TITLE: If every continuous $f:X\to X$ has $\text{Fix}(f)\subseteq X$ closed, must $X$ be Hausdorff?
QUESTION [35 upvotes]: Given a function $f:X\to X$, let $\text{Fix}(f)=\{x\in X\mid x=f(x)\}$.
In a recent comment, I wondered whether
$X$ is Hausdorff $\iff$ $\text{Fix}(f)\subseteq X$ is closed for every continuous $f:X\to X$ (the forwards implication is a simple, well-known result). At first glance it seemed plausible to me, but I don't have any particular reason for thinking so. I'll also repost Qiaochu's comment to me below for reference:

I would be very surprised if this were true, but it doesn't seem easy to construct a counterexample. Any counterexample needs to be infinite and $T_1$, but not Hausdorff, and I don't have good constructions for such spaces which don't result in a huge collection of endomorphisms...

Is there a non-Hausdorff space $X$ for which $\text{Fix}(f)\subseteq X$ is closed for every continuous $f:X\to X$?

REPLY [3 votes]: This is a remark related to the question.
Theorem : let $X$ be a topological space. Suppose that for every space $Y$ and pair of maps $f,g : Y \rightarrow X$, the subset $\{ y \in Y : f(y)=g(y) \}$ is closed. Then $X$ is Hausdorf.
It is easy to prove by considering the $2$ projections $X \times X \rightarrow X$. In algebraic geometry, a (pre)variety which satisfies this axiom is called separated (see for example Milne's note on page 61 http://www.jmilne.org/math/CourseNotes/ag.html). This notion was introduced because algebraic varieties are not Hausdorff, and if we work with separated varieties, then maps are determined on dense subset.


TITLE: Minimal Polynomial of a scalar multiple of a Matrix
QUESTION [2 upvotes]: I got the following problem:
Let $A$ be a square matrix of order $n$ over field $\mathbb{F}$
and let $M_A$ be the minimal polynomial of $A$ of degree $k\in\mathbb{N}$ and let $0\neq c\in\mathbb{F}$ be any nonzero scalar in $\mathbb{F}$ then the minimal polynomial of $cA$ denoted by $M_{cA}$ is equal to
$c^nM_A({\frac tc})$ or $c^kM_A({\frac tc})$
I tried to prove it but didn't manage to continue that much.
 Thanks.

REPLY [2 votes]: Indeed your guests are in the right direction: $M_A(\frac{t}{c})$ annihilates $cA$, which is the first property the minimal polynomial of $cA$ must fulfil.
The second one is that it must be monic; that is, its leading coefficient must be equal to one. Hence you need to multiply it by $c^k$.
Finally, you need to prove that, if $P(t)$ is such that $P(cA) = 0$, then $M_{cA}\ | \ P$. And this is the case, since $P(cA) = 0$ means that $P(ct)$ annihilates $A$. Hence, $M_A (t)\ | \ P(ct)$. Now, we do a little change with the name of the variable: put $u = ct$ and you get $M_A(\frac{u}{c}) \ | \ P(u)$. That is, $M_{cA} \ | \ P$.


TITLE: prove that C = (C \D)∪(C ∩D).
QUESTION [0 upvotes]: Let C and D be sets. Use element chasing to prove that C = (C \D)∪(C ∩D). We can only use deﬁnitions and logical deduction. You can not use properties of sets. I am having trouble figuring out where to start on this proof. I know for element chase we have to do a case by case proof.

REPLY [1 votes]: If $C = \varnothing$ you are done. If $C \neq \varnothing$, pick an element of $C$, let's say, $c$. Now, note that, $c$ satisfy one, and only one, of the following two conditions:

$c$ is an element of $D$,
$c$ is not an element of $D$.

If $c$ satisfies the first, we then have that $c \in C \cap D$, and hence $c \in (C \setminus D) \cup (C \cap D)$. In the other case, we have that $c \in C \setminus D$ and again, $c \in (C \setminus D) \cup (C \cap D)$.
In either case, we see that $c \in (C \setminus D) \cup (C \cap D)$, showing that any element of $C$ is also an element of $(C \setminus D) \cup (C \cap D)$. So, $C \subseteq (C \setminus D) \cup (C \cap D)$.
Can you show the other inclusion?


TITLE: If the current is increased, is there more charge flowing or is it moving quicker?
QUESTION [8 upvotes]: Problem
Current is the amount of charge that is flowing through a component per unit of time.
For a given voltage, Ohm's law tells us that if we increase the resistance, then the current must decrease.
But what's actually happening to decrease the current?
My reasoning so far - is it correct?
More resistance (if we're talking about something of the same size for simplicity) is more 'stuff' in the way (higher resistivity), so more collisions. More collisions means it takes longer for the charge to 'get through' the component. The charge is moving slower and so the current is lower.
Or, is it that the speed is always the same and that somehow if we have more resistance then it means there is just less charge able to flow, hence a lower current. If so, why?

REPLY [0 votes]: For a given voltage, Ohm's law tells us that if we increase the resistance, then the current must decrease.
But what's actually happening to decrease the current?

If we increase the potential, the current increases, rest all the parameters are constant - which are resistance, material, temperature etc. Drift velocity of the charges, here, have increased due to increased potential.
Drift velocity is $v_d = \frac{e.V.d.\tau}{m}$, where
$e$ - charge
$V$ -  Potential
$d$ - distance moved by the charge in time $\tau$
$m$ - mass of the charge
So here we can say that Charges move with greater speed now
Now to change the resistance, let us just change resistivity i.e changing the material and rest all the parameters are same except Free Charge Density of the material, it depends on the material.
Current, $I = n.e.A.v_d$, here $(n.e)$ is the charge density, the effective charge that is flowing per unit volume.
So here the amount freely flowing charge changes and also the drift velocity changes because $\tau$, the effective time between consecutive collisions, may also change.

REPLY [0 votes]: For an Ohmic conductor (not a semiconductor) the number of conduction electrons is constant. It is the drift velocity that increases as a larger potential difference means a larger acceleration between scattering events. You can change the resistivity by changing the temperature. This will change the scattering probability, hence the time interval of acceleration hence the drift velocity.
Of course you can also change to a different material, which will have a different resistivity for example because it has a different conduction electron density and mobility.


TITLE: Is that proof correct? I'm proving that the set of injective Linear transformations in open.
QUESTION [0 upvotes]: Let $T$ an inyective linear transfomation from $\mathbb{R}^n$ to $\mathbb{R}^m$. Then there exist an $\alpha>0$ such that $$||T(x)||\geq \alpha||x||$$ for all $x$. Let $S$ a linear transformation such that
$$||T-S||<\alpha$$
So $$||T(x)||\geq \alpha||x||$$
   $$||T(x)-S(x)+S(x)||\geq \alpha||x||$$
$$||S(x)||+||T(x)-S(x)||\geq \alpha||x||$$
$$||S(x)||+||(T-S)(x)||\geq \alpha||x|| $$
$$||S(x)||+||T-S||\cdot||x||\geq \alpha||x|| \ \ (*)$$
$$||S(x)||\geq (\alpha-||T-S||)\cdot||x||$$
for all $x$. So the constant that work in this case is $\alpha-||T-S||>0$.
I have a dude in $(*)$ i can do this?? If not how can i do it in the right way?
||T|| is the Frobenius norm of a matrix.

REPLY [2 votes]: Nicely done. For the $*$ there is no problem because $T-S$ is a bounded linear map(or bounded operator,or continuous map and many more) ,because $\|T-S\|<a$. So $\|(T-S)(x)\|\leq \|T-S\|\|x\|$


TITLE: Is U union W necessarily a vector space?
QUESTION [1 upvotes]: Let U and w be two subspaces of the same vector space V.Is U union W necessarily a vector space?

REPLY [0 votes]: Let $V = R^3$. Choose the subspaces to be generated by two non parallel vectors, respectively. Then the sum of the two vectors is not contained in the union.


TITLE: Simplify the radical expression
QUESTION [0 upvotes]: Simplify the following:
  $$\sqrt{48 a^2 b^7}$$

All I have so far is 16  I think i forgot how to do this, can anyone help?

REPLY [2 votes]: $$\begin{align}48&=\underbrace{2\times 2}\times\underbrace{2\times2}\times3\\a^2&=\underbrace{a\times a}\\b^7&=\underbrace{b\times b}\times \underbrace{b\times b}\times \underbrace{b\times b}\times b\end{align}$$
When you take the square root of them:
$$\begin{align}\sqrt{48}&=\underbrace{\sqrt{2\times 2}}_2\times\underbrace{\sqrt{2\times2}}_2\times\sqrt3\\a^2&=\underbrace{\sqrt{a\times a}}_{a}\\b^7&=\underbrace{\sqrt{b\times b}}_{b}\times \underbrace{\sqrt{b\times b}}_{b}\times \underbrace{\sqrt{b\times b}}_{b}\times \sqrt{b}\end{align}$$
So you're left with:
$$4ab^3\sqrt{3b}$$
Note that:
$$(4ab^3)^\frac22\sqrt{3b}=\sqrt{16a^2b^6}\sqrt{3b}=\sqrt{48a^2b^7}$$


TITLE: Show a function is decreasing
QUESTION [1 upvotes]: I'm having some trouble finding a way to prove that the following function is decreasing
$f(x) = \dfrac{\left(\dfrac{3^m-2}{3^m-1} + \dfrac{1}{(3^m-1)(3^m)^x}\right)^s}{\dfrac{6^m-5}{6^m-2} + \dfrac{3}{6^m-2}\left(\dfrac{2}{6^m}\right)^x}$
where $m\in\mathbb{N},m \geq 2, x\geq 1$ and $s=\log_36$.
Taking the derivative is rather complicated (I did it in Wolfram Mathematica) and I haven't found a good way to go from there to show that the resulting expression is negative.
Definition of decreasing seems a little less painful but I run into the issue where the numerator of $f(x)$ and denominator of $f(x)$ are both decreasing and I'm not sure if there's a way to analyze that further to show the whole function is decreasing.
Thanks for any help.

REPLY [1 votes]: I think I made a mistake in my previous answer. Since my answer can't be deleted, I update it to clarify the argument. My point is that this function can't be decreasing.
Let $f(x) = \dfrac{(a+ \frac{1}{kb^{x}})^s}{c+ \lambda d^x}$, where $a,c,d,k,\lambda <1; b>1$ are some positive constants (compare these letters with terms depending on $m$ in the original function. For a fixed $m$ & $s$, these can be considered constants).
Note that, {$1/kb^{x}$} decreases since $b>1$. Hence, {$a+ \frac{1}{kb^{x}}$} is decreasing as well.  Moreover  $a+ \frac{1}{kb^{x}}< 1$ & $s>1$; therefore,
{$(a+\frac{1}{kb^{x+1}})^s$} is decreasing as well.
Similarly,{$c+ \lambda d^x$} decreases since $d<1$.
Now , let $p(x)$ & $q(x)$ be the numerator & the denominator of $f(x)$ , respectively. So, $p(x) > p(y)$ & $q(x) > q(y)$ for all $x<y$. Thus, $1/q(y) > 1/q(x)$.
$f(y)=\dfrac{p(y)}{q(y)} > \dfrac{p(y)}{q(x)}$
That's all we can conclude.  It's not possible to prove the desired result.


TITLE: induction with remainders
QUESTION [2 upvotes]: $$a_n=
\sum_{i=0}^5 (n+i)^2$$
I have to prove by induction that for every natural $n$ (including $0$ in this case) that if you divide $a_n$ by $12$ you will always get a remainder of $7$. so i proved the base case which was n=0
Then I started my induction step which was to take $a_n+1 - a_n$ (i got help there because i don't understand why to subtract for my induction step)
It all equals $(n+6)^2-n^2 = 12n+36 = 12(n+3)$
so my problem is that i have to show that this will give me a remainder of $7$ when divided $12$ but it won't because its all a multiple of $12$. So where did i go wrong??

REPLY [0 votes]: The key idea is that a sequence $\,a_n\,$ is constant $\,a_n\equiv a_0\equiv 7\,$ if it $\color{#c00}{\text{never changes value.}}$
Lemma $\ $ The sequence $\,a_n\,$ satisfies $\,a_n \equiv  a_0\,$ for all $\,n\ge 0\,$ if $\,\color{#c00}{a_{n+1} \equiv a_{n}}\,$ for $\,n\ge 0$
Proof $\ $ The base case is clear, and $\,a_n\equiv a_0\, \Rightarrow\, \color{#c00}{a_{n+1}\equiv a_n}\equiv a_0\,$ is the induction step.
You have proved that $\ 7\mid a_{n+1}-a_n,\ $ i.e. $ $  that $\ \color{#c00}{a_{n+1}\equiv a_n}\pmod{7},\,$ so the Lemma applies.

Remark $\ $ The proof is a special case of telescopy since we can write $\,a(n) = a_n\,$ as a telescoping sum of its successive differences $\displaystyle \, a(n)\, =\, a(0) + {\sum_{k=0}^{n-1}\, \overbrace{a(n\!+\!1)-a(n)}^{\large\color{#c00}0}}\ $ i.e.
$$ \color{#c00}{a(0)}\, =\,  \underbrace{\color{#c00}{a(0)}\phantom{-a(0)}}_{\Large\color{#0a0}0}\!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{-\,a(0) +\!\phantom{a(1)}}^{\Large\!\!\!\!\! \ \ \ \,  \color{#c00}{0}} \!\!\!\!\!\!\!\!\!\! \underbrace{a(1) -a(1)}_{\Large\color{#0a0}0}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{\phantom{-a(1)}\!+ a(2)}^{\Large \!\!\!\!\!\!\!\!\!\!\! \quad\ \ \  \color{#c00}{0} }\!\!\!\!\!\!\!\!\!\!\underbrace{\phantom{a(2)}-a(2)}_{\Large\color{#0a0}0}\!+\: \overbrace{\underbrace{\cdots\phantom{I_{I_I}\!\!\!\!\!\!\!\!}}_{\Large\color{#0a0}0}+\,\color{#0a0}{a(n)}}^{\Large \!\!\!\!\! \ \ \ \color{#c00}{0}}\ =\ \color{#0a0}{a(n)} $$
The induction essentially rebrackets the sum from the green bracketing $\,\color{#0a0}{0+\cdots0 +a(n)}\,$ into the red bracketing $\,\color{#c00}{a(0)+0+\cdots0},\, $ using the associativity of addition. For further details on this ubiquitous telescopic form of induction,  see this answer and its links.


TITLE: Rotations in 3D space: order of concatenated rotations.
QUESTION [0 upvotes]: I am currently doing an online course in Udemy to clarify some things for a project I'm currently doing on solid rigid motion and simulation in University, although understanding some of the things regarding the use of rotation matrixes has confused for some time already.
Given a coordinate system $\{XYZ\}_I$, and a sequence of rotations in $Z$, $Y$, $X$ represented by the matrices $R_z$,$R_y$,$R_x$ respectively performed over the original axis and not the one after each rotations that rotate $\{XYZ\}_I$ to $\{XYZ\}_B$ , and given $u_B \in \mathbb{R}^3$, I do not understand completely the way to calculate $u_I$ is:
\begin{equation}
      u_I = R_xR_yR_zu_B
\label{eq1}
\end{equation}
What I understand of Rotation matrices is that a rotation from a frame $\{XYZ\}_A$ to an rotated frame $\{XYZ\}_B$ can be described as a rotation matrix $R$ that maps a vector from the rotated frame $B$ to an original frame $A$ the following way:
$$u_A = R u_B$$
Following that intuition, by calculating the rotations as in the first equation it would imply that the $Rz$ maps the vector $u_B$ to an intermediate frame (let's say $B_z$), then $R_y$ maps from $B_z$ to another intermediate frame (let's say $B_y$), and finally, $R_x$ rotates from $B_y$ to the inertial and original body frame $I$.
According to the course, that is not like that because the rotations in this case are described as rotations from the inertial frame to each of the intermediate body frames. That would mean that $Rz$ maps from the first intermediate frame to the inertial frame (instead of maping form the final body frame to the first intermediate), and then, the next rotation would be mapping from the inertial frame to the inertial frame again (?) since the vector that was returned in the previous transformation is to the inertial frame, but that is a contradiction unless all the rotations left are the identity.
I know there is something I'm not getting, and I know there is also the possibility of post-multiplying instead of pre-multiplying, which I guess that implys doing the opposite rotaiton, although I'm still a bit lost and don't totally understand this elements.
Also, even if I can kind of see how it works if we are rotating from a given frame to another frame, and a sequence of rotations, how would we define a sequence of rotations if they are all respect to the world, inertial frame?

REPLY [0 votes]: If $x $ is a vector in an initial coordinate system $B_i$ , the corresponding the world coordinates is the vector $y$ where
$ y = B_i \ x $
If you now apply $3$ consecutive rotations to vector $y$ about the world axes $z$, $y$, $x$ in that order, then the final image of $y$ is the vector $y'$ where
$y' = R_x\ R_y\ R_z\ y = R_x\ R_y\ R_z\ B_i\ x  $
which means that the final coordinate system is
$B_f = R_x\ R_y\ R_z\ B_i $
So that the relation between a vector in frame $B_i$ and its corresponding coordinate vector $y'$ after applying the rotations, is
$ y' = B_f\ x $


TITLE: Solve the differential equation:$\frac{\,dx}{mz-ny}=\frac{\,dy}{nx-lz}=\frac{\,dz}{ly-mx}$
QUESTION [2 upvotes]: QUESTION: Solve the differential equation: $$\frac{\,dx}{mz-ny}=\frac{\,dy}{nx-lz}=\frac{\,dz}{ly-mx}$$

MY ATTEMPT: I tried out to proceed by using
$$\frac{\,dx}{mz-ny}=\frac{\,dy}{nx-lz}=\frac{\,dz}{ly-mx}=k(\text{say, a constant})$$
Then I got $$dx=k(mz-ny)$$ $$dy=k(nx-lz)$$ $$dz=k(ly-mx)$$
And then I could by elimination of variables, create diffential equations with $2$ variables, like $$l\,dx+n\,dz=k(lmz-nmx)$$ Or $$l\,dx+knmx=klmz-n\,dz$$
But then I could not proceed any further since I have no idea on how to integrate expressions like $l\,dx+knmx$ where $x$ and $\,dx$ are separated by a $+$ sign.
Can anyone help.

REPLY [0 votes]: What if you consider $\frac {dx}{mz-ny}=\frac{dy}{nx-lz}$ , $\frac{dy}{nx-lz}=\frac{dz}{ly-mx}$ and $\frac{dx}{mz-ny}=\frac{dz}{ly-mx}$separately. Yielding $\int{(nx-lz)dx}=\int{(mx-ny)dy}$ , $\int{(ly-mx)dy}=\int{(nx-lz)dz}$, and $\int{(ly-mx)dx}=\int{(mz-ny)dz}$. Giving Three equations $\frac{nx^2}{2}=-\frac{ny^2}{2}$, $\frac{ly^2}{2}=-\frac{lz^2}{2}$, and $\frac{nx^2}{2}=-\frac{lz^2}{2}$.
Yielding $x^2=-y^2$, $y^2=-z^2$, and $x^2=-z^2$.
or
$x^2=-y^2=-z^2$.


TITLE: Show that $\lim_{x\rightarrow \infty}f(x)=\infty.$
QUESTION [1 upvotes]: The function $f:(0,\infty)\rightarrow \mathbb{R}$ is twice differentiable in its domain and satisfies the following:
i) $f(2)=-1,$
ii) $f'(2)>0,$
iii) $f''(x)\geq 0, \ \text{in} \ [2,\infty).$
Show that:
a) $\lim_{x\rightarrow \infty}f(x)=\infty.$
b) $f$ has at most one root in $(2,\infty).$
c) $f$ has at least one root in $(2,\infty).$

Attempt:
a) We know that $f''\geq 0 $ in $[0,\infty)\Longrightarrow f'$ is strictly increasing in $(2,\infty)$ and that $f$ is convex in the same interval.
This implies that $f(x)\rightarrow\infty$ as $x\rightarrow\infty$ and a) is shown.
b)+c)
Since the above holds and since $f(2)=-1<0,$ we know that there exists a $c\in\mathbb{R}$ such that $f(c)>0.$ Thus, according to the intermediate value theorem there exists at least one $\xi\in(2,c)$ such that $f(\xi)=0$
Is this "proof" (it's a proof to me until i'm disproven) correct or am I really off the track here?

REPLY [0 votes]: You wrote

a) We know that $f''\geq 0 $ in $[0,\infty)\Longrightarrow f'$ is strictly increasing in $(2,\infty)$ and that $f$ is convex in the same
  interval.

This is false. For instance, if $f'(x) = 1$ for all $x \in (2, \infty)$, then $f''(x) = 0 \ge 0$ for all $x$...but $f'$ is not strictly increasing. There's a real difference between "less than" and "less than or equal to".
You're certainly on the right track; you just need to be more careful.


TITLE: Prove for $\gcd$ of two integers $a,b$, that $(a^n,b^n) = (a,b)^n$
QUESTION [3 upvotes]: Let $d=(a^n,b^n), d_1=(a,b)^n$; need show $d\mid d_1$ & $d_1\mid d$ to prove.
For the general case of $(a,b)\ne 1, (a,b)= (d_1)^\frac{1}{n}$, else $(a,b)=d_1$.
Let, $d_2= (a,b)=d_1^\frac{1}{n}\implies a=a'd_2, b = b'd_2\implies a^n= (a')^nd_2^n, b^n= (b')^nd_2^n$
Unable to progress further.

REPLY [1 votes]: It is easily reduced to the coprime case $\rm\,\color{#c00}{(a,b)=1\,\Rightarrow\,(a^{\large n},b^{\large n})=1}\,$ proved here . Namely
to show $\rm\ (A^{\large n},B^{\large n}) = (A,B)^{\large n}\,$ let $\rm\,d=(A,B),\,$ so $\rm\,A,B=ad,bd\,$ with $\rm\,\color{#c00}{(a,b)=1}.\,$ Then
$\qquad\quad\,\ \rm (A^{\large n},B^{\large n})=((ad)^{\large n},(bd)^{\large n})=\color{#c00}{(a^{\large n},b^{\large n})}\,d^n = d^{\large n} = (A,B)^{\large n}$
where we applied the GCD Distributive Law $\rm\ (x,y)z = (xz,yz)$
The same idea works for homogeneous forms to reduce to the case of coprime arguments.


TITLE: Solving equation in Matlab
QUESTION [0 upvotes]: I want to solve this equation for several values of y
y=4 exp(0,034x)

y= [60 14 63 34 21 12 11];

Thanks

REPLY [0 votes]: This is not mainly a site for programming specific help, so don't be surprised if people sometimes vote down questions that are too specific about computer language. You can use the fsolve command and an anonymous function
f = @(x)[60 14 63 34 21 12 11]' - 4*exp(0.034.*x);
x_solution = fsolve(f,rand(1,7)');


TITLE: Edit distance with move operations
QUESTION [14 upvotes]: Motivation: A coauthor edits a manuscript and I would like to see a clear summary of the edits. All "diff"-like tools tend to be useless if you are both moving text around (e.g., re-organising the structure) and doing local edits. Is it really so hard to get it right?

Definitions: I would like to find the minimum edit distance, where the permitted operations are:

"cheap" operations: add/change/delete a single character (the usual Levenshtein operations),
"expensive": operations: move a substring to a new location ($abcd \mapsto acbd$ for any strings $a$, $b$, $c$, $d$).

Given two strings $x$ and $y$ and integers $k$ and $K$, I would like to solve the following problem:

can you transform $x$ into $y$ using at most $k$ cheap operations and at most $K$ expensive operations?

Questions:

Does this problem have a name? (It sounds like a very standard question in the context of sequence alignment.)
Is it hard?
If it is hard, is it fixed-parameter tractable with $K$ as the parameter?
Are there efficient approximation algorithms? (E.g., find a solution with at most $2k$ cheap and $2K$ expensive operations if a solution with $k$ cheap and $K$ expensive operations exist.)

I tried to have a look at the string metrics listed in Wikipedia, but none of them looked right.

REPLY [5 votes]: The problem becomes easier, if we consider long deletions and substring copying instead of transpositions. Assume that we are using the standard dynamic programming algorithm for edit distance computation, and that an expensive operation of length $k$ increases the distance by $ak+b$, for some constants $a,b \ge 0$. These constants may be different for long deletions and substring copying.
A long deletion is the deletion of an arbitrary substring from $x$. Supporting them is easy, if we break them down into two kinds of simple operations: deleting the first character (cost $a+b$) and extending the deletion by one character (cost $a$). In addition to the standard array $A$, where $A[i,j]$ is the edit distance between prefixes $x[1 \dots i]$ and $y[1 \dots j]$, we use another array $A_{d}$ to store the edit distance, when the last operation used was a long deletion. With this array, we only have to look at $A[i-1,j]$, $A[i-1,j-1]$, $A[i,j-1]$ and $A_{d}[i-1,j]$ when computing $A[i,j]$ and $A_{d}[i,j]$, allowing us to do it in $O(1)$ time.
Substring copying means the insertion of an arbitrary substring of $x$ into the edited string. As with long deletions, we break the operation down into two simple operations: inserting the first character and extending the insertion by one character. We also use array $A_{s}$ to store the edit distance between prefixes, provided that the last operation used was substring copying.
Doing this efficiently is more complicated than with long deletions, and I am not sure whether we can get to amortized $O(1)$ time per cell. We build a suffix tree for $x$, which takes $O(|x|)$ time, assuming a constant-size alphabet. We store a pointer to the current suffix tree node in $A_{s}[i,j-1]$, allowing us to check in constant time, whether we can extend the insertion by character $y[j]$. If that is true, we can compute $A[i,j]$ and $A_{s}[i,j]$ in constant time.
Otherwise $zy[j]$, where $z$ is the inserted substring that was used to compute $A_{s}[i,j-1]$, is not a substring of $x$. We use the suffix tree to find the longest suffix $z'$ of $z$, for which $z'y[j]$ is a substring of $x$, in $O(|z|-|z'|)$ time. To compute $A_{s}[i,j]$, we now need to look at cells $A[i, j-|z'|-1]$ to $A[i,j-1]$. Finding suffix $z'$ requires just amortized $O(1)$ time per cell, but computing $A_{s}[i,j]$ with a brute-force approach takes $O(|z'|)$ time. There is probably some way to do this more efficiently, but I cannot find it right now.
In the worst case, the algorithm takes $O(\min(|x| \cdot |y|^{2}, |x|^{2} \cdot |y|))$ time, but a better analysis should be possible. The resulting edit distance with long deletions and substring copying is not symmetric, but that should not be a problem. After all, it is usually easier to reach the empty string from a nonempty one than the other way around.


text:
\section{Conclusion}
In this paper, a novel all-digital self-interference cancellation technique for full-duplex systems is proposed. The proposed technique uses an auxiliary receiver chain to obtain a digital-domain copy of the transmitted RF self-interference signal including all transmitter impairments. The self-interference signal copy is then used in the digital-domain to cancel out both the self-interference signal and the transmitter impairments. In order to alleviate the receiver phase noise effect, a common oscillator is shared between the auxiliary and ordinary receiver chains. In addition, a nonlinearity estimation and suppression technique is proposed to mitigate the receiver nonlinearity effects. A thorough analytical and numerical analysis for the effect of the transmitter and receiver impairments on the cancellation capability of the proposed technique is presented. The analyses show that the proposed technique significantly mitigates the transceiver phase noise and nonlinearity effects. The overall full-duplex system performance using a combination of the proposed digital cancellation technique and practical passive suppression techniques is numerically investigated. The results show that, the proposed technique significantly mitigates the self-interference signal to $\sim$3dB higher than the receiver noise floor, which results in up to 76\% rate improvement compared to conventional half-duplex systems at 20dBm transmit power values.

\begin{figure}[!ht]
\begin{center}
\noindent
  \includegraphics[width=6.5in]{figure8.pdf}
  \caption{Full-duplex and Half-duplex acheivable rates at different transmit power and SNR values: (a) $1^{st}$ scenario, (b) $2^{nd}$ scenario, (c) $3^{rd}$ scenario.\label{Fig8Label}}
\end{center}
\end{figure}

\clearpage

\begin{table}[ht]
\caption{Average full-duplex rate improvement compared to half-duplex system in different operating scenarios}
\label{Table1Label}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
 & \multicolumn{2}{c|}{$1^{st}$ Scenario} & \multicolumn{2}{c|}{$2^{nd}$ Scenario} & \multicolumn{2}{c|}{$3^{rd}$ Scenario} \\
& \multicolumn{2}{c|}{(25dB passive suppression)} & \multicolumn{2}{c|}{(45dB passive suppression)} & \multicolumn{2}{c|}{(60dB passive suppression)} \\
 \hline
 & Proposed & Conventional & Proposed & Conventional & Proposed & Conventional \\
& Technique & DC & Technique & DC & Technique & DC \\
 \hline
 5dBm Transmit Power& 14\%  &   $-$58\%  &  61\%  & 11\%  &  76\%  & 72\% \\
\hline
 20dBm Transmit Power&  $-$58\% & $-$98\% & 23\% &  $-$63\% &  67\% & 14\%  \\
\hline
\end{tabular}
\end{table}

config: arxiv


text:
\section{Inconsistent Systems}\label{inconsistent}
In this section, we show that the distributed Kaczmarz algorithm with admissible relaxation parameters converges regardless of the consistency of the system and that the limit point is an approximation of a weighted least-squares solution when the system of equations is inconsistent. We first develop the relevant theory by following Successive Over-Relaxation (SOR) analysis of the Kaczmarz algorithm as developed in \cite{Natterer2001}.

Let $\ell \in \mathcal{L}$, and suppose $r = u_1 \to u_2 \to \cdots \to u_{p-1} \to u_p = \ell$, the path from $r$ to $\ell$. We denote the initial estimate at $r$ by $\vec{x}_{u_0}$. Then, from the Kaczmarz update, we recursively attain $\vec{x}_{u_j}$, the relaxed projection of $\vec{x}_{u_{j-1}}$ onto the hyperplane given by $\vec{a}_{u_j}^*\vec{x} = b_{u_j}$,
\begin{align}\label{relaxedprojection}
\vec{x}_{u_j} = Q_{u_j}^{\vec{\omega}}\vec{x}_{u_{j-1}} = \vec{x}_{u_{j-1}} + \omega_{u_j}\frac{b_{u_j} - \vec{a}_{u_j}^*\vec{x}_{u_{j-1}}}{\|\vec{a}_{u_j}\|^2}\vec{a}_{u_j}.
\end{align}
Hence, there exist complex scalars $\{c_k\}_{k=1}^p$ such that, for all $j$,
\begin{align}\label{sumrepresentation}
\vec{x}_{u_j} = \vec{x}_{u_0} + \sum_{k=1}^jc_k\vec{a}_{u_k}.
\end{align}
Substituting Equation \ref{sumrepresentation} into Equation \ref{relaxedprojection},
\begin{align}\label{coefficients}
c_j = \omega_{u_j}\dfrac{b_{u_j} - \vec{a}_{u_j}^*\vec{x}_{u_0} - \sum\limits_{k = 1}^{j-1}c_k\vec{a}_{u_j}^*\vec{a}_{u_k}}{\|\vec{a}_{u_j}\|^2}.
\end{align}
We can then consolidate Equation \ref{coefficients} for all $j$ into the matrix equation
\begin{align}\label{coefficientsvector}
D_\ell\vec{c} = \Omega_\ell(\vec{b}_\ell - A_\ell\vec{x}_{u_0} - L_\ell\vec{c})
\end{align}
where $\vec{c} = (c_1,c_2,...,c_p)^T$ and $D_\ell$, $\Omega_\ell$, $\vec{b}_\ell$, $L_\ell$ and $A_\ell$ are as follows:
\[
D_\ell = \begin{pmatrix}
\|\vec{a}_{u_1}\|^2 & 0 & \hdots & 0 \\
0 & \|\vec{a}_{u_2}\|^2 & \hdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \hdots & \|\vec{a}_{u_p}\|^2
\end{pmatrix},\,\Omega_\ell = \begin{pmatrix}
\omega_{u_1} & 0 & \hdots & 0 \\
0 & \omega_{u_2} & \hdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \hdots & \omega_{u_p}
\end{pmatrix},
\]
\[
\vec{b}_{\ell} = \begin{pmatrix}
b_{u_1} \\
b_{u_2} \\
\vdots \\
b_{u_p}
\end{pmatrix},\,L_\ell = \begin{pmatrix}
0 & 0 & 0 & \hdots & 0 & 0 \\
\vec{a}_{u_2}^*\vec{a}_{u_1} & 0 & 0 & \hdots & 0 & 0 \\
\vec{a}_{u_3}^*\vec{a}_{u_1} & \vec{a}_{u_3}^*\vec{a}_{u_2} & 0 & \hdots & 0 & 0 \\
\vec{a}_{u_4}^*\vec{a}_{u_1} & \vec{a}_{u_4}^*\vec{a}_{u_2} & \vec{a}_{u_4}^*\vec{a}_{u_3} & \hdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
\vec{a}_{u_p}^*\vec{a}_{u_1} & \vec{a}_{u_p}^*\vec{a}_{u_2} & \vec{a}_{u_p}^*\vec{a}_{u_3} & \hdots & \vec{a}_{u_p}^*\vec{a}_{u_{p-1}} & 0
\end{pmatrix},\,A_\ell = \begin{pmatrix}
\vec{a}_{u_1}^* \\
\vec{a}_{u_2}^* \\
\vdots \\
\vec{a}_{u_p}^*
\end{pmatrix}.
\]
Altogether, from Equations \ref{sumrepresentation} and \ref{coefficientsvector}, respectively, we may express the iterate $\vec{x}_\ell^{(n)}$ at the leaf $\ell$ given the initial vector $\vec{x}^{(n)}$ at the root in terms of the scalar vector $\vec{c}$,
\begin{align*}
\vec{x}_\ell^{(n)} &= \vec{x}^{(n)} + A_\ell^*\vec{c}, \\
\vec{c} &= (D_\ell + \Omega_\ell L_\ell)^{-1}\Omega_\ell\left(\vec{b}_\ell - A_\ell\vec{x}^{(n)}\right).
\end{align*}
We eliminate the scalar vector and attain
\begin{align*}
\vec{x}_\ell^{(n)} = (I - A_\ell^*(D_\ell + \Omega_\ell L_\ell)^{-1}\Omega_\ell A_\ell)\vec{x}^{(n)} + A_\ell^*(D_\ell + \Omega_\ell L_\ell)^{-1}\Omega_\ell\vec{b}_\ell.
\end{align*}

We then aggregate the leaf operators as follows:
\[
D = \begin{pmatrix}
D_{\ell_1} & 0 & ... & 0 \\
0 & D_{\ell_2} & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & D_{\ell_t}
\end{pmatrix}, \Omega = \begin{pmatrix}
\Omega_{\ell_1} & 0 & ... & 0 \\
0 & \Omega_{\ell_2} & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & \Omega_{\ell_t}
\end{pmatrix},
\]
\[
\vec{\mathfrak{b}} = \begin{pmatrix}
\vec{b}_{\ell_1} \\
\vec{b}_{\ell_2} \\
\vdots \\
\vec{b}_{\ell_t}
\end{pmatrix}, L = \begin{pmatrix}
L_{\ell_1} & 0 & ... & 0 \\
0 & L_{\ell_2} & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & L_{\ell_t}
\end{pmatrix}, \mathcal{A} = \begin{pmatrix}
A_{\ell_1} \\
A_{\ell_2} \\
\vdots \\
A_{\ell_t}
\end{pmatrix},
\]
\[
W = \begin{pmatrix}
w(r,\ell_1)I_{\text{dim}(\Omega_{\ell_1})} & 0 & ... & 0 \\
0 & w(r,\ell_2)I_{\text{dim}(\Omega_{\ell_2})} & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & w(r,\ell_t)I_{\text{dim}(\Omega_{\ell_t})}
\end{pmatrix}.
\]
The estimate obtained from the pooling stage of the $n$th iteration can be expressed in terms of these matrices,
\begin{align}\label{groupiterate}
\vec{x}^{(n+1)} = \sum_{\ell \in \mathcal{L}}w(r,\ell)\vec{x}_\ell^{(n)} = B^{\vec{\omega}}\vec{x}^{(n)} + \vec{b}^{\vec{\omega}}
\end{align}
where
\begin{align}
B^{\vec{\omega}} &= I - \mathcal{A}^*(D + \Omega L)^{-1}W\Omega \mathcal{A}, \\
\vec{b}^{\vec{\omega}} &= \mathcal{A}^*(D + \Omega L)^{-1}W\Omega\vec{\mathfrak{b}}.
\end{align}

Note that there exists a vector $\vec{h}$ such that $Q^{\vec{\omega}}\vec{x} = P^{\vec{\omega}}\vec{x} + \vec{h}$ for every $\vec{x}$. Then, from Equation \ref{groupiterate} and the linearity of $P^{\vec{\omega}}$ and $B^{\vec{\omega}}$, we have $B^{\vec{\omega}} = P^{\vec{\omega}}$ and $\vec{b}^{\vec{\omega}} = \vec{h}$.

\begin{proposition}\label{eigenvalues}
Suppose $B^{\vec{\omega}}\vec{x} = \lambda\vec{x}$ for some $\vec{x} \neq \vec{0}$. Then $\lambda = 1$ or $|\lambda| < 1$, and
\begin{enumerate}
    \item $\lambda = 1$ if and only if $\vec{x} \in \mathcal{R}(A^*)^\perp$,
    \item $|\lambda| < 1$ if and only if $\vec{x} \in \mathcal{R}(A^*)$.
\end{enumerate}
\end{proposition}

\begin{proof}
Suppose $P^{\vec{\omega}}\vec{x} = \lambda\vec{x}$ for some $\vec{x} \neq \vec{0}$. By Lemma \ref{contractions}, we note that $|\lambda| \leq 1$. Let $P$ be the orthogonal projection onto $\mathcal{R}(A^*)^\perp$. Then we find
\[
\lambda P\vec{x} + \lambda (I - P)\vec{x} = \lambda\vec{x} = P^{\vec{\omega}}\vec{x} = P^{\vec{\omega}}P\vec{x} + P^{\vec{\omega}}(I - P)\vec{x} = P\vec{x} + P^{\vec{\omega}}(I - P)\vec{x}.
\]
By uniqueness of the decomposition in $\mathcal{R}(A^*)\oplus\mathcal{R}(A^*)^\perp$, we have
\begin{align*}
    P\vec{x} &= \lambda P\vec{x}, \\
    P^{\vec{\omega}}(I - P)\vec{x} &= \lambda(I - P)\vec{x}.
\end{align*}
If $\lambda \neq 1$, then $P\vec{x} = \vec{0}$ so that $\vec{x} = (I - P)\vec{x} \in \mathcal{R}(A^*)$. From this observation and Proposition \ref{inconsistentlemma1}, we find that $|\lambda| < 1$. Now suppose $\lambda = 1$. Then, by Proposition \ref{inconsistentlemma1}, $(I - P)\vec{x} \in \mathcal{R}(A^*)^\perp$ so that $\vec{x} = P\vec{x} \in \mathcal{R}(A^*)^\perp$. The sufficient statement of (1) is straightforward, and (2) follows.

\end{proof}

\begin{lemma}\label{Convergence}
Let $\vec{x}^{(0)} \in \mathcal{R}(A^*)$. The sequence $\{\vec{x}^{(n)}\}$ converges to the fixed point of the mapping $\vec{x} \in \mathcal{R}(A^*) \mapsto B^{\vec{\omega}}\vec{x} + \vec{b}^{\vec{\omega}}$. Precisely, the sequence converges to
\[
(I - B^{\vec{\omega}})|_{\mathcal{R}(A^*)}^{-1}\vec{b}^{\vec{\omega}} = \sum_{j=0}^\infty(B^{\vec{\omega}})^j\vec{b}^{\vec{\omega}}.
\]

\end{lemma}

\begin{proof}
Throughout the proof, we assume that every operator is restricted to $\mathcal{R}(A^*)$. From Proposition \ref{eigenvalues}, there exists an induced matrix norm $\|\cdot\|$ such that $\|B^{\vec{\omega}}\| < 1$. Note that, with respect to this norm, $(B^{\vec{\omega}})^n$ converges to the zero matrix and $(B^{\vec{\omega}})^{n-1} + ... + B^{\vec{\omega}} + I$ converges to the matrix $(I - B^{\vec{\omega}})^{-1}$. Then
\[
\vec{x}^{(n)} = (B^{\vec{\omega}})^n\vec{x}^{(0)} + ((B^{\vec{\omega}})^{n-1} + ... + B^{\vec{\omega}} + I)\vec{b}^{\vec{\omega}} \to (I - B^{\vec{\omega}})^{-1}\vec{b}^{\vec{\omega}} =: \vec{z}.
\]
Note that $\vec{z} \in \mathcal{R}(A^*)$ and that $\vec{z} = B^{\vec{\omega}}\vec{z} + \vec{b}^{\vec{\omega}}$, as desired.

\end{proof}

\begin{remark}
We observe that, in general, the sequence $\{\vec{x}^{(n)}\}$ converges to
\begin{align}\label{limit}
\vec{y} = \sum_{j=0}^\infty(B^{\vec{\omega}})^j\vec{b}^{\vec{\omega}} + P\vec{x}^{(0)}
\end{align}
where $P$ is the orthogonal projection onto $\mathcal{N}(A)$. Hence, it is novel to choose $\vec{x}^{(0)} \in \mathcal{R}(A^*)$ (e.g., $\vec{x} = \vec{0}$) so that the norm of the vector in Equation \ref{limit} is minimized.

\end{remark}

\begin{theorem}\label{mainthm}
Let $\vec{x}^{(0)} \in \mathcal{R}(A^*)$. The distributed Kaczmarz algorithm with admissible relaxation parameters converges to the vector $\vec{y}$ in Equation \ref{limit}. If the system is inconsistent and $\Omega = s\Omega_1$ where $s \in (0,1]$, then $\vec{y} = \vec{y}_M + o(s)$ where $\vec{y}_M$ minimizes the functional
\[
\vec{x} \in \mathcal{R}(A^*) \mapsto \langle D^{-1}W\Omega_1(\vec{\mathfrak{b}} - \mathcal{A}\vec{x}),\vec{\mathfrak{b}} - \mathcal{A}\vec{x}\rangle.
\]
\end{theorem}

\begin{proof}
With Lemma \ref{Convergence}, the proof is similar to the proof of Theorem V.3.9. in \cite{Natterer2001}. Nonetheless, we provide a self-contained proof for clarification of our adaptation. First, by Lemma \ref{Convergence}, we have that the sequence $\{\vec{x}^{(n)}\}$ converges to the vector $\vec{y}$ satisfying $\vec{y} = B^{\vec{\omega}}\vec{y} + \vec{b}^{\vec{\omega}}$, that is
\begin{align}\label{unique}
\mathcal{A}^*(D + \Omega L)^{-1}W\Omega\mathcal{A}\vec{y} = \mathcal{A}^*(D + \Omega L)^{-1}W\Omega\vec{\mathfrak{b}}.
\end{align}
Note that $\vec{y}_M$ minimizes $\|D^{-1/2}W^{1/2}\Omega_1^{1/2}(\vec{\mathfrak{b}} - \mathcal{A}\vec{x})\|$ if and only if
\[
(D^{-1/2}W^{1/2}\Omega_1^{1/2}\mathcal{A})^*(D^{-1/2}W^{1/2}\Omega_1^{1/2}\mathcal{A})\vec{y}_M = (D^{-1/2}W^{1/2}\Omega_1^{1/2}\mathcal{A})^*D^{-1/2}W^{1/2}\Omega_1^{1/2}\vec{\mathfrak{b}}
\]
(see Theorem 1.1 of IV.1 in \cite{Natterer2001}), that is
\begin{align}\label{eq1}
\mathcal{A}^*D^{-1}W\Omega_1\mathcal{A}\vec{y}_M = \mathcal{A}^*D^{-1}W\Omega_1\vec{\mathfrak{b}}.
\end{align}
Substituting $\Omega = s\Omega_1$ into Equation \ref{unique}, we have
\begin{align}\label{eq2}
\mathcal{A}^*(D + s\Omega_1L)^{-1}W\Omega_1\mathcal{A}\vec{y} = \mathcal{A}^*(D + s\Omega_1L)^{-1}W\Omega_1\vec{\mathfrak{b}}.
\end{align}
From Equations \ref{eq1} and \ref{eq2}, we observe that $\vec{y} = \vec{y}_M + o(s)$, as desired.

\end{proof}

\begin{remark}
The minimizer of the functional in Theorem \ref{mainthm} is the weighted least-squares solution of
\[
\vec{x}\in\mathcal{R}(A^*)\mapsto\sum_{v \in \mathcal{V}}(\Omega_1)_v\left(\sum_{\ell\,:\,v\preceq\ell}w(r,\ell)\right)\dfrac{|b_v - \vec{a}^*_v\vec{x}|^2}{\|\vec{a}_v\|^2}.
\]
We note that there is a trade-off between the convergence rate of the algorithm and the approximation error; that is, the algorithm converges more slowly as $s$ approaches zero.

\end{remark}


TITLE: $[F(a):F] \leq [\mathbb{Q}(a):\mathbb{Q}]$ - proof?
QUESTION [0 upvotes]: Let $a \in \mathbb{C}$ be algebraic over $\mathbb{Q}$ and $F \subseteq \mathbb{C}$ be a subfield of $\mathbb{C}$.
I'm trying to prove that $[F(a):F] \leq [\mathbb{Q}(a):\mathbb{Q}]$ (where $F(a) := \mathbb{Q}(F\cup\left\{ {a}\right\}))$ and deduce from this that the set of complex numbers which are algebraic over the rationals is a subfield of $\mathbb{C}$.
For the first part, I guess $a$ has to be algebraic to ensure the two degrees are finite. Now as $F$ is a field, certainly $\mathbb{Q} \subseteq F$. Then I've tried thinking in terms of minimal polynomials but I get completely confused and dont really see how to provide a rigorous proof... Also, I don't really see how I could use such a result to show the last part.
I've been going round and round about this and got no result, could you hint me please? It would really help me as I'm only starting to read about field extensions and am not familiar with them!
Thank you!

REPLY [1 votes]: Let $\;P_F(x)\,,\,\,P_{\Bbb Q}(x)\;$ be the minimal polynomial of $\;a\;$ over $\;F\,,\,\,\Bbb Q\;$ , resp., so that
$$[F(a):F]=\deg P_F(x)\;\;,\;\;\;[\Bbb Q(a):\Bbb Q]=\deg P_{\Bbb Q}(x)$$
But we certainly have that also
$$\;P_{\Bbb Q}(x)\in F[x]\;\implies\;\; P_F(x)\mid P_{\Bbb Q}(x)\implies \deg P_F(X)\le\deg P_{\Bbb Q}(x)$$
adn we're done.


TITLE: Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes
QUESTION [2 upvotes]: Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes when $b_1<b_2\le 4$, where $b_1,b_2$ are the numbers of objects in boxes $1$ and $2$, respectively.
I understand that boxes $3, 4$, and $5$ will all just have the fundamental exponential generating function and that their combined generating function will be $e^{3x}$. What I don't understand is the first two boxes with the equality parameter given.
Can someone explain how to do this?

REPLY [3 votes]: Here is a very painstaking approach that may help you to see exactly what’s going on.
The possible values of $b_1$ are $0,1,2$, and $3$, so far starters we try
$$1+x+\frac{x^2}2+\frac{x^3}6$$
to account for $b_1$. Similarly, the possible values of $b_2$ are $1,2,3$, and $4$, so we try
$$y+\frac{y^2}2+\frac{y^3}6+\frac{y^4}{24}$$
to account for $b_2$. I’m using different indeterminates for now, because at this point I still need to keep the $b_1$ and $b_2$ contributions separate.
The product of these polynomials is
$$\begin{align*}
&y+\frac{y^2}2+\frac{y^3}6+\frac{y^4}{24}+\\
&xy+\frac{xy^2}2+\frac{xy^3}6+\frac{xy^4}{24}+\\
&\frac{x^2y}2+\frac{x^2y^2}4+\frac{x^2y^3}{12}+\frac{x^2y^4}{48}+\\
&\frac{x^3y}6+\frac{x^3y^2}{12}+\frac{x^3y^3}{36}+\frac{x^3y^4}{144}\;;
\end{align*}$$
however, we don’t want the terms in $x^ky\ell$ with $k\ge\ell$, since they correspond to having $b_1\ge b_2$. After we throw them away, we have
$$y+\frac{y^2}2+\frac{y^3}6+\frac{y^4}{24}+\frac{xy^2}2+\frac{xy^3}6+\frac{xy^4}{24}+\frac{x^2y^3}{12}+\frac{x^2y^4}{48}+\frac{x^3y^4}{144}\;.$$
Now replace $y$ by $x$, collect terms, and adjust the denominators to match the exponents to get
$$\frac{x}{1!}+\frac{x^2}{2!}+\frac{4x^3}{3!}+\frac{5x^4}{4!}+\frac{15x^5}{5!}+\frac{15x^6}{6!}+\frac{35x^7}{7!}\;,$$
which is the egf for boxes $1$ and $2$ combined. Multiply this by $e^{3x}$, and you’re done.
(And now that I’ve written this, I see that Markus has given you the abbreviated version of it.)


TITLE: How to calculate this mass.
QUESTION [1 upvotes]: Determinate the mass of portion of spherical surface given by the equation $x^2+y^2+z^2=2a^2$ who is inside of the cylindrical surface given by the equation $(x^2+y^2)=2a^2(x^2-y^2)$, knowing that the density at the point $P(x,y,z)$ is given by $\rho(x,y,z) = k(x^2+y^2+z^2)$.
I don't know how to find the integration limits, because the graphic is so strange :( If anyone could help me I would really appreciate it.

REPLY [2 votes]: As you are interested, we are on the way of finding the proper limits. First of all, note that there is a nice symmetric throughout the equations so, I restrict myself to consider the $x>0,y>0,z>0$. If you have a nice imagine, you'll find out that we should multiply the result by $8$. Working on $$(x^2+y^2)=2a^2(x^2-y^2)$$ gives us $$y/x=\sqrt{\frac{2a^2-1}{2a^2+1}}=k$$ so we found a nice range for $\theta$: $$\theta|_0^{\arctan{k}}$$ Other limits are easy to find. Indeed, $$\rho|_0^{\sqrt{2}a},~~\phi|_0^{\pi}$$


text:
\begin{document}

\title{$p$-torsion elements in local cohomology modules II}

\author{Anurag K. Singh}

\address{Department of Mathematics, University of Utah, 155 S. 1400 E., Salt
Lake City, \quad UT 84112, USA \newline E-mail: {\tt singh@math.utah.edu}}

\thanks{February 14, 2001 \\
This manuscript is based on work supported in part by the National Science
Foundation under Grant No. DMS 0070268}

\maketitle

\section{Introduction}

Gennady Lyubeznik conjectured that if $R$ is a regular ring and ${\mathfrak a}$
is an ideal of $R$, then the local cohomology modules $H_{\mathfrak a}^i(R)$
have only finitely many associated prime ideals, \cite[Remark 3.7 (iii)]{Ly1}.
While this conjecture remains open in this generality, several results are now
available: if the regular ring $R$ contains a field of prime characteristic $p
> 0$, Huneke and Sharp showed in \cite{HS} that the set of associated prime
ideals of $H_{\mathfrak a}^i(R)$ is finite. If $R$ is a regular local ring
containing a field of characteristic zero, Lyubeznik showed that $H_{\mathfrak
a}^i(R)$ has only finitely many associated prime ideals, see \cite{Ly1} and
also \cite{Ly2, Ly3}. Recently Lyubeznik has also proved this result for
unramified regular local rings of mixed characteristic, \cite{Ly4}.

In \cite{Hu} Craig Huneke first raised the following question: for a Noetherian
ring $R$, an ideal ${\mathfrak a} \subset R$, and a finitely generated
$R$-module $M$, is the number of associated primes ideals of $H_{\mathfrak
a}^i(M)$ always finite? For some of the work on this problem, we refer the
reader to the papers \cite{BL, BRS, He} in addition to those mentioned above.
In \cite{Si} we constructed an example of a hypersurface $R$ for which a local
cohomology module $H_{\mathfrak a}^3(R)$ has $p$-torsion elements for every
prime integer $p$, and consequently has infinitely many associated prime
ideals. Since this is the only known source of infinitely many associated prime
ideals so far, it is worthwhile to investigate whether similar techniques may
yield an example of a regular ring $R$ for which a local cohomology module
$H_{\mathfrak a}^i(R)$ has $p$-torsion elements for every prime integer $p$.
This leads to some very intriguing questions as we shall see in this paper. Our
results thus far support Lyubeznik's conjecture that local cohomology modules
of all regular rings have only finitely many associated prime ideals.

Let $R$ be a polynomial ring over the integers and $F_i,G_i$ be elements of $R$
for which
$$
F_1G_1 + F_2G_2 + \dots + F_nG_n = 0.
$$
Consider the ideal ${\mathfrak a} = (G_1, \dots, G_n)R$ and the local
cohomology module
$$
H_{\mathfrak a}^n(R) = \varinjlim \frac{R}{(G_1^k, \dots, G_n^k)R}
$$
where the maps in the direct limit system are induced by multiplication by the
element $G_1 \cdots G_n$. For a prime integer $p$ and prime power $q=p^e$, let
$$
\lambda_q = \frac{(F_1G_1)^q + \dots + (F_nG_n)^q}{p}.
$$
Note that has $\lambda_q$ has integer coefficients, i.e., that
$\lambda_q \in R$. Consider
$$
\eta_q = [\lambda_q + (G_1^q, \dots, G_n^q)R] \ \in \ H_{\mathfrak a}^n(R)
= \varinjlim \frac{R}{(G_1^k, \dots, G_n^k)R}.
$$
It is immediately seen that $p\eta_q=0$ and so if $\eta_ q$ is a nonzero
element of $H_{\mathfrak a}^n(R)$, then it must be a $p$-torsion element.
Hence if the local cohomology module $H_{\mathfrak a}^n(R)$ has only finitely
many associated prime ideals then, for all but finitely many prime integers
$p$, the elements $\eta_q$ as constructed above must be zero i.e., for some
$k \in \mathbb N$, we have
$$
\lambda_q (G_1 \cdots G_n)^{k} \in (G_1^{q+k}, \dots, G_n^{q+k})R.
$$

This motivates the following conjecture:
\begin{conj}
Let $R$ be a polynomial ring over the integers and let $F_i,G_i\in R$ such that
$$
F_1G_1 + \dots + F_nG_n = 0.
$$
Then for every prime integer $p$ and prime power $q=p^e$, there exists
$k \in \mathbb N$ such that
$$
\frac{(F_1G_1)^q + \dots + (F_nG_n)^q}{p} (G_1 \cdots G_n)^{k} \in
(G_1^{q+k}, \dots, G_n^{q+k})R.
$$
\label{conj}
\end{conj}

We shall say that Conjecture~\ref{conj} holds for $G_1,\dots,G_n \in R$, if it
holds for all relations $\displaystyle{\sum_{i=0}^n F_i G_i =0}$ with
$F_1,\dots,F_n \in R$.

\begin{remark}
The above conjecture is easily established if $n=2$ since in this case we have
$F_1G_1 + F_2G_2 = 0$ and so
$$
\frac{(F_1G_1)^q + (F_2G_2)^q}{p} = \frac{(F_1G_1)^q + (-F_1G_1)^q}{p} =
\begin{cases}
(F_1G_1)^q & \text{if } \ p=2, \\
0 & \text{if } \ p \neq 2,
\end{cases}
$$
which is an element of $(G_1^q, G_2^q)R$.
\end{remark}

\begin{example}{\bf The hypersurface example.}
Conjecture~\ref{conj} is false if the condition that $R$ is a polynomial ring
over the integers is replaced by the weaker condition that the ring $R$ is a
hypersurface over the integers. To see this, let
$$
R={\mathbb Z}[U,V,W,X,Y,Z]/(UX+VY+WZ)
$$
and ${\mathfrak a}$ denote the ideal $(x,y,z)R$. (We use lowercase letters here
to denote the images of the corresponding variables.) Consider the relation
$$
ux+vy+wz=0
$$
where, in the notation of Conjecture~\ref{conj}, $F_1=u, \ F_2=v, \ F_3=w$ and
$G_1=x, \ G_2=y, \ G_3=z$. By means of a multi-grading argument, it is
established in \cite[\S 4]{Si} that
$$
\frac{(ux)^p + (vy)^p + (wz)^p}{p}(xyz)^k \notin
(x^{p+k}, \ y^{p+k}, \ z^{p+k})R \ \text { for all } \ k \in {\mathbb N}.
$$
Consequently Conjecture~\ref{conj} does not hold here even for the choice of the
prime power $q=p$. This is, of course, the example from \cite{Si} of a local
cohomology module with infinitely many associated prime ideals: more precisely,
for every prime integer $p$,
$$
\left[ \frac{(ux)^p + (vy)^p + (wz)^p}{p} + \ (x^p, y^p, z^p)R \right] \
\in \ H^3_{(x,y,z)}(R)
$$
is a nonzero $p$-torsion element.
\end{example}

\begin{conj}
We record another formulation of Conjecture~\ref{conj}. Let $F_i,G_i$ be
elements of a polynomial ring $R$ over the integers such that
$$
F_1G_1 + \dots + F_nG_n = 0.
$$
Let ${\mathfrak a}$ denote the ideal $(G_1, \dots, G_n)$ of $R$. For an
arbitrary prime integer $p$ and prime power $q=p^e$, we have
$$
(F_1G_1)^q + \dots + (F_nG_n)^q \equiv 0 \mod p
$$
which is a relation on the elements $\overline{G}_1^q, \dots, \overline{G}_n^q
\in R/pR$, where $\overline{\phantom{G}}$ denotes the image of an element of
$R$ in the ring $R/pR$. This relation may be viewed as an element $\mu_q \in
H_{\mathfrak a}^{n-1}(R/pR)$. Conjecture~\ref{conj} is equivalent to the
conjecture that this element $\mu_q$ is in the image of the natural
homomorphism $H_{\mathfrak a}^{n-1}(R) \to H_{\mathfrak a}^{n-1}(R/pR)$.

To see the equivalence of these conjectures, suppose
$$
\mu_q \in \image\Big( H_{\mathfrak a}^{n-1}(R)
 \to H_{\mathfrak a}^{n-1}(R/pR) \Big).
$$
Then the relation $(\overline{F}_1^q, \dots, \overline{F}_n^q)$ on the elements
$\overline{G}_1^q, \dots, \overline{G}_n^q \in R/pR$ lifts to a relation in
$H_{\mathfrak a}^{n-1}(R)$, i.e., there exists an integer $k$ and elements
$\alpha_i \in R$ such that
\begin{align*}
& \alpha_1 G_1^{q+k}+ \dots + \alpha_n G_n^{q+k}=0 \qquad \text{ and} \\
& \alpha_i \equiv F_i^q (G_1 \cdots G_{i-1} G_{i+1} \cdots G_n)^k \mod p
\quad \text{ for all } \quad 1 \le i \le n.
\end{align*}
Hence we have
\begin{multline*}
\Big( (F_1G_1)^q + \dots + (F_nG_n)^q \Big)(G_1 \dots G_n)^k \\
= (F_1^q G_2^k \cdots G_n^k -\alpha_1)G_1^{q+k} + \dots +
        (F_n^q G_1^k \cdots G_{n-1}^k -\alpha_n)G_n^{q+k} \\
\in \Big( p G_1^{q+k}, \dots, p G_n^{q+k} \Big)R,
\end{multline*}
and so
$\displaystyle{
\frac{(F_1G_1)^q + \dots + (F_nG_n)^q}{p} (G_1 \cdots G_n)^{k} \in
\Big( G_1^{q+k}, \dots, G_n^{q+k} \Big)R}$.

The proof of the converse is similar.
\end{conj}

\begin{remark}
We next mention a conjecture due to Mel Hochster. While this was shown to be
false in \cite{Si}, our entire study of $p$-torsion elements originates from
this conjecture.

Consider the polynomial ring over the integers $R={\mathbb Z}[u,v,w,x,y,z]$
where ${\mathfrak a}$ is the ideal generated by the size two minors of the
matrix
$$
M = \begin{pmatrix}
u & v & w \\
x & y & z \\
\end{pmatrix},
$$
i.e., ${\mathfrak a} = (\Delta_1, \Delta_2, \Delta_3)R$ where $\Delta_1 =
vz-wy$, $\Delta_2 = wx-uz$, and $\Delta_3 = uy-vx$. In the ring $R$ we have
$u\Delta_1 + v\Delta_2 + w\Delta_3 = 0$. The relation
$$
(u\Delta_1)^q + (v\Delta_2)^q + (w\Delta_3)^q \equiv 0 \mod p
$$
may be viewed as an element $\mu_q \in H_{\mathfrak a}^2(R/pR)$. Hochster
conjectured that for every prime integer there exists a choice of $q=p^e$ such
that
$$
\mu_q \notin \image \Big(H_{\mathfrak a}^2(R) \to H_{\mathfrak a}^2(R/pR)\Big),
$$
and consequently that the image of $\mu_q$ in $H_{\mathfrak a}^3(R)$ is a
nonzero $p$-torsion element of $H_{\mathfrak a}^3(R)$. In \cite{Si} we
constructed an equational identity which provided us with an element of
$H_{\mathfrak a}^2(R)$ that maps to $\mu_q \in H_{\mathfrak a}^2(R/pR)$.
While we refer the reader to \cite{Si} for the details of the construction, we
would like to provide a brief sketch.

Consider the following equational identity:

\begin{align*}
&\Delta_1^{2k+1} u^{k+1} \sum_{n=0}^k \binom{k}{n} x^n \sum_{i=0}^n (-1)^i
\binom{k+i}{k} \binom{k+n-i}{k} w^i v^{n-i} \Delta_2^{k-i}\Delta_3^{k-n+i} \\
+&\Delta_2^{2k+1} v^{k+1} \sum_{n=0}^k \binom{k}{n} y^n \sum_{i=0}^n (-1)^i
\binom{k+i}{k} \binom{k+n-i}{k} u^i w^{n-i} \Delta_3^{k-i}\Delta_1^{k-n+i} \\
+&\Delta_3^{2k+1} w^{k+1} \sum_{n=0}^k \binom{k}{n} z^n \sum_{i=0}^n (-1)^i
\binom{k+i}{k} \binom{k+n-i}{k} v^i u^{n-i} \Delta_1^{k-i}\Delta_2^{k-n+i} \\
&=0.
\end{align*}

As it is a relation on the elements $\Delta_i^{2k+1}$ this identity gives us,
for every $k \in \mathbb N$, an element $\gamma_k \in H_{\mathfrak a}^2(R)$.
Using $k=q-1$ and examining the binomial coefficients above $\mod p$, we obtain

$$
\Big((u\Delta_1)^q+(v\Delta_2)^q+(w\Delta_3)^q\Big)
(\Delta_1 \Delta_2 \Delta_3)^{q-1} \equiv 0 \mod p.
$$

Consequently $\gamma_{q-1} \mapsto \mu_q$ under the natural homomorphism
$$
H_{\mathfrak a}^2(R) \to H_{\mathfrak a}^2(R/pR).
$$
\end{remark}

\begin{prop}
If Conjecture~\ref{conj} is true for the relations
$\displaystyle{\sum_{i=0}^n E_i G_i =0}$ and
$\displaystyle{\sum_{i=0}^n F_i G_i =0}$ where
$E_i, F_i, G_i \in R={\mathbb Z}[X_1,\dots,X_m]$, then it is also true for
the relation $\displaystyle{\sum_{i=0}^n (sE_i+tF_i) G_i =0}$ in
$S={\mathbb Z}[X_1,\dots,X_m,s,t]$. More precisely, if for a prime
power $q=p^e$, there exists $k_1, k_2 \in \mathbb N$ such that
\begin{multline*}
\left[ \sum_{i=0}^n \frac{(E_iG_i)^q}{p} \right] (G_1 \cdots G_n)^{k_1} \in
   \Big( G_1^{q+k_1}, \dots, G_n^{q+k_1} \Big) R \ \text{ and } \\
\left[ \sum_{i=0}^n \frac{(F_iG_i)^q}{p} \right] (G_1 \cdots G_n)^{k_2} \in
  \Big( G_1^{q+k_2}, \dots, G_n^{q+k_2} \Big) R, \ \text{ then } \\
\left[ \sum_{i=0}^n \frac{(sE_i+tF_i)^q G_i^q}{p} \right] (G_1 \cdots G_n)^{k}
 \in \Big( G_1^{q+k}, \dots, G_n^{q+k} \Big) S

\end{multline*}
for $k = \max\{k_1,k_2\}$.
\label{sum}
\end{prop}

We leave the proof as an elementary exercise.

\section{A special case of the conjecture}

In Theorem~\ref{reg} below we prove what is perhaps the first interesting case
of Conjecture~\ref{conj}.

\begin{theorem}
Let $R$ be a polynomial ring over the integers and $F_i, G_i$ be elements of $R$
such that $F_1, F_2, F_3$ form a regular sequence in $R$ and
$$
F_1G_1 + F_2G_2 + F_3G_3 = 0.
$$
Let $q=p^e$ be a prime power. Then for $k=q-1$ we have
$$
\frac{(F_1G_1)^q + (F_2G_2)^q + (F_3G_3)^q}{p} (G_1 G_2 G_3)^k \ \in \
(G_1^{q+k}, G_2^{q+k}, G_3^{q+k})R.
$$
\label{reg}
\end{theorem}

\begin{remark}
In Hochster's conjecture discussed earlier, the relation under consideration is
$u\Delta_1 + v\Delta_2 + w\Delta_3 = 0$ and the elements $u,v,w$ certainly
form a regular sequence in the polynomial ring $R={\mathbb Z}[u,v,w,x,y,z]$.
\end{remark}

\begin{proof}[Proof of Theorem~\ref{reg}]

Since $F_3G_3 \in (F_1,F_2)R$ and $F_1, F_2, F_3$ form a regular sequence, there
exist $\alpha, \beta \in R$ such that $G_3 = \alpha F_1 + \beta F_2$. In
\cite[\S 2, 3]{Si} we showed that in the polynomial ring $\mathbb{Z}[A,B,T]$
$$
\frac{(A+B)^q+(-A)^q+(-B)^q}{p} \Big[ (A+B)AB \Big]^k
$$
is an element of the ideal
\begin{multline*}
(A+B)^{q+k}(T,A)^k (T,B)^k \ + \ A^{q+k}(T,B)^k (T+B, A+B)^k \\
+ \ B^{q+k}(T,A)^k (T-A, A+B)^k
\end{multline*}
when $k=q-1$. We shall use this fact with $A=-F_2G_2$, \ $B=-F_3G_3$ \ and \
$T=\beta F_2F_3$. Note that this gives $A+B = F_1G_1, \ T+B = -\alpha F_1 F_3$,
\ and \ $T-A = -F_1 G_1 - \alpha F_1 F_3$. Consequently
$$
\frac{(F_1G_1)^q+(F_2G_2)^q+(F_3G_3)^q}{p} \Big[ F_1G_1 F_2G_2 F_3G_3 \Big]^k
$$
is an element of the ideal
\begin{multline*}
(F_1G_1)^{q+k} (F_2F_3, \ F_2G_2)^k (F_2F_3, \ F_3G_3)^k \\
+ (F_2G_2)^{q+k} (F_2F_3, \ F_3G_3)^k (F_1 F_3, \ F_1G_1)^k \\
+ (F_3G_3)^{q+k} (F_2F_3, \ F_2G_2)^k (F_1 G_1 + \alpha F_1 F_3, \ F_1G_1)^k \\
\subseteq \ (F_1F_2F_3)^k(G_1^{q+k}, \ G_2^{q+k}, \ G_3^{q+k}).
\end{multline*}
The required result follows from this statement.
\end{proof}

\section{A Pl\"ucker relation}

One of the main goals of \cite{Si} was to settle Conjecture~\ref{conj} for the
relation
$$
u(vz-wy) + v(wx-uz) + w(uy-vx) = 0 \qquad \qquad (1)
$$
in the polynomial ring ${\mathbb Z}[u,v,w,x,y,z]$. This was accomplished by
establishing that for every prime integer $p$ and $q=p^e$ we have, for $k=q-1$,
\begin{multline*}
\frac{1}{p} \left[ u^q(vz-wy)^q+v^q(wx-uz)^q+w^q(uy-vx)^q \right] \\
\times \left[ (vz-wy)(wx-uz)(uy-vx) \right]^k \\
 \in \ \Big( (vz-wy)^{q+k}, \ (wx-uz)^{q+k}, \ (uy-vx)^{q+k} \Big).
\end{multline*}
The syzygy of the matrix
$\displaystyle{\begin{pmatrix}vz-wy & wx-uz & uy-vx \end{pmatrix}}$ is
$$
\begin{pmatrix}
x & u\\
y & v\\
z & w\\
\end{pmatrix}
$$
but, by symmetry, Conjecture~\ref{conj} also holds for the relation
$$
x(vz-wy) + y(wx-uz) + z(uy-vx) = 0. \qquad \qquad (2)
$$
In the light of Proposition~\ref{sum}, this gives us the following:

\begin{theorem}
Conjecture~\ref{conj} holds for
$$
\Delta_1=vz-wy, \ \Delta_2=wx-uz, \ \Delta_3=uy-vx \ \in \
R={\mathbb Z}[u,v,w,x,y,z],
$$
i.e., if \ $F_1,F_2,F_3 \in R$ \ satisfy \
$F_1\Delta_1+F_2\Delta_2+F_3\Delta_3=0$ \
then for any prime power $q=p^e$ we have, for $k=q-1$,
$$
\frac{(F_1\Delta_1)^q + (F_2\Delta_2)^q + (F_3\Delta_3)^q}{p}
(\Delta_1\Delta_2\Delta_3)^{k}
\in (\Delta_1^{q+k}, \ \Delta_2^{q+k}, \ \Delta_3^{q+k})R.
$$
\end{theorem}

\bigskip

By taking a combination of $(1)$ and $(2)$ above, we get the relation
$$
(vz-wy)(ut-xs) + (wx-uz)(vt-ys) + (uy-vx)(wt-zs) = 0
$$
in $R={\mathbb Z}[s,t,u,v,w,x,y,z]$. This is, of course, the Pl\"ucker relation
$$
-\Delta_{34}\Delta_{12}+\Delta_{24}\Delta_{13}-\Delta_{23}\Delta_{14}
$$
where $\Delta_{ij}$ is the size two minor formed by picking rows $i$ and $j$ of
the matrix
$$
\begin{pmatrix}
s & u & v & w \\
t & x & y & z \\
\end{pmatrix}.
$$
The syzygy of the matrix
$\displaystyle{\begin{pmatrix}ut-xs & vt-ys & wt-zs \end{pmatrix}}$ is
$$
\begin{pmatrix}
vz-wy & 0 & -wt+zs & vt-ys \\
wx-uz & -wt+zs & 0 & -ut+xs \\
uy-vx & vt-ys & ut-xs & 0
\end{pmatrix}.
$$
Since the Koszul relations are easily treated, the following theorem shows
that Conjecture~\ref{conj} holds for $ut-xs, \ vt-ys, \ wt-zs$.

\begin{theorem}
\label{plucker}
Consider the Pl\"ucker relation
$$
(vz-wy)(ut-xs) + (wx-uz)(vt-ys) + (uy-vx)(wt-zs) = 0
$$
in the polynomial ring $R = {\mathbb Z}[s,t,u,v,w,x,y,z]$. Then for $k=q-1$,
we have

\begin{multline*}
\frac{1}{p} \left[ (vz-wy)^q(ut-xs)^q + (wx-uz)^q(vt-ys)^q + (uy-vx)^q(wt-zs)^q
\right] \\
\times [(ut-xs)(vt-ys)(wt-zs)]^k \\
\in \ \Big( (ut-xs)^{q+k}, \ (vt-ys)^{q+k}, \ (wt-zs)^{q+k} \Big)R.
\end{multline*}

\end{theorem}

Towards the proof of this theorem, we first record some identities with
binomial coefficients. These identities can be proved using Zeilberger's
algorithm (see \cite{PWS}) and the Maple package {\tt EKHAD}, but we include
proofs for the sake of completeness.

When the range of a summation is not specified, it is assumed to extend over
all integers. We set $\displaystyle{\binom{k}{i}=0}$ if $ i<0$ or if $k<i$.

\begin{lemma}
\label{binomiden}
\begin{alignat*}2

& \sum_{n} (-1)^n \binom{m+s-r}{m-n}\binom{k-n}{k-r} =
\begin{cases}
(-1)^{r-s} \binom{k-m}{s}, & \text{if } \ m \le k-s, \\
(-1)^r \binom{m+s-k-1}{s}, & \text{if } \ k+1 \le m, \\
0 & \text{else.}
\tag 1
\end{cases} \\

& \sum_{r} (-1)^r \binom{2k-r}{m-1} \binom{m}{r-s} =
0 \qquad \text{if} \qquad 1 \le m \le 2k-s.
\tag 2

\end{alignat*}
\end{lemma}

\begin{proof}
(1) Let
$$
F(s,n) = (-1)^n \binom{m+s-r}{m-n}\binom{k-n}{k-r}, \qquad H(s)=\sum_n F(s,n),
$$
and $\displaystyle{G(s,n) = (-1)(k+1-n)F(s,n)}$. It is easily verified that
$$
\frac{G(s,n+1)}{G(s,n)} = \frac{(-1)(m-n)(r-n)}{(k+1-n)(s-r+n+1)},
\quad \frac{F(s+1,n)}{F(s,n)} = \frac{m+s+1-r}{n+s+1-r},
$$
and these can then be used to obtain the relation
$$
(s+1)F(s+1,n) - (m+s-k)F(s,n)= G(s,n+1) - G(s,n).
$$
Summing the above equation with respect to $n$ gives
$$
(s+1)H(s+1) - (m+s-k)H(s) = 0,
$$
and using this recurrence relation for $H(s)$ we get
$$
H(s) = \frac{(m-k+s-1) \cdots (m-k)}{s \cdots 1} H(0).
$$
The result now follows since $H(0) = (-1)^r$.

\bigskip

(2) Let $\displaystyle{
F(r) = (-1)^r \binom{2k-r}{m-1} \binom{m}{r-s}
}$
and $G(r) = (2k+1-r)(r-s) F(r)$. It is a routine verification that
$$
G(r) - G(r+1) = m(2k+1-m-s) F(r).
$$
Consequently if $1 \le m \le 2k-s$ then
$$
\sum_{r} F(r) = \frac{1}{m(2k+1-m-s)} \sum_{r} \big(G(r) - G(r+1)\big) = 0.
$$
\end{proof}

We use the above results to establish the following equational identity:

\begin{lemma}
\label{asym}
\begin{multline*}

\sum_{r=0}^k \sum_{n=0}^r \binom{2k+1-r}{n}\binom{k-n}{r-n}
\frac{T^{k-r}}{2k+1-r} (T-Y)^n (Y-Z)^{2k+1-r-n}Y^rZ^r \\
= \sum_{s=0}^k (-1)^s \binom{k}{s} \frac{T^{k-s}}{2k+1-s}
\big(Y^{2k+1}Z^s- Z^{2k+1}Y^s\big),

\end{multline*}
\end{lemma}

\begin{proof}
Comparing the coefficients of $T^{k-s}$, we need to show that
\begin{multline*}

\sum_{r=s}^k \sum_{n=r-s}^r \frac{(-1)^{n-r+s}}{2k+1-r}
\binom{2k+1-r}{n} \binom{k-n}{r-n} \binom{n}{r-s} \\
\times \ (Y-Z)^{2k+1-r-n}Y^{n+s} Z^r
\ = \ \frac{(-1)^s}{2k+1-s} \binom{k}{s} \big(Y^{2k+1}Z^s - Z^{2k+1}Y^s\big),

\end{multline*}
i.e., that
\begin{multline*}

\sum_{r=s}^k \sum_{n=r-s}^r \frac{(-1)^{n-r}}{2k+1-r}
\binom{2k+1-r}{n} \binom{k-n}{r-n} \binom{n}{r-s} \\
\times \ (Y-Z)^{2k+1-r-n}Y^n Z^{r-s} \
= \ \frac{1}{2k+1-s} \binom{k}{s} \big( Y^{2k+1-s} - Z^{2k+1-s} \big).

\end{multline*}
The coefficient of $Y^mZ^{2k+1-s-m}$ in the expression on the left hand side is
\begin{multline*}

\sum_{r,n} \frac{(-1)^{n+m+1}}{2k+1-r} \binom{2k+1-r-n}{m-n} \binom{2k+1-r}{n}
\binom{k-n}{r-n} \binom{n}{r-s} \\
=\sum_r \frac{(-1)^{m+1}}{2k+1-r} \binom{m}{r-s} \binom{2k+1-r}{m}
\sum_n (-1)^n \binom{m+s-r}{m-n} \binom{k-n}{r-n}.
\end{multline*}
which we denote by $\gamma_{m,s}$. Note that $\gamma_{m,s}=0$ unless
$m \le k-s$ or $m \ge k+1$ since
$\displaystyle{
\sum_n (-1)^n \binom{m+s-r}{m-n} \binom{k-n}{r-n} = 0
}$ for such $m$ by Lemma~\ref{binomiden}~(1).

\bigskip

We next consider the case $m \le k-s$. It follows from Lemma~\ref{binomiden}~(1)
that
$$
\gamma_{m,s} = \sum_r \frac{(-1)^{m+1+r-s}}{2k+1-r} \binom{m}{r-s}
\binom{2k+1-r}{m} \binom{k-m}{s}.
$$
If $m=0$, then the only term that contributes to this sum is when $r=s$, and
we get $\displaystyle{\gamma_{0,s}=\frac{-1}{2k+1-s}\binom{k}{s}}$. We may
now assume $1 \le m \le k-s$ and then
$$
\gamma_{m,s}= \frac{(-1)^{m+1-s}}{m} \binom{k-m}{s}
\sum_r (-1)^r \binom{m}{r-s} \binom{2k-r}{m-1} =0
$$
by Lemma~\ref{binomiden}~(2).

\bigskip

For $m \ge k+1$, Lemma~\ref{binomiden}~(1) gives
\begin{align*}
\gamma_{m,s} & = \sum_r \frac{(-1)^{m+1+r}}{2k+1-r} \binom{m}{r-s}
\binom{2k+1-r}{m} \binom{m+s-k-1}{s} \\
& = \frac{(-1)^{m+1}}{m} \binom{m+s-k-1}{s}
\sum_r (-1)^r \binom{m}{r-s} \binom{2k-r}{m-1}.
\end{align*}
By Lemma~\ref{binomiden}~(2), this is zero if $m \le 2k-s$. If $m=2k+1-s$ the
only term that contributes to this sum is when $r=s$, and
$\displaystyle{\gamma_{2k+1-s,s}=\frac{1}{2k+1-s}\binom{k}{s}}$.
\end{proof}

This lemma enables us to establish the following crucial identity:

\begin{lemma}
\label{iden}
\begin{align*}

\sum_{r=0}^k \sum_{n=0}^r \frac{k+1}{2k+1-r} & \binom{2k+1-r}{n} \binom{k-n}{r-n}
\ T^{k-r} \\
\times \bigg[ & X^{2k+1}(T-Y)^n (Y-Z)^{2k+1-r-n}Y^rZ^r \\
+ & Y^{2k+1}(T-Z)^n (Z-X)^{2k+1-r-n}Z^rX^r \\
+ & Z^{2k+1}(T-X)^n (X-Y)^{2k+1-r-n}X^rY^r \bigg] = 0

\end{align*}
\end{lemma}

\begin{proof}
Using Lemma~\ref{asym}, the left hand side in the equation equals
\begin{multline*}

\sum_{s=0}^k (-1)^s \binom{k}{s} \frac{k+1}{2k+1-s}T^{k-s}
\times \bigg[
X^{2k+1}(Y^{2k+1}Z^s- Z^{2k+1}Y^s) \\ \ + \
Y^{2k+1}(Z^{2k+1}X^s- X^{2k+1}Z^s) \ + \
Z^{2k+1}(X^{2k+1}Y^s- Y^{2k+1}X^s)
\bigg] \ = \ 0.

\end{multline*}
\end{proof}

\begin{proof}[Proof of Theorem~\ref{plucker}]
Since $k=q-1$, we have $(k+1)/(2k+1-r)=q/(2q-1-r)$. For $0 \le r \le q-2$, if
$q/(2q-1-r) = a/b$ for relatively prime integers $a$ and $b$, then $p$ divides
$a$ since $q+1 \le 2q-1-r \le 2q-1$. Consequently there exists an integer
$d \in {\mathbb Z}$ such that $d$ is relatively prime to $p$ and
$$
\frac{dq}{2q-1-r} \in {\mathbb Z} \quad \text{ for all } \quad 0 \le r \le q-1.
$$
After replacing $d$ by a suitable multiple, if necessary, we may
assume that $d \equiv 1 \mod p$. Using Lemma~\ref{iden} with
$$
T=\frac{t}{s}, \qquad X=\frac{t}{s}-\frac{x}{u},
\qquad Y=\frac{t}{s}-\frac{y}{v}, \qquad Z=\frac{t}{s}-\frac{z}{w},
$$
we get
\begin{align*}

\sum_{r=0}^k & \sum_{n=0}^r \frac{k+1}{2k+1-r} \binom{2k+1-r}{n}
\binom{k-n}{r-n} \ \left( \frac{t}{s} \right)^{k-r} \\
\times \bigg[
& \left( \frac{ut-xs}{su} \right)^{2k+1} \left(\frac{y}{v} \right)^n
 \left( \frac{vz-wy}{vw} \right)^{2k+1-r-n} \left(\frac{vt-ys}{sv} \right)^r
 \left( \frac{wt-zs}{sw} \right)^r \\
& \left( \frac{vt-ys}{sv} \right)^{2k+1} \left(\frac{z}{w} \right)^n
 \left( \frac{wx-uz}{wu} \right)^{2k+1-r-n} \left(\frac{wt-zs}{sw} \right)^r
 \left( \frac{ut-xs}{su} \right)^r \\
& \left( \frac{wt-zs}{sw} \right)^{2k+1} \left(\frac{x}{u} \right)^n
 \left( \frac{uy-vx}{uv} \right)^{2k+1-r-n} \left(\frac{ut-xs}{su} \right)^r
 \left( \frac{vt-ys}{sv} \right)^r \bigg] \\
& = 0.

\end{align*}
Multiplying by $ds^{4k+1}(uvw)^{2k+1}$ clears denominators, and gives us the
equational identity (with integer coefficients):
\begin{multline*}

\sum_{r=0}^k \sum_{n=0}^r \frac{d(k+1)}{2k+1-r} \binom{2k+1-r}{n}
\binom{k-n}{r-n} \ (st)^{k-r} \\ \times \bigg[
(ut-xs)^{2k+1} (wy)^n (vz-wy)^{2k+1-r-n} (vt-ys)^r (wt-zs)^r \\ + \
(vt-ys)^{2k+1} (uz)^n (wx-uz)^{2k+1-r-n} (wt-zs)^r (ut-xs)^r \\ + \
(wt-zs)^{2k+1} (vx)^n (uy-vx)^{2k+1-r-n} (ut-xs)^r (vt-ys)^r
\bigg] = 0.

\end{multline*}
Note that for $0 \le r \le k=q-1$ and $0 \le n \le r$,
$$

\frac{d(k+1)}{2k+1-r} \binom{2k+1-r}{n} \binom{k-n}{r-n} \equiv
\begin{cases}
1 \mod p, & \text{if } \ r=k \text{ and } n=0, \\
0 \mod p, & \text{else}.
\end{cases}

$$
Consequently we get
\begin{align*}

 & (ut-xs)^{2k+1} (vz-wy)^{k+1} (vt-ys)^k (wt-zs)^k \\
+ \ & (vt-ys)^{2k+1} (wx-uz)^{k+1} (wt-zs)^k (ut-xs)^k \\
+ \ & (wt-zs)^{2k+1} (uy-vx)^{k+1} (ut-xs)^k (vt-ys)^k \\
& \qquad \qquad \in \ \Big( p(ut-xs)^{2k+1}, \ p(vt-ys)^{2k+1}, \
p(wt-zs)^{2k+1} \Big) R.

\end{align*}
Finally, this shows that

\begin{multline*}
\frac{1}{p}[(vz-wy)^q(ut-xs)^q + (wx-uz)^q(vt-ys)^q + (uy-vx)^q(wt-zs)^q] \\
\times [(ut-xs)(vt-ys)(wt-zs)]^k \\
\in \ \Big( (ut-xs)^{q+k}, \ (vt-ys)^{q+k}, \ (wt-zs)^{q+k} \Big)R.
\end{multline*}

\end{proof}

\section*{Acknowledgments}

I am indebted to Mel Hochster for valuable discussions regarding this material.
The use of the Maple package {\tt EKHAD} is gratefully acknowledged. I also
want to express my appreciation to Gennady Lyubeznik and Xavier G\'omez-Mont
Avalos for organizing the {\it Conference on Local Cohomology}.


TITLE: What causes acceleration of particles in the expansion section of a De Laval nozzle?
QUESTION [11 upvotes]: A De Laval nozzle has a compression section, where the propellant is compressed (and thereby accelerated) as it moves towards a narrow section (the throat).  After the throat, the nozzle widens out again.  Here's a reference image.

According to what I've been reading, the propellant is moving at subsonic speeds before the throat, and is accelerated to supersonic speeds as it passes the throat.  Furthermore, the propellant continues to accelerate as the nozzle expands again.  What I don't understand is: what force continues to accelerate the propellant after it passes through the throat?  I've been trying to imagine a single molecule of propellant and thinking about the forces from the particles around it, but can't come up with a force "pushing" from upstream (because it's moving at supersonic speeds), or "pulling" from downstream (because at the minimum there would be 0 opposing force in a vacuum, but certainly not a negative attractive force from downstream).  Any ideas?

REPLY [1 votes]: Basic Bernoulli principle for a compressible flow can be stated as :
$$ {\frac {v^{2}}{2}} + H={\text{const}} $$
where $H$ is specific enthalpy. Expressing enthalpy in terms of gas temperature, gives :
$$ {\frac {v^{2}}{2}} + c_p \,T={\text{const}}$$
where $c_p$ is specific heat capacity. Now as gas goes down through the valve, it expands, while cooling down in temperature. This can be best seen by the temperature distribution of the rocket nozzle wall (red line) :

As per equation, as gas cools down and it's enthalpy is decreasing - exhaust gas speed must increase. This explanation may not be the best or accurate, but should give the basic idea.


TITLE: On the convergence of $\sum \mu(n)/n^s$
QUESTION [3 upvotes]: I arrived at something during my maths ponderings which is really exciting for me.
It is clearly stated in the book on Riemann Hypothesis by Borwein that the convergence of $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$ for $\Re(s) > 1/2$ is necessary and sufficient for RH. This is ofcourse valid since, $\sum \mu(n)/n^s = 1/\zeta(s)$ for $\Re(s) > 1$
Having said that, I have reached a point where I got, for $\Re(s) > 1/2$
$$ \left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| < \infty $$
$\eta(s)$ is the Dirichlet eta function.
My question is,

What do I interpret out of this formula. Does this result imply RH, or falls short of it?

I believe the second option might be more correct, because this result does not say anything about the zeros of the eta function. But at least it is clear that if $\sum \mu(n)/n^s$ blows up then $\eta(s)$ also must have a zero to bring it down.
Any elaborated answer will be highly appreciated, because it is a current work in progress. :)

REPLY [3 votes]: Let me make sure I understand what you mean by the formula
$$ \left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| < \infty. $$
Do you actually mean that you have proven that this sum is convergent for $s$ in the critical strip and that, for such $s$, this inequality holds? Because, if you have shown that this sum converges in that range, you have already shown RH.
I can think of other interpretations of your statement, but I will wait for clarification before elaborating on them. Here is a basic point to remember: The statement that $\sum_{n=1}^{\infty} a_n$ converges is a statement about the partial sums $\sum_{n=1}^N a_n$. (Namely, that they form a Cauchy sequence.) If you never say anything about these partial sums, it is unlikely you have proved convergence.


TITLE: Why do we equate an indefinite integral to a specific value?
QUESTION [6 upvotes]: Suppose we want to obtain a displacement vector defined as $\mathbf s(t) = x(t)\mathbf i + y(t)\mathbf j + z(t)\mathbf k$ from the components of a velocity vector $\mathbf v(t) = \dot x(t)\mathbf i + \dot y(t)\mathbf j + \dot z(t)\mathbf k=\mathbf 0$. According to my notes, this can be done by equating each scalar component of the displacement vector to the indefinite integral of the corresponding scalars of the velocity vector, i.e.
$$
\mathbf s(t)=\begin{pmatrix}
x(t)=\int \dot x\ dt \\
y(t)=\int \dot y\ dt  \\
z(t)=\int \dot z\ dt
\end{pmatrix}
$$
But, as $\int f(x)\ dx=\{F(x): \frac {dF}{dx}=f(x)\}$, this should be syntactically wrong, because we're implying that a number is equal to an infinite set of numbers, or am I missing something?
Moreover, this also leads to a weird equation when solving the integral; for example, by taking into consideration the $x$-component of $\mathbf s$, we would have that
$$
x(t)=\int \dot x\ dt=c_1
$$
Which is correct, but it would also mean that the $x$-component of the velocity could be equal to any value belonging to $\mathbb R$. Because of that, we substitute $c_1$ with the initial condition and we equate it to zero, giving it a specific value: $c_1=0$. But, to me, this sounds like a break of the definition of indefinite integrals, as
$\int \dot x\ dt=c_1=0$ would basically mean that an indefinite integral is one, specific function.
I know this may be a very stupid question, and maybe it has to do with the same shortcuts that make us not specify "$\forall c \in \mathbb R$" when adding the constant $c$ in the solutions of an indefinite integral, but this doubt is really challenging me and I still don't understand whether I'm missing some point or it should actually be written $x(t)=c_1=0 \in \int \dot x\ dt$. Thanks a lot in advance!

REPLY [21 votes]: In physics we frequently leave off the limits of the integral when the limits can be figured out from the context. So, in the first case, the actual relation is:
$$x(t) = x(0) + \int_0^t \dot{x} \operatorname{d}t'.$$
Most often, though, when the limits are left off the implied limits are over all possible values of the dummy variable. For example:
$$ Q = \int \rho(\mathbf{x}) \operatorname{d}x^3$$
is understood to be the integral over all of space.

REPLY [12 votes]: You are being confused by the shortcut that people took who wrote that expression. They mean for the integral to be taken between definite limits.
It would be more proper to say
$$x(t) = x(0) + \int_0^t \dot{x} dt$$
But that gets longwinded. Most people, when seeing the expression as you gave it, will understand it to mean what I wrote. But technically, it's not the same.


TITLE: k! perfect matches
QUESTION [4 upvotes]: Let $G=($ $ A \cup B $ , $E$ $)$ be a bipartite graph with perfect matching. Denote $|A| = n$. Prove that if every vertex in A has degree $\geq$ $k$ then G has at least $k!$ perfect matches.

I tried the following solution ,but It failed... I think I am just missing something really easy...I tried induction on both $|A|$ and $k$ . The base of the induction is trivial (|A| = 1 , then there is a perfect matching and we must have $k$=1, for general |A| if k=1 then there is a matching...). Lets move to the general case.
Let $a$ be some vertex of degree $k$ (if all the vertices are of degree $>k$ then take some vertex and remove edges that are incident to it and do not participate in the perfect matching. After we finish removing them we have a graph such that all the vertices on one side have degree at least $k$ and we have a vertex of degree k, plus G still contains the same perfect matching we had before removing edges). If we have that for each edge $e=(a,b)$ the graph $G'= G$-$\{a,b\}$(deleting both $a$ and $b$ from $G$) we still have perfect matching then by induction we have in $G'$ $(k-1)!$ perfect matches . By adding to each matching the edge $e$ we get a matching of $G$ , since $deg(a)=k$ we get $k!$ matchings.
O.w. we have that there exist an edge such that $G'$ doesn't have perfect matching. Using Konig's theorem (the size of maximum matching in bipartite graph is equal to the minimum size of vertex cover) we get that in $G'$ the minimum vertex cover is of size at most $n-2$. If it has a vertex cover of size $n-3$ then by joining $a$ and $b$ to the vertex cover we would get a vertex cover of size $n-1$ for G , in contradiction to the assumption that the minimal vertex cover is $n$ (again , by Konig's theorem).
So we get that there is a vertex cover $T$ of size $n-2$ in $G'$, and it is minimal. If all the neighbours of $a$ or $b$ reside in $T$, then by joining only one vertex from $\{a,b\}$ we would get a vertex cover of size $n-1$ ,which is a contradiction.
So we can assume that there are vertices $a_2 , b_2$, both not in T , such that $(a_2,b),(a,b_2) \in E$. But here I got stuck. I also tried using Hall's theorem , but again , I got nowhere. I'll be happy if you could give me an hint about what to do.

REPLY [1 votes]: I provide a complete answer here. I try to give them in a "hint"-ish order, so that you can just read a prefix of my answer for a hint.
Suppose the minimum degree is $k$. I will show that there must exist a node $a \in A$ such that there exists at least $k$ different nodes $b \in B$, such that there exists a perfect matching in $G$ that includes $(a, b)$. Thus, you can obtain $k!$ perfect matching by induction: from the graph $G$, pick $a \in A$ as described above. For every $b \in B$ such that $(a, b)$ is a member of at least one perfect matching, consider the graph $G' = (A \cup B - a - b, E)$ (that is, we remove nodes $a$ and $b$ and all edges adjacent to them from $G$). Clearly $G'$ has a perfect matching and all of its nodes have degree at least $k-1$, so it contains $(k-1)!$ perfect matching by the inductive hypothesis. Since there exists at least $k$ such node $b$, the total number of matchings is $k \times (k-1)! = k!$.
Now I show that there must exist a node $a \in A$ such that there exists at least $k$ different nodes $b \in B$, such that there exists a perfect matching in $G$ that includes $(a, b)$. I prove by contradiction. Find the smallest $k$ for which this may not hold. Clearly $k > 1$. In this case, for every node $a' \in A$, there must exists $b' \in B$ such that $(a', b') \in E$ and that $(a', b')$ cannot be a member of a perfect matching. This is true since the degree of $a'$ is at least $k$, but there can only be at most $k-1$ different nodes $b \in B$ such that there exists a perfect matching in $G$ that includes $(a', b)$. Define the function $f: A \to B$ by $f(a') = b'$ as defined above (if there are more than one such $b'$, this function just pick one arbitrarily).
Let $P \subset E$ be any perfect matching of $G$. Define the function $g: B \to A$ such that $g(b) = a$ iff $(a, b) \in P$ (that is, $g$ maps each member of $B$ to its match in $A$ according to the matching $P$). Let $a \in A$ be any node in $A$.
Consider the following sequence $S$:

$s_1 = a$
$s_i = g(f(s_{i-1}))$ for $i > 1$

$S$ will be an infinite sequence of nodes in $A$. Since $A$ is finite, there will exist $1 \le i < j$ such that $s_i = s_j$ (the node repeats). Pick the smallest such $i$, and then the smallest such $j$. Let $O = \{(g(f(s_k)), f(s_k)) | i \le k < j\}$ and $O' = \{(s_k, f(s_k)) | i \le k < j\}$. Clearly $O$ and $O'$ contains the same number of elements, and contains the exact same set of nodes. In addition, $O \subseteq P$ and $O' \cap P = \emptyset$. Thus, $P - O + O'$ is a perfect matching of $G$. But this contradicts the fact that none of the members of $O'$ can be included in any perfect matching. This completes the proof.


TITLE: Is it true that for every $n \times n$ matrix, $\text{det}(A^k) = (\text{det} A)^k$?
QUESTION [0 upvotes]: I'd say yes, because
$\det(A^k) = \det(A) \det(A) \cdots \det(A) = (\det(A))^k$
But I might be missing something?

REPLY [1 votes]: Yes, because $\det AB = \det A \det B$ for all $n \times n$ matrices $A,B$.
$\det(A^k) = (\det(A))^k$ follows by induction on $k$.


TITLE: How does escape velocity relate to energy and speed?
QUESTION [11 upvotes]: I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11 \text{km s}^{-1} $ but I am talking about exact escape speed.   That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$.
Wikipedia says and I quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible?
As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right?
A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

REPLY [7 votes]: Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a negative value. Since energy is conserved, if you can sum the KE (a positive value) and the PE (a negative value) and the result is $0$, that means there is enough KE for the projectile to reach $r\to\infty$, where $PE=0$ just as $KE=0$. At that point, the object never returns. Thus that KE defines the escape speed. Also, $KE=\frac{1}{2}mv^2$ as you wrote, but $PE=-\frac{GMm}{r}$. The equation $PE=mgh$ is only valid near Earth's surface and it represents the change in potential energy from the surface, not the total potential energy (which is why it is positive, not negative).
If $KE+PE=0$ then at $r=\infty$, the speed will be zero. However, in a more practical sense, you are correct. An object at escape speed in a universe with just one gravity source and the object will never reach zero speed, it will simply move slower and slower forever.
At $r=\infty$, the PE goes to zero. The KE then becomes the only contributor to the total energy. so whatever the total energy of the object is, that's its KE. $r=0$ is a complicated case. For a point source of gravity, the PE would become infinite, but (other than a black hole, which isn't covered by Newtonian mechanics), there is no such thing as a point source of gravity. In the usual case, the Force of gravity disappears at $r=0$, but since moving away from that position would still constitute a gain in PE (as you'd experience an opposing force), it can get more complicated to talk about the PE at or near $r=0$. Generally, beneath the surface of an object, the PE becomes dependent on the distribution of the mass of the object. The KE is again dependent on the total energy. But there is nothing very special at $r=0$. In fact, there are no particularly interesting points anywhere. At every point, the total energy is $E=PE+KE$, PE is a negative value that approaches 0 as $r\to\infty$, and KE is whatever is left over such that total energy is conserved.

REPLY [5 votes]: Potential energy for a point mass (and also for a sphere) is not $PE = mgh $ (this is special case for a uniform field) but rather: $$ PE = - \frac{GMm}{r}$$
where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere)
Can you see the answer to your question yourself now?


TITLE: How many valid expressions for countdown numbers round?
QUESTION [4 upvotes]: If you haven't seen Countdown before watch a quick numbers round. For the more detailed constraints please read. I'll paraphrase the rules below.
Given a list $A$ of $n$ positive integers, how many unique sequences of calculations are possible? Each sequence is subject to:

It only uses the four basic operations of addition, subtraction, multiplication and division.
Not all $n$ integers have to be used to be considered a sequence.
A number may not appear more times than it is provided in $A$.
Division can only be performed if there is no remainder.
Only positive integers may be obtained as a result at any stage of the calculation.
Associative calculations should only be counted once e.g. $(4 + 2) - 5$ is considered the same as $(2 + 4) - 5$.

My observation:

For any pair of integers there are only 4 possible permutations not 8. Because addition and multiplication are associative so we only count them once leaving us with 6. One of the permutations will result in a fraction, leaving us with 5. One of the permutations will result in a negative number leaving us with 4. The last two observations hold where the numbers are different but the operations will be redundant if the numbers are identical so we still end up with 4.

Following this I came up with

$$
total = n!4^n + (n-1)!4^{n-2} + ... + 1!4^{0}
$$
$$
total = \sum_{i=1}^{n} i!4^{i-1}
$$
So for $n=6$ as in the game there are 769,641?
This is a decent upper bound but there will still be whole branches of calculations which will be void because there could be some fraction, zero division or negative number.
Any way to get a better approximation?

REPLY [1 votes]: I have created a python3 script to generate all solutions to a countdown numbers game, available on github: https://github.com/armchaircaver/Countdown-numbers-game
The program produces some analysis for a few random selections of cards.  A typical set of 6 cards, either 5 small + 1 large or 4 small+2 large, produces up to 40000 (approx) valid expressions that are in the range 101 to 999, the target range for the countdown game.
Without the constraint that the expression has a value in the range 101-999, there are typically up to 100,000 valid expressions for a set of cards.
Additional comments 10/8/2021:
Here are results of a larger statistical test, using samples of 100 for each type of card combination:
Random combinations of 2 top + 4 bottom, counting all expressions
mean: 58451.15 , median: 50650 , st dev: 21933.5
minimum: 15046 ,maximum: 110308
Random combinations of 1 top + 5 bottom, counting all expressions
mean: 51877.96 , median: 47105 , st dev: 21859.7
minimum: 9461 ,maximum: 125137
The variabilility of the number of expressions and the sensitivity of the number of valid expressions to the cards selected suggest that it would be difficult to obtain a mathematical formula for the number of solutions, and that a statistical analysis is the only realistic approach.


TITLE: Field Extensions and Properties of Algebraic Numbers
QUESTION [0 upvotes]: Problem
Let $F \leq E$ be an extension of fields, let $a \in E$ and let $b \in F$.
Show that $a$ is algebraic over $F$ if and only if $a + b$ is algebraic over $F$.
Thoughts
Is the best approach here to try and explicitly construct some polynomial $f(x) \in F[x]$ that satisfies $f(a+b)=0$ (to show the $\Rightarrow$ direction), or is it better to try and argue for the finiteness of $F(a+b)$? [I understand these are effectively the same thing].
And for the reverse direction ($\Leftarrow$), is it best to argue that, since $b \in F$, $F(b)=F$ and so $F(a+b)=F(a)$ (I'm not sure if this can simply be stated without justification) meaning that $|F(a):F|=|F(a+b):F|=n$, where n is the degree of the minimmum polynomial for $a+b$, which exists by assumption.
Thinking about another problem
How can we change approach to deal with the equivalent problem but with $ab$ rather than $a+b$. Does the $\iff$ relation still hold? (We should probably add the assumption that $b \neq 0 $ for this case).
Any help is very appreciated. Thanks!

REPLY [2 votes]: $F(a+b)$ is the smallest subfield of $E$ that contains both $F$ and the element $a+b$. Since $b\in F$ it follows that $-b\in F$ and so $b=(a+b)-b\in F(a+b)$. It follows that $F(a+b)$ also contains $a$ and so: $F(a)\subseteq F(a+b)$, since $F(a)$ is the smallest subfield of $E$ containing both $F$ and $a$. Similarly, since $a\in F(a)$ and $b\in F$ it follows that $a+b\in F(a)$, and so $F(a+b)\subseteq F(a)$. So $F(a)=F(a+b)$ and the conclusion about algebraicity follows. A similar argument will work with multiplication instead of addition, as long as you are careful with the case $b=0$.
In general, find explicit minimal polynomials is hard. The trick of considering a minimal field instead usually simplifies a lot of proofs.


TITLE: Proof or counterexample : Supremum and infimum
QUESTION [1 upvotes]: If $($An$)_{n \in N}$ are sets such that each $A_n$ has a supremum and $∩_{n \in N}$$A_n$ $\neq$ $\emptyset$ , then $∩_{n \in N}$$A_n$ has a supremum.
How to Prove This.

REPLY [1 votes]: Let $x \in \cap_{n \in \mathbb N} A_n$. Since $x$ belongs to each $A_n$, $x \le sup\ A_n$ for any $n \in \mathbb N$. Then we see that $\{x : x \in \cap_{n \in \mathbb N} A_n\}$ is bounded above (by $sup\ A_1$ for example) and so by the Completeness Property of $\mathbb R$, $\cap_{n \in \mathbb N} A_n$ has a supremum.


TITLE: Finding binomial Coefficient using expansion
QUESTION [1 upvotes]: If ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$, prove that $a_0=a_{2n},a_1=a_{2n-1},....a_n=a_{n+1}$
My approach is as follow ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$
${\left( {1 + x + {x^2}} \right)^n} = {\left( {1 + X} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{X^r}}  = \sum\limits_{r = 0}^n {{}^n{C_r}{{\left( {x + {x^2}} \right)}^r}}  = \sum\limits_{r = 0}^n {{}^n{C_r}\sum\limits_{g = 0}^r {{}^r{C_g}{x^{r - g}}.{x^{2g}}} }  = \sum\limits_{r = 0}^n {{}^n{C_r}\sum\limits_{g = 0}^r {{}^r{C_g}{x^{r + g}}} } $
How do we proceed from here

REPLY [1 votes]: $$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k$$                                                                                Change $x$ to $1/x$ in this udentity, then
$$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^{2n-k}=\sum_{k=0}^{2n} A_{2n-k} ~ x^k,$$ $k \to 2n-k$ used in above. Finally we get $A_{2n-k}=A_k$.