问:
First order system in polar coordinates : finding the trap region and showing it has a limit cycle
I have the following first order EDOs system,

$$
\begin{cases}
  \dot{x} = y + ax(1-b-x^2-y^2) \\
  \dot{y} = -x + ay(1-x^2-y^2)
\end{cases}
$$

where $a \in ]0,1]$ and $b \in [0,1[$

Switching to polar coordinates $(r, \theta)$, I find the following system,

$$
\begin{cases}
  \dot{r} = ar^2(1-r) - abr\cos^2(\theta) \\
  \dot{\theta} = a\cos(\theta)\sin(\theta)b - 1
\end{cases}
$$

I would like to show that this system has a limit cycle. How could I do that ? I was thinking about showing that some surface $S$ in the plane is a trapping region. Therefore I could show using Pointcaré-Bendixson theorem that inside there would be a periodic orbit (limit cycle).

However, I'm wondering how I could find this trapping region ?

I can see that
$$
0 < \cos^2(\theta) < 1
$$

And that
$$
ar^2(1-r) > 0 \quad , r \in ]0,1[
$$

$$
ar^2(1-r) < 0 \quad, r \in]1, +\infty[
$$

But then there is this $abr\cos^2(\theta)$ that is so annoying.

答:
The polar form of the equations are
\begin{align}\dot{r}&=ar(1-r^2)-ab\,r\cos^2\theta\\
\dot{\theta}&=ab\sin\theta\cos\theta-1
\end{align}

Let $D$ be the annulus defined by $\frac{1}{2}\sqrt{1-b}<r<2$. Then

it does not contain any equilibrium points, in fact $\dot{\theta}<0$,

on the outer boundary, $\dot{r}<0$,

on the inner boundary, $\dot{r}=ar(1-r^2)-abr\cos^2\theta\ge\frac{3}{4}(1-b)ar>0$

Thus $D$ is a bounded trapping region without equilibrium points, so must contain a non-trivial closed orbit by the Poincare-Bendixson theorem.


问:
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.

答:
We use the identity $(u+v)^2=u^2+v^2.$ From $$x^4+x^2+x=0$$ we see that
$$x=x^4+x^2=(x^2)^2+x^2=(x^2+x)^2$$
And we can construct the root of an arbitrary polynomial
$$ax^3+bx^2+cx+d=x(ax^2+c)+bx^2+d=(x^2+x)^2(ax+c)^2+(bx+d)^2=((x^2+x)(ax+c)+(bx+d))^2=(ax^3+(a+c)x^2+(b+c)x+d)^2$$
So we see that every polynomial has exactly one square root.

ContentLicense: CC BY-SA 4.0


问:
Sum ${\sum_{n=1}^\infty \frac{4}{n(n+3)}}$
Determine if the series converges or diverges. If it converges what does it converge to?

${\sum_{n=1}^\infty \frac{4}{n(n+3)}}$

Here's what I have been able to figure out

By the limit comparison test ${\sum\frac{1}{x^2}}$

${\lim_{x\to\infty}\frac{4n^2}{n^2 + 3n} = 4\Rightarrow}$ Convergence

But I can't figure out how to tell what it converges to, it's not an alternating series, or a geometric series that I can tell. And we haven't yet covered taylor series. It might be a power series, but I don't see any x value to tell what it converges to.

答:
Observe that it can be showed that, in general (it's quite evident why this happens):
$$\sum_{i = 1}^{\infty} \left(\frac{1}{i} - \frac{1}{(i+p)} \right) = \sum_{i = 1}^{p} \frac{1}{i}$$
But see that:
$$\sum_{n = 1}^{\infty} \frac{4}{(n(n+3))} = 4 \sum_{n = 1}^{\infty} \frac{1}{3n} - \frac{1}{3(n+3)} = \sum_{n = 1}^{\infty} \frac{4}{3}\left(\frac{1}{n} - \frac{1}{n+3} \right)$$
In this way, $p = 3$, then we have:
$$\sum_{n = 1}^{\infty} \frac{4}{3}\left(\frac{1}{n} - \frac{1}{n+3} \right) = \sum_{n = 1}^{3} \frac{4}{3}\left(\frac{1}{n} - \frac{1}{n+3} \right) = 73/45$$
Logo a sequência converge.


问:
Convergence in $L^2(]0,T[,v)$
if I have a sequence $(u_n)_n$ that converges to $u$ in $L^2(]0,T[,v)$ , then why $ ||u_n(.)||^2_v \rightarrow ||u(.)||^2_v$ in $L^1(]0,T[)$?
Remark that $L^2(]0,T[,v)$ is the space of measurable functions $t \rightarrow ||f(t)||_v$ in $L^2(]0,T[)$ with the norm $||f||_{L^2v}=$$(\int_0^T ||f(t)||^2_v\,dt)^\frac{1}{2}$ , $v$ is a Hilbert space.

I tried to use the assumption that  in general I can use : $| ||x||^2-||y||^2 | \le ||x-y||^2$ , but as answered below, it is not true.

答: No. Let $||\cdot|| $ be the usual absolute value in $ \mathbb R$ and take $x=2$ and $y=3.$

Title: Convergence in $L^2(]0,T[,v)$


问:
Rank-Induction Principle (following Van Dalen's Logic and Structure, 4th edition)
This post contains questions on understanding Van Dalen's proof of the Rank-Induction principle and questions concerning its wording and presentation. Please don't feel obliged to answer everything. Anyways, we begin with the theorem's statement:

Theorem 1.1.8 (Induction on rank-Principle) If for all $\varphi$, [$A(\psi)$ for all $\psi$ with rank less than $r(\varphi)$] $\Rightarrow A(\varphi)$, then $A(\varphi)$ holds for all $\varphi \in PROP$.

I assume here that the theorem is simply a statement of a strong mathematical induction principle on the rank of propositions. Please correct me if this is the wrong interpretation of the theorem's statement. Assuming this is the case, why must we show that strong (mathematical) induction on rank is equivalent to  the previously defined structural induction (Theorem 1.1.3)? In other words, what is the significance of this equivalence? Moving on, we continue with the author's proof:

Let us show that induction on $\varphi$ and induction on the rank of $\varphi$ are equivalent. First we introduce a convenient notation for the rank-induction: write $\varphi \prec \psi$ ($\varphi \preceq \psi$) for $r(\varphi) < r(\psi)$ ($r(\varphi) \leq r(\psi)$). So $\forall\psi\preceq\varphi A(\psi)$ stands for "$A(\psi)$ holds for all $\psi$ with rank at most $r(\varphi)$". The Rank-Induction Principle now reads $$\forall\varphi(\forall\psi\prec\varphi A(\psi) \Rightarrow A(\varphi)) \Rightarrow \forall\varphi A(\varphi).$$
We will now show that the rank-induction principle follows from the induction principle. Let $\forall\varphi(\forall\psi\prec\varphi A(\psi) \Rightarrow A(\varphi))$ be denoted by ($\dagger$). In order to show $\forall\varphi A(\varphi)$ we have to indulge in a bit of induction loading. Put $B(\varphi) := \forall\psi\preceq\varphi A(\psi)$. Now show $\forall\varphi B(\varphi)$ by induction on $\varphi$.

It appears that this part of the proof is showing that rank induction follows from structural induction. But, later in the proof Van Dalen writes

For the converse we assume the premises of the induction principle.

This made me further confused as to the organization of the proof. It seems as if the first section is using structural induction to show $\forall\varphi B(\varphi)$, but then the second part starts off by assuming the premises of structural induction. Are these not proving the same direction (i.e. rank induction follows from structural induction)? Moreover, what motivates the definition of $B(\varphi)$? Doesn't strong induction only require that we show a property holds for all propositions with rank strictly less than the rank of the current proposition $\varphi$? Yet the author uses $\preceq$ when stating the property $B$.

If someone could either carefully trace through the author's proof or provide an alternative proof, it would be greatly appreciated.

答:
Van Dalen's approach is explained in the previous paragraph:

"The introduction of the rank function above is not a mere illustration of the ‘definition by recursion’, it also allows us to prove facts about propositions by means of plain complete induction (or mathematical induction). We have, so to speak, reduced the tree structure to that of the straight line of natural numbers."

Th.1.1.3 is structural induction because it uses the construction tree.

Th.1.1.8 (Induction on rank-Principle) is strong induction based on rank that is a natural number: thus, it is the "usual" mathematical induction.

The equivalence is showed in two steps: the first one concludes that $A(\varphi)$ holds for all $\varphi \in \text {PROP}$ assuming that

"if $A$ holds for all formulas $\psi$ with rank less than that of $\varphi$, then $A$ holds of formula $\varphi$".

This is the antecedent of Th.1.1.8 ["iff for all $\varphi$ [$A(\psi)$ for all $\psi$ with rank less than ..."] and we have to prove the consequent: "then $A(\varphi)$ holds for all ...".

To derive it, the author defines a new property $B$ and shows that $B$ holds of atomic formulas, and that it "propagates" along the construction tree (the two basic cases with connectives $\lnot$ and $\square$).

Thus, using Th.1.1.3 [structural induction], it follows that "$∀\varphi B(\varphi)$, and as a consequence $∀\varphi A(\varphi)$", which is the conclusion of Th.1.1.8 [rank induction].

The second part proves Th.1.1.3 using Th.1.1.8.

So, again, we assume the antecedent of Th.1.1.3: "$A$ holds for atomic formulas and propagates with connectives" and we have to prove the consequent: "$A(\varphi)$ holds for all $\varphi \in \text {PROP}$".

To do this, the author shows that from the assumptions above it follows that the formula ["dagger"] holds. This formulas expresses the antecedent of Th.1.1.8 [rank-induction]; thus, from it and modus ponens we conclude with the consequent of the theorem, that is the consequent of Th.1.1.3 [structural induction].


问:
If for every sequence $(x_n)$, $x_n \to x_0 \implies (f(x_n))$ is a Cauchy sequence, then is $f$ continuous?
Given $f:D \subseteq \mathbb{R} \to \mathbb{R}$, if for every sequence $(x_n)$ in $D$ such that $x_n \to x_0$ is true that $(f(x_n))$ is a Cauchy sequence, then is $f$ continuous?

This is what I know:

Every Cauchy sequence is a convergent sequence.

$f$ is continuous if and only if $x_n \to x_0 \implies f(x_n) \to f(x_0)$ for every sequence $(x_n)$.

However, I only know that $f(x_n) \to a \in \mathbb{R}$; how can I see that $a = f(x_0)$ and then have the continuity of $f$?

If the result is false, please show a counterexample, i.e., a discontinuous function such that for every $x_n \to x_0$ I have $(f(x_n))$ Cauchy.

答: The statement is correct. To see this, let $x_0\in D$ and $(x_n)$ a sequence in $D$ with $x_n\to x_0$. We want to show that $f(x_n)\to f(x_0)$. Define the sequence $(y_n)=(x_1,x_0,x_2,x_0,x_3,x_0,\dots)$. Then $y_n\to x_0$, so by assumption $f(y_n)$ is a Cauchy-sequence. It has the constant convergent subsequence $f(x_0)$, hence $f(y_n)$ converges to $f(x_0)$ (A Cauchy-sequence with convergent subsequence is itself convergent with the same limit). As $f(x_n)$ is a subsequence of $f(y_n)$ it follows that $f(x_n)\to f(x_0)$


问:
If $\Omega$ is open and $D$ is a discrete subset of $\Omega$, is $\Omega\setminus D$ open?
Let $(E,\tau)$ be a topological space and $\mathcal N_\tau(x):=\{N\subseteq E:N\text{ is a }\tau\text{-neighborhood of }x\}$.

If $B\subseteq E$, then $x\in E$ is called

$\tau$-accumulation point of $B$ if $$\forall N\in\mathcal N_\tau(x):B\cap N\setminus\{x\}\ne\emptyset;\tag1$$

$\tau$-isolated point of $B$ if $$\exists N\in\mathcal N_\tau(x)=B\cap N=\{x\}\tag2.$$

Now, in a Wikipedia article it is mentioned, that if $B\subseteq E$ and $x\in E$ is not $\tau$-accumulation point of $B$, then $x$ is an $\tau$-isolated point of $B$. However, I think there is a subtle issue with this: If $x$ is not an $\tau$-isolated point of $B$, then $$\forall N\in\mathcal N_\tau(x):B\cap N\ne\{x\}\tag3.$$ From this, the desired claim is obvious, as long as $x\in B$. But without assuming $x\in B$, I don't see how the claim follows. We need at least to assume that $B$ is nonempty, since otherwise it is obviously wrong.

Question 1: So, is the claim only correct when $x\in B$?

Builindg up on that, $D\subseteq E$ is called $\tau$-discrete if $$\forall x\in E:\exists N\in\mathcal N_\tau(x):D\cap N\subseteq\{x\}\tag4.$$

Question 2: Now, if $\Omega\in\tau$ and $D\subseteq\Omega$ is $\left.\tau\right|_\Omega$-discrete$^1$, can we show that $\Omega\setminus D\in\tau$?

$\left.\tau\right|_\Omega:=\{U\cap\Omega:U\in\tau\}=\{U\subseteq\Omega:U\in\tau\}$.

答:
Firstly I did not see that statement in your linked page. What is mentioned and correct is that if $S$ is a subset of some space and $x \in S$ then either $x$ is an isolated point of $S$ or $x$ is a limit point ($\tau$-accumulation point as you call it) of $S$. So each set has two kinds of points.

To come to the question in the title: No. In $\Bbb R$ we can take $\Omega=\Bbb R$ or $(-1,1)$ and $D=\{\frac{1}{n}: n\in \Bbb N^+\}$ which is discrete but $0$ is not an interior point of $\Omega\setminus D$.

A discrete set need not be closed, as illustrated by this example. If we define a discrete set as a set $D$ such that for all $x \in X$, there is an open neighbourhood $O_x$ such that $O_x \cap D$ has at most one point (which can be $x$ if $x \in D$ or empty), then $D$ is indeed closed and the result holds. One sees such usage in some complex analysis books e.g. Be careful with how discrete is defined. For a topologische it means the subspace topology is the discrete topology, but in analysis the meaning is sometimes the stronger one.


问:
How to show that a formula is not a norm in function space
I have to show that the formula $||g|| = |g(c)|, c \in [a, b]$ ($c$ is fixed) does not define a norm in the space $\mathcal{C}([a, b]$) of continuous functions on the interval $[a, b]$. I find proofs in vector spaces easy, as then the formulas are usually defined in terms of the components of the vectors. But I do not know how to make the jump to function spaces.

Namely here, it is pretty obvious that unless we have a zero constant function, $||g|| \neq 0$. Also the scalar multiplication seem pretty obvious, as for $t \in [a, b]$ we have that $||t\cdot g|| = |tg(c)| = |t||g(c)|$. So the only thing left would be to show that somehow the triangle inequality does not hold with this formula. But expanding $||g + f|| = |(g + f)(c)| = |g(c) + f(c)| \leq |g(c)| + |f(c)|$. So where am I going wrong?

答:
If $g$ is not the zero function, but $g(c)=0$, then we have $||g||=0$.

If we asumme that $|| \cdot||$ defines a norm, then we would obtain $g=0,$ a contradiction.


问:
Planar graph with $|V|\ge3$ has the average degree of vertices smaller than $6$
$G$ is a planar graph with $|V|\geq 3$. How to show that $G$ has an average degree of vertices smaller than 6?

Would it be enough to note that if $G$ has less then 6 nodes the average cannot be over 5 and the existence of connected components further decreases the $\frac{\sum_{v\in V}deg(v)}{|V|}$, so w.l.g. it can be proved for a connected graph (component) and extended to all connected components if needed?

Proof: If $G$ has no cycles it is a tree. $|E|=|V|-1$ and that is smaller than $|V|\leq |V|+2|V|-6 = |3|V-6$ so $$|E|\leq |3|V-6$$
If $G$ has cycles: we note that each face is bounded by 3 edges and the sum of the degrees of the faces has to be at least three times number of all faces: $$\sum_{f\in F}deg(f)\geq3|F|$$
Furthermore, each edge bounds at most two faces so $$\sum_{f\in F}deg(f)\leq2|E|$$

We would have $3|F|\leq2|E|$. Using Euler's formula we get $$3|F|=3|E|-3|V|-6$$ after multiplying it by scalar and combining both we get: $2|E|\geq 3|E|-3|V|-6$ and  $$|E|\leq 3|V|-6$$

Let's suppose that the degree of each node in $G$  is at least 6 (contraposition).
$$\frac{\sum_{v\in V}deg(v)}{|V|} \geq 6 \\  2|E|\geq 6|V| \\ |E|\geq 3|V|$$
But we have just shown that $|E|\leq 3|V|-6$, so $2|E|\leq 6|V|-12$. This two cannot be true at the same time for graph/connected component with $|V|\geq 6$.
If $G$ isn’t connected, the result still holds, because we can apply our proof to each connected component individually.

答:
Your proof is pretty much correct, and on the Wikipedia page for planar graphs , there is a brief explanation which is completely captured in your answer : first we show that $2|E| \geq 3 |F|$ , and then substitute in Euler's formula to get $|E| \leq 3|V|-6$. Then we just multiply by $\frac 2{|V|}$ on both sides, and rearrange to get that the average degree is at most $6 - \frac{12}{|V|}$, and therefore strictly less than $6$.

Now this is true for each connected component, but the average degree of the whole graph is at most the average degree of any connected component (using the pigeonhole principle), so the whole graph has average degree at most $6 - \frac {12}{|V|}$, where $|V|$ is the number of vertices.

Interestingly, we can study the average degree of a random planar graph. Although an upper bound on the degree is $6$, it was shown that if you uniformly sample, from all planar graphs with $n$ vertices, a random graph, then the number of edges of this graph is with high probability concentrated around $2.21n$ (at least for large enough $n$). This implies that the average degree of a uniformly randomly chosen planar graph is very highly concentrated around $4.42$, and $6$ is not really a very tight bound for a lot of planar graphs. On the other hand, planar graphs are also not trees, which have an average degree of at most $2$. So an average planar graph lies somewhere in the middle.


问:
Understanding the set $M_1 \cap M_2$
Let $ M_1=  \left\lbrace \left( \begin{matrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & f \end{matrix} \right), a,d,f \in i\mathbb{R}, b = \overline{c} \right\rbrace $ be a subspace of skew-hermitian matrices.

Let $ M_2 = \lbrace g \lambda g ^{-1}, g \in SU(3)\rbrace $ such that $\lambda = \operatorname{diag}(i\lambda_1,i\lambda_2, i\lambda_3), \lambda_i \in \mathbb{R}.$

My first question is how can we describe the intersection of $M_1$ and $M_2$, in other words what are the matrices $g\in SU(3)$ s.t $g \lambda g ^{-1} \in M_2 $.

My second question is what are the connected components of $M_1 \cap M_2 $ ?

答:
I am assuming $b = -\overline{c}$, so that the matrices in $M_1$ are indeed skew-Hermitian.

The matrices in $M_1$ are precisely the skew-Hermitian matrices with eigenvector $(0, 0, 1)$. Clearly, $(0, 0, 1)$ is an eigenvector of every matrix in $M_1$ (with eigenvalue $f$, as per the form in $M_1$'s definition). Conversely, skew-Hermitian matrices have orthogonal eigenspaces, and so if $\operatorname{span}(0, 0, 1)$ is invariant, then so is $\operatorname{span}((1, 0, 0), (0, 1, 0)) = \operatorname{span}(0, 0, 1)^\perp$. This gives us the required block-diagonal form.

So, we are looking for matrices $g\lambda g^{-1}$ in $M_2$ where one of the columns of $g$ is $(0, 0, 1)^\top$. We can permute the columns of $g$ provided we correspondingly permute the diagonal entries of $\lambda$ and we maintain an even permutation (to keep $\det g = 1$), hence we can assume without loss of generality that the third column of $g$ is $(0, 0, 1)^\top$.

The columns of $g$ are orthonormal, hence the other two columns will be orthogonal to the third column, giving us the familiar form
$$g = \begin{pmatrix} p & q & 0 \\ r & s & 0 \\ 0 & 0 & 1 \end{pmatrix},$$
where $\begin{pmatrix} p & q \\ r & s\end{pmatrix} \in SU(2)$ (note that the determinant must be $1$, by simple calculation, so that $\det g = 1$). This seems like a fairly complete characterisation of the matrices $g$.

As for the connected components, the set is connected. The map
\begin{align*}
\Bbb{R}^3 \times SU(2) &\to M_1 \cap M_2 \\
(\lambda, h) &\mapsto \begin{pmatrix} h & \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 & 0 \end{matrix} & 1 \end{pmatrix} \begin{pmatrix} i\lambda_1 & 0 & 0 \\ 0 & i\lambda_2 & 0 \\ 0 & 0 & i\lambda_3\end{pmatrix}\begin{pmatrix} h & \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 & 0 \end{matrix} & 1 \end{pmatrix}^{-1}
\end{align*}
is continuous and surjective, making $M_1 \cap M_2$ connected.

Title: Understanding the set $M_1 \cap M_2$


问:
Harmonic Series Divergence Proof
I was given a proof of the divergence of the harmonic series, but I can't fully understand it (I found other ones that I find exhaustuve, so the problem is not trying to understand why the series diverges)

Here is the proof:

Having: $\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\gt\frac{3}{x}$ for $x\gt1$

Let be $$S=\sum_{n=1}^\infty \frac1n\lt\infty$$

$$S=1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+...$$
$$S = 1+\frac{1}{3-1}+\frac{1}{3}+\frac{1}{3+1}+\frac{1}{6-1}+\frac{1}{6}+\frac{1}{6+1}+...$$

Now, using the previous inequality

$$S\gt1+3(\frac13+\frac16+\frac19+...) =1+1+\frac12+\frac13+...=1+S$$

So we get at the end $S\gt1+S$, as a consequence S can't be finite.

My question is if it is possible to treat an infinite sum like that.

Thanks in advance.

答:
Though clever, this manipulation is not rigorous. A series converges if the limit of the partial sums converges. If we consider the $7^{\text{th}}$ partial sum, like you did,
we do have

$$1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17>1+\frac33+\frac36=1+1+\frac12$$

but this is not

$$S_7>1+S_7.$$ Instead, it establishes $$S_7>1+S_2$$ which generalizes to
$$S_{3n+1}>1+S_n$$
and this not enough to prove divergence.

The situation is different with the classical proof by groups of doubling sizes,

$$1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18>1+\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\frac18
\\=1+\frac12+\frac12+\frac12$$

or

$$S_9>1+\frac{3}2,$$ which generalizes to $$S_{2^n+1}>1+\frac{n}2$$

or

$$S_m>1+\frac{\log_2(m-1)}2.$$

Title: Harmonic Series Divergence Proof


问:
Focal chords of ellipse
The feet of the perpendiculars from the foci S and S' ,on to the tangent to an ellipse at a point P,are Y and Y', and YQ and Y'Q' are drawn to touch the ellipse in Q and Q'. To show that QS and Q'S' meet the ellipse again at the end of the diameter through P.
I know that the feet of the tangents lie on the auxiliary circle of the ellipse, but cannot use this to prove the result.Could we use affine tramsformations here? Any help would be appreciated.</p>

答:
Let $U,V$ be the intersections of the ellipse with $PS$ and, respectively, $QS$ and $X$ be the pole of $UV$.

The quadrilater $PQUV$ is inscribed in the ellipse, hence the points
\begin{align}
PQ\cap UV\\
PU\cap QV\\
PV\cap QU
\end{align}
are conjugated each other respect to the ellipse.
Consequently, the line $(PU\cap QV)(PQ\cap UV)$ is the polar of $PV\cap QU$.
Moreover, $S=PU\cap QV$ and $R=PQ\cap UV$ is the pole of $XY$, hence  $S$ belongs to $XY$ and the lines $PY$, $VX$ and $SR$ meets at the pole $T$ of $PV$.
Since $S$ is a focus for the ellipse, we have $SR\perp SY$, hence $SR$ and $PY$ are parallel, hence $T$ is at infinity.
This proves that $PV$ is a diameter of the ellipse.

Title: Focal chords of ellipse


问:
Some calculus Prove/Disprove question
I have 2 questions that I would like to make sure I did the first one right, and get some tips for the next one :)

Prove or disprove:

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function. If the equation $f'(x)=0$ has exactly one solution, then the equation $f(x)=0$ has at least two solutions.

I think this one is false, because I can take $f(x)=x^2$. It is clear that the function is differentiable, and $f'(x)=2x$, so the equation $f'(x)=0$ has only one solution as needed, when $x=0$.

And the equation $f(x)=0$ has only one solution, when $x=0$, which is in contradiction to the fact that $f(x)=0$ has at least two solutions.

--

Let $f:[a,b) \to \mathbb{R}$ be a differentiable function such that $\lim_{x \to b^-}f(x)$ does not exist in the extended sence.
Then  there exists an $x_{0} \in [a,b)$ for which $f'(x_{0})=0$.

So I have tried to think for functions that I know which for their limits does not exist, like $\cos (\frac{1}{x})$, and started playing with them, but no matter what I did, $f'$ always had a point $x_{0}$ which for $f'(x_{0})=0$. So I think that this statement is true, but if you guys can please enlighten me, I will very appreciate that!

Thanks a lot!

答:
Hint: if $f'(x) \neq 0$ on $[a,b),$ then $f'(x)>0$ or $f'(x)<0$ on $[a,b).$
What can you conclude from this information?

(We don't implicitly require $f'$ to be continuous here! See theorem 5.12 in Rudin's Principles.)

Edit:
Suppose that $f'(x)>0$ on $[a,b).$ Then by the mean value theorem, $f$ is increasing. It follows that $\lim_{x \to b^-} f(x) = \sup_{x \in [a,b)} f(x),$
so the limit exists in the extended sense. The case $f'(x) < 0 $ is similar. This proves the contrapositive of the statement.

Title: Some calculus Prove/Disprove question


问:
Validity of Jordan's lemma
I'm trying to calculate the following integral

$$ \int_0^\infty dx \, e^{ix}$$

with contour integration. By choosing a quarter circle contour $\gamma$, by the residue theorem we get

\begin{align}
0 =\oint_\gamma dz \, e^{iz} =& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i\int_0^\infty dy \, e^{-y} \\
=& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i
\end{align}

therefore, if the integral over the arc is zero, the original integral becomes $i$.

Can Jordan's lemma, of a variation thereof, be applied in this case?

Edit: the parametrization would be $z=R e^{i\theta}$ with $\theta \in [0, \frac{\pi}{2}]$.

Aside: it seems to me that the original integral can be interpreted as a distribution as the Fourier transform of the theta function at $k=1$, which would give $i$ as well.

答:
The answer to my question is that, as pointed out by Nico Terry, the integral does not converge. However, it can be assigned a value by analytic continuation as follows

$$ f(z)=\int_0^\infty dt \, e^{zt} = -\frac{1}{z}, \qquad \mathrm{for\, Re}(z) < 0 $$

so we can analytically continue $f(z)$ over all $\mathbb C -\{0\}$. Hence $f(i)=i$ means that we can assign the value $i$ to the original integral.

Title: Validity of Jordan's lemma


问:
proof that the sequence satisfies weak law of large numbers
strong textHow to check if this sequence satisfies the weak law of large numbers?
$$P(X_n=1)= {{1}\over{3^n}} ,P(X_n=-1)= {{1}\over{3^n}}, P(X_n=0)= {1 -{{2}\over{3^n}}}  $$
I calculated that:
$$E(X_n)=0 $$
so according to the definition of this law: $${\sum_{n=1}^{N}{(X_n-0)}}\over{N} $$and from this one we can deduce that this expression converges to 0 so the sequence satisfies the weak law of large numbers. Is that correct?

答:
Simply using the definition of WLLN

$$\lim\limits_{n\to +\infty}\mathbb{P}[|X_n|>\epsilon]=\frac{2}{3^n}=0$$

this means that

$$X_n\xrightarrow{\mathcal{P}}0$$

This if $\epsilon$ is chosen $|\epsilon|\leq1$. If it is chosen >1 the probability is exactly zero. Anyway the sequence converges.


问:
exponential functions and calculus
Assume the rate of change of a quantity T is proportional to the difference between the quantity T and some fixed constant A.
That is
dT/dt = K(T-A)
show that T=A+Ce^(kt) (where C is a constant) is a solution to this differential equation.
I dont really know what the question is asking me to even try and begin to attempt it, if anyone can help me out that would be great thanks :) </p>

答:
Rewrite $\frac{dT}{dt}=K(T-A)$ as follows:

$$\frac{dT}{(T-A)}=K dt$$

Or equivalently,

$$d(\ln (T-A))=K dt$$

Now, integrate both sides to get the desired result.

ContentLicense: CC BY-SA 3.0

Title: exponential functions and calculus


问:
Big list: collecting results of diagram chasing
I plan giving a homework problem on linear algebra for my students
and my idea was to collect many results of diagram chasing.

Could you help me and give me some results of diagram chasing you know? They can be very basic or they can be more difficult. I think of results such as the "Five lemma".

Remarks

PS:
So far, they know all the basic stuff of linear algebra (vector space, linear maps, kernels, images, etc.). They don't know what is a finite-dimension space but they know what is a free/spanning finite families and finite basis.

PPS: They don’t know what is the quotient $E/F$... but maybe this can be overpassed by using subspaces $G$ such that $E = F \oplus G$.

答:
It's a bit late, but here's another interesting one:

If in the short five lemma, you replace "is an isomorphism" by "is $0$", you get a wrong statement, but if you put two short exact sequences on top of one another and only ask for the composite to be $0$, then it becomes true. More precisely :

Suppose we have a commutative diagram $$\require{AMScd}\begin{CD}0 @>>>A @>>> B @>>> C @>>> 0\\
@VVV @VVV@VVV@VVV@VVV\\
0 @>>>D @>>> E @>>> F @>>> 0\\
@VVV @VVV@VVV@VVV@VVV\\
0 @>>>H @>>> K @>>> L @>>> 0\end{CD}$$
with exact lines, and suppose $A\to D, C\to F, D\to H$ and  $F\to L$ are $0$. Then the composite $B\to E\to K$ is zero. Find an example where $B\to E$ isn't.

(Note : naively you would expect there to be no counterexamples if the short exact sequences split, so you would expect it to hold with just one map on vector spaces, but even then it doesn't)

It's an instructive diagram chasing argument, and finding a counterexample is instructive as well - moreover, it is related to deep mathematics.


问:
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.

Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$

Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:

\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}

\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}

\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}

However, I am not sure if I see a pattern here that can help solve the question.

Body:
$$
f(x) = x^{f(x)}\\
\ln(f(x))=f(x) \ln(x)\\
\frac{1}{f(x)}f'(x) = \frac{f(x)}{x} + \ln(x)f'(x),\\
f'(x)=\left(\frac{1}{f(x)}-\ln(x)\right)^{-1}\frac{f(x)}{x},\\
f'(x)=\left(\frac{f(x)}{1-\ln(x)f(x)}\right)\frac{f(x)}{x}.
$$


答:
$$x^{x^{x^\cdots}}= y \implies x^y = y \implies1 = yx^{-y} = ye^{-y\log (x)}$$

which means

$$y = \frac{-W(-\log (x))}{\log (x)}$$

where $W$ is the principal branch of the Lambert-W function. Now, the problem of the successive derivatives is "much" simpler (have a look here for the derivatives of $W(t)$).

Title: Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.

Body:
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$

Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:

\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}

\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}

\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}

However, I am not sure if I see a pattern here that can help solve the question.


答:
Use the method of first order linear differential equations:

$$\frac{dT}{dt}+p(t)T=q(t)$$

$$T(t)=\frac{\int u(t)q(t)dt+C}{u(t)}$$

where

$$u(t)=e^{\int p(t)dt}$$

Title: exponential functions and calculus

Body:
Assume the rate of change of a quantity T is proportional to the difference between the quantity T and some fixed constant A.
That is
dT/dt = K(T-A)
show that T=A+Ce^(kt) (where C is a constant) is a solution to this differential equation.
I dont really know what the question is asking me to even try and begin to attempt it, if anyone can help me out that would be great thanks :) </p>


问:
Finding the smallest $n$ that satisfies $\left(\frac{1-i}{\sqrt{2}}\right)^n=1$
Maybe I'm just stupid, but right now I'm in a dilemma so please help. I had the question

$$\left(\frac{1-i}{\sqrt{2}}\right)^n=1$$ where I need to find the smallest number for $n$ ($n> 0$) that would satisfy the problem.

My first step was to $$\left[\left(\frac{1-i}{\sqrt{2}}
\right)^2\right]^\frac{n}{2}=1$$ and than $$(-i)^\frac{n}{2}=1$$ after that just ignore the minus because its powered by a even number and so on so forth $$i^\frac{n}{2}=1$$ $$\frac{n}{2}=4$$the  answer is $$n=8$$. But here is where the dilemma came if $$-i=i^3$$ than it would be $$\left(i^3\right)^\frac{n}{2}=1$$ which turns to $$i^\frac{3n}{2}=1$$ and the $$\frac{3n}{2}=4$$. This seems alright in the head but than the answer turns out to be $n=\frac{8}{3}$.

I used latex. Sorry if bad latex i'ts my first time.

If you know the answer please please tell me.

By the way I'm 17 and dumb. So please don't hate

答:
Recall how two complex numbers are multiplied: the length of the product is the product of the length as the product's argument, (https://en.wikipedia.org/wiki/Argument_(complex_analysis), is the sum of the factors' arguments.

Considering $z=(1-i)/\sqrt2$; it has length $1$, any power of $z$ has length $1$ two.  As the argument of $z$ is $-45^\circ$ we'll need the eighth power of $z$ to get $1$.


Body:
Let $f_1(x)=x$ and $f_n(x)=x^{f_{n-1}(x)}=x^{x^{\ldots^{x}}}$ ($n$ $x's$ on the exponent).
Then $\ln(f_n(x))=f_{n-1}(x)\cdot\ln(x)$ so: $$\frac{f'_n(x)}{f_{n}(x)}=f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\implies f'_{n}(x)=f_{n}(x)\cdot\left(f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\right).$$


问:
Transpose of a linear map
Q. Let $T:V \to U$ be linear and $y\in U$. If $f\in V^*$ such that $f(y)=1$, then $T^t(f)=\rule{1.5cm}{0.15mm}$

This question was asked in my assignment test.

My Doubt-
I know that $T:V \to U$ is a linear transformation then the transpose is $T^t: U^* \to V^*$,where $U^*$ and $V^*$ are dual spaces of vector spaces $U$ and $V$ respectively.
So $T^t$ should take values from $U^*$ but in the question $T^t(f)$ is required. I guess the question is wrong. Can someone please verify if the question is correct?

答: You are right.  $T^t$ is defined on $U^*$. If $U^* \ne V^*$, then $T^t(f)$ makes no sense for $f \in V^*.$

Title: Transpose of a linear map


问:
Uncountable Disjoint Union of $\mathbb{R}$ and Local Euclideanness
I'm working on Lee: Introduction to Smooth Manifolds problem 1-2 which asks to prove that the uncountable disjoint union of copies of $\mathbb{R}$ is locally euclidean, and just wanted to verify/disprove a thought I had regarding the question.

I've thought about the construction for some time and have concluded that it's equivalent to $\mathbb{R}$ equipped with the disjoint topology $\times$ $\mathbb{R}$ equipped with the usual topology.

From that construction, would it be safe to assume that at each point, we could construct a local open neighborhood that is homeomorphic to an open subset of $\mathbb{R}^2$? My reasoning is that the disjoint topology implies that each disjoint point is an open set, and then an open neighborhood of a point $(i, r)$ would be akin to $(i, (r-\epsilon, r+\epsilon))$ (where $i$ belongs to the copy of $\mathbb{R}$ equipped with the disjoint topology and $r$ belongs to $\mathbb{R}$ equipped with the usual topology). This can then be mapped by a homeomorphism to an open neighborhood of $\mathbb{R}^n$ by way of bases (not 100% sure how to deal with $i$ in this case though, as it is a singleton).

Is the idea correct, or have I made an error somewhere?

答:
From that construction, would it be safe to assume that at each point, we could construct a local open neighborhood that is homeomorphic to an open subset of $\mathbb{R}^2$?

No.

Consider $X=\bigsqcup_{i\in I}X_i$ for some (uncountable or not) set $I$, where each $X_i\equiv\mathbb{R}$ is just a copy. Formally $X_i=\{i\}\times\mathbb{R}$ and $X=\bigcup_{i\in I}X_i$. The topology on disjoint union is defined as follows: $U\subseteq X$ is open if and only if $U\cap X_i$ is open in $X_i$ for every $i\in I$. Formally, the topology on $X_i$ is given by the product topology and thus $X_i$ is homeomorphic to $\mathbb{R}$ via $(i,r)\mapsto r$.

In particular each $X_i$ is open in $X$. Therefore we can construct a straight forward atlas: each point $x\in X$ belongs to exactly one $X_i$ and thus the homeomorphism we are looking for is the trivial $X_i\to\mathbb{R}$ homeomorphism:
$(i,r)\mapsto r$. This shows that $X$ is of dimension $1$, even though it can be arbitrarly big as a set.


问:
Let $f: X \to Y$ be a function, and let A be a subset of X, B is a subset of Y. Find examples which show that the following statements are false.
I have a couple of questions on elementary set theory.
Q1: Let $f: X \to Y$ be a function, and let $A$ be a subset of $X$, $B$ is a subset of $Y$. Find examples which show that the following statements are false.

a. If $B$ is not equal to an empty set, then $f^{-1}(B)$ is not equal to an empty set.

b. $f^{-1}(f(A))=A$

c. $f(f^{-1}(B))=B$

d. $f(X)=Y$

i am having a problem from a.
A solution says, if $f(x)=1$, $B=\{2\}$, $f(B)=1$. For all $x$, $f^{-1}(B)=\emptyset$.
why $f^{-1}(B)$ is an empty set?

Q2:
Find $f^{-1}(\{0,1\})$, $f^{-1}(\{-3,3\})$, and $f^{-1}(\{1,2,3\})$ from below function (venn diagram).
To descrbie the venn diagram in words..

(domain) Set $X={x,y,z,v,u}$

(range) Set  $Y={a,b,c,d,e}$

$\{x\} \to \{d\} \\
\{y\} \to \{e\} \\
\{z\} \to \{a\} \\
\{u,v\} \to \{b\}$

I am having trouble finding the original elements.
this is what I found but i am not sure if they are right:
$f^{-1}(\{0,1\})=\{\{\pm \sqrt{2}\}, \{\pm \sqrt{3}\}\}\\
f^{-1}(\{-3,3\})=\{\{\pm \sqrt{5}\}\}\\
f^{-1}(\{1,2,3\})=\{\{\pm \sqrt{3}\}, \{\pm 2\}, \{\pm \sqrt{5}\}\}$

Q3: Let $X=\{x,y,z\}$, and $Y=\{1,2,3\}$. Which of the following constitute functions from $X$ to $Y$? If they do not, give the reason.

d) $i=\{(x,1),(x,2),(y,1),(z,3)\}$
It is not a function from $X$ to $Y$ because there two elements in the domain that have different ranges (different y-values).  (would it be correct to say this? Also, if the two elements $(x)$ had the same $y$ value, let's say 1, --> $(x,1) \ \text{&} \  (x,1)$,
can i say that this is a function from X to Y?

Thank you very much for your insights!

答:
To just do the first:

Let $X=\{a,b,c\}$, $Y=\{1,2,3\}$, and let the map be $a \to 1, b\to 1, c \to 2$.

Then $f[X]=\{\{1,2\} \neq Y$.

For $A=\{a\}$, $f[A]=\{1\}$ but $f^{-1}[f[A]]= f^{-1}[\{1\}]= \{a,b\} \neq A$.

For $B=\{3\}$ we have that $f^{-1}[B]=\emptyset$

For that same $B$ we also have that $f[f^{-1}[B]]=\emptyset \neq B$ as well.
Also for $B=Y$ we have that $f[f^{-1}[B]]=f[X]\neq B$.

Hope this helps.


答:
All the other answers are for the case where there are infinite x's. However, there are finite ($n$) x's in this problem!

Let's first denote the function as $f_n(x)=x^{x^{\cdots^x}}$ with $n$ x's in the exponent. For example, $f_1(x)=x^x, f_0(x)=x.$

Take the logarithm of both sides, we get $\ln f_n(x)=f_{n-1}(x)\ln x$. Taking the derivative gives $$\frac{f'_n(x)}{f_n(x)}=f'_{n-1}(x)\ln x+\frac{f_{n-1}(x)}{x}$$

$$f'_n(x)=f_n(x) f'_{n-1}(x)\ln x+\frac{f_n(x) f_{n-1}(x)}{x}$$

That's just a basic recurrence relation. We know that $$a_n=Ca_{n-1}+D\implies a_n=c C^{n-1} + \frac{D(C^n-1)}{C-1}$$ where $c$ is an arbitrary constant.

Similarly, we may get $$f'_n(x)=c(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$

To get the value of the parameter $c$, let $n=1$, then $f'_1(x)=c+x^x$. Hence, $c=x^x \ln x$. Substituting in, we get the final answer (in a rather cumbersome form) $$f'_n(x)=x^x \ln x(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$


问:
Check the following statements.
Let $X=\{(x_1,x_2,\dots):x_i\in\Bbb{R}$ and finitely many $x_i$'s are non zero} and $d:X\times X\to\Bbb{R}$ be a metric on $X$ defined by $d(x,y)=\text{sup}_{i\in\Bbb{N}}|x_i-x_j|,x=(x_1,x_2,\dots),y=(y_1,y_2,\dots) $in $X$.

Consider the following statements:

$(X,d)$ is complete metric spaces.

The set $\{x\in X:d(\overline{0},x)\leq 1\}$ is compact, where $\overline{0}$is the zero elements of $X$.

I have to check that whether the statements are true or false.

consider the sequence $e_n,n\geq 1$, where $e_1=(\delta/2,0,0,...), e_2=(0,\delta/2,0,0,...)$ and so on.
clearly this sequence is Cauchy, but how to show that this question is not convergent? I am having difficulty for writing in my notebook.
help me with both statements.
Thanks|

答:
Consider the sequence $(e_n)_{n\in\Bbb N}$ of elements of $X$ defined by:

$e_1=(1,0,0,0,0,\ldots)$;

$e_2=\left(1,\frac12,0,0,0,\ldots\right)$;

$e_3=\left(1,\frac12,\frac13,0,0,\ldots\right)$;

and so on. Then$$d(e_m,e_n)=\begin{cases}0&\text{ if }m=n\\\frac1{\min\{m,n\}+1}&\text{ otherwise,}\end{cases}$$and therefore it is a Cauchy sequence. But it doesn't converge in $X$. If it did and if $e=(a_1,a_2,a_3,a_4,\ldots)$ was it s limit, then, for each $m\in\Bbb N$, $a_m$ would be the limit of the sequence of the $m$th terms of the sequence $(e_n)_{n\in\Bbb N}$, which means that $a_m=\frac1m$. But then $e\notin X$.

And $\left\{x\in X\,\middle|\,d\left(\overline0,x\right)\leqslant1\right\}$ is not compact, since it contains a sequence without a convergent subsequence, which is $(e_n)_{n\in\Bbb N}$.

Title: Check the following statements.


问:
Natural numbers to prime exponents
I am currently teaching a class for middle school students who are interested in mathematics and we were talking about Hilbert's Hotel and how you could fit $\mathbb{N}\times\mathbb{N}$ guests into the Hotel, i.e. the goal is to show that $\mathbb{N}\times\mathbb{N}$ is countable.
One of the students came up with the idea of defining the following map:
$$ f: \mathbb{N}\times\mathbb{N}\to\mathbb{N},\; (n,m)\mapsto n^{p_m}$$
where $p_m$ is the m-th prime number. While this is not injective on the whole domain, I have the feeling that it should be injective when restricting to $\{n\in\mathbb{N}\;:\; n\ge2\}\times\mathbb{N}$, which with some extra work should be enough to prove countability.
Since I am not very good at number theory, I still failed to show this restricted injectivity.
It is easy to see that for pairs $(n,m)$ and $(k,m)$ the map is injective and the same holds for pairs $(n,m)$ and $(n,l)$ but for arbitrary pairs $(n,m),\; (k,l)$ one would have to show that
$$n^{p_m} = k^{p_l}\Rightarrow (n,m) = (k,l).$$ Is there an easy way to show this?

答:
Suppose $n^{p_m} = k^{p_l} \tag{1}$

If we assume there exists a prime number $q$ such that $q$ divides one of $n, k$ but not the other, then obviously (1) cannot be true.

So it follows that $n,k$ have the same prime factors $q_1, q_2, ... q_s$

Let their respective powers in $n,k$ be $a_1, a_2, ..., a_s$ and $b_1, b_2, ..., b_s$

$$n = q_1^{a_1}q_2^{a_2}...q_s^{a_s} \tag{2}$$

$$k = q_1^{b_1}q_2^{b_2}...q_s^{b_s} \tag{3}$$

Then from $(1)$ we get that $a_1 \cdot p_m = b_1 \cdot p_l \tag{4}$

But from here it does not follow that $a_1 = b_1$, we can take e.g. $a_1 = p_l$ and $b_1 = p_m$ and satisfy $(4)$.

So... counter example:

$n=2^{p_{15}} 3^{p_{15}}$

$k=2^{p_{20}} 3^{p_{20}}$

$m=20$

$l=15$

Now $(n, 20)$ and $(k, 15)$ map to the same number.
But these two pairs are not equal.

So... it seems to me this mapping that the student defined is not injective.

Please verify the logic of this proof because I came up with it and wrote it quite quickly. But yeah, I believe it's valid, I don't see a flaw in it.

Title: Natural numbers to prime exponents


问:
Given integers $a$ and $b$ prove that if there exist integers $x$ and $y$ for which $ax + by = (a, b)$, then $(x, y) = 1$.
I know that the answer involves Bezout's theorem in some way.
I tried this:

Let $(a, b) = d$
Let $a = dk_1 and b = dk_2$
So, $ax + by = d$ becomes $dk_1x + dk_2x = d$
Dividing by d, we get:
$k_1x + k_2x = 1$
So, $(x,y) = 1$
Is this sufficient?

答:
I would prove it in the following way:

Let $(a,b)=d$. By definition, there exists $q_1,q_2$ such that $a=dq_1, b=dq_2$. Then, by hypothesis, $ax + by = dq_1x + dq_2y = d$. As we are in an integral domain, we can simplify the expression to $q_1x + q_2y=1$. Now, notice that you directly concluded that (x,y)=1. This is only true in the case that a linear combination of integers is equal to 1 (or unity, if we were working with polynomials in $\Bbb K[x]$).

By definition of gcd, we say $d'=(x,y)$. Then, $d$ has to divide 1, because $q_1x + q_2y=1$. Hence, the only possibility is $(x,y)=1$.


问:
check for incompressibility in sin and cos
Given $\mathbf{u}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}$ check that it satisfies the condition:

$\nabla\cdot\mathbf{u}=0$

I tried approaching it likeso:

$\frac{dx}{dy}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}=\frac{dx}{\sin{x}\cos{x}}=\frac{dy}{\cos{x}+\sin{y}}$

Am I calculating this correctly? I'm trying to check whether $\mathbf{u}$
is incompressible?

答:
Given $\mathbf{u}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}$ check that it satisfies the condition:
$\nabla\cdot\mathbf{u}=0$.

Note that $$\color{blue}{\boxed{\nabla:=\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right)}}$$
and $$\mathbf{u}=(\sin(x)\sin(y),\cos(x)\cos(y))$$
Then,
\begin{eqnarray*}
\nabla\cdot \mathbf{u}&=&\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right)\cdot (\sin(x)\sin(y),\cos(x)\cos(y))\\
&=&\frac{\partial}{\partial \color{red}{x}}\left(\color{red}{\sin(x)}\sin(y) \right)+\frac{\partial}{\partial \color{red}{y}}\left(\cos(x)\color{red}{\cos(y)} \right)\\
&=&\cos(x)\sin(y)\color{red}{-}\sin(y)\cos(x)\\
&=&0
\end{eqnarray*}

Therefore, $\nabla \cdot \mathbf{u}=0$.


问:
A variant of Russian Roulette
Rules: When it is your turn you can either shoot yourself or shoot your opponent and then shoot yourself. When your turn ends you spin the cylinder and hand the gun to the other player. The cylinder holds 6 rounds.

How should you play the game so that you have the greatest chance of survival?

This is my attempt at the problem:

I’m going to cover only some cases.

Case 1: You start and only shoot yourself but your opponent always shoots you then himself when it’s his turn.

Then at the end of each round you have a 1 - (5/6)^2 = .306 probability of death and your opponent has a 1/5 = .2 probability of death (you shoot yourself (1/6) then opponent spins cylinder and shoots you (1/6) then opponent shoots himself (1/5)).

In this case your opponent has a greater chance of “beating you” at the game.

Case 2: Your opponent starts and you only shoot yourself but your opponent always shoots you then himself when it’s his turn.

The round-end probabilities here are the same as in Case 1. Your opponent has an advantage still, but less of an advantage than in Case 1 because he could shoot himself at the start of the game on the first shot.

Case 3: You start and always shoot opponent then yourself. Your opponent always shoots you and then himself.

Then at the end of each round the probability of death is the same for you as it is for your opponent, so you have an advantage since you could kill your opponent in round one when you start the game by shooting him first.

Case 4: Opponent starts and always shoots you then himself when it’s his turn. You always shoot opponent then yourself when it’s your turn.

Round-end probabilities are the same and you have a disadvantage since your opponent could shoot you on the first shot in the first round.

So it seems that it is always in your best interest to shoot your opponent and then yourself when it is your turn. Using that method, you could only be more likely to lose than your opponent IF your opponent starts off the game. This would be a great question for mathstackexchange. I might ask it there and get back to you because now I’m curious.

答:
Let $P_n$ be the probability of the first person surviving, when there are $n$ bullets, if both participants always shoot at the opponent first. Then $P_0=P_1=1$.

For $n\ge2$,

$$P_n=\frac{1}{n}+\frac{n-2}{n}(1-P_{n-2})$$

and so $$nP_n+(n-2)P_{n-2}=n-1.$$

From this it is easy to see that $nP_n$ is the non-decreasing sequence $0,1,1,1,2,3,3.$

If the first deviation from this policy occurs with $n$ bullets remaining then the person making that choice changes their probability of survival from

$$\frac{1}{n}+\frac{n-2}{n}(1-P_{n-2})$$ to
$$\frac{n-1}{n}(1-P_{n-1}).$$

This is a change of
$$\frac{1}{n}((n-2) P_{n-2}-(n-1) P_{n-1}  ).$$

Since this change is never positive, there is no (mathematical) reason to not shoot at the opponent first.

The general case

This analysis applies irrespective of how many bullets are initially used. In particular, the pattern $$0,1,1,1,2,3,3,3,4,5,5,5,6,7,...$$ means that the probability of survival for the first person is precisely $\frac{1}{2}$ when $n$ is even. For an odd number of bullets the probability alternates between being above and below $\frac{1}{2}$ and tends to $\frac{1}{2}$ as $n$ tends to infinity.

Title: A variant of Russian Roulette


问:
$M:=\{(x,y)\in \Bbb R^2: x\cdot y=0\}$ is not a submanifold
I observe the following set:
$$M:=\{(x,y)\in \Bbb R^2: x\cdot y=0\}$$
$M$ is nothing else than the intersection of the $x$ and the $y$ axis.The question is, can this set be a submanifold of $\Bbb R^2$?  I guess no for the point in the origine i.e. but I can't help me to give some better explanation. As hint I could look at the tangent space but I still don't get.

答:
Firs of all, that set is not the intersection of the $x$-axis and the $y$-axis; its their union.

If $V$ is a neighborhood of $(0,0)$ on $M$, then $V\setminus\{(0,0)\}$ has, at least, four connected components. If $M$ was a manifold, then it would either be $1$-dimensional, in which case $(0,0)$ would have a neighborhood such that, if you remove $(0,0)$ from it, you get two connected components, or it would be $n$ -dimensional with $n>1$, in which case $(0,0)$ has a neighborhood such that, if you remove $(0,0)$ from it, you still get a single connected component.


问:
Big Oh: T(n) = O(f(n)), G(n) = O(h(n)). Is T(G(n)) = O(h(f(n)))?
$T(n)= O(f(n))$ , $G(n)= O(h(n))$

How would I prove or disprove:

$  T(G(n))= O(h(f(n))$

I think this is false, because it should be $O(f(h(n)))$ instead of $O(h(f(n)))$, since G is applied before T is applied, I tried substituting polynomial functions for T and G, I think the order matters, ($n^2)!$ is not equal to ($n!)^2$ , but I am not sure if this reasoning is correct?

Body:
This is false.
Let:
$$T(n) = 2^n = O (2^n) \\ G(n) = \lg (n!) = O(n \lg n)$$

$$T(G(n)) = 2^{\lg (n!)} \equiv n! \neq O(2^n \lg 2^n) = O(n \cdot 2^n)$$


问:
prove that inf $f\geq 0$ if and only if $ae^{1+b}\geq 1$.
Ive got a function $f(x)=a+bx+xlnx$ for all $x\in(0,\infty)$ and I need to prove that inf $f\geq 0$ if and only if $ae^{1+b}\geq 1$.

I have already proved that $f$ is convex and have found a formula for inf, inf $f=$max$\{b\in \mathbb{R}: v\leq x$ for all $x\in f\}$ but I am unsure how to implement the question to show what is required. Any help will be great.

答: We want $(b+\ln x)x\geq -a$ for all $x >0$  which means  $\inf \{(b+\ln x)x: x>0\} \geq -a$. By differentiation we see that the infimum is attained at $x=e^{-1-b}$. [The function is decreasing up to this point and increasing after that]. Thus the condition  we need is $(b+(-1-b)) e^{-1-b} \geq -a$ which says $ae^{1+b } \geq 1$.


问:
Calculate this limit without L'Hopital's rule
I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:
$$
\lim_{x \to 0} f(x) = (1+\frac{x}{2})^{\frac{1}{x}}
$$
I know how to make it using L'Hopital and that the result is $ e^{\frac{1}{2}} $ ,but I'm getting nowhere when I try without it. Any advice?

Body:
If $$L=\lim_{x \to a} f(x)^{g(x)}\to 1^{\infty}$$
then
$$L=\exp[\lim_{x\to a} [g(x)(f(x)-1))]$$
So here
$$L=\exp[\lim_{x\to 0}\frac{1}{x}(1+\frac{x}{2}-1)]=e^{1/2}.$$


答:
This is a standard $1^{\infty}$ indeterminate form.
A useful result for such limits is:
$$ \lim_{x \to a} ~(1+f(x))^{g(x)} = e^{\lim_{x \to a}f(x) \cdot g(x)} ~ \text{if} ~ \lim_{x \to a}f(x) = 0 ~\text{and} \lim_{x \to a}g(x)=\infty $$
In your question, $f(x) = \dfrac{x}{2}$ and $g(x)=\dfrac1x$
$$ \therefore \lim_{x \to a} \left(1+\dfrac{x}{2}\right)^{\frac1x} = e^{\lim_{x \to a} \frac{x}{2}\cdot\frac1x} = e^{\frac12}$$

Title: Calculate this limit without L'Hopital's rule

Body:
I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:
$$
\lim_{x \to 0} f(x) = (1+\frac{x}{2})^{\frac{1}{x}}
$$
I know how to make it using L'Hopital and that the result is $ e^{\frac{1}{2}} $ ,but I'm getting nowhere when I try without it. Any advice?


问:
Changes in the price of Apples and Bananas 2

The change in the price of Apples each month is normally distributed with a standard deviation of 10 cents. The change in the price of Bananas each month is normally distributed with a standard deviation of 20 cents. You can assume that there is only a single Apple price that all apples follow, and there is a single Banana price that all bananas follow. These monthly price changes of Apples and Bananas have correlation of 1.0. Assuming these facts hold true going forward, if the price of Apples goes up by 10 cents, what would the change in the difference in price between Bananas and Apples in cents?

My first reaction to this is how can you tell exactly? Because the price difference will be distributed like

$$B_t - A_t \sim N(\mu_X+\mu_Y, 0.1^2+0.2^2)$$

So it must have something to do with the correlation being equal to 1, which gives us this

$$0.02 = Cov(A_t,B_t)$$

I cannot see where to go from here. I dont see how we can that we are certain that the price of Bananas would go up 20 cents just because the correlation is 1. Even if we know that the correlation is 1 we dont know by how much the variable will go up.

答:
With a perfect correlation of $1$, with probability $1$  you can say there is a linear relationship where  $B_t=c+dA_t$ for some $c$ and $d$.

From the ratio of the standard deviations, you can say $d=\frac{20}{10}=2$.

You can then say this implies $B_t-A_t=c+(d-1)A_t$ and that $d-1=1$

This suggests that if the price of Apples goes up by $10$ cents, then the difference in price between Bananas and Apples also goes up by $10$ cents.


问:
What are the applications of nilpotent elements/nilpotent ideals?
As I am doing exercises related to group and ring theory I constantly see questions regarding nilpotent elements/ideals/groups. However, I have yet to see any practical use of them in theory, but I imagine they are used for something since I constantly do exercises about them. What would be some useful applications of the idea of nilpotent elements/ideals/groups?</p>

答:
Let $R$ be a ring, $a\in R$ nilpotent with degree of nilpotency $n$. Then we have the extremely nice property that whenever we have any polynomial expression $\sum_{k=0}^m c_ka^k$ in $a$, we can ignore all terms where $k\geq n$. That is, polynomials in $a$ effectively have degree at most $n-1$, if you allow my sloppy language. This means that to calculate all polynomial expressions in $a$, we only have to really calculate $n$ elements $a^0,a^1,a^2,\dots,a^{n-1}$, which we can then insert into any polynomial we want, ignoring all higher order terms.

For an application, look at the Jordan normal form of an endomorphism (endomorphisms over a field are a ring!). From a more abstract point of view, the Jordan normal form of an endomorphism $T$ gives us a decomposition $T=N+D$, where $N$ is nilpotent, $D$ is diagonal, and $N$ and $D$ commute. This allows us to simplify polynomial expressions in $T$ as well: We have $T^n=(N+D)^n=\sum_{k=0}^n\binom{n}{k}N^kD^{n-k}$. Since $N$ is nilpotent (let's say with degree of nilpotency $m$), we only need to consider the terms where $k<m$, because all other terms contain higher powers of $N$, which vanish due to nilpotency. And since $D$ is diagonal, powers of $D$ are easy to compute: Just take the powers of the diagonal elements. That is, effectively we only need to calculate $N^0,N^1,\dots,N^{m-1}$ to efficiently calculate all polynomial expressions in $T$. And this goes beyond just polynomial expressions. Limits of polynomial expressions can also be considered, power series for instance. This way, we have it way easier to calculate, for instance, the matrix exponential of $T$, which is
$$\exp(T)=\sum_{k=0}^\infty \frac{T^k}{k!},$$
and contains only polynomial expressions in $T$ - which are reasonably easy to calculate due to the decomposition of $T$ into commuting nilpotent and diagonal matrices/endomorphisms.

In short, nilpotent elements make arbitrary polynomial expressions easy to handle.


问:
What is the difference between interpretations of the following predicates?
I need to translate 2 sentences into predicate logic using the following definitions:-

```<code>A(x, y): x admires y
B(x, y): x attended y
P(x): x is a professor
S(x): x is a student
L(x): x is a lecture
m: mary
</code>```

Now I want to translate the following 2 sentences into predicate logic:

No student attended every lecture.

No lecture was attended by every student.

My translations:

$ \neg \exists x \space (S(x) \land (\forall y \space L(y) \rightarrow B(x, y) ) ) $

$ \neg \exists x \space (L(x) \land (\forall y \space S(y) \rightarrow B(y, x) ) ) $

Problem:

First of all, I'm not sure if my translations are correct(although I'm convinced that they are).

Second, I am not able to deduce why the following translations are wrong:

$ \neg \exists x \forall y \space (S(x) \land L(y) \rightarrow B(x, y)) $

$ \neg \exists x \forall y \space (L(x) \land S(y) \rightarrow B(y, x)) $

  or,

$ \neg \exists x \space (S(x) \rightarrow (\forall y L(y) \rightarrow B(x, y))) $

$ \neg \exists x \space (L(x) \rightarrow (\forall y S(y) \rightarrow B(y, x))) $

  I know that they are all different, that much is evident by converting each translation into clause form and comparing. But I'm not able to put a finger on what is wrong.
 For the last set of translations, the only reasoning I'm able to use is by translating the predicate into a non-English language and finding that the sentence constructed is not very idiomatic.

Lastly, as a result of 3, I find that using an implication in $\exists$ just seems wrong(as I said, the sentence formed are not very idiomatic when translated into a non-English language(I tried 2 other languages)). Are there any cases where an implication makes sense in an $\exists$.

答:
(a) The bracketing is wrong, for a start.

Take the first example in stages, using LogLish (a mix of English and logical symbolism) as a halfway house!

How would you render "x attended every lecture"? $\forall y(Ly \to Bxy)$.

How would you render 'No student is $\varphi$" using a negated existential? $\neg\exists x(Sx \land \varphi x)$

Now you want the case where $\varphi x = x$-attended-every-lecture = $\forall y(Ly \to Bxy)$. So plugging that in, we get

$\neg\exists x(Sx \land \forall y(Ly \to Bxy))$

Do you see why the $Bxy$ needs to be inside the scope of the existential quantifier?

The second example is similar.

(b) On the general issue of translating restricted quantifiers, see e.g. Restricted quantifiers - Logic (and for more, see the reference to Teller's Primer given there).


问:
For a probability measure $\mu$, does $|1-\hat{\mu} (\xi)| \geq c \xi^2$ hold near zero?
Let $\mu$ be a probability measure on $\mathbb{R}$, with $\mu \neq \delta_0$.

Does there exists $c>0$ and a neighborhood of $0$ on which $|1-\hat{\mu} (\xi)| \geq c \xi^2$?

This is true if $\mu$ has a finite second moment (then $\hat{\mu}$ is $\mathcal{C}^2$ with non-zero second derivative at $0$), but I neither have a general argument nor a counter-example.

答: Yes, this is true. $|1-\hat {\mu} (\xi)| \geq \int_{-r}^{r} [1-\cos \xi x] d\mu (x)$. We can choose $s$ so that $1-\cos u \geq \frac {u^{2}} 3$ for $|u| \leq s$. So we get $|1-\hat {\mu} (\xi)| \geq c\xi^{2}$ for $|\xi| \leq \frac  s r$ where $c=\frac 1 3 \int_{-r}^{r} x^{2} d\mu(x)$ which cannot be $0$ for every $r$ unless $\mu = \delta_0$.


问:
Find a 3x3 Matrix given its Kernel, an element in its Image and one of its Eigenvectors
The question asks to find a 3x3 matrix, knowing:

Its Kernel is the span of $\begin{bmatrix}{1\\1\\1}\end{bmatrix}$

The vector $\begin{bmatrix}{1\\2\\-1}\end{bmatrix}$ is in  its Image (column space)

$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$ is one of its eigenvectors

I don't know the definite way to do problems like this, so how I did it requires a bit of guessing. From everything I know so far, here's my train of thoughts:

So, first of all, the Kernel is the span of one 3D vector aka. a 3D line through the origin, the Kernel has the dimension of 1, the rank-nullity theorem suggests the Image or column space of that matrix has dimension 2. Thus the column space is a plane going through the origin, and the vector $\begin{bmatrix}{1\\2\\-1}\end{bmatrix}$ is on the plane. Also, the matrix is not in full rank, because its rank is 2. This implies its determinant must be zero, so 0 is one of its eigenvalues. We can imply from the first clue that $\begin{bmatrix}{1\\1\\1}\end{bmatrix}$ is also an eigenvector corresponding to the eigenvalue 0, together with $\begin{bmatrix}{1\\0\\1}\end{bmatrix}$ (but I'm not sure if these two vectors are sufficient to form an eigenbasis, later on it turns out to be not).

Suppose $\begin{bmatrix}{1\\0\\1}\end{bmatrix}$ has a non zero corresponding eigenvalue. Because a matrix only scales its eigenvector if the eigenvalue is not zero, $\begin{bmatrix}{1\\0\\1}\end{bmatrix}$ is in its Image plane. Choose $\begin{bmatrix}{1\\0\\1}\end{bmatrix}$ and $\begin{bmatrix}{1\\2\\-1}\end{bmatrix}$, which are linearly independent, as the basis of that Image plane, put that into our matrix, we got two of its column. The third must be some linear combination of the first two.

Here's when the guessing comes. First, I guess it's this matrix:
$\begin{bmatrix}1 & 1 & -2\\0 & 2 & -2\\1 & -1 & 0\end{bmatrix}$

It has the right column space and kernel. But the eigenvectors are found to be: $\begin{bmatrix}{1\\1\\1}\end{bmatrix}$, $\begin{bmatrix}{3\\2\\1}\end{bmatrix}$ and $\begin{bmatrix}{1\\1\\0}\end{bmatrix}$.

So I switched the first two columns:
$\begin{bmatrix}1 & 1 & -2\\2 & 0 & -2\\-1 & 1 & 0\end{bmatrix}$

This time the matrix' eigenvectors are correct: $\begin{bmatrix}{1\\1\\1}\end{bmatrix}$, $\begin{bmatrix}{1\\0\\1}\end{bmatrix}$ and $\begin{bmatrix}{1\\1\\0}\end{bmatrix}$, also its kernel and column space satisfy the problem. We can say this is one answer.

However, I'm pondering: Is there any method to do this without guessing? And is that matrix the only answer to this question?

答:
First, I will say that I prefer your method.

That said, here's a way to do things systematically.  If we select a matrix so that its eigenvectors are $(1,1,1),(1,2,-1),(1,0,1)$ with associated eigenvalues $0,\lambda_1,\lambda_2$ (where $\lambda_1$ and $\lambda_2$ are both non-zero), then the resulting matrix will have the required properties.

If we take
$$
D = \pmatrix{0 \\ & \lambda_1 \\ && \lambda_2}, \quad S = \pmatrix{1&1&1\\1&2&0\\1&-1&1},
$$
then we can take the matrix $A = SDS^{-1}$. If you want to try this for yourself, the inverse of $S$ is given by
$$
S^{-1} = \frac 12 \pmatrix{-2 & 2 & 2\\ 1 & 0 & -1\\ 3 & -2 & -1}.
$$
With the help of block-matrix multiplication, we can also take advantage of the zero in the diagonal matrix $D$ to write this as
$$
A = SDS^{-1} = \frac 12 \pmatrix{1&1\\2&0\\-1&1}
\pmatrix{\lambda_1\\ & \lambda_2}
\pmatrix{1&0&-1\\3&-2&-1}.
$$
For instance, taking $\lambda_1 = \lambda_2 = 1$ yields the answer
$$
A = \pmatrix{2 & -1 & -1\\
1 & 0 & -1\\
1 & -1 & 0}.
$$


问:
Which of the following Maps is Bilinear?
Problem:

Show which of the following maps are bilinear ?

$\phi_1 : \left( \left(\begin{array}{c} v_1 \\ v_2  \end{array}\right) , \left(\begin{array}{c} w_1 \\w_2  \end{array}\right)\right) \mapsto v_1w_2 + v_2w_1 $

$\phi_2 : \left( \left(\begin{array}{c} v_1 \\ v_2 \\v_3 \end{array}\right) , \left(\begin{array}{c} w_1 \\w_2 \\ w_3 \end{array}\right)\right) \mapsto v_1v_2 + 2w_1w_2 $

$\phi_3 : \left( \left(\begin{array}{c} v_1 \\ v_2  \end{array}\right) , \left(\begin{array}{c} w_1 \\w_2  \end{array}\right)\right) \mapsto v_1 + v_2 + w_1 + w_2 $

Questions:

I have already shown that $\phi_1$ is bilinear, though i want to know if i was right with my assumption.
Though right now what i am struggling with is to show that $\phi_2$ and $\phi_3$ are bilinear, due to lack of knowledge of transforming the following results of $\phi_2$ and $\phi_3$ to see if it is even possible for them to be bilinear. If there is a piece of advice on seeing quickly if it's bilinear, i would love to hear it. Other than that i thank you in advance for the following pieces of advise and awnsers !

答:
Remark: Remember that a $\color{red}{bilinear form}$ on a vector space $V$ over a field $F$ is a $\color{blue}{map}$ $$\color{blue}{\phi: V \times V \to F}$$
such that:

$\forall v_{1},v_{2},w\in V$, we have that $$ \color{blue}{\boxed{\phi(v_{1}+v_{2},w)=\phi(v_{1},w)+\phi(v_{2},w)}}$$

$\forall v,w_{1},w_{2}\in V$, we have that  $$\color{blue}{\boxed{\phi(v,w_{1}+w_{2})=\phi(v,w_{1})+\phi(v,w_{2})}}$$

$\forall v,w\in V, a\in F$, we have that $$\color{blue}{\boxed{ \phi(av,w)=a\phi(v,w)}}$$

$\forall v,w\in V, a\in F$, we have that $$\color{blue}{\boxed{\phi(v,aw)=a\phi(v,w)}}$$

So you should try to verify each of these properties (at least when you first start studying a definition in mathematics it is best not to skip steps but to be as detailed as possible).

Now, note that for $\phi_{2}$ we have
\begin{eqnarray*}
\phi_{2} \left( \begin{pmatrix} v_{1}\\v_{2}\\ v_{3} \end{pmatrix},\begin{pmatrix} w_{1}\\ w_{2}\\ w_{3} \end{pmatrix} +\begin{pmatrix} u_{1}\\u_{2}\\ u_{3} \end{pmatrix}\right)&=&\phi_{2}\left( \begin{pmatrix} v_{1}\\v_{2}\\ v_{3} \end{pmatrix},\begin{pmatrix} w_{1}+u_{1}\\ w_{2}+u_{2}\\ w_{3}+u_{3} \end{pmatrix} \right)\\
&=&v_{1}v_{2}+2(w_{1}+u_{1})(w_{2}+u_{2})\\
&=&v_{1}v_{2}+2(w_{1}w_{2}+w_{1}u_{2}+u_{1}w_{2}+u_{1}u_{2})\\
&=&v_{1}v_{2}+2w_{1}w_{2}+2w_{1}u_{2}+2u_{1}w_{2}+2u_{1}u_{2}\\
&\not=&\color{blue}{v_{1}v_{2}+2w_{1}w_{2}}+\color{red}{v_{1}v_{2}+2u_{1}u_{2}}\\
&=&\color{blue}{\phi_{2}\left(\begin{pmatrix} v_{1}\\ v_{2} \\ v_{3}\end{pmatrix},\begin{pmatrix} w_{1}\\ w_{2} \\ w_{3}\end{pmatrix}  \right)}+\color{red}{\phi_{2}\left( \begin{pmatrix} v_{1}\\ v_{2} \\ v_{3}\end{pmatrix},\begin{pmatrix} u_{1}\\ u_{2} \\ u_{3}\end{pmatrix}  \right)}
\end{eqnarray*}
with $\begin{pmatrix} v_{1}\\v_{2}\\ v_{3} \end{pmatrix},\begin{pmatrix} w_{1}\\ w_{2}\\ w_{3} \end{pmatrix},\begin{pmatrix} u_{1}\\u_{2}\\ u_{3} \end{pmatrix}\in \operatorname{Dom}(\phi_{2})$.

Therefore, $\phi_{2}$ it does not satisfy condition $\color{red}{2}$ and that is enough to affirm that $\phi_{2}$ it is not a bilinear form.

I will leave it to you to conclude with literal $3$. Many successes in your studies.


问:
I am finding a contradiction in my vector proof. What am I doing wrong?
Prove the midpoint of the hypotenuse of a right-angled triangle is equidistance from the three vertices by vector methods.

I have drawn out a triangle with the right angle at A, then going clockwise, vertices B, and C.

By definition of perpendicular vectors, $(\vec{AM} + \vec{MC})\cdot(\vec{AM}-\vec{BM})=0$. Then use the fact that $\vec{BM} = \vec{MC}$ to conclude the proof.

However, lets try to define $\vec{AC}$ = a and  $\vec{AB}$ = b

$\vec{BC}$ = a $-$ b  so $\vec{BM} = \vec{MC} = \frac{1}{2}($a$-$b)

But then for $\vec{AM} =$ a $- \vec{MC} = \frac{1}{2} ($a$+$b) which is obviously not equal to BM or MC. Why is that happening?

答: What you are asked to prove is that $\vec{AM}$, $\vec{BM}$, and $\vec{CM}$ have the same norm (or length if you prefer), not that they are equal as vectors.


答:
There is no contradiction: by equidistant we mean that $|\vec {AM}| = |\vec {BM}| = |\vec {CM}|$.

Indeed we have $\vec {AM} = \frac12(\mathbf a + \mathbf b)$. Thus:

$$|\vec {AM}| = \frac12\sqrt{(\mathbf a + \mathbf b)\cdot (\mathbf a + \mathbf b)}=\frac12\sqrt{\mathbf a \cdot \mathbf a+\mathbf a \cdot \mathbf b+2\mathbf a \cdot \mathbf b} = \frac12\sqrt{|\mathbf a|^2+|\mathbf b|^2} = \frac12|BC|$$

so the length of $AM$ is equal to $BM$ and $MC$.

Title: I am finding a contradiction in my vector proof. What am I doing wrong?

Body:
Prove the midpoint of the hypotenuse of a right-angled triangle is equidistance from the three vertices by vector methods.

I have drawn out a triangle with the right angle at A, then going clockwise, vertices B, and C.

By definition of perpendicular vectors, $(\vec{AM} + \vec{MC})\cdot(\vec{AM}-\vec{BM})=0$. Then use the fact that $\vec{BM} = \vec{MC}$ to conclude the proof.

However, lets try to define $\vec{AC}$ = a and  $\vec{AB}$ = b

$\vec{BC}$ = a $-$ b  so $\vec{BM} = \vec{MC} = \frac{1}{2}($a$-$b)

But then for $\vec{AM} =$ a $- \vec{MC} = \frac{1}{2} ($a$+$b) which is obviously not equal to BM or MC. Why is that happening?


问:
proof that the sequence satisfies weak law of large numbers
strong textHow to check if this sequence satisfies the weak law of large numbers?
$$P(X_n=1)= {{1}\over{3^n}} ,P(X_n=-1)= {{1}\over{3^n}}, P(X_n=0)= {1 -{{2}\over{3^n}}}  $$
I calculated that:
$$E(X_n)=0 $$
so according to the definition of this law: $${\sum_{n=1}^{N}{(X_n-0)}}\over{N} $$and from this one we can deduce that this expression converges to 0 so the sequence satisfies the weak law of large numbers. Is that correct?

答: $\sum P(X_n \neq 0)=\sum \frac 2 {3^{n}}<\infty$. By Borel Cantelli Lemma $P(\lim \sup (X_n \neq 0))=0$. Hence, with probaility $1$, $X_n=0$ for all $n$ suffciently large. But this implies that the averages $\frac 1n \sum\limits_{k=1}^{n} X_k$ tends to $0$ with probability $1$. Thus, we actually have strong law for this sequence.


答:
So, you have shown:

$\vec{BM} = \vec{MC} = \frac{1}{2}(\mathbf{a}-\mathbf{b})$

and

$\vec{AM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$

All that remains is to show AM = BM (not that the vectors are equal, simply their magnitude).  Here, we can use the observation that $\mathbf{a} \cdot \mathbf{b} = 0$, and the result that $|\mathbf{v}|^2=\mathbf{v}\cdot\mathbf{v}$

$BM^2 = (\frac12)^2(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=(\frac12)^2(a^2 -2\mathbf{a}\cdot\mathbf{b}+b^2)=(\frac12)^2(a^2 +b^2)$

$AM^2 = (\frac12)^2(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})=(\frac12)^2(a^2 +2\mathbf{a}\cdot\mathbf{b}+b^2)=(\frac12)^2(a^2 +b^2)$

So AM = BM.


问:
A bag contains toys marked 1,2,3,...n.Two tags are chosen at random.
A bag contains toys marked 1,2,3,...n.Two tags are chosen at random.

Find P such that the markers will be consecutive integers if the tags are chosen

1)without replacement

2)with replacement

I am having a solution that goes like this:

For case 1:(Without replacement)

There are $nC2$ ways of choosing an element.

Take n=5, you get (1,2)(2,3)(3,4)(4,5) Generalising you can get (n-1) cases of getting consecutive tags.

So the required probability stands $(n-1)/nC2$

For case 2:(With replacement)

There is a sample space of $nC1 * nC1$ or $n^2$.

Here my instructor says that the number of choices will be 2(n-1).

Therefore, the probability of choosing consecutive tags is $\frac{2(n-1)}{n^2}$.

But I don't understand why it's $2*(n-1)$.

Is it simply because we now take both the tags without order for example: 1,2 in two ways (1,2) and (2,1) and they still remain conseceutive.

答:
In the first case, the space of possible picks is counted as unordered pairs (or distinct sorted pairs, which might be easier to imagine and easier to write down). So the pick $(1,3)$ and the pick $(3,1)$ are not counted as distinct, and together constitute a single possible pick. Thus, when you subsequently count the number of good picks, you must stay with the same regime.

If you want, you can change this to counting ordered pairs, with a total of $_nC_1\cdot{}_{n-1}C_1$ possible picks and $2n-2$ good picks.

On the other hand, in case 2, the space of possible picks is counted as ordered pairs. So the pick $(1,3)$ and the pick $(3,1)$ are counted as distinct, and together constitute two possible picks. Thus, when you subsequently count the number of good picks, you must stay with the same regime.

It is difficult to change this into an unordered approach, as the possibility of picking the same toy twice makes it non-uniform.


问:
If for every sequence $(x_n)$, $x_n \to x_0 \implies (f(x_n))$ is a Cauchy sequence, then is $f$ continuous?
Given $f:D \subseteq \mathbb{R} \to \mathbb{R}$, if for every sequence $(x_n)$ in $D$ such that $x_n \to x_0$ is true that $(f(x_n))$ is a Cauchy sequence, then is $f$ continuous?

This is what I know:

Every Cauchy sequence is a convergent sequence.

$f$ is continuous if and only if $x_n \to x_0 \implies f(x_n) \to f(x_0)$ for every sequence $(x_n)$.

However, I only know that $f(x_n) \to a \in \mathbb{R}$; how can I see that $a = f(x_0)$ and then have the continuity of $f$?

If the result is false, please show a counterexample, i.e., a discontinuous function such that for every $x_n \to x_0$ I have $(f(x_n))$ Cauchy.

答:
Correct me if wrong.

Perhaps a bit of nitpicking.

1)If $x_0 \in D;$

Then consider $x_n:=x_0$ and you are done

2)Assume $x_0\not \in D$, and $x_0$ a limit point of $D,$ then

$f(x_n) \rightarrow f(a);$ i.e. $f(x_n)$ has a limit;

Continuity does not follow.

Example:

$D_L=$ { $x|x<0$ };

$f:(x)=0$ for $x<0;$

Define $D'=$ { $0$ }:

$f(0)=1.$

$D_R$ defined similarly.


问:
Why is it not obvious that the Zermelo hierarchy is a hierarchy?
In Sets, Logic and Categories, Cameron writes

First, [we show that the Zermelo hierarchy] really is a hierarchy: the sets get larger as we progress. (This is not obvious; each set is the power set of its predecessor, and most sets $X$ don't satisfy $X\subseteq\mathcal{P}X$.)

I'm stuck on the parenthetical sentence. What does he mean by "each set is the power set of its predecessor"? (Is this referring to ordinals, or to sets in general?) And what does he mean by "most sets $X$ don't satisfy $X\subseteq\mathcal{P}X$"? (Most sets = most ordinals? Certainly this holds for finite sets, no? But not ordinals?)

答:
For "each set is the power set of its predecessor", it refers to the hierarchy itself. If $V_\alpha$ is the stage $\alpha$ of the hierarchy, then $V_{\alpha+1}=\mathcal{P}(V_\alpha)$.

For "most sets $X$ don't satisfy $X⊆\mathcal{P}(X)$", you can check that indeed finite sets do not satisfy this except for very special cases. For instance, if $X$ is a singleton, then $\mathcal{P}(X)=\{X,\emptyset\}$, and unless $X=\{X\}$ (impossible with the foundation axiom) or $X=\{\emptyset\}$, you don't get $X\subset \mathcal{P}(X)$. (The good news is that $V_1$ is indeed $\{\emptyset\}$, so there is no contradiction.)


问:
Simplify $\sum_{i=1}^n\frac{a}{\theta-x_i}$ where $\{x_i\}\in(0,\theta)$ and $a\in(-1,0]$
I have this problem where I need to solve $-\frac{n(1+a)}{\theta}+\sum_{i=1}^n\frac{a}{\theta-x_i}=0 $ for $\theta$ (the solution is allowed to be in terms of the mean of $x_1,..., x_n$), provided  that $\{x_i\}\in(0,\theta)$ and $a\in(-1,0]$.

I really have no idea how to simplify the summation. Maybe I'm just missing something simple?

答:
I think there is no solution. Let's take the simple case where $n=1$. Then we have the following :
$$\frac{1+a}{\theta} = \frac{a}{\theta - x}$$
$$\frac{\theta}{\theta - x } = \frac{1+a}{a}$$
Subtracting one from both sides,
$$\frac{x}{\theta -x} = \frac{1}{a}$$
$$x = \frac{\theta}{1+a}$$
Now, since $a \in (-1,0]$, the denominator is less than one, therefore $x$ is larger then $\theta$ for any allowed value of $a$.


问:
Variance estimation for uncorrelated variables
Consider a sequence of uncorrelated variables $(X_n)$. If these variables are bounded in $L^2$ and have common mean $\mu$, a well-known variant of the law of large numbers states that
$$
  \frac{1}{n}\sum_{i=1}^n X_i \stackrel{\mathrm{a.s.}}\longrightarrow \mu.
$$
This allows for estimation of parameters when observations are not i.i.d., a situation occurring for example for non-uniformly spaced observations of continuous-time processes.

My question is about the variance analogue of the above. Again, assume that $(X_n)$ is a sequence of uncorrelated variables. Assume that the variables have common variance $\sigma^2$. We do not assume that the variables have common mean. Is there an asymptotic result similar to the above yielding $\sigma^2$ as an almost sure limit?

I would imagine something like
$$
  \frac{1}{n}\sum_{i=1}^n (X_i - \bar{X}_n)^2 \stackrel{\mathrm{a.s.}}\longrightarrow \sigma^2,
$$
where $\bar{X}_n$ is the average of $X_1,\ldots,X_n$, but this result is not
obvious to me. Assume boundedness in $L^p$ for any $p\ge 1$ necessary.

For completeness, here is an explicit motivation for the above example. Assume that $S_t = \exp(X_t)$ for some Levy process $X$ and $t\ge0$. Here, $S$ denotes an asset price, and this is then an instance of the exponential Levy market model. In this model, continuous returns of period $\Delta$ are given by $\log S_{t+\Delta} / S_t$ and have mean $\Delta \alpha$ and variance $\Delta \beta$ for some $\alpha$ and $\beta$. Making observations $S_{t_1},\ldots,S_{t_n}$, I'm interested in estimating $\beta$. Here, it holds that the variables
$$
  \frac{1}{\sqrt{t_i-t_{i-1}}}\log\frac{S_{t_i}}{S_{t_{i-1}}}
$$
are uncorrelated (in fact, independent) and have the same variance $\beta$. If something like the above almost sure result for convergence to the variance holds, I would then be able to use the above transformed variables to estimate $\beta$.

答:
Consider the example where $E[X_i] = +m>0$ when $i$ is even and $E[X_i]=-m$ when $i$ is odd, where $\mathrm {Var}(X_i)=\sigma^2$ and the fourth moment is bounded

Then $\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i \stackrel{\mathrm{a.s.}}\longrightarrow 0$ while $\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X}_n)^2 \stackrel{\mathrm{a.s.}}\longrightarrow \sigma^2 +m^2 \not=\sigma^2$ .


问:
$\lim_{k\to\infty} X^{(k)}=A^{-1}$ if and only if $ρ(I − AX^{(0)}) < 1$
A is $n × n$  nonsingular matrix, and $X^{(0)}$ is n × n matrices and sequence is defined

$X^{(k+1)} = X^{(k)} + X^{(k)}(I − AX^{(k)}),$ $k=0,1,2.....$

I want to show that $\lim_{k\to\infty} X^{(k)}=A^{-1}$ if and only if $ρ(I − AX^{(0)}) < 1$ where  $ρ$ is spectral radius.

I'm stuck on the problem. How can I solve it?

答: Hints. Prove by mathematical induction that $I-AX^{(k)}=(I-AX^{(0)})^{2^k}$. Now the RHS tends to zero as $k\to\infty$, because $\rho(I-AX^{(0)})<1$.


答:
Note that $\vec{BM}$ and $\vec{MC}$ point in same direction (along the hypotenuse). While $\vec{AM}$ points in a different direction (from right vertex to midpoint of hypotenuse). So $\vec{AM}$ must be different from $\vec{BM}$ or $\vec{MC}$.

But when you calculate their lengths, you'll obtain $$|\vec{AM}|=|\vec{BM}|=|\vec{MC}|$$

Things would have been easier had you also defined position vector of vertex $A$ as origin. Then position vector of $M$ could easily be seen as $$\vec{M}=\frac{\mathbf{a}+\mathbf{b}}{2}$$
that is
$$\vec{AM}=\vec{M}-\vec{0}=\frac{\mathbf{a}+\mathbf{b}}{2}$$


问:
Parametric equation of ellipse
Find the curvature and the radius of curvature for (f) $x = 2 \cos t$ and $y = 3 \sin t$, $0 < t < 2\pi$ at point $(2, 0)$ and $(0, 3)$, where the parametric equation given is a ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ . May i know how to use the point given? I forgot the formula . thanks

Title: Parametric equation of ellipse

Body:
$$\rho(x,y)=\left|\frac{(1+y'^2)^{3/2}}{y''(x)}|_{(x,y)}\right|$$

Here $$y=3\sqrt{1-x^2/4} \implies y'(x,y)=\frac{3x}{2\sqrt{1-x^2/4}}'$$
$$y''(x,y)=\frac{-6}{(4-x^2)^{3/2}}$$
so $$\rho(x,y)=\frac{(16+5x^2)^{3/2}}{48}$$
So $$\rho(2,0)=\frac{(16+20)^{3/2}}{48}=\frac{9}{2},$$
$$\rho(0,3)=\frac{4}{3}$$


问:
Existence of closed ball outside of closed convex set.
Let $H$ be a Hilbert space, and $K\subseteq H$ a closed convex subset, with $p\in \partial K$. Is there $B$, a closed ball, with $B \cap K=\emptyset$ and $\text{conv}(B,p)\cap K=\{p\}$?

答:
This is not true. It should be clear from looking at the drawing you made. Try moving the point $p$ (in your drawing  ! )along the boundary of $K$ and you will clearly see that for some $p$ the cone $Conv(B,p)$ intersects $K$ at more than one point.

Another counterexample :

Consider $H = \mathbb R^2$,  $K = [0,1]^2$ , $p = (0,0)$ and $B_\epsilon$ the closed ball centered at $(2,2)$ with radius $\epsilon$.

For $\epsilon$ small enough $B_\epsilon \cap K = \varnothing $ but Conv$(B_\epsilon,p) \cap K$ will always contain the diagonal $[(0,0),(1,1)]$ of the square K.


问:
Solve the equation $\sqrt{3x-2} +2-x=0$

Solve the equation: $$\sqrt{3x-2} +2-x=0$$

I squared both equations $$(\sqrt{3x-2})^2  (+2-x)^2= 0$$

I got $$3x-2 + 4 -4x + x^2$$

I then combined like terms $x^2 -1x +2$

However, that can not be right since I get a negative radicand when I use the quadratic equation.

$x = 1/2  \pm \sqrt{((-1)/2)^2 -2}$

The answer is 6

答:
$$\sqrt{3x-2} +2-x=0$$

Isolating the radical:$$\sqrt{3x-2} =-2+x$$

Squaring both sides:$$\bigg(\sqrt{3x-2}\bigg)^2 =\bigg(-2+x\bigg)^2$$

Expanding $(-2+x)^2$ and gathering like terms: $$3x-2=-2(-2+x)+x(-2+x)$$

$$3x-2=4-2x-2x+x^2$$

Set x equal to zero:$$3x-2=4-4x+x^2$$

Gather like terms:$$0=4+2-3x-4x+x^2$$

Factor the quadratic and find the solutions:$$0=x^2-7x+6$$

$$0=(x-6)(x-1)$$

$$0=x-6\implies\boxed{6=x}$$
$$0=x-1\implies\boxed{1=x}$$

Checking 6 as a solution:

$$\sqrt{3(6)-2} +2-(6)=0$$

$$\sqrt{16} +2-6=0$$

$$4+2-6=0$$

$$6-6=0$$

Checking 1 as a solution:

$$\sqrt{3(1)-2} +2-(1)=0$$

$$\sqrt{1} +2-1=0$$

$$2\neq0$$

The solution $x=1$ does not equal zero and therefore is not a solution.

The solution $x=6$ does equal zero and therefore is our only solution to this equation.

Title: Solve the equation $\sqrt{3x-2} +2-x=0$


问:
Differential of the product of two matrices
From Muirhead, Aspects of Multivariate Statistical Theory, pages 56-57.

For any matrix $X$, $\text{d}X$ denotes the matrix of differential ($\text{d}x_{ij}$). It's easy to check that if $X$ is a $n\times m$ and $Y$ is $m\times p$ then
\begin{equation}\text{d}(XY)=X.\text{d}Y+\text{d}X.Y\end{equation}

I'm trying to figure out what kind of product is involved in the previous expression. In order to do that, I've tried to see what happen in the simple case $n\triangleq m\triangleq p\triangleq 2$.

example

If $n\triangleq m\triangleq p\triangleq 2$ then
\begin{equation}X=\begin{bmatrix}
x_{11} & x_{12} \\
x_{21} & x_{22} \\
\end{bmatrix} \qquad \text{d}X=\begin{bmatrix}
\text{d}x_{11} & \text{d}x_{12} \\
\text{d}x_{21} & \text{d}x_{22} \\
\end{bmatrix}\quad Y=\begin{bmatrix}
y_{11} & y_{12} \\
y_{21} & y_{22} \\
\end{bmatrix} \qquad \text{d}Y=\begin{bmatrix}
\text{d}y_{11} & \text{d}y_{12} \\
\text{d}y_{21} & \text{d}y_{22} \\
\end{bmatrix}\end{equation}
I start from
\begin{equation}
XY=\begin{bmatrix}
x_{11}y_{11}+x_{12}y_{21} & x_{11}y_{12}+x_{12}y_{22} \\
x_{21}y_{11}+x_{22}y_{21} & x_{21}y_{12}+x_{22}y_{22}
\end{bmatrix}\end{equation}
then, I compute the differential of $XY$ element-wise. By recalling that for a scalar function $f(x)$ with $x=[x_1, \dots, x_n]'$, its differential in $x$ is defined as
\begin{equation}\text{d}f(x)\triangleq \sum_{k=1}^n \frac{\partial\,f(x)}{\partial\, x_k}\text{d}x_k\end{equation}
follows that for the first term of $\text{d}(XY)$ (element in position $i,j=1,1$) holds
\begin{equation}\begin{aligned}
\text{d}(x_{11}y_{11}+x_{12}y_{21})&=\text{d}(x_{11}y_{11})+\text{d}(x_{12}y_{21})\\
&=(y_{11}\text{d}x_{11}+x_{11}\text{d}y_{11})
+(y_{21}\text{d}x_{12}+x_{12}\text{d}y_{21})\\
&=y_{11}\text{d}x_{11}+x_{11}\text{d}y_{11}
+y_{21}\text{d}x_{12}+x_{12}\text{d}y_{21}
\end{aligned}\end{equation}
so, with the same reasoning applied to the others 3 elements, follows
\begin{equation}
\text{d}(XY)=\begin{bmatrix}
y_{11}\text{d}x_{11}+x_{11}\text{d}y_{11}
+y_{21}\text{d}x_{12}+x_{12}\text{d}y_{21}
&
y_{12}\text{d}x_{11}+x_{11}\text{d}y_{12}
+y_{22}\text{d}x_{12}+x_{12}\text{d}y_{22}\\
y_{11}\text{d}x_{21}+x_{21}\text{d}y_{11}
+y_{21}\text{d}x_{22}+x_{22}\text{d}y_{21}
&
y_{12}\text{d}x_{21}+x_{21}\text{d}y_{12}
+y_{22}\text{d}x_{22}+x_{22}\text{d}y_{22}
\end{bmatrix}\end{equation}
which can be written, by separating the differentials, as
\begin{equation}\text{d}(XY)=
\begin{bmatrix}
y_{11}\text{d}x_{11}+y_{21}\text{d}x_{12} &
y_{12}\text{d}x_{11}+y_{22}\text{d}x_{12} \\
y_{11}\text{d}x_{21}+y_{21}\text{d}x_{22} &
y_{12}\text{d}x_{21}+y_{22}\text{d}x_{22}
\end{bmatrix}+
\begin{bmatrix}
x_{11}\text{d}y_{11}+x_{12}\text{d}y_{12} &
x_{11}\text{d}y_{12}+x_{12}\text{d}y_{22} \\
x_{21}\text{d}y_{11}+x_{22}\text{d}y_{21} &
x_{21}\text{d}y_{12}+x_{22}\text{d}y_{22}
\end{bmatrix}\end{equation}
In conclusion, it seems to me that
\begin{equation}\text{d}X.Y=\begin{bmatrix}
y_{11}\text{d}x_{11}+y_{21}\text{d}x_{12} &
y_{12}\text{d}x_{11}+y_{22}\text{d}x_{12} \\
y_{11}\text{d}x_{21}+y_{21}\text{d}x_{22} &
y_{12}\text{d}x_{21}+y_{22}\text{d}x_{22}
\end{bmatrix}\end{equation}
\begin{equation}
\text{d}Y.X=\begin{bmatrix}
x_{11}\text{d}y_{11}+x_{12}\text{d}y_{12} &
x_{11}\text{d}y_{12}+x_{12}\text{d}y_{22} \\
x_{21}\text{d}y_{11}+x_{22}\text{d}y_{21} &
x_{21}\text{d}y_{12}+x_{22}\text{d}y_{22}
\end{bmatrix}
\end{equation}
In theese two expressions I don't recognise any type of matricial product between $\text{d}X$, $Y$ or between $\text{d}Y$, $X$ (not the usual product "rows times columns", nor the Hadamard or Kronecker products).

答:
Well, the condition is
$$\mathrm{d}(\bf{A} \bf{B}) = \mathrm{d}\bf{A} .\bf{B}+\bf{A}.\mathrm{d}\bf{B}  $$

So,
\begin{align}
\text{LHS}_{ij} &=\mathrm{d} \sum_k A_{ik}B_{kj} \\
&=  \sum_k \mathrm{d} (A_{ik}B_{kj} + A_{ik}B_{kj}) \\
&= \sum_k  (\mathrm{d}A_{ik}B_{kj} + A_{ik}\mathrm{d}B_{kj}) \\
&= \sum_k \mathrm{d} A_{ik}B_{kj} + \sum_k A_{ik}\mathrm{d}B_{kj} \\
&= \text{RHS}_{ij}
\end{align}


问:
Problem with some Bayesian posterior pdf functions.
I've been given a problem and I'm not entirely sure how to approach it, I'm aware of the basics like $Posterior ∝ Likelihood × Prior$ etc. but I'm not sure what steps to take here to solve the below problem.

Any help would be greatly appreciated,

Thanks,

MathewG

The exponential distribution is often used to model a continuous positive random
variable measured on a time scale. Assume that the random variable $y$ follows
an exponential distribution with rate parameter $θ$. Hence, $y ∼ Exp(θ)$ with
probability density function

$$f (y|θ) = θ\exp(−θy),\ y > 0\, θ > 0.$$

In order to perform inference about $θ$ in the Bayesian framework, a researcher
has adopted a Gamma prior distribution for $θ, θ ∼ Gamma(a, b)$, with
probability density function

$$π(θ) = b^a/\Gamma(a)*θa−1exp (−b θ),\ θ > 0,$$

in which $a$ and $b$ are known positive constants and $Γ(a)$ is the Gamma function.
It can be shown that $E [θ] = a/b$ and $Var [θ] = a/b^2$

Using this information how would you determine $π(θ | y)$, the posterior probability density function of θ and then show that $$θ | y ∼ Gamma(a + 1, b + y)$$

答:
It is very simple.

I'm aware of the basics like Posterior∝Likelihood×Prior

That is all you have to do to finish your exercise:

Prior density (skip all the quantities not depending on $\theta$ and substitute = with $\propto$):

$$\pi(\theta)\propto \theta^{a-1}e^{-b \theta}$$

model density:

$$p(y|\theta)=\theta e^{-\theta y}$$

Posterior density:

$$\pi(\theta|y)\propto \theta^a e^{-\theta(b+y)}$$

where you immediately recognize the kernel of a $Gamma(a+1;b+y)$$

Just for the sake of completeness, you can write the exact posterior density, with the normalization constant required

$$\pi(\theta|y)=\frac{(b+y)^{a+1}}{\Gamma(a+1)}\theta^{(a+1)-1}e^{-\theta(b+y)}$$


问:
The arcsine is "not really a function"
My math teacher said that later on in life, if you pursue math, you would see that the arcsine function isn't actually a function. I do not see how it's not a function.</p>

答:
Following clarification from the OP that the arcsine function had been referred to: the arcsine is "not really a function" because there are infinitely many solutions to (for example) $\sin x=0$, so $\arcsin0$ could be any one of infinitely many values, whereas a function must map one input to one output. $\arcsin$ is therefore, stricto sensu, a multifunction.

Implementations of the arcsine in programming languages must choose one principal value out of infinitely many to serve as the aforementioned $x$, which leads to branch cuts over which that principal value is not continuous. For the arcsine's case, those branch cuts most often are $(-\infty,-1)$ and $(1,\infty)$.

Title: The arcsine is "not really a function"


答:
As a starter for $10$, iIf you have a sampling density of the form
$$f(x|θ,ν)=\dfrac{θ^νx^{ν−1}}{Γ(ν)}\exp\{−θx\}\,,$$
the conjugate prior is $\theta\sim\mathcal{G}(\alpha,\beta)$ with posterior
$$\theta|x\sim \mathcal{G}(\alpha+\nu,\beta+x)$$

(where throughout $\mathcal{G}(\alpha,\beta)$ is the gamma distribution).

Title: Problem with some Bayesian posterior pdf functions.

Body:
I've been given a problem and I'm not entirely sure how to approach it, I'm aware of the basics like $Posterior ∝ Likelihood × Prior$ etc. but I'm not sure what steps to take here to solve the below problem.

Any help would be greatly appreciated,

Thanks,

MathewG

The exponential distribution is often used to model a continuous positive random
variable measured on a time scale. Assume that the random variable $y$ follows
an exponential distribution with rate parameter $θ$. Hence, $y ∼ Exp(θ)$ with
probability density function

$$f (y|θ) = θ\exp(−θy),\ y > 0\, θ > 0.$$

In order to perform inference about $θ$ in the Bayesian framework, a researcher
has adopted a Gamma prior distribution for $θ, θ ∼ Gamma(a, b)$, with
probability density function

$$π(θ) = b^a/\Gamma(a)*θa−1exp (−b θ),\ θ > 0,$$

in which $a$ and $b$ are known positive constants and $Γ(a)$ is the Gamma function.
It can be shown that $E [θ] = a/b$ and $Var [θ] = a/b^2$

Using this information how would you determine $π(θ | y)$, the posterior probability density function of θ and then show that $$θ | y ∼ Gamma(a + 1, b + y)$$


问:
Can Green's theorem be used in a plane other than the xy-plane?
In the following 2D case, Green's theorem solves the following problem:

$$\vec{F}=\langle{xy+\ln{(\sin{e^{x})},x^2+e^{y^2}}}\rangle$$

$$\oint_C\vec{F}\cdot{d\vec{r}}=\iint_Dx\space{dA}$$

where C is the unit circle $x^2+y^2=1$, and D is the unit disk $x^2+y^2\le{1}$.

But what if instead the problem were:

$$\vec{F}=\langle{\frac{\sqrt{2}}{2}(xy+\ln{(\sin{e^x}}),x^2+e^{y^2},-\frac{\sqrt{2}}{2}(xy+\ln{(\sin{e^x}})}\rangle$$

$$\oint_C\vec{F}\cdot{d\vec{r}}$$

And $C$ is instead the unit circle $\vec{r}(t)=\langle{\frac{\sqrt{2}}{2}\cos{t},\sin{t},-\frac{\sqrt{2}}{2}\cos{t}}\rangle$.

This is the same problem as the first one, but rotated 45 degrees about the y-axis. I know I could solve the second problem by simply rotating the path and the vector field back 45 degrees. But is there any way to apply Green's theorem to the second problem directly?

答:
Green's theorem is just a special case of Stokes' theorem (which is just a special case of the generalized Stokes' theorem).

$$\oint_{\partial M} d\ell \cdot F = \int_M dA \cdot (\nabla \times F)$$


问:
Alternative definitions of neighbourhood of a point
There are some text like Munkres and Rudin that defines neighbourhood of a point x to be an open set that contains x, while other text defines neighbourhood to be any set that contains an open set that contains x. I was wondering which definition is the one that math journals actually use?</p>

答: Both are used, and most authors specify which version they use. I prefer the general one, as it makes for some shorter expressions of concepts. In this question I also give some examples in my answer.</p>


问:
Proof on sequence of finite-rank operators
I had difficulties figuring out the solutions to the following problem:

Let $H$ be a separable infinite dimensional Hilbert space. Let
{${L_n}$} be a sequence of linear finite rank operators from $H$ to
into itself such that {${L_n}x$} converges in $H$ for any $x \in H$.
Prove that:

i) there exist $L: H \rightarrow H$ linear and bounded such that $L_nx
 \rightarrow Lx$ for any $x \in H$.

ii) $L$ can be either compact or not by showing and discussing two
examples.

As an idea (maybe wrong idea) I propose my solution concerning point (i):

Fixed $x$, we have that {${L_n}x$} is convergent; it follows that $||{L_n}x|| < \infty$ for any $n$.  Consider now n being fixed. We have that  $||{L_n}x|| < \infty$ for any $x \in H$, hence it follows that $L_n$ is linear and bounded for any n (is this correct?). The solution to point one follows by applying the uniform boudedness principle (or Banach-Steinhaus theorem) using the fact that $lim_n  (L_nx)$ is well defined for all $x \in H$.

Concerning point (ii): I would figure out a $L$ bounded using the argumentation of the point (i), and on the other hand I would construct a $L$ compact by using the fact that we can find a compact operator by summing $L_n$ (using the approximation propriety of compact operators). Probably this answer is a total mess.

Thanks!

答:
Yes, Uniform Boundedness Principle give the first part.

Take $L_n$ independent of $n$ to get an example where the limit is compact.

Let $(e_n)$ be an orthonormal basis for $H$. Let $ P_n$ be the projection on the span of $\{e_1,e_2,...,e_n\}$. Then $P_nx \to Ix$  for every $x$ and the identity operator $I$ is never compact on an infinite dimensional space.


答:
(i) $L_n$ are finite rank, so by definition are bounded. Since $L_nx$ converges, let $L(x):=\lim_{n\to\infty}L_nx$. That $L$ is linear is easy to verify.

Also, $\|L_nx\|\le c_x$ since $L_nx$ is a convergent sequence. Thus by the uniform boundedness theorem, $\|L_n\|\le c$ independent of $n$. Taking the limit of $L_nx\to L(x)$ then shows $L$ is bounded $$\|Lx\|=\lim_{n\to\infty}\|L_nx\|\le c\|x\|$$

(ii) For an example where $L$ is compact, take $H=\ell^2$ with $L_N(a_n):=(a_n/n)$ for $n\le N$ and $0$ otherwise. It converges to a compact operator.

For an example where $L$ is not compact take $L_N(a_n):=(a_n)$ for $n\le N$ and $0$ otherwise. It converges pointwise to the identity operator $L=I$ which is not compact.

Title: Proof on sequence of finite-rank operators

Body:
I had difficulties figuring out the solutions to the following problem:

Let $H$ be a separable infinite dimensional Hilbert space. Let
{${L_n}$} be a sequence of linear finite rank operators from $H$ to
into itself such that {${L_n}x$} converges in $H$ for any $x \in H$.
Prove that:

i) there exist $L: H \rightarrow H$ linear and bounded such that $L_nx
 \rightarrow Lx$ for any $x \in H$.

ii) $L$ can be either compact or not by showing and discussing two
examples.

As an idea (maybe wrong idea) I propose my solution concerning point (i):

Fixed $x$, we have that {${L_n}x$} is convergent; it follows that $||{L_n}x|| < \infty$ for any $n$.  Consider now n being fixed. We have that  $||{L_n}x|| < \infty$ for any $x \in H$, hence it follows that $L_n$ is linear and bounded for any n (is this correct?). The solution to point one follows by applying the uniform boudedness principle (or Banach-Steinhaus theorem) using the fact that $lim_n  (L_nx)$ is well defined for all $x \in H$.

Concerning point (ii): I would figure out a $L$ bounded using the argumentation of the point (i), and on the other hand I would construct a $L$ compact by using the fact that we can find a compact operator by summing $L_n$ (using the approximation propriety of compact operators). Probably this answer is a total mess.

Thanks!


答:
This equation is a variant of the so-called Underwood equation (have a look at my answer here).

Taking into account the specificity of this one, there are $n$ solutions each of them between a pair of vertical asymptotes corresponding to the $x_i$.

Now, if your condition is that you look for $0 <\theta < x_i^{min}$, then, instead of considering the function
$$f(\theta)=-\frac{n(1+a)}{\theta}+\sum_{i=1}^n\frac{a}{\theta-x_i}$$ consider
$$g(\theta)=\theta(\theta - x_i^{min})f(\theta)$$ which removes the first asymptotes.

Now, for an approximation, use Newton method with $\theta_0=0$ and you will see all $x_i$'s appearing in $x_1$.

This is a problem I worked for decades and I have published may papers on this topic. If you need more, just tell.

Edit

Since it is the first time I face a problem with an asymptote at exactly $\theta=0$, I reworked the equations.

$x_1$ is the samllest of the list and no other $x_i$ is equal to it.

What we have is
$$g(0)=(a+1) n x_1 > 0 \qquad \text{and} \qquad g(x_1)=a x_1 <0$$
$$g'(0)=a-(1+a)n+a x_1 \sum_{i=2}^n \frac 1 {x_i}$$
$$g''(0)=2a \Big[x_1 \sum_{i=2}^n \frac 1 {x_i^2} -\sum_{i=2}^n \frac 1 {x_i}\Big]$$

The first iterate of Newton method is then
$$\theta_1^{(2)}=\frac{(a+1) n x_1}{(1+a)n-a-a x_1 \left(\sum _{i=2}^n \frac{1}{x_i}\right)}$$

Let us try for $n=5$, $a=-\frac 1 2$, $x_i=\left\{\frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{1}{2}\right\}$. This gives $\theta_1^{(2)}=\frac 1 {10}=0.1000$ while the exact solution is $0.1127$.

Using Halley's method, the first iterate would be $\theta_1^{(3)}=\frac 5 {47}=0.1064$ while the exact solution is $0.1127$.

Using Householder's method, the first iterate would be $\theta_1^{(4)}=\frac {47} {430}=0.1093$ while the exact solution is $0.1127$.

Notice that using the secant method, we should have
$$\theta_1^{(1)}=\frac{ (1+a)n x_1}{(1+a)n-a}$$ which totally hides the contribution of the other $x_i$'s. For the worked case, this would give $\theta_1^{(1)}=\frac {5} {36}=0.1389$ while the exact solution is $0.1127$.

We could continue with higher-order method which, as a function of their order, would generate the sequence
$$\left\{\frac{1}{10},\frac{5}{47},\frac{47}{430},\frac{2150}{19407},\frac{19407}{173
   918},\frac{173918}{1552301},\frac{1552301}{13822834},\frac{69114170}{614604843}\right\}$$ which are closer and closer to the solution (the last in the table is $0.1125$).

Title: Simplify $\sum_{i=1}^n\frac{a}{\theta-x_i}$ where $\{x_i\}\in(0,\theta)$ and $a\in(-1,0]$

Body:
I have this problem where I need to solve $-\frac{n(1+a)}{\theta}+\sum_{i=1}^n\frac{a}{\theta-x_i}=0 $ for $\theta$ (the solution is allowed to be in terms of the mean of $x_1,..., x_n$), provided  that $\{x_i\}\in(0,\theta)$ and $a\in(-1,0]$.

I really have no idea how to simplify the summation. Maybe I'm just missing something simple?