问:
Which polygon tile grids allow convex polygons to be formed from multiple tiles?
If I have a grid made of equilateral triangles, I can easily form larger convex polygons as a set of tiles in that grid. I believe this holds for some (but not all) tilings of non-equilateral triangles.

The same for quadrilaterals - it's obvious for squares, rectangles and parallelograms, but I believe it also holds for some other (but not all) tilings of quadrilaterals.

I'm pretty sure tiling a flat area purely with convex pentagons is impossible, so skip that case.

With hexagons, a tiled grid is possible, but the only way to make a convex polygon from those convex hexagonal tiles seems to be to use only one tile - no multi-tile convex polygons seem possible.

My speculation is that it's only possible to form a multi-tile convex polygon from convex polygon tiles if all those tiles have interior angles at all vertices of 90 degrees or less, at least for those vertices that are at a vertex or edge of that larger convex polygon.

Is that speculation correct? Is there a proof or disproof?

答:
Sorry - I figured it out within a minute of posting. Should have thought longer before asking.

If two polygons are joined such that such that a vertex and edge are on the same point and line, there are two cases...

If the sum of the two interior angles is greater than 180 degrees, this is a concavity. It's only valid as part of a larger convex polygon if that concavity is filled by more tiles.

If the sum of the two interior angles is less than or equal to 180 degrees, there is no concavity. This vertex may be at the vertex or edge of a larger convex polygon formed from these tiles.

So the issue is the sum of the angles, not the interior angle at the vertex of one particular tile.


问:
Use the Axiom of Regularity to prove $(A,B)\neq A$ and $(A,B)\neq B$
Attempted proof: Let $(A,B)=\{ \{A \},\{A,B \}\}$ be the ordered pair of A and B. Using double contention, for A to equal (A,B), there must happen that $A\subseteq(A,B)$ and $(A,B)\subseteq A$. But the second contention would imply $\{A\} \in A$ and thus $\{A\}\cap A \neq \emptyset$, which contradicts the axiom of regularity.

And that would be it. I'm confused by the notation: is the last part a correct implication of $\{A\} \in A$? What approach could there be for it to improve? Where can I start to get the second part going since $\{A\}\not\in B$? Should I use $\{A,B\}\in B$ and disprove it by the Axiom of Regularity?

Thanks a lot.

答:
The axiom of regularity implies that for any two sets $X,Y$, if $X\in Y$, then $Y\not\in X$ (can you show this?).  This is the simplest fact to use here.  Given that result, proceed to prove the desired inequalities by contradiction:

First, suppose that
$(A,B)\subseteq A$.
Then every element of $(A,B)$ is an element of $A$.  In particular, $\{A\}\in A$; since $A\in\{A\}$ as well, this is a contradiction.  We conclude that $(A,B)\not\subseteq A$ and hence that $(A,B)\neq A$.

Second, suppose that
$(A,B) \subseteq B$.
Then every element of $(A,B)$ is an element of $B$. In particular, $\{A,B\}\in B$; since $B\in\{A,B\}$ as well, this is also a contradiction.  We conclude that $(A,B)\not\subseteq B$ and hence that $(A,B)\neq B$.

ContentLicense: CC BY-SA 4.0


问:
About partition the real number
I know this a very simple. I want to make sure. Let $\mathcal P=\{P_\xi\colon \xi<\mathfrak c\}$ be a set of pairwise disjoint perfect sets such that $$D:=\bigcup\mathcal P$$ is meager set. So, $R\setminus D$ is co-meager and has cardinality continuum. Coonisder $$\mathcal F=\mathcal P \cup\{\{x\}\colon x\in\mathbb R\setminus D\}$$ note that $\mathcal F$ is a partition of $\mathbb R.$ So, we can enumerate $\mathcal F$ as follows $$\mathcal F=\{ A_\eta\colon \eta<\mathfrak c\}$$ Here is my easy question , every set in $\mathcal F$ even is a perfect set or a singleton. Is that right?

答: Yes, you have a partition of $\mathbb{R}$ into continuum-many perfect sets and continuum-many singletons, where the union of the perfect sets involved is meager.

Title: About partition the real number


问:
Poisson distribution question: Company XYZ provides a warranty on a product that it produces...

Company XYZ provides a warranty on a product that it produces. Each year, the number of warranty claims follows a Poisson distribution with mean c. The probability that no warranty claims are received in any given year is 0.60.

Company XYZ purchases an insurance policy that will reduce its overall warranty claim payment costs. The insurance policy will pay nothing for the first warranty claim received and 5000 for each claim thereafter until the end of the year.

Calculate the expected amount of annual insurance policy payments to Company XYZ.

My attempt:

Let $Y$ be the RV denoting the amount of annual insurance policy payments to Company XYZ and $X$ be the number of annual warranty claims. Then,
$$Y=5000(X-1) \;\; \text{if}\;\; X>1$$

and $0$ otherwise.
$$\begin{align}
E[Y] &= 5000 \cdot E[X-1\mid X>1] \\
&= 5000 \cdot \sum_{x=2}^\infty (x-1) \cdot P[X=x-1]\\
&=5000 \cdot \sum_{x=1}^\infty (x) \cdot P[X=x]\\
&=5000[E(X)-0\cdot P[X=0]]\\
&=5000[E(X)]
\end{align}$$

Note that $P(X=0) = 0.6 \implies c= -\ln 0.6$. So, $E[Y]=5000[-\ln 0.6].$
This is not the correct answer, however. Can someone please point out where I went wrong?

答:
Number of claims is $X \sim \mathsf{Pois}(c),$ so
$P(X=0) = e^{-c} = 0.6,$ which implies $c = -\log_e(.6) = 0.511.$ So $E(X) = 0.5108.$

```<code>c =-log(.6); c
[1] 0.5108256
</code>```

Because of XYZ can't collect on the first claim,
you can't multiply $c$ by \$5000 to get the average
annual payout as $5000c.$

There may be some years with only
one claim so XYZ collects nothing, and there  may be years with several valid claims so thqt
XYZ collects from the insurance company.

Here is a strictly computational method.
Technically, Poisson probabilities extend from $0$ to $\infty.$ But terms decay rapidly to 0, Let's use the total number of claims for illustration.
You can come
extremely close to $E(X)$ by summing only the first 100 terms of the infinite series,

```<code>k = 1:99;  s = sum(k*dpois(k, -log(.6)))
s
[1] 0.5108256  # exactly E(X) to many places
</code>```

Now we find the expected number $Y$ of claims with payouts. We seek $E(Y) = 0.1108$ so that we can give the expected annual payout of $5000\,E(Y)\approx \$554$.

We let the 100-vector j represent the number
of paid claims and the 1000-vector i be the number of claims.

```<code>j = c(0,0,1:98);  i = 0:99
  5000*s
[1] 0.1108256  # E(Y) = avg nr. payable claims
[1] 554.1281   # 5000E(Y) = avg annual payout
</code>```

Note: in case you're wondering whether it is sufficient to sum the first 100 terms of the
series, the first 50 would have been more than enough:

```<code>j = c(0,0.1:48); i = 0:48
sum(j*dpois(i, -log(.6)))
[1] 0.1508256
</code>```

Almost all of the probability of the distribution $\mathsf{Pois}(-log(.6))$ is below 6.

```<code>qpois(.9999, -log(.6))
[1] 5
</code>```

Body:
Number of claims is $X \sim \mathsf{Pois}(c),$ so
$P(X=0) = e^{-c} = 0.6,$ which implies $c = -\log_e(.6) = 0.511.$ So $E(X) = 0.5108.$

c =-log(.6); c
[1] 0.5108256

Because of XYZ can't collect on the first claim,
you can't multiply $c$ by \$5000 to get the average
annual payout as $5000c.$

There may be some years with only
one claim so XYZ collects nothing, and there  may be years with several valid claims so thqt
XYZ collects from the insurance company.

Here is a strictly computational method.
Technically, Poisson probabilities extend from $0$ to $\infty.$ But terms decay rapidly to 0, Let's use the total number of claims for illustration.
You can come
extremely close to $E(X)$ by summing only the first 100 terms of the infinite series,

k = 1:99;  s = sum(k*dpois(k, -log(.6)))
s
[1] 0.5108256  # exactly E(X) to many places

Now we find the expected number $Y$ of claims with payouts. We seek $E(Y) = 0.1108$ so that we can give the expected annual payout of $5000\,E(Y)\approx \$554$.

We let the 100-vector j represent the number
of paid claims and the 1000-vector i be the number of claims.

j = c(0,0,1:98);  i = 0:99
  5000*s
[1] 0.1108256  # E(Y) = avg nr. payable claims
[1] 554.1281   # 5000E(Y) = avg annual payout

Note: in case you're wondering whether it is sufficient to sum the first 100 terms of the
series, the first 50 would have been more than enough:

j = c(0,0.1:48); i = 0:48
sum(j*dpois(i, -log(.6)))
[1] 0.1508256

Almost all of the probability of the distribution $\mathsf{Pois}(-log(.6))$ is below 6.

qpois(.9999, -log(.6))
[1] 5


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

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

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

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

答:
Don't rush to square equations to get rid of square roots.  It's almost always more work and you end up missing opportunities for simplification.  Instead eliminate square roots by substitution.  Here, let $a\ge0$ and $b\ge0$ be the values of the two square roots:

$$
\begin{cases}
x+y=23\\
a+b=33\\
a^2 = x^2 + 12y\\
b^2 = y^2 + 12x
\end{cases}
$$

While this turns your two equations into four equations, it makes simplifications easier to spot.  Subtracting the last two equations will let you factor $a^2 - b^2$ and $x^2 - y^2$ and make use of the fact that $x+y$ and $a+b$ are known:
$$33(a-b) = 23(x-y) - 12(x-y)$$
$$3(a-b) = x-y$$

Now this becomes very simple.  Let $z = a-b$ and solve for $z$.   For example, express $x$, $y$, $a$ in terms of $z$ and substitute into $(a-x)(a+x) = 12y$.  Everything will neatly cancel out and you will get $z^2 = 1$ or $a=b\pm1$ and $x=y\pm3$.  The two solutions are $x=10, y=13$ and $x=13,y=10$


问:
On algebraic integers and minimal polynomials with roots that satisfy $\prod_{i=1}^n (1-\alpha_i^s) = \pm 1$
This question in inspired on a particular case of this post, which is:

If $A$ is an integer coefficient matrix such that $I -A^s \in \mathsf{GL}_n(\mathbb{Z})$ for all $s$, is $A$ nilpotent?

As stated there, if $f = \chi_A = \prod_{i=1}^n(X-\alpha_i)$ the hypothesis is precisely that

$$
\prod_{i=1}^n(1-\alpha_i^s)  = \pm 1 \tag 1
$$

for all $s \geq 1$. Or noting $f_m = \prod_{i=1}^n(X-\alpha_i^m)$, that $f_s(1) = \pm 1 $ for all $s$.

In general, we know that for any $g \in \mathbb{Z}[X]$ the coefficients of $g_m$ are integers, hence factorizing $f$ in $\mathbb{Z}[X]$ into irreducible factors $p_1,\ldots, p_r$ we get $f_m = (p_1)_m \cdots (p_r)_m$ and thus $(p_j)_s(1) = \pm 1$ for all $j$ and $s$.

Hence we can assume $f$ to be irreducible; $f = m(\alpha, \mathbb{Q})$ for some algebraic integer $\alpha$, recovering the question in the title:

If $\alpha \in \mathbb{C}$ is an algebraic integer and $f =  m(\alpha, \mathbb{Q})$ is such that $f_s(1) \in \mathbb{Z}^\times$ for all $s \geq 1$, is $f = X$?

In the linked post, an argument using Dirichlet approximation and taking limits for a subsequence of $(1)$ shows (if correct) that $f = X$, but I am asking whether there is an algebraic proof of this, and more importantly, if there is an available reference for such a result.

答:
Let $K = \mathbf{Q}(\alpha)$ where $\alpha$ is a root of $f(x)$.
By assumption $f(x) \ne x$ so $\alpha \ne 0$. Let $\mathcal{O}$ be the ring of integers of $K$.  Since $\alpha \ne 0$, it only lies in finitely many prime ideals of $\mathcal{O}$.  Let $\mathfrak{p}$ be a prime ideal in $\mathcal{O}$ not containing $\alpha$ of residue characteristic $p$, and let $\mathbf{F}_q = \mathcal{O}/\mathfrak{p}$ be the residue field. Since $\alpha \ne 0 \bmod \mathfrak{p}$, it follows that $\alpha^{q-1} \equiv 1 \bmod \mathfrak{p}$ and so $1-\alpha^m \equiv 0 \bmod \mathfrak{p}$ if $m$ is divisible by $q-1$. But then

$$N_{K/\mathbf{Q}}(1 - \alpha^m) = \prod (1 - \alpha^m_i) \equiv 0 \bmod p$$

for any $m$ divisible by $q-1$.


问:
Relationship between the modified Bessel function of the second kind and the Psi function?
While reading this paper this paper, I was stumble upon a strange relationship between the modified Bessel function of the second kind and the Psi function that is:

$c{K_1}\left( c \right) \approx 1 + \frac{{{c^2}}}{2}\left( {\ln \left( {\frac{c}{2}} \right) + {{\text{C}}_0}} \right) + \frac{{{c^4}}}{{16}}{{\text{C}}_1}$

Where ${C_0}$= −φ(1) /2 − φ(2) /2 and C1 = −φ(2)/2 − φ(3)/2

The paper said that φ(•) denotes the Psi function and point a citation to

. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals Series and Products, New York, NY, USA:Academic, 2007.

Which yield the following definition and integral representation

Please help me understand where this relationship come from ?

Also, is it possible to extent this approximation for a modified Bessel function of the second kind of order $M$ like this ${K_M}\left( {2\sqrt x } \right)$ ?

答:
The DLMF (eq. 10.3.1) provides the following sum formula for $K_n(x)$ with nonnegative integer $n$:

\begin{align}
K_n(x)
&=\frac12 \left(\frac{x}{2}\right)^{-n}\sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(-\frac{x^2}{4}\right)^k+(-1)^{n+1}\ln\left(\frac{x}{2}\right)I_n(x)\\
&\qquad\quad\quad +(-1)^n \frac12 \left(\frac{x}{2}\right)^n \sum_{k=0}^\infty\left[\psi(k+1)+\psi(n+k+1)\right]\frac{(x^2/4)^k}{k!(n+k)!}
\end{align}

For the case of $n=1$, this reduces to
$$
K_1(x)
=\frac1x+\ln\left(\frac{x}{2}\right)I_1(x)
 - \frac{x}{4} \sum_{k=0}^\infty\left[\psi(k+1)+\psi(n+k+1)\right]\frac{(x^2/4)^k}{k!(k+1)!}
$$

Since $I_1(x)\approx x/2$ for small $x$, we find

\begin{align}
x K_1(x) &\approx
 1+\ln\left(\frac{x}{2}\right)\cdot \frac{x^2}{2}-\frac{1}{4}(\psi(1)+\psi(2))x^2-(\psi(2)+\psi(3))\frac{x^4}{32}\\
&=1+\frac{x^2}{2}\left(\ln\left(\frac{x}{2}\right)-\frac12 \psi(1)- \frac12 \psi(2)\right) -\left(\frac12 \psi(2)+\frac12 \psi(3)\right)\frac{x^4}{16}
\end{align}
which is the stated result.


问:
How to show that $(\Bbb Q, <)$ contains an order isomorphic copy of $\epsilon_0$ which is the least ordinal satisfies $\omega^\epsilon = \epsilon$?
We define $\epsilon_0 = \sup\{\omega, \omega^\omega,\omega^{\omega^\omega},\omega^{\omega^{\omega^\omega}},\ldots\}$. How to find a subset of $ \Bbb Q$, $(A, <_{\Bbb Q})$ that is isomorphic to $\epsilon_0$?

So far, I'm only able to show that there's an order isomophic copy of $\omega^\omega$ in $\Bbb Q$ and don't know how to proceed.

答:
Any countable totally ordered set $(X,<)$ is isomorphic to a subset of $(\mathbb Q,<)$ :

Pick any bijection $f : \Bbb N \to X$, and define $\phi : X \to \Bbb Q$ blindly by picking for each $n$, any value for $\phi(f(n))$ that is consistent with the previously chosen values (no matter what you do, there will always be enough room, because $\Bbb Q$ with $n-1$ points removed is $n$ open intervals, so $n$ copies of $\Bbb Q$)

Here you could have some almost explicit map if you have an order-preserving map $f : \Bbb Q^\omega \to (0 ; 1)$ by setting $\phi(k) = 1-1/(k+2)$ for $k \in \omega$, and if $\alpha = (a_1, \ldots, a_n) \in \beta^\omega$, $\phi(\alpha) = f((\phi(a_1),\ldots,\phi(a_n)) + \max \lfloor \phi(a_i) \rfloor + 1\in \Bbb Q^\omega)$

In turn, you can make such a map $f: \Bbb Q^\omega \to \Bbb Q$ using the same construction with map a $\mathbb Q^2 \to (0;1)$.

ContentLicense: CC BY-SA 3.0


问:
Prove or disprove that $f(x)=ax^{(2b+1)}$ is injective by using the contrapositive.
I'm just getting into proofs, so a clear and intuitive explanation here would be most appreciated. I'm not sure if this is the correct notation either. How would I go about proving $$(\forall x, a \in \mathbb R)(\forall b \in \mathbb N)\{f(x)=ax^{2b+1}: x_1 \neq x_2 \Rightarrow f(x_1)\neq f(x_2)\}, a \neq0$$ I am essentially concerned with figuring out if there are any odd-powered functions that are not injective. Thank you!

Body:
Note that $$f(x_2)-f(x_1)=a(x_2^{2b+1}-x_1^{2b+1})=a(x_2-x_1)(x_2^{2b}+x_2^{2b-1}x_1+\cdots+x_2x_1^{2b-1}+x_1^{2b}).$$
First, by assumption, we know that $x_2-x_1\neq 0$. Denote $$S:=x_2^{2b}+x_2^{2b-1}x_1+\cdots+x_2x_1^{2b-1}+x_1^{2b}.$$

Without loss of generality, we assume $x_1<x_2$.

If $x_1=0$, then $S=x_2^{2b}>0$. Similarly, if $x_2=0$, then $S=x_1^{2b}>0$.

If $x_1>0$, then $x_2>0$ and clearly $S>0$.

If $x_2<0$, then $x_1<0$. Suppose $x_1=-x_1'$ and $x_2=-x_2'$ where $x_1',x_2'>0$. Then $$x_2^ix_1^{2b-i}=(-x_2')^{i}(-x_1')^{2b-i}=(-1)^{2b}\cdot(x_2')^i(x_1')^{2b-i}=(x_2')^i(x_1')^{2b-i}>0.$$ Hence $S>0$ as well.

In the first three cases, we have $f(x_2)-f(x_1)\neq 0$, so $f(x_1)\neq f(x_2)$. As for the last case,

Finally, assume $x_1<0<x_2$. Observe that $$f(x_1)f(x_2)=a^2(\underbrace{x_1x_2}_{<0})^{2b+1}<0,$$ so $f(x_1)\neq f(x_2)$. If they are equal, then we must have $f(x_1)f(x_2)\geq 0$.


问:
A question about book <Introduction to Topological Manifolds> by John Lee
The author first proved the following important theorem - "Unique Lifting Property of the Circle"

Then, the author proved the following corollary.

My question is: if corollary 8.6's assumption underlined by a red line is wrong. When two lifts $\tilde{f_0},\tilde{f_1}$
have the same initial point, they agree at the point $0$ of $[0,1]$. Then they should be identically equal by Theorem 8.3!

答: I think there is no issue. Theorem 8.3 describes lifts of the same path, whereas Corollary 8.6 describes lifts of distinct paths.


问:
is treating first order differential equations as results of chain rule valid?
The entire treatise of differential equations seems obscure to me, it doesn't really have a foundation from which you can work from, as differentiation or integration do (that I know of--at least), like the difference quotient definition of a derivative, and infinite series definition of the Riemann integral.
But nonetheless, I tried to make sense of how you could work out a differential equation given only standard knowledge of differentiation/integration, and it follows as such:

The question asks to find $y$ such that:

$$ \frac{dy}{dx} = (2x + 3)\sqrt{y}$$

Now, this is a simple question, but if you notice that the RHS product could resemble a product of the chain rule, i.e. something of the form $[f(g(x))]' = g'(x)f'(g(x))$, now pick $g'(x) = \sqrt{y}$ and $f'(g(x) = 2x + 3$, then immediately you can solve $g(x) = \frac{2\sqrt{y^3}}{3} + c_0$, but then we have:

$$ \begin{align}
f'(g(x)) = f'\left(\frac{2\sqrt{y^3}}{3}\right) &= 2x + 3 \\
\therefore f(g(x)) = f\left(\frac{2\sqrt{y^3}}{3}\right) &= x^2 + 3x + c_1
\end{align}
$$

But this is not the answer, and resources don't seem to be abundant on this particular way of reasoning about ODEs, but I guess that the obstacle is to do with the fact that I am integrating with respect to $x$ in the second step on a function whose input is not necessarily dependent (?) on $x$, though I'm stuck.

If this method is aimless, then my question is to ask whether any related methods exist?

答:
There seem to be major conceptual flaws in your working. For instance, how do you go from:

$\displaystyle g'(x) = \sqrt{y}$

to $\displaystyle g(x) = \frac{2\sqrt{y^3}}{3} + c_0$ ?

You are integrating wrt $\displaystyle x$ on the LHS, but wrt $\displaystyle y$ on the RHS.

Basically, this equation is immediately solvable by separation of variables. What you must realise is that separation of variables is already based on the chain rule!

The whole point of ordinary differential equations is to try to get $\displaystyle y$ as a function of $\displaystyle x$, basically to find the function $h$ where $\displaystyle y=h(x)$.

So if you're given a simple ODE like $\displaystyle \frac{dy}{dx} = \frac{f(x)}{g(y)}$

You can then write $\displaystyle \frac{d}{dx}h(x) = \frac{f(x)}{g(y)}$

and proceed $\displaystyle h'(x) = \frac{f(x)}{g(y)}$ (from this point on, the prime notation denotes differentiation wrt $\displaystyle x$).

$\displaystyle h'(x)g(y) = f(x)$

$\displaystyle h'(x)g(h(x)) = f(x)$

At this point, by the (reverse of) chain rule, you should observe the LHS is the derivative of $\displaystyle G(h(x))$ wrt $\displaystyle x$, where $\displaystyle G'(x) = g(x)$ giving:

$\displaystyle h'(x) G'(h(x)) = f(x)$

$\displaystyle (G(h(x))'= f(x)$

$\displaystyle G(h(x)) = F(x) + c$

where $\displaystyle F'(x) = f(x)$.

and finally $\displaystyle y = h(x) = G^{-1}(F(x)+c)$.

assuming $\displaystyle G$ is invertible.

So separation of variables is already rooted in the application of chain rule. Our "normal" working by separating $\displaystyle y$ and $\displaystyle x$ to opposite sides is simply a shorthand representation that works.


答:
Proceed by induction.

Suppose that $A_\alpha$ is a subset of $\Bbb Q$ of order type $\alpha$, and that $A_\alpha$ is bounded. Let $q>\sup A_\alpha$, then $A_{\alpha+1}=A_\alpha\cup\{q\}$. Clearly $A_{\alpha+1}$ has order type $\alpha+1$ and is bounded.

Now suppose $\delta$ is a limit ordinal, and for all $\alpha<\delta$ we have $A_\alpha$ defined. Pick an increasing sequence $\alpha_n$, and write $\delta=\bigcup[\alpha_n,\alpha_{n+1})$, and let $\beta_n$ be the order type of the $n$-th interval. Then $\beta_n<\delta$, and therefore we can find $A_{\beta_n}$ as a subset of $\Bbb Q$.

We know that $(0,1)\cap\Bbb Q$ is order-isomorphic to $\Bbb Q$, so embed $A_{\beta_n}$ into $(n,n+1)$, and let $A'_\delta$ be the union of those embedded copies. It is easy to see that $A'_\delta$ has order type $\delta$. Since we want it to be bounded (so the induction can continue), we use the order isomorphism mentioned above to map $A'_\delta$ into an isomorphic copy of itself in $(0,1)\cap\Bbb Q$.

Title: How to show that $(\Bbb Q, <)$ contains an order isomorphic copy of $\epsilon_0$ which is the least ordinal satisfies $\omega^\epsilon = \epsilon$?

Body:
We define $\epsilon_0 = \sup\{\omega, \omega^\omega,\omega^{\omega^\omega},\omega^{\omega^{\omega^\omega}},\ldots\}$. How to find a subset of $ \Bbb Q$, $(A, <_{\Bbb Q})$ that is isomorphic to $\epsilon_0$?

So far, I'm only able to show that there's an order isomophic copy of $\omega^\omega$ in $\Bbb Q$ and don't know how to proceed.


问:
Isometric linear operators on Hilbert spaces
Is it possible that an isometric (bounded) operator ${\rm T}: \mathcal H_1 \to \mathcal H_2$ with norm $\|\rm T\| = 1 $ is not unitary, i.e.  $\rm T^*T = I \neq TT^*$?

If yes, could you please provide some examples?

答:
Yes. The canonical example is the unilateral shift.

Take any separable Hilbert space $H$, and choose some orthonormal basis $\{e_n\}_{n\in\mathbb N}$ for it. Then define
$$
S\sum_kc_ke_k=\sum_kc_ke_{k+1}.
$$


问:
Constructing a function discontinuous precisely in a given closed/open set.
I have to find a real function which is discontinuous in a given closed set $F.$

I defined $f : \mathbb{R} \to \mathbb{R}$ by $$f(x) = \begin{cases}
 0&\text{if}\, x\in \mathbb{R}\setminus F\\
 1&\text{if}\, x \in F\cap{\mathbb{Q}}, \;\mathbb{Q}\;\text{being the set of the rationals}\\
      2&\text{if}\, x\in F\cap{\mathbb{I}}, \;\mathbb{I}\;\text{being the set of the irrationals}
\end{cases}$$
This function is discontinuous on $F.$

Also, I have to find another function that is discontinuous on a given open set $O.$
So I replaced $F$ with $O$ in the above construction.

Am I right? Is my construction okay as I don't use the closed or open property of the sets?
Thank you.

答: Let $U$ be the open set. Set $f(x)=d(x,\mathbb R \setminus U)\cdot\chi_{U\cap\mathbb Q}(x).$


答:
The form of the chain rule you need to be using is that which you learned in multivariable calculus, an exact differential:

\begin{align}\tag{1}
\frac{\mathrm d}{\mathrm dx}\lambda(x,y)=\frac{\partial \lambda}{\partial y}\frac{\mathrm dy}{\mathrm dx}+\frac{\partial \lambda}{\partial x}.
\end{align}
You can read about it on the wiki page for chain rule or just google "multivariable chain rule". Rewriting your ODE gives
\begin{align}\tag{2}
\frac{\mathrm dy}{\mathrm dx}-(2x+3)\sqrt y=0,
\end{align}
which is similar in form to Eqn. (1), so we'll begin with the ansatz that Eqn. (2) is indeed an exact differential, so then it must be that
\begin{align}
1&=\frac{\partial \lambda}{\partial y},\ \ \ \ \text{and}\\
-(2x+3)\sqrt y&=\frac{\partial \lambda}{\partial x}.
\end{align}
If we integrate the first Eqn. we see that
\begin{align}
\lambda(x,y)=y+f(x),
\end{align}
where $f(x)$ is an arbitrary function picked up from integrating with respect to $y$. Taking a derivative with respect to $x$ and setting it equal to the other equation we get that
\begin{align}
\frac{\mathrm df}{\mathrm dx}=-(2x+3)\sqrt y.
\end{align}
But $f(x)$ is only a function of $x$, therefore the ansatz that Eqn. (2) is a perfect differential was wrong.

That being said, multiplying the entire equation by a integrating factor $m(x,y)$ sometimes makes your ODE a perfect differential, you can read about that by googling "exact differential equation integrating factor"

Title: is treating first order differential equations as results of chain rule valid?

Body:
The entire treatise of differential equations seems obscure to me, it doesn't really have a foundation from which you can work from, as differentiation or integration do (that I know of--at least), like the difference quotient definition of a derivative, and infinite series definition of the Riemann integral.
But nonetheless, I tried to make sense of how you could work out a differential equation given only standard knowledge of differentiation/integration, and it follows as such:

The question asks to find $y$ such that:

$$ \frac{dy}{dx} = (2x + 3)\sqrt{y}$$

Now, this is a simple question, but if you notice that the RHS product could resemble a product of the chain rule, i.e. something of the form $[f(g(x))]' = g'(x)f'(g(x))$, now pick $g'(x) = \sqrt{y}$ and $f'(g(x) = 2x + 3$, then immediately you can solve $g(x) = \frac{2\sqrt{y^3}}{3} + c_0$, but then we have:

$$ \begin{align}
f'(g(x)) = f'\left(\frac{2\sqrt{y^3}}{3}\right) &= 2x + 3 \\
\therefore f(g(x)) = f\left(\frac{2\sqrt{y^3}}{3}\right) &= x^2 + 3x + c_1
\end{align}
$$

But this is not the answer, and resources don't seem to be abundant on this particular way of reasoning about ODEs, but I guess that the obstacle is to do with the fact that I am integrating with respect to $x$ in the second step on a function whose input is not necessarily dependent (?) on $x$, though I'm stuck.

If this method is aimless, then my question is to ask whether any related methods exist?


问:
Find the expected value for these random variables.
Let $X_{1}, X_{2}, ..., X_{n}$ be independent and identically distributed random variables with finite expected value. Find
$\mathbb{E}[X_{1}|(X_{1}+X_{2}+ ... + X_{n})=x]$.
Assuming that $f_{(X_{1}+X_{2}+ ... + X_{n})}(x)\neq 0$, I must have the expected value is $\frac{x}{n}$. ¿Could you give me a hint to do it, please?

答:
By symmetry,
$$\mathbb E[X_1\mid X_1+\dots+X_n]=\mathbb E[X_2\mid X_1+\dots+X_n]=\dots=\mathbb E[X_n\mid X_1+\dots+X_n].$$
But by linearity
$$\sum_{i=1}^n\mathbb E\left[X_i\,\middle|\,\sum_{j=1}^n X_j=x\right]=\mathbb E\left[\sum_{i=1}^nX_i\,\middle|\,\sum_{j=1}^nX_j=x\right]=x,$$
so we have $\mathbb E[X_i\mid X_1+\dots+X_n=x]=\frac{x}{n}$.


问:
Multiplication Operator on $L^2[(0,1])$
Let $$T: L^2([0,1]) \mapsto L^2([0,1]),\quad T(f) = tf(t) $$

I have read that for  $z\in \mathbb{C}$ follows $T - z\text{id}$ is invertable iff $z \notin [0,1]$. We didnt have spectral theory yet so i am struggling to show it. Any help is appreciated

答:
You have $[(T-z I)f](t)=(t-z)f(t)$.

If $z\not\in[0,1]$, then $g(t)=\frac1{t-z}$ is continuous on $[0,1]$, so in $L^2$. This shows that $Sf=gf$ defines an inverse for $T-z I$.

Conversely, if $z\in[0,1]$, then $T-z I$ is not surjective. For instance, the constants are not in its range. For if you had $(t-z)f(t)=1$ for some $f$, then $f(t)=\frac1{t-z}$, which is not in $L^2[0,1]$

Title: Multiplication Operator on $L^2[(0,1])$


问:
Are pointed CW complexes for which the Yoneda embedding restricted to finite CW complexes reflects isomorphisms forced to be connected?
I'm currently reading Edgar H. Brown's paper, Abstract Homotopy Theory, which proves a categorical version of the Brown Representability Theorem. I've come across a statement that I don't really how to verify (if it is in fact true).

First the notation and definitions. Let $\mathscr{C}$ be the homotopy category of pointed CW complexes. Let $\mathscr{C}_0$ be (as I understand it) a countable (though small is apparently sufficient) full subcategory of the subcategory of finite pointed CW complexes that contains a representative of every isomorphism class of finite pointed CW complexes.

Now we define $\newcommand\oln\overline\oln{\mathscr{C}_0}$ to be the set of all objects $Y\in\mathscr{C}$ such that
for all homotopy classes of maps $f:Y\to Y'$, if we have that for all $X\in\mathscr{C}_0$, $f_* : [X,Y]\to [X,Y']$ is an isomorphism, then we have that $f$ is an isomorphism. In other words, $\oln{\mathscr{C}_0}$ consists of the objects $Y$ for which isomorphisms out of $Y$ can be detected by restricting the Yoneda embedding to finite CW complexes.

Brown claims that the set $\oln{\mathscr{C}_0}$ is the set of all connected CW complexes. I can see why the Whitehead theorem implies that the CW complexes are a subset of $\oln{\mathscr{C}_0}$, but I can't find any example of a space not in $\oln{\mathscr{C}_0}$, although it appears such a space must exist, due to a claim that there is no small full subcategory of the homotopy category of spaces that reflects isomorphisms in this manner appearing in this arxiv paper. It doesn't seem to really matter for the purposes of Brown's paper, but I do want to find some resolution to this question for my own personal curiosity.

To that end, Brown appears to be claiming that for any space $Y$ with at least 2 connected components, one can construct a space $Y'$ and a map $f:Y\to Y'$ which induces bijections $f_*:[X,Y]\to [X,Y']$ for all finite CW complexes $X$. If this is in fact true, I'd appreciate any reference to a proof of this fact, or even just a proof of this for one space, like $S^0$. Alternatively, a disproof of the claim would also resolve my question. Perhaps this is an easy question, but my first thoughts seem to have failed.

Actually, since I know that not all spaces can be in $\oln{\mathscr{C}_0}$, I know the following must be wrong, but I can't figure out where. I can't see why we can't apply the Whitehead theorem in the disconnected case by taking $X=S^0$ and $X=S^n_+$ (unpointed $S^n$ plus a disjoint base point). The first forces $f$ to induce a bijection on $\pi_0$, and the others appear to force $f$ to induce bijections on homotopy classes of maps from $S^n$ to the paired up components of $Y$ and $Y'$, which seems like it would imply that the induced map of homotopy groups for each component is an isomorphism. I would also appreciate if someone could explain where I'm going wrong.

答:
Let $f:A\to B$ be a map of non-pointed CW complexes which looks like an isomorphism to the finite complexes in the homotopy category. (For example as in the paper you link we might take $A=B$ the classifying space of the group of permutations of a countable infinite set with all but finitely many points fixed, and $f$ the map induced by the “suspension” homomorphism, which shifts the action of a permutation up by one.) Then $f_+:A_+\to B_+$ has the same properties in the pointed category. Indeed the maps $X\to A_+$ in the pointed homotopy category are just the maps from subsets of the connected components of $X\setminus X_0$ to $A$ in the unpointed category ($X_0$ the component containing the basepoint).

The fact that unpointed homotopy rears its head in the pointed but not connected category away from the basepoint also explains the error in your last paragraph. Maps from $S^n_+$ in the pointed homotopy category are free homotopy classes of maps from $S^n$, which is not what $\pi_n$ is. In case $\pi_1$ is being annoying, this difference ruins everything.


问:
Can two 3-dimensional vectors spans a $\mathbb{R}^2$
I have done some similar questions, for example the following:

Verify if three 3-dimensional vectors (v1:[...],v2:[...],v3:[...]) will span $\mathbb{R}^3$.

All I have to do is to put three vectors into a 3 by 3 matrix and perform elementary row operations, and to check if there are 3 pivots.

However, this makes me think if I have two 3-dimensional vectors and it does have a reduced row-echelon form with 2 pivots. Can I say that they span $\mathbb{R}^2$

More generally, does the dimension of vectors limit the space that the set of these vectors can span? (Ex: Can three 4-dimensional vectors span $\mathbb{R}^3$)

答:
The first sentence after the colon is incomplete.  However, if I get what you're driving at, the answer is yes that $k$ vectors in $\Bbb R^n$ will, if they are independent, span a $k$-dimensional subspace.

Recall that the number of vectors in a basis is an invariant; and that this number is called the dimension.

Moreover, any two $k$-dimensional vector spaces are isomorphic.  So if we have $k$ linearly independent vectors, they span a space isomorphic to $\Bbb R^k$.


问:
Understanding this proof of the reflection principle
I was reading the Wikipedia article about the reflection principle for Wiener processes, which can be written as :

Lemma (Reflection Principle) : If $(W(t) : t \ge 0)$ is a Wiener process and $a > 0$ is a threshold, then $\forall t\ge 0$ $$ \mathbb{P}\left(\sup_{0\le s\le t} W(s) \ge a \right) = 2\mathbb{P}(W(t) \ge a)$$

And there are some parts of the proof that I can't seem to understand. After defining $\tau_a := \inf \{t : W(t) \ge a\}$ and $X(t) := W(t + \tau_a) - a$, the following points are unclear
:

The proof relies on the fact that events $\{\sup_{0\le s\le t} W(s) \ge a \}$ and $\{X(t - \tau_a) < 0\}$ are independent conditional on $\mathcal{F}_{\tau_a}^W$. I understand that by the strong Markov property, $X(t)$ is a Brownian Motion independent of $\mathcal{F}_{\tau_a}^W$, but to me that only holds for $t \ge 0$. However as far as I know there are no reasons for $ t - \tau_a $ to be nonnegative.

The point above raises a second concern : if $ t - \tau_a < 0 $, then  $X(t - \tau_a)$ doesn't make sense if you think of $X$ as a Brownian Motion, since they are only defined for nonnegative times. I thus fail to understand how the probability $\mathbb{P}\left(\sup_{0\le s\le t} W(s) \ge a, X(t - \tau_a)\right) $ is well defined.

Any help on this would be greatly appreciated.

Bonus question (if this is inappropriate I will create a separate question) : Why is the lemma as written above equivalent to $\mathbb{P}(\tau_a \leq t) = 2 \mathbb{P}(W(t) \geq a)$ ?

答: I think answering the bonus question likely answers the two main questions as well.  The reason is that $\{\tau_a \le t \} = \{ \sup_{0 \le s \le t} W(s) \ge a\}$, i.e. $W$ hits $a$ at a time before $t$ if and only if the max up to time $t$ is at least $a$.  Therefore on the event $\{ \sup_{0 \le s \le t} W(s) \ge a\}$ we have that $t \ge \tau_a$, so $t - \tau_a \ge 0$ and $X(t-\tau_a)$ is well defined on that event.


答:
The things that you mentione are false, The problem is your definition of basis.

Let a finite vectorial space $V$ over some field $\mathbb{K}$, a subset $\mathcal{B}=\lbrace v_1,v_2, \cdots v_n\rbrace \subset V$ is a basis for $V$ if for every $v\in V$ then $v$ is linear combination of the elements  of  $\mathcal{B}$(it is $v=\sum_{j=1}^{n}\alpha_jv_j $ for $\alpha_j\in \mathbb{K}) $ and if $\mathcal{B}$ is a set linearly independent(it means if $\sum_{j=1}^{n}\alpha_j v_j=0$ then should occur that $\alpha_j=0 \forall j$.
If the set $\mathcal{B}$ is finite we say that $V$ have a finite basis and the number of elements of $B$(unique) the dimention of $V$

Now in the particular case of $\mathbb{R^n}$ if you have a set of vectors $\mathcal{B}=\lbrace (1,0,0, \cdots 0),(0,1,0, \cdots, 0)\rbrace$ then $\mathcal{B}$ spann a subespace of $\mathbb{R^n}$ in particular spann vectors of the form $(x,0,0, \cdots,0)+(0,y,0, \dots,0) :x,y\in \mathbb{R}$ which in any case spann $\mathbb{R^{2}}$.

Title: Can two 3-dimensional vectors spans a $\mathbb{R}^2$

Body:
I have done some similar questions, for example the following:

Verify if three 3-dimensional vectors (v1:[...],v2:[...],v3:[...]) will span $\mathbb{R}^3$.

All I have to do is to put three vectors into a 3 by 3 matrix and perform elementary row operations, and to check if there are 3 pivots.

However, this makes me think if I have two 3-dimensional vectors and it does have a reduced row-echelon form with 2 pivots. Can I say that they span $\mathbb{R}^2$

More generally, does the dimension of vectors limit the space that the set of these vectors can span? (Ex: Can three 4-dimensional vectors span $\mathbb{R}^3$)


问:
A group of students weigh on average 70 Kg, with a standard deviation of 5 Kg...
If a group of 50 students are sampled at random, the probability that their total weight does not exceed 3600 Kg is:

A. 0.9961
B. 0.9977
C. 0.9989
D. 0.9952
E. 0.9931

I know that you must find z-score; (3600-3500)/5

But the official answers say that the standard deviation must be written as so, σS= sqrt(50) × 5 = 35.35

The part where I get confused is when you multiply 5 by sqrt (50) to reflect the new standard deviation, as I think I should instead be writing 5/sqrt 50

Any help would be appreciated!

答:
Let $X_i$ be a random variable with mean $\mu_x$ and variance $\sigma_x^2$ for all $i = 1, 2, ..., n$. Assume $X_i$ are independent and identically distributed.

Let $Z = \sum_i X_i $.

Then, the mean of $Z$, $\mu_z$, is given by

$$ \mu_z = E[Z] $$
$$ = E\left[\sum_i X_i\right] $$
$$ = \sum_i{E\left[X_i\right]} $$
$$ = nE\left[X_i\right] $$
$$ = n\mu_x $$

The variance of $Z$, $\sigma_z^2$, is given by

$$ \sigma_z^2 = V[Z] $$
$$ = V\left[\sum_i X_i\right] $$
$$ = \sum_i{V\left[X_i\right]} $$
$$ = nV\left[X_i\right] $$
$$ = n\sigma^2_x $$

So, the standard deviation of $Z$ is $\sigma_z = \sqrt{n}\sigma_x$.

Now, with $X_i$ normally distributed, we further have that $Z$ is also normally distributed - the sum of normally distributed random variables is itself normally distributed.

In this case, the mean of the sum is

$$ \mu_z = n\mu_x = 50\cdot 70 = 3500 . $$

That's the 3500 you're comparing 3600 against in your $Z$-score.

The standard deviation of the sum is

$$ \sigma_z = \sqrt{n}\sigma_x = \sqrt{50}\cdot 5 . $$

That's what you need to divide by in your $Z$-score.


问:
Taylor series of $f(x)=\frac {e^x-1}{x}$
I am asked to expand $f(x)=\frac {e^x-1}{x}$ centered at 0 using the known Talyor series of functions.
How to simplify the function so that it can be expanded more easily?</p>

答:
If you expand $e^x$ and subtract 1, then you get something divisible by $x$. You should find
$$\displaystyle\sum_{k=0}^{+\infty}\frac{x^k}{(k+1)!}$$</p>


答:
The short answer to the title of your question is no, two vectors, independent of the number of dimensions they have, can only generate at most, a 2d space.

This is because you need at least as many vectors as dimensions for the generated space.

If you have only one vector, you can spawn a line, in any higher dimensional space, you can have a 6d vector, but because is only one, it can only generate a line, 1d space, which is the set of all linear combinations of it.

The dimensionality of your generator vectors only describe their geometry in the containing, source space, but in the inner space generated by their linear combinations, you will have as many dimensions as independently linear vectors, as they can't be linearly represented by each other, and thus spawn an internal space taking each vector as a coordinate axis (just a way of seeing it), and this space is isomorphic to $R^n$.

Imagine the columns of your matrix as vectors in space, and maybe you can see more easily the relationship between the source space (normally $R^n$) and the target space.


问:
Solving $\int_0^{\pi/2}x\cot x\,\mathrm{d}x$ while running into a $0\times\infty=0$ problem
While i have been trying to solve the integral  $ \int_0^{\pi/2} x\cot x \, \mathrm{d}x $ i have noticed that by trying integrating by parts using $u = x$ and $\mathrm{d}v = \cot x \, \mathrm{d}x$, i get that:

$$ \int_0^{\pi/2} x\cot x \, \mathrm{d}x = x\ln\sin x\Big|_0^{\pi/2} - \int_0^{\pi/2} \ln\sin x\,\mathrm{d}x $$

Plugging the first integral in Wolfram Alpha and after that the resulting integral in the right hand side i noticed that they have the same value! So the evaluation of $u\cdot v$ from $0$ to $\pi/2$ must be $0$, but how is this possible? I mean $\sin 0= 0$ and $\ln 0= -\infty$, so, how is that the term $x\sin x\Big|_0^{\pi/2}$ can be $0$? Or how am i wrong?

Ps: I am trying to integrate by parts first because i am trying to solve the integral by differentiation under the integral sign, if you have any suggestions for this i would also appreciate it!

答:
For your main concern, it's been in the comments but I'll copy it again into the answer, we have

$$\lim_{x\to 0^+}x\ln(\sin x)=0$$

by L'Hospital's rule, for example. Therefore it does hold that
$$x\ln(\sin x)\vert_{0}^{\pi/2}=0-0=0$$

I believe there are sources online that shows you how to do this integral (maybe not MSE, but somewhere), but I will try to answer it using Leibniz's rule as you specifically required. Consider for $0\le a\le 1$,
$$I(a)=\int_0^{\frac{\pi}{2}}\arctan\left(a\tan x\right)\cdot \cot x\,dx$$
Then
\begin{align*}
I'(a)&=\int_0^{\frac{\pi}{2}}\frac{\tan x}{a^2\tan^2 x+1}\cdot \cot x\,dx\\
&=\int_0^{\frac{\pi}{2}}\frac{1}{a^2\tan^2 x+1}\,dx\quad\left[\text{let}\ t=\tan x\right]\\
&=\int_0^\infty\frac{1}{a^2t^2+1}\cdot\frac{dt}{t^2+1}\quad\left[\text{just ask Wolfram}\right]\\
&=\frac{\pi}{2\left(a+1\right)}
\end{align*}
Thus we conclude that $I(a)=\frac{\pi}{2}\ln\left(a+1\right)+C$ where $I(0)=0$, so that
$$I(1)=\int_0^{\frac{\pi}{2}}x\cot x\,dx=\frac{\pi}{2}\ln(2)$$
as desired.


问:
Formula for radius of circles on vertices of regular polygon
TLDR: Given a square, a number of circles, a rotation angle, and some spacing factor, how can I determine the largest possible radius of circles arranged on the vertices in a regular polygon such that the circles do not exceed the bounds of the square?

I am trying to determine a formula or algorithm for calculating the radius of circles situated on the vertices of regular polygons with 1 to n sides.

I have tried the following with inconsistent results.

```<code>2 * a * tan(π / n) / x
</code>```

Where x is a sizing factor and a is the apothem. For example say I wanted the spacing between circles to be such that you could fit another circle on the midpoint of the side, x could be 4. Or if I wanted the circles to be touching x could be 1. But for n = 3 the results look quite different than say for 6 in that the spacing between the circles is too small/big respectively.

As for the apothem, it's not clear how I would calculate that.

For n = 1 or 2 I would have to handle it differently. For 1 I'd just want a circle in the centre of the available space, and for 2 they would be arranged in a line. Is there a more general formula I could use?

Something like the circles on the vertices of the hexagon in the image below, where I could adjust the radius of the circles as needed.

Further complication is that I calculate the coordinates of the vertices based on a given square and an angle to offset the resulting vertices, but I reduce the size of the resulting polygon based on the radius of the circles to make sure the resulting circles do not exceed the bounds of the square.

Here's a few imperfect drawings I made for 1–3:

The circles don't have to fill the square necessarily; I want to be able to adjust the spacing between them by reducing the radius.

I'm a bit stuck on a circular dependency (no pun intended) since I think I need the positions of the vertices to calculate the circle radius, but I need the circle radius to determine the positions of the vertices so that the circles do not exceed the size of the square.

答:
I believe this becomes easier to think about if we bound the circles inside of a larger circle.

Labeled diagram of an example setup with four circles:

With five circles:

Below is some trigonometry that relates the size of the square and the number of circles to the radius of the circles, their distance from the center of the square, and the angle of displacement between the circles. Let $n$ be the number of circles. The other labels are described in the diagram(s).

\begin{align*}
    e &= \frac{2\pi}{2n} \ \text{rad}=\frac{\pi}{n} \ \text{rad} \\
    d &= \pi-\frac{\pi}{2}-e=\frac{\pi}{2}-e \\
    b &= a \cdot \sec{d} \\
    c &= b+a \\
    f &= 2c
  \end{align*}

Now, to solve for $a$ in terms of $f$ and $n$:

\begin{align*}
  f &= 2(b+a) \\
  f &= 2(a \cdot \sec(d) + a) \\
  f &= 2a(\sec(d) + 1) \\
  a &= \frac{f}{2(\sec(d) + 1)} \\
  a &= \frac{f}{2(\sec(\frac{\pi}{2}-e) + 1)} \\
  a &= \frac{f}{2(\sec(\frac{\pi}{2}-\frac{\pi}{n}) + 1)} \\
  \end{align*}

And for $b$ in terms of $f$ and $n$:

\begin{align*}
    b &= a \cdot \sec(d) \\
    b &= \frac{f \cdot \sec(d)}{2(\sec(\frac{\pi}{2}-\frac{\pi}{n}) + 1)} \\
    b &= \frac{f \cdot \sec(\frac{\pi}{2}-\frac{\pi}{n})}{2(\sec(\frac{\pi}{2}-\frac{\pi}{n}) + 1)} \\
  \end{align*}

This works for any number of circles greater than $n=1$.

Bonus eight circle example


问:
Theorem 46.8 Munkres Topology
In the proof of Theorem 46.8 Munkres Topology following is claimed (not exact copy but exact interpretation)

If A is any compact subset of a metric space Y and V is any  open subset of Y containing A then there is a $\epsilon>0 $ such that the $\epsilon-$neighborhood of A is contained in V.

The mentioned claim is true provided $d(A, X-V) >0$ is positive. 1- But how to prove that? And 2- how any open set containing a compact set is a proper superset?

答:
As Brian M. Scott stated, the function $f: A \to \mathbb R$ sending $x$ to $d(x, Y \setminus V)$ is continuous on $A$ and $A$ is compact, so $f$ attains its minimum on $A$, at say $x \in A$. But if $d(x, Y \setminus V) = 0$, this means there exists a sequence of elements $\{y_n\}_{n \in \mathbb N}$ in $Y \setminus V$ satisfying $d(x,y_n) \to 0$. Since $Y \setminus V$ is closed, this implies $x \in Y \setminus V$, a contradiction since $A \subseteq V$. Therefore, we can pick $\varepsilon = \frac{d(x,Y \setminus V)}2$ to get an $\varepsilon$-neighborhood $U$ of $A$ contained in $V$. By definition, $d(U, Y \setminus V) = \frac{d(x, Y \setminus V)}2 > 0$, so $U$ is contained in $V$.

You won't be able to prove that $A \subsetneq U$ because it's not true in some cases. Take the example of $Y$ finite and discrete; $\varepsilon$-neighborhoods of $A$ are just $A$ for $\varepsilon$ small enough (smaller than any distance between any two distinct points, for example).

Hope that helps,

Title: Theorem 46.8 Munkres Topology


问:
Prove, if $L$ real symmetric matrix over $S$, then $M(L)$ representing $L$ with base change over an orthonormal base of $S$ is also symmetric
im new here, and im having trouble checking some old theorems i didnt prove in their time. This is my first post, and also English is not my native language, so please be patient with me, i will update my question and fix any mistake or add any other requirements y may be missing. I currently only need to prove this, which as been a bit confusing:

Given $L$ a linear transform over $S$ subspace of $R^n$, and $B$ orthonormal base of $S$.
Prove the matrix $M(L)$ representing the transform $L$ after a base change over $B$ (from and into) is also symmetric iff $L$ symmetric.

In other words:

$L$ symmetric <=> $M(L)$ symmetric

PD: i can't directly use $B^⊤$ = $B^{-1}$ if $B$ the orthonormal matrix generated by the base $B$.

答:
If $B$ is the orthonormal matrix given by sending the standard basis to your orthonormal basis, then $B$ satisfies $B^{\top} = B^{-1}$, and in particular, if $M_B(L)$ denotes the transformed $L$ with matrix $B$, then
$$
M_B(L)^{\top} = (BLB^{-1})^{\top} = (BLB^{\top})^{\top} = (B^{\top})^{\top} L^{\top} B^{\top} = BLB^{\top} = M_B(L).
$$
Since $L = M_{B^{-1}}(M_B(L))$, you can repeat the argument by substituting $(B,L)$ by $(B^{-1}, M_B(L))$ to prove the equivalence.

Hope that helps,


问:
How can you prove Nakayama's lemma over nonunital rings using the unitization?
For a nonunital commutative ring $A$, an $A$-module $M$ is called finitely generated over $A$ if there is a finite set of elements $x_1,...,x_n\in M$ such that every element of $M$ is of the form $a_1x_1+...+a_nx_n$ for some $a_1,...,a_n\in A$. Under this definition, notice that the ring $A$ may not be finitely generated over itself, for example $A=2\Bbb Z$.

Now let us consider a equivalent form of Nakayama's Lemma below:

For a nonunital commutative ring $A$ and a finitely generated module $M$ over $A$, there is an element $a\in A$ such that $am=m$ for all $M$.

The proof goes as usual: Write the generator $x_i=a_{i,1}x_1+...+a_{i,n}x_n$ and notice the determinant of the matrix $\phi=\det(a_{i,j}-\delta_{i,j}\Bbb{1}_M)$ vanishes on every element of $M$ while $\phi$ is of the form $\Bbb{1}_M+k\Bbb{1}_M$ for some $k\in A$(just expand out the $\det$.)

To obtain the usual version of Nakayama's lemma, simply replace $A$ by an ideal $\mathfrak a$ in a commutative unital ring with $\mathfrak aM=M$.

I am requesting another proof of this fact using unitalization, as I was told the category of all $A$-modules is indeed isomorphic to the category of all $U(A)$-modules, where $U(A)$ is the unitalization of $A$. I really cannot come up with another categorical proof of the nonunital version of Nakayama's Lemma, although this "nonunital" version involves less restriction while seems indeed fundamental in its form. I will be appreciated if another angle is possible to attack this result.

答:
To obtain the usual version of Nakayama's lemma, simply replace $A$ by an ideal $\mathfrak a$ in a commutative unital ring with $\mathfrak aM=M$.

You can just reverse this argument.  $A$ is an ideal in its unitization $U(A)$, and considering $M$ as a $U(A)$-module, it satisfies $AM=M$.  So, the usual version of Nakayama's lemma gives an element $a\in A$ such that $am=m$ for all $m\in M$.


问:
Are the statements about the correlation coefficient true or false?
Let us define the correlation coefficient as
$\rho(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$.

Are the following statements true or false?

If $\rho(X,Y)=\rho(Y,Z)=0$ then $\rho(X,Z)=0$

If $\rho(X,Y)>\rho(Y,Z)>0$ then $\rho(X,Z)>0$

If $\rho(X,Y)<\rho(Y,Z)<0$ then $\rho(X,Z)<0$

I think they are false, but I can't find counterexamples. Could you help me?

答: They are indeed all false.  For the first one, you can take $X=Z$ as a counterexample, and have $Y$ be independent of $X$.  For the second, you can take $X$ and $Z$ to be iid $N(0,1)$ random variables and $Y:= X + Z$.  Basically the same counterexample works for the third, but with $Y := -X-Z$ instead.


问:
Consecutive consecutive sums of equal value
Given the list of counting numbers, what is the largest amount of consecutive consecutive sums of equal value that can be found? Is there a limit?

For example,

[1, 2] and [3] are two consecutive consecutive numbers that both add up to 3.

(If somebody understands the question and has a better way of phrasing, please do help me rephrase this.)

Body:
Consider
$$\quad N=x+(x+1)+(x+2)+\cdots +(x+n)\\
\implies N=\frac{(n^2 + 2 n x + n + 2 x)}{2}
= \frac{(2x+n)(n+1)}{2}\\
$$
N is the sum of consecutive numbers if and only if $N$ is factorable as
$\space 2N=(2x+n)(n+1).$

For example
$$N=7 \rightarrow
14=\big(2(x)+n\big)\big(n+1\big)=\big(2(3)+1\big)\big(1+1\big),
\quad 7=3+4$$

Now we solve for $x$ and, try values of $n$ where
$\quad\sum_{1}^{n} k<N\quad $ and the highest integer solution where
$\quad (1+2+3+\cdots+n)<N\quad$ indicates $n$ as the maximum possible number of consecutive integers sums equal to $N$.

$$N=\frac{(n^2 + 2 n x + n + 2 x)}{2}
   =\frac{2(n+1)x+n(n+1)}{2}
\implies x=\frac{2N-n(n+1)}{2(n+1)}$$

The number $n$ will sometimes be greater than the number of sums equal to $N$ but it will never be less. Here is a table showing the results for
$\quad 1\le N\le 25$

\begin{align*}
(N,n)\qquad &\text{  consecutive integer sums}\\
(2,0)\qquad &2\\
(3,1)\qquad &3 = 1+2\\
(4,0)\qquad &4\\
(5,1)\qquad &5 = 2+3\\
(6,2)\qquad &6 = 1+2+3\\
(7,1)\qquad &7 = 3+4\\
(8,0)\qquad &8\\
(9,2)\qquad &9 = 4+5\quad = 2+3+4\\
(10,3)\qquad &10 = 1+2+3+4\\
(11,1)\qquad &11 = 5+6\\
(12,2)\qquad &12 = 3+4+5\\
(13,1)\qquad &13 = 6+7\\
(14,3)\qquad &14 = 2+3+4+5\\
(15,4)\qquad &15 = 7+8 \quad  = 4+5+6 \quad = 1+2+3+4+5\\
(16,0)\qquad &16\\
(17,1)\qquad &17=8+9\\
(18,4)\qquad &18=5+6+7\quad =3+4+5+6\\
(19,1)\qquad &19=9+10\\
(20,4)\qquad &20 =2+3+4+5+6\\
(21,6)\qquad &21=1+2+3+4+5+6\quad =6+7+8=10+11\\
(22,3)\qquad &22=4+5+6+7+8\\
(23,1)\qquad &23=11+12\\
(24,1)\qquad &24=7+8+9\\
(25,4)\qquad &25=3+4+5+6+7\quad =12+13\\
\end{align*}


问:
Intermediate field of $\mathbb F_9(x)/$ $\mathbb F_3(x^2)$.
The title question was a question asked in interview examination in some university.
I want to find all of Intermediate field of $\mathbb F_9(x)/$ $\mathbb F_3(x^2)$.

$\mathbb F_9(x)$＝$\mathbb F_3(x,$$\sqrt{2}$) is OK, but I cannnot proceed from here.
Do I need to rewrite the extension as simple extension?
For the beginning,I cannot specify the extension degree.

Thank you for your help.

答:
Note that $T^2 - x^2 \in \Bbb F_3(x^2)[T]$ and so $[\Bbb F_3(x) : \Bbb F_3(x^2)] = 2$. On the other hand, we have $\Bbb F_9(x) = \Bbb F_3(\sqrt{2})(x)$ and $T^2-2 \in \Bbb F_3[T] \subset \Bbb F_3(x)[T]$, so $[\Bbb F_9(x) : \Bbb F_3(x)] = 2$.

Thus
$$
[\Bbb F_9(x) : \Bbb F_3(x^2)] = [\Bbb F_9(x) : \Bbb F_3(x)] \cdot [\Bbb F_3(x) \colon \Bbb F_3(x^2)] = 4.
$$

All polynomials considered are separable as we are in characteristic $3$, and the extension is normal because $\Bbb F_9(x) = \Bbb F_3(x^2)(x,\sqrt{2})$ is a splitting field, so the original extension is Galois and

$$
|\mathrm{Gal}(\Bbb F_9(x)/\Bbb F_3(x^2))| = 4.
$$

Then $\mathrm{Gal}(\Bbb F_9(x)/\Bbb F_3(x^2))$ is either the Klein four group of the cyclic group of order $4$, but since any automorphism of $\Bbb F_9(x) = \Bbb F_3(x^2)(x,\sqrt{2})$ is determined by the images of $x$ and $\sqrt{2}$ and those must be roots of their respective minimal polynomials, a direct computation shows that

$$
\mathrm{Gal}(\Bbb F_9(x)/\Bbb F_3(x^2)) \simeq \Bbb Z_2 \oplus \Bbb Z_2.
$$

By the Galois corresponence, intermediate fields correspond to subgroups of the Klein group. I'll leave the rest to you.


问:
Applications of Double/Triple Integrals
This is the question that I need to solve using mathematica:

The concentration of an air pollutant at a point $(x,y,z)$ is given by: $$p(x,y,z) = x^2y^4z^3 \text{ particles}/m^3$$ We're interested in studying the air quality in a region in 3-space which satisfies $x\ge0$ and is bounded by the $xy$-plane, the surface $z=\sqrt{4-x^2-y^2}$, the plane $y=-2x$, and the $xz$-plane.  Find the total amount of pollutant in this region.

I don't want a full answer to this question, obviously (I need to work it out myself), but I'm just having trouble getting started.  How do I use the bounds here to set up an integral for the problem?  What exactly do the bounds mean/represent in the first place?

答:
It's a bit of a puzzle to work out the region of integration and a nice way to express it using limits of integration.  A priori the fact that a bunch of "bounds" are thrown out by a problem does not tell us whether the region is actually finite (a finite volume in this case), and even if so you might not be able to package up the region with a single set of nested integral signs and their associated limits of integration.

Without a picture to guide us, I'd probably start from the beginning of the list, assuming some good faith on the part of the problem's constructor.  Now $x \ge 0$ is an easy one to express, by itself.  If we put $x$ as the variable for the outer integration, then the two nested integrals can have limits that depend on the $x$ value referenced in the outer integration.

"Bounded by the $xy$-plane" is a little ambiguous.  The $xy$-plane is where $z=0$, so as a boundary this means either restricting to $z \ge 0$ or $z \le 0$, one side or the other.  Let's keep an open mind as we parse the remaining pieces of the boundary puzzle.

Aha! The surface $z = \sqrt{4 - x^2 - y^2}$ clarifies things.  For one, the square root sign here explicitly means nonnegative values of $z$ on the boundary, so evidently it's the upper half space $z \ge 0$ we need.  Also this is a hemisphere, half the surface of a ball, so it does restrict integration to a finite volume.

I'll let you pursue the remaining pieces of the puzzle, but what we ultimately hope to do is express the region using nested limits of integration something like this:

$$ \int_0^c \int_{f(x)}^{g(x)} \int_{u(x,y)}^{v(x,y)} p(x,y,z) dz dy dx $$

where the functional notation I've used should be replaced by some expressions, and I've assumed $x$ for the outer integration, $z$ for the innermost integration.

I've left you a particularly tricky point to work out, perhaps with the aid of drawing a picture.  The problem says $y = -2x$ is part of the boundary, but which side of this plane is the region on?  Is $y$ supposed to be above or below $-2x$?  The final boundary piece is described as the $xz$-plane, which is where $y = 0$.  How does that fit with the other pieces?

If the order of integration were as I've suggested above, we'd have to find a constant $c$ that is the maximum of the range for $x$, and I've filled in $0$ as the minimum of that range.  The limits of integration on $y$ can then depend on the particular value of $x$, and further the limits of $z$ on both $x$ and $y$.

Evaluating the integral is then a matter of working from the innermost one out, doing the integration with respect to $x$ (or so I've assumed) last.

Title: Applications of Double/Triple Integrals


Title: Solving $\int_0^{\pi/2}x\cot x\,\mathrm{d}x$ while running into a $0\times\infty=0$ problem

Body:
While i have been trying to solve the integral  $ \int_0^{\pi/2} x\cot x \, \mathrm{d}x $ i have noticed that by trying integrating by parts using $u = x$ and $\mathrm{d}v = \cot x \, \mathrm{d}x$, i get that:

$$ \int_0^{\pi/2} x\cot x \, \mathrm{d}x = x\ln\sin x\Big|_0^{\pi/2} - \int_0^{\pi/2} \ln\sin x\,\mathrm{d}x $$

Plugging the first integral in Wolfram Alpha and after that the resulting integral in the right hand side i noticed that they have the same value! So the evaluation of $u\cdot v$ from $0$ to $\pi/2$ must be $0$, but how is this possible? I mean $\sin 0= 0$ and $\ln 0= -\infty$, so, how is that the term $x\sin x\Big|_0^{\pi/2}$ can be $0$? Or how am i wrong?

Ps: I am trying to integrate by parts first because i am trying to solve the integral by differentiation under the integral sign, if you have any suggestions for this i would also appreciate it!

Body:
$$I=\int_{0}^{\pi/2} x \cot x dx$$
Integrate by parts
$$I=x \ln \sin x-|_{0}^{\pi/2}-\int_{0}^{\pi/2} \ln \sin x$$
Using $\lim_{x \to 0}$ by L'Hospitalof the first part, we get
$$I=-\int_{0}^{\pi/2}\sin x dx~~~~(1)$$
Use the property $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$
Get $$I=-\int_{0}^{\pi/2}\ln \cos x dx~~~~(2)$$
add (1) and (2), then
$$2I=-\int_{0}^{\pi/2} \ln \sin x \cos x dx=\int_{0}^{\pi/2} \ln \sin 2x dx -\frac{\pi}{2}\ln 2~~~~(3)$$
Take $2x=t$, then
$$2I=-\frac{1}{2}\int_{0}^{\pi} \ln \sin t dt+\frac{\pi}{2}\ln 2~~~~(4)$$
Use the property: $\int_{0}^{a} f(x) dx=\int_{0}^{a/2}[f(x)+f(a-x)] dx,$ then
$$2I=-\int_{0}^{\pi/2} \ln \sin t dt +\frac{\pi}{2}\ln 2~~~(5)$$
Using (1), we get $$2I=I+\frac{\pi}{2}\ln 2 \implies I=\frac{\pi}{2}\ln 2$$


问:
Understanding a proof of the uniqueness lemma for bases
I'm having trouble understanding this proof about the uniqueness of bases, specifically, a line in the forward direction (highlighted in the image). I feel like I'm missing something obvious, and any clarification would be appreciated.

答:
Since $S$ is linearly independent, the linear dependence relation would normally force all those scalars to be zero; however, by assumption, these scalars are all nonzero. So, $a_1 = b_l$ for some $l$.

Another thing that follows, for the same reason, is that $m = k$. This should’ve been mentioned in the proof, as it is not obvious.


答:
Suppose $v = a_1u_1+...+a_nu_n = b_1w_1+...+b_kw_k$ with $u_i,w_i \in S$ and $a_i,b_i \in \mathbb{F}\setminus\{0\}$. Then we have that
$$
a_1u_1+...+a_nu_n -b_1w_1-...-b_kw_k = 0
$$
Suppose $a_1 \neq b_l$ and $u_1 \neq w_l$ for all values of $l$. This implies the existence of a nontrivial relation, since $a_1 \neq 0 $ by assumption. Since $u_1,...,u_n, w_1,...,w_k \in S$ and $S$ is linearly independent, this yields a contradiction. This lets you 'pair off' $a_1u_1$ with $b_lw_l$ for some value of $l$, also implying that $n = k$. As with so many things, you can formalize this with induction.

Title: Understanding a proof of the uniqueness lemma for bases

Body: I'm having trouble understanding this proof about the uniqueness of bases, specifically, a line in the forward direction (highlighted in the image). I feel like I'm missing something obvious, and any clarification would be appreciated.


问:
Principal normal vector of a parabolic path is not orthogonal
I have a parameterized equation of a path that looks like a parabola, $\textbf{r}(t) = \langle t, t^2 \rangle $. The principal normal vector is $\textbf{N}(t) = \frac{\textbf{T}'(t)}{|\textbf{T}'(t)|}$. Without normalizing, which doesn't change the direction of the vector, we can find the tangent vector $\textbf{T}(t) \propto \langle 1, 2t \rangle$, and the normal vector $\textbf{N}(t) \propto \langle 0, 2 \rangle$. At any time t, the normal vector $\textbf{N}(t)$ and the tangent vector $\textbf{T}(t)$ are not orthogonal to each other. I read somewhere that it has to do with curvature. Is there an easier way to understand why this simple path's tangent and normal vectors are not orthogonal to each other?

The theorem in the Briggs Calculus textbook states:

Let $\textbf{r}$ describe a smooth parameterized curve with unit tangent vector $\textbf{T}$ and principal unit normal vector $\textbf{N}$. Then, $\textbf{T}$ and $\textbf{N}$ are orthogonal at all points of the curve; that is, $\textbf{T} \cdot \textbf{N} = 0$ at all points where $\textbf{N}$ is defined.

I would argue that this path satisfies the conditions to this theorem, so not sure why the two vectors wouldn't be orthogonal.

答:
I say this without having seen your computations, but I suspect your issue was subbing the velocity vector $\textbf{v}(t) = \textbf{r}'(t)$ for the unit tangent vector $\textbf{T}(t) = \textbf{v}(t)/\|\textbf{v}(t)\|.$

If I try to compute $$\textbf{N}(t) = \frac{\textbf{v}'(t)}{\|\textbf{v}'(t)\|} = \frac{\textbf{a}(t)}{\|\textbf{a}(t)\|},$$ instead of correctly using $$\textbf{N}(t) = \frac{\textbf{T}'(t)}{\|\textbf{T}'(t)\|},$$ because I mistakenly assume $\textbf{T}(t) = c\textbf{v}(t)$ for some fixed $c$ and all $t$, then of course my resulting $\textbf{N}(t)$ is going to fail to be orthogonal to $\textbf{T}(t)$.

This is because when $\|\textbf{v}(t)\|$ is non-constant (i.e. I'm traversing the curve with changing speed), the acceleration vector $\textbf{a}(t) = \textbf{v}'(t)$ has components both in the direction of $\textbf{v}(t)$ (changing the speed) and in a direction orthogonal to $\textbf{v}(t)$ (changing the direction). You can find this equation in most textbooks on vector calculus:

If the acceleration vector is written as $$\textbf{a} = a_{T} \textbf{T} + a_{N} \textbf{N},$$ then $$a_T = \frac{d}{dt} \| \textbf{v} \| \text{ and } a_N = \kappa \| \textbf{v} \|^2.$$

Because the unit tangent vector's length never changes, $\textbf{T}'(t)$ solely represents the change in direction of $\textbf{T}(t)$, which is the same as the change in direction of $\textbf{v}(t)$. Normalizing then gives us a unit normal vector $\textbf{N}(t)$. Another  user has already posted the correct equations for the unit tangent and normal in the specific case of the parabola, as well as verifying that they're orthogonal.


问:
When is a comparison test inconclusive and when is a limit comparison test inconclusive?
Apologies if the question is too broad or if it has been answered before, but I understand what makes a comparison test convergent and divergent, but what makes a comparison test inconclusive? I know that if the smaller series is divergent, then the larger one is also divergent. And if the larger series is convergent, then the smaller one is too. But what makes a comparison test inconclusive?</p>

答:
Any test is a comparison with some known series (or class of series). So for example the ratio test is about checking if your series is below a convergent geometric series or above a divergent geometric series. When the limit in that test is $1$, it means that your series is, in some sense,  in the gap between all convergent geometric series and all divergent geometric series, and you need a more sensitive test to draw a conclusion.

So for example Raabe's test can be conclusive when $\lim a_{n+1}/a_n=1$ if the difference $\lim a_{n}/a_{n+1}-1$ is of the order of $1/n$, and identifies a generalized harmonic series $\sum 1/n^p$ that behaves asymptotically as the given series. There are many other such tests.

The key idea here is that given a convergent series, there is always another convergent series that converges more slowly, and given a divergent series, there is another series that diverges, its terms decaying faster.


问:
model fuzzy assignment problem
i wanna ask something about fuzzy assignment, why in fuzzy assignment problem specially in min z form didn't using $tilde{x}$ in $x_{ij}$ meanwhile it is used in $c_{ij}$ and also $z$.

\begin{align*}
 \min \widetilde{z}=&{\sum^{n}_{i=1}}{\sum^{n}_{j=1}}\widetilde{c}_{ij}x_{ij}
\end{align*}

答: The $x$ has no tilde because $x$ is not fuzzy. Instead, $x$ is a binary decision variable whose value is to be chosen to minimize the objective function subject to the constraints.

Title: model fuzzy assignment problem


问:
Why is this statement of a limit not necessarily true?

I feel as though I understand the solution of (b) [algebraic manipulation correct?]

But what is going on with (a)? What is the brace trying to convey? Where does the 5 come from? Why is one not equal to zero but the other is? Does this have anything to do with the conditions of the quotient law of limits?

Thank you

答: The braces denote a piecewise expression; $5$ could be replaced by any non-zero number. The point of this exercise is that the limit says nothing about the value of the function at the point being approached; it only deals with behaviour in a neighbourhood of that point.


问:
Is the zero map unique?
If I have a linear transformation $T$, from a vector space $V$ to $V’$, and this transformation is a zero map, i.e. $T(v)=0$, $\forall$ $ v \in V$, is $T$ unique? Is there any other transformation $T_o$, say, which is also a zero map?

I guess this is not possible since,
$T(v)=0=T_o(v)$
$\implies T(v)=T_o(v)$
Since $v \in V$ is arbitrary, the maps $T$ and $T_o$ are one and the same and thus the zero map must necessarily be unique. Is this the right way to prove this?

答: Yes, this correct. Concisely, a function $f:X\to Y$ between sets is specified uniquely by its values on the elements of $X$.

Title: Is the zero map unique?


问:
About the definition of a singular and non-invertible matrix
In Wolfram Mathworld, they defined a singular matrix as:

A square matrix that does not have a matrix inverse.

So my questions regard this definition are:

Why is the definition of singular matrices restricted to square matrices? Aren't all rectangular matrices singular and non-invertible since they don't have inverses?

When talking about matrix inverse, does it only refer to two-sided inverses and not one-sided inverses?

Can the terms singular and non-invertible be used interchangeably?

The invertible matrix is defined as having a determinant not equal to zero. Do rectangular matrices also have determinants and can this rule be applied to rectangular matrices as well?

答:
You are correct that all non-square matrices are non-invertible. This is why the term "singular" is reserved for the square case: the colloquial meaning of "singular" is "unusual" and non-invertibility is unusual for square matrices but not for non-square matrices. Similarly, we want to be able to say things like "a (square) matrix is singular iff its determinant is nonzero" and we need the matrix to be square to talk about determinants.

"Inverse" always means "two-sided inverse." It happens that a square matrix has a left inverse iff it has a right inverse iff it has an inverse but this isn't true in more general contexts.

Yes, keeping in mind the restriction to square matrices.

An invertible matrix is a matrix with an inverse. It's a theorem that this is equivalent to having a nonzero determinant, for a square matrix. Non-square matrices don't have determinants so this theorem does not apply to them. (However, they do have minors, and you can state a generalization of "invertible iff determinant nonzero" which describes the rank of an arbitrary matrix in terms of the vanishing and nonvanishing of its minors.)


问:
Does $ \sum_1^\infty {(-1)^n \over n \ \log_x(p_n)}$ converge for all $x$?
Does
$$ \sum_1^\infty {(-1)^n \over n \ \log_x(p_n)}$$
converge for all x, where $p_n$ is the nth prime?

答: Since $\frac{1}{n\log_x(p_n)} \geq 0$, and decreases monotonically to zero, your series converges by the alternating series test.


问:
Convergence of $\sum a_n$ (quasi-Cauchy condensation test)
Let $a_n\geq a_{n+1}$ for $\forall n\in\mathbb{N}$.

Is There any example of $a_n$ which satisfies $$\sum a_n<\infty \nLeftrightarrow\sum 2^na_{2^n}<\infty$$

I found an example of $a_n$ which satisfies $$\sum a_n<\infty \nLeftrightarrow\sum 2^na_{2^n}<\infty$$when decreasing condition deleted from cauchy condensation conditions, but It's hard to generate example in this case.

I think $a_n$ must have negative terms but I don't have any idea to construct $a_n$.

So, I need any hints for constructing $a_n$ which satisfies this or proving there's no $a_n$ for this.

Thank you

答: If you are using $<\infty$ for convegence (which should not be done!) then there is no such example. If $ \sum a_n$ converges then $a_n \to 0$. Since $a_n$ is decreasing it follows that $a_n \geq 0$ for all $n$. If $\sum 2^{n}a_{2^{n}}$ converges then $2^{n}a_{2^{n}} \to 0$ which implies $a_{2^{n}} \to 0$. But then $a_n \geq 0$ for all $n$.


问:
Permutation-invariant embedding preserving as much information as possible
Suppose there are 2 discrete random variables $x$ and $y$ that each can take $n$ possible values, and $n$ is not too large, at most 10. Then their joint PMF $p \in \mathbb{R}^{n^2}$ is a histogram of $n^2$ bins, which are all non-negative and sum up to 1. I am interested in finding a map $f: \mathbb{R}^{n^2} \rightarrow \mathbb{R}^m$ which satisfies the following:

It is symmetric under exchange of the random variables
$$f(p[x, y]) = f(p[y, x])$$

It preserves as much information about its argument as possible. That is, given $f(p)$ is should be possible to reconstruct as much of $p$ as possible. I'm not quite sure how exactly to formalize this statement, I welcome suggestions.

The dimension $m$ is a free parameter and can be chosen sufficiently large to satisfy the above constraints. Smaller $m$ are preferable if constraints are satisfied. The goal is to find an algorithm that implements any solution to this problem and demonstrate that it satisfies the conditions. In case optimal solution is intractable, an approximate solution may also be of interest.

My attempts:

$f(p)$ = const. For sure satisfies first condition but not the second.

$m = n^2$, two indices go over the values of $x$ and $y$ respectively and

$$
f_{ij} = \begin{cases}
  \min(p_{ij}, p_{ji}) & i < j \\
  \max(p_{ij}, p_{ji}) & i \geq j
\end{cases}$$

This looks like a good candidate solution, but I don't know how to check if it is optimal or if one can do better. I strongly suspect it is not optimal, because, for example, given $p$ I can test if $x$ and $y$ are independent, but given this version of $f$ I can't, and this information should not depend on their order.

Edit: Let me make the independence example more clear. If $x$ and $y$ are statistically-independent, then $p[x,y] = p[x]p[y]$, where $p[x] = \sum_y p[x,y]$. The test for statistical independence is $\Delta = p[x,y] - p[x]p[y] = 0$. $\Delta$ is invariant of the order of $x$ and $y$, and thus an optimal $f(p)$ should preserve that information. It is easy to verify that my attempt 2. does not preserve this information, so it is not optimal

Edit 2: There was a request to know the purpose of this exercise. It is as follows. I have a recording of many variables (e.g. 50) over many trials (e.g. 1000). I want to explore what sort of pairwise and triple-wise statistical relations are likely among these variables. For this purpose, I can compute all pairwise and triple-wise empirical PMF's. Now I would like to perform some classification/clustering on those PMF's. A standard tool to use a low-dimensional embedding, e.g. tSNE. However, to do the embedding, the dimensions of the PMFs have to have the same meaning for all PMFs. This is where the problem comes in - if I can succeed to filter out the permutation-asymmetric part from the PMFs, the resulting vectors will all have the same meaning, and can be clustered. For this reason, there are other properties of $f$ that are desirable for good result. Namely, it would be good if $f$ does not experience discontinuous jumps due to small perturbations. Generally, it would be good if $f$ does not amplify small differences and does not dampen large differences.

答:
Here is a standard tool which does something like this, although it satisfies a modified version of your first condition. Namely, you can think of the joint distribution in $\mathbb{R}^{n \times n}$ as an $n \times n$ matrix, compute the $k$ largest singular values $\sigma_1 \ge \dots \ge \sigma_k$ in its singular value decomposition $P = U \Sigma V^T$, and take the truncated SVD

$$P_k = U_k \Sigma_k V_k^T$$

where only the $k$ largest singular values and the corresponding singular vectors are kept and the others are discarded. By the Eckhart-Young theorem this is the best approximation of $P$ by a matrix of rank $\le k$, with respect to both the operator / spectral norm and the Frobenius / Hilbert-Schmidt norm. The accuracy of the approximation is controlled by the next largest singular value $\sigma_{k+1}$ (at least in the operator norm) so if this is small then the approximation is good.

This is like trying to compute the best approximation of the joint distribution by a mixture of $k$ distributions where $x$ and $y$ are independent (these correspond to rank-$1$ matrices) except that $P_k$ is not guaranteed to be non-negative or to sum to $1$. It's an interesting question what is the actual best approximation by a mixture of $k$ distributions where $x$ and $y$ are independent; I don't know the answer.

Exchanging the random variables corresponds to taking transposes, and the truncated SVD has the property that its transpose

$$P_k^T = V_k \Sigma_k U_k^T$$

is the truncated SVD of $P^T$; in other words it is not invariant with respect to transpose but is instead naturally equivariant. You can force it to be invariant by identifying $P_k$ and $P_k^T$ but this doesn't seem to me to be at all a natural thing to do.


问:
The nature of variables within proofs
I am confused about how best to view certain kinds of variables that show up in the course of an elementary proof. For example, to prove that the square of every odd integer is odd, I would write:

Let $m \in \mathbf{Z}$ be arbitrary. Suppose $m$ is odd. Choose $k \in \mathbf{Z}$ such that $m = 2k + 1$. Then $m^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1$, so since $2k^2 + 2k \in \mathbf{Z}$, it follows that $m^2$ is odd.

The symbol $m$ is introduced as part of the technique of universal generalization; Velleman [1] describes this step as "introducing a new variable [$m$] ... to stand for an arbitrary object." The symbol $k$ is introduced as part of the technique of existential instantiation; [1] describes this as "introducing a new variable [$k$] to stand for an object for which [the predicate $m = 2k + 1$] is true."

In what sense are $m$ and $k$ variables? They do not seem to be bound variables, as they are in the scope of no quantifier. They also can't be free variables, because if they were, then, for example, the sentence being proved true after the introduction of arbitrary integer $m$ -- namely that if $m$ is odd then $m^2$ is odd -- would be a predicate and not even have a truth value; I am also obviously not free to substitute values into $m$ and $k$, if for no other reason than $m$ and $k$ are dependent.

Stoll [2] says a variable is free if it is not bound, so there is no third option.

Is it more correct to think of the phrases "let $m \in \mathbf{Z}$ be arbitrary" and "choose $k \in \mathbf{Z}$ such that ..." as introducing new constants rather than variables, as their values are fixed despite being unspecified ($m$) or unknown ($k$)? More generally, the idea of a symbol that stands for a particular but arbitrary element of some set seems incompatible with the typical description of a variable as a placeholder as seen in, e.g., Epp [3].

[1] Velleman, D. (2006). How to Prove It: A Structured Approach (2nd ed.). Cambridge: Cambridge University Press.

[2] Stoll, Robert Roth (1963). Set Theory and Logic. W.H. Freeman.

[3] Epp, Susanna S. Variables in Mathematics Education. https://condor.depaul.edu/~sepp/VariablesInMathEd.pdf

答: $m$ is a variable in that its value is not fixed, but due to the relation between $m$ and $k$ once you fix one of them you determine the value of the other (and $m$ must be odd). You are not proving the statement for particular constants but for arbitrary ones, which is to say, variables.

Title: The nature of variables within proofs


答:
In what sense are $m$ and $k$ variables? They do not seem to be bound variables, as they are in the scope of no quantifier.

Actually every time you say something like "let $m \in \mathbb{Z}$ be arbitrary" the word "arbitrary" is code for an implicit universal quantifier $\forall m \in \mathbb{Z}$, in this case part of the statement

$$\forall m \in \mathbb{Z} : \left( \exists k \in \mathbb{Z} : 2k+1 = m \right) \Rightarrow \left( \exists k' \in \mathbb{Z} : 2k'+1 = m^2 \right)$$

you are trying to prove. So it is a bound variable, bound to a quantifier which is being left implicit. This is a very common convention in mathematics which is usually not described explicitly. Then we appeal to a general principle about how to prove a result involving a universal quantifier, which I guess is called universal generalization: to prove $\forall m, P(m)$ you introduce $m$ as an arbitrary variable and write down a proof where you assume nothing about $m$ except whatever premises might be in the statement $P(m)$ and see if you can deduce the desired conclusion.

Title: The nature of variables within proofs

Body:
I am confused about how best to view certain kinds of variables that show up in the course of an elementary proof. For example, to prove that the square of every odd integer is odd, I would write:

Let $m \in \mathbf{Z}$ be arbitrary. Suppose $m$ is odd. Choose $k \in \mathbf{Z}$ such that $m = 2k + 1$. Then $m^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1$, so since $2k^2 + 2k \in \mathbf{Z}$, it follows that $m^2$ is odd.

The symbol $m$ is introduced as part of the technique of universal generalization; Velleman [1] describes this step as "introducing a new variable [$m$] ... to stand for an arbitrary object." The symbol $k$ is introduced as part of the technique of existential instantiation; [1] describes this as "introducing a new variable [$k$] to stand for an object for which [the predicate $m = 2k + 1$] is true."

In what sense are $m$ and $k$ variables? They do not seem to be bound variables, as they are in the scope of no quantifier. They also can't be free variables, because if they were, then, for example, the sentence being proved true after the introduction of arbitrary integer $m$ -- namely that if $m$ is odd then $m^2$ is odd -- would be a predicate and not even have a truth value; I am also obviously not free to substitute values into $m$ and $k$, if for no other reason than $m$ and $k$ are dependent.

Stoll [2] says a variable is free if it is not bound, so there is no third option.

Is it more correct to think of the phrases "let $m \in \mathbf{Z}$ be arbitrary" and "choose $k \in \mathbf{Z}$ such that ..." as introducing new constants rather than variables, as their values are fixed despite being unspecified ($m$) or unknown ($k$)? More generally, the idea of a symbol that stands for a particular but arbitrary element of some set seems incompatible with the typical description of a variable as a placeholder as seen in, e.g., Epp [3].

[1] Velleman, D. (2006). How to Prove It: A Structured Approach (2nd ed.). Cambridge: Cambridge University Press.

[2] Stoll, Robert Roth (1963). Set Theory and Logic. W.H. Freeman.

[3] Epp, Susanna S. Variables in Mathematics Education. https://condor.depaul.edu/~sepp/VariablesInMathEd.pdf


答:
universal generalization: "for all odd $m \in \Bbb{Z}$".

existential instantiation: "there exists a $k \in \Bbb{Z}$ such that $m = 2k+1$".

These are both quantified and they are quantified in the way you recite.

It is more accurate to think of variables as variables.  A statement with a universally quantified variable may be specialized, removing the universal quantifier, by replacing the occurrences of the variable with a particular member of the set from which that variable is drawn.  For instance, in your proof, an equally valid set of statements is produced by replacing every instance of $m$ by "$3$" (an odd integer).  Existentially quantified variables can be specialized to the value that satisfies their constraints.

When $m$ is generic (free to range over the odd integers), $k$ must be generic, since the value of $k$ is dependent on the value of $m$.  (In a more pedantic setting, the dependence of $k$ on $m$ would be made more apparent by writing, for instance $k(m)$, $k_m$, or some other syntax to indicate dependency.  This dependency enforces an order on the quantifiers.  Equivalently, the order of the quantifiers, $\forall m \text{ odd}, \exists k \in \Bbb{Z} \dots$.)  Even though $k$ is generic, in this case, we can solve for $k$ in terms of $m$: $k = \frac{m-1}{2}$.  Consequently, throughout, we may specialize the variable $k$ to the value $\frac{m-1}{2}$.

When $m$ is specialized to a value, we may replace $k$ with the value that is half of one less than the value of $m$.  We still end up with a valid set of statements -- this valid set of statements is the explicit proof of the oddness of the square of whatever value $m$ has taken.  This last should make it clear: $m$ and $k$ are variables.  For each specialization of $m$ and $k$ to (valid) values, we obtain a concrete proof.  If we do not specialize, we obtain a proof that is universally quantified over $m$ and existentially quantified over $k$ (which is dependent on $m$).

Title: The nature of variables within proofs


问:
Solving a PDE by method of characteristics
I solved the following:
\begin{align}
\frac{dy}{y}&=\frac{dy'}{2y'+wy}\\
\ln y + \ln C &= \frac{1}{2} \ln \left(2y'+wy\right)\\
2\ln y + \ln C &= \ln \left(2y'+wy\right)\\
C &= \frac{2y'}{y^2} + \frac{w}{y}\\
\end{align}

I isolated the constant since the purpose of solving is to find the differential invariant. However the answer given is:
$$
\frac{y'}{y^2}+\frac{w}{y}
$$

where am I going awry?

答:
Your mistake is, $y \ \& \ y'$ aren't independent variables, so you can't suppose them as constant when you integrate with respect to the other ones! You must follow Exact Differential Equations method instead. By inspiration from given answer, Write it as follows, to see its exactness:
$$\frac{dy}{y} = \frac{dy'}{2y'+\omega y} \Longrightarrow (2y'+\omega y)dy = ydy' \Longrightarrow ydy' - (2y'+\omega y)dy = 0 \Longrightarrow$$
$$\Longrightarrow \frac{1}{y^2}dy' + (\frac{-2y'}{y^3}+\frac{-\omega}{y^2})dy = 0 \Longrightarrow d(\frac{y'}{y^2}+\frac{\omega}{y}) = 0 \Longrightarrow \frac{y'}{y^2}+\frac{\omega}{y} = C$$


问:
Binomial theorem and a congruence modulo $2^{2m+2}$
I'm interested in the congruence problem $$(1 + 2^m)^{2x} \equiv 1+ 2^{m+1} \pmod{2^{2m+2}}\,,\quad m \geq 2\,.\quad(\dagger)$$

Though I can solve it for various small values of $m$, I still haven't found a proof that a solution exists for every such $m$.

For context, I'm reading a paper and thinking about the possibilities of extending its results. The author himself claims that extending the theorem in a certain direction is trivial, and part of this is because he claims that $(\dagger)$ always has a solution for each $m \geq 2$.

I have tried so far to expand $(1 + 2^m)^{2x}$ using the binomial theorem and reducing the problem to the quadratic equation $$2^mx^2 + (1-2^{m-1})x-1\equiv 0 \pmod{2^{m+1}}\,,\quad m\geq 2\,.\quad (\ddagger)$$ but that didn't helped much either.

I'm looking for hints to prove that $(\dagger)$ or equivalently $(\ddagger)$ aways has a solution $x$ for every such value of $m$.

答:
If we choose $x$ such that
$$
(1-2^{m-1})x\equiv 1+2^m\;(\text{mod}\;2^{m+1})
$$
then $2^mx^2$ and $(1-2^{m-1})x-1$ are both odd multiples of $2^m$, hence
$$
2^mx^2 + \bigl((1-2^{m-1})x-1\bigr)
$$
is a multiple of $2^{m+1}$.


问:
Cardinality of Center of p Groups $ |Z(G)| \neq p^{n-1}$
Question

Let G be a p group, such that $ |G|=p^n$. Prove that $ |Z(G)|\neq p^{n-1}$.

Beginning of Proof

So, I began by writing out $ p^n=|G|=\sum_{i\in I}[G:Stab_G(x_i)]+|Z(G)|=\sum_{i\in I}[G:Stab_G(x_i)]+p^{n-1}$, when $ \left\{x_i\right\}_{i\in I}$ are representatives in the following way $ \cup_{i\in I}x_iGx_i^{-1}\cup Z(G)=\cup_{x\in G}xGx^{-1}$, when $ \forall i\in I, |x_iGx_i^{-1}|\geq p$ (because otherwise $x_i \in Z(G)$). Therefore, $ \sum_{i\in I}[G:Stab_G(x_i)]=p^n(p-1)$.

Then, we can say that $ \forall i\in I, \exists j_i\in [1,n-1]: [G:Stab_G(x_i)]=p^{j_i}$, according to Lagrange's theorem. Since p-1 is even (if $ p\neq 2$), then in order for this sum of odd numbers to be even, then |I| must be even. Therefore, we can write that $|I|=2k$, when $ 2k \in [2,\frac{p^n(p-1)}{2}]$. $2k\neq p^n(p-1)$, because then, we'll have $p^n(p-1)$ numbers in the sum, and therefore every number in the sum must be 1, such that $\forall i\in I, |Stab_G(x_i)|=|G| \Rightarrow x_i\in Z(G)$. Therefore, $k\in [1,\frac{p^n(p-1)}{4}]$, and therefore, $p-1\equiv 0 (mod 4) \Rightarrow p \equiv 1 (mod 4)$. Therefore, $ |Z(G)|\neq p^{n-1}$, for $p: p \equiv 3(mod 4)$.

I haven't been able to prove this for all other primes p. Thanks a lot for the help!!!

答: Hint: If $|G| = p^n$ and $|Z(G)| = p^{n-1}$, $|G/Z(G)| = p$. Now take a look at If $G/Z(G)$ is cyclic, then $G$ is abelian. What can you conclude now?</p>


问:
$p$ prime , $n\in \mathbb{N_0},f= \sum_{j=0}^{n}a_jx^{jp}\in\mathbb{Z}_p[x]$, show that $f=g^p,g\in\mathbb{Z}_p[x]$ exists
I could prove that if it exists it is unique but I could not show the existence for such a function

My uniquencess proof draft

Assume $a^p=g^p$, then $a^{p}-g^p=0$ thus if $g=\sum_{k=0}^{m}a_kx^k$ and $g=\sum_{k=0}^{l}g_kx^k $

$(a_0+\dotsb+a_mx^m)^p=\sum_{k_0+\dotsb+k_m=p}{p\choose k_0,\dotsc,k_m}\cdot a_0^{k_1}\dotsm a_mx^{{m}^{k_m}}$

and

$(g_0+\dotsb+g_{l}x^l)^p=\sum_{k_0+\dotsb+k_l=p}{p\choose k_0,\dotsc,k_l}\cdot a_0^{k_1}\dotsm a_lx^{{l}^{k_l}}$

(I still have not proved this part) but the only way the difference can be zero is when the both polynomials are exactly the same

答: Hint: note that in $\Bbb Z_p$ we have $a_i= a_i^p$, and $(x+y)^p = x^p+y^p$.


问:
Prove that $f(x)=0$ for all $x∈\Bbb R$ if all rationals are mapped to $0$ using only topology
Prove that if all rationals are mapped to $0$ and f is continuous then f(x) = 0 for all x in R.

I have done this proof before using epsilon delta definition. I was wondering if this proof can be done using only topology and open and closed sets.

答: Suppose that there is an $x\in\Bbb R$ such that $f(x)\ne 0$. Let $a=f(x)$, let $U$ be an open nbhd of $a$ such that $0\notin U$, and let $V=f^{-1}[U]$. Then $V$ is an open nbhd of $x$, and $0\notin f[V]$, so $V\cap\Bbb Q=\varnothing$, which is absurd.


答:
Since $f$ is continuous, $f^{-1}(0)$ is closed.

But the rationals are dense in $\mathbb{R}$, hence $f^{-1}(0)=\mathbb{R}$.

Title: Prove that $f(x)=0$ for all $x∈\Bbb R$ if all rationals are mapped to $0$ using only topology

Body:
Prove that if all rationals are mapped to $0$ and f is continuous then f(x) = 0 for all x in R.

I have done this proof before using epsilon delta definition. I was wondering if this proof can be done using only topology and open and closed sets.


问:
Laurent series expansion of holomorphic/meromorphic functions
I'm not familiar with complex analysis, but I need to catch up a basic fact about holomorphic and meromorphic functions.

Let $\Omega\subseteq\mathbb C$, $f:\Omega\to\mathbb C$ and $z_0\in\Omega$.

Am I correct that

if $f$ is holomorphic at $z_0$, then there is a $\varepsilon>0$ and $(a_n)_{n\in\mathbb N_0}\subseteq\mathbb C$ with $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\;\;\;\text{for all }z\in B_\varepsilon(z_0)\tag1;$$

if $f$ is meromorphic at $z_0$, then there is a $\varepsilon>0$, $n_0\in\mathbb N_0$, $a_{-n_0},\ldots,a_{-1}\in\mathbb C$ and $(a_n)_{n\in\mathbb N_0}\subseteq\mathbb C$ with $$f(z)=\sum_{n=-n_0}^\infty a_n(z-z_0)^n\;\;\;\text{for all }z\in B_\varepsilon(z_0)\tag2.$$

If so, does the converse hold as well? If not, how do we need to correct the claims? In any case, I'd highly appreciate if someone could confirm whether I'm right or wrong and point me to a reference which provides rigorous proofs.

答:
Claim 1. is correct. This is a very famous result of complex analysis, namely that holomorphic functions are analytic. Claim 2. on the other hand is a little strange. If $f$ is holomorphic, no terms of negative exponent is needed in its Laurent series. On the other hand, if it is, the function is said to have a pole at this point, which means it is not holomorphic there. More precisely:

If $f$ is holomorphic on $r < |z-z_0| < R$ and its Laurent series in this domain is given by

$$f(z)= \sum_{n=-m}^{\infty} a_n(z-z_0)^n,$$

with $m>0$, $f$ is said to have a pole of order $m$ at $z_0$.

When it comes to the converse, you are really asking if series are holomorphic, and the answer is a little subtle. Uniformly convergent series are holomorphic, but more specifically series that "look like" Laurent series, are in fact Laurent series for a holomorphic function.


Title: Solving a PDE by method of characteristics

Body:
I solved the following:
\begin{align}
\frac{dy}{y}&=\frac{dy'}{2y'+wy}\\
\ln y + \ln C &= \frac{1}{2} \ln \left(2y'+wy\right)\\
2\ln y + \ln C &= \ln \left(2y'+wy\right)\\
C &= \frac{2y'}{y^2} + \frac{w}{y}\\
\end{align}

I isolated the constant since the purpose of solving is to find the differential invariant. However the answer given is:
$$
\frac{y'}{y^2}+\frac{w}{y}
$$

where am I going awry?

Body:
$$\frac{dy}{y}=\frac{dy'}{2y'+wy}$$
$$(2y'+wy){dy}-y{dy'}=0$$
$$wy{dy}+2y'dy-y{dy'}=0$$
Multiply by $y$
$$wy^2{dy}+y'dy^2-y^2{dy'}=0$$
Divide by $y^4$:
$$\dfrac w {y^2}{dy}+\dfrac {y'dy^2-y^2{dy'}}{y^4}=0$$
$$\dfrac w {y^2}{dy}-d\dfrac {y'}{y^2}=0$$
Integrate:
$$-\dfrac w y- \dfrac {y'}{y^2}=k$$
$$\dfrac w y+ \dfrac {y'}{y^2}=C$$


问:
Rank of a module over an integral domain
Suppose $A$ be an integral domain. The rank of an $A$-module $M$ is defined to be the maximal number of $A$-linear independent elements of $M$.

Let $S=A-\{0\}$. Then $S^{-1}A:=k$ is a field and it can be proved that rank $M=$ the $k$-vector space dimension of $M\otimes_{A}k$. Hence rank is well defined.

Now suppose $B\supset A$, where $B$ is also a domain. Let $T=B-\{0\}$ and $L=T^{-1}B$. It is also given that $L/k$ is finite algebraic extension. Let $[L:k]=r$.

I want to prove that $\operatorname{rank}B$ as an $A$-module is equals to $r$. Clearly $k\subset S^{-1}B\subset L$. Hence $\operatorname{rank}B\leq r$.

How to prove that rank is exactly equals to $r$?

Thank you in advance.

答:
There is a canonical map of $Q(A)$-vector spaces $\iota: B \otimes_A Q(A) \to Q(B) \otimes_{Q(A)} Q(A)$ where $Q(A)$ denotes the fraction field of a domain $A$. The map just sends $b \otimes \frac as$ to $\frac b1 \otimes \frac as$. This map induces an isomorphism of $Q(B)$-modules
$$
Q(B) \otimes_B (B \otimes_A Q(A)) \simeq Q(B) \otimes_B (Q(B) \otimes_{Q(A)} Q(A))
$$
because both are naturally isomorphic to $Q(B)$ by the multiplication map, and this natural isomorphism commutes with $Q(B) \otimes_B \iota$. If $\iota$ had a non-zero kernel or cokernel, then by exactness of localization so would $Q(B) \otimes_B \iota$, a contradiction (this is an exercise: use the existence of an exact sequence of the form $0 \to B \to \ker \iota \to B \otimes_A Q(A)$ of $B$-modules to derive the contradiction). It follows that $\iota$ is an isomorphism.

This is in some way the algebraization of the proof that says that a basis of $Q(B)$ over $Q(A)$ gives a maximally $A$-linearly independent subset of $B$ by clearing denominators; tensoring with $Q(B)$ over $B$ is what allows us to clear denominators.

Hope that helps,


问:
Is the set $U=\{ x \in \mathbb{R} : \sin (x) = \frac{1}{2} \}$ closed or open set?
I tried to simplify the set definition by solving for different values of $x$'s and ended up with this set.
$U = \{ ... , -\pi-\frac{\pi}{6} , \frac{\pi}{6} , \pi-\frac{\pi}{6}, ... \}$
I know that a set is said to be closed if all of its cluster points belong to the set.
I don't know how to go ahead with the problem.
Thanks for the help in advance
Edit 1: From the comments although I don't know exactly why the set is closed but yes the set U is closed. Then π/6 is a cluster point i.e every neighborhood of π/6 has atleast one element in U leaving π/6 . How is that true?

答:
Take any arbitrary $x$=nπ + $(-1)^n$$\frac{π}{6}$.

Now let $\epsilon$ < min {[(n+1)π + $(-1)^{(n+1)}$$\frac{π}{6}$] - [nπ + $(-1)^n$$\frac{π}{6}$], [nπ + $(-1)^n$$\frac{π}{6}$] - [(n-1)π + $(-1)^{(n-1)}$$\frac{π}{6}$] }.

Although it looks a bit complex but it's not at all complex. If you see the format of any three consecutive numbers in U, you'll understand what I have written in that minimum part. Now let's say $x_1$, $x$, $x_2$∈U such that $x_1$ < $x$ < $x_2$. Now what I have done is I have taken $\epsilon$<min{$x$-$x_1$, $x_2$-$x$}.(You can take equality but < makes the argument stronger and more explicit.) Then the neighbourhood N($x$,$\epsilon$) contains no point of U.This shows that there exists a neighbourhood of $x$ such that it doesn't contain any point of U. Hence, x is not a limit point. (Remember the definition of limit point). Since x is arbitrary, none of the points of U is a limit point. Hence, derived set U$'$ = {$\phi$} $\subset$ U and therefore, U is a closed set.


答:
Simply we can understand the Meromorphic functions is that:- Only singularity at poles i.e., a function which is analytic everywhere in the finite plane except at a finite number of poles.
Simply, Holomorphic functions are analytic functions which satisfies Laplace equation.</p>

Title: Laurent series expansion of holomorphic/meromorphic functions

Body:
I'm not familiar with complex analysis, but I need to catch up a basic fact about holomorphic and meromorphic functions.

Let $\Omega\subseteq\mathbb C$, $f:\Omega\to\mathbb C$ and $z_0\in\Omega$.

Am I correct that

if $f$ is holomorphic at $z_0$, then there is a $\varepsilon>0$ and $(a_n)_{n\in\mathbb N_0}\subseteq\mathbb C$ with $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\;\;\;\text{for all }z\in B_\varepsilon(z_0)\tag1;$$

if $f$ is meromorphic at $z_0$, then there is a $\varepsilon>0$, $n_0\in\mathbb N_0$, $a_{-n_0},\ldots,a_{-1}\in\mathbb C$ and $(a_n)_{n\in\mathbb N_0}\subseteq\mathbb C$ with $$f(z)=\sum_{n=-n_0}^\infty a_n(z-z_0)^n\;\;\;\text{for all }z\in B_\varepsilon(z_0)\tag2.$$

If so, does the converse hold as well? If not, how do we need to correct the claims? In any case, I'd highly appreciate if someone could confirm whether I'm right or wrong and point me to a reference which provides rigorous proofs.


问:
Showing $e^{-x^2}$ small when $x$ is large just from series definition
This is a sort of open ended question. Suppose
$$F(x):=\sum_{L=0}^{\infty} (-1)^L\frac{x^{2L}}{L!}.$$
One can show that the series is absolutely convergent for every fixed $x$, by noting that $x^{2L} \le (L-2)!$ for all large enough $L$. So, $F(x)$ is well defined. I want to show $F(x)$ is very small for $x$ large enough.

I DON'T want to use the fact that $F(x)$ is actually $e^{-x^2}$. Actually, in my research, I am dealing with a similar problem where I have an alternating series quite similar to exponential (but no closed form), where I need to show its small when $x$ is large. So, I was trying with this toy problem.

答:
Here is a method that might work even if you don't have the function in closed form.

First it is seen that since $$F(x):=\sum_{k=0}^{\infty}{\frac{(-1)^kx^{2k}}{k!}}$$

Is a power series with infinite radius of convergence, it has derivatives of all orders that can be obtained by differentiating the terms of the series. In particular,$$F'(x)=\sum_{k=1}^{\infty}{\frac{(-1)^kx^{2k-1}2k}{k!}}=2\sum_{k=1}^{\infty}{\frac{(-1)^kx^{2k-1}}{(k-1)!}}=-2xF(x)$$

for all $x\in \mathbb{R}$. Which can be seen with a shift of index.
$$\\$$
Next we want to show that $1\geq F(x)>0$ on $[0,\infty)$

$F(0)=1>0$. Suppose that $F(x_0)<0$ for some $x_0>0$. Since $F$ is continuous, there is some $c\in(0,x_0)$ such that: $F(c)=0$. We can also assume that there is a largest such $c$, because otherwise $c$ could get arbitrarily close to $x_0$, which would imply that $F(x_0)=0$ via the continuity of $F$.

Since $F<0$ on $(c,x_0)$, then by the relation between $F$ and its derivative, we have that $F'>0$ on $(c,x_0)$. Thus $F$ is increasing on $(c,x_0)$ which implies that $F(x)>F(c)=0$ on $(c,x_0)$. Which is a contradiction.

It can also be shown that $F(x)\neq 0$ on $[0,\infty)$ - I'll leave this to you -
Thus, $0<F(x)\leq 1$ on $[0,\infty)$.
$$\\$$
Lastly,
$$F''(x)=-2xF'(x)-2F(x)=F(x)\left(4x^2-2\right)$$
$$$$
So $x_1=\frac{1}{\sqrt{2}}$ is the only point on $[0,\infty)$ where $F''=0$.

If $0\leq x \leq \frac{1}{\sqrt{2}}\implies F''\leq 0 \implies F'(\frac{1}{\sqrt{2}})\leq F'(x)\leq 0$

If $x\geq \frac{1}{\sqrt{2}}\implies F''\geq 0\implies F'(\frac{1}{\sqrt{2}})\leq F'(x)\leq 0$

Thus $|F'|$ is bounded on $[0,\infty)$.
$$$$
Since, $|F(x)|=\left|\frac{F'(x)}{2x}\right|$, we can now see that $F$ is small if $x$ is large.