问:
What is the least number of cuts that should be made to obtain the maximum pieces possible in a parallelogram?
The problem is as follows:

The figure from below represents a piece of wood that has the shape of
a parallelogram. whose sides are of equal length. If we want to cut
piece of wood in the maximum number of equilateral triangles possible
with a side of $5$ cm. How many straight cuts at least are necessary
to make using a circular saw to obtain these pieces?

The alternatives which do appear in my book are:

$\begin{array}{ll}
1.&\textrm{3 cuts}\\
2.&\textrm{4 cuts}\\
3.&\textrm{5 cuts}\\
4.&\textrm{6 cuts}\\
\end{array}$

Gee for this problem I have no idea where to begin. What I thought was to construct equilateral triangles inside the figure. But this did not helped that much.

In other words I've attempted to fold the figure by its corners but this approach did not yielded the desired results and I ended up with no answer.

Thus what should be the way to solve this problem?. Is there any trick here or what?.
It would help me a lot an answer which would indicate with some arrows or lines where should the cuts be made or the justification for these cuts because I'm lost with this.

答:
It’s not hard to check that the parallelogram can be divided into $32$ equilateral triangles with $5$ cm sides. Clearly you will have to stack the pieces after each cut if you are to get anything like that number. Note that each cut can at most double the number of pieces that you currently have, and $32=2^5$, so it will take at least $5$ cuts to get the maximum possible number of triangles. And it can in fact be done in $5$ cuts.

HINT: Make the first cut parallel to the base, producing two identical parallelograms, and stack them in the natural way before making your next cut.


问:
Eigenvalue preserving projection
Given a symmetric $m \times m$ matrix $X$, and say we know that the smallest eigenvalue of $X$ has eigenvector $u$. Is there a way to use this information to construct a $(m-1)\times (m-1)$ matrix $Y$, such that $Y$ has the same first $m-1$ eigenvalues as $X$?

答: Hint: Rotate $X$ to a coordinate system where $u$ is parallel to one of the axes. Since the eigenvectors of $X$ must be orthogonal, what does this new matrix look like? Can you find $Y$ from there?

ContentLicense: CC BY-SA 4.0

Title: Eigenvalue preserving projection


问:
Necessary and sufficient condition on a set for a continuous function $f$ to be bounded by the subsets
Question: Find  a  necessary  and  sufficient  condition  on  the  set $A \subset \mathbb{R}$ such that $\forall B \subset A, f(B)$ is bounded, where $f$ is any continuous function $f:A \rightarrow \mathbb{R}$.

My attempt: I think the answer should be $A$ must be closed in $\mathbb{R}$. Suppose A is not closed, then we want to show that there exist a function $f:A \rightarrow \mathbb{R}$ and a subset $B$ such that $f(B)$ is unbounded. Since $A$ is not closed, there exist $a_1, a_2, ... , a_n \in A$ such that $a_n \rightarrow a$ and $a \notin A$. Consider the function $f(x)=\frac{1}{x-a}$ which is continuous on $A$. We can find an $\epsilon > 0$ such that $B_a(\epsilon) \cap A$ is a subset of $A$. Then $f(B_a(\epsilon) \cap A)$ is clearly unbounded. However, I don't know how to prove the other direction ($A$ is closed $\implies \forall B \subset A, f(B)$ is bounded). Any help would be greatly appreciated.

答:
We show that $A$ is closed and bounded (compact) is a NS-condition.

If $A$ is closed and bounded, $B \subset A$, and $f: A \to \Bbb R$ is continuous, since $f(B) \subset f(A)$ and $f$ is continuous we get the result.

Suppose $\forall B \subset A, f(B)$ is bounded where $f$ is continous. Take $f=\text{id}$. We see that $A$ is bounded. You already showed that it's closed.


答:
I get $5$ cuts:

Cut along the middle, parallel to the $x$ axis.  Then stack the top half exactly atop the bottom half.

Cut along the middle, parallel (again) to the $x$ axis, giving four "strips."  Stack them atop one another, perfectly.

Cut the stack in half (by the only way possible).  Restack the pieces to be in a $4 \times 1$ parallelogram.

Cut this stack in half (by the only way possible).  Restack the pieces to be in a $2 \times 1$ parallelogram.

Cut this stack in half (by the only way possible).

As @Brian M. Scott shows:  $5$ cuts is the minimum.  So this must be an optimal solution (likely not unique).

Title: What is the least number of cuts that should be made to obtain the maximum pieces possible in a parallelogram?

Body:
The problem is as follows:

The figure from below represents a piece of wood that has the shape of
a parallelogram. whose sides are of equal length. If we want to cut
piece of wood in the maximum number of equilateral triangles possible
with a side of $5$ cm. How many straight cuts at least are necessary
to make using a circular saw to obtain these pieces?

The alternatives which do appear in my book are:

$\begin{array}{ll}
1.&\textrm{3 cuts}\\
2.&\textrm{4 cuts}\\
3.&\textrm{5 cuts}\\
4.&\textrm{6 cuts}\\
\end{array}$

Gee for this problem I have no idea where to begin. What I thought was to construct equilateral triangles inside the figure. But this did not helped that much.

In other words I've attempted to fold the figure by its corners but this approach did not yielded the desired results and I ended up with no answer.

Thus what should be the way to solve this problem?. Is there any trick here or what?.
It would help me a lot an answer which would indicate with some arrows or lines where should the cuts be made or the justification for these cuts because I'm lost with this.


问:
If $f:S\rightarrow \mathbb{R}^m$ is a map such that each $f|_{S_i}:S_i\rightarrow \mathbb{R}^m$ is differentiable, prove that $f$ is differentiable.
QUESTION 1: Let $f, g: S\rightarrow \mathbb{R}^m$ be differentiable vector-valued functions and let $\lambda\in \mathbb{R}$. Prove that the function $(f+g):S\rightarrow \mathbb{R}^m$ is also differentiable.
PROOF: Let $X:U\rightarrow S$ be a parametrization of $S$. Then $$(f+g)\circ X= f\circ X + g\circ X$$ which is differentiable, as a sum of differentiable functions.

MY DOUBT IN QUESTION 1: How can I show in a step by step way that the sum above is distributive with the composition maps?

QUESTION 2: Suppose that a surface $S$ is a union $S=\displaystyle\bigcup_{i\in I} S_i$, where each $S_i$ is open. If $f:S\rightarrow \mathbb{R}^m$ is a map such that each $f|_{S_i}:S_i\rightarrow \mathbb{R}^m$ is differentiable, prove that $f$ is differentiable.

MY ATTEMPT: We have already proved that open sets in surfaces are also surfaces. Defining $f|_{S_i}:S_i\rightarrow \mathbb{R}^m$ as $f_i:S_i\rightarrow \mathbb{R}^m$ such that $f=\sum f_i$, considering $X:U\rightarrow S$ a parametrization of S (which is a surface), then we can write $(f_1+\cdots +f_n)\circ X= f_1\circ X+\cdots +f_n\circ X$ is differentiable, this is $f$ is differentiable.

MY DOUBT IN QUESTION 2: I'm not sure if I can interpret the union as a sum. Would you help me?

答:
This is correct. Distributivity comes pretty much directly from definition, try calculating $(f+g)\circ X(p)$. Recall that $(f+g)(q) := f(q)+g(q)$.

This is wrong. Recall that differentiability is a local property. That means if we want to prove $f$ is differentiable at $p$, we can take $X:U\to S$ with $X(S)$ small enough to be contained in $S_i$ (where $p\in X(S)\cap S_i$). Compatibility of parametrizations takes over the rest.


问:
Does the column space of a matrix have the same dimension of a column space of that matrix in echelon form?
Suppose we have two matrices $A$, $B$ that are row-equivalent. Matrix $B$ is in echelon form. Would $\operatorname{Col}(A)$ and $\operatorname{Col}(B)$ have the same dimension? I know that the contents of the column spaces themselves aren't always the same, but would the dimensions still be the same?

答:
Yes, if $A$ and $B$ are row-equivalent then their columns spaces have the same dimension (i.e. rank).

This is because $A$ and $B$ have the same null space. To see this, note that there is an invertible matrix $M$ such that $B = MA$. Suppose $x$ is in the null space of $A$. Then $Bx = MAx = M(Ax) = 0$, so $x$ is in the null space of $B$. Conversely, if $x$ is in the null space of $B$, then $Bx = 0$, hence $MAx = 0$. But $M$ is invertible, so this implies $Ax = M^{-1}0 = 0$, so $x$ is in the null space of $A$.

Now let $n$ denote the number of columns of $A$ (and hence also of $B$). The rank-nullity theorem tells us that
$$\dim(\operatorname{col}(A)) + \dim(\operatorname{null}(A)) = n$$
and
$$\dim(\operatorname{col}(B)) + \dim(\operatorname{null}(B)) = n$$
Since $\dim(\operatorname{null}(A)) = \dim(\operatorname{null}(B))$, this means that $\dim(\operatorname{col}(A)) = \dim(\operatorname{col}(B))$.


答:
Question 1. The definition of sum of functions $f$ and $g$ is a function $(f+g):S\rightarrow \mathbb{R}^m$, which can be described with a formula $(f+g)(s)=f(s)+g(s)$. It is correct since $\mathbb{R}^m$ is a vector space. To show that functions $(f+g)\circ X$ and $f\circ X+g\circ X$ you need to show that they are pointwisely equal on $U$. Indeed, \begin{equation*}((f+g)\circ X)(u)=(f+g)(X(u))=f(X(u))+g(X(u))=
(f\circ X)(u)+(g\circ X)(u)=(f\circ X+g\circ X)(u),\end{equation*}
where I used the definition of composition of functions. Here you have no more than definitions.

Body:
QUESTION 1: Let $f, g: S\rightarrow \mathbb{R}^m$ be differentiable vector-valued functions and let $\lambda\in \mathbb{R}$. Prove that the function $(f+g):S\rightarrow \mathbb{R}^m$ is also differentiable.
PROOF: Let $X:U\rightarrow S$ be a parametrization of $S$. Then $$(f+g)\circ X= f\circ X + g\circ X$$ which is differentiable, as a sum of differentiable functions.

MY DOUBT IN QUESTION 1: How can I show in a step by step way that the sum above is distributive with the composition maps?

QUESTION 2: Suppose that a surface $S$ is a union $S=\displaystyle\bigcup_{i\in I} S_i$, where each $S_i$ is open. If $f:S\rightarrow \mathbb{R}^m$ is a map such that each $f|_{S_i}:S_i\rightarrow \mathbb{R}^m$ is differentiable, prove that $f$ is differentiable.

MY ATTEMPT: We have already proved that open sets in surfaces are also surfaces. Defining $f|_{S_i}:S_i\rightarrow \mathbb{R}^m$ as $f_i:S_i\rightarrow \mathbb{R}^m$ such that $f=\sum f_i$, considering $X:U\rightarrow S$ a parametrization of S (which is a surface), then we can write $(f_1+\cdots +f_n)\circ X= f_1\circ X+\cdots +f_n\circ X$ is differentiable, this is $f$ is differentiable.

MY DOUBT IN QUESTION 2: I'm not sure if I can interpret the union as a sum. Would you help me?


问:
Uniform continuous with simple properties
$f$ is a map from $f:\mathbb{R} \rightarrow \mathbb{R}$. Let $f$ be continous and assume $f$ is both uniformly contiuous on $(-\infty,0]$ and $[0,\infty)$ . Show that $f$ is uniformly continuous on all of $\mathbb{R}$.

I would've thought that the function $f$ is actually uniformly continuous over all the real numbers since it is in both the intervals $(-\infty,0]$ and $[0,\infty)$. I guess the "problem" arises on $f(0)$. Do I have to use that $f(x)$ and $f(y)$ for $x<0<y$ somehow and $f(0)$?

How would such a proof be done?

答:
We wish to show, for all $\varepsilon > 0$, there exists some $\delta > 0$ such that, for all $x, y \in \Bbb{R}$ (might as well assume $x < y$),
$$|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon.$$
As you seem to be aware, there are three cases of note: either $x < y \le 0$, $0 \le x < y$, or $x < 0 < y$. We need a $\delta$ small enough to cover whichever case it happens to be. The tricky one seems to be the third case: $x < 0 < y$, which we can cover using continuity at $0$.

Because $f$ is continuous at $0$, there exists some $\delta_1 > 0$ such that
$$|x| < \delta_1 \implies |f(x) - f(0)| < \frac{\varepsilon}{2}.$$
If we have $|x - y| < \delta_1$ with $x < 0 < y$, then
$$|x| = -x < y - x = |x - y| < \delta,$$
and similarly,
$$|y| = y < y - x = |x - y| < \delta.$$
Therefore, for such $x$ and $y$ we have
$$|f(x) - f(0)| < \frac{\varepsilon}{2} \text{ and } |f(y) - f(0)| < \frac{\varepsilon}{2}.$$
Thus,
$$|f(x) - f(y)| \le |f(x) - f(0)| + |f(0) - f(y)| < \varepsilon.$$
This covers the $x < 0 < y$ case. To complete the full proof, let $\delta_2 > 0$ such that
$$|x - y| < \delta_2 \text{ and } x, y \le 0 \implies |f(x) - f(y)| < \varepsilon$$
and $\delta_3 > 0$ such that
$$|x - y| < \delta_3 \text{ and } x, y \ge 0 \implies |f(x) - f(y)| < \varepsilon.$$
Then
$$|x - y| < \min\{\delta_1, \delta_2, \delta_3\} \implies |f(x) - f(y)| < \varepsilon,$$
as required.


问:
Find the limit of the sequence $(x_n)$, where $x_1=1$ and $x_{n+1}=(2+x_n)^{1/2}$
I have been given this hint in the problem: In order to determine the limit, it may be helpful to use the fact that if $y_n \rightarrow y$ and $y_n > 0$ for every $n$,then $\sqrt{y_n} \rightarrow \sqrt{y}$. You can use this without proving it. You can also use without proof the fact that if $0\le a \le b,$ then $\sqrt{a} \le \sqrt{b}$.

答:
My attempt :

In the first, is easy to showing $x_n$ is incremental, just pose $f(x) =\sqrt{2+x}$ and use recurrence for showing $x_n<x_{n+1}$

And  you can see that $x_2=x_3=x_4=.......=2$

So $x_n\leq 2$ and for showing that also you can use recurrence

For $n=1$ $x_1\leq2$$\Rightarrow $$1\leq 2$ that is correct

Suppose $x_n\leq 2$

Let's show $x_{n+1}\leq 2$

We know $x_n\leq 2$

So $f(x_n) \leq f(2)$

so $x_{n+1}\leq 2$ because f is incremental

we see $x_n\leq 2$ and $x_n$ is incremental so $x_n $ is convergent and $\lim x_n=l$

So. $l=\sqrt{2+l} $$\Rightarrow $$ l^2-l-2=0$$\Rightarrow l=2$ or $l=-1 $but this is impossible because $l\geq 0$

So finally

$\lim x_n=2. $


问:
Series equal to an integral
If I have
\begin{eqnarray*}
S = \sum_{n\geq 0}{\frac{(-1)^{n}}{3n+1}}.
\end{eqnarray*}

Prove that
\begin{eqnarray*}
S = \int_{0}^{1}{\frac{1}{1+x^{3}}}\,\mathrm{d}x.
\end{eqnarray*}

I know that serie is convergent but I don't know how to prove this, I think in some theorem of uniformly convergence and integration where
\begin{eqnarray*}
f(x)=\sum_{n=1}^{\infty}{f_{n}(x)}
\end{eqnarray*}
but I don't know how to do this. Can you give me some advice to prove it? Thank you.

Title: Series equal to an integral

Body: We find $$\int_0^1 \frac{1}{1+x^3} dx = \int_0^1  \sum_{n \ge 0} (-x)^{3n} \ dx = \sum_{n \ge 0} (-1)^n \int_0^1 x^{3n} dx = \sum_{n \ge 0} \frac{(-1)^n}{3n+1}.$$


问:
Fixed Points in Mobius Transformation
Let $$w(z)=\frac{e^z-1}{e^z+1}; \ z=x+iy$$ and $$w=u+iv$$ Then show that the image of the $y$-axis in the domain is $v$-axis in the co-domain. Does this contradict the result that a Mobius transformation with at least $3$ fixed points is a identity mapping ? Why ?

I have already proved the fact that $y$-axis is mapped onto $v$-axis by showing that
$$w(iy)=i\frac{\sin y}{1+\cos y}$$

Now I need to show that this mapping does not contradict the property of a Moebius transformation. For that I would either have to show that $$\frac{\sin y}{1+\cos y}=y$$ has less than $3$ solutions or that $w(z)$ is not a Moebius transformation. Which one is it and how do I prove that ?

答: A mobius transformation $w(z)=\frac{az+b}{cz+d}, ad-bc\ne 0$ has a unique simple pole at $z=-d/c$ whereas in your problem there are infinite poles $z=(2k+1)i\pi, k=0,1,2,...$ so their equivalence in $\mathbb C$ is impossible.

Title: Fixed Points in Mobius Transformation


答: Question 2. You do not need to extend $f$ on $S$ (it is an answer to your comment to another answer) because $f$ is already defined on $S$, it is already a normal function. You have just to show that it is differentiable. But differentiability is a local property, in other words, it is enough to check it in any neighborhood of every point of $S$. Since $S$ is the union of open sets $S_i$, then for each point $s\in S$ there is an index $i(s)$ such that $s\in S_{i(s)}$. Since $f |_{S_{i(s)}}$ is differentiable on $S_{i(s)}$, then $f$ is differentiable at point $s$, because $f$ and $f|_{S_{i(s)}}$ are equal on $S_{i(s)}$. You do not even need a parameterization of $S$.

Title: If $f:S\rightarrow \mathbb{R}^m$ is a map such that each $f|_{S_i}:S_i\rightarrow \mathbb{R}^m$ is differentiable, prove that $f$ is differentiable.


问:
Nice Identity, Logarithm, Euler Constant
I would like to prove the following identity:
For $x\to 1$ :

$$\sum_{k=1}^{+\infty} \ln\left(1+\frac{x^{k}}{k}\right) = - \ln(1-x) - \gamma +o(1)$$

where $o(1)=\varepsilon (x)$ with $\lim_{x \to 1} \varepsilon (x)= 0$ and $\gamma=\lim_{n \to +\infty} (\sum_{k=1}^{n} \frac{1}{k} - \ln(n))$.

I tried to use that for $t \to 0$:

$$\ln(1+t)=t- \frac{t^{2}}{2} +o(t^{2})$$

For $x \in (0,1)$ fixed, let $u_{n}(x)=\frac{x^{n}}{n}$ . We easily have that $u_{n}(x) \to_{n \to +\infty} 0$. Hence

$$\ln(1+u_{n}(x))= u_{n}(x)- \frac{u_{n}(x)^{2}}{2} + o(u_{n}^{2}(x))$$ and then I am trying to prove that : $\lim_{x \to 1} \sum_{n=1}^{+\infty} \frac{u_{n}(x)^{2}}{2}  = \gamma $  If I can do that maybe the estimation $x-\frac{x^{2}}{2} \leq \ln(1+x) \leq x-\frac{x^{2}}{2}+\frac{x^{3}}{3} $ can help to prove the identity.

答: $$-\ln(1-x)-\sum_{k=1}^\infty\ln\left(1+\frac{x^k}{k}\right)=\sum_{k=1}^\infty\left[\frac{x^k}{k}-\ln\left(1+\frac{x^k}{k}\right)\right].$$ We can take $\lim\limits_{x\to1^-}$ termwise, since $t\mapsto t-\ln(1+t)$ is increasing (for $t>0$). And $$\sum_{k=1}^n\left[\frac1k-\ln\left(1+\frac{1}{k}\right)\right]=\sum_{k=1}^n\frac1k-\ln(n+1)$$ tends to $\gamma$ as $n\to\infty$, because $\ln(n+1)-\ln n$ tends to $0$.


问:
Fill in a $5 \times 5$ square to make equal products in rows and columns
Let's say you fill in a $5 \times 5$ square with $1, 2,\dots, 25$. Is there a way to fill it so that the product of the first row is equal to the product of the first column, the product of the second row is equal to the product of the second column, etc?

What about generalization for a $n \times n$ square filled with $1, 2,\ldots, n^2$?

Edit: Here are some updates on progress. For $n=3$, it is a possibility:

$$\begin{matrix} 5 & 1 & 6\\
                 2 & 7 & 4\\
                 3 & 8 & 9\end{matrix}$$

I'm not sure how to do $n=4, 5$. I know that if there are $>n$ prime number between $n^2/2$ and $n^2$ then $n$ doesn't work. I do not know how to proceed.

答:
Here's one solution:
$$
\matrix{
7 & 18 & 14 & 10 & 5 \\
24 & 13 & 11 & 6 & 15 \\
21 & 22 & 17 & 12 & 4 \\
1 & 20 & 9 & 19 & 8 \\
25 & 3 & 16 & 2 & 23
}
$$
The row and column products are 88200, 308880, 376992, 27360, and 55200.  You can easily modify this by applying an arbitrary permutation to both the rows and the columns, applying another arbitrary permutation to the diagonal entries, and/or taking the transpose.

How I found it:  The primes 13, 17, 19, and 23 only appear once each, so they must all go on the diagonal.  I made the last diagonal entry 7 because $25!$ equals $7 \cdot 13 \cdot 17 \cdot 19 \cdot 23$ times a square, although I don't think this was actually necessary.  (For example, the three multiples of 7 could appear in positions $(1, 2), (2, 3)$, and $(3, 1)$, instead of putting one on the diagonal and the other two symmetrically around it.)  Then, by trial and error, I found a way to place the multiples of 5 so that each row and corresponding column are divisible by 5 with the same multiplicity.  After this, I placed the 1 and 3 near the bottom left to make the row-product-to-column-product ratios nice, and did the rest by more trial and error.

Update:  Here's a solution for the 4x4 version, found by a similar method (but with much less casework):
$$
\matrix{
1 & 9 & 8 & 4 \\
3 & 5 & 7 & 12 \\
6 & 14 & 11 & 10 \\
16 & 2 & 15 & 13
}
$$
I'll let you attempt the cases $n = 6, 7, 8$, and $10$, which seem to be the only other cases not ruled out by primes between $n^2/2$ and $n^2$.


问:
Is Axiom of Choice related to Indeterminism
Is it fundamentally wrong to think if the Axiom of Choice (AC) as a logical codification of indeterminism? Could it not be related to matching distributions with a random number generator?</p>

答:
Yes. It's wrong.

The axiom of choice is not choosing "random elements". That's what existential instantiation is doing, in some sense. The axiom of choice is best understood in some sense in a constructive framework: if you can choose an element from each set in a family of sets, then you're supposedly able to do it uniformly.

Of course, in classical mathematics this is not true, but the axiom of choice lets you overcome that issue.

The "non-constructive part" of the axiom of choice is really an artifact of existential instantiation, if you think about it, that's the part moving from "there is a choice function" to "$f$ is a choice function".


问:
Help with my solution to an integral of Erf functions
I have been trying to solve this integral, but numerical evaluations between my solution and the integral do not match. Could someone please let me know where I went wrong?

$$\int_{-\infty}^{\infty}\exp(-ax^2)\DeclareMathOperator{\erf}{Erf}\erf(bx)\erf(\sqrt{a}x)\mathrm dx,\qquad a,b\in\mathbb{R}^{+}$$

I began with partial differentiation,

$$\frac{\partial}{\partial b}\int_{-\infty}^{\infty}\exp(-ax^2)\DeclareMathOperator{\erf}{Erf}\erf(bx)\erf(\sqrt{a}x)\mathrm dx = \frac{2}{\sqrt{\pi}}\int_{-\infty}^{\infty}2x\exp\Big[-(a+b^2)x^{2}\Big]\erf(\sqrt{a}x)\mathrm dx$$

I was able to solve this integral, and checked it numerically:

$$\frac{2}{\sqrt{\pi}}\int_{-\infty}^{\infty}2x\exp\Big[-(a+b^2)x^{2}\Big]\erf(\sqrt{a}x)\mathrm dx = \frac{2}{\sqrt{\pi}}\Bigg[\frac{\sqrt{a}}{(a+b^2)\sqrt{2a+b^2}}\Bigg]$$
In order to retrieve the original integral, I found the antiderivative wrt. $b$

$$\frac{2}{\sqrt{\pi}}\int\frac{\sqrt{a}}{(a+b^2)\sqrt{2a+b^2}}\mathrm db = \frac{2}{\sqrt{\pi a}}\cdot\arctan\Bigg(\frac{b}{\sqrt{2a+b^2}}\Bigg)+\mathcal{Q}(a)$$

Here, $\mathcal{Q}(a)$ is a random function. In order to solve for $\mathcal{Q}(a)$, I took limits:
$$\lim\limits_{b\to\infty^{+}}\int_{-\infty}^{\infty}\exp(-ax^2)\DeclareMathOperator{\erf}{Erf}\erf(bx)\erf(\sqrt{a}x)\mathrm dx \longrightarrow 0$$
$$\lim\limits_{b\to\infty^{+}}\frac{2}{\sqrt{\pi a}}\cdot\arctan\Bigg(\frac{b}{\sqrt{2a+b^2}}\Bigg)\longrightarrow \frac{\sqrt{\pi}}{2\sqrt{a}} = -\mathcal{Q}(a)$$

Thus, I'd surmise that

$$\int_{-\infty}^{\infty}\exp(-ax^2)\DeclareMathOperator{\erf}{Erf}\erf(bx)\erf(\sqrt{a}x)\mathrm dx = \frac{2}{\sqrt{\pi a}}\cdot\arctan\Bigg(\frac{b}{\sqrt{2a+b^2}}\Bigg) - \frac{\sqrt{\pi}}{2\sqrt{a}}$$

However, when I evaluate the integral numerically, it doesn't match with my closed-form solution. There must be something I did wrong after taking the partial derivative, but I can't figure out what. Please let me know your opinion!

答:
For two copies of $\operatorname{erf}$ I'm unsure if this has a closed form, but if $a=0$ or $b=0$ there is a nice identity
$$\int_{\mathbb{R}}\operatorname{erf}(ax+b)\frac{\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\sqrt{2\pi\sigma^2}}\mathrm{d}x=\operatorname{erf}\left(\frac{a\mu+b}{\sqrt{1+2a^2\sigma^2}}\right)$$


问:
How do calculate the constant c of the Midpoint algorithm?
I hope this is a question for this community.

I consider the Midpoint algorithm for a line as known.

We have the following funktion:
$g(x+1,y+\frac{1}{2})=\Delta y*(x+1)-\Delta x*(y+\frac{1}{2}) + c$, $c=\Delta y*x_1-\Delta x*y_1$

Where $\Delta x = x_2-x_1$ and $\Delta y = y_2-y_1$

Lets have a line between points $P_1(x_1,y_1)=P_1(100,5)$ and $P_2(x_2,y_2)=P_2(107,9)$ the goal is to calculate the rasterization or better the pixels between does two point on the line.

If $g(x+1,y+\frac{1}{2}) > 0$ then we draw $pixel(x+1,y+1)$ else $pixel(x+1,y)$.

But if we take a look at the constant $c$ then I don't see how we ever get $g(...)\leq 0$, because $c$ would be
$c=\Delta y*x_1-\Delta x*y_1 = 4*100 - 7*5 = 365$ way to big.

So the question is, where am I wrong, because i have to be worong?

答:
The function is
$$\begin{aligned}
g(x,y) & = \Delta_y x - \Delta_x y + c \\
c & = \Delta_y x_1 - \Delta_x y_1 \\
\Delta_x & = x_2 - x_1 \\
\Delta_y & = y_2 - y_1 \\
\end{aligned}$$
Expanding it we get
$$\begin{aligned}
g(x,y) & = (y_2 - y_1) x - (x_2 - x_1) y + (y_2 - y_1) x_1 - (x_2 - x_1) y_1 \\
     ~ & = (y_2 - y_1) (x - x_1) - (x_2 - x_1) (y - y_1) \\
\end{aligned}$$
Note how $g(x_1, y_1) = 0$ and $g(x_2, y_2) = 0$?

If we use $0 \le \lambda \le 1$ to interpolate linearly between $(x_1, y_1)$ and $(x_2, y_2)$, i.e. the line between the two points, using
$$\left\lbrace\begin{aligned}
x = (1 - \lambda) x_1 + \lambda x_2 &= x_1 + \lambda (x_2 - x_1) \\
y = (1 - \lambda) y_1 + \lambda y_2 &= y_1 + \lambda (x_2 - x_1) \\
\end{aligned}\right.$$
we get
$$\begin{aligned}
g(\lambda) &= (y_2 - y_1) \lambda (x_2 - x_1) - (x_2 - x_1) \lambda (y_2 - y_1) \\
~          &= (x_2 - x_1) \lambda (y_2 - y_1) - (x_2 - x_1) \lambda (y_2 - y_1) \\
~          &= 0 \\
\end{aligned}$$

Let's examine the case where $(x_1, y_1) = (100, 5)$ and $(x_2, y_2) = (107, 9)$.  Now,
$$\begin{aligned}
\Delta_x = x_2 - x_1 = 107 - 100 &= 7 \\
\Delta_y = y_2 - y_1 = 9 - 5 &= 4 \\
c = \Delta_y x_1 - \Delta x y_1 &= 365 \\
\end{aligned}$$
Note that $\Delta_x \gt \Delta_y$.  This means the line is closer to horizontal than vertical, and at every step $x$ increases by one. The rule for mostly horizontal lines, $x_2 \ge x_1$, and $x_2 - x_1 \ge y_2 - y_1 \ge 0$, is
$$\text{if } g\left(x + 1, y + \frac{1}{2}\right) \ge 0, \text{ draw } (x + 1, y), \text{ else draw } (x + 1, y + 1)$$

First pixel is $(x_1, y_1)$, and we always draw that.

Because $g\left(x_1 + 1, y_1 + \frac{1}{2}\right) = -\frac{1}{2}$, we draw $(x_1 + 1, y_1 + 1)$.

Because $g\left(x_1 + 2, y_1 + 1 + \frac{1}{2}\right) = \frac{5}{2}$, we draw $(x_1 + 2, y_1 + 1)$.

Because $g\left(x_1 + 3, y_1 + 1 + \frac{1}{2}\right) = -\frac{3}{2}$, we draw $(x_1 + 3, y_1 + 2)$.

Because $g\left(x_1 + 4, y_1 + 2 + \frac{1}{2}\right) = \frac{3}{2}$, we draw $(x_1 + 4, y_1 + 2)$.

Because $g\left(x_1 + 5, y_1 + 2 + \frac{1}{2}\right) = -\frac{5}{2}$, we draw $(x_1 + 5, y_1 + 3)$.

Because $g\left(x_1 + 6, y_1 + 3 + \frac{1}{2}\right) = \frac{1}{2}$, we draw $(x_1 + 6, y_1 + 3)$.

Because $g\left(x_1 + 7, y_1 + 3 + \frac{1}{2}\right) = -\frac{7}{2}$, we draw $(x_1 + 7, y_1 + 4)$.

We realize that $x_1 + 7 = x_2$ (and $y_1 + 4 = y_2$), so we arrived where we should have.  Line drawn.

All eight octants (divided by lines $x = 0$, $y = 0$, $y = x$, and $y = -x$) have their own rule, but if we swap the line endpoints when necessary, we only need to deal with four of them.  The logic is as follows:

If $x_1 = x_2$ and $y_1 = y_2$, the line is only a single point.

Otherwise,

If $y_1 = y_2$, the line is horizontal.  Draw from $x = \min(x_1, x_2)$ to $x = \max(x_1, x_2)$ at $y = y_1 = y_2$.

Otherwise,

If $x_1 = x_2$, the line is vertical.  Draw from $y = \min(y_1, y_2)$ to $y = \max(y_1, y_2)$ at $x = x_1 = x_2$.

Otherwise,

If $\lvert x_2 - x_1 \rvert \ge \lvert y_2 - y_1 \rvert$, the line is mostly horizontal.

Otherwise, the line is mostly vertical.

When the line is mostly horizontal, make sure $x_1 \lt x_2$ (swapping the endpoints if necessary; remember to update $c$, $\Delta_x$, and $\Delta_y$). Then, if $y_2 \gt y_1$, the rule is
$$\text{if } g\left(x + 1, y + \frac{1}{2}\right) \ge 0, \text{ draw } (x + 1, y), \text{ else draw } (x + 1, y + 1)$$
Otherwise, for a mostly horizontal line, $y_2 \lt y_1$, and the rule is
$$\text{if } g\left(x + 1, y - \frac{1}{2}\right) \ge 0, \text{ draw } (x + 1, y), \text{ else draw } (x + 1, y - 1)$$

When the line is mostly vertical, make sure $y_1 \lt y_2$ (swapping the endpoints if necessary; remember to update $c$, $\Delta_x$, and $\Delta_y$). Then, if $x_2 \gt x_1$, the rule is
$$\text{if } g\left(x + \frac{1}{2}, y + 1\right) \ge 0, \text{ draw } (x, y + 1), \text{ else draw } (x + 1, y + 1)$$
Otherwise, for a mostly vertical line, $x_2 \lt x_1$, and the rule is
$$\text{if } g\left(x - \frac{1}{2}, y + 1\right) \ge 0, \text{ draw } (x, y + 1), \text{ else draw } (x - 1, y + 1)$$

In both mostly horizontal line cases and both mostly vertical line cases, we start with $x = x_1$ and $y = y_1$, and draw pixel $(x, y)$.

Mostly horizontal lines loop for as long as $x \lt x_2$, and mostly vertical lines loop for as long as $y \lt y_2$.


问:
Cosine of complex variable

Show that $|\cos z|\leq2$  for all $z \in B(0,r)\subset C$ for some $r>0$ . Can we have $|\cos z|\leq 1$  for all $z \in B(0,r)\subset C$ for some $r>0$ ?

I have found the answer for the first part but can't think of a solution for the second question. For the first part

$$|\cos z|=\biggl|\frac{e^{iz}+e^{-iz}}{2}\biggr|$$
$$ =\biggl|\frac{e^{ix}*e^{-y}+e^{-ix}*e^{y}}{2}\biggr|$$
$$ \leq\biggl|\frac{e^{ix}*e^{-y}}{2}\biggr|+ \biggl|\frac{e^{-ix}*e^{y}}{2}\biggr| $$
$$ \text{But }\:|e^{-ix}|=|e^{ix}|=1$$
$$ =\biggl|\frac{e^{-y}}{2}\biggr|+ \biggl|\frac{e^{y}}{2}\biggr|$$
$$= \cosh y$$
$$\text{So }\: |\cos z|\leq \cosh  y$$
and since $\cosh 0=1$ we can find a neighborhood around the point $0$ such that $|\cos z|\leq 2$.

Since $\cosh x$ is a strictly increasing function I can't use it to prove that $|\cos z|\leq 1 $ for some neighborhood. Another method I can think of is taking partial derivative of $|\cos z|$ along $x$ and $y$-axes to check the direction in which it increases or decreases. But is there a simpler way for checking whether such a neighborhood exists ?

答:
For $r>0$, consider a disc $B(0,r)$. Then at some point on imaginary axis $(x=0)$, you have

$f(y)=\cos (iy)=\cos hy=\frac{e^y+e^-y}{2}$

Since $f(y)$ attains its minimum value $1$ at $y=0$ and is an increasing function $( f'(y)>0)$ for $y>0$ hence $|f(y)|\ge 1$ (even for $y<0$). Thus $|\cos z|\le 1$ for all $z\in B(0,r)$ is impossible regardless of the choice of $r$.

Title: Cosine of complex variable


问:
Probability of second ball being blue
There are $3$ boxes. One box has $2$ blue balls. One box has $2$ red. The last has one blue and one red.

You randomly pick a box and take $1$ ball out. You see that it's blue. What is the probability that the other ball in the box is blue.

The answer is $\frac{2}{3}$ and I confirmed this with Bayes rule. However, I am confused why the answer is not $0.5$. I know that if we aren't given the information that the first ball taken out is blue, then for sure, the second ball being blue has a probability of $0.5$.

However, without referring to Bayes rule, if I account for the fact that the first ball is blue, I can essentially eliminate the box with $2$ reds from my consideration. Because it is surely not possible that we chose that box. That leaves us with $2$ boxes. Then my thought process was that well, we could remove a blue ball from each box (because we just drew one and no longer need to consider it), so now the $2$ boxes each has $1$ ball. Blue in $1$ box and red in another, with each having the same probability of being the second ball, which would make the answer $0.5$, but apparently that's not correct. What is wrong with this thought process?

One thing that I am pondering is, if we know the first ball is blue, then we can remove the box with $2$ reds, but I think knowing this information might also make it so that there's a higher probability that we chose the box with $2$ blues, than the box with $1$ blue and $1$ red. In these $2$ boxes, there are $3$ blue balls total, but because $2$ of these belong in a single box, then that box should have a $2/3$ chance of being chosen conditioned on us knowing that we observed $1$ blue already. Is this right? If so, then I think that answers my previous question.

答:
This is a variant of the infamous Monty Hall Problem.

The problem you are having is identifying what events are equally likely.  If you have six balls in all, three of them blue, and you pull out a blue ball, then there are three possibilities, not two.  You could have taken the one ball from the mixed, or you could have taken the first blue ball from the box with two, OR you could have taken the SECOND blue ball from the box with two blue.  That's three possibilities, not two.

Title: Probability of second ball being blue


问:
How could the math for this combinatorics solution be simply explained?
I am faced with organizing a team of 5 speakers: $A,B,C,D,E$.

I must figure out how many permutations I can organize the five speakers, such that $Speaker\,A$ presents before $Speaker\,B$.

A valid order of speakers, which fulfills the above condition, is $A,B,C,D,E$.

Through methods used in other online forums, I stumbled upon the (possible) solution for this answer. From what I can tell, it uses the rule of sum and indirect method calculations to deduce how many permutations are possible ($n$) in these conditions:

$n= 4!+(4!-3!)+(4!-2(3!))+(4!-3(3!))$

$n=4(4!)-6(3!)$

$n=60$

I have tried to look online for potential matches to my technique, such as in falling/rising factorials, although the summations and algebra looks very different from what my solution provides.

Should my solution be correct, what is the mathematical technique being used here and how does it actually work to solve the above problem?

答:
The simplest is to note that there are $5!$ ways to order the speakers.  You can pair up each order where $A$ precedes $B$ with a matching one where $B$ precedes $A$, so the number where $A$ precedes $B$ is $\frac 12 \cdot 5!=60$

Another way is that there are ${5 \choose 2}=10$ ways to choose the slots for $A$ and $B$.  Then you must put $A$ in the earlier one.  There are $3!=6$ ways to put the other people in the $3$ remaining slots.


问:
Find a condition for the unilateral forward shift
Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward shift with $Te_n= t_n e_{n+1}$, where  $(t_n) \in \ell^{\infty}$

Let $\mathcal{K}(H)$ denote the set of compact operators acting on $H$.

Let $\pi:\mathcal{B(H)}\rightarrow\mathcal{B(H)}/\mathcal{K(H)}$ be the canonical quotient map,

Note that $\pi(T^*)= \pi(T)^*$.

Find a necessary condition on the sequence $(t_n)$ so that $\pi(T^*)\pi(T)=\pi(T)\pi(T^*)$.

I suspect that the condition will have something to do we the normality and compactness of $T$. But I'm not sure about this.

Any ideas or help will be appreciated!

答:
Since $\pi$ is multiplicative, the right point of view is to look at $\pi(T^*T)=\pi(TT^*)$. Even better, looking at $\pi(T^*T-TT^*)=0$, the condition required is that $T^*T-TT^*$ is compact.

In terms of the matrix units $\{E_{kj}\}$ coming from $\{e_n\}$,
$$
T=\sum_k t_kE_{k+1,k},\qquad\qquad T^*=\sum_k\overline{t_k}\,E_{k,k+1}.
$$
Then
$$
T^*T=\sum_k|t_k|^2\,E_{kk},\qquad\qquad TT^*=\sum_k|t_k|^2\,E_{k+1,k+1},
$$
and
$$
T^*T-TT^*=|t_1|^2\,E_{11}+\sum_{k>1}(|t_k|^2-|t_{k-1}|^2)\,E_{kk}.
$$
This is a diagonal operator, and it is compact if and only if $\lim|t_k|^2-|t_{k-1}|^2=0$.

Edit: Working on the basis elements

$$
\langle T^*Te_n,e_m\rangle=\langle Te_n,Te_m\rangle=t_n\overline{t_m}\langle e_{n+1},e_{m+1}\rangle=\langle |t_n|^2e_n,e_m\rangle,
$$
so $T^*Te_n=|t_n|^2e_n$. For $n,m>1$,
$$
\langle TT^*e_n,e_m\rangle=\langle T\overline{t_{n-1}}e_{n-1},e_m\rangle=\langle |t_{n-1}|^2e_n,e_m\rangle.
$$
So,
$$
(T^*T-TT^*)e_1=T^*Te_1=|t_1|^2e_1,
$$
and
$$\tag{$*$}
(T^*T-TT^*)e_n=(|t_n|^2-|t_{n-1}|^2)e_n,\qquad\qquad n>1.
$$


答:
The only difficulty here is justifying switching the order of integration and the summation in

$$\int_0^1 \frac{dx}{1+x^3} = \int_0^1 \sum_{n=0}^\infty (-x^3)^n\, dx = \sum_{n=0}^\infty \int_0^1 (-x^3)^n\, dx = \sum_{n=0}^\infty  \frac{(-1)^n}{3n+1}$$

We don't have uniform convergence of the series on the interval $[0,1]$ since the series diverges at $x=1$.

One way to justify is to note that

$$\sum_{n=0}^N(-x^3)^n = \frac{1 - (-x^3)^{N+1}}{1+x^3},$$

and

$$\int_0^1\sum_{n=0}^N(-x^3)^n\, dx  - \int_0^1 \frac{dx}{1+x^3} = (-1)^{N+2}\int_0^1 \frac{x^{3N+3}}{1+x^3}\, dx$$

Since we can switch the integral and a finite sum , we have

$$\int_0^1\sum_{n=0}^N(-x^3)^n\, dx =\sum_{n=0}^N \int_0^1(-x^3)^n\, dx = \sum_{n=0}^N \frac{(-1)^n}{3n+1},$$

and it follows that

$$\left|\sum_{n=0}^N \frac{(-1)^n}{3n+1}- \int_0^1 \frac{dx}{1+x^3}\right| = \int_0^1 \frac{x^{3N+3}}{1+x^3}\, dx \leqslant \int_0^1 x^{3N+3}\, dx \\= \frac{1}{3N+4} \underset{N \to \infty}\longrightarrow 0$$

Thus,

$$\lim_{N \to \infty}\sum_{n=0}^N \frac{(-1)^n}{3n+1} = \int_0^1 \frac{dx}{1+x^3}$$

Title: Series equal to an integral

Body:
If I have
\begin{eqnarray*}
S = \sum_{n\geq 0}{\frac{(-1)^{n}}{3n+1}}.
\end{eqnarray*}

Prove that
\begin{eqnarray*}
S = \int_{0}^{1}{\frac{1}{1+x^{3}}}\,\mathrm{d}x.
\end{eqnarray*}

I know that serie is convergent but I don't know how to prove this, I think in some theorem of uniformly convergence and integration where
\begin{eqnarray*}
f(x)=\sum_{n=1}^{\infty}{f_{n}(x)}
\end{eqnarray*}
but I don't know how to do this. Can you give me some advice to prove it? Thank you.


问:
Prove directly that a regular function on $\mathbb{P}^n$ is a constant map.
Exercise 3 of section 1.5.5 of Shavarevich's Basic Algebraic Geometry I asks us to prove directly that regular functions on $\mathbb{P}^n$ are constant. I understand how one proves this statement for irreducible projective varieties using the theorem that the image of a projective variety under a regular map is closed. However I am not sure how to prove it directly, which I assume means without using that theorem. Does anyone have any useful hints or tips?

答: For $n=1$, $\mathbb P^1$ can be obtained by gluing two open sets $U_1$ and $U_2$ which are both isomorphic to $\mathbb A^1.$ So a regular function on $\mathbb P^1$ would mean a regular function $f_1$ on $U_1$ and a regular function $f_2$ on $U_2$ whose restriction to $U_1 \cap U_2$ is identical. Let's write $f_1 = p(x)$ and $f_2= q(y).$ Then we need $f_1$ and $f_2$ to satisfy $p(T) = q(T^{-1})$ in the ring $k[T, T^{-1}],$ where $k$ is the base field. This implies that $p$ and $q$ should both have degree $0$ as polynomials and in fact should agree as constants. The general case follows a similar idea.


问:
Proving that the outer measure of an arbitrary union of intervals is equal to the infinite sum of the outer measures of the intervals.
I found in the solution to this question that

$$m^{*}\left( \bigcup_{n=1}^{\infty}\left( a_n+\frac{1}{2^{n}}\varepsilon, b_n-\frac{1}{2^{n}}\varepsilon \right) \right)= \sum_{n=1}^{\infty} m^{*}\left( \left( a_{n}+\frac{1}{2^{n}}\varepsilon,
 b_{n}-\frac{1}{2^n}\varepsilon\right) \right)$$
where $m^\ast$ denotes outer measure.

There is a proof here for the $n=2$ case. However, that proof cannot be used to justify the equality above, of course. This proof also cannot be extended to the infinite case using mathematical induction, as far as I understand. Could someone please prove a proof for the equality above?

答: Actually, you can use proof for a case with $n=2$. Indeed, for each $N\in \mathbb{N}$ it is wright that $m^*(\bigcup\limits_{n=1}^{+\infty} I_n(\varepsilon))\geq m^*(\bigcup\limits_{n=1}^{N} I_n(\varepsilon))$, where $I_n(\varepsilon)=(a_n+\frac{\varepsilon}{2^n},b_n-\frac{\varepsilon}{2^n})$, because $m^*$ is monotonic as an outer measure. With the induction you can show that $m^*(\bigcup\limits_{n=1}^{N} I_n(\varepsilon))=\sum\limits_{n=1}^N m^*(I_n(\varepsilon))$. So for each $N\in \mathbb{N}$ you have that $m^*(\bigcup\limits_{n=1}^{+\infty} I_n(\varepsilon))\geq \sum\limits_{n=1}^N m^*(I_n(\varepsilon))$. Then you can take a limit as $N$ goes to $+\infty$. You will get that $m^*(\bigcup\limits_{n=1}^{+\infty} I_n(\varepsilon))\geq \sum\limits_{n=1}^{+\infty} m^*(I_n(\varepsilon))$. But converse is true because of countable subadditivity of $m^*$. So the equality holds.


问:
Why is probability divided by time = annual rate? (Force of Mortality)
In force of mortality, one divides the probability of failure in a time interval $(x, x+\Delta x)$ by $\Delta x$. Apparently, this gives an annualised instantaneous rate of failure.

In this article, a post says "If we divide this by h (in our case, $\Delta x$), we get the annual rate at which the person is dying over that single day."

Paraphrasing the example provided there: If the probability of death for an individual over one day is $0.00004$, then the annual rate at which the individual dies is $\frac{0.00004}{1/365}=0.0146$ pa.

I don't quite understand the units for this rate. Is it "Number of deaths / year"? Because I'm not sure how probability divided by time leads to such a rate. Or is it a different type of rate?

答:
The dimensions are probability per time.

The units are $year^{-1}$, but could have been expressed in other units of time.

It is not number per year, but probability per year.  If you multiply by the size of the population, you get number per year.


问:
Find a holomorphic function which is also injection
Find a holomorphic function which is also injection such that $$\{z\in \mathbb C: \text{Re}z>0, \text{Im}z>0\} \mapsto \{z\in \mathbb C: |z|<1, \text{Im}z>0\}.$$

I think that I need two transformations: $$\{z\in \mathbb C: \text{Re}z>0, \text{Im}z>0\} \mapsto \{z\in \mathbb C: |z|<1, \text{Re}z>0, \text{Im}z>0\} \mapsto\{z\in \mathbb C: |z|<1, \text{Im}z>0\},$$ but I don't know how to designate them.

答:
If $f(z) = \frac{z+1}{z-1}$ then by computation $f^{-1} = f$ and you can see that it sends the upper unit semi-disc to the third quadrant of $\mathbb{C}$ by checking on the boundary: $f(-1)=0, f(1)=\infty, f(0) =-1, f(i) = -i$.  You can prove that it's injective by a direct computation.

So if $g(z) = -z$ then $g$ maps the first quadrant of the plane to the third, and thus $f^{-1} \circ g = f \circ g = f(-z) = \frac{z-1}{z+1}$ is a conformal automorphism from your first domain to the other.


问:
complete the squares to find expression of quadratic forms
I don't understand what is the intuition/ strategy to obtain $q(u) = \frac{1}{4}(u_1 + u_2 + 2u_3)^2 - \frac{1}{4}(u_1-u_2)^2-u_3^2$ from $q(u) = u_1u_2 + u_1u_3 + u_2u_3.$

I understand the result of course but I don't see how what motivated the choice of grouping $u_1$, $u_2$ and $u_3$ together, and why we have the factor $2$ in front of $u_3$, ...

Actually when we have squares with only two element I understand it very well but when it comes to 3 elements inside the square I become lost.

Thank you for your help !

答:
The idea here is to rewrite $q(u) = u_1u_2 + u_1u_3 + u_2u_3$ as $$q(v) = v_1^2 \pm v_2^2 \pm v_3^3$$
where $v$ is some linear transformation of $q$. It should be obvious that there are multiple solutions. In particular, from the solution you've given, two more are immediately apparent, since $q$ is symmetric in $u_1, u_2, u_3$:
$$q(u) = \frac14(2u_1 + u_2 + u_3)^2 - \frac14(u_2-u_3)^2-u_1^2\\q(u)= \frac14(u_1 + 2u_2 + u_3)^2 - \frac14(u_1-u_3)^2-u_2^2$$

So we are not looking for a single solution, which gives us the freedom to make simplifying choices, which is exactly what someone did here.

Start off by letting $v_1 = a_1u_1 +a_2u_2 + a_3u_3$. Then $$v_1^2 = \left[a_1^2u_1^2 + a_2^2u_2^2 + a_3^2u_3^2\right] + (2a_1a_2)u_1u_2 + (2a_1a_3)u_1u_3 + (2a_2a_3)u_2u_3$$
That gives us all the mixed products, but we also pick up the squared terms that we don't want. So we need to subtract them back out with $v_2, v_3$.

We cannot get rid of all three squares with $v_2$ alone. If we set $v_2 = b_1u_1 +b_2u_2 + b_3u_3$, then $$v_1^2 - v_2^2 = (a_1^2 - b_1^2)u_1^2 + (a_2^2 - b_2^2)u_2^2 + (a_3^2 - b_3^2)u_3^2 + \text{ mixed terms}$$
If this were to only consist of mixed terms, we would need $b_i = \pm a_i$ for each $i$. But if we choose the same sign for all three, $v_1^2 = v_2^2$ and we just get $0$. We can assume wlog that $b_1 = a_1$. If $b_2 = a_2$ as well, then the $u_1u_2$ terms in $v_1^2$ and $v_2^2$ will also be the same and $v_1^2 - v_2^2$ would have no $u_1u_2$ term. So instead we take $b_2 = -a_2$. But then $b_3 = a_3$ would give the same problem, leaving no $u_1u_3$ term. And the other choice of $b_3 = -a_3$ leaves $v_1^2 - v_2^2$ with no $u_2u_3$ term. Thus we cannot rewrite $q$ as $q(v) = v_1^2 - v_2^2$ alone. $v_3$ is needed.

So if we cannot get rid of all three squared terms with just $v_2$, can we get rid of two of them? That can be done. Since we can't get rid of $u_3^2$, just ignore it for now and let $b_3 = 0$. Getting rid of $u_1^2$ and $u_2^2$ without also getting rid of $u_1u_2$ requires $b_1 = a_1$ and $b_2 = -a_2$ (or vice versa), so we set
$$v_2 = a_1u_1 - a_2u_2$$
which leaves
$$v_1^2 - v_2^2 = a_3^2u_3^2 + (4a_1a_2)u_1u_2 + (2a_1a_3)u_1u_3 + (2a_2a_3)u_2u_3$$
Now if we include $u_1$ or $u_2$ in $v_3$, it would bring back $u_1^2$ and $u_2^2$ terms, which we don't want. So $v_3 = cu_3$ for some $c$. And obviously $c = \pm a_3$, if $v_1^2 - v_2^2 - v_3^2$ is to have no $u_3^2$ term. It doesn't matter which sign we pick, so we might as well just use $c=a_3$.

So now $$v_1^2 - v_2^2 - v_3^2 = (4a_1a_2)u_1u_2 + (2a_1a_3)u_1u_3 + (2a_2a_3)u_2u_3$$
which means $$4a_1a_2 = 1\\2a_1a_3 = 1\\2a_2a_3 = 1$$
Thus $$a_1 = a_2 = \frac 1{2a_3}$$
and $$4\frac 1{2a_3}\frac 1{2a_3} = 1\\a_3^2 = 1\\a_3 = \pm 1$$
Arbitrarily choose the positive value, and we get $$a_1 = \frac 12, a_2 = \frac 12, a_3 = 1$$
which is exactly the solution you gave.

We arrived at it by making some arbitrary choices, two of which are trivial, while the other two locked us into this solution.

$b_1$ is the same sign as $a_1$ (though changing this choice does not truly give a different solution - it only flips the sign of $v_2$, which doesn't matter because we are only concerned with its square).

$b_3 = 0$.

$v_1^2 - v_2^2$ should have no $u_1^2$ or $u_2^2$ terms.

$a_3 > 0$ (changing this choice flips the sign of all three of $v_1, v_2, v_3$, which does not matter).


问:
$\operatorname{Ext}^2(A, \mathbb{Z})$ for $A$ abelian
Why for all abelian groups $A$ the second ext-group $\operatorname{Ext}^2(A, \mathbb{Z})$ vanishes? Saw this 'fact' several times without any proof attached to.

答:
Every abelian group $A$ has a free resolution of the form $0 \to \mathbb{Z}^{(n)} \to \mathbb{Z}^{(m)} \to A \to 0$ (i.e. a presentation). Here, $\mathbb{Z}^{(n)}$ means "the" free abelian group on a set of $n$ generators ($n$ may be an infinite cardinal). This comes from the fact that any subgroup of a free abelian group is free abelian: here is a proof from Wikipedia.

Then we compute $\operatorname{Ext}^i(A,\mathbb{Z})$ by applying $\operatorname{Hom}({-},\mathbb{Z})$ to the deleted resolution $0 \to \mathbb{Z}^{(n)} \to \mathbb{Z}^{(m)} \to 0$ and taking the $i$th homology group. There are only two nonzero terms in this deleted resolution, so $\operatorname{Ext}^i(A,\mathbb{Z})$ is automatically $0$ for $i \geq 2$!


问:
How do I find $\operatorname{det} T_Q$?

Let $S$ be the space of all $n \times n$ real skew symmetric matrices and let $Q$ be a real orthogonal matrix. Consider the map $T_Q: S \to S$ defined by $$T_Q(X) = QXQ^T.$$ Find $\operatorname{det} T_Q$.

I thought about diagonalizing $Q$, but I don't think we know it is real diagonalizable. I can show it is an isometry using the Hilbert-Schmidt inner product, but I can't really relate it to the determinant of $Q$ (I've seen posts here that say the determinant should be $\operatorname{det}Q^{n-1}$). So all I know is that $\operatorname{det} T_Q = \pm1$. How would I find $\det T_Q$?

答:
Partial Answer: I will focus on the case that $n$ is even. The odd case can be handled similarly.

As a consequence of the block-diagonalizability of skew-symmetric matrices and the fact that $Q = \exp(P)$ holds for some skew-symmetric matrix $P$, we can show that there exists an orthogonal matrix $W$ such that
$$
WQW^T = D:= \pmatrix{A_1\\ & \ddots \\ && A_k}, \quad A_k = \pmatrix{a_k & -b_k\\ b_k & a_k}.
$$
Note that $T_Q = T_W \circ T_D \circ T_W^{-1}$, so that $\det(T_Q) = \det(T_D)$. Decompose $D$ into a product $D = D_1 \cdots D_k$ where
$$
D_1 = \pmatrix{A_1 \\ & I \\ & & \ddots \\ &&& I}, \dots, \quad D_k = \pmatrix{I \\ & \ddots \\ & & I \\ &&& A_k}.
$$
Note that $T_D = T_{D_1} \circ \cdots \circ T_{D_k}$. With that, it suffices to determine $\det T_{D_k}$.

Another approach is to use the fact that there exists a skew-symmetric matrix $M$ for which $Q = \exp(M)$ and that $T_Q = \exp(C_M)$, where
$$
C_M(X) = MX - XM.
$$
From there, we can use the fact that
$$
\det(T_Q) = \det(\exp(C_M)) = \exp(\operatorname{tr}(C_M)).
$$
Once we show that $\operatorname{tr}(C_M) = (n-1)\operatorname{tr}(M)$, it follows that
$$
\begin{align}
\det(T_Q) &= \exp(\operatorname{tr}(C_M)) = \exp((n-1)\operatorname{tr}(M)) = \exp(\operatorname{tr}(M))^{n-1}
\\ & = \det(\exp(M))^{n-1} = \det(Q)^{n-1}.
\end{align}
$$

Title: How do I find $\operatorname{det} T_Q$?


问:
Monomial into power sum
This might be a routine problem but I'm unable to express monomial symmetric functions $m_\lambda$ into power sum symmetric functions $p_\mu$'s. Could you please help me with it?

答: There is a formula, but it is not pretty. See Mike Zabrocki's symmetric function notes, page 38. The definition of $z_\lambda$ is elsewhere, I believe it is $z_\lambda=1^{m_i}m_i!\cdot 2^{m_2}m_2!\cdots$, where $m_i$ is the multiplicity of $i$. They reference a chapter not in that document for an idea of the proof, but it is included here, page 43=44.

Title: Monomial into power sum


问:
How to show $t{B^{2}_t}+t^2$ is $\mathcal{F}_t$-addapted process?
How to show $t{B^{2}_t}+t^2$ is $\mathcal{F}_t$-adapted process? Here $B_t$ is Brownian Process.

Please Help

答: As $B_t$ is $\mathcal F_t$-measurable, $B_t^2$ is also (note that continuous functions $\mathbb R \to \mathbb R$ are measurable), so is $tB_t^2 + t^2$. </p>

ContentLicense: CC BY-SA 3.0


问:
how to make a venn diagram of the following
I need to do a venn diagram of $$X^\complement \cap  (Y\cup Z) $$
I know that I start with the union of $Y$ and $Z$, having all of circles $Y$ and $Z$ colored, then I take the intersection with $X$ which would only be the part of the circle that touch $X$ with the circles $X$ and $Y$ and to take the complement (part where I am confused) would it be the rest of circle $X$ that has previously not drawn and the rest of the universe without counting the previous work of $Y$ and $Z$?

答: Take the complement before the intersection.  It should be the parts of $Y$ and $Z$ that are not in $X$.


问:
Help with polynomials give zeros
My teacher hasn’t covered this type of material yet and I’m really struggling.
All I know is that it’s a third degree polynomial and the roots are $x=2$ and $x=-3i$. Also given is that the polynomial equals $60$ when $x=-1$.

What do I do whit this information and how can I construct the expanded polynomial with just this?

答:
I accept the coefficients of your polynomial are all real.

If the all coefficients are real, then you have

$$x_1=2, x_2=-3i, x_3=3i$$

Your polynomial is equivalent to

$$P(x)=A(x-2)(x-3i)(x+3i)=A(x-2)(x^2+9)$$

$$P(-1)=-30A=60\Longrightarrow A=-2$$

So, your polynomial equals to

$$P(x)=-2(x-2)(x^2+9)$$

Title: Help with polynomials give zeros


答:
Well, you can make use of this fact:
\begin{gather*}
\text{For integers}\; a,b\ \text{and}\ c.\\
\text{if}\ ax^{2} +bx+c=0\ \text{has}\ x=di\ \text{as a root},\ \\
\text{then the other root of the equation must be}\ x=-di.
\end{gather*}
Utilizing this fact, with $d=-3$ from your question, the set of quadratic equations which have these roots are:
\begin{equation*}
c( x-3i)( x+3i) =c\left( x^{2} +9\right)
\end{equation*}
where $c$ is any rational number except $0$.
Similarly, the cubic equation that has roots $-2, 3i$ and $-3i$ is
\begin{equation*}
f( x) =c( x-2)( x-3i)( x+3i) =c( x-2)\left( x^{2} +9\right)
\end{equation*}
From your question $f(-1)= 60$, so
\begin{gather*}
c( -1-2)\left(( -1)^{2} +9\right) =60\\
c( -3)( 10) =60\\
c=\frac{60}{-30} =-2
\end{gather*}
Can you now complete the answer?

Title: Help with polynomials give zeros

Body:
My teacher hasn’t covered this type of material yet and I’m really struggling.
All I know is that it’s a third degree polynomial and the roots are $x=2$ and $x=-3i$. Also given is that the polynomial equals $60$ when $x=-1$.

What do I do whit this information and how can I construct the expanded polynomial with just this?


问:
How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$
$\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.</p>

Body:
Idea: You can unwrap like this:

$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$

if

$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<2^2-1$$

if

$$\sqrt{3+\cdots+\sqrt{2006}}<(2^2-1)^2-2$$

and so on, so we want to show

$$2006 < (((2^2-1)^2-2)^2-\cdots)^2-2005$$

might as well prove it by induction for all $n$ rather than just 2006, so we need to show that

$$n+1 < (((2^2-1)^2-2)^2-\cdots)^2-n$$

implies

$$n+2 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2-(n+1)$$

but thats just

$$2n+3 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2$$

which holds since

$$2n+3 < (n+1)^2$$

for all $n>1$.


问:
Proving Expected Value Formula for Discrete Random Variable Involving Infinite Series
I am currently trying to prove the following formula. Let Y be a discrete random variable that assigns positive probabilities to only positive integers. Show that:

$$
E(Y) = \sum_{k=1}^{\infty }P(Y \geq k)
$$
where k is a positive integer.

My initial foray into this involved abstracting it into two series, with the inner series having y go from k to infinity, and the outer series going from k = 1 to infinity, as so:
$$
\sum_{k=1}^{\infty}\sum_{y=k}^{\infty}P(Y=y)
$$

I am unsure what to do next. I had discussed this with a professor but his explanation remained confusing to me. I am still confused as to
what the integer k represents. Any help and insight would be highly appreciated!

答:
As a next step, you can swap the summations. Your double summation is over pairs $(k,y)$ with $y \ge k$, which we can also obtain as
$$
\sum_{y=1}^\infty \sum_{k=1}^y \mathbb{P}(Y=y).
$$
Note that the summand does not depend on $k$, so we simply get
$$
\sum_{y=1}^\infty y \cdot \mathbb{P}(Y=y),
$$
which, because $Y$ only takes positive integer values is by definition equal to $\mathbb{E}[Y]$.


问:
Erroneous reference to EGA I 6.9.17 and Deligne's formula
I'm attempting to track down the original EGA proof of Deligne's formula that for a noetherian ring $A$, ideal $\mathfrak{a}\subset A$, and $A$-module $M$ we have $$\varinjlim \operatorname{Hom}_A(\mathfrak{a}^n,M) \cong \widetilde{M}\left(\operatorname{Spec} A \setminus V(\mathfrak{a})\right)$$ which I've heard cited as EGA I prop 6.9.17. Unfortunately, the version of EGA hosted by Numdam here (warning, large file) does not even contain a section 6.9. What's going on?

答:
I think they are citing the 1971 edition of EGA I published by Springer.

Grothendieck, Alexandre; Dieudonné, Jean (1971). Éléments de géométrie algébrique: I. Le langage des schémas. Grundlehren der Mathematischen Wissenschaften (in French). 166 (2nd ed.). Berlin; New York: Springer-Verlag. ISBN 978-3-540-05113-8.

In this edition, Proposition 6.9.17 is the following (my translation).

Proposition (6.9.17) Let $X$ be a noetherian scheme, $\newcommand{\I}{\mathscr{I}} \I$ a quasicoherent $\newcommand{\O}{\mathscr{O}} \O_X$-module, $Y$ the closed subscheme of $X$ defined by $\I$, and $U$ the open set $X \setminus Y$. Let $\newcommand{\F}{\mathscr{F}} \F$ be a coherent $\O_U$-module, $\newcommand{\G}{\mathscr{G}} \G$ a quasicoherent $\O_X$-module, $\F'$ a coherent $\O_X$-module such that $\F'|_U = \F$ (6.9.8). Then we have a canonical bijection
\begin{equation}
\DeclareMathOperator{\Hom}{Hom}
\varinjlim_n \Hom_{\O_X}(\I^n \F', \G) \overset{\sim}{\to} \Hom_{\O_U}(\F, \G|_U) \, . \tag{6.9.17.1}
\end{equation}


问:
Hermitian Product Orthogonality
In Lang’s Linear Algebra, after defining the Hermitian product for vector spaces over complex fields, he mentioned that all previously mentioned notions involving orthogonality with scalar products (including theorems) held true for the Hermitian product. I’m not sure if this theorem is though:

For all $v,w\in V$, where $V$ is a vector space over a complex field and $||a||$ is the norm with respect to a Hermitian product $(s,t)$:
$$||v-w||^2=||v+w||^2 \Leftrightarrow (v,w) =0$$.

I tried to prove this and got to the following:
$$||v-w||^2=||v+w||^2 $$$$\Leftrightarrow (v-w,v-w)=(v+w,v+w)$$$$ \Leftrightarrow (v-w,v)-(v-w,w)=(v+w,v)+(v+w,w)$$$$\Leftrightarrow \overline{(v, v-w) - (w, v-w)} = \overline{(v, v+w)+(w,v+w)}$$$$\Leftrightarrow (v,v) - (v,w) - (w,v) + (w,w) = (v,v) + (v,w) + (w,v) + (w,w)$$$$\Leftrightarrow (v,w) + (w,v)=0 $$$$ \Leftrightarrow (v,w) + \overline{(v,w)} = 0.$$

My trouble is with the fact that for all complex numbers $z$, $z+\overline{z}=0$ does not imply that $z$ must be $0$ - take $z=i$ for instance. Is this just not what Lang meant by “all other properties work the same?” I mean this property wouldn’t really affect the proofs of the others anyways, but I’m still curious.

Of course, any input would be appreciated.

答:
Indeed, this statement is not true. Take $v=1,w=i$ in your above notation as you suggest:
$$
\lVert 1-i\rVert^2=2=\lVert1+i\rVert^2
$$
but $\langle 1,i\rangle=i\ne 0.$ What Lang means really is that the notion of orthogonality we get using this definition of Hermitian inner product allows us to (re)prove the standard results that one proves for inner products over $\Bbb{R}$. For instance, given $W\subseteq V$, $\dim W+\dim W^\perp=\dim V$. We also get a pairing of $V$ with $V^*$ but you should be careful as this is in general only $\Bbb{R}-$linear.

Title: Hermitian Product Orthogonality


问:
Gluing locally defined smooth maps to get a globally defined smooth map
Please find the discussion underlined with red, which is to be clarified. $\tilde{f_p}$ is a smooth vector-valued function defined on an open subset $W_p$ of a smooth manifold $M$, and $\psi_p$ is a smooth real-valued function defined on the whole $M$ with $\mathrm{supp}(\psi_p)\subseteq W_p$. My question is how to extend the scalar product $\psi_p\tilde{f_p}$ defined on $W_p$ to a globally defined smooth function. I don't understand what John M. Lee, the author, means when he says

if we interpret it to be zero on $M\setminus\mathrm{supp}(\psi_p)$

What exactly is "it"? I have no idea what's happening to $\tilde{f_p}$ around $\partial W_p$. Is there any easy way to understand the trick Lee is using? Thank you.

答:
For a real-valued function $f: U \to \mathbb{R}$ and a vector-valued function $g : V \to \mathbb{R}^k$, their product is the vector-valued function $fg : U\cap V \to \mathbb{R}^k$ given by $(fg)(x) = f(x)g(x)$; note, we had to restrict the domain of $fg$ to be the intersection of the domains of $f$ and $g$ for this to make sense.

In this case we have $\psi_p : M \to \mathbb{R}$ and $\tilde{f_p} : W_p \to \mathbb{R}^k$, so the product is $\psi_p\tilde{f_p} : W_p \to \mathbb{R}^k$; note, the domain is $M\cap W_p = W_p$, not $M$. What Lee is saying is that we can extend this to a smooth function on $M$ by defining $(\psi_p\tilde{f_p})(x) = 0$ if $x \in M\setminus W_p$.


问:
Error in book's definition of open sets in terms of neighborhoods?
The following is copied verbatim from a book (I. Protasov, Combinatorics of numbers, p. 14):

Suppose that to each point $x$ of a set $X$ a collection $\mathcal{B}(x)$ of subsets of $X$, which are called neighborhoods of $x$, is assigned so that the following conditions are satisfied:

(B1) $x\in U$ for every neighborhood $U \in \mathcal{B}(x)$;
(B2) if $U \subseteq V, U \in \mathcal{B}(x)$, then $V\in \mathcal{B}(x)$;
(B3) if $U_1, \dots, U_n \in \mathcal{B}(x)$, then $U_1 \cap \dots \cap U_n\in \mathcal{B}(x)$;
(B4) if $U\in \mathcal{B}(x)$, then there is a neighborhood $V\in\mathcal{B}(x)$ such that $U\in \mathcal{B}(y)$ for every $y\in V$.

A subset $A\subseteq X$ is defined to be open, if $A$ is a neighborhood of each of its points, i.e. $A\in\mathcal{B}(x)$ for every $x\in A$.  Evidently, open sets satisfy the following properties:

(O1) $X, \varnothing$ are open sets;
(O2) if $U_1, \dots, U_n$ are open sets, then $U_1 \cap \dots \cap U_n$ is an open set;
(O3) if $U_\alpha, \alpha\in J$, is a collection of open sets, then $\bigcup\{U_\alpha : \alpha \in J\}$ is an open set.

I don't see why (O1) is true.  More generally, I don't see what guarantees that $\mathcal{B}(x) \neq \varnothing$ for any $x \in X$.  In fact, if we set $\mathcal{B}(x) = \varnothing$ for all $x \in X$, conditions (B1)-(B4) are all satisfied vacuously.  In this case there could be no open sets, so (O1) could not hold.

Am I missing something?

If not, there must be some error in the book, and I'd like to know how to fix it such that what results agrees with the standard way of defining open sets in terms of neighborhoods.

答:
Your observation appears to be correct, and you don't appear to be missing anything.

Note that simply adding the condition that $\mathcal{B}(x) \neq \emptyset$ for all $x \in X$ (or, equivalently, $X \in \mathcal{B}(x)$ for all $x \in X$) will yield the usual definition of a topology.  (The sets in $\mathcal{B}(x)$ will be those sets whose interior contains $x$.)


问:
Metric space union proof
Suppose $(A, d)$ is a metric space and let $D, E \subseteq A$ be subsets. Show that $\partial(D \cup E) \subseteq \partial(D) \cup \partial(E)$. Then, if $\overline{D} \cap \overline{E} = \emptyset$, show that $\partial(D \cup E) = \partial(D) \cup \partial(E)$.

Note: For a set $B$, $\partial B$ refers to the frontier of that set; that is, $\partial B = \overline{B} \cap \overline{B^c}$ where $B^c$ is the complement of $B$.

My attempt:

Suppose $x \in \partial(D \cup E)$. Then for all $\epsilon > 0$ we have $B_0(x, \epsilon) \cap (D \cup E) \neq \emptyset \land B_0(x, \epsilon) \cap (D \cup E)^c \neq \emptyset$. Since $B_0(x, \epsilon) \cap (D \cup E)^c \neq \emptyset$, the sets $B_0(x, \epsilon)$, $D^c$ and $E^c$ have a non-empty intersection. So $x \notin (D \cup E)^c$ and then $x \in D \land x \in E$. Hence at least one of the intersections $B_0(x, \epsilon) \cap D$ or $B_0(x, \epsilon) \cap E$ is not empty. So we get that for all $\epsilon > 0$

$$(B_0(x, \epsilon) \cap D \neq \emptyset \land B_0(x, \epsilon) \cap D^c \neq \emptyset) \lor (B_0(x, \epsilon) \cap E \neq \emptyset \land B_0(x, \epsilon) \cap E^c \neq \emptyset)$$

We may conclude that $x \in \partial(D) \cup \partial(E)$.

QED.

Is this proof correct? I am not sure how $\partial(D) \cup \partial(E) = \partial(D \cup E)$ follows if $\overline{D} \cap \overline{E} = \emptyset$. Any assistance is appreciated.

答:
Suppose $\bar D \cap \bar E =\emptyset. $ It suffices to consider $ x \in \partial D$ and show it is in the boundary of $D \cup E$. Clearly $x$ is in the closure of that union, so we have to show it is also in the closure of $(D \cup E)^c=D^c \cap E^c$. Every ball $B(x,r)$ centered at $x$ intersects $D^c$. Moreover, since $x \in \bar D$ we infer that $x \notin \bar E$. Thus if $r$ is small enough, then $B(x,r)$ is disjoint from $E$ so
$B(x,r) \cap E^c \cap  D^c = B(x,r)   \cap  D^c \neq \emptyset. $

Title: Metric space union proof


问:
Question on Homotopy and Homology Groups of Spaces $X$ and $X \cup_{S^{n-1}} D^n$
I have a general question about the differnce of homology and (simplicial) homotopy groups
of a topological space in degrees $X$ and the space $Y := X \cup_{S^{n-1}} D^n$ obtained by glueing
$D^n$-ball to $X$ along continuous boundary map $f: S^{n-1} \to X$ and obtain in that way the new space $Y:= X \cup_{S^{n-1}} D^n$.

What can we say about the change of homotopy and homology groups of degree $0 \le k <n$?
I conjecture that for these $k$ the answer should be that homotopy and homology doesn't
change, ie $\pi_k(X)= \pi_k(Y)$ and $H^k(X, \mathbb{Z})= H^k(Y, \mathbb{Z})$ but
I would like to know I a 'geometrical' argument for it. Now I would like to present my efforts and I would be thankful if somebody could give feedback if my arguments work:

What I tried on $\pi_k(Y)$ : Let $[s] \in [S^k, Y]$ and we have to show that $[s]$ is homotopy
equivalent to a map $S^k \to X$. Choose a representant $s: S^k \to Y=  X \cap_{S^{n-1}} D^n$
and compose it with collapse map $c:X \cup_{S^{n-1}} D^n \to S^n$ which collapses $X$ to a
point. By Sard's theorem the composition of $s$ with $c$ cannot be surjective, therefore
there exist a point $p \in S^n$ with hasn't a preimage and $c \circ s$ factorizes over
$S^n \backslash p \simeq \mathbb{R}^n$.

So we conclude that $c \circ s$ is hotopic to m map which factorizes over contractable space
$\mathbb{R}^n$, and therefore $c \circ s$ should be homotopic to constant map, right?

About homology: We can always assume that $X$ is a CW-complex and let
$C^X_m \to C^X_{m-1} \to ... \to C^X_k \to ... \to C^X_1$ it's cell complex
with differentials $d_i^X: C^X_i \to C^X_{i-1}$ described in every book on algebraic topology and where
$C^X _i = \bigoplus_{c_i \text{ i-cell }} \mathbb{Z} \cdot c_i $. The resolution
of $Y$ differs from that one of $X$ only by an additional $n$-cell, ie
$C^Y_i = C^X_i$ for $i \neq n$ and $C^Y_n = C^X_n \oplus \mathbb{Z} \cdot c_n$. Making this
identification between $i$-cells and using the canonical
inclusion map $C^X_n \subset C^Y_n$ we obtain a map between chain complexes of $X$ and $Y$.
The $i$-th homology is by definition $H^i()=ker(d_i)/im(d_{i+1})$. For all $i < n -1$ we have equal
differentials $d^X_i=d^Y_i$ by construction of our cell complexes $X$ and $Y$, the $d^X_n $
and $d^Y_n$.

So I think that from this we can always conclude $H^i(X) = H^i(Y)$ for $ i < n-2$, right?
Up to now, are my reasonings correct?
If yes, then similary, $H^i(X) = H^i(Y)$ for $i >n$, as $C^X_n \subset C^Y_n$ is injective.
About $i = n, n-1$ in general we cannot say anything.

Are my 'proofs' 1 and 2 ok?

答:
In general, this will be heavily dependent on your choice of map $f$, as well as the dimension of $D^n$. For example, say $X$ is a circle, so $\pi_1(X)\simeq\mathbb{Z}$, and all higher homotopy groups vanish.

Now, let $Y$ be obtained from $X$ by attaching a $2$-cell by a map of degree $k$. Van Kampen's theorem asserts that the fundamental group of $Y$ is now cyclic of order $k$, and now the higher homotopy groups need not vanish (e.g. $k=2$ yields $\mathbb{R}P^2$). However, if instead we attach the $2$-cell via a constant map, then $Y$ is now $S^1\wedge S^2$, which also has different homotopy groups than $X$.

You can find more examples in Chapter 4 of Hatcher's Algebraic Topology, available for free in his own website.

P.D.: although I didn't mention homology in my answer, you can also see that (in the second example) letting $f$ be a constant map adds a $2$-cycle that wasn't there before, so the homologies also differ.


问:
Prove that the sequence $(x_n)$ is monotone increasing, for $x_1=1, x_{n+1}=(2+x_n)^{1/2}$
I proved that the sequence is bounded and that its limit is 2, but how do I prove by induction that it is monotone increasing? i.e. that $x_{n+1} \ge x_n$ for every $n \in \mathbb{N}$

答: Hint: The function $f(x) = \sqrt{x+2}$ has derivative $f’(x) > 0$.


Title: Prove that the sequence $(x_n)$ is monotone increasing, for $x_1=1, x_{n+1}=(2+x_n)^{1/2}$

Body: I proved that the sequence is bounded and that its limit is 2, but how do I prove by induction that it is monotone increasing? i.e. that $x_{n+1} \ge x_n$ for every $n \in \mathbb{N}$

Body:
As you said$,\quad x_{n}< 2$
$$x_{n+ 1}^{2}- x_{n}^{2}= 2+ x_{n}- x_{n}^{2}> 2+ x_{n}- 2x_{n}> 0$$


答:
The statement is indeed not true.  The true statement is

If $\lVert v+\varepsilon w\rVert$ is a constant (independent of $\varepsilon$) for all $\lvert\varepsilon\rvert=1$ then $(v,w)=0$.

Over $\mathbb{R}$ there is only $\varepsilon=\pm 1$.  Over $\mathbb{C}$ there is a whole circle of them, but it turns out you only have to check 4 of them $\pm 1,\pm i$.  This is because of the polarisation identity
$$
(v,w)=\frac1{2\pi i}\int_{\mathbb{T}}\lVert v+\zeta w\rVert^2\,\mathrm{d}\zeta=\frac1{2\pi}\int_{0}^{2\pi}e^{i\theta}\lVert v+e^{i\theta}w\rVert^2\,\mathrm{d}\theta=\frac14\sum_{n=1}^4 i^n\lVert v+i^nw\rVert^2.
$$

Title: Hermitian Product Orthogonality

Body:
In Lang’s Linear Algebra, after defining the Hermitian product for vector spaces over complex fields, he mentioned that all previously mentioned notions involving orthogonality with scalar products (including theorems) held true for the Hermitian product. I’m not sure if this theorem is though:

For all $v,w\in V$, where $V$ is a vector space over a complex field and $||a||$ is the norm with respect to a Hermitian product $(s,t)$:
$$||v-w||^2=||v+w||^2 \Leftrightarrow (v,w) =0$$.

I tried to prove this and got to the following:
$$||v-w||^2=||v+w||^2 $$$$\Leftrightarrow (v-w,v-w)=(v+w,v+w)$$$$ \Leftrightarrow (v-w,v)-(v-w,w)=(v+w,v)+(v+w,w)$$$$\Leftrightarrow \overline{(v, v-w) - (w, v-w)} = \overline{(v, v+w)+(w,v+w)}$$$$\Leftrightarrow (v,v) - (v,w) - (w,v) + (w,w) = (v,v) + (v,w) + (w,v) + (w,w)$$$$\Leftrightarrow (v,w) + (w,v)=0 $$$$ \Leftrightarrow (v,w) + \overline{(v,w)} = 0.$$

My trouble is with the fact that for all complex numbers $z$, $z+\overline{z}=0$ does not imply that $z$ must be $0$ - take $z=i$ for instance. Is this just not what Lang meant by “all other properties work the same?” I mean this property wouldn’t really affect the proofs of the others anyways, but I’m still curious.

Of course, any input would be appreciated.


问:
What is the least amount of straight cuts that should be made to get squares from a rectangle?
The problem is as follows:

Rachel has a rectangular piece of cloth which measures $2$ meters
large and $0.2$ meters wide. She is to use a guillotin which can only
makes cuts of $60$ cm of maximum in length. Another constrain in this
guillotin is that it can only cut one layer of this cloth. Using this
information, how many straight cuts minimum can Rachel make in order
to get from such fabric $10$ square pieces of $20\,cm$ from each
side?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{3 cuts}\\
2.&\textrm{1 cut}\\
3.&\textrm{2 cuts}\\
4.&\textrm{4 cuts}\\
\end{array}$

I'm not sure how to solve this question regarding cuts. But I believe that a way to reduce the number of cuts will require to fold the piece. But in this case this seems not to be possible because there is a condition in the problem which is forbidding to do such approach.

My initial approach was to make the necessary cuts to make some sort of grid which could I guess reduce the number of cuts but 10 is not a perfect square. Therefore what sort of logic should be here?.

I've went in circles for long on this question so I will really appreciate someone could give me a hand with this.

Therefore what should be the way to go here?. Can someone help me here please?. I'm confused.

答:
To cut the strip of cloth, it only takes 9 cuts of 20 cm, and each cut of the guillotin cuts through 60 cm at most, so at least 3 cuts are necessary.

Number the cuts $1$ to $9$ sequentially, and the resultant squares $0$ to $9$.

Align cut $1$ under the guillotin, then fold two right angled triangles on the strip around cuts $2$ and $3$ so that the strip goes back under the guillotin, and align cut $4$ under the guillotin next to cut $1$.

(Fold twice like (B) in this random image found in Google:

)

Do the same thing to align cut $7$ under the guillotin. With cut $1, 4, 7$ all aligned side by side, one guillotin cut of 60 cm would do all the 3 20-cm cuts.

Now squares 1-3, squares 4-6, squares 7-9 are on separated pieces of cloth, and they can be aligned freely to make the remaining 6 cloth-cuts in 2 guillotin cuts.

The optimal answer is 3 guillotin cuts.


问:
Product of two functions converging in $L^1(X,\mu)$

Let $f_n\to f$ in $L^1(X,\mu)$, $\mu(X)<\infty$, and let $\{g_n\}$ be a sequence of measurable functions such that $|g_n|\le M<\infty\ \forall n$ with some constant $M$, and $g_n\to g$ almost everywhere. Prove that $g_nf_n\to gf$ in $L^1(X,\mu)$.

This is a question from one of my past papers, but unfortunately there is no solution provided.

Here is as far as I have gotten:

$$\int|gf-g_nf_n|=\int|(f-f_n)g_n+(g-g_n)f|\le\int|f-f_n||g_n|+\int|g-g_n||f|$$
$$\le M\int|f-f_n|+\int|g-g_n||f|$$

We know that $f_n\to f$ in $L^1$, so $\int|f-f_n|\to 0$, and by Lebesgue's bounded convergence theorem it follows that $\int|g-g_n|\to 0$. But I am unsure whether this also implies $\int|g-g_n||f|\to0$.

答: Hint: Observe that $2M|f|$ is an integrable bound for $|g_n - g|\cdot |f|$ and the latter converges a. e. to $0$. Now apply the bounded convergence theorem.


问:
Use Lagrange multipliers to find the maximum(s) and minimum(s)
Use Lagrange multipliers to find the maximum(s) and minimum(s) of

$$f(x,y)=2x^2+3y^2−4x−5 \text{ subject to } x^2+y^2=16$$

So far I've taken the partials of f and g:

\begin{align}
f_x &= 4x-4 \\
f_y &= 6y \\
g_x &= 2x \\
g_y &= 2y
\end{align}

Then I set them equal to each other and multiplied by $\lambda$:
\begin{align}
4x-4 &= 2x\lambda \\
6y &= 2y\lambda
\end{align}

I'm kind of stuck here where I need to solve for $\lambda$.

答:
In this problem, the constraint curve $ \ x^2 \ + \ y^2 \ = \ 16 \ $ has four-fold symmetry about the origin, but the function $ \ f(x,y) \ = \ 2x^2 \ + \ 3y^2 \ − \ 4x \ − \ 5 \ $ is only symmetrical about the $ x- $ axis.  This indicates that we can expect critical points on the constraint circle to appear in pairs $ \ (x,y) \ $ and $ \ ( x, -y) \ $ .

For your Lagrange equations, the one involving $ \ x \ $ does not have a convenient factorization after being rearranged, but the one with $ \ y \ $ does:

$$ 4x \ - \ 4 \ = \ \lambda·2x \ \ \rightarrow \ \ (2 \ - \ \lambda)·x \ = \ 2 \ \ \ , \ \ \ 6y \ = \ \lambda·2y \ \ \rightarrow \ \ (3 \ - \ \lambda)·y \ = \ 0 \ \ . $$

As discussed already in the comments above, we have either

$ \mathbf{y = 0 :  } \quad x^2 \  = \  16  \ \ \Rightarrow \ \ x \  = \  \pm 4 \ \ , $  giving us the two function values

$ \ f(4,0) \ = \ 2·4^2 \ − \ 4·4 \ − \ 5 \ \ = \ \ 11 \ \ \ \text{or} \  \ \  f(-4,0) \ = \ 2·(-4)^2 \ − \ 4·(-4) \ − \ 5 \ \ = \ \ 43 \ \ ; $

or, applying your Lagrange equation for $ \ x \ $ ,

$ \mathbf{\lambda = 3 : }  \quad (2 \ - \ 3)·x \ = \ 2  \ \ \Rightarrow \ \ x \  = \  -2 \ \ \Rightarrow \ \ (-2)^2 \ + \ y^2 \  =  \ 16 \ \ \Rightarrow \ \ y^2 \ = \ 12 \ \ \Rightarrow \ \ y \ = \ \pm 2\sqrt{3}  $
$$ f(-2 \ , \ \pm 2\sqrt{3}) \ = \ 2·(-2)^2 \ + \ 3·12 \  − \ 4·(-2) \ − \ 5 \ \ = \ \ 47 \ \ , $$

this latter result illustrating the function symmetry mentioned above.

The main aim here in summarizing the earlier work is to show the geometric interpretation of the problem, which also allows two of the results to be found with relatively little  calculation.  If we consider level curves of our function $ \ 2x^2 \ + \ 3y^2 \ − \ 4x \ − \ 5 \ \ = \ \ c \ \ ,  $ we can re-write this equation as

$$ 2x^2  \ − \ 4x \ + \ 3y^2  \ \ = \ \ 5 \ + \ c \ \ \rightarrow \ \ 2·(x^2  \ − \ 2x \ + \ 1 ) \ + \ 3y^2  \ \ = \ \ 5 \ + \ 2 \ + \ c  $$
$$  \rightarrow \ \ \frac{(x \ − \  1 )^2}{\left( \frac{7+c}{2} \right)} \ + \ \frac{y^2}{\left( \frac{7+c}{3} \right)}  \ \ = \ \ 1 \ \ . $$

These level curves are ellipses centered on $  \ (1,0) \ $ with their major axes on the $ \ x-$ axis; the graph below shows the ellipse for $ \ c \ = \ 0 \ $ and the constraint circle.

Critical values of $ \ f(x,y) \ = \ c \ $ occur when the endpoints of the major axis are tangent to the circle, that is, when either the right endpoint lies at $  \ (4,0) \ $ or the left endpoint becomes $ \ (-4,0) \ $ .  In the first case, the distance from the center of the ellipse to the right-hand intersection of the circle with the $ \ x-$ axis is $ \ 3 \ $ , so the semi-major axis of the ellipse must be $ \ \sqrt{\frac{7+c}{2}} \ = \ 3 \ \Rightarrow \ 7 + c \ = \ 2·9 \ \Rightarrow \ c \ = \ 11 \ . $  The left $ \ x-$ intercept of the circle is $ \ 5 \ $ units from the ellipse's center, so we obtain $ \ \sqrt{\frac{7+c}{2}} \ = \ 5 \ \Rightarrow \ 7 + c \ = \ 2·25 $
$ \Rightarrow \ c \ = \ 43 \ . $  The corresponding graphs are displayed below.

This does not seem quite complete, however, as it appears that if the ellipse were made slightly larger than the $ \ c \ = \ 43 \ $ level curve, we would have the circle circumscribed by the ellipse.

EDIT:  We can locate the remaining tangent points by inserting the constraint condition into the level curve equation:

$$  2x^2 \ + \ 3·(16-x^2) \ - \ 4x \ - \ 5 \ \ = \ \ c \ \ \Rightarrow \ \ -x^2 \ - \ 4x \ \ = \ \ c \ - \ 43 \ \ $$
$$  \Rightarrow \ \ (x \ + \ 2)^2 \ \ = \ \  47 \ - \ c \ \ \Rightarrow \ \ x \ \ = \ \ -2 \ \pm \ \sqrt{47-c} \ \ . $$

What would be four intersection points of the level curve and constrain circle "coalesce" into two tangent points for $ \ c \ = \ 47 \ . $  We thus have the result $ \ x \ = \ -2 \ $ for the tangent points that we obtained from the $ \ \lambda \ = \ 3 \ $ case.


问:
Solving $Ax \approx b$ with $0 < x_i < 1$?
I have a known matrix $A \in \mathbb{R}^{n, m}$ and an input vector $b \in \mathbb{R}^n$ and I'm looking for an algorithm to solve for $x \in \mathbb{R}^m$ such that:

$$
Ax \approx b
$$

but each $x_i$ must be between 0 and 1.

(n is 6, and m is in the range 1-100)

Minimizing square error would be fine.

Any advice on how to approach this?

答:
Formulate it as a quadratic programming problem.

Choose $\epsilon > 0$ to be small.

$$\min \left\|Ax - b \right\|^2$$

subject to

$$\epsilon  \le x_i \le 1-\epsilon , \forall i \in \{1, \ldots, n\}.$$


答:
Imagine the Venn diagram of the sets $X, Y, Z$ is represented as below.

Then $Y\cup Z$ is represented as

and $X^C$ is represented as

And finally, $X^C \cap (Y\cup Z)$ is the intersection of the two diagrams above and is represented as

Title: how to make a venn diagram of the following

Body:
I need to do a venn diagram of $$X^\complement \cap  (Y\cup Z) $$
I know that I start with the union of $Y$ and $Z$, having all of circles $Y$ and $Z$ colored, then I take the intersection with $X$ which would only be the part of the circle that touch $X$ with the circles $X$ and $Y$ and to take the complement (part where I am confused) would it be the rest of circle $X$ that has previously not drawn and the rest of the universe without counting the previous work of $Y$ and $Z$?


问:
∀x(A→ B) ├ ¬∃x(A ∧ ¬B) prove with natural deduction
Please help me prove the following
    $$∀x(A→ B) ├ ¬∃x(A ∧ ¬B)$$
using natural deduction.

This is what I have done so far:

$∀x(A→ B)$

$∃x(A ∧ ¬B)$ assumption

n   $(A ∧ ¬B)[n/x]$ assumption

$A[n/x] ∧ ¬B[n/x]$

$A[n/x]$  $∧e,4$

$B[n/x]$  $∧e,4$

How should I continue?

答: $$\def\fitch#1#2{~~~\begin{array}{|l} #1\\\hline #2\end{array}}\fitch{~~1.~~\forall x~(A\to B)}{\fitch{~~2.~~\exists x~(A\land\lnot B)}{\fitch{\hspace{-4ex}\boxed n~~~~3.~~(A\land\lnot B)[n/x]}{~~4.~~A[n/x]\;\land\lnot B[n/x]\\~~5.~~A[n/x]\hspace{15ex}{\land}\mathsf e~4\\~~6.~~\lnot B[n/x]\hspace{13.5ex}{\land}\mathsf e~4\\[1ex]~~7.~~(A\to B)[n/x]\hspace{8ex}\forall\mathsf e~1\\~~8.~A[n/x]\to B[n/x]\\~~9.~~B[n/x]\hspace{15ex}{\to}\mathsf e~8,5\\10.~~\bot\hspace{20ex}\lnot\mathsf e~9,6}\\11.~~\bot\hspace{23ex}\exists\mathsf e~2,3{-}10}\\12.~~\lnot\exists x~(A\land\lnot B)\hspace{13.5ex}\lnot\mathsf i~2{-}11}$$


问:
Rational solutions of quadratic forms
Is there an algorithm or a method that one can use to determine whether an equation of the form $(\text{E})$:
$$ax^2+by^2+cz^2+dt^2=0$$
has a solution $(x,y,z,t)$ in whole numbers. In other words, given the tuple $(a,b,c,d)\in\mathbb{Z}^4$, how to determine whether there is a non-trivial tuple $(x,y,z,t)\in\mathbb{Z}^4$ that satisfies $(\text{E})$?

In case when one of $a,b,c$ or $d$ is zero, one can use Legendre's theorem on ternary quadratic forms. This paper shows the full process of how it's done. Is there anything similar for the case of four variables?

答:
I actually realized some time after my comment above that, because we're trying to hit zero, at least one member of $(a,b,c,d) < 0$. If we assume WLOG that $a \geq b \geq c \geq d$, then at least $d$ is negative. If we set $x = t$ and $a+d = m$, then we can simplify to

$$mx^2 + by^2 + cx^2 = 0$$

But now one of those coefficients still has to be negative if we want to hit $0$. Either $|a| < |d|$ and $m$ is negative, or $b$ and/or $c$ are negative. Again, WLOG we can assume $c$ is negative. Then setting $y = z$ and $b+c-n$, we get:

$$mx^2 + ny^2 = 0$$

And either $m$ or $n$ has to be negative, so we really have

$$mx^2 = ny^2$$

I have to imagine there are solutions for this that don't have $m = n$ and $x = y$. But I'm not sure if there's a generalized form.

Edit: Oh, right. Either $m = n, x = y$, or both $m$ and $n$ must be squares for there to be integer solutions, because (for instance) $2x^2 = 5y^2$ can't have an integer solution, since both $x$ and $y$ have prime factorizations where all the exponents are $2$.

Title: Rational solutions of quadratic forms


问:
Why are Fuchsian groups interesting?
I am recently reading the book "Fuchsian groups" by Katok and now on Chapter $2$. I am curious about why Fuchsian groups are interesting. I look it up online and find answers here. Those are great answers. However, I am at an undergrad level now and I don't get a very clear picture. Hyperbolic geometry is interesting since there are some great properties, like hyperbolic Gauss-Bonnet. Would someone like to provide an example, e.g. rubric cube when studying group theory?

Appreciate any resources and helps!

Addendum: Fuchsian groups are discrete subgroups of PSL($2$, $\mathbb{R}$) or automorphism of the upper-half plane.

答:
Well, there might be other answers, but I think a very clear motivation is the study of surfaces.

Namely, consider a closed orientable surface of genus $1$, i.e. a torus. You probably know that the torus can be obtained by taking a square and identifying its opposite sides adequately. Somewhat more precisely, we can pair the sides of a square via Euclidean isometries to obtain a torus.

Naturally, we want to know whether this can be done for surface of higher genus (with more holes), and it turns out that hyperbolic polygons provide the answer. As you will learn in Katok's book, not every side-pairing of a polygon will work (for example, at the end of day you'd like your surface to be closed, and this imposes a restriction on the sum of the angles of your polygon).

Going back to hyperbolic polygons, consider (for simplicity) the closed orientable surface of genus $2$. It is not so hard to see that you may glue the sides of an (hyperbolic) octagon to obtain such a surface. Now, how do we describe this side-pairing? In the case of the torus, this is achieved via translations (which are discrete Euclidean isometries), and it occurs that for hyperbolic polygons, this side-pairing is achieved precisely by discrete subgroups of the isometry group of the hyperbolic plane; in other words, Fuchsian groups!

This means that understanding Fuchsian groups yields an understanding of compact orientable surfaces, although I must say that Fuchsian groups are interesting in their own right : )

Title: Why are Fuchsian groups interesting?


问:
Finding the singularity type at $z=0$ of $\frac{1}{\cos(\frac{1}{z})}$
I have the following homework problem:

What kind of singular point does the function
  $\frac{1}{\cos(\frac{1}{z})}$ have at $z=0$ ?

What I tried:

We note (visually) that $z_{0}$ is the same type of singularity for
both $f,f^{2}$ hence the type of singularity of $f(z)=\frac{1}{\cos(\frac{1}{z})}$
have at $z=0$ is the same type of singularity $f^{2}(z)=\frac{1}{\cos^{2}(\frac{1}{z})}$.
We recall
$$
\frac{1}{\cos^{2}(z)}=1+\tan^{2}(z)
$$

We also note that for any constant $z_{0}\in\mathbb{C}$ the singularity
of $f,f+z_{0}$ are the same, this can be proved by noting the Laurent
expansions of both functions are the same, up to an additive constant.

It remains to determine the singularity type at the origin of
$$
\tan(\frac{1}{z})
$$

This is where I'm stuck, we didn't study what the Taylor series of
$\tan(z)$.

I also know that type of singularity $$\cos(\frac{1}{z})$$ have, but
I don't know how to connect this with the singularity type of $$\frac{1}{\cos(\frac{1}{z})}$$

Can someone please hint me in the right direction ?

答: The singularity is not an isolated singularity, as $\cos(1/z)$ has a sequence of zeroes approaching $0$ (namely $z_n=1/(n\pi+\pi/2)$). In particular, $0$ cannot be an essential singularity of the function.


问:
Obtain the coefficient of $x^2$ in the expansion of $1+\frac{6}{2x+1}+\frac{5}{2-3x}$
Hello so this is a 2 part question and I managed to express that praction as a partial fraction which was equaled to $$1+\frac{6}{2x+1}+\frac{5}{2-3x}$$ I will add my work below I tried lot to Obtain the coefficient of $x^2$ but cant get to the answer of 237/8 and help is much appreciated. Thank you!

Body:
This is how to do it the hard way.

Let's suppose that, for $x$ close enough to $0$,

$$1 + \frac{6}{2x+1} + \frac{5}{2-3x} = a_0 + a_1x + a_2x^2 + \dots$$

$$(2x+1)(2-3x)\left(1 + \frac{6}{2x+1} + \frac{5}{2-3x}\right) = 19 - 7x - 6x^2 + \dots$$

$$(2x+1)(2-3x)(a_0 + a_1x + a_2x^2)
   = 2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1 + 2 a_2)x^2 + \dots$$

Comparing $19 - 7x - 6x^2$  to $2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1 + 2 a_2)x^2$

\begin{align}
   2a_0 &= 19 \\
   a_0 &= \frac{19}{2} \\
\hline
   a_0 + 2a_1 &= -7 \\
   \frac{19}{2} + 2a_1 &= -7 \\
   a_1 &= -\frac{33}{4} \\
\hline
   -6 a_0 + a_1 + 2 a_2 &= -6 \\
   -57 - \frac{33}{4} + 2a_2 &= -6 \\
   a_2 &= \frac{237}{8}
\end{align}

So the coefficient of x^2 is $\frac{237}{8}$.


问:
How to prove that $X \sim N(\sqrt\theta,1)$,then $X^2\sim\chi^2_1(\theta)$.
How to prove that $X\sim N(\sqrt\theta,1)$,then $X^2$ has non-central chi-square $\chi_1^2(\theta)$.

I can obtain the pdf of $X$, $f(x)=\frac{1}{\sqrt {2\pi}}\exp\left(-\frac{(x-\sqrt \theta)^2}{2}\right)$, I don't know how to get the pdf of $\chi^2_1(\theta)$ and how to get this result.

答:
The statement is not true. In particular for $\theta=1$, we can see that $X\sim N(1,1)$, and $X^2\nsim \chi^2(1)$. On the other hand if $X\sim N(0,1)$, then $X^2\sim \chi ^2(1)$. To show that the statement is false, here are plots for $\theta=1$ and $\theta = 2:$

In the above plot, the red curve is the pdf of $\chi ^2(1)$, and the blue curve is the pdf of $X^2$ where $X\sim N(1,1)$ . Now consider $\theta=2:$

In this case the red curve is the pdf of $\chi ^2(2)$, and the blue curve is the pdf of $X^2$ where $X\sim N(\sqrt 2,1)$.

To prove that they are indeed different, we derive the pdf of $X^2$. Suppose we have $X\sim N(\sqrt \theta,1)$. Then for $x\ge 0$, \begin{align*}
\mathbb P(X^2<x)&=\mathbb P(-\sqrt x<X<\sqrt x)=\mathbb P\left(\frac{-\sqrt{x}-\sqrt\theta}{1}<Z<\frac{\sqrt x-\sqrt \theta}{1}\right)\\
&=\Phi(\sqrt x-\sqrt \theta)-\Phi(-\sqrt x-\sqrt \theta)
\end{align*}
where $\Phi(.)$ is the cdf of $N(0,1)$. Differentiating, we get \begin{align*}
\frac{d}{dx}\mathbb P(X^2<x)&=\phi(\sqrt x-\sqrt \theta)\frac{d}{dx}(\sqrt x-\sqrt\theta)-\phi(-\sqrt x-\sqrt \theta)\frac{d}{dx}(-\sqrt x-\sqrt\theta)\\
&= \frac{\phi(\sqrt x-\sqrt\theta)}{2\sqrt x}+\frac{\phi(-\sqrt x-\sqrt \theta)}{2\sqrt x}=\frac{\phi(\sqrt x-\sqrt\theta)+\phi(-\sqrt x-\sqrt\theta)}{2\sqrt x}
\end{align*}
where $\phi(x)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)$ is the pdf of $N(0,1)$. Now we get \begin{align*}
\frac{\phi(\sqrt x-\sqrt\theta)+\phi(-\sqrt x-\sqrt\theta)}{2\sqrt x}&=\frac{\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(\sqrt x-\sqrt \theta)^2}{2}\right)+\frac{1}{\sqrt {2\pi}}\exp\left(-\frac{(-\sqrt x-\sqrt \theta)^2}{2}\right)}{2\sqrt x}\\
&=\frac{\exp\left(-\frac{x-2\sqrt\theta\sqrt x+\theta}{2}\right)+\exp\left(-\frac{x+2\sqrt\theta\sqrt x+\theta}{2}\right)}{2\sqrt{2\pi}\cdot \sqrt x}\\
&= \frac{\exp\left(-\frac x2+\sqrt\theta\sqrt x-\frac\theta 2\right)+\exp\left(-\frac x2-\sqrt\theta\sqrt x-\frac\theta 2\right)}{2\sqrt{2\pi}\sqrt x}
\end{align*}
which is not the pdf of $\chi ^2(\theta)$.


答: The claim simply is not true.  If $X$ is normal with mean $\mu = \sqrt{\theta}$ and variance $\sigma^2 = 1$, then $X^2$ is not chi-square distributed with $\theta$ degrees of freedom.  You can immediately see this by choosing $\theta = 1$:  then we can say that if $Z \sim \operatorname{Normal}(0,1)$, i.e., $Z$ is standard normal, then $Z^2$ is chi-squared with $\theta = 1$ degree of freedom.  But $X \sim \operatorname{Normal}(1,1)$ has mean $1$, so that would imply that $X^2$ has the same distribution as $Z^2$, which is absurd, since $Z = X-1$.  We can immediately see, for instance, that $$\Pr[X^2 < 1] = \Pr[-1 < X < 1] = \Pr[-2 < X - 1 < 0] = \Pr[-2 < Z < 0],$$ whereas $$\Pr[Z^2 < 1] = \Pr[-1 < Z < 1],$$ and these are obviously not equal.

Title: How to prove that $X \sim N(\sqrt\theta,1)$,then $X^2\sim\chi^2_1(\theta)$.

Body:
How to prove that $X\sim N(\sqrt\theta,1)$,then $X^2$ has non-central chi-square $\chi_1^2(\theta)$.

I can obtain the pdf of $X$, $f(x)=\frac{1}{\sqrt {2\pi}}\exp\left(-\frac{(x-\sqrt \theta)^2}{2}\right)$, I don't know how to get the pdf of $\chi^2_1(\theta)$ and how to get this result.


问:
Are there nonstandard $\mathsf{PA}$ models without $\Delta^1_1$ cuts?
My question is the following:

Is there a nonstandard model $\mathcal{M}\models\mathsf{PA}$ such that $\mathcal{M}$ has no $\Delta^1_1$-with-parameters-definable nonempty proper successor-closed initial segments ("cuts")?

Here "$\Delta^1_1$" is meant in the sense of the standard semantics of second-order logic - so a $\Delta^1_1$ subset of $\mathcal{M}$ doesn't need to be "internal" to $\mathcal{M}$ in any nice sense.

If we replace $\Delta^1_1$ with $\Pi^1_1$ the answer is trivially negative since the cut of standard naturals is $\Pi^1_1$; no parameters are needed here. If we replace $\Delta^1_1$ with $\Sigma^1_1$ this answer again becomes negative since for each nonstandard $a\in\mathcal{M}$ the set of elements infinitely below $a$ is $\Sigma^1_1(a)$ over $\mathcal{M}$. (See here.) However, I don't see a way to get a $\Delta^1_1$ cut in a nonstandard model of $\mathsf{PA}$. On the other hand, I don't see how to build a nonstandard model without a $\Delta^1_1$ cut. In particular, a natural hope might be to look at a nontrivial ultrapower of $\mathbb{N}$, but while $\Sigma^1_1$ formulas are preserved by taking ultrapowers, $\Delta^1_1$-ness (= an equivalence between a $\Sigma^1_1$ formula and a $\Pi^1_1$ formula) doesn't obviously need to be.

答:
I have posted a complete answer to this question on MathOverflow.

The general idea of the solution is that sufficiently saturated models will always fail to have $\Delta^1_1$ definable cuts, even with parameters. Computably saturated countable models should also be sufficient, which avoids set theoretic assumptions.


问:
A finitely based monoid whose semigroup reduct is not finitely based, and vice versa
Does there exist a monoid $(M,*,1)$ which is finitely based, but whose $\{*\}$ reduct is not finitely based? Also, does there exist a monoid $(M,*,1)$ which is not finitely based, but whose $\{*\}$ reduct is finitely based?

答:
Case 0. (Relevant, but not the question asked!)
It is known that there is a finitely based semigroup $S$ such that the monoid $S^1$ is not finitely based (Peter Perkins, 1969). It is also known that there is a nonfinitely based semigroup $T$ such that the monoid $T^1$ is finitely based (Edmund Lee, 2013).

But you are not asking about formally adjoining a unit to a semigroup. Let's turn to the question asked.

Case 1.
$(M,*,1)$ is a monoid satisfying some identity of the form $x^n=1$, $n>0$.

In this case, if $\Sigma$ is a finite semigroup basis for $(M,*)$, then $\Sigma\cup \{x^n=1\}$ is a finite monoid basis for $(M,*,1)$. Conversely, if $\Gamma$ is a finite monoid basis for $(M,*,1)$, then a finite semigroup basis for $(M,*)$ is $\{x^ny=y, yx^n=y\}\cup \Gamma'$ where $\Gamma'$ is obtained from $\Gamma$ by replacing all instances of $1$ with $x^n$.

What seems to be the hard case is

Case 2. $(M,*,1)$ satisfies no identity of the form $x^n=1$.

In this case, we still have the following: if  $\Sigma$ is a finite semigroup basis for $(M,*)$, then a finite monoid basis for $(M,*,1)$ is $\Sigma\cup \{x*1=x, 1*x=x\}$.

I don't see how to argue "conversely" in Case 2. That is,
I don't see how to rule out the possibility that $(M,*)$ is nonfinitely based and $(M,*,1)$ is finitely based when $(M,*,1)$ satisfies no identity of the form $x^n=1$.


问:
Hao Wang's $\mathfrak S$ system/$\Sigma$ system: a "transfinite type" theory that avoids the Goedel's theorems.
Long ago, while I was reading a book ($*$) about the various way to build set theories (Zermelo-Freankel, Von Neumann–Bernays–Gödel, and type theories), I read about a variant of type theory with weird metalogics properties called Hao Wang's $\mathfrak S$ system, and presented for the first time in the year 1954.

The weird properties were something about his ability to avoid the Goedel's theorems.

Now I can't find that book anymore, and on internet I can't find nothing about this. Someone can explain me in easy words that theory's special properties and help me finding some free readings and introductions about?

Edit: The thory allowed the types of the variables to have transfinite value like $\omega$ and can be extended to even bigger ordinals.

Update 05-28-13 I remember the title of the chapter, and was about "Cumulative orders and cumulative types".

($*$)The book was Ettore Casari - Questioni di filosofia della matematica. Milan: Feltrinelli 1964

答:
The following is a list of the papers written by Hao Wang on the topic and published on journals:

1) The Formalization of Mathematics, Journal of Symbolic Logic 19, 241-266;

2) Ordinals Number and Predicative Set Theories, Zeitschrift fur mathematische Logik und Grundlagen der Mathematik 5, 216-239.

There is another paper that has the title "Some formal details on Predicative Set Theories" (3). Interestingly, all those papers can be found in the following Hao Wang's book:

A Survey of Mathematical Logic, Amsterdam: North Holland, 1963.

[Reprinted as Logic, Computers and Sets, New York: Chealsea, 1970]

In particular, we have that paper (1) (the introductory one and the most philosophical, with an interesting introduction on the method of mathematics) is on chapter XXIII [pp.559-584], paper (3) is on chapter XXIV [pp.585-623], and paper (2) is on chapter XXV [pp.624-651].

Last piece of information about a presentation of the theory, that is about introducing the constructive approach in mathematics,
is this review by Mueller (unfortunately, the other two pages are not visible, but they can be easily downloaded with a JSTOR account).

For the sake of completeness, it is possible to find some information in the "Handbook of Philosophy of Mathematics" edited by Andrew Irvine in the part on Constructivism in Mathematics under the section Predicativsm [pp.333-337].

I hope it helps.


问:
Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection
Clearly, the total number of subsets possible is $2^5$

For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.

Now there are far too many cases (IMO) for me to manually curate, so is their a shorter way, complementary probability perhaps?

Even in complimentary method there seem to be a lot of subsets, but here is what I got

$$\binom 51(\binom 41 +\binom 42 +\binom 43 +\binom 44) + \binom 52 (\binom 31 + \binom 32 +\binom 33).....+\binom 54 (\binom 11)$$
$$=5(2^4-1) +10 (2^3-1) +10(2^2-1) +5 (2-1)$$
$$=180$$

Which comes out to be greater than $2^5$, which isn’t possible. Where am I going wrong?

Edit: My reasoning

I found the cases where one set has an element, and then the other set does not have that element, although I now realise I missed the cases where they have 1 element common or 3 or 4 or 5, which effectively makes my attempt useless.

答:
Note: $2^5$ is the count of whether each element is in a subset or not.  However, you are selecting 2 such subsets.  So we must count whether each element is included in each subset: which is $4^5$ outcomes in the space.

For the favored case, we wish a selection of 2 from the five elements to be in both, and each of the remaining 3 elements have but 3 options (in one, in the other, or in neither).  So that is $\binom 52 3^3$ favored outcomes.

As all outcomes are unbiased, the probability is: $$\dbinom 5 2\cdot \dfrac{3^3}{4^5} =\dfrac{135}{512}$$


问:
Find a finite extension of $\mathbb{Q}$ in which all primes split
Dear all, I would be grateful if someone could provide a solution to the following problem (using decomposition and inertia groups):

Find a finite extension of $\mathbb{Q}$ in which all primes split.

[Hint: Use the fact that a prime splits if and only if its decomposition group is
not the full Galois group (and that the decomposition group is cyclic for
all unramified primes)]

Many thanks,
Mohammad.

答: It may be possible to do this without using the hint. The polynomial $(x^2-13)(x^2-17)(x^2-221)$ has a root mod $p$ for all $p$; does that imply that every $p$ splits in ${\bf Q}(\sqrt{13},\sqrt{17})$? Or do I have my wires crossed?</p>


答:
There are $\binom52=10$ two-element subsets of $\{1,2,3,4,5\}$, and every pair $\langle A,B\rangle$ of subsets of $\{1,2,3,4,5\}$ whose intersection has exactly $2$ elements has one of them as its intersection. $A\cap B=S$ if and only if $A\setminus S$ and $B\setminus S$ are disjoint subsets of $\{1,2,3,4,5\}\setminus S$, so to count the pairs $\langle A,B\rangle$ such that $A\cap B=S$, we need to count the pairs of disjoint subsets of a $3$-element set.

The number of pairs $\langle C,D\rangle$ of disjoint subsets of a set $X$ of $n$ elements can be computed as follows. For $k=0,\ldots,n$ there are $\binom{n}k$ ways to choose $C$ of cardinality $k$, and there are then $2^{n-k}$ ways to choose $D$ disjoint from $C$, so by the binomial theorem there are altogether

$$\sum_{k=0}^n\binom{n}k2^{n-k}=\sum_{k=0}^n\binom{n}k1^k\cdot2^{n-k}=(1+2)^n=3^n$$

such pairs.

Alternatively, and even more simply, each of the $n$ elements of $X$ can be assigned arbitrarily to $C$, $D$, or $X\setminus(C\cup D)$, and this assignment can be made in $3^n$ ways.

In our case $|X|=|\{1,2,3,4,5\}\setminus S|=3$, so there are $3^3=27$ pairs of disjoint subsets of $\{1,2,3,4,5\}\setminus S$. Thus, there are $10\cdot27=270$ ordered pairs of subsets of $\{1,2,3,4,5\}$ that have $2$-element intersections. If you want unordered pairs, you need only cut this number in half.

Title: Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection

Body:
Clearly, the total number of subsets possible is $2^5$

For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.

Now there are far too many cases (IMO) for me to manually curate, so is their a shorter way, complementary probability perhaps?

Even in complimentary method there seem to be a lot of subsets, but here is what I got

$$\binom 51(\binom 41 +\binom 42 +\binom 43 +\binom 44) + \binom 52 (\binom 31 + \binom 32 +\binom 33).....+\binom 54 (\binom 11)$$
$$=5(2^4-1) +10 (2^3-1) +10(2^2-1) +5 (2-1)$$
$$=180$$

Which comes out to be greater than $2^5$, which isn’t possible. Where am I going wrong?

Edit: My reasoning

I found the cases where one set has an element, and then the other set does not have that element, although I now realise I missed the cases where they have 1 element common or 3 or 4 or 5, which effectively makes my attempt useless.