问:
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano

Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$

Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.

Also, can we prove that there is only one real solution without using the discriminant?

答:
When you say "full Cardano" are you referring to using a formula? Because there is a process for solving a cubic that is constructive. But maybe you do not want this because it is essentially how to construct Cardano's formula.

Substituting $x=y+3$ leads to:

$$\begin{align}
x^3-9x^2-15x-6&=0\\
(y+3)^3-9(y+3)^2-15(y+3)-6&=0\\
y^3-42y-105&=0
\end{align}$$

Now introduce parameters $u$ and $v$ such that their sum is a solution $y$:
$$\begin{align}
(u+v)^3-42(u+v)-105&=0\\
u^3+v^3+(3uv-42)(u+v)-105&=0
\end{align}$$
We have freedom to choose $u$ such that $3uv-42=0$. That implies $v=\frac{14}{u}$ and simplifies the equation above:
$$\begin{align}
u^3+\left(\frac{14}{u}\right)^3-105&=0\\
u^6-105u^3+14^3&=0
\end{align}$$
This is quadratic in $u^3$:
$$\begin{align}
u^3&=\frac{105\pm\sqrt{105^2-4\cdot14^3}}{2}\\
u^3&=\frac{105\pm7}{2}\\
\end{align}$$
You may assume the "$+$" for $u^3$, since the "$-$" would give the symmetric $v^3$.
$$\begin{align}
u^3&=\frac{105+7}{2}=56\\
u&=2\sqrt[3]{7}
\end{align}$$
From which $v=\frac{14}{2\sqrt[3]{7}}=\frac{7}{\sqrt[3]{7}}=\sqrt[3]{7^2}$.

And now $x=y+3=u+v+3=2\sqrt[3]{7}+\sqrt[3]{7^2}+3$.


问:
Every distance squared between two points in $\mathbb{Z}^n$ is clearly all $\in \mathbb{N}$. Is the converse true (upto isomorphism)?

Every distance squared between two points in $\mathbb{Z}^n$ is clearly all $\in \mathbb{N}$. Is the converse true (upto isomorphism)?

The title was rather informal, so clarifying the question:

Is every countably infinite set $S \subset \mathbb{R}^n$ such that

$$\forall v_1, v_2 \in S,  v_1 \neq v_2  \rightarrow d(v_1, v_2)^2 \in \mathbb{N}$$

subset of $L + \delta$, where  $L$ is some lattice in $\mathbb{R}^n$ and $\delta \in \mathbb{R}^n$
?

I suspect it would be true, here is some approach I've took.

Fix a point v. translate $S$, so that $v = 0$. Also transform so that $\dim S = n$. Define $S^*$ as maximal superset of $S$ that also satisfies such property. It would be enough to show that $S^*$ is a lattice.

Fix some set of $k$ points, and denote them as $T$. Denote set of points $p$ in $\mathbb{R}^n$ such that $d^2(p, t) \in \mathbb{N}$ for every $t \in T$ as $P_T$. By definition, $S^* \subset P_T$.

(choosing nice $k$ points). Since $S^*$ is countable, set of every size $k$ subset of $S^*$ is also countable. Therefore we select $k$ points $T$ that has maximal rank with respect to $S$. Formally this would mean selecting $k$ points that $\dim (Span(T) \cap Span(S^*))$ is maximum. Here dim is used in lattice sense.

It would be enough to prove that for $k = n$, such $T$ satisfies $P_T = S^*$.

There are close problems, but non exact search results...
https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Ulam_problem

Seems to be lots of researches with "integral distances" but my google search skills limits me with finding non about distance squared being integral.

https://link.springer.com/article/10.1007/s00454-003-0014-7

答:
Following your approach, we translate so that $0\in S$, and define $S^*$ as a maximal superset of $S$ satisfying the requested property.

To prove that $S^*$ is a lattice, it suffices to show that it is a discrete subgroup of $\mathbb{R}^n$, i.e. that for any points $a,b \in S^*$, their sum and difference $a\pm b$ are also in $S^*$ (discreteness is obvious since each point is at least $1$ unit of distance away from the others).

First, we note that for any $x,y \in S^*$, we have

$$|x|^2 = d(x,0)^2 \in \mathbb{N}\cup\{0\}$$

and

$$2\,x\cdot y = d(x,0)^2+d(y,0)^2-d(x,y)^2 \in \mathbb{Z},$$

where $\cdot$ and $|\:\:|$ denote the standard Euclidean inner product and norm respectively.

Next, given an arbitrary $c \in S^*$ we compute

$$d(a\pm b,c)^2=|a\pm b-c|^2 = |a|^2+|b|^2+|c|^2\pm 2\,a\cdot b-2\,a\cdot c \mp 2\,b \cdot c \in \mathbb{N}\cup\{0\}.$$

We conclude that $a\pm b \in S^*$ by maximality.

Finally, we prove that the rank of the lattice $S^*$ must be $n$: if it were lower, say $m<n$, then any unit-length vector perpendicular to the hyperplane $S^*\otimes \mathbb{R} \simeq \mathbb{R}^m$ would have integral squared distance to any vector in $S^*$ by the Pythagorean theorem, contradicting maximality.

ContentLicense: CC BY-SA 4.0


问:
Need a basic reminder regarding a simple vector calculus question
Hello I was recently working through the https://www.deeplearningbook.org/ and the author wrote the following;

In the context of a section on Principal Component Analysis,

Suppose $x=\{x^{(1)},x^{(2)},...x^{(m)}\}$ with each $ x \in \mathbb{R}^{n}$
For each $ x^{(i)} \in \mathbb{R}^{n}$ we want a corresponding code vector $c \in \mathbb{R}^{l}$  with $l \le n$. We want an encoder, $f(x)=c$ and a decoder $g$ so that $x$ approximates $g(f(x))$ We also constrain the columns of D to be orthogonal.

For $D \in \mathbb{R}^{n \times l}$, $c \in \mathbb{R}^{l}$ let $g(c)=Dc$ define the decoding

$$\nabla_{c}(-2x^{T}Dc+c^{T}c)=0$$

implies $c=D^{T}x$

My question is how does it go from $-2x^{T}Dc$ to $D^{T}x$ instead of $x^{T}D$

答: Notice that $x$ is not a matrix, $x$ is a set of vectors and so $D^T x$ and $x^T D$ are the same set of vectors.  Pedantically, in one case you have a set of column vectors and in the other a set of row vectors.  However, being a row or column vector is usually only relevant when you are thinking of the set as a matrix.  Vectors qua vectors don't care about row-ness or column-ness.


问:
Prove that $x \in c=\{x=(x_k)_{k=1}^\infty \in l^\infty: \exists \lim_{k \to +\infty} x_k \in \mathbb{C}\}$ and $x_n \in c, x_n \to (x_k)_{k=1}$
I am working on this problem
Let $c=\{x=(x_k)_{k=1}^\infty \in l^\infty: \exists \lim_{k \to +\infty} x_k \in \mathbb{C}\}$, let $x_n \in c, x_n \to (x_k)_{k=1}^\infty$ in the sup norm (that is in $l^{\infty})$. Prove that $x \in c$.

My try:

Since $x_n \in C : \lim_{k \to +\infty} x_{n,k}=l_n \in \mathbb{C}$ that is  $ \forall \varepsilon > 0, |x_{n,k}-l_n|< \varepsilon \;\forall k>K_{\varepsilon}(n) ....(\theta)$

Since  $x_n \to x :$ for $\varepsilon > 0 $ $|x_{n,k}-x_k| \leq ||x_n-x||_{\infty} <\varepsilon \;\forall n> N_{\varepsilon }....(\beta)$

Taking the limit for $k \to \infty$:$$\left|\lim_{k \to +\infty}x_{n,k}-\lim_{k \to +\infty}x_k\right| \leq \varepsilon $$  $$ \left|l_n-\lim_{k \to +\infty}x_k\right| \leq \varepsilon$$
which means that, calling $l :=\lim_{k \to +\infty}x_k...(\zeta)$, $$ |l_n-l| \leq \varepsilon...(\gamma)$$ so that $$l :=\lim_{n \to +\infty}l_n$$

then $|x_k -l|=|x_k-x_{n,k}+x_{n,k}-l_n+l_n-l|\leq |x_k-x_{n,k}|+|x_{n,k}-l_n|+|l_n-l|$

and using $(\theta),(\beta)$ and $(\gamma)$:
$$|x_k -l|<3\varepsilon, \text{for } k> K(N_{\varepsilon})$$ which proves that  $x_k \to l$ and so $x \in c$.

I think it is wrong because at $(\zeta)$,I am defining $\lim_{k \to +\infty}x_k =l$, which is precisely what I wanted to prove. Can someone tell me if this or somewhere else is wrong and help me fixed it?

答:
You correctly identified the mistake, we have to first prove that $(l_n)_n$ converges and then that $(x_k)_k$ converges to $\lim_{n\to\infty} l_n$.

To prove this, we can show that $(l_n)_n$ is a Cauchy sequence. Pick $\varepsilon > 0$. Sequence $(x_n)_n$ is converges in $\|\cdot\|_\infty$ so it is in particular Cauchy. Therefore we can pick $n_0 \in \Bbb{N}$ such that $m,n \ge n_0 \implies \|x_m-x_n\|_\infty \le \frac\varepsilon3$.

Now for $m,n\ge n_0$ pick an index $k \in \Bbb{N}$ such that $|x_m(k)-l_m|, |x_n(k)-l_n|\le\frac\varepsilon3$. Then
$$|l_m-l_n| \le |l_m-x_m(k)| + \underbrace{|x_m(k)-x_n(k)|}_{\le \|x_m-x_n\|_\infty} + |x_n(k)-l_n| \le \frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.$$

Hence $(l_n)_n$ converges to some $l \in \Bbb{C}$.

Now it remains to prove that $(x_k)_k$ converges to $l$, which you already did in the second part of your proof.


问:
Transforming overdetermined system of equations to normal system of equations
Consider the following overdetermined linear system

$$Ax=b$$

where $A$ is a given tall $n \times k$ matrix ($n>k$), $x$ is a $k$-vector of unknowns, and $b$ is a given $n$-vector. Suppose further that matrix $A$ has full column rank, i.e., $\mbox{rank} (A)=k$. Is it legitimate to delete linearly dependent rows?

If so, then we obtain another system of linear equations which has the same number of equations as the number of unknowns. Hence, the conclusion is that any overdetermined system of linear equations can be transformed to one where the number of equations is equal to the number of unknowns. Where is the problem in my reasoning? Please guide me!

答:
Whether it is a consistent or inconsistent system of equations, if you want to use analytical methods, you can delete any dependent row which you can establish to be a valid linear combination of any subset of the other undeleted rows. However you must also join the left hand side values as a column of your matrix with value: -b. Otherwise you will struggle with inconsistent results in your calculation. The negation of that column is essential.

Special Note: make sure you do your row deletions, only one row at a time. If you delete more than one row at a time, you will inadvertently delete a linearly essential row. Guassian reduction remains the best way to identify such rows even when N > M. It also exposes the row which will never combine with any set of the rest, or the cause of the inconsistency, if the system is inconsistent.


问:
Verbal Subgroups of $\mathbb{Z}^2$
So I want to try to show that all the verbal subgroups of an abelian group $G$ are of the form $G(X^n), n \geq 1$. I want to start with $\mathbb{Z}$ and $\mathbb{Z}^n$. So I worked with $\mathbb{Z}$, I know that all the subgroups of $\mathbb{Z}$ are $m\mathbb{Z} = $ $\langle {m} \rangle$ $ = \mathbb{Z}(X^m)$, so then all the subgroups of $Z$ are verbal.

Now if I look at $\mathbb{Z}^2$, the subgroups of this are:

$(0,m) \mathbb{Z} = \langle (0,m) \rangle$

$(m,0) \mathbb{Z} = \langle (m,0) \rangle$

$(a,b) \mathbb{Z} = \langle (a,b) \rangle$ $ a,b \in \mathbb{Z}-0$

$(a,b) \mathbb{Z} + (c,d) \mathbb{Z} = \langle (a,b) \rangle + \langle (c,d) \rangle$

Now I know that a verbal subgroup must be fully characteristic, so i know the first 3 subgroups are not verbal under the endomorphism $f(x,y) = (y,x)$. In the third case if $a = b$ then it fails under the endomorphism $g(x,y) = (x+1,y+2)$.

The last case, I'm not really sure about, I could show it fails in general with $f$ as above but perhaps in certain cases it would work? But then how do I show it is verbal , I'm not even sure how to begin to write any of these subgroups as $\mathbb{Z}(X^n)$, it doesn't seem like you can, at least not as easily as we did for $\mathbb{Z}$.

Am I missing a subgroup of $\mathbb{Z}^2$? How could I extend this to a general $\mathbb{Z^n}$?

Edit: Definition of verbal subgroup. Let $W$ be a set of words, then the verbal subgroup $G(W)$ of a group G, is the subgroup of G generated by the words $w \in W$, written with elements of G.

i.e. $G(X^2) = \langle g^2 | g \in G\rangle$ or $G(X^2, XYX) = \langle g^2, ghg | g,h \in G\rangle$

答:
Any word $w(x_1,\ldots,x_n)$ is equivalent to a power word $x^a$ and a commutator word $c$; that is, the verbal subgroup determined by $w$ is equal to the verbal subgroup determined by $x^a$ and $c$.

Indeed, this can be verified through the collection process. Consider an arbitrary word $w$. The first occurrence of $x_1$ will look something like
$$w_1 x_1^a w_2$$
where $w_1$ and $w_2$ are words, and $w_1$ does not contain any instances of $x_1$. But we can rewrite this as
$$w_1x_1^aw_2 = x_1^aw_1[w_1,x_1^a]w_2 = x_1^aw_1w_2[w_1,x_1^a][[w_1,x_1^a],w_2] = x_1^aw_1w_2c,$$
where $c\in [F_n,F_n]$, the commutator subgroup of the free group in $n$ letters. Then we can do the same thing with the next occurence of $x_1$, which will be in $w_2$, and so on until we move the last original occurrence of $x_1$ all the way to the left. We end up with a word of the form
$$x_1^{a_1}w_3c_1$$
where $w_3$ is a word that does not contain $x_1$, and $c_1$ is an element of the commutator subgroup of the free group $F_n$. We can then repeat the process with $x_2$, $x_3$, etc. In the end, we obtain that the original word is equal to the word
$$w(x_1,\ldots,x_n)=x_1^{a_1}\cdots x_n^{a_n}c$$
where $c\in [F_n,F_n]$.

Among the elements in $w(G)$ (the verbal subgroup of $G$ determined by $w$) are the elements $w(g,e,\ldots,e)$, $w(e,g,e,\ldots,e),\ldots,w(e,\ldots,e,g)$. For each of these words we have that $c$ evaluates to the trivial element, since the value lies in the subgroup $\langle g\rangle$ which is abelian. Thus, the verbal subgroup $w(G)$ includes all values of the form $g^{a_1},\ldots,g^{a_n}$.

If $d=\gcd(a_1,\ldots,a_n)$, then we know that $d=\alpha_1a_1+\cdots+\alpha_na_n$ for integers $\alpha_1,\ldots,\alpha_n$, hence the elements $g^d$ are also all in the verbal subgroup. And clearly with the elements $g^d$ we also obtain all elements $g^{a_1},\ldots,g^{a_n}$.

Thus, the verbal subgroup determined by the words $x_1^d$ and $c$ (the commutator word from the rewritten $w$) is equal to $w(G)$: because the latter is generated by the values $g_1^{a_1}\cdots g_n^{a_n}c(g_1,\ldots,g_n)$, and since we also have $g_1^{a_1},\ldots,g_n^{a_n}\in w(G)$, we also have $c(g_1,\ldots,g_n)$ separately. Thus, $w(G)$ is contained in the verbal subgroup $W(G)$ with $W=\{x_1^d,c\}$, and conversely.

From this we conclude that any verbal subgroup is equal to the verbal subgroup determined by a single power word $x^a$, plus a set of commutator words $\{c_i\}_{i\in I}$. This holds in any group.

(This argument can be found near the beginning of Hanna Neumman’s book, Varieties of groups, and is due to Bernhard Neumman.)

Now if $G$ is abelian, then all commutator words evaluate to the trivial element. Thus, any verbal subgroup of an abelian group is of the form $G^n = \{g^n\mid g\in G\}$ for some positive integer $n$ (technically, the subgroup generated by these elements, but in an abelian group, the set of $n$th powers is already a subgroup).

In particular, this holds for $\mathbb{Z}^2$.

Title: Verbal Subgroups of $\mathbb{Z}^2$


问:
On the sheaf of discontinuous sections
Let $X$ be a topological space and $\mathcal F$ be a sheaf. Godement defined the sheaf of discontinuous sections $\mathcal G(\mathcal F)$ by putting, for an open set $U\subseteq X$, $\mathcal G(\mathcal F)=\prod_{x\in U}\mathcal F_x$, with the natural restriction maps. If $X_{\text{disc}}$ is the set $X$ with the discrete topology, this Wikipedia's article says that we have $\mathcal G(\mathcal F)\simeq p_*p^{-1}\mathcal F$, where $p:X_{\text{disc}}\to X$ is the identity, $p_*$ is the direct image of $p$ and $p^{-1}$ is the inverse image of $p$. I have tried to prove this directly computing the stalks but failed.

How can one prove the isomorphism $\mathcal G(\mathcal F)\simeq p_*p^{-1}\mathcal F$?

答:
For any sheaf on any topological space $X$, if $U=\coprod U_i$ is a disjoint union of open sets then $\mathcal{F}(U)=\prod\mathcal{F}(U_i)$ as it can easily be seen from the gluing condition of a sheaf. Thus on a discrete space $X$, we have $U=\coprod_{x\in U}\{x\}$ and so $\mathcal{F}(U)=\prod_{x\in U}\mathcal{F}(\{x\})$. Moreover $\mathcal{F}(\{x\})=\mathcal{F}_x$ since the collection of neighborhood of $x$ has a smallest element, namely $\{x\}$. Hence over a discrete space $X$, for any sheaf $\mathcal{F}$, we have $\mathcal{F}(U)=\prod_{x\in U}\mathcal{F}_x$.

Now let us compute the stalks of $p^{-1}\mathcal{F}$. If $x\in X$, denote $i_x:\{x\}\to X$ the inclusion, then $$(p^{-1}\mathcal{F})_x=i_x^{-1}p^{-1}\mathcal{F}=(pi_x)^{-1}\mathcal{F}=i_x^{-1}\mathcal{F}=\mathcal{F}_x$$
In other words, the stalks of $p^{-1}\mathcal{F}$ are the same as the stalks of $\mathcal{F}$.

Combining 1. and 2. together, we find that
$$(p^{-1}\mathcal{F})(U)=\prod_{x\in U}(p^{-1}\mathcal{F})_x=\prod_{x\in U}\mathcal{F}_x$$

Finally, take the pushforward of $p^{-1}\mathcal{F}$ and remember that $p$ is just the identity over the underlying sets :
$$(p_*p^{-1}\mathcal{F})(U)=(p^{-1}\mathcal{F})(p^{-1}(U))=(p^{-1}\mathcal{F})(U)=\prod_{x\in U}\mathcal{F}_x=\mathcal{G(F)}(U)$$
Therefore, we have an isomorphism $\mathcal{G(F)}\simeq p_*p^{-1}\mathcal{F}$.

Title: On the sheaf of discontinuous sections


问:
Smoothness of the associated G-valued map of a bundle automorphism of a principal G-bundle
Let $\pi: P\rightarrow M$ be a principal $G$-bundle and $f:P\to P$ be a bundle automorphism. Define $\sigma_f: P\to G$ by $f(p)=p\cdot \sigma_f(p)$. It is claimed that $
\sigma_f$ is a G-valued map, i.e. a smooth map and $\sigma_f(p\cdot g)=g^{-1}\sigma_f(p)g$. I can prove the second condition, but why is $\sigma_f$ smooth?

答:
Let $\phi: P_U \to U\times G$ be a local trivialization associated with a local gauge (section) $s: U \to P$, i.e. $\phi(s(x)\cdot g) = (x,g)$. Then $\phi$ can be written as $\phi(p) = (\pi(p),\psi(p))$ where $\psi: P_U \to G$ is smooth.

Therefore, any $p\in P_U$ can be written as $p=s(\pi(p))\cdot \psi(p)$, and $f(p)=s(\pi(f(p)))\cdot \psi(f(p))=s(\pi(p))\cdot  \psi(f(p))$ as $f$ preserves fibres.

By the properties of free and transitive actions, and since $f(p)=p\cdot \sigma_f(p)$, we have $\sigma_f(p)=(\psi(p))^{-1}\psi(f(p))$. As the multiplication map, inverse map, $\psi$ and f are all smooth, so is $\sigma_f$ on $P_U$. As smoothness is a local property, $\sigma_f$ is smooth on the whole of $P$.


问:
Integration involving roots
$$\int\frac{dx}{(1+x^\frac{1}{4})x^\frac{1}{2}}$$
This is my work:
$$u^4=x$$
$$4u^3=dx$$
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u^3du}{(1+u)u^2}=-4(1+x^\frac{1}{4})^{-1}+2(1+x^\frac{1}{4})^{-2}+C$$
But wolframAlpha gives quite a different answer. So, have I done a mistake? Or somehow both answers are correct?</p>

答:
Check this calculations. They give the same result as wolfram alpha
$$
\int\frac{4u}{1+u}du=
4\int\left(1-\frac{1}{u+1}\right)du=
4\int du-4\int\frac{1}{u+1}du=
4u-4\ln(1+u)=\\
4x^{1/4}-4\ln(1+x^{1/4})
$$</p>

ContentLicense: CC BY-SA 3.0

Title: Integration involving roots


问:
Using the precise definition of limits to prove a limit
Question

Let $a$ be a positive real number. Use the $\epsilon-\delta$ definition of limits to prove that $$\lim\limits_{x\to a}\ \frac {3x^2 + 2ax + a^2} {2x + a} = 2a\ .$$

My working

Let $\epsilon > 0$ be given.

For $|x - a| < \delta$, choose $\delta = \frac 1 3 \epsilon - 2a$.

\begin{align}
|x + a| & =
|x - a + 2a|
\\[5 mm] & \leq
|x - a| + 2a
\\[5 mm] & <
\frac 1 3 \epsilon
\end{align}

\begin{align}
\left|\frac {3x^2 + 2ax + a^2} {2x + a} - 2a\right| & =
\left|\frac {3x^2 - 2ax - a^2} {2x + a}\right|
\\[5 mm] & =
\left|\frac {(3x + a)(x - a)} {2x + a}\right|
\\[5 mm] & <
\left|\frac {3(x + a)(x - a)} {2(x - a)}\right|
\\[5 mm] & =
\frac 3 2 |x + a|
\\[5 mm] & <
\frac 1 2 \epsilon
\\[5 mm] & <
\epsilon
\end{align}

May I know if I have chosen a suitable $\delta$ and if my proof is logical? Any comments/suggestions to improve my working will be greatly appreciated :)

Edit

Thank you everyone for your inputs! I have learnt a lot :)

答:
There are at least two major errors.

The first is where you try to choose $\delta$ in such a way to ensure that
$$|x - a| + 2a \stackrel?< \frac13 \epsilon. \tag1$$
Leaving aside the problems with your method of choosing $\delta,$
just take a good long look at Inequation $(1)$.

Now recall that $a$ is some given positive integer and that absolute values are never negative. Therefore the left hand side is never less than the positive number $2a.$
So if we just consider the case $\epsilon = a,$ it should be clear that
$\frac13 \epsilon = \frac13 a < 2a \leq |x - a| + 2a,$ and therefor Inequation $(1)$ is false for this particular choice of a positive number $\epsilon$.
Now do you recall that part of the definition of a limit where it says
"for all $\epsilon > 0$"?
It means that whatever you do, you have to have an answer for every positive $\epsilon.$ If there is even one positive $\epsilon$ for which your proof doesn't work, your proof doesn't work. And that's how it will go if you try to base a proof on Inequation $(1)$.

The second major error I found is here:
$$
\left|\frac {(3x + a)(x - a)}{2x + a}\right| \stackrel?<
\left|\frac {3(x + a)(x - a)}{2(x - a)}\right|. \tag2
$$

Note that the right-hand side is not even defined unless $x - a\neq 0,$
but in the context of the definition of a limit we actually have
$0 < |x-a| < \delta$, not just $|x-a| < \delta$, so that's OK.
And given that $x - a \neq 0,$ we can cancel a factor of $|x - a|$ on both sides, which means you actually need this to be true:
$$
\left|\frac {3x + a}{2x + a}\right| \stackrel?<
\left|\frac {3(x + a)}{2(x - a)}\right|.
$$

So let $x - a = t$, where $0 < |t| < \delta.$ Then $x = a + t$ and
$$
\left|\frac {3x + a}{2x + a}\right|
= \left|\frac {4a+3t}{3a + 2t}\right|.
$$
For very small values of $t$ the right-hand side is approximately $\frac 43.$

On the other hand,
$$
\left|\frac {3(x + a)}{2(x - a)}\right|
= \left|\frac {6a + 3t}{2t}\right|,
$$
which gets very large as $t$ gets very small.
For example, if $0 < t < \frac1{100}a,$
the right-hand side is greater than $200.$
So in order to show that Inequation $(2)$ is true, you need to show that at least some numbers greater than $200$ are approximately equal to $\frac 43.$
That is not true even for the largest possible value of $\frac43,$ so a proof that relies on Inequation $(2)$ won't work.


问:
Homology group of a topological n-manifold
Let M be an n dimensional topological manifold where $n \geq 0$ and $x \in M$. Then, $H_n(M,M \setminus \{ x\}) \cong \mathbb{Z}$.

I am familiar with singular homology and I already proved the claim for every $n \neq 1$ (by using excision).

I am having trouble proving the claim for $n=1$. I already proved that $H_1(M,M \setminus \{ x\}) \cong H_1(\mathbb{R},\mathbb{S}^0)$  and I'm wondering how to prove that $H_1(\mathbb{R},\mathbb{S}^0) \cong \mathbb{Z}$ where $\mathbb{S}^0= \{\pm1\}$ to finish the proof.

I also know that the sequence $H_1(\mathbb{R}) \rightarrow H_1(\mathbb{R},\mathbb{S}^0) \rightarrow H_0(\mathbb{S}^0) \rightarrow H_0(\mathbb{R}) \rightarrow H_0(\mathbb{R},\mathbb{S}^0)$ is exact-this is a part of the long exact sequence of the pair $(\mathbb{R},\mathbb{S}^0)$-  and that $H_1(\mathbb{R}) \cong 0,H_0(\mathbb{S}^0) \cong \mathbb{Z } \oplus \mathbb{Z}, H_0(\mathbb{R}) \cong \mathbb{Z}$ in case this can help.

答:
Use instead the long exact sequence of reduced homology, which ends
$$H_1(\mathbb R) \mapsto H_1(\mathbb R,\mathbb S^0) \mapsto \widetilde H_0(\mathbb S^0) \mapsto \widetilde H_0(\mathbb R)
$$
Since $\text{rank}(\widetilde H_0(X)) = \text{rank}(H_0(X))-1$ for any $X$, the above sequence becomes
$$0 \mapsto H_1(\mathbb R,\mathbb S^0) \mapsto \mathbb Z \mapsto 0
$$
(You can also do it with the sequence in your question, if you work out the actual formulas for the homomorphisms).


问:
Let $X$ be the disjoint union of copies of a subspace of $\mathbb{R}$. Is the quotient of $X$ by identifying accumulation points first countable?
Sorry the title isn't super clear, I ran out of characters.

From Willard's General Topology, 16A.5:

For each $n \in \mathbb{N}$, let $X_n$ be a copy of the subspace
$\{0\} \cup \{\frac{1}{m}: m = 1, 2, \ldots \}$ of $\mathbb{R}$. Let $X$ be the disjoint union of the $X_n$. Is the quotient $Y$ of $X$ obtained by identifying all accumulation points of $X$ first
countable?

I'm really having a hard time with this question. I'm struggling more generally with both the notions of the disjoint union topology, and the quotient topology, so this is really throwing me for a loop.

The relevant information that I know is:

If $X$ is a topological space and $a \in X$, then we say $a$ is an accumulation point of a set $A$ iff each nbhood of $a$ meets $A$ in some point other than $a$,

The set $X_n'$ of all accumulation points of $X_n$ (defined above) is $X_n' = \{0\}$.

But I'm not really sure what $X$, the disjoint union of these $X_n$, "looks" like. Is the set of accumulation points of $X$ the union of the accumulation points of the $X_n$? I'm really just not sure how to proceed.

Any help is appreciated.

答:
$X$ is homeomorphic to $\Bbb Z^+\times X_1$ with the product topology, where $\Bbb Z^+$ has the discrete topology. $\Bbb Z^+\times X_1$, and therefore also $X$, looks something like the diagram below: the points, symbolized by labelled bullets, are ordered pairs $\left\langle k,\frac1n\right\rangle$ for $k\in\Bbb Z^+$ and $\frac1n\in X_1$.

$$\begin{array}{rrr}
\langle 1,1\rangle\;\bullet&\langle 1,1\rangle\;\bullet&\langle 1,1\rangle\;\bullet&\langle 1,1\rangle\;\bullet&\cdots\\
\left\langle 1,\frac12\right\rangle\bullet&\left\langle 1,\frac12\right\rangle\bullet&\left\langle 1,\frac12\right\rangle\bullet&\left\langle 1,\frac12\right\rangle\bullet&\cdots\\
\left\langle 1,\frac13\right\rangle\bullet&\left\langle 1,\frac13\right\rangle\bullet&\left\langle 1,\frac13\right\rangle\bullet&\left\langle 1,\frac13\right\rangle\bullet&\cdots\\
\left\langle 1,\frac14\right\rangle\bullet&\left\langle 1,\frac14\right\rangle\bullet&\left\langle 1,\frac14\right\rangle\bullet&\left\langle 1,\frac14\right\rangle\bullet&\cdots\\
\vdots\,&\vdots\,&\vdots\,&\vdots\,\\
\langle 0,0\rangle\;\bullet&\langle 0,0\rangle\;\bullet&\langle 0,0\rangle\;\bullet&\langle 0,0\rangle\;\bullet&\cdots
\end{array}$$

To get $Y$, just glue all of the $0$ points together in a single point: you end up with a sort of hedgehog, with countably infinitely many spikes — the sets $X_n\setminus\{0\}$, which are columns in the picture above — sticking out of the central point where all of the $0$s are glued together. Call that central point $p$.

HINT: You should be able to check easily that points of $Y\setminus\{p\}$ are isolated, so $Y$ is first countable at each of those points.

If $\sigma=\langle k_n:n\in\Bbb Z^+\rangle$ is any sequence of positive integers, let $$B(\sigma)=\{p\}\cup\bigcup_{n\in\Bbb Z^+}\left\{x\in X_n:x\le\frac1{k_n}\right\}\,.$$

Check that each of these sets $B(\sigma)$ is an open nbhd of $p$ and that in fact they form a local base at $p$. Then show that if $\mathscr{V}$ is any countable family of open nbhds of $p$, there is a $B(\sigma)$ that does not contain any member of $\mathscr{V}$, so $\mathscr{V}$ cannot be a local base at $p$. This shows that $Y$ is not first countable at $p$.


问:
Commutator ideal of $\mathfrak{gl}(V)$ for $V$ infinite-dimensional
It is well-known that the commutator ideal of $\mathfrak{gl}(V)$ is $\mathfrak{sl}(V)$, the Lie algebra of traceless endomorphisms, if $V$ is a finite-dimensional vector space. I am now wondering what is known about $[\mathfrak{gl}(V), \mathfrak{gl}(V)]$ if the vector space $V$ is infinite-dimensional. I am in particular wondering about the the codimension of $[\mathfrak{gl}(V), \mathfrak{gl}(V)]$ in $\mathfrak{gl}(V)$.

答:
The codimension is zero. That is, $\mathfrak{gl}(V)$ is a perfect Lie algebra.

Indeed, first use that $V\simeq W\oplus W$ for some vector space $W$. Under this "block" decomposition, every element of $\mathfrak{gl}(V)$ can be written as a square block matrix
$$\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&0\\0&0\end{pmatrix}+\begin{pmatrix}0&B\\0&0\end{pmatrix}+\begin{pmatrix}0&0\\C&0\end{pmatrix}+\begin{pmatrix}0&0\\0&D\end{pmatrix}.$$
It is enough to prove that each of these matrices is a commutator (thus every element of $\mathfrak{gl}(V)$ is a sum of 4 commutators; 4 is probably not optimal). Indeed,
$$\begin{pmatrix}0&B\\0&0\end{pmatrix}=\Big[\begin{pmatrix}I&0\\0&0\end{pmatrix},\begin{pmatrix}I&B\\0&0\end{pmatrix}\Big],$$
and similarly $\begin{pmatrix}0&0\\C&0\end{pmatrix}$ is a commutator.

Next, to deal with $f=\begin{pmatrix}A&0\\0&0\end{pmatrix}$, we use that $W$ is isomorphic to $W^{(\mathbf{N})}$ and thus write $V$ as $\bigoplus_{n\in\mathbf{Z}} W_n$, where each $W_n$ is given with an isomorphism with $W$, and $W_0$ is identified to the left-hand $W$ in the previous $W\oplus W$ decomposition. Hence, in this new decomposition, $f$ is the block-diagonal matrix whose diagonal is $(\dots 0,0,A,0,0\dots)$ where $A$ is in position zero. For $w\in W$, write $w[n]$ as the element $w$ viewed in the $n$-th copy $W_n$ (through the given identification). Define $g,h\in\mathfrak{gl}(V)$ as follows: $g(w[n])=0$ if $n\le 0$, $g(w[n])=w[n-1]$ if $n>0$; $h(w[n])=(Aw)[n+1]$ for all $n\in\mathbf{Z}$. Then
$$[g,h](w[0])=gh(w[0])-hg(w[0])=g((Aw)[1])=(Aw)[0];$$
$$[g,h](w[n])=gh(w[n])-hg(w[n])=0-0=0 \quad (n<0);$$
$$[g,h](w[n])=gh(w[n])-hg(w[n])=g((Aw)[n+1])-h(w[n-1])=$$
$$= (Aw)[n]-(Aw)[n])=0  \quad (n\ge 0).$$

That is, $[g,h]=f$, so $f$ is a commutator. Similarly, $\begin{pmatrix}0&0\\0&D\end{pmatrix}$ is a commutator.


答:
Often a substitution can simplify the calculations or clarify the issues. Let $f(x)=\frac {3x^2+2ax+a^2} {2x+a}$ when $2x+a\ne 0.$ Let $x=y+a.$ When $2x+a\ne 0$ we have $$|f(x)-2a|=|y|\cdot\frac {|a+3y|}{|3a+2y|}.$$ Now if $|y|<|a|$ then (i)... $|a+3y|\le |a|+3|y|<4|a|=4a\;$ and (ii)... $|3a+2y|\ge |3a|-|2y|>|a|=a>0,\;$ so (iii)... $\frac {|a+3y|}{|3a+2y|}<4,\;$ so $$|f(x)-2a|\le 4|y|.$$
So, given $\epsilon>0,$ let $\delta=\min(\epsilon/4, a).$ Now if $|x-a|<\delta$ then  $f(x)$ exists (because if $|x-a|0$ then $2x+a\ne 0$) and $$|f(x)-2a|\le 4|y|=4|x-a|<4\delta\le\epsilon.$$

Title: Using the precise definition of limits to prove a limit

Body:
Question

Let $a$ be a positive real number. Use the $\epsilon-\delta$ definition of limits to prove that $$\lim\limits_{x\to a}\ \frac {3x^2 + 2ax + a^2} {2x + a} = 2a\ .$$

My working

Let $\epsilon > 0$ be given.

For $|x - a| < \delta$, choose $\delta = \frac 1 3 \epsilon - 2a$.

\begin{align}
|x + a| & =
|x - a + 2a|
\\[5 mm] & \leq
|x - a| + 2a
\\[5 mm] & <
\frac 1 3 \epsilon
\end{align}

\begin{align}
\left|\frac {3x^2 + 2ax + a^2} {2x + a} - 2a\right| & =
\left|\frac {3x^2 - 2ax - a^2} {2x + a}\right|
\\[5 mm] & =
\left|\frac {(3x + a)(x - a)} {2x + a}\right|
\\[5 mm] & <
\left|\frac {3(x + a)(x - a)} {2(x - a)}\right|
\\[5 mm] & =
\frac 3 2 |x + a|
\\[5 mm] & <
\frac 1 2 \epsilon
\\[5 mm] & <
\epsilon
\end{align}

May I know if I have chosen a suitable $\delta$ and if my proof is logical? Any comments/suggestions to improve my working will be greatly appreciated :)

Edit

Thank you everyone for your inputs! I have learnt a lot :)

Body:
Often a substitution can simplify the calculations or clarify the issues. Let $f(x)=\frac {3x^2+2ax+a^2} {2x+a}$ when $2x+a\ne 0.$ Let $x=y+a.$ When $2x+a\ne 0$ we have $$|f(x)-2a|=|y|\cdot\frac {|a+3y|}{|3a+2y|}.$$ Now if $|y|<|a|$ then (i)... $|a+3y|\le |a|+3|y|<4|a|=4a\;$ and (ii)... $|3a+2y|\ge |3a|-|2y|>|a|=a>0,\;$ so (iii)... $\frac {|a+3y|}{|3a+2y|}<4,\;$ so $$|f(x)-2a|\le 4|y|.$$
So, given $\epsilon>0,$ let $\delta=\min(\epsilon/4, a).$ Now if $|x-a|<\delta$ then  $f(x)$ exists (because if $|x-a|<a>0$ then $2x+a\ne 0$) and $$|f(x)-2a|\le 4|y|=4|x-a|<4\delta\le\epsilon.$$


问:
Integral involving gamma density function
I am trying to numerically calculate the value of $I$ using R's "integrate" function.

$$
 I = \int_{t_1}^{t_2}\frac{b^{a(t-t_1)}}{\Gamma(a(t-t_1))}(y_2-y_1)^{a(t-t_1)-1}e^{-b(y_2-y_1)}(t_2-t)dt
$$
Note that, this is a gamma pdf multiplied with $(t_2-t)$ where $a(t-t_1)$ is the shape parameter. As per my understanding, I am multiplying the probability of jumping $(y_2-y_1)$ length in $(t-t_1)$ time period with $(t_2-t)$. I am expecting to get expected time between $t_1=0$ to $t_2$ that it jumps over $(y_2-y_1)$. Therefore my expectation is to get a result between $0$ to maximum $t_2$.

I used following parameters and got an answer 628.97 which does not make sense. It should be between $t_1$ and $t_2$.

```<code>  y1 <- 28
  y2 <- 30
  b <- 1
  a <- .2
  t1 <- 0
  t2 <- 140
</code>```

I am rather doubtful whether I am implementing the integration in R wrong or my formulation is wrong. Can you help me verify if my answer is correct. In that case, I can be sure that my formulation is wrong. Also any suggestion about formulation will be appreciated.

Edit:

Here I have a stochastic process where the process makes jumps as time progresses. Any jump between a time interval follows a gamma distribution with shape parameter $a(t-t_1)$ and scale parameter $b$ as depicted in the sketch. I am trying to calculate the expected length of time the process spends over $y_2$.

答:
With the parameters you have given, the integrand is
$$  \frac{2^{\frac{t}{5}-1} (140-t)}{e^2 \Gamma \left(\frac{t}{5}\right)}  $$

Plotting that integrand,

which area we can estimate to be aorund $20 \times 30 = 600$, compatible with your numerical result.  Asking Mathematica to perform the integration, we obtain $628.9792121\dots$, which is compatible with your R calculation.

I note that the gamma distribution with shape parameter $a(t-t_1)$ and scale parameter $b$ (a guess, since you do not specify both parameters to your gamma distribution) gives PDF
$$  \frac{e^{-\frac{t}{b}} t^{a (t-t_1)-1} b^{-a (t-t_1)}}{\Gamma (a(t-t_1))}  \text{,}  $$
which partially matches your integrand.  The difference might be your choice of scale parameter or may be something else.


问:
Statement of the DeFinetti's representation theorem
I'm using a standard textbook in statistics [1], and am confused with the statement of the DeFinetti's representation theorem.

Theorem 1.49 (DeFinetti's representation theorem). Let $(S,A,\mu)$ be a probability space, and let $(\mathcal{X},\mathcal{B})$ be a Borel space. For each $n$, let $X_n: S\to \mathcal{X}$ be measurable. The sequence $\{X_n\}_{n=1}^\infty$ is exchangeable if and only if there is a random probability measure $\mathbb{P}$ on $(\mathcal{X},\mathcal{B})$ such that, conditional on $\mathbb{P} = P$, the $X_n$'s are IID with distribution $P$. Furthermore, if the sequence is exchangeable, then the distribution of $\mathbb{P}$ is unique, and $\mathbb{P}_n(B)$ converges to $\mathbb{P}(B)$ almost surely for each $B \in \mathcal{B}$.

More precisely, I'm a confused with what a "random probability measure $\mathbb{P}$ on
$(\mathcal{X},\mathcal{B})$" means. Does it mean a random variable that takes value in
the space $M_{1}(\mathcal{X},\mathcal{B})$ of probability measure on $(\mathcal{X},\mathcal{B})$? This seems to
make the most sense because it later says

... the $X_n$ are IID with distribution P.

Here each $X_{n}$ is a measurable function from $S$ to $\mathcal{X}$, and thus its
distribution is the pushforward measure $X_{n,\star}(\mu)$, which is
an element in $M_{1}(\mathcal{X},\mathcal{B})$.

But if that's the case, then what's the domain of $\mathbb{P}$? I guess the domain is $S$.. it does not say it clearly. If the domain is indeed $S$, can $\mathbb{P}$ be written out in general in terms of the $X_{n}$?

Reference

[1] Theory of Statistics-[Mark Schervish]

答:
$\mathbb P$ is a map from $S \times \mathcal B$ to $[0,1]$ such that

a) $E \mapsto \mathbb P (\omega,E)$ is a probability measure on $\mathcal  B$ for each $\omega \in S$

b) $\omega  \mapsto \mathbb P (\omega,E)$ is a measurable function for each $E \in \mathcal B$

Conditioned on $\mathbb P=P$ means conditioned on the event $\{\omega \in S: \mathbb P (\omega,E) =P(E) \forall E \in \mathcal B\}$.


问:
Calculate the norm of vectors $(1,0)$ and $(0,1)$
Calculate the norm of vectors $(1,0)$ and $(0,1).$ We are on $\mathbb{R}^2$ where $\langle x,y\rangle=2x_1y_1+x_2y_2$. I know that I need to calculate the norm by $||x||=\sqrt{\langle x,x \rangle}$, but I'm having difficulty using that.

Looking at the first one, some of my classmates are computing it as
$\sqrt{2\cdot 1\cdot1+0\cdot 0}$, but I don't understand where that comes from.

Personally, I was thinking it would be $\sqrt{2\cdot 1\cdot 1 + 1\cdot 1}$, but honestly I'm not super confident there.

Any thoughts?

答: The norm of $x=(1,0)$ is  $\sqrt{\langle x,x\rangle}=\sqrt{2x_1x_1+x_2x_2}=\sqrt{2\cdot1\cdot1+0\cdot0}$.  Now can you compute the norm of $(0,1)$?


问:
Statistics Workshop for High School Students
We are going to hold an introductory workshop about the statistics. The participants will be students who have just finished their 8th or 9th grade. The workshop consists of 10 two-hour sessions. The students have a very limited background about the statistics: They can compute arithmetic average of a set of data, create pie charts, etc.

So we’re looking for some good educational material to introduce the students to some of the main statistical ideas, methods and concepts as well as to illustrate the importance of statistics in daily life.

Your suggestions would be of great help.

Thanks.

答:
Maybe you can look into:

Share My Lesson

High School Statistics Resources

Statistics and Probability

Books

Activities and Projects for High School Statistics Courses by Ron Millard, John C. Turner

See this list of books on Google

Teaching Stats in High School

Search out other such books and resources

You might also want to try some of the Opencourseware, search out more books and visit your local college library and peruse their book collection for ideas and materials.


Title: Calculate the norm of vectors $(1,0)$ and $(0,1)$

Body:
Calculate the norm of vectors $(1,0)$ and $(0,1).$ We are on $\mathbb{R}^2$ where $\langle x,y\rangle=2x_1y_1+x_2y_2$. I know that I need to calculate the norm by $||x||=\sqrt{\langle x,x \rangle}$, but I'm having difficulty using that.

Looking at the first one, some of my classmates are computing it as
$\sqrt{2\cdot 1\cdot1+0\cdot 0}$, but I don't understand where that comes from.

Personally, I was thinking it would be $\sqrt{2\cdot 1\cdot 1 + 1\cdot 1}$, but honestly I'm not super confident there.

Any thoughts?

Body:
$$\begin{align}||(1,0)|| & = \sqrt{\langle (1,0),(1,0)\rangle}\\ & = \sqrt{2 \cdot 1 \cdot 1 + 0 \cdot 0}\\ & = \sqrt{2}\end{align}$$

$$\begin{align}||(0,1)|| & = \sqrt{\langle(0,1),(0,1) \rangle}\\ & = \sqrt{2 \cdot 0 \cdot 0 + 1 \cdot 1} \\ & = \sqrt{1} \\ & = 1\end{align}$$


问:
Minimum value of exponential order
Find the smallest value of $\alpha$ such that $f(t)=6^{t+1}$ is of exponential order $\alpha$.
($f$ is said to be of eponential order $\alpha>0$ if there exists $M>0$ suc that $|f(t)|\leq Me^{\alpha t}$ )

We have $f(t)=6\cdot 6^t=6 \cdot e^{(\ln 6) \cdot t}$, Can $\alpha$ be determined from this inspiration?

答:
Say you have $0<\alpha < \ln 6$ then you must have that there exists $M>0$ and $T>0$ such that

$$6^{t+1} = 6 e^{(\ln 6) \cdot t} \leq M e^{\alpha \cdot t}$$

for all $t > T$.

This means $6 e^{t(\ln 6-\alpha)} \leq M $, but this is impossible because $\ln 6-\alpha >0 $ so $6 e^{t(\ln 6-\alpha)}$ becomes larger and larger . If we want to find a counterexample then taking $t = \frac{|\ln M|}{\ln 6-\alpha}+T$  suffices to negate  $ 6 e^{t(\ln 6-\alpha)} \leq M $ for any $M$ and $T$ .

Instead, $\alpha = \ln 6$ follows the definition so it is the minimum value we were looking for.

Title: Minimum value of exponential order


问:
Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$
I am working on this problem:

$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$

After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have

$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$,  $b' =\sqrt[3]{b}$ and  $c' =\sqrt[3]{c}$

So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.

答:
Rewrite the equation  $x^3 +3x^2-24x +1 = 0$ as
$$(x+1)^3-27x=0$$
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $ a^2+ab+b^2>0$ to obtain
$$x-3\sqrt[3]x+1=0
$$
which is a cubic equation in $\sqrt[3]x$. Thus, given its roots $\sqrt[3]{a}, \sqrt[3]{b} $ and $\sqrt[3]{c} $, the Vieta’s formulas produces
$$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
and
$$\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} =-3<0$$
Note that the latter is actually not needed to prove the former.


问:
What is the meaning of "independent events " and how can we logically conclude independence of two events in probability?

What is the meaning of  "independent events " in probability

For eg: Two events (say A and B)are independent , what I understand is the occurrence of A is not affected by occurrence of B .But I am not comfortable with this understanding , thinking this way every events I meet upon are independent!, does there exist a more mathematical definition( I don't want the formula associated with it)

Another thing I want to know in a real case or a problem how do we
  understand(logically) if two events are independent?

That is without verifying that $P(A\cap B) = P(A)P(B) \Large\color{blue}{\star}$  how do we conclude that two events $A$ and $B$ are independent or dependent?.

Take for eg an example by "Tim"

Suppose you throw a die. The probability you throw a six(Event $A$) is $\frac16$ and the
  probability you throw an even number(Event $B$) is $\frac12$.

And event $C$ such that $A$ and $B$ both happen would mean $A$ should happen(As here $A$ is a subset of $B$) hence it's too $\frac16$

My thought:

The above example suggest something like this "Suppose I define two events $A$ and $B$ and if one of the even is a subset of the other then the events are not independent". Of course such an argument is not a sufficient condition for dependency , for eg: consider an event $A$ throwing a dice and getting an odd number and event $B$ getting 6. They aren't independent ($\Large\color{blue}{\star}$ isn't satisfied) . So again I improve my suggestion "Suppose I define two events $A$ and $B$ and if one of them is a subset of the other or their intersection is a null-set then the events are not independent $\Large\color{red}{\star}$"

So at last is $\Large\color{red}{\star}$ a sufficient condition?. Or does there exist a suffcient condition (other than satisfying the formulas $\Large\color{blue}{\star}$? And what is the proof, I can't prove my statement as the idea for me is not that much mathematical .

答:
Philosophically, two events $A$ and $B$ are independent for an observer iff gaining the information that event $A$ occurred or did not occur does not alter the probability that $B$ occurred for that observer. Thus independence is observer dependent - I might say the events are independent, you might say they're dependent; and the version of me in the future with more information might agree with you that they're dependent, thereby disagreeing with his past self.

Now there are events that most people will agree are dependent. The event that a coin shows a heads and the event that the coin shows a tails are almost certainly dependent; if I see a heads, my probability that the coin is showing a tails suddenly plumets to zero. But even in this example, if I already knew the coin was biased such that heads always shows, then I always knew the probability of tails would show is/was $0$, and thus my probability of seeing tails wouldn't change upon noting that heads is showing. So even here, it's possible that, for some observers, heads and tails will be independent events.

Also, at any given moment in time, there are events that most observers situated at that moment in time will regard as independent. Like if people are gambling at a craps table, then the numbers that show on the die the next time they're rolled will tend to be viewed as independent of the numbers that show on the die the time after that. Of course, this isn't entirely true. The dice may have a non-negligible bias, after all; in which case, the numbers that show after the first throw can be used to make a better prediction on the second throw. But dice tend to be pretty fair, and most sensible player's will tend to model the dice as such, and thus the numbers showing on each roll will tend to be treated as independent from one another. Note that we're thinking about idealized players here - these are players free from human flaws e.g. they certainly don't believe in the law of averages.

Finally, a word about cause and effect. If we flip two coins simultaneously, well neither can have a causal effect on the other, so we'd likely model the outcome of one flip as being independent of the outcome of the other. But, what if we knew they both came from the same factory, either factory X or factory Y, and factory X's coins always come up heads and factory Y's coins always come up tails? Well, the coins still have no causal effect on one another - but, lo and behold, the outcome of one coin allows us to predict the outcome of the other. That is, from the point of view of an observer that knows that both coins come from the same factory, either X or Y, and that those factories make systematically biased coins, the outcomes of the two coins are utterly dependent, even though there is no cause/effect relationship between them.

So, do not confound causal independence with probabilistic independence! Yes, these concepts do have some bearing on one another, but they're distinct concepts.


问:
What is this property in model theory called?
Take three L-structures: $\cal{M}_0 \subset\cal{M}_1 \subset\cal{M}_2$. If $\cal{M}_0 \prec \cal{M}_2$ and $\cal{M}_1 \prec \cal{M}_2$, then we have $\cal{M}_0 \prec \cal{M}_1$.

Does this property have a name?

答: This is one of the axioms of abstract elementary classes, and in that context it is called "coherence".</p>


问:
About a case in a proof that every subgroup of a cyclic group is cyclic.
In the following proof I do not understand why it is assumed that $\{e\}$ is cyclic and why we treat the case $h = \{e\}$ separately.

Let $G$ be a cyclic group generated by $a.$

Let $H$ be a subgroup of $G$.

If $H=\{e\},$ then $H$ is a cyclic group subgroup generated by $e$.

Let $H\neq\{e\}.$

By definition of cyclic group, every element of $G$ has the form $an$.

Then as $H$ is a subgroup of $G, an\in H$ for some $n\in\Bbb Z$.

Let $m$ be the smallest (strictly) positive integer such that $am\in H.$

Consider an arbitrary element $b$ of $H$.

As $H$ is a subgroup of $G, b=an$ for some $n$.

By the Division Theorem, it is possible to find integers $q$ and $r$ such that $n=mq+r$ with $0\le r<m.$

It follows that:

$$an=amq+r=(am)qar$$
and hence:

$$ar=(am)−qan$$

Since $am\in H$ so is its inverse $(am)^{−1}.$

By Group Axiom G0: Closure, so are all powers of its inverse.

Now $an$ and $(am)−q$ are both in $H$, thus so is their product $ar,$ again by Group Axiom G0: Closure.

However:

$m$ was the smallest (strictly) positive integer such that $am\in H$
and:

$$0\le r<m$$
Therefore it follows that:

$$r=0$$
Therefore:

$$n=qm$
and:

$$b=an=(am)q.$$
We conclude that any arbitrary element $b=an$ of $H$ is a power of $am$.

So, by definition, $H=\langle am\rangle$ is cyclic.

答: There's no real need to consider $\{e\}$ separately (as far as I am aware): the subgroups of $G=\langle a\rangle$ correspond to multiples (or "powers") of $a$, and we say, by convention, that $a^0=e$, the zero-th power of $a$; perhaps the additive nature of abelian groups (and hence cyclic groups) calls to question how one defines $0a$, which is the additive way of taking the zero-th power - the zero-th multiple of $a$ (which is still $e$ by convention) - and so, here, for pedagogical reasons, it is put to one side conceptually; and, besides, if $G$ is infinite, $\{e\}$ is its only finite subgroup.


答:
Probability with dice and coin tosses seems like a standard starting point. Discussion of normal distributions (percentiles, standard deviation) also is pretty fundamental.

The great thing about both of those is that they're easy to illustrate. Probability is great to discuss in terms of games of chance (and while talking about dice and coin tosses is great to introduce the idea, most students aren't going to find those too interesting, so once you've taught the concept, move on to more interesting examples, e.g. Poker). The distribution of heights of students in a class is a standard example for introducing normal distributions, probably pretty good for introducing the idea and getting students to conceptually understand the concept of a normal distribution.

However, that won't be enough to really rouse the interest of the students. You'll want to show them that normal distributions are everywhere (that's a lie, I really mean Gaussian distributions) and can be used to model a great number of really interesting real-life distributions. Show them lots of these examples, and try to make them relevant or appealing to the lives of the students. Try to get them to visualise "bell curves" when thinking about any sort of population with a quantifiable characteristic.

Perhaps look at this Udacity course for some inspiration: https://www.udacity.com/course/st101

Title: Statistics Workshop for High School Students

Body:
We are going to hold an introductory workshop about the statistics. The participants will be students who have just finished their 8th or 9th grade. The workshop consists of 10 two-hour sessions. The students have a very limited background about the statistics: They can compute arithmetic average of a set of data, create pie charts, etc.

So we’re looking for some good educational material to introduce the students to some of the main statistical ideas, methods and concepts as well as to illustrate the importance of statistics in daily life.

Your suggestions would be of great help.

Thanks.


问:
Prove that $L$ is closed if and only if for any converging sequence $(x_n)$ with $x_n\in L$, the limit $x=\lim x_n$ is also an element of $L$
Could somebody please point me in the right direction to prove $\Leftarrow$?

Here is my proof for $\Rightarrow$:

Let $L\subseteq R$ be closed and $(x_n)$ be a converging sequence with $x_n\in L$.

Show that (1) $L$ is closed gives (2) $\lim x_n=x\in L$: suppose by contradiction that (2) does not hold given (1), i.e. $\lim x_n=x\notin L$; for any arbitrary $\epsilon>0$, there exists $N^\prime\in N$ such that for $n\geq N^\prime,|x_n-x|<\epsilon\Leftrightarrow -\epsilon<x_n-x<\epsilon\Leftrightarrow x-\epsilon<x_n<x+\epsilon$; $x_n\in L$ but $(x-\epsilon,x+\epsilon)\subseteq L^c\Rightarrow$ contradiction; thus, (1) gives (2).

答:
The previous answer is perfect but i will try to give an explanation with concepts.

Cosider those two things:

L is closed if and only if every point that is arbitrarily near to L is in L.

The limit point of a sequence is arbitrarerly near to the set of points on the sequence.

Your theorem states that every point that is arbitrarely near to L it is also arbitrarely near to a sequence in L.

So the key is that the definition of limit point links 1) with 2). In fact if L were the set of points on a convergence sequence you get the equivalence. So your question states that in some sense if L is closed it is the union of all his convergent sequences and its limits points.
I hope this helps you


答:
Rewrite the inequality as

$$\sqrt[3]{abc\over a}+\sqrt[3]{abc\over b}+\sqrt[3]{abc\over c}\lt0$$

but now note that $abc=-1$, so this simplifies to

$$\sqrt[3]{1\over a}+\sqrt[3]{1\over b}+\sqrt[3]{1\over c}\gt0$$

Next, note that $1/a$, $1/b$, and $1/c$ are roots of $x^3-24x^2+3x+1=0$. Calling this polynomial $P$, we see that $P(-1)\lt0$, $P(0)\gt0$, and $P(1)\lt0$, so its three roots lie in $(-1,0)$, $(0,1)$, and $(1,\infty)$. Their cube roots lie in the same intervals, and so the sum of the cube roots is clearly positive, since the one in $(1,\infty)$ more than offsets the one in $(-1,0)$.

Title: Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$

Body:
I am working on this problem:

$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$

After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have

$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$,  $b' =\sqrt[3]{b}$ and  $c' =\sqrt[3]{c}$

So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.


问:
Translation invariance of cross product.
For two vectors $a, b \in \mathbb{R}^3$. Their cross product $a \times b$ is a vector that orthogonal to the plane spanned by them. If both $a, b$ was translated by a vector $c\in \mathbb{R}^3$, would the cross product subject to a simple translation?

$(a+c)\times(b+c) ?= (a\times b)+c$

答:
The cross product is distributive, and $c\times c=0$, so

$$ (a+c)\times(b+c)= a\times b+a\times c+c\times b. $$

The only way this equals $a\times b+c$ is if

$$ a\times c+c\times b=c. $$

This isn't possible if $c\ne 0$ though; the left side is perpendicular to the right side!


问:
Calculate $\sup \{r \in \mathbb{R} : \lim_{n \to \infty} \frac{r^n n!}{n^n} = 0\}$
Determine $\alpha$, where
$$
\alpha \equiv \sup \left\{r \in \mathbb{R} : \lim_{n \to \infty} \frac{r^n n!}{n^n} = 0 \right\}
$$

Some Observations

We clearly have $\alpha \geq 1$ since
$$
\lim_{n \to \infty} \frac{n!}{n^n} = 0
$$
Moreover, Wolframalpha tells me
$$
\lim_{n \to \infty} \frac{4^n n!}{n^n} = \infty
$$
so $\alpha \leq 4$.

A basic estimate shows that
$$
0 \leq \frac{r^n n!}{n^n} \leq \frac{r^n}{2^n} \left(1 + \frac{1}{n} \right)^n
$$
which may be some kind of clue.

This is reminiscent of Stirling's approximation, which says
$$
\lim_{n \to \infty} \frac{e^n n!}{n^{n + \frac{1}{2}}} = \sqrt{2\pi}
$$
But the $n + \frac{1}{2}$ in the exponent in the denominator prevents us from simply concluding that $\alpha = e$

Playing around in Wolframalpha, it seems like $\alpha = e$ is a good conjecture, but I haven't been able to confirm it. I also have no idea how to prove it.

答:
Your guess is correct, and your problem can indeed be tackled by the Stirling's approximation:

$$\frac{r^n n!}{n^n} \sim \sqrt{2\pi n}\,\frac{r^n n^n e^{-n}}{n^n}=\sqrt{2\pi n}\,\left(\frac{r}{e}\right)^n.$$

Focusing on $r>0$, this limit converges if $r<e$ and diverges if $r\geq e$.

In fact, we may borrow the idea from the Ratio Test and come up with a simpler argument. If $a_n$ denotes the expression inside of the limit, then

$$\frac{a_{n+1}}{a_n}=\frac{r}{(1+\frac{1}{n})^n}.$$

Now by noting that $(1+\frac{1}{n})^n \uparrow e$ as $n\to\infty$,

$$ \frac{a_{n+1}}{a_n} \geq \frac{r}{e} \qquad\text{and}\qquad \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{r}{e}. $$

This can be used to prove that $a_n \to 0$ if $0 < r < e$ and $a_n \not\to 0$ if $r \geq e$.


问:
Proving surjectivity of a floor function.
So the function given is $$ g{_r}: \mathbb{Z}\rightarrow\mathbb{Z}, \: x \mapsto \big\lfloor\dfrac{x}{r}\big\rfloor, \quad \quad \text{while} \: r\in\mathbb{N}.$$
The original question was to prove or disprove the injectivity or surjectivity of this function. I have already proven that for $r=1$, this function is injective and for $r>1$ it is not injective.

Now I have to check if the function is surjective for $r>1$.
 My thought is that we assume that the function is surjective, then we have to show that for every $\left\lfloor\dfrac{x}{r}\right\rfloor\in\mathbb{Z}$ exists an $x \in\mathbb{Z}$. How can I prove (or disprove) this? Are there some transformations that I can do to the floor function?

答: Hint:  $\lfloor\frac{rt}{r}\rfloor=\lfloor t\rfloor$


问:
construction of the Witt group
I've seen a couple of constructions of the so-called Witt group: it seems that most authors start with the commutative monoid of isometry classes of quadratic spaces under direct sum, pass to the Grothendieck group, and then quotient out the subgroup generated by the hyperbolic plane (sometimes called the split quadratic spaces).

In Bump's Automorphic Forms and Representations, however, it is claimed that one need only quotient out the submonoid generated by the hyperbolic plane and the monoid thus obtained is already a group. So we can avoid the Grothendieck construction entirely. The content of this statement is that given a quadratic space, there exists another such that the direct sum of the two is split (i.e. a sum of copies of the hyperbolic plane). But I don't see why. Does anyone have a reference or an explanation?

答:
Yes, this is true.  (Or almost true: one doesn't take isometry classes of all quadratic forms, but only the nonsingular ones.)

If $q = \langle a_1,\ldots,a_n \rangle$ is a nonsingular diagonal quadratic space, then take $q' = \langle -a_1,\ldots,-a_n \rangle$.  Then

$q \oplus q' \cong \langle a_1, -a_1 \rangle \oplus \ldots \oplus \langle a_n, - a_n \rangle$,

and it is a standard early exercise to show that for any $a \in K^{\times}$, $\langle a, - a \rangle$ is isomorphic to the hyperbolic plane. (For instance, every nonsingular isotropic quadratic form $Q$ is of the form $Q' \oplus \langle 1,-1 \rangle$ for some quadratic form $Q'$.  Or see Corollary 11 in these notes.) So $q \oplus q'$ is a totally hyperbolic space and gets identified with $0$ in your quotient monoid, which is therefore a group.

If I am remembering correctly, this is indeed much closer to Witt's original take on the matter than the more modern Grothendieck group approach.

Title: construction of the Witt group


问:
Equivalent norms and isometries
Let $X$ be a vector space, $\|\cdot\|_1$ and $\|\cdot\|_2$ two equivalent norms on $X$. Under what further assumptions can we prove there is an isometry between $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$?

In particular, are $(\mathbb{R}^2,\|\cdot\|_1)$ and $(\mathbb{R}^2,\|\cdot\|_2)$ isometric (where the two norms are the maximum norm and the usual euclidean norm respectively)?

答: These two spaces are not isometric. Unit sphere of the first space contains straight segment while unit sphere of second space doesn't. You can draw them to see this. But property of unit sphere to contain segments preserved under linear isometries, so there is no isometry between these two normed spaces. You can also check similar question asked earlier.</p>

Title: Equivalent norms and isometries


问:
Sum of subspaces is the smallest containing subspace
In sheldon Axler's book,

Suppose $U_1+\cdots+U_n$ are subspaces of $V$. Then $U_1+\cdots+U_n$ is the
smallest subspace of V containing $U_1+\cdots+U_n$

I can't understand two parts of this definition

$U_1,\ldots,U_n\subseteq U_1+\cdots+U_n$

$U_1+\cdots+U_n$ is the smallest subspace in $V$

答: The sum $U_1 + \cdots + U_n$ is the smallest subspace of $V$ that contains $U_1,\dots,U_n$ in the sense that if $W$ is some other subspace of $V$ that contains $U_1,\dots,U_n$, then $U_1 + \cdots + U_n \subseteq W$. Compare the $\subseteq$ with the $\leq$!


问:
derivative of product matrix
Let $$f(\alpha)= \alpha^{3} A(B+\alpha I)^{-3}C,$$
where A , B and C are square real matrices that do not depend on the scalar $\alpha$ and I is the identity matrix. I need help with computing the derivative of f with respect tot $\alpha$.

答:
For the majority of steps you can use the product rule. Do keep in mind that matrices in general do not commute, so don't alter their order (unless you multiply by a scalar like $\alpha$). Before doing this $f(\alpha)$ can be written into an equivalent form, to which it is easier to find the intermediate derivatives when applying the product rule

$$
f(\alpha) = \alpha^{3} A\,(B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1} C. \tag{1}
$$

Therefore, the derivative of $(1)$ with respect to $\alpha$ can be expressed as

$$
\frac{d\,f(\alpha)}{d\alpha} = A \left(3\,\alpha^{2}(B + \alpha\,I)^{-3} + \alpha^{3} \frac{d\,(B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1}}{d\alpha} \right) C, \tag{2}
$$

with

$$
\frac{d\,(B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1}}{d\alpha} = \frac{d\,(B + \alpha\,I)^{-1}}{d\alpha} (B + \alpha\,I)^{-2} + (B + \alpha\,I)^{-1} \frac{d\,(B + \alpha\,I)^{-1}}{d\alpha} (B + \alpha\,I)^{-1} + (B + \alpha\,I)^{-2} \frac{d\,(B + \alpha\,I)^{-1}}{d\alpha}. \tag{3}
$$

In general the derivative of an expression of the form $(X+\alpha\,Y)^{-1}$ with respect to $\alpha$ can be obtained by applying the product rule to $(X+\alpha\,Y)^{-1} (X+\alpha\,Y) = I$, resulting in

$$
\frac{d\,(X+\alpha\,Y)^{-1} (X+\alpha\,Y)}{d\alpha} = \frac{d\,(X+\alpha\,Y)^{-1}}{d\alpha} (X+\alpha\,Y) + (X+\alpha\,Y)^{-1} \frac{d\,(X+\alpha\,Y)}{d\alpha} = 0, \tag{4}
$$

where it is hopefully clear that the derivative of $X+\alpha\,Y$ with respect to $\alpha$ should equal $Y$. Therefore, solving $(4)$ for the derivative of $(X+\alpha\,Y)^{-1}$ with respect to $\alpha$ yields

$$
\frac{d\,(X + \alpha\,Y)^{-1}}{d\alpha} = -(X+\alpha\,Y)^{-1} Y\,(X+\alpha\,Y)^{-1}. \tag{5}
$$

When using $(5)$ for the derivative of $(B + \alpha\,I)^{-1}$ with respect to $\alpha$ yields

$$
\frac{d\,(B + \alpha\,I)^{-1}}{d\alpha} = -(B + \alpha\,I)^{-1} I\,(B + \alpha\,I)^{-1} = -(B + \alpha\,I)^{-2}. \tag{6}
$$

In this case the matrix $Y$ from $(5)$ is the identity matrix, which always commutes. However, if $Y$ would have been a different matrix it would not simplify as neatly. Substituting $(6)$ in $(3)$ therefore gives

$$
\frac{d\,(B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1} (B + \alpha\,I)^{-1}}{d\alpha} = -3\,(B + \alpha\,I)^{-4}. \tag{7}
$$

Lastly, substituting $(7)$ in $(2)$ yields

$$
\frac{d\,f(\alpha)}{d\alpha} = 3\,A \left(\alpha^{2}(B + \alpha\,I)^{-3} - \alpha^{3} (B + \alpha\,I)^{-4} \right) C. \tag{8}
$$

Title: derivative of product matrix


问:
How do I approach this complex integral?
I want to compute the following

$$\oint_{\delta D(0,1)} \frac{e^{\alpha z}}{2z^2-5z+2}dz.$$

My attempt:

\begin{align*}
\oint_{\delta D(0,1)} \frac{e^{\alpha z}}{2z^2-5z+2}dz &=\oint_{\delta D(0,1)} \frac{e^{\alpha z}}{(z-2)(2z-1)}dz \\
&= \frac{1}{3}\left( \oint_{\delta D(0,1)} \frac{e^{\alpha z}}{z-2}dz - \oint_{\delta D(0,1)}\frac{e^{\alpha z}}{z-\frac{1}{2}}dz \right) \\
&= \frac{1}{3} \left(\oint_{\delta D(0,1)} \frac{e^{\alpha z}}{z-2} dz -2\pi i e^{\frac{\alpha}{2}} \right)
\end{align*}

where the later integral was computed through Caychy's Integral Formula.

Here is where I am stuck, I don't know how to integrate $\oint_{\delta D(0,1)} \frac{e^{\alpha z}}{z-2}dz$. Any advice on how to proceed?

答:
In fact
$$\oint_{\delta D(0,1)} \frac{e^{\alpha z}}{z-2}dz=0$$
since $\frac{e^{\alpha z}}{z-2}$ is analytic in $\bar{D}(0,1)$.


问:
Determine the length and angle of a hypotenuse starting from a rectangle at a different angle
Lumber support for Plywood
I'm trying to orient a piece of lumber over a piece of plywood as shown. The dimensions of the plywood and the width of the lumber are known. The lumber needs to be oriented and cut as a parallelogram. I have to determine the length of the lumber (l), the distance from the end to trim off (t), and the angle of the cut in degrees (d). I think I can figure out the trim off length but I'm not sure.

Referring to the diagram:
a, o, h - is known
w - is known
l, t, d - is unknown

I need a formula to determine:

the length (l) of the (second) rectangle created from filling out the parallelogram

the length (t) of a triangle created from filling out the parallelogram

the parallelogram's lesser angle in degrees (d)

My math isn't too good and I don't understand Greek characters. I would GREATLY appreciate a reply using something like Excel functions.

答:
Diameter $h=AB$ across the lumber.

Purple angle $\beta$ is marked. By the Pythagoras theorem

$$ \text{ length l}= \sqrt{a^2+o^2-w^2}$$

$$ \beta= \sin^{-1}\dfrac{BC}{BA}-\sin^{-1}\dfrac{w}{BA}=\sin^{-1}\dfrac{a}{h}-\sin^{-1}\dfrac{w}{h}$$

$$\text{trim}={w} \tan {\beta}$$

$$ d= \beta+ 90^{\circ}$$


答:
Normally, one proves that two normed linear  spaces are isometric in a constructive™ way, by exhibiting an isometry between them. One common exception: since it's widely known that any two separable Hilbert spaces are isometrically isomorphic to each other, an explicit isometry is unnecessary in this case.

One elementary way to show that two spaces $X,Y$ are not isometrically isomorphic is to find a subset of $X$ that cannot be isometrically embedded into $Y$. For example, consider the subset $$A=\{(\pm 1,0), (0,\pm 1)\}\subset (\mathbb R^2,\|\cdot \|_1)$$
Any two points of $A$ are at distance $2$ from each other. Such a configuration is impossible in $(\mathbb R^2,\|\cdot \|_2)$. Indeed, three points at distance $2$ from one another must lie at the vertices of an equilateral triangle, and there is nowhere to put the fourth one.

Title: Equivalent norms and isometries

Body:
Let $X$ be a vector space, $\|\cdot\|_1$ and $\|\cdot\|_2$ two equivalent norms on $X$. Under what further assumptions can we prove there is an isometry between $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$?

In particular, are $(\mathbb{R}^2,\|\cdot\|_1)$ and $(\mathbb{R}^2,\|\cdot\|_2)$ isometric (where the two norms are the maximum norm and the usual euclidean norm respectively)?


问:
Variations on $x^x=y$
Pretty sure at least a few people know how to solve for x in an equation of the form $x^x=y$.

One way (with 2 as an example):
$x^x=2$

$x=2^{1/x}$

$x=({e^{ln(2)}})^{1/x}$

Multiplying both sides by $ln(2)/x$:

$ln(2)=ln(2)/x*{e^{{ln(2)}/x}}$

Which can then be solved using the Lambert W function on both sides (I don't understand it completely but I believe it to be something like the inverse of $f(z)=z*e^z$) to yield:

$W(ln(2))=ln(2)/x$

And so $x=ln(2)/W(ln(2))$.

That makes sense, but it relies on the base and exponent being equal to utilise the Lambert W function. Say I were to attempt something similar using $x^{6*x^{12}}=2$, the same process doesn't work exactly as it doesn't fit the form required (of $x*e^x$), even though there's still only one variable. Any ideas on how to get it into this form? Or perhaps use another method entirely? I would prefer if the answer is in exact form, so any iterating equations or graphical methods wouldn't help much.

Thanks and much appreciated :)

答: Square $x^{6 x^{12}} = 2$ to get $x^{12 x^{12}} = 4$. Can you take it from here?

Title: Variations on $x^x=y$


问:
Searching/Sorting Algorithm
I just started studying algorithms and data structures and came across this problem:

Given $x \in \mathbb{N}$ and two integer Arrays $A_1$ and $A_2$ each of the length $n$. Write an algorithm in pseudocode that has a time complexity of $O(n.log(n))$ or better to determine the set $$X = \{ (a,b) |  a \in A_1  \land  b \in A_2  \land  a + b = x \}$$

My idea was to first sort the two arrays, loop through the first array and then perform a binary search through the second one and then, if such a couple (a,b) exists, add the result to a 2D Array(?). But i'm not sure if the time complexity is $O(n.log(n))$ or better.

Thanks in advance!

答:
We can do this in linear time. In one pass, we can place the elements of $A_2$ in a hash set which has constant time access. Call the hash set $H$.

Then we get

```<code>for num in A_1
   if H.contains(x-num) then add (num, x-num) to X
</code>```

This makes a one time pass through $A_1$ and for each value executes a constant time access.

Therefore, this algorithm is $O(n)$

Title: Searching/Sorting Algorithm

Body:
We can do this in linear time. In one pass, we can place the elements of $A_2$ in a hash set which has constant time access. Call the hash set $H$.

Then we get

for num in A_1
   if H.contains(x-num) then add (num, x-num) to X

This makes a one time pass through $A_1$ and for each value executes a constant time access.

Therefore, this algorithm is $O(n)$


问:
Let $X$ be the disjoint union of copies of a subspace of $\mathbb{R}$. Is the quotient of $X$ by identifying accumulation points first countable?
Sorry the title isn't super clear, I ran out of characters.

From Willard's General Topology, 16A.5:

For each $n \in \mathbb{N}$, let $X_n$ be a copy of the subspace
$\{0\} \cup \{\frac{1}{m}: m = 1, 2, \ldots \}$ of $\mathbb{R}$. Let $X$ be the disjoint union of the $X_n$. Is the quotient $Y$ of $X$ obtained by identifying all accumulation points of $X$ first
countable?

I'm really having a hard time with this question. I'm struggling more generally with both the notions of the disjoint union topology, and the quotient topology, so this is really throwing me for a loop.

The relevant information that I know is:

If $X$ is a topological space and $a \in X$, then we say $a$ is an accumulation point of a set $A$ iff each nbhood of $a$ meets $A$ in some point other than $a$,

The set $X_n'$ of all accumulation points of $X_n$ (defined above) is $X_n' = \{0\}$.

But I'm not really sure what $X$, the disjoint union of these $X_n$, "looks" like. Is the set of accumulation points of $X$ the union of the accumulation points of the $X_n$? I'm really just not sure how to proceed.

Any help is appreciated.

答:
This is a very similar question as Some topological properties of "countable lines with one origin".

Let $L  = \{0\} \cup \{\frac{1}{m}: m \in \mathbb N \}$. Then we may take $X = L \times \mathbb N$, where $\mathbb N$ has the discrete topology (which is the subspace topology inherited from $\mathbb R \supset \mathbb N$). The set of accumulation points of $L$ is $L' = \{0\}$. Hence the set of accumulation points of $X$ is $X' = L' \times \mathbb N = \{0\} \times \mathbb N$. Certainly no $(x,n) \in X$ with $x \ne 0$ is an accumulation point of $X$ because $\{(x,n)\} = \{x\} \times \{n\}$ is open in $X$. Now consider $(0,n) \in X$ and an open neigborhood $U$ of $(0,n)$ in $X$. There exist open neigborhoods $U_n$ of $0$ in $X$ and $V_n$ of $n$ in $\mathbb N$ such that $U_n \times V_n \subset U$. The set $U_n$ contains some $p \in L$ with $p \ne 0$. Hence $(p,n) \in U$ and $(p,n) \ne (0,n)$.

This means that $Y = X/X'$. Let $p : X \to Y$ denote the quotient map and $*$ the common equivalence class of the points $(0,n)$. For $M \in \mathbb N$ let $L_M = \{0\} \cup \{\frac{1}{m}: m \ge M \}$ which is an open neigborhood of $0$ in $L$.

Let $\{U_k\}$ be any countable family of open neigborhoods of $*$ in $Y$. Then $p^{-1}(U_k)$ is an open neigborhood of $(0,n)$ in $X$, thus there exists $M(n)$ such that $L_{M(n)} \times \{n\} \subset p^{-1}(U_k)$. The set  $V = p(\bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\})$ is an open neighborhood of $*$ in $Y$ because $p^{-1}(V)  = \bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\}$ which is open in $X$. For each $n$ we have $p^{-1}(U_n) \not\subset p^{-1}(V) = \bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\}$ because $(0,M(n)) \in p^{-1}(U_n)$, but $(0,M(n))\notin \bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\}$. This means $U_n \not\subset V$, thus $\{U_k\}$ cannot be a neighborhood base of $*$.


问:
Incidence matrix proof.

I don't fully understand the proof in this link on Springer of Lemma 2.2. Why is it that the $i$th and $j$ components of $x$ are equal when $i\sim j$? Also, what does $i\sim j$ even mean? Does it mean there is a path from $i$ to $j$?

I get that if $G$ is connected and components of $x$ are equal if their corresponding vertices are joined by a path, then $x$ must have all components equal. But is there a formal justification for why the components have to be equal if their corresponding vertices are joined by a path (e.g. there's a justification that the diagonal entries of the Laplacian matrix of a graph $G$ are the vertex degrees)?

答:
The notation $i\sim j$ means that vertices $i$ and $j$ are connected by an edge, say $e_k$. Let $q_k$ be the $k$-th column of $Q$. If $i\sim j$, the only two non-zero entries in $q_k$ are those in rows $i$ and $j$; call them $q_{k,i}$ and $q_{k,j}$, respectively. One of them is $1$, the other is $-1$, and

$$[0]=x'q_k=x_iq_{k,i}+x_jq_{k,j}=q_{k,i}(x_i-x_j)\,,$$

so $x_i=x_j$.

Now suppose that there is a path $i=i_0\sim i_1\sim\ldots\sim i_\ell=j$ from $i$ to $j$. Then

$$x_i=x_{i_0}=x_{i_1}=\ldots=x_{i_\ell}=x_j$$

by repeated applications of the previous result. $G$ is connected, so there is a path between any two vertices, and therefore $x_1=x_2=\ldots=x_n$. Thus, the left null space of $Q$ is either trivial or one-dimensional, and hence $\operatorname{rank}Q=n$ or $\operatorname{rank}Q=n-1$, respectively. On the other hand, the $n$ rows of $Q$ are sum to a zero vector and are therefore linearly dependent, so $\operatorname{rank}Q\le n-1$. Combining results, we see that $\operatorname{rank}Q=n-1$.

Title: Incidence matrix proof.


问:
Statement of the DeFinetti's representation theorem
I'm using a standard textbook in statistics [1], and am confused with the statement of the DeFinetti's representation theorem.

Theorem 1.49 (DeFinetti's representation theorem). Let $(S,A,\mu)$ be a probability space, and let $(\mathcal{X},\mathcal{B})$ be a Borel space. For each $n$, let $X_n: S\to \mathcal{X}$ be measurable. The sequence $\{X_n\}_{n=1}^\infty$ is exchangeable if and only if there is a random probability measure $\mathbb{P}$ on $(\mathcal{X},\mathcal{B})$ such that, conditional on $\mathbb{P} = P$, the $X_n$'s are IID with distribution $P$. Furthermore, if the sequence is exchangeable, then the distribution of $\mathbb{P}$ is unique, and $\mathbb{P}_n(B)$ converges to $\mathbb{P}(B)$ almost surely for each $B \in \mathcal{B}$.

More precisely, I'm a confused with what a "random probability measure $\mathbb{P}$ on
$(\mathcal{X},\mathcal{B})$" means. Does it mean a random variable that takes value in
the space $M_{1}(\mathcal{X},\mathcal{B})$ of probability measure on $(\mathcal{X},\mathcal{B})$? This seems to
make the most sense because it later says

... the $X_n$ are IID with distribution P.

Here each $X_{n}$ is a measurable function from $S$ to $\mathcal{X}$, and thus its
distribution is the pushforward measure $X_{n,\star}(\mu)$, which is
an element in $M_{1}(\mathcal{X},\mathcal{B})$.

But if that's the case, then what's the domain of $\mathbb{P}$? I guess the domain is $S$.. it does not say it clearly. If the domain is indeed $S$, can $\mathbb{P}$ be written out in general in terms of the $X_{n}$?

Reference

[1] Theory of Statistics-[Mark Schervish]

答:
Under the hypotheses of the theorem, the sequence $(X_n)_{n=1}^\infty$ converges almost surely. The probability measure $\mathbb P$ is the marginal probability distribution of the random variable $\lim\limits_{n\to\infty} \overline X_n= \lim\limits_{n\to\infty}(X_1+\cdots+X_n)/n.$ The conditional distribution of the whole sequence $(X_n)_{n=1}^\infty$ given the value of $\mathbb P$ is they are i.i.d. and the conditional distribution of each $X_i$ given that value of $\mathbb P$ is that value of $\mathbb P.$

Maybe an example will make things clearer. Suppose $(X_1,X_2,X_3,\ldots)$ is random and takes values in $\{0,1\}^n.$ If the distribution of this whole sequence is exchangeable, then there is some random variable $P$ taking values in $[0,1]$ such that the conditional distribution of $X_1,X_2,X_3,\ldots$ given $P=\text{(some particular number $p\in[0,1]$)}$ is expressed by saying they are are i.i.d. and each is equal to $1$ or $0$ with respective probabilities $p$ and $1-p.$ In this case the random probability measure is a measure on the sigma-algebra of all (four) subsets of $\{0,1\}.$ And $\Pr\left( \lim\limits_{n\to\infty} X_n =P \right) = 1.$

Notice that this means $X_1,X_2,X_3,\ldots$ are positively correlated unless the distribution of $P$ is degenerate, a delta distribution, in which latter case the correlation is zero.

(Exchangeable sequences with negative correlation between terms of the sequence cannot be extended to infinitely long exchangeable sequences.)


问:
Infinite field extensions and linear transformations
I have some issues proving this statement, which is an exercise in Jacobson's Lectures in Abstract Algebra (vol. 3):

Let $P$ be an arbitrary extension field of $\Phi$. Show that, if $\alpha \in P$ satisfies $\alpha_R \circ f = f \circ \alpha_R$ for all $f \in hom_{\Phi} \left ( P, P \right )$ then $\alpha \in \Phi$, where: $\alpha_R$ is the defined as $\alpha_R:P \rightarrow P, \alpha_R\left ( p  \right ) = p.\alpha$, and $hom_{\Phi}\left ( P, P \right )$ represents the set of $\Phi$-linear transformations from $P$ to $P$.

In other words, that the only elements of $P$ that can be "taken out" from every linear transformation are the elements of $\Phi$. Now, as a consequence of a theorem given in the aforementioned book ("Jacobson-Bourbaki theorem"), the statement is true for finite extensions. However, i don't seem to be able to deduce the general case from the finite one. Any help would be appreciated.

答:
Assume that $\alpha\in P\setminus\Phi$, that is, $\alpha$ is linearly independent from $1$ in the $\Phi$-vector space $P$.

Then (e.g. by extending $(1,\alpha)$ to a basis) we can define a $\Phi$-linear map $f:P\to P$ that satisfies $f(1)=1,\ f(\alpha)=0$.

This will violate the condition, as
$$\alpha_R(f(1))=\alpha\ \ne\ 0=f(\alpha_R(1))\,.$$


答: The general solution of $x^{ax^b}= c$  is $$x = \left(\dfrac{b}{a} \log(c)\cdot \dfrac1{ W\left(\frac{b}{a} \log(c)\right)}\right)^{1/b}$$

Title: Variations on $x^x=y$

Body:
Pretty sure at least a few people know how to solve for x in an equation of the form $x^x=y$.

One way (with 2 as an example):
$x^x=2$

$x=2^{1/x}$

$x=({e^{ln(2)}})^{1/x}$

Multiplying both sides by $ln(2)/x$:

$ln(2)=ln(2)/x*{e^{{ln(2)}/x}}$

Which can then be solved using the Lambert W function on both sides (I don't understand it completely but I believe it to be something like the inverse of $f(z)=z*e^z$) to yield:

$W(ln(2))=ln(2)/x$

And so $x=ln(2)/W(ln(2))$.

That makes sense, but it relies on the base and exponent being equal to utilise the Lambert W function. Say I were to attempt something similar using $x^{6*x^{12}}=2$, the same process doesn't work exactly as it doesn't fit the form required (of $x*e^x$), even though there's still only one variable. Any ideas on how to get it into this form? Or perhaps use another method entirely? I would prefer if the answer is in exact form, so any iterating equations or graphical methods wouldn't help much.

Thanks and much appreciated :)


答: Your approach is $O(n \log n)$ assuming you use a sort of that order.  Given that the sort is that order, you then are going through $n$ elements of the first array and doing a binary search, which is of order $\log n$, for each one.  Each step is of order $n \log n$ so the whole thing is.

Title: Searching/Sorting Algorithm

Body:
I just started studying algorithms and data structures and came across this problem:

Given $x \in \mathbb{N}$ and two integer Arrays $A_1$ and $A_2$ each of the length $n$. Write an algorithm in pseudocode that has a time complexity of $O(n.log(n))$ or better to determine the set $$X = \{ (a,b) |  a \in A_1  \land  b \in A_2  \land  a + b = x \}$$

My idea was to first sort the two arrays, loop through the first array and then perform a binary search through the second one and then, if such a couple (a,b) exists, add the result to a 2D Array(?). But i'm not sure if the time complexity is $O(n.log(n))$ or better.

Thanks in advance!


问:
Evaluate a limit
I've been stuck on this limit $$\lim_{x\to 0} \frac{e^{-1/x²}}{x²}$$ i've tried evaluating from both sides, and got $\lim_{x\to 0^-}\frac{1}{e^{1/x²}}\times \frac{1}{x²}$ which actually gives 0 times something that tends to infinity. Same thing happens with the right hand limit, not sure from where to go from here, i don't see any possible cancellations or anything to get rid of 1/x².

Title: Evaluate a limit

Body:
$$L=\lim_{x\to 0} \frac{e^{-1/x^2}}{x^2}$$
Let $x=1/t$, then
$$L=\lim_{t\to \infty} t^2e^{-t^2}=\lim_{t \to \infty} \frac{t^2}{e^{t^2}}\to \frac{0}{0}$$
Use L-Hospital, then
$$L=\lim_{t \to \infty} \frac{2t}{2t e^{t^2}}=\lim_{t\to \infty} e^{-t^2}=0$$


问:
Pushout of injective is injective.

Let $f:A\to X$ be a continuous injective map of spaces. Let $g:A\to Y$ be continuous. Prove that the map $i_Y:Y\to X\cup_A Y$ is injective. Here, $X\cup_A Y:=X\coprod Y/\sim$, where $\sim$ is the finest equivalence relation such that $f(a)\sim g(a)$. This is called the pushout of $X$ and $Y$ along $A$. Also, the map $i_Y$ is defined as the composition of the inclusion $X\to X\coprod Y$ with the quotient map $X\coprod Y\to X\coprod Y/\sim$.

I'm new to the idea of pushouts, so I'm having a hard time getting started here. I know that we have the "diagram commutativity" relation $i_X\circ f=i_Y\circ g$ where the definition of $i_X$ is analogous to the definition of $i_Y$, but I'm not sure how to use this. My first thought was to just jump in and try to prove this using the definition of injective as follows.

Pick $y_1,y_2\in Y$ such that $i_Y(y_1)=i_Y(y_2)$. A first case I tried considering was when $y_1=g(a_1)$ and $y_2=g(a_2)$ for some $a_1,a_2\in A$. Then we have (using $[-]$ to denote eq. class),
$$
[f(a_2)]=[g(a_2)]=i_Y(y_2)=i_Y(y_1)=[g(a_1)]=[f(a_1)],
$$
and then the injectivity of $f$ implies $a_1=a_2$? I'm not sure here. Also, what about the cases when $y_1$ and $y_2$ are not in the image of $g$?

Any suggestions? Am I approaching this right?

答:
One nice thing about the category of all topological spaces is that you can make your favorite set-theoretic function continuous by either making the codomain have the indiscrete topology, or make the domain have the discrete topology. This means that you have a lot of liberty for choosing when you have a property that guarantees the existence of a map whenever “there exists a topological space $Z$ and continuous functions with properties blah.” This is a consequence of the fact that the underlying set functor from $\mathsf{Top}$ to $\mathsf{Set}$ has both a left and a right adjoint, given precisely by the discrete topology functor and the indiscrete topology functor, respectively.

So... for this problem. Since we want maps into a “well-chosen $Z$”, we probably want to endow $Z$ with the indiscrete topology; then we won’t have to worry about continuity and can just define the maps any way we want, to make sure they satisfy the relevant properties. Here goes.

If $Y$ is empty, then $A$ is empty and there is nothing to do: any map with empty domain is injective. So assume $Y$ is not empty and let $y_0\in Y$.

Let $Z$ be the topological space with underlying set $Y$ and the indiscrete topology. Define the map $\phi_2\colon Y\to Z$ to be the identity on underlying sets. Define the map $\phi_1\colon X\to Z$ as follows:

If $x\in X$ and there exists $a\in A$ with $x=f(a)$, then let $\phi_1(x) = \phi_2(g(a))$ (really just $g(a)$, since the underlying set of $Z$ is the same as the underlying set of $Y$). This is well defined, because $f$ is injective. If $x\notin f(A)$, then let $\phi_2(x)=y_0$.

Since any map into an indiscrete space is continuous, both $\phi_1$ and $\phi_2$ are continuous. Moreover, if $a\in A$, then $\phi_1(f(a)) = \phi_2(g(a))$ by definition. Thus, by the universal property of the pushout, there exists a unique continuous map $\Phi\colon X\cup_A Y\to Z$ such that $\phi_1=\Phi\circ i_X$ and $\phi_2=\Phi\circ i_Y$. But $\phi_2$ is a bijection, so $\Phi\circ i_Y$ is a bijection. Therefore, $\Phi$ is surjective and $i_Y$ is injective. This shows that $i_Y$ is injective.

Title: Pushout of injective is injective.


问:
Why are the 1-skeleton graphs of the Regular Polytopes distance transitive?
A graph $G$ is distance transitive if for all vertices $u,v,w,x$ of $G$ such that $D_G(u,v) = D_G(w,x) $ implies that there exists a graph automorphism of $G$, $\psi \in \Gamma(G)$ such that $\psi(u) = w$, $\, \psi(v) = x$.

If we have a regular polytope, $\mathcal{P}$, (spherical, euclidean or hyperbolic) then if we take the graph whose vertices are the vertices of $\mathcal{P}$ and whose vertices are adjacent if and only if they share an edge in $\mathcal{P}$.

I should be clear that I do NOT know that this s true for all regular polytopes. But checking the few I can they seem to be so.

答:
As stated by Chris Godsil in the comments, not all Platonic solids are distance-transitive: the 24-cell, 120-cell ans 600-cell are not distance-transitive.

Here is a complete list of polytopes that are distance-transitive (i.e. not only their edge-graph is distance-transitive, but also their Euclidean symmetry group acts distance-transitively on the skeleton):

regular polygons,

the regular dodecahedron and regular icosahedron,

the $d$-dimensional crosspolytope,

cartesian powers of regular simplices (this includes, of course, the regular simplices, but also the regular $d$-dimensional cube as the power $[0,1]\times\cdots\times[0,1]$),

hyper-simplices,

demi-cubes,

the $2_{21}$-polytope,

the $3_{21}$-polytope.

See also Theorem 5.10 in my recent preprint:

M. Winter, "Eigenpolytopes, Spectral Polytopes and Edge-Transitivity." arXiv preprint arXiv:2009.02179 (2020).

However, the true reason for why exactly those polytopes are distance-transitive is more complicated. Chris Godsil should indeed be the perfect person to explain this, as it was essentially proven in his paper

C.D. Godsil, "Eigenpolytopes of distance regular graphs." Canadian Journal of Mathematics 50.4 (1998): 739-755.


答:
Here's what I've got for you:

https://drive.google.com/file/d/1kxf3H83sP6BI3BP7CBGj5u3EMs4RGY8z/view?usp=sharing

$$ l^2 + w^2 = h^2 \\
l = \sqrt{h^2 - w^2}$$

Knowing $l$, you can use similar triangles XYZ, NMZ to show,

$$ t = h' \text{ cos}(d) \\
t = \frac{w}{\text{ tan}(d)} \\
a' = l \text{ cos}(d) \\
\frac{t}{a'} = \frac{w}{o} \\
\text{therefore, } h' = \frac{wl}{o}$$

Knowing $h'$ we can solve for $d$, and consequently $t$.

$$w = h' \text{ sin}(d) \\
d = \text{ arcsin}(\frac{w}{h'}) \\
t = h' \text{ cos}(d)$$

I may have made some mistake here having done this so quickly, but this can be easily implemented into excel for calculations by varying $a,o,$ and $w$.

Title: Determine the length and angle of a hypotenuse starting from a rectangle at a different angle

Body:
Lumber support for Plywood
I'm trying to orient a piece of lumber over a piece of plywood as shown. The dimensions of the plywood and the width of the lumber are known. The lumber needs to be oriented and cut as a parallelogram. I have to determine the length of the lumber (l), the distance from the end to trim off (t), and the angle of the cut in degrees (d). I think I can figure out the trim off length but I'm not sure.

Referring to the diagram:
a, o, h - is known
w - is known
l, t, d - is unknown

I need a formula to determine:

the length (l) of the (second) rectangle created from filling out the parallelogram

the length (t) of a triangle created from filling out the parallelogram

the parallelogram's lesser angle in degrees (d)

My math isn't too good and I don't understand Greek characters. I would GREATLY appreciate a reply using something like Excel functions.


问:
Without calculating limit directly show that it is equal to zero
$$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\frac{1}{3^n}=0$$
I am not really sure what it means by "without calculating limit" and I don't really have ideas how to do it.</p>

答:
$\left(1+\dfrac 1 n\right)^n < e \,\,$so $\left(1+\dfrac 1 n\right)^{n^2} < e^n$
so $\left(1+\dfrac 1 n\right)^{n^2} \dfrac 1{3^n} < \dfrac{e^n}{3^n}
= \left(\dfrac e 3 \right)^n \to 0$.</p>


答:
This is a slightly nontrivial result saying that a finite sum of subspaces is again a subspace, and that it is in fact the smallest subspace containing each of the summands.

So there will be two things to check.

You can use the subspace criterion for the first part.   That is, a nonempty subset of a vector space is a subspace if it is closed under addition and scalar multiplication.

For the second you need to prove that any subspace containing each of the summands contains the sum.  Another way of saying it is that it is the intersection of all the subspaces containing it.

Title: Sum of subspaces is the smallest containing subspace

Body:
In sheldon Axler's book,

Suppose $U_1+\cdots+U_n$ are subspaces of $V$. Then $U_1+\cdots+U_n$ is the
smallest subspace of V containing $U_1+\cdots+U_n$

I can't understand two parts of this definition

$U_1,\ldots,U_n\subseteq U_1+\cdots+U_n$

$U_1+\cdots+U_n$ is the smallest subspace in $V$


问:
Evaluating predicate logic formula
× denotes the multiplication operation.

$ (\forall x)(\forall z)(\exists y)(x × y = z) $

Domain: $\mathbb{Q}$

(Q.1) The answer I have says this formula would evaluate to true for this domain, which I don't understand.

If x was 0 ($\frac{0}{1}$) and z was 5 ($\frac{5}{1}$), then there exists no rational number y, such that 0 × y = 5

$\therefore$ $ (\forall x)(\forall z)(\exists y)(x × y = z) $ is false under this interpretation.

Am I correct or am I missing something here?

Domain: $\mathbb{R}$

(Q.2) Essentially my argument is the same for why this formula is false for this domain, but for brevity:

If x was 0.0 and z was 5.0, then there exists no real number y, such that 0.0 × y = 5.0

$\therefore$ $ (\forall x)(\forall z)(\exists y)(x × y = z) $ is false, under this interpretation.

Am I correct or am I missing something here?

答:
With the help of the users in the comment section, I can conclude the arguments made are correct.

The formula was changed to:

$ (\forall x) (\forall z) (x \ne 0 \rightarrow (\exists y)(x \times y = z))$

Title: Evaluating predicate logic formula


问:
A triple integral over the interior of a shifted ellipsoid
Suppose I want to compute the integral

$$ \iiint_E f(x,y,z)\ dV $$

where $E$ is given by

$$ \frac{(x-1)^2}{4^2} + y^2+z^2\leq 1 $$

and

$$ f(x,y,z) = e^{xy+z} $$

I am thinking that I want to use spherical coordinates but I'm a little uncertain if my reasoning is correct about how to apply that to a shifted ellipsoid.  I believe that I should use

$$ (x-1)/2 = \rho \sin\varphi\cos\theta $$

$$ y = \rho \sin\varphi\sin\theta $$

$$ z = \rho\cos\varphi $$

with

$$ 0\leq \rho\leq 1 $$
$$ 0\leq \varphi \leq \pi $$
$$ 0\leq \theta < 2\pi $$

and then treat this like a change of coordinates, so that I therefore compute the Jacobian and perform the change of variables in the integral.

But I'm not sure that my change of coordinates faithfully preserves the shape, or that it's the right one, or that there isn't a simpler way to do this.  Typically when parameterizing any ellipsoid I have only parameterized the surface and therefore I can specify a scalar associated with the length along each axis.  But here since I need a parameter $\rho$ I'm a little less confident in this approach, and I'm also a little concerned over whether the shift needs some other sort of treatment.

答:
Traditionally with ellipsoids and triple integrals, some form of transformation of the form $(x,y,z) \to (au,bv,cw)$ is done first. If you're given an ellipsoid of the form:

$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} + \frac{(z-\ell)^2}{c^2}$$

Then a transformation would first look something like:

$$x-h = au, y-k = bv, z-\ell = cw$$.

A bit of manipulation shows that:

$$x = au+h, y = bv + k, z = cw + \ell$$

The Jacobian matrix of THIS is $abc$. Put in the proper values to turn your integral into $u,v,w$, and then now you're ready to turn this into the unit sphere, in other words:

$$ u = \rho \sin\varphi\cos\theta $$

$$ v = \rho \sin\varphi\sin\theta $$

$$ w = \rho\cos\varphi $$

This is the same way as you did it, but you've done it as one transformation instead of two. This should hopefully alleviate your concern that the transformation somehow isn't proper. You've done it correctly, but there's a little bit of extra machinery in the background that you've actually done in your head. You've gone straight from the ellipsoid to a spherical integral, but REALLY what you should be doing is going from the ellipsoid to the sphere, then using regular spherical coordinates.

My point is that both the methods are the same. But the method I'm using will hopefully outline a little bit better some of the things that are happening behind the scenes. Rather than going straight from Ellipsoid -> Spherical integral, you should go Ellipsoid -> Sphere -> Spherical integral.

I would even argue my way is easier, because you don't have to calculate all the nasty partial derivatives and do the determinant, and simplifying everything out. You can just use that the first jacobian is $abc$, then when you transform into regular spherical, you get a jacobian of $\rho^2\sin\varphi$ as expected. Let me know if you have any questions. Just a different way of explaining what's going on.


答:
Without L'Hospital or Taylor:

$e^x=\underset{n\to \infty}\lim\left(1+\frac xn\right)^n$. It's easy to show that the sequence is increasing so $e^x\ge \left(1+\frac xn\right)^n\ge  1+x.$ So, $1+x\le e^x$ and so $0\le e^{-x}\le \frac{1}{1+x}.$ Thus,  $0\le e^{-x^2}\le \frac{1}{1+x^2}\Rightarrow 0\le x^2e^{-x^2}\le \frac{x^2}{1+x^2}$ which implies that $x^2e^{-x^2}\to 0$ as $x\to 0.$ But this is exactly the limit you want, once you make the substitution $t=1/x.$

Title: Evaluate a limit

Body: I've been stuck on this limit $$\lim_{x\to 0} \frac{e^{-1/x²}}{x²}$$ i've tried evaluating from both sides, and got $\lim_{x\to 0^-}\frac{1}{e^{1/x²}}\times \frac{1}{x²}$ which actually gives 0 times something that tends to infinity. Same thing happens with the right hand limit, not sure from where to go from here, i don't see any possible cancellations or anything to get rid of 1/x².


问:
Are "non-zero eigenspaces of a matrix" the same as "eigenspaces corresponding to non-zero eigenvalues"?
Sorry for the long title. In essence, I'm wondering if non-zero eigenspaces are the same as eigenspaces corresponding to non-zero eigenvalues. That is, are non-zero eigenspaces = $E$$\lambda$ for $\lambda$ $\neq$ $0$ ?

答:
Yes.  The former would be eigenspaces that are not the zero vector space, from the sounds of it.  But, in fact, eigenvectors are always restricted to being nonzero.

The eigenspace for eigenvalue zero would be another way of referring to the kernel.


问:
incidence matrix with full rank

I don't understand certain steps in the accepted answer for this question on Mathoverflow that the rows of the incidence matrix of a graph are linearly independent iff no connected component is bipartite. In particular, for the converse, how would one show that the incidence matrix of $H$ is invertible if $H$ is an odd cycle?

Also, why is it that if there's at least one edge not in the cycle, there will be a vertex of valency one? And why would $H\backslash x$ be connected and not bipartite? Is there some theorem for this or is there a simple justification?

How does one get that $H$ is invertible if $H\backslash x$ is invertible?

答:
Here's a more concrete argument.

If $H$ is connected and not bipartite (hence not a tree), it must have at least as many edges as vertices, so "full rank" means "full row rank": we want to show that the rows are linearly independent. Equivalently, if $M_H$ is the incidence matrix of $H$, then we want to show $M_H \mathbf x = \mathbf 0 \implies \mathbf x = \mathbf 0$.

Pick any $\mathbf x \in \mathbb R^n$, where $n = |V(H)|$; this assigns a real number $x_v$ to every vertex $v$. Then the vector $M_H \mathbf x \in \mathbb R^m$ (where $m = |E(H)|$) assigns each edge $\{v,w\}$ the value $x_v + x_w$. If $M_H \mathbf x = \mathbf 0$, this means that whenever vertices $v$ and $w$ are adjacent, $x_v = -x_w$. Because $H$ is connected, this means that every entry of $\mathbf x$ is the same up to sign.

But now, find an odd cycle $v_1, v_2, \dots, v_{2k+1}$ in $H$. The edge condition above says that
$$
   x_{v_1} = -x_{v_2} = x_{v_3} = -x_{v_4} = \dots = x_{v_{2k-1}} = -x_{v_{2k}} = x_{v_{2k+1}} = -x_{v_1}
$$
and $x_{v_1} = -x_{v_1}$ means $x_{v_1} = 0$. From connectivity, we conclude $\mathbf x = \mathbf 0$.

Title: incidence matrix with full rank


问:
Understanding $ \forall x \exists y \forall z P(x, y, z) $
$ \forall x \exists y \forall z P(x, y, z) $

Domain $D$: {1, 2}

For this formula to evaluate to true, would it be correct to say that the predicate $P$ would need to evaluate to true for the following tuples of terms?

$ (1, e_1, 1) $

$ (1, e_1, 2) $

$ (2, e_2, 1) $

$ (2, e_2, 2) $

Where $ e_1 \in D $ and $ e_2 \in D $

答:
Yes, you are right.

You have to show that such $e_1$ and $e_2$ exists to make them all true.


问:
Prove that two bases for a vector space are equinumerous
If $B_{1},B_{2}$ are two bases for a vector space $V$, prove that there exists a bijection $f:B_{1}\rightarrow B_{2}$.

It is known to all that the statement is true for finite-dimensional vector spaces, with quite a lot of elegant proofs. I suppose there at least should be one or two similarly elegant proofs for the infinite-dimensional version, assuming that the Axiom of Choice is correct.

答:
Clearly, if any one basis is finite then they are both finite. So we may assume they are both infinite.

For each vector $v$ in $B_1$ there is a finite subset $S_v$ of $B_2$ that suffices to express $v$ as a linear combination. Moreover, every vector in $B_2$ must occur in one of the $S_v$, lest a proper subset of $B_2$ span $B_1$ and hence $V$. Therefore,
$$|B_2| = \left|\bigcup_{v\in B_1}S_v\right|\leq \sum_{v\in B_1}|S_v| \leq \sum_{v\in B_1}\aleph_0 = \aleph_0|B_1| = |B_1|.$$
Symmetrically, $|B_1|\leq |B_2|$.


问:
Find the general solution to the differential equation $y' + 12 x^{11} y = x^{12}$
$y' + 12 x^{11} y = x^{12}$

Use the variable $I = \int e^{x^{12}} dx$ where it occurs in your answer.

I attempted this one and found the variable $I$ also works as the integrating factor. Then I arrived at $ ye^{x^{12}} = \int x^{12} e^{x^{12}}dx $. I'm stuck here.

Body:
You wish to simplify:

$$I_{2} = \int x^{12}e^{x^{12}}\ dx$$

Integrating by parts with $u = \frac{x}{12}, v = e^{x^{12}}$, so $v'= 12x^{11}e^{x^{12}}$:

$$I_{2} = \int uv'\ dx = uv - \int u'v \ dx$$

$$I_{2} = \frac{xe^{x^{12}}}{12} - \int\frac{e^{x^{12}}}{12}\ dx = \frac{xe^{x^{12}} - (I + C)}{12}$$

Thus:

$$y = \frac{I_{2}}{e^{x^{12}}} = \boxed{\frac{x}{12} - \frac{I}{e^{x^{12}}} + \frac{C}{e^{x^{12}}}}$$

Here, $C$ can be any constant so we do not need to show its negation or division by $12$.