TITLE: Principal disjunctive normal form conversion problem.
QUESTION [2 upvotes]: I am solving a question where I have to convert $(\neg P \ \lor Q)$ this equation into principal disjonction normal form. so far I know that to convert it I have to multiply 1 with it. ie; $(\neg P \ \lor Q) \land T$, where $T$ is true.
now in the solution of the question after multiplying the equation with 1 it is written like this.
$(\neg P \ \lor Q) \land (\neg P \ \lor P)$.
so my question is why $(\neg P \ \lor P)$ is coming why can't we write $(P \ \lor \neg Q)$ instead.

REPLY [0 votes]: $(P \ \lor \neg P)$ is exactly true. I guess the solution wants to express the normal form uniquely with elements of the original logical proposition (i.e $P$ and $Q$), and avoids using other symbols (such as $T$). This is why they are using $(P \ \lor \neg P)$.
Moreover, note that $(\neg P \ \lor Q) \land (P \ \lor \neg Q) \neq (\neg P \ \lor Q)$. For example, take $P$ False and $Q$ True.


TITLE: Almost sure convergence and lim sup
QUESTION [5 upvotes]: I'm struggling to understand the beginning of the solution to the following exercise:

Let $(X_n)_{n\geq 1}$ and $X$ be random variables. Prove that $X_n \to X$ almost surely if and only if for every $\epsilon>0$
$$P(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\})=0$$

Solution:
Notice that $\omega \in\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\}$ iff there exist a subsequence $(n_k)_{k\geq 1}$ such that $|X_{n_k}-X|\geq \epsilon$, so we have
$$\{\limsup\limits_{n\to \infty}|X_n-X|> \epsilon\}\subset \limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\} \subset \{\limsup\limits_{n\to \infty}|X_n-X|\geq \epsilon\}\quad (*)$$
and so on..

I don't understand what this means.

First what subsequence are we referring to? A subsequence that does converge? Why do we need this exactly?
What's the difference between those 3 sets in line $(*)$. Why does it matter if the $\limsup$ is within the brackets and why does it matter whether $>$ or $\geq$? I really don't understand what the difference are.

I'm happy if someone could explain it to me.

REPLY [5 votes]: Note the definition of lim sup for a sequence $(A_n)$ of sets:
$$ \limsup_{n\to\infty} A_n=\bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n.$$
This translates to:
$$ \omega\in\limsup_{n\to\infty} A_n \Leftrightarrow \forall k\, \exists n\ge k\colon \omega\in A_n.$$
Or, in everyday language: There are arbitrarily large $n$ for which $\omega\in A_n$. If you collect all such $n$ in a set and number them consecutively, you get an increasing sequence $(n_k)$ with $\omega\in A_{n_k}$ for all $k$.
That is the sequence mentioned in the proof, where $A_n=\{|X_n-X|\ge\epsilon\}$.
It matters whether lim sup is inside or outside the sets, because the two kinds of lim sup are entirely different beasts! But clearly, related.
The difference between sharp and non-sharp inequalities arise for $\omega$ such that $\limsup|X_n-X|=\epsilon$ exactly. Such an $\omega$ belongs in the set on the right end of $(*)$, but not in the left set. And it may or may not be in the middle set, depending on whether $|X_n-X|\ge\epsilon$ infinitely often or not. Both are possible, under the current assumption.


TITLE: Components of four-force
QUESTION [2 upvotes]: Temporal component of the four-accelaration is:
$$\mathbf{A}_t = \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c}$$
that, multipliying by the rest mass, should give a value of the temporal component of the four-force of:
$$\mathbf{F}_t = m_0 \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c} = \gamma_u^4\frac{\mathbf{f}\cdot\mathbf{u}}{c}$$
where I replaced $ m_0 \mathbf{a} = \mathbf{f} $.
I reach same value if I take derivate respect proper time of the temporal component of the four-momentum $m_0 \gamma_u c$:
$ { d \gamma_u \over dt } = {d \over dt} \frac{1}{\sqrt{ 1 - \frac{\mathbf{v} \cdot \mathbf{v}}{c^2} }} = \frac{1}{\left( 1 - \frac{\mathbf{v} \cdot \mathbf{v}}{c^2} \right)^{3/2}} \, \, \frac{\mathbf{v}}{c^2} \cdot \, \frac{d \mathbf{v}}{dt} \, = \, \frac{\mathbf{a \cdot u}}{c^2} \frac{1}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \, = \, \frac{\mathbf{a \cdot u}}{c^2} \, \gamma_u^3 $
$ { d \gamma_u \over d\tau } = { d \gamma_u \over dt }{ dt \over d\tau } = \frac{\mathbf{a \cdot u}}{c^2} \, \gamma_u^3 \, \gamma_u $
$ \mathbf{F}_t = { d \mathbf{P}_t \over d\tau } = m_0 c { d \gamma_u \over d\tau } = m_0 \frac{\mathbf{a \cdot u}}{c} \, \gamma_u^4 =\frac{\mathbf{f \cdot u}}{c} \, \gamma_u^4  $
However, wikipedia gives as correct value:
$ \mathbf{F}_t  = \frac{\mathbf{f \cdot u}}{c} \, \gamma_u  $
I'm making an error of $\gamma_u^3$ and I can not find where it is .

REPLY [3 votes]: The relation $\:\mathbf f \boldsymbol{=} m_0 \mathbf a\:$ is not valid. Instead of it use this
\begin{equation}
\mathbf f \boldsymbol{=}\gamma_u m_0 \mathbf a\boldsymbol{+}\gamma^3_u m_0 \dfrac{\left(\mathbf a \boldsymbol{\cdot}\mathbf u\right)}{c^2}\mathbf u \quad \boldsymbol{\Longrightarrow} \quad
\boxed{\:\:\mathbf f\boldsymbol{\cdot}\mathbf u \boldsymbol{=}\gamma^3_u m_0 \left(\mathbf a \boldsymbol{\cdot}\mathbf u\right)\vphantom{\dfrac{a}{b}}\:\:}
\tag{A-01}\label{A-01}
\end{equation}
To reach that combine
\begin{equation}
\mathbf f \boldsymbol{=}\dfrac{\mathrm d\mathbf p}{\mathrm d t} \boldsymbol{=}\dfrac{\mathrm d\left(\gamma_u m_0 \mathbf u\right)}{\mathrm d t} \boldsymbol{=}\cdots
\tag{A-02}\label{A-02}
\end{equation}
with yours
\begin{equation}
\dfrac{\mathrm d\gamma_u}{\mathrm d t} \boldsymbol{=}\cdots
\tag{A-03}\label{A-03}
\end{equation}


TITLE: Ergodicity under transformation
QUESTION [2 upvotes]: Suppose $\Omega := [0,1]^{\mathbb Z}$ is equipped with the product topology and endowed with the Borel $\sigma$-algebra $\mathcal B(\Omega)$ and there is a probability measure $\mathbb P$ on $(\Omega,\mathcal B(\Omega))$ such that the shift $T:\Omega \to \Omega$, $$T(\omega)(k) := \omega(k+1),\quad \omega\in\Omega,k\in \mathbb Z$$ is measure preserving, i.e. $\mathbb P = \mathbb P \circ T^{-1}$ on $\mathcal B(\Omega)$, and ergodic, i.e. $A=T^{-1}(A)$ implies $\mathbb P (A)\in\{0,1\}$ for any $A\in\mathcal B(\Omega)$.
Now let $f:[0,1]^3\to[0,1]$ a measurable function and $U:\Omega \to \Omega$ the transformation defined by $$ U(\omega)(k) := f(\omega(2k-1),\omega(2k),\omega(2k+1)),\quad \omega\in\Omega,k\in\mathbb Z.$$ We consider the probability measure $\widetilde {\mathbb P}:= \mathbb P\circ U^{-1}$ where $U^{-1}$ denotes the preimage.
Then, by $T\circ U= U\circ T^2$, it holds that $(\Omega,\mathcal B(\Omega), \widetilde {\mathbb P},T)$ is still a measure-preserving dynamical system. Is it also ergodic?
Edit: What are examples of probability measures $\mathbb P$ on $\mathcal B(\Omega)$ and sets $A\in\mathcal B(\Omega)$ such that $T^{-2}(A)=A$ but $\mathbb P(A)\notin \{0,1\}$ (and hence necessarily $T^{-1}(A)\neq A$)?

REPLY [2 votes]: The answer is negative: Let \begin{align*} \mathbb P &:= \frac 1 2 \left(\delta_{\left(\mathbb 1_{2 \mathbb Z}(k)\right)_{k\in \mathbb Z}} + \delta_{\left(\mathbb 1_{2 \mathbb Z+1}(k)\right)_{k\in \mathbb Z}} \right), \\ A &:= \left\lbrace (1)_{k\in\mathbb Z} \right\rbrace ,\\ f(x,y,z) &:= y. \end{align*}
The probability $\mathbb P$ corresponds to the irreducible Markov chain on state space $\{0,1\}$ with transition matrix $P = \begin{pmatrix} 0 & 1\\ 1& 0\end{pmatrix}$ which has unique stationary distribution $(\frac 1 2 \,\,\, \frac 1 2)$. In light of the answer to this math.SE question the dynamical system $(\Omega,\mathcal B(\Omega),\mathbb P, T)$ is measure-preserving and ergodic (but not mixing). Now, $T^{-1}(A)=A$ but $$U^{-1}(A) = \prod_{k\in \mathbb Z} \begin{cases} \{1\},&\quad k\in 2 \mathbb Z \\ [0,1],&\quad k\in 2\mathbb Z+1\end{cases}, $$ whence $\widetilde{\mathbb P}(A) = \frac 1 2 $.


TITLE: Find the number of integer solutions to $x_1+x_2+x_3+x_4= 30$ where $0\leq x_n <10$ for $1\leq n \leq 4$
QUESTION [2 upvotes]: I'm reviewing for my final coming up and I'm not sure about my solution to this problem.

Find the number of integer solutions to $$x_1+x_2+x_3+x_4= 30$$ where $0\leq x_n <10$ for $1\leq n \leq 4$.

My first thought was to find the number of solutions where at least one of the x's is greater than ten and subtracting that from the total, which is equivalent to finding the number of integer solutions to
Find the number of integer solutions to $$x_1+x_2+x_3+x_4= 20.$$
That's a problem I know how to solve, and the answer is $nCr(23,20)=1771$.
Since there are $nCr(33,30)=5456$ total solutions, the final answer is $$5456-1771=3685.$$
Is this right?

REPLY [0 votes]: HINT:
I got $$
(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)^4 = \\
x^{36}+4 x^{35}+10 x^{34}+20 x^{33}+35 x^{32}+56 x^{31}+84 x^{30}+120 x^{29}+165 x^{28}+220 x^{27}+282 x^{26}+348 x^{25}+415 x^{24}+480 x^{23}+540 x^{22}+592 x^{21}+633 x^{20}+660 x^{19}+670 x^{18}+660 x^{17}+633 x^{16}+592 x^{15}+540 x^{14}+480 x^{13}+415 x^{12}+348 x^{11}+282 x^{10}+220 x^9+165 x^8+120 x^7+84 x^6+56 x^5+35 x^4+20 x^3+10 x^2+4 x+1$$
( cf. hint of @dREaM: ) so the number is $84$.


TITLE: Natural Deduction Proof: A ↔ B |- (C → A) → (C → B)
QUESTION [1 upvotes]: In an attempt to prove the formula, I tried setting a hypothesis
C -> A

like the following
1   A <-> B             Premise
2   |C -> A         Hypothesis
3   |A -> B         1 (<->E)
4   |C -> B         1, 2, 3 (?????)

, but I don't know what rule to apply to support the forth line
|C -> B

, even though I do know that I can reach this conclusion by combining 1, 2, 3.
What rule goes in to the ????? in the parenthesis above?
Also, I assume that I can prove
(C → A) → (C → B)

if I prove
A -> B
C -> A

, but how do I add
C ->

part to A and B?
I know that I can prove like if
A

is true, then
A V B

is true as well (VI rule)
Also, if
A

is true and
B

is true, then
A Ʌ B

is true as well (ɅI rule).
and in this proof I want to add
C ->

to A and B so it will be
(C -> A) -> (C -> B)

, but how can I do it? There are introduction rules like
(VI rule)
(ɅI rule)

, but it seems there is no rule like
(->I rule)

REPLY [1 votes]: An alternative proof using mainly (1) conditional proof and (2) contraposition.
It also makes use of the replacement rule : (X --> Y) :: ~ ( X & ~Y)
as definition of the material implication operator.
The strategy is as follows :
(1) since the desired conclusion is a conditional statement ( having itself conditional statements as antecedent and consequent, but that does not matter), and since a conditional statement is always equivalent to its contrapositive statement ( the contrapositive of X --> Y is  ~ Y --> ~ X) , one can prove a conditional by proving its contrapositive ;
(2) the contrapositive of the desired conclusion is :
                 ~ ( C --> B)  -->  ~ ( C --> A)

(3) in order to prove this contrapositive, we use conditional proof, by assuming the antecedent, and deriving the consequent under this assumption.
(4) finally, we use  contraposition to recover the desired conclusion ( for the contrapositive of the contrapositive of X is X itself, or, in other words, the relation " being the contrapositive of " is a symmetric relation).


TITLE: Bifurcation Help - Fixed and Varied Constants
QUESTION [1 upvotes]: Question
The dynamic system is two dimensional. The variables $\alpha$, $\beta$, $A$ and $B$ are real constants. However both $A$ and $B$ are nonzero.
$$f(x,y) = A(y-x^2)$$
$$g(x,y) = B(x-\alpha)(y-\beta x+x^2)$$
(a) Show that Hopf bifurcation cannot exist in this system.
(b) Show this dynamic system can have all three other types of local bifurcations as $\alpha$ varies and $\beta$ is fixed and vice versa.
To show that the Hopf bifurcation cannot exist, I have founded all the fixed points and the corresponding eigenvalues to which none have imaginary parts. From this I draw the conclusion that a Hopf bifurcation is impossible. However, I am not really sure if this is actually the case.
I am completely stuck on (b). I have founded fixed points before, I do not see how I can transform my system into any of the bifurcations.
Results:
Let $f = g = 0$. Solving the equations and labelling the new function $h$ gives
$$h(x) = x(x-\alpha)(2x-\beta)$$
Taking the roots of this function gives the fixed points $x^* = 0$, $\alpha$ and $\beta/2$.

REPLY [1 votes]: To show that the Hopf bifurcation cannot exist, solve the following system:
$$
\begin{align}
f(x,y)&=0,\\
g(x,y)&=0,\\
\text{tr}(J(x,y))&=0,
\end{align}
$$
for $(x,y,A)$ and with this solution evaluate the determinant of Jacobian Matrix $J(x,y)$. Then, the same for $(x,y,B)$, $(x,y,\alpha)$ and $(x,y,\beta)$. In all cases, the determinant is negative, therefore the Hopf bifurcation cannot exist in your system.


TITLE: What is the exact gravitational force between two masses including relativistic effects?
QUESTION [26 upvotes]: Is there a closed-form formula for the force between two masses $m_1$ and $m_2$ if relativistic effects are included? My understanding is that the classic formula $G \frac{m_1 m_2}{r^2}$ is just an approximation (which is probably good enough even for going to the Moon), but what would the correct formula be according to the general theory of relativity? Does a closed formula even exist? For example for an idealized situation of two uniform spherical masses?

REPLY [2 votes]: The expression that is actually used by JPL to calculate/approximate the relativistic effects under given promises comes from the "post-Newtonian expansion" at the 1PN-level and, including the classical Newtonian acceleration, it looks like:
$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$
So they are basically introducing one component that is "repulsive inverse cube gravity" and two velocity dependent terms. This expression is also availabe at the 3PN-level, see this post. Note that this is only an approximation that is able to reproduce the "anomalous perihelion shift" in the weak field limit. The extra repulsive inverse cube term causes very strange behaviour if you try to use it in the strong field limit. Here are a few simulations I made where the green circle denotes the Schwarzschild radius and the red circle denotes the radial distance to the "innermost stable circular orbit." It is easy to see the bouncing caused by the repulsive inverse cube term:

Note: This is the expression for the case of infinite mass ratio. If you cannot use the infinite mass ratio approximation, the expression would be more complicated I think.


TITLE: Constructing a Simple Group with a Specific Number of Sylow $p$-subgroups
QUESTION [4 upvotes]: I've been studying for my final exams, and I came across the following question:

If possible, give an example of a simple group $G$ with $n>1$ Sylow $p$-subgroups such that the order of $G$ does not divide $n!$. If not possible, briefly explain why.

Now, it didn't take me long to think of the following "numerical" example: a group of order $60=2^2\cdot3\cdot5$ where there are $3$ Sylow $2$-subgroups. My question is whether or not such a group exists, and if it does, is there any simple way to describe these groups based on knowing how the Sylow structure is built?
In trying to describe the above group, I think I showed that it can't exist. This is because there either have to be $4$ Sylow 3-subgroups or $10$ Sylow 3-subgroups. The second case leads to a contradiction since then there would have to be $k$ Sylow 5-subgroups such that $k(5-1)=30$, and likewise, in the first case, we find we'd have to have 13 Sylow 5-subgroups, also a contradiction. Is all of this correct and is there an easy way to determine whether such a group exists or not?

REPLY [2 votes]: The answer is no.
Let $C$ be the set of sylow $p$ subgroups of $G$. It is easy to verify that the map $G\times C\rightarrow C$ that sends $(x,H)$ to $xHx^{-1}$ is an action of $G$ on $C$. By the orbit stabilizer theorem, we know that:
$$\forall H\in C[|C||N(H)|=|G|]$$
Since $1<|C|$, we deduce that $|N(H)|<|G|$ for all $H\in C$....(1)
By sylow's theorem we know that the function $T_x:C\rightarrow C$ that sends $H$ to $xHx^{-1}$ is a permutation of $C$. Now consider the homomorphism $\phi:G\rightarrow S_C$ that sends $x$ to $T_x$.
Claim:  $\phi$ is an injection.
Proof: We know that $\operatorname{Im}\phi\cong G/\ker\phi$. Thus, it suffices to show that $|\ker\phi|=1$. This is easy to see because $\ker\phi\unlhd G$, since $G$ is simple, therefore either $\ker\phi=\{e\}$ or $\ker\phi=G$. Let $H\in C$, it is easy to verify that $\ker\phi \leq N(H)$. Thus, if $\ker\phi=G$, we would conclude that $N(H)=G$. However, we know from (1) that $N(H)$ is a proper subgroup of $G$. Therefore, we finally deduce that $\ker\phi=\{e\}$.
Thus, $G$ is isomorphic to a subgroup of $S_C$ (The group of permutations of $C$). Hence, $|G|\,\Big|\,|S_C|!=n!$


TITLE: How to solve: $4^x=3^x+1$
QUESTION [0 upvotes]: For $x \in \mathbb{R}$, solve for $x$:
$$
4^x=3^x+1
$$
By inspection, it's easy to see that $x=1$ is a solution. How would I go about showing that it is the only solution? If I look at the graph it's quite obvious, but I don't see how I can do it formally.

REPLY [3 votes]: If $x>1$, then $4^x >4\cdot 3^{x-1} = (3+1)3^{x-1}=3^x + 3^{x-1}\geq 3^x+1$.
Let me know if any of the steps are unclear.


TITLE: Prime divisors of two large integers which are close.
QUESTION [1 upvotes]: $\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\N}{\mathbb N}$
For a positive integer $x$ let us write $A_x$ to denote the set of all the prime factors of $x$.

Conjecture.
For any given positive integer $n$ there is $N$ large enough such that whenever $x$ is a positive integer with $x\geq N$ we have $A_x\neq A_{x+n}$.

Suppose $x$ is such that $A_{x}=A_{x+n} = \set{p_1,\ldots, p_k}$.
Then each $p_i$ must divide $n$.
Write
$$
x=p_1^{\alpha_1} \cdots p_k^{\alpha_k},\quad
x+n= p_1^{\beta_1} \cdots p_k^{\beta_k}
$$
Then the equation $(x+n) - x= n$ gives
$$
p_1^{\beta_1} \cdots p_k^{\beta_k} - p_1^{\alpha_1} \cdots p_k^{\alpha_k} = n
$$
This seems to suggest the following conjecture.
Let $S=\set{p_1,\ldots, p_k}$ be a set of primes.
For $\alpha = (\alpha_1,\ldots, \alpha_k)\in \N^k$, we write $p^\alpha$ to mean the number $p_1^{\alpha_1} \cdots p_k^{\alpha_k}$.
Then

Conjecture.
For any given natural number $n$, there are only finitely many distinct elements $\alpha, \beta\in \N^k$ such that $|p^\alpha-p^\beta|\leq n$.

(Cancelling out from LHS and RHS in the equation $p^\alpha-q^\beta = \pm n$, we may assume that $\sum_i \alpha_i\beta_i=0$.)
In particular, the conjecture above, if true, implies that for any given $n$, the gap $|2^\alpha-3^\beta|$ is less than $n$ for only finitely many pairs $(\alpha, \beta)\in \N^2$, which seems intuitively reasonable but I am not able to make any progress.

REPLY [2 votes]: R. Tijdeman, On integers with many small prime factors, Compositio Mathematica 26 (1973) 319-330, says:
Let $p$ be a prime, $p\ge3$, and let $n_1=1<n_2<\cdots$ be the sequence of all positive integers composed of primes $\le p$. [...] Thue derived from his famous result on the approximation of algebraic numbers by rationals that $\lim_{i\to\infty}(n_{i+1}-n_i)=\infty$.
This is already enough to prove the "Conjecture". Tijdeman goes on to improve on Thue's result. The Thue reference is Bermerkungen uber gewisse Naherungsbruche algebraischer Zahlen, Skrift. Vidensk. Selsk. Christ., 1908, Nr. 3.


TITLE: If $X$, $Y$ are i.i.d and mutually exclusive, then is $\Pr(\min(X,Y)=n)=\Pr(X=n)+\Pr(Y=n)$?
QUESTION [0 upvotes]: I was wondering if $X$, $Y$ are independent and mutually exclusive, then is $\Pr(\min(X,Y)=n)=\Pr(X=n)+\Pr(Y=n)$?
 $\Pr(\min(X,Y)=n)=\Pr(X=n \text{ or }Y=n)=\Pr(X=n)+\Pr(Y=n)$. Is this correct or am I missing something?

REPLY [3 votes]: $$Pr(min(X,Y)=n)=Pr(X=n\cap Y=n)+Pr(X=n\cap Y\gt n)+Pr(Y=n\cap X\gt n)\ ,$$ since these events are mutually exclusive.
Using independence $Pr(min(X,Y)=n)=Pr(X=n)Pr(Y=n)+Pr(X=n)Pr(Y\gt n)+Pr(Y=n)Pr(X\gt n)$.


TITLE: Show that sets are equal
QUESTION [0 upvotes]: Let $X=\{x_1,x_2,...,x_n\}$ and $Y=\{y_1,y_2,...,y_n\}$ be sets over a finite field $F$ with $p=char(F)>2$. Assume
$$x_1^k+x_2^k+...+x_n^k=y_1^k+y_2^k+...+y_n^k,\ 1\leq k\leq n$$
I wanna show that $X=Y$.

REPLY [2 votes]: Let me prove that if for two sets of $n$ distinct numbers $X=\{x_1,\ldots,x_n\}$ and $Y=\{y_1,\ldots,y_n\}$ the sums of $k$th powers for $k=1,\ldots,2n-1$ are the same, then $X=Y$, irrespectively of the ground field. This shows that the example of @zeb given in his answer (for $n=p$ over a field of characteristic $p$) is, in a sense, optimal.
For that, let us take set $X$ and consider the Newton identities number $n$, \ldots, $2n-1$:
\begin{aligned}
ne_n&=\sum_{k=1}^n (-1)^{k-1}p_k e_{n-k},\\
 0&=\sum_{k=1}^{n+1} (-1)^{k-1}p_k e_{n+1-k},\\
 \ldots \\
 0&=\sum_{k=1}^{2n-1} (-1)^{k-1}p_k e_{2n-1-k},\\
\end{aligned}
with the usual convention $e_0=1$, $e_m=0$ for $m>n$. Given the power sums $p_1, p_2, \ldots$, these formulas are $n$ linear equations with $n$ unknowns $e_1, \ldots, e_n$, whose matrix of coefficients is easily seen to have the entry $i,j$ equal to $(-1)^{i+j-1}p_{i+j-2}$, where $p_i$ is the $i$th power sum of $x_1,\ldots,x_n$, including $p_0=n$ (this accounts for the term $ne_n$ in the first equation). Changing the variables $x_i\to -x_i$ and multiplying the matrix by $-1$, we get the matrix whose entry $i,j$ equal to $p_{i+j-2}$. This matrix is, by a direct inspection, equal to $VV^T$, where $V$ is the Vandermonde matrix, so its determinant is $\prod_{i<j}(x_i-x_j)^2$. Thus, for pairwise distinct $x_i$, this system has the only solution for $e_1,\ldots,e_n$. Clearly, the elementary symmetric functions determine the set $X$ uniquely.


TITLE: Weaking the path test for multivariable limits
QUESTION [3 upvotes]: The multivariable limit $\lim_{(x,y)\to (x_0,y_0)} f(x,y)$ exists and is equal to some scalar $L$ if and only if the limit $\lim_{t\to 1} f(r(t))$ exists and is equal to $L$ for all functions $r:\mathbb{R}\to \mathbb{R}^2$ such that $\lim_{t\to 1} r(t) = (x_0, y_0)$. It is common to use this fact as a way to conclude that a certain limit does not exist, by checking that taking the limit along different paths leads to different limits or by showing that the limit along a certain path does not exist. I wonder if this characterizarion can be weakened in the following way:
The multivariable limit $\lim_{(x,y)\to (x_0,y_0)} f(x,y)$ exists and is equal to some scalar $L$ if and only if $\lim_{y\to y_0} f(x_0,y)=L$ and $\lim_{x\to x_0} f(x,g(x)) = L$ for all functions $g:\mathbb{R}\to \mathbb{R}$ such that $\lim_{x\to x_0} g(x) = y_0$.
That is, it the limit exists and is the same going through all graphs of functions $g(x)$ and through the $y$ axis, can I conclude the existence of the multivariable limit?
EDIT: I substituted my initial condition of $\lim_{y\to y_0} \lim_{x\to x_0} f(x,y)=L$ by $\lim_{x\to x_0} f(x,g(x)) = L$ since, as José Carlos Santos noted in the comments, formalizes better the idea of approaching the limit vertically.

REPLY [5 votes]: Yes, you can.
Suppose that, for some $L\in\Bbb R$, you don't have $\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L$. Then there is some $\varepsilon>0$ such that, for every $\delta>0$, there is some $(x,y)\in D_f$ such that$$\|(x,y)-(x_0,y_0)\|<\delta\quad\text{and that}\quad|f(x,y)-f(x_0,y_0)|\geqslant\varepsilon.$$In particular, for every $n\in\Bbb N$, there is some $(x_n,y_n)\in D_f$ such that$$\|(x_n,y_n)-(x_0,y_0)\|<\frac1n\quad\text{and that}\quad|f(x_n,y_n)-L|\geqslant\varepsilon.$$At least one of the following assertions holds:

We have $x_n=x_0$ for infinitely many $n$'s.
We have $x_n>x_0$ for infinitely many $n$'s.
We have $x_n<x_0$ for infinitely many $n$'s.

If the first assertion holds, we can assume without loss of generality that $x_n=x_0$ for each $n\in\Bbb N$. But then, since $\lim_{n\to\infty}(x_n,y_n)=(x_0,y_0)$, $\lim_{n\to\infty}y_n=y_0$, and so we cannot have $\lim_{y\to y_0}f(x_0,y)=L$ (because $|f(x_0,y_n)-L|\geqslant\varepsilon$ for each $n\in\Bbb N$).
If the second assertion holds, we can assume without loss of generality that the sequence $(x_n)_{n\in\Bbb N}$ is strictly decreasing. Consider the only function $g\colon[x_0,x_1]\longrightarrow\Bbb R$ such that:

$(\forall n\in\Bbb Z_+):g(x_n)=y_n$;
the restriction of $f$ to each interval $[x_{n-1},x_n]$ is affine.

Then, since $\lim_{n\to\infty}x_n=x_0$, $g$ is continuous. But, since $(\forall n\in\Bbb N):\bigl|f(x_n,g(x_n))-L\bigr|\geqslant\varepsilon$, we cannot have $\lim_{x\to x_0}f(x,g(x))=L$.
What can be done under the third assumption is similar to what was done with the second one.


TITLE: What are the sufficient conditions for a strongly convex function to have a unique minimum if it exists?
QUESTION [0 upvotes]: What are the sufficient conditions for a strongly convex function to have a unique minimum if it exists? I think continuity is necessary? And have a counter example but just wanted to be sure?

REPLY [0 votes]: It's always true. A strongly convex function is also strictly convex; a strictly convex function reaches its global minimum (if it exists) in exactly one point. Note also that any local minimum of a convex function is also a global minimum.


TITLE: Invert transform of a generating function
QUESTION [2 upvotes]: A055887's generating function is given as a transform of another generating function $\frac{1}{1-P(x)}$. It's clear from the definition that this is $\sum_{n=0}^{\infty} \left [ \sum_{i+j+k+...=n} P(i)P(j)P(k)... \right ] x^n$ but I'm struggling to derive this.
I've seen $x$ in $\frac{1}{1-x} = 1+x+x^2+x^3$ be replaced $x$ with $a x$, $x^2$, $x^m$, $-x$ but I only proved these by working backwards. Starting with the result and showing
$S - Sx = 1$ then concluding $S = \frac{1}{1-x}$.
This was called "invert transform" in A067687 but that term doesn't seem to be defined or commonly used outside of OEIS.

REPLY [0 votes]: Hint: A proof of
\begin{align*}
\color{blue}{\left(1-A(z)\right)^{-1}=\sum_{n=0}^{\infty}A^n(z)}
\end{align*}
with constant term of $A(z)$ equal to zero is given here.


TITLE: Solve for $x$: $\sqrt{ x + 4} = −2$
QUESTION [2 upvotes]: Solve for $x$: $\sqrt{ x + 4} = −2$
I did a lot of positive numbers.
I did negative exponents.
I did everything.
It doesn't seem possible or is it? It might be a trick question...

REPLY [3 votes]: (Answer to original question, which was mis-formatted):
You get $\sqrt{x} = -2-4 = -6.$
Since there are no real numbers whose square root is negative (the $\sqrt{\cdot}$ notation means "the positive square root or zero"), the solution set is the empty set (by which I mean that there is no solution to the problem -- there's no real number $x$ that makes the equation true).
It's possible that your book was hoping that you'd square both sides to get
$x = 36$ and conclude that $x = 36$ is an answer, which is wrong (i.e., they were trying to fool you into making a mistake.)

OP says that the actual equation to be solved is
$$
\sqrt{x + 4} = -2.
$$
Since $\sqrt{\cdot}$ always (by definition) produces a non-negative number, there's no solution. But your book may have hoped that you'd write
\begin{align}
\sqrt{x + 4} &= -2\\
(x + 4) &= (-2)^2\text{, by squaring both sides}\\
x + 4 &= 4\text{, because $-2 \cdot -2 = 4$}\\
x  &= 0\text{, by subtracting $4$ from both sides}\\
\end{align}
and conclude that $x = 0$ is a solution, which is not true: if you plug $x = 0$ into the left half, you get $\sqrt{0+4} = \sqrt{4} = 2$, and this is not the same as the right hand side.
The failure was in the first line of the deduction: the solutions to an equation $A = B$ are not the same as the solutions to $A^2 = B^2$; put less formally, "it's not OK to square both sides of an equation when you're trying to solve it, or it you do, you have to check at the end that the solutions you got really do work in the original equation."