问:
Testing whether forecasting models differ significantly in their forecasts
I am not finding my question on here which I sort of find it hard to believe unless I am simply on the wrong stack board. If it is on this board, kindly direct me and I will be grateful. Otherwise, I have two models used to forecasting the same data set. I want to find out if my models are producing significantly different forecasts or not. How would I go about doing this?

Null: Models do not differ.
Alt: Models do differ from one another.

Models are used to forecast demand.

答:
Comments are correct that this question is quite vague, somewhat
less so in view of your response. But I will try to give an answer
that shows various statistical procedures that may be useful.

Suppose you have 100 paired forecasts as in vectors x1 and x1 below;

How well do they agree?
We can look at descriptive statistics of their differences:

```<code>d = x2 - x1
summary(d); length(d);  sd(d)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-0.2626  1.3748  2.0057  1.9440  2.5832  5.0812
[1] 100        # number of differences
[1] 0.9689753  # standard deviation

hist(d, prob=T, col="skyblue2")
</code>```

So it appears that forecasts $X_2$ are mostly larger than
forecasts $X_1.$ However, the two forecasts are highly correlated,
so it may not make much difference which forecast is used.

```<code>cor(x1,x2)
[1] 0.9989661

plot(x1, x2);  abline(0, 1, col="green2")
</code>```

Nevertheless, we do know that the forecasts are not exactly the same,
and we can do a formal test whether the two are significantly
different in a statistical sense. The histogram of differences
looks roughly normal, so we use a paired t test. The tiny P-value
very near $0$ shows that the two forecasts are significantly different.

```<code>t.test(x1, x2, pair=T)

       Paired t-test

data:  x1 and x2
t = -20.063, df = 99, p-value < 2.2e-16
alternative hypothesis:
  true difference in means is not equal to 0
95 percent confidence interval:
 -2.136304 -1.751773
sample estimates:
mean of the differences
              -1.944039
</code>```

A nonparametric Wilcoxon paired (signed rank) test also shows a hightly
significant difference with a P-value very near $0:$

```<code>wilcox.test(x1, x2, pair=T)$p.val
[1] 6.407145e-18
</code>```

However, statistical significance is not the same thing as practical importance.
If the fact that $X_2$ forecasts average about $2$ higher then
$X_1$ forecasts is not important, and because the two never seem to be
far apart, then it may not make any practical difference which
forecast is used.

By contrast, if a difference of $2$ is of practical importance, we
should look at the record of past performance and use the forecasting
method that has been most often correct.

Note: The following R code was used to simulate x1 and x2:

```<code>set.seed(121)
x1 = rgamma(100, 5, .1)
x2 = x1 + rnorm(100, 2, 1)
</code>```

Body:
Comments are correct that this question is quite vague, somewhat
less so in view of your response. But I will try to give an answer
that shows various statistical procedures that may be useful.

Suppose you have 100 paired forecasts as in vectors x1 and x1 below;

How well do they agree?
We can look at descriptive statistics of their differences:

d = x2 - x1
summary(d); length(d);  sd(d)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-0.2626  1.3748  2.0057  1.9440  2.5832  5.0812
[1] 100        # number of differences
[1] 0.9689753  # standard deviation

hist(d, prob=T, col="skyblue2")

So it appears that forecasts $X_2$ are mostly larger than
forecasts $X_1.$ However, the two forecasts are highly correlated,
so it may not make much difference which forecast is used.

cor(x1,x2)
[1] 0.9989661

plot(x1, x2);  abline(0, 1, col="green2")

Nevertheless, we do know that the forecasts are not exactly the same,
and we can do a formal test whether the two are significantly
different in a statistical sense. The histogram of differences
looks roughly normal, so we use a paired t test. The tiny P-value
very near $0$ shows that the two forecasts are significantly different.

t.test(x1, x2, pair=T)

       Paired t-test

data:  x1 and x2
t = -20.063, df = 99, p-value < 2.2e-16
alternative hypothesis:
  true difference in means is not equal to 0
95 percent confidence interval:
 -2.136304 -1.751773
sample estimates:
mean of the differences
              -1.944039

A nonparametric Wilcoxon paired (signed rank) test also shows a hightly
significant difference with a P-value very near $0:$

wilcox.test(x1, x2, pair=T)$p.val
[1] 6.407145e-18

However, statistical significance is not the same thing as practical importance.
If the fact that $X_2$ forecasts average about $2$ higher then
$X_1$ forecasts is not important, and because the two never seem to be
far apart, then it may not make any practical difference which
forecast is used.

By contrast, if a difference of $2$ is of practical importance, we
should look at the record of past performance and use the forecasting
method that has been most often correct.

Note: The following R code was used to simulate x1 and x2:

set.seed(121)
x1 = rgamma(100, 5, .1)
x2 = x1 + rnorm(100, 2, 1)


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

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

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

My Approach:

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

Case 1:

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

Case 2:

taking $y=1 \Longrightarrow$

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

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

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

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

答:
Given
$\qquad x+y=2\space\land\space y+z=4\qquad$
we can start by eliminating $y$.
$$
y+z=4\implies z=4-y\\
(x+y=2)\implies x=2-y\\
z-x=(4-y)-(2-y)=2\qquad\longrightarrow (z-x=2)\\
(x+z=4)-(z-x=2)\implies 2x=2\implies x=1\\
(x+z=4)+(z-x=2)\implies 2z=6\implies z=3\\
\therefore\quad x+z=1+3=4
$$
At this point we can also see that $y=1$ but who cares? Ha ha

ContentLicense: CC BY-SA 4.0

Title: Is there an other way for it?


问:
How we can know the ramification ideals geometrically?
Let $L/\mathbb{Q}$ be a finite Galois extension of degree n, let $\mathcal{O}_{L}$ be the ring of integers of $L$,
By Dedekind lemma we have that

$\mathfrak{p}=\mathfrak{b}_{1}^{e}...\mathbb{b}_{g}^{e}$
where $ \mathfrak{b}_{i} \in \mathcal{O}_{L}$ and $\mathfrak{p}$ $\in $$\mathbb{Z}$
Since this extension is Galois we have
$gfe=[L:\mathbb{Q}]=n$ where $e$ is the ramification index and $f$ is the inertia degree which is given by $f_{i}=[\mathcal{O}_{L}/\mathfrak{b} :\mathbb{Z}/\mathfrak{p}]$

we know that $\mathfrak{p}$ is ramified if $e >1 $,
I am asking is there a geometric way to know whether a prime ideal is ramified or not? if it exists can you apply it on this picture (give any example)
Also, I want to know what is the geometric interpretation of the inertia degree and the ramification index ?

答:
Note that if $k'/k$ is any finite separable extension of fields, then $k' \otimes_k \bar{k} \cong \bar{k}^{[k' : k]}$. So, given an extension of number fields $L/K$, a nonzero prime ideal $\mathfrak{p} \subseteq \mathcal{O}_K$ with residue field $k = \mathcal{O}_K/\mathfrak{p}$ is unramified in the extension $\mathcal{O}_L/\mathcal{O}_K$ if and only if the geometric fiber $\operatorname{Spec}(\mathcal{O}_L/\mathfrak{p} \otimes_{k} \bar{k})$ (that is, the base change of the fiber, which is a $k$-scheme, to the algebraic closure $\bar{k}$) has exactly $[L : K]$ closed points.

This exactly mirrors the situation for branched coverings of Riemann surfaces, where a branch point is exactly one where the fiber has fewer points than the degree of the covering map. The difference is that since we're dealing with non-algebraically-closed fields, we might have to extend the residue field in order to "see" the full size of the fiber.

From this perspective, here are geometric interpretations of inertia degree and ramification index of a prime $\mathfrak{b} \in \operatorname{Spec} \mathcal{O}_L$:

The inertia degree is the number of geometric points lying above $\mathfrak{b}$. In other words, the inertia degree tells you how many points a given prime splits into under extension of the base field.

The ramification index, conversely, is essentially the multiplicity with which points occur in the geometric fiber. So, for example, if $\mathfrak{p}$ is totally ramified, then the geometric fiber above $\mathfrak{p}$ will be of the form $\operatorname{Spec} A$, where $A$ is a $\bar{k}$-algebra that is $[L : K]$-dimensional as a $\bar{k}$-vector space, and we can think of this as a "point with multiplicity $[L : K]$".


问:
Number of pairs $(A, B)$ such that $A \subseteq B$ and $B \subseteq \{1, 2, \ldots, n\}$
I need to determine the number of pairs $(A,B)$ of sets such that $A \subseteq B$ and $B \subseteq \{1,2,\ldots,n\}$

My solution:

How $A \subseteq B$ and $B \subseteq \{1,2,\ldots,n\}$, we have:

$$A :=  B := \sum_{i = 0}^{n} \left(\begin{array}{c} n \\ i \end{array} \right)$$

in this way, we have that the number of pairs $(A,B)$ of sets are:

$$\sum_{i=1}^{n} \left( \begin{array}{c} n \\ i \end{array}\right) + \frac{\sum_{i=1}^{n} \left( \begin{array}{c} n \\ i \end{array}\right)}{2} \cdot \left( \begin{array}{c} n \\ 0 \end{array}\right) + \left( \begin{array}{c} n \\ 0 \end{array}\right) \\ 2^n + 2^{n-1} + 1 $$

Here is a little example.

Suppose $n = 2$

Then $A \subseteq B$ and $B \subseteq \{1,2\}$. With this, we have:

$$(A, B) = (\emptyset,\emptyset), (\emptyset,{1}), (\emptyset,{2}), (\emptyset, {1,2}), ({1}, {1}), ({1}, {1,2}), ({1,2}, {1,2})$$

答:
We use $[n]$ to denote the set of natural numbers $\{1,2,\ldots,n\}$.

We obtain
\begin{align*}
\color{blue}{\sum_{{(A,B)}\atop{A\subseteq B\subseteq[n]}}1}
&=\sum_{B\subseteq [n]}\sum_{A\subseteq B}1\\
&=\sum_{j=0}^n\sum_{{B\subseteq [n]}\atop{|B|=j}}\sum_{k=0}^j\sum_{{A\subseteq B}\atop{|A|=k}}1\tag{1}\\
&=\sum_{j=0}^n\binom{n}{j}\sum_{k=0}^j\binom{j}{k}\tag{2}\\
&=\sum_{j=0}^n\binom{n}{j}2^j\tag{3}\\
&\,\,\color{blue}{=3^n}\tag{4}
\end{align*}

Comment:

In (1) we rearrange the sum according to terms with increasing size of subsets $A$ and $B$.

In (2) we use the fact that the number of subsets of size $q$ of a finite set $X$ is $\binom{|X|}{q}$.

In (3) and (4) we apply the binomial theorem.

Example: $n=2$

We have the following $3^2=9$ pairs $(A,B)$ when considering $[2]=\{1,2\}$:

\begin{align*}
&(\emptyset,\emptyset),\\
&(\emptyset,\{1\}),\,(\{1\},\{1\}),\\
&(\emptyset,\{2\}),\,(\{2\},\{2\}),\\
&(\emptyset,\{1,2\})\,(\{1\},\{1,2\}),\,(\{2\},\{1,2\}),\,(\{1,2\},\{1,2\})
\end{align*}


问:
Proof of the Extended Reeb's Theorem for dimensions less than 7
Reeb's Theorem states that every compact smooth manifold which admits a Morse funtion with exactly two critical points is homeomorphic to n-sphere.

I have heard an extension of this theorem for dimensions less than 7, as homeomorphism replaced by diffeomorphism and couldn't found any proof yet.

One might say it is a direct consequence of Smooth Poincare Conjecture for dimensions less than 7 which states the equivalency of being homeomorphic to standart n-sphere and being diffeomorphic to standart n-sphere.

However, since the reason I am asking this question is that this theorem does constitutes a half of the proof of the theorem which states that exotic 4-spheres, if any, have at least 4 Morse critical points, I wonder if there exist a Poincare Conjecture independent proof of this extended Reeb's Theorem, since Smooth Poincare Conjecture in dimension 4 is still an open problem.

答:
This question amounts showing that for all integer $1\leqslant n\leqslant 6$, the twisted $n$-sphere group (denoted by $\Gamma_n$) is trivial, where $\Gamma_n$ fits in the following short exact sequence:
$$\pi_0\operatorname{Diff}^+(D^n)\to\pi_0\operatorname{Diff}^+(S^{n-1})\to\Gamma_n\to 0,$$
where $S^{n-1}$ is seen as the boundary of $D^n$ and the first map is induced by the restriction, then $\Gamma_n$ basically measures the obstruction for a diffeomorphism of $S^{n-1}$ to extend to a diffeomorphism of $D^n$.

Let me explain why this is true. Let $M^n$ be a closed manifold and $f\colon M\to\mathbb{R}$ be a Morse function on $M$ with two critical points. Since $M$ is compact, the critical points of $f$ are a minimum $x_-$ and a maximum $x_+$, thus the stable manifold of $x_-$ and the unstable manifold of $x_+$ (for any Riemannian metric on $M)$ are $n$-dimensional disks. The proof of Reeb's theorem now boils down to showing that $M$ is obtained by gluing these disks along their boundaries, but this gluing map is an orientation preserving diffeomorphism of $S^{n-1}$. If is isotopic to the identity, then $M$ is diffeomorphic to the standard $n$-sphere.

It is easy to show that $\Gamma_1$ and $\Gamma_2$ are trivial (the extension is explicit and straightforward), however, the higher dimensional cases are more subtle:

for $n=3$, this is a result due independently to Munkres, 1960 (the proof is elementary and uses the Poincaré-Bendixson theorem) and Smale, 1959,

for $n=4$, this is a hard result due to Cerf, 1968 (see also this wonderful contact geometric proof),

for $n=5$ or $6$, this is a result due to Kervaire and Milnor, 1963.


问:
Regex with at least three even digits and odd length
As the title said, I need to make a regex for a number with at least three even digits and odd length.
As an example, the number 248 should match.
Thank you!</p>

答:
If you are allowed to create capturing and non-capturing groups (e.g. python) then it is possible to reduce the regex by having an AND operation performed rather than having to list all combinations like in Brian's answer.

```<code>(?=oddcount)(?:evendigits)

oddcount = ^[0-9]([0-9][0-9])+$

evendigits = ^[0-9]*[02468][0-9]*[02468][0-9]*[02468][0-9]*$
</code>```

Note: I enclosed expression inside ^...$ to match the whole string, not just some part of it.

Test it here: https://regex101.com/r/dgbngY/1

Body:
If you are allowed to create capturing and non-capturing groups (e.g. python) then it is possible to reduce the regex by having an AND operation performed rather than having to list all combinations like in Brian's answer.

(?=oddcount)(?:evendigits)

oddcount = ^[0-9]([0-9][0-9])+$

evendigits = ^[0-9]*[02468][0-9]*[02468][0-9]*[02468][0-9]*$

Note: I enclosed expression inside ^...$ to match the whole string, not just some part of it.

Test it here: https://regex101.com/r/dgbngY/1


答:
First we count the possibilities for $B$.  So if $B$ has $k$ elements, $0 \leq k \leq n$, then there are ${n \choose k}$ possibilities. For each of these possibilities for $B$, there are
$$ \sum_{i=0}^{{n \choose k}} {{n \choose k} \choose i} $$
possibilities for $A$. So to conclude, if $B$ has $k$ elements, then there are
$$ {n \choose k} \cdot \sum_{i=0}^{{n \choose k}} {{n \choose k} \choose i} $$
possible pairs of sets $(A,B)$.

Since we can choose any $0 \leq k \leq n$, sum over all values of $k$:
$$ \sum_{k=0}^{n} \left( {n \choose k} \cdot \sum_{i=0}^{{n \choose k}} {{n \choose k} \choose i} \right)$$

Title: Number of pairs $(A, B)$ such that $A \subseteq B$ and $B \subseteq \{1, 2, \ldots, n\}$

Body:
I need to determine the number of pairs $(A,B)$ of sets such that $A \subseteq B$ and $B \subseteq \{1,2,\ldots,n\}$

My solution:

How $A \subseteq B$ and $B \subseteq \{1,2,\ldots,n\}$, we have:

$$A :=  B := \sum_{i = 0}^{n} \left(\begin{array}{c} n \\ i \end{array} \right)$$

in this way, we have that the number of pairs $(A,B)$ of sets are:

$$\sum_{i=1}^{n} \left( \begin{array}{c} n \\ i \end{array}\right) + \frac{\sum_{i=1}^{n} \left( \begin{array}{c} n \\ i \end{array}\right)}{2} \cdot \left( \begin{array}{c} n \\ 0 \end{array}\right) + \left( \begin{array}{c} n \\ 0 \end{array}\right) \\ 2^n + 2^{n-1} + 1 $$

Here is a little example.

Suppose $n = 2$

Then $A \subseteq B$ and $B \subseteq \{1,2\}$. With this, we have:

$$(A, B) = (\emptyset,\emptyset), (\emptyset,{1}), (\emptyset,{2}), (\emptyset, {1,2}), ({1}, {1}), ({1}, {1,2}), ({1,2}, {1,2})$$


问:
simplify equation by removing double summation
Is it possible to simplify the following function by removing the double summation?

$$f(x) = \sum_{n=1}^{x-1} \sum_{m=n+1}^x a_{n}b_{m}$$

Or is there no way of removing the sigmas?

Thanks in advance

答:
There's not much you can do, but if you like you could write

\begin{align}
f(x)
&=
\sum_{n=1}^{x-1}\sum_{m=n+1}^xa_nb_m
\\
&=
\sum_{n=1}^{x-1}\sum_{m=n}^{x-1}a_nb_{m+1}
\\
&=
\sum_{{\scriptstyle m,n=1}\atop{\scriptstyle m\ge n}}^{x-1}a_nb_{m+1}
\end{align}

ContentLicense: CC BY-SA 3.0


问:
Continuity in the Mean Value Theorem and the Implicit Function Theorem
Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be a $\mathcal{C}^2$ function of $x\in\mathbb{R}^n$. Then we can use the mean value theorem (MVT) to formulate the following equality:
$$f(x_1)-f(x_2)=\frac{\partial f}{\partial x}(x_2+\lambda(x_1-x_2))(x_1-x_2),$$
with $\lambda\in(0,1)$. This equality is point-wise true by the MVT for every pair $(x_1,x_2)$. Furthermore, it is easy to show with a simple example that for a pair $(x_1,x_2)$, there might be multiple $\lambda$ that yield the equality.

I'm currently looking in the continuation of $\lambda$ when $x_1,x_2$ vary continuously. There are also quite simple examples to formulate that show that continuous variation of $x_1,x_2$ yield discontinuous continuations of $\lambda$. (Check e.g. this example, and move point B to 4)

Now I want to show that for every pair $(x_1,x_2)$ you can find a neighborhood $\mathfrak{B}$ around $(x_1,x_2)$ for which there exists a continuous function $\lambda(x_1,x_2)$ such that for all $(x_1,x_2)\in\mathfrak{B}$, we have that $f(x_1)-f(x_2)=\frac{\partial f}{\partial x}(x_2+\lambda(x_1,x_2)(x_1-x_2))(x_1-x_2)$ and for any continuous trajectory that takes values in $\mathfrak{B}$, the function $\lambda(x_1,x_2)$ is continuous. Trying to counterexample this with multiple examples in geoabgebra (and others) did not help. Up until now I found that this must be true in my setting, I just don't know how to prove this...

What I tried:
I tried to show this with the implicit value theorem by analyzing the implicit function
$$\mathscr{F}(x_1, x_2, \lambda)= f(x_1)-f(x_2)-\frac{\partial f}{\partial x}(x_2+\lambda(x_1-x_2))(x_1-x_2),$$
However, the classic version requires a differentiability condition: $\frac{\partial^2 f}{\partial x^2}\neq 0$ at the point where $\mathscr{F}=0$, which I cannot guarantee perse - plus this seems like a very restrictive assumption for my case. I found a variation of the theorem in [1]. But this theorem requires a one-to-one relationship, which is for me not the case, as there might be multiple lambda's. This was the closed I could get, but no concrete results yet...

Can anybody help me with this problem?

[1] Jittorntrum (1978) - TECHNICAL NOTE: An Implicit Function Theorem

答:
Too long for a comment.

However, the classic version requires a differentiability condition: $\frac{\partial^2 f}{\partial x^2}\neq 0$ at the point where $\mathscr{F}=0$, which I cannot guarantee perse - plus this seems like a very restrictive assumption for my case. I found a variation of the theorem in [1]. But this theorem requires a one-to-one relationship, which is for me not the case, as there might be multiple lambda's.

Unfortunately, some one-to-one conditions are necessary for the continuity of $\lambda$. For instance, suppose that $n=1$ and $f’(x_2)\ne \frac{f(x_1)-f(x_2)}{x_1-x_2}$. For each $x$ such that $\lambda(x)$ is defined put $h(x)=x+\lambda(x_1,x)(x_1-x)$. Then $f’(h(x))=\frac{f(x_1)-f(x)}{x_1-x}$ is monotonic (and so one-to-one) on some open neighborhood $U$ of $x_2$. Since $h$ is a continuous one-to-one map on a non-empty open subset $U$ of $\Bbb R$, by the invariance of domain $h(U)$ is an open subset of $\Bbb R$ and $h$ is a homeomorphism between $U$ and $h(U)$. Moreover, $f’$ is one-to-one on $h(U)\ni h(x_2)$.


问:
Liu's "Algebraic Geometry and Arithmetic Curves" - Proposition 4.4. (Ch. 7) and Exercises 5.1.29 and 5.1.30
In Liu's book, Chapter 7, Proposition 4.4, the question is about closed embeddings (or: closed immersions) coming from very ample divisors. The theorems are quite well-known when the ground field $k$ is algebraically closed (e.g., Hartshorne's Proposition 7.3 and Sec. 7, Chap. 2 there, in general). But Liu does not assume $k$ to be algebraically closed because of the Exercises 5.1.29 and 5.1.30. and not assuming $k$ to be algebraically closed is also a major difference to various other texbooks!

Here it is said that the statements in the exercises could be incorrect. Can one prove them (at least) in special cases?

Ex. 5.1.29: Let $\pi: X'\rightarrow X$ be a faithfully flat morphism of schemes, $F$ quasi-coherent sheaf on $X$.

(a) There is a morphism $\phi: \mathcal{O}_{X}^{(I)} \rightarrow F$ such that $\phi(X)$ is surjective. Further, $F$ is generated by global sections iff $\phi$ is surjective.

(b) $F$ generated by global sections iff $\pi^{\ast}F$ generated by global sections.

(c) Now $X,X'$ quasi-compact, $L$ invertible sheaf on $X$. If $\pi^{\ast}L$ is ample, then $L$ is ample.

Ex. 5.1.30 (light version): Let $X$ be aproper scheme over $\mathbb{R}$, $L$ invertible sheaf on $X$, $\pi: X_{\mathbb{C}} \rightarrow X$ projection (to base change $\mathbb{R}$ to $\mathbb{C}$) and let $\pi^{\ast}L$ be very ample.

(a) Show that $L$ is ample and generated by some global sections $s_0, \ldots , s_n$. Let $f: X \rightarrow Y:=\mathbb{P}^{n}_\mathbb{R}$ be the morphism associated to $L$ and the $s_i$. Show that $f_\mathbb{C}: X_\mathbb{C} \rightarrow Y_\mathbb{C} $ is a closed immersion.

(b) Let $y \in Y$ and let $y' \in Y_\mathbb{C}$ be a point lying over $y$. Show that the fiber of $X_\mathbb{C}$ over $y'$ is isomorphic to $X_y \times_{Spec(k(y))} Spec(k(y')) $. Deduce from this that $f^{-1}(y)$ consists of at most one point, and that $f$ is a topological closed immersion.

(c) Show that $\mathcal{O}_Y \rightarrow f_{\ast}\mathcal{O}_X$ is surjective and that $f$ is a closed immersion.

答:
Exercise 5.1.29(b) and (c) are just plain wrong. Mohan's counterexamples are completely correct, and some further exposition of the general case of a cyclic cover of a curve can be found here. The key issue is that there is no good way to go from a surjective map $\mathcal{O}_{X'}^I\to \pi^*\mathcal{F}$ to a surjective map $\mathcal{O}_{X}^I\to \mathcal{F}$.

On the other hand, when we are in the situation of exercise 5.1.30, everything works out. The big idea that lets us bypass the issue from exercise 5.1.30 is that $H^0(X,\mathcal{L})\otimes_A B \cong H^0(X_B,\mathcal{L}_B)$ by flat base change (ref Stacks 02KH, for instance). So pick a generating set $\{s_i\}_{i\in I}$ for $H^0(X,\mathcal{L})$, construct the obvious morphism $g:\mathcal{O}_X^I\to \mathcal{L}$, and then note that $\{s_i\otimes 1\}_{i\in I}$ is a generating set for $H^0(X,\mathcal{L})\otimes_A B = H^0(X_B,\mathcal{L}_B)$. Thus $\pi^*g:\pi^*\mathcal{O}_X^I\to\pi^*\mathcal{L}$ is a map $\mathcal{O}_{X_B}^I\to\mathcal{L}_B$ which is surjective on global sections. So $\mathcal{L}_B$ is globally generated iff $\pi^*g$ is surjective, which happens if and only if $g:\mathcal{O}_X^I\to\mathcal{L}$ is surjective by the argument in Niven's post addressing 5.1.29(b). This fixes all of the issues in Niven's outline of a solution to exercise 5.1.30.


问:
Extension of a continuous function $f: \mathbb{Q}\to\mathbb{R}$ to a continuous function $g:\mathbb{R}\to\mathbb{R}$.
Prove or disprove that for all continuous functions $f:\mathbb{Q}\to\mathbb{R}$ there is a unique continuous function $g:\mathbb{R}\to\mathbb{R}$ with $g(x)=f(x),\ \forall x\in\mathbb{Q}$.

Now, this is a task consisting of 2 subtask and this is the latter. In the first one (I think) I have proved that this statement is true for uniformly continuous functions. Now I will try to disprove this statement.

Let us choose $x,y\in\mathbb{R}$ with $d(x,y)<\delta/3$ where $x$ is fixed. Now let us choose $x_n\to x\wedge y_n\to y,\ x_n,y_n\in\mathbb{Q}$. Thereby $d(x_n,y_n)<\delta$. In my other proof, I now used the property for uniformly continuous functions and said that $d(f(x_n),f(y_n))<\varepsilon/2$ and followed that $g$ is uniformly continuous and unique. However, I can't use that premise here, which is why I think that this statement is false as $d(f(x_n),f(y_n))$ can be anything because of $x_n$ not being fixed (and we can't apply the continuity).

How do I go from here / is this the right path in the first place?

Any help would be greatly appreciated.

答:
Is false

Let
$$
f:\Bbb Q\to\Bbb R\\
f(x) =
\begin{cases}
1,  & \text{if $x<\sqrt2$} \\
0, & \text{if $x>\sqrt2$}
\end{cases}
$$
The function is continuos but
$$
\lim_{x\to \sqrt2^{+}}f=0\\
\lim_{x\to \sqrt2^{-}}f=1
$$
So it cant be extended to a continuos function in $\Bbb R$.


问:
Difficult integral $\int_{\nu+\epsilon}^{2\nu}\frac{x^{2}}{(x-\nu)^{4}+b}dx$
As part of breaking up a larger integral to solve, I have ended up with the following integral:
\begin{equation*} \int_{\nu+\epsilon}^{2\nu}\frac{x^{2}}{(x-\nu)^{4}+b}dx \end{equation*}
where $\nu,b>0$ are fixed constants and $\epsilon\in (0,\nu)$ is small as I want. I have tried everything I can think of (i.e. various substitutions, breaking the integral apart even more, wolframalpha, etc..) Maybe I have missed something or maybe there is some way to do this with partial fraction decomposition, but I am not sure.

I have a solution to this integral if I assume instead that $b=0$, but in this case letting $\epsilon\to 0$ makes the answer blow up and I need it to stay bounded. In any case, I would be very grateful for any help/hints in solving the integral if $b>0$.

Edit: If an explicit solution to this integral is difficult, it would suffice for my purposes to show that the integral is bounded above by a positive constant that is independent of $\nu$ and $\epsilon$.

Body:
Rescale the integral with $t= \frac{x-v}{b^{1/4}}$, $a=\frac{v}{b^{1/4}}$ and $\delta=\frac{\epsilon}{b^{1/4}}$

\begin{align*}
& \int_{v +\epsilon}^{2v}\frac{x^{2}}{(x-v)^{4}+b}dx
=\frac1{b^{1/4}} \int_{\delta}^{a}\frac{t^{2}+2at +a^2}{t^{4}+1}dt\\= &\frac1{b^{1/4}} \int_{\delta}^{a}\left( \frac{1+a^2}2 \frac{t^{2}+1}{t^{4}+1}
 + \frac{1-a^2}2 \frac{t^{2}-1}{t^{4}+1}+ \frac{2at}{t^{4}+1}  \right)dt \\= &\frac{1+a^2}{2\sqrt2 b^{1/4}} \left( \tan^{-1}\frac{a^2-1}{\sqrt2a} - \tan^{-1}\frac{\delta^2-1}{\sqrt2\delta} \right) \\
&\hspace{2mm} -  \frac{1-a^2}{2\sqrt2 b^{1/4}} \left( \coth^{-1}\frac{a^2+1}{\sqrt2a} - \coth^{-1}\frac{\delta^2+1}{\sqrt2\delta} \right)
+ \frac a{b^{1/4}} \tan^{-1} \frac{a^2-\delta^2}{1+a^2\delta^2}
\end{align*}


问:
Continuity and composition
Let $A \subset X$, where $X$ is a topological space and $\iota_A :A \to X$ the inclusion map. Let now also $Y$ be a topological space.

I wanna show that $f:Y \to A$ is continuous iff $\iota_A \circ f: Y \to X$ is continuous.

Well one side is easy: If $f$ is continuous, so also $\iota_A \circ f$ is continuous as a composition of continuous function. But I can't proof the other one.

答:
An open set of $A$ is the intersection of $A$ and an open set of $X$. Let $U$ be open in $X$, one has
\begin{equation}
f^{-1}(A\cap U) = (\iota_A \circ f)^{-1}(U)
\end{equation}
whence the equivalence.

Title: Continuity and composition


问:
A continuous and derivable decreasing function s.t. the derivative is $0$ for $t>t_0$?
The question may be silly, but I am stucked with this thought...

I ask for a nonconstant continuous and derivable nondecreasing function s.t. the derivative approaches $0$ for $t\to\infty$, but more than that I might have $f'(t_0)=0$ for some $t_0$ and also for all $t>t_0$.

I really cannot think this example. For instance, $f(t)=\text{log}(t)$ does not mind, because $f'$ approaches to $0$, but $f'(t)\neq0\forall t$. And so on.

But because this question?

Because I have an equation on a model of form $\dfrac{d f}{d t}=g(t)$ with $g(t)$ nonpositive and I might have $g(t)=0$ for all $t>t_0$ for some $t_0>0$, but I cannot realize this.

Thank you so much.

Body: Let $f(x)=-x^2$ when $x\le 0$ and $f(x)=0$, when $x>0$.


问:
Prove that $\sqrt{1+\pi^2}, \hspace{2pt}\pi-\sqrt\pi,\hspace{2pt} \pi^2+\pi+\sqrt{1+2\pi}$ are not algebraic
Considering the fact that $\pi$ is not an algebraic number, I need to prove these expressions are not algebraic :

$$\sqrt{1+\pi^2},\hspace{10pt} \pi-\sqrt\pi,\hspace{10pt} \pi^2+\pi+\sqrt{1+2\pi}$$

Thanks

Some explanations will be appreciated ! Thank you all for your time .

答:
If $\alpha_1$ is algebraic, then so is $\alpha_1^2-1$.

If $\alpha_2$ is algebraic, then so is $\left(\sqrt{\alpha_2+\frac14}+\frac12\right)^2$.

Can you see a similar reason for $\alpha_3$?


问:
Example of dense subset in topological space that does not have a metric
We just went over the Zariski topology in my algebraic geometry class and we see the zero ideal is dense in $\text{Spec}(\mathbb{Z})$. This made me realize that I cant think of any other good examples of a dense subset in a topological space that isnt just a metric space. Does anyone have any good examples?

答: For instance, on $\Bbb R$ consider the topology$$\tau=\{\emptyset,\Bbb R\}\cup\{(-\infty,a)\mid a\in\Bbb R\}.$$Then $\Bbb Z$ is dense on $(\Bbb R,\tau)$. Actually, every subset of $\Bbb R$ without an upper bound is dense in $(\Bbb R,\tau)$.


问:
Matrices which determinant is obvious.
At least in my current Linear Algebra course, exercises concerning the determinant of a matrix of dimension $n\geq 5$ more often than not require one to notice some property of the matrix so that its determinant becomes obvious and one needs not compute it in the usual way. I'm trying to find a simple way to realize that the determinant of
$$\begin{pmatrix} 0 & -1 & -1 & -1 & -1
\\
1 & 0 & -1 & -1 & -1
\\
1  & 1 & a & -1 & -1
\\
1 & 1 & 1 & 0 & -1
\\
1 & 1 & 1 & 1 & 0\end{pmatrix}$$

is equal to $a$, and I would appreciate anyone who points me in the right direction.

Also, since these are the types of questions that might go to my (as well as other students') final, I'm open to users giving other examples of matrices which determinants are obvious without the need to compute them, here are a few:

$$\det \begin{pmatrix}
-4 & 1 & 1 & 1 & 1
\\
1 & 1 & 1 & -4 & 1
\\
1  & 1 & -4 & 1 & 1
\\
1 & -4 & 1 & 1 & 1
\\
1 & 1 & 1 & 1 & -4\end{pmatrix}
=
\det\begin{pmatrix}
-4 & 1 & 1 & 1 & 1
\\
1 & 1 & 1 & -4 & 1
\\
0  & 0 & 0 & 0 & 0
\\
1 & -4 & 1 & 1 & 1
\\
1 & 1 & 1 & 1 & -4
\end{pmatrix}=0
$$

by adding rows $1,2,4,5$ to row $3$.

$$\det \begin{pmatrix}
1 & 2 & 3 & 4 & 5
\\
2 & 3 & 4 & 5 & 6
\\
3  & 4 & 5 & 6 & 7
\\
4 & 5 & 6 & 7 & 8
\\
5 & 6 & 7 & 8 & 9\end{pmatrix}
=\det \begin{pmatrix}
1 & 2 & 3 & 4 & 5
\\
1 & 1 & 1 & 1 & 1
\\
2  & 2 & 2 & 2 & 2
\\
4 & 5 & 6 & 7 & 8
\\
5 & 6 & 7 & 8 & 9\end{pmatrix}=0$$

by subtracting row $1$ from row $2$ & $3$.

$$\det \begin{pmatrix}
0 & -a & -b & -d & -g
\\
a & 0 & -c & -e & -h
\\
b  & c & 0 & -f & -i
\\
d & e & f & 0 & -j
\\
g & h & i & j & 0\end{pmatrix}=0$$

since $(-1)^5\det (A)=\det(A^T)=\det(A)$. All of these can be generalized to other dimensions, with the exception of the last one for which the dimension of the matrix must be odd.

答:
Using cofactor expansion, the determinant is equal to $aD_4+b$ where $D_4$ and $b$ are fixed, and $a$ is the unknown entry.

If $a=0$, then the matrix is skew-symmetric of odd-dimension, which means its determinant is zero. That implies that $b = 0$. So the determinant is $aD_4$.

$D_4$ is the determinant of an even-dimensional skew-symmetric matrix with $-1$s above the diagonal. This will always be $1$ (by an admittedly nontrivial induction argument). That means the determinant is $a$.

Title: Matrices which determinant is obvious.


问:
Computing Probabilities with Several Normally Distributed Random Variables
I encountered the following problem.

$X$, $Y$ and $Z$ are independent normally distributed random variables with expected values $2$, $1$, and $0$, respectively. They all have the same variance $2$. Compute $P(4X - 3Y > 5Z )$.

I don't really know how to approach this problem. I tried defining a new variable as $U = 4X - 3Y$, which would then have and expected value of $5$ and a variance of $14$, but then I don't know how to deal with this new variable U, given that we have still $5Z$ on the other side of the inequality. I never get to the correct solution of $0.6915$. Also, $Z$ is not yet standardized either, since its variance is $2$.

答:
Hint.

You need the "joint density" $f(x,y,z)$ and
$$
P(4X-3Y>5Z)=\iiint_{A}f(x,y,z)dxdydz
$$
where $A=\{(x,y,z):4x-3y>5z\}$

But since $X,Y,Z$ are independent, the joint density is given by $f(x,y,z)=h(x)g(y)l(z)$, where $h,g,l$ are the density functions for $X$, $Y$ and $Z$ respectively.

The approach above is a general one. But your problem has special properties.

@user6247850's answer is very nice and here is a remark about it.

In general, any linear combination of independent normal-distributed random variables is of normal distribution.

In particular, if $X_i\sim N(\nu_i,\sigma_i^2)$ are independent, then
$$
\sum a_iX_i\sim N(\sum a_i\mu_i,\sqrt{\sum a_i^2\sigma_i^2})
$$

See for instance this set of lecture notes: http://www.stat.ucla.edu/~nchristo/introstatistics/introstats_normal_linear_combinations.pdf

So once you know the mean $\mu$ and variance $\sigma^2$ for the random variable $4X-3Y-5Z$, you get
$$
P(4X-3Y-5Z>0)=1-P(\frac{4X-3Y-5Z-\mu}{\sigma}\le -\mu)=1-\Phi(-\mu)
$$
where $\Phi$ is the CDF for the standard normal.


答:
To each of the elements of $\{1,2,\ldots,n\}=[n]$, we assign a $a,b$ or $c$, independently. We have $3^n$ ways to do this.

For each assignment we make a pair $(A,B)$ as follows: if $i \to a$ we put $i$ in $A$ and also in $B$, if $i \to b$ we put $i$ in $B$ and if $i \to c$ we put it in neither $A$ nor $B$. Then $(A,B)$ is a good pair and to each such pair there is also a unique such assignment as above (based on the partition $A$,$B \setminus A$, $[n]\setminus (A \cup B)$, etc. So there are also $3^n$ such pairs.


问:
Riemann integral and Lebesgue integral
$f:R\rightarrow [0,\infty)$ is a Lebesgue-integrable function. Show that
$$
\int_R f \ d m=\int_0^\infty m(\{f\geq t\})\ dt
$$
where $m$ is Lebesgue measure.

I know the question may be a little dump.

答:
We have, using Fubini and denoting by$\def\o{\mathbb 1}\def\R{\mathbb R}$ $\o_A$ the indicator function of a set $A \subseteq \R$
\begin{align*}
  \int_\R f(x)\, dx &= \int_\R \int_{[0,\infty)}\o_{[0,f(x)]}(t)\, dt\,dx\\
        &= \int_{[0,\infty)} \int_\R \o_{[0,f(x)]}(t)\, dx\, dt\\
        &= \int_{[0,\infty)} \int_\R \o_{\{f \ge t\}}(x)\, dx\, dt\\
        &= \int_{[0,\infty)} m(\{f\ge t\})\, dt.
\end{align*}
For the third line note that
\begin{align*}
  \o_{[0,f(x)]}(t) = 1 &\iff 0 \le t \le f(x)\\
                       &\iff f(x) \ge t\\
                       &\iff x \in \{f\ge t\}\\
                       &\iff \o_{\{f\ge t\}}(x) = 1
\end{align*}
and hence $\o_{[0,f(x)]}(t) =  \o_{\{f \ge t\}}(x)$ for all $(x,t) \in \R \times [0,\infty)$.</p>

Title: Riemann integral and Lebesgue integral


答:
Denote your determinant by $f(a)$ so we get a function $f : \Bbb{R} \to M_5(\Bbb{R})$. Notice that $f(0)$ is a skew-symmetric $5 \times 5$ matrix and hence has determinant $0$ so $f(0)=0$.

The derivative of a determinant of a matrix function can be calculated by differentiating every column of the matrix and summing them up:
$$f'(a) = \begin{vmatrix} 0 & -1 & 0 & -1 & -1\\1 & 0 & 0 & -1 & -1\\1  & 1 & 1 & -1 & -1\\1 & 1 & 0 & 0 & -1\\1 & 1 & 0 & 1 & 0\end{vmatrix} = \begin{vmatrix} 0 & -1  & -1 & -1\\1 & 0  & -1 & -1\\1 & 1 & 0  & -1\\1 & 1 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 0 & -1  & -1 & -1\\1 & 0  & -1 & -1\\0 & 0 & 0  & -1\\0 & 0 & 1 & 0 \end{vmatrix}.$$
by adding the second and subtracting the first row from the third and fourth. The latter matrix is block upper-triangular so
$$f'(a) = \begin{vmatrix} 0 & -1  & -1 & -1\\1 & 0  & -1 & -1\\0 & 0 & 0  & -1\\0 & 0 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 0 & -1 \\ 1 & 0\end{vmatrix}\cdot \begin{vmatrix} 0 & -1 \\ 1 & 0\end{vmatrix} = (-1)\cdot(-1)=1.$$
Therefore $f(a)= a+C$ for some constant $C \in \Bbb{R}$. From $f(0)=0$ we get $C=0$ and hence $f(a)=a$.

Title: Matrices which determinant is obvious.

Body:
At least in my current Linear Algebra course, exercises concerning the determinant of a matrix of dimension $n\geq 5$ more often than not require one to notice some property of the matrix so that its determinant becomes obvious and one needs not compute it in the usual way. I'm trying to find a simple way to realize that the determinant of
$$\begin{pmatrix} 0 & -1 & -1 & -1 & -1
\\
1 & 0 & -1 & -1 & -1
\\
1  & 1 & a & -1 & -1
\\
1 & 1 & 1 & 0 & -1
\\
1 & 1 & 1 & 1 & 0\end{pmatrix}$$

is equal to $a$, and I would appreciate anyone who points me in the right direction.

Also, since these are the types of questions that might go to my (as well as other students') final, I'm open to users giving other examples of matrices which determinants are obvious without the need to compute them, here are a few:

$$\det \begin{pmatrix}
-4 & 1 & 1 & 1 & 1
\\
1 & 1 & 1 & -4 & 1
\\
1  & 1 & -4 & 1 & 1
\\
1 & -4 & 1 & 1 & 1
\\
1 & 1 & 1 & 1 & -4\end{pmatrix}
=
\det\begin{pmatrix}
-4 & 1 & 1 & 1 & 1
\\
1 & 1 & 1 & -4 & 1
\\
0  & 0 & 0 & 0 & 0
\\
1 & -4 & 1 & 1 & 1
\\
1 & 1 & 1 & 1 & -4
\end{pmatrix}=0
$$

by adding rows $1,2,4,5$ to row $3$.

$$\det \begin{pmatrix}
1 & 2 & 3 & 4 & 5
\\
2 & 3 & 4 & 5 & 6
\\
3  & 4 & 5 & 6 & 7
\\
4 & 5 & 6 & 7 & 8
\\
5 & 6 & 7 & 8 & 9\end{pmatrix}
=\det \begin{pmatrix}
1 & 2 & 3 & 4 & 5
\\
1 & 1 & 1 & 1 & 1
\\
2  & 2 & 2 & 2 & 2
\\
4 & 5 & 6 & 7 & 8
\\
5 & 6 & 7 & 8 & 9\end{pmatrix}=0$$

by subtracting row $1$ from row $2$ & $3$.

$$\det \begin{pmatrix}
0 & -a & -b & -d & -g
\\
a & 0 & -c & -e & -h
\\
b  & c & 0 & -f & -i
\\
d & e & f & 0 & -j
\\
g & h & i & j & 0\end{pmatrix}=0$$

since $(-1)^5\det (A)=\det(A^T)=\det(A)$. All of these can be generalized to other dimensions, with the exception of the last one for which the dimension of the matrix must be odd.


问:
Determine if $f(x) = \frac{100}{1+2^{-x}}$ is one-to-one
I currently have $$f(x) = \frac{100}{1+2^{-x}}$$ and I'm trying to determine if it is one-to-one. I have looked at the other posts however I'm not sure as to how to deal with the $-x$.

答:
It is because $y=2^{-x}$ is.  An exponential function to any positive base is.  You could take the derivative, and check that it's strictly decreasing.

Also, $y=\dfrac{100}{1+x}$ is one-to-one.  So we have the composition of two one-to-one functions.


答: Without using the joint density, you could define the new variable $V:= 4X-3Y-5Z$ which is a normal random variable with expected value $5$ and variance $32+18+50=100$ (the variances add even if the random variables are subtracted).  Then you just need $P(4X-3Y > 5Z) = P(V > 0)$, and $P(V>0)$ can be computed by standardizing $V$.

Title: Computing Probabilities with Several Normally Distributed Random Variables

Body:
I encountered the following problem.

$X$, $Y$ and $Z$ are independent normally distributed random variables with expected values $2$, $1$, and $0$, respectively. They all have the same variance $2$. Compute $P(4X - 3Y > 5Z )$.

I don't really know how to approach this problem. I tried defining a new variable as $U = 4X - 3Y$, which would then have and expected value of $5$ and a variance of $14$, but then I don't know how to deal with this new variable U, given that we have still $5Z$ on the other side of the inequality. I never get to the correct solution of $0.6915$. Also, $Z$ is not yet standardized either, since its variance is $2$.


问:
How can you know you are performing a legal operation?
There are many examples in math where you have fake proofs of something because you do something extraneous that you were not allowed to do. For example:

Let $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \log 2,$$ but $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{5} + \dots = \frac{3}{2}\log 2$$ (rearranging the series). In elementary calculus, we are taught that we cannot rearrange series precisely for this reason: We may change the value that the series equals. Another example would be the equation

$$\large{x^{x^{x^{x^{\dots}}}}} = 2.$$ Then we use the property that it repeats to see that $x^2 = 2,$ which implies that $x = \sqrt{2}$ for positive $x$. But then, if we wish to solve $$\large{x^{x^{x^{x^{\dots}}}}} = 4,$$ then using the same trick we see that $$x^4 = 4 \implies \sqrt[4]{x^4} = \sqrt[4]{4} \implies x = \sqrt{2}.$$ Once again, we did something we are not allowed to do.

My question is therefore: how do we know that we aren't doing something illegal when exploring mathematics? When we try to solve problems, or discover patterns or learn, how can we tell when we do something extraneous?

答:
These problems always surface when you try to work with infinite expressions. As was already pointed out in comments, infinite expressions like $x^{x^{x^{x^{\dots}}}}$ are not part of standard mathematical notation, so when dealing with them one needs to define what they mean. In this case, you probably mean something like
$$ x^{x^{x^{x^{\dots}}}} := \lim_{n \to \infty} a_n \qquad \text{where} \qquad
a_0 = x \quad\text{and}\quad a_{n + 1} = x^{a_n} \quad\text{for $n \geq 1$}. $$
But it is possible to imagine other definitions, like taking a different $a_0$, setting $a_{n+1} = x^{x^{a_n}}$, or any number of other modifications. Are these equivalent? Maybe. If so, this needs to be proven using theorems about limits. If not, we need to carefully distinguish them.

A famous example is the equation
$$ 1 + 2 + 3 + \cdots = \sum_{n = 1}^\infty n = -\frac{1}{12}. $$
Clearly this does not make sense with the usual definition of the infinite sum
$$ \sum_{n = 1}^\infty n = \lim_{t \to \infty} \sum_{n = 0}^t n, $$
which would give $\infty$. But mathematicians have other, less standard, ways of defining what infinite sums are, which they find useful, which share properties of the finite sum that they consider imporant, and which indeed make the sum equal to $-1/12$.

So as long as you know exactly what every piece of notation means, you should know exactly what you may and may not do with each mathematical object. If you don't, you are not reasoning formally, but are in the process of exploring or inventing new mathematics (which is a valid and important activity for mathematicians).


问:
Binomial determinants related to Hoggatt triangles
Let $d$ be a positive integer and let $0\leq{k}\leq{n}$ be non-negative integers.

Let $a_d(n)=\binom{n+d-1}{d}$ and $\binom{n}{k}_{d}=\prod_{j=0}^{k-1} \frac{a_d(n-j)}{a_d(k-j)}.$

(The matrix with entries $\binom{n}{k}_{d}$ is sometimes called Hoggatt triangle of order $d$. For $d=1$ it coincides with Pascal's triangle and for $d=2$ with the triangle of Narayana numbers).

Consider the matrix $A_d(n,k)$ whose $i,j-$ entry is $\binom{n-i+j+d-1}{n-1}$ for $0\leq {i,j}\leq{k-1}.$

Computer experiments suggest that
$$\det{A_d(n,k)}=\binom{n}{k}_{d}.$$
Any idea how to prove this? Is it possible to see directly from the matrix that the determinant satisfies $\binom{n}{k}_{d}=\binom{n}{n-k}_{d}?$

答:
The Jacobi-Trudi identities give

$$s_{\lambda} = \det(h_{\lambda_i + j - i})_{i,j=1}^k$$

where $\lambda = (\lambda_1, \dots \lambda_k)$ is a partition with $k$ parts, $s_{\lambda}$ is a Schur function, and $h_i$ is a complete homogeneous symmetric function. If we work with $n$ variables and set them all equal to $1$ then

$$h_{\lambda_i + j - i}(\underbrace{1, 1, \dots 1}_{n \text{ times}}) = {\lambda_i + j - i + n - 1 \choose n - 1}$$

so comparing to the desired determinant we see that we want to set $\lambda_i = d$. (Note that it doesn't matter whether we index $i, j$ starting from $1$ or $0$ since only the difference between them matters anyway.) This gives that the desired determinant evaluation is

$$s_{d^k}(\underbrace{1, 1, \dots 1}_{n \text{ times}})$$

which is the number of semistandard Young tableaux with shape $d^k$ (a box with $d$ columns and $k$ rows) and entries in $\{ 1, \dots n \}$. These can be counted with the semistandard hook length formula, which gives

$$s_{d^k}(\underbrace{1, 1, \dots 1}_{n \text{ times}}) = \prod_{1 \le i \le k, 1 \le j \le d} \frac{n - i + j}{(k - i) + (d - j) + 1} = \prod_{1 \le i \le k} \frac{(n - i + 1)^{\overline{d}}}{(k - i + 1)^{\overline{d}}}$$

where $x^{\overline{d}}$ denotes the rising factorial $x(x + 1) \dots (x + (d-1))$. Writing $a_d(n) = \frac{n^{\overline{d}}}{d!}$ and $\frac{a_d(n-j)}{a_d(k-j)} = \frac{(n-j)^{\overline{d}}}{(k-j)^{\overline{d}}}$, and reindexing the sum above by $1$, shows that this is equivalent to the desired result.

This argument might be slight overkill and one or more of the methods in Krattenthaler's Advanced Determinant Calculus might be easier. It is maybe also worth saying that this argument can be rephrased using the Lindstrom-Gessel-Viennot lemma, since that lemma proves the Jacobi-Trudi identities and can be specialized to this case.


答:
An obvious example would be any non empty subspace of a space with the trivial topology, or any infinite set in a space with the finite complement topology.

An example with more separation axioms would be the successor ordinals in the first uncountable ordinal with the order topology.

As a last example, $\mathbb{Q}$ is dense in the sorgenfrey line, which is not metrizable.

Title: Example of dense subset in topological space that does not have a metric

Body: We just went over the Zariski topology in my algebraic geometry class and we see the zero ideal is dense in $\text{Spec}(\mathbb{Z})$. This made me realize that I cant think of any other good examples of a dense subset in a topological space that isnt just a metric space. Does anyone have any good examples?


答:
If $f(x)=\dfrac{100}{1+2^{-x}}$ then $2^{-x}=\dfrac{100}{f(x)}-1$ so $x=-\log_{2}\dfrac{100}{f(x)}=\log_2 f(x)-\log_2 100$.

$f$ is injective, as if $f(x_1)=f(x_2)$ then $x_1=-\log_{2}\dfrac{100}{f(x_1)}=-\log_{2}\dfrac{100}{f(x_1)}=x_2$, because $\log_2$ is injective. Whether $f$ is surjective or not depends on the Range you choose for $f$.

Title: Determine if $f(x) = \frac{100}{1+2^{-x}}$ is one-to-one

Body: I currently have $$f(x) = \frac{100}{1+2^{-x}}$$ and I'm trying to determine if it is one-to-one. I have looked at the other posts however I'm not sure as to how to deal with the $-x$.


问:
Why has the equation $ax + by = (a + b)z$ always a solution with the given constraints?
Consider the equation $ax + by = (a+b)z$  with $a < z$ and $b > z$. The variables a, b, z are fixed and one can vary x and y. Why is there always a solution to this equation? It might be obvious but I somehow don't see it... I maybe should also add that $a, b, z, a, x, y \in \mathbb{N}$

答:
Suppose $a = 0$.

Then you need to show that

$0*x + by = (0+b)z$ has a solution.  It obviously does if $x = anything$ and $y = z$.

Then $0*anything + bz = (0 + b)z$.  Which is true.

Suppose $b = 0$ then we need to show that

$ax + 0*y = (a+0)z$ has a solution.  It obviously does if $x = z$ and $y = anything$.

The $az + 0*anything =(a + 0)z$. Which is true.

And if $a\ne 0; b\ne 0$ then

$ax + by = (a+b)z$

$ax = (a+b)z-by$

$x = \frac {(a+b)z- by}a = z + \frac ba(z -y)$

If $a$ divides into $b$ evenly then $\frac ba$ is an integer.

If we let $y = anything$ and $x = z + \frac ba(z-y)$ we have a solution.

$a(z + \frac ba(z-y))+by=$

$(az + bz -by)+by = az+bz = (a+b) z$.

Otherwise if $\frac ba$ reduces to the fraction $\frac {b'}{a'}$ (which means $a'b=ab'$).

Then if we let $r = $ the remainder of $z$ when divided by $a'$ (so that there is a $q$ so that $z = qa' +r$) then if we let $y = ka'+r$ for any $k$ and let $x = z +b'q -b'rk$

Then we have a solution

$ax + by = $

$a(z +b'q -b'rk) +b(ka' + r)=$

$az + ab'q - ab'rk + a'bk + br=$

$az + a'bq+br = az + b(a'q+r) =az + bz = (a+b)z$


问:
How to calculate cumulative probabilities for a discrete random variable?
I want to understand the difference between the cumulative distribution function and probability density function for a discrete random variable. I thought I understood the difference, but then I attempted this question:

The number of accidents resulting in claims per month is modeled by the random variable $N$ with $F(n) = 1-\frac{1}{n}$. What is the probability that $3, 4,$ or $5$ claims are received in a given month?

My approach was to differentiate $F(n)$, which gave me $f(n)$, the probability density function, and I then found $f(3)+f(4)+f(5)$, by using the definition of PDF's of R.V.s. This approach turned out to be incorrect (the correct approach was $F(5)-F(2)$). Could someone please explain why this is incorrect? Wouldn't this approach be correct if $N$ were a continuous  random variable? Is my understanding of PDFs of R.V.s incomplete?

答:
For discrete random variables, there's only CDF. There are no PDFs for them. The naming is very telling, "cumulative" meaning when you plug in a value into $F(n)$, that's the probability, than the random varible is less than that value. Let's say you want to calculate $P(\xi \leq 5)$. Since $\xi$ is discrete, you add up all the probabilites until $5$, meaning $P(\xi\leq 5)=P(\xi=0)+P(\xi=1)+P(\xi=2)+P(\xi=3)+P(\xi=4)+P(\xi=5)=F(5)$.

Now, in your case, you want to calculate $P(3\leq \xi \leq 5)$. You do the the same, however since the CDF is cumulative, you added everything up, until $5$. But you don't need those before $3$. So you subtract the unnecessary part. Hence $F(5)-F(2)$. Or more precisely, $P(\xi=3)+P(\xi=4)+P(\xi=5)$.

If $\xi$ is absolute continous, and you want to use the PDF, then you have to use integration.$$\int\limits_3^5 f(x)\; dx$$where $f(x)$ is the PDF, which you obtain by differentiating $F(x)$, the CDF.

Note: $\xi$ or $X$ is the standard notation for a random variable, in your case, this is $N$.


问:
find angle x in isosceles triangle
Considering the attached image, is that possible to compute for $\angle x$ in the isosceles triangle below, with no further known values?

$\\$I have found $\angle B$ and $\angle C$ are $50^o$. So $\angle B$ is divided by $\overline{BP}$ to $10^o$ and $40^o$. But I can't find how the $\angle C$ is divided by $\overline{CP}$. Also its clear that $\overline{BP}$ is perpendicular to leg $\overline{AC}$.

答:
Consider the figure below.

$\\$Thanks to @saulspatz answer, and based on Trigonometric Form of Ceva's Theorem we may write the following:
$$Sin(\angle B1).Sin(\angle C1).Sin(\angle A1) = Sin(\angle B2).Sin(\angle C2).Sin(\angle A2)$$
$$\\$$so we have:
$$\frac{Sin(\angle C1)}{Sin(\angle C2)} = \frac{Sin(40^o)*Sin(20^o)}{Sin(10^o)*Sin(60^o)}$$
$\\$Also we have $\angle C1 + \angle C2 = 50^o$, and by solving for $\angle C2$:

$$\angle C2 = 20^o$$

$\\$Hence:
$$\angle x = 100^o$$

Title: find angle x in isosceles triangle


问:
What is the relationship between an existential statement $\exists(x)\varphi(x)$ and the phrase "can be calculated"?
I recently read a proof that made me question the relationship between an existential claim $\exists x \varphi(x)$ and "can be calculated". The exact proof can be found as the answer to the post here --> Direct Proof: $A\subseteq\bigcap_{n\lt\omega}f^n[A]\cup\bigcup_{n\lt\omega}f^n[A]\setminus f^n[B]\cup \bigcup_{n\lt\omega} f^n[B]\setminus f^{n+1}[A]$, but I will summarize the part that specifically led to the question I will ask in this current post.

Without going into all of the detail, there was a starting set $A$ and a nested chain of subsets that worked as follows:

$A \supseteq B \supseteq f[A] \supseteq f[B] \supseteq f^2[A] \supseteq f^2[B] \supseteq...$

Importantly, this chain spanned to infinity (i.e. $n \lt \omega$).

In this chain, $f$ is an injective function $f:  A \xrightarrow{1-1} B$, and the notation $f^n$ simply refers to $n$ iterations of the function $f$ (with $f^0$ representing $I_A$).

In the aforementioned post, the author develops a function $N$ that takes an arbitray element $x$ in $A$ and maps it to the last such subset that does NOT contain it. For example, let's say that $f^n[A]$ happens to be the first appearance of the element $x$ (i.e. $x$ is not an element of any subset of $f^n[A]$). Then $N(x) = f^n[B]$

...or, alternatively, let's say that $f^n[B]$ happens to be the first appearance of the element $x$. Then $N(x)=f^{n+1}[A]$.

In order to prove that such a function exists (it does), I confirmed a formula with the following basic structure: $\forall x \in A \exists ! y \varphi(x,y)$.

When I thought a little more about this, though, I realized that if someone handed me an actual number that was in the function $N$'s domain (e.g. $2$), and asked me, "Where does $N$ take $2$?", I wouldn't know how to respond.

It seems like my only option would be to start at set $A$ and begin my decent down the infinitely nested chain of subsets until I spotted the first occasion of $2$'s disappearance. I guess with enough time, such a method could be carried out...so perhaps this is "calculatable".

Conversely, however, I can imagine some variant of the above proof where I am not working with $n \lt \omega$ (which has the perk of a minimal element) but instead $\mathbb Z$...or maybe a transfinite ordinal "bigger" than $\omega$. In these other cases, it seems like there may not be a calculation that can be provided, at least with any guaranteed chance of working.

So what exactly does "$\exists x \varphi(x)$ tell me? Is this FOL statement synonymous with "Depending on the circumstance, the thing may be calculatable" without leaving any inkling about how it can be calculated?

答:
It says nothing more than that there is at least one object with the desired property. This object may or may not be constructible (in whatever sense), but all "$\exists x\varphi(x)$" asserts is bare existence.

Depending on what you want from logic this may feel unsatisfying. But note that being able to express bare existential claims does not prevent us from also being able to express more complicated constructibility claims (although again one would need to specify what "construct" means - is the halting problem "constructible"?). I would argue that in fact it's the other way around: more loaded language has more difficulty expressing finer distinctions.

This is how I personally think of classical logic most of the time: it's a drastically, almost unnaturally, bare framework in which each assertion means the bare minimum that it has to. This makes it perfect for developing really complicated loaded ideas  because there's no danger of "contamination" from background intentionalities. For example, I would rather develop a theory of "constructive truth" within a classical framework than develop a theory of "classical truth" within a constructive framework, since in the latter approach I would be worried that my background constructivity was interfering with my notion of classicality. I'm not saying that the latter can't be done - it absolutely can - just that at the beginning I would have a worry which would not be present, at least for me, if the former approach were being followed.

(Of course the previous paragraph can be severely deconstructed, and the words "I" and "most" were carefully chosen in its first sentence. But I really do think that there's a lot to it.)


问:
Multiplicity of an eigenvalue is equal to $\dim V_{\lambda}$
I am trying to prove that multiplicity of an eigenvaliue $\lambda$ = $\dim V_{\lambda}$ and I have problems with this inequality:

$\dim V_{\lambda} \le $ multiplicity $\lambda$.

I know that characyeristic polynomial of $W_f(t)=f|_{V_{\lambda}}$ (where $f$ has Jordan form) is $\pm(t-\lambda)^{dimV_{\lambda}}$ because there are only $\lambda$'s on the matrix's diagonal.

I also need to prove that $\pm(t-\lambda)^{dimV_{\lambda}}$ divides $W_f(t)$, but I don't know how. Could you help me with that?

$V_{\lambda} = \bigcup_{k=0} ^{\infty} V_{\lambda} ^{k}$, $V_{\lambda} ^{k} = ker((f-\lambda)^k)$

By multiplicity I mean algebraic multiplicity.

答:
Hint: Over the complex numbers you have both,

$$\sum_\lambda \dim V_\lambda=n$$

and

$$\sum_\lambda mult(\lambda)=n$$

where $n$ is the size of the matrix.


问:
Infinity of Natural Numbers
How can I prove the infinity of the Natural Numbers just be using this definition: (Bartle Analysis Book) "A set B is finite if it is empty or if there is a one-one function with domain B and range in an initial segment of N"

Using this definition I need to prove that is not possible to form a bijection between N and the initial segment of N, however, I do not really know how to start solving this.

Thank you in advance

答: If the range of the function is an initial segment of $\Bbb N$, there must be some number $c$ that is greater than any element of the range.  Now look at the image of the numbers from $0$ to $c$ (assuming you accept $0$ as a natural).  There are $c+1$ of them and only $c$ numbers in the range.

Title: Infinity of Natural Numbers


问:
Sampling from an arbitrary distribution on Polish spaces
Let $U\sim \text{Unif}(0,1)$, and let $\mu \in \mathcal{P}(\mathbb{R})$ be an arbitrary probability measure on $\mathbb{R}$. Then from $\mu$, we can derive an associated CDF $F(x) = \mu((-\infty,x])$. We consider the following inverse of $F$:

$$F^{-1}(u) = \inf\{x \in \mathbb{R}: F(x) \geq u\}.$$

Then it's easy to show that $F^{-1}(U)\sim \mu$. In other words, $F^{-1}(U)$ is a sample of a random variable distributed like $\mu$.

Is there a similar construction when we now assume that $\mu \in \mathcal{P}(S)$ is a probability distribution on some arbitrary Polish space $S$? That is, does there exist a random element $X$ taking values in some other Polish space $T$ and a mapping $\psi: \mathcal{P}(S) \to (T \to S)$ from probability measures on $S$ to measurable functions from $T$ to $S$ such that $\psi(\mu)(X)\sim \mu$ for all $\mu \in \mathcal{P}(S)$?

I would be interested in proofs that such a mapping exists (or even more interesting, doesn't), but it would be absolutely amazing if the proof was constructive. I'd also be interested in references to the literature where this problem may have already been addressed. Thank you!

答:
By "measurable", here, I presume you mean "Borel measurable" and that your probability measures are all Borel probability measures. If so, the answer is yes. In fact, you can always take $\ X\sim \text{Unif}(0,1)\ $.

Here's a theorem I proved in Differential Games with No Information, SIAM J. Control Optim., 15(2), p.242 (Theorem 5.1):

If $\ S\ $ is a complete separable metric space and $\ \mu\ $ a Borel probability measure on $\ S\ $, then there exists a Borel measurable function $\ \phi:(0,1)\rightarrow S\ $ which has $\ \mu\ $ as its distribution with respect to Lebesgue measure on $\ (0,1)\ $. That is, $\ \text{m}(\{\alpha\in(0,1); \phi(\alpha)\in A\})=$$\,\mu(A)\ $ for any Borel subset $\ A\ $ of $\ S\ $.

Here, I have replaced the symbols "$Z$" and "$\Pi$" which originally appeared in my statement of the theorem with "$S$" and "$\mu$", respectively.

All the heavy lifting in the theorem's proof comes from Section 1.2, The Isomorphism Theorem, in K.R.Parthasarathy, Probability Measures on Metric Spaces.  The key results from this book are Theorem 2.8, from which it follows that either $\ S\ $ is countable or has the same cardinality as $\ \mathbb{R}\ $, and the isomorphism theorem itself (Theorem 2.14).  The latter implies that when $\ S\ $ has the same cardinality as $\ \mathbb{R}\ $, there exists a Borel measurable bijection $\ \zeta:\mathbb{R}\rightarrow S\ $ such that $\ \zeta^{-1}\ $ is Borel measurable. Thus, if $\ \mu\ $ is a Borel probability measure on $\ S\ $, then $\ \mu_\mathbb{R}(A)=$$\mu(\zeta(A))\ $ defines a Borel probability measure on $\ \mathbb{R}\ $, and you already know that if $\ F\ $ is the distribution function of $\ \mu_\mathbb{R}\ $ then  $\ F^{-1}(X)\sim\mu_\mathbb{R}\ $ (where $\ X\sim \text{Unif}(0,1)\ $). It then follows that $\ \zeta    F^{-1}(X)\sim\mu\ $.

The proof of the theorem for the case when $\ S\ $ is countable is straightforward.


问:
$\lim_{|z| \to 1 } |f(z)| = \infty$ for $f(z) = \sum_{k=1}^\infty z^{k!}$, a strange series manipulation
I'm dealing with a rather complicated series:

$$f(z) = \sum_{k=1}^\infty z^{k!}.$$

I know that the radius of the convergence of it is 1 but here is another problem:

$$\lim_{|z| \to 1 } |f(z)| = \infty.$$

Here is my try:

Put $z=Re^{i\theta}$ then $|z| = R$ and the limit reduces to $\lim_{R \to 1 } |f(Re^{i\theta})| $:

$$|f(Re^{i\theta})| = |\lim_{n \to \infty} \sum_{k=1}^n (Re^{i\theta})^{k!}|$$

I'm not sure that I can exchange absolute with the limit but I proceed

$$\lim_{n \to \infty}| \sum_{k=1}^n (Re^{i\theta})^{k!}|\leq \lim_{n \to \infty}\sum_{k=1}^n| R^{k!}|$$

And at this point, I have no idea how to continue.

Can anyone help me withhow to tackle it down? Maybe that's a bad approach and there is an elegant one.

答:
There is no holomorphic function $f$ in $\mathbb D$ such that $\lim_{|z|\to 1}|f(z)| =\infty.$

Proof: If there were, this $f$ would have only finitely many zeros. Let $B$ be the finite Blashke product with these zeros. Then $g=f/B$ is holomophic and nonzero in $\mathbb D,$ and $\lim_{|z|\to 1}|g(z)| =\infty.$ This implies $1/g$ is holomophic in $\mathbb D$ with $\lim_{|z|\to 1}1/g(z) =0.$ Thus $1/g$ extends to be continuous on $\overline {\mathbb D}$ with boundary values $0.$ This violates the maximum modulus theorem, to say the least, and we're done.


问:
$\exists f \in L(V,V)$ such that $\ker(f ) = W_1$ and $\operatorname{Im} (f ) = W_2$ $\iff \dim W_1 + \dim W_2 = \dim V$
Let $W_1$ et $W_2$ two vector subspaces of $V$ a $K$-vector space of finite dimension. Show that it exists $f \in L(V,V)$ such that $\ker(f ) = W_1$ and $\operatorname{Im} (f ) = W_2$ $\iff$ $\dim W_1 + \dim W_2 = \dim V.$

For $\implies$, by rank theorem we have that $\dim V = \dim \ker f + \dim \operatorname{Im} f = \dim W_1 + \dim W_2$.

For the other direction, I am not sure of what I am doing : we have that $\dim W_1 + \dim W_2 = \dim V$ so $$\dim (W_1+W_2) + \dim(W_1 \cap W_2) = \dim V$$ but I don't really know how it can be useful.

答: Let $\{e_1,\ldots,e_n\}$ be a basis of $V$ such that $\{e_1,\ldots,e_k\}$ is a basis of $W_1$. Let $\{f_1,\ldots,f_{n-k}\}$ be a basis of $W_2$. Now take $f\colon V\longrightarrow V$ such that $f(e_i)=0$ if $i\in\{1,2,\ldots,k\}$ and that $f(e_i)=f_{i-k}$ otherwise. Then $\ker f=W_1$ and $\operatorname{Im}f=W_2$.


问:
"Spread-Out" Partitions
Firstly, the given definition of spread-out is

"A partition is said to be 'spread-out' if all the parts are distinct and no two parts are consecutive numbers."

I am trying to work out how many spread-out partitions of 21 are there into 4 parts? I believe the answer is 7, but I only achieved this by brute force, and hence wondering if there is a easier way and more general formula of working this out.

答:
For admissible partitions of $n$ into $k$ elements we choose the first
element and a sequence of $k-1$ differences between consecutive
elements, with these being at least two. The first element contributes to
all $k$ elements. The first difference contributes to $k-1$ values, the
second  to $k-2$ and so on. This yields the OGF

$$G_k(z) =
\frac{z^k}{1-z^k} \times \prod_{q=1}^{k-1} \frac{z^{2q}}{1-z^q}
= z^{k^2} \prod_{q=1}^k \frac{1}{1-z^q}.$$

We are interested in $g_k(n) = [z^n] G_k(z).$ Differentiate to get

$$G'_k(z) = k^2 z^{k^2-1} \prod_{q=1}^k \frac{1}{1-z^q}
+ z^{k^2} \prod_{q=1}^k
\frac{1}{1-z^q} \sum_{q=1}^k \frac{qz^{q-1}}{1-z^q}.$$

Extracting the coefficient on $[z^n]$ yields

$$(n+1) g_k(n+1) = k^2 g_k(n+1)
+ \sum_{q=1}^k q [z^{n+1-q}] \frac{1}{1-z^q} G_k(z)
\\ = k^2 g_k(n+1)
+ \sum_{q=1}^k q
\sum_{p=0}^{\lfloor (n+1)/q \rfloor -1} g_k(n+1-(p+1)q).$$

This gives the recurrence

$$g_k(n+1) = \frac{1}{n+1-k^2}
\sum_{q=1}^k q
\sum_{p=0}^{\lfloor (n+1)/q \rfloor -1} g_k(n+1-(p+1)q)$$

or alternatively

$$\bbox[5px,border:2px solid #00A000]{
g_k(n) = \frac{1}{n-k^2}
\sum_{q=1}^k q
\sum_{p=1}^{\lfloor n/q \rfloor} g_k(n-pq).}$$

with the boundary conditions being $g_k(n) = 0$ when $n\lt k^2$ and
$g_k(k^2) = 1.$

 This will indeed produce $g_4(21) = 6.$

Title: "Spread-Out" Partitions


问:
Calculate control points of cubic bezier curve approximating a part of a circle
I'm not mathematically inclined, so please be patient with my question.

Given

$(x_0, y_0)$ and $(x_1, y_1)$ as the endpoints of a cubic Bezier curve.

$(c_x, c_y)$ and r as the centerpoint and the radius of a circle.

$(x_0, y_0)$ and $(x_1, y_1)$ are on the circle.

if it makes the calculation simpler, it's safe to assume the arc is less than or equal to $\frac{\pi}{2}$.

How do I calculate the two control points of the Bezier curve that best fits the arc of the circle from $(x_0, y_0)$ to $(x_1, y_1)$?

答:
As a different way to choose where the error falls, one could pick the control points such that the enclosed area matches that of the circle.

If we consider the first quarter of a circle (the same as the first image in the other answer), then the area should be $\pi/4$, so we need the integral

$$
\begin{align}
\pi/4
  &= \int_{0}^{1} y(t)x'(t)dt \\
  &= \int_{0}^{1} ( (1-t)^3 + 3(1-t)^2t + 3(1-t)t^2c )( 3(1-t)^2c + 6(1-t)t(1-c) ) dt \\
  &= -\frac{3c^2}{20}+\frac{3c}{5} + \frac12 \\
c &= 2 - \sqrt{\frac13(22-5\pi)} \\
  &≈ 0.5517784778...
\end{align}
$$

That's slightly smaller than the other two answers, as it attempts to even out the error between over- and under-shooting the circle, rather than minimizing the maximum error.

Aside: Edits appreciated to add a solution for angles other than 90°.  My maths are rusty.


问:
Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$.
Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$.

Attempt: Does this approach true?

$(\Rightarrow)$
Let $S$ be a group. Denote $e_S$ be an identity element in $S$. Clearly, $e_S$ is idempotent since $e_S^2 = e_S$.
Suppose that there is another idempotent in $S$, say $x$. Then, $x^2 = x$. So,
\begin{equation*}
x^2=x \Leftrightarrow x^{-1}x^2 = x^{-1}x \Leftrightarrow x = e_S.
\end{equation*}
Hence, $x=e_S$. Thus, the only idempotent in $S$ is $e_S$.

$(\Leftarrow)$
First, to show for any $x \in S$, there exist $n \in \Bbb N$ such that $x^n$ is idempotent. I found this proof in the ProofWiki. Next, since the only idempotent in $S$ is $e_S$, then $x^n$ must equal to $e_S$. I have proved this already. But, I am still confused. If $n=2$, then $a^2 = e_S$ i.e. the inverse of $a$ is itself.
But, what about $n=3$? $a^3=e_S$, does it implies that the inverse of $a$ is $a^2$?

答:
Let $x\in S$. Since $S$ is finite, there exist $a$, $d> 0$ such that
$$x^a= x^{a+d}$$

Now you can show easily by induction that
$$x^b = x^{b+ m d}$$

whenever $b\ge a$, and $m\ge 0$ integer. Let us choose $b\ge a$, and $m d$ such that they are both equal.  ( for instance, take $b = \operatorname{lcm}(a,d)$). Then we get
$$x^{b} = x^{b+b}$$
or $x^b= (x^b)^2$.  Therefore, $x^b$ is idempotent, so it must be $e$.
We conclude $x^{b-1} = x^{-1}$.


答:
First note that
$$
\int_0^{f(x)}1\,{\rm d}t
~=~
f(x)
$$
since $f(x)$ is constant with respect to variable $t$, so
$$
\int_{\mathbb R}f(x)\,{\rm d}m
~=~
\int_{\mathbb R}
\left(
  \int_0^{f(x)}1\,{\rm d}t
\right){\rm d}m
$$
Now let $\chi_A(x)$ be the characteristic function of the set $A$, i.e.
$$
\chi_A(x) =
\begin{cases}
1 & \text{if }x\in A \\
0 & \text{if }x\notin A
\end{cases}
$$
so that the previous expression turns out to be equal to
$$
\int_{\mathbb R}f(x)\,{\rm d}m
~=~
\int_{\mathbb R}
\left(
  \int_0^{+\infty}\chi_{[0,f(x)]}(t)\,{\rm d}t
\right){\rm d}m
$$
Now the crucial observation is that since $f(x)\geq 0$ then
$$
\chi_{[0,f(x)]}(t)
~=~
\chi_{\{f\geq t\}}(x)
\quad \forall t\geq 0
$$
Now, going back to the integral,
$$
\int_{\mathbb R}f(x)\,{\rm d}m
~=~
\int_{\mathbb R}
\left(
  \int_0^{+\infty}\chi_{\{f\geq t\}}(x)\,{\rm d}t
\right){\rm d}m
~=~
\int_{\mathbb R}
m\left(
 \{f\geq t\}
\right){\rm d}t
$$</p>

Title: Riemann integral and Lebesgue integral

Body:
$f:R\rightarrow [0,\infty)$ is a Lebesgue-integrable function. Show that
$$
\int_R f \ d m=\int_0^\infty m(\{f\geq t\})\ dt
$$
where $m$ is Lebesgue measure.

I know the question may be a little dump.


问:
How to compute $a \ast b \ast c$ or more generally $a_1^{b_1} a_2^{b_2} a_3^{b_3} \cdots$?
I'm currently a self-learner on Abstract Algebra and Group theory. The book that I'm currently reading is A First Course In Abstract Algebra by John B. Fraleigh. So I know this question might truly be a basic question which every student would be taught in class and there is confusion about it. However, maybe it is me who is not careful in reading the book that raises in my mind this question. So given a binary operation $\ast $ on a set $S$, how can we compute:

$$a \ast b \ast c$$
Actually, on page 23 of the book, the author does mention this but I still can not get it. I don't know if we should compute this from left to right or reversely or we will just leave it there if $\ast$ is not associative. Moreover, if this binary operation is associated, is it true to define that:
$$a \ast b \ast c = \left(a \ast b \right)\ast c$$
Then, after learning about Cyclic group there is a problem of which can be stated as below:

Let $a$ and $b$ be elements of a group $G$. Show that if $ab$ has finite order $n$, then $ba$ also has order $n$.

I know that we should show that $(ba)^n = e$ which can be driven from $(ab)^n = e$. Unfortunately, since $G$ might not be an Abelian group so what I had gone so far before looking for a solution is just manipulating $(ab)^n = e$ by using the fact that $(ab)^{-1} = b^{-1} a^{-1}$ since I don't know how to move further after splitting $(ab)^n = (ab)(ab)^{n-1}$. I also get confused whether $(ab)^n = (ab)(ab)^{n-1}$ or $(ab)^n = (ab)^{n-1} (ab)$. So I checked up for the solution and this is the way they manipulate it that I found: Show that if ab has finite order n, then ba also has order n. - Fraleigh p. 47 6.46.
So I completely have no idea about this can be infered:

\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies
\color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e
\end{align}
I also tried to check another link in order to understand the solution: same question And the asker states that:
$$(ab)^n=\underbrace{(ab)(ab)(ab)\cdots(ab)}_{n~\text{copies}}$$
$$=a\underbrace{(ba)(ba)(ba)\cdots(ba)}_{n-1~\text{copies}}b$$
$$=a(ba)^{n-1} b$$
It is here that the number of terms which we're dealing with is $n$ now. I don't have any idea how could he manipulates from the first line to the second line so I think that I need some helps on this. In summary, how can we compute binary operation of more than 3 elements? Pleas explain the problems that I state above and give some other examples if possible. Thank you so much.

答:
Write

$$(ab)^n = \underbrace{ababab \ldots ab}_{n {\text{ times}}}$$

Then

$$(ba)^n = \underbrace{bababab \ldots ba}_{n {\text{ times}}} =  (\underbrace{bababab \ldots ba}_{n {\text{ times}}})bb^{-1} = b(\underbrace{ababab \ldots ab}_{n {\text{ times}}})b^{-1} = b(ab)^nb^{-1}.$$

If $(ab)^n = e$, then the line above gives

$$(ba)^n = b(ab)^nb^{-1} = beb^{-1} = e.$$

Suppose we were to replace $e$ with any element $z \in Z$ that commutes with any other element of $G$. Then what would we have.


问:
What are sets made up of in topology?
I am looking for a very basic answer, I am in 10th grade and studying it for a project. I don't understand what the sets are referring to. Is it nodes? Edges? Axioms? Vertices? Or does it just depend?</p>

答:
A set, in mathematics, is just a bunch of things. A set could be a bunch of numbers like $\{1, 2, 5, 1000\}$, or it could be a bunch of points on the surface of a sphere like $\{(x, y, z) : x^2 + y^2 + z^2 = 1\}$, or it could be a bunch of words like $\{\mbox{cat},\mbox{dog},\mbox{Ouagadougou}\}$.

Perhaps most confusingly, a set can be a bunch of other sets - for example, $\left\{
\left\{1, 2, 5, 1000\right\}, \left\{(x, y, z) : x^2 + y^2 + z^2 = 1\right\}, \left\{\mbox{cat},\mbox{dog},\mbox{Ouagadougou}\right\} \right\}$ is a valid, if not particularly useful, set.

A subset is a bunch of things inside a larger[1] set. For example, some of the subsets of $\{1, 2, 5, 1000\}$ are $\{1, 5\}$, $\{2\}$ and $\{1, 2, 5, 1000\}$ itself. Additionally, the set containing nothing, $\{\}$, sometimes written as $\varnothing$, is also a subset of that set - and, indeed, of every set in existence.

A topology is a way of defining which elements of a given set are somehow "close to" or "next to" or "linked to" each other, by listing out the subsets of the set that define "neighbourhoods" of associated points. The only rules for a topology are:

It must contain the empty set, as well as the original set.

If you take any combination of subsets from the list, and find all the elements they have in common (called the intersection of the sets), then the resulting set is also in the list.

If you take any combination of subsets from the list, and find all the elements they cover in total (called the union of the sets), then the resulting set is also in the list.

The simplest topology, and one that exists for any set, is the trivial topology consisting of the empty set and the set itself. So $\left\{ \varnothing, \left\{ \mbox{cat},\mbox{dog},\mbox{Ouagadougou}\right\} \right\}$ is a valid topology for the set $\left\{\mbox{cat},\mbox{dog},\mbox{Ouagadougou}\right\}$.

The other simple topology that exists for any set is the discrete topology, which is one that lists all possible subsets. For example, the discrete topology on $\{A, B, C\}$ is $\left\{ \varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\right\}$.

If we take a more geometric view, then the elements of a topology are called the "open sets", and if your starting set is $\mathbb{R}$, which is the set of real numbers (or essentially the number line), then you can define your open sets as being the open intervals (i.e. ranges $(x, y)$ with $x \leq y$ that don't contain their endpoints) and saying that open sets are those that can be constructed from unions of those intervals. Similarly, you can define topologies on 2D or 3D space by starting with circles and spheres and building from there.

Somewhat nicely, we can look at some topological spaces (that is, sets with topologies on them) and say that they're equivalent - if we map the elements of one set to the elements of the other, then their topologies map in the same way. This is where the whole "a donut is the same as a coffee cup" stuff comes from - by defining the right topology, we can assign characteristics to certain types of sets, especially ones that have some kind of geometry involved, along with certain "valid" ways to reshape those sets, that lets you group them together by those common characteristics. For example, you can formalise the idea of a shape with a hole in it versus one without, such that we would say that any shapes with a single hole are somehow "topologically equivalent" to each other. We can even go so far as to start talking about joining topologies together or cutting them apart  or "gluing" points together and looking at how their properties behave (for example, if you take the closed interval $[0, 1]$ and just say that for your purposes the points at $0$ and $1$ are joined to each other, you have a shape that is topologically equivalent to a circle).

Which then gets to graphs - in mathematics, a graph is a set of points (the nodes or vertices) along with a set of pairs of points (the edges). It doesn't necessarily have geometric properties - a database of Facebook users (nodes) and information on which ones are friends (edges) is a graph, as is a database that links people (nodes) to websites (also nodes) via their user accounts (edges). However, we can add some geometry by  treating the edges as line segments, and that lets us look at how they relate to other things. For example, you can represent polyhedra as graphs (literally just taking their vertices and edges), and hence you can define a bunch of topologically equivalent polyhedra based on them having the same graph. So when you're working in this kind of space, the set that's of interest (topologically speaking) is a combination of the vertices and their edges, not as physical points but as discrete identities and information about which ones are joined to each other (and often also some information about the order in which they're joined, because that helps define the faces of the shape and some other handy properties).

tl;dr Yes?

Title: What are sets made up of in topology?


问:
Prove that for every $\varepsilon>0$ there exists $n \in \mathbb{N}$ such that $|\cos(x-2^n)-1|<\varepsilon$
Let $x \in \mathbb{R}$ and $\varepsilon>0$. I would like to know if there is a natural number $n$ such that $|\cos(x-2^n)-1|<\varepsilon$.

I was trying to prove that there exists $k \in \mathbb{Z}_+$ such that $x-2^n \in \left(2k \pi, (4k+1)\frac{\pi}{2}\right)$, but I couldn't prove it. This would only imply that $0<\cos(x-2^n)<1$. I couldn't find a correct way to solve the problem. Note that I'm not asking for the existence of a natural number $n_0$ such that $|\cos(x-2^n)-1|<\varepsilon$ for all $n \geq n_0$.

答:
Translate into what it means for the argument of $\cos$

$$\|\frac{x - 2^n}{2\pi}\| < \delta$$

where $\|\cdot\|$ denotes the distance from the closest integer.  Or,
$$\frac{x}{2\pi} - 2^n\frac{1}{2 \pi}$$ is very close to an integer, for some $n$.

Let's write everything in base $2$. $x' \colon = \frac{x}{2\pi}$, $\eta= \frac{1}{2 \pi}$. The statement would be that you can guarantee that there always exists a chunk of the binary expansion of $\eta$ of a given form.  For instance, there exists in the binary expansion of $\frac{1}{2\pi}$ a sequence of a trillion $0$'s followed by a trillion $1$'s.  Now, everybody believes that this is true, but we cannot prove it yet.
See  normal numbers.


问:
Convergence in $L^1$ of martingale related to a Poisson process
I was solving Problem 3.21 b) of Karatzas and Shreve's Brownian Motion and Stochastic Calculus. Essentially, the problem says the following:

Let $N_t$ be a Poisson random variable with parameter $\lambda t$. Define the process
$$X_t=\exp(N_t-\lambda t(e-1)).$$
Does $X$ converges in $L_1$?

I know $(X_t)_t$ is a martingale, and I tried to prove that $(X_t)_t$ is uniformly integrable, but I couldn't do it. Any hisnts would be appreciated. Thanks!

答: By the law of large numbers we have $N_t \sim \lambda t$ as $t \to \infty$, thus $\lim_{t \to \infty} X_t =0$ almost surely. Since $X$ is a martingale, $\mathbb{E}[X_t] = \mathbb{E}[X_0] = 1$ so $X_t$ cannot converge in $L^1$.


问:
Inequality for determinant of sum of two orthogonal matrices
My goal is to show that :

$$\forall A,B\in O_n(\mathbb{R}), |{\rm det}(A+B)| \le 2^n.$$

How can I access the determinant of the sum of two matrices ? I don’t know many ways to establish inequalities in Algebra, except with Bessel’s inequality and Cauchy Schwarz inequality, but they don’t seem to be very helpful here.

答:
Since $AA^T = I$, we have
$$\det (A+B)= \det A(I+A^TB) = \det A \,\det(I+A^TB).$$
Now, $|\det A| = 1$ since $A$ is orthogonal, and $A^TB$ is orthogonal as well so all its eigenvalues $\lambda_1, \ldots, \lambda_n$ are on the unit sphere in $\Bbb{C}$. Therefore
$$|\det(A+B)| = |\det(I+A^TB)| = |(1+\lambda_1)\cdots(1+\lambda_n)| \le (1+|\lambda_1|)\cdots(1+|\lambda_n|) \le 2^n.$$


问:
effects on the Radon Nikodym derivative by restriction of the measures
Let $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0},\mathbb P)$ a filtered probability space satisfying the usual conditions and $\mathbb Q$ a further probability measure with $\mathbb P\ll\mathbb Q$. By the Radon-Nikodym theorem we know of the existence of a $\mathcal F$-measureable function $$\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$$
so that $\mathbb P(A)=\int_A \frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}d\mathbb Q$ for $A\in\mathcal F$. My question is, if we now restict the measures for each $t$ to the sub-$\sigma$-algebras $\mathcal F_t$ of the filtration, what can we say about the Radon Nikodym derivatives
$$S_t:=\frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}.
$$
My guess is, that the Radon Nikodym derivative $S_t$ has to be 'very similar' to the 'original one' $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$, since I know from this post that if $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ is already $\mathcal F_t$-measureable, it wouldn't change by the restiction. But what can we say in genereal? Are there some interesting results regrading this? Also I would realy like to have a picturesque description of how the restriction effects on the Radon Nikodym derivative. And can we conclude from the 'usual conditions' that the function $t\mapsto S_t$ is càdlag?

EDIT: Tanks to the comment of Kavi Rama Murthy, I know that $S_t$ is the conditional expectation of $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ given $\mathcal F_t$ - to be exactly the conditional $\mathbb Q$-expectation. Here is my proof:

For all $A\in\mathcal F_t$ holds
$$\int_A \frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\mathrm d\mathbb Q=\mathbb P(A)=\mathbb P|_{\mathcal F_t}(A)=\int_A \frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}\mathrm d\mathbb Q|_{\mathcal F_t}=\int_A \frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}\mathrm d\mathbb Q.
$$
Thus we know that $S$ is a uniformly integrable $\mathbb Q$-Martingale, i.e.
$$\forall 0\leq s\leq t:\mathbb E_{\mathbb Q}\left(\left.S_t\right|\mathcal F_s\right)=S_s\qquad\text{and}\qquad \mathbb E_{\mathbb Q}\left(\left.\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\right|\mathcal F_t\right)=\frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}.
$$
This answers pretty good the question in which kind $S$ and $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ are similar. Also by this we can assue $S$ to be càdlàg. But this brings me up to a new question: Is $S$ also a uniformly integrable $\mathbb P$-Martingale?

Here are my thoughts about it: If we had the additional assumption $\mathbb Q\ll\mathbb P$ (i.e. the measures are equivalent) and $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ is $\mathbb P$-integrable (or equivalent $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\in L^2(\mathbb Q)$), the statement would hold since $\mathbb E_{\mathbb Q}\left(\left.\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\right|\mathcal F_t\right)=\mathbb E_{\mathbb P}\left(\left.\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\right|\mathcal F_t\right)$ by using the Baye's rule for conditional expectations.

It is clear that the condition $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ is $\mathbb P$-integrable always is requiered, as it would be implied by the martingale property anyway. But I think this requirement is not sattisfied in general, is it? The other requirement $\mathbb Q\ll\mathbb P$ is very restrictive. So is it possible to proof anything without this?

答: $S_t$ is the conditional expectation of $\frac {d\mathbb P} {d\mathbb Q}$ given $\mathcal F_t$ (w.r.t the probability space $(\Omega, \mathcal F, \mathbb Q)$). This is an easy consequence of the definition of conditional expectation.


问:
Smooth Julia Set
My textbook tells me that for a polynomial $P$, its Julia set $J(P)$ is the nuit circle iff $P(z)=az^n$,where $\lvert a\rvert =1$,and $n\geq 2$.

So I want to know whether there is a Julia set of a polynomial is the boundary of a square. Or more generally, whether there is such a rational map.

I found a paper stating that under certain conditions, the Julia set of rational mappings is a Jordan curve. But for a specific square, whether or not there is a corresponding rational map, that's my puzzle.

答:
For a polynomial, the answer is no. By a theorem of Zdunik, any connected Julia set either is a fractal (in the sense that its Hausdorff dimension is strictly larger than one) or it's a segment or the unit circle (and in both cases, we know what polynomials those are).

I am fairly sure the unit square is not the Julia set of any rational map but I don't have any argument in mind.

Title: Smooth Julia Set


问:
Deriving Jacobi triple from the Weyl-Kac denominator formula of affine $\mathfrak{sl}_2$
The Weyl-Kac denominator formula states

$$\prod_{\alpha\in\Phi^+}(1-e(-\alpha))^{mult(\alpha)}=\sum_{w\in W}\epsilon(w)e(w(\rho)-\rho))$$
Where the product is taken over the positive roots and the sum is taken over the Weyl group. Here $\epsilon(w)=(-1)^{l(w)}$ where $l(w)$ denotes the length of $w$ and $e(\alpha)$ is the formal exponential.

I'm trying to apply this to the affinization of $\mathfrak{sl}_2$ with simple roots $\{\alpha_0,\alpha_1\}$. In this case the $\rho=\frac{\alpha_1}{2}+2\Lambda_0$ seems to work. Here $\Lambda_0$ is defined on the coroots st. $\Lambda_0(\alpha_i^\vee)=\delta_{i0}$.

The Weyl group of the affinization of $\mathfrak{sl}_2$ is generated by $s_0,s_1$ st. $s_0^2=s_1^2=1$ (it's the infinite dihedral group). The generators works on $\rho$ in the following way
$$s_0(\rho)=\rho-\rho(\alpha_0^\vee)\alpha_0=\rho+\alpha_0=\rho-\alpha_1+\delta$$
$$s_1(\rho)=\rho-\rho(\alpha_1^\vee)\alpha_1=\rho-\alpha_1$$

Now for the formula. The positive roots are exactly $\alpha_1+k\delta$ for $k\geq 0$, $-\alpha_1+k\delta$ for $k\in\mathbb{N}$ and $k\delta$ for $k\in\mathbb{N}$. Here $\delta=\alpha_0+\alpha_1$. Writing $e(-\delta)=q$ and $e(-\alpha_1)=r$ then the left side is
$$\prod_{k>0}(1-q^k)(1-q^{k-1}r)(1-q^kr^{-1})$$
Turning to the right side, is where I get in trouble. I've seen it should equal $\sum_{k\in\mathbb{Z}}(-1)^kr^kq^{\frac{k(k-1)}{2}}$, so working my way backwards then $\sum_{w\in W}\epsilon(w)e(w(\rho)-\rho)$ should equal
\begin{align*}
\sum_{k\in\mathbb{Z}}(-1)^ke(-k\alpha_1)e(-\frac{k(k-1)}{2}\delta) &= \sum_{k\in\mathbb{Z}}(-1)^ke(-k\alpha_1-\frac{k(k-1)}{2}\delta) \\
&=\sum_{m\in\mathbb{Z}}e(-2m\alpha_1-\frac{2m(2m-1)}{2}\delta)\\
&-e(-(2m+1)\alpha_1-\frac{(2m+1)((2m+1)-1)}{2}\delta) \\
&=\sum_{m\in\mathbb{Z}}e(-2m\alpha_1-m(2m-1)\delta)\\
&-e(-(2m+1)\alpha_1-(2m+1)m\delta)
\end{align*}
Which is where I'm stuck.

答:
+) Your $s_0$ should be $s_0(\rho)=\rho-\langle \rho,\alpha_0^{\vee}\rangle \alpha_0=\rho-\alpha_0=\rho+\alpha_1-\delta$.

+) The Weyl group $W\cong \mathbb{Z} \rtimes S_2$ is generated by $1\in \mathbb{Z}$ (which is $s_0s_1$) and $s\in S_2$ (which is $s_1$).
In other words, its elements are of the form $\{s_1(s_0s_1)^m,(s_0s_1)^m\}$.
One can show $(s_0s_1)^m:\rho \mapsto \rho+2m\alpha_1-(2m^2+m)\delta$ and $s_1(s_0s_1)^m:\rho\mapsto \rho-(2m+1)\alpha_1-(2m^2+m)\delta$.

+) Summing over all elements in the Weyl group we get the desired answer.


问:
Generalized Counting Measure
Let $E$ be an uncountable set. Fix a specific countable subset $M$ from $E$, namely
$$M:=\{x_1,x_2,\ldots\}\subset E.$$
Fix a sequence $\{c_i\}_{i=1}^\infty\in\ell_1(\mathbb{N})$. For any subset $S$ of $E$, we define the measure as
$$\mu(S):=\sum\limits_{x_i\in S}|c_i|.$$
So is this meansure well-defined? If it is, for any function $f$ on $E$, the integration with the measure above should be
$$\int_E fd\mu:=\sum\limits_{n=1}^\infty f(x_i)\cdot |c_i|.$$
Is this correct? Thanks.

答: The answer to both questioins is YES and this is stanadrd in Probailtiy Theory. Counatble addiitvity of $\mu$ follows from the fact that we can always interchange two infinite sums when the terms are non-negative. The last part holds whenver $f$ is non-negative or integrable w.r.t. $\mu$. (The measure $\mu$ is defined on the power set and every function is measurable).

Title: Generalized Counting Measure


问:
Either-or condition for equality constraints
Consider the following optimization problem
\begin{equation*}
\begin{split}
\text{min} & \sum_{j\in J} c_jx_j \\
& \quad \sum_{j\in J} a_{1j}x_j \leq b_1 \\
& \quad \sum_{j\in J} a_{2j}x_j \leq b_2 \\
& \quad x_j \geq 0 \quad \forall j\in J
\end{split}
\end{equation*}
It is well known that, if we need to impose that just one of these two constraints must be satisfied, then we can use introduce a binary variable $y\in\{0,1\}$ and two upper bounds $M_1$ and $M_2$ so that
\begin{equation*}
\begin{split}
\text{min} & \sum_{j\in J} c_jx_j \\
& \quad \sum_{j\in J} a_{1j}x_j \leq b_1 + M_1y\\
& \quad \sum_{j\in J} a_{2j}x_j \leq b_2 +M_2(1-y)\\
& \quad x_j \geq 0 \quad \forall j\in J
\end{split}
\end{equation*}

Imagine that, in the initial problem, the constraints are equalities, that is to say,
\begin{equation*}
\begin{split}
\text{min} & \sum_{j\in J} c_jx_j \\
& \quad \sum_{j\in J} a_{1j}x_j = b_1 \\
& \quad \sum_{j\in J} a_{2j}x_j = b_2 \\
& \quad x_j \geq 0 \quad \forall j\in J
\end{split}
\end{equation*}
How can we impose that just one of these constraints holds with binary variables?

Any help would be appreciated.

答:
You can do this by reformulating each equality constraint as two inequality constraints:
\begin{align}
& \quad \sum_{j\in J} a_{1j}x_j \leq b_1 +M_1y \\
& \quad \sum_{j\in J} a_{1j}x_j \geq b_1 -M_1y \\
& \quad \sum_{j\in J} a_{2j}x_j \leq b_2 +M_2(1-y) \\
& \quad \sum_{j\in J} a_{2j}x_j \geq b_2 -M_2(1-y)
\end{align}


答:
Yes, the measure $\mu$ is well-defined in the $\sigma$-algebra $2^E$. An easy way to see this is to note that, for any $S \subseteq E$,
$$\mu(S)=\mu(S \cap M) = \sum\limits_{x_i\in S \cap M}|c_i|$$
It is then immediate to see that $\mu(\emptyset)=0$ and that $\mu$ is countably additive.

Since the $\sigma$-algebra is $2^E$, it follows that any function $f$ defined on $E$ is measurable.

Now, note that $\mu(E\setminus M)=0$ and, for all $x_i \in M$,  $\mu({x_i}) =|c_i|$.  It follows that

$$\int_E fd\mu=\int_{E \setminus M} fd\mu+\int_M fd\mu=\int_M fd\mu=\sum\limits_{n=1}^\infty f(x_i)\cdot \mu({x_i})=\sum\limits_{n=1}^\infty f(x_i)\cdot |c_i|.$$

Title: Generalized Counting Measure

Body:
Let $E$ be an uncountable set. Fix a specific countable subset $M$ from $E$, namely
$$M:=\{x_1,x_2,\ldots\}\subset E.$$
Fix a sequence $\{c_i\}_{i=1}^\infty\in\ell_1(\mathbb{N})$. For any subset $S$ of $E$, we define the measure as
$$\mu(S):=\sum\limits_{x_i\in S}|c_i|.$$
So is this meansure well-defined? If it is, for any function $f$ on $E$, the integration with the measure above should be
$$\int_E fd\mu:=\sum\limits_{n=1}^\infty f(x_i)\cdot |c_i|.$$
Is this correct? Thanks.


答:
Let $(\delta, \varpi_1, \Lambda_0)$ be the basis of $\mathfrak{h}^*$ which is dual to the basis $(d, \alpha_1^\vee, c)$ of $\mathfrak{h}$, where $c$ is the canonical central element and $d$ is the derivation. The element $\varpi_1$ is a lift of the fundamental weight of $\mathfrak{sl}_2$ to the affine algebra, but is not a fundamental weight of the affine algebra: instead the element $\Lambda_1 = \varpi_1 + \Lambda_0$ is. In this basis the missing roots and coroots are
$$ \alpha_1 = 2 \varpi_1, \quad \alpha_0 = \delta - 2 \varpi_1, \quad \alpha_0^\vee = c - \alpha_1^\vee.$$
The reason we use this basis is because it shows the structure of the Weyl group a bit better: abstractly, the affine Weyl group (for untwisted affine Lie algebras) is the semidirect product of the finite coroot lattice and the finite Weyl group. In terms of the basis $(\delta, \varpi_1, \Lambda_0)$ of $\mathfrak{h}^*$, the matrices (with zeros omitted) of the simple reflections are
$$ s_0 = \begin{pmatrix} 1 & 1 & -1 \\ & -1 & 2 \\ & & 1 \end{pmatrix}, \quad s_1 = \begin{pmatrix} 1 & & \\ & -1 & \\ & & 1\end{pmatrix}. $$
We can see the semidirect product structure by defining $t = s_0 s_1$, then we have $s_1 t^k = t^{-k} s_1$, and the Weyl group is the semidirect product of $\mathbb{Z}$ (generated by $t$) and $\mathbb{Z}_2$ (generated by $s_1$). In terms of matrices, we have
$$ t^k = \begin{pmatrix} 1 & -k & -k^2 \\ & 1 & 2k \\ & & 1 \end{pmatrix}.$$

A little computation shows that
$$ t^k \rho - \rho = (-k - 2k^2)\delta + 4k \varpi_1, \quad s_1 t^k \rho - \rho = (-k -2k^2) \delta + (-4k - 2) \varpi_1.$$
Therefore letting $q = e^{-\delta}$ and $r = e^{\alpha_1} = e^{2 \varpi_1}$ we get
$$ \begin{aligned}
\sum_{w \in W} (-1)^w e^{w \rho - \rho}
&= \sum_{k \in \mathbb{Z}}(e^{t^k \rho - \rho} - e^{s_1 t^k \rho - \rho}) \\
&= \sum_{k \in \mathbb{Z}}q^{k + 2k^2}r^{2k} - \sum_{k \in \mathbb{Z}}q^{k + 2k^2}r^{-2k-1} \\
&= \sum_{l \in 2\mathbb{Z}} q^{(l^2 + l)/2} r^l - \sum_{l \in 2\mathbb{Z} + 1}q^{(l^2 - l)/2} r^{-l} \\
&= \sum_{l \in 2\mathbb{Z}} q^{(l^2 + l)/2} r^l - \sum_{l \in 2\mathbb{Z} + 1}q^{(l^2 + l)/2} r^{l} \\
&= \sum_{l \in \mathbb{Z}} (-1)^l q^{(l^2 + l)/2} r^l.
\end{aligned}$$

The first equality is using the semidirect product fact, the third  equality is reindexing the first sum over even integers $l = 2k$ and the second sum over odd integers $l = 2k + 1$, and the fourth equality is using the $l \mapsto -l$ symmetry in the second sum. The final answer is equivalent to yours because I used $r = e^{\alpha_1}$ rather than $r = e^{- \alpha_1}$.

Title: Deriving Jacobi triple from the Weyl-Kac denominator formula of affine $\mathfrak{sl}_2$

Body:
The Weyl-Kac denominator formula states

$$\prod_{\alpha\in\Phi^+}(1-e(-\alpha))^{mult(\alpha)}=\sum_{w\in W}\epsilon(w)e(w(\rho)-\rho))$$
Where the product is taken over the positive roots and the sum is taken over the Weyl group. Here $\epsilon(w)=(-1)^{l(w)}$ where $l(w)$ denotes the length of $w$ and $e(\alpha)$ is the formal exponential.

I'm trying to apply this to the affinization of $\mathfrak{sl}_2$ with simple roots $\{\alpha_0,\alpha_1\}$. In this case the $\rho=\frac{\alpha_1}{2}+2\Lambda_0$ seems to work. Here $\Lambda_0$ is defined on the coroots st. $\Lambda_0(\alpha_i^\vee)=\delta_{i0}$.

The Weyl group of the affinization of $\mathfrak{sl}_2$ is generated by $s_0,s_1$ st. $s_0^2=s_1^2=1$ (it's the infinite dihedral group). The generators works on $\rho$ in the following way
$$s_0(\rho)=\rho-\rho(\alpha_0^\vee)\alpha_0=\rho+\alpha_0=\rho-\alpha_1+\delta$$
$$s_1(\rho)=\rho-\rho(\alpha_1^\vee)\alpha_1=\rho-\alpha_1$$

Now for the formula. The positive roots are exactly $\alpha_1+k\delta$ for $k\geq 0$, $-\alpha_1+k\delta$ for $k\in\mathbb{N}$ and $k\delta$ for $k\in\mathbb{N}$. Here $\delta=\alpha_0+\alpha_1$. Writing $e(-\delta)=q$ and $e(-\alpha_1)=r$ then the left side is
$$\prod_{k>0}(1-q^k)(1-q^{k-1}r)(1-q^kr^{-1})$$
Turning to the right side, is where I get in trouble. I've seen it should equal $\sum_{k\in\mathbb{Z}}(-1)^kr^kq^{\frac{k(k-1)}{2}}$, so working my way backwards then $\sum_{w\in W}\epsilon(w)e(w(\rho)-\rho)$ should equal
\begin{align*}
\sum_{k\in\mathbb{Z}}(-1)^ke(-k\alpha_1)e(-\frac{k(k-1)}{2}\delta) &= \sum_{k\in\mathbb{Z}}(-1)^ke(-k\alpha_1-\frac{k(k-1)}{2}\delta) \\
&=\sum_{m\in\mathbb{Z}}e(-2m\alpha_1-\frac{2m(2m-1)}{2}\delta)\\
&-e(-(2m+1)\alpha_1-\frac{(2m+1)((2m+1)-1)}{2}\delta) \\
&=\sum_{m\in\mathbb{Z}}e(-2m\alpha_1-m(2m-1)\delta)\\
&-e(-(2m+1)\alpha_1-(2m+1)m\delta)
\end{align*}
Which is where I'm stuck.


问:
Can't understand the answer to this combinatorics problem
```<code>Consider a deck of 40 cards (10 separate cards for each of the four suits).
Alice, Bianca, and Carlo draw from the deck respectively 5, 4, and 3 cards.
How many are the possible extractions in which Alice and Bianca have not drawn swords, while Carlo has
extracted at least 2?
</code>```

The answer is: $30\choose{5}$$\times$$25\choose{4}$$\times$$($$10\choose{2}$$\times$$21\choose{1}$+$10\choose3$$)$

But I can't understand why $10\choose3$ is inside the parentheses, something like:
$30\choose{5}$$\times$$25\choose{4}$$\times$$($$10\choose{2}$$\times$$21\choose{1}$$)$+$10\choose3$ it seemed more logical to me, because I'm choosing $A+B\iff A \ or \ B \iff (exactly\ 2 \ drawn \ swords) \ or \ (at\ least \ 2\ swords)$

However it's a little different than I thought.

答:
I think, in the original answer, it is basically a 3 term product. Alice's draw, times Bianca's draw, times Carlo's draw. There are Two possibilites for Carlo's draw. He either draws 2 swords and 1 other, or he draws 3 swords and 0 other. All three people draw, so their respective terms are multiplied, however, Carlo's draw, can only be one of the aforementioned possibilites, hence you have addition between them, since you're counting all the possibilites. You have to be careful with ands and ors.

Your answer reads like this:

$\bigg($Alice draws 5 cards AND Bianca draws 4 cards AND Carlo draws exactly 2 swords$\bigg)$ OR Carlo draws 3 swords.

The given answer reads like this:

Alice draws 5 cards AND Bianca draws 4 cards AND $\bigg($Carlo draws exactly 2 swords OR Carlo draws 3 swords.$\bigg)$

The main point here is, that you have to treat Carlo's draw, as ONE thing. In your answer, the last part stands on it's own, meaning Carlo draws 3 swords, and the other two doesn't do anything, hence you don't take account for a lot of outcomes, hence it is less, than the given answer. It can help, if you look at the different events as sets, like $A,B,C$ etc. and $+,\times$ as $\cup,\cap$ respectively. You applied the associative property of sets incorrectly. It only holds, if there's only one of the two set operators present.


问:
What is the relation between linear transformations with the same matrix on different bases?
Given two linear transformations $$Tx=Ax$$ and $$T'y=Ay$$ where $x$ and $y$ are two different bases of a vector space $V$, how are those linear transformations related?

Suppose that $$y=Px$$ where $P$ is the change of basis matrix, then I have $$T'x=APx$$
I can't see how $T=Ax$ and $T'=APx$ relate to each other, perhaps if it was true that $AP=PA$ then $$T'=APx=PAx=PTx$$ but $AP=PA$ is not necessarily true, so what is their relations?

答:
Let $\mathcal A:=(a_1,\dots,a_n)$ and $\mathcal B=(b_1,\dots,b_n)$ be two bases of $V$, and let $P:V\to V$ be the base change map $a_i\mapsto b_i$.

The condition that linear map $T$ has the same matrix in basis $\mathcal A$ as map $T'$ in basis $\mathcal B$ should rather be translated as follows:

for every index $i$, we have $[Ta_i]_{\mathcal A}\ =\ [T'b_i]_{\mathcal B}$, where $[x]_{\mathcal A}$ denotes the coordinate vector of $x$ in basis $\mathcal A$.

By the base of change formula, we have $[x]_{\mathcal A}=[Px]_{\mathcal B}$ for every $x\in V$. (It's obvious for the basis elements and we can use linearity.)
So, the above condition is equivalent to $PTa_i=T'b_i$, or, using $b_i=Pa_i$, to
$$PTa_i=T'Pa_i\,.$$
And thus, it holds if and only if $PT=T'P$, or, put another way, $T'=PTP^{-1}$.

Since any invertible map can be viewed as a base-of-change map, we obtained that the answer is that $T$ and $T'$ are conjugate.