问:
What is wrong with this proof that inaccessible cardinals are inconsistent
Disclaimer: I believe this proof is wrong, but I'm asking because I can't find what's wrong with it, which means I must have some basic misunderstanding of the concepts involved.

First, some definitions. Recursively, we define an ordinal $\alpha$ to be $\eta$-unboundedly elementary:

$\alpha$ is $0$-unboundedly elementary iff $\alpha$ is an ordinal

$\alpha$ is $\eta$-unboundedly elementary iff for all $\beta<\eta$, for all $\gamma$, there exists $\delta>\gamma$, such that $V_\alpha\prec V_\delta$ and $\delta$ is $\beta$-unboundedly elementary. That is, $V_\alpha$ is elementary in unboundedly many $V_\delta$ for $\delta$'s lower in this hierarchy.

Some examples: $1$-unboundedly elementary ordinals are those $\alpha$'s such that $V_\alpha\prec V_\beta$ for unboundedly many ordinals $\beta$. In this blogpost they are called totally otherworldly cardinals. $2$-unboundedly elementary ordinals are those $\alpha$'s such that $V_\alpha$ is elementary in unboundedly many totally otherworldly cardinals.

An interesting feature of these unboundedly elementaries is that they exhibit increasing degree of correctness (where $\kappa$ is $\Sigma_n$-correct iff $V_\kappa\prec_{\Sigma_n}V$). Let us first see that $1$-unboundedly elementary ordinals are $\Sigma_3$-correct:

That they are $\Sigma_2$ correct is proven in the blogpost linked above. To see that they are in fact $\Sigma_3$-correct, suppose $\exists x~\varphi(x)$ is a true $\Sigma_3$ statement with parameters in $V_\alpha$. Fix a witness $a$ such that $\varphi(a)$ is true and fix $\beta$ large enough with $a\in V_\beta$ and $V_\alpha\prec V_\beta$. Then since $\beta$ must be a $\beth$-fixed point (this follows from the fact that, for any ordinals $\alpha,\beta$, $V_\alpha\prec V_\beta$ implies $V_\beta\vDash \mathsf{ZFC}$), $V_\beta\prec_{\Sigma_1}V$. Then the truth of the $\Pi_2$ statement $\varphi(a)$ is preserveed downward to $V_\beta$. This means $V_\beta\vDash \exists x~\varphi(x)$, and by elementarity, so does $V_\alpha$. This shows that $\alpha$ is $\Sigma_3$-correct.

Essentially the same trick can give us that $2$-unboundedly elementary ordinals are $\Sigma_4$ correct: this is because there are unboundedly many $\Sigma_3$-correct ordinals that they are elementary in, and if $\exists x~\varphi(x)$ is a true $\Sigma_4$ statement, then we can find some target $V_\beta$ large enough to include a witness, then proceed the same way as in the last paragraph.

Now here's the problem: the recursive definition of $\eta$-unboundedly elementary ordinals is $\Pi_3$. This means ``there exists a $2$-unboundedly elementary ordinal" a $\Sigma_4$ statement. So the least $2$-unboundedly elementary ordinal must be below the least $\Sigma_4$-correct cardinal. But this contradicts the fact we've just shown, that $2$-unboundedly elementary ordinals are $\Sigma_4$-correct!

Nevertheless, assuming there is an inaccessible cardinal, then the usual Löwenheim–Skolem argument shows the consistency of essentially any degree of unboundedly elementary ordinals, and in particular the consistency of $2$-unboundedly elementary.

Taking stock, we have: $\mathsf{ZFC}\vdash ``\text{there exists an inaccessible cardinal}"\rightarrow$ Con($2$-unboundedly elementary), and also $\mathsf{ZFC}\vdash``2\text{-unboundedly elementary ordinals are inconsistent}"$. Putting things together, this shows that $\mathsf{ZFC}$ refutes the existence of inaccessible cardinals. What went wrong?

Some possibilities I can think of: 1) the proof of their increasing degree of correctness is wrong, 2) the complexity calculation of the recursive definition wrong (or maybe it's not recursive after all, that there is some metamathematical issues), 3) inaccessible cardinals doesn't show that these unboundedly elementary ordinals are consistent.

答:
The recursive definition of $\eta$-unbounded elementary which you propose cannot actually be carried out in ZFC.  Recall that the way recursive definitions work is that you prove by induction on $\eta$ that there is a unique function $F_\eta$ defined on $\eta$ which takes each $\alpha<\eta$ to the $\alpha$th stage of your definition.  This only works if the $\alpha$th stage of your definition is a set, so that this function $F_\eta$ actually is a set (so you can quantify over such functions).  In your case, you would want to be defining the class of all $\alpha$-unbounded elementary ordinals as the $\alpha$th stage, which you cannot do.  (Note that for variant notions like that of $\eta$-inaccessibility, you can define the set of $\eta$-inaccessible cardinals less than $\kappa$ for any fixed $\kappa$ recursively, since $\eta$-inaccessibility of a cardinal depends only on the cardinals smaller than it.  This doesn't work for $\eta$-unbounded elementary ordinals, though, since to determine whether an ordinal is $\eta$-unbounded elementary, you need to know the entire proper class of $\beta$-unbounded elementary ordinals for each $\beta<\eta$.)

Incidentally, you can carry out this definition in MK, where class quantification is allowed (and can be used in Separation, and thus in proofs by induction where you want to define "the set of counterexamples" to prove it is empty).  However, your $\Sigma_4$-correctness statement then only works for statements using set quantifiers, and the recursive definition of $2$-unboundedly elementary would need to use class quantifiers (or, if you did it with set quantifiers by brute force instead of by recursion, it would need to have higher complexity, as described in Noah's answer).


问:
How can I compute closure/morphology and dilution for an image using math?
Suppose we have image A and the structuring element B from figure 2. How can I compute A dilated by B and A closed(or can be called morphology) by B? How can I apply math on paper for this?

答:
To dilate image A by structuring element B, place the center of (a copy of) B over each pixel in A, and set all of the pixels under all of (each copy of) B.

To compute the closure of A by B, first dilate A by B, then erode the result by B (actually, by the transpose of B, but as this B is symmetrical, it amounts to the same thing.  If, e.g, the bottommost pixel were removed from B, then it would matter).  To erode an image by B, place (a copy of) B -completely- inside the image at every pixel where it will fit like that, and mark the center of (each copy of) B.

ContentLicense: CC BY-SA 4.0


问:
Solve $(-3 - i)^{\frac{5}{7}}$ using de Moivre formula
$$\text{I tried the following: } |2|^{5/7}(\cos(5/7(-\pi/6 + 2k\pi))+ i\sin(5/7(-\pi/6+2k\pi)))
 $$

I don't know how to complete from here. I am also having trouble with "2k$\pi$" as a lot of times I don't know if I should multiply with exponent of the complex number or not, and if there is denomenator underneath it or not, it confuses me.

答: What is the meaning of a "fractional exponent" in case of a complex number as its base? Whosoever gave you that question in THAT form ought to scratch behind his ears. Take for example $\sqrt{i}$. There is no unique solution to this either, but we DO have 2 solutions to $x^2=i$. So in your case, we are essentially interested in $x^7=(-3-i)^5$. Now this can be done with DeMoivre. Finding out the 5th power isn't to bad and then extracting 7 roots is also just some annoying algebra. And so yes, as a 7th degree equation we expect 7 solutions. Do you know how to do $x^7=12-316i$? (absolute value, argument, etc) The algebra isn't pretty


问:
When does a matrix $A$ with ones on and above the diagonal have $\det(A)=1$?
What conditions, if they're even necessary, must be placed on $\star$ so that the matrix

$$ \begin{pmatrix} 1 & & \huge{1} \\ & \ddots & \\ \huge{\star} & & 1 \end{pmatrix}, $$

so that $\det{(A)}=1$, where $\huge{1}$ denotes "all entries above the diagonal are $1$'s," and $\huge{\star}$ is just some arbitrary scalar entries from a general field $F$? My intuition tells me that that no matter what it's $1$.

EDIT:

More specifically, disregard my call for conditions, but assume rather that the lower diagonal entries are $\color{red}{\text{guaranteed to be less than $1$}}$.

答:
Let $$A = \begin{pmatrix} 1 & & &\huge{1} \\ a_1 & \ddots \\ & \ddots & \ddots \\ \huge{\star}& & a_{n-1} & 1 \end{pmatrix}$$ be your matrix. I claim that $\det(A) = \prod_{i=1}^{n-1}(1-a_i)$.

To see why this is true, note that by expanding the determinant in the first column, we get $$\det(A) = \begin{vmatrix}1 & & &\huge{1} \\ a_2 & \ddots \\ & \ddots & \ddots \\ \huge{\star}& & a_{n-1} & 1\end{vmatrix} - a_1\begin{vmatrix}1 & & &\huge{1} \\ a_2 & \ddots \\ & \ddots & \ddots \\ \huge{\star}& & a_{n-1} & 1\end{vmatrix}+0+0+\cdots+0,$$
since the first two rows are equal except for their first entries.

Hence $\det(A) = (1-a_1)\begin{vmatrix}1 & & &\huge{1} \\ a_2 & \ddots \\ & \ddots & \ddots \\ \huge{\star}& & a_{n-1} & 1\end{vmatrix}$ and we get our result by induction.

Thus the restriction you are looking for is $\prod_{i=1}^{n-1} (1-a_i) = 1$, where $a_1,\ldots,a_{n-1}$ is the lower diagonal of your matrix.

ContentLicense: CC BY-SA 3.0


问:
Distance between sets of $\mathbb {R^n}$
Distance between sets of $\mathbb {R^n}$.

Definition: Let $A, B \subseteq \mathbb{R^n}$. We define the distance $d(A, B)$ between the sets $A$ and $B$, as:

$d(A, B)= \inf$ {$|x-y|: x \in A, y \in B$}

Now consider the sets $A = ${$(x, 0): x \in \mathbb{R}$} and $B=$ { $(x, 1/x): x >0$}, which are disjoint. Show that $d (A, B) = 0$

答:
For $x>0$, the distance between the points $p=(x,0)$ and $q=(x,1/x)$ is exactly $|p-q|=1/x$. Thus
$$d(A,B)=\inf\{|p-q|:p\in A, q\in B\}\le\inf\{1/x:x>0\}=\lim_{x\to\infty}\frac1x=0$$


问:
Deducing $L^1$ convergence
Let $X_i$ be iid, non-negative with mean $1$ and $M_0=1$, $M_n=X_1 X_2\cdots X_n$ an martingale with respect to the natural filtration. If I know that $E[\mathbb{1}_{\{max_{1\leq k \leq n}M_k \geq c\}}]\leq \frac{1}{c}$, that $E[M_k]=1$ for all $k$, and that $M_n$ converges to some finite $M_{\infty}$ almost surely, can I conclude that $(M_n)$ converges to it in $L^1$?

答:
No, you cannot conclude this without a stronger condition on $M$.  Doob's inequality will always give that $\mathbb{P}(\max_{1 \le k \le n} M_k \ge c) \le \frac 1c \mathbb{E}[M_n 1_{\max_{1 \le k \le n} M_k \ge c}] \le \frac 1c \mathbb{E}[M_n] = \frac 1c$ whenever $M$ is a martingale of this form.

For a specific counterexample, let $X_i$ be i.i.d. with $\mathbb{P}(X_i = 2) = \mathbb{P}(X_i = 0) = \frac 12$.  Then $M_n = X_1 X_2 \cdots X_n$ converges to $M_{\infty} = 0$ almost surely because $\mathbb{P}(X_i = 2 \text{ for all }i \in \mathbb{N}) = 0$, and since $\mathbb{E}[M_n] = 1$ for all $n$ we can't have $(M_n) \rightarrow M_{\infty}$ in $L^1$

Title: Deducing $L^1$ convergence


问:
Find the linear transformation of a matrix knowing 4 linear transformations
I'm stuck with an exercise where they give me 4 linear transformations $T:M_{2\times2}\to\Bbb R$ for 4 matrices 2x2 and then they ask me for the linear transformation of a fifth matrix.

$$ T
  \begin{pmatrix}
    1 &0\\
    0 & 0\\
  \end{pmatrix}
=3,
T\begin{pmatrix}
    0 &1\\
    1 & 0\\
  \end{pmatrix}
=-1,
T
  \begin{pmatrix}
    1 &0\\
   1 & 0\\
  \end{pmatrix}
=0,
T
  \begin{pmatrix}
    0 &0\\
    0 & 1\\
  \end{pmatrix}
=0,
T
  \begin{pmatrix}
    a &b\\
    c & d\\
  \end{pmatrix}
=?$$

and I have worked with this situation for $T:\Bbb R^n \to\Bbb R^m$ but never with matrices so I have the doubt that how is supposed to find 4 constants that will multiply de 4 matrices I know to compose the fifth matrix to finally can find the linear transformation of this last one. (I have tried adding up the 4 matrices as follows)

$$
  \begin{pmatrix}
    1 &0\\
    0 & 0\\
  \end{pmatrix}
+
\begin{pmatrix}
    0 &1\\
    1 & 0\\
  \end{pmatrix}
+
  \begin{pmatrix}
    1 &0\\
   1 & 0\\
  \end{pmatrix}
+
  \begin{pmatrix}
    0 &0\\
    0 & 1\\
  \end{pmatrix}=
  \begin{pmatrix}
    a &b\\
    c & d\\
  \end{pmatrix}
$$

\begin{cases}
2=a  &  \\
1=b & \\
2=c\\
1=d\\
\end{cases}

and then use the coefficients
$$
  2\begin{pmatrix}
    1 &0\\
    0 & 0\\
  \end{pmatrix}
+
1\begin{pmatrix}
    0 &1\\
    1 & 0\\
  \end{pmatrix}
+
  2\begin{pmatrix}
    1 &0\\
   1 & 0\\
  \end{pmatrix}
+
  1\begin{pmatrix}
    0 &0\\
    0 & 1\\
  \end{pmatrix}=
  \begin{pmatrix}
    a &b\\
    c & d\\
  \end{pmatrix}
$$

and I just want to know if I continue in this way or I have committed any mistakes, thanks for your time <3

答:
What is given can be translated as follows. Denote, in a standard way,
$$E_1=\begin{pmatrix}1&0\\0&0 \end{pmatrix},\quad E_2=\begin{pmatrix}0&1\\0&0 \end{pmatrix}, \quad E_3=\begin{pmatrix}0&0\\1&0 \end{pmatrix},\quad E_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$
We have $\:M=\begin{pmatrix}a&b\\c&d \end{pmatrix}=aE_1+bE_2+cE_3+dE_4$, so
$$T(M)=aT(E_1)+bT(E_2)+cT(E_3)+dT(E_4).$$

Now, we are given that $T(E_1)=3,\enspace T(E_2)+T(E_3)=-1,\enspace T(E_1)+T(E_3)=0, \enspace T(E_4)=0$. Can you proceed?


问:
transformation for my self
I am trying to prove this equation of integral transformation with Borel measure. But so far have no idea. Can you help me how to start.
Suppose $(X, A, \mu)$ is a $\sigma$-finite measure space and $f: X\to [0,\infty]$ an $A$-measurable function. Let $\nu$ be a Borel measure on $[0,\infty)$ such that $\nu(K)<\infty$ for any compact set $K$ in $[0,\infty)$. Show that

$$\int_X \nu([ 0 ,f(x))\;\mu(dx) = \int_0^\infty\mu(\{\ x\in X \ | f(x)>t\})\;\nu(dt).$$

答: The question has been answered, however I'd like to ask how to prove, that $1_{[0,f(x)[} (t)$ is measurable since this is required for using Tonelli and therefore prove that given equality. We want to show that for every $C\in B[0,\infty]$ following holds: $(1_{[0,f(x)[})^{-1} (C) \in A $ x $B[0,\infty[$..but I'm struggling with this..

Title: transformation for my self


问:
Parameterization of a curve (line & circle)
Let the curve C be the boundary of the region bounded by $y=2$ and $x^2+y^2=9$.

a) How would C look like?

(I think it would be circle segment?)

b) Which of the following characteristics does C have?

smooth

piece wise smooth

Jordan curve

closed

(I think it is closed and smooth and also a Jordan curve?) But how can I prove this mathematically?

c) I need to find a parameterization of the curve, and here I do not know how to start, do I parameterize the line and the circle arc separately?

the intersection points would be $P(\sqrt{5} | 2), Q(-\sqrt{5} | 2)$

答:
Rather than giving an explicit answer, I'll give you a plan on as how to think of this yourself.

The first thing I would recommend is drawing the two boundaries, you will earn lots of intuition about the problem.

Once you've done that, you will see the curve is a circle segment AND the segment joining the two ends since the radius is 3 > 2. Therefore it's not smooth, but piecewise-smooth since it fails on those two points (you will see it graphically), it is closed and a Jordan Curve.

As for the parametrization, draw the boundaries, and parametrize those two individually and then join them continuously. With that parametrization you can actually prove (b)


问:
How can I figure out the non differentiable values of this function?
Basically, I am trying to answer question B, but I am confused on what it's asking. I know that a function is differentiable at a point when there's a defined derivative at that point, but I'm confused on the non differentiable part. I have graphed the derivative by finding out the function's slope at each point, but I am not sure if that helps. Could anyone help me explain and solve part B? Thanks.

答:
Intuitively, a function is not differentiable at a point if:

The function isn't even defined there (think $f(x)=1/x$ at $x=0$)

The function has a "sharp/angled point" there (think $f(x)=|x|$ at $x=0$) -- as opposed to a smooth one (compare with $g(x) = x^2$ at $x=0$). The former means you could easily draw multiple lines tangent to the function through that same point. In particular what this often means is that there is a "jump" discontinuity in the graph of the derivative.

The derivative "blows up" to infinity at that point (the tangent becomes vertical).

For instance, some examples:

In this example, the function $f$ is not differentiable at $x=0$.

In this example, the function $f$ is not differentiable at $x=-2$ or $x=1$

In this example, $f$ is not differentiable at $x=0$. This is because, not of a jump in the derivative, but $f$ not being defined there:

$$f(x) = \operatorname{sign}(x) = \begin{cases}
1 & x > 0 \\
-1 & x < 0
\end{cases}$$

Sometimes it's preferable to say that $f'(0) = \delta(0)$ in this case, where $\delta$ represents the Dirac delta function. You can probably say the same of the previous example as well, when I think about it.

Another example: here, non-differentiability is at $x=-1$ and $x=2$. (Note these are vertical asymptotes of $f$.)

This example is problematic because of the vertical tangent as $x=0$. Notice how the derivative blows up to infinity.

A few cases combined here: non-existence of $f$ on $[-1,1]$, and the fact that towards each of $-1$ and $1$, the tangent becomes vertical and thus $f'$ blows up to infinity.

With these examples and the intuition of "sharp points, vertical tangents, and discontinuities being trouble spots" in mind, you should be able to get the idea of how to solve the problem, especially if you already have a graph of your derivative.


问:
Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$
If $n=7+4\sqrt3$,then what is the simplified value of
$$\sqrt n+\frac {1}{\sqrt n}$$
I was taking LCM but how to get rid of  $\sqrt n$ in denominator</p>

答: Square $2 + \sqrt{3}$, and you will know what to do.</p>


问:
If set cardinality and power sets are not functions, then what are they?
Lets say you have a (finite) set $S$ and want to know its cardinality $|S|$. Then this seems like a function
$$|\cdot|: \mathbb{S} \to \mathbb{N}$$
which certainly cannot exist, given that there is no such thing as "$\mathbb{S}$", which would be the set of all sets. The same goes for the power set $\mathcal{P}(S)$, which would look like
$$\mathcal{P} : \mathbb{S} \to \mathbb{S}$$
which also certainly makes no sense. However, these still seem like functions: they take in an input (a set) and spit out an output. So since these are not functions, what are they?

答:
This is a great question, and the answer in ZFC set theory lies in the notion of a "class function". Formalizations of ZFC view it as a first-order theory in the language $\mathcal{L}=\{\in\}$, which has a single binary predicate meant to express the idea of "membership". The idea of "classes" and "proper classes" can then be formalized through the idea of $\mathcal{L}$-formulas; I recommend reading a bit about first-order logic to understand how to formalize the ideas written below.

$\mathcal{L}$-formulas are powerful enough to express the majority of constructions we want in our set theory; for instance, the following formula expresses that $X$ is a subset of $S$: $$\sigma(X,S)\equiv\forall x(x\in X\rightarrow x\in S).$$ We can then use this to create a formula expressing that $P$ is the power set of $S$: $$\eta(S,P)\equiv\forall X(X\in P\leftrightarrow\sigma(X,S)).$$ With this tool under our belt, we can codify the idea of a "class" as intuitively being the collection of all sets that satisfy a particular $\mathcal{L}$-formula. For instance, the "class" of all sets is cut out by the formula $\psi(X)\equiv X=X$. Since every set is equal to itself, every set will satisfy $\psi$, so $\psi$ "cuts out" the proper class of all sets, in some sense. Formally, we can think then of classes as equivalence classes of $\mathcal{L}$-formulas, where two $\mathcal{L}$-formulas are equivalent when one holds if and only if the other does.

With some work, it is possible in ZFC to create a class $\textbf{Card}$ of all the sets we want to consider "cardinal numbers", satisfying the properties we expect of cardinality. (If we are interested only in the cardinalities of finite sets, then we can just take $\textbf{Card}$ to be the set $\mathbb{N}$, but this construction is much more general. Look up ordinal numbers and well-orderings if you want to see the tools needed for it.) $\textbf{Card}$ is "too large" to be a set – ie, there is no set that contains every cardinal number. If there were, we would be able to recreate a Russell-like paradox. However, the property of "being a cardinal number" (when suitably defined) is expressable by a (complicated) $\mathcal{L}$-formula, and this justifies referring to $\textbf{Card}$ as a "class". For convenience, let's suppose $\phi(X)$ is a formula expressing "$X$ is a cardinal number". Let's also take $\psi(X)$ as above, so that $\psi(X)$ cuts out the class $\textbf{Set}$, consisting of all sets.

With classes under our belt, we can work on defining "class functions", which codify in some sense the idea of functions between classes. For instance, a "function" from the class $\textbf{Set}$ to the class $\textbf{Card}$ is a formula $\theta(X,Y)$ with the property that, for any $S$, if $S$ lies in the class $\textbf{Set}$ – ie if $\psi(S)$ holds – then there exists a unique $T$ such that $T$ lies in the class $\textbf{Card}$ – ie, such that $\phi(T)$ holds – and such that $\theta(S,T)$ holds. If $\theta$ further satisfies that there exists a bijection between $S$ and the unique cardinal number $T$ such that $\theta(S,T)$ holds, we then denote this unique $T$ by $|S|$ and consider $\theta(X,Y)$ to define the "cardinality" operation that you mention.

Similarly, the "power set" operation that you mentioned is a class function from $\textbf{Set}$ to $\textbf{Set}$. In fact, the formula $\eta(X,Y)$ that I've given above defines the power set class function! The fact that, for each set $S$, there is a unique set $P$ satisfying $\eta(S,P)$ is one of the axioms of ZFC, and this justifies our claim that $\eta$ is in fact a class function. (Actually, to be pedantic, the existence of $P$ is one of the axioms of ZFC, and uniqueness follows from a different axiom called "extensionality", which says that two sets are equal if and only if they have the same elements.)

These ideas lie at the root of the deep field of set theory, and are covered in any good introductory resource on the subject. If you find them interesting I very much recommend learning them in detail! For instance, here is a good set of lecture notes that should hopefully be accessible, which covers all of the ideas above (and many more) in detail.


答:
As pointed out in the previous answer, it is false.

I am posting this just to add a counterexample where $X_n>0$.

Let $(W_t)_{t\geq0}$ be a Brownian Motion and let $X_n = e^{W_n-W_{n-1}-\frac{1}{2}}$. They are iid with mean $1$.

Then $M_n = e^{W_n - \frac{n}{2}}$ and it's well known that $M_n \stackrel{a.s.}{\to} 0$.

Once again, this implies the impossibility of the $L^1$ convergence because $\mathbb{E}[M_n] = 1$.

Title: Deducing $L^1$ convergence

Body: Let $X_i$ be iid, non-negative with mean $1$ and $M_0=1$, $M_n=X_1 X_2\cdots X_n$ an martingale with respect to the natural filtration. If I know that $E[\mathbb{1}_{\{max_{1\leq k \leq n}M_k \geq c\}}]\leq \frac{1}{c}$, that $E[M_k]=1$ for all $k$, and that $M_n$ converges to some finite $M_{\infty}$ almost surely, can I conclude that $(M_n)$ converges to it in $L^1$?


问:
Is there a dissection tool available online?
I want to write some pages on a website exploring geometric dissections (initially 2d, e.g. Dudeney's, Haberdasher's Problem, Archimedes' Loculus, tangrams and pentominoes, etc., but eventually moving up to 3d: Piet Hein's Soma cubes etc.)

I can spend a lot of time hacking around in e.g. GeoGebra to draw the shapes, but in such an environment I am limited more-or-less to static constructions, and there does not exist the capability to drag, rotate, reflect, pivot around a point, and all that stuff.

There are some fairly sophisticated graphics out there which must have been constructed using some sort of tool (unless they have been crafted from raw source code from a graphics package in whatever programming language). But I've done a search and nothing comes to light (and the search results tend also to be clogged with tools allowing visualisations of biological dissections.

Anyone able to help?

Many thanks.

答:
BurrTools offers the ability to construct 3D puzzles with a given set of initial and final shapes, and will search for possible solutions. One can of course make 2D puzzles by creating pieces and targets which are only one layer thick.

The puzzles are fixed to be on a given grid (cubic, triangular prisms, a few other options), so the flexibility is not infinite, but it should cover many of the use cases you have in mind.


答: Let $\varepsilon>0$. By convergence of $\frac{1}{x}\to 0$ when $x$ to infinity, exists $A>0$ such that for $x>0, |x|>A$ implies $\frac{1}{x}<\varepsilon$. By these argument, choosing one  $x$ with  the property above, $d(A,B)\leq |(x,0)-(x,1/x)|=|(0,1/x)|=1/x<\varepsilon$. By arbitrariety of $\varepsilon $, we  can conclude that $d(A,B)=0$.

Body:
Distance between sets of $\mathbb {R^n}$.

Definition: Let $A, B \subseteq \mathbb{R^n}$. We define the distance $d(A, B)$ between the sets $A$ and $B$, as:

$d(A, B)= \inf$ {$|x-y|: x \in A, y \in B$}

Now consider the sets $A = ${$(x, 0): x \in \mathbb{R}$} and $B=$ { $(x, 1/x): x >0$}, which are disjoint. Show that $d (A, B) = 0$


问:
No polynomials of degree > 0 are such that $p(t)^2 + q(t)^2 =1$
I'm trying to prove here that ellipses can't be represented by a parametric polynomial and got stuck on this part of the problem:

No real polynomials of degree $> 0$ are such that $$p(t)^2 + q(t)^2 =1$$

I want to prove the above statement.

答:
Real polynomials that are not constants go to infinity at infinity (just factor out the term of higher degree, other terms are negligible).

Squaring them make them both positive, i.e. $p(t)^2\to+\infty$ and $q(t)^2\to+\infty$ so their sum cannot be bounded.


问:
Limit in Big-O in the proof of Poisson limit theorem
while I was studying the proof of the Poisson limit theorem (One can find it here), I've got confused with one thing:
$$
\lim_{n \to \infty} \frac{n^k + O\left(n^{k-1}\right)}{k!}\frac{\lambda^k}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k} = \lim_{n\to\infty} \frac{\lambda^k}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k}.
$$
How to show that
$$
\lim_{n \to \infty} \frac{n^k + O\left(n^{k-1}\right)}{k!} = 1?
$$
Thanks in advance!

答:
You've made an error. It should be

$$
\lim_{n \to \infty} \frac{n^k + O\left(n^{k-1}\right)}{k!}\frac{\lambda^k}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k} = \lim_{n\to\infty} \frac{\lambda^k}{k!}\left(1-\frac{\lambda}{n}\right)^{n-k}.
$$

Which is not difficult.


问:
V.I. Arnold's Topological Proof of the Fifth Degree
I have been playing around with some topological ideas in Galois Theory, and I find that it is easy to force a swapping of the roots in a quadratic equation if I slowly drag one of the coefficients around the origin, returning it to where I started. But I can't easily do the same with a cubic. I can take an equation like x^3 + 7x -11 = 0, I can drag either the 7 or the 11 in a big circle about the origin, but I can't thereby swap the two complex roots (or any other pair of roots). Sometimes I can force a cyclic permutation, and sometimes the roots meander around and return to where I began (depending on what cubic equation I start with). But I haven't been able to swap two roots. Is there some reason for this?</p>

答:
We can write out quite explicitly what a path through the space of cubics with distinct roots that transposes two roots looks like, even a path that maintains a zero quadratic term: take the three roots to be $r_1(t) = e^{it}, r_2(t) = - e^{it}, r_3(t) = 0$ where $t \in [0, \pi]$ (so that $r_1(0) = r_2(\pi) = 1, r_2(0) = r_1(\pi) = -1$). This gives the cubic polynomial

$$(x - r_1)(x - r_2)(x - r_3) = x^3 - e^{2it} x$$

which has a linear term going in a circle around the origin and a zero constant term.

Getting an example where the constant term is nonzero is a bit more annoying if you insist on keeping the quadratic term zero (it's easy without this, just translate the example above); this amounts to requiring that $\frac{1}{r_1(t)} + \frac{1}{r_2(t)} + \frac{1}{r_3(t)} = 0$ which makes the formulas less nice. We can take

$$\frac{1}{r_1(t)} = 2 + e^{it}$$
$$\frac{1}{r_2(t)} = 2 - e^{it}$$
$$\frac{1}{r_3(t)} = -4$$

which gives the cubic polynomial

$$(x - r_1)(x - r_2)(x - r_3) = x^3 - \left( \frac{4}{4 - e^{2it}} - \frac{1}{4} \right) x - (16 - 4e^{2it}).$$

So the constant term travels in a circle. The linear term does also, because it's an inversion of a circle. If you take the path to be $t \in \left[ \frac{\pi}{2}, \frac{3 \pi}{2} \right]$ you even get an example where the two roots being swapped start out complex.

I don't know what happens if you only make the constant and linear terms go around in circles separately. What might be going on there is the following: if you move both terms freely you are working in the space of (monic) cubics with distinct roots and zero quadratic term, which can be identified with the complement in $\mathbb{C}^2$ of the zero locus of the discriminant, which for a cubic polynomial $x^3 + px + q$ with zero quadratic term takes the form $\Delta = - 4p^3 - 27 q^2$. I believe this complement is homotopy equivalent to the complement of the trefoil knot; in any case there's a somewhat complicated locus of points you can't cross, and if you don't take sufficiently complicated loops you may not be able to generate the entire fundamental group (which is famously the braid group $B_3$, also the fundamental group of the entire configuration space $C_3(\mathbb{C})$, also known as the space of (monic) cubics with distinct roots).

I just found a quite nice page that explains the connection to configuration spaces and the trefoil knot (with pictures and a video!) here.


问:
Prove that $\dim(T(H))\leq \dim(H)$.

Let $V$ and $W$ be finite-dimensional vector spaces and $T$ be a linear transformation $T:V\to W$.

Let $H$ be a non-zero subspace of $V$, and let $T(H)$ be the set of images of vectors in $H$. Prove that $\dim(T(H))\leq \dim(H)$.

Proof:

Let $B=\{b_1,...,b_n\}$ be a basis for $H$.

Because $B$ is a basis and $T$ is linear, for any $x\in T(H)$, we have

$$x=T(c_1b_1+...+c_nb_n)=c_1T(b_1)+...+c_nT(b_n)$$

Therefore, $\text{span}\{T(b_1)...T(b_n)\}=T(H)$.

If $\{T(b_1)...T(b_n)\}$ is linearaly independent, then $\text{dim}(T(H))=\text{dim}(H)$ and we are done.

Otherwise, $\{T(b_1)...T(b_n)\}$ is linearaly dependent, and there exist $T(b_i)$ that can be expressed as a linear combination of the other vectors in this set.

So let $T(b_i)=d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)$.

Now let $y\in T(H)$,then
$$y=c_1T(b_1)+..+c_iT(b_i)+..+c_nT(b_n)=c_1T(b_1)+...+d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)+...+c_nT(b_n)=(c_1+d_1)T(b_1)+..+(c_{i-1}+d_{i-1})T(b_{i-1})+(c_{i+1}+d_{i+1})T(b_{i+1})+...+(c_n+d_n)T(b_n)$$

So, $\text{span}\{T(b_1)...T(b_n)\}=\text{span}\{T(b_1)...T(b_{i-1}),T(b_{i+1}),...,T(b_n)\}=T(H)$

If this new set is linearly independent then $\text{dim}(T(H))=n-1<\text{dim}(H)$.

Otherwise, we are linearly dependent and a vector can be removed without changing the span. If we get down to one remaining vector and that vector is not the $0$ vector then  $\text{dim}(T(H))=1<\text{dim}(H)$. Otherwise, it is the $0$ vector and $\text{dim}(T(H))=0<\text{dim}(H)$.

Thank you for reading the proof. Please provide feedback on the validity of my proof. Have I assumed anything I shouldn't have assumed? Any leaps in logic? I appreciate any tips suggestions and insights.

答:
I think you are correct. Essentially you go from the space you have control to the space you do not have control.

 Now, In my humble opinion, this is too much suffering. I would argue that is easier to do the opposite i.e., Suppose $dim(H)<dim(T(H)),$ and consider a basis in $T(H)$ like $\{T(b_i)\}_{i\in [n]}.$ Then $\{b_i\}_{i\in [n]}$ has to be dependent in $H$ so WLOG $b_1=\frac{1}{c_1}\sum _{i>1}c_ib_i,$ where $c_i$ are scalars. Applying $T$ to the last equation,you get the contradiction.(Notice that essentially this is the same as your proof, but sometimes is easier to arrive to contradictions).

Title: Prove that $\dim(T(H))\leq \dim(H)$.


问:
Spivak Rising Sun Lemma
Suppose that all points of $(a,b)$ are shadow points, that is $\forall x \in (a,b) \space \exists y\space(y>x \mbox{ and } f(y)>f(x))$.

For all $x\in(a,b)$, prove that $f(x)\leq f(b)$.

let $A=\{y: x\leq y \leq b \mbox{ and } f(x)\leq f(y)\}$. Then in the solutions manual he proceeds:

Notice that we must have $f(x)\leq f(\sup A)$ because $f$ is continuous at
$sup A$ and there are points $y$ arbitrary close to $\sup A$ with
$f(x)\leq f(y)$. (A simple $\varepsilon - \delta$ argument is being
suppressed).

How does $f(x)\leq f(\sup A)$ follow from the continuity of $f$? It would've been very helpful if he had not suppressed the argument.

答:
This is really not difficult and it is better that you understand this without using any $\epsilon, \delta$.

You should note that if $y\in A$ then $f(y) \geq f(x) $. Let us suppose on the contrary that (P) $f(\sup A) <f(x) $ then by continuity (Q) there is a neighborhood of $\sup A$ where values of $f$ are less than $f(x) $ and thus this entire neighborhood contains no points of $A$. This clearly contradicts the definition of $\sup A$ (every neighborhood of supremum must contain some point of the set).

You should be able to grasp why the continuity of $f$ allows you to deduce the conclusion (Q) from assumption (P). In simple language by continuity we can find a neighborhood of $\sup A$ such that the values of $f$ are much nearer to $f(\sup A) $ than to $f(x) $ and hence these values are all less than $f(x) $.

To simplify it further consider the obvious. If we take numbers which are too close to $1$ rather than to $2$ then they will be less than $2$. And the statement remains true if $2$ is replaced by $1.5$ or $1.1$ or $1.001$ or any specific number which is greater than $1$.

The $\epsilon, \delta $ stuff is just a translation of the obvious things explained above in symbols.

Title: Spivak Rising Sun Lemma


问:
Is the solution to the Poisson equation smooth?
In my lecture notes this well known theorem is presented in the following way:

Let $f\in C_{0}^{\infty}(\mathbb{R}^n)$
and define
$$u(x) = (f*Φ)(x) =
\int_{\mathbb{R}^n}
f (y)Φ(x− y)d y,$$
where Φ is the fundamental solution of the Laplace equation. Then $u\in C^2(\mathbb{R}^n)$
and the Poisson equation: $−∆u = f$ holds in the whole space $\mathbb{R}^n$.

Just to put it here for clarification, the fundamental solution is the function $\Phi:\mathbb{R}^n\setminus \{0\}\to \mathbb{R}$ such that:
$$\Phi(\boldsymbol  x)=\begin{cases}
{\displaystyle -\frac{1}{2\pi}\ln\left(\left\vert\boldsymbol x\right\vert\right) \;,
\quad n=2}
\\[2mm]
{\displaystyle\frac{1}{n\left(n - 2\right)\alpha\left(n\right)}
\frac{1}{\left\vert\boldsymbol x\right\vert^{n-2}}\;,\quad n\geq 3}\\
\end{cases}$$
Here $\alpha(n)$ is just the volume of the unit ball in $n$ dimensions.

My question: Is it not true that the solution $u$ is not just in $C^2$, but is also smooth ($u\in C^{\infty}(\mathbb{R}^n)$)? If so how would I prove this?

答:
Yes, the solution is indeed smooth. Did you prove the $C^2$ regularity in your lectures? The $C^\infty$ regularity is proved in a similar way. You change variables in the integration to get
$$u(x) = \int_{\mathbb{R}^n} f(x-y)\Phi(y)\, dy,$$
and then you differentiate under the integral to get
$$D^\alpha u(x) = \int_{\mathbb{R}^n} [D^\alpha f(x-y)]\Phi(y)\, dy$$
for any multi-index $\alpha$. The compact support and smoothness of $f$ is used to justify exchanging the derivative and integral (with, say, the dominated convergence theorem).


问:
Expressing $\sqrt{n +m\sqrt{k}}$
Following this answer, is there a simple rule for determining when:
$$\sqrt{n +m\sqrt{k}}$$
Where $n,m,k \in \mathbb{N}$, can be expressed as:
$$a + b\sqrt{k}$$
For some natural $a,b$?

This boils down to asking for what $n,m,k \in \mathbb{N}$ there exist $a,b\in \mathbb{N}$ such that:
$$2ab= m,\ \ \text{and}\ \ a^2+b^2k = n$$

答:
Hint: Please see this link. To translate it to your problem, note that $ b = m^2k $. From here, one of the square roots on the right hand side of their expression must reduce.

In the case that $ a + \sqrt{a^2 - b} = 2p^2, p \in \mathbb{N} $, $ p^2 = \frac{a + \sqrt{a^2 - b}}{2} $. For the second case $ q^2 = \frac{a - \sqrt{a^2 - b}}{2} $. Hence, if $ \frac{a \pm \sqrt{a^2 - b}}{2} $ is a perfect square then the nested square root is reducible.

For example, let $ a = 30, b = 896 $. In this case, $ \frac{a + \sqrt{a^2 - b}}{2} = 16 $ so $ \sqrt{30 + \sqrt{896}} $ is reducible, being equal to $ 4 + \sqrt{14} $. Also, note that if $ a = 30, b = 756 $, it is reducible because $ \frac{a + \sqrt{a^2 - b}}{2} = 9 $, being equal to $ 3 + \sqrt{21} $.

Title: Expressing $\sqrt{n +m\sqrt{k}}$


问:
closure of image of polynomial map from affine variety
Let $K$ be an algebraic closed field. Let $A$ be an affine variety of $K^{n}$, and $f:A\rightarrow K^{m}$ be a polynomial map. Prove that closure of $f(A)$ equal to $V(\langle I(A),Y_{1}-f_{1},...,Y_{m}-f_{m}\rangle_{K[X_{1},...X_{n},Y_{1},...Y_{m}]}\cap K[Y_{1}...Y_{m}]):=V(B\cap C)$. Where $f_{i}$ are the components of $f$,$V(I)$ denote zero set of ideal $I$, $I(X)$ denote vanishing ideal of set $X$.

I am stuck at both directions. I want to to show that $I(f(A))=\langle B\cap C\rangle$, but what I can show is only $(A,\overline{f(A)})\subset V(B\cap C)$, this is clearly not enough. Can anyone help or give a hint?

答:
First, note that $V(I(A),Y_1-f_1,\cdots,Y_m-f_m)\subset K^{n+m}$ is the graph $(a,f(a))$ of $f$, which is isomorphic to $A$ via the projection to $K^n$. So you're really trying to understand why if $X\subset K^{n+m}$ is a variety then the closure of $\pi(X)$, the projection of $X$ to $K^m$, is the variety cut out by $I(X)\cap K[Y_1,\cdots,Y_m]$.

For one direction, any polynomial in the $Y_i$ which vanishes on $\pi(X)$ will vanish on $X$ since it doesn't depend on the $X$-coordinates, so $I(\pi(X))\subset I(X)\cap K[Y_1,\cdots,Y_m]$. On the other hand, if there is a polynomial $f$ in the $Y_i$ which vanishes on $X$, then it must vanish on $\pi(X)$, since $f$ does not depend on the $X$ coordinates. So $I(X)\cap K[Y_1,\cdots,Y_m]\subset I(\pi(X))$. So $V(I(X)\cap K[Y_1,\cdots,Y_m])=V(I(\pi(X)))=\overline{\pi(X)}$ and we're done.


问:
The set of isotropic vectors contains the kernel of a Bilinear form
I'm studying linear algebra, specifically bilinear forms. I found many sources where it's said that
$$ker(g)\subseteq I(g)$$
where $I$ is the set of all isotropic vectors and $g$ is a bilinear form.

I can find some examples, but I fail to understand why this is true in every case, specifically, why the set of isotropic vectors can be bigger and always contains the kernel.

Let $V$ be a finite dimensional vector space.

I define the kernel of a bilinear form as

$$ker(g)= [v\in V | g(v,w)=0 \forall w \in V] $$

and I define the set of all isotropic vectors as

$$ I= [v \in V | g(v,v)=0 \forall v\in V]$$

答: If $v\in\ker(g)$, then, for each vector $w\in V$, $g(v,w)=0$. In particular, $g(v,v)=0$. In other words, $v\in I(g)$.


答:
Well, it is clear that if $g(v,w)=0$ for all $w\in V$, then a fortiori $g(v,v)=0$; therefore $\ker g\subseteq I(g)$. For a strict inclusion, you can think about $$g:\Bbb R^2\times\Bbb R^2\to\Bbb R\\ g(v,w)=v_1w_1-v_2w_2=v^T\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}w$$

for which $\ker g=\{(0,0)\}$ but $I(g)=\operatorname{span}\{(1,1)^T\}\cup \operatorname{span}\{(1,-1)^T\}$.

Title: The set of isotropic vectors contains the kernel of a Bilinear form

Body:
I'm studying linear algebra, specifically bilinear forms. I found many sources where it's said that
$$ker(g)\subseteq I(g)$$
where $I$ is the set of all isotropic vectors and $g$ is a bilinear form.

I can find some examples, but I fail to understand why this is true in every case, specifically, why the set of isotropic vectors can be bigger and always contains the kernel.

Let $V$ be a finite dimensional vector space.

I define the kernel of a bilinear form as

$$ker(g)= [v\in V | g(v,w)=0 \forall w \in V] $$

and I define the set of all isotropic vectors as

$$ I= [v \in V | g(v,v)=0 \forall v\in V]$$


问:
$\int_{\Omega}|fg|d \mu \leq ||f||_2||g||_2$
Let $(\Omega, \mathcal{M}, \mu)$ be a measure space. Let $f,g \in \mathcal{L}^2(\Omega, \mu)$.

We have that $\int_{\Omega}|fg|d \mu \leq \|f\|_2\|g\|_2$

I would like to prove it, without applying directly Cauchy–Schwarz inequality.

I would proceed in the following way:

Show that $\int_{\Omega}|fg|d \mu \leq \|f\|_2\|g\|_2$ is true for

$\|f\|_2=0$ or $\|g\|_2=0$.

$\|f\|_2=1 = \|g\|_2$.

$\|f\|_2 \neq 0$ and $\|g\|_2 \neq 0$.

Can someone show me how to do one case? Hopefully I can do the rest alone then.

Body:
Case $1$: For any $h \in \mathcal{L}^2(\Omega, \mu)$, if $\|h\|_2=0$ then $\int |h|^2 d\mu =0$, so $|h| = 0$ $a.e.$, so $h=0$ $a.e.$.

So we have that, if $\|f\|_2=0$ or $\|g\|_2=0$ then $f=0$ $a.e.$ or $g=0$ $a.e.$. So $fg=0$ $a.e.$, so $| fg| =0$ $a.e.$,
$$\int_{\Omega}|fg|d \mu =0 \leq 0=\|f\|_2\|g\|_2$$

Case $2$:  If $\|f\|_2=1 = \|g\|_2$, the we have $\int |f|^2 d\mu =1$ and $\int |g|^2 d\mu =1$.

Note that $(|f|-|g|)^2 \geq 0$. So
$$\int_\Omega (|f|-|g|)^2 d\mu \geq 0$$
So, we have,
$$\int_\Omega |f|^2 d\mu +  \int_\Omega |g|^2 d\mu -2 \int_\Omega |fg|d\mu = \int_\Omega (|f|-|g|)^2 d\mu \geq 0$$
Since $\int |f|^2 d\mu =1$ and $\int |g|^2 d\mu =1$, we have
$$2 -2 \int_\Omega |fg|d\mu \geq 0$$
So
$$\int_\Omega |fg|d\mu \leq 1 = \|f\|_2 \|g\|_2$$

Case $3$: If $\|f\|_2 \neq 0$ and $\|g\|_2 \neq 0$. Then let $\hat{f} =\frac{f}{\|f\|_2}$ and $\hat{g} =\frac{g}{\|g\|_2}$. Clearly, $\hat{f}, \hat{g} \in \mathcal{L}^2(\Omega, \mu)$ and $\|\hat{f}\|_2=1 = \|\hat{g}\|_2$. So, by Case $2$, we have

$$\int_\Omega |\hat{f}\hat{g}|d\mu \leq 1 = \|\hat{f}\|_2 \|{g}\|_2$$

But, since $\hat{f} =\frac{f}{\|f\|_2}$ and $\hat{g} =\frac{g}{\|g\|_2}$, we have

$$\frac{1}{\|f\|_2\|g\|_2 }\int_\Omega |fg|d\mu=\int_\Omega |\hat{f}\hat{g}|d\mu \leq 1 $$
So
$$\int_\Omega |fg|d\mu \leq  \|f\|_2 \|g\|_2$$

Remark: From Case $1$ and Case $3$ we have that $$\int_\Omega |fg|d\mu \leq  \|f\|_2 \|g\|_2$$  is valid for all  $f, g \in \mathcal{L}^2(\Omega, \mu)$.


问:
How does this difference equation (discrete function map) behave?
Consider the iterated map

$x_{n+1} = rx_n(1 - x_n^2)$

As this is non-linear, I'm a bit confused on how to approach it's analysis. I'd like to find out/justify the following things I found out with plotting and numeric testing:

Is it true that $[x_n \in (0,1) \implies x_{n+1} \in (0,1)], \  \forall r \in (0,\frac{3\sqrt{3}}{2})$. Why?

This iterated map has two fixed points, which are $x_0=0$ and $x_0=\sqrt{1-\frac{1}{r}}$. Why does this second fixed point only exist when $r \gt 1$? How do I determine it's stability, which depends on r?

How do I show that $x_0=0$ is an attractive fixed point?

What happens when $r \gt 2$?

How must I procede? Either hints or solutions are welcome.

答:
Clearly, $x_{n+1}>0$. On the other hand, by the inequality between geometric and quadratic means,
$$x_{n+1}<\frac {3\sqrt 3}{2}x_n(1-x_n^2)=\frac {3\sqrt 3}{2\sqrt{2}}\cdot \sqrt{2}x_n\cdot\sqrt{1-x_n^2}\cdot  \sqrt{1-x_n^2}\le $$ $$\frac {3\sqrt 3}{2\sqrt{2}}\left(\frac{2x_n^2+1-x_n^2+1-x_n^2} 3 \right)^{3/2}=
\frac {3\sqrt 3}{2\sqrt{2}}\left(\frac23 \right)^{3/2}=1.$$

Fixed point of the iterated map are exactly real roots of the equation $x=rx(1-x^2)$. For any $r$ the equation has a root $0$. Since $f’(0)=r$, this point is attractive when $|r|<1$ and repelling when $|r|>1$. Moreover, when $|r|\le 1$ and $|x_n|\le 1$ then $|x_{n+1}|\le |x_n|$, so the sequence $|\{x_n\}|$ is non-increasing and thus it converges to some number $a$. Since $a=ra(1-a^2)$, we have $a=0$.

It is easy to check that two other roots exist iff $1/r<1$ and the are equal $\pm \sqrt{1-\frac{1}{r}}$. The value of $f’(x_0)$ at any of these roots $x_0$ equals $3-2r$, so $x_0$ is attractive when $1<r<2$ and repelling when $r<0$ and $r>2$. When $r=2$ then put $f(x)=rx(1-x^2)$.

Since $f(x_0)=x_0$ and $f’(x_0)=0$, by Taylor's formula in Lagrange’s form we have $$x_{n+1}-x_0=f(x_n)-x_0=f(x_0)+(x_n-x_0)f’(x_0)+\frac{f’’(c)}{2}(x_n-x_0)^2-x_0=$$ $$\frac{f’’(c)}{2}(x_n-x_0)^2=-3c(x_n-x_0)^2,$$ for some $c$ between $x_0$ and $x_n$. Since $x_0=\pm \frac{1}{\sqrt{2}}$, we have $|x_{n+1}-x_0|\le |x_n-x_0|/2$ provided $|x_n-x_0|\le\frac 16$, so the point $x_0$ is attractive.


问:
How many groups of order n are possible?
I'm wondering what total number of possible groups of order n would be if we are counting isomorphisms separately. I mean that if we were to write out the multiplication tables for elements $a_1, a_2, ... , a_n$, what is the maximum number of such tables? Can we make the bound tighter by noting that the table must satisfy the group axioms, or do we need to know something about n?

答:
This is A000001 on the OEIS. The answer is quite complicated and depends delicately on the prime factorization of $n$; only asymptotic results are known for general $n$. It's a result due to Higman and Sims that if $n = p^k$ then there are asymptotically $p^{\frac{2}{27} k^3 + O(k^{8/3})}$ groups of order $n$. This gives that if $n = \prod p_i^{k_i}$ then there are at least

$$\prod p_i^{\frac{2}{27} k_i^3 + O(k_i^{8/3})}$$

groups of order $n$ (this is a count of the nilpotent groups of order $n$, which are the products of their Sylow $p$-subgroups), and I'd guess this is an accurate asymptotic (up to that huge error in the exponent) but I don't know if this is known.

These results suggest, if you work with them for a bit, that almost all finite groups are finite $2$-groups (having order a power of $2$); this is an open problem to prove rigorously, though, as far as I know. As experimental evidence for this, it's known that over 99% of the first $50$ billion groups have order $1024 = 2^{10}$: there are known to be exactly

$$49487365422 \approx 4.9 \times 10^{10}$$

groups of this order.

A much easier special case to think about is the number of abelian groups of order $n$: by the structure theorem for finite abelian groups, if $n = \prod p_i^{k_i}$ then this is exactly $\prod p(k_i)$ where $p$ is the partition function.

In the opposite direction, there are a special set of positive integers $n$, including the primes, for which the only group of order $n$ is the cyclic group $C_n$; see this exposition by Keith Conrad for more, and also A003277 on the OEIS.


问:
Cutting an equilateral triangle into $n$ equal pieces
We have an equilateral triangle and we want to cut it into $n$ equal triangular pieces.

For which $n$ is it possible?

My Attempt: I found these possible numbers: $2,3,4,6$. I also proved every $n$ of the form $4^n$.

Note: I don't mean equal areas, I mean equal triangles.

答:
Michael Beeson has a series of papers on tiling triangles by smaller congruent triangles listed here, of which a recent one dealing with the equilateral case is here.

In it, it is shown that an equilateral triangle can be tiled by $N$ congruent triangles when $N$ is of the form $m^2, 2m^2, 3m^2,$ or $6m^2$. Shockingly, these are not the only possibilities! Among others, the case $N=109345=3^7\cdot 5$ is possible, and shown in the paper; for convenience, I'll reproduce the diagram below.

However, it is proven that no prime $N$ beyond $N=2,3$ are possible.

The paper also shows that the question of whether $N$ tiles are possible is computable, though the given algorithm is not efficient enough to make a search over large $N$ tractable.


问:
Given any sequence of real numbers $a_n$, show that there exists a subsequence $a_{n_k}$ of $a_n$....
Given any sequence of real numbers $a_n$, show that there exists a subsequence $a_{n_k}$ of $a_n$ and a real number $b$ such that $\sum\limits_{k\geq1}\mid b-\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}\mid$ is convergent.

I tried like this. For any subsequence  $a_{n_k}$ of $a_n$ , the sequence $\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}$ is always a bounded sequence. Hence it contains a convergent subsequence. Now from this I got stuck. please help

答:
let $a_{n} > 0$ then $a_{n}+1 > 1$ $\Rightarrow$ $\frac{1}{a_{n}+1}<1$

$\dfrac{a_{n}}{1+\mid a_{n}\mid}=\dfrac{a_{n}}{1+a_{n}}=1-\dfrac{1}{1+a_{n}}<1$  which means it is bounded so there exists a subsequence $a_{n_k}$

$a_{n_k}$ is convergent $\Rightarrow$ $\dfrac{a_{n}}{1+\mid a_{n}\mid}$ is convergent

$\dfrac{a_{n}}{1+\mid a_{n}\mid}$ is convergent  $\Rightarrow$  $\sum\limits_{k\geq1}\mid b-\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}\mid$ is convergent

let $a_{n} < 0$ then $1-a_{n}>1$ $\Rightarrow$ $-1<\frac{1}{1-a_{n}}<0$

$\dfrac{a_{n}}{1+\mid a_{n}\mid}=\dfrac{a_{n}}{1-a_{n}}=\dfrac{1}{1-a_{n}}-1>-1$ $\Rightarrow$ $\mid \dfrac{a_{n}}{1+\mid a_{n}\mid}\mid <1$ which means it has a convergent subsequence $a_{n_k}$

$a_{n_k}$ is convergent $\Rightarrow$ $\mid \dfrac{a_{n}}{1+\mid a_{n}\mid} \mid$ is convergent

$\mid \dfrac{a_{n}}{1+\mid a_{n}\mid} \mid$ is convergent $\Rightarrow$ $\sum\limits_{k\geq1}\mid b-\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}\mid$ is convergent

NOTE: I also found  an article that may help you If $(a_n)$ is any real sequence , then $(\frac{a_n}{1+|a_n|})$ has a convergent subsequence


问:
Where is the flaw in my proof claiming that if $P$ is a prime minimal over $\operatorname{ann}M$, then $P=\operatorname{rad}(\operatorname{ann}M)$.
Let $R$ be a Noetherian ring, and $M$ a finitely generated $R$-module. If $P$ is prime minimal over $\operatorname{ann}M$, then $P=\operatorname{rad}(\operatorname{ann}M)$.

$\textit{Proof:}$ Since $\operatorname{ann}M \subset P$ is minimal, if we localize at $P$, then $(\operatorname{ann}M)R_P \subset PR_P$ and $PR_P$ is the unique prime ideal containing $(\operatorname{ann}M)R_P$. Hence $PR_P = \operatorname{rad}((\operatorname{ann}M)R_P)$. Since radical commutes with localization, we have $$PR_P = \operatorname{rad}((\operatorname{ann}M)R_P) = (\operatorname{rad}(\operatorname{ann}M))R_P.$$
So, $P = \operatorname{rad}(\operatorname{ann}M)$.

A counter example is $R=\mathbb Z$, $P=(2)$, $M = \mathbb Z/6\mathbb Z$ where we have $\operatorname{rad}(\operatorname{ann}M)=(6) \ne (2)$.

答:
You obtain
$$PR_P=(\operatorname{rad}(\operatorname{ann} M)R_P$$ and conclude from this that $P=\operatorname{rad}(\operatorname{ann} M)$. But this implication is not true; indeed, in your example where $R=\mathbb{Z}$, $P=(2)$ and $\operatorname{rad}(\operatorname{ann} M)=(6)$, you have $(2)R_P=(6)R_P$, as $3$ is a unit in $R_P$, but of course $(2) \ne (6)$ in $\mathbb{Z}$.


问:
Lebesgue-measure function
We are looking at $\mathbb{R}$ with the Lebesgue-measure. Let $h \colon [0,1] \to \mathbb {R}$ be a measurable function and let $f(x,t) = h(t) \sin (x-t)$. For $x \in \mathbb {R}$ we are looking at $F(x) =$ $\int _{[0,1]} f(x,-)\,d\lambda$  (might not be defined, depending on $h$). Which of these following statements is true?

a. $F$ is defined and continuous on $[0,1]$, without further requirements to $h$.

b. If $h$ is integrable on $[0,1]$, then $F$ is defined and continuous.

c. Let $F$ be defined. If $h$ is not continuous, then $F$ is also not continuous.

I think that a is false and that b and c are true, is that correct?

答:
a. False. A counterexample is $h(t) = \frac{1}{t}$ for $t \in (0,1]$ and $h(0)=0$. Then if $x \notin \pi \mathbb{Z}$, then $\sin(x) \neq 0$. Then $\frac{\sin(x-t)}{t} \sim_{t \to 0} \frac{\sin(x)}{t}$ so $F(x)$ is $\mbox{sign}(\sin(x)) \cdot \infty$.

b. True. If $h$ is integrable, then for any $x$, for all $t$, $|h(t)\sin(x-t)| \le |h(t)|$, and for any $t$, $x \mapsto \sin(x-t)$ is continuous, so the dominated convergence theorem yields that $F$ is continuous (you can actually show that $F$ is $\mathcal{C}^{\infty}$.

c. False. A trivial counterexample would be $h(t) = 1_{\{0\}}(t)$. $h$ is not continuous but $F = 0$. Another counterexample is $h(t) = 1_{t < \frac{1}{2}}$. Indeed, $F(x) = \displaystyle{\int_0^{\frac{1}{2}}} \sin(x-t)dt = \cos\big(x-\frac{1}{2}\big)-\cos(x)$, which is continuous.

Title: Lebesgue-measure function


问:
Let $a_1/b_1,\ldots,a_n/b_n$ be rational numbers in lowest terms. If $M={\rm lcm}(b_1,\ldots,b_n)$, prove $\gcd(Ma_1/b_1,\ldots,Ma_n/b_n)=1$.

Problem. Let $a_1/b_1,\ldots,a_n/b_n$ be rational numbers in lowest terms. If $M={\rm lcm}(b_1,\ldots,b_n)$, prove that the gcd of $Ma_1/b_1,\ldots,Ma_n/b_n$ is $1$.

I think this statement is false. For example, consider $$\frac{2}{3}\quad{\rm and}\quad \frac{2}{9}.$$ Both of them are in the lowest form, but $M={\rm lcm}(3,9)=9$ where $$9\cdot\frac{2}{3}=6\quad{\rm and}\quad 9\cdot\frac{2}{9}=2$$ are not relatively prime.

Nonetheless, if I add the assumption that $\gcd(a_1,\ldots,a_n)=1$, will the assertion be true? I tried writing $1$ as a linear combination of those numbers, but it seems not working.

答:
It is true when $\,d := \gcd\{a_i\} = 1.\,$ More generally it is the special case $\,d = 1,\,m = \ell\,$ of the Theorem below, which is the integral-scaled version of the gcd formula for reduced fractions $\frac{a_i}{b_i}$ (putting $\,m=1\,$ below gives the $k$-ary extension of this formula), which says the the gcd of reduced fractions is the gcd of their numerators over the lcm of their denominators (and dually for lcm).

Theorem $\, $ If $\,m,a_i,b_i\in\Bbb Z,\,$ $\,b_i\mid m a_i,\,$   $\,\color{#c00}{\gcd(b_i,a_i/d)\!=\!1},\,$ $\,d\! =\! \gcd\{a_i\},\,$ $\,\ell\! =\! {\rm lcm}\{b_i\}\,$ then

$\qquad\qquad\quad \gcd\left(\!\dfrac{ma_1}{b_1},\ldots,\dfrac{ma_k}{b_k}\!\right)\, =\, \dfrac{m\gcd(a_1,\ldots,a_k)}{{\rm lcm}(b_1,\:\!\ldots,\:\!b_k)\!\!\!\!\!\!}\,=\,\dfrac{md}{\ell}$

$\begin{align}{\bf Proof}\
\ \ c\,&\mid \gcd\{ma_i/b_i\}\\[.1em]
\iff \, \ c\,&\mid {ma_i}/{b_i},\,\forall i, \ \rm by\ gcd\ \color{#90f}{universal}\ property\\[.3em]
\iff\! cb_i&\mid ma_i,\, \forall i\\[.3em]
\iff\ \color{#c00}{b_i}&\mid ma_i/c={md}/c\,(\color{#c00}{a_i/d}),\, \forall i,\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\smash{\overbrace{\&\ \ md/c\in\Bbb Z}^{\textstyle c\mid ma_i\Rightarrow c\mid\gcd\{ma_i\}= md\!\!\!\!\!\!\!\!\!\!\!}}}\\[.1em]
\iff\ b_i&\mid {md}/{c},\,\forall i,\ \  {\rm by\ } \color{#c00}{\gcd(b_i,a_i/d)\!=\!1}\ \text{&  Euclid's Lemma}\\[.1em]
\iff\ \ell\ &\mid {md}/c,\ \ \ \ \ \ \ \ \rm by\ lcm\ \color{#90f}{universal}\ property\\[.1em]
\iff\:\! \ell c&\mid md\\[.1em]
\iff\ c\ &\mid {md}/\ell
\end{align}$

We used the gcd distributive law in $\,\gcd\{ma_i\} = m\gcd\{a_i\} = md,\,$ and Euclid's Lemma and the  gcd & lcm $\rm \color{#90f}{universal}\ properties$.

Remark $ $ Since the proof used only laws valid in every gcd domain, the theorem remains true if we replace $\Bbb Z$ by any gcd domain, e.g. any UFD.


问:
Connection between quadratic characters and quadratic twists of elliptic curves
So apparently there is some kind of a connection between quadratic characters and twists of elliptic curves that I can't understand.

I am well aware of quadratic twisting of an elliptic curve $E$, we get a curve $E^d$, and the curves are isomorphic over $\mathbb{Q}(\sqrt{d})$.

I also know about quadratic characters associated to fields $\mathbb{Q}(\sqrt{d})$, given by
$$\chi=\chi_{\mathbb{Q}(\sqrt{d})}:\mathbb{Z}_{>0} \rightarrow \mathbb{C},$$
$$ \chi(n)=\left( \frac{disc(\mathbb{Q}(\sqrt{d}))}{n} \right). $$

In literature I often see twists of elliptic curves $E^d$ being denoted by $E^{\chi}$. Why is that so? what is the connection? Am I even using the right character here?

答:
There are a few ways to intemperate this question. As @Mathmo123 says in their comment there is a natural bijection between quadratic characters and twists of $E$ (when the $j$-invariant is not $0$ or $1728$).

From a more elementary point of view we can define a quadratic twist of $E : y^2 = f(x)$ by $E^d : dy^2 = f(x)$. Notice that this is determined by $d$ up to squares - hence is determined by the quadratic extension $F = \mathbb{Q}(\sqrt{d}) / \mathbb{Q}$. We obtain a quadratic character
$$ \chi : G_{\mathbb{Q}} \to \langle \pm 1 \rangle$$
of the absolute galois group $G_{\mathbb{Q}}$ by restricting $\sigma \in G_{\mathbb{Q}}$ to $F$ (note that $\operatorname{Gal}(F/\mathbb{Q}) \cong C_2$) (conversely we obtain a quadratic extension of $\mathbb{Q}$ as the fixed field of the kernel of such a character). Thus there is a bijective correspondence between quadratic twist of $E$ and quadratic characters of the absolute galois group.

A useful application of this thinking is the following

Lemma We have
$$E^d(\mathbb{Q}) = \{P \in E : \sigma(P) = \chi(\sigma)P \}$$

Proof: Notice that we have $P \in E^d(\mathbb{Q})$ if and only if $P = (x', \sqrt{d}y')$, where $P' = (x', y') \in E(\mathbb{Q})$. Thus $\sigma(P) = P$ if $\sigma$ fixes $F$, or $\sigma(P) = -P$ otherwise, which is precisely the statement.


问:
Let $G$ be a group. Prove that $G'$ is central iff ${\rm Inn}(G)$ is abelian.
This is Exercise 4.11 of Roman's "Fundamentals of Group Theory: An Advanced Approach." According to this search and Approach0, it is new to MSE.

The Details:

On page 33 of Roman's book,

Definition: A subgroup $H$ of a group $G$ is central if $H$ is contained in the centre of $G$.

On page 120 ibid., we have

Theorem 4.13 (1): [. . .] $${\rm Inn}(G)\cong G/Z(G).$$

The Question:

Let $G$ be a group. Prove that $G'$ is central if and only if ${\rm Inn}(G)$ is abelian.

Thoughts:

I think Theorem 4.13 (1) is important here. However, I find it hard to articulate why.

Suppose $G'\subseteq Z(G)$. Consider $\gamma_g, \gamma_h\in{\rm Inn}(G)$ for $g, h\in G$. We have for any $x\in G$ that

$$\begin{align}
(\gamma_g\circ\gamma_h)(x)&=\gamma_g(\gamma_h(x))\\
&=\gamma_g(hxh^{-1})\\
&=ghxh^{-1}g^{-1}\\
&=(gh)x(gh)^{-1}\\
&=\gamma_{gh}(x).
\end{align}$$

Now, I just follow my nose. Something tells me I'm heading in the right direction.

Consider $\gamma_{gh}(x)x^{-1}$; that is, $[gh, x]$. This commutes with all elements of $G$ by assumption.

Where do I go from here?

Please help :)

答: This is simpler. Since $\mathrm{Inn}(G)\cong G/Z(G)$, then $\mathrm{Inn}(G)$ is abelian if and only if $G'\subseteq Z(G)$: recall that for a group $G$, a normal subgroup $N$ contains $G'$ if and only if $G/N$ is abelian.


问:
How to show that the distribution of $Y_{1}, Y_{2}, \ldots, Y_{n-1}$ is uniform on the simplex in $\mathbb{R}^{n-1}$
I need to show that the distribution of $Y_{1}, Y_{2}, \ldots, Y_{n-1}$ is uniform on the simplex in $\mathbb{R}^{n-1}$, that is, the pdf is $(n-1)!$ in a suitable support.

Here, for $i = 1, \ldots, n-1$, $Y_{i}$ is defined as
$$
Y_{i}=\frac{X_{i}}{X_{1}+\ldots+X_{n}}, i=1, \ldots, n-1,
$$
where $X_{1}, X_{2}, \ldots, X_{n}$ are independent distributed random variables, each with a standard
exponential distribution.

答:
To prove that result, you can integrate your distribution over $T=\sum_1^n{X_i}$

$$
f(y_1,y_2,...,y_{n-1})dy_1dy_2...dy_{n-1}=\int_{t=0}^\infty{g(x_1,x_2,...,x_n)}dx_1dx_2...dx_{n-1}dt
$$

where g is the distribution over $X_1,...,X_n$

you can notice that $dx_1dx_2...dx_{n-1}=t^{n-1}dy_1dy_2...dy_{n-1}$

and $g(x_1,x_2,...,x_n) = \lambda^n.e^{-\lambda.t}$

so
$$
f(y_1,y_2,...,y_{n-1})=\int_{t=0}^\infty{t^{n-1}\lambda^ne^{-\lambda.t}dt}
$$

with $z=\lambda t$, you get
$$
f(y_1,y_2,...,y_{n-1})=\int_{t=0}^\infty {z^{n-1}e^{z}dz}=(n-1)!
$$


问:
Probability of generating a sequence of numbers between 1 and n
Let's say we want to generate a random number between 1 and n, n-times,then the probability that every integer between 1 and n appears once in the generated sequence of random numbers is $\frac{n!}{n^n}$. So if we were to generate a number between 1 and 10, 10-times, the probability that every number between 1 and 10 appears once in the generated sequence of numbers is $\frac{10!}{10^{10} }$, which is about 1 in 3000. So we should expect that if we do this process of generating 10 numbers 3000 times we should get a sequence of numbers where all numbers between 1 and 10 apeear once only one time. But when I tried this on my calculator(it has a button for generating integers) I got that pattern of all integers between 1 and 10 appearing without repeating after my 15th trial! So does that mean my calculator dose not actually generate random numbes?

Or in another words: what is the probability that after generating 10 numbers (between 1and 10) randomly 15 times you get a sequence where all integers between 1 and 10 appear once without repeating?

答:
The probability that after generating 10 numbers (between 1 and 10) randomly 15 times you get a sequence where all integers between 1 and 10 appear at least once is

$$
1-\left(1-\frac{10!}{10^{10}}\right)^{15}\approx0.00543.
$$


问:
Converse of $(A\rightarrow B)\rightarrow((B\rightarrow C)\rightarrow(A\rightarrow C))$
I have a basic question about the following proposition.

$(A\rightarrow B)\rightarrow((B\rightarrow C)\rightarrow(A\rightarrow C))$

I can prove it in intuitionistic logic. But I wonder if we have the converse in intuitionistic logic? Or in classical logic?

$((B\rightarrow C)\rightarrow(A\rightarrow C))\rightarrow(A\rightarrow B)$

Thanks!

答:
You can show that the formula evaluates to False under some assignments by picking a falsifying assignment and then check if it indeed falsifies the formual. This can be done easily, you only have $2^3 = 8$ assignments so it won't take you long to go over all assignments and look for a falsifying one (note that when you have $n$ variables, you have an exponential number of assignments so this is not always a good approach). So if you meet more complicated formulas, it can be helpful to first simplify the formula using distribution and de Morgan rules as follows. In propositional logic, every formula of the form $\varphi_1 \to \varphi_2$ is equivalent to the formula $\neg \varphi_1 \lor \varphi_2$. Therefore,
$$\begin{align*}
((B\to C) \to (A\to C)) \to (A\to B)&\equiv ((\neg B \lor C) \to (\neg A \lor C)) \to (\neg A \lor B) \\
&\equiv  ((\neg(\neg B \lor C)) \lor (\neg A \lor C)) \to (\neg A \lor B)\end{align*}$$
Using de Morgan laws, the above formula is equivalent to
$$\begin{align*}
&\equiv ((\neg\neg B \wedge \neg C) \lor (\neg A \lor C)) \to (\neg A \lor B)\\
&\equiv (( B \wedge \neg C) \lor (\neg A \lor C)) \to (\neg A \lor B)\\
&\equiv \neg(( B \wedge \neg C) \lor (\neg A \lor C)) \lor (\neg A \lor B)\end{align*}$$
Again, by de Morgan
$$\begin{align*}
 &\equiv (\neg( B \wedge \neg C) \wedge \neg(\neg A \lor C)) \lor (\neg A \lor B)\\
&\equiv (( \neg B \lor \neg \neg C) \wedge (\neg \neg A \wedge \neg C)) \lor (\neg A \lor B)\\
 &\equiv (( \neg B \lor  C) \wedge ( A \wedge \neg C)) \lor (\neg A \lor B)\end{align*}$$
By the distribution rule
$$ \equiv ( (\neg B \lor C) \lor (\neg A \lor B) ) \wedge ( (A \wedge \neg C) \lor (\neg A \lor B))$$

Since $B\lor \neg B \equiv \text{True}$, the above formula is equivalent to
$$\begin{align*}
 &\equiv ( \text{True} \lor C \lor \neg A ) \wedge ( (A \wedge \neg C) \lor (\neg A \lor B))\\
&\equiv \text{True} \wedge ( (A \wedge \neg C) \lor (\neg A \lor B))\\
&\equiv  (A \wedge \neg C) \lor (\neg A \lor B)
\end{align*}$$
This formula is a bit easier than the original one. When $A$ is True, the above formula is equivalent to $\neg C \lor B$ which is obviously not equivalent to True. Indeed, if you follow the assignment in the comment section and assign $A = \text{True}$, $C = \text{True}$, $B = \text{False}$, you get that the formula evaluates to $\neg \text{True} \lor \text{False} = \text{False}$.


问:
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$

Find all matrices $A$ of order $2 \times 2$ that satisfy the equation

$$
A^2-5A+6I = O
$$

My Attempt:

We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$

This implies that $A\in\{3I,2I\} = \left\{\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix},
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}\right\}$.

Are these the only two possible values for $A$, or are there other solutions?If there are other solutions, how can I find them?

Body:
In fact there are infinitely many solutions.

Similar to the question $1$ in http://hk.knowledge.yahoo.com/question/question?qid=7010022700078:

Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ ,

Then $\begin{pmatrix}a&b\\c&d\end{pmatrix}^2-5\begin{pmatrix}a&b\\c&d\end{pmatrix}+6\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$

$\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}-\begin{pmatrix}5a&5b\\5c&5d\end{pmatrix}+\begin{pmatrix}6&0\\0&6\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$

$\begin{pmatrix}a^2+bc-5a+6&ab+bd-5b\\ac+cd-5c&bc+d^2-5d+6\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$

$\begin{pmatrix}a^2-5a+bc+6&b(a+d-5)\\c(a+d-5)&d^2-5d+bc+6\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$

$\therefore\begin{cases}a^2-5a+bc+6=0~......(1)\\b(a+d-5)=0~.........(2)\\c(a+d-5)=0~.........(3)\\d^2-5d+bc+6=0~......(4)\end{cases}$

From $(2)$ and $(3)$ ,

$\begin{cases}b=0\\c=0\\a+d-5=k_1\end{cases}$ or $\begin{cases}b=k_1\\c=k_2\\a+d-5=0\end{cases}$ , $k_1,k_2\in\mathbb{C}$

Case $1$: $\begin{cases}b=0\\c=0\\a+d-5=k_1\end{cases}$

Put it into $(1)$ and $(4)$ :

$a^2-5a+6=0$

$(a-2)(a-3)=0$

$a=2$ or $3$

$d^2-5d+6=0$

$(d-2)(d-3)=0$

$d=2$ or $3$

They are automatically satisflied $a+d-5=k_1$

$\therefore A=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}2&0\\0&3\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&3\end{pmatrix}$

Case $2$: $\begin{cases}b=k_1\\c=k_2\\a+d-5=0\end{cases}$

$(1)-(4)$ :

$a^2-d^2-5a+5d=0$

$(a+d)(a-d)-5(a-d)=0$

$(a+d-5)(a-d)=0$

$\therefore\begin{cases}a+d-5=0~......(5)\\a-d=2k_3~........(6)\end{cases}$ , $k_3\in\mathbb{C}$

$(5)+(6)$ :

$2a-5=2k_3$

$a=2.5+k_3~......(7)$

$(5)-(6)$ :

$2d-5=-2k_3$

$d=2.5-k_3~......(8)$

Put $(7)$ into $(1)$ and $(8)$ into $(4)$ :

$\begin{cases}(2.5+k_3)^2-5(2.5+k_3)+k_1k_2+6=0\\(2.5-k_3)^2-5(2.5-k_3)+k_1k_2+6=0\end{cases}$

$\begin{cases}6.25+5k_3+k_3^2-12.5-5k_3+k_1k_2+6=0\\6.25-5k_3+k_3^2-12.5+5k_3+k_1k_2+6=0\end{cases}$

$\begin{cases}k_3^2+k_1k_2-0.25=0\\k_3^2+k_1k_2-0.25=0\end{cases}$

$k_3^2+k_1k_2-0.25=0$

$k_3^2=0.25-k_1k_2$

$k_3=\pm\sqrt{0.25-k_1k_2}$

$\therefore A=\begin{pmatrix}2.5\pm\sqrt{0.25-k_1k_2}&k_1\\k_2&2.5\mp\sqrt{0.25-k_1k_2}\end{pmatrix}$ , $k_1,k_2\in\mathbb{C}$

Hence $A=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}2&0\\0&3\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&3\end{pmatrix}$ or $\begin{pmatrix}2.5\pm\sqrt{0.25-k_1k_2}&k_1\\k_2&2.5\mp\sqrt{0.25-k_1k_2}\end{pmatrix}$ , $k_1,k_2\in\mathbb{C}$


问:
Exterior Algebra and linearly independent vectors
Suppose that $v_1,\cdots,v_r$ are linearly independent vectors in some vector space $V$. I want to try and show that for any $w \in \bigwedge^p(V)$ that
$$
w = \sum_{i=1}^{r} v_i \wedge \psi_i
$$
for some $\psi_i \in \bigwedge^{p-1}(V)$ if and only if
$$
v_1\wedge v_2\wedge \cdots \wedge v_r\wedge w = 0.
$$

The forward direction is trivial by writing $w$ as the sum and extending the wedge product linearly. It's the second implication that's giving me some trouble.

If we assume that $v_1\wedge v_2\wedge \cdots \wedge v_r\wedge w = 0$, then I want to conclude that I can write $w$ in the appropriate form by examining well-chosen alternating, multi-linear forms from $V^{p+r}$ into some vector space so that I can use the universal property of $\bigwedge^{p+r}(V)$, and evaluate the induced map at $v_1\wedge v_2\wedge \cdots \wedge v_r\wedge w$ and get $0$.

The problem I'm having is that $w$ is not necessarily an elementary wedge product, so I don't have a canonical way of thinking of it as an element of $V^p$.
Any ideas for this backward direction would be greatly appreciated.

答:
Let $\{e_1,\ldots, e_k\}$ be a basis of $V$ such that $v_i=e_i$ for $1\le i\le r$. $w\in \bigwedge^p(V) \implies$

$$w = \sum_{\alpha\in P}f_{\alpha}e_{\alpha_1}\wedge\ldots \wedge e_{\alpha_s}$$
Where $P = \{(i_1,\ldots, i_s) \mid 1 \le  i_1 < i_2 < \cdots < i_s \le k, s\leq p\}$ and I will use $|\alpha|$ to denote the number of elements in the tuple. Clearly
$$v_1\wedge \cdots \wedge v_r =  e_{1}\wedge\cdots \wedge e_{r}$$
So
\begin{align*}
&v_1\wedge  \cdots \wedge v_r\wedge w = 0\\
\implies&  e_{1}\wedge\cdots \wedge e_{r}\wedge  \sum_{\alpha\in P}f_{\alpha}e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_s} = 0\\
\implies& \forall \alpha\in P, f_\alpha \neq 0 \implies \exists l_\alpha \leq |\alpha|, \alpha_{l_\alpha} \leq r \text{ (Let $l_\alpha$ denote the least such value)}\\
\implies& w = \sum_{\alpha\in P, f_\alpha \neq0}f_{\alpha}e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_m}\wedge e_{l_\alpha}\wedge e_{\alpha_n}\wedge \cdots \wedge e_{\alpha_s} \space\space(\alpha_m<l_\alpha<\alpha_n)\\
\implies& w = \sum_{\alpha\in P, f_\alpha \neq0}f_{\alpha}(-1)^m e_{l_\alpha}\wedge e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_m}\wedge e_{\alpha_n}\wedge \cdots \wedge e_{\alpha_s} \space\space(\alpha_m<l_\alpha<\alpha_n)\\
\implies& w = \sum_{i=1}^rv_{i}\wedge\sum_{\alpha\in P, f_\alpha \neq0,l_\alpha = i}f_{\alpha}(-1)^m\wedge e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_m}\wedge  e_{\alpha_n}\wedge \cdots \wedge e_{\alpha_s} \space\space(\alpha_m<l_\alpha<\alpha_n)
\end{align*}
Might have made a mistake somewhere but the idea should be clear. If you have a notation that you suggest I use for clarity, please feel free to comment!


问:
Characteristic method for function of three variables
I have a function of three variables, and I'm wondering if the method of characteristics can be used to solve the PDE.
Specifically, let $u(t, x, y)$ be a function of three variables satisfying the following PDE:
$$
\frac{\partial u}{\partial t}= f(x,y) \frac{\partial u}{\partial x}+g(x,y)\frac{\partial u}{\partial y},
$$
where $f(x,y)$ and $g(x,y)$ are given functions that do not depend on $t$.

Given boundary data and a characteristic curve $(t, x(t), y(t))$ such that $u(t,x(t),y(t))$ is constant, can an explicit solution for $u$ be found?

答:
$$
\frac{\partial u}{\partial t}= f(x,y) \frac{\partial u}{\partial x}+g(x,y)\frac{\partial u}{\partial y},
$$
$$u_t-f(x,y)u_x-g(x,y)u_y=0$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dt}{1}=\frac{dx}{-f(x,y)}=\frac{dy}{-g(x,y)}=\frac{du}{0}$$
A first characteristic equation comes from solving $\frac{dx}{-f(x,y)}=\frac{dy}{-g(x,y)}$
$$\frac{dy}{dx}-\frac{g(x,y)}{f(x,y)}=0$$
Without knowing explicitly the functions $f(x,y)$ and $g(x,y)$ it isn't possible to say if this ODE is analytically solvable or not.

Supposing that the solution of the ODE can be found on the form :
$$\varphi(x,y)=c_1$$
with $\varphi$ a function explicitly known, then one could continue.

A second characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{-f(x,y(x,c_1))}$

$y(x,c_1)$ supposes that one can solve $\varphi(x,y)=c_1$ for $y$.
$$t+\int \frac{dx}{f(x,y(x,c_1))}dx=c_2$$

Supposing that one can explicitly integrate, then the second caracteristic equation would be :
$$t+\phi(x,c_1)=t+\phi(x,\varphi(x,y))=c_2$$
with $\phi$ a function explicitly known.

If all the above calculus can be carried out, the general solution of the PDE would be $u=F(c_1,c_2)$ :
$$u(t,x,y)=F\big(\varphi(x,y)\:,\:(t+\phi(x,\varphi(x,y)))\big)$$
in which $\varphi$ and $\phi$ are known functions and in which $F$ is an arbitrary function of two variables.

If some valid conditions (initial and/or boundary) are specified, one expect to determine the function $F$ in order to satisfy those conditions.

Thus the answer to your question about the possibility to solve the PDE thanks to the method of characteristics is : It is theoretically possible but not explicitly for all kind of functions $f(x,y)$ and $g(x,y)$.

It is not possible to definitively answer without knowing explicitly what are the functions $f(x,y)$ and $g(x,y)$.


问:
Matrix $BA\neq$$I_{3}$
If $\text{A}$ is a $2\times3$ matrix and $\text{B}$ is a $3\times2$ matrix, prove that $\text{BA}=I_{3}$ is impossible.

So I've been thinking about this, and so far I'm thinking that a homogenous system is going to be involved in this proof. Maybe something about one of the later steps being that the last row of the matrix would be $0\neq \text{a}$, where a is any real number. I've also been thinking that for a $2\times3$ matrix, there is a (non-zero) vector $[x,y,z]$ such that $\text{A}[x,y,z]=[0,0]$ because the dot product could possibly yield $0$. I'm not sure if that's helpful at all though.

Trouble is I'm not really too sure how to continue, or even begin.

Any help would be appreciated.

答: Consider the possible dimension of the columnspace of the matrix $BA$. In particular, since $A$ has at most a two-dimensional columnspace, $BA$ has at most a two-dimensional columnspace. Stated more formally, if $A$ has rank $r_a$ and $B$ has rank $r_b$, then $BA$ has rank at most $\min\{ r_a, r_b \}$.</p>

Title: Matrix $BA\neq$$I_{3}$


问:
symmetric difference function
Let $X$ be a given set and let $A$ be its subset. Define $D$ as a map from the power set of $X$ to itself such that $D(B)=(A \setminus B) \cup (B\setminus A)$. I have already proved that $D$ is injective and I can't seem to conclude if the following statements are true (and if not, one needs to provide a counter example)

D is surjective for all subsets $A$ of $X$.

D is bijective for all subsets $A$ of $X$.

答:
Let $A \triangle B$ denote the symmetric difference of $A$ and $B$. You should convince yourself that $A\triangle \emptyset=A$ and $A \triangle A =\emptyset$, that $A\triangle B= B \triangle A$ and that $A\triangle (B\triangle C)=(A\triangle B) \triangle C$. With these properties, it is clear that $C = A \triangle (A \triangle C)$ for any sets $A$ and $C$. Hence, in the language of your question, for a fixed $A$, Let $D(B)=A\triangle B$. To show that $D$ is surjective, let $C$ be an arbitrary subset of $X$. Then $D(D(C))=C,$ so every $C$ is in the image of $D$.
Bijectivity is just injectivity together with surjectivity.

Title: symmetric difference function


问:
There is no inner product that induces the infinity norm in the space of continuous complex functions in [a,b]
There is no inner product that induces the infinity norm $\|f\|_{\infty}:=\sup\{｜f(x)｜: x \in [a,b] \}$ in the space of continuous COMPLEX functions of domain $[a,b]$, where $[a,b]$ is a closed interval of the real numbers.

I couldn't prove this and I couldn't find counterexamples for the parallelogram rule. Can you provide a proof of this statement?

答: You can take$$f_1(x)=\begin{cases}\frac{\frac{a+b}2-x}{b-a}&\text{ if }x\leqslant \frac{a+b}2\\0&\text{ if }x>\frac{a+b}2\end{cases}\quad\text{and}\quad f_2(x)=\begin{cases}0&\text{ if }x\leqslant \frac{a+b}2\\\frac{x-\frac{a+b}2}{b-a}&\text{ if }x>\frac{a+b}2.\end{cases}$$Then all of the functions $f_1$, $f_2$, $f_1+f_2$, and $f_1-f_2$ have norm $\frac12$. Therefore, the parallelogram law doesn't hold.


问:
On an evaluation of a limit
I want to evaluate
$$\lim_{n \to \infty} n^{3/2}\int_0^1 \frac{x^2}{(1+x^2)^n}\ dx$$

All that I needed is an intergrable control function $g(\cdot)$ independent of $n \in \mathbb{N}$ such that $n^{3/2} \frac{x^2}{(1+x^2)^n}\leq g(x)$, but I do not find direct control function anyway....

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{n^{3/2}
\int_{0}^{1}{x^{2} \over \pars{1 + x^{2}}^{n}}\,\dd x}}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{3/2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n - 1}} -
n^{3/2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{3/2}\int_{0}^{1}\expo{-\pars{n - 1}\ln\pars{1 + x^{2}}}\,\,\,\dd x -
n^{3/2}\int_{0}^{1}\expo{-n\ln\pars{1 + x^{2}}}\,\,\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{3/2}\int_{0}^{\infty}\expo{-\pars{n - 1}x^{2}}\,\,\,\dd x -
n^{3/2}\int_{0}^{\infty}\expo{-nx^{2}}\,\,\,\dd x}\
\pars{\substack{\ds{Laplace's}\\[0.5mm] \ds{Method}}}
\\[5mm] = &\
\lim_{n \to \infty}\pars{%
n^{3/2}\,{\root{\pi} \over 2}{1 \over \root{n - 1}} -
n^{3/2}\,{\root{\pi} \over 2}{1 \over \root{n}}}
\\[5mm] = &\
{\root{\pi} \over 2}\lim_{n \to \infty}\pars{%
n^{3/2}\,{\root{n} - \root{n - 1} \over \root{n - 1}\root{n}}}
\\[5mm] = &\
{\root{\pi} \over 2}\lim_{n \to \infty}\bracks{%
{n^{3/2} \over
\root{n - 1}\root{n}\pars{\root{n} + \root{n - 1}}}}
\\[5mm] = &\
{\root{\pi} \over 2}\lim_{n \to \infty}\bracks{%
{1 \over
\root{1 - 1/n}\pars{1 + \root{1 - 1/n}}}}
\\[5mm] = &\
\bbx{\root{\pi} \over 4} \approx 0.4431 \\ &
\end{align}

Title: On an evaluation of a limit


问:
Why is the multiplicative group of a field an algebraic group?
Reading from here: https://en.wikipedia.org/wiki/Algebraic_torus

In particular, the paragraph under the heading "Multiplicative group of a field."

So, in my mind, a multiplicative group of a field is denoted $F^\times$ and is just the group $(F \backslash \{0\}, \times)$ where $\times$ is the multiplicative operation in $F$. So, this multiplicative group is an algebraic group? That means it is also an affine variety: a set of solutions of a system of polynomial equations. What is this system of polynomial equations, is it $\{f-x : f \in F^\times \}$?

Also, the next sentence says that the multiplicate group of $F$ is such that for any field extension $E \backslash F$ the $E$-points are isomorphic to the group $E^\times$.

Are $E$-points those points that are in $E \backslash F$? What does it mean that the $E$-points are isomorphic to $E^\times$?

I'm realizing I asked a few different questions here... If anyone could answer anything it would be greatly appreciated!! Thanks, I don't know what I'd do without this website!

答:
The system of polynomial equations is $xy = 1$ in $\mathbb{A}^2$. The system you wrote down has no solutions as soon as $F^{\times}$ has at least two elements.

There are several different ways of thinking about algebraic varieties, at different levels of abstraction and requiring different levels of background, and the differences between them matter when the ground field is not algebraically closed. The simplest way to say it is that if you have a variety defined by some equations over a field $F$, then you can consider the solutions to those equations over any field extension $E$ of $F$; these are called the $E$-points of the variety. We need to do this sort of thing because a variety over a non-algebraically closed field $F$ need not have any $F$-points; consider, for example, the variety $x^2 + y^2 + 1 = 0$ over $\mathbb{R}$, which has no $\mathbb{R}$-points but does have $\mathbb{C}$-points. This is also a very natural thing to do in number theory, taking $F = \mathbb{Q}$ and $E$ some number field for example.

If the variety is an algebraic group $G$ then the solutions to the defining equations over any field extension $E$ form a group denoted $G(E)$. If $G$ is the multiplicative group, then $G(E) \cong E^{\times}$, since the set of solutions to $xy = 1$ in $\mathbb{A}^2(E)$ is the set of pairs of points $(x, x^{-1})$ where $x \in E^{\times}$ (and the multiplication is pointwise so it's the same as the ordinary multiplication in $E^{\times}$).

There's a more abstract and more powerful way to say this using the language of group schemes and functors of points also, which has the benefit of not requiring a specific choice of equations and generalizing from fields to arbitrary commutative rings.


问:
What are the current research areas in Differential Topology?
I am currently in my 4th year of a 5-year BS-MS course. I am majoring in Mathematics and I am quite interested in differential topology and would really appreciate if someone could tell me about possible areas of research in the field (With resources like books, survey papers etc. if possible).

As far as my background is concerned, I have done courses on Point-set topology, Calculus on Manifolds and Algebraic Topology and have dabbled a bit with 3-manifolds (Schultens' book titled "Introduction to 3-Manifolds") as well. I have studied "Differential Topology" by Guillemin and Pollock and was thinking of going towards Morse Theory after this.

答:
Here are some:

Symplectic topology

Contact topology

Low-dimensional topology (dimensions 3 and 4)

Gauge theory (primarily in low dimensions) and gauge-theory inspired invariants.

But one needs an advisor familiar with these fields in order to succeed in any of these areas. (Yes, there are exceptions to this rule, but very few.)


问:
Using Fast Fourier Transform (FFT) to calculate a polynomial and its derivatives at a specific point $x_0$
Those are my homeworks so i prefer a hint and not a full answer

I need to use FFT to calculate a polynomial and its derivatives at a certain point $x_0$ at time complexity $O(n\log n)$

Now, i saw this answer:
https://cs.stackexchange.com/questions/24314/how-to-evaluate-all-derivatives-of-a-polynomial-at-a-point-with-fft

Yet, i dont understand it.

It says that i can conclude from the fourier transform of the polynomial, on the fourier transform of its derivatives.

Therefore i tried to take fourier transform of a polynomial: $P(x) = -1 + 2x + 5x^2 - 4x^3$ and its $4$ derivatives.

I get: (Did it using Octave)

I dont see any correlation between the FFT of the original polynom and the FFT of its derivatives.

And i dont know how to proceed with the question, I'm stuck.

Thank you.

答:
try to get in $\Theta(n\,log n)$ the convolution of  $(1,x,\frac{1}{2!} x^2,\frac{1}{3!} x^3,...,\frac{1}{n!} x^n )$ and $(n!a_n,(n-1)! a_{n-1},...,a_0)$ when  $a_i$ is the $i-th$ coefficient of the Polynomial.

example:

for $P(x)= −1+2x+5x^2−4x^3$ when $x=1$ you'll get $(1,1,\frac{1}{2!} ,\frac{1}{3!}) (-24,10,2,-1)=(-24,-14,0,2,...)$ and those are the derivatives of $P(x)$ at $1$.


答:
To see how I found the assignment that falsifies the converse,
$$\Bigl((B\to C)\to (A\to C)\Bigr)\to (A\to B),$$
recall that an implication $P\to Q$ is false when $P$ is true and $Q$ is false. So to (try to) falsify this implication, we want $A\to B$ to be false, which means that we must have $A$ true and $B$ false.

Once you have $A$ true and $B$ false, the implication $(B\to C)$ will be true regardless of the truth value of $C$, so to make the antecedent true we just need $A\to C$ to be true as well, which we can achieve by letting $C$ be true. Thus, $A$ and $C$ true and $B$ false will make this proposition formula false: the consequent $A\to B$ is false; the antecedent is true because its consequent $A\to C$ is true (because its consequent $C$ is true).

Title: Converse of $(A\rightarrow B)\rightarrow((B\rightarrow C)\rightarrow(A\rightarrow C))$

Body:
I have a basic question about the following proposition.

$(A\rightarrow B)\rightarrow((B\rightarrow C)\rightarrow(A\rightarrow C))$

I can prove it in intuitionistic logic. But I wonder if we have the converse in intuitionistic logic? Or in classical logic?

$((B\rightarrow C)\rightarrow(A\rightarrow C))\rightarrow(A\rightarrow B)$

Thanks!


问:
Types of possible moves of tiling puzzle pieces
This is a question about tiling puzzle jargon. What are the types of moving a tiling piece? I mean the single word or a phrase for rearrangement move of a piece. I would be grateful for proper mathematical names as well as contemporary English names. Suppose we are talking about 2D puzzle that lay on the table.

Sliding the item along the plane on the table without picking it up off the table, without rotating it. Translate.

Rotate the piece clockwise or counterclockwise without picking it up from the table Rotate or Twist.

Flip the piece to the other side, just like a pancake. It is necessary to take the piece off the table to do that. Here I found some inconsistency. It is also called Twist as in twist-hinged dissection. Flip sounds good but I have not found it in use. Instead, I found Reflect - reflecting the item in the plane as if being viewed in a mirror.

Scale up or down. This transformation is called dilation although I am not sure if both ways up and down.

I especially ask for the third type of move.

Reference:

http://www.mathpuzzle.com/Tessel.htm

Move a polygon to a specified position using only allowed rotations, reflections, and dilations

https://www.cs.purdue.edu/homes/gnf/book2.html

答:
If you wish for the pieces to stay the same size and shape (which precludes the dilation), then the motion taking one piece to another is an isometry of the plane. These can be classified into four types:

Translation, as you mentioned.

Rotation about a point, again pretty self-explanatory.

Reflection about a line, which roughly corresponds to the "flip" you describe.

Glide reflection, which consists of a reflection followed by a translation along the line of reflection, and is shown in this diagram from Wikipedia:

A similar classification can be performed in higher dimensions, though the number of possible motions increases; a list for three-dimensional Euclidean space is given here.


问:
How to identify characters
Suppose I have a compact connected Lie group $G$, and a smooth class function $f : G \to \mathbb{C}$. Suppose that $\| f \|_{L^2(G)} = 1$ and $\Delta f = - \kappa_\pi f$ where $\Delta$ is the Laplace-Beltrami operator on $G$, $\kappa_\pi$ is the Casimir of the irreducible representation $\pi$. I would like to conclude that $f = \chi_\pi$ (the character of the irreducible representation $\pi$). However there is a problem: it may be that there are inequivalent irreducible representations $\pi \neq \pi^\prime$ where $\kappa_\pi = \kappa_{\pi^\prime}$. Then any linear combination $f = a  \chi_\pi + b \chi_{\pi^\prime}$ for which $|a|^2 + |b|^2 = 1$ would satisfy the above requirements but $f$ would not be a character. One could also consider more general linear combinations. In certain cases like $U(1)$ and $SU(2)$ I believe the eigenvalues of $\Delta$ are non-degenerate and so this issue doesn't arise, however I don't think this is true in general.

Is there any analytical property that can be used to identify characters? (By "analytical" I mean something you could apply purely at the level of functions.) One thing you could do is evaluate at the identity. If $f(\mathrm{Id})$ isn't a non-negative integer then it can't be a character. However this probably isn't sufficient.

答:
I wonder if this might be of any help, but characters may be characterized as follows.

A necessary and sufficient set of conditions for a function $\phi:G\to\mathbb C$ to be a nonegative multiple of a character is:

$\phi$ is a continuous class function,

$\phi$ is positive semi-definite,

If $\psi$ is any continuous class function such that
$$0\leq \psi \leq \phi,$$
in the sense that $\psi$ and $\phi-\psi$ are positive semi-definite, then there exists a nonnegative constant $\lambda$ such that $\psi=\lambda \phi$.

EDIT.  Below I will try to outline a justification of the assertion made above.

If $G$ is a compact Lie group (actually,  for the time being, all we  need is that  $G$ is a locally compact topological group) then $C_c(G)$,  that
is,  the space of all compactly supported,  continuous, complex valued functions on $G$, is a complex $^\ast$-algebra with
convolution for multiplication, namely
$$
  (f\star g)(t) = \int_Gf(s)g(s^{-1}t)\, ds
  $$
(the integral being taken with respect to Haar measure)
and involution given by
$$
  f^\ast(t) = \overline{f(t^{-1})}.
  $$

This algebra has a canonical norm, and the completion, denoted $C^\ast(G)$, is called the group C$^\ast$-algebra of $G$.

For every (strongly continuous) representation $\pi $ of $G$ on a Hilbert space $H$, one can define the so called integrated
form of $\pi $, which is  a representation $\Pi$ of $C^\ast(G)$ on the same Hilbert space $H$.  The description of
$\Pi$ is pretty explicit on $C_c(G)$:  for every $f$ in this dense subalgebra   one has
$$
  \Pi(f) = \int_G f(t)\pi(t)\, dt,
  $$
where the integral is taken with respect to the weak operator topology.  Once $\Pi$ is defined on $C_c(G)$,  as above,
it may be extended to  $C^\ast(G)$ by continuity.

A key result of Harmonic Analysis states that every representation of $C^\ast(G)$ arises as the integrated form of some
representation of $G$ and in fact  the correspondence $\pi\to\Pi$ is a bijection between the set of
representations of $G$ and those of $C^\ast(G)$.

Regarding characters, states, and all that, if $\phi:G\to \mathbb C$ is a positive semi-definite function, one may also  define an
integrated form
$$
  \Phi:C_c(G) \to \mathbb C,
  $$
by the formula
$$
  \Phi(f) = \int_G f(t)\phi(t)\, dt.
  \tag {$\dagger$}
  $$
One may prove that $\Phi$ extends to $C^\ast(G)$ by continuity, providing a positive linear functional,  namely one that
satisfies
$\Phi(a^\ast a)\geq 0$,
for all $a$ in $C^\ast(G)$.

As before, ($\dagger$) provides a  bijective correspondence between positive semi-definite functions on $G$ and positive linear functionals on
$C^\ast(G)$.

The correspondence $\phi\to\Phi$  may be shown to map class functions to traces (a trace on $C^\ast(G)$ is a linear
functional $\Phi$ satisfying $\Phi(ab)=\Phi(ba)$).

When $G$ is compact, the Peter-Weyl Theorem essentially says that
$$
  C^\ast(G)=\bigoplus_{\pi\in\hat G}M(\pi),
  $$
(a $c_0$-direct sum) where $\hat G$ is the set of equivalence classes of irreducible representations of $G$, and each
$M(\pi)$ is the algebra of $n\times n$ complex matrices, $n$ being the dimension of $\pi$.

If $\phi$ is the character of $G$ associated to a given irreducible representation $\pi$, that is $\phi(g)=\text{tr}(\pi (g))$,
then one may show that the corresponding positive linear functional $\Phi$ is given by
$$
  \Phi\big ((a_\rho )_{\rho \in \hat G}\big ) = \text{tr}(a_\pi ).
  $$

Addressing the characterization proposed in the first part of this answer,
assume  that $\phi$ and $\Phi$ are as above, and that
$\psi$ is a positive semi-definite continuous class function on $G$, such that $\psi\leq\phi$.  Letting $\Psi$ be the
integrated form of $\psi$, we then have that $\Psi$ is a positive trace on $C^\ast(G)$ with $\Psi\leq\Phi$.
As $\Phi$ vanishes on every $M(\rho)$, with $\rho\neq\pi$, the same must be true for $\Psi$, and hence $\Psi$ factors through
a trace on $M(\pi)$.  However,  as we all know, matrix algebras admit a unique trace up to scalar multiples, so  it follows
that
$\Psi$ is a multiple of $\Phi$ which in turn implies  that   $\psi$ is a multiple of $\phi$, as desired.

Conversely, assuming that $\phi$ satisfies conditions (1--3) above, its  integrated form $\Phi$ is a trace,  so its
restriction  to each component
$M(\pi)$ must be a nonnegative scalar multiple,  say $\lambda_\pi$, of the standard trace on $M(\pi)$.  Therefore
$$
  \Phi\big ((a_\pi )_{\pi \in \hat G}\big ) = \sum_{\pi\in\hat G}\lambda_\pi\text{tr}(a_\pi ).
  $$

Picking any $\pi$ such that $\lambda_\pi\neq0$, observe that the character $\phi_\pi$
associated to $\pi$ satisfies
$\lambda_\pi\phi_\pi\leq\phi$ as a consequence of the fact that  their integrated forms obey that inequality.  The conclusion then
follows from point (3).

Title: How to identify characters


问:
"Product" topology of $\mathbb N\times \mathbb R$
Background: Consider a set that is a finite or countable set (for example $\mathbb N$) endowed with discrete topology

Question: What topology is endowed with $\mathbb N\times \mathbb R$? Is $\mathbb N\times\mathbb R$ connected?

Is it possible to define a reasonable topology such that $\mathbb N\times\mathbb R$ is connected?

Motivation: "Connectness" is a useful property that is a necessary condition for many theorems. In order to apply those theorems in discrete space $\mathbb N$, one may want try to "connectify" the discrete space, for example, by "put it together" with $\mathbb R$.

My guess: $\mathbb N\times\mathbb R$ is not connected by its standard product topology. Consider set $\{1\}\times\mathbb R$ and $(\mathbb N\setminus \{1\})\times\mathbb R$, both of those sets are open by the standard product topology as $\mathbb R$ and $\{1\}$ are both open sets.

In order to make $\mathbb N\times\mathbb R$ connected, can we force define that the whole set $\mathbb R$ to be neither closed nor open? Intuitively, take a closed interval $[0,1)$, then $\mathbb N\times [0,1)$ can only be connected?

Additionally, I am not sure if things like $\mathbb N\times \mathbb R$ is connected. Any comments will help!

答: The easiest way to make N×R connected is to give N the indiscrete topology.</p>


问:
Prove that running time T(n)=$(n + a)^b$ is $O(n^b)$ with a and b > 0
Image of the question

I'm studying asymptotic notation, but i'm stuck at this question, can anyone help me?

答:
$$T(n)=(n + a)^b \leqslant (n + n)^b=2^bn^b$$
when $\lfloor a \rfloor+1<n$, so we can take any $C\geqslant n^b$ and $N=\lfloor a \rfloor+1$.


问:
Prove that $\dim(T(H))\leq \dim(H)$.

Let $V$ and $W$ be finite-dimensional vector spaces and $T$ be a linear transformation $T:V\to W$.

Let $H$ be a non-zero subspace of $V$, and let $T(H)$ be the set of images of vectors in $H$. Prove that $\dim(T(H))\leq \dim(H)$.

Proof:

Let $B=\{b_1,...,b_n\}$ be a basis for $H$.

Because $B$ is a basis and $T$ is linear, for any $x\in T(H)$, we have

$$x=T(c_1b_1+...+c_nb_n)=c_1T(b_1)+...+c_nT(b_n)$$

Therefore, $\text{span}\{T(b_1)...T(b_n)\}=T(H)$.

If $\{T(b_1)...T(b_n)\}$ is linearaly independent, then $\text{dim}(T(H))=\text{dim}(H)$ and we are done.

Otherwise, $\{T(b_1)...T(b_n)\}$ is linearaly dependent, and there exist $T(b_i)$ that can be expressed as a linear combination of the other vectors in this set.

So let $T(b_i)=d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)$.

Now let $y\in T(H)$,then
$$y=c_1T(b_1)+..+c_iT(b_i)+..+c_nT(b_n)=c_1T(b_1)+...+d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)+...+c_nT(b_n)=(c_1+d_1)T(b_1)+..+(c_{i-1}+d_{i-1})T(b_{i-1})+(c_{i+1}+d_{i+1})T(b_{i+1})+...+(c_n+d_n)T(b_n)$$

So, $\text{span}\{T(b_1)...T(b_n)\}=\text{span}\{T(b_1)...T(b_{i-1}),T(b_{i+1}),...,T(b_n)\}=T(H)$

If this new set is linearly independent then $\text{dim}(T(H))=n-1<\text{dim}(H)$.

Otherwise, we are linearly dependent and a vector can be removed without changing the span. If we get down to one remaining vector and that vector is not the $0$ vector then  $\text{dim}(T(H))=1<\text{dim}(H)$. Otherwise, it is the $0$ vector and $\text{dim}(T(H))=0<\text{dim}(H)$.

Thank you for reading the proof. Please provide feedback on the validity of my proof. Have I assumed anything I shouldn't have assumed? Any leaps in logic? I appreciate any tips suggestions and insights.

答:
Your proof looks correct to me. Here is a much shorter proof.

Let $\{\xi_1, \dots, \xi_n\}$ be a basis of  $H$. Then $\{T\xi_1, \dots, T\xi_n\}$ spans $T(H)$. Hence, $\dim T(H) \leq  n = \dim(H)$.

Basically, my proof relies on the following fact, which you probably know:

If the vector space $V$ is linearly spanned by a set of vectors $E\subseteq V$, then
there exists a subset $F \subseteq E$ such that $F$ is a basis for
$V$.

Title: Prove that $\dim(T(H))\leq \dim(H)$.


问:
Function satisfying CR equations but not differentiable
Consider $f(z)=||z|^2 -1|^2$. Prove that $f$ is differentiable at 0 and 1 only.

I find that the Cauchy Riemann equations are satisfied at the origin and at all points on the unit circle. I am able to prove differentiability at the origin. But I am not able to prove differentiability at 1 and non differentiability at all points $z\ne1$ on the unit circle. Please help.

答:
As comments have already pointed out, the the function you wrote is $\mathbb{C}$-differentiable at all points on the unit circle. This is fairly straight-forward to check using the limit definition. Let $w$ be a point on the unit circle, so $|w|=1$. Then $|z-w|\geq ||z|-|w||=||z|-1|$ by reverse-triangle inequality. Thus
$$\frac{||z|-1|}{|z-w|}\leq 1.$$
Since $f(z) = |(|z|-1)(|z|+1)|^2$ and $f(w) = 0$ we have
$$\left|\frac{f(z)-f(w)}{z-w}\right| = \frac{|(|z|-1)(|z|+1)|^2}{|z-w|}\leq ||z|-1|\cdot(|z|+1)^2.$$
Since the right-hand side tends to $0$ as $z\to w$, we conclude by squeeze theorem that $f$ is $\mathbb{C}$-differentiable at $w$ and $f'(w)=0$.


答:
Why forcing properties on the topology of $\mathbb R$ when $\mathbb R$ is already connected?

The obstruction in $\mathbb N\times \mathbb R$ connection is clearly the fact that $\mathbb N$ itself is not connected (which is what you used to prove that the product is not connected, i.e., being $\{1\}$ open in $\mathbb N$, you conclude that $\{1\}\times\mathbb R$ is open in $\mathbb N\times\mathbb R$), therefore I'd say that the easiest way to get to a connected topology on $\mathbb N\times \mathbb R$ is requiring a connected $\mathbb N$ (for instance, giving it the trivial topology, i.e. $\{\varnothing,\mathbb N\}$).

If your question requires us to keep the discrete topology on $\mathbb N$, I'm afraid that changing the topology of $\mathbb R$ won't be much of a help. You could always take a clopen set of $\mathbb N$ (e.g., $\{1\}$), and its product with any other topological space will still be a clopen in the product topology.

Title: "Product" topology of $\mathbb N\times \mathbb R$

Body:
Background: Consider a set that is a finite or countable set (for example $\mathbb N$) endowed with discrete topology

Question: What topology is endowed with $\mathbb N\times \mathbb R$? Is $\mathbb N\times\mathbb R$ connected?

Is it possible to define a reasonable topology such that $\mathbb N\times\mathbb R$ is connected?

Motivation: "Connectness" is a useful property that is a necessary condition for many theorems. In order to apply those theorems in discrete space $\mathbb N$, one may want try to "connectify" the discrete space, for example, by "put it together" with $\mathbb R$.

My guess: $\mathbb N\times\mathbb R$ is not connected by its standard product topology. Consider set $\{1\}\times\mathbb R$ and $(\mathbb N\setminus \{1\})\times\mathbb R$, both of those sets are open by the standard product topology as $\mathbb R$ and $\{1\}$ are both open sets.

In order to make $\mathbb N\times\mathbb R$ connected, can we force define that the whole set $\mathbb R$ to be neither closed nor open? Intuitively, take a closed interval $[0,1)$, then $\mathbb N\times [0,1)$ can only be connected?

Additionally, I am not sure if things like $\mathbb N\times \mathbb R$ is connected. Any comments will help!