问:
Prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$.
If I want to prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$, I need to show that:

$\exists\varepsilon>0$ $\forall\delta>0$  $\exists{x,y}\in\mathbb{R}\ : |x-y|<\delta$ and $|\sin(x^2) - \sin(y^2)|\geq\varepsilon$.

So let's take $\varepsilon = 1$. Then I want $|\sin(x^2)-\sin(y^2)|\ge1$. That's the case if $\sin(x^2)=0$ and $\sin(y^2)=\pm1$. Thus $x^2=n\pi$ and $y^2=n\pi + \frac{1}{2}\pi$. Now I'm stuck on expressing x and y, which I want to express in $\delta$, to ensure that $|x-y|<\delta$.

Thanks in advance for any help.

答: You can also say that $f$ is uniformly continuous if and only if for any $x_n$ and $y_n$ such that $(x_n - y_n)\to 0 $ implies $|f(x_n) - f(y_n)| \to 0$ thus you can choose for example $x_n=n$ and $y_n=n-\frac{1}{n}$, which we see $x_n-y_n$. However we have that $f(x_n)-f(y_n) = 2cos(\frac{...}{2})sin(\frac{2-n^2}{2}) \not\to 0$. So we see that the condition for $f$ being uniformly continuous fails.


问:
If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ be a module homomorphism, then $\mathbb{Z}^m/\operatorname{im}(f)$ is a finite abelian group

If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ is an injective group homomorphism, then is $\mathbb{Z}^m/\operatorname{im}(f)$ a finite abelian group?

I think yes. Let the homomorphism be given by a $m\times m$ matrix. Then, by existence of smith normal form over $\mathbb{Z}$, I think $\mathbb{Z}^m/\operatorname{im}(f)$ would be a direct product of cyclic groups of orders given by the entries in the diagonal of the smith normal form. Is this right? If so, how do we prove it? Any hints? Thanks beforehand.

答: The Smith normal form works fine (and with $f(x)=Ax$ it proves that $|\Bbb{Z}^n/Im(f)| = |\det(A)|$) but it is easier to find a non-zero matrix $B\in M_n(\Bbb{Z})$ such that $AB=d\ I$.

ContentLicense: CC BY-SA 4.0


问:
Dice rolling: When rolling n>3 die, probability for getting any combination from a fixed set of random 3-element-combinations.
Good afternoon,

fyi: I am currently working to balance a dice-based combat system for tabletop rpg, which heavily relies on selecting certain combinations of die from a larger pool to outbid your opponent.

Using Excel, I've worked my way through determining the number of possible combinations for rolling n 6 sided die, as well as the probability of getting a certain combination from a larger dice pool. However, I am stuck now. Simply put, I want to know the probability for getting the (say) 6 highest ranking 3d6-combinations out of a nd6 dicepool.

In particular: Say I have a given set S of several (not all) 3-element-combinations (order doesn't matter) from rolling 3 6-sided-die like

S = {{1,1,6}, {6,6,6}, {1,1,5}, {5,5,5}, {1,1,4}, {4,4,4}, {1,1,3}, {3,3,3}}

Given S, I want to count all 4-element-permutations (now from rolling 4 6-sided-die) of wich at least one element of S is a subset. So basically, I want to know how likely I am to get a combination from S when rolling 4 die. Preferably, I would like to generalize the method for any number >1 of die.

Is there a "mathmatical" way to do this without using brute force (that would be counting done by Excel)?

答:
There is a mathematical way to do it. First, you need to figure out the probability of hitting any single combination, like $\{1,1,6\}$. This can be done recursively; for each possible result of the first die, you add up the probability that the remaining $n-1$ die hit the combination with one instance of the first roll removed. There are ways to speed this up by avoiding recalculating repeated cases.

Then, given a set $S$ of combinations, you need to find the probability of hitting at least one. This is can be done with the principle of inclusion/exclusion. That is, add up the probabilities of hitting each combination in $S$, then subtract the probability of hitting the combinations in each pair of $S$, add in the triplets, etc.

I made a program that does this algorithm at the link below. It works for any number of dice, and any subset $S$ of combinations to hit at least one of, and you can check it against a simulated probability. The program will be too slow if $S$ has $20$ or more combinations.

https://repl.it/@mearnest/Tabletop-game-probability


问:
Can it be proven in ZF that $\mathbb{N}$ is a universal size comparator?
This is a follow-up to my previous question several years ago about whether every finite set is a universal size comparator. Now I am asking, can it be proven in ZF without the axiom of choice that the set $\mathbb{N}$ of natural numbers is a universal size comparator, meaning, it can be compared with every set in terms of cardinality?

答:
No, it is consistent that there are infinite Dedekind finite sets.

Such set can easily shown to admin no injective to or from $ℕ$:

Let $D$ be that set, if $f:ℕ→D$ is injective, then $|D|=|D\setminus {f(0)}|$ by fixing a bijection between $f''ℕ\setminus\{0\}$ to $f''ℕ$, hence $ℕ\not≤D$.

If $g:D→ℕ$ is injective, then either $g''D$ is countable infinite or $g''D$ is finite, the first case implies it is Dedekind infinite, the second case implies it is finite, both cases lead to contradiction. ⇒ $D\not≤ℕ$

That being said, we cannot prove that there exists a set $D$ that is not comparable with $ℕ$ from $ZF$(otherwise we could prove it exists in $ZFC$ as well, where we know all sets are comparable)


问:
What natural integers satisfy the relation $\frac{3}{7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
I would like to find all natural triplets $(a, b, c)$ that satisfy the  relation
$\frac{3}{7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\,.$

Body:
The solutions are the following
$$
\begin{array}{rrr}
 3 & 11 & 231\\
 3 & 12 & 84 \\
 3 & 14 & 42 \\
 3 & 15 & 35 \\
 3 & 21 & 21 \\
 4 & 6 & 84 \\
 4 & 7 & 28 \\
 5 & 5 & 35 \\
 7 & 7 & 7 \\
\end{array}
$$


问:
Uncenter version of the Besicovitch covering lemma
It is wellknown that in the classical Besicovitch covering theorem, the point being covered should be the center of some ball. However in Javier Duoandikoetxea's book, Fourier Analysis, it says that the point $x$ need not to be the center of the ball or cube but must be uniformly close to the center. For more details, you can see the following picture

Where I don't understand is the meaning of "uniformly close to the center". Does it mean that
$$\sup_{x}|x-c_{B_x}|<\infty$$
or the ratio of $|x-c_B|$ and $r_B$ is not very large, e.g.
$$\sup_x\frac{|x-c_{B_x}|}{r_{B_x}}\leq 1/2 ?$$
I have seen the proof of the Besicovitch covering lemma, but have no progress. I wonder if there are some related references or papers? Thanks for your warm help.

答:
I haven't been able to find a reference that addresses this specific point, but I'm farily sure that "uniformly close to the center" has the second meaning from the question, i.e.
$$
\sup_{x \in A} \frac{|x - c_{B_x}|}{r_x} < 1.
$$

Here's an adaptation of a proof of the centered Besicovitch covering lemma from some notes that handles the uncentered case for balls.

Step 0: Parameters

We begin by defining a few parameters that will be important as we go along. First, let
$$
\gamma = \sup_{x \in A} \frac{|x - c_{B_x}|}{r_x},
$$
so that $\gamma < 1$. Select $\alpha$ satisfying $(1 + \gamma)/2 < \alpha < 1$, and further define $\delta = (\alpha - \gamma) / (\alpha + 1)$, so that $0 < \delta < 1$. Finally, let $R = \sup_{x \in A} |x|$.

Step 1: Selecting the cover

Now we iteratively select a sequence of balls from our collection. Define $R_1 = \sup_{x \in A} r_x$. We can assume that $R_1$ is finite, as otherwise we can find a single ball in the collection which covers $A$. (Since we're allowing uncentered balls, the uniformity condition comes into play here.) Pick the first ball $B_1 = B(c_1, r_1)$, associated with the point $x_1 \in A$, so that $r_1 > \alpha R_1$. Having selected $B_1, B_2, \ldots, B_m$ (and assuming they do not cover $A$), let
$$
R_{m+1}
= \sup_{x \in A \setminus \bigcup_{k=1}^m B_k} r_x
$$
and select $B_{m+1} = B(c_{m+1}, r_{m+1})$ associated to $x_{m+1} \in A \setminus \bigcup_{k=1}^m B_k$ so that $r_{m+1} > \alpha R_{m+1}$. Note that for $i < j$, we have
$r_i > \alpha R_i \geq \alpha R_j \geq \alpha r_j.$

The balls $\{ B'_k = B(c_k, \delta r_k) \}$ are then disjoint: assuming $i < j$, we have $x_j \notin B(c_i, r_i)$ and as a result
\begin{align*}
|c_i - c_j|
&\geq |c_i - x_j| - |x_j - c_j| \\
&\geq r_i - \gamma r_j \\
&> \delta r_i + (1 - \delta) \alpha r_j - \gamma r_j \\
&= \delta r_i + \delta r_j.
\end{align*}
We can also say that every $B'_k$ is contained in the ball $B(0, 2R_1 + R)$ since $|c_k| \leq |c_k - x_k| + |x_k| \leq R_1 + R$. It then follows that the radii $r_k \to 0$, and so $R_k \to 0$ as well. We can in turn conclude that the main balls $\{ B_k = B(c_k, r_k) \}$ cover $A$, as otherwise we obtain a contradiction to $R_k \to 0$.

Step 2: Bounding intersections, small radii

Going forward, we let $M$ denote a (potentially large) parameter whose value we defer selecting until later. (Its value will ultimately depend on the values of $\gamma$ and $\alpha$, but otherwise there won't be more specific dependence on the starting collection of balls.) We now show that for any $m$, there are at most $(2M + 1)^n \alpha^{-n} \delta^{-n}$ of the balls $B_k$, $k = 1, 2, \ldots, m-1$ which both intersect $B_m$ and satisfy $r_k \leq M r_m$.

To see this, fix an $m$ and suppose that $B_{k_1}, \ldots, B_{k_{N}}$ satisfy the given criteria. Then the associated dilated balls $B'_{k_1}, \ldots, B'_{k_{N}}$ are disjoint (as showed in Step 1) and are further contained in $B(c_m, (2M+1) r_m)$: for $x \in B(c_{k_i}, \delta r_{k_i})$ we have
\begin{align*}
|x - c_m|
&\leq |x - c_{k_i}| + |c_{k_i} - c_m| \\
&\leq \delta r_{k_i} + r_{k_i} + r_m \\
&\leq 2 M r_m + r_m.
\end{align*}
Measure considerations now give
\begin{align*}
(2M + 1)^n r_m^n
\geq \delta^n \sum_{i = 1}^{N} r_{k_i}^n
\geq \delta^n \sum_{i = 1}^{N} \alpha^{n} r_m^n
= N \alpha^{n} \delta^n r_m^n,
\end{align*}
since $r_{k_i} > \alpha r_m$ as showed in Step 1. This establishes the claimed bound.

Step 3: Bounding intersections, large radii

Next suppose that $i < j < m$ and that the balls $B_i$ and $B_j$ both intersect $B_m$, while also $r_i, r_j > M r_m$. We establish a lower bound for the angle made by the rays from $x_m$ to each of $c_i$ and $c_j$.

Since $x_m \notin B_i$ by construction, we first have the lower bound $| x_m - c_i | \geq r_i$, and likewise $| x_m - c_j | \geq r_j$. We also have the upper bound
$$
| x_m - c_i |
\leq | x_m - c_m | + | c_m - c_i |
\leq (1 + \gamma) r_m + r_i,
$$
and likewise replacing $i$ by $j$. Third, we have a lower bound
\begin{align*}
| c_i - c_j |
&\geq | c_i - x_j | - | x_j - c_j | \\
&\geq r_i - \gamma r_j \\
&\geq (\alpha - \gamma) r_j \\
&> 0,
\end{align*}
relying on the fact that $x_j \notin B_i$.

Letting $\theta$ denote the angle between the rays, the law of cosines now gives
\begin{align*}
\cos \theta
&= \frac{|x_m - c_i|^2 + |x_m - c_j|^2 - |c_i - c_j|^2}{2 |x_m - c_i| |x_m - c_j|} \\
&\leq \frac{[(1 + \gamma) r_m + r_i]^2 + [(1 + \gamma) r_m + r_j]^2 - [r_i - \gamma r_j]^2}
{2 r_i r_j} \\
&= \gamma
+ \frac{1 - \gamma^2}{2} \frac{r_j}{r_i}
+ (1 + \gamma)
\left(
\frac{r_m}{r_i} + \frac{r_m}{r_j}
\right)
+ (1 + \gamma)^2 \frac{r_m}{r_i} \frac{r_m}{r_j}.
\end{align*}
We further have $r_j / r_i < 1 / \alpha$ since $i < j$, and by hypothesis both $r_m / r_i$ and $r_m / r_j$ are less than $1/M$. Therefore
$$
\cos \theta
\leq \gamma
+ \frac{1 - \gamma^2}{2 \alpha}
+ \frac{2 (1 + \gamma)}{M}
+ \frac{(1 + \gamma)^2}{M^2}.
$$
Our initial choice of $\alpha$ ensures that
$$
\gamma
+ \frac{1 - \gamma^2}{2 \alpha}
< 1,
$$
and so by now selecting $M = M(\gamma, \alpha)$ sufficiently large we can also bound $\cos \theta$ away from $1$, thereby bounding $\theta$ away from $0$. That is, we can say that there is an $\epsilon = \epsilon(\gamma, \alpha)$ such that $\theta \geq \epsilon$.

To conclude this step, suppose that $m$ is fixed and that among the balls $B_k$, $k = 1, 2, \ldots, m-1$, there are $N$ which both intersect $B_m$ and satisfy $r_k > M r_m$. The angular separation condition just established then implies that there is a constant $C = C(n, \gamma, \alpha)$ such that $N \leq C$.

Step 4: Wrapping up

Putting together Steps 2 and 3, we can say that there is a constant $C' = C'(n, \gamma, \alpha)$ so that for any $m$, at most $C'$ of the balls $B_1, B_2, \ldots, B_{m-1}$ intersect $B_m$.

Finally, it follows that
$$
\sum_{k} \chi_{B_k}(x) \leq C' + 1
$$
for all $x$. For otherwise, we can select $x$ and $m$ so that $x \in B_m$ and
$$
\sum_{k = 1}^{m} \chi_{B_k}(x) > C' + 1.
$$
So then $x$ lies in more than $C'$ of the balls $B_1, \ldots, B_{m-1}$, meaning that more than $C'$ of these balls intersect $B_m$.

EDIT:

Here's a 1-dimensional counterexample to show that a Besicovitch-type result isn't ensured by the condition $\sup_x |x - c_{B_x}| < \infty$.

Let $A = (0, 1)$, and for $x \in A$ let $B_x = (0, (1+x)/2)$. Note that $|x - c_{B_x}| \leq 1/2$ for every $x$. It's straightforward to see that no finite collection of these balls (intervals) covers $A$. But the interval $(0, 1/2)$ is contained in each of the balls, so if $A' \subseteq A$ is such that $\{ B_x : x \in A' \}$ covers $A$ then $\sum_{x \in A'} \chi_{B_x}$ is infinite on $(0, 1/2)$.


问:
Building a Function with saddle point that has Certain Properties
I'm working on a problem and I'd like to compute a couple examples. For one example I'd like to create a function from $\mathbb{R}^2 \rightarrow \mathbb{R}$ with the following properties:

It has a saddle at zero

It goes to positive infinity on the x AND y axis (so maybe cross sections along these axes are upward opening parabolas)

It goes to negative infinity along the line $x = y$ AND $x = -y$ (so maybe cross sections along these axes are downward opening parabolas).

What is an example of a function with such properties?

Thanks!

答:
Following the commentor, I searched Monkey Saddles. I found more complicated saddles: I guess these types of surfaces are called n-th order saddles.

Formulas for them can be found here:
https://rivix.com/download/Docs/Peckham_2011_Saddles_Final.pdf

For my particular problem:
$f(x,y) = x^4 - 6x^2y^2 +y^4$


问:
Find all entire functions $f$ such that $|f|$ is harmonic.
I'm trying to find all entire functions $f$ such that $|f|$ is harmonic. My attempt is as follows.

Because $f$ is entire, we may write $f(z) = f(x + iy) = u(x,y) + iv(x,y)$ where $u$ and $v$ have continuous first-order partial derivatives and satisfy the Cauchy-Riemann equations. Furthermore, $u$ and $v$ are harmonic. Now $|f| = \sqrt{u^2 + v^2}$ so I thought maybe to look at $|f|^2$ first. Then,
$$(|f|^2)_{xx} = 2(u_x)^2 + 2(v_x)^2 + 2(u\cdot u_{xx} + v \cdot v_{xx})$$
and
$$(|f|^2)_{yy} = 2(u_y)^2 + 2(v_y)^2 + 2(u\cdot u_{yy} + v \cdot v_{yy}).$$
I need $(|f|^2)_{xx} + (|f|^2)_{yy} = 0$ in order for $|f|^2$ to be harmonic. So adding these equations and using that $u$ and $v$ are harmonic, I obtain
$$(u_x)^2 + (u_y)^2 + (v_x)^2 + (v_y)^2 = 0.$$
By the Cauchy-Riemann equations, I can simplify this to
$$(u_x)^2 + (v_x)^2 = 0.$$
I'm a little stuck on how to proceed from here. I want to somehow conclude $|f|^2$ is constant but I'm not sure how.

答:
You have demonstrated that constant functions are the only entire functions for which $|f|^2$ is harmonic.

Now let us investigate the case where $f$ is entire and $ |f|$ is harmonic. We can assume that $f$ is not identically zero.

It is surely possible to calculate $\Delta |f|$ directly, but one can simplify the work by showing first that $f$ has no zeros, so that $f=e^g$ for some entire function $g$: Assume that $f(z_0) = 0$, then
$$
 0 = |f(z_0)| = \frac{1}{2\pi} \int_0^{2\pi} |f(z_0+re^{it}| \, dt
$$
for all $r > 0$, which implies that $f$ is identically zero, in contrast to the assumption.

So we have $|f| =|e^g| = e^{\operatorname{Re} g} = e^v$ where $v =  \operatorname{Re} g$ is harmonic. Then
$$
 0 = \Delta |f| = \Delta(e^v) = e^v \left(v_{xx} + (v_x)^2 + v_{yy} +  (v_y)^2\right)
 = e^v \left( (v_x)^2 +  (v_y)^2\right)
$$
so that $v_x$ and $v_y$ are identically zero. It follows that $v$ is constant. Consequently, $|f|$ is constant, which implies that $f$ is constant.


问:
Sequence of continuous functions, integral, series convergence
Let $f_k$ be a sequence of continuous functions on $[0,1]$ such that $\int _0 ^1 f_k(x)x^ndx  = \int _0^1 x^{n+k} dx$ for all $n \in \mathbb{N}$.

Is $\sum _{k=1} ^{\infty}f_k(x)$ convergent?

Could you tell me how to solve this? I would appreciate all the hints.

Thank you.

答:
No. Hint: The right-hand-side $\int_0^1 x^{n+k} dx=c_k$ is a constant for each $k$. You can make $f_k(x)$ completely arbitrary on the interval $[0,9/10]$, and then adjust it on the interval $[9/10,1]$ to give the integral $\int_0^1 f_k(x)x^n dx$ the value $c_k$. Choose $f_k$ on the interval $[0,9/10]$ in such a way that the series diverges for any $0\leq x\leq 9/10$. If instead of always "adjusting" $f_k$ on the interval $[9/10,1]$, and you instead adjust it somewhere else for different $k$, you can construct a sequence $f_k$ which makes the series diverge for all $x\in[0,1]$.

EDIT I'll flesh out the hint. For the first case: Let $f_k(x)=1$ for $x\in[0,9/10]$, and let $f_k(x)$ on the interval $[0,9/10]$ be a straight line from the point $(9/10,1)$ to a point $(1,a_k)$. If $a_k$ were $1$, then $f_k$ would be identically 1 on the entire interval, and we would get $\int_0^1 f_k(x) x^n dx = \int_0^1 x^n dx > \int_0^1 x^{n+k} dx=c_k$. On the other hand, if we let $a_k\to-\infty$, then the integral $\int_0^1 f_k(x) x^n dx$ will tend to $-\infty$, which is less than $c_k>0$, so by the intermediate value theorem, there must exist some $a_k$ such that $\int_0^1 f_k(x) x^n dx$ is equal to $c_k$. Now the series $\sum_{k=1}^\infty f_k(x)$ will diverge for all $x\leq 9/10$ since it will just be the series $\sum 1=\infty$.

If we want the series to diverge for all $x$, then we could repeat the idea above, but instead of letting $f_k$ be $1$ on $[0,9/10]$, we could let it be $1$ on $[0,1-1/k]$ instead. Then the series would diverge for all $x<1$, since for each such $x$, all but finitely many terms of the series will be $1$. For $x=1$ one gets the series $\sum a_k$, and one needs to check that this series necessarily diverges. I will leave that as an exercise to you.

ContentLicense: CC BY-SA 3.0


问:
How to prove that a quotient group is isomorphic to a group of matrices
Consider the group $(\mathbb{R},+)$ and the subgroup

$$H=\{2n \pi |n\in \mathbb{Z}\}$$

Let $G$ denote the group of matrices of the form
\begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}
Where $\theta$ is any real number and the group operation is matrix multiplication. Prove that $\mathbb{R}/H$ is isomorphic to $G$ by first giving an explicit formula for a map $\phi:\mathbb{R}/H\to\ G$, and then checking that your map is well-formed, bijective, and respects the group operation.

I know that for the map to be well defined, the mapping has to be element to element, and that I need to ensure that its one-to-one and onto. But I am not making any progress with the proof nor able to give a formula for the map.
Can someone please help me out!

答:
Problem asks for an explicit map $\phi: R/H \to G$ but probably, the best approach here is to use the First Isomorphism Theorem. We will try to define a homomorphism $\phi: \mathbb{R} \to G$ such that $\phi$ is surjective and $\text{Ker}(\phi) = H$.

Another thing is if $T:\mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation that rotates a vector $\theta$ radians around origin clockwise, and $B$ is the ordered standard basis for $\mathbb{R}^2$, then $[T]_B^B = \begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}$. So, we can consider $G$ as group of rotations in $\mathbb{R}^2$ (I am writing this part just for intuition).

Now, define $\phi: \mathbb{R} \to G$ such that $\phi(\theta) = \begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}, \forall \theta \in \mathbb{R}$. Then, $\phi$ is well-defined and a homomorphism (I leave this part to you. To prove it is a homomorphism, you can use the intuition part). Also $\phi$ is clearly surjective and,

$$\text{Ker}(\phi) = \bigg\{\theta \in \mathbb{R} \mid \phi(\theta) = \begin{bmatrix}
    1 & 0 \\
    0 & 1
  \end{bmatrix}\bigg\} = \{\theta \in \mathbb{R} \mid \cos(\theta) = 1, \sin(\theta) = 0\} = H$$

(Note that we could also find this by using the intuition part above. Because kernel of this map is simply the angles in radians that when we rotate a vector that much, we get the same vector)

Then, by using the proof of First Isomorphism Theorem, we can easily obtain a map $\psi: \mathbb{R}/H \to G$ such that $\psi(\theta H) = \phi(\theta) = \begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}$. I am leaving it to you to check that $\psi$ is well-defined, homomorphism and bijective.


问:
Nested intervals and $\lim_{n \to \infty} |M| = 1$?
Considering a sequence $(M_n)_{n\in \mathbb{N}}$ of non-empty and compact subsets of $\mathbb{C}$ such that $M_1 \supset M_2 \supset M_3 \supset \ldots$. Furthermore define

\begin{align}
d(M) := \text{max}\{|z-w| : z,w \in M\}.
\end{align}

The sequence above fulfills $\lim_{n \to \infty} d(M_n) = 0$.

Does this alone imply that
\begin{align}
\lim_{n \to \infty} |M_n| = 1,
\end{align}
where $|M_n|$ denotes the cardinality of $M_n$?

答:
Let $M=\bigcap_{n\in\Bbb Z^+}M_n$. Your hypotheses imply that $|M|=1$, but it’s entirely possible that $|M_n|=\mathfrak{c}$ for each $n\in\Bbb Z^+$, in which case $\lim\limits_{n\to\infty}|M_n|=\mathfrak{c}$.

The sets $M_n$ are compact, so the Cantor intersection theorem ensures that $M\ne\varnothing$. If $x,y\in M$, and $x\ne y$, the hypotheses imply that $\operatorname{diam}M_n<|x-y|$ for some $n\in\Bbb Z^+$, contradicting the fact that $x,y\in M_n$. Thus, $M=\{x\}$ for some $x\in\Bbb C$. But if

$$M_n=\left\{z\in\Bbb C:|z|\le 1+\frac1n\right\}$$

for each $n\in\Bbb Z^+$, then $|M_n|=|\Bbb C|=\mathfrak{c}$ for each $n\in\Bbb Z^+$, and $\lim\limits_{n\to\infty}|M_n|=\mathfrak{c}$.


问:
Proof by induction, $2^{n} > n^{2} - 2$
So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.

I start of with the "basecase" for n = 3:

RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$

Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$.
To prove this inequality, we have to proof that it holds for n = k + 1.

$2^{k} > k^{2} - 2$

We multiply both sides with 2:

$2^{k+1} > 2(k^{2} - 2)$

$2^{k+1} > 2k^{2} - 4$

$2^{k+1} > k^{2} + k^{2} - 2 - 2$

According to our assumption,  $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $

Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:

$2^{k+1} > k^{2} + 2k + 1 - 2$

Thus:

$2^{k+1} > (k+1)^{2} - 2$

QED.

How does my proof look, does it have any deficits, and how would you prove it?

Thank you!

答: The inductive step uses $2k^2-4\ge(k+1)^2-2$, or equivalently $(k-1)^2\ge4$, which holds for $k\ge3$. With quadratic inequalities, it's always wise to complete the square.

Title: Proof by induction, $2^{n} > n^{2} - 2$


问:
Curve adaptation with a finite number of measured values
I consider a function $f(t)=\alpha\cdot 2^{-t}+\beta\cdot 2^{-2t}$ and a finite number of measured values $f_0,...,f_n$. The coefficients $ \alpha, \beta $ are calcualted by the least squares method which means
$$ \sum\limits_{i=0}^n |f(t_i)-f_i|^2$$
is minimized.

Then $$\sum\limits_{i=0}^n 2^{-t_i}\cdot f(t_i)=\sum\limits_{i=0}^n 2^{-t_i}\cdot f_i.\quad (*)$$

My idea:

I consider the expressions

$$ \begin{aligned}f_0&\stackrel{!}{=}f(t_0)=2^{-t_0}+\beta\cdot 2^{-2t_0}\\f_1&\stackrel{!}{=}f(t_1)=2^{-t_1}+\beta\cdot 2^{-2t_1}\\\vdots\\f_n&\stackrel{!}{=}f(t_n)=2^{-t_n}+\beta\cdot 2^{-2t_n}\end{aligned}$$

which can be reprensented by $$ \underbrace{\begin{pmatrix}2^{-t_0}&2^{-2t_0} \\2^{-t_1}&2^{-2t_1} \\\vdots&\vdots\\2^{-t_n}&2^{-2t_n} \end{pmatrix}}_{=:A}\cdot \underbrace{\begin{pmatrix}\alpha\\\beta \end{pmatrix}}_{=:x}=\underbrace{\begin{pmatrix}f_0\\f_1\\\vdots\\f_n \end{pmatrix}}_{=:f} $$

Then I want to minimize

$$ F((\alpha,\beta)):=\|A\cdot x-f\|_2^2=\sum\limits_{i=0}^n |f(t_i)-f_i|^2=\sum\limits_{i=0}^n \Big(\alpha\cdot 2^{-t_i}+\beta\cdot 2^{-2t_i}-f_i \Big)^2 $$

I calculate the gradiant of $ F $:

$$ \nabla F((\alpha,\beta))= 2\cdot A^T\cdot (A\cdot x-f)\stackrel{!}{=}0$$

which means I have to solve $$ A^T\cdot A\cdot x=A^T\cdot f $$

At first I have
$$ A^T\cdot A=\begin{pmatrix}\sum\limits_{i=0}^n 2^{-4t_i}&\sum\limits_{i=0}^n 2^{-3t_i}\\\sum\limits_{i=0}^n 2^{-3t_i}&\sum\limits_{i=0}^n 2^{-2t_i} \end{pmatrix}=:\begin{pmatrix}a&b\\b&c \end{pmatrix} $$
and

$$ A^T\cdot f=\begin{pmatrix}\sum\limits_{i=0}^nf_i\cdot 2^{-2t_i}\\\sum\limits_{i=0}^nf_i\cdot 2^{-t_i} \end{pmatrix}=:\begin{pmatrix}b_1\\b_2 \end{pmatrix} $$

The solution is $$ \alpha=\frac{a\cdot b_2-b\cdot b_1}{a\cdot c-b^2}\qquad \beta=\frac{c\cdot b_1-b\cdot b_2}{a\cdot c-b^2} $$

And here comes the problem. These expressions of the solution are soo complicated that I'm not able to handle them to show the upper identity (*).

Is there annother way to show or what could I do else(*)?

答:
Given that $$ A^T\cdot A=\begin{pmatrix}\sum\limits_{i=0}^n 2^{-2t_i}&\sum\limits_{i=0}^n 2^{-3t_i}\\\sum\limits_{i=0}^n 2^{-3t_i}&\sum\limits_{i=0}^n 2^{-4t_i} \end{pmatrix}=:\begin{pmatrix}a&b\\b&c \end{pmatrix}, $$
note that $$(A^TA)^{-1} = \frac{1}{ac - b^2}\begin{pmatrix} c & -b \\ -b & a \end{pmatrix}.$$
Let $d$ be the first column of $A$ (unfortunate nomenclature...). Then $$d^TA = \begin{pmatrix} \sum_{i=0}^n2^{-2t_i} & \sum_{i=0}^n2^{-3t_i} \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix},$$ and $$d^TA(A^TA)^{-1} = \frac{1}{ac - b^2}\begin{pmatrix} ac - b^2 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \end{pmatrix}$$ -- et voila! Since now $$d^TA(A^TA)^{-1}A'f = d^Tf,$$
which had to be shown. Note that ($\ast$) can be expressed as $d^TA\hat x = d^Tf$, where $\hat x = (A^TA)^{-1}A^Tf$ is the LS estimate of $x$.


答:
I think it is correct. For me, it is better to write (although simple) why we have $k^2-2\geq 2k-1$.

I write the proof as follows. We need to prove that $2^{k+1} > (k+1)^2-2$.

Indeed, (as you said) from $2^k > k^2-2$, we have $2^{k+1} > 2k^2-4$. We now prove that $2k^2-4\geq(k+1)^2-2$. After expanding and reducing, it leads to prove that $(k+1)(k-3)\geq 0$. This is correct for $k\geq 3$.

I present this way because I do not want to surprise the reader why we need statement $k^2-2\geq 2k-1$. I mean your statement "Because ..." is not natural in a mathematical proof.

Title: Proof by induction, $2^{n} > n^{2} - 2$

Body:
So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.

I start of with the "basecase" for n = 3:

RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$

Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$.
To prove this inequality, we have to proof that it holds for n = k + 1.

$2^{k} > k^{2} - 2$

We multiply both sides with 2:

$2^{k+1} > 2(k^{2} - 2)$

$2^{k+1} > 2k^{2} - 4$

$2^{k+1} > k^{2} + k^{2} - 2 - 2$

According to our assumption,  $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $

Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:

$2^{k+1} > k^{2} + 2k + 1 - 2$

Thus:

$2^{k+1} > (k+1)^{2} - 2$

QED.

How does my proof look, does it have any deficits, and how would you prove it?

Thank you!


问:
Limits And Asymptotes Question
If $f$ has an asymptote as $x \to \infty$, then $\lim_{x\to\infty} \frac{f(x)}{x}$ exists,
is true.

If $\lim_{x\to\infty} \frac{f(x)}{x}$ exists, then $f$ has an asymptote as $x\to\infty$,
is false.

Asymptote means that there exists some $m$ and $b$ as $x\to\infty$, $[f(x) − (mx + b)] = 0$.
Why is claim 1 true and claim 2 false?

答:
If $ f $ has the line $ y=mx+n $ as an asymptote at $ +\infty $, then

$$\lim_{x\to+\infty}(f(x)-(mx+b))=0$$

and
$$\lim_{x\to+\infty}\frac{f(x)-(mx+b))}{x}=0$$

which gives

$$\lim_{x\to+\infty}(\frac{f(x)}{x}-m)=0$$
or
$$\lim_{x\to+\infty}\frac{f(x)}{x}=m$$

But,
$$\lim_{x\to+\infty}\frac{\sin(x)}{x}=0$$
and
$x\mapsto \sin(x) $ has NO asymptote at $ +\infty$.

Title: Limits And Asymptotes Question


问:
How to deduce the Vieta's substitution $y = z-\frac {p}{3z}$ for cubic equation $y^3+py +q=0$
In deriving the formula of cubic equations, Vieta substituted the following
$$y = z-\frac {p}{3z}$$

for the depressed cubic equation

$$y^3+py +q=0$$

and transformed it into a quadratic one.

My question: How did he get that substitution, or how did he know that by substituting $y = z-\frac {p}{3z}$, he could turn cubic into quadratic?

Please help me!

答:
The quadratic equation in $u$ has the equivalent form below
$$u+ \frac a{u}+ b=0$$
Determine below the substitution that transforms the depressed cubic equation $y^3+p y+q=0$ into a quadratic one, i.e.

\begin{align}
y^3+p y +q &= u+\frac a{u}+ b\\
& = \left(u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13}}\right)\left( u^{\frac23 }+\frac {a^{\frac23}}{u^{\frac23}} -a^{\frac13}\right)+b\\
&= \left(u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13}}\right)
\left( \left(u^{\frac13 }+\frac {a^{\frac13}}{u^{\frac13}}\right)^2 - 3a^{\frac13}\right)+b\\
&= \left(u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13}}\right)^3
- 3a^{\frac13} \left(u^{\frac13 }+\frac {a^{\frac13}}{u^{\frac13}}\right)+b\\
\end{align}
Compare the two sides to get $b=q$, $a^{\frac13}=-\frac p3$  and $y = u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13} }$. Then, let $u(z)= z^3$ to obtain the Vieta’s substitution $y= z - \frac p{3z}$.


答:
$\lim_\infty f(x)/x$ only means the tangent line at a point takes a given direction as the point goes to $\infty$, whereas the existence of an asymptote adds that the curve gets closer and closer to a line  with this direction.

For instance, if $\lim_\infty\frac{f(x)}x=0$, the tangent line to the curve gets more and more horizontal. This is the case for, say $f(x)=\sqrt x$. If the curve has a horizontal asymptote, the function has a finite limit at $\infty.

I'll some words about the terminology: if $\lim_\infty\frac{f(x)}x=m$, but there's no asymptote, one says that the representative curve has a parabolic branch in the direction with slope $m$.

Title: Limits And Asymptotes Question

Body:
If $f$ has an asymptote as $x \to \infty$, then $\lim_{x\to\infty} \frac{f(x)}{x}$ exists,
is true.

If $\lim_{x\to\infty} \frac{f(x)}{x}$ exists, then $f$ has an asymptote as $x\to\infty$,
is false.

Asymptote means that there exists some $m$ and $b$ as $x\to\infty$, $[f(x) − (mx + b)] = 0$.
Why is claim 1 true and claim 2 false?


问:
A question on limsup
Let $a_n>0$. Prove that $$\varlimsup_{n\to\infty}n\left(\frac{1+a_{n+1}}{a_n}-1\right)\geq 1.$$

I argue by contradiction. If it is not ture, then $$\exists\ N,\ \forall\ n\geq N, n\left(\frac{1+a_{n+1}}{a_n}-1\right)<1\Rightarrow 1+a_{n+1}-a_n<\frac{a_n}{n}.$$
What contradiction shall I get then....

答:
Suppose to the contrary that for $n\ge N$,
$$
n\left(\frac{1+a_{n+1}}{a_n}-1\right)\lt1
$$
Then, for $n\ge N$,
$$
\frac{a_{n+1}}{n+1}\lt\frac{a_n}{n}-\frac1{n+1}
$$
It can be shown by induction that
$$
\frac{a_{n+k}}{n+k}\lt\frac{a_n}n-\sum_{j=1}^k\frac1{n+j}
$$
Since the harmonic series diverges, at some point $\dfrac{a_{n+k}}{n+k}<0$, contrary to the hypothesis that $a_n\gt0$.</p>

Title: A question on limsup


问:
According to what rigorous definition of "polynomial" is $\sum_{k=0}^n a_k D^k$ a polynomial, where $D$ is the derivative operator?
Let $Df$ be the derivative of $f\in C^{\infty}$.
\begin{align}
D^k\colon C^{\infty}&\to C^{\infty}\\
f&\mapsto f^{(k)}=\begin{cases}f&\text{ if }k=0\\
Df^{(k-1)}&\text{ if }1\leq k
\end{cases}
\end{align}
I was told that if $(a_0,\ldots,a_n)\in\mathbf{C}^{n+1}$, $\sum_{k=0}^n a_k D^k$ is a polynomial and can thus be factorized. That is, there is a $(b_1,\ldots,b_n)\in\mathbf{C}^n$ such that
\begin{equation}
\sum_{k=0}^n a_k D^k = a_n \prod_{i=1}^{n}(D^1 - b_iD^0).
\end{equation}

According to what rigorous definition of "polynomial" is that true and where can I find more information?

Unfortunately, I didn't find the Wikipedia article helpful and I'm not sure if the definition as sequences in some ring is what I'm looking for, as I have no background in abstract algebra.

答:
As you probably know from school, polynomial functions can be factorized and that result can be used to show that any polynomial (with complex coefficients) can be factorized.

To make this precise, let's introduce some notation: For all $n\in\mathbf{N}_0$, let's define
\begin{align}
\tau_n\colon\mathbf{C}&\to\mathbf{C}\\
x&\mapsto x^n=\begin{cases}1&\text{if }n=0\\
x^{n-1}\cdot x&\text{else}
\end{cases}
\end{align}
Then
\begin{equation}
V:=\left\{\sum_{n=0}^{\infty}a_n\tau_n:a_n=0\text{ for all almost all }n\right\}
\end{equation}
is the set of all polynomial functions (it's a vector space and $\{\tau_n:n\in\mathbf{N}_0\}$ is a basis).
Using the fundamental theorem of algebra and the division algorithm (Theorem 17.6 in [1] or theorem 3.46 in [2]), one can show that any polynomial function can be factorized - this means that there is $(b_1,\ldots,b_n)\in\mathbf{C}$ such that
\begin{equation}
\sum_{k=0}^n a_k\tau_k = a_n \prod_{i=1}^{n}\tau_1 - b_i\tau_0.
\end{equation}
(Here are more details.)

To prove
\begin{equation}
\sum_{k=0}^n a_k X^k = a_n \prod_{i=1}^{n}X^1 - b_iX^0,
\end{equation}
we can introduce the linear map
\begin{align}
\phi\colon V&\to\mathbf{C}[X]\\
\sum_{k=0}^{\infty}c_k\tau_k&\mapsto\sum_{k=0}^{\infty}c_kX^k
\end{align}
You can easily prove that for all $A,B\in V$, $\phi(A\cdot B)=\phi(A)\cdot\phi(B)$ and therefore
\begin{align}
\sum_{k=0}^n a_k X^k=\phi\left(\sum_{k=0}^n a_k\tau_k\right)=\phi\left(a_n \prod_{i=1}^{n}\tau_1-b_i\tau_0\right)\\=a_n\prod_{i=1}^{n}\phi\left(\tau_1 - b_i\tau_0\right)=a_n \prod_{i=1}^{n}X^1 - b_iX^0
\end{align}

One further comment: Let $R$ be a ring.
\begin{equation}
R[X]:=\left\{\sum_{n=0}^{\infty}a_nX^n:a_n=0\text{ for all almost all }n\right\}
\end{equation}
If nothing is known about $X$, we have to define
\begin{equation}\tag{1}
\sum_{n=0}^{\infty}a_nX^n+\sum_{n=0}^{\infty}b_nX^n=\sum_{n=0}^{\infty}(a_n+b_n)X^n
\end{equation}
and
\begin{equation}\tag{2}
\sum_{n=0}^{\infty}a_nX^n\cdot\sum_{n=0}^{\infty}b_nX^n=\sum_{n=0}^{\infty}\left(\sum_{k+l=n}a_kb_l\right)X^n
\end{equation}
to make $R[X]$ a ring. But if $A$ is an algebra and $X\in A$, you can define $X^0:=1\in A$ and $X^{n+1}:=X^n\cdot X$ and prove $(1)$ and $(2)$.

In the case of my question, $R=\mathbf{C}$, $A=L\left(C^{\infty},C^{\infty}\right)$ is the set of linear maps from $C^{\infty}$ to $C^{\infty}$, $X=D$ and $\sum_{k=0}^na_kD^k\in\mathbf{C}[D]$.

[1] Thomas W. Judson. Abstract Algebra. 2019.

[2] Joseph J. Rotman. A first course in abstract algebra. 3rd edition.


问:
Matrix for rotation about axis in $\mathbb{R^3}$
The following are the matrices that represent rotations anticlockwise about the $x$, $y$ and $z$ axis respectively:
$$\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\theta & -\sin\theta \\
0 & \sin\theta & \cos\theta
\end{pmatrix}$$

$$\begin{pmatrix}
\cos\theta & 0 & \sin\theta \\
0 & 1 & 0 \\
-\sin\theta & 0 & \cos\theta
\end{pmatrix} $$

$$\begin{pmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1
\end{pmatrix}
$$
My question is, when we say anticlockwise what exactly do we mean?

Do we mean anticlockwise when facing in the positive direction of the given axis or anticlockwise when facing in the negative direction of the given axis?

Thank you very much for your help.

答:
Assuming a “standard” right-handed coordinate system, it means if you stick the thumb of your right hand in the positive direction of the axis, the rotation is in the direction of the curl of your fingers.

Were we using the alternative left-handed system, the rotation would be in the other direction.

Edit: We say that the system is right-handed if, when your right thumb is pointing straight up in the positive z direction, and your right fingers are pointing in the positive x direction, your fingers curl in the direction of positive y. When you use your left hand to determine positive y, you get a left-handed system.


问:
How to prove that $DK=DL$

In $\triangle ABC$ the cevians $AD$,$BE$ and $CF$ are concurrent at the point $O$. $DL\parallel  EF$ and $BE$ intersects $DL$ at $K$. How to prove that $DK = DL$.

I applied menelaus theorem twice in the $\triangle KEL$ and got a relation $\frac{DL^2} {DK^2}=\frac{BE} {BK}\cdot \frac{CL} {CE} \cdot \frac{AL} {AE} \cdot \frac{EO} {KO}$ and also  tried to use the ratio of similar triangles but no result. any ideas  on utilizing the ratio of similar triangles.

答:
Following timon92, we are going to prove $\dfrac{DA}{AJ}=\dfrac{DO}{OJ}$.

Consider the figure (only add point $M$, which is an intersection of lines $EF$ and $BC$), we focus on $\triangle MDJ$, and apply Menelaus's theorem 4 times:
\begin{align}
\color{blue}{\frac{MB}{BD}}\color{red}{\frac{DA}{AJ}}\frac{JF}{FM}&=1;\tag{1}\\
\color{blue}{\frac{MB}{BD}}\color{orange}{\frac{DO}{OJ}}\color{green}{\frac{JE}{EM}}&=1;\tag{2}\\
\color{purple}{\frac{MC}{CD}}\color{orange}{\frac{DO}{OJ}}\frac{JF}{FM}&=1;\tag{3}\\
\color{purple}{\frac{MC}{CD}}\color{red}{\frac{DA}{AJ}}\color{green}{\frac{JE}{EM}}&=1.\tag{4}
\end{align}
Combining (1) and (4), and combining (2) and (3), respectively, we obtain
\begin{align}
\left(\color{red}{\frac{DA}{AJ}}\right)^2
=\left(\color{blue}{\frac{MB}{BD}}\frac{JF}{FM}\color{purple}{\frac{MC}{CD}}\color{green}{\frac{JE}{EM}}\right)^{-1}
=\left(\color{orange}{\frac{DO}{OJ}}\right)^2.
\end{align}

Title: How to prove that $DK=DL$


问:
An ${\rm Aut}(\mathbb{Z}/n\mathbb{Z}) \rightarrow(\mathbb{Z}/n\mathbb{Z})^\times$ construction
i would like to understand how to construct a morphism between 2 groups $${\rm Aut}(\mathbb{Z}/n\mathbb{Z}) \rightarrow (\mathbb{Z}/n\mathbb{Z})^\times$$ and prove that it is an isomorphism.

The only thing that I need is how to construct it.

I'm still not good  in manipulating Automorphisms so I have a difficulty. Then, to prove that it is an isomorphism I'm sure it's pretty easy to prove it's surjection and injection.

I would probably consider a generator of $(\mathbb{Z}/n\mathbb{Z})^\times$ and construct an image of $\phi([r]_n)$, where $\phi \in{\rm Aut}(\mathbb{Z}/n\mathbb{Z})$ but I'm not sure how to manipulate all of this.

Thanks in advance

答:
Hint: Consider the map $\Phi : (\Bbb Z/n \Bbb Z)^\times \to \mathrm{Aut}(\Bbb Z/n\Bbb Z)$ defined by
$$
[\Phi([r]_n)]([s]_n) = [rs]_n.
$$
I think that the trickiest part from there should be proving that $\Phi$ is surjective.

If you insist on constructing a map in the opposite direction, then you could consider the map $\phi \mapsto \phi(1)$, as is discussed in the comments to this answer.


问:
Decide if image or prototype of a connected component is a connected component in case of continuous map
Decide if image or prototype of a connected component is a connected component in case of continuous map

X->Y , Y is a component if X is.

答: The continuous image of a connected component of $X$ will be connected, but it need not be a component of $Y$. The continuous pre-image of a connected component of $Y$ need not even be connected. Let $X=\Bbb Z$ and $Y=\Bbb R$, both with the usual topology, and let $f:X\to Y$ be the constant function $f(n)=0$; this will give you both counterexamples.


答:
Hint:

Check that the map
\begin{align}
\mathbf R&\longrightarrow G \\
x &\longmapsto \begin{bmatrix}
    \cos x & -\sin x\\
    \sin x & \phantom{-}\cos x
  \end{bmatrix}
\end{align}
is a homomorphism from the additive group $(\mathbf R, +)$ onto the multiplicative  group $G$, and dtermine its kernel, to apply the $1^\text{st}$ isomorphism theorem.

Title: How to prove that a quotient group is isomorphic to a group of matrices

Body:
Consider the group $(\mathbb{R},+)$ and the subgroup

$$H=\{2n \pi |n\in \mathbb{Z}\}$$

Let $G$ denote the group of matrices of the form
\begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}
Where $\theta$ is any real number and the group operation is matrix multiplication. Prove that $\mathbb{R}/H$ is isomorphic to $G$ by first giving an explicit formula for a map $\phi:\mathbb{R}/H\to\ G$, and then checking that your map is well-formed, bijective, and respects the group operation.

I know that for the map to be well defined, the mapping has to be element to element, and that I need to ensure that its one-to-one and onto. But I am not making any progress with the proof nor able to give a formula for the map.
Can someone please help me out!


问:
If $\lambda$ is a characteristic root of a non-singular matrix $A$, then prove that $\det(A)/\lambda$ is a characteristic root of $\mathrm{Adj}(A)$.
Let $A$ be a $n\times n$ matrix. Then the characteristic equation is $\mathrm{det}(A-\lambda I)=0$, with possibly characteristic roots of $A$ $\lambda_1,\dots,\lambda_n$. Let $A$ be regular, i.e. $\det(A)\neq0$. Now $\mathrm{Adj}(A)=\det(A)\mathrm{Inv}(A)$. I don't know how to find its characteristic roots.

答:
$$Au= \lambda u \Rightarrow \mbox{Adj}(A)Au = \mbox{Adj}(A)\lambda u  \Rightarrow  \det(A) u = \mbox{Adj}(A)\lambda u \\
\Rightarrow  \frac{\det(A)}{\lambda} u = \mbox{Adj}(A) u$$

Alternately note that
$$\det(xI-\mbox{adj}(A))= \det(x\frac{1}{\det(A)}A \mbox{adj}(A)-\mbox{adj}(A))\\= \det(x\frac{1}{\det(A)}A -I) \det(\mbox{adj}(A))= (\frac{x}{\det(A)})^n \det(A - \frac{\det(A)}{x}I) \det(\mbox{adj}(A))\\= (\frac{1}{y})^n \det(A - yI) \det(\mbox{adj}(A))$$
where $y=\frac{\det(A)}{x}$.


问:
What natural integers satisfy the relation $\frac{3}{7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
I would like to find all natural triplets $(a, b, c)$ that satisfy the  relation
$\frac{3}{7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\,.$

答:
WLOG assume $a \ge b \ge c$ then $3 \le a \le 7$.

$\frac 1b + \frac 1c=\frac{2}{21}, \frac{5}{28}, \frac{8}{35}, \frac{11}{42}, \frac{2}{7}$.

Now $\frac 1b + \frac 1c = \frac mn \iff (mb-n)(mc-n)=n^2$

Can you take it from here?


问:
Strong induction for natural number object
From Lawvere's paper about ETCS we know that we can do induction for natural number object. My aim is to prove the strong induction principle for an NNO in a topos by translating this argument https://math.ou.edu/~nbrady/teaching/f14-2513/LeastPrinciple.pdf ((I) $\to$ (SI)) into an abstract version, so it works in any category satisfies ETCS. In particular, the axioms of ETCS make sure that we have a Boolean topos.

I think the statement of strong induction to prove in ETCS is as the following.

For a subobject $p: P \to N$, if for an arbitary element $n: 1 \to N$ of the NNO, "all member $n_0: 1\to N$ such that $- \circ \langle n_0, n\rangle = o$ factorises through $P$" implies "$s \circ n: 1 \to N \to N$ factors through P", then $P\cong N$ where $p$ is an isomorphism.

Here $N$ is the NNO, $s: N \to N$ is the successor map  the $\le$ relation on NNO is given as the pullback of $o: 1\to N$, as the element $0$ of natural number, along the map truncated subtraction $-: N\times N \to N$, as in Sketch of an elephant by Johnstone (page 114, A2.5).

I am not sure what to do with the $Q$ though, and I am not sure if constructing such a $Q$ need the comprehension axiom for ETCS (which is not in Lawvere's original paper). Hence I have trouble translating the proof in the link to also work for NNO. I think if we translate the proof, then we should reduce the task of $P\cong N$ into the task of proving $Q \cong N$. I am now asking about how to construct such a $Q$, which corresponds to the predicate $Q$ in the link. Any help, please?

If anything I said is wrong, or if there is a better approach, thanks a lot for pointing out!

答:
The construction you want definitely works in any elementary topos (with an NNO), even those which are not Boolean, and in fact it will also work in any arithmetic universe ("predicative toposes, toposes without power objects").

I'm in a hurry, there are surely more elegant ways to construct $Q$, but here is one:

Construct $List(P)$, the object of (finite) lists of values from $P$.

Construct the map $N \to List(N)$ thought as mapping $n$ to $[0,1,\ldots,n-1]$.

Intersect $List(P)$ with the image of that map (as subobjects of $List(N)$).

The image of the resulting object under the map $length : List(N) \to N$ is the desired object $Q$.


问:
Substitution of Complex Variable in integral
I am trying to evaluate the integral

$$\int_{0}^{\frac{\pi}{2}} \frac{\log(e^{ix}+e^{-ix})]-\log{2}}{e^{2ix}-1}e^{ix} \,dx$$

I am wondering if a complex variable substitution the likes of $iu=e^{ix}$, $\,du=e^{ix}\,dx$ is justifiable? In my mind. when $x=0, u=0$ and when $x=\pi/2, u=1$ which would make the integral
$$-\int_{0}^{1} \frac{\log(iu-i/u)-\log(2)}{1+u^2}\,du$$

but it is not numerically equivalent. Is it possible to introduce a complex variable substitution to solve the integral, and if so how do I go about it?

答:
With some rearranging the integral is equal to

$$\frac{1}{2i}\int_0^{\frac{\pi}{2}}\frac{\log\cos x}{\sin x}dx = \frac{1}{2i}\int_0^1\frac{\log t}{1-t^2}dt$$

by the substitution $t = \cos x$. Then expand the denominator as a geometric series

$$\sum_{n=0}^\infty \frac{1}{2i} \int_0^1 \log t \cdot t^{2n}dt=\sum_{n=0}^\infty \frac{-1}{2i(2n+1)^2} = \frac{i\pi^2}{16}$$

by integration by parts and then using the sum of the odd terms of the Basel problem.


问:
Why is there a right adjoint and a left adjoint?
In my category theory course, we saw the principle of adjunction and there is something that I really do not understand. Let me explain it with an example.

We were studying the hom-set relation, i.e. we define a map
$$\theta_{N, Z}: {}_A\text{Hom}_B(M\otimes_B N, Z) \to {}_B\text{Hom}_C(N, {}_A\text{Hom}(M,Z))$$
where $A, B, C$ are $k$-algebra (for $k$ a field) and $M,N,Z$ are modules in ${}_A\text{Mod}_B, {}_B\text{Mod}_C, {}_A\text{Mod}_C$ respectively. Explicitly, we get
$$\theta_{N,Z}(f)(n)(m) = f(m\otimes_B n)$$
for any $f \in {}_A\text{Hom}_B(M\otimes_B N, Z)$. This function is in fact a bijection and the inverse in given by
$$\theta_{N, Z}^{-1}(g)(m\otimes_Bn) = g(n)(m).$$
From this fact, my teacher deduce that the functor $M \otimes_B -$ is the left adjoint of the functor ${}_A\text{Hom}(M,-)$. But here I don't really understand why we make a difference between left and right because the choice of "the sens" of $\theta_{N,Z}$ seems arbitrary, we could very well have chosen $\tilde{\theta}_{N, Z} = \theta_{N, Z}^{-1}$ and now $M \otimes_B -$ becomes the right adjoint. What's wrong with my argument? Why could we not take $\tilde{\theta}_{N, Z} = \theta^{-1}_{N, Z}$ ?

答:
Be careful, the notion of "left" or "right" adjoint does not refer to the direction of the natural transformation, but rather where the functor is standing in the hom.

An adjunction between $F$ and $G$ is a natural isomorphism $\hom(F(X),Y)\cong \hom(X,G(Y))$. No matter the direction of the natural transformation, $F$ is the left adjoint because it is on the left in the hom (in $\hom(a,b)$, $a$ is on the left, $b$ is on the right) and $G$ the right adjoint because it is on the right in the hom.

So if you choose $\theta^{-1}$, $M\otimes_B-$ is still left adjoint.

Contrary to the direction of $\theta$, the position in the hom is not arbitrary and makes this notion slightly asymmetric. For instance left adjoints always preserve colimits, but not necessarily limits, whereas right adjoints always preserve limits but not necessarily colimits, so there is an actual distinction.

Note: it does happen that the tradition is often to write it as a natural isomorphism $\hom(F(X),Y)\overset\cong\to \hom(X,G(Y))$, so that's probably why you were confused, but  again, the direction $\to$ does not really matter.


问:
How do I find the coordinates of symmetric matrices after finding the eigenvalues, eigenvectors and eigenbases?
I've been playing around with Symmetric matrices and orthogonal bases of said Symmetric matrices, but I cannot figure out how to find the coordinates.

So, let's say that I have a quadratic function : $6x^{2}_1 + 4x_1x_2+3x^2_2$.

Well now I know that this forms  a matrix of $~~~A=~\begin{pmatrix}6 & 2 \\ 2 & 3\end{pmatrix}$.

So now I have to find the eigenvalues of this matrix which are $\lambda_1 = 7 , \lambda_2 = 2$ .

and now I have two Eigenbases of $E_7 = \mathrm{span}\left(\begin{pmatrix}-2\\-1\end{pmatrix}\right), E_2 = \mathrm{span}\left(\begin{pmatrix}~~~1\\-2\end{pmatrix}\right)$.

and the orthogonal bases of these two are just the lengths multiplied by the matrices, so

$$
\vec{w_1} = \frac{1}{\sqrt5}\begin{pmatrix}-2\\-1\end{pmatrix} \qquad \qquad \vec{w_2} = \frac{1}{\sqrt5}\begin{pmatrix}~~~1\\-2\end{pmatrix}
$$

But the problem I'm having is that I need an equation $q(\vec{w1} ~c_1 +\vec{w2}~c_2) = c_1^2+c_2^2$.

But how do I find these coordinates? I read my book and it wasn't clear on what I'm supposed to be doing to obtain them.

答:
The quadratic form is written

$$v^TMv$$ and after diagonalization

$$v^TP\Lambda P^Tv$$ or

$$(P^Tv)^T\Lambda (P^Tv).$$

So with

$$w:=P^Tv$$ the form reduces to

$$w^T\Lambda w.$$

Finally

$$\lambda_1w_1^2+\lambda_2w_2^2=(\sqrt{\lambda_1}w_1)^2+(\sqrt{\lambda_2}w_2)^2=c_1^2+c_2^2.$$


答:
It turns out that for every $a \in \mathbb Z/n$, there is a unique map $\mathbb Z/n \longrightarrow \mathbb Z/n$ sending $1 \mapsto a$. This is just multiplication by $a$, and it is unique as $1$ generates $\mathbb Z/n$. On the other hand, if $\phi: \mathbb Z/n \longrightarrow \mathbb Z/n$ then $\phi$ is totally determined by where $1$ is sent. Indeed, $\phi(m)=m \phi(1)$ so $\phi$ is just the multiplication by $\phi(1)$ map. Thus, we have the following map: $\Phi: End(\mathbb Z/n) \longrightarrow \mathbb Z/n$ via $\phi \mapsto \phi(1)$. If you are unfamiliar with this notation, $End(\mathbb Z/n)$ is the set of endomorphisms of $\mathbb Z/n$, i.e. group homomorphisms $\mathbb Z/n \longrightarrow \mathbb Z/n$.

This map $\Phi$, it turns out, will restrict to the desired map $Aut(\mathbb Z/n)\longrightarrow (\mathbb Z/n)^\times$, and this is a group isomorphism. I'd like to point out that in addition, $\Phi: End(\mathbb Z/n)\longrightarrow \mathbb Z/n$ is actually an isomorphism of rings, where multiplication in $End(\mathbb Z/n)$ is composition. The isomorphism you're looking for arises as the induced map on the unit groups of these rings.

Ring theory aside, here are some ideas to guide you to proving that $\Phi$ is your desired isomorphism.

As discussed, if $\Phi(\phi)=a$ then $\phi$ is the multiplication by $a$ map. If $\phi$ is an automorphism, why must $a$ be invertible in $\mathbb Z/n$?

You can try to prove that $\Phi$ is an isomorphism by constructing an inverse map. How can you take an element of $\mathbb Z/n$ and get a map $\mathbb Z/n \longrightarrow \mathbb Z/n$?

Title: An ${\rm Aut}(\mathbb{Z}/n\mathbb{Z}) \rightarrow(\mathbb{Z}/n\mathbb{Z})^\times$ construction

Body:
i would like to understand how to construct a morphism between 2 groups $${\rm Aut}(\mathbb{Z}/n\mathbb{Z}) \rightarrow (\mathbb{Z}/n\mathbb{Z})^\times$$ and prove that it is an isomorphism.

The only thing that I need is how to construct it.

I'm still not good  in manipulating Automorphisms so I have a difficulty. Then, to prove that it is an isomorphism I'm sure it's pretty easy to prove it's surjection and injection.

I would probably consider a generator of $(\mathbb{Z}/n\mathbb{Z})^\times$ and construct an image of $\phi([r]_n)$, where $\phi \in{\rm Aut}(\mathbb{Z}/n\mathbb{Z})$ but I'm not sure how to manipulate all of this.

Thanks in advance


答:
I think you are on the wrong track. I would look at
$$
\log(e^{ix}+e^{-ix})-\log2
=\log\left(\frac{e^{ix}+e^{-ix}}2\right)
=\log(\sin x)
$$
and
$$
\frac{e^{ix}}{e^{2ix}-1}
=\frac1{e^{ix}-e^{-ix}}
=\frac1{2\cos x}
$$
so your integral turns into
$$
\int_0^{\pi/2}\frac{\log(\sin x)}{2\cos x}\;dx.
$$
Can you work on this?

Title: Substitution of Complex Variable in integral

Body:
I am trying to evaluate the integral

$$\int_{0}^{\frac{\pi}{2}} \frac{\log(e^{ix}+e^{-ix})]-\log{2}}{e^{2ix}-1}e^{ix} \,dx$$

I am wondering if a complex variable substitution the likes of $iu=e^{ix}$, $\,du=e^{ix}\,dx$ is justifiable? In my mind. when $x=0, u=0$ and when $x=\pi/2, u=1$ which would make the integral
$$-\int_{0}^{1} \frac{\log(iu-i/u)-\log(2)}{1+u^2}\,du$$

but it is not numerically equivalent. Is it possible to introduce a complex variable substitution to solve the integral, and if so how do I go about it?


问:
Shortcut for value of $f(1)$ where $f(x) = \int e^x \left(\arctan x + \frac {2x}{(1+x^2)^2}\right)\,dx$
If $$f(x) = \int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx$$ and $f(0)=0$ then value of $f(1)$ is?

This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a shortcut to finding this result quickly? It seems very complicated. The answer is $e(\pi/4-(1/2)). $

答:
Use the result $$\int e^x\left(g(x)+g'(x)\right)dx=e^xg(x)+c$$ to prove that

$$\int e^x\left(g(x)-g''(x)\right)dx=e^x\left(g(x)-g'(x)\right)+c$$

Put $g(x)=\arctan x$ to obtain $g'(x)=\frac{1}{1+x^2}$ and then $g''(x)=-\frac{2x}{(1+x^2)^2}$

and we are done


问:
Start with PCA and multiple regression or Start with multiple regression and PCA
I would like to know something easy but very important.

Imagine I have a database with 0 NA, a perfect database who has been clean. And  I have to do a PCA on this database.  This datebase got a lot of individuals and variables ( 95 individuals and 10 variables)

I have to do a multiple regression and a PCA.

I must start per my multiple regression and eventually delete somme individuals who has been a Cook's distance > at the limit.  And after I do my PCA on " new data base"

OR  I must start per my PCA on my complete database,  and after I do my multiple regression.

In conclusion, I must do :

```<code>- PCA
- multiple Regression
</code>```

or

```<code>-multiple Regression
-PCA
</code>```

Ty for helping me !

答: Regression should be the final step, not the first one. By using PCA you can reduce dimension (i.e., number of explanatory variables) by discarding "unimportant" (that is, with small variance) variables. You can use PCA to perform whitening, i.e., eliminating autocorrelation or heteroscedsticity (inhomogeneous variance) in your data (or future model's residuals).  Note that if you are interested in point prediction or R square, regressing on the original features yield the same results as on the principal components. Namely, you should use the PCA for further reduction and tiding of your data (if possible), and not just for the sake of doing PCA itself.</p>


问:
What K value minimizes this function
I got a function $$ nK^{\frac{-1}{\beta}}\log{K} + ZK\log{n} $$ where $K$ and $n$ are integers and $Z > 0$.

I followed the typical way of differentiating the function and equating to zero.  But I couldn't find a closed form.  I got the following result

$$
\frac{Z\log{n}}{n} = \frac {\frac{\log{K}}{\beta}-1}{K^{\frac{1+\beta}{\beta}}}
$$

I am not sure how can I go from here.

答: K and n are constant integers. $\beta$ is constant real. Being in form a + b Z as a linear relation.. minimum when Z=0.

Title: What K value minimizes this function


问:
A mixed strategy Nash equilibrium uses all possible states
This is a problem I've been discussing with a friend just for fun. Neither of us has formally studied game theory so it may be that this is a well-known fact and we've just been lacking the appropriate terminology to successfully search for the answer.

Consider a two-player, zero-sum, symmetric game, represented by a finite matrix $R$, where $r_{ij}$ is player 1's reward when playing strategy $i$ against player 2's strategy $j$ (so the assumptions imply $r_{ij}=-r_{ji}$).
Suppose the game is in reduced form, so that no row of $R$ is equal to, or dominated by, any convex combination of other rows.

Conjecture: any Nash equilibrium for this game involves players using a mixed strategy with strictly positive probability for every state.

There seem to be two directions to approach a proof of the claim but we haven't made progress with either:

If you claim to have a Nash strategy which doesn't use $i$, perhaps I can show that adding a little bit of state $i$ to your strategy (and removing something else) yields a better strategy. Formally, this claim is more or less that there exists a convex combination of other pure strategies which is dominated by strategy $i$. This seems plausible because $i$ is not itself dominated by any convex combination of the other pure strategies, but is not automatic from the definition.

If I know you won't be using state $i$, I should be able to choose a strategy which highly weights those states $j$ which lose to $i$, since I don't need to worry about that outcome. If my strategy then beats yours, that shows you can't be playing the Nash strategy. However, you may have options available for beating each state which loses to $i$ even without using $i$ itself (e.g. in a game with 5 strategies, where each strategy wins vs two and loses vs two, forbidding yourself one option still leaves you able to beat anything I choose).

It feels like induction on the number of strategies is a natural approach, but inductive arguments are hurt by the fact that even though $R$ is in reduced form, removing a row needn't leave it in reduced form (e.g. in rock paper scissors, removing rock leaves scissors as the dominant strategy).

Proofs, counterexamples, or references related to the conjecture are welcome!

答:
The name used for a mixed equilibrium strategy which involves every state is a "completely mixed" strategy, so the key terminology here is "completely mixed (Nash) equilibria".

A precise condition under which the exists a completely mixed Nash equilibrium is given in the paper "Completely mixed equilibria in bimatrix games", Igal Milchtaich and Tadeusz Ostrowski (see equation 8; the paper is written for games which are not necessarily symmetric, so the condition on the matrix B can be ignored, with $R$ in the notation from the question here substituted for $A$ in the paper).

However, it's not clear to me whether $R$ being in reduced form implies the condition of equation 8 holds.


问:
Curve adaptation with a finite number of measured values
I consider a function $f(t)=\alpha\cdot 2^{-t}+\beta\cdot 2^{-2t}$ and a finite number of measured values $f_0,...,f_n$. The coefficients $ \alpha, \beta $ are calcualted by the least squares method which means
$$ \sum\limits_{i=0}^n |f(t_i)-f_i|^2$$
is minimized.

Then $$\sum\limits_{i=0}^n 2^{-t_i}\cdot f(t_i)=\sum\limits_{i=0}^n 2^{-t_i}\cdot f_i.\quad (*)$$

My idea:

I consider the expressions

$$ \begin{aligned}f_0&\stackrel{!}{=}f(t_0)=2^{-t_0}+\beta\cdot 2^{-2t_0}\\f_1&\stackrel{!}{=}f(t_1)=2^{-t_1}+\beta\cdot 2^{-2t_1}\\\vdots\\f_n&\stackrel{!}{=}f(t_n)=2^{-t_n}+\beta\cdot 2^{-2t_n}\end{aligned}$$

which can be reprensented by $$ \underbrace{\begin{pmatrix}2^{-t_0}&2^{-2t_0} \\2^{-t_1}&2^{-2t_1} \\\vdots&\vdots\\2^{-t_n}&2^{-2t_n} \end{pmatrix}}_{=:A}\cdot \underbrace{\begin{pmatrix}\alpha\\\beta \end{pmatrix}}_{=:x}=\underbrace{\begin{pmatrix}f_0\\f_1\\\vdots\\f_n \end{pmatrix}}_{=:f} $$

Then I want to minimize

$$ F((\alpha,\beta)):=\|A\cdot x-f\|_2^2=\sum\limits_{i=0}^n |f(t_i)-f_i|^2=\sum\limits_{i=0}^n \Big(\alpha\cdot 2^{-t_i}+\beta\cdot 2^{-2t_i}-f_i \Big)^2 $$

I calculate the gradiant of $ F $:

$$ \nabla F((\alpha,\beta))= 2\cdot A^T\cdot (A\cdot x-f)\stackrel{!}{=}0$$

which means I have to solve $$ A^T\cdot A\cdot x=A^T\cdot f $$

At first I have
$$ A^T\cdot A=\begin{pmatrix}\sum\limits_{i=0}^n 2^{-4t_i}&\sum\limits_{i=0}^n 2^{-3t_i}\\\sum\limits_{i=0}^n 2^{-3t_i}&\sum\limits_{i=0}^n 2^{-2t_i} \end{pmatrix}=:\begin{pmatrix}a&b\\b&c \end{pmatrix} $$
and

$$ A^T\cdot f=\begin{pmatrix}\sum\limits_{i=0}^nf_i\cdot 2^{-2t_i}\\\sum\limits_{i=0}^nf_i\cdot 2^{-t_i} \end{pmatrix}=:\begin{pmatrix}b_1\\b_2 \end{pmatrix} $$

The solution is $$ \alpha=\frac{a\cdot b_2-b\cdot b_1}{a\cdot c-b^2}\qquad \beta=\frac{c\cdot b_1-b\cdot b_2}{a\cdot c-b^2} $$

And here comes the problem. These expressions of the solution are soo complicated that I'm not able to handle them to show the upper identity (*).

Is there annother way to show or what could I do else(*)?

答:
$$\min_{\alpha,\beta}\sum_{i=0}^n(f(t_i)-f_i)^2=\min_{\alpha,\beta}\sum_{i=0}^n(\alpha\,2^{-t_i}+\beta\,2^{-2t_i}-f_i)^2$$ is minimized by

$$\begin{cases}\displaystyle\sum_{i=0}^n2^{-t_i}(f(t_i)-f_i)=0,\\
\displaystyle\sum_{i=0}^n2^{-2t_i}(f(t_i)-f_i)=0.\end{cases}
$$

The first equation is the relation you are after.


问:
proof related to Fundamental Theorem of Calculus $f$ is not necessarily continuous on an interval
I want to solve the problems from the book "A Course of Modern Analysis" (P 65) by E.T. Whittaker & G.N. Watson. The notation in this book is a little old-fashioned and I don't really like their notation.

If $f(x)$ is integrable when $a_1 \le x \le b_1$, when $a_1 \le a < b < b_1,$ we write
$$
\int_{a}^bf(x)dx = \phi(a,b),
$$
and if $f(b+0)$ exists, then
$$
\lim_{\delta \to 0^+} \frac{\phi(a, b+\delta)-\phi(a, b)}{\delta} = f(b+0).
$$

Here is my attempt for the problem:

\begin{align*}
\lim_{\delta \to 0^+} \frac{\phi(a, b+\delta)-\phi(a, b)}{\delta}  &= \lim_{\delta \to 0^+} \frac{\int_{a}^{b+ \delta}f(x)dx -\int_{a}^bf(x)dx }{\delta} \\
&=  \lim_{\delta \to 0^+} \frac{\int_{b}^{b + \delta}f(x)dx}{\delta}.
\end{align*}
Then I got stuck here.

答:
\begin{align*}
& \lim_{\delta \to 0^+} \frac{\phi(a, b+\delta)-\phi(a, b)}{\delta} \\
&= \lim_{\delta \to 0^+} \frac{\int_{a}^{b+ \delta}f(x)dx -\int_{a}^bf(x)dx }{\delta} \\
&= \lim_{\delta \to 0^+} \frac{\int_{b}^{b + \delta}f(x)dx}{\delta} \\
&= \lim_{\delta \to 0^+} \frac{\lim\limits_{n\to+\infty}\sum\limits_{i=1}^{n} f(b+i\delta/n) \, \delta/n}{\delta} \tag{$f$ is integrable on $[a,b_1]$} \\
&= \lim_{\delta \to 0^+} \lim_{n\to+\infty} \frac1n \sum_{i=1}^{n} f(b+i\delta/n) \\
&= f(b+) \tag{claim to be proved}
\end{align*}

To prove the claim, let $\epsilon > 0$.  Using the existence of $f(b+)$ to establish $\delta > 0$ such that $|f(x)-f(b+)| < \epsilon$ whenever $x \in (b,b+\delta)$.

\begin{align}
\left(\frac1n \sum_{i=1}^{n} f(b+i\delta/n)\right) - f(b+) &= \left[\frac1n\sum_{i=1}^{n-1} (f(b+i\delta/n)-f(b+))\right] + \frac{f(b+\delta)-f(b+)}{n} \\
\left \lvert \left(\frac1n \sum_{i=1}^{n} f(b+i\delta/n)\right) - f(b+) \right \rvert &< \frac{n-1}{n} \,\epsilon + \left \lvert \frac{f(b+\delta)-f(b+)}{n} \right \rvert \\
&< \epsilon + \left \lvert \frac{f(b+\delta)-f(b+)}{n} \right \rvert
\end{align}

Take $N > \left\lceil\dfrac{f(b+\delta) - f(b+)}{\epsilon}\right\rceil$.  Whenever $n \ge N$, then
$$\left \lvert \left(\frac1n \sum_{i=1}^{n} f(b+i\delta/n)\right) - f(b+) \right \rvert < 2 \epsilon.$$


问:
Finding Tangent Planes to Paraboloid Containing a Specified Line
Find all tangent planes to the paraboloid $z = 100 - x^2 - y^2$ that contain the line $[x,y,z] = [10,5,40] + t[3,-1,-4]$. </p>

<p>I have looked around and have not been able to find a clear response for solving this type of problem. </p>

答:
$z = 100 - x^2 - y^2$ that contain the line $[x,y,z] = [10,5,40] + t[3,-1,-4]$.

The parametric equations of the line are $r:(10+3t,5-t,40-4t)$.

The points $A(10,5,40)$ and $B(13,4,36)$ lie on $r$

The plane tangent to the paraboloid $z=f(x,y)=100-x^2-y^2$ in the point $(a,b)$ has equation
$$z-f(a,b)=\frac{\partial f}{\partial x}(a,b)(x-a)+\frac{\partial f}{\partial y}(a,b)(y-b)$$
$$\frac{\partial f}{\partial x}=-2x;\;\frac{\partial f}{\partial y}=-2y$$
is in this case we have
$$\pi:z-(100-a^2-b^2)=-2a(x-a)-2b(y-b)$$
Conditions

$A\in\pi$ and $B\in\pi$
$$
\begin{cases}
a^2+b^2-60=-2 (10-a) a-2 (5-b) b\\
a^2+b^2-64=-2 (13-a) a-2 (4-b) b \\
\end{cases}
$$
subtract the two equations
$ b=3a-2$ and substitute in the first
$$a^2+(3 a-2)^2-60=-2 (10-a) a-2 (7-3 a) (3 a-2)\to 5 a^2-31 a+42=0$$
Solutions are $a=2;\;b=4$ and $a=\frac{21}{5};\;b=\frac{11}{5}$

So we have two tangent planes
$$\pi_1:z=-4 x-8 y+120;\;\pi_2:z=-\frac{42 x}{5}-\frac{22 y}{5}+\frac{3062}{25}$$


问:
If all $f_n(x) \to 0$ as $|x| \to \infty$ and $f_n \to f$ pointwise, does $f \to 0$ as $|x| \to \infty$?
It sounds like it should be a standard result but I can't find a statement or proof of it. Does anyone have a reference or counterexample? Thanks.</p>

答:
I don't think this is true.
$$f_n(x) = \min\{1,\max\{0,n-x^2\}\}$$
should be a counterexample


问:
Serious applications of Colouring proofs
Are there any research-level applications of proofs by colouring?

This is the kind of proof you use to show that you can't cover a mutilated chessboard with 31 dominoes. Afaik, this technique chiefly finds a market in IMO preparation classes; see  Arthur Engel's book or  this handout .

答: Conway's Soldiers is a proof-by-coloring with an infinity of colors.</p>


问:
Formulas for calculating pythagorean triples
I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is:

$$\begin{align}
&a = m^2-n^2 \\
&b = 2mn\\
&c = m^2+n^2
\end{align}$$

With numerous parameters.

So are there other formulas/methods?

答:
The one I was taught in my lectures was:

Assume that $(x, y, z)$ is a Pythagorean triple in which $x$ is odd, so that $y$ is even and $z$ is odd. For similar reasons, we assume that $p$ and $q$ are coprime $(x,y,z,p,q \in \mathbb{Z})$. The theory of Pythagorean triples then tells us that there are nonzero integers $p$ and $q$ (with $p > q$)such that

$$x + iy = (p + iq)^2 \hspace{1.5cm} z = |p + iq|^2 = p^2 + q^2.$$

If $x$ is odd then one of $p$ and $q$ must be even and the other is odd.

In case it isn't obvious, the way it works is this: let's first check with the smallest three triples. Let's pick two small numbers satisfying the criteria for $p$ and $q$, i.e take $p = 2$ and $q = 1$ (one of them is odd, both are co-prime and $p > q$). First, to find our $z$ value, we simply do $p^2 + q^2 = 1^2 + 2^2 = 5$. Now, let's calculate our $x$ and $y$ values:

$$x + iy = (p + iq)^2 = (2 + i)(2 + i) = 4 + 4i - 1 = 3 + 4i.$$

From here, we see that $x = 3$ and $y = 4$, which we know to be true as $3^2 + 4^2 = 5^2$/

This might not give you ALL the Pythagorean triples but its another way of doing it, seeing as you only need to pick $p,q$ and the rest are worked out for you.

The coding I've ever done is Matlab so I can only really think how to put this in terms of Matlab (and even then it might be wrong), so I'm not sure if it will work for your program or whatever you're making, but here is another way anyway.


问:
$ \operatorname{rank} \left( A - A^{-1} \right) = n $ for every invertible $n \times n$ matrix $A$?
Given invertible matrix $A$ of size $n \times n$. Is it true that:

$$ \operatorname{rank} \left( A - A^{-1} \right) = n $$

if $A$ has no eigenvalue of magnitude 1, i.e., $|\lambda_i| \neq 1$ for all $i$. It's clear for diagonalizable matrices and it seems to hold for Jordan normal form as well.

答:
Yes, it is true. Note that
$$
\operatorname{rank}(A - A^{-1}) =
\operatorname{rank}(A^{-1}[A^2 - I]) =
\operatorname{rank}(A^2 - I).
$$
From there, it suffices to note that $A^2$ has full rank if and only if $A^2$ does not have $1$ as an eigenvalue, which holds if and only if $A$ has neither $1$ nor $-1$ as an eigenvalue. So, if $A$ does not have any eigenvalues of magnitude $1$, then $A^2 - I$ has full rank, which means that $A - A^{-1}$ has full rank.


答: Suppose that $A-A^{-1}$ is not invertible, there exists $x$ such that $(A-A^{-1})(x)=0$, this implies that $A(x)=A^{-1}(x)$ and $A^2(x)=x$. If $A(x)=cx$, $A^2(x)=c^2x=x$ implies that $|c|=1$. If $x,A(x)$ are independent,$A(x+A(x))=x+A(x)$. Contradiction.

Title: $ \operatorname{rank} \left( A - A^{-1} \right) = n $ for every invertible $n \times n$ matrix $A$?

Body:
Given invertible matrix $A$ of size $n \times n$. Is it true that:

$$ \operatorname{rank} \left( A - A^{-1} \right) = n $$

if $A$ has no eigenvalue of magnitude 1, i.e., $|\lambda_i| \neq 1$ for all $i$. It's clear for diagonalizable matrices and it seems to hold for Jordan normal form as well.


问:
Is $\cot x = \tan (π/2 - x) $ true for any angle $x$?
Is $$\cot x = \tan \Big(\frac{π}{2} - x\Big)$$ true even when $x$ is not an acute angle ?

答: It is an identity by definition.. yes, it is true for all arguments.</p>


答:
If $J$ is Jordan matrix for $A$ then $A= P^{-1}JP$ for some invertibile $P$.

Then $$A-I = P^{-1}(J-I)P$$ and $$A+I = P^{-1}(J+I)P$$

Let $X= A-A^{-1}$ then $$XA = A^2-I =  P^{-1}(J-I)(J+I)P$$

so $$\det X \det A = \det (J-I)\cdot \det (J+I) \ne 0$$ and thus $\det X \ne 0$ so the claim is true if all $\lambda _i \neq \pm 1$.


问:
How to derive formulas for sequences that converge to a specific value using simple functions
If I have some value $L\in \mathbb{R}$ such that a sequence $a_n$
$$\lim_{n\rightarrow\infty}{a_n}=L$$
If I state the properties of the sequence such as monotonically increasing or decreasing is it possible to find the terms in the sequence in the form
$a_{n+1}=A(a_n)$ where $A$ is a function that requires only simple functions, where simple means exactly calculable over a finite proccess. We can express this as a guess
$$|L-x_n|<\epsilon$$
And we want to rearrange it to be
$$|L-A(x_n)|<B(\epsilon)$$
And $A(x_n)$ is a better approximation if $B(\epsilon) < \epsilon$. However, we also want $A(x_n) > x_n$ since we want it to be monotonically increasing.
For example, if I want to find
$$\log_2(b)$$
We can first express it as
$$|\log_2(b)-x|<\epsilon$$
But how can I express it as $$|L-A(x_n)|<B(\epsilon)$$ where B is composed of only simple functions (i.e. exactly calcuable) and using the property that $A(x_n) > x_n$

答:
I'm sorry that I don't have a complete answer for you, but I'll attempt to help in two ways: (i) by rephrasing the question and (ii) by providing a widget that you can use to experiment with various choices of $A$ and see if inspiration strikes you.

(i) Here is my interpretation of the problem:

You want fixed point iterative method (i.e., $x_{n+1}=A(x_n)$), subject to the additional constraint that the sequence is increasing (i.e., $A(x_n) \geq x_n$). Essentially, you want a fixed point iterative method that starts with a guess $x_0 < L$ and then never overshoots.

This seems like a tall order. There exist numerous iterative methods, but I don't know of any offhand that will give the monotonicity property you want.
Bisection method might give you monotonicity, but isn't of the form you asked for.

For instance, to solve the equation $f(x)=0$, Newton's method asks us to iterate
$$
    x_{n+1} = A(x_n)
$$
where
$$
    A(x) = x - \frac{f(x)}{f'(x)}.
$$
We want to solve $x = \log_2(b)$.
Squaring both sides, this is equivalent to $2^x = b$, so $2^x-b = 0$
and thus we can try $f(x)=2^x-b$ in Newton's method.
So,
$$
     A(x)
=
    x - \frac{f(x)}{f'(x)}
=
     x - \frac{2^x-b}{\ln(2)2^x}
$$
However, as one can check (I did, it took more time than I'd like to admit), this function $A$ has the property that if $y \leq \log_2(b)$ then $y \leq \log_2(b) \leq A(y)$.
So, approximation from below cannot be achieved with this $A$ (note: if you want your sequence $(x_n)$ to be monotonically increasing and converge to $\log_2(b)$, then it is a necessary condition that $x_n \leq \log_2(b)$ for every $n$).

(ii) You (or anyone) can use the following R code to test various A

```<code>#
# you want to solve for the true value of a function "true" at a point "b",
# we you want to do this by iterating a function # A
#

true <- function(b) log(b,2)                  # the true value to test against
A <- function(x,b) x - (2^x-b)/(log(2)*2^x)   # Newton's method for computing log_2(b)

Iterate <- function( f, b=3, N=8, initial=NULL ) {
    if (is.null(initial)) initial <- 0.3     # small enough to be \leq \log_2(b)
    n <- 1:N
    a <- rep( initial, N )
    for (j in 2:N) { a[j] <- f(a[j-1],b) }
    plot( n, a, las=1 )
    abline( h=true(b), lty=3 )
    a
}

Iterate(A)
#
</code>```

Edit: to run that code online, you might be able to use https://rweb.webapps.cla.umn.edu/Rweb/Rweb.general.html the author of whom once taught a class I took at the U of M.

Sorry for not answering your question completely, but I hope this was at least helpful.

Body:
I'm sorry that I don't have a complete answer for you, but I'll attempt to help in two ways: (i) by rephrasing the question and (ii) by providing a widget that you can use to experiment with various choices of $A$ and see if inspiration strikes you.

(i) Here is my interpretation of the problem:

You want fixed point iterative method (i.e., $x_{n+1}=A(x_n)$), subject to the additional constraint that the sequence is increasing (i.e., $A(x_n) \geq x_n$). Essentially, you want a fixed point iterative method that starts with a guess $x_0 < L$ and then never overshoots.

This seems like a tall order. There exist numerous iterative methods, but I don't know of any offhand that will give the monotonicity property you want.
Bisection method might give you monotonicity, but isn't of the form you asked for.

For instance, to solve the equation $f(x)=0$, Newton's method asks us to iterate
$$
    x_{n+1} = A(x_n)
$$
where
$$
    A(x) = x - \frac{f(x)}{f'(x)}.
$$
We want to solve $x = \log_2(b)$.
Squaring both sides, this is equivalent to $2^x = b$, so $2^x-b = 0$
and thus we can try $f(x)=2^x-b$ in Newton's method.
So,
$$
     A(x)
=
    x - \frac{f(x)}{f'(x)}
=
     x - \frac{2^x-b}{\ln(2)2^x}
$$
However, as one can check (I did, it took more time than I'd like to admit), this function $A$ has the property that if $y \leq \log_2(b)$ then $y \leq \log_2(b) \leq A(y)$.
So, approximation from below cannot be achieved with this $A$ (note: if you want your sequence $(x_n)$ to be monotonically increasing and converge to $\log_2(b)$, then it is a necessary condition that $x_n \leq \log_2(b)$ for every $n$).

(ii) You (or anyone) can use the following R code to test various A

#
# you want to solve for the true value of a function "true" at a point "b",
# we you want to do this by iterating a function # A
#

true <- function(b) log(b,2)                  # the true value to test against
A <- function(x,b) x - (2^x-b)/(log(2)*2^x)   # Newton's method for computing log_2(b)

Iterate <- function( f, b=3, N=8, initial=NULL ) {
    if (is.null(initial)) initial <- 0.3     # small enough to be \leq \log_2(b)
    n <- 1:N
    a <- rep( initial, N )
    for (j in 2:N) { a[j] <- f(a[j-1],b) }
    plot( n, a, las=1 )
    abline( h=true(b), lty=3 )
    a
}

Iterate(A)
#

Edit: to run that code online, you might be able to use https://rweb.webapps.cla.umn.edu/Rweb/Rweb.general.html the author of whom once taught a class I took at the U of M.

Sorry for not answering your question completely, but I hope this was at least helpful.


问:
If $x \leq g(x) \leq x^2-x+1$ where $x \in [0,2]$, can we say that $g(x)$ is continuous at $x=1$?
If $x \leq g(x) \leq x^2-x+1$ where $x \in [0,2]$, can we say that $g(x)$ continuous at $x=1$ ?

Is $g(x)$ continuous in $[0,2]$?

答: Hint: For the first part, you know that your function can't be "further away" from 1 than either of the bounding functions are. For the second part, you should be able to just draw a counterexample. Start by sketching the two bounding functions.


答:
$f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$.

Take this as a particular example of the class of polynomials $$f_d(x)=x^d+2x^{d-1}+3x^{d-2}+\cdots+dx+(d+1)$$ where $d$ is the degree. Here we have the values of the $d$ elementary polynomials in the roots $x_1,x_2,\cdots, x_d$ (Vieta's coefficients) which are $e_k=e_k(x_1,x_2,\cdots, x_d)=(-1)^k(k+1)$ and we have five answers to choose the correct:
$$a)\space -(d+1)\hspace1cm b)\space -d\hspace1cm c)\space -(d-1)\hspace1cm d)\space -(d-2)\hspace1cm e)\space -(d-3)$$
The problem being to calculate $$\sum_{n=0}^{d-1}(x_1^n+x_2^n+\cdots+x_d^n)$$

We use Newton's identities to solve the problem for general $d$ degree.

Noting $P_i=x_1^i+x_2^i+\cdots+x_d^i$, Newton's identities give for the five first values
$$\begin{cases}P_1=e_1(x_1,x_2,\cdots, x_d)=e_1\\P_2=e_1^2-2e_2\\P_3=e_1^3-3e_1e_2+3e_3\\P_4=e_1^4-4e_1^2e_2+4e_1e_3+2e_2^2-4e_4\\P_5=e_1^5-5e_2e_1^3+5e_3e_1^2+5e_2^2e_1-5e_4e_1-5e_3e_2+5e_5\end{cases}$$ Calculation gives now a constant value for each $P_i$
$$P_i=-2\text{ for } 1\le i\le 5$$ We assume as a consequence and without proof that $P_i=-2$ for all $i$ which we have also verified for $P_6$ but it is too long to write.
Thus $$\sum_{n=0}^{d-1}(x_1^n+x_2^n+\cdots+x_d^n)= (d)+\underbrace{(-2)+(-2)+\cdots+(-2)}_{(d-1)\space\text{ times}}=d-2(d-1)=-(d-2)$$ Thus in this particular problem the answer is $\boxed{-2013}$ (the option d).

Body:
(From OMITS Indonesian $2015$)

Known that $f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$, if $x_1, x_2, x_3,..., x_{2015}$ is the roots of $f(x)$, Find :

$\sum_{n=0}^{2014}$ $\sum_{k=1}^{2015}$ $(x_k)^n$

$a. -2016$

$b. -2015$

$c. -2014$

$d. -2013$

$e. -2012$

What should i do? and where i can learn more about it?


问:
What is a real world application of polynomial factoring?
The wife and I are sitting here on a Saturday night doing some algebra homework.  We're factoring polynomials and had the same thought at the same time: when will we use this?

I feel a bit silly because it always bugged me when people asked that in grade school.  However, we're both working professionals (I'm a programmer, she's a photographer) and I can't recall ever considering polynomial factoring as a solution to the problem I was solving.

Are there real world applications where factoring polynomials leads to  solutions? or is it a stepping-stone math that will open my mind to more elaborate solutions that I actually will use?

Thanks for taking the time!

答:
Ever since middle school, I used to wonder about 'real-life' applications of algebra in general; except using them for groceries.

My professor once answered the question brilliantly. Ever since, the constant bugging of this question has faded massively.

"When imaginary numbers were acknowledged, no one ever thought that complex numbers would be fundamental in electricity!"

Furthermore, I could argue that when graph theory was discovered, no one every thought that it would have strides in social sciences and criminology. Examples abound.

As mathematicians, we accumulate our knowledge in our field to draw a logical and cohesive conclusion/result that might prove to be extremely useful to humanity. :)


问:
Need to prove or disprove an alternative group definition.
Let be the non empty set $G$ with the operation $*$ satisfy the following 3 conditions:

$a(bc) = (ab)c$, for all $a, b, c \in G$ (associative law).

For every $a,b$ there is $c$ such that $ac = b$. ($c$  is the "path" from $a$ to $b$).

For every $a,b$ there is $d$ such that $bd = a$. ($d$ is the "path back" from $b$ to $a$).

Prove or disprove: $G,*$  is a group.

Asociativity:

Proof:

from 1)

Identity element:

Proof:

According to 2) For every $a,a$ there is $e$ such that $ae = a$.

Let proof $ea=a$.

According to 2) For every $e,e$ there is $X$ such that $eX= e$.

Let's check the two possibilites:

$X=a,$

$X\neq a$

If 1) then $ea=a$ proved.

If 2): $eX=e$ then what?

Every try failed. Someone can do it?

答:
If the third condition is written as

For every $a,b \in G$ there is $d \in G$ such that $db=a$.

then the answer is that $G$ is a group and the existence of the identity element can shown as follows:

Fix $a \in G$. Then there exists $x_a$ and $y_a$ in $G$ such that $ax_a=a=y_aa$. We will prove that $gx_a=g=y_ag$ for any other $g \in G$. Indeed, if $g \in G$, since we know that there exists $x,y \in G$ with $ax=g=ya$, then $$gx_a = (ya)x_a = y(ax_a) = ya = g$$ and similarly $y_ag = g$. In particular $x_a=y_a$ (why?) and then $e := x_a (=y_a)$ is the identity element.


问:
Compact operator $T_{A,B}$
Let $A,B \in \mathcal{B}(X)$ where $X$ is a Banach space. Define $T_{A,B}: \mathcal{B}(X) \rightarrow \mathcal{B}(X)$ by $T_{A,B}(S)=ASB$.

Prove that if $A$ and $B$ are compact then $T_{A,B}$ is compact.

I consider a bounded sequence $\{S_n\}$ in $\mathcal{B}(X)$ but I can't find a subsequence of $T_{A,B}(S_n)$ such that is convergent. Any hint?

答:
In the special case of Hilbert spaces, there is a much simpler proof of this result based on the fact that every compact
operator is the limit of finite rank ones (this in fact characterizes the Approximation Property for
Banach spaces).

Here is the proof:

If  $y\in X$  and $f\in  X'$,  consider the operator
$U_{f,y}\in \mathcal B(X)$
given by
$$
  U_{f,y}(x) = f(x)y, \quad \forall x\in  X.
  $$
Evidently $U_{f,y}$ has rank-one and it is easy to see that every rank-one operator is of this form.

In case $A=U_{f,y}$ and $B= U_{g,z}$ then
$$
  T_{AB}(S)(x) = ASBx =  $$$$ =
  f(SBx)y = f(S(g(x)z))y = $$$$ =
  f(Sz)g(x)y = f(Sz)U_{g, y}(x),
  $$
which is to say that   $T_{AB}(S) = f(Sz)U_{g, y}$.  So  the range of $T_{A,B}$ is the one-dimensional space
spanned by $U_{g, y}$, whence $T_{A,B}$ has rank-one.

If  $A$  and $B$ have finite rank, then both $A$ and $B$ may be writen as finite sums of rank-one operators, from where
it follows that $T_{A,B}$ has finite rank (actually this proves that $\text{rank}(T_{A, B})\leq  \text{rank}(A)\text{rank}(B)$).

Assuming that $X$ has the Approximation Property,  such as if $X$ is a Hilbert space,  then the given compact operators
$A$ and $B$ may be written as
$$
  A=\lim_{n\to\infty }A_n,  \quad \text{and}\quad   B=\lim_{n\to\infty }B_n,
  $$
where $A_n$ and $B_n$ have finite rank,  so it easily follows
that $T_{A, B}$ is the operator-norm limit of $T_{A_n, B_n}$,  hence proving that $T_{A, B}$ is compact.

Title: Compact operator $T_{A,B}$


问:
For every prime of the form $2^{4n}+1$, 7 is a primitive root.
What I want to show is the following statement.

For every prime of the form $2^{4n}+1$, 7 is a primitive root.

What I get is that

$$7^{2^{k}}\equiv1\pmod{p}$$
$$7^{2^{k-1}}\equiv-1\equiv2^{4n}\pmod{p}$$
$$7^{2^{k-2}}\equiv(2^{n})^2\pmod{p}$$
Thus $(\frac{2^n}{p})=(\frac{7^{2^{k-2}}}{p})=1$.

I think that $7$ is important because $7$ is a primitive root but I don't know how to use $7$.

答:
Assume that $p=2^{4n}+1$ is a prime. We know that the order of $7$ is a factor of $p-1$, so it is a power of two. The claim is equivalent to saying that the order is exactly $p-1$. Assume that this is not the case. Then the order is a factor of $(p-1)/2$ meaning that
$$
\left(\frac7p\right)\equiv 7^{(p-1)/2}\equiv 1\pmod p.
$$
The case $n=0$ is easy, so we can assume that $n>0$. Then $p\equiv1\pmod4$, so the law of quadratic recpirocity tells that
the claim is equivalent to
$$
\left(\frac{p}7\right)=1.
$$
As $2^3\equiv1\pmod7$, $p\equiv 2^n+1\pmod7.$ The residue class of $2^n+1$ modulo $7$ can be either $2,3$ or $5$, when $n\equiv 0,1,2\pmod3$ respectively. Of these, only $2$ is a quadratic residue modulo $7$. This means that we must have $3\mid n$. I leave it to you to prove that in that case $p$ cannot be a prime unless $n=0$ and $p=2$.</p>


问:
interesting set of integrals
When integrating functions $\dfrac {\sin x}{\sin x}$, as well as $\dfrac {\sin3x}{\sin x}$ , $\dfrac {\sin 5x}{\sin x}$, $\dfrac {\sin 7x}{\sin x}$ etc. on interval from $0$ to $\dfrac {\pi}{2}$, the answer turns out to be $\dfrac {\pi}{2}$. How can this be proven?</p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi/2}{%
\sin\pars{\bracks{2n + 1}x} \over \sin\pars{x}}
\,\dd x\,\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq\ 0}}} \\[5mm] = &\
\Im\int_{0}^{\pi/2}{\expo{\ic\pars{2n + 1}x}\,\,\, -\ 1 \over \sin\pars{x}}\,\dd x
\\[5mm] = &\
\left.\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\,\,
{z^{2n + 1} -1 \over \pars{z - 1/z}/\pars{2\ic}}
\,{\dd z \over \ic z}\,\right\vert_{\,z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\,\,
{1 - z^{2n + 1} \over 1 - z^{2}}
\,\dd z\,\right\vert_{\,z\ =\ \exp\pars{\ic x}}
\end{align}
Hereafter, I'll "close" the integration around a quarter circle in the complex plane first quadrant: Thetre isn't any pole inside such contour.

Namely,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi/2}{%
\sin\pars{\bracks{2n + 1}x} \over \sin\pars{x}}
\,\dd x\,\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq\ 0}}}
\\[5mm] = &\
-2\,\Im\int_{1}^{0}\,\,
{1 - \ic^{2n + 1}\,\,y^{2n + 1}\,\, \over 1 - \ic^{2}\,y^{2}\,}
\,\ic\,\dd y\ -\
\pars{\substack{\mbox{An integral, along the}\ \ds{x}\mbox{-axis}
\\[1mm]
\mbox{segment}\ \ds{\pars{0,1}},\ \mbox{which}
\\[1mm] \ds{vanishes\ out}\ \mbox{because}
\\[1mm]\mbox{the integrand}\ \ds{\in \mathbb{R}.}}}
\\[5mm] = &\
2\int_{0}^{1}{\dd y \over 1 + y^{2}} = \bbx{\pi \over 2} \\ &
\end{align}

Title: interesting set of integrals


问:
Is there simple proof that no finite polynomial with rational coefficients can generate only primes?
There are many proofs, suitable for beginners and new students, that there is no polynomial that generates only primes - with the constraint that the polynomial has finite terms, is non-constant, and has positive integer coefficients.

However, searching the internet and this maths stack exchange - there does not appear to be a simple proof that loosens the requirements to positive rational coefficients.

The question has been discussed in comments to other questions, but no proof set out, or linked to.

I would appreciate a short simple proof, suitable for beginners (not trained at university maths).

Also can the proofs be extended to negative coefficients, integer or rational?

答:
A proof very easy was given by the great Alan Baker for all polynomial $f(x)$ with integer coefficients:
$$f(mf(n)+n)\text{ is divisible by $f(n)$ for all integer $m$ }.$$


问:
Expected number of pieces of a chessboard
If n squares are randomly removed from a $p \ \cdot \ q$ chessboard, what will be the expected number of pieces the chessboard is divided into?

Can anybody please provide how can I approach the problem? There are numerous cases and when I go through case consideration it becomes extremely complex.

答:
2 =         = 2 piece
3 = (3+1)/2 = 2 piece
4 =         = 2 piece
5 = (5+1)/2 = 3 piece
6           = 3 piece
7 = (7+1)/2 = 4 piece
8 =         = 4 piece

Series = (n+1)/2 pieces

Regards,
Yuvaraj


问:
Functoriality of the local Artin reciprocity
Let $k$ be a finite extension of $\mathbb{Q}_p$. Then the local Artin reciprocity tells us that
$$\text{Gal}(\bar{k}/k)^{\text{ab}}\cong \hat{k^\times}.$$
If we have a finite extension $\bar{k}/l/k$, then we have a natural inclusion $\text{Gal}(\bar{k}/l)^{\text{ab}}\to \text{Gal}(\bar{k}/k)^{\text{ab}}$. I would like to understand the corresponding map $\hat{l^\times}\to \hat{k^\times}$. For any extension $l'/l$ the map
$$\text{Gal}(l'/l)\cong l^\times/N_{l'/l}(l'^\times)\longrightarrow \text{Gal}(l'/k)\cong k^\times/N_{l'/k}(l'^\times)$$
is induced by the norm $N_{l/k}$. It would seem natural that the continuous map $\hat{l^\times}\to \hat{k^\times}$ is given by the completion of the norm $l^\times \to k^\times$. But that doesn't make sense, no? Because this map isn't injective, and the inclusion is. Clearly I'm missing something stupid...

答: There is a natural homomorphism $\text{Gal}(\bar{l}/l)^{\text{ab}}\to \text{Gal}(\bar{k}/k)^{\text{ab}}$ it is not injective, the natural guess is that it corresponds to $N_{l/k}: l^\times\to k^\times$


问:
Vector space span over real numbers
Let the set be $S=\{(a,b,c)∈\mathbb{C}^{3}|b=\bar a\}$ over $\mathbb{R}$.

I am asked to find the linear span on that set. My attempt was to define the following set member: $(w,v,z)∈S$, from here follows accoring to the definition of the set: $(w,\bar w,z)∈S$.

Now I don't know if it suits the definition of span, or there is some additional work to be done.

答:
Your answer is correct, but perhaps incomplete depending on the context you're working in. Certainly any $\mathbb{R}$-linear combination of vectors in $S$ would be of the form
$$
a_1(w_1,\overline{w_1},z_1) + ... + a_n(w_n,\overline{w_n},z_n) = (a_1w_1+...a_nw_n,\overline{a_1w_1+...a_nw_n},z_1+...+z_n) = (w,\overline{w},z)
$$
For some $a_1,...,a_n \in \mathbb{R}$, $w_1,...,w_n,z_1,...,z_n \in \mathbb{C}$.

So we have that
$$
\text{span}(S) = \left\{ (w,\overline{w},z)\mid w,z \in \mathbb{C}\right\}
$$
Which is as you say, although perhaps put a little more explicitly. In order to give a fuller description of the span of $S$ as a vector space over $R$, it might be helpful/more descriptive to also find a basis.

Title: Vector space span over real numbers


问:
Change of variable to show $\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$
Which change of variable can be used to show
$$\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$$
(direct integration isn't allowed, must use change of variable)

答:
To show the equivalent result $\int_1^\infty\frac{dx}{1+x^2}=\int_0^1\frac{dx}{1+x^2}$, use $x\mapsto1/x$.

Harder version, requiring no rewriting of the problem in an equivalent form: to show $\int_0^\infty\frac{dx}{1+x^2}=\int_0^1\frac{2dx}{1+x^2}$, use $x\mapsto\exp(-|\ln x|)$.


问:
Limit as $x$ approaches $1$ from the right of $\frac{1}{\ln x}-\frac{1}{x-1}$
$$
\lim_{x\rightarrow 1^+}\;\frac{1}{\ln x}-\frac{1}{x-1}
$$

So I would just like to know how to begin to solve this limit, or what topic does this problem fall under so that I can search for examples online or in text.

I know the answer is 1/2 and I have tried plugging in numbers  smaller or greater than one but as I assumed that was the wrong way to go.

Body:
If you need some fun, you may solve like this: Since

$$ \frac{1}{\log x} + \frac{1}{1-x} = \int_{0}^{1} \frac{1-x^{t}}{1-x} \, dt, $$

we have

$$ \lim_{x\to 1^+} \left( \frac{1}{\log x} + \frac{1}{1-x} \right) = \int_{0}^{1} \lim_{x\to 1^+} \frac{1-x^{t}}{1-x} \, dt = \int_{0}^{1} t \, dt = \frac{1}{2}. $$


Title: Change of variable to show $\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$

Body:
Which change of variable can be used to show
$$\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$$
(direct integration isn't allowed, must use change of variable)

Body:
Alternative

$$\tan(u) ~=~ x ~\implies~$$

$$\text{[informally]}
~\sec^2(u)du ~=~ dx ~~~\text{and}$$

$$~\frac{1}{1 + x^2} ~=~ \cos^2(u) ~\implies $$

$$\int_0^\infty \frac{dx}{1+x^2} = \int_0^{\pi/2} du
= 2\int_0^{\pi/4} du = 2\int_0^1 \frac{dx}{1+x^2}.$$


问:
Proving $6+12+18+24+...+6n=3n(n+1)$ by induction
I'm learning prove by induction, but for some reason I just can't figure out the following example. I'll just work out the example and tell you guys where I'm stuck.

prove the following by induction $$6+12+18+24+...+6n=3n(n+1)$$

prove basis step (n=1).
$$
6 = 6*1 = 3*1(1+1) = 6
$$

induction step, assume Sk is true, n = k
$$
6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)
$$

induction step n = k + 1

$$
6 + 12 + 18 + 24 + ... + 6(k+1) = 3(k+1)(k+2)
$$

bring Sk to both sides
$$
3k(k+1)+6(k+1) = 3(k+1)(k+2)
$$

Simplify, I really have no idea how these equal eachother, what are the simplifying steps taken in this example? Any help would be greatly appreciated!

答:
$6+12+18+24+\dots+6n=6(1+2+3+4+\dots+n)=6n(n+1)/2=3n(n+1)$.

The key is the well-known Gauß sum.  He did it as a school boy by pairing the first and last term, second and second to last, etc...


问:
The complete orthonormal basis on $L^2(-\infty,+\infty)$ about Hermite polynomials.
Prove that $\{\varphi_n(t)=(2^n n! \sqrt{\pi})^{-\frac{1}{2}}e^{-\frac{t^2}{2}}H_n(t)\}_{n=0}^{\infty}$ are the complete orthonormal basis on $L^2(-\infty,+\infty)$ where $H_n(t)$ are the Hermite polynomials.

I have aleady proved $\{\varphi_n(t)\}$ are orthonormal by the trick "Integration by parts". But I have difficulties on proving the completeness. I want to use the denseness of the continuous functions with a compact supported set in $L^2(-\infty,+\infty)$ and use polynomials to approximate those continuous functions, and the use the properties of integral $\int_{-\infty}^{+\infty} e^{-\frac{t^2}{2}}p(t)dt$ where $p(t)$ is a polynomial. But I faied.

I want to know if I am on the right track. And if it isn't correct, what's the right way to do it?

Any kind of help is appreciated and thank you very much in advance!

答:
There is a nice machinery for proving that polynomials are dense on certain weighted $L^2$-spaces, which involves some beautiful complex analysis tools. In this particular case, the proof is rather simple and follows from standard Hilbert space theory on $L^2(\mathbb{R})$.

Let $\mathcal{P}$ denote set of polynomials with complex coefficients. According to the comments, we came to agreement that it suffices to prove that
$$ S = \left\{e^{-t^2 /2}p(t): p \in \mathcal{P} \right\}.
$$
forms a dense subspace in $L^{2}(\mathbb{R})$. This is equivalent to proving that the orthogonal complement of $S$ in $L^2(\mathbb{R})$, denoted by $S^\perp$ only consists of the zero function, that is $S^\perp = \{0\}$. Let $f\in S^\perp$, that is
$$ \int_\mathbb{R} f(t)\,t^n \, e^{-t^2 /2}dt = 0 \qquad ,\forall n\geq 0.
$$
Consider the Fourier transform of $f(t)e^{-t^2 /2}$, defined by
$$
F(\xi) := \int_{\mathbb{R}}f(t) e^{-t^2 /2 } \, e^{-it\xi} dt \qquad, \, \xi \in \mathbb{R}.
$$
A simple series expansion (easily justified by the dominated convergence theorem, for instance) shows that
$$
F(\xi) = \sum_{n=0}^{\infty} \frac{(-i\xi)^n}{n!} \underbrace{\int_{\mathbb{R}}f(t)\,t^n\, e^{-t^2 /2} dt}_{=0} = 0 \qquad, \, \forall \xi \in \mathbb{R}.
$$
Since $\varphi(t)=f(t)e^{-t^2 /2} \in L^2$, it follows that by Parseval's theorem that
$$
\int_{\mathbb{R}} \left| f(t)e^{-t^2/2} \right|^2 dt = \int_{\mathbb{R}} \left|F(\xi) \right|^2 d\xi = 0.
$$
This implies that $f(t) \equiv 0$, thus we conclude that $S$ is dense in $L^2(\mathbb{R})$.


答:
If $I$ is an intervall and $f,h\colon I\to\mathbb R$ are continuous functions with $f(x)\le h(x)$ for all $x\in I$. Then

If $f(a)=h(a)$ for some $a\in I$, then any function $g\colon I\to\mathbb R$ with $f(x)\le g(x)\le h(x)$ for all $x\in I$ is continuous at $a$.

If $f(a)\ne h(a)$, there exists $g\colon I\to \mathbb R$ with $f(x)\le g(x)\le h(x)$ for all $x\in I$ that is not continuous at $a$.

For the first part note that for given $\epsilon>0$ there are $\delta_f,\delta_h>0$ such that for $x\in I$ with $|x-a|<\delta_f$ we have $|f(x)-f(a)|\le \epsilon$ and if $|x-a|<\delta_h$ then $|h(x)-h(a)|<\epsilon$.
Conclude that with $\delta:=\min\{\delta_f,\delta_h\}$ we have $|g(x)-g(a)|\le\epsilon$ if $|x-a|<\delta$.

For the second part, consider $g(x)=\begin{cases}f(x)&\text{if }x\in\mathbb Q,\\h(x)&\text{if }x\notin \mathbb Q.\end{cases} $

Title: If $x \leq g(x) \leq x^2-x+1$ where $x \in [0,2]$, can we say that $g(x)$ is continuous at $x=1$?

Body:
If $x \leq g(x) \leq x^2-x+1$ where $x \in [0,2]$, can we say that $g(x)$ continuous at $x=1$ ?

Is $g(x)$ continuous in $[0,2]$?


问:
Differentiable structures on Unit circle same
Let $\mathbb{S}^1$ have the manifold structure induced by it being a regular submanifold of $\mathbb{R}^2$. Let $\{$ $U_i$,$\phi_i$ $\}_{i=1}^4$ be the differentiable structure on $\mathbb{S}^1$ where $U_i$ are the open semicircles and each $\phi_i$ projects $U_i$ to an axis.

I would like to show that these smooth structures determine the same smooth structure. To do so, I must show that their union is also a smooth atlas.

Let $\{$ $(V_{\alpha}\cap \mathbb{S}^1,\psi_{\alpha}) \}_{\alpha}$ be the smooth structure given by the adapted charts. On $V_{\alpha}\cap \mathbb{S}^1 \cap U_i$ , we have

$\phi_{i}(\psi_{\alpha}^{-1}(\psi(p))=\phi_{i}(\psi_{\alpha}(x^1(p))=\phi_{i}(p)=p=\psi_{\alpha}^{-1}(\psi_{\alpha}(p))$

Moreover, $\psi_{\alpha}({\phi^{-1}(\phi(p))}=\psi_{\alpha}(p)$

Similarly for other charts.

Is this correct? How do I fix this?

答:
Let us see which differentiable structure is induced on $S^1$ as a submanifold of $\mathbb R^2$. Define
$$\phi : (-1,1) \times \mathbb R \to (-1,1) \times \mathbb R, \phi(x,y) = (x, y - \sqrt{1-x^2}).$$
This map is a diffeomorphism (with inverse $\phi^{-1}(x,y) = (x, y + \sqrt{1-x^2})$) such that

$\phi((-1,1) \times (0,\infty)) = (-1,1) \times [0,\infty) \cup$ open lower half disk of $D^2$ = $W_1$.

$\phi(U_1) = (-1,1) \times \{0\}$, where $U_1$ is the open upper half circle of $S^1$.

Thus $\phi : (-1,1) \times (0,\infty) \to W$ is a chart on $\mathbb R^2$ such that $\phi(S^1 \cap  (-1,1) \times (0,\infty)) = W \cap (\mathbb R \times \{0\})$.

This gives us the chart
$$\phi_1 : U_1  \stackrel{\phi}{\to}  W \cap (\mathbb R \times \{0\}) \stackrel{proj}{\to} (-1,1) $$
on $S^1$ as a submanifold of $\mathbb R^2$. You see that this is nothing else than the map in your question. For the other $U_i$ you can apply the same construction.


问:
Prove F is closed in $\mathbb{R}$ but F is not open in $\mathbb{R}$
I am using Topology Without Tears by Morris

Theorem

If $F$ is non-empty countable subset of $\mathbb R$, prove $F$ is closed in $\mathbb R$
but $F$ is not open in $\mathbb R$

My Best Effort to date

Part 1

It suffices  to show that $\mathbb R\setminus F$ is open in $\mathbb R$

Since $F$ being countable and closed in $\mathbb R$,then $\mathbb R\setminus F $ is open and infinite
and $S=\{s_1…,s_n \}$

where $f:S\mapsto \mathbb N$ is a one-to-one correspondence and $s_n=f(n)$ $\forall n\in N$

So $\mathbb R\setminus  F$ is open and denumerable.

Part 2

Suppose X$\setminus$(1/n)${n \in N}$ is open then there is a point p such that (X$\setminus$(1\n)${ n\in N}$ -p ,X$\setminus$(1/n)$ {n\in N}$ +p)

But this means ((1/n)${n \in N}-p,(1/n){n \in N}+p)$$\notin$ A

So $p \notin A$ and is not open.

Any help would be appreciated..

答:
It is easy to show every not empty open set of R is uncountable.

Thus F is not open.

If F = {0}, then F is closed.

If F = Q, then F is not closed.