
and an arc $I$ with $r$ as one of its endpoints such that $I\cap [a,b]=\{r\}$. So if we take two distinct points $c,c^{\prime}\in (a,b)$,
we can find two disjoint arcs one has $c$ as one of its endpoints and the other has $c^{\prime}$ as one of its endpoints and each one of them
intersects $[a,b]$ only in $c$ and $c^{\prime}$ respectively. We conclude the existence of a dense sequence $(r_n)_{n}$ of points in $[a,b]$
and a pairwise disjoint arcs $(I_n)_{n}$, each of them has $r_n$ as one of its endpoints with $I_{n}\cap [a,b]=\{r_n\}$.
Since $X$ is hereditarily locally connected, then by (\cite{ku}, Theorem 50.IV.9), the arcs $(I_n)_{n}$ is a null family; i.e. for any
$\varepsilon >0$, only finitely many $I_{n}$ have diameter $> \varepsilon$. Therefore $\underset{n\to +\infty}\lim \textrm{diam}(I_{n}) = 0$.
We conclude that the set $Z:= [a,b]\cup \underset{n\in\mathbb{N}}\cup I_n$ is homeomorphic to the dendrite $D_{0}$.
\medskip
\smallskip

\textbf{(b)} Second, we need the following proposition.

\begin{prop}[\cite{sp}, proposition 11]\label{p:4}
 Let $X$ be a hereditarily locally connected continuum and let $Y$ be a locally connected continuum. Then any
continuous map $f: M \longrightarrow Y$ defined on a closed subset $M$ of $X$ can be extended to a continuous map $F : X\longrightarrow Y$
defined on $X$.
\end{prop}
\medskip

Now, we complete the proof of Theorem \ref{thm2} as follows. From the Example 3.1, there exist a point $x\in Z$ and a
continuous map $F: Z\longrightarrow Z$ defined on $Z$ such that $\omega_F(x)= [a,b]= \textrm{Fix} F$.
So, by Proposition \ref{p:4}, the map $F$ can be extended to a continuous map $\tilde{F}$ defined on $X$ into $X$
with $\omega_{\tilde{F}}(x)=
[a,b]\subset \textrm{Fix}\tilde{F}$. Therefore $X$ does not admit
the $\omega$-FTP property. \qed
\medskip

\section{\bf The $\omega$-FTP property in higher dimension}
\medskip

\begin{prop}\label{prop51} For each integer $n\geq 2$, there exists a homeomorphism $f$ of $B_n$ onto itself such that
$S^{n-1} \cup \{0\}\subset \mathrm{Fix}f$ and $\omega_f(x)$ is a circle included in
$S^{n-1}$, for some $x\in B_n\setminus (S^{n-1}\cup\{0\})$.
\end{prop}
\medskip
\medskip

\begin{proof}
 We shall construct such homeomorphism $f$ by induction on $n$.
\medskip

 \textbullet \ The case $n=2$. In this case $B_{2}$ is the unit disk, denoted by $D= \{z\in \mathbb{C}:  |z|\leq 1\}$. Let
$z_{0}= r_{0}e^{i\frac{1}{1-r_{0}^{2}}}\in D$ where $0< r_{0}< 1$ is a real. We shall construct a
$C^{\infty}$-diffeomorphism $f$ on $D$ such that $\omega_{f}(z_{0})= \partial D = S^{1}$
with Fix$(f) =  \partial D\cup \{0\}$. Such a diffeomorphism will be the time-one map of a $C^{\infty}$-flow on $D$. Let $X$ be the vector fields defined on $D$ by
$$X(z)= \begin{cases} e^{\dfrac{i}{1-|z|^{2}}}\left (|z|^{2}+
2i(1-|z|^{2})|z|^{4}\right)\varphi (z), & \textrm{ if } |z|< 1\\
0, & \textrm{ if } |z|=1
\end{cases} $$ where
$$\varphi(z) = \begin{cases} e^{\dfrac{-1}{1-|z|^{2}}},& \textrm{ if } |z|< 1\\
                                0, & \textrm{ if } |z|= 1
                                                 \end{cases}$$

 The map $\varphi$ is $C^{\infty}$ on $D$. Indeed, set $h(x)=  e^{\dfrac{-1}{1-x^{2}}}$, $x\in [0,1[$. The derivative of $h$ of order $n$,
($n\in \mathbb{Z}_{+}$)
 is: $h^{(n)}(x)= P_{n}(x)\dfrac{e^{{\dfrac{-1}{1-x^{2}}}}}{(1-x^{2})^{2n}}$, where $P_{n}$ is a polynomial function. Hence
$\underset{x\mapsto 1}\lim h^{(n)}(x) =0$ and so $h$ is $C^{\infty}$ on $1$. We conclude that $\varphi$ is $C^{\infty}$ on $D$ and so on $X$. The set of singular points of $X$ is $\{0\}\cup S^{1}$. Let $(\phi_{t})_{t\in \mathbb{R}}$ denote the flow on $D$ associated to $X$.
Write $\phi_{t}(z_{0}) := z(t)= r(t)e^{i\theta(t)}, \ t\in \mathbb{R}$ where $z(0)= z_{0}$, be the solution of the differential
equation $z^{\prime}(t)= X(z(t))$ with  $z(0)= z_{0}$. Then $r$ is a solution of the real differential equation
$r^{\prime}(t)= r^{2}(t)\varphi(r(t))$ with $r(0)=  r_{0}$ and $\theta(t) = \dfrac{1}{1-r^{2}(t)}$.
 The function $r$ satisfies  $\underset{t\to +\infty}\lim r(t)=1$,
$\underset{t\to -\infty}\lim r(t)=0$. The function $\theta$ is increasing and satisfies $\underset{t\to +\infty}\lim
\theta(t)= +\infty$ and $\underset{|t|\to +\infty}\lim\theta^{\prime}(t)
= \underset{|t|\to +\infty}\lim \dfrac{2r^{3}\varphi(r(t))}{(1-r^{2}(t))^{2}}= 0$.
We let $f= \phi_{1}$ be the time one map of the flow. Then $f$ is a $C^{\infty}$-diffeomorphism of $D$.
Let us show that $\omega_{f}(z_{0})= S^{1}$. One has $f^{n}(z_{0})= \phi_{n}(z_{0})$, so $\omega_{f}(z_{0})\subset S^{1}$.
Conversely, if $z\in S^{1}$, there is a sequence
$(t_{n})_{n}\subset \mathbb{R}$ with $t_{n}\to +\infty$ such that $\phi_{t_{n}}(z_{0})\to z$. As
$\underset{n\to +\infty}\lim r(n)=1$ and $\underset{n\to +\infty}\lim
\theta(n+1)-\theta(n) =0$, it follows that $\underset{n\to +\infty}\lim |f^{[t_{n}]}(z_{0})-\phi_{t_{n}}(z_{0})|=0$ where $[t_{n}]$ is
the integer part of $t_{n}$. Therefore $\underset{n\to +\infty}\lim f^{[t_{n}]}(z_{0})=z$ and so $S^{1}\subset \omega_{f}(z_{0})$.
We conclude that $\omega_{f}(z_{0})= S^{1}$.
\bigskip

\textbullet \ Suppose that for $n\geq 2$, there exists a homeomorphism $f_n$ of $B_{n}$ onto itself satisfying the required properties.
Let $f_{n+1}$ denote the map given by:
$$f_{n+1}(x) = \begin{cases}

 \left(\sqrt{1-x_{n+1}^{2}}~f_n(\frac{1}{\sqrt{1-x_{n+1}^{2}}}(x_1,x_2,\dots,x_n)),x_{n+1}\right), & \textrm{if } x_{n+1}\neq \pm 1\\
(0,\dots, \pm 1),  & \textrm{if } x_{n+1} = \pm 1
\end{cases}$$ where $x=(x_1,x_2,\dots,x_n,x_{n+1})\in B_{n+1}$. It is plain that $f_{n+1}$ is continuous and bijective, hence $f_{n+1}$ is a
homeomorphism of $B_{n+1}$ onto itself. In addition, $S^{n}\cup \{0\}\subset \textrm{Fix}f_{n+1}$. For any $x\in (x_1,x_2,\dots,x_n,x_{n+1})\in B_{n+1}$, we have
$$\omega_{f_{n+1}}(x) = \sqrt{1-x_{n+1}^{2}}~\omega_{f_n}\left (\frac{1}{\sqrt{1-x_{n+1}^{2}}}(x_1,x_2,\dots,x_n)\right)\times\{x_{n+1}\}.$$
We conclude that there is a point $x_{0}\in B_{n+1}\setminus (S^{n}\cup \{0\})$ such that $\omega_{f_{n+1}}(x_{0})\subset S^{n}$ and
 is homeomorphic to the circle $S^1$.
\end{proof}
\medskip

\textit{Proof of Theorem \ref{t4}}. Let $V$ be a free topological $n$-ball in $X$. Its boundary $\partial V$
 is homeomorphic to the sphere $S^{n-1}$. By Proposition \ref{prop51},
there exists a homeomorphism $f: V\to V$ and a point $c\in V$ such that $\partial V \cup \{c\}\subset \textrm{Fix}f$ and for some
point $x\in V\setminus\{c\}$, $\omega_f(x)\subset \partial V$ is homeomorphic to the circle $S^1$. In this way, the map $f$ can be easily extended to a continuous map, say $\widetilde{f}$ defined on the
whole space $X$ by fixing all points in $X\setminus V$. Hence $\widetilde{f}$ is homeomorphism of $X$ onto itself, where
$\omega_{\widetilde{f}}(x)$ is infinite and consisting of fixed points.
We conclude that $X$ does not admit the $\omega$-FTP property. \qed
\bigskip

\textit{Proof of Proposition \ref{p3}}. Let $(x_{1},\dots, x_{n})\in X_{1}\times \dots \times X_{n}$ and assume that
$\omega_{f}((x_{1},\dots, x_{n}))\subset P(f)$. Let $i\in\{1,\dots,n\}$ and  $a_{i}\in \omega_{f_{i}}(x_{i})$, then there exists a sequence
$(n_{k}^{(i)})_{k}\to +\infty$ such that
$f_{i}^{n_{k}^{(i)}}(x_{i})\to a_{i}$. Now the sequence  $(f^{n_{k}^{(i)}}((x_{1},\dots, x_{n}))_{k} = (f_{1}^{n_{k}^{(i)}}(x_{1}), \dots,
f_{n}^{n_{k}^{(i)}}(x_{n}))_{k}$ has a subsequence that converges to a point $(b_{1}^{(i)}, \dots, a_{i},\dots, b_{n}^{(i)})\in P(f)$.
 As $P(f)\subset P(f_{1})\times \dots \times P(f_{n})$ then $a_{i}\in P(f_{i})$. We conclude that $\omega_{f_{i}}(x_{i})\subset P(f_{i})$.
By Theorem \ref{thm2}, the $\omega_{f_{i}}(x_{i})$ ($1\leq i\leq n$) are
finite. As $\omega_{f}(x_{1},\dots, x_{n})\subset \omega_{f_{1}}(x_{1})\times \dots \times \omega_{f_{n}}(x_{n})$,
hence $\omega_{f}(x_{1},\dots, x_{n})$ is finite. \qed
\bigskip

Recall that a Peano continuum can be defined as a continuous image of $[0,1]$. In view of Theorem \ref{thm2},
 one may be tempted to ask the following question.
\medskip

\textbf{Question.} Is it true that a Peano continuum has the
$\omega$-FTP property if and only if it is completely regular ?

\bibliographystyle{amsplain}
\bigskip

config: arxiv


TITLE: how to prove $(-1)\cdot(-1)=1$ based only on the field axioms?
QUESTION [3 upvotes]: How do I prove $(-1)\cdot(-1)=1$ (or in other words that $-1$ is the multiplicative inverse of itself), based only on the field axioms?
Thanks

REPLY [1 votes]: $(-1) \cdot (-1)=(-1) \cdot (-1)+(-1) \cdot (1)-(-1) \cdot 1=(-1) \cdot (-1+1)-(-1)=1.$


TITLE: Justification for Interchange of integral and sum
QUESTION [1 upvotes]: Let $\mu$ be a probability measure and $t\in\mathbb{R}$. I would like to write this equality
$$\int_{\mathbb{R}}e^{ixt}d\mu(x)=\sum_{n\geq0}\frac{(it)^{n}}{n!}\int_{\mathbb{R}}x^{n}d\mu(x).$$
This is a part of an exercise from Probability giving together the Fourier transform and moments of a measure.
I want to ask, what is the right (rigorous!) justification for the interchange of the sum and the integral in this case? Thanks.

REPLY [3 votes]: You would need a rather special probability measure on $\mathbb{R}$ (for example, $e^{-x^2/2}dx/\sqrt{2\pi})$ in order to guarantee the convergence of the sequence of integrals on the right hand side of your equation, and even in that case you would need to examine their rate of growth in order to allow for integrating the sum term by term. In general the formula is not correct.


TITLE: Find matrix representation of following linear transformation
QUESTION [1 upvotes]: Consider subspace $V=span\{f_1,f_2\}$ of the complex vector space of complex valued functions with domain $\mathbb{R}$, where $f_1(x)=\cos x $ and $f_2(x)=\sin x$ for all $x \in \mathbb{R}$. $D: V \to V$ is a linear operator defines as the derivative $D(g)=g'$ for any function $g \in V$

Find the matrix representation of D with respect to the ordered basis $\{f_1,f_2\}$
Find the matrix representation of the vector $h$ with respect to the ordered basis $\{f_1,f_2\}$ where $h(x)=3e^{ix}-2e^{-ix}, \:$ for all $x \in \mathbb{R}$
Calculate derivative of $h(x)=3e^{ix}-2e^{-ix}$ using the matrix representations you have found in parts 1 and 2.

This is an exercise in my book.
My thoughts:
For $1$:
$V=span(\cos x, \sin x)=a \cos x+b \sin x,$ for all $a,b \in \mathbb{C}$
So this gives:
$D(span(\cos x, \sin x))=D(a \cos x+b \sin x)=(a \cos x+b \sin x)'=- a\sin x + b \cos x = (- a)(\sin x) + (b )(\cos x ) = \begin{bmatrix} - a\\ b \end{bmatrix}$
I am not sure whether it is correct or not. If I could do first part this might give me some idea about part 2 and 3.
Thank you in advance for your help.

REPLY [1 votes]: Your matrix isn't correct. Remember, if you're representing a linear transformation as a matrix, the columns of the matrix are the images of the basis vectors under the linear transformation. The derivative of the span is a vector - there's no reason that should be the correct matrix.
The correct approach here is to determine $D(\cos(x))$ as a vector $\begin{bmatrix} a \\ b \end{bmatrix}$, and $D(\sin(x))$ as a vector $\begin{bmatrix}c \\ d \end{bmatrix}$, then you have $D = \begin{pmatrix}a & c \\ b & d \end{pmatrix}$. This is the definition of representing the linear transformation as a matrix. Once you have that, for part (b) and (c), recall that $e^{ix} = \cos(x) + i\sin(x)$, and they should follow quickly.


TITLE: Given two substances of different specific heats, how do you find the final temperature and these phase related things
QUESTION [1 upvotes]: Given two substances of different specific heats, how do you find the final temperature of them both, phase of either substance if there was a phase change, and how much either substance has changed if not all of it has changed? I would like a general formula and index of which variables are which.
I tried this and this for example, but the examples assume the same specific heat capacity for both substances.

REPLY [1 votes]: There isn't a single formula because the process is an iterative one. The guiding principle is energy conservation, but you need to incorporate additional forms of energy transfer if a phase change occurs.
Example: Substance 1 (with mass $m_1$ and temperature $T_\mathrm{1,\,initial}$) starts off as a liquid, and substance 2 (with mass $m_2$ and temperature $T_\mathrm{2,\,initial}$) starts off as a gas. The total amount of energy gained by all substances must be zero according to energy conservation. Initially assume that no phase change occurs; thus, the final temperature $T_\mathrm{final}$ must satisfy
$$m_1C_1(T_\mathrm{1,\,initial}-T_\mathrm{final})+m_2C_2(T_\mathrm{2,\,initial}-T_\mathrm{final})=0$$
for temperature-independent (but material- and phase-dependent) heat capacities $C$. (If the heat capacities vary non-negligibly with temperature within a single phase, then you need to numerically or analytically integrate them.)
If the final temperature is above the liquid's boiling temperature or below the gas's boiling temperature, then the initial assumption is violated; some amount of phase change has occurred. Assume that it is partial. You need to revise the equation by adding $mxL$, where $x$ is the extent of the phase change (i.e., from 0 to 1, representing a lack of to a complete phase change, respectively) and $L$ is the latent heat of vaporization. The final temperature is the phase change temperature that was violated earlier. Now re-solve the equation.
If the extent $x$ is greater than 1, then your most recent assumption is violated. Set $x$ equal to 1 (because the phase change was completed) and add yet another term that represents a temperature change in the new phase. If the liquid was found to completely boil away, for example, then you have
$$m_1\left[C_\mathrm{1,\,liquid}(T_\mathrm{1,\,initial}-T_\mathrm{vaporization})+L_1+C_\mathrm{1,\,gas}(T_\mathrm{1,\,vaporization}-T_\mathrm{final})\right]+m_2C_2(T_\mathrm{2,\,initial}-T_\mathrm{final})=0$$
Solve again.
In general, you would conduct this iterative process with additional phase change temperatures and phase-dependent heat capacities until your most recent assumption is validated. Then you have obtained the equilibrium temperature.


TITLE: Cover $\mathbb{R}^3$ with skew lines
QUESTION [2 upvotes]: Is there a covering of $\mathbb{R}^3$ with pairwise skew lines?

REPLY [4 votes]: For each $\omega = x_0 +  y_0 i\in \mathbb{C} \simeq \mathbb{R}^{2}$. Consider the set $l_{\omega} \subset \mathbb{R}^3$ defined by:
$$l_{\omega} = \left\{ (x,y,z) \in \mathbb{R}^{3} : \frac{x + iy}{1 + iz} = \omega \right\}$$
Since the map $(x,y,z) \mapsto \frac{x+iy}{1+iz}$ is defined over all $\mathbb{R}^3$, we have:
$$\mathbb{R}^3 = \cup_{\omega \in \mathbb{C}}\;l_{\omega}\quad\text{ and }\quad
l_{\omega_1} \cap l_{\omega_2} = \emptyset \;\text{ for distinct }\;\omega_1, \omega_2 \in \mathbb{C}$$
Furthermore, each $l_{\omega}$ is a straight line as it admits a linear parametrization:
   $$\mathbb{R} \ni t \mapsto ( x_0 - y_0 t,\, y_0 + x_0 t,\, t ) \in l_{\omega}$$
With this parametrization, one can see that its tangent vector is along the direction
proportional to $(-y_0, x_0, 1)$.
This means $l_{\omega_1} \not\parallel l_{\omega_2}$ for distinct $\omega_1, \omega_2 \in \mathbb{C}$.
In short, $\{ l_{\omega} : \omega \in \mathbb{C} \}$ is a partition of $\mathbb{R}^3$ into a family of lines in which all pairs are skew.


TITLE: Multivariable calculus: what principle is this step based on?
QUESTION [0 upvotes]: The background is that I was asked to solve the following problem using Green's formula

$L$ is a Jordan curve (smooth and closed) which encloses the origin point in $xOy$ plane. Caculate this integral
  $$I=\oint_L \frac{e^x}{x^2+y^2}\left[(x\sin y-y\cos y)dx+(x\cos y+y\sin y)dy \right]$$
  in the counter-clockwise direction.

Let $P,Q$ denote the $x,y$ components respectively, and it is easy to see $-P'_{y}+Q'_{x}=0$. However, since the origin is a singularity point, we can't apply Green's formula directly on the area enclosed by $L$. So we consider a small circle $\Gamma:x^2+y^2=\epsilon^2$ which is enclosed by $L$. After some simple adjustment we have
$$I=\oint_\Gamma \frac{e^x}{x^2+y^2}\left[(x\sin y-y\cos y)dx+(x\cos y+y\sin y)dy \right]\quad\text{(in the clockwise direction)}$$
I'll skip all the complicated calculations here. Finally I got
$$I=\int_0^{2\pi}e^{-\epsilon\cos\theta}\cos(\epsilon\sin\theta)d\theta$$
and I couldn't proceed any further. However when I looked it up in the answerbook, it said, without explanation, that I should let $\epsilon\to 0^+$ and hence
$$I=\int_0^{2\pi}1d\theta=2\pi$$
It really surprised me a great deal, not for its simplicity, but because I had never seen such a trick taught in my textbook before!
I want to know what mathematical principle this trick is based on (continuity? I suspect), and on what condition we are able to use it.
Best regards.

I have been pondering on this problem for a while, and just now I had a thought as follows.
By Green's formula, $I$ is independent of $\epsilon$ (and even independent of the shape of $\Gamma$ as long as $\Gamma$ is a Jordan curve enclosing the origin and enclosed by $L$). Therefore we can regard $I$ as a constant function w.r.t. $\epsilon\in\Bbb R^{+}$. Then if $\displaystyle\exists\lim_{\epsilon\to 0^+}I(\epsilon)$, it will be the desired result. So all I need to do is prove the existence of this limit. Am I right?

REPLY [1 votes]: Your integral comes from the complex integral
\begin{align}
&\int_0^{2\pi } {e^{ - \varepsilon \cos \theta } \cos \left( {\varepsilon \sin \theta } \right)d\theta }  + i\int_0^{2\pi } {e^{ - \varepsilon \cos \theta } \sin \left( {\varepsilon \sin \theta } \right)d\theta }  \\
  &= \int_0^{2\pi } {e^{ - \varepsilon \cos \theta } \left[ {\cos \left( {\varepsilon \sin \theta } \right) - i\sin \left( {\varepsilon \sin \theta } \right)} \right]d\theta }  \\
  &= \int_0^{2\pi } {e^{ - \varepsilon \cos \theta } e^{ - i\varepsilon \sin \theta } d\theta }  \\
  &= \int_0^{2\pi } {e^{ - \varepsilon \left( {\cos \theta  + i\sin \theta } \right)} d\theta }  \\
  &= \int_0^{2\pi } {e^{ - \varepsilon e^{i\theta } } d\theta }  \\
  &=  - i\int_0^{2\pi } {\frac{{e^{ - \varepsilon e^{i\theta } } }}{{\varepsilon e^{i\theta } }}i\varepsilon e^{i\theta } d\theta }  \\
  &=  - i\int_L {\frac{{e^{ - z} }}{z}dz}  \\
  &=  - i\left( {2\pi i} \right) \\
  &= 2\pi
\end{align}
So that $$
\int_0^{2\pi } {e^{ - \varepsilon \cos \theta } \cos \left( {\varepsilon \sin \theta } \right)d\theta }
=2\pi$$  and $$\int_0^{2\pi } {e^{ - \varepsilon \cos \theta } \sin \left( {\varepsilon \sin \theta } \right)d\theta } =0$$


TITLE: Prove that $\forall x\in\mathbb{N}\ \text{ there always exists a prime }p\equiv1 \pmod 6 \text{ s.t. }p|(2x)^2+3;$
QUESTION [2 upvotes]: I want to prove the following:

$$\forall x\in\mathbb{N}\  \text{ there always exists a prime }p\equiv1 \pmod 6 \text{ s.t. }p|(2x)^2+3;$$$\  \text{i.e. } (2x)^2\equiv-3 \pmod p$ where $p$ is some prime sufficing $p\equiv1 \pmod 6$ has to be true for all $x\in\mathbb{N}$.

I want to represent $2x$ as a product of 2 and primes $p_i$ of the form $6k_i+1$; i.e. $p_i\equiv1(\mod 6)$ and $2x=2\prod_{i=1}^{n}p_i$. Note that $x\in\mathbb{N}$ is chosen arbitrarily beforehand. Is this even possible?
What I've tried so far:
I've prove with the Legendre symbol that $\big(\frac{-3}{p}\big)=1$ when $p\equiv1(\mod6)$ which clarifies the existence of an element $y\in\mathbb{Z}/(p)$ s.t. $y^2\equiv-3(\mod p)$. But since I want to show that it holds for all $y=2x$ where $x$ is chosen, this is useless i.m.o..
So maybe brainless trial and error will get me somewhere:
$x=1$:   $2^2+3\equiv0(\mod p)$ where we choose $p=7\equiv1(\mod6)$ prime.
$x=2$:   $4^2+3=19\equiv0(\mod p)$ where we choose $p=19\equiv1(\mod6)$ prime.
...
$x=a$:   $4a^2+3\equiv0(\mod p)$ for some $p\equiv1(\mod6)$ iff (?) $4a^2+3\equiv1(\mod6)$.
This seems a dead end.
I also was considering $\big(\frac{-3\cdot4^{-1}}{p}\big)$, but this is difficult to compute since we don't know the inverse of 4 explicitly modulo $p$. Of course there's an algorithm for it, but I doubt it will bring succes in computing the Legendre symbol.
Still, $x\in\mathbb{N}$ is fixed, hence we cannot just manipulate $x$. Clearly, I'm stuck. Could anyone get me on the right track? All help is appreciated!
EDIT:
First title that is now deleted: For every $x\in\mathbb{N}$ write $2x=2\prod_{i=1}^{n}p_i$ with $p_i=6k_i+1$ primes

REPLY [2 votes]: In fact something stronger is true. All of the prime divisors of $(2x)^2+3$ ($x\in\mathbb{N}$) are either equal to $3$ or are congruent to $1$ mod $6$. (And then since they can't all be equal to $3$, there is always some prime divisor congruent to $1$ mod $6$.)
Suppose $p$ is some odd prime dividing $(2x)^2+3$. Note that either $p=3$, or $p\equiv1$ or $p\equiv5$ mod $6$. Assume that $p\equiv5$ mod $6$. Then $$(2x)^2\equiv-3\mod{p}$$ So $-3$ is a square mod $p$. So using Legendre symbols and quadratic reciprocity:$$
\begin{align}
1&=\left(\frac{-3}{p}\right)\\
&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\
&=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)(-1)^{(p-1)(3-1)/4}\\
&=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)(-1)^{(p-1)/2}\end{align}$$
Since $p\equiv5$ mod $6$, then $p\equiv2$ mod $3$. So $p$ is not a square mod $3$. So continuing:
$$
1=\left(\frac{-1}{p}\right)(-1)^{(p+1)/2}
$$
It is "well-known" that $-1$ is a quadratic residue mod $p$ if and only if $p$ is congruent to $1$ mod $4$. So if $p$ is congruent to $1$ mod $4$, the above says:
$$
1=(1)(-1)
$$
and if $p$ is not congruent to $1$ mod $4$, the above says:
$$
1=(-1)(1)
$$
So it is a contradiction either way. We conclude that all primes dividing $(2x)^2+3$ are either equal to $3$ or are $1$ mod $6$. There is the possibility that $(2x)^2+3$ is a power of $3$, but we can eliminate that easily. Mod $9$, $6$ is not a square. So $(2x)^2+3\not\equiv0$ mod $9$ no matter what $x$ is. So $(2x)^2+3$ is sometimes divisible by $3$ but never divisible by $9$.
So $(2x)^2+3$ is always the product of some primes that are congruent to $1$ mod $6$, sometimes with a single factor of $3$ thrown in the mix. But $(2x)^2+3$ always has prime factors congruent to $1$ mod $6$.


TITLE: Are all exotic affine spaces count equivalent to affine space?
QUESTION [10 upvotes]: This question is inspired by this MO question; indeed it is a special case on which to focus.
An exotic affine space is an affine variety $V$ whose $\mathbb{C}$-points are diffeomorphic to $\mathbb{R}^{2n}$ yet $V$ is not algebraically isomorphic to $\mathbb{A}^n$.
Say that two varieties are count equivalent if they are both polynomial count varieties with the same counting polynomial.
As shown in the comments here, the Russell Cubic is count equivalent to $\mathbb{A}^3$ although it is not isomorphic to $\mathbb{A}^3$.

Question: Are all exotic affine spaces count equivalent to affine space?

REPLY [11 votes]: For all but finitely many primes, yes. Any such $V$ has $V_{\mathbb C}$ smooth and has $H^i(V_{\mathbb C}, \mathbb Q_\ell)=0$ for $i\neq 0$ and $=\mathbb Q_\ell$ for $i=0$. Both these properties are known to be constructible, so they hold for $V_{\mathbb F_q}$ for all but finitely many $q$.
For any such $q$, the action of $\operatorname{Frob}_q$ on $H^0(V_{\overline{\mathbb F}_q}, \mathbb Q_\ell)$ must be trivial as it is one-dimensional and contains at least one invariant class. So its cohomology, with Frobenius action, matches affine space. Applying Poincare duality (using smoothness), the same is true for its cohomology with compact supports. Applying the Grothendieck-Lefschetz formula, its number of points matches affine space.