问:
Find the limit of $(n^3+2n^2)^{1/3}-n$
Find the limit of sequence.I tried and get this
$$
\lim_{n\to \infty} \frac{((n^3+2(n)^2)^{1/3}-n)\cdot((n^3+2 n ^2)^{1/3}+n)}{(2(n)^2+n^3)^{1/3}+n}
$$ I am getting limit 0 but this is wrong answer can you help?

答:
Find the $\lim_{n\to \infty} \left(n^3+2n^2\right)^{1/3} ~-~ n$

You can regard $a_n = \left(n^3+2n^2\right)^{1/3} ~-~ n$

and then use the epsilon definition of the limit of a sequence to prove that the limit $~= (2/3).$

For all $n \in \mathbb{Z^+}, (n^3 + 2n^2) < (n + [2/3])^3 \implies$

$(n^3 + 2n^2)^{1/3} < (n + [2/3]) \implies$

$|~~(n^3 + 2n^2)^{1/3} - n - [2/3]~~| ~~=~~ (n +[2/3]) - (n^3 + 2n^2)^{1/3}.$

Further, you have the identity that

$(A - B) = \frac{A^3 - B^3}{A^2 + AB + B^2} < \frac{A^3 - B^3}{A^2} ~:~$ when $A,B > 0.$

Therefore, the problem reduces to establishing that for any fixed $\epsilon > 0$, you can identify $N \in \mathbb{Z^+}$ such that for all $n \geq N$,

$$\frac{(n + [2/3])^3 - (n^3 + 2n^2)}{(n + [2/3])^2} < \epsilon.$$

$$(n + [2/3])^3 - (n^3 + 2n^2) = [4/3]n + [8/27] < [4/3](n + [2/3]).$$

Therefore,

$$\frac{(n + [2/3])^3 - (n^3 + 2n^2)}{(n + [2/3])^2} < \frac{[4/3]}{n + [2/3]}.$$

Therefore, the problem reduces to demonstrating that for any $\epsilon > 0, N \in \mathbb{Z^+}$ can be found such that for all $n \geq N,$

$$\frac{[4/3]}{n + [2/3]} < \epsilon.$$

Simply choose $N \in \mathbb{Z^+}$ such that

$$\frac{[4/3]}{\epsilon} < N + [2/3].$$

Then you have that

$$n \geq N \implies (n + [2/3]) \geq (N + [2/3]) \implies
\frac{[4/3]}{n + [2/3]} < \epsilon.$$

Title: Find the limit of $(n^3+2n^2)^{1/3}-n$


问:
Topological manifold example
$\theta(x,x^2)=x$

$\Bbb X =${$(x,x^2)| x$ in $\Bbb R$}

And V is subset of $\Bbb R$

$dim\Bbb X=1$

My instructor said that this is topological manifold.

Why?

Please can you explain me? This example is from my notebook. But there is No explanation except for this. Thank you for help.

答:
We know that : The Graph of a smooth function is a manifolds.

For a subset of $A ⊂ \mathbb R^n$ and a function
$f : A→\mathbb R^m$, the graph of $f$ is defined to be the subset
$$G( f ) = \{(x, f (x)) ∈ A×\mathbb R^m\}$$
If $U $is an open subset of $\mathbb R^n$ and $f : U →\mathbb R^n$ is $C^{\infty}$, then the two maps
$$f : G( f )→U$$   $$(x, f (x)) \mapsto  x$$
and
$$(1, f ) : U →G( f )$$$$ x \mapsto (x, f (x))$$
are continuous and inverse to each other, and so are homeomorphisms. The graph
$G( f )$ of a $C^{\infty}$ function $f : U →\mathbb R^m$ has an atlas with a single chart $(G( f ),f )$, and is
therefore a $C^{\infty}$ manifold. This shows that many of the familiar surfaces of calculus,
for example your question are maniflods .

ContentLicense: CC BY-SA 3.0

Title: Topological manifold example


问:
Computing all possible transition scenarios
thanks for reading. I have the following problem:

We have data on the progression of the health status of 52 patients over two time periods (Table 1). The patients only progressed to a worse health status, but we do not know if during this period of time the patients have progressed 1, 2 or 3 health statuses. What are all the transition posibilities?

Table 1.

```<code>|Health state| Time 0 | Time 1 |
|------------|--------|--------|
|Best        |20      |10      |
|Good        |10      |1       |
|Bad         |20      |19      |
|Death       |2       |22      |
</code>```

As result, I'm looking for all scenarios that fit in Table 2.

Table 2.

```<code>|↓Time 0\Time 1→|Best    |Good    |Bad    |Death |Total|
|---------------|--------|--------|-------|------|-----|
|Best           |    a   |    b   |   c   |   d  |20   |
|Good           |    -   |    e   |   f   |   g  |10   |
|Bad            |    -   |    -   |   h   |   i  |20   |
|Death          |    -   |    -   |   -   |   j  |2    |
|Total          |   10   |    1   |  19   |  22  | 52  |
</code>```

All numbers must be integers and positive and, as you can see, a=10.
Is there any general method to solve this? Is there any software that can help me with this?
Thank you! :)

ContentLicense: CC BY-SA 4.0

Body:
With $11$ unknowns and $8$ data points, there isn't enough information given to decide what values $a,\ldots, k$ have.

$$\begin{pmatrix}1&1&1&1&0&0&0&0&0&0&0\\0&0&0&0&1&1&1&0&0&0&0\\0&0&0&0&0&0&0&1&1&1&0\\0&0&0&0&0&0&0&0&0&0&1\\1&0&0&0&0&0&0&0&0&0&0\\0&1&0&0&1&0&0&1&0&0&0\\0&0&1&0&0&1&0&0&1&0&0\\0&0&0&1&0&0&1&0&0&1&1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\\f\\g\\h\\i\\j\\k\end{pmatrix}=\begin{pmatrix}20\\10\\20\\2\\10\\1\\19\\22\end{pmatrix}$$

Usually one would use the method of least squares, but in this case this gives negative values for the unknown values. Instead, perform Gaussian elimination to get

$$\begin{pmatrix}1&0&0&0&0&0&0&0&0&0&0\\0&1&0&0&0&-1&-1&0&-1&-1&0\\0&0&1&0&0&1&0&0&1&0&0\\0&0&0&1&0&0&1&0&0&1&0\\0&0&0&0&1&1&1&0&0&0&0\\0&0&0&0&0&0&0&1&1&1&0\\0&0&0&0&0&0&0&0&0&0&1\\0&0&0&0&0&0&0&0&0&0&0\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\\f\\g\\h\\i\\j\\k\end{pmatrix}=\begin{pmatrix}10\\-29\\19\\20\\10\\20\\2\\0\end{pmatrix}$$

Keeping the positivity constraints, this implies $$\begin{matrix}a=10,& b=0 \textrm{ or }1,& c+d=9\textrm{ or }10,&
e=0\textrm{ or }1,\\ f+g=9\textrm{ or }10,& h=0\textrm{ or }1,& i+j=19\textrm{ or }20,& k=2.\end{matrix}$$
with the additional constraints $f+i\le 19$, $g+j\le 20$.


问:
Different descent methods to bound the rank of an elliptic curve
I'm trying to find out as much as possible about different descent methods to bound the rank of an elliptic curve.

Currently I understand and am able to apply the 'Descent via 2-isogeny' method.

I've also heard of a '$\textit{complete}$ 2-isogeny' method (?) and a 'descent via 3-isogeny' method.

My question is, firstly, does anyone know of any links that explain these two other methods preferably with concrete examples of the application?

Secondly, are there any other descent methods? I believe there are but I have struggled to find a comprehensive list.

Any links to help me understand any of these or any other way of bounding the rank of an elliptic curve would be much appreciated.

答:
The second method you have heard of is "complete $2$-descent". There are examples of this in Chapter X of Silverman's The Arithmetic of Elliptic Curves (see Prop 1.4 and the following example).

When $E$ has a $2$-torsion point, one may use descent by $2$-isogeny which appears in $\S 14$ "A $2$-isogeny" of Cassels' Lectures on Elliptic Curves (also as Proposition 4.9 of Silverman chapter X).

The more general method of $n$-descent exists. Searching google for "$n$-descent" for some fixed $n$ probably gives a reference. There is a paper of Stoll called Descent on Elliptic Curves which discusses the general case, and also a paper of Schaefer and Stoll How to do a $p$-Descent on an Elliptic Curve.

To understand this you should be familiar with Chapter X of Silverman, and be comfortable with complete $2$-descent. It is a useful exercise to specialise to $2$-descent (which encompasses "complete $2$-descent") and compute some examples within this framework.


问:
Solve $A\mathbf{x}=\mathbf{v}(\mathbf{x}^TD\mathbf{v})$ for $A$
Is there a way to simplify
$$f(\mathbf{x})=\mathbf{x}-\mathbf{v}(\mathbf{x}^TD\mathbf{v})$$

to the form

$$f(\mathbf{x})=(I-A)\mathbf{x}$$

where
$\mathbf{x}$ and $\mathbf{v}$ are column-vectors and $D$ is a matrix?

In particular, I seek to extract the variable $\mathbf{x}$ as a common factor so that it appears only once in the expression for $f(\mathbf{x})$.

I could write $\mathbf{x}-\mathbf{v}(\mathbf{v}^TD^T\mathbf{x})$ but then I still don't know how to get that into the form $(...\text{constant matrix expression }...)\mathbf{x}$.

答:
This expression is linear in $x$ so it can be reduced to such form:

$f(\mathbf{x})=M \mathbf{x}$

where explicitely the components of the matrix $M$ are $M_{a,b}=\delta_{a,b}-v_a (D\mathbf{v})_b$

We can put it in a more vector form, if needed:

$v_a (D\mathbf{v})_b=\sum_c v_a D_{b,c}v_c=\sum_c (\mathbf{v}\mathbf{v}^+)_{a,c}D_{b,c}=[(\mathbf{v}\mathbf{v}^+) D^+]_{a,b}$


问:
Taking Velocity Derivative wrt X
I was looking at some examples in my Fluid Mechanics textbook, and it showed that
$$\frac{du(x)}{dx} = \frac{d}{dx}(v_0 + v_0\frac{R^3}{x^3}) = -3\frac{v_0^2}{R}\frac{1+(x/R)^3}{(x/R)^4} $$
where $v_0$ is the initial velocity of the fluid and $R$ is the radius of a sphere where the fluid traverses across. Both constants. When I tried to derive it by myself, I got the following:
$$\frac{du}{dx} = -3v_0\frac{R^3}{x^4} $$
I do not understand how to reach the form that the book reaches, please help.

答:
If $v_0$ and $R$ are constants, your book cannot be correct. One can easily check that without writing out the calculations.

You need more information on those parameters. Perhaps, you can read more context around that formula.

Title: Taking Velocity Derivative wrt X


问:
calculate $\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$
How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get

$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
and this doesn't get me further. Any ideas?

答:
$$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}\frac{\sqrt{1-\cos(\pi t)}}{\sqrt{1-\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)\sqrt{1-\cos(\pi t)}}{\sqrt{\sin^2(\pi t)}}$$

P.S. Pay attention to the sign of $\sin(\pi t)$ .


问:
Subspace topology and initial topology
Let $f_i:X \to X_i$, $i \in I$ be a family of functions in topological spaces $X_i$, $A \subset X$ a subset and $\iota_A:A \to X$ the inclusion map. Suppose that $X$ has the initial topology with respect to the family $\{f_i\}_{i \in I}$.

I wanna show that the subspace topology of $A$ with the initial topology with respect to the family  $\{f_i \circ \iota_A\}_{i \in I}$ the same is but I got stuck.

Some help would be really nice!

答:
This is a consequence of the transitive law of initial topologies, that I showed in a general form here, to show that $\prod_i A_i$ as a subspace of the product has the same topology as the product topology of the subspace topologies.

I'll give a direct proof for your case:

The initial topology on $X$ wrt $\{f_i \mid i \in I\}$ has as its standard subbase

$$\mathcal{S} = \{f_i^{-1}[O]\mid i \in I, O \subseteq X_i \text{ open }\}$$

so the subspace topology on $A$ has subbase

$$\mathcal{S}_A= \{f_i^{-1}[O] \cap A\mid i \in I, O \subseteq X_i \text{ open }\}$$

Now note that $(f_i \circ \iota_A)^{-1}[O] = f_i^{-1}[O] \cap A$ as well, so that $\mathcal{S}_A$ is exactly the standard subbase for the initial topology on $A$ wrt the maps $\{f_i \circ \iota_A \mid i \in I\}$, as required.

The explicit description by subbases of these topologies is helpful here.

Title: Subspace topology and initial topology


答: Let $\tau_A$ denote the subspace topology on $A$ and $\tau_F$ denote the initial topology given by $F=\{f_i \circ \iota_A: i \in I\}$. Since for all $i$ we have that $f_i \circ \iota_A \colon (A,\tau_A) \to X_i$ is continuous, (because composition of continuous gives continuous) then this shows that the initial topology $\tau_F$ is coarser than $\tau_A$ (i.e. $\tau_F \subseteq \tau_A$). Now take $U$ open in $\tau_A$ and  $p \in U$. Then there exists an open set of $X$, say $O$ such that $U= A\cap O$. Since $p \in O$ there exists some open sets $U_{j_k} \subseteq X_{j_k}$, with $j_k \in I$ for all $k\in \{1,...,n\}$ such that $p \in f_{j_1}^{-1}[U_{j_1}] \cap \ldots \cap f_{j_n}^{-1}[U_{j_n}] \subseteq O$. Finally, note that $p \in (f_{j_1}\circ \iota_A)^{-1}[U_{j_1}] \cap \ldots \cap (f_{j_n}\circ\iota_A)^{-1}[U_{j_n}] \subseteq O\cap A$. This shows that $p$ is an interior point of $U$ with respect to the initial topology and hence $\tau_A \subseteq \tau_F$.

Title: Subspace topology and initial topology

Body:
Let $f_i:X \to X_i$, $i \in I$ be a family of functions in topological spaces $X_i$, $A \subset X$ a subset and $\iota_A:A \to X$ the inclusion map. Suppose that $X$ has the initial topology with respect to the family $\{f_i\}_{i \in I}$.

I wanna show that the subspace topology of $A$ with the initial topology with respect to the family  $\{f_i \circ \iota_A\}_{i \in I}$ the same is but I got stuck.

Some help would be really nice!


问:
Elementary methods for checking double roots
How can I check if a large polynomial has any double roots? I was trying to see if the polynomial
$$x^{2021} + x^{3} + 1$$
had any double roots, but I had no idea what to do. After some research, I saw that that the polynomial should never be equal to it's derivative, but unfortunately I am unfamiliar with calculus. I tried seeing if I could find some roots and perhaps discover anything interesting, but it would take too long.

Is there an elementary method of verifying this?

答:
I shall admi that we are in the real domain
$$f(x)=x^{2021}+x^3+1 \implies f'(x)=x^2 \left(2021 x^{2018}+3\right)$$ $x=0$ is not a root of $f(x)$ and the next term is always positive.

This case is easy but, in a general manner, ???


答:
The powers $2021$ and $3$ are both odd, hence the terms $x^{2021}$ and $x^3$ are both monotonically increasing, hence so is their sum.  Adding $1$ merely shifts the sum up a bit.  No double roots.

Incidentally, the test for multiple roots is not 'is the polynomial equal to its own derivative' (never true for polynomials), it is 'does the polynomial share a factor with its own derivative'.

Title: Elementary methods for checking double roots

Body:
How can I check if a large polynomial has any double roots? I was trying to see if the polynomial
$$x^{2021} + x^{3} + 1$$
had any double roots, but I had no idea what to do. After some research, I saw that that the polynomial should never be equal to it's derivative, but unfortunately I am unfamiliar with calculus. I tried seeing if I could find some roots and perhaps discover anything interesting, but it would take too long.

Is there an elementary method of verifying this?


问:
A doubt on Lie subgroups of Lie groups
The following theorem can be found in Lee's ``Introduction to Smooth Manifold".

Suppose $G$ is a Lie group with Lie algebra $\mathfrak g.$ If $\mathfrak h$ is a Lie subalgebra of $\mathfrak g$ then there is a unique connected Lie subgroup $H$ of $G$ with Lie algebra $\mathfrak h.$

Are the following statements true? Given the hypothesis of the above theorem, there is a subgroup $H$ of $G$ such that $H$ can be given a topology and a smooth structure such that the natural inclusion map $i:H\to G$ is a Lie group homomorphism and $Di(e)T_{e}H=\mathfrak h.$

My question is in which topology $H$ is connected? Does uniqueness mean that if there is another subgroup $H^\prime$ with a topology and smooth structure, connected in its own topology and $Di(e)T_{e}H=\mathfrak h$ then $H=H^\prime$ and the identity map from $H$ to $H^\prime$ is a Lie group isomorphism?

Can there be some natural counterexamples violating the asumptions?

Definition: Let $M$ be a smooth manifold. Then $S\subseteq M$ is said to be a submanifold if $S$ has its own topology and smooth structure with respect to the inclusion map is a smooth immersion.
Definition: Let $G$ be a Lie group. A subgroup $H\subseteq G$ is called a Lie subgroup if $H$ is a submanifold and Lie group with respect to the smooth structure of being a submanifold.

答:
I don't have a copy of the book accessible to me at the moment. I think it is a good question. Let me explain a standard "pathological" example first.

Let $V = \mathbb{R}^2$ and $\mathbb{Z}^2 \subset \mathbb{R}^2$ be the integer lattice inside $V$. Then $T^2 := V/\mathbb{Z}^2 \simeq U(1) \times U(1)$ is a (abelian) two-dimensional Lie group, where the Lie group "product" is induced by addition of vectors in $V$.

At $\mathbf{0} \in V$, consider a tangent line $l \subset T_\mathbf{0}(\mathbb{R}^2) = \mathbb{R}^2$. If the slope of $l$ is rational, then $l \mod \mathbb{Z}^2$ defines a Lie subgroup of $T^2$ with the subspace topology (it is a closed submanifold of $T^2$). But if the slope of $l$ is irrational, then $l \mod \mathbb{Z}^2$ defines a Lie subgroup of $T^2$ which is not a closed submanifold.

It is due to pathological examples such as this one that many authors define the notion of a Lie subgroup without requiring that a Lie subgroup have the induced topology.

I don't know the precise definitions that Lee is using in his book, so I cannot discuss further before knowing the precise definitions he uses for a "Lie subgroup" and for a "submanifold".

If the OP wishes to discuss further, please add these definitions to your post, and then I may be able to discuss in more detail.

Title: A doubt on Lie subgroups of Lie groups


问:
If $A\cap B= \emptyset \Rightarrow \bar{A}\cap\dot{B} = \emptyset$
$A\cap B= \emptyset \Rightarrow \bar{A}\cap\overset{\,\circ}{B} = \emptyset$

This should be a really simple question but I cannot seem to work it out. I suppose there is an element $x \in \bar{A}\cap\overset{\,\circ}{B}$ then it must also be $x \in \bar{A}$ and $x\in B$. So I now only have to show that it must be in $A$ also. I suppose it is not, then it must be in $\partial A$. But I cannot seem to arrive at a contradiction.

答:
$A \cap B = \emptyset$ implies $$A \subseteq B^\complement$$ (complement in the space $X$)

so $$\operatorname{int}(A) \subseteq \operatorname{int}(B^\complement)$$ follows by taking interior on both sides, which preserves inclusion.

By a standard duality identity we have $$\operatorname{int}(B^\complement)=(\overline{B})^\complement$$

so $$\operatorname{int}(A) \subseteq (\overline{B})^\complement$$

or $$\operatorname{int}(A) \cap \overline{B}= \emptyset$$

as required.


问:
The proof of $\log(1+x) < x.$
I want to prove $\log(1+x) < x.$

Of course, I can prove this by letting $f(x):= x - \log(1+x),$ and calculating $f'(x)$, ... .

But I wonder if I can prove this using Maclaurin's expansion.

$\log(1+x)=x-\dfrac{1}{2}x^2+\dfrac{1}{3} x^3-\cdots.$

The right-hand side includes the term "$x$" so it seems that I can use this equality in order to prove $\log(1+x) < x$, but I cannot.

I would like you to give me some ideas.

答:
Take $0<x<1$, then if we let $a_k(x) = (-1)^{k+1}\frac{1}{k}x^k$ we see that
$$\log (1+x)-x = \sum_{k=2}^{\infty}a_k(x).$$

Now note that in the specified interval we have that $|a_k(x)|>|a_{k+1}(x)|$ and that $\mathrm{sgn}\, a_k(x) = -\mathrm{sgn}\, a_{k+1}(x)$ so that the sum is alternating. These two properties implies that $a_{2k}(x)+a_{2k+1}(x)<0$ and therefore
$$\log(1+x)-x= \sum_{k=2}^{\infty}a_k(x) = \sum_{k=1}^{\infty}\underbrace{a_{2k}(x)+a_{2k+1}(x)}_{<0}<0$$

Title: The proof of $\log(1+x) < x.$


Title: Find the limit of $(n^3+2n^2)^{1/3}-n$

Body:
Find the limit of sequence.I tried and get this
$$
\lim_{n\to \infty} \frac{((n^3+2(n)^2)^{1/3}-n)\cdot((n^3+2 n ^2)^{1/3}+n)}{(2(n)^2+n^3)^{1/3}+n}
$$ I am getting limit 0 but this is wrong answer can you help?

Body:
Just standard rearranging gives:

\begin{eqnarray*} (n^3+2n^2)^{1/3}-n
& = & (n^3+2n^2)^{1/3}-(n^3)^{\frac 13}\\
& \stackrel{a^3-b^3 = (a-b)(a^2+ab+b^2)}{=} & \frac{n^3+2n^2 - n^3}{(n^3+2n^2)^{2/3} + (n^3+2n^2)^{1/3}n + n^2}\\
& = & \frac 2{(1+\frac 2n)^{2/3} + (1+\frac 2n)^{1/3} + 1}\\
& \stackrel{n\to\infty}{\longrightarrow} & \frac 23
\end{eqnarray*}


答:
For me the simplest proof is to look for the minimum value of $$f(x)=x- \log (1+x)$$
$$f'(x)=1-\frac 1{1+x} \qquad \text{and} \qquad f''(x)=\frac 1{x^2} \quad > 0~~\forall x$$  So the extremum is at $x=0$ and the second derivative test shows that this is a minimum. So, in real domain, $\forall x$
$$x \geq \log(1+x).$$

Title: The proof of $\log(1+x) < x.$

Body:
I want to prove $\log(1+x) < x.$

Of course, I can prove this by letting $f(x):= x - \log(1+x),$ and calculating $f'(x)$, ... .

But I wonder if I can prove this using Maclaurin's expansion.

$\log(1+x)=x-\dfrac{1}{2}x^2+\dfrac{1}{3} x^3-\cdots.$

The right-hand side includes the term "$x$" so it seems that I can use this equality in order to prove $\log(1+x) < x$, but I cannot.

I would like you to give me some ideas.


问:
Norm of a vector product
I'm looking into the norm of a rank-1 matrix $A$.

Because $A$ is a rank-1 matrix, we know that $A=b c^T$, for a $b,c\in\mathbb{R}^n$. For the norm we then have \begin{equation}
\left\Vert A\right\Vert = \max_{x\ne 0}\frac{\left\Vert Ax\right\Vert}{\left\Vert x\right\Vert} = \max_{x\ne 0}\frac{\left\Vert bc^Tx\right\Vert}{\left\Vert x\right\Vert} = \frac{\left\Vert bc^Tc\right\Vert}{\left\Vert c\right\Vert}.
\end{equation}

I think I can further simplify this, but my question is how? Can I write $\left\Vert bc^Tc\right\Vert = \left\Vert b\right\Vert\left\Vert c^Tc\right\Vert$ for example?

答:
Note that $\Vert ax \Vert = |a|\Vert x \Vert$  for all $a \in \mathbb{R}$ and all $x \in \mathbb{R}^n$.

$c^\top c$ is a scalar.

Title: Norm of a vector product


问:
Elementary proof of the fact that any orientable 3-manifold is parallelizable
A parallelizable manifold $M$ is a smooth manifold such that there exist smooth vector fields $V_1,...,V_n$ where $n$ is the dimension of $M$, such that at any point $p\in M$, the tangent vectors $V_1(p),...,V_n(p)$ provide a basis for the tangent space at $p$. Equivalently, a manifold is parallelizable if its tangent bundle is trivial.

There is a theorem that states that any compact orientable 3-manifold is parallelizable, and
there is a proof of this result which uses $spin^c$ structures and the Steifel-Whitney class.

I am wondering whether there exists a more elementary, perhaps more straightforward proof. Otherwise, I would be grateful for some intuition on why this is true.
Also, it the theorem still true without the compactness assumption? If so, is there a relatively simple proof in that case?

答:
Benedetti and Lisca wrote a paper on this same question. They write “the answers given [on this question] until July 18, 2018 use
the same tools employed in the proofs mentioned above”, and give three more-elementary answers.
https://arxiv.org/abs/1806.04991</p>


问:
Properties of homeomorphic function
Is this function homeomorphic and most importantly why?

$$f(x)=x^2$$

What are the properties of homeomorphic functions? When is the function continuous and $1-1$?

Best regards,
Sergey

答:
A function $f : X\to Y$ ($X$ and $Y$ topological spaces) is a homeomorphism iff $f$ is bijective, continuous and has an inverse function $f^{-1}$ that is continuous.
(https://en.wikipedia.org/wiki/Homeomorphism)

Hence, to study whether $f:x\mapsto x^2$ is a homeomorphism, you have to study if it is bijective and continuous then see if its inverse function is continous.

Then, what is a bijection? Here is a definition for you :

$f:I\to J$ is bijective iff for every element $y$ of $J$, there is an unique element $x$ of $I$ verifying $f(x)=y$.

So, at first, is $x^2$ bijective over $\bf{R}$? Then, on which domain can $x^2$ be a homeomorphism?

I hope this can help you start, otherwise don't hesitate to ask specific questions so I can help you.

Title: Properties of homeomorphic function


问:
A problem with a application of SLLN and CLT and distribution convergence

Let $Y_{i}\overset{i.i.d}{\sim} N(0,\sigma^{2})$ with $\sigma^{2}$ unknow. I'm trying to prove that $$
    \sqrt{n}\left(\overline{Y}_{n}^{2}- \sigma^2 \right) \overset{D}{\to}  N\left(0,2\sigma^4\right)$$

My approach: I proved that $\overline{Y}_{n}^{2}\overset{D}{\to} \sigma^{2}$ using the strong law large number and using the central limit theorem for random variables i.i.d we have that $\sqrt{n}(\overline{Y}_{n}-\mathbb{E}X_{i})\overset{D}{\to} N(0,\sigma^{2})$, then I was trying use that fact that $f(x)=x^{2}$ and apply the Delta's method, but I can't continue from here because $f(x)=[\mathbb{E}X_{i}]^{2}=0$ and I obtained $\sqrt{n}\overline{Y}_{n}^{2}\overset{D}{\to}N(0,0)$ but its not
desired. 1) How can I solve this problem from here? and can I use that result for conclude that $\overline{Y}_{n}^{2}$ is consistent for $\sigma^{2}$? Note: I proved that $\overline{Y}_{n}^{2}$ is consistent without that result above, but also I'm trying to see if I can conclude the same using the result above.

答:
there are various ways to approach the problem.

The first I get in mind is the more statistical one.

Just to avoid misunderstandings, let's set

$$Y_i \sim N(0;\theta)$$

where $\theta=\sigma^2$

First observe that $T=\overline{Y}_n^2$ is the maximum likelihood estimator for $\sigma^2$ thus is is well known (it is one important property of MLE that you can check in any basic Statistics Textbooks) that

$$\sqrt{I_n(\theta)}\Big(\overline{Y}_n^2-\theta\Big)\dot{\sim}N[0;1]$$

Where $I_n(\theta)$ is the Fischer Information.

$$I_n(\theta)=-nE\Bigg\{\frac{\partial^2}{\partial \theta^2}\log f(x,\theta)\Bigg\}$$

Calculations...

$$f(x,\theta)\propto \theta^{-1/2}e^{-y^2/(2\theta)}$$

$$\log f(x,\theta)=-\frac{1}{2}\log\theta-\frac{y^2}{2\theta}$$

$$\frac{\partial}{\partial\theta}\log f(x,\theta)=-\frac{1}{2\theta}+\frac{y^2}{2\theta^2}$$

$$\frac{\partial^2}{\partial\theta^2}\log f(x,\theta)=\frac{1}{2\theta^2}-\frac{y^2}{\theta^3}$$

$$\mathbb{E}\Bigg[\frac{\partial^2}{\partial\theta^2}\log f(x,\theta)\Bigg]=\frac{1}{2\theta^2}-\frac{\theta}{\theta^3}=-\frac{1}{2\theta^2}$$

$$-n\mathbb{E}\Bigg[\frac{\partial^2}{\partial\theta^2}\log f(x,\theta)\Bigg]=\frac{n}{2\theta^2}=\frac{n}{2\sigma^4}$$

This means that

$$\overline{Y}_n^2\dot{\sim}N\Bigg(\sigma^2;\frac{2\sigma^4}{n}\Bigg)$$

Your are all set!


答:
A followup to the answer by Claude Leibovici.

This is a quick way of seeing it, but it might not be elementary enough:

By the Eisenstein criterion the polynomial $2021x^{2018}+3$ is irreducible over $\Bbb Q$. Hence $\gcd(f,f')\ne1$ if and only if $2021x^{2018}+3\mid f$. But this is impossible as $x^{2018}+\frac{3}{2021}\notin\Bbb Z[x]$.

(In general if $R$ is a UFD and $f\in R[x]$ monic with $f=gh$, $g,h\in (\operatorname{Frac}(R))[x]$ monic then $g,h\in R[x]$)

Therefore $\gcd(f,f')=1$, i.e. $f$ has no multiple roots.


答: Since you talk about series expansions, one could exponentiate the inequality (the exponential function is monotonically increasing, so this is equivalent to the original inequality) to obtain $e^x > 1 + x$, which is easy to prove with the series expansion of the exponential function: we have $$e^x = 1 + x + \underbrace{x^2 + \ldots}_{> 0} > 1 + x$$ for all $x > 0$.

Title: The proof of $\log(1+x) < x.$


问:
Asymptotes of a conicoid
What do we mean by asymptotes of a conicoid ?

The asymptotes of below conicoid
$$ax^2+by^2+cz^2=1$$
is the cone
$$ax^2+by^2+cz^2=0$$

We can derive this by the general definition of a asymptotes . Mathematics seems correct here.
However , i cannot visualize how a cone can meet a fixed surface(in this case a conicoid) at infinity ?
What do we mean by asymptote of a cone ,cylinder etc. ?

答:
Perhaps this will help:

The outer surface is the hyperboloid $x^{2} + y^{2} - z^{2} = 1$, and the inner surface is the cone $x^{2} + y^{2} - z^{2} = 0$. You can see that the surfaces get closer together as $z$ goes to $\pm\infty$.

Title: Asymptotes of a conicoid


Body:
You may apply the classic techniques in formal derivatives whose definition doesn't rely on calculus.  Let $f(x) = x^{2021} + x^3 + 1$.  Then $f'(x) = 2021x^{2020} + 3x^2 = (2021x^{2018}+3) x^2$, which has

a double root $x = 0$.

$2018$ $4036$-th roots of unity $\omega_{4036}^{2k+1}$ with $k \in \{0, 1, \dots, 2017\}$ multiplied by $\sqrt[2018]{3/2021}$.

It suffices to verify that these $2019$ roots of $f'(x)$ aren't a zero of $f(x)$.

\begin{aligned}
&f(\omega_{4036}^{2k+1}) \\
=& (\omega_{4036}^{2k+1})^3 (-3/2021) + (\omega_{4036}^{2k+1})^3 + 1 \\
=& 1+2018\omega_{4036}^{3({2k+1})}/2021
\end{aligned}

Note that $\mathrm{Im}(f(\omega_{4036}^{2k+1})) = 2018 \sin(3(2k+1)\pi/2018)/2021 \ne 0$ as $3(2k+1)/2018 \notin \mathbb{Z}$ for all $k \in \mathbb{Z}$.


问:
What are the correct interpretations for these basic probability questions?
I have this homework on basic probability for my engineering mathematics course, but I don't understand these two sub-questions.

Here is the text.

The probabilities of the monthly snowfall exceeding 10 cm at a
  particular location in the months of December, January and February
  are 0.20, 0.40 and 0.60, respectively.

Here are the questions that I do not know how to interpret.

(For a particular winter:)

What is the probability of receiving at least 10 cm snowfall in a
  month, in at least two of the three months of that winter?

Find the expected number of months in which monthly snowfall does not exceed 10 cm.

Given and my assumptions:

There are three consecutive months, each with a corresponding probability of exceeding 10 cm snowfall.

A month of exceeding snowfall does not affect the amount of snowfall for the next month (not to be confused with the months' relationship in a tree diagram, they are still related)

Sorry to ask about this. English is not my main language.

答:
a)  The probability of the monthly snowfall NOT exceeding $10 $ cm is $0.8, 0.6$, and $0.4$. Since these are assumed to be independent (not really a safe assumption), the probability of all three being less than $10$ cm is:
\begin{align*}
0.8 \times 0.6 \times 0.4 = 0.19200
\end{align*}

b)

All $3 = .2 \times .4 \times .6 = 0.04800$

Dec & Jan & not Feb  $ =0.2 \times 0.4 \times 0.4 = 0.03200$

Dec, not Jan, Feb $= 0.2 \times 0.6 \times 0.6 = 0.07200$

not Dec, Jan, Feb $= 0.8 \times 0.4 \times 0.6 = 0.19200 $

Now add up these four numbers: $0.34400 $

c)
Consecutive
Dec-Jan-not Feb $= 0.03200$

Jan-Feb-not Dec $= 0.19200$

$0.19200 + 0.03200 = 0.22400$
Not Consecutive

Dec - Feb-not Jan $= 0.07200$

$0.07200 + 0.19200 + 0.03200 = 0.29600$

probability of consecutive $= 0.244/0.296 = 0.756$


问:
Determine the integral $\int_{C} xy^2 \ dx + x^2y \ dy$ when $C$ denotes the circle $x^2+y^2=a^2, a>0.$

Determine the integral $$\int_{C} xy^2 \ dx + x^2y \ dy$$ when $C$ denotes the circle $x^2+y^2=a^2, a>0.$

I have that $F(x,y) = (xy^2, x^2y)$ which is a conservative vector field and the circle is a closed loop so one would assume that the integral would be $0$?

I got that $$\iint_D 2xy-2xy \ dA = \iint_D 0 \ dA = 0$$ I'm not sure if my reasoning here is correct?

答:
That is correct, by Green's theorem.

Being "conservative" also means that the line integral is path independent.


问:
$\cos (\tan \alpha)=\sin \alpha $
I can't find where is my mistake so if someone can enlighten me...

$\tan \alpha=\arccos (\sin \alpha) $

$\tan \alpha=\arccos (\cos (\alpha -\frac{\pi }{2}))= \alpha -\frac{\pi }{2}$

$\frac{d}{d\alpha }\tan \alpha=\frac{d}{d\alpha }(\alpha -\frac{\pi }{2})  $

$\frac{1}{\cos (\alpha)^{2}}=1  $

$\alpha =\pi n$

there is something wrong and I don't really know.

答:
The fact that $\alpha$ satisfies $F(\alpha)=G(\alpha)$ does not imply that $F'(\alpha)=G'(\alpha)$. For instance, $x=0$ is a solution to $x=x^2$, but not a solution to $1=2x$.

Added: There are of course a few other issues of purely algebraic nature, namely that $\cos t=c$ does not imply $t=\arccos c$, and that $\arccos \cos c\ne c$.

Title: $\cos (\tan \alpha)=\sin \alpha $


问:
Algebraic numbers and Number theory
I would like to know good references for undergraduate students on this topic. I went to wiki page, but it seems most of them are for graduate level. Anyways, one of questions on this topic is the following: Let's suppose we have

$a+b = A$

in which A is algebraic. I suppose we can conclude the following, assuming that the algebraic numbers is a closed field:

If $b$ is algebraic, then so is $a$. In addition, $a$ can be either rational or irrational.

If $b$ is transcedental, then so is $a$.

Are these conclusions and the assumption correct?

答:
Yes. It is a standard fact that the sum of algebraic numbers is algebraic.

In the first case $a = A - b$, so $a$ is algebraic since $A$ and $b$ are by assumption. For rational $a$ take $A = \sqrt{2} + 1$ and $b = \sqrt{2}$, for irrational $a$, take $b = 1$.

In the second case, suppose $a$ is algebraic. Then $A - a = b$ is algebraic. Hence if $b$ is transcendental, then so is $a$.

Title: Algebraic numbers and Number theory


问:
Finding a calculation formula to convert different units
A solution to my problem definitely must exist somewhere online, but I definitely don't know the right terminology to find it.
Essentially I am making a quick script to convert measuring units of oil (cst and wt) both of which are brand specific so are difficult to calculate the 'weight' of the oil.
I have this basic information for the two oil units. Is it possible to find a formula to convert between these two units based on this information?

\begin{align*}
1000\mathrm{cst} &= 80\mathrm{wt} \\
650\mathrm{cst} &= 50\mathrm{wt} \\
200\mathrm{cst} &= 20\mathrm{wt} \\
100\mathrm{cst} &= 10\mathrm{wt} \\
\end{align*}

答:
I don't think it is possible to know exactly how the amounts in cst units are converted to the amounts in wt units.
Either the relationship is non-linear or there is some rounding going on.

If the relationship is based on $1\,\mathrm{cst} = k\,\mathrm{wt}$
but the result of the calculation is then rounded to the nearest multiple of $10,$
there are various values of $k$ that might fit the given results.
An obvious choice would be $k = 0.08$, which is exactly for the first conversion and also works out for the other three conversions with rounding to the nearest multiple of $10.$
But a factor $k = 0.082$ would also work.

I don't think there is any way to reconcile the given figures with a linear factor with the result rounded to the nearest multiple of $5.$
If the rule for conversion had that form, the first equation says $k$ must be less than $0.0825$ (otherwise the linear conversion would be from $1000$ to at least $82.5$ which then would round to at least $85$), but the third equation says $k$ must be at least $0.0875$, since otherwise the linear conversion would give less than $17.5$, which would round to $15$ or less.

Conversion in the other direction would represent only a guess as to what the weight in cst is, since rounding causes multiple possible cst weights to be represented by the same wt weight.


问:
Decomposing an abelian extension into cyclic extensions
I have a question regarding abelian and cyclic extensions on number fields.

Let $L/K$ be an extension of number fields which is abelian (i.e., its Galois group is commutative). By the fundamental theorem of abelian groups, we can decompose Gal$(L/K)$ into a product of cyclic groups, namely

$$\mbox{Gal}(L/K) \cong C_1 \times C_2 \times ... \times C_r,$$

for some $C_i$ cyclic. (Note that all of these groups are finite of course).

How can i decompose $L/K$ into a compositum of cyclic extensions over $K$?

My guess: We know that $L = L^{C_1}L^{C_2}...L^{C_r}$ is the compositum of all the fixed fields by these cyclic groups. However, the extension $L^{C_i}/K$ is not cyclic (actually $L/L^{C_i}$ is) and so this doesn't work. I am considering now the following decomposition:

$$L = L^{\{id\}\times C_2\times...\times C_r}L^{C_1\times \{id\}\times...\times C_r}\cdots L^{C_1\times C_2\times...\times C_{r-1}\times \{id\}}$$

It is clear then that each of these fields in the decomposition are cyclic extensions of $K$ with Galois group $C_i$. Is this decomposition true? Namely, does this last equality really hold? And if not, how can i construct a decomposition of $L$ into cyclic extensions of $K$?

Thanks in advance!

答:
To be a bit more general, we have the following theorem:

Let $L/F$ be a finite Galois extension whose Galois group $G$ may be written as $G= G_1\times G_2\times\cdots \times G_n$, and let $L_i$ denote the fixed field of $G_1\times\cdots\times \{1\}\times\cdots \times G_n$, where the $\{1\}$ occurs in the i-th position of the direct product. Then each $L_i/F$ is Galois, $L_{i+1}\cap (L_1\cdots L_i)=F,$ and $L=L_1\cdots L_n$.

Mercio already gave a proof to the last assertion. Note that the original assumption that $C_i$ are cyclic wasn't used, so there are only notation changes, namely, $G_i$ becomes $C_i, L_i$ becomes $L^{G_i}$, and $F$ becomes $K$ in Mercio's answer.

For the first assertion, notice that $(G_1\times\cdots\times \{1\}\times\cdots \times G_n )\lhd (G_1\times\cdots\times G_n)$ regardless of the position of $\{1\}$ and therefore by the fundamental theorem of Galois theory, $L_i/F$ is Galois for each $i$.

Similarly to Mercio's answer we have that $L_1\cdots L_i$ corresponds to the intersection $\{1\}\times\cdots\times\{1\}\times G_{i+1}\times\cdots\times G_n$ , and  $L_{i+1}\cap (L_1\cdots L_i)$ corresponds to the smallest subgroup generated by $\{1\}\times\cdots\times\{1\}\times G_{i+1}\times\cdots\times G_n$ and $G_1\times G_2\times\cdots\times\{1\}\times G_{i+2}\cdots \times G_n$, which is obviously $G_1\times G_2\times\cdots \times G_n=G$. Then since the fixed field of $G$ is $F$, we get $L_{i+1}\cap (L_1\cdots L_i)=F$, which proves our second assertion.


问:
Exercise: The measure of the boundary of a level set of a Holder continuous function is positive

Let $M$ be a compact manifold in $R^3$ and consider a non negative
function $f\in C^{1,\alpha}(M)$ for some $0<\alpha<1$. I am interested
in showing that $|\partial \{f>0\} |>0$, where $|\cdot|$ stands for
the Hausdorff measure and $\partial A$ denotes
the boundary of $A$.

My Attempt:

By definition, $x^* \in M$ is a boundary point of $\{f>0\}$, if for every $r>0$ it holds that $B_r(x^*) \cap \{f>0\} \neq \emptyset$ and $B_r(x^*) \cap \{f=0\} \neq \emptyset$. In other words, if for every ball with center at $x^*$, I can find not only points of $\{f>0\}$ but also of its complement. Now, it is also true that $M=\{f>0\}\cup \{f=0\} $. Hence, if I could find/construct a dense subset of $\{f>0\}\cup \{f=0\}$, let's say $D$, I could obtain that $B_r(x^*) \cap D \neq \emptyset$ by density. This in particular would yield that $B_r(x^*) \cap (\{f>0\}\cup \{f=0\})=(B_r(x^*) \cap \{f>0\})\cup (B_r(x^*) \cap\{f=0\}) \neq \emptyset$. At this point, two issues arise:

The latter does not imply that both intersections must be non empty, but only at least one between them.

I am not sure if such a dense set exists.

DISCLAIMER: I have seen some similar question here in which the answer is given by a Cantor set. However, since I am not that familiar with topology, I would prefer to prove this (if possible) using the "slightest" tools.

I feel that density is a key part here but I am not able to tackle this problem properly. Any help or hint will be much appreciated!

Thanks in advance.

答:
I think I figured out the missing keys so I will post the whole argument here as an answer. In case, there are still mistakes, please correct me.

We know that $f \in C^{1,\alpha}(M)$ for some $0<\alpha<1$. Hence, $f \in C^{1,\beta}(M)$ for some $0<\beta<\alpha<1$. We can find a sequence of smooth functions $g_n$ converging to $f$ such that $g_n(x)>0$ for all $x\in B_{\delta_n}(x)$ and $g_n(x)=0$ otherwise (here $\delta_n$ is chosen arbitrarily small). Moreover, since $M$ is a compact manifold, there exists a countable dense subset $\{ x_n, n \in \mathbb N\}$ in $M$. Passing to a subsequence $\hat x_n$ we obtain by density that $\hat x_n \to x^* \in M$.

We can write $M=M \setminus \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) \cup \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big)$. We recall that $\big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) \neq \emptyset$ since $\{x_n\}$ is dense. At this point, we distinguish cases. If $x^*\in M \setminus \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) $ then $f(x^*)=0$. If $x^* \in \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big)$ then due to $g_n$ and by passing to a subsequence, we obtain that $f(x^*)>0$. This completes the argument.


问:
Subspace topology and initial topology
Let $f_i:X \to X_i$, $i \in I$ be a family of functions in topological spaces $X_i$, $A \subset X$ a subset and $\iota_A:A \to X$ the inclusion map. Suppose that $X$ has the initial topology with respect to the family $\{f_i\}_{i \in I}$.

I wanna show that the subspace topology of $A$ with the initial topology with respect to the family  $\{f_i \circ \iota_A\}_{i \in I}$ the same is but I got stuck.

Some help would be really nice!

答:
The other answers are good, but I personally prefer to prove things by using the universal property when possible, so here is another proof. There are two things to show:

The subspace topology on $A\subset X$ is the initial topology for the inclusion map $i_A:A\to X$.

In general, if you are given functions $f_i: Y\to Y_i$ and $f_{i,j}: Y_i\to Y_{i,j}$ where the $Y_{i,j}$ are topological spaces, then the initial topology on $Y$ defined by the family $(f_{i,j}\circ f_i: Y\to Y_{i,j})_{i,j}$ (topology $(i)$) is the same as the initial topology defined by $(f_i: Y\to Y_i)_i$ where each $Y_i$ has the initial topology defined by $(f_{i,j}: Y_i\to Y_{i,j})_j$ (topology $(ii)$); this is the transitivity property evoked by Henno Brandsma.

The first point can be taken as the definition of the subspace topology, but it is easy to see that it is equivalent to the more usual definition using open subsets (it is just saying that the subspace topology on $A$ is characterized by the fact that any map $B\to A$ is continuous if and only if it is continuous as a map $B\to X$).

For the second point, let $g:B\to Y$ be a function, where $B$ is a topological space. Then it is continuous for the topology $(i)$ iff each $f_{i,j}\circ f_i\circ g$ is continuous, by definition of the initial topology. And it is continuous for the topology $(ii)$ iff each $f_i\circ g$ is continuous, and by assumption on the topology of $Y_i$ this is equivalent to each $f_{i,j}\circ f_i\circ g$ being continuous. So the two conditions are the same, which means the two topologies coincide.

Title: Subspace topology and initial topology


问:
Two triangles of equal area

Claim. Given a convex quadrilateral $ABCD$ that isn't parallelogram. The Newton line intersects sides $AD$ and $BC$ of the quadrilateral at the points $H$ and $G$ , respectively . Then $Area(\triangle HGD)=Area(\triangle HBG)$ and $Area(\triangle HGC)=Area(\triangle HAG)$.

GeoGebra applet that demonstrates this claim can be found here.

Proof. Observe that diagonal $HG$ of the quadrilateral $HBGD$ lie on the Newton line of the quadrilateral $HBGD$. If we apply Anne's theorem on point $H$ and quadrilateral $HBGD$ we can write $Area(\triangle HGD)+0=Area(\triangle HBG)+0$ , hence $Area(\triangle HGD)=Area(\triangle HBG)$. Similarlly we can show that $Area(\triangle HGC)=Area(\triangle HAG)$.

Question. Is this proof acceptable? Can we use Anne's theorem in the case when the point lies on the side of the quadrilateral? Can you provide an alternative proof?

答:
Drop the perpendicular from $D$ onto $\overline{GH}$ at $E$.

Drop the perpendicular from $B$ onto $\overline{GH}$ at $F$.

Since $I$ bisects $\overline{BD}$, we have $\overline{ID}=\overline{IB}$. Furthermore, $\angle DIE=\angle BIF$ since they are vertically opposite. Thus, $\triangle DIE=\triangle BIF$. Therefore, the altitude $\overline{DE}$ is the same as the altitude $\overline{BF}$ and thus, $|\triangle HGD|=|\triangle HGB|$.

Drop the perpendicular from $C$ onto $\overline{GH}$ at $E$.

Drop the perpendicular from $A$ onto $\overline{GH}$ at $F$.

Since $J$ bisects $\overline{AC}$, we have $\overline{JA}=\overline{JC}$. Furthermore, $\angle AJH=\angle CJG$ since they are vertically opposite. Thus, $\triangle AFJ=\triangle CEJ$. Therefore, the altitude $\overline{AF}$ is the same as the altitude $\overline{CE}$ and thus, $|\triangle HAG|=|\triangle HCG|$.

Title: Two triangles of equal area


问:
Is $\begin{bmatrix} a &b \\0 &1 \end{bmatrix}$ a cyclic group?
I appreciate all the help I can get with this task.
$$
G=\left \{ \begin{bmatrix}
a &b \\0
 &1
\end{bmatrix},a,b\in \mathbb{Z}_3, a\neq 0 \right \}
$$

Is G a cyclic group?

Does a subgroup with 3 elements exists?

All elements in G:
$$\begin{bmatrix}
1 &0 \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &1  \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &2 \\0
 &1
\end{bmatrix},\begin{bmatrix}
2 &0 \\0
 &1
\end{bmatrix},\begin{bmatrix}
2 &1 \\0
 &1
\end{bmatrix},\begin{bmatrix}
2 &2 \\0
 &1
\end{bmatrix}$$

What do I do now?

EDIT: Thanks for your comments.

Is this one subgroup of order 3? $a=1$ generates 3 matrices.
$$\begin{bmatrix}
1 &0 \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &1  \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &2 \\0
 &1
\end{bmatrix}$$

From my book

A group is said to be cyclic if it contains an element x such that every member of G is a power of x.

Can $x$ be one of the six elements above?

EDIT: Thanks for all the help! I appreciate it very much.

答:
$\begin{pmatrix}
a &b \\0
 &1
\end{pmatrix}\begin{pmatrix}
c &d \\0
 &1
\end{pmatrix}=\begin{pmatrix}
ac &ad+b \\0
 &1
\end{pmatrix}$ $\Rightarrow$ $\begin{pmatrix}
a &b \\0
 &1
\end{pmatrix}^n=\begin{pmatrix}
a^n &b(a^n+a^{n-1}+...+1)\\0
 &1
\end{pmatrix}$

which means $a \neq 1$ if there exists an element such that $\langle x\rangle=G$

$b=0 \Rightarrow \begin{pmatrix}
a &b \\0
 &1
\end{pmatrix}^n=\begin{pmatrix}
a^n &0 \\0
 &1
\end{pmatrix}$ which implies $b \neq 0$ if $\begin{pmatrix}
a &b \\0
 &1
\end{pmatrix}$ is a generator.

$a=2 \Rightarrow$ $\begin{pmatrix}a^n &b(\frac{a^{n+1}-1}{a-1}) \\0
 &1
\end{pmatrix}=\begin{pmatrix}2^n &b(2^{n+1}-1) \\0
 &1
\end{pmatrix}$

$2^{n+1}=1$ or $2^{n+1}=2 \Rightarrow b(2^{n+1}-1)=0$ or $b(2^{n+1}-1)=b$

if you choose $b=1$ then it will be impossible to find a $n$ such that $\begin{pmatrix}
a &b \\0
 &1
\end{pmatrix}^n$=$\begin{bmatrix}
2 &2 \\0
 &1
\end{bmatrix}$

Also it will not be possible to find a $n$ such that $\begin{pmatrix}
a &b \\0
 &1
\end{pmatrix}^n$=$\begin{bmatrix}
2 &1 \\0
 &1
\end{bmatrix}$ if $b$ is chosen as $2$

so G is not cyclic


问:
Finding $\mathbb{P}(\max_{t\leq 1} (W_t+t)\geq 1)$
I was trying to find $\mathbb{P}(\max_{t\leq 1}  (W_t+t)\geq 1),$ where $W_t$ is a one-dimensional Brownian motion for $t\in [0,1].$ I first applied the Girsanov's theorem to get $$\mathbb{P}\Big(\max_{t\leq 1}  (W_t+t)\geq 1\Big)=\mathbb{E}\Big[\mathbb{1}_{\max_{t\leq 1}  (W_t)\geq 1} \exp\Big(W_1-\frac{1}{2}\Big)\Big],$$ where $\mathbb{1}_A$ is the indicator function of set $A$.

A given hint here is to use $\mathbb{P}(\max_{t\leq 1} W_t\geq 1, W_1\leq b)$ and use the fact that if $\mathbb{P}(\xi\leq a ,\eta\leq b)=\int_{-\infty}^b f(x) \ dx$ for every $b$, then $\mathbb{E}[g(\eta)\mathbb{1}_{\xi\leq a}]=\int_\mathbb{R}g(x)f(x) \ dx.$
Hence, I've found
$$
\mathbb{P}(\max_{t\leq 1} W_t\geq 1, W_1\leq b) = \left\{
        \begin{array}{ll}
            \mathbb{P}(W_1\geq 2-b) & \quad b \leq 1 \\
            2\mathbb{P}(W_1\geq 1)-\mathbb{P}(W_1\geq b) & \quad b \geq 1
        \end{array}
    \right.
$$
and we have $g(x)=\exp(x-\frac{1}{2})$ (using the same notation as the hint), since we want to compute $\mathbb{E}\Big[\mathbb{1}_{\max_{t\leq 1}  (W_t)\geq 1} \exp\Big(W_1-\frac{1}{2}\Big)\Big].$

Thus, by using the above result, we get for $b\leq 1$ $$\mathbb{P}(\max_{t\leq 1} W_t\geq 1, W_1\leq b)=\int_{-\infty}^b \frac{1}{\sqrt{2\pi}} \ e^{-(2-x)^2/2} \ dx.$$
However, I could not find an integral of this form for the case $b\geq 1$. Can someone offer me guidance on how to proceed from here?

The answer, by the way, is $$ \mathbb{P}(\max_{t\leq 1}  (W_t+t)\geq 1)=\int_1^\infty e^{x-\frac{1}{2}}\frac{1}{\sqrt{2\pi}} \ e^{-x^2/2} \ dx + \int_{-\infty}^1 e^{x-\frac{1}{2}} \frac{1}{\sqrt{2\pi}} \ e^{-(2-x)^2/2} \ dx.$$

答:
First, as you noticed, $\tilde{W}_t:=W_t+t$, $t\in[0,1]$, is a Brownian motion under $\mathsf{Q}$ s.t. $d\mathsf{P}=\exp(\tilde{W}_1-1/2)d\mathsf{Q}$.
Then, letting $A:=\{\max_{0\le t\le 1}\tilde{W}_t\ge 1\}$,
$$
\mathsf{P}(A)=\mathsf{E}_{\mathsf{Q}}[1_{A}\exp(\tilde{W}_1-1/2)]=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{x-\frac{1}{2}}e^{-\frac{(|x-1|+1)^2}{2}}\,dx
$$
because for a $\mathsf{Q}$-Brownian motion $\{B_t\}$ and $y\ge 0$,
$$
\mathsf{Q}\!\left(\max_{0\le s\le t}B_s\ge y, B_t\in dx\right)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(|x-y|+y)^2}{2t}}\, dx.
$$
(See, for example, Eq. 1.1.8 on page 154 here)

The last formula can be derived as follows. For $\epsilon>0$ and $y\le x-\epsilon$,
$$
\mathsf{Q}\!\left(\max_{0\le s\le t}B_s\ge y, x-\epsilon< B_t< x+\epsilon\right)=\mathsf{Q}\!\left(x-\epsilon< B_t< x+\epsilon\right),
$$
and for $x+\epsilon<y$, using the strong Markov property,
$$
\mathsf{Q}\!\left(\max_{0\le s\le t}B_s\ge y, x-\epsilon< B_t< x+\epsilon\right)=\mathsf{Q}\!\left(2y-x-\epsilon< B_t< 2y-x+\epsilon\right).
$$
Sending $\epsilon\to 0$ yields the result.


答:
$$\begin{aligned} \cos( \tan ( \alpha)) = \sin( \alpha) &\\ t \rightarrow \frac{\pi}{2}- \alpha &\\ \cos( \tan ( \frac{\pi}{2} -t)) = \cos(t) &\\ \tan( \frac{\pi}{2} -t) = t &\\ \cot(t) = t &\\\end{aligned}$$
Which does not have an analytical solution

Title: $\cos (\tan \alpha)=\sin \alpha $

Body:
I can't find where is my mistake so if someone can enlighten me...

$\tan \alpha=\arccos (\sin \alpha) $

$\tan \alpha=\arccos (\cos (\alpha -\frac{\pi }{2}))= \alpha -\frac{\pi }{2}$

$\frac{d}{d\alpha }\tan \alpha=\frac{d}{d\alpha }(\alpha -\frac{\pi }{2})  $

$\frac{1}{\cos (\alpha)^{2}}=1  $

$\alpha =\pi n$

there is something wrong and I don't really know.


问:
Solve the following Dif. Equation
The Dif. Equation is $$(\tan y-2)dx+\left(x\sec^2y+\frac{1}{y}\right)dy=0$$ with the initial condition $y(0)=1$

As it is a non linear equation I think its better to try to do change of variable. Can someone please suggest anyrhing

答:
Hint:
$$(\tan y-2)dx+(x\sec^2 y+\frac{1}{y})dy=0$$
Change the variables this way:
$$(\tan y-2)dx+x d(\tan y)+\dfrac{dy}{y}=0$$
$$d(x\tan y)-2dx+\dfrac{dy}{y}=0$$
Integrate both sides.

Title: Solve the following Dif. Equation


答: $$\frac{dx}{dy}+\frac{x\sec^2y}{\tan y -2}=-\frac1{y(\tan y-2)}$$is linear ODE in $x$.

Title: Solve the following Dif. Equation

Body:
The Dif. Equation is $$(\tan y-2)dx+\left(x\sec^2y+\frac{1}{y}\right)dy=0$$ with the initial condition $y(0)=1$

As it is a non linear equation I think its better to try to do change of variable. Can someone please suggest anyrhing


问:
Proving the limit of a fraction at infinity by replacing the fraction by a larger fraction.
To prove $$\lim_{x \to -\infty} \frac{x^3}{5x^3 - 2x^2 - 4} = \frac{1}{5},$$ one can identify an $N$ such that $x < N$ implies that $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| < \epsilon$$ by working backwards. That is $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| = \left|\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}\right|.$$ It would be nice to simplify this rather messy expression; something you can do as long as you do not introduce changes that prevent the final inequality from holding. In this case, the $4$ term in the numerator of $\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}$ is an inconvenience, and it would be nice to remove it. Simply removing this positive term would make the absolute value of the fraction smaller when what is needed is to make the fraction larger. The same thing happens when the expression $-2x^2 - 4$ is removed from the denominator of our fraction. How should I replace this fraction with a simpler fraction that is clearly larger?

答:
I just continue where you stopped and I assume $|x|\geq 2$:

\begin{eqnarray*}\left|\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}\right|
& \stackrel{||a|-|b||\leq |a-b|}{\leq} & \frac{2x^2 + 4}{|5|x|^3 -|2x^2 + 4| |}\\
& \stackrel{2x^2+4\leq 3x^2 \leq 3|x|^3}{\leq} & \frac{3x^2}{5|x|^3 - 3|x|^3} \\
& = & \frac 3{2|x|}
\end{eqnarray*}


问:
Property of continuous map from compact space onto Hausdorff space
I am self studying Topology, and I am having difficulty in proving the following statement.

Let $f$ be a continuous mapping defined on a compact space $X$ onto a Hausdorff space $Y$. Suppose $y\in Y$ and $U$ is an open subset of $X$ that contains $f^{-1}[y]$. Show that there exists an open neighborhood $V$ of $y$ such that $f^{-1}[V] \subset U$.

I know that

$Y=f[X]$ is compact, and thus, being Hausdorff, it is also normal;

$f$ is a closed map;

$f^{-1}[y]$ is closed and compact.

Can you give me some hints on how to use these facts (or other consequences of the hypothesis) to prove the theorem?

答:
We know $f^{-1}[\{y\}] \subseteq U$. We know that $f$ is a closed map (continuous  from a compact space to a Hausdorff space) so $f[X\setminus U]$ is closed in $Y$ and so $V:= Y\setminus f[X\setminus U]$ is open in $Y$.

I claim that $y \in V$ and $f^{-1}[V] \subseteq U$.

Proof: if $y \notin V$ then $y \in f[X\setminus U]$ so $y=f(x)$ for some $x \notin U$, but this implies $x \in f^{-1}[\{y\}] \nsubseteq U$ a contradiction. So $y \in V$.

If $x \in f^{-1}[V]$ then $f(x) \in V$ by definition and if $x \notin U$, we'd have $x \in X\setminus U$ so $f(x) \in f[X\setminus U]$ or $f(x) \notin V$, a contradiction, and hence $x \in U$ and the inclusion holds.


问:
find b when $\frac{36^{3b}}{36^{2b}}=216^{2-b}$
I am to find b when $\frac{36^{3b}}{36^{2b}}=216^{2-b}$. I got $b=\frac{12}{7}$ whereas my textbook says the solution is $\frac{6}{5}$

My working:

$$\frac{36^{3b}}{36^{2b}}=216^{2-b}$$
$$36^b=36^{6(2-b)}$$
$$b=6(2-b)$$
$$b=12-6b$$
$$7b=12$$
$$b=\frac{12}{7}$$

Where did I go wrong and how can I arrive at $\frac{6}{5}$?

The online textbook is here. It's exercise 9 at the end of that page. The solution is here under section exercises for chapter 6.6, exercise 9.

Body:
Note that $216 = 6^3$,

Hence we have $$6^{2b}=6^{3(2-b)}$$

$$2b=3(2-b)$$

$$5b=6$$


问:
The proof of $\log(1+x) < x.$
I want to prove $\log(1+x) < x.$

Of course, I can prove this by letting $f(x):= x - \log(1+x),$ and calculating $f'(x)$, ... .

But I wonder if I can prove this using Maclaurin's expansion.

$\log(1+x)=x-\dfrac{1}{2}x^2+\dfrac{1}{3} x^3-\cdots.$

The right-hand side includes the term "$x$" so it seems that I can use this equality in order to prove $\log(1+x) < x$, but I cannot.

I would like you to give me some ideas.

答:
The definition of natural log in How can we come up with the definition of natural logarithm? allows a proof without words.

interactive graph made from Desmos

\begin{align}
\mathrm{LHS} &:= \int_1^{1+x} \frac{\mathrm{d}t}{t} = \text{area under the curve} \\
\mathrm{RHS} &= x = \text{area of the rectangle}
\end{align}

The curve $y = 1/x$ is below $y = 1$ on $x \ge 1$.

Exercise: complete this proof for the case $x \in (-1,0)$.

Title: The proof of $\log(1+x) < x.$


问:
Meaning of $\lim_{p\to 0^+}$
I would like to know the interpretation of limit expression which calculates the number of non-zero entries in vector $x$ in the expression below:
$$\lVert x\rVert_0 = \#\{i \mid x[i] \neq 0\} = \lim_{p\to 0^+} \left(\sum_{i=1}^{m} |x[i]|^p\right),$$
where $x[i]$ is the $i$th coordinate of $x$ in $\mathbb{R}^m$ and $p \geq 1$</p>

答:
As Qiaochu Says $\lim_{x \to 0+}$ means that $x$ approaches to $0$ from the Positive side. It shall be clear once you see an example.

Example so that you can understand better. Consider $f: \mathbb{R} \to \mathbb{R}$ Let $f(x) = \lfloor{x\rfloor}$. We prove that the point at which $f$ is not continuous at $2$. Let us consider the right hand limit $\lim_{h \to 0+} f(2)$ and the left hand limit $\lim_{h \to 0-} f(2)$.

Right hand Limit: $\displaystyle\lim_{h \to 0+} f(2) = \displaystyle\lim_{h \to 0} f(2+h) = \lim_{h \to 0}\lfloor{2+h\rfloor} =2$.

Left Hand Limit: $\displaystyle\lim_{h \to 0-} f(2)= \displaystyle\lim_{h \to 0} f(2-h) =\lim_{h \to 0} \lfloor{2-h\rfloor} = 1$.

Hence $f$ is not continuous at $x =2$.

Title: Meaning of $\lim_{p\to 0^+}$


问:
Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.
Interesting number theory question, which I feel should be reasonably straight forward, but I can't seem to crack it.

Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.

Now, I know that a quadratic residue $a$ (mod m) means that there exists $x$ such that $x^2\equiv a$(modm).
So if 7 is a quadratic residue mod 28k+1 or 28k+3 then $x^2\equiv 7$ (mod 28k+1 or 28k +3).  But I am unsure of how to tackle this problem.
Perhaps, I could go on to say that for the first case, $x^2\equiv 6$ (mod 28k) $\equiv$ 3 (mod 14k)??

Or perhaps I am going about this wrong.

I also know the laws of quadratic reciprocity, so I need to show $(\frac{7}{28k+1})$ =1 and likewise for 28k +3

Help would be greatly appreciated.  Thank you

Body: Hint: If $p=28k+1$ then $\frac{7-1}{2}\cdot \frac{p-1}{2}$ is even. If $p=28k+3$ then $\frac{7-1}{2}\cdot\frac{p-1}{2}$ is odd.


问:
What are examples of discrete functions that are monotonically increasing/decreasing?
What are examples of discrete functions that are monotonically increasing/decreasing?
Can monotonically increasing/decreasing  be defined over discrete functions?</p>

答: Consider the function $f:\mathbb{Z} \mapsto \mathbb{Z}$ defined by $f(x)=x$. This function is monotonically increasing, since for every $x,y$ in the domain of $f$, $f(x) \geq f(y)$ whenever $x \geq y$.


问:
partial derivative of transpose matrix-matrix multiplication
I came across some problems that are related to partial derivative but I haven't learnt this yet. And I looked up many online resources but couldn't find answers to my doubts. Really hope someone can help me.

Here is my problem. $y=A^TB$, where A, B are two matrices. Now I want to know what $\frac{\partial y}{\partial A}$,$\frac{\partial y}{\partial B}$ are.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$Let $\,(\alpha,\beta)\,$ be fourth-order tensors
with components
$$\eqalign{
\alpha_{ijk\ell} &= \delta_{ik}\,\delta_{j\ell} \\
\beta_{ijk\ell} &= \delta_{i\ell}\,\delta_{jk} \\
}$$
and properties with respect to the matrices $(F,G,H)$
$$\eqalign{
\alpha:H &= H:\alpha = H \\
\beta:F &= F:\beta = F^T \\
HFG &= H\alpha G^T:F \\
}$$
where a colon denotes a double-contraction product, i.e.
$$\eqalign{
\left(\alpha:H\right)_{ij} &= \sum_k\sum_\ell\alpha_{ijk\ell}\,H_{k\ell} \\
\left(F:\beta\right)_{k\ell} &= \sum_i\sum_jF_{ij}\,\beta_{ijk\ell} \\
}$$
and juxtaposition implies a single-contraction product
$$\eqalign{
\left(H\alpha\right)_{mjk\ell} &= \sum_i H_{mi}\,\alpha_{ijk\ell} \\
\left(\alpha G^T\right)_{ijkm} &= \sum_\ell\alpha_{ijk\ell}\,G^T_{\ell m} \\
}$$

With these tensors, the posted question can be answered as follows
$$\eqalign{
Y &= A^TB \\&= A^T\alpha:B
\quad&\implies\quad\p{Y}{B} &= A^T\alpha \\
Y  &= \alpha B^T:A^T \\
 &= \alpha B^T:\beta:A
\quad&\implies\quad\p{Y}{A} &= \alpha B^T:\beta \\
}$$
So the gradients in question are seen to be fourth-order tensors.

An approach which avoids higher-order tensors, is to transform the relationship into a vector equation using Kronecker products.
$$\eqalign{
{\rm vec}(Y) &= (I\otimes A^T)\;{\rm vec}(B)
\quad&\implies\quad \p{{\,\rm vec}\,Y}{{\,\rm vec}\,B} = (I\otimes A^T) \\
 &= (B^T\otimes I)K\;{\rm vec}(A)
\quad&\implies\quad \p{{\,\rm vec}\,Y}{{\,\rm vec}\,A} = (B^T\otimes I)K \\
}$$
where $K$ is the commutation matrix associated with vectorization.

Another approach is to use component-wise derivatives
$$\eqalign{
\p{Y}{A_{ij}} &= E_{ij}^TB
\qquad\quad
\p{Y}{B_{ij}} &= A^TE_{ij}
\\
}$$
where $E_{ij}$ is a matrix with all components equal to zero,
except the $(i,j)$ component which equals one. And any matrix with
independent components satisfies the identity
$$\eqalign{
\p{G}{G_{k\ell}} &= E_{k\ell} \qquad\iff\qquad
\p{G^T}{G_{k\ell}} &= E_{k\ell}^T  \\
}$$
Finally, to bring things full circle
$$\eqalign{
\p{G_{ij}}{G_{k\ell}} &= \alpha_{ijk\ell} \qquad\iff\qquad
\p{G_{ij}^T}{G_{k\ell}} &= \beta_{ijk\ell}  \\
}$$


问:
The number of solutions of $ax^2+by^2\equiv 1\pmod{p}$ is $ p-\left(\frac{-ab}{p}\right)$
What I need to show is that

For $\gcd(ab,p)=1$ and p is a prime, the number of solutions of the equation $ax^2+by^2\equiv 1\pmod{p}$ is exactly $$p-\left(\frac{-ab}{p}\right)\,.$$

I got a hint that I have to use Legendre symbol from the answer.

I think that I may count a solution one by one.

What I did :

$$(ax)^2 \equiv a-aby^2 \pmod{p}$$
It suffices to count $y$ such that $(\frac{a-aby^2}{p})=1$.

I tried to use the complete residue system or a primitive root but it didn't work.

The factorization also didn't work.

I think that the pigeonhole principle may not work because it just says the existence.

Thanks in advance.

答:
Let $k = -b/a$ and $c = 1/a$ (both are nonzero), so that you can rewrite the equation as $x^2-ky^2 = c$.

Call $N(k,c)$ the number of solutions $(x,y)$ to $x^2-ky^2 = c$. It is easy to check that this number only depends on $c$ and wether $k$ is a square, a nonsquare, or $0$, and that $N(0,c) = p(1 + \binom c p)$

Call $M(y,c)$ the number of solutions $(x,k)$ to $x^2-ky^2 = c$. If $y=0$ then again, $M(0,c) = N(0,c)$. And if $y \neq 0$, we clearly have $k = (x^2-c)/y^2$, hence $M(y,c) = p$ (one value of $k$ for each value of $x$)

The total number $T(c)$ of triplets $(x,y,k)$ satisfying the equation is $T(c) = \sum N(k,c) = \sum M(y,c) = M(0,c) + (p-1)p$, hence $\sum_{k \neq 0} N(k,c) = p^2-p$.

Now, if $k$ is a square $u^2$ you can factor the left-hand side to get $(x-uy)(x+uy) = c$. Since $p>2$, the change of variable $(x,y) \to (x-uy,x+uy) = (s,t)$ is invertible, and you get $st = c$, of which there are $p-1$ solutions for $c \neq 0$, (and $2p-1$ for $c=0$).

This leaves you, when $c \neq 0$, with $(p^2-p)-(p-1)^2/2 = (p-1)(p+1)/2$ triplets with $k$ nonsquare, hence $p+1$ solutions for each nonsquare $k$.

And when $c = 0$, with $(p^2-p)-(p-1)(2p-1)/2 = (p-1)/2$ triplets, hence $1$ solution (the trivial $x=y=0$ one) for each nonsquare $k$.


问:
Solve $A\mathbf{x}=\mathbf{v}(\mathbf{x}^TD\mathbf{v})$ for $A$
Is there a way to simplify
$$f(\mathbf{x})=\mathbf{x}-\mathbf{v}(\mathbf{x}^TD\mathbf{v})$$

to the form

$$f(\mathbf{x})=(I-A)\mathbf{x}$$

where
$\mathbf{x}$ and $\mathbf{v}$ are column-vectors and $D$ is a matrix?

In particular, I seek to extract the variable $\mathbf{x}$ as a common factor so that it appears only once in the expression for $f(\mathbf{x})$.

I could write $\mathbf{x}-\mathbf{v}(\mathbf{v}^TD^T\mathbf{x})$ but then I still don't know how to get that into the form $(...\text{constant matrix expression }...)\mathbf{x}$.

答:
Checking with 2 by 2 matrix,
$$f(x)=(I-vv^TD^T)x$$


问:
$\cos (\tan \alpha)=\sin \alpha $
I can't find where is my mistake so if someone can enlighten me...

$\tan \alpha=\arccos (\sin \alpha) $

$\tan \alpha=\arccos (\cos (\alpha -\frac{\pi }{2}))= \alpha -\frac{\pi }{2}$

$\frac{d}{d\alpha }\tan \alpha=\frac{d}{d\alpha }(\alpha -\frac{\pi }{2})  $

$\frac{1}{\cos (\alpha)^{2}}=1  $

$\alpha =\pi n$

there is something wrong and I don't really know.

答:
The function
$$f(\alpha)=\cos (\tan (\alpha))-\sin (\alpha)$$ will not show explicit roots (remember that is already the case of $x=\cos(x)$). SO, you will need a numerical method, so a reasoble guess for the solution you are looking for since there is an infinity of roots.

If you look for the ones close to zero, compose Taylor series to get
$$f(\alpha)=1-\alpha -\frac{\alpha ^2}{2}+O\left(\alpha ^3\right)$$
Solving the quadratic will give two roots
$$a_-=-1-\sqrt{3} \sim -2.732 \qquad \text{and} \qquad a_+=-1+\sqrt{3}\sim 0.732$$ while the "exact" solutions (given by Newton method) are
$$a_-\sim -1.287 \qquad \text{and} \qquad a_+\sim 0.710$$

Edit

Staring from @logichtech's answer, we need to find the zero's of function
$$f(t)= \cot(t) -t$$ which is not very well conditioned because of the dicontinuities induced by the cotangent.

It is then much better to consider function
$$g(t)=\cos(t)-t \sin(t)$$ the roots of which being closer and closer to $n\pi$ because of the sine.

So, let expand as a Taylor series function $g(t)$ around $t=n\pi$. This will give
$$g(t)=\sum_{k=0}^\infty a_k (t-n\pi)^k$$ where the coefficients make the sequance
$$\left\{1,-\pi ,-\frac{3}{2},\frac{\pi }{2},\frac{5}{24},-\frac{\pi
   }{24},-\frac{7}{720},\frac{\pi }{720},\frac{1}{4480},-\frac{\pi
   }{40320},-\frac{11}{3628800}\right\}$$ Now, using series reversion
$$t=k+\frac{1}{k}-\frac{4}{3 k^3}+\frac{53}{15 k^5}-\frac{1226}{105
   k^7}+\frac{13597}{315 k^9}-\frac{1531127}{8960 k^{11}}+O\left(\frac{1}{k^{12}}\right)$$ where $k=n \pi$.

Using the above, compare the approximation and the solution given by Newton method for the $n^{th}$ root
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{solution} \\
 2 & 3.42544789218750 &  3.42561845948170 \\
 3 & 6.43729815226868 &  6.43729817917194 \\
 4 & 9.52933440521558 &  9.52933440536196 \\
 5 & 12.6452872238531 &  12.6452872238566 \\
 6 & 15.7712848748157 &  15.7712848748159 \\
 7 & 18.9024099568600 &  18.9024099568600
\end{array}
\right)$$

For the very first root, the same procedure would lead to
$$t=\frac{200683117717}{95234227200 \sqrt{6}}\sim  0.860285$$ while the "exact" value is $0.860334$.

Title: $\cos (\tan \alpha)=\sin \alpha $


问:
Dirac delta function?
I am reading the paper http://web.mit.edu/2.14/www/Handouts/Convolution.pdf with the goal of learning about convolutions. In the paper, the unit function is defined as
$$
\delta_T (t) = \begin{cases}
      0 & t \leq 0 \\
      1/T & 0 < t \leq T \\
      0 & t > 0
   \end{cases}
$$
and then defines the Dirac Delta function as $\delta(t) = \lim_{T \rightarrow 0} \delta_T(t)$.

What then confuses me is that he says that

An impulse occurring at $t = a$ is $\delta(t-a)$

I cannot see why this is true with the definition of the unit function above. For example if $t = a \leq T$, wouldn't $\delta(t) = 1/T$ and $\delta(t-a) = 0$? Thus, $\delta(t) \neq \delta(t-a)$ as he claims?

Wiki is using the same definition of unit function(as far as I can see) https://en.wikipedia.org/wiki/Dirac_delta_function

答:
$$
\delta_T (t-a) = \begin{cases}
      0 & t-a \leq 0 \\
      1/T & 0 < t-a \leq T \\
      0 & t-a > T.
   \end{cases}
$$

If you prefer,

$$
\delta_T (t-a) = \begin{cases}
      0 & t \leq a \\
      1/T & a < t \leq a+T \\
      0 & t > a+T.
   \end{cases}
$$

Title: Dirac delta function?


问:
Probability of $-\frac{1}{4}\leq \sin (a x)\leq \frac{1}{2}$?
We know that probability of having $ \sin (a x)>0$ for a random $x$ is $\frac{1}{2}$.

Can we say something about the probability of the following condition?
$$-\frac{1}{4}\leq \sin (a x)\leq \frac{1}{2}$$

Here, $a$ and $x$ are continuous and $x>0$ and $a>0$.

答:
The function
$$x\mapsto\sin(a x)\qquad(-\infty<x<\infty)$$
is periodic with period $T:={2\pi\over a}$. When we assume that $x$ is uniformly distributed modulo $T$ then we may assume as well that $a=1$, and that $x$ is uniformly distributed modulo $2\pi$.

Draw the sin curve in the $x$-interval $[0,2\pi]$ representing a full period, and measure the total length $L$ of the subintervals where $-{1\over4}<\sin x<{1\over2}$. The probability $p$ you are after then is ${L\over2\pi}$. In this way you obtain
$$p={L\over2\pi}={2\arcsin{1\over2}+2\arcsin{1\over4}\over2\pi}=0.247097\ .$$


问:
The Denseness of Q
I am answering the question:

Consider $a, b ∈ R$ where $a < b$. Use Denseness of $\mathbb Q$ to show there are infinitely many rationals between $a$ and $b$.

I have chosen to answer this thing using induction.
I know $P_1$ is a true assertion since by the denseness of $\mathbb Q$ there exists a rational, $r_1$ such that $a<r_1<b$. I can then assume $P_n$ is true and that there are $n$ distinct rationals between $a$ and $b$ of the form

$a<r_n<r_{n-1}<r_{n-2}<\cdots<r_2<r_1<b$

This is where I'm stuck but I know I want to use the denseness of $\mathbb Q$ again to say since $a<r_n$, I can find a rational $r_{n+1}$. At the same time, I don't know what it is about $\mathbb Q$ that allows me to say it.

答: You're basically there: there's more than any finite number of rationals between $a$ and $b$ because 1. given any finite list of them you can generate more and 2. $a$ and $b$ themselves compose such a finite list; therefore, the number of rationals between $a$ and $b$ is infinite.

Title: The Denseness of Q


问:
Leaves of a tree made up of internal nodes with either 3 or 2 children
I'm currently struggling with solving the following problem: given a tree T made up of n internal nodes, where n-k nodes have 2 children and k nodes have 3 children, demonstrate that T has n+k+1 leaves.

Can anyone help me or point me to the right direction? Thank you in advance!

答:
Ok, after a while I managed to solve this.

Being m the number of leaves, and n the number of internal nodes, we know that the size of the tree equals the total number of nodes, which is then m+n.

Now we know that the sum of the degrees of all nodes tot_deg = size-1, so tot_deg = m+n-1. We can write tot_deg as 2(n-k) + 3k (number of children per each node multiplied by number of nodes).

We then solve the equation 2(n-k) + 3k = m+n-1 obtaining m, demonstrating that it's equal to n+k+1.


答:
Yes they can.

Go look at your definition of what it means to be monotonic - all that's required is that some sense of "$\lt$" be defined on both the domain and range of your function.

Also, note that if you already have a monotonic function on a large, non-discrete set, and restrict its domain to a smaller, discrete subset, the restricted function is still monotonic. The gives you a huge set of examples by taking monotonic functions from  $\mathbb R \rightarrow \mathbb R$ and restricting their domain to $\mathbb Z$.

Body:
What are examples of discrete functions that are monotonically increasing/decreasing?
Can monotonically increasing/decreasing  be defined over discrete functions?</p>


问:
Prove: $ T ◦ S $ is surjective iff $ $ T is surjective and $ ImS + KerT = V $
Let $  S: U \rightarrow V $ and $ T: V \rightarrow W $ be linear transformations over field $ F $.

Prove that $ T ◦ S $ is surjective iff $  $ T is surjective and $ ImS + KerT = V $

My attempt:

$ \rightarrow $ : Suppose that $ T ◦ S $ is surjective. Let $ w \in W $ be arbitrary. Then there exists $ u \in U $ such that $ w=T(S(u)) $. Since $ S(u) \in V $, denote $ v = S(u) $ and so $ w = T(v) $ and so we immediately see that $ T $ is surjective.
Next, by dimension theorem, we know that,

$ dim U=dimKer(T ◦ S) + dimIm(T ◦ S) $ , $ dim U = dimKer(S) + dimIm(S)  $ ,  $ dim V = dimKer(T) + dimIm(T)  $,

Since $ T ◦ S $ is surjective and we showed  $ T $ is surjective, then: $ dimIm(T ◦ S) =  dimW $  , $ dimImT = dimW $

[ Basically after this I'm pretty much stuck. I've reached the equation $ dimV = dimKer(T) + dimImS +  dimKerS  -  dimKer(T ◦ S)  $ , so In order to proceed I wanted to show that $ dimKerS  =  dimKer(T ◦ S) $ I tried to do this since It is immediate that $ ker(S) \subseteq Ker(T ◦ S) $ but it is left to show that $ Ker(T ◦ S)  \subseteq ker(S) $ and this part stumbled me   ]

$ \leftarrow $ : Suppose T is surjective and $ ImS + KerT = V $. So  $ dimImT = dimW $   [ I got a few more equations on dimensionality but they didn't help, so I got stuck here. ]

答:
I think it might be the easiest to verify the equality by direct computation via elements instead of dimension (the latter approach is actually a bit problematic if the vector spaces are no longer finite-dimensional)

For $\Rightarrow$: Let $v\in V$ be arbitrary. We wish to express it as a sum of elements in $\ker T$ and $\operatorname{im} S$. For this notice that since $TS$ is surjective there is some $u\in U$ with $T(v)=T(S(u))$, so we have $v-S(u)\in\ker T$. Now $$v=(v-S(u))+S(u)$$ is the desired decomposition.

The reverse implication is similar. Take $w\in W$. By assumption there is some $v\in V$ with $T(v)=w$. Furthermore we have $v=x+y$ with $x\in\ker T,y\in \operatorname{im} S$, i.e. $x=S(u)$ for some $u\in U$. Check that $T(S(u))=w$.

(I left out some details in the answer, which you might need to fill in)


答:
For the "$\Leftarrow$" part: Let $T$ be surjective and let's assume $V = \operatorname{Ker}(T) + \operatorname{Im}(S)$. Let $w \in W$. We need to find $u \in U$ such that $T(S(u))=w$.

Since T is surjective, $\exists v \in V$ such that $w=T(v)$. Since $V = \operatorname{Ker}(T) + \operatorname{Im}(S)$, there are some $v_1 \in \operatorname{Ker}(T), v_2 \in \operatorname{Im}(S)$ such that $v=v_1+v_2$. Can you continue from here?

Title: Prove: $ T ◦ S $ is surjective iff $ $ T is surjective and $ ImS + KerT = V $

Body:
Let $  S: U \rightarrow V $ and $ T: V \rightarrow W $ be linear transformations over field $ F $.

Prove that $ T ◦ S $ is surjective iff $  $ T is surjective and $ ImS + KerT = V $

My attempt:

$ \rightarrow $ : Suppose that $ T ◦ S $ is surjective. Let $ w \in W $ be arbitrary. Then there exists $ u \in U $ such that $ w=T(S(u)) $. Since $ S(u) \in V $, denote $ v = S(u) $ and so $ w = T(v) $ and so we immediately see that $ T $ is surjective.
Next, by dimension theorem, we know that,

$ dim U=dimKer(T ◦ S) + dimIm(T ◦ S) $ , $ dim U = dimKer(S) + dimIm(S)  $ ,  $ dim V = dimKer(T) + dimIm(T)  $,

Since $ T ◦ S $ is surjective and we showed  $ T $ is surjective, then: $ dimIm(T ◦ S) =  dimW $  , $ dimImT = dimW $

[ Basically after this I'm pretty much stuck. I've reached the equation $ dimV = dimKer(T) + dimImS +  dimKerS  -  dimKer(T ◦ S)  $ , so In order to proceed I wanted to show that $ dimKerS  =  dimKer(T ◦ S) $ I tried to do this since It is immediate that $ ker(S) \subseteq Ker(T ◦ S) $ but it is left to show that $ Ker(T ◦ S)  \subseteq ker(S) $ and this part stumbled me   ]

$ \leftarrow $ : Suppose T is surjective and $ ImS + KerT = V $. So  $ dimImT = dimW $   [ I got a few more equations on dimensionality but they didn't help, so I got stuck here. ]


问:
$\int_0^\infty\int_x^\infty\left(\frac{1}{y}\right)e^{\frac{-y}{2}}dydx$
The value of the integral
$$\int_0^\infty\int_x^\infty\left(\frac{1}{y}\right)e^{\frac{-y}{2}}dydx$$
is ______

I tried to solve it using substitution but did not succeed as I didn't find any favourable substitution to make above Integral easier one,I am totally stuck how to approach this type of problem in which limits of integration are involving infinity in case of multiple integrals.

答:
This integral is a classic example where the order of integration is important.

Your integral can be easily evaluated by rewriting it as

$$\int_0^{\infty}\frac{e^{-y/2}}{y}\Bigg[\int_0^y dx  \Bigg]dy$$


答:
There is a typo in the lecture notes:

$$
\delta_T (t) = \begin{cases}
      0 & t \leq 0 \\
      1/T & 0 < t \leq T \\
      0 & \color{red}{t > T}
   \end{cases}
$$

(You can find this typo by looking at the picture in Figure 1(a) of your linked notes.)

Now, if you shift this function, you get
$$
\delta_T (t-a) = \begin{cases}
      0 & t \leq a \\
      1/T & a < t \leq T \\
      0 & \color{red}{t > T+a}
   \end{cases}
$$

Now by definition,

$$
\delta(t-a)=\lim_{T\to 0}\delta_T(t-a)
$$

Note: you read the definition of $\delta$ incorrectly:
$$
\delta(t)=\lim_{\color{red}{T}\to 0}\delta_T(t)
$$

Title: Dirac delta function?

Body:
I am reading the paper http://web.mit.edu/2.14/www/Handouts/Convolution.pdf with the goal of learning about convolutions. In the paper, the unit function is defined as
$$
\delta_T (t) = \begin{cases}
      0 & t \leq 0 \\
      1/T & 0 < t \leq T \\
      0 & t > 0
   \end{cases}
$$
and then defines the Dirac Delta function as $\delta(t) = \lim_{T \rightarrow 0} \delta_T(t)$.

What then confuses me is that he says that

An impulse occurring at $t = a$ is $\delta(t-a)$

I cannot see why this is true with the definition of the unit function above. For example if $t = a \leq T$, wouldn't $\delta(t) = 1/T$ and $\delta(t-a) = 0$? Thus, $\delta(t) \neq \delta(t-a)$ as he claims?

Wiki is using the same definition of unit function(as far as I can see) https://en.wikipedia.org/wiki/Dirac_delta_function


问:
Showing that this equation for z holds
Assuming that the function $z = z(x, y)$ has continuous partial derivatives and $\phi = \phi(s)$
is continuously differentiable, show that the following holds
$$x^2z_x − xyz_y + y^2 = 0, \text{where } z = y^2/3x + \phi(xy)$$
I'm not too sure where to start with the function ϕ within this question.

答:
This is just straightforward computation. Since $\phi$ is a function of a single variable,
$$\frac{\partial }{\partial x} \phi(xy) = \phi_x(xy) \cdot y = y \cdot \phi'(xy)$$
So just use this to compute $z_x$ and $z_y$ and plug it in. It works out. Whatever the mysterious $\phi$ function is, the terms containing it cancel out.

Title: Showing that this equation for z holds


问:
An intuitive explanation for the ecological fallacy
I believe that it is called the ecological fallacy. People say that one cannot apply population-wide statistics to individuals of that population. So, just because some trait exists in a higher proportion in population A than it does in B, that doesn't mean that an individual from pop. A is more likely to have that trait than an individual from pop. B.

Now, this doesn't make sense to me. An individual coming from a population where a certain trait is more common, is more likely to have that trait than an individual coming from a population where said trait is less common, right?

I guess this depends on how one mathematically defines "common". The mean and the average are a bit scary statistical quantifications. I see how "the average" is problematic, but a weighted average is surely quite illuminatory of the real likelihoods?

答:
An individual coming from a population where a certain trait is more common, is more likely to have that trait than an individual coming from a population where said trait is less common, right?

Perhaps.

Let's assume that 14% of all men work in manufacturing, and 7% all women work in manufacturing. Does this mean that if I select a random man from the population (using a uniformly random selection method), I have a 14% chance of selecting someone who works in manufacturing? Yes, it does.

Now, it just so happens that last month, I was working in a General Motors factory. While I was there, I met an employee named Chris. Chris is a man. Does this mean that there's a 14% chance that Chris works in manufacturing?

The answer is no, of course, because I didn't meet Chris by selecting a random man using a uniformly random selection method, and because I have additional information about Chris. As a matter of fact, there is a 100% chance that Chris works in manufacturing.


问:
Zeta in summations
Does anyone know if there is any reference that discuss the following kind of summations?

$$S = \displaystyle\sum_{n=0}^\infty a_n \zeta(n+2)$$

I have read Srivastava's article https://www.sciencedirect.com/science/article/pii/0022247X88900133. But none deals with such series. Especially when:

$a_n = \frac{2}{(n+2)(n+3)}$. Which leads to

$S = \frac{\zeta(2)}{3} + \frac{\zeta(3)}{6} + \frac{\zeta(4)}{10} + \cdots$

Have you seen this one before?

Title: Zeta in summations

Body:
Thanks to a CAS,
$$S_m=\sum_{n=0}^\infty  \frac{2 _,\zeta(n+m)}{(n+m)(n+m+1)}$$ have closed forms
$$S_2=\log (2 \pi )-\gamma$$
$$S_3=\log (2 \pi )-\gamma -\frac{\zeta(2)}{3}$$
$$S_4=\log (2 \pi )-\gamma -\frac{\zeta(2)}{3}-\frac{\zeta (3)}{6}$$
$$S_5=\log (2 \pi )-\gamma -\frac{\zeta(2)}{3}-\frac{\zeta (3)}{6} -\frac{\zeta (4)}{10}$$
$$S_6=\log (2 \pi )-\gamma-\frac{\zeta(2)}{3}-\frac{\zeta (3)}{6} -\frac{\zeta (4)}{10}-\frac{\zeta (5)}{15} $$


答:
Since
$$
\zeta (s) = {1 \over {\Gamma (s)}}\int_0^\infty  {{{x^{\,s - 1} } \over {e^{\,x}  - 1}}dx}
$$
then
$$
\eqalign{
  & \sum\limits_{n = 0}^\infty  {a_{\,n} \zeta (n + 2)}
  = \sum\limits_{n = 0}^\infty  {a_{\,n} {1 \over {\left( {n + 1} \right)!}}\int_0^\infty  {{{x^{\,n + 1} }
 \over {e^{\,x}  - 1}}dx} }  =   \cr
  &  = \int_0^\infty  {\left( {\sum\limits_{n = 0}^\infty  {a_{\,n} } {{x^{\,n + 1} }
 \over {\left( {n + 1} \right)!}}} \right){1 \over {e^{\,x}  - 1}}dx}  =   \cr
  &  = \int_0^\infty  {{{I(x)} \over {e^{\,x}  - 1}}dx}  \cr}
$$
if we admit that the integral and sum can be exchanged.

Here
$$
I(x) = \sum\limits_{n = 0}^\infty  {a_{\,n} } {{x^{\,n + 1} } \over {\left( {n + 1} \right)!}}
 = \int_0^x {\left( {\sum\limits_{n = 0}^\infty  {a_{\,n} } {{t^{\,n} } \over {n!}}} \right)dt}
$$
is the integral of the e.g.f of the $\{a_n\}$ sequence.

Title: Zeta in summations

Body:
Does anyone know if there is any reference that discuss the following kind of summations?

$$S = \displaystyle\sum_{n=0}^\infty a_n \zeta(n+2)$$

I have read Srivastava's article https://www.sciencedirect.com/science/article/pii/0022247X88900133. But none deals with such series. Especially when:

$a_n = \frac{2}{(n+2)(n+3)}$. Which leads to

$S = \frac{\zeta(2)}{3} + \frac{\zeta(3)}{6} + \frac{\zeta(4)}{10} + \cdots$

Have you seen this one before?


问:
What is the $n^{th}$ digit of this sequence?
This is from a closed question:
What is the $n$th digit of this sequence?: What's the $n$th digit of the sequence where all natural numbers are repeated the time equal to itself:

$$122333\cdots1010101010101010101011\cdots$$

My results so far:
First we compute how many digits are there from $10^{k-1}$ to $10^k-1, \forall k=1, 2, \ldots$

$$
a_k = k(10^{k-1}  + \cdots + (10^k-1))=k\frac{10^{k-1}+10^k-1}{2} \cdot (9\cdot 10^{k-1})
$$

Then we compute the partial sum (define $S_0=0$)
$$
S_m=\sum_{k=1}^m a_k = \sum_{k=0}^m \frac{9k}{2} \left(10^{2k-1}+10^{2k-2}-10^{k-1} \right)\\
=\frac{1}{198}(10^m-1)(99m \cdot 10^m - 10^m+10)$$
$$=5(10^m-1)\left(m\cdot 10^{m-1}-\frac{10^{m-1}-1}{10^2-1}\right)\tag 1
$$

I admit I cheated with WA.

For any $n \in \mathbb N$ we need to find $m$ such that $S_{m-1} < n \le S_m$, then we'd know the digit we are looking for is $$\text{the } \left(m- ((S_{m-1}-n) \pmod m ) \right)^{th} \text{ digit of the (} m \text{-digit number)
 }10^{m-1}-1 + \lceil \frac{n-S_{m-1}}{m}\rceil \tag 2$$

My questions:

a) Any single  formula to get the digit in terms of $n$? I doubt there is one because getting $m$ out of $S_m$ from $(1)$ looks difficult but who knows.

b) If a) is not possible, any better alternative to $(2)$ given $m$?

答:
COMMENT.- No problem with the nine digits, I mean with the number
$$N=122333444455555\cdots999999999$$
because its position is clearly given
$$2\text { in the } 2^a\text{and }3^a\\3\text { in the } 4^a\text{and }6^a\\4\text { in the } 7^a\text{and }10^a\\5\text { in the }11^a\text{and }15^a$$
The pattern is manifest and thus $6$ is between $15 + 1 = 16$ and $15 + 6 = 21$ positions and following this model we have that $8$ is between positions $29$ and $36$ and $9$ is between positions $37$ and $45$.

But starting with $10$, which will be between positions $45 + 1 = 46$ and $45 + 10 = 55$, there is no longer a fixed digit but there is an interleaving of $1$ and $0$ and the matter will be further complicated with other numbers.


问:
Topological manifold example
$\theta(x,x^2)=x$

$\Bbb X =${$(x,x^2)| x$ in $\Bbb R$}

And V is subset of $\Bbb R$

$dim\Bbb X=1$

My instructor said that this is topological manifold.

Why?

Please can you explain me? This example is from my notebook. But there is No explanation except for this. Thank you for help.

答: Check that $\theta:X \to\mathbb{R}$ it's an homeomorphism, where $\mathbb{R}$ has the euclidean topology; then you conclude since the codomain is surely a one-dimensional topological manifold.</p>

Title: Topological manifold example


问:
Equality in Jensen's inequality for non convex functions
Jensen's inequality, $\mathbb{E}[\phi(X)]\geq\phi(\mathbb{E}[X])$, holds for any convex function $\phi$ and random variable $X$.

Given that $\mathbb{E}[\phi(X)]=\phi(\mathbb{E}[X])$ for some non-degenerate random variable $X$, what can we say about $\phi$? any why? (assuming that it is not necessarily convex). What about the case in which $\phi$ is differentiable?
I wonder if there is an interesting universal property on all function $\phi$ that lead to such equality.

答: I don't think there is much that can be said without more restrictions. If $X$ is some random variable such that $P(X=E[X]) = 0$, then for any function $\phi$ you can define $\tilde{\phi}(t) = \begin{cases} \phi(t) & t \ne E[X] \\ E[\phi(X)] & t = E[X] \end{cases}$ to get $\tilde{\phi}(E[X]) := E[\phi(X)] = E[\tilde{\phi}(X)]$.


问:
If $\sin(x)/\sin(y) = \sin(9)/\sin(18)$, can we say that $y = 2x$?
I am not really sure whether trigonometric functions work like that, and if we can say that, how would you prove that?</p>

答:
The general principle is that you can not "cancel" function symbols, i.e., in general,

$$
\frac{f(x)}{f(y)}\color{red}{\ne}\frac{x}{y}\tag{1}
$$

Your particular case is $f(u)=\sin(u)$.

Try more simple examples to convince yourself:
$$
f(u)=2u+1,\quad f(u)=u^2,\quad \cdots
$$


问:
Given $x,y,z\in \mathbb{Z}$ such that $x^3+y^3-z^3$ is multiple of $7$ prove that at least one is divisible by $7$

Given $x,y,z\in \mathbb{Z}$ such that $x^3+y^3-z^3$ is multiple of $7$ prove that at least one is divisible by $7$

Given $n\in \mathbb{Z}$ then $n\in \lbrace [0],[1],[2],[3],[4],[5],[6]\rbrace$, of the same way
$n^3\in \lbrace[0],[1],[6] \rbrace$.

Since $7\mid x^3+y^3-z^3$ then $x^3+y^3-z^3\in [0]$ there be two possibilities

$x^3,y^3,z^3\in [0]$ and then $x,y,z\in[0]$ and the result is true.

$x^3\in [6],y^3\in [1],z^3\in[0]$ and then $z\in [0]$ therefore the results follows.

$x^3\in [0], y^3 \in [1], z^3 \in [1]$ and then $x\in [0]$and the afirmation is too true

The other possibilities only switch the order of the variables with the classes.

答:
If it's a more optimized proof you're looking for, these exist. Again, we notice that
$$a^3=a^{(7-1)/2}\equiv\left(\frac a7\right)\in\{-1,0,1\}\pmod7,$$
which we can also see by direct casework. We note that $a^3\equiv0$ if and only if $a\equiv0.$

The advantage of saying $\{-1,0,1\}$ is that we can say the sum $x^3+y^3-z^3$ lives in $[-3,3],$ and further, if all are nonzero, then this sum is $1+1+1\equiv1\pmod2$; i.e., the sum is odd. In particular, if all are nonzero, then $x^3+y^3-z^3$ is not divisible by $7,$ and we finish by contraposition.


答:
Suppose $x = 2y$ then
$$\frac{\sin(x)}{\sin(2x)} = \dfrac{\sin(9)}{\sin(18)}$$

Which means, by using $\sin(2x) = 2\sin(x)\cos(x)$,

$$x = \arccos\left(\dfrac{\sin(18)}{2\sin(9)}\right) \approx 2.716$$

Hence

$$y = \dfrac{\sin(x)\sin(18)}{\sin(9)} = -0.75234$$

From which you can see $y \neq 2x$

Title: If $\sin(x)/\sin(y) = \sin(9)/\sin(18)$, can we say that $y = 2x$?

Body: I am not really sure whether trigonometric functions work like that, and if we can say that, how would you prove that?</p>


问:
Thoughts about the use of $\epsilon$-$\delta$ language in proofs involving limits
The limit of a function $f: \mathbb{R} \to \mathbb{R}$ at $x_{0}$ is defined as follows:

$\lim_{x \to x_{0}} f\left(x\right) = L \iff \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon$.

In proofs of limits using the definition, we typically fix an arbitrary constant $\epsilon > 0$, and as long as we are able to prove
\begin{equation}
\exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon,
\end{equation}
we can conclude that $\lim_{x\to x_{0}}f\left(x\right) = L$. The key is to construct such a $\delta$ so that the above relation holds. However, as I saw in some textbooks, for an arbitrary $\epsilon > 0$, if we are able to find a $\delta > 0$ such that
\begin{equation}
\forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < g\left(\epsilon\right),
\end{equation}
where $g\left(\epsilon\right)$ is a term containing $\epsilon$ ($g\left(\epsilon\right) = 3\cdot\epsilon$, for example),
then we can still claim that $\lim_{x\to x_{0}}f\left(x\right) = L$.

I am wondering if there exists a simple proof for using the general $g\left(\epsilon\right)$ in proofs of limits?

答:
If you have $|A|<3\epsilon$ for any $\epsilon>0$, you can conclude that $|A|<\epsilon$ for any $\epsilon>0$, which implies that $|A|=0$.

In general, if you have the estimates of the form $|A|<g(\epsilon)$ for all $\epsilon>0$, where $\displaystyle\lim_{y\to 0+}g(y)=0$, you can conclude that $|A|=0$.

Suppose you have shown that
$$
\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<g(\epsilon)\tag{1}
$$
where $\displaystyle \lim_{y\to 0+}g(y)=0$,
you can then show that
$$
\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon\tag{2}
$$

Proof.

Let $\epsilon>0$. Since $\displaystyle \lim_{y\to 0+}g(y)=0$, there exists $r>0$ such that $g(r)<\epsilon$. By (1), there exists $\delta>0$ such that

$$
|x-x_0|<\delta\Rightarrow |f(x)-L|<g(r)<\epsilon.
$$

Now you have (2).


问:
Solve the equation $\tan(x-45 °)+\cot x =2$ for $0°\le x\le180°$
This is A levels trigonometry I am doing self studying so I have no teacher if someone could help me that would be great.</p>

答:
As an A Level student myself, I really recommend you get these textbooks if you are doing the Edexcel course; they contain excellent explanation of the course's content as well as hundreds of practice questions.

As for your question:

HINT

Use the addition formulae for $\tan$ to give yourself an equation entirely in terms of $\tan x$. If you require any more help please don't hesitate to ask.


答:
As long as $g(\epsilon)$ goes infinitely close to $0$ while staying positive, as $\epsilon \to 0$, i.e. $\lim_{\epsilon\to 0} g(\epsilon) = 0$ and $g(\epsilon) > 0$, any such $g$ will do because

$\forall \zeta > 0,\ \exists \gamma > 0, \ 0 < \epsilon < \gamma \implies g(\epsilon) < \zeta$

So, if using an $\epsilon-\delta$ technique we conclude that $\exists$ a suitable $\delta$ such that $|f(x) - L| < g(\epsilon)$, then $\forall \zeta > 0$, we choose $\epsilon < \gamma$ and $|f(x) - L| < g(\epsilon) < \zeta$ whenever $0 < |x-x_0| < \delta$.

Title: Thoughts about the use of $\epsilon$-$\delta$ language in proofs involving limits

Body:
The limit of a function $f: \mathbb{R} \to \mathbb{R}$ at $x_{0}$ is defined as follows:

$\lim_{x \to x_{0}} f\left(x\right) = L \iff \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon$.

In proofs of limits using the definition, we typically fix an arbitrary constant $\epsilon > 0$, and as long as we are able to prove
\begin{equation}
\exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon,
\end{equation}
we can conclude that $\lim_{x\to x_{0}}f\left(x\right) = L$. The key is to construct such a $\delta$ so that the above relation holds. However, as I saw in some textbooks, for an arbitrary $\epsilon > 0$, if we are able to find a $\delta > 0$ such that
\begin{equation}
\forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < g\left(\epsilon\right),
\end{equation}
where $g\left(\epsilon\right)$ is a term containing $\epsilon$ ($g\left(\epsilon\right) = 3\cdot\epsilon$, for example),
then we can still claim that $\lim_{x\to x_{0}}f\left(x\right) = L$.

I am wondering if there exists a simple proof for using the general $g\left(\epsilon\right)$ in proofs of limits?


答: Hint: with $t:=\tan x$, the equation is $\frac{t-1}{t+1}+\frac1t=2$, which you can rearrange to a quadratic in $t$.

Title: Solve the equation $\tan(x-45 °)+\cot x =2$ for $0°\le x\le180°$

Body: This is A levels trigonometry I am doing self studying so I have no teacher if someone could help me that would be great.</p>


问:
Wasserstein distance of linear combination
Given three measures $\mu,\nu,\xi$ on a common space and $d$ be a distance on such space. Let also pick an $\alpha \in (0,1)$

is there some result connecting $W_p(\alpha\mu + (1-\alpha)\nu, \xi)$ and $W_p(\mu,\xi)$ and $W_p(\nu,\xi)$, were $W_p$ denotes the $p-$Wasserin distance, i.e. $W_p(\mu,\nu)=\min_\pi\mathbb{E}_{\pi}[d^p(X,Y)]$, where $X\sim\mu$, $Y\sim\nu$ and the minimization is over couplings of $\mu$ and $\nu$?
I could not find anything anywhere, but maybe there's a quick trick I fail to see?

答:
Nice question. I don't think there is, but I don't have a full answer. $W_p$ makes the space of probability measures into a geodesic metric space, and it looks like your question is about the curvature of this space. More precisely, if you put $\phi_\alpha = W_p(\alpha \mu + (1-\alpha) \nu,\xi)$ for the interpolation between $\phi_0=\nu$ and $\phi_1=\mu$, then you have a "triangle" with vertex $\xi$ and opposite side $\phi_\alpha$, $0 \leq \alpha \leq 1$, and the it seems like you're for the curvature of this opposite side. I don't know of general results for curvature in the Wasserstein space. Also, this vector space interpolation may have very little to do with the Wasserstein geodesic from $\mu$ to $\nu$. So I can't see how this is possible.

Perhaps there is something if you include $W_p(\mu,\nu)$, or a geometric condition on the underlying metric space (like hyperbolicity) ... but maybe that's just pointing back to triangle inequality.

Letting $\mu,\nu,\xi$ be translations of a simple measure seems to rule out any of the basic relations that occur to me. I'd be very interested to hear if there is a neat relation here that I'm not seeing.


问:
"Translating" one value $- \infty$ to $+ \infty$ to another ($+ \infty$ to $ \gt 0$)
Well even if I need to use the following in a computer game this is a math question. I have a world map which I can scroll with a scroll velocity $(f(x))$ with my mouse. And I have a zoom factor $(x)$ which is $0$ in its initial state and can be positive infinite and negative infinite.

So I'm searching for a function where:

$x$ can range from $- \infty$ to $+ \infty$

$f(x)$ should be $\gt 0$

$f(x)$ should tend to $0$ for $x$ tending to $+ \infty$

$f(x)$ should tend to $+ \infty$ for $x$ tending to $- \infty$

I bet the solution is somewhat so obviously easy I'm just to blind to see in the moment.

答:
Any function in the form $f(x)=a^{-x}$ where $a>1$ will satisfy your four conditions.

The closer $a$ is to 1, the more slowly $f(x)$ will approach its asymptotes.


问:
ENR sets and homotopy between constant maps and homotopies
I'm stumped by this homework question:
"Given an ENR space $X$, then for every points $x_0 \in X$ and open set $U \ni x$ there exists a neighborhood $V$ with $x_0 \in V \subset U$ such that the maps $i:V\rightarrow U$ and $x_0:V \rightarrow U$ are homotopic".

By definition this seems to need a lot of simple results that my mind can't memorize, I probably need to use this corollary to Borsuk's extension theorem but I'm probably missing some background to get to fulfill the hypothesis:

"If $M$ is a contractible metric space, $A$ a closed subset of $M$ and $X$ ENR, then $f:A \rightarrow X$ has an extension to $M$ if and only if it's homotopic to a constant"

Of course if I can find a closed subset $A$ of $U$ and an open set $V\subset A$ then the result follows but having only the retraction doesn't seem to cut it to find such $A$.

答:
I learned this fact from A. Dold's book Lectures on Algebraic Topology. If you are interested, you will find there a much more general statement.

With the ENR $X$ given, fix an embedding $j:X\hookrightarrow\mathbb{R}^n$, and a retraction $r:N\rightarrow X$, from an open neighbourhood $N$ of $X$ in $\mathbb{R}^n$. Now let $U\subseteq X$ be open and $x_0\in U$. By means of the embedding $j$ we understand $U$ to be a subset of $\mathbb{R}^n$. Consider the set
$$V=\{x\in U\mid (1-t)j(x)+tj(x_0)\in N\;\forall\, t\in I\}.$$
Clearly $x_0\in V$. In fact $V$ is open too, since it is the inverse image of the open set $N$ under the continuous function $U\times I\rightarrow \mathbb{R}^n$, $(x,t)\mapsto (1-t)j(x)+tj(x_0)$.

To complete we define a homotopy $H:V\times I\rightarrow U$ by putting
$$H(x,t)=r\left((1-t)j(x)+tj(x_0)\right).$$


问:
Gauss theorem and potential
Let $V(r) = qe^{-\mu r}/r$

if R is any region containig the origin, show that:

$\int_{dR} \nabla V dS = \mu^2 \int_{R} Vd^3r - 4\pi q$

dr is the boundary of R

I applied Gauss theorem here, so that
$\int_{dR} \nabla V dS = \int \Delta V d³r = \mu² \int_{R} Vd³r$

Now i can guess that the difference between my answer and the book answer is due the singularity at origin, but how to deal with this? In case like that i can't apply Gauss theorem?

答:
Yes it is to do with the singularity at the origin. The way you have calculated the divergence of $\nabla V$, it will ignore the fact that there is singularity at the origin.

$V(r) = \displaystyle \frac{q \ e^{-\mu r}}{r}, \nabla V = \displaystyle (- \frac{q \ e^{-\mu r}}{r^2} - \frac{q \ \mu \ e^{-\mu r}}{r}) \hat{r}$

$\nabla^2 V = \displaystyle \nabla \cdot \big(- \frac{q \ e^{-\mu r}}{r^2} \hat{r}\big) + \nabla \cdot \big( - \frac{q \ \mu \ e^{-\mu r}}{r} \hat{r}\big)$

$ = \displaystyle - q \ e^{-\mu r} \ \nabla \cdot \big(\frac{\hat{r}}{r^2}\big) - \frac{q }{r^2} \hat{r} \cdot  \ \nabla \big(e^{-\mu r}\big) + \frac{1}{r^2}\frac{\partial}{\partial r} \big( - q \ r \mu \ e^{-\mu r}\big)$

$ = \displaystyle - q \ e^{-\mu r} \ \nabla \cdot \big(\frac{\hat{r}}{r^2}\big) + \frac{q \ \mu \ e^{-\mu r}}{r^2} - \frac{q \ \mu \ e^{-\mu r}}{r^2} + \frac{q \ \mu^2 \ e^{-\mu r}}{r}$

$ = \displaystyle - q \ e^{-\mu r} \ \nabla \cdot \big(\frac{\hat{r}}{r^2}\big) + \mu^2 \ V$

Now the second term is straightforward that is in your working too. Coming to the first term, we cannot calculate divergence of vector field $\frac{\hat{r}}{r^2}$ by just multiplying by $r^2$ and taking derivative (which works out to be zero) due to singularity at the origin. It turns out that the divergence of the vector field $\frac{\hat{r}}{r^2}$ is in fact zero everywhere but the origin. You have to use Dirac Delta function to calculate the divergence -

$\nabla \cdot \big(\frac{\hat{r}}{r^2}\big) = 4 \pi \ \delta^3(r - \varepsilon)$ and $\int_\Omega f(r) \ \delta^3(r-\epsilon) \ dV = f(\epsilon), \epsilon \to 0$

So, $\int_\Omega - q \ e^{-\mu r} \ \nabla \cdot \big(\frac{\hat{r}}{r^2}\big) \ dV = - 4 \pi  q \ \big($as $ \ f(r) = e^{-\mu r} \big)$

Title: Gauss theorem and potential