问:
Importance of Axiom of Extension in Set Theory
As far as I understand, one importance of Axiom of Extension is the following:

Let $A=\{1,2,2,2,3,3\}$ and $B=\{1,2,3\}$. then due to Axiom of Extension we can conclude $A=B$.

Is there any other importance of Axiom of Extension apart from the one I specified?

答: Well, it establishes the connection between the statement "$a=b$" (identity of sets) and a statement in terms of the symbol "$\in$". Otherwise you could hardly ever, if at all, derive identity of sets from their properties as sets.


问:
How can I merge these functions?
Given Data:

There are three functions

f(x)=sin(πx/2)

g(x)=sin(πx/3)

h(x)=sin(πx/4)

and a point A(6,0).

To find:

Which of the given functions pass via point A.

My Approach:

As the  y coordinate of pt.A is 0, we know that any function which passes through pt.A will definitely pass via x axis at that point. Also, if a function passes through  A(6,0), this implies that for that function, at x=6 , the  y value, or f(6)  should equal 0.
Plugging in x =6, we get:

f(6)=g(6)= 0 ;h(6)≠0

⇒ The functions f(x)  & g(x)  pass via point A(6,0).

My question:

Is there any method by which I can merge/combine the equations of the given functions so that I get a single equation which serves the whole purpose (instead of plugging values in individual equations, can I merge all equations to get a single equation which works for all)? Is it possible?

答:
It depends on what you mean.  If you're looking for a way to have three yes/no answers from the same process, that would be multivalued, and typically we define functions to only output a single value.

If you're fine with losing information in the process, there are some simple options that might give you a useful single value, depending on your application.  For example, if you are trying to test for at least one of these functions passing through (6,0), then you could use:

$$
q(x) = f(x) \cdot g(x) \cdot h(x) \\
$$

which would return 0 if and only if at least one of them were 0.

Or if you were only interested in the case where all the functions are 0, you could use:

$$
q(x) = (f(x))^2 + (g(x))^2 + (h(x))^2
$$

which (assuming your functions output real numbers and not imaginary numbers) will output zero if and only if all the functions are 0.

ContentLicense: CC BY-SA 4.0

Title: How can I merge these functions?


问:
Calculate the determinant of a multilinear operator
How to calculate the determinant of a multilinear operator? Is it something different  from the determinant of the linear operator? Thanks.</p>

答:
As far as I'm aware, it doesn't make sense to talk about the determinant of a multilinear operator. The usual determinant requires that the linear operator "start" and "end" at the same vector space: specifically, given a linear operator $T\colon V\to V$, where $V$ is an $n$-dimensional vector space, applying the $n$th exterior power functor $\Lambda^n$ produces a new linear operator $(\Lambda^nT)\colon \Lambda^n V\to\Lambda^nV$. Because $\Lambda^nV$ is a one-dimensional vector space, any linear operator from it to itself is necessarily given by multiplication by some scalar, and this scalar is the determinant of $T$. Observe that if we had started with a linear operator $T$ starting at a vector space $V$ and ending at a different vector space $W$, it would be impossible to identify this scalar; if $W$ is of a different dimension than $V$, there's no hope, because $\Lambda^n V$ and $\Lambda^n W$ will be of different dimensions, but even if $W$ is of the same dimension as $V$, it would be necessary to choose bases of $V$ and $W$ so that they (and hence also their exterior powers) could be identified in a consistent way.

In contrast, an $m$-linear operator (for $m>1$) looks like
$$V_1\times\cdots\times V_m\to W.$$
for some vector spaces $V_1,\ldots,V_m,W$, and the above process wouldn't be applicable.

ContentLicense: CC BY-SA 3.0


问:
Find dimension of subspace with condition
How can I find the dimension of this subspace rapidly? So far, I've tried putting the columns in reduced row echelon form and simply taking out the last one but that won't work.
$$V=\left\{\left[\begin{array}{c} v_1\\ v_2\\ v_3\\ v_4\end{array}\right]\in\text{Span}\left\{\left[\begin{array}{c} 1\\ 1\\ 0\\ 1\end{array}\right],\left[\begin{array}{r} 0\\ -1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c} 0\\ 2\\ 0\\ 1\end{array}\right],\left[\begin{array}{c} 0\\ 0\\ 2\\ 1\end{array}\right]\right\}\ \text{such as}\ v_4=0\right\}.$$

答:
Let $v=(v_1,v_2,v_3,v_4)^t$, and $e_1= (1,1,0,1)^t$, $e_2=(0,-1,1,0)^t$, $e_3=(0,2,0,1)^t$, $e_4=(0,0,2,1)^t$.

$v$ is in the span, so for some $a_i \in \mathbb{F}$, $v=a_1e_1+a_2e_2+a_3e_3+a_4e_4$. Looking at the fourth coordinate implies $a_4=-a_1-a_3$. So, if we let $f_1=e_1-e_3$, $f_3=e_3-e_4$ we have $v=a_1f_1+a_2e_2+a_3f_3$. Moreover, any such vector is in your set $V$.

So, $V$ is the span of $f_1$, $e_2$, $f_3$. See how many of them are linearly independent to obtain your dimension.


问:
exponential generating functions containing an even/odd number of cycles
How to derive the exponential generating functions that having an even/odd number of cycles?
And how to define a bijection between them? Is there any example of this? Thanks in advance!</p>

答:
Minor correction to the answer by joriki, If you are counting permutations with an even number of cycles, then there is one for $n=0$ but not for $n=1$, and for odd number of cycles it is the opposite. Therefore for the even number of cycles you get
$$
  \frac12(1-x+\frac1{1-x}) = \frac12\times\frac{2-2x+x^2}{1-x}
 =1+\frac12(\frac{x^2}{1-x}),
$$
and for odd numbers of cycles you get similarly
$$
  \frac12(-1+x+\frac1{1-x}) = \frac12\times\frac{2x-x^2}{1-x}
 =x+\frac12(\frac{x^2}{1-x}) .
$$</p>


问:
How is that possible to be $X$ discrete random variable?

Suppose that the cumulative distribution function of discreate random variable $X$ is given by,
$$F(x) =
\begin{cases}
0  & \text{$x$ < 0 } \\[1.5ex]
\dfrac{x}{4} & \text{$0 \leq x<1$}\\[1.5ex]
\dfrac{1}{2}+\dfrac{x-1}{4} & \text{$1 \leq x<2$}\\[1.5ex]
\dfrac{11}{12}   &\text{$2 \leq x<3$}\\[1.5ex]
1 &\text{$3 \leq x$}\\
\end{cases}$$
Find $P(X=i),i=1,2,3$.

My problem is, how is that possible to be $X$ discrete random variable as I think it is mixed random variable(both discrete and continuous) since there are $x$ in $F(X)$. Can anyone help me to understand this?

答:
Well answered by Kavi sir.
Let's see it visually.

You can see from graph its neither discrete nor continuous it's mixed of both.

Let's zoom a little bit and look at the required probabilities.


问:
Universal set as a collection of objects
The universal set $U$ can be defined as $U = \{x : x=x\}$, which is equivalent to $\forall x[x \in U \leftrightarrow x = x]$. Since $x=x$ is a tautology, then we infer $\forall x \in U$. This definition tells us that every free variable in the scope of the universal quantifier is an element of $U$. Since $x \in U$ means $\{x\} \subseteq U$, then $\forall x \in U$ could be rewritten as $\forall x[\{x\} \subseteq U ]$ or $\forall X[X \subseteq U]$. Therefore, every set $X$ is a subset of the universal set $U$.

However, I've seen in some proofs and commentaries cases which define the universal set as a finite collection of objects, for example $U = \{ x_0, x_1\}$. This formula is equivalent to $\forall x[x \in U \leftrightarrow x \in \{x_0, x_1\}]$ and to $\forall x[x \in U \leftrightarrow x = x_0 \lor x=x_1\}]$. Compatibilization problems with the classical definition $\forall x [x \in U \leftrightarrow x=x]$ arise when we substitute $x$ by a free variable that is not $x_0$ and $x_1$. For example, $x_2 \in U \leftrightarrow x_2=x_0 \lor x_2=x_1$ returns $x_2 \notin U$, while $x_2 \in U \leftrightarrow x_2=x_2$ is equivalent to $x_2 \in U$. Thus, we have an apparent contradiction between the standard definition of universal set and its definition as a finite collection of objects.

I was inclined to think that my mistake would lie on the impossibility of universally instantiating the formula $\forall x[x \in U \leftrightarrow x = x_0 \lor x=x_1\}]$ with a variable $x_2$ that is not in the universal set $\{ x_0, x_1\}$. To formalize such notion, we would have to explicit that the domain of this formula itself is "limited" to the defined universal set, thus considering it equivalent to $\forall x \in U[x \in U \leftrightarrow x = x_0 \lor x=x_1\}]$. Therefore, we would have by universal instantiation $x_2 \in U \rightarrow [x_2 \in U \leftrightarrow x_2=x_0 \lor x_2=x_1]$, which is a tautology, since $x_2 \notin U$. But then again, from the formula $\forall x[x \in U \leftrightarrow x=x]$, we would conclude $x_2 \in U$, so the contradiction remains.

Is the universal set not definable as a collection of finite objects?

答:
Long comment

Consider formulas in the first-order language of arithmetic, with $0,+,\times, \le$.

The "natural" domain of interpretation is the set $\mathbb N = \{ 0,1,2,\ldots \}$ of natural numbers.

We can prove e.g. that $\forall x (x \le 1 \to P(x)) \leftrightarrow (P(0) \land P(1))$, but we cannot prove that $\forall x (x \ge 0 \to P(x)) \leftrightarrow (P(0) \land P(1) \land \ldots)$ because the infinite-long conjunction to the right is not a syntactically correct formula: in FOL, formulas are expression of finite length.

We can prove instead:  $\forall x (x \ge 0 \to P(x)) \rightarrow P(n)$, for every $n$.

But you can consider so-called Infinitary Logic, i.e. logic based on

"languages which can be obtained in a straightforward manner from first-order languages by allowing conjunctions, disjunctions and, possibly, quantifier sequences, to be of infinite length."

Regarding the "universal set" $U$, we can mimic it in predicate logic with an universal predicate $U(x) \leftrightarrow (x=x)$, in which case the formula $∀x[x=x ↔ U(x)]$ becomes a tautology.


问:
Prove that $L^{\infty} [0,1]$ is not separable.

Prove that $L^{\infty} [0,1]$ is not separable.

I know that $\ell^{\infty} = L^{\infty} (\Bbb N, \mathcal P (\Bbb N), \mu_c)$ is not separable where $\mu_c$ is the counting measure on $\mathcal P (\Bbb N).$ The way I proved that is to construct sequences in $\ell^{\infty}$ whose range set is contained in $\{0,1\}.$ Then such collection is uncountable and any two sequences are $1$ unit apart with respect to $\ell^{\infty}$-norm and then if we consider open balls of radius $\frac {1} {2}$ around each of the sequences then they are all disjoint and hence any dense subset of $\ell^{\infty}$ has to contain at least one element from each of these disjoint open balls. Since the number of such open balls is uncountable it follows that any dense subset of $\ell^{\infty}$ contains uncountably many elements and hence $\ell^{\infty}$ cannot be separable. How do I argue the same for $L^{\infty} [0,1]\ $?

I think of a collection of functions $\{f_{\alpha}\}_{\alpha \in I},$ for some indexing set $I$ such that $f_{\alpha} \left (\frac {1} {n} \right ) = 0\ \text {or}\ 1$ and $f_{\alpha} (x) = 0,$ for $x \not\in \left \{\frac {1} {n}\ \big |\ n \in \Bbb N \right \}.$ Then I think this collection will work here. Am I right?

答:
Functions in $L^{\infty}$ are defined only up to a set of measure $0$. The value at $\frac 1  n$ is not well defined. So your argument does not work.

Define $T:\ell^{\infty} \to L^{\infty} ([0,1])$ by $T(a_n)=\sum a_n \chi_{(\frac 1{n+1}, \frac 1n)}$ Check that $T$ is an isometric isomoprphism. Can you finish?


问:
Is the union of two circles homeomorphic to an interval?
Let $Y$ be the subspace of $\Bbb R^2$ given by $Y=\{(x,y): x^2+ y^2=1\}\cup \{(x,y): (x−2)^2+ y^2=1\}$. Is $Y$ homeomorphic to an interval?

I have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $Y$. Basically if we remove a point from $Y$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?

答:
A cutpoint of a connected space $X$ is a $p \in X$ such that $X\setminus\{p\}$ is disconnected.

If $f:X \to Y$ is a homeomorphism of connected spaces $X$ and $Y$ and $p$ is a cut point of $X$ then $f(p)$ is a cutpoint of $Y$ (and vice versa).

If we take $X$ to be an interval, then $X$ has at most two non-cutpoints. (the endpoints in the case of a closed interval). $Y$ on the other hand has infinitely many non-cutpoints (all points except $(1,0)$). So there can be no homeomorphism between them by the observations in the second paragraph.


问:
Infinite imprimitive non abelian group?
My new question is

Is there an infinite, imprimitive and non abelian group?

Thank you for the further answers.

答: Consider the subgroups $A={\rm Sym}(2{\bf Z})$ and $B={\rm Sym}(1+2{\bf Z})$ sitting inside $G={\rm Sym}({\bf Z})$. So $G$ is the set of bijections from the set of integers to itself, $A$ is the set of permutations of the even integers or equivalently the permutations which fix all odd integers, and $B$ the set of permutations of the odd integers or equivalently the permutations which fix all even integers. Let $h:x\mapsto x+1$ be the simple forward translation map. Then $H=\langle A,B,h\rangle$ is an infinite group which acts transitively on the integers and preserves the nontrivial partition ${\bf Z}=2{\bf Z}\cup(1+2{\bf Z})$.</p>

Title: Infinite imprimitive non abelian group?


问:
Find the indicial equation of $x^{3}y''+(\cos 2x-1)y'+2xy=0$
Find the indicial equation of $x^{3}y''+(\cos 2x-1)y'+2xy=0$
(where $y''$ stands for the second derivative)

The reference answer is $m(m-1)-2m+2=0$

I really don't know how to ... please help. thanks in advance!!

答:
$$x^3y''+(\cos 2x-1)y'+2xy=0$$ near $x=0$, we use $\cos 2x-1\approx 1-2x^2-1=-2x^2$, then the ODE is
$$x^2y''-2xy'+2y=0$$
Let $y=x^m$, then
$$m(m-1)x^m-2mx^m+2x^m=0\implies m^2-3m+2=0 \implies m=1,2$$
So trhe indicial roots are $m=1,2$.


问:
Is there an other way for it?
Consider two equations:

$x+y=2$
$y+z=4$

Find the value of $(x+z)$.
($x,y,z$ all are positive real numbers)

My Approach:

$\because x+y=2 \Longrightarrow y=0$ or $y=1$.

Case 1:

taking $y=0$$ \Longrightarrow$
 $x=2$ ; $z=4$.
$\therefore x+z = 2+4 = 6.$

Case 2:

taking $y=1 \Longrightarrow$

$x=1$ ; $z=3$.

$\therefore x+z = 1+3 = 4.$

$x+z=6$
or
$x+z=4$

I want to know if my approach is correct or is there more better way to evaluate this ?

答: Adding the equations gives $x+2y+z= 6$ or $x+z=6-2y$. So for every $y$ you choose you'll get a different sum and all values $\le 6$ can be assumed. So taking these two specific ones proves or shows nothing. You can just say that $x+z \le 6$ as $y\ge 0$.

Title: Is there an other way for it?


答: Another use of extension is to allow us to define sets like $\{x|\phi(x)\}$ for those $\phi$ for which our other axioms imply such a set's existence, e.g. those of the form $\psi(x)\land x\in y$ for some set $y$. So instead of merely saying $\exists S\forall x(x\in S\iff\phi(x))$, which would be consistent with multiple such $S$, we can with extension not only know this $S$ is unique but also refer to such an $S$.

Title: Importance of Axiom of Extension in Set Theory

Body:
As far as I understand, one importance of Axiom of Extension is the following:

Let $A=\{1,2,2,2,3,3\}$ and $B=\{1,2,3\}$. then due to Axiom of Extension we can conclude $A=B$.

Is there any other importance of Axiom of Extension apart from the one I specified?


问:
Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$
The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $x = 0$). This attempt failed because firstly it really is true but after that we get opposing inequality. Also, I have noticed that $\lim \frac{x}{1 + x \cdot \frac{2}{\pi}} = \lim \arctan(x) = \frac{\pi}{2}$ as $x \rightarrow + \infty$ but it didn't lead me to solution. I also tried to apply Taylor's formula but it didn't help much either.

So, what are available ways to solve this problem?

答:
Consider the function $f$ defined by
$$
f(x): = \arctan x - \frac{x}{{1 + \frac{2}{\pi }x}}
$$
for all $x>0$. It is easy to check that
$$
f'(x) = \frac{{x((4 - \pi ^2 )x + 4\pi )}}{{(x^2  + 1)(2x + \pi )^2 }}.
$$
If $0 < x \le \frac{{4\pi }}{{\pi ^2  - 4}} = 2.14 \ldots$, then $f'(x) \geq 0$, i.e., $f$ is increasing on this range. Thus,
$$
f(x) > \mathop {\lim }\limits_{x \to 0 + } f(x) = 0  \Leftrightarrow \arctan x > \frac{x}{{1 + \frac{2}{\pi }x}}
$$
for $0 < x \le \frac{{4\pi }}{{\pi ^2  - 4}}$. If $x > \frac{{4\pi }}{{\pi ^2  - 4}}$, then $f'(x) < 0$, i.e., $f$ is decreasing on this range. Hence,
$$
f(x) > \mathop {\lim }\limits_{x \to  + \infty } f(x) = 0 \Leftrightarrow \arctan x > \frac{x}{{1 + \frac{2}{\pi }x}}
$$
for $x > \frac{{4\pi }}{{\pi ^2  - 4}}$.


问:
Underdetermined system has infinitely many solutions
To prove that there can be at most n linearly independent vectors in R^n, I have to show, that a matrix equation $$ A_{n \times (n +1)}x=b $$ has infinitely many solutions. I have been looking for a proof for that but so far most of the proofs rely on some kind of other "obvious" result or  intuition. For example some simply state that a system with more variables than equations has infinitely many solutions. Even though obvious, I can't prove this result.

So could someone please present an exact proof? Without simply stating that for sure the matrix can be transformed to some kind of other form etc?

答:
This proof is lifted from "Linear Algebra and It's Applications" by David C. Lay

Let $A=[v_1...v_p]$ where each column is condensed into vectors of dimension $n$, then $A$ is $n \times p$, and the equation $Ax=0$ corresponds to a system of $n$ equations in $p$ unknowns. If $p>n$, there are more variables than equations, there must be be a free variable*. Hence $Ax=0$ has a nontrivial solution and the columns of $A$ are linearly dependent.

*You can see this statement is true: When a matrix is transformed into it's row reduced echelon form (which is always possible: see here), each row must contain a pivot point (basic variable) or be completely composed of zeros (free variable). The number of pivot points can never exceed the number of columns (because each column can only have, at most, one pivot point), so naturally, if the rows exceed the number of columns there must be a free variable.


问:
Identity for a weighted sum of sines / sines with different amplitudes
I'm trying to simplify the following sum of sines with different amplitudes
$$
a \sin(\theta) + b \sin(\phi) = ???  \,\,\,\,\, (1)
$$

I know that
$$
a \sin(\theta) + a \sin(\phi) = a\cdot2\sin\left(\frac{\theta+\phi}{2}\right)\cos\left(\frac{\theta-\phi}{2}\right)
$$

But how do I do the same thing in the case where both sines have different amplitudes, as in (1).

Thanks!

答:
I'll try to give both geometrical solution and a solution using complex numbers.

But perhaps you might have a look at Wikipedia first, their explanation is probably better than mine. (They certainly have nicer pictures.)

Wikipedia

You can find very similar formulas in Wikipedia's List of trigonometric identities. (Here is also link to the current revision of the Wikipedia article. I hope I have not made mistake somewhere and their formulas are equivalent to mine.)

A link to Phasor addition (current revision) is also given there. This Wikipedia article might help to visualize the whole thing.

Geometry

We have situation as in the following picture:

Note that the $y$-coordinates of the two vectors are precisely the values you want to add together. So we would like to know the length and the angle for the sum of these two vectors.

We simply rotate the situation. We denote $\alpha=\phi-\theta$.

We want to find the values $c$ and $\varkappa$. Once we know them, we get
$$a\sin\theta+b\sin\phi= c\sin(\theta+\varkappa).$$

For $c$ we can simply use law of cosines:
$$c^2=a^2+b^2+2ab\cos\alpha$$
(Note that the angle in the bottom triangle is $\pi-\alpha$.)

We can calculate $\varkappa$ using law of sines:
$$\frac{\sin(\pi-\alpha)}{\sin\varkappa} = \frac cb \qquad \Rightarrow \qquad \sin\varkappa=\frac{b\sin\alpha}c$$

Note that this does not determine $\varkappa$ uniquely, but you can find out whether $\varkappa>\pi/2$ or $\varkappa<\pi/2$ by checking the sign of $x$-coordinate of the sum of the two vectors, which is $a+b\cos\alpha$.

Complex numbers

We want to get
$$ae^{i\theta}+be^{i\phi}=ce^{i(\theta+\varkappa)}\tag{1}$$
(If we get such an expression then the imaginary part is a formula for sines.)

This is equivalent to
$$a+be^{i(\phi-\theta)}=ce^{i\varkappa}.\tag{2}$$

To find $c$ we simply compute the absolute value of both sides. On the LHS we get $(a+be^{i(\phi-\theta)})(a-be^{i(\phi-\theta)})=a^2+b^2+2ab\cos(\phi-\theta)$, hence
$$c^2=a^2+b^2+2ab\cos(\phi-\theta).$$

We also have
$$e^{i\varkappa}=\frac ac+\frac bce^{i(\phi-\theta)}\\
\cos\varkappa+i\sin\varkappa = \frac ac+\frac bc\cos(\phi-\theta)+i\frac bc\sin(\phi-\theta)$$
Comparing the imaginary parts yields $$\sin\varkappa=\frac bc\sin(\phi-\theta).$$

The remark that we have to do a little more to find which of the two possible values of $\varkappa$ we should choose applies here, too.


问:
Show that there does not exist $g \in L^1[0,1]$ such that $\varphi (f) = \int fg\ d\mu.$

Let $0 \neq \varphi \in (L^{\infty} [0,1])^*$ such that $\varphi (f) = 0,$ for all $f \in C[0,1].$ The existence of such a $\varphi$ is guaranteed by Hahn-Banach theorem since $C[0,1]$ is not dense in $L^{\infty} [0,1].$ Show that there does not exist any $g \in L^{1} [0,1]$ such that $\varphi (f) = \int fg\ d\mu,$ for all $f \in L^{\infty} [0,1],$ where $\mu$ is the Lebesgue measure.

How do I prove that? Can anybody please give me some hint?

Thanks for your time.

答: Consider $\phi (e^{2\pi i nx})=\int e^{2\pi i nx} g(x)d\mu (x)$. [You can split this into real and imaginary parts if you don't want to use complex valued fuctions].  The hypothesis implies that all the Fourier coefficients of the intergable function $g$ are $0$. Hence $g=0$ a.e. which forces $\phi $ to be $0$.


问:
Find the slope of the line go through a point such that the area between the graph and the line is minimum

If $a$ is the slope of $(L)$ that go through point $(-1,2)$ and $f(x)=x^2$ then find $a$ such that the area between the graph and the line is minimum.

For what I thought, the area is minimum if the line tangent to the graph because that mean the area is $0$. For the above problem, the point $(-1,2)$ is inside the graph so there can't be a line that tangent to it ( I think ). But still for what I did, I take $a=f'(-1)$ which is $a=-2$. I dunno if what I did is right or wrong at all. Please help

答:
Please note I am using notation $m$ for the slope instead of $a$.

If you have two points $A$ and $B$ with $x-$coordinates $x = a$ and $x =b$ ($b \gt a$) on the parabola $y=x^2$, area of parabola under the straight line between these two points can be found as below.

For points on the line segment $AB$, $y -a^2 = \frac{b^2-a^2}{b-a} (x-a) \implies y = (a+b)x - ab$

So the area is $A = \int_a^b ((a+b)x-ab - x^2) \ dx = \frac{1}{6}(b-a)^3$ ...(i)

Now for a line with slope $m$ through point $(-1, 2)$, the equation of the line will be

$y-2 = m(x+1)$ and $x-$coordinates of its intersection with parabola $y=x^2 \ $ is given by,

$x^2 - mx - (m+2) = 0 \implies x = \frac{m \pm \sqrt{m^2+4m+8}}{2}$

Plugging in $(i)$, $A = \frac{1}{6}(m^2+4m+8)^{3/2}$

So if we minimize $(m^2+4m+8)$, we minimize the area. We can rewrite $m^2+4m+8$ as $(m+2)^2+4$ which shows that the minimum area occurs at $m = -2$.


问:
Calculation of a limit, using left eigenvectors, an eigenvector and a positive matrix
I've tried to calculate the limit of 1 over the Spectral Radius of a positive Matrix A times the Matrix A itself, the whole thing to the power of k, but it went wrong somewhere. My attempt is in the picture below, I hope this is alright. Many thanks in advance for your help

my attempt

答:
I just noticed this, and a fair amount of time has gone by. You clearly recognize what you did wrong and you are actually asking how to go about the proof correctly. There are many approaches. The most obvious approach that comes to mind (without assuming that $A$ is disgonalizable) is the following.  Try calculating
$$ \lim_{u \to \infty} A^u v $$
for any vector $v$. Then write
$$ v = a x + b v_\perp $$
where the second term is the part perpendicular to $x$. Then consider the Frobenius Perron theorem and what this says about $r(A)$ versus $|A v_\perp|$ by considering induced matrix norms.


问:
Is it possible to rewrite $\frac{m\cdot \ln2}{\frac{n(1-(1-p)^r)}{p}}$ as $\frac{p\cdot \ln2}{(1-(1-p)^r)\frac{n}{m}}$?
Is this rewriting possible? If so, how?

$$\frac{m\cdot \ln2}{\dfrac{n(1-(1-p)^r)}{p}} = \frac{p\cdot \ln2}{(1-(1-p)^r)\dfrac{n}{m}}$$

答:
One way to proceed is to factor out $p$ from denominator & try to cancel out with $p$ in numerator...but that will make it look messy .

Otherwise, $$(p.m.ln2)/(n(1-(1-p)^r)$$ is the simplest form.


问:
Similarity of a complex matrix with a triangular matrix with small elements besides the diagonal
Let $A$ be a $n \times n$ matrix with complex coefficients and let $r>0$. I know that there exist a nonsingular matrix $C$ and an upper triangular matrix $T$ such that $A=C^{-1}TC$.
I wish to prove the existence of nonsingular $C$ and upper triangular $T$ such that that equality holds and aditional all elements $t_{ij}$ of $T$ over the main diagonal satisfy  $|t_{ij}|<r$.

Thanks

答:
Lemma: Let $N$ be a $k\times k$ matrix with $1$'s on the superdiagonal and $0$ elsewhere. The $N$ and $\epsilon N$ are similar for any $\epsilon\not=0$.

Proof: By looking at the ranks of $(\epsilon N)^s$  we see that the Jordan form of $\epsilon N$ is $N$.

Now $A$ is similar to its Jordan Form, and we can apply the lemma to $J_{p}(\lambda)-\lambda$ for each Jordan block.


问:
Uniform convergence of sum to constant, plus $L^2$ convergence of summands to constant, implies uniform convergence of summands?
If one has $f_N,g_N \geq 0$ continuous, defined over some compact sets (assumed of measure one for simplicity), such that:

$||f_N+g_N||_{L^1} = 1$;

$f_N + g_N \to C$ in $L^\infty$;

$f_N \to C_f$ in $L^2$;

$g_N \to C_g$ in $L^2$;

is it true that $f_N$ (resp $g_N$) $\to C_f$ in $L^\infty$?

答: Let $h_n(x)=nx$ for $0 \leq x \leq \frac 1 n$ and $0$ for $x >\frac 1 n$. Let $C=1,f_n=1-h_n$ and $g_n=h_n$. [Note that $0\leq h_n \leq 1$ and $\|h_n\|_{\infty}=1$ for all $n$].


问:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$

Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$

The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.

The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)

答:
Hint:

Let $\sqrt{x^2+12y}-\sqrt{y^2+12x}=1+2a$

$$\implies\sqrt{x^2+12y}=\dfrac{23+1+2a}2,\sqrt{y^2+12x}=11-a$$

$\implies11\ge a\ge-12\ \  \ \ (1)$

$$(12+a)^2-(11-a)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$

Replace the value of $x+y$ to find $x-y$

Consequently we find $x,y$ in terms of $a$

But we have $$x^2+12y=(12+a)^2$$

Replace the values of $x,y$ in terms of $a$ to from a quadratic equation $a$

Solve and remember $(1)$


Title: $\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$

Body:
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$

The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.

The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)

Body:
$$\left(\sqrt{x^2+12y}+\sqrt{y^2+12x}\right)^2=(x+y+10)^2$$

$$\implies x^2+12y+y^2+12x+2\left(\sqrt{(x^2+12y)(y^2+12x)}\right)=x^2 + y^2 + 100+ 2xy+20x+20 y $$

$$\implies \sqrt{(x^2+12y)(y^2+12x)}=xy+4(x+y)+50 $$

$$\implies(x^2+12y)(y^2+12x)=(xy+142)^2 $$

$$\implies x^3 - \dfrac{35 x y}{3}+ y^3 - \dfrac{5041}{3}=0$$

$$ \implies (x+y)^3-3xy(x+y)-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$

$$\implies 23^3-69xy-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$

Finally, we have

$$\begin{cases} x+y=23 \\ xy=130 \end {cases}$$

$$\implies x=13, y=10 $$ and

$$\implies x=10, y=13.$$


问:
What is the Perron-Frobenius theorem for non-negative matrices?
Let $M\succeq 0$ (i.e. $M_{ij}\geq 0$ for all $i,j$). No further conditions on $M$ such as irreducibility, aperiodicity, or what not. What is the formulation of the theorem in this case?

I believe that still the eigenvectors belonging to the largest eigenvalue are all non-negative. However the algebraic multiplicity of $\lambda_1$ is no longer 1, but rather I believe $n$. What is the statement here?

Edit:

Refined and more constrained question in a new post. It's okay for me if this one is put on hold.

答: This is discussed in some detail on the Wikipedia page. Each irreducible diagonal block of the $M$ satisfies the Perron-Frobenius theorem separately and the results of the theorem mostly carry over to the corresponding components of the eigenvectors. As you suspect, irreducible nonnegative matrices can have Perron-Frobenius roots with multiplicities  other than one if they belong to different blocks. Within the same block, the magnitude can be the same but the phase can be different, and they are always roots of unity (see this section titled “Perron–Frobenius theorem for irreducible non-negative matrices”).


问:
Optimal control of the gradient type PDE
I recently encountered the following optimal control problem.

We have an object described by a quasi-PDE of a gradient type with the ability to apply an input control action $u$:

$\frac{dx}{dt}=\frac{df}{dx}+u$

where $F=\frac{1}{(x-x_*)^2+1}$, and $x_*$ - constant, at which the maximum function is reached.

The purpose of the control is to find the value of the controlled parameter $x$, at which the maximum of the function $F$.

The task is complicated by the fact that the transition process of the transition from the initial point $x(0)$ to the final one $x=x_*$ has a hard form due to the nonlinearity of the static map $F$).

We have two variants:

It is necessary to make sure that the transition process from any initial state $x(0)$ to an extremum point occurs exponentially, i.e.use one of the cost functions $J$ suggested in the answer, then the solution will look like this, where blue is the existing transient and orange is the desired one.

It is necessary that the solution of the differential equation be bounded from above and below by an exponential signal, i.e:

Blue and orange are the upper and lower boundaries. Red - a transient process that does not go beyond their limits and converges to an extremum.

Problem:  Build such a control system so that the required cost function $J$ so that the transient process from $x(0)$ to $x_*$ in the system is exponential, or make sure that the transient process does not go beyond the required boundaries

We know and are available for measurement: $x$,$x^{'}$,$F$,$\frac{dF}{dx}$.

I am new to optimal control of partial differential equations. I can't seem to figure out how to find an approach to this problem, so any help would be appreciated. I thank all the helpers.

答:
The answer is trivial and I am not sure if it is what you want.

Define $e=x-x_*$, then the dynamics of the error is
$$
\dot{e} = \frac{-2e}{\left(1+e^2\right)^2}.
$$
Note that if $e_0=e(0)$ is bounded then the convergence is exponential, where the rate of the convergence depends on $e_0$.

If you want $e$ to converge to zero exponentially with the rate $\kappa$ then one possible cost function is
$$
J(e) = \int_0^\infty \left(e(\tau)-e_0\exp(-\kappa \tau)\right)^2d\tau.
$$
Another $J$ ensuring the exponential convergence is
$$
J(e,\dot{e}) = \int_0^\infty \left(e(\tau)+\kappa\dot{e}(\tau)\right)^2d\tau.
$$
You see that both $J$ are defined with respect to $e$, i.e., with respect to $x$ and $x_*$.

Control

To get the exponential convergence with the constant rate $\kappa$ you choose
$$
u = -\frac{df}{dx} -\kappa e.
$$
Note that to implement is you must know your $x_*$. If it is not the case, then I do not see how to get a constant rate of convergence. However, recall that for any initial condition your system is exponentially converging as it is bounded by
$$
|e(t)| \le e_0 \exp\left(-\left(\frac{\sqrt{2}}{1+e_0^2}\right)^2t\right).
$$


问:
Exterior power of lie algebra representation in coordinates
Suppose I have a representation $V = \mathbb{C}^3$ of a group $G$ where elements are of the form $M = \begin{bmatrix}m_1^1 & m_1^2 & m_1^3 \\ m_2^1 & m_2^2 & m_2^3 \\m_3^1 & m_3^2 & m_3^3 \end{bmatrix}$ in some basis $v_1,v_2,v_3$. If I wish to compute the representation $\Lambda^2 V \subset V \otimes V$, I could look at the action $g \cdot (v \wedge u) := (gv) \wedge (gu)$ and see that I will get a matrix $N$ of minors of $M$, i.e. $n_{i,j} = $minor obtained by removing $i$th row and $j$th column.

My question is as follows: In the case of a representation of a Lie algebra $\mathfrak{g}$, is there an analogous interpretation when considering an exterior power of the representation?

As an example, using the Lie algebra action $g(v \wedge u) := (gv) \wedge u + v \wedge (gv)$, with basis $v_1 \wedge v_2$, $v_1 \wedge v_3$, and $v_2 \wedge v_3$, I obtain the following matrix for $\begin{bmatrix}m_1^1+m_2^2 & m_2^3 & -m_1^3 \\ m_3^2 & m_1^1+m_3^3 & m_1^2 \\-m_3^1 & m_2^1 & m_2^2 + m_3^3 \end{bmatrix}$. Any insight would be appreciated!

答:
Consider a linear operator $T \colon V \to V$. This induces a linear operator $\wedge^k T\colon \wedge^k V \to \wedge^k V$ on the wedge power, and hence we get a map between Lie groups
$$ \varphi \colon \operatorname{GL}(V) \to \operatorname{GL}(\wedge^k V).$$
This map is polynomial (each coordinate on the right is a determinant of a matrix minor on the left), and therefore can be differentiated at the identity $I \in \operatorname{GL}(V)$ to a linear map
$$ d \varphi_I: \operatorname{End}(V) \to \operatorname{End}(\wedge^k V).$$
It is relatively straightforward (in principle) to compute $d \varphi_I$: let $M = [m_i^j]$ be a matrix of indeterminates, then the derivative should be the first-order term in the expansion $$d \varphi_I(I + tM) = I + t d\varphi_I(M) + t^2(\text{higher order terms...}).$$

For example, to compute the top-left entry of $d \varphi_I(M)$ in your example, we compute the determinant
$$ \det_{1, 1} (I + tM) = \det \begin{pmatrix} 1 + t m_1^1 & tm_1^2 \\ tm_2^1 & 1 + tm_2^2 \end{pmatrix}  = 1 + t(m_1^1 + m_2^2) + t^2(...),$$
and therefore the result is $m_1^1 + m_2^2$. We could try another entry:
$$ \det_{1, 3} (I + tM) = \det \begin{pmatrix} t m_2^1 & 1 + tm_2^2 \\ tm_3^1 & tm_3^2 \end{pmatrix} = t(-m_3^1) + t^2(m_2^1 m_3^2 - m_3^1 m_2^2),$$
so the result is $-m_3^1$ (note that we don't expect a $1+$ at the start here, since $\det_{1, 3}(I) = 0$).

We can work this more carefully into a general rule. Say we are working with an $n \times n$ matrix $M$ and calculating in the $k$th exterior power. The matrix $\varphi(M)$ is indexed by pairs $J, K$ of $k$-element subsets of $\{1, \ldots, n\}$, where $\varphi(M)_J^K = \det M_J^K$, where $M_J^K$ is the $k \times k$ submatrix of $M$ with rows $J$ and columns $K$. This determinant can be written more explicitly as
$$ \det M_J^K = \sum_{\sigma : J \to K} (-1)^\sigma \prod_{j \in J} M_j^{\sigma(j)},$$
where $\sigma$ ranges over all bijections $J \to K$, and $(-1)^\sigma$ is $(+1)$ if the diagram of $\sigma$ has an even number of crossings, and $(-1)$ for an odd number of crossings. Now consider the determinant $\det(I + tM)_J^K$: the only summands that produce a $t^1$ term are either if:

$\sigma(j) = j$ for all $j \in J$, in which case $J = K$, and $\det(I + tM)_J^J = 1 + t \operatorname{tr}(M_J^J) + t^2(...)$. This is like the first example above.

$\sigma(j) = j$ for all but one $j \in J$, in which case we have $|J \cap K| = k - 1$, and $\det(I + tM)_J^K = t (-1)^\sigma M_j^k + t^2(...)$, where $j$ is the unique element of $J \setminus K$, $k$ is the unique element of $K \setminus J$, and $\sigma$ is the identity on $J \cap K$ and sends $j$ to $k$.

All other terms are zero: in particular, if $|J \cap K| < k - 1$ then $d \varphi_I(M)_J^K = 0$. To recap, we have

$\operatorname{tr}(M_J^K)$ if $J = K$,

$(-1)^\sigma M_j^k$ if $|J \cap K| = k - 1$, where $\sigma: J \to K$ fixes $J \cap K$ and $j \in J \setminus K$, and $k \in K \setminus J$,

$0$ if $|J \cap K| < k - 1$.

Finally, an example for $\wedge^2 M$ where $M$ is a $4 \times 4$ matrix. There are six two-element subsets of $\{1, 2, 3, 4\}$, and we get:

$$
      \begin{array}{@{} c c c c c c c @{}}
& \{1, 2\} & \{1, 3\} & \{1, 4\} & \{2, 3\} & \{2, 4\} & \{3, 4\} \\
\{1, 2\}&        m_1^1 + m_2^2 & m_2^3& m_2^4& -m_1^3& -m_1^4& 0\\
\{1, 3\}&        m_3^2 & m_1^1 + m_3^3 & m_3^4& m_1^2& 0& -m_1^4\\
\{1, 4\}&        m_4^2& m_4^3& m_1^1 + m_4^4 & 0 & m_1^2& m_1^3\\
\{2, 3\}&        -m_3^1& m_2^1& 0& m_2^2 + m_3^3 & m_3^4& -m_2^4\\
\{2, 4\}&        -m_4^1& 0 & m_2^1& m_4^3& m_2^2 + m_4^4 & m_2^3\\
\{3, 4\}&        0& -m_4^1& m_3^1& -m_4^2 & m_3^2& m_3^3 + m_4^4
      \end{array}
$$

The structure is rather simple, but the signs (as usual) make coordinate descriptions annoying. Even so, it is in some ways simpler than the matrix of minor determinants: most of the reason that Lie algebras are so useful is that representations become linear (rather than polynomial on a group).

The transpose-symmetry (ish) you can see in the matrix for $\wedge^2 \mathbb{C}^4$ is kind of special because this space is self-dual, in general there should be some kind of symmetry between the operators for $\wedge^k \mathbb{C}^n$ and $\wedge^{n - k} \mathbb{C}^n$, by complementing the sets indexing the rows/columns inside $\{1, \ldots, n\}$.


答:
$A$ corresponds to a linear map $T_A: \Bbb R^{n+1} \to \Bbb R^n$.

By one of the fundamental theorems on linear maps (see here e.g.):

$$n+1 = \dim(\Bbb R^{n+1}) = \dim(\operatorname{Im}(T_A)) + \dim(\operatorname{Ker}(T_A))$$

As $\dim(\operatorname{Im}(T_A)) \le \dim(\Bbb R^n)=n$ it follows that $\dim(\operatorname{Ker}(T_A)) \ge 1$ and it follows that if $Ax=b$ has one solution $x_0$ it has infinitely many solutions $x_0 + y$, where $y \in \operatorname{Ker}(T_A)$ and there are infinitely many such $y$ as the dimension is positive and $\Bbb R$ is infinite. It is perfectly possible that there is no solution as well. So $0$ or infinitely many solutions whenever the dimension of the domain exceeds that of the codomain.

Title: Underdetermined system has infinitely many solutions

Body:
To prove that there can be at most n linearly independent vectors in R^n, I have to show, that a matrix equation $$ A_{n \times (n +1)}x=b $$ has infinitely many solutions. I have been looking for a proof for that but so far most of the proofs rely on some kind of other "obvious" result or  intuition. For example some simply state that a system with more variables than equations has infinitely many solutions. Even though obvious, I can't prove this result.

So could someone please present an exact proof? Without simply stating that for sure the matrix can be transformed to some kind of other form etc?


问:
Is this Sequence Bounded Above?

I know the Sequence is increasing as $\Big(a_{n+1}-a_n> 0\Big)$ and it is bounded below as $a_1 = \frac{1}{2}$

But how do we know whether the sequence is bounded above or not .

Title: Is this Sequence Bounded Above?

Body:
$$
\sum_{k=1}^n\frac{1}{n+k}<\sum_{k=1}^n\frac{1}{n+1}=\frac{n}{n+1}<1.
$$


问:
Is the truncated exponential series for matrices injective?
If $k$ is a field of characteristic $p$, we can define a map $\exp:\mathfrak{gl}_n(k)\to GL_n(k)$ by:

$$\exp(A)=\sum_{i=0}^{p-1}\frac{A^i}{i!}$$

In the answer to this question, we see that if $A^p=B^p=0$, and if $\exp(A)=\exp(B)$, then $A=B$.  So if $p>n$, $\exp$ is injective when restricted to nilpotents.  I'd like to know whether or not $\exp$ is injective on all of $\mathfrak{gl}_n(k)$.

答:
I am not sure if there is any nice characterisation of injectiveness, but the truncated matrix exponential is not injective in all cases. Let $p=3>n=2$ and $A=\pmatrix{1&\ast\\ 0&0}$. Then $A^m=A$ for every $m\ge1$ and
$$\exp(A) = I + A + \frac12 A^2 = I + A + 2A = I = \exp(0).$$</p>


答:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$,$x+y=23$

The obvious way to solve it is by substituting for one variable.

No, cart before the horse.  The first step is to move one of the radicals to the RHS and square both sides.  This gives:

$$(x^2 + 12y) = 1089 + (y^2 + 12x) - 66\sqrt{y^2 + 12x}.\tag1$$

At this point, there are two key factors:

When you square both sides, you will generate an equation such that some of its solutions may not satisfy the original equation.

You are looking for some elegant way of utilizing that $(x + y) = 23.$

I experimented with various tries and then decided on the following:

$(x^2 + 12y) = x^2 + 12[23 - x] = x^2 - 12x + 276.$

$(y^2 + 12x) = [23-x]^2 + 12x = 529 - 46x + x^2 + 12x = x^2 - 34x + 529.$

Therefore, equation (1) above simplifies to

$$(x^2 - 12x + 276) = 1089 + (x^2 - 34x + 529) - 66\sqrt{x^2 - 34x + 529} \implies $$

$$(1342 - 22x) = 66\sqrt{x^2 - 34x + 529} \implies $$

$$(61 - x) = 3\sqrt{x^2 - 34x + 529} \implies $$

$$x^2 - 122x + 3721 = 9[x^2 - 34x + 529] \implies $$

$$(8x^2 - 184x + 1040) = 0 \implies$$

$$x = \left(\frac{1}{16}\right) \times
\left[184 \pm \sqrt{33856 - \left(4 \times 8 \times 1040\right)}\right]$$

$$= \left(\frac{1}{16}\right) \times
\left[184 \pm \sqrt{576}\right]$$

$$= \left(\frac{1}{16}\right) \times
\left[184 \pm 24\right] \implies x \in \{10, 13\}.$$

Here, you have to check each candidate value of $x$ against the original equation.  Doing so, you realize that there are in fact two solutions: $(x=10, y=13)$ and $(x = 13, y = 10).$


问:
Induced Isomorphism between Homology Groups
How can I show the following isomorphism
$$H_n(\mathbb{R}^n, \mathbb{R}^n-A) \cong H_n(\mathbb{R}^{n+1}, (\mathbb{R}^n-A) \times (-1,1) )$$

What I am trying for this problem is to use the following sequence, but I am not sure if this will help.

$$H_n(\mathbb{R}^n, \mathbb{R}^n-A) \to H_n(\mathbb{R}^{n} \times \{0\}, (\mathbb{R}^n-A) \times \{0\} ) \to H_n(\mathbb{R}^{n+1}, (\mathbb{R}^n-A) \times (-1,1) )$$

Is it true that $\to$ are isomorphism between the homology groups

答: Consider the map of pairs $j : (\mathbb{R}^n, \mathbb{R}^n \setminus A) \to (\mathbb{R}^{n+1}, (\mathbb{R}^n \setminus A) \times (-1,1)), j(x) = (x,0)$. Both the "absolute part" $j_{abs} : \mathbb{R}^n \to \mathbb{R}^{n+1}$ and the "relative part" $j_{rel} : \mathbb{R}^n \setminus A \to (\mathbb{R}^n \setminus A) \times (-1,1)$ are homotopy equivalences. Now consider the long exact sequences of the pairs $ (\mathbb{R}^n, \mathbb{R}^n \setminus A)$ and $(\mathbb{R}^{n+1}, (\mathbb{R}^n \setminus A) \times (-1,1))$ which are connected by the homomorphisms $j_* : H_k(\mathbb{R}^n, \mathbb{R}^n \setminus A) \to H_k(\mathbb{R}^{n+1}, (\mathbb{R}^n \setminus A) \times (-1,1))$, $(j_{abs})_* : H_k(\mathbb{R}^n) \to H_k(\mathbb{R}^{n+1})$ and  $(j_{rel})_* : H_k(\mathbb{R}^n \setminus A) \to h_k((\mathbb{R}^n \setminus A) \times (-1,1))$. The last two of them are isomorphisms in all dimensions $k$ and therefore the five lemma shows that also $j_*$ is an isomorphism in all dimensions $k$.


问:
Is the union of two circles homeomorphic to an interval?
Let $Y$ be the subspace of $\Bbb R^2$ given by $Y=\{(x,y): x^2+ y^2=1\}\cup \{(x,y): (x−2)^2+ y^2=1\}$. Is $Y$ homeomorphic to an interval?

I have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $Y$. Basically if we remove a point from $Y$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?

答:
Here's an alternative argument that covers wide range of spaces, regardless of cut points.

Let $X$, $Y$ be any topological spaces. Consider a homeomorphism $f:X\to Y$. Now let $Z$ be any topological space and $\alpha:Z\to X$ be a continuous function. Then $\alpha$ is injective if and only if $f\circ\alpha$ is injective. Which is easy to see by applying $f$ to $\alpha$ and $f^{-1}$ to $f\circ\alpha$.

In particular $X$ admits an injective map $Z\to X$ if and only if $Y$ admits an injective map $Z\to Y$.

This shows that your $Y$ (or more generally any space containing $S^1$ as a subspace) cannot be homeomorphic to the interval $[0,1]$. Because there is an obvious injective continuous map $S^1\to Y$ while there is no continuous injective map $S^1\to [0,1]$. That's because every map $S^1\to [0,1]$ arises from a map $[0,1]\to [0,1]$ with the same values at endpoints and so by the intermediate value property it cannot be injective on $(0,1)$, which then "lifts" to $S^1$.


问:
Find the argument of a complex number $z$
$z$ is a complex number $z=2+\sqrt 3+ i$

I found that $\tan(a)=\frac{1}{(2+\sqrt 3 )}$
What am i supposed to do next?
Without using calculator, please

答:
As Matti P. already hinted: $\frac{1}{2+\sqrt{3}} = 2 - \sqrt{3}$ which is $\tan(\frac{\pi}{12})$ or $\tan(15^\circ)$ if you prefer degrees over radians. As $|z|^2 = (2+\sqrt{3})^2 + 1^2 = 8 + 4\sqrt{3}$ we get that

$$z = 2\sqrt{2 + \sqrt{3}}e^{i \frac{\pi}{12}}$$

in polar representation.


问:
Positive Definite Quadratic Form
I am following the solution of a problem which has the following matrix
$$\begin{bmatrix} 2 & 0 & 0\\ 0  & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$$ along with $dz = -dx - dy$ and also corresponding quadratic form is $2dx^2 + 2dy^2$. Looking at the matrix there is clearly an eigenvalue of $0$, yet the problem I am following says using the identity $dz = -dx - dy$ it is possible to "eliminate" $dz$ and show that the quadratic form is actually positive. How is $\lambda = 0$ discarded?

答: If $dz=-dx-dy$ then $[dz]^TAdz=(-1,-1,0)A(-1,-1,0)^T$ and because $A$ is positive definite with respect to $x$ and $y$ its apparent positive semi-definiteness is actually properly seen as positive definite. You're eliminating $dz$ as a variable from your space, which amounts to expressing it in the basis $dx,dy$ as that identity already does. If it had a $-1$ in the lower-right corner, then this simplification would be impossible (and the identity would be wrong because of the quadratic form associated with the matrix).

Title: Positive Definite Quadratic Form


问:
An Easter game related to the geometric distribution
Consider the following game. You have a bowl of 10 eggs. With your 10 eggs, you want to compete in an egg-braking contest for Easter. In other words, you take your egg and bang it against your opponents’ egg. For simplicity, let us only consider only one hit on one side with probability of a tie equal to zero. Each of your eggs has a probability of a loss $p_k,\quad k=\overline{1,10}$. This is an averaged probability, it does not change with change in the opposing egg. Let $\xi_i$ be the random variable "the number of eggs you broke with the i-th egg.

Let $$\xi=\min_{i=\overline{1,10}}\xi_i$$
What is the distribution of $\xi$?

My idea is that for each i, $\xi_i$ is geometrically distributed, but what can we tell about $\xi$? Can it be done with the generating functions?

答:
For each $i$, $\xi_i\sim Ge(p_i)\Rightarrow\xi=\min_i\xi_i\sim Ge(1-\prod_i(1-p_i))$.
This follows by induction and the equality $$\min_{i=\overline{1,n}}x_i=\min\{\min_{i=\overline{1,n-1}}x_i,x_n\}$$ For the base of the induction look at this answer.</p>


问:
A "Fourier Series" for $\sin$?

Consider a series $\sum_{k=2}^\infty a_n \sin(kx)$, where $\{a_n\}$ is a sequence of reals. Is it possible to construct $\{a_n\}$ such that  $\sum_{k=2}^\infty a_n \sin(kx)$ converges uniformly to $\sin(x)$ on $[0,\pi]$?

If there were indeed an expression
$$
\sin(x) = \sum_{k=2}^\infty a_n \sin(kx)
$$
then it implies
$$
0 = \sin(x) - \sum_{k=2}^\infty a_n \sin(kx)
$$
saying that $0$ has a Fourier sine series, which is nonsense. But I can't think up a proof to show the problem, so is it really impossible to construct one?

答: If it is true you get $\int_0^{\pi}\sin^{2}xdx=\sum_k a_k \int_0^{\pi} \sin x \sin (kx)dx=\sum a_k 0=0$ which is a contradiction.

Title: A "Fourier Series" for $\sin$?


问:
unit ball of $(C([0, 1], \mathbb{R}), \mathrm{d}\infty)$ is not compact
I am trying to show that the unit ball of $(C([0, 1], \mathbb{R}), \mathrm{d}\infty)$ is not compact.
thanks

答:
The constant functions $f_t(x)=t$ for all $x$ give an uncountable compact subset of $C[0,1]$.

The sequence $\{1,x,x^{2},...\}$ has no convergent subsequence , so the unit ball is not compact.


问:
Continuous functions on $(0,1)$ that have no fixed point
I need an example of a continuous function that does not have a fixed point, a function $f:(0,1)\to(0,1)$ with domain $(0,1)$ and image $(0,1)$.
Thanks.

Body:
$$
f(x)=x^2
$$
Also, $f(x)=x^a$, for every $a\ne 1$, $a>0$.


问:
Nondegenaracy of differential mapping for differmorphism?
Let $\Psi : U_p \to (-\delta,\delta)^2 \subseteq \mathbb{R}^2 $ be a differmorphism, the differential mapping is defined as
$$
 D \Psi = \left(\begin{matrix}
\frac{\partial \Psi_1}{\partial x_1} & \frac{\partial \Psi_2}{\partial x_1} \\
\frac{\partial \Psi_1}{\partial x_2} & \frac{\partial \Psi_2}{\partial x_2}
\end{matrix}\right).
$$

It is always needed to verify the nondegeneracy of differential mapping for $ \Psi $ to be a differmorphism. On the contrary, given a differmorphism $ \Psi $, how can we prove that $ D \Psi $ is nondegenerated?
Thanks in advance!

答: If you have a diffeomorphism $\Psi$ you have that there exists a smooth inverse $\Psi^{-1}$ and so we have that $D\Psi \circ D\Psi^{-1}=D(\Psi\circ \Psi^{-1})=D(Id)=Id =D(\Psi^{-1}\circ \Psi)=D\Psi^{-1}\circ D\Psi$.


问:
one 1-form on the Riemann surface of an algebraic function
Given the many-valued algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$,  we can get a Riemann surface  $S$ that is topologically equivalent to a torus. I am wondering whether the $1$-form
$ \displaystyle\frac{dz}{w}$ is a holomorphic form on the torus $S$.

On the surface we know that

$$ \frac{dz}{w}=\frac{dz}{\sqrt{(z-1)(z-2)(z-3)}},$$

I think that $z=1,2,3$ are poles of the above 1-form, so it is a meromorphic 1-form on the $z-$plane. Is it a holomorphic 1-form on the torus $S$ determined by the algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$?

I just know a little about Riemann surface theory. I will appreciate any suggestions and comments. Thanks.

答: Given the algebraic function $w^2=(z-1)(z-2)(z-3)$, differentiate each side, we get $$2wdw=[(z-2)(z-3)+(z-1)(z-3)+(z-1)(z-2)]dz .$$ Therefore we have $$\frac{dz}{w}=\frac{2dw}{(z-2)(z-3)+(z-1)(z-3)+(z-1)(z-2)}.$$ It is easy to see that the RHS is holomorphic locally at (1,0), (2,0),(3,0) respectively.


问:
How to prove the use of boundedness may be the key to this problem
Let $f$ be a bounded convex function on $(0,+\infty)$,prove that $f$  is decreasing and
$$\displaystyle\lim_{x\to+\infty}xf'_+(x)=0.$$
Geometrically speaking, the significance of the first problem is obvious, but I don't know how to analyze and prove it.I try to use the following inequality$$\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\leqslant\dfrac{f(x_3)-f(x_1)}{x_3-x_1}\leqslant\dfrac{f(x_3)-f(x_2)}{x_3-x_2}.$$I don't know how to take advantage of boundedness...

答:
For the first part we need that $f$ is bounded above, e.g. by $M$: For $0 < x < y < z$ we have
$$
 f(y) \le \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x}f(z) \le \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x} M \, .
$$
Now take the limit for $z \to \infty$ to conclude that $f(y) \le f(x)$, i.e. $f$ is decreasing.

For the second part we need that $f$ is also bounded below: $f$ is  decreasing which implies that $L = \lim_{x \to \infty} f(x)$ exists. Now write the convexity condition for $0 < x/2 < x < y$ as
$$
 \frac{f(x)-f(x/2)}{x-x/2} \le  \frac{f(y)-f(x)}{y-x} \le 0 \, .
$$
Taking the limit $y \to x^+$ gives
$$
 \frac{f(x)-f(x/2)}{x-x/2} \le f'_+(x) \le 0 \\
\implies 2 \bigl(f(x) - f(x/2) \bigr) \le x f'_+(x) \le 0 \, .
$$
For $x \to \infty$ the left-hand side converges to $2 (L-L) = 0$, and it follows that $\lim_{x \to \infty} x f'_+(x) = 0$.


问:
Find all prime ideals and maximal ideals of $\mathbb{Z}_{36}$
I'm trying to solve this problem from my abstract algebra course:

Find all prime and maximal ideals of $\mathbb{Z}_{36}=\mathbb{Z}/36\mathbb{Z}$.

I've seen other posts about this topic, but I still don't get it since they have orders so much lower that $36$. I think in this case the prime and the maximal ideals are the same, and I guess that's because $\mathbb{Z}_{36}$ is a principal ideal domain (if this were the case, how can I also prove this? In a general case $\mathbb{Z}_n$ assuming every $\mathbb{Z_n}$ verifies that it is a principal ideal domain).

After some research I found out that the maximal ideals of $\mathbb{Z}_{36}$ are $k\mathbb{Z}/36\mathbb{Z}$ for $k\in\{1,\dots,35\}$, but these are so much posibilities to check to not exist some easier way to see it, like checking only the coprimes to $36$ (or the non-coprimes, I'm not sure about what of them should I check, that's why I'm asking). How can I act in these kind of problems?

Thanks in advance, any help will be appreciated.

答:
$\Bbb Z_{36}$ is not a principal ideal domain, however it is a principal ideal ring. In general for a commutative ring $R$ with ideal $I$ the ideals of $R/I$ correspond bijectively to the ideals of $R$ containing $I$. This correspondence preserves prime/maximal/... ideals. So in this case the ideals in $\Bbb Z$ containing $36\Bbb Z$ are exactly $$\Bbb Z,2\Bbb Z,3\Bbb Z,4\Bbb Z,6\Bbb Z,9\Bbb Z,12\Bbb Z,18\Bbb Z,36\Bbb Z$$
(All divisors of $36$) By quotienting out $36\Bbb Z$ we get the corresponding ideals of $\Bbb Z/36\Bbb Z$. I will let you figure out which of these are maximal.


问:
Real Analysis Differentiation
Real analysis question about differentiation

Assume that $f$ and $f'$  are differentiable on $\mathbf R$ and that for every x in $\mathbf R$,  $f(x) + f''(x) = 0 $. Show that $g (x)= f^2(x) + (f'(x))^2$ is a constant.

Title: Real Analysis Differentiation

Body:
Solution of ode $f+f''=0$ is $f=c_1\cos x+c_2\sin x$. Then
$$f^2+f'^2=c_1^2+c_2^2$$


问:
For two linear maps, does having the same matrix when expressed in different bases imply they are equal?
I am working on some image filters and I have come across this issue. I can express every point $u$ of the image (the space for all that matters) in two distinct packets (bases for the space) $\{x_i\}, \{y_i\}$. That is for all $u$,

$$u=\sum u_i x_i, \ u=\sum v_i y_i$$
where $u_i,v_i$ are scalars.

Applying the filter amounts to considering two linear maps on the space, call them $B$ and $C$. I managed to express these on the two distinct bases and they have the same matrix $a_{ij}$:
$$
Bx_i=\sum_j a_{ij}x_j, \ Cy_i=\sum_j a_{ij}y_j
$$

My question is, how are $B$ and $C$ related in this context?

Am I justified to say that they are equal? Is there a theorem I can invoke?

Thanks a lot.

答:
Edit: the discussion is relevant only when the domain and the range of the linear map are the same, otherwise we cannot talk about similarity, regards to Christoph for the remark.

In general case - no, they are not equal. Let us declare a linear transformation $f: R^2 \to R^2$ and $g: R^2 \to R^2$.
Let us fix 2 bases in $R^2: B_{1} = \{(1, -1)^T, (2, 3)^T\}$ and $B_{2} = \{(4, 6)^T, (3, 1)^T\}$ (they constitute a basis, because they are linearly independent and there are 2 of them, which is $dim(R^2)$).

Let us say that linear transformations $f$ and $g$ have the same matrix with respect to bases $B_{1}$ and $B_{2}$ respectively:

$[f]_{B_{1}} = [g]_{B_{2}}= \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$

We would need the following property on which you can read more in the Change of basis topic (or here) while discussing linear maps. This is a corollary of a more general theorem for the linear maps which have the same domain and range:

Assume that $f : R^n \to R^n$ is a linear map and let $B=\{v_{1},...,v_{n}\}$ and $C=\{v'
_{1},...,v'_{n}\}$ be bases of $R^n$. Let $g : R^n \to R^n$ be the uniquely determined linear map for which $g(v_{j}) = v'_{j}$ holds for every $1 ≤ j ≤ n.$ Then $[f]_{C} = [g]^{−1}_{B}[f]_{B}[g]_{B}$. Likewise $[f]_{B} = [g]_{B}[f]_{C}[g]_{B}^{-1}$.

We will show that the linear maps defined above have different matrices with respect to the standard basis $\{(1,0)^T, (0, 1)^T\}$ in $R^2$.

$[f]$

We need matrix of linear map $h$ which maps standard basis in $R^2$ to the basis $B_{1}$

$[h] = \begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$

$[h]^{-1} = \frac{1}{5}*\begin{bmatrix}3 & -2\\1 & 1\end{bmatrix}$

$[f] = [h][f]_{B_{1}}[h]^{-1} = \frac{1}{5}*\begin{bmatrix}7 & -3\\8 & -7\end{bmatrix}$

$[g]$

As in the first case, we need matrix of linear map $h$ which maps standard basis in $R^2$ to the basis $B_{2}$

$[h] = \begin{bmatrix}4 & 3\\6 & 1\end{bmatrix}$

$[h]^{-1} = \frac{1}{14}*\begin{bmatrix}-1 & 3\\6 & -4\end{bmatrix}$

$[g] = [h][g]_{B_{2}}[h]^{-1} = \frac{1}{2}*\begin{bmatrix}3 & -1\\5 & -3\end{bmatrix}$

As we can see, matrices of linear maps $f$ and $g$ w.r.t. the standard basis are not equal, that is, these maps are not the same.


问:
"Flat chain complex"?
Given a flat module $F$ and an exact sequence $A_\bullet$, a tensor product of chain complexes
$$F\otimes A_\bullet$$
is again exact, where $\otimes$ is defined as in Tensor product of chain complexes
and $F$ regarded as a chain complex concentrated in degree $0$.

My question is, is it true that for any chain complex $F_\bullet$ whose components are all flat and an exact sequence $A_\bullet$, a tensor product of chain complexes
$$F_\bullet\otimes A_\bullet$$
is again exact?

Moreover, how about the inner-hom $[F_\bullet, A_\bullet]$?

答:
If there are no boundedness conditions on the complexes then this need not be true. For example, let $F_\bullet=A_\bullet$ be the unbounded complex
$$\cdots\xrightarrow{\times 2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times 2}\cdots$$
of $\mathbb{Z}/4\mathbb{Z}$-modules (so it is both exact and a complex of flat $\mathbb{Z}/4\mathbb{Z}$-modules).

Then each component of $F_\bullet\otimes_{\mathbb{Z}/4\mathbb{Z}}A_\bullet$ is isomorphic to
$$\bigoplus_{i\in\mathbb{Z}}\mathbb{Z}/4\mathbb{Z},$$
the kernel of the differential is
$$\bigoplus_{i\in\mathbb{Z}}2\mathbb{Z}/4\mathbb{Z},$$
and the image of the differential is
$$\left\{(a_i)_{i\in\mathbb{Z}}\in\bigoplus_{i\in\mathbb{Z}}2\mathbb{Z}/4\mathbb{Z}
\,\middle\vert\,\sum_{i\in\mathbb{Z}}a_i=0\right\}.$$
So $F_\bullet\otimes_{\mathbb{Z}/4\mathbb{Z}}A_\bullet$ has homology isomorphic to $2\mathbb{Z}/4\mathbb{Z}$ in each degree, and in particular is not exact.

However, with suitable boundedness conditions, it is true.

If $F_\bullet$ is bounded, then an induction argument on the length of $F_\bullet$ shows that $F_\bullet\otimes A_\bullet$ is exact. From this one can prove that $F_\bullet\otimes A_\bullet$ is exact if

(a) either $F_\bullet$ or $A_\bullet$ is bounded, or

(b) both $F_\bullet$ and $A_\bullet$ are bounded above, or

(c) both $F_\bullet$ and $A_\bullet$ are bounded below.

[Alternatively, spectral sequence arguments can be used.]

Similar results hold for the internal Hom, but conditions (b) and (c) are replaced by the condition that one of $F_\bullet$ and $A_\bullet$ is bounded below, and the other is bounded above.

Title: "Flat chain complex"?


答: Any continous function will do as long as its graph does not intersect the diagonal line $y=x$. So for example the function $$x\mapsto \frac x2$$ will work, since $   \frac x2<x$ in the region of interest.  Similarly $$x\mapsto\frac {x+1}2$$ or $$ x\mapsto \min\bigg(\frac x3, \frac14\bigg)$$ or what have you.

Title: Continuous functions on $(0,1)$ that have no fixed point

Body:
I need an example of a continuous function that does not have a fixed point, a function $f:(0,1)\to(0,1)$ with domain $(0,1)$ and image $(0,1)$.
Thanks.


问:
Why are linear functions the natural analogue of exponential functions in a tropical semiring?
I was reading a blog post on the Fourier transform and the Legendre transform as being the same thing over different semirings, in which the author says

It's not obvious how to interpret the exponential function in
  (R',min,+) but it turns out that the natural choice is to consider the
  ordinary linear functions (in the conventional sense) to be the
  correct analogue.

So why are ordinary linear functions a natural analogue for the exponential function in a tropical semiring?

(Note: an intuitive answer is fine. I don't know any tropical geometry and only basic algebraic geometry, so not necessarily looking for a very technical response, though that would be awesome as well.

答:
You haven't logged in in years, but seeing as someone just edited the tags on this question maybe other people will find it.

I think "it's not obvious" is being a little too pessimistic. Insofar as exponentiation is repeated multiplication, and multiplication in a tropical semiring is addition, it makes sense to define the tropical exponential "$x^n$" to be the linear function $nx$, at least for integer $n$. Then you extend to rational exponents $\frac{p}{q}$ by taking unique roots; that is, you ask that the "tropical exponential" satisfy the property that "$\left( x^{\frac{p}{q}} \right)^q$" is equal to "$x^p$" which forces the exponential to be $\frac{p}{q} x$. And then, just as for the ordinary exponential, the result for real exponents is determined by continuity.

The main conceptually tricky thing about this argument, to my mind, is that you are forced to make clear distinctions between the base and the exponent: while the base is an element of the tropical semiring the exponent is an ordinary integer or rational number or real number, regarded as a ring with the usual ring operations. This interpretation is necessary to make the exponent rules continue to be true: for example, we have "$x^{n+m} = x^n x^m$" but while the addition on the LHS is ordinary the multiplication on the RHS is tropical, since this identity really says

$$(n+m)x = nx + mx$$

and similarly we have "$(x^n)^m = x^{nm}$" which says

$$m(nx) = mnx.$$

A closely related argument that comes to the same conclusion is the following. You can pass from ordinary addition and multiplication to tropical addition and multiplication as follows: for a real parameter $t$ consider the family of operations

$$x +_t y = \frac{\log (e^{tx} + e^{ty})}{t}$$
$$x \times_t y = \frac{\log (e^{tx} e^{ty})}{t}$$

obtained by conjugating ordinary addition and multiplication by the map $x \mapsto e^{tx}$. The second operation is just always equal to tropical multiplication $x + y$. The first operation approaches tropical addition $\text{min}$ as $t \to - \infty$. Intuitively speaking this says that the tropical operations describe the asymptotic behavior of ordinary addition and multiplication applied to exponentials with very large (negative in this case, but you can use a different tropical semiring with $\text{max}$ corresponding to the positive case) exponents.

So tropical exponentiation ought to be given by conjugating ordinary exponentiation $x^r$ by the same map. This gives

$$\frac{\log e^{txr}}{t} = rx$$

as expected (and as with multiplication we don't even need to take the $t \to - \infty$ limit). And again we see the key feature that the base $x$ is treated as an element of the tropical semiring but the exponent $r$ is not.


问:
How to minimize $\Big( \frac{2}{a^2}+\frac{1}{b^2}\Big)(a^2+b^2)$
I've tried completing the square but I can't however i think there's something called a cauchy schwarz inequality that could help me solve this. Help?

But how $\frac2{a^2}$ is not a square number for the inequality $$(++)^2\le(^2+^2+^2)(^2+^2+^2)$$ do you split it into $\frac1{a^2}+\frac1{a^2}$?

答:
As you stated

$$\left( \left( \frac{\sqrt2}a\right)^2 + \left(\frac1b\right)^2\right)\left( a^2+b^2\right) \ge \left(\sqrt2 + 1 \right)^2$$

Now, let's check if it can be attained.

If $b=1$ and $a=\sqrt[4]2$,

$$\left( \left( \frac{\sqrt2}{\sqrt[4]2}\right)^2 + \left(\frac11\right)^2\right)\left( \sqrt[4]2^2+1^2\right) = \left(\sqrt2 + 1 \right)^2$$


问:
Change of variable in univariate polynomials
Given $ f(x) = a_{0} + a_{2} x^{2} +  \cdots + a_{n}  x^n $.

Define $ g(t) $ as $ f(15t -7) $.

How can I show that

$$ g(t) = b_{0} + b_{2} t^{2} +  \cdots + b_{n}  t^n $$

Do I need calculus for this?

答:
By the binomial theorem (or by the comment of Pedro),

$$(15t-7)^k$$ is a sum of powers of the two terms, with total degree $k$. Hence it is a polynomial in $t$ of degree $k$.

Now the original expression is a linear combination of such terms, hence it is also a polynomial in $t$, of maximum degree $n$.

By a similar reasoning,

$$f(g(t))$$ where $f$ is a polynomial of degree $d_f$, and $g$ a polynomial of degree $d_g$, is a polynomial of degree $d_fd_g$.


问:
Prove that $\{x\} \times B \in R^{n} \times R^{m} $ is compact, when $B \in R^{m} $ is compact.
Prove that $\{x\} \times B \subseteq \Bbb{R}^{n} \times \Bbb{R}^{m} $ is compact, when $B \subseteq \Bbb{R}^{m} $ is compact.

My attempt:
Let $\{x\} \times B \subset \cup G_{a} $, where $\cup G_{a} = \cup G_{a1} \times \cup G_{a2}$ such that $\{x\} \subset \cup G_{a1}$ and $B \subset \cup G_{a2}$. $( G_{a}  , G_{a1} $ and $G_{a2}
$
are open sets.$) $

So there exists an open set such that  $\{x\} \subset O_{a1} \subset \cup G_{a1}$ and a finite open cover of $B$ such that $ B \subset \bigcup_{a2=1}^{n} G_{a2} \subset
\cup G_{a2}$

This implies $\{x\} \times B \subset O_{a1} \times \bigcup_{a2=1}^{n} G_{a2}$

So, $\{x\} \times B$ is compact.
Is this proof correct$? $
If it's not right then please help me out with right one using the definition of compact set (without using heine borel theorem.)

答:
Not every open set of $\Bbb R^m \times \Bbb R^n$ is of the form $U\times V$, with $U\subseteq \Bbb R^m$, $V\subseteq \Bbb R^n$ open. Take for example the interior of the unit ball $B(1,0)\subseteq \Bbb R^2$. I understand your definition of your cover in that way...

Yet, it is the right idea. As the open sets of the form $U \times V$ form a basis of the topology on $\Bbb R^m \times \Bbb R^n$, if we are given a cover
$$X \subseteq \bigcup \limits_{i \in I} W_i \;\;\;(1)$$
of a subspace $X \subseteq \Bbb R^m \times \Bbb R^n$, we immediately obtain another cover
$$X\subseteq \bigcup \limits_{i \in I} \bigcup \limits_{j \in J_i} U_{ij}\times V_{ij}\;\;\;(2)$$
writing the opens $W_i$ with respect to the basis.

Now assume that $X = \{x\} \times K$ for some $x\in \Bbb R^m$ and compact $K \subseteq \Bbb R^n$. Then the cover (2) yields an open cover
$$K \subseteq \bigcup \limits_{i \in I, j \in J_i} V_{ij}.$$
By compactness of $K$ finitely many $V_{ij}$ suffice, with pairs of indices denoted  $i_1j_1,...,i_kj_k$. Note that every $U_{ij}$ has to contain $\{x\}$. Hence we find
$$\{x\}\times K \subseteq \bigcup \limits_{t=1}^k U_{i_tj_t}\times V_{i_tj_t} \subseteq \bigcup \limits_{t=1}^k W_{i_t}$$
which is a finite subcover of our original cover (1).


问:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$

Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$

The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.

The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)

Body:
WLOG

$$\sqrt{x^2+12y}=33\cos^2t\text{ and }\sqrt{y^2+12x}=33\sin^2t$$

$$(33\cos^2t)^2-(33\sin^2t)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$

$$x+y=23\implies x-y=99(\cos^2t-\sin^2t)=99(2\cos^2t-1)$$

$$\implies x=99\cos^2t-38\implies y=61-99\cos^2t$$

$$\implies(33\cos^2t)^2=x^2+12y=(99\cos^2t-38)^2+12(61-99\cos^2t)$$

Let $33\cos^2t=b$  $$\implies0=b^2-33b+272=(b-16)(b-17)$$

Hope you can take it from here?


答:
Just expand to find $$2+1+\dfrac{2b^2}{a^2}+\dfrac{a^2}{b^2}=3+\left(\dfrac{\sqrt2b}a-\dfrac ab\right)^2+2\sqrt2\ge3+2\sqrt2$$

the equality occurs if $\dfrac{\sqrt2b}a-\dfrac ab\iff\sqrt2b^2=a^2$ with $a,b$ non zero finite real

Title: How to minimize $\Big( \frac{2}{a^2}+\frac{1}{b^2}\Big)(a^2+b^2)$

Body:
I've tried completing the square but I can't however i think there's something called a cauchy schwarz inequality that could help me solve this. Help?

But how $\frac2{a^2}$ is not a square number for the inequality $$(++)^2\le(^2+^2+^2)(^2+^2+^2)$$ do you split it into $\frac1{a^2}+\frac1{a^2}$?


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

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

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

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

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

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

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

The answer is 6

答:
You can't square the equation the way that you did.

What you did was essentially
$$
\sqrt{3x-2}^2+(2-x)^2 = 0
$$
but it should be
$$
\sqrt{3x-2}^2 = (x-2)^2
$$
which becomes
$$
3x-2 = x^2-4x+4
$$
Solve from there.

And generally, if you have an equation, you can't apply operations to only part of one side. The rule is that "what you do to one side, you do to the other". So if you square the left side, you square the right side. If you were to square the left side of your original equation, it would become
$$
(\sqrt{3x-2}+2-x)^2=0^2
$$
which won't get you where you want to go (at least, not as easily - you can get there by using the original equation to substitute out the square root from that point).

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


问:
Modified heat equation - solving PDE
Conisder the following two PDEs:

$(1)$:$\;$$\displaystyle\frac{\partial \omega_1'}{\partial t}+A x_2 \frac{\partial \omega_1'}{\partial x_1}=\nu\left(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2}\right)$

and

$(2)$:$\;$
$
\begin{cases}
\displaystyle\frac{\partial \omega_1'}{\partial t}+A  \frac{\partial u_3'}{\partial x_1}+A x_2 \frac{\partial \omega_1'}{\partial x_1}=\nu\left(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2}\right)\\ \\
\displaystyle\omega_1'=\frac{\partial u_3'}{\partial x_2}-\frac{\partial u_2'}{\partial x_3}\\ \\
\displaystyle\frac{\partial u_2'}{\partial x_2}+\frac{\partial u_3'}{\partial x_3}=0
\end{cases}
$

The first PDE $(1)$ highly resembles the heat equation, aswell as the first equation in $(2)$.

How can I solve for $\omega_1'$ in both cases?

I know how to solve the heat equation using the method of separation of variables but I do not know how to apply it on these PDEs.

I am very grateful for any advice.

答:
$$\displaystyle\frac{\partial \omega_1'}{\partial t}+A x_2 \frac{\partial \omega_1'}{\partial x_1}=\nu\left(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2}\right)$$
They are four variables : $x_1\:,x_2\:,x_3\:,t.$ The function has to be separated into four functions :
$$\omega_1'=f(x_1)g(x_2)h(x_3)v(t)$$
$$fghv'+Ax_2f'hgv=\nu(fg''hv+fgh''v)$$
$$\frac{v'}{v}+Ax_2\frac{f'}{f}-\nu\frac{g''}{g}-\nu\frac{h''}{h}=0$$
A function of one variable can be equal to a function of a different variable any values of the variables only if both functions are equal to a common constant. Thus :
$$\begin{cases}
\frac{v'}{v}=a \qquad\text{function of}\quad t\quad\text{only.}\\
\frac{f'}{f}=b \qquad\text{function of}\quad x_1\quad\text{only.}\\
\frac{h''}{h}=c \qquad\text{function of}\quad x_3\quad\text{only.}\\
a+bAx_2-\nu\frac{g''}{g}-c\nu=0 \qquad\text{function of}\quad x_2\quad\text{only.}
\end{cases}$$
$g$ is obtained in solving the ODE : $\quad \nu g''=(a+bAx_2-c\nu)g\quad$ which is an Airy ODE.

$a,b,c$ are arbitrary real or complex constants.
$$\begin{cases}
v=c_1e^{at}\\
f=c_2e^{bx_1}\\
h=c_3e^{\pm\sqrt{c}\:x_3}\\
g=c_4\text{Ai}\left(z\right)+c_5\text{Bi}\left(z\right)\quad ;\quad z=\frac{a-c\nu+bAx_2}{(bA)^{2/3}\nu^{1/3}}
\end{cases}$$
$\text{Ai}(z)$ and $\text{Bi}(z)$ are the Airy functions.

$c_1,c_2,c_3,c_4,c_5$ are are arbitrary constants.

Independant solutions of the PDE :
$$\omega_1'=C_1e^{at}e^{bx_1}e^{\pm\sqrt{c}\:x_3}\text{Ai}\left(\frac{a-c\nu+bAx_2}{(bA)^{2/3}\nu^{1/3}}\right)+C_2e^{at}e^{bx_1}e^{\pm\sqrt{c}\:x_3}\text{Bi}\left(\frac{a-c\nu+bAx_2}{(bA)^{2/3}\nu^{1/3}}\right)$$

Any linear combination of those independant solutions with different arbitrary constants $C_1,C_2,a,b,c$ is solution of the PDE.

HINT :

In the case of the system of three functions $\omega_1',u'_1,u'_2$ proceed on the same manner with :
$$\begin{cases}
\omega'_1=f_1(x_1)g_1(x_2)h_1(x_3)v_1(t)  \\
u'_2=f_2(x_1)g_2(x_2)h_2(x_3)v_2(t)  \\
u'_3=f_3(x_1)g_3(x_2)h_3(x_3)v_3(t)  \\
\end{cases}$$
Separate the functions of different variables as shown above and find the 12 independant functions each one of one variable only.

I let for you this awfully booring task. Good luck !

Title: Modified heat equation - solving PDE


问:
$y = \beta_0 + \beta_1 x_1 + \dots + \beta_p x_p$, unknowns $(x_1, \dots, x_p, y)$: Describes a hyperplane of affine space $\mathbb{R}^{p + 1}$?
Let's say we have an equation $y = \beta_0 + \beta_1 x_1 + \dots + \beta_p x_p$, where the unknowns are $(x_1, \dots, x_p,y)$. It is said that this describes a hyperplane $H$ of the affine space $\mathbb{R}^{p + 1}$. How is this a hyperplane of the affine space $\mathbb{R}^{p + 1}$? What is it about the mathematics of affine spaces that makes this so?

答:
Affine:

Affine is a term to say it not a linear function, but shifted/translated( do not confuse with a  nonlinear function either). Example $y = ax$  is linear as it includes the $0$ point. It(Linear) accepts the principle of superposition and homogeneity, whereas $y = ax + b$ is not a linear function but an affine function.

For your function the constant term $\beta_{0}$ makes it an affine function.

Next Hyperplane,

Hyperplane or a Hypersurface is a subspace of $n$ - dimensional space with one dimension less. Normally its something that splits an $n$ dimensional space.

General examples that are shown are.

"a point" is a hyperplane in 1-D space $\mathbb{R}^{1}$. It splits the line.

"a line" is a hyperplane in 2-D space $\mathbb{R}^{2}$. It splits the plane.

"a plane" is a hyperplane in 3-D space $\mathbb{R}^{3}$.

also an unit sphere is a "hypersurface" or a 2-D object in a 3-D space

Other way to see may be is , how many variables you require to define a point in a  surface(in general)/plane. Always it is one dimension less.

So if a plane is defined as $Ax +By + Cy +d$ then $x,y,z$ is the space $\mathbb{R}^{3}$ plus a shift/translation/a constant term that is "real value" makes it to $\mathbb{R}^{3+1}$ which is an affine space(not $\mathbb{R}^{4}$,  . So what above describes is a "hyperplane" a plane of one dimension less in that affine space.

So in general an affine space is defined by $\mathbb{R}^{n+1}$

If you don't have the constant term, then it is a plane containing the origin and you can relax mentioning it as an affine space.


问:
minimal graphs converge to a harmonic function
Suppose $\{u_i\}$ is a sequence of positive $C^2$-functions on the unit ball $B,$ satisfying the minimal surface equation, i.e.,
$${\rm div}\left(\frac{Du_i}{\sqrt{1+|Du_i|^2}}\right)=0$$
for all $i.$ Define another sequence
$$w_i:=\frac{u_i}{u_i(0)}.$$
The problem is as follows. If $u_i$ converges to $0$ uniformly to $0$ on any compact subset of $B,$ with uniform $C^2$-bound (that is, $|u_i|+|Du_i|+|{\rm Hess}_{u_i}|<\varepsilon$ for all $i$), will a subsequence of $w_i$'s converge to a "harmonic" function?

Since the minimal surface equation for $u$ could be written as
$$\sum_{i,j}\left(\delta_{ij}-\frac{D_iuD_ju}{1+|Du|^2}\right)D_{ij}u=0,$$
I think it suffices to prove that $|Du_i|$ converges to $0$ also. However, I even don't know if this is correct.

Any suggestion or comments are appreciated!

答:
Finally I somehow get a satisfactory explanation. The equation could be written as
\begin{equation}\label{MSE}
\sum_{i,j}\left(\delta_{ij}-\frac{D_iuD_ju}{1+|Du|^2}\right)D_{ij}u=0.
\end{equation}
Thus we attempt to show a subsequence of $Du_i$ converges to $0.$ For any $\varphi\in C^\infty_0(B),$  we have
$$0=\int \varphi u_i \text{div}\frac{Du_i}{\sqrt{1+|Du_i|^2}}=\int\left\langle \frac{Du_i}{\sqrt{1+|Du_i|^2}},u_iD\varphi+\varphi Du \right\rangle.$$
Thus
$$\int \frac{\varphi|Du_i|^2}{\sqrt{1+|Du_i|^2}}\le \int\frac{|u_i||Du_i||D\varphi|}{\sqrt{1+|Du_i|^2}}\le \varepsilon $$
for any $i=i(\varphi,\varepsilon)$ sufficiently large. Take $\varphi=1$ on $\overline{B_r(x)},$ and we derive (based on the a priori bound of $Du_i$)
$$C\sup_{\overline{B_r(x)}}|Du_i|^2\le \varepsilon,$$
so $Du_i\to 0$ as $i\to\infty.$

Finally we show that $w_i$'s converge subsequentially (to a harmonic function by above). The the equation is uniformly elliptic since $|Du|$ is counded. Since $w_i$ is positive and $w_i(0)=1,$ the Harnack inequality implies
$$\|w_i\|_{C^0(K)}\le C(K)$$
for any $K\subset\subset B,$ and the Schauder estimate implies
$$\|w_i\|_{C^{2,\alpha}(K)}\le C(K)$$
also. The Arzela-Ascoli theorem then implies the subsequential convergence.


答:
If you meant

$$\sqrt{3x-2}+2-x=0\implies \sqrt{3x-2}=x-2\implies3x-2=(x-2)^2=x^2-4x+4\implies$$

$$\implies x^2-7x+6=0$$

Now just check that $x^2-7x+6=(x-6)(x-1)$ ...and remember to check at the end whether both solutions of this quadratic are actually solutions of your original equation, since when squaring some mess can happen there. For example, one of the soltuions of the quadratic is not a solution of your equation.

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

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

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

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

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

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

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

The answer is 6


问:
Do we have $f=\mathrm{Id} +c$ if $f'=1$ almost everywhere and $f$ is 1-Lipschitz?
Assume that we have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that is $1$-Lipschitz, and so that $f'(x)=1$ almost everywhere.

Can we show that there exist a unique constant $c$ so that $f(x)=x+c$ ?

The main problem of this question is that the constant $c$ that we find depends on the connexe component of the set $R\setminus A=R\setminus \{x \vert f'(x)=1\}$ with $\lambda(A)=1$.
The goal is to find that it is the same constant for any of these components using 1-Lipschitzness of $f$, but I couldn't make it.

答: Lipschitz implies absolute continuity. So $f(x)=f(0)+\int_0^{x} f'(t)dt=f(0)+x$. [See Theorem 7.20 of Rudin's RCA].


问:
Evaluate $\int_{-\infty}^{-1}e^{\lambda t}\operatorname{Log}\left(\frac{\lambda+1}{\lambda}\right)\Bbb d\lambda$
This question is still related to my question 3 days ago about inverse laplace transform. Here is the link: enter link description here.
Now, i'm working on path $L_1$ which led me to this integral when i set the radius around branch point goes to $0$. Now forget about my question 3 days ago, and my question here is how to evaluate this improper integral:

$$\int_{-\infty}^{-1}e^{\lambda t}\operatorname{Log}\left(\frac{\lambda+1}{\lambda}\right) \Bbb d\lambda$$.

$\lambda$ is now real variable instead of complex variable since my path goes close to real axis. I need to compute this integral but the problem is $t$ is unkown and i can't approximate it using geogebra. How to process this? I haven't tried to taylor it then integrating term by term cz i'm not sure if the given answer will be as my desired result.

答:
Integral in terms of $\boldsymbol{\Gamma(0,t)}$

$$
\begin{align}
&\int_{-\infty}^{-1}e^{\lambda t}\log\left(\frac{\lambda+1}\lambda\right)\,\mathrm{d}\lambda\\
&=e^{-t}\int_0^\infty e^{-tx}\log\left(\frac{x}{x+1}\right)\,\mathrm{d}x\tag1\\
&=\frac{e^{-t}}t\int_0^\infty e^{-x}\log\left(\frac{x}{x+t}\right)\,\mathrm{d}x\tag2\\
&=\frac{e^{-t}}t\int_0^\infty e^{-x}\log(x)\,\mathrm{d}x-\frac{e^{-t}}t\int_0^\infty e^{-x}\log(x+t)\,\mathrm{d}x\tag3\\
&=-\frac{\gamma e^{-t}}t-\frac1t\int_t^\infty e^{-x}\log(x)\,\mathrm{d}x\tag4\\
&=-\frac{\gamma e^{-t}}t-\frac{e^{-t}\log(t)}t-\frac1t\int_t^\infty\frac{e^{-x}}x\,\mathrm{d}x\tag5\\
&=-\frac{\gamma+\log(t)}te^{-t}-\frac1t\Gamma(0,t)\tag6
\end{align}
$$
Explanation:

$(1)$: substitute $\lambda\mapsto-x-1$

$(2)$: substitute $x\mapsto x/t$

$(3)$: $\log(a/b)=\log(a)-\log(b)$

$(4)$: $\int_0^\infty e^{-x}\log(x)\,\mathrm{d}x=-\gamma$, see this answer or $(4)$ from this answer

$\phantom{\text{(4):}}$ substitute $x\mapsto x-t$ in the right-hand integral

$(5)$: integrate by parts

$(6)$: apply the Upper Incomplete Gamma function

Asymptotic Expansion for $\boldsymbol{\Gamma(0,t)}$

We can integrate by parts multiple times to get
$$
\begin{align}
\int_t^\infty\frac{e^{-x}}x\,\mathrm{d}x
&=e^{-t}\sum_{k=0}^{n-1}\frac{(-1)^kk!}{t^{k+1}}+(-1)^n\overbrace{n!\int_t^\infty\frac{e^{-x}}{x^{n+1}}\,\mathrm{d}x}^{\le n!\frac{e^{-t}}{t^{n+1}}}\tag7\\
&\sim e^{-t}\left(\frac1t-\frac1{t^2}+\frac2{t^3}-\frac6{t^4}+\dots\right)\tag8
\end{align}
$$
Note that the contribution by the integral is no more than the next term in the series.

According to $(7)$, the right-most term in $(6)$
$$
\frac1t\Gamma(0,t)\le\frac{e^{-t}}{t^2}\tag9
$$
is much smaller than the preceding terms as $t\to\infty$.

Series Expansion for $\boldsymbol{\Gamma(0,1)-\Gamma(0,t)}$
$$
\begin{align}
\int_1^t\frac{e^{-x}}{x}\,\mathrm{d}x
&=\int_1^t\left(\frac1x+\sum_{k=1}^\infty(-1)^k\frac{x^{k-1}}{k!}\right)\mathrm{d}x\tag{10}\\
&=\log(t)+\sum_{k=1}^\infty(-1)^k\frac{t^k-1}{k\,k!}\tag{11}
\end{align}
$$
For $k\ge et$, the terms are decreasing in absolute value and so the alternating series test says that the error from truncating the series at that point is at most the next term in the series.

Combining the Series and Asymptotic Expansions

The error when using the sum in $(7)$ is $\le\frac{n!\,e^{-t}}{t^{n+1}}$. For a given $t$, we don't want to use $n\gt t$ since the error gets bigger. For $t=n$, we get an error of $\sim\sqrt{\frac{2\pi}n}\,e^{-2n}$. Since $\frac2{\log(10)}\ge\frac67$, using $t\ge n\ge\frac76d$ will give at least $d$ digits of precision.

To get a similar error in $(11)$, we want to use $n\ge\frac{d\log(10)}{\operatorname{W}\left(\frac{d}{et}\log(10)\right)}$.

To compute $\Gamma(0,1)$ to $30$ places, we want to use $t=n=35$ in $(7)$, then $n=150$ in $(11)$ which gives
$$
\Gamma(0,1)=0.219383934395520273677163775460\tag{12}
$$
Having the value of $\Gamma(0,1)$ allows us to compute $\Gamma(0,t)$ with $(11)$ when that can give a more precise result than $(7)$.


问:
Multiplicity of Irreducible Representations for Faithful Representation of Group Algebra, $\mathbb{C}G$.
Let $G$ be a finite group and $\mathbb{C}G$ its associated group algebra. The regular representation $\mathbb{C}G\subset B(\ell^2(G))$ defined by:

$$\delta^g(e_{g_2})=e_{g_1g_2},$$

contains each irreducible representation $\mathbb{C}G\subset B(V_\alpha)$ with multiplicity equal to its degree.

In the case of $\mathbb{C}S_3$, where $V_0$ is associated to trivial representation, $V_1$ to the sign representation, and $V_2$ to the degree two irreduciblee representation, the regular representation is a faithful representation of $\mathbb{C}S_3$ on $\mathbb{C}^6\cong V_0\oplus V_1\oplus V_2\oplus V_2$.

However there is also a faithful representation of $\mathbb{C}S_3$ that just uses each irreducible representation once.

Question: Do we always have a faithful representation when we just use every irreducible representation just once?

答:
HINT:

If $\rho$ contains all the irreducible representations, then $k(G)\subset \rho^{\oplus n}$ for some $n$. If $\rho$ were not faithful, then so would be $\rho^{\oplus n}$, contradiction.


问:
Union of deletion Cantor sets
I know that it is possible to have deletion Cantor set which are of non-zero measure (fat Cantor sets) furthermore it must never of measure one since it would result in a contradiction (as Cantor is nowhere dense but full measure implies dense in $[0,1]$).

I have found that there exists a partition of $[0,1]$ by Cantor sets. Now if it is countable (not necessarily finite) it holds that the Lebesgue measure of the union is the sum of the measure of each set.

My question is; can we construct such union with fat Cantor sets and such that it is countable?

If so, can we "add" enough such disjoint fat Cantor sets so that the measure of the union is one (i.e. the union has full measure in $[0,1]$)?

Any help is appriciated!

答: That would violate Baire Category Theorem. So we need uncountable many fat Cantor sets to cover $[0,1]$.

Title: Union of deletion Cantor sets


问:
Is the cardinal of a subset of random variable a random variable?
Here is my problem :
$Z_i$  a sequence of i.i.d. random variables on the same probability space with median k.

$M_n = |{i = 1 . . . n : Z_i ≤ k}|$ and $P_n = |{i = 1 . . . n : Z_i > k}|$.

Can we describe $L_n$ as a sum of i.i.d. random variables?
Is $M_n$  a random variable? If yes how can we describe it with the $Z_i$?
Does $M_n/n$ converge as $n → ∞$ ?
Does $n^{−1/2}(M_n − P_n)$ converge in law as $n → ∞$ ?
thanks for the help

答:
As a partial answer to get you started. You can describe $M_n$ and $P_n$ as sums of iid indicator random variables: For instance, if we set $I_i = \cases{1 {\text { if } } Z_i\le k\\0{\text { otherwise } }}$, then $M_n = I_1 + \cdots + I_n$. This is a common construction, for example the binomial distribution is a sum of iid Bernoulli random variables, which each itself can be thought as an indicator random variable. We have $M_n/n$ converge in probability to the expected value of $I_i$, which is $E(I_i)=P(Z_i\le k)$. See weak law of large numbers (https://en.wikipedia.org/wiki/Law_of_large_numbers).

Also note, $M_n + P_n = n$, so $M_n - P_n = n- 2P_n$, this may be helpful to analyze your other question.


问:
$\mathbb{Z}_p$ with $p$ prime is integral domain
I was trying to solve some exercise from my abstract algebra course, and at some point, I assumed $\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}_2$ is an integral domain, because I thought to remember that $\mathbb{Z}_p$ is integral domain iif $p$ is a prime number.

Am I right? If I am, how can I prove it to add it to my execise? If not, could you give me a counter example? Any help will be appreciated, thanks in advance.

答:
Being an integral domain means: if $a,b\in R$ are such that $ab=0$ then $a=0$ or $b=0$.

In case of $\mathbb{Z}/n\mathbb{Z}$ this translates to: if $a,b\in\mathbb{Z}$ are such that $n|ab$ then $n|a$ or $n|b$. Then the classical number theory shows that this works only for prime $n$.


问:
Calculate surface area of $f(x+y+z)$

Calculate the following surface integral:
$$\iint_{\sigma}(x+y+z)\mathrm{d}S$$
When $\sigma$ is the plane $x+y=1$ (x,y positive) between $z=0$ and $z=1$.

For some reason I don't manage because I can't express $z$ in terms of $x$ and $y$.

答:
There are two simplifications to be made. Since we are on the surface $x+y=1$ we have that

$$\iint\limits_\sigma x+y+z\:dS = \iint\limits_\sigma 1 + z\:dS$$

Second, the surface is a rectangle with side lengths $1$ and $\sqrt{2}$. Since the function only varies in the $z$ direction, we can say that

$$\iint\limits_\sigma 1+z\:dS = \sqrt{2}\int_0^11+z\:dz = \frac{3}{\sqrt{2}}$$

Title: Calculate surface area of $f(x+y+z)$


答:
You're right. $\Bbb Z_p$ is an integral domain if $p$ is prime. Otherwise, it's not.

First, the units of $\Bbb Z_p$ are the elements $a$ with $gcd(a,p)=1$ and so are the nonzero elements of $\Bbb Z_p$. For the proof use Euclid's algorithm and Bezout.

Moreover, units are not zero divisors, since if $a$ is a unit with $ba=0$, the

$b= b1 = b(aa^{-1}) = (ba)a^{-1} = 0 a^{-1}=0$.

Title: $\mathbb{Z}_p$ with $p$ prime is integral domain

Body:
I was trying to solve some exercise from my abstract algebra course, and at some point, I assumed $\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}_2$ is an integral domain, because I thought to remember that $\mathbb{Z}_p$ is integral domain iif $p$ is a prime number.

Am I right? If I am, how can I prove it to add it to my execise? If not, could you give me a counter example? Any help will be appreciated, thanks in advance.


答:
What you did was not squaring both sides of the equation. If we have an equation
$$a+b=0,$$
then squaring both sides produces
$$\begin{align*}
(a+b)^2&=0^2\\a^2+2ab+b^2&=0.
\end{align*}$$
In your case, you have $a=\sqrt{3x-2}$ and $b=2-x$. But when you squared the left side of the equation, what you've written is instead $a^2+b^2$. But
$$(a+b)^2=a^2+2ab+b^2\neq a^2+b^2.$$
Thus, when correctly squaring both sides of the equation you've been given, the result is
$$(\sqrt{3x-2})^2 +2(\sqrt{3x-2})(2-x)+(2-x)^2=0.$$
However, this is not a productive place to start solving the question. A better approach would be to move the things not inside the radical to the other side, and then square, like this:
$$\begin{align*}
\sqrt{3x-2} +2-x&=0\\\\
\sqrt{3x-2} &=x-2\\\\
\qquad\qquad\qquad(\sqrt{3x-2})^2&=(x-2)^2\qquad \leftarrow \text{squared both sides}
\end{align*}$$
Can you take it from here?

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


Body: Let $c=a/b.$ Then $$\left(\frac {2}{a^2}+\frac {1}{b^2}\right)(a^2+b^2)\,=\,3+\frac {2b^2}{a^2}+\frac {a^2}{b^2}=$$ $$=3+\frac {2}{c^2} +c^2=$$ $$=3+ \left(\frac {\sqrt 2}{c}-c\right)^2+2\sqrt 2\ge$$ $$\ge 3+2\sqrt 2$$ with equality iff $\frac {\sqrt 2}{c}-c=0,$ that is, iff $\frac {a^2}{b^2}=c^2=\sqrt 2.$


问:
Interpretation of Breslow Estimator for Baseline hazard
I am trying to understand the intuition behind Breslow's estimator of baseline hazard for a survival model. The Estimator is defined as follows:

$$\lambda_{0}(t_j)=\frac{d_j}{\sum_{k\in \mathbb{R}(t_j)}\text{exp}(\beta^{T}Z_k)}$$

Note: $\mathbb{R}(t_j)$ is the collection of observations that are at risk at time $t_j$, $d_j$ are sample that have experienced event, T is a time to event random variable,  and $Z$ are covariates.

The derivation is based on this similarity:

$$d_j\approx\sum_{k\in \mathbb{R}(t_j)}P(t_j<T_k<t_j + \Delta t|T_k>t_j)$$

I am having trouble making sense of this connection. Can someone explain the intuition behind this.

答:
The significance of the hazard function $\lambda(t)$ for a survival model is that it is sufficient to describe the distribution of the lifetime through $P(T > t) =: S(t) = e^{-\int^t_0 \lambda(s) \,\mathrm{d} s}$. (For example, the hazard function can be defined by $\lambda(t) = -\frac{S'(t)}{S(t)}$.) In the proportional hazards model, we assume that the hazard function takes a semi-parametric form: $\lambda(t) = \lambda_0(t) \exp(\beta^T Z_k)$.

To come up with an estimator for the baseline hazard $\lambda_0(t)$, we approximate the number of events $d_j$ that happen at time $t_j$ with the expected number of events that happen based on our survival model, which depends on the pool of risks $\mathbb{R}(t_j)$ still active at $t_j$. That is, $d_j$ is approximately the number of events that happen in the interval $[t_j, t_j + \Delta t]$, and the occurrence of the event for risk $k$ is approximately Bernoulli($p_k$), where
$$
p_k = P(t_j < T_k < t_j + \Delta t \,|\, T_k > t_j) = \frac{S_k(t_j) - S_k(t_j + \Delta)}{S_k(t_j)}
$$
Since $\lim_{\Delta t \to 0} \frac{S_k(t_j) - S_k(t_j + \Delta)}{\Delta t \, S_k(t_j)} = \lambda_k(t)$, we can approximate $p_k \approx \Delta t \cdot \lambda_k(t_j)$. Further approximating with $\Delta t = 1$ and putting all this in together with the proportional hazards assumption, we have
$$
d_j \approx \sum_{k \in \mathbb{R}(t_j)} P(t_j < T_k < t_j + \Delta t \,|\, T_k > t_j) \approx \sum_{k \in \mathbb{R}(t_j)} \lambda_k(t) = \lambda_0(t) \sum_{k \in \mathbb{R}(t_j)} \exp(\beta^T Z_k).
$$
Rearranging gives the estimator $\lambda_0(t) = d_j / \sum_{k \in \mathbb{R}(t_j)} \exp(\beta^T Z_k)$. Thus, $\lambda_0(t)$ can be interpreted as describing the "instantaneous probability" that the event happens at a particular time (recall that a lot of classical probability is counting the ratio of favourable outcomes to all possible outcomes; with this estimator, we assume the baseline hazard is discrete at the times $t_j$ when the event has actually happened from the data).

As for the correctness of this, I think looking at an actual proof would be better, but hopefully this gives some intuition to accompany the rigour.


问:
Proof that sigmoid function is not homogeneous.
How can we prove that the sigmoid function is not homogeneous?

For the sigmoid function $$S(x) = \frac{1}{1+e^{-x}} $$ I want to show that it does not satisfy $$ \forall (\alpha, x) \in \mathbb{R}^+ \times \mathbb{R}, S(\alpha x) = \alpha S(x) $$

I can get to $$ \frac{\alpha}{1+e^{-x}} = \frac{1}{1+e^{-\alpha x}} $$ which obviously is true for $\alpha = 1$ but it is trivial to find a counterexample e.g. $\alpha = 2, x = 1$  ... but is there a nicer way to show that it doesn't hold?

答:
Another easy choice of parameter is to let $x=0$, and you can let $\alpha$ be any positive number not equal to $1$.

Alternatively, you can rearrange them such that $$\alpha\left( 1 + e^{-\alpha x}\right)=1+e^{-x}$$

and we can see that the LHS is dependent on $\alpha$ while the right hand side is not. If you let $\alpha \to 0 $, the LHS goes to $0$, but the right hand side is at least $1$. We can also let $\alpha \to \infty$ to get a contradiction as well.


问:
Chebyshev's Inequality proof explanation - squared values of probability
In the proof of Chebyshev's Inequality we do the following:

Claim: for some random variable $Y$ and some value $a > 0$

$Pr[|Y - E[Y]| \geq a] = \frac{E[(Y - E[Y])^2]}{a^2}$

Let's refer to $E[Y]$ as $\mu$

\begin{align*}
Pr[|Y - \mu| \geq a] =&\  Pr[|Y - \mu|^2 \geq a^2] \\
=& \ Pr[X \geq a^2]\ \ \ \ if \ X = (Y - \mu)^2\\
\leq& \ \frac{E[X]}{a^2}\ \ \ \ by \ Markov's \ inequality\\
=& \ \frac{E[(Y - E[Y])^2]}{a^2}
\end{align*}

My question is, how can we just assert:

$Pr[|Y - \mu| \geq a] =\  Pr[|Y - \mu|^2 \geq a^2]$

What is the intuition or line of thought behind this?

答:
Another way to see it is that
$$\mathbb{P}(|X|\geq a)=\mathbb{E}\left[1_{|X|\geq a}\right]$$
Now look at the indicator function:
$$1_{|X|\geq a}=\begin{cases}
1 & \text{if }|X|\geq a\\
0 & \text{if}\,|X|<a
\end{cases}=\begin{cases}
1 & \text{if }|X|^{2}\geq a^{2}\\
0 & \text{if}\,|X|^{2}<a^{2}
\end{cases}$$
As it preserves inequalities in this setting. Meaning: $1_{|X|\geq a}=1_{|X|^{2}\geq a^{2}}$. It is the same analogy as Epiousios, but just specifying that the indicator function is equal in both cases.


答:
Hint $\ $ The inductive step is $\rm\bmod 11\!:\ 10\equiv -1,\ 10^n\equiv (-1)^n  \Rightarrow\, 10^{n+1}\equiv (-1)^{n+1}$

This is simply a special case of the following ubiquitous rule for multiplying congruences

Congruence Product Rule $\rm\ \ A\equiv\: a,\ B\:\equiv\: b\ \ \Rightarrow\ \ AB\equiv ab\ \ \ (mod\ m)$

Proof $\rm\ \ m\ |\ A-a,\: B-b\ \ \Rightarrow\ \ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $

Remark $ $ If in the proof we put $\rm\,A = 10,\ b = -1,\ B = 10^n,\ b = (-1)^n\ $ then it shows

$$\rm  11\mid \underbrace{10^n\!-(-1)^n}_{\Large f_n}\Rightarrow\, 11\mid \underbrace{11(10^n)-(10^n\!-(-1)^n)}_{\textstyle (A-a)B\ +\ a(B-b)}= \underbrace{10^{n+1}\!-(-1)^{n+1}}_{\Large f_{n+1}}\qquad $$

i.e. $\rm\,11\mid f_n\Rightarrow\, 11\mid f_{n+1},\,$ which is the inductive step that is usually pulled out of a hat like magic in many proofs. But by above we see we can use math (Product Rule) vs. magic to discover a proof. But it is more conceptual arithmetically to not give a direct inductive proof using divisibility but instead proceed as in the hint using modular arithmetic and the Product Rule.

Abstracting a bit we can easily deduce the following more general

Congruence Power Rule $\rm\ \ \color{}{A\equiv a}\ \Rightarrow\ \color{}{A^n\equiv a^n}\ \  (mod\ m)\ \ $ for all naturals $\rm\,n.$

Remark $ $ If congruences are not familiar then  we can still use the product rule in an equivalent divisibility form - see here, where we explain these ideas at much further length (including showing in detail how a couple posted inductive proofs are precisely special cases of the Product Rule).

You can find much further discussion of this fundamental product rule in many of my prior posts. For example, this post explains how the derivative product rule is closely related to the congruence product rule. See also this post which explains the relationship between arithmetic divisibility rules and subring structure. Many of these diverse topics will be unified when you learn about quotient rings, a ubiquitous prototype being the ring $\rm\:\mathbb Z/m\:$ of integers modulo $\rm\:m\:.$

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Counting Cosets of $\langle\tfrac12\rangle$, in $\Bbb{R}$ and in $\Bbb{R}^{\times}$
Describe the cosets of the subgroups described:

The subgroup $\langle\frac{1}{2}\rangle$ of $\mathbb{R}^{\times}$, where $\mathbb{R}^{\times}$ is the group of non-zero real numbers with multiplication.

The subgroup $\langle\frac{1}{2}\rangle$ of $\mathbb{R}$, where $\mathbb{R}$ is the group of real numbers with addition

答:
If I understand correctly, you mean the subgroups
$$H_1=\langle\tfrac{1}{2}\rangle\subset\mathbb{R}^{\times}\qquad\text{ and }\ H_2= \langle\tfrac{1}{2}\rangle\subset\mathbb{R}^+.$$
So as sets, these subgroups look like
$$H_1=\left\{2^k:\ k\in\mathbb{Z}\right\}\qquad\text{ and }\qquad H_2=\left\{\tfrac{k}{2}:\ k\in\mathbb{Z}\right\}.$$
Can you now determine what the cosets of these subgroups look like?
Can you describe a set of representatives for the cosets?


问:
Non-principal Ultrafilter on Measurable Subsets of $\mathbb R$
Let $\Omega$ be the set of all measurable subsets of $\mathbb R$ ordered by inclusion. Is it possible to construct a non-principal ultrafilter on $\Omega$?

答: Take a maximal filter generated by all complements of null sets in the collection of all Lebesgue measurable sets, by Zorn.</p>