问:
Classification of Groups of Order $p^2$
The book I am reading contains contains the following claim:

Every group of order $p^2$, where $p$ is a prime, is isomorphic to $\mathbf{Z}_{p^2}$ or $\mathbf{Z}_p \oplus \mathbf{Z}_p$.

I tried to do the proof myself and ended up with the following. Is the following sketch correct? --

We can show all groups of order $p^2$ are abelian by examining the center of the group. https://proofwiki.org/wiki/Group_of_Order_Prime_Squared_is_Abelian

We are left to show, that such an abelian group of order $p^2$ is of the two forms given above. Examine arbitrary subgroup $\langle a \rangle \subset G$. By lagrange's theorem it's order can be $p^2$, $p$, or $1$.

If $|\langle a \rangle|=p^2$ we obtain an abelian subgroup order $p^2$ -- i.e. $\langle a \rangle$ generates G. So $\mathbf{Z}_{p^2} = \langle a \rangle = G$ (the first isomorphism established by sending $k \mapsto a^k$).

If $|\langle a \rangle|=p$ we have an abelian subgroup of $G$ order of $p$. We can see $\mathbf{Z}_{p} = \langle a \rangle$ by again sending $k \mapsto a^k$.  Construct $G/\langle a \rangle$ (Can check $\langle a \rangle$ is normal in $G$). $|G/\langle a \rangle| = p$ by using coset definition and counting. By using Lagrange's theorem, one can show any group of order $p$ is cyclic and abelian. Thus we can construct the isomorphism: $\mathbf{Z}_{p} = G/\langle a \rangle$. We note: $\langle a \rangle = \mathbf{Z}_{p} = G/\langle a \rangle$. And so: $G/\langle a \rangle \oplus \langle a \rangle=\mathbf{Z}_p \oplus \mathbf{Z}_p$. (How do I show $G/\langle a \rangle \oplus \langle a \rangle$ = $G$? I think this is still missing)

If $|\langle a \rangle|=1$ then $G/\langle a \rangle =G$. So we can refer to 1 and 2.

答:
None of the answers so far shows the hardest part - why $\langle a\rangle$ is normal.

So assume that $G$ is of order $p^2$ and all non-trivial elements have order $p$. Let $e \neq a \in G$. If $\langle a \rangle$ is not normal, then there's some $x \in G$ s.t. $xax^{-1} \notin \langle a \rangle$. Now it's easy to prove that for any $b \notin  \langle a \rangle$, $G = \langle a \rangle \langle b \rangle$ i.e. every element of $G$ is expressible as $a^ib^j$. So by choosing $b = xax^{-1}$ we can express $x^{-1}$ this way:

$$x^{-1} = a^ib^j = a^ixa^jx^{-1}$$

So $e = a^ixa^j$, which implies that $x = a^{-i-j}$ $\to$ contradiction, because then $xax^{-1} \in \langle a \rangle$.

Title: Classification of Groups of Order $p^2$


问:
What is the cardinality of set of functions $F:\mathbb{N} \longrightarrow \{0,1,2\}$?
Actually I have tried it but I am unable to create a mapping to see the reality as I wants to know that what is the value of 3 to the power aleph null?</p>

答: This is the cardinality of the set of infinite sequences of characters over the alphabet $\{0,1,2\}$. Each such sequence is the decimal expansion in the ternary number system of a real number in the closed interval $[0,1]$. So the cardinality is $\aleph_1$. A similar argument can be used to show that the cardinality of any function space $\mathbb{N} \rightarrow S$ is $\aleph_1$ if $S$ is a finite, non-empty set.

ContentLicense: CC BY-SA 4.0


问:
Does this number belong to the set of real numbers?
Suppose we build a number in this way:
we put the natural numbers one after the other. For example, for the first 5 numbers:
$n_1=1,n_2=2,n_3=3,n_4=4,n_5=5$ we obtaine a new number $N_5=12345$. For the first 20 numbers we have in the same way:
$$N_{20}=1234567891011121314151617181920$$ and so on.
If we build a new number using all the natural numbers in this way, is this new element a real number? I suppose yes, because it seems to be uncountable, but I'm not sure. Can anyone suggest a proof? Thanks.</p>

答:
Every integer has a finite number of decimal digits, because it is a finite sum of $1$ (or $-1$ for negative integers).

While real numbers can have infinitely decimal digits, those can only appear in the fractional part. Your construction, if so, is not a real number. It's an infinite string of digits.

ContentLicense: CC BY-SA 3.0


问:
(convolution) If f*g=g then g is a trigonometric polynomial
The problem:

Let $g$ be a continuous and $2\pi$ periodic, such that there exists a continuous and $2\pi$ periodic function $f$ which sustains $f*g=g$. Prove that $g$ is a trigonometric polynomial.

Ok. So I know that I need to show somehow that $g(x)=\sum_{n=-\infty}^{\infty}a_ne^{inx}$.

I tried this approach but got stuck:

According to the convolution theorem, we get that the fourier coefs are equal: $$\widehat{(f*g)}(n)=\hat{f}(n)\cdot\hat{g}(n)$$ and from that we get $$\hat{f}(n)\cdot\hat{g}(n)=\hat{g}(n)$$

and that's pretty much where I got stuck.

Any help would be appreciated.

答: That's it, is it not? $\left(\widehat f(n)\right)_{n\in\Bbb Z}$ is a square-summable sequence by Bessel, and therefore it can't be frequently equal to $1$ as $\lvert n\rvert\to\infty$. Therefore, $\left(\widehat g(n)\right)_{n\in\Bbb Z}$ must be eventually $0$ as $\lvert n\rvert\to\infty$.


问:
Confused on how the MGF of the exponential distribution works
How can the MGF of the exponential dist with parameter $\lambda$ be $E[X^n] = \frac{n!}{\lambda^n}$ if the first and second moments are $\frac{1}{\lambda}$ and $\frac{1}{\lambda^2}$?

Where is the n! going that we get a one on top?

答: Let $M(t)$ be the moment generating function of a random variable.  The power series at $t=1$ is $M(1)=\sum_0^\infty \frac{m_n}{n!}$ where $m_n$ is the $n^{th}$ moment.  I hope this clarifies your concern.


问:
Find the volume inside the curve $x^2+4z^2+8y=16$ and on the positive side of $xz$-plane.
Find the volume inside the curve $x^2+4z^2+8y=16$ and on the positive side of $xz$-plane.

What will be the volume of this as I'm getting wrong answer?

As per me the volume is this.

Please help me as I'm unable to get it.

答:
@player3236
my answer is in the image, please let me know where i went wrong, Thank you in advance
$$v = \int_0^2 \int_0^\sqrt{16-4z^2} 1/8(16-x^2 +4z^2) \,dx\,dz$$

$$v = \int_0^2 \int_0^\sqrt{16-4z^2} (2-x^2/8 +z^2/2) \,dx\,dz$$

My solution for the above question


问:
monomorphism from $\mathrm{GL}(n,q^2)$ to $\mathrm{GL}(2n,q)$
Construct an explicit monomorphism from $GL(n,q^2)$ to $GL(2n,q)$. Will $GL(n,q^m)$ always have a subgroup isomorphic with $GL(mn,q^2)$ for every $m,n\in \mathbb{N}$, and prime power $q$?</p>

答:
Can you see how $\mathrm{GL}_n(\mathbb{C})$ is naturally a subgroup of $\mathrm{GL}_{2n}(\mathbb{R})$? You may apply the same idea to the finite fields.
Hint: A $\mathbb{C}$-linear operator is a priori $\mathbb{R}$ linear.

I wonder if there is some typo in the second part of the question. As it stands now, it doesn't make much sense, since $\mathrm{GL}_{mn}(\mathbb{F}_{q^2})$ has larger cardinality than $\mathrm{GL}_{n}(\mathbb{F}_{q^m})$.


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

What is the meaning of  "independent events " in probability

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

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

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

Take for eg an example by "Tim"

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

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

My thought:

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

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

答:
The mathematical definition is very easy.  Two events $A$ and $B$ are independent if and only if $$P(A\cap B) = P(A)P(B).$$

In "pure" probability theory there's no interpretation of this, it's just a definition.  It's a purely mathematical statement I can make about two events and a probability distribution.

To explain what it "means" you have to explain what probability means. There's no acceptable answer to this question. It's a big philosophical problem that mathematicians avoid by writing down some equations and solving them.

The motivation comes from the idea of conditional probability.

Suppose you throw  a die.  The probability you throw a six is $\frac 16$ and the probability you throw an even number is $\frac 12$.  You can check with the formula above that the two events are not independent.

To get an idea of why suppose you throw a die but don't look at it. You want to get a six. I tell you whether or not it's even and you decide whether to keep it or roll it again.  If I tell you it's odd then you know it's not a six and you roll it again. If I tell you it's even then there are only three numbers it could be and one of them is a six.  So the probability that you got a six is now one in three.  So you'd be crazy to throw again.

In maths we define conditional probabilities as follows$$P(A|B) = \frac{P(A\cap B)}{P(B)}.$$
Again in "pure" maths there's no interpretation of this, it's just a formula.
But in the real world $P(A|B)$ is the probability that $A$ happens if you already know that $B$ happened.

So the interpretation of independence is that $A$ and $B$ are independent if and only if $P(A|B) = P(A)$ if you know that $B$ happened it doesn't affect the probability that $A$ happened.

This concept makes intuitive sense to people.  If my team is winning at half time it's more likely to win the game than if it wasn't, so not independent. If my team is winning at half time it doesn't make it less likely that it's going to rain tomorrow, so independent.

It's worth noting though that independence is an assumption that I might be wrong about. If my team happens to play well in the rain and they're winning at half time it's more likely to be raining during the game. This might mean it's more likely to be raining tomorrow. So they might not be independent.

So in fact a better definition of independence would be an assumption I make to simplify my model which is usually wrong, but hopefully not that wrong.


问:
Evaluate $\lim_{x\to \infty}(x-(\ln x)^3))$
Here is my approach:
$\lim_{x\to \infty}(x-(\ln x)^3))=\lim_{x\to\infty}\frac{x^2-(\ln x)^6}{x+(\ln x)^3}=\lim_{x\to \infty}\frac{x^2}{x}=\infty$

I used the fact that $x^n$ goes faster to $\infty$ than $\ln x$.
Is my approach right?

答:
It is correct, although it is simpler to write
$$x-\ln(x)^3=x\left(1-\dfrac{\ln(x)^3}{x}\right)$$
and say that the fraction tends to $0$ for the reason you've mentioned.


答: Yes, your method works with $n=1/3$. But in case it's not obvious any $n>0$ achieves $\ln x\in o(x^n)$, I'll flesh this out. Since $\int_0^\infty y^3e^{-y}dy=6<\infty$ has non-negative continuous integrand, $\lim_{x\to\infty}\frac{\ln^3x}{x}=\lim_{y\to\infty}y^3e^{-y}=0$, i.e.$$\ln^3x\in o(x)\implies x-\ln^3x\sim x.$$So your limit is $\infty$.

Title: Evaluate $\lim_{x\to \infty}(x-(\ln x)^3))$

Body:
Here is my approach:
$\lim_{x\to \infty}(x-(\ln x)^3))=\lim_{x\to\infty}\frac{x^2-(\ln x)^6}{x+(\ln x)^3}=\lim_{x\to \infty}\frac{x^2}{x}=\infty$

I used the fact that $x^n$ goes faster to $\infty$ than $\ln x$.
Is my approach right?


问:
$\|x+\mathcal{M}\| = \inf\{\|x+y\|: y \in \mathcal{M}\}$ defines a norm
Taken From Conway's A course in Functional Analysis chapter 3 Section 4 problem 1:

Problem Statement:
Show that if $\mathcal{M}\leq \mathcal{H}$, then (4.1) defines a norm on $\mathcal{X}/\mathcal{M}$.

I wanted to see if what I had below was complete. Any feedback on rigor is appreciated!
To prove that $\|x+\mathcal{M}\| = \inf\{\|x+y\|: y \in \mathcal{M}\}$ defines a norm, we check that it satisfies the three axioms. We first prove $\|x+\mathcal{M}\| \Rightarrow x \in \mathcal{M}$. We have $\inf\{\|x+y\|: y \in \mathcal{M}\} = 0$, which can only be true if $x\in \mathcal{M}$.

Now we hope to prove that $\|\alpha(x+\mathcal{M})\| =|\alpha|\|(x+\mathcal{M})\|$. We have

\begin{align}
\|\alpha(x + \mathcal M)\| &= \|\alpha x + \mathcal M\|
\\ & = \inf\{\| \alpha(x+y)\| : y \in \mathcal M\}
\\ & = |\alpha| \inf\{\|x+y\|:y \in \mathcal M\} =
|\alpha|\cdot \|(x + \mathcal M)\|.
\end{align}

Finally, we prove the triangle inequality, which can be verified with: $\|x+z\| = \inf\{\|x+z+y\|: y \in \mathcal{M}\}= \inf\{\|x+y_1+z + y_2\|: y_1,y_2 \in \mathcal{M}\}\leq \inf\{\|x+y_1\|+\|z + y_2\|: y_1,y_2 \in \mathcal{M}\} =  \inf\{\|x+y_1\|: y_1 \in \mathcal{M}\} +  \inf\{\|z + y_2\|:y_2 \in \mathcal{M}\}$.

答:
I'm not sure about notation, but based on the problem context I'd wager that $\mathcal{M}\leq\mathcal{H}$ might be supposed to mean $\mathcal{M}$ is a closed vector subspace of $\mathcal{H}$? Assuming this is the case, here are some hints/comments on the 3 parts of your proof.

I think you're missing an "$=0$" in the $\|x+\mathcal{M}\|\Rightarrow x\in \mathcal{M}$ part. Also, the argument you've stated is a tautology. Hint: What does it actually mean for the infimum to be zero? Also, is $\mathcal{M}$ closed? If so, this could be quite useful in proving this part.

The positively homogeneous part (involving $\alpha$) is only true if $\mathcal{M}$ is positively homogeneous. For instance, if $\mathcal{M}$ is a vector subspace then this would be satisfied, but you may want to state the reasoning.

The triangle inequality part looks OK to me, however I would also note that you may want to write a justification (in words) as to why you can express an arbitrary $y\in\mathcal{M}$ as $y_1+y_2$ where $y_1,y_2\in\mathcal{M}$. Again, if $\mathcal{M}$ is a vector subspace then this is true. However, we have to be careful, because this is not true for arbitrary sets.


问:
When will level sets be closed/Jordan curves?
Are there any conditions on a $C^{1}$ function $f$, for example from $\mathbb{R}^2$ to $\mathbb{R}$, and constant $c$ which would guarantee that the level set $\left\{(x,y) \in \mathbb{R}^2: f(x,y)=c\right\}$ is a closed curve? By "closed curve" I mean a curve which is a closed loop (homeomorphic to the circle), rather than merely a curve which, seen as a set of points, is a closed set (as discussed in comments here: Do the curves of a level set of a continuous function have to be closed?).

Similarly, what about conditions for the level set to be a simple (non-self-intersecting) closed curve, i.e. a Jordan curve in the plane?

答:
If you haven't studied manifolds yet then this answer is going to be a little rough, in that it will use some of the more sophisticated tools about smooth submanifold of $\mathbb R^n$ including Morse Theory.

Here goes.

One list of condidions that guarantees your level set is a Jordan curve is:

$f$ is smooth,

$f$ is proper,

$f$ has a unique critical point $(x_0,y_0) \in \mathbb R^2$. I'll denote
$$b = f(x_0,y_0)
$$

$c \in \text{image}(f) - \{b\}$.

The critical point $(x_0,y_0)$ is a Morse singularity of $f$. This just means that the matrix of second partial derivatives of $f$ at $(x_0,y_0)$ has nonzero determinant.

Here's a few details of proof.

The Regular Value Theorem (aka the Submersion Theorem), together with 2, 3, and 4, implies that $f^{-1}(c)$ is a 1-dimensional submanifold of $\mathbb R^2$ for all $c \in \text{image}(f) - \{b\}$. Combining this with 1 it follows that $f^{-1}(c)$ is a compact 1-dimensional submanifold, which in turn implies that $f^{-1}(c)$ is a union of a disjoint collection of Jordan curves. Let's denote the number of components of $f^{-1}(c)$ number by $N(c)$. A further application of the Regular Value Theorem implies that $N(c)$ is locally constant: for each $c$ there exists $\epsilon>0$ so that if $c-\epsilon < c' < c+\epsilon$ then $N(c')=N(c)$.

So it remains to check that $f^{-1}(c)$ is just a single Jordan curve, in other words that $N(c)=1$ for all $c \in \text{image}(f) - \{b\}$.

The Morse Lemma, applied to $f$ and $b$, implies that $N(c)=1$ if $c$ is close to $b$.

Applying the Intermediate Value Theorem to $N(c)$ together with the fact that $N(c)$ is locally constant, it follows that $N(c)$ is constant. Knowing that $N(c)=1$ at values close to $b$, it follows that $N(c)=1$ for all values of $c$.


问:
Computing PDF of Products of Two Random Variables
I've been stuck on this problem for some days. I'm hoping someone would help by chipping in a few comments. I have two i.i.d. r.v.:

$$
f_X(x)=\frac{\left(1-e^{-\frac{x}{\alpha }}\right)^{\tilde{r}-1} exp({-\frac{x \left(K-\tilde{r}+1\right)}{\alpha })}}{\alpha }
$$
and
$$
f_Y(y)=\frac{\left(1-e^{-\frac{y}{\beta }}\right)^{r-1} exp({-\frac{y (K-r+1)}{\beta })}}{\beta }
$$
Since they are i.i.d, the joint pdf is given as:
$$
f_{XY}(x,y)=\frac{\left(1-e^{-\frac{y}{\beta }}\right)^{r-1} \left(1-e^{-\frac{x}{\alpha }}\right)^{\tilde{r}-1} e^{-\frac{x \left(K-\tilde{r}+1\right)}{\alpha }-\frac{y (K-r+1)}{\beta }}}{\alpha  \beta }
$$

Goal is to find the the PDF of Z = XY. Suppose z>0, I'm using the product relation given in Henry Stark's, (page 137, also here on wiki http://en.wikipedia.org/wiki/Product_distribution)

$$
f_Z(z)=\int_{-\infty }^{\infty } \frac{1}{\left| y\right| } f_{XY}\left(\frac{z}{y}, y\right) \, dy
$$
Subtituting, I have,

$$
f_Z(z)=\int_{-\infty }^{\infty } \frac{\left(1-e^{-\frac{y}{\beta }}\right)^{r-1} \left(1-e^{-\frac{z}{\alpha  y}}\right)^{\tilde{r}-1} \exp \left(-\frac{z \left(K-\tilde{r}+1\right)}{\alpha  y}-\frac{y (K-r+1)}{\beta }\right)}{\alpha  \beta  y} \, dy
$$

From here, I don't seem to know how to proceed to obtain a closed form equation. Almost at my wits end but hoping I might get some guidance here.

答:
As explained in comments, neither $f_X$ nor $f_Y$ is a PDF in general. For example, using the change of variable $t=\mathrm e^{-x/\alpha}$ in $f_X$ yields $\mathrm dt=-t\mathrm dx/\alpha$, hence
$$
\int_0^{+\infty}f_X(x)\mathrm dx=\int_0^1(1-t)^{\tilde{r}-1} t^{K-\tilde{r}}\mathrm dt=\mathrm{Beta}(K-\tilde{r}+1,\tilde{r}).
$$
For each $\bar r\gt0$, there is exactly one value of $K$ such that $f_X$ is a PDF (for example, if $\bar r=2$ then $K=\frac12(1+\sqrt5)$).</p>


问:
$\det\left[ \begin{matrix} A & B \\ B & A \end{matrix} \right] \ = \ \det\left( A + B \right) \det( A - B )$ for $AB \neq BA$
For two square matrices $A$ and $B$ of the same size, we have the identity
$$
\det\left[ \begin{matrix} A & B \\ B & A \end{matrix} \right] \ = \ \det\left( A + B \right) \det( A - B ) \ .
$$
This is supposed to hold even when the matrices are non-commuting ie. $A B \neq BA$.

In the case that $AB=BA$, I can see how the above follows from the identity $$
\det\left[ \begin{matrix} A & B \\ C & D \end{matrix} \right] \ = \ \det(D) \det\left( A - B D^{-1} C \right)
$$
by taking $(C,D) \to (B,A)$ and mutiplying the argument of the determinants together, leads to
$$
\det\left[ \begin{matrix} A & B \\ C & D \end{matrix} \right] \ \to \ \det(A) \det\left( A - B A^{-1} B \right) = \det\left( A^2 - B A^{-1} B A \right) = \det\left( A^2 - B^2 \right) =  \det\left( A - B \right)\det\left( A + B \right)
$$
but the above proof relies heavily on $AB=BA$.

How do you see this identity is true in the case that $AB\neq BA$?

答:
By elementary operations over the columns and the rows of the matrix, you have$$\det\left[ \begin{matrix} A & B \\ B & A \end{matrix} \right] \ = \det\left[ \begin{matrix} A+B & B \\ B+A & A \end{matrix} \right] \ = \det\left[ \begin{matrix} A+B & B \\ 0 & A-B \end{matrix} \right] $$

which gives the formula.


问:
Injection of ordinal $\alpha$ into $\beta$ means $\alpha \subseteq \beta$
I'm starting reading about cardinals (Shoenfield, page 253) and showing $x \subseteq y \implies \alpha = Card(x) \le Card(y) = \beta$. I notice that there is a injection $\alpha \to \beta$, so I want to conclude $\alpha \subseteq \beta$. Injection between ordinals $\alpha \hookrightarrow \beta$ asks the same but with an order-preserving injection. So maybe this property is false. I was wondering what is an easy counterexample.

My try

I examine first what can be the proof of Injection between ordinals $\alpha \hookrightarrow \beta$. As shown there, we can built an order preserving injection $h:\beta \to \beta$. Therefore, $\beta \cong h(\beta)$. Now, I'm proving that $h(\beta)$, which is clearly a set of ordinals (using axiom of replacement), is transitive. I believe $x \in h(\beta) \implies h^{-1}(x) \in \beta \implies h^{-1}(x) \subseteq \beta \implies x \subseteq h(\beta)$ where I use transitivity of $\beta$. So, since every transitive set of ordinals is an ordinal, we conclude $h(\beta)$ is an ordinal. But $\beta \cong h(\beta)$ implies $\beta = h(\beta)$ which contradicts the assumption $h(\beta) \subset \beta$ (since $\beta \subset \alpha$ so there are elements of $\alpha$ that are not mapped in the image.

So we see that it was needed to have an isomorphism, not just an injection.

答: There is an injection $f: \omega + 1 \to \omega$ defined by $f(\omega)=0, f(n)=n+1$ for $n \in \omega$, but it's not the case that $\omega \cup \{ \omega \} = \omega + 1 \subseteq \omega$.  More generally, there's an injection from any countable ordinal into $\omega$.  That's how you know the ordinal is countable.


问:
Calculating the spectrum of a specific coordinate ring
So, let $F$ be a field and $b_1,b_2,b_3,t_1^{\pm},t_2^{\pm}$ be formal variables satisfying $t_it_i^{-1}=1$.

Let $I$ be the ideal generated by the equations $b_1+b_3+b_1b_2b_3=t_1^{-1}$ and $1=t_1^{-1}t_2^{-1}$

I need some help figuring out what $\operatorname{Spec} (\frac{F[b_1,b_2,b_3,t_1^{\pm},t_2^{f\pm}]}{I})$ is?

If there is a series flaw inherent in this question, please let me know.

I know that $\operatorname{Spec} (\frac{F[b_1,b_2,b_3,t_1^{\pm},t_2^{f\pm}]}{I})$ is going to be an affine variety. What are the defining equations for this variety? Sorry if this is a stupid question, but I'd really appreciate some clarification from one of the legit math Einsteins that I know hang out here... Thanks

edit:

If $t_1^{-1}=1$, then we would have:

$b_1+b_3+b_1b_2b_3-1=0$ and $t_2^{-1}=1$.

Just looking at the first equation I see that:

if $b_1=0$ then $b_3=1$ and $b_2$ could be anything.

if $b_3=0$ then $b_1=1$ and $b_2$ could be anything.

if $b_1 \neq 0 \neq b_3$ then $b_2 = (-b1-b3)(b_1^{-1}b_3^{-1})$, and so $b_1$ and $b_3$ can be any invertible invertible element.

This leads me to conclude that $\operatorname{Spec} (\frac{F[b_1,b_2,b_3,t_1^{\pm},t_2^{f\pm}]}{I}) \cong F \amalg F \amalg (F^*)^2$

Why is this incorrect?

答:
Edit: The answer that was previously here was incorrect. First, the second equation lets you eliminate $t_2$ so we're left with a single equation

$$F[b_1, b_2, b_3, t_1^{\pm 1}]/(b_1 + b_3 + b_1 b_2 b_3 = t_1^{-1}).$$

This single equation just says that $f(b_1, b_2, b_3) = b_1 + b_3 + b_1 b_2 b_3$ must be invertible; equivalently, the above is a localization

$$F[b_1, b_2, b_3][f^{-1}]$$

which geometrically means it corresponds to the open subvariety of $\mathbb{A}^3$ given by the complement of $V(f) = \{ (x_1, x_2, x_3) \in \mathbb{A}^3 : x_1 + x_3 + x_1 x_2 x_3 = 0 \}$. Probably more can be said but it depends on what you want to know.

If we set $t_1^{-1} = 1$ then we instead get the closed subvariety of $\mathbb{A}^3$ given by $V(f - 1) = \{ (x_1, x_2, x_3) \in \mathbb{A}^3 : x_1 + x_3 + x_1 x_2 x_3 = 1 \}$. Your case analysis is correct but it doesn't imply the disjoint union decomposition you wrote because the three cases are not disconnected; in fact $V(f - 1)$ is irreducible.


问:
Benford's law not working?
So I recently came across Benford's law and immediately tried to code it out but the answer I got was rather confusing. I think my code is correct I'm pretty sure it is but the result is just not the one others get. Here's the code:

```<code>int a;
float[] flag=new float[10];

  for(int i=0;i<10000;i++)
  {

    a=floor(random(0,99999));
      while(a>=10)
       a/=10;

      flag[a]++;
  }

</code>```

But I get roughly the same flag[k] (around 1100) but according to Benford's law that should not be the case.

I think it is because my upper limit is set so I tried modifying my code as follows :

```<code>int a;
float p,q,r;
float[] flag=new float[10];

  for(int i=0;i<10000;i++)
  {
    r=random(0,99999)
    q=random(0,r);
    p=random(0,q);
    a=floor(random(0,p));
      while(a>=10)
       a/=10;

      flag[a]++;
  }

</code>```

And this gives really satisfactory results, the problem is that this creates a very different type of distribution, it's certainly not random, so the problem I'm running into is that: is this how Benford's law work? Since we don't know the "Cutoff" it makes 1s significantly more likely than 9s and other numbers, Since they would require the "Cutoff" to be high which reduces the probability of them occuring in a random "Cutoff" scenario?

答: Benford's law does not apply to data with a uniform distribution, be it continuous or discrete. It applies to data whose logarithms' fractional parts have a continuous uniform distribution.</p>

Title: Benford's law not working?


问:
Characterization of a virtually nilpotent group
Let $G$ be a group that has an increasing sequence of subgroups $G_i \le G_{i+1}$ satisfying the following properties.

(a) $G=\bigcup_{i\ge 1} G_i$.

(b) Each $G_i$ contains a nilpotent subgroup $N_i$ such that $[G_i:N_i] \le k$ for some constant $k$ independent of $i$.

Can we find a nilpotent subgroup $N \le G$ such that $[G:N]<\infty$?

答:
Sorry, despite getting four upvotes, my previous answer was wrong, so I have deleted it. The result is false even when $k=1$.

As counterexample, for each $i \ge 1$, let $H_i$ be a finite nilpotent group of class $i$. For example, we could choose $H_i$ to be the dihedral group of order $2^{i+2}$.

Now let $G_i = H_1 \times H_2 \times \cdots \times H_i$ for all $i$, with the natural embedding of $G_i$ into $G_{i+1}$. Then $G := \cup_{i \ge 1} G_i = \times_{i \ge 1} H_i$ is not nilpotent, because it has subgroups that are nilpotent of class $i$ for each $i$.

But we need to prove that $G$ does not have a nilpotent subgroup of finite index. A subgroup of finite index would have to project onto $H_i$ for all but finitely many $i$, and so again it could not be nilpotent.


问:
How to compute $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}/ \langle(2,1,2)\rangle$?
I am trying to compute the factor group  $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}/ \langle(2,1,2)\rangle$. Here is what i did:

First i expand  $\langle(2,1,2)\rangle$, it is equal to {(2,1,2),(0,0,4),(2,1,6),(0,0,0)}. Then, since order of $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}$ is 64, we have 64/4=16. So the order of factor group is 16 and so it must be isomorphic to one of the following: $\mathbb{Z}_{16}$,  $\mathbb{Z}_{2} \times\mathbb{Z}_{8}$,  $\mathbb{Z}_{4} \times\mathbb{Z}_{4}$,  $\mathbb{Z}_{2} \times\mathbb{Z}_{2}  \times\mathbb{Z}_{4}$ or $\mathbb{Z}_{2} \times\mathbb{Z}_{2}  \times\mathbb{Z}_{2} \times\mathbb{Z}_{2}$. But from here on i do not know how to continue. Can anyone help? How can i eliminate  $\mathbb{Z}_{16}$ etc.?

Edit: Actually, i have a solution, saying that consider 4(a,b,c)=(4a,4b,4c)=(0,0,4c). 4c is either 0 or 4, in both cases both (0,0,0) and (0,0,4) are in $\langle(2,1,2)\rangle$ so any coset in the factor group has order at most 4. I do not understand why this is the case. Can anyone explain it?

Thanks

答:
The original group was generated by $\langle1,0,0\rangle,
\langle0,1,0\rangle, $ and $\langle0,0,1\rangle$, with orders 4, 2, and 8, respectively.  A good place to begin would be to investigate the orders of these elements of the new group, and to see if they generate the entire new group.</p>


问:
Multiplicative Functions Proof
I don't know if this question was asked before, but I could not find it on this forum: a function $f(n)$ is multiplicative if $f(mn)=f(m)f(n)$, where m and n are coprime and positive. If $d(n)$ is the number of positive divisors of n, how can I prove $d(n)$ is multiplicative? Thanks.

答:
Welcome to Math SX!

Note that if two numbers are coprime, they do not have common divisors (except for $1$), therefore every combination $pq$ (where $p$ is divisor of $m$ and $q$ is a divisor of $n$) is a divisor of $mn$. Using this, the result will follow (I hope you can see it).

Title: Multiplicative Functions Proof


问:
Change of basis-linear algebra
I see there are other change of basis posts, but it seems like they are not asking the same question as me, or perhaps my understanding is so poor, I do not understand they are asking the same questions.

Linear algebra was going so well until this concept, which I cannot wrap my head around. I honestly need someone to work through this practice problem( not for a grade) to explain how to solve it. There is an example in my book, but for some reason when I solve it as they do, I do not get the answer...

Part a is not the problem, it is part b

Instructions: In Exercises 37–42, find T(x) by using (a) the standard matrix and (b) the matrix relative to $B$ and $B'$.

$T: \mathbb{R}^3 \to \mathbb{R}^3, T(x, y, z) = (x + y +z, 2z- x, 2y-z)$

$v = (4, -5, 10)$

$B = \{(2, 0, 1), (0, 2, 1), (1, 2, 1)\}$

$B' = \{(1, 1, 1), (1, 1, 0), (0, 1, 1)\}$

$A=\begin{pmatrix}1&1&1\\-1&0&2\\0&2&-1\end{pmatrix}$

$P=\begin{pmatrix}2&0&1\\0&2&2\\1&1&1\end{pmatrix}$

$Q^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\-1&1&0\end{pmatrix}$

答:
Let $A$ be the matrix of $T$ in the standard basis.

If we denote $X'$ the column matrix of a vector in the domain and if $P$ is the change of basis matrix from the standard basis to basis $\mathcal B$, we have the relation $$X=PX'.$$
Similarly, if   $X'$ denotes the column matrix of a vector in the codomain and if $Q$ is the change of basis matrix from the standard basis to basis $\mathcal B'$, we have the relation $$Y=QY'.$$

Therefire the relation $Y=AX$ becomes
$$QY'=A(PX')\iff Y'=(Q^{-1}AP)X',$$
and as a consequence, the matrix of $T$ relative to bases $\mathcal B$ and $\mathcal B'$ is
$\:\color{red}{Q^{-1}AP}$.

Title: Change of basis-linear algebra


问:
Euler's formula question
I saw this formula in my book:

$$2i \sin(w)=e^{iw}-e^{-iw}$$

but I don't understand why this is true? isn't $\,\sin(w)=\frac{e^{iw}-e^{-iw}}{2}$?

答:
No. It's$$\sin(w)=\frac{e^{iw}-e^{-iw}}{2i}.\tag1$$Therefore, that formula from your book is correct.

If you want to know why we have $(1)$, note that$$e^w=1+\frac w1+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots$$So,$$e^{iw}=1+i\frac w1-\frac{w^2}{2!}-i\frac{w^3}{3!}+\cdots$$and$$e^{-iw}=1-i\frac w1-\frac{w^2}{2!}+i\frac{w^3}{3!}+\cdots$$and so\begin{align}e^{iw}-e^{-iw}&=2i\left(\frac w1-\frac{w^3}{3!}+\frac{w^5}{5!}-\cdots\right)\\&=2i\sin(w).\end{align}

Title: Euler's formula question


问:
Solving systems of ordinary differential equations when you can't work out constants from given initial conditions.
Not sure if this is the right place to ask but I am trying to solve systems of ODE's using eigenvectors and eigenvalues.

$$
x'(t) = \begin{bmatrix}5 & 4\\-2 & 1\end{bmatrix}  \begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix}, \ \  x_1(0)=0, x_2(0)=1
$$

I have worked out the eigenvectors and eigenvalues to be $ λ_1 = 3 +2j $ and $ V_1 =  \begin{bmatrix} -1-j \\ 1\end{bmatrix}$ and $ λ_2 = 3 - 2j $ and $ V_2 =  \begin{bmatrix} -1+j \\ 1\end{bmatrix}$

Subbing them into the general formula gives:

$$
x(t)= e^{3t}(d_1(\cos2t\begin{bmatrix} -1 \\ 1\end{bmatrix} -\sin2t\begin{bmatrix} -1 \\ 1\end{bmatrix})+d_2(\cos2t\begin{bmatrix} -1 \\ 1\end{bmatrix}+\sin2t\begin{bmatrix} -1 \\ 1\end{bmatrix}))
$$

However when trying to work out the values of the d_1 and d_2 I always get an error.

Can anyone help me understand where i've gone wrong, thank you.

答:
You have two eigenvectors $v_\pm = v_1\pm jv_2$ with
$$
v_1 = \begin{bmatrix}
-1\\
1
\end{bmatrix}
$$
and
$$
v_2 =
\begin{bmatrix}
-1\\
0
\end{bmatrix}
$$
for eigenvalues $\lambda_\pm = a\pm jb$.
Recall that the general solution to the ode is of the form
$$
x(t) = e^{at}\left(\ \  d_1(v_1\cos(bt) - v_2\sin(bt)) + d_2(v_1\sin(bt) + v_2\cos(bt))\ \ \right)
$$
(see more details in eq (2) here). Or equivalently
$$x(t) = e^{at}\left(\ \  d_1\left(\begin{bmatrix}
-1\\
1
\end{bmatrix}\cos(bt) - \begin{bmatrix}
-1\\
0
\end{bmatrix}\sin(bt)\right) + d_2\left(\begin{bmatrix}
-1\\
1
\end{bmatrix}\sin(bt) + \begin{bmatrix}
-1\\
0
\end{bmatrix}\cos(bt)\right)\ \ \right)$$

Compare this with your result. You basically used $v_1=v_2$


问:
Rank preservation of Hankel matrix by adding constrained sample
Let some $x_i \in \mathbb{R}$ for every $i$ such that the Hankel matrix
$$H_0=\begin{bmatrix} x_0 & x_1 & x_2 & x_3\\
x_1 & x_2 & x_3 & x_4\\
x_2 & x_3 & x_4 & x_5
\end{bmatrix} $$
is full rank, with the rank equal to 3. Show that there exists some $y\in [-1,1]$ such that the following Hankel matrix is full rank too
$$H_1=\begin{bmatrix}  x_1 & x_2 & x_3 & x_4\\
 x_2 & x_3 & x_4 & x_5\\
 x_3 & x_4 & x_5 & y
\end{bmatrix} $$
The matrix $H_1$ is obtained by deleting the first column of $H_0$ and adding a new column as $[x_4 \ x_5 \ y]^\top$.

I tried to solve this by showing that there exists some $a_0,a_1,a_2,a_3\in \mathbb{R}$ with $a_0\neq 0$ and some $y_0\in [-1,1]$ such that
\begin{equation}
\begin{bmatrix}
x_4\\ x_5\\ y_0
\end{bmatrix}= \begin{bmatrix} x_0 & x_1 & x_2 & x_3\\
x_1 & x_2 & x_3 & x_4\\
x_2 & x_3 & x_4 & x_5
\end{bmatrix} \begin{bmatrix}
a_0\\ a_1 \\ a_2 \\ a_3
\end{bmatrix}
\end{equation}
But still couldn't come up with a formal proof for it. Do you have any suggestion?

答:
We denote
$$
P =  \pmatrix{
 x_1&x_2&x_3\\
 x_2 & x_3 & x_4\\
  x_3 & x_4 & x_5}, \quad a = \pmatrix{x_0\\x_1\\x_2},
\quad b = \pmatrix{x_4\\x_5\\y}.
$$
In the case that $P$ is invertible, $H_1$ would have full rank for all $y$, so suppose that this is not the case. From the fact that $H_0 = [a \ \  P]$ has full rank, we see that $P$ must have rank $2$.

Write $b$ in the form $b(y) = b_0 + ye_3$, where $e_3 = (0,0,1)$. Our goal is to show that there exists a value of $y$ for which $[P\ \ b(y)]$ has rank greater than $P$, which is to say that the line parameterized by $b(y)$ does not intersect the plane spanned by the columns of $P$.

Suppose for contradiction that this is not the case. Then it must hold that the line $b(y)$ is parallel to to the plane spanned by the columns of $P$.  That is, there exists a vector $v$ for which $Pv = e_3$. Let $w$ denote a non-zero solution to $Pw = 0$; one exists because $P$ is not invertible. Because $P$ is symmetric, $w$ must be orthogonal to $Pv$ for all $v$, which means that we must have $e_3^Tw = 0$. In other words, there is a non-zero vector $w = (w_1,w_2,0)$ for which
$$
Pw = 0 \implies w_1 \pmatrix{x_1\\x_2\\x_3} = -w_2 \pmatrix{x_2\\x_3\\x_4}.
$$
If neither column is zero, then let $\alpha = -w_1/w_2$. We have
$$
P = x_1 \cdot \pmatrix{1 & \alpha & \alpha^2\\ \alpha & \alpha^2 & \alpha^3\\ \alpha^2 & \alpha^3 & \alpha^4} + \pmatrix{0&0&0\\0&0&0\\0&0&x_5-x_1\alpha^4}.
$$
Because $P$ has rank $2$, the second matrix in this sum can't be zero. So, $x_5 \neq \alpha x_4$, since $\alpha x_4 = x_1 \alpha^3$. This means that $(x_4,x_5)$ is not a multiple of $(1,\alpha)$, which means that $Pv = b_0$ has no solution.  This means that taking $y = 0$ gives us the desired outcome. (In fact, because the line is parallel to the column space, taking any $y$ yields the desired outcome).

If the first column is zero, then we have $x_1 = x_2 = x_3 = 0$, but we must have $x_4 \neq 0$ since the rank of $P$ is greater than $1$. However, this means that reversing the columns of $H_1$ gives us a triangular matrix, which means that $H_1$ is invertible for all $y$.

If the second column is zero, something similar happens (where we note that this time, we must have $x_1 = 0$). Again, $H_1$ has full rank for all $y$.


问:
Can I know the Fourier coefficients of a function that uniformly converges? if so why is it?
For example the function

$$f(x) = \sum\frac{1}{2^n}\sin(2^nx)$$
we can see that the function uniformly converges

Can I know that the Fourier coefficients are $$\hat{f}(x)=\frac{1}{2^n}$$

if so why is that possible?

答:
$$f(x) \sim \sum_k b_k \sin k x$$ is a Fourier sine series of an odd function with period $2\pi.$

We have  $$f(x) \sim \sum_n \frac{1}{2^n} \sin 2^n x,$$

so

$$\bbox[5px, border: 1.25pt solid green]{b_k = \left\{ \begin{aligned} \frac{1}{k}, &\quad \text{when }\exists n \text{ s.t. }k=2^n,\\
0, &\quad \text{otherwise.} \end{aligned}\right.}$$

$$f(x) \sim \sin x + \frac{1}{2} \sin 2x + \frac{1}{4} \sin 4x + \frac{1}{8} \sin 8x + \cdots$$

We could also check that $b_k$ arises from the formula

$$b_k = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin k x \, dx.$$

This checks because the sinusoids are orthogonal:

$$ \frac{1}{\pi}\int_{-\pi}^\pi \sin k x \sin l x \, dx = \delta_{kl}.$$


问:
Finding the residue at $z=z_0$ of $f(z)$ given...
i need to resolve this problem:

Given $f(z)=\frac{1}{\big(g(z)\big)^2}$, show that if $g(z_0)=0$ and $g'(z_0)\neq 0$ then $f$ has a second order pole at $z_0$ and $\text{Res}_{z=z_0}f(z)=-\frac{g''(z_0)}{\big(g'(z_0)\big)^3}$

It's easy to show that $f$ has a second order pole at $z_0$ however i have some difficulties finding de residue, since $z_0$ is a pole of order $2$ then we have that

$$\text{Res}_{z=z_0}f(z)=\lim_{z\to z_0}\frac{d}{dz}\big((z-z_0)^2f(z)\big)\\
=\lim_{z\to z_0}\frac{d}{dz}\Big(\frac{z-z_0}{g(z)}\Big)^2=\lim_{z\to z_0}2\Big(\frac{z-z_0}{g(z)}\Big)\Big(\frac{g(z)-(z-z_0)g'(z)}{g(z)^2}\Big)$$

By L'Hôpital's rule I get the correct answer, but i would like to compute this limit without the aid of L'Hôpital's rule, just with the properties/theorems of residue calculus.
Thanks.

Body:
Assuming that $g$ is analytic with $g(z_0)=0$ and $g'(z_0)\ne0$, we can write

$$\begin{align}
\lim_{z\to z_0}\frac{d}{dz}\left(\frac{z-z_0}{g(z)-g(z_0)}\right)^2&=2\lim_{z\to z_0}\left(\frac{z-z_0}{g(z)-g(z_0)}\right)\frac{d}{dz}\left(\frac{z-z_0}{g(z)-g(z_0)}\right)\\\\
&=\frac2{g'(z_0)}\lim_{z\to z_0}\left(\frac{g(z)-g(z_0)-g'(z)(z-z_0)}{(g(z)-g(z_0))^2}\right)\\\\
&=\frac2{g'(z_0)}\lim_{z\to z_0}\left(\frac{\color{red}{\frac{g(z)-g(z_0)-g'(z_0)(z-z_0)}{(z-z_0)^2}}-\color{blue}{\frac{(g'(z)-g'(z_0))(z-z_0)}{(z-z_0)^2}}}{\frac{(g(z)-g(z_0))^2}{(z-z_0)^2}}\right)\\\\
&=\frac2{(g'(z_0))^3}\left(\color{red}{\frac12 g''(z_0)}-\color{blue}{g''(z_0)}\right)\\\\
&=-\frac{g''(z_0)}{(g'(z_0))^3}
\end{align}$$


问:
Prove the existence of a minimal non-negative solution to a linear Diophantine equation
How can I show that there is a unique integer solution $(x_1, y_1)$ to the Diophantine equation $ax+by=c$ such that $0\leq x_1< b/\gcd(a,b)$?

I am given $a,b,c\in\mathbb{Z}^{+}$ such that $\gcd(a,b)|c$. I gave it a try first. I made $\gcd(a,b)=d$ and $d$ is a positive integer for sure. Then, we know $d|a$, $d|b$, and $d|c$. By divisibility rules, $d|(ax+by)$, so $d|c$. I've come full circle.

The I conjectured, since $ax+by=c$, by Bezout's Lemma $\gcd(a,b)=c$? If $\gcd(a,b)=c$, then $\gcd(a,bc)|c$ is $c|c$. So that doesn't do anything. This question is a bit challenging. Many thanks for the help.

答:
The equation $ax+by=c$ is equivalent to $\frac{a}{(a,b)}x+\frac{b}{(a,b)}y=\frac{c}{(a,b)},$ so it suffices to treat the case where $a$ and $b$ are coprime.

Let the equation be $mx+ny=N$ where $m,n$ are positive integers such that $(m,n)=1.$ By Bezout's lemma, there exist integers $(x_0,y_0)$ such that $mx_0 +ny_0=N.$ The problem is that we don't know the signs of $x_0$ and $y_0.$ By the classification of all solutions to a linear Diophantine equation, all solutions are given by $(x_0-kn,y_0+km)$ over all integers $k.$

We want there to exist a unique integer $k$ such that $0\le x_0-kn<n.$ By the Euclidean division of $x_0$ by $n,$ there exists a unique non-negative remainder $x_1$ and a unique quotient $k$ such that $$x_0-kn=x_1,0\le x_1<n.$$ So $x_1$ is the unique solution that we seek. You can actually then show that the corresponding $y_1=y_0+qm$ is also non-negative, but since you did not ask for it, I'll leave that to you.


答:
If you're not sure you can do the proof:
We know that : $e^{iw} =\cos{w} +i\sin{w}$ and $e^{-iw} =\cos{w} -i\sin{w}$

If you do the difference of those two terms, you'll find the first formula.

Title: Euler's formula question

Body:
I saw this formula in my book:

$$2i \sin(w)=e^{iw}-e^{-iw}$$

but I don't understand why this is true? isn't $\,\sin(w)=\frac{e^{iw}-e^{-iw}}{2}$?


答:
All of the change of base posts are basically the same.  The key is that any invertible matrix can be interpreted as the change of basis matrix from the standard basis to the basis consisting of its columns.

The standard matrix for $T$ is found by applying it to the standard basis vectors:  $T(1,0,0)=(1,-1,0),T(0,1,0)=(1,0,2)$ and $T(0,0,1)=(1,2,-1)$.  Thus we get $\begin{pmatrix}1&1&1\\-1&0&2\\0&2&-1\end{pmatrix}$.

The matrix rel $B$ and $B'$ is gotten by multiplying the matrix in the standard basis by $P=\begin{pmatrix}2&0&1\\0&2&2\\1&1&1\end{pmatrix}$ on the right and the matrix $Q^{-1}$ on the left, where $Q=\begin{pmatrix}1&1&0\\1&1&1\\1&0&1\end{pmatrix}$.

(Notice the columns of $P$ and $Q$ are the elements of $B$ and $B'$ respectively.)

Title: Change of basis-linear algebra

Body:
I see there are other change of basis posts, but it seems like they are not asking the same question as me, or perhaps my understanding is so poor, I do not understand they are asking the same questions.

Linear algebra was going so well until this concept, which I cannot wrap my head around. I honestly need someone to work through this practice problem( not for a grade) to explain how to solve it. There is an example in my book, but for some reason when I solve it as they do, I do not get the answer...

Part a is not the problem, it is part b

Instructions: In Exercises 37–42, find T(x) by using (a) the standard matrix and (b) the matrix relative to $B$ and $B'$.

$T: \mathbb{R}^3 \to \mathbb{R}^3, T(x, y, z) = (x + y +z, 2z- x, 2y-z)$

$v = (4, -5, 10)$

$B = \{(2, 0, 1), (0, 2, 1), (1, 2, 1)\}$

$B' = \{(1, 1, 1), (1, 1, 0), (0, 1, 1)\}$

$A=\begin{pmatrix}1&1&1\\-1&0&2\\0&2&-1\end{pmatrix}$

$P=\begin{pmatrix}2&0&1\\0&2&2\\1&1&1\end{pmatrix}$

$Q^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\-1&1&0\end{pmatrix}$


问:
Prove that this block matrix is positive definite
I'm trying to prove that this symmetric block matrix is positive definite
$$
    \begin{bmatrix}
    B & -I & 0 & \dots& 0 \\
    -I & B & -I & \ddots & \vdots\\
    0 & -I & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -I\\
    0 & \dots & 0& - I & B \\
    \end{bmatrix}
$$

with $I$ the identity matrix

$$B = \begin{bmatrix}
    4 & -1 & 0 & \dots& 0 \\
    -1 & 4 & -1 & \ddots & \vdots\\
    0 & -1 & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -1\\
    0 & \dots & 0& - 1 & 4 \\
    \end{bmatrix}$$

by proving that this term $x^T A x$ is positive, but given the fact that it is a block matrix I'm stuck. Could you help me please?

答:
Another way to get the desired result is via Perron-Frobenius Theory and with a little cleverness we get Taussky's refinement of Gerschgorin Discs as a corollary.

$A:=\begin{bmatrix}
    B & -I & 0 & \dots& 0 \\
    -I & B & -I & \ddots & \vdots\\
    0 & -I & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -I\\
    0 & \dots & 0& - I & B \\
    \end{bmatrix}$

notice that $A$ is irreducible / represents a single communicating class in an underlying (possibly weighted and directed) graph

Direct Argument

consider $A':=-\big(A-4I\big)$

now $A'$ has maximum row sum of $4$ and minimum row sum of $2$. Since $A'$ is irreducible and real non-negative we may apply Perron-Frobenius Theory to observe that $2\lt \lambda_\text{Perron} \lt 4$

Being symmetric, $\lambda_\text{Perron} = \sigma_{max} = \Big\Vert A'\Big\Vert_2= \Big\Vert -A'\Big\Vert_2$

and since $-A' = A - 4I\implies A=-A'+4I$, then for any $\big \Vert \mathbf x\big \Vert_2=1$ we have

$\min \mathbf x^TA\mathbf x$

$=  \min \Big\{\mathbf x^T\big(-A'+4I\big)\mathbf x\Big\}$

$=  \min\Big\{ \mathbf x^T\big(-A'\big)\mathbf x+\mathbf x^T\big(4I\big)\mathbf x\Big\}$

$\geq  \min\Big\{ \mathbf x^T\big(-A'\big)\mathbf x\Big\}+ \min\Big\{ \mathbf x^T\big(4I\big)\mathbf x\Big\}$

$= -  \max\Big\{ \mathbf x^T\big(A'\big)\mathbf x\Big\}+ 4$

$\gt -4 +4$

$=0$

$\implies A \succ \mathbf 0$

Optional Second

$\text{Gerschgorin Discs & Perron-Frobenius Theory} \implies \text{Taussky's refinement  of G-Discs}$

In words, the refinement says  for any Weakly Diagonally Dominant irreducible $A$ where the dominance is strict in at least one row

$\det\big(A\big ) \neq 0$

arguing by contradiction, suppose $\det\big(A\big )=0$.

Now left multiply by invertible diagonal matrix $D$ where $d_{k,k} = -(a_{k,k})^{-1}$  (and we know $a_{k,k}\neq 0$, why?).  In other words $D:=-\big(I\circ A\big)^{-1}$.  Note that $D$ preserves rank, irreducibility and the diagonal dominance structure.

Since $(DA)$ has its diagonal homogenized to consist solely of $-1$'s, let us write $C-I=(DA)$

$\det\big(C-I\big)=\det\big(DA\big)=\det\big(A\big)=0\implies$  there is some $\mathbf x \neq \mathbf 0$ such that

$\big(C-I\big)\mathbf x = \mathbf 0\implies C\mathbf x = \mathbf x \implies \lambda_\text{max modulus}\Big(C\Big)\geq 1$

but

$\lambda_\text{max modulus}\Big(C\Big) \leq \lambda_\text{max modulus}\Big(\big \vert C\big \vert \Big) \lt 1$

which is a contradiction.

notes:

(i.)  $\big \vert C\big \vert$ denotes $C$ with each element replaced by its modulus.  The inequality is proven here $\rho(B) \leq \rho(|B|)$, where $\rho$ is the spectral radius

(ii.) The second inequality follows because $\big\vert C\big \vert$ is irreducible and substochastic and strictly so in at least one row.  So $\big \vert C\big \vert$ cannot have an eigenvalue of 1, by standard Markov chain arguments or directly applying Perron-Frobenius Theorem.


问:
trace of $M \mapsto PM+MP$ with P a matrix of projection
let $n \in \mathbb{N} \:\: n\geq2$ and $P \in \mathcal{M}_n(\mathbb{R})$ a matrix of projection

$\begin{array}{ccccc}
Φ & : & \mathcal{M}_n(\mathbb{R}) & \to & \mathcal{M}_n(\mathbb{R}) \\
 & & M & \mapsto & PM + MP \\
\end{array}$

Express $Tr(Φ)$ as a function of $Tr(P)$

答:
Consider it in $\mathcal M_n(\mathbb C)$.

Suppose $P(e_{ij})=e_{ij}$ for $i,j\leq k$ and $P(e_{ij})=0$ for $i,j>k$.

$\text{Tr}(\Phi)=\sum_{ij} \langle \Phi(e_{ij}),e_{ij}\rangle=k^2=\text{Tr}(P)^2$


问:
Solving the simultaneous equations
I need to solve the following simultaneous equations for the constants $A$ and $B$:

$$zA + (1-z)B = 1$$
$$\hspace{1.8cm} z^2A + (1-z)B = z^2 - z + 1$$

where$z$ is just a variable. What I did was this:

$$zA = 1 - (1-z)B \implies z^2A = z - (z - z^2)B$$

and so subbing this into the second equation gives me

$$z - (z - z^2)B + (1-z)B = z^2 -z + 1$$

Rearanging gives me

$$(z^2 - 2z + 1) = (1 - z)^2 = \{(1-z) - z(1 - z)\}B $$

which can then be simplified to get

$$(1 - z) = (1 - z)B \implies B = 1$$

which is wrong beacuse in my answers it says $B = \frac{1-z}{1 - 2z}$. I think I will get $A$ once I have $B$ but I can't see my mistake right now.

答:
You system has infinitely many solutions if $A=B=1$ (for instance, substract the second from the first, and see what happens to $A$). Then, any $z$ satisfies it.

If $A\neq1$, then $z$ must be either $0$ or $1$. But plugging $z=1$ in the equations yields $A=1$, so the only solution is $z=0$.

Title: Solving the simultaneous equations


问:
Decompose a motion into its screw motion (mozzi-chasles' theorem)
The theorem says that any rigid body movement can be described (for each point) as a set translation along a set axis and a rotation (by a set angle) around it.

It's easy to imagine what the movement looks like if only a rotation or only a translation takes place around a set axis.

However, I fail to imagine where would the axis be and which way pointing, and the rotation angle and the amount of translation, for one particular movement (I chose a slightly more complex one than those allowed in the trivial cases above).

The movement is as follows:

Looking from the top, I have a cube on the table, lying flat on one of its faces.

I simply stand it on one of its corners reaching unstable equilibrium (by a rotation along one axis), leaving one corner on the table exactly where it was.

Along the vertical axis, I rotate it clockwise by for example 30°.

Where is the screw axis of this movement? Is there any translation. Any rotation?

Being this a more complex movement, I suspect that there is both a non-zero translation (that is applied to every point) and none-zero rotation (along the screw axis). How come it cancels out at the corner point, that does not move?

答:
In short, Euler's rotation theorem says "any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point."

That means, that if I choose the center point of the body as fixed for example and I rotate it many times around various axes (going through center), I could achieve the same displacement with a single rotation along an axis.

In the question cube example, the fixed point was the corner (both in rotation 1. and 2.). So, according to Euler, there exists an axis going through the corner and if I rotated the cube around it, I would achieve the displacement in the question. All of that without any translation along the rotation axis. In fact, if it was a true screw motion (with the translational component), indeed no point would be fixed.

For imagining the direction of the screw axis (we now know that just rotation axis):
You can try grabbing a cube not by opposing corners but slide your fingers a little closer, on the opposing edges, forming a rotation axis, that way you can achieve similar (however not sure if the same) displacement to that in question with exception the center is fixed (if you managed to form an axis going through the center) -- just to illustrate compound rotations along two different axes might be substituted by a single rotation.

More formally, the results in the following link imply that chained rotations actually have a single equivalent rotation:

A proof that rotations in 3d form a group (and have at least one axis). (I would say that proves the Euler's theorem I cited above),  https://www.math.purdue.edu/~arapura/algebra/algebra13.pdf

It is important that the rotation in proof above, in linear algebra sense, is meant the vectors rotate around an axis always going through the origin -- zero-vector. Fortunately, the twice rotated cube in question is always rotated around the same point -- corner --, that would be the zero-vector if we applied the theory on the example.


问:
Enclosing Circle for a Set of 'n' points
Given a set of 'n' points, I need to construct a circle enclosing the points(might not be the smallest enclosing circle) given that minimum of 2 points lie on the circle. I want to find the result in order of 'n' time complexity.

What I tried:
I tried by first finding the maximum and minimum x-coordinate and finding the distance between them and then repeating the same for y-coordinate. Then I found out the maximum distance out of the two(between x_min and x_max and between y_min and y_max). For the pair for which I got the maximum distance, I found the euclidean distance and took that as the diameter, and constructed the circle around it. For eg. if I found the maximum distance between x_min and x_max, I calculated the euclidean distance between (x_min,y1) and (x_min,y2) and considered it as the diameter.

But that seems to not work !!

答:
find the leftmost point in $O(n)$ operations (this is a point of the convex hull),

find the leftmost point "around the first" in $O(n)$ operations (this is the next point on the hull),

compute the radii of all circles by these two points and the remaining ones, in $O(n)$ operations. Keep the circle with the largest radius and you are done.

(The first two steps correspond to the start of a "Jarvis walk" https://en.wikipedia.org/wiki/Gift_wrapping_algorithm.)


Title: Solving the simultaneous equations

Body:
I need to solve the following simultaneous equations for the constants $A$ and $B$:

$$zA + (1-z)B = 1$$
$$\hspace{1.8cm} z^2A + (1-z)B = z^2 - z + 1$$

where$z$ is just a variable. What I did was this:

$$zA = 1 - (1-z)B \implies z^2A = z - (z - z^2)B$$

and so subbing this into the second equation gives me

$$z - (z - z^2)B + (1-z)B = z^2 -z + 1$$

Rearanging gives me

$$(z^2 - 2z + 1) = (1 - z)^2 = \{(1-z) - z(1 - z)\}B $$

which can then be simplified to get

$$(1 - z) = (1 - z)B \implies B = 1$$

which is wrong beacuse in my answers it says $B = \frac{1-z}{1 - 2z}$. I think I will get $A$ once I have $B$ but I can't see my mistake right now.

Body:
Hints:

From eq. (1):

$$(1-z)B=1-Az$$

From eq. (2):

$$(1-z)B=(1-A)z^2-z+1$$

so

$$ (1)=(2)\implies (1-A)z^2-(1-A)z=0\implies A=1\;\; \text {or} \;\,\,z(z-1)=0\;\;\ldots$$


问:
Method to solve this non-linear differential equation
I have a differential equation as follows and have got no idea how to solve it. Any help would be appreciated.

$$ \frac{d^2t}{dx^2} -2 (\frac{dt}{dx} )^2\frac{1}{t} = 0 $$

答:
$$t''-2\frac{(t')^2}{t}=0 \iff \frac{t''}{t'}=2\frac{t'}{t}$$
Now you can integrate in both side and get
$$\log|t'|=2\log|t|+\tilde C\iff t' = C\cdot t^2$$
and you can finish from here.


问:
How many possible Barcodes are there of width $12$?
I was doing a past paper for a competition and I came across this question:

A barcode of the type shown in the two examples is composed of alternate strips of black and white, where the leftmost and rightmost strips are always black. Each strip (of either colour) has a width of $1$ or $2$. The total width of the barcode is $12$. The barcodes are always read from left to right. How many distinct barcodes are
possible?

This is the solution given:

Any code will start with a black strip and a white strip followed by a shorter barcode. Let $C (m)$ be the number of distinct barcodes of width  $m$. Those codes which start with $BW$ will be followed by a code of width $m-2$; so there will be $C(m-2)$ of these. Likewise, there will be $C(m-3)$ codes starting $BBW$, the same number starting $BWW$, and $C(m-4)$ starting $BBWW$; and that exhausts the
possibilites. So it follows that
$$C (m) = C (m − 2) + 2C (m − 3) + C (m − 4)$$
When $m\le4$, it is simple to list all possible barcodes; namely $B$, $BB$, $BWB$ and $BBWB$,
$BWBB$, $BWWB$. Therefore $$C (1) = C (2) = C (3) = 1 ,~~C (4) = 3$$ We can now calculate for $m>4$.
Thus $C(5) = C(3) + 2C(2) + C(1) = 1 + 2 + 1 = 4$, and continuing like this, we get $$C (6) = 6,~ C (7) = 11,~ C (8) = 17,~ C (9) = 27,~ C (10) = 45,~ C (11) = 72,~
C (12) = 116$$

There's a couple of points I don't understand.

Won't the value of $C(m-3)$ be different for different starting points? More specifically, if the starting points are $BBW$ or $BWW$ why won't the value of $C(m-3)$ be different? Surely if it starts with
$BWW$ then there's only one choice for the colour of the next strip?

Aren't the number of barcodes starting with $BWW$ included in those starting with $BW$? If so, haven't they been counted twice in the solution?

答: There are certain unstated assumptions lurking here. One is that each strip, be it black or white, is of "width" $1$ or $2$ only; that is, $BB$ and $WW$ are allowed, but not $BBB$ or $WWW$. Another is that the bar code both starts and stops with a black strip. The latter assumption explains why we have $C(1)=C(2)=1$ and the two assumptions together explain why we have $C(3)=1$ and $C(4)=3$. Finally, the no-width-$3$ assumption explains the recursion: The first pair of strips in a barcode of length greater than $4$ are either $BW$, $BBW$, $BWW$, or $BBWW$, followed by a bar code (starting with a new $B$) of the appropriate shorter length.


问:
If $y_1$ and $y_2$ fundamental set of solutions, then no common point of inflection, unless $p(t)$ and $q(t)$ vanish simultaneous there
For $y''+p(t)y'+q(t)y=0$, prove that if $y_1$ and $y_2$ are a fundamental set of solutions, then they cannot have a common point $t^*$ of inflection unless $p$ and $q$ vanish simultaneously there.

We know that if $y_1$ and $y_2$ are two solutions of the ODE, then their Wronskian $W[y_1,y_2](t)$ satisfies $W' + p(t)W = 0;$; lemma 1 [1]. From this we can assert that $p(t)$ has to vanish at $t^*$, since we have $W'(t^*) = 0$ there.

How can we make an implication on $q(t)$?

Context: this is problem 2.1.17 of Braun's Differential Equations and Their Applications. [1]

答:
Since $t^*$ is a common point of inflection of $y_1(t),y_2(t)$, one has
$$ y_1''(t^*)=y_2''(t^*)=0. $$
Putting $y_i$ into the equation and evaluating at $t=t^*$, one has
$$ \left\{\begin{array}{ll}p(t^*)y_1'(t^*)+q(t^*)y_1(t^*)=0, \\
p(t^*)y_2'(t^*)+q(t^*)y_2(t^*)=0.
\end{array}\right.
$$
Now the determinant of coefficient matrix regarding to $p(t^*),q(t^*)$ is the Wronskian $W(y_1,y_2)(t^*)$ which is not zero since $y_1,y_2$ are fundamental solutions and hence one must have $p(t^*)=q(t^*)=0$.


问:
Need Help with Modulus Algebra
Non-mathematician here.  I need help solving a formula with a modulus term.

First some background:  I am trying to calculate the hole positions for a dividing head plate, which is a tool used in the field of machining when cutting gears.  The idea is that the user will crank an input handle a certain amount, which will precisely rotate an output shaft holding a metal blank disk to be cut into a gear.  In the photo below, there are 9 concentric rings.  The correct ring is chosen depending on the number of teeth to cut in the blank, and input arm is advanced a calculated number of holes for each gear tooth.  These dividing heads are very expensive, and I am making a home-made one. Thus I have to figure out the positions of the holes.  (Continued below photo)

In my case, 144 complete turns of the input crank will cause the output to rotate 360 degrees, or 2.5 output degrees per 1 input turn.  I want to create a gear with 42 teeth, which means 8.571428571 degrees/tooth and 3.428571429 turns of the input crank.  This means that I will need to crank 3 full turns, and 0.428571429 additional turns, which is ~154.29 degrees. This means the input arm advances ~154.29 degrees on the dividing plate each time.

The way this usually works is that the dividing head machine has 2 sector arms on the face of the dividing plate (see the bronze arms in the photos).  These would be set to the angle that needs to be advanced each time, ~154.29 degrees in my case. The user starts at the position of one of the arms, and stops at the position of the 2nd arm.

In my example, I created a spreadsheet here that calculates that the angle of the input arm (and thus the location for the hole) for the first 8 teeth are as follows.  Notice that tooth #8 wraps, and input arm is back to 0 degrees:

```<code>    tooth    Input arm Degrees    Output Degrees
    1        0.0                  0.00
    2        154.3                8.57
    3        308.6                17.14
    4        102.9                25.71
    5        257.1                34.29
    6        51.4                 42.86
    7        205.7                51.43
    8        0.0                  60.00
</code>```

And if I graph these out, I get:

So what I REALLY want, is the angle between the red dots in the photo.  When I have this, I can generate the dividing plate I need.  So I am going to try to write all this in some formulas.  But here I am getting lost...

The formula for input turns I (units: turns) to achieve output degrees O (units: degrees) in a machine with a conversion C (units: turns/degree)  =

```<code>O = I / C

O is Output, in degrees
I is Input, in full turns
C is Conversion factor, in turns/degree
</code>```

Since the input arm is wrapping, this would involve a modulus operation. In my example, I calculate 7 different angles before the 8th operation returns the input handle back to position of 0 degrees.  Thus there are 7 red dots on the image above, and the angle between each dot is:

```<code>Angle between red dots = 360 / 7
</code>```

But how to determine this value of 7 without using an iterative algorithm such as a spreadsheet??

$$\frac{360\ output\ deg}{42\ tooth} * \frac{1\ input\ turn}{2.5\ output\ deg} * \frac{360\ input\ deg}{1\ input\ turn} = 1234.28571\ \frac{input\ deg}{tooth}$$

$$Modulus(1234.28571\ deg,\ 360\ deg) = 154.285714285714286\ deg$$

$$Modulus(154.285714285714286*N,\ 360) = 0$$

In my case N = 7.  But how to calculate this, i.e for different cases when cutting gears with different number of teeth?

tl;dr

if Modulus(C*n, 360) = 0, how to calculate value of N.  I understand that there are multiples of N that will also work.  I just want the first value.

Thanks in advance.

答:
Note that $154.\overline{285714} = \frac{1080}{7}$. Then, we wish to find values of $N$ such that:

$$\text{Modulus}(\frac{1080N}{7}, 360) = 0$$

This happens if and only if $\frac{1080N}{7} = 360k$ for some (positive in your application) integer $k$. Then:

$$N = \frac{7\cdot 360k}{1080} = \frac{7}{3}k$$

Thus, the smallest value of $N$, which happens at $k=1$, is $\boxed{\frac{7}{3}}.$

In general, the smallest value of $N$ is $\frac{360}{C}$.

Title: Need Help with Modulus Algebra


问:
if $\sin24^\circ = p$ what is $\cos24^\circ$?
Let $p=\sin 24^\circ$

Then what would $\cos (24^\circ)$ be in terms of $p$?

What would $\sin (168^\circ) \cdot  \sin(-78^\circ)$ be in terms of $p$?

I'm not sure how to approach these as we have only started this section.

Thanks.

答:
My favorite method

$$\sin \theta = \cfrac{\text {opposite}}{\text{hypothenus}}=\cfrac p1,\quad \quad  \cos\theta = \cfrac{\text {adjacent}}{\text{hypothenus}}$$
What is "adjacent"? By Pthagoras', we have  $\quad \text{adjacent}^2+\text{opposite}^2=\text{hypothenus}^2$
so $$ \text{adjacent}^2 + p^2=1 \implies \text{adjacent}=\sqrt{1-p^2}$$
et voila! $$\cos\theta = \cfrac{\text {adjacent}}{\text{hypothenus}}=\cfrac{\sqrt{1-p^2}}{1}=\sqrt{1-p^2}$$


问:
Prove $H(x,y)=\alpha\ln(y)-\beta y+\gamma\ln(x)-\delta x$ is constant
Given:

$x'=\alpha x-\beta xy,y'=-\gamma y +\delta xy$
where $\alpha,\beta,\gamma,\delta>0$
Prove:

1.For the solution $(x(t),y(t))$ find the value of $y(t)$ where $x(t)$ attains maximum/minimum.

2.(H=consnt) is a closed curve in $\mathbb{R}^2$ where $H(x,y)=\alpha\ln(y)-\beta y+\gamma\ln(x)-\delta x$ is from voltka-lotera

Any hint for the second part?

答:
It is easy to see $\frac{d}{dt}H(x,y)=0$. So
$$ H(x,y)(t)=C. $$
Note that the system has a non-trivial equilibrium point $(\frac{\gamma}{\delta},\frac{\alpha}{\beta})$. Using
$$ \ln x=\ln a+\frac{x-a}{a}-\frac{(x-a)^2}{2a^2}+O((x-a)^3) $$
one has
\begin{eqnarray}
\ln x&=&\ln\bigg(\frac{\gamma}{\delta}\bigg)+\frac{\delta}{\gamma}\bigg(x-\frac{\gamma}{\delta}\bigg)-\frac{\delta^2}{2\gamma^2}\bigg(x-\frac{\gamma}{\delta}\bigg)^2+O\bigg(\bigg(x-\frac{\gamma}{\delta}\bigg)^3\bigg),\\
\ln y&=&\ln\bigg(\frac{\alpha}{\beta}\bigg)+\frac{\beta}{\alpha}\bigg(y-\frac{\alpha}{\beta}\bigg)-\frac{\beta^2}{2\alpha^2}\bigg(y-\frac{\alpha}{\beta}\bigg)^2+O\bigg(\bigg(x-\frac{\alpha}{\beta}\bigg)^3\bigg).
\end{eqnarray}
So $H(x,y)=C$
$$ -\frac{\beta^2}{2\alpha}\bigg(y-\frac{\alpha}{\beta}\bigg)^2-\frac{\delta^2}{2\gamma}\bigg(x-\frac{\gamma}{\delta}\bigg)^2+\alpha\ln\bigg(\frac{\alpha}{\beta}\bigg)+\gamma\ln\bigg(\frac{\gamma}{\delta}\bigg)-\alpha-\gamma+h.o.t=C $$
which is a circle.


答:
Suppose the contrary that $\operatorname{rank}(H_1)\le2$ for all $y\in[-1,1]$. Let $S$ be the square submatrix consisting of the first three columns of $H_1$. Then $S$ must be singular. Since it is also comprised of the last three columns of $H_0$, we have $3=\operatorname{rank}(H_0)\le\operatorname{rank}(S)+1\le\operatorname{rank}(H_1)+1\le3$. Therefore $\operatorname{rank}(S)=\operatorname{rank}(H_1)=2$.

It follows that $\operatorname{span}\{(x_4,x_5,y)^T: y\in[-1,1]\}$ must lie inside the column space of $S$. In particular, both $(x_4,x_5,0)^T$ and $(0,0,1)^T$ lie inside the column space of $S$. Therefore the augmented matrix
$$
A=\begin{bmatrix}
x_1 & x_2 & x_3 & x_4 & 0\\
x_2 & x_3 & x_4 & x_5 & 0\\
x_3 & x_4 & x_5 & 0 & 1
\end{bmatrix}
$$
has rank $2$. By elementary column operations, we can reduce $A$ to
$$
B=\begin{bmatrix}
x_1 & x_2 & x_3 & x_4 & 0\\
x_2 & x_3 & x_4 & x_5 & 0\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}.
$$
Hence $\operatorname{rank}(B)=2$ too and the first two rows of $B$ must be linearly dependent. It follows that the last two rows of $H_0$ are also linearly dependent. Now we arrive at a contradiction because $H_0$ has full row rank. Thus we conclude that $H_1$ must have full row rank for some $y\in[-1,1]$.

Title: Rank preservation of Hankel matrix by adding constrained sample

Body:
Let some $x_i \in \mathbb{R}$ for every $i$ such that the Hankel matrix
$$H_0=\begin{bmatrix} x_0 & x_1 & x_2 & x_3\\
x_1 & x_2 & x_3 & x_4\\
x_2 & x_3 & x_4 & x_5
\end{bmatrix} $$
is full rank, with the rank equal to 3. Show that there exists some $y\in [-1,1]$ such that the following Hankel matrix is full rank too
$$H_1=\begin{bmatrix}  x_1 & x_2 & x_3 & x_4\\
 x_2 & x_3 & x_4 & x_5\\
 x_3 & x_4 & x_5 & y
\end{bmatrix} $$
The matrix $H_1$ is obtained by deleting the first column of $H_0$ and adding a new column as $[x_4 \ x_5 \ y]^\top$.

I tried to solve this by showing that there exists some $a_0,a_1,a_2,a_3\in \mathbb{R}$ with $a_0\neq 0$ and some $y_0\in [-1,1]$ such that
\begin{equation}
\begin{bmatrix}
x_4\\ x_5\\ y_0
\end{bmatrix}= \begin{bmatrix} x_0 & x_1 & x_2 & x_3\\
x_1 & x_2 & x_3 & x_4\\
x_2 & x_3 & x_4 & x_5
\end{bmatrix} \begin{bmatrix}
a_0\\ a_1 \\ a_2 \\ a_3
\end{bmatrix}
\end{equation}
But still couldn't come up with a formal proof for it. Do you have any suggestion?


问:
Ito formula for Jump-processes
Hey here I found Ito formula for jump-processes (Proposition 20.13). It states that for Ito process of the form
$$X_t+\int_0^tu_sdB_s+\int_0^tv_sds+\int_0^t\eta_sdY_s$$

where $Y_t$ is a compound Poisson process, we have
$$f(X_t)=f(X_0)+\int_0^tv_sf'(X_s)ds+\int_0^tu_sf'(X_s)dB_s+\frac{1}{2}\int_0^t f''(X_s)|u_s|^2ds+\int_0^t(f(X_s)-f(X_{s-}))dN_s.$$

What we have to assume about processes $u_s, v_s, \eta_s$ and function $f$? Can we take function of two arguments $t$ and $x$ or it must be afucntion of $x$ only? Because I would like to make this statement precisely

答:
You can deduce some of the properties directly from the statement about $f(X_t)$.

Firstly, $f$ has to be twice differentiable (otherwise $\frac{1}{2}\int_0^t f''(X_s)|u_s|^2ds$ would be undefined).

Secondly, $u_s$ has to be square-integrable (again, otherwise the integral above would be undefined).

Thirdly, $v_s$ has to be integrable (otherwise $\int_0^tv_sf'(X_s)ds$ would be undefined).

In addition, $u_t$ has to be measurable with respect to the filtration generated by $B_t$ (otherwise the integral $\int_0^tu_sdB_s$ would be undefined), and $\eta_t$ has to be measurable with respect to the filtration generated by $Y_t$ (otherwise the integral $\int_0^t\eta_sdY_s$ would be undefined).

Title: Ito formula for Jump-processes


问:
Is there a locally compact space which is not a k-space
Definitions:

A locally compact space is a space where every point has a local base of compact neighborhoods.

A $k$-space $X$ has its topology generated by maps from compact Hausdorff spaces, i.e. $C$ is closed iff for every compact Hausdorff space $K$ and every continuous function $f: K \to X$, $f^{-1}[C]$ is closed in $K$. Strickland's notes call this compactly generated.

By compact I mean not necessarily Hausdorff.

The reason I ask is that standard constructions of non $k$-spaces for example the square of the one-point compactification of $\mathbb{Q}$ and the product $\mathbb{R}\setminus \{1,\frac{1}{2},\frac{1}{3}\} \times \mathbb{R}/\mathbb{Z}$ where the second quotient means identifying $\mathbb{Z}$ to one point, are usually not locally compact.

I'm looking for a locally compact space which is not a $k$-space.

答:
The answer is that, assuming AC, there are locally compact spaces which are not compactly generated. (That is, do not have topologies which are final with respect to all maps with compact Hausdorff domains. i.e. do not belong to the monocoreflective hull of the class of compact Hausdorff spaces as computed in $Top$.)

The construction of such a space is due to J. Isbell, A Distinguishing Example in k-Spaces, Proc. Am. Math. Soc. 100 (1987), 593-594. Below is my attempt at his details.

Let $X=\omega_1+1$, where $\omega_1$ is the first uncountable ordinal, and topologise $X$ by giving it the base of closed sets generated by $(i)$ countable initial segments $[0,\alpha]$, and $(ii)$ the singleton $\{\omega\}$. The space $X$ is our counterexample.

$X$ is locally compact.

Proof: Indeed, more is true: every subspace of $A\subseteq X$ is compact. For $A$, if nonempty, contains a least element, every neighbourhood of which contains all of $A$ except for possibly $\omega_1$. $\square$

$X$ is not compactly generated.

Proof: The claim is that there is a non-closed subspace $A\subseteq X$ with the property that whenever $f:K\rightarrow X$ is a continuous function with $K$ a compact Hausdorff space, the preimage $f^{-1}(A)$ is a closed subset of $K$. We will show that the open subspace $A=[0,\omega_1)$ has this property.

So take $K$ and $f$ as above and suppose $f^{-1}[0,\omega)\subset K$ is not closed. Fix a countable ordinal $\alpha_1$ and consider the closed set $f^{-1}[0,\alpha_1]\subseteq K$. This is disjoint from the closed set $K\setminus f^{-1}[0,\omega_1)$, so by normality of $K$ there are disjoint open sets $U_1,V_1\subset K$ with $f^{-1}[0,\alpha_1]\subseteq U_1\subseteq X\setminus V_1\subseteq f^{-1}[0,\omega_1)$. Because we are assuming that $f^{-1}[0,\omega_1)$ is not closed we have $X\setminus V_1\neq f^{-1}[0,\alpha)$, so in particular the inclusion $U_1\subseteq f^{-1}[0,\omega_1)$ is proper. Thus there is a countable $\alpha_2>\alpha_1$ for which there is a proper inclusion $U_1\subseteq f^{-1}[0,\alpha_2]$.

Now iterate the process to obtain a sequence of proper inclusions
$$f^{-1}[0,\alpha_1]\subseteq U_1\subseteq f^{-1}[0,\alpha_2]\subseteq U_2\subseteq\dots$$
Since $\bigcup_\mathbb{N}\alpha_i$ is countable, $\bigcup f^{-1}[0,\alpha_i]=f^{-1}(\bigcup [0,\alpha_i])$ is closed in $K$ and hence compact. On the other hand it is covered by the $U_i$, but by no finite subfamily of them. Thus there is a contradiction. It must be that $f^{-1}[0,\omega_1)$ is closed in $K$. $\square$

Remark: $(i)$ All that was needed of $K$ was its compactness and normality. A more general statement from which the proof flows is contained in Isbell's paper. $(ii)$ We quietly used the axiom of choice to ensure that $\bigcup\alpha_i$ as countable.


问:
Factorial of infinity
So, I've read in this article that:

$$\zeta'(0) = \log\sqrt\frac 1 {2\pi}$$

And that:
$$e^{-\zeta'(0)} = 1\cdot2\cdot3\cdots \infty = \infty! = \sqrt{2\pi}$$

I found this result very strange and tried to simplify if but i couldn't. Can somebody help me?

Thanks!

答:
Let me try to explain what is going on with a simpler example. Consider geometric series
$$F(s)=\sum_{n=0}^{\infty}s^n,$$
which converges inside the unit circle $|s|<1$ to the function $F(s)=\frac{1}{1-s}$. Note, however, that $F(s)$ can be analytically continued to the whole complex plane with excluded point $s=1$, which is a simple pole of $F(s)$. This point was in fact the origin of radius of convergence equal to $1$. So we are tempted to write identities like
$$ F(2)=1+2+4+8+\ldots=\frac{1}{1-2}=-1.$$
It is important to understand that they do not "really" hold. The logic is the following: the initial series defines some function in a "small" domain, but then sometimes this function can be (uniquely!) continued to a larger domain where the original series no longer makes sense. In general, this continuation will have singularities.

Sometimes - actually, almost always - one cannot construct a continuation to the whole complex plane with isolated singular points because of specific singularities called natural boundaries. An example of this is encountered if we consider instead of $F(s)$ the series
$$G(s)=\sum_{n=0}^{\infty}s^{n^2},$$
which converges inside the unit circle even better than $F(s)$ but cannot be continued elsewhere: $|s|=1$ is a dense set of singularities for $G(s)$.

Now let us go back to initial question. If we consider the series
$$\zeta(s)=\sum_{n=1}^{\infty}n^{-s},$$
it converges to some function $\zeta(s)$ in the halfplane $\mathrm{Re}\,s>1$. It turns out that this function can be (again, uniquely!) continued to the whole complex plane except the point $s=1$ where it has simple pole, similarly to our previous example. This continued function, as well as its derivatives, can be evaluated for any $s\neq1$ (for example, at $s=-1$ and $0$). One of the ways to do it is to systematically use integral representation
$$\Gamma(s)\zeta(s)=\int_0^{\infty}\frac{x^{s-1}dx}{e^x-1},$$
which for $\mathrm{Re}\,s>1$ reproduces the original series definition.

Title: Factorial of infinity


问:
Can we apply L'hopital Rule when we have $\frac{-\infty}{+\infty}$?
I saw in a book, it evaluated $\lim_{x\to0^+}\cfrac{(\ln x)^3}{\frac1x}$  with L'hopital Rule. the fraction is $\frac{-\infty}{+\infty}$ actually. so can I use L'hopital rule safely every time I see something like $\frac{-\infty}{+\infty}$ ?

答:
hint

Write

$$\frac{f(x)}{g(x)}=\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}$$

to transform $ \frac{\infty}{\infty} $ into $ \frac 00$.


答: Sure. That's so because, if $\lim_{x\to a}f(x)=-\infty$ and $\lim_{x\to a}g(x)=\infty$, then\begin{align}\lim_{x\to a}\frac{f(x)}{g(x)}&=-\lim_{x\to a}\frac{-f(x)}{g(x)}\\&=-\lim_{x\to a}\frac{(-f)'(x)}{g'(x)}\\&=-\lim_{x\to a}\frac{-f'(x)}{g'(x)}\\&=\lim_{x\to a}\frac{f'(x)}{g'(x)}.\end{align}

Title: Can we apply L'hopital Rule when we have $\frac{-\infty}{+\infty}$?

Body: I saw in a book, it evaluated $\lim_{x\to0^+}\cfrac{(\ln x)^3}{\frac1x}$  with L'hopital Rule. the fraction is $\frac{-\infty}{+\infty}$ actually. so can I use L'hopital rule safely every time I see something like $\frac{-\infty}{+\infty}$ ?


问:
Lower bound for minimum vertices in tournament
A tournament on a set $V$, with $|V| = n$ is an orientation of the edges of the complete graph on $V$.

A tournament has property $S_k$ if for every set of $k$ edges, there is one who dominates them all. ($x$ dominates $y$ iff $(x,y)$ is an edge, and note that then  $(y,x)$ is not).

In the book 'The Probabilistic Method' by Alon and Spencer we find this proposition:

below which we find:

Can you explain how it follows that $f(k) <= k^2s^kln(2)(1+o(1))$? Specifically I'm looking to understand how they came up with this bound, and not merely why it's true that if $n$ is larger than it then there is such a tournament.

答:
With the two bounds $\binom nk < (\frac{en}{k})^k$ and $(1 - 2^{-k})^{n-k} < e^{-(n-k)/2^k}$, we can rewrite Theorem 1.2.1: its conclusion holds if $(\frac{en}{k})^ke^{-(n-k)/2^k} < 1$, or (by taking the log)
$$
   k \log \frac{en}{k} - \frac{n-k}{2^k} < 0.
$$
Solving for $n$ is difficult, but if we do a kind of half-assed job of solving for $n$, we get
$$
    n > k 2^k (1 + \log n - \log k) + k.
$$
To get an upper bound on $f(k)$, we want to get an upper bound on the least $n$ for which this inequality holds.

First, observe that the inequality holds (for sufficiently large $k$) if we set $n = k^3 2^k$, because then the right-hand side is only $O(k^2 \log k)$. So for sufficiently large $k$, the conclusion of Theorem 1.2.1 holds provided that
$$
    n > k 2^k(1 + \log (k^3 2^k) - \log k) + k
$$
and this simplifies to $ (1 + o(1))k^2 2^k \log 2$.


问:
Determine the values of k that make the equation a circle.
Given the equation: $$(2x^2) + (2y^2) + 4kx - 2y = 2k - 8$$

Find the values of k that make the equation a circle.

My question is what is the procedure for reaching this conclusion:

Answer: $k \in (-\infty, -5/2) \cup (3/2, +\infty)$

Body:
$$(2x^2) + (2y^2) + 4kx - 2y = 2k - 8\\
x^2+y^2+2kx-y=k-4\\
(x+k)^2-k^2+(y-\frac12)^2-\frac14=k-4\\
(x+k)^2+(y-\frac12)^2=k^2+\frac14+k-4\\f(x,y)>0\\
k^2+\frac14+k-4>0$$


问:
Maximal ideals correspond to G-orbits
Let $\mathbb{F}$ be a field and $\overline{\mathbb{F}}$ its algebraic closure. I denote with $G = \text{Aut}_\mathbb{F}(\overline{\mathbb{F}})$ the automorphism group.
I have to show that
$$
\text{Max} \; \mathbb{F}[x_1,\cdots,x_n] = \overline{\mathbb{F}^n}/G.
$$
By Max I denote the maximum spectrum, thus all maximal ideals of the polynomial ring. Further I would like to construct examples to understand this fact better using $\mathbb{F} = \mathbb{R}$ and $\overline{\mathbb{F}} = \mathbb{C}$. Thanks in advance!

答:
I ask for patience, as I am a beginner.
I almost got it.
We define
$$\mathcal{O}(\mathbb{F}[x_1,\cdots,x_n]) := \{f \vert \mathbb{F}[x_1,\cdots,x_n]: \mathbb{F}[x_1,\cdots,x_n] \rightarrow K \; \vert \; \text{just for all elements in the polynomial ring} \}.$$

These are the regular functions. Then a maximal ideal would be a polynomial on this set, which evaluates to zero, such that:
$$
\mathfrak{m} := \{f \in \mathcal{O}(\mathbb{F}[x_1,\cdots,x_n]) \; \vert \; f(x) = 0\}.
$$
The maximal spectrum is just all of these ideals, as I said. I understand now, that I seek $\theta: \overline{\mathbb{F}^n} \rightarrow \text{Max} \; \mathbb{F}[x_1,\cdots,x_n]$, as on the right side I have all polynomials that evaluate to zero on the entire ring, thus the kernel of the evaluation, which eduard mentioned in the comments.
Now a polynomial in $n$-variables looks like $\sum_{k = (k_1,\ldots,k_n) \in \mathbb{N}^n_0}a_k x^{k_1}_1 \cdots x^{k_n}_n$. When does this guy evaluate to zero then? If all summand evaluate to zero. Thus it is enough that all $a_i$ are zero. I assume that every summand is zero by some translation. As we are in a commutative group, right and left translation are the same thing and we see, that this is the factor group of $\overline{F}^n/\text{Aut}_{F}(\overline{\mathbb{F}})$.

Is this correct?

How can this be written down in a proper form?

Title: Maximal ideals correspond to G-orbits


问:
maximal tree by Zorn's lemma
In using the Zorn's lemma to show that every connected graph contains a spanning tree, we let $\{T_{\lambda}: \lambda \in \Lambda \}$ be a family of trees contained in $X$ which is totally ordered by inclusion. But how to show that $\bigcup\limits_{\lambda \in \Lambda} T_\lambda$ is a also a tree in $X$?</p>

答:
The argument here is one of the common arguments when using Zorn's lemma.

If $G$ is a graph which is not a tree then it has a finite subgraph which witnesses that.

If the chain was finite, then it has a maximal element. Therefore the union of the chain is that maximal element, and we are done. In fact we don't even care that the chain was finite, just that it had a maximal element.

If the chain doesn't have a maximal element, and the union is not a tree then there is a finite subset which witnesses that. But that finite subset must have been added somewhere along the way. This counterexample, if so appears in some $T_\lambda$ in our chain. But we assumed that all the $T_\lambda$ are trees, so that is impossible.

Note that this is the same argument that the increasing union of linearly independent sets is linearly independent; that the increasing union of filters is a filter; and that the increasing union of ideals is an ideal, etc. The key is if the union wasn't a such object, it would mean that somewhere along the union we added a counterexample -- which contradicts our assumptions.

Title: maximal tree by Zorn's lemma


问:
How many boxes of cookies did each woman buy?
I'm stuck on this problem from the 2019 Grade 8 International Math Contest, which I'm using to help prepare my team for the Mathcounts competition:

Mrs Munch and Mrs. Crunch each bought the same number of identical cookie boxes from their local troop. Both women took either $2$ or $3$ days to finish each box of cookies. It took Mrs. Munch $43$ days to empty every last one of her cookie boxes; It took Mrs. Crunch $62$ days to empty all of hers. How many boxes of cookies did each woman buy?

I feel like I should be using a system of equations to solve this, but if that's correct, I can't figure out how to set it up.

答:
I solved this immediatly by figuring out what the minimal number of boxes (finishing every box in 3 days) and the maximal number (finishing every box in 2 days) is.

Maximal:

Ms. Munch 21 boxes (eating 20 in 2 days and 1 in 3 days)

Ms. Crunch 31 boxes (eatin 31 in 2 days)

Minimal:

Ms. Munch 15 boxes (eating 13 in 3 days, and 2 in 2 days)

Ms. Crunch 21 boxes (eating 20 in 3 days and 1 in 2 days)

Which leads to one solution, that every person had 21 boxes.

Also this would lead to a general solving method, but of course it is here luck that it turns out this well.


问:
Given a position vector for a particle, determine when the velocity vector and acceleration vector are perpendicular
Problem

Given the position vector of a particle measured in $t$ seconds

$$\vec r(t) = [4t+2, \ 2t-5t^2]$$

find out when the velocity vector and acceleration vector are perpendicular to each other.

My thoughts

Velocity vector: $\vec v(t) = [4, \ 2-10t]$

Acceleration vector: $\vec a(t) = [0, \ -10]$

We know that the dot product needs to be 0 at this point. Thus we solve

$$4\cdot 0 + (2-10t)(-10) = 0 \ \Rightarrow \ \boxed{t = 5 \text{ seconds}}$$

Is this correct?

答:
It's all correct except for

$$ t = \dfrac{1}{5} s $$


问:
Let $a, b ∈ Z$ with $gcd(a, b) = 1$, and let $c ∈ Z$. Show the equivalence: $ab | c ⇔ a | c$ and $b | c$.
I'm not sure how to do this question. Please help.

All I know that $gcd(a,b)=1$

But what can I do with this extra information?

"Hint: To prove the implication “$⇐$”, start from $a | c$ and use the fact, due to $gcd(a, b) = 1$, that for any $k ∈ Z$, if $b | ka$ then $b | k$."

b)  Find integers $a, b, c ∈ Z$ with $gcd(a, b)$ does not equal to 1 where the equivalence in ($a$) does not hold.

^ I am so lost from what the question asking

答: Integers $x, \, y$ exist with $xa+yb=1$, so $k=xka+ybk$ is a multiple of $b$.


问:
Linear bijection between a normed vector space and its proper subspace
Let $X,Y$ be normed vector spaces such that $X$ is a proper subspace of $Y$ (i.e., $X$ is a subspace of $Y$ and $X\neq Y$), and $X$ is dense in $Y$. Is there a linear bijection $T$ from $X$ to $Y$? If it does exist, can $T$ and $T^{-1}$ both be bounded?

I only know that in finite dimensional spaces, this is impossible. But I have no idea how to come up with an example with such $T$ even in indefinite dimensional spaces (or disprove the existence).

答:
First question. In general the desired bijection does not exist. Consider dense subspace $X$ of countable dimension and compete separable Banach space $Y$ (for example $X=c_{00}$, $Y=\ell_2$), then cardinality of Hamel basis of $X$ is $\aleph_0$, of $Y$ is $\mathfrak{c}$. Since cardianality of Hamel basis is preserved under linear bijections, then there is no bijective linear operator between $X$ and $Y$.

Second question. I don't know the answer but the following can be stated. If such $T$ exists then $Y$ is incomplete. Indeed $T$ is an isomorphism, so if $Y$ is complete then $X$ is complete too. Then you can easily show that $X=Y$. So counterexamples (if they exist) are among incomplete normed spaces.


问:
If $u:[0,\infty)\times\mathbb R\to\mathbb R$ is a $C^\infty$ solution of $u_t(t,x)-u_{xx}(t,x)=0$ then $\frac{d}{dt}\int_\mathbb R u(t,x)\,dx\equiv0$
Problem: Suppose $u:[0,\infty)\times\mathbb R\to\mathbb R$ is a smooth solution to $u_t(t,x)-u_{xx}(t,x)=0$ such that $u$ and its derivatives are integrable. Show that
$$\frac{d}{dt}\left[\int_\mathbb R u(t,x)\,dx\right]\equiv0.$$

My Attempt: Differentiating under the integral and applying the hypothesis we get
$$\frac{d}{dt}\left[\int_\mathbb R u(t,x)\,dx\right]=\int_\mathbb{R}u_t(t,x)\,dx=\int_\mathbb{R}u_{xx}(t,x)\,dx=u_x(t,x)\bigg\vert_{x\to-\infty}^{x\to\infty}.$$
But I cannot see how the above limits converge to zero. I also thought about applying some energy estimate, using the energy
$$E(t)=\int_{\mathbb R}u(t,x)^2\,dx,$$
but I could not see a way to go with this either.

Could anyone point me in the right direction, with a hint only.

Thank you for your time and appreciate any help.

答:
Hint: if $f$ and $f'$ are both integrable then
$$\lim_{x\rightarrow\pm\infty} f(x)=0$$
(use the mean value theorem)


问:
The proof that a permutation and its conjugate have the same cycle structure from Cooper's notes
I've been idly reading "Cooper's Notes" on graph theory the past few days.

Following one example, he states (see $\S2.10$ Conjugates) that

"the permutation and its conjugate have the same cycle structure."

where a conjugate of a permutation $a$ is defined as $b^{-1}ab$ for an arbitrary permutation $b$ on the same set.

How interesting! I think. It's followed by a theorem:

Theorem 2: The conjugate of $a=(x_1\:x_2\:\dots\:x_n)\dots$ by $b$ is $c=a^b=(b(x_1)\:b(x_2)\dots\:b(x_n))\dots$

no problem yet. Then the proof begins:

Proof: We'll show that $ab=bc$ from which it follows that $b^{-1}ab=c$.

Wait a minute. First of all, $c$ is defined as $b^{-1}ab$. Shouldn't it be the other way around? Show that $b^{-1}ab=c$ means $ab=bc$?

Second, and more importantly, all that this shows is that $ab$ has the same cycle structure as $bc$. Why should $ab=bc$ mean that $c$ has the same cycle structure as $a$?

The proof itself is short, which I reproduce here:

Now $ab(x_1)=b(a(x_1))=b(x_2)$ and $bc(x_1)=c(b(x_1))=b(x_2)$ Thus
$ab$ and $bc$ have the same effect on the symbol $x_1$. Similarly they
have the same effect on any symbol in the cycle notation for $a$. If
$z$ is any other symbol then it's fixed by $a$ so
$ab(z)=b(a(z))=b(z)$. Since $z$ is not present in the cycle notation
for $c$ it's fixed by $c$ and so $bc(z)=c(b(z))=b(z)$. We've thus
shown that $ab$ and $bc$ behave identically on all symbols so $ab=bc$.

What am I missing? Why does this proof have anything at all to do with the Theorem? Note that while I'd be interested in seeing a different proof, I'm more interested in why this specific proof works.

答:
This is admittedly not an answer to your question, but just another proof of the claim.

The cycle structure of a permutation $\sigma \in S_n$ shows itself when we consider the action of $\langle \sigma \rangle$ as a group of permutations on the set $X:=\{1,\dots,n\}$. By the Orbit-Stabilizer Theroem, we get:

$$|O_\sigma(i)||\operatorname{Stab}_\sigma(i)|=o(\sigma), \space\forall i\in X \tag 1$$

where:

$$\operatorname{Stab}_\sigma(i):=\{\sigma^k\mid \sigma^k(i)=i\} \le \langle\sigma\rangle, \space\forall i\in X \tag 2$$

Now, given $\tau\in S_n$, it is $(\tau\sigma\tau^{-1})^k=\tau\sigma^k\tau^{-1}$ (induction on $k$), so we get:

\begin{alignat}{1}
\operatorname{Stab}_{\tau\sigma\tau^{-1}}(i)&=\{\tau\sigma^k\tau^{-1}\mid (\tau\sigma^k\tau^{-1})(i)=i\} \\
&=\{\tau\sigma^k\tau^{-1}\mid \tau(\sigma^k(\tau^{-1}(i)))=i\} \\
&=\{\tau\sigma^k\tau^{-1}\mid \sigma^k(\tau^{-1}(i))=\tau^{-1}(i)\} \\
&=\{\tau\sigma^k\tau^{-1}\mid \sigma^k \in \operatorname{Stab}_\sigma(\tau^{-1}(i))\} \\
&=\tau\operatorname{Stab}_\sigma(\tau^{-1}(i)) \tau^{-1}, \space\forall \tau\in S_n,\forall i\in X\\
\tag 3
\end{alignat}

whence:

$$|\operatorname{Stab}_{\tau\sigma\tau^{-1}}(i)|=|\operatorname{Stab}_\sigma(\tau^{-1}(i))|, \space\forall \tau\in S_n,\forall i\in X \tag 4$$

But since $\forall \tau\in S_n, o(\tau\sigma\tau^{-1})=o(\sigma)$, we get that $(4)$ implies (again by the Orbit-Stabilizer Theorem):

$$|O_\sigma(\tau^{-1}(i))|=|O_{\tau\sigma\tau^{-1}}(i)|, \space\forall \tau\in S_n, \forall i\in X
\tag 5$$

Therefore, for every $\tau\in S_n$, the orbits induced by $\langle \tau\sigma\tau^{-1}\rangle$ and $\langle \sigma\rangle$ are pairwise of equal size. Moreover, if we denote by $\mathcal{O}$ the set of orbits, we have:

\begin{alignat}{1}
|\mathcal{O}_{\tau\sigma\tau^{-1}}| &= \frac{1}{o(\tau\sigma\tau^{-1})}\sum_{i=1}^{n}|\operatorname{Stab}_{\tau\sigma\tau^{-1}}(i)| \\
&=\frac{1}{o(\sigma)}\sum_{i=1}^{n}|\operatorname{Stab}_\sigma(\tau^{-1}(i))| \\
&=\frac{1}{o(\sigma)}\sum_{i=1}^{n}|\operatorname{Stab}_\sigma(i)| \\
&=|\mathcal{O}_\sigma| \\
\tag 6
\end{alignat}

So, given $\sigma\in S_n$, for every $\tau\in S_n$ the natural actions of $\langle\sigma\rangle$ and $\langle\tau\sigma\tau^{-1}\rangle$ on $X$ induce the same number of orbits of the same size in pairs, namely $\sigma$ and $\tau\sigma\tau^{-1}$ have the same cycle structure.


问:
Finding a tricky quotient set
We will denote by $\mathbb{K}$ one of the fields $\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$. On $\mathbb{K}\times \mathbb{K}$ we define the following equivalence relation:
$$(a,b)\equiv (a', b') \iff \exists (q,\alpha)\in \mathbb{K}^{*}\times \mathbb{K} \text{ such that } \begin{cases} a=q^2a'+\alpha^2-b\alpha \\ b=qb'+2\alpha \end{cases}.$$
We wish to determine the quotient set $\mathbb{K} \times \mathbb{K}/\equiv$ $\space$ for all $\mathbb{K}$s.

This problem was an extra problem in my abstract algebra class (don't worry, this isn't an attempt to cheat, I am posting this a week after the solution was due to be sent) and I kind of got stuck when it comes to $\mathbb{K}=\mathbb{Q}$.

For $\mathbb{K}=\mathbb{C}$, the things were nice and easy, because the original question asked me to prove that $\mathbb{C}\times \mathbb{C}/\equiv$ is equal to $\{\hat{(0,0)}, \hat{(0,1)}\}$ and this can be checked through (tedious) direct computations.

For $\mathbb{K}=\mathbb{R}$, a friend came up with the idea of expressing $\alpha$ from the second equation and then substituing it in the first one. This gives us the following equivalent characterisation of the equivalence relation:
$$(a,b)\equiv (a', b') \iff \exists q\in \mathbb{K}^{*} \text{ such that } 4a+b^2=q^2(4a'+b'^2) \space (*).$$
(notice that this works for all $\mathbb{K}$s, the case $\mathbb{K}=\mathbb{C}$ can be solved much easier by using this, but I didn't really need to think that much for that one since direct computations worked in my context)

For real numbers, this rewrites as $(a,b)\equiv (a', b') \iff \operatorname{sgn}(4a+b^2)=\operatorname{sgn}(4a'+b'^2)$.

As a result, there will be three equivalence classes: the parabola $4x+y^2=0$, its interior and its exterior. A representative for each of these are, respectively, $(0,0), (-1,0)$ and $(0,1)$, so $\mathbb{R}\times \mathbb{R}/\equiv \space = \{\hat{(0,0)}, \hat{(-1,0)}, \hat{(0,1)}\}$.

For $\mathbb{K}=\mathbb{Q}$, the things get pretty nasty by this approach. In this case, $(*)$ rewrites as $(a,b)\equiv (a',b') \iff \sqrt{\frac{4a+b^2}{4a'+b'^2}}\in \mathbb{Q}$ and I haven't been able to make any further progress.

答:
Consider the map $\varphi\colon K\times K\rightarrow K,\,(a,b)\mapsto4a+b^2$. This map is surjective, since $\varphi(a/4,0)=a$ for all $a\in K$. Next, consider the projection map $\pi\colon K\rightarrow K/(K^{\times})^2$, where $K/(K^{\times})^2$ is the set of equivalence classes under the equivalence relation $a\sim b:\Leftrightarrow\exists q\in K^{\times}\colon a=q^2b$ for $a,b\in K$. This map is surjective by definition. Your equation $(\ast)$ now says that
$$(a,b)\sim(a^{\prime},b^{\prime})\Leftrightarrow(\pi\circ\varphi)(a,b)=(\pi\circ\varphi)(a^{\prime},b^{\prime}),\qquad\forall (a,b),(a^{\prime},b^{\prime})\in K\times K.$$
Now, the universal property of the quotient set tells us that $\pi\circ\varphi\colon K\times K\rightarrow K/(K^{\times})^2$ factors as a bijection $\widetilde{\pi\circ\varphi}\colon K\times K/\sim\rightarrow K/(K^{\times})^2$, sending the equivalence class of $(a,b)$ in $K\times K$ to the equivalence class of $4a+b^2$ in $K/(K^{\times})^2$. (This holds true for any field $K$ of characteristic $\neq2$.)

In the specific case $K=\mathbb{Q}$, there is also a somewhat natural system of representatives of $K/(K^{\times})^2$. The equivalence class $[0]=\{0\}$ is represented by $0$ and every other equivalence class contains a unique square-free integer.

Title: Finding a tricky quotient set


问:
Calculating limit of sequence using only $1/n\to0$
First of all, thank you for visiting this thread, and thanks for whom will try helping.
So, the teacher explained to us how to calculate the limit of the sequence by only using one fact: $1/n$ goes to $0$.
However, I can't find the solution for this question, using only this fact.
In red, I wrote in Hebrew "Goes to zero".
In addition, there are about 4 or 5 more students who shared their way to solve this question, and tried the same way as I wrote. But I think we're missing some rule here.

Is it even possible to calculate the sequence's limit using only "$1/n$ goes to $0$"?
Again, thanks in advance. :)

答:
What's critical here is that the coefficients of the expressions on the right add up to $0$. The best way to handle the problem is to use that fact that
$$\sqrt{1+x} = 1+\frac x2 + \epsilon(x),$$
where $\epsilon(x)/x \to 0$ as $x\to 0$. This follows from the definition of the derivative of $\sqrt{1+x}$.

These problems are usually more mundane and handled — at the beginning of a calculus course — by a multiplication by the "conjugate" trick. For example, to handle
$$\lim_{n\to\infty} \sqrt{n+3} - \sqrt{n-1},$$
you multiply by $\dfrac{\sqrt{n+3}+\sqrt{n-1}}{\sqrt{n+3}+\sqrt{n-1}}$ and use basic algebra, together with the $1/n\to 0$ fact to which you've alluded (in this case, you need $1/\sqrt n\to 0$).


问:
Probability: pairs of Shoes problem
Given n different pair of shoes,a total of 2n shoes,we put these 2n shoes together and randomly divided into n groups,with each group having 2 shoes.If 2 shoes in a group are just a pair,them we call this group the "pair group".Let X be the number of "pair groups".Calculate the expectation of x.

For this problem,I figured out that 2n shoes divided into n groups that has $C{2 \choose 2n}$ * $C{2 \choose 2n-2}$*... * $C{2 \choose 2}$ and the probability of X = n is $ \frac{2^n*n!}{2n!} $

X can be 0,1,2,...,n.So what is the probability of X = 0,X = 1,...,X=n-1,and how to calculate the E(X)?

答:
Approaching the problem by Linearity of Expectation -

First to find, probability of a group of two shoes be a pair -

The number of ways to arrange $2n$ shoes into $n$ groups of two $= \displaystyle \frac{(2n)!}{2^n} \times \frac{1}{n!}$

(which comes from ${2n \choose 2} {2n-2 \choose 2} ...{4 \choose 2}$ and we divide by $n!$ to account for overcounting).

If a group is a pair, the number of ways to arrange $(2n-2)$ remaining shoes into $(n-1)$ groups of two

$= \displaystyle \frac{(2n-2)!}{2^{(n-1)}} \times \frac{1}{(n-1)!}$

So the probability of a group being a pair $P(X_1) = \displaystyle \frac{ 1}{2n-1}$ (which makes sense as we are choosing one particular shoe from $(2n-1)$ to make a pair).

To find expected number of pairs, we will sum their individual expected values.

So expected number of pairs, $E(\mathbb{X}) = P(X_1) \times 1 + P(X_2) \times 1 + ... + P(X_n) \times 1 = \displaystyle \frac{n}{2n-1}$

Title: Probability: pairs of Shoes problem


问:
If $\forall x \in \mathbb{R}^n, \exists y \in A, d(x,y) = \inf\{d(x,a) \mid a \in A \}$, A is a closed set in $\mathbb{R}^n$?
Let $\mathbb{R}^n$ has Euclidean distance $d: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$.

(a) $\mathrm{A} \neq \emptyset$ is a closed set in $\mathbb{R}^n$

(b) $\forall x \in \mathbb{R}^n, \exists y \in A, d(x,y) = \inf\{d(x,a) \mid a \in A\} $

I trying to prove that (a) and (b) are equivalent.

the proof of (a) to (b) is suggested in other questions, but I think (b) to (a) is not proved.

please give me some ideas to prove (b) to (a)...

答: If $A$ is not closed, take $x\in\overline A\setminus A$. Then $d(x,A)=0$, but there is not $a\in A$ such that $d(x,a)=0$.


问:
Is it a general truth that when two equations can be described with the same sentence, only varying intonation they are equivalent?
My question is the effect of something that I have noticed during some of the courses I am taking as a first year maths bachelor, that in tutorials we will be asked to proof that two statements are equal but that when I try to describe the left and right hand side I end up with one and the same sentence, I will give three examples:
\begin{equation}
    (A^m)^{-1} = (A^{-1})^m
\end{equation}
Both the left and right hand side are described by the sentence "the inverse of matrix A to the power of m"
\begin{equation}
f^{-1}(A\cap B) = f^{-1}(A) \cap f^{-1}(B)
\end{equation}
With both A and B being supsets of the real numbers, both the left and right hand side are described by the sentence "the values that when the function f is applied to them end up inside set A and also inside set B", something similar can be done replacing the intersect with a union and the 'and' with an 'or'.
\begin{equation}
    (AB)C = A(BC)
\end{equation}
This example is discussed in three blue one browns series on linear algebra, both the left and right hand side are described as "first apply transformation C, then transformation B, and finally transformation A"

Of course, the fact that two sides are described the same is not a rigorous proof since it could be that sneaky business is applied in words with double meanings - although often that should be easy enough to spot - but I do think this is a rather interesting patern and I would absolutely love to know if anyone can come up with an example where both the left and right hand side are described the same and still not equal.

答:
Mathematica has a Speak function, where (for instance)

Speak[x^2]

will produce the aural utterance "x squared."

Speak[(x y) z] and Speak[x (y z)] each give the identical utterance "x times y times z," where of course these need not be equal in non-associative algebras.