问:
Describing $\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(x, n) \; \mathrm{d}n$ as a contour integral.
My question essentially has to do with the derivative of a Contour Integral's parameterized curve.
$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$ to be exact. Where $\partial \Omega(x)$ is a Jordan curve which is differentiable for any $x \in \mathbb{C}$, and $f(n, x): \mathbb{C}^2 \to
\mathbb{C}$ integrable around the curve $\partial \Omega(x)$ in respect to $n$. Define $\gamma$ as the parameterized curve of $\partial \Omega$, and the terminology $f_x(n, x) = \frac{\partial f(n, x)}{\partial x}$ is used.

My work has essentially gotten down to these steps.

STEP 1: Turning the contour integral into the usual integral.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \frac{\partial}{\partial x} \int_{0}^{2\pi} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 2: Using the Liebniz rule.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \int_{0}^{2\pi} \frac{\partial}{\partial x} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 3: Taking the derivative.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) + \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) + \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 4: Separating the integrals.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta$$
$$+ \int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \mathrm{d}\theta$$
$$+ \int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 5: Simplifying the first integral into a contour integral.

$$\int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta = \oint_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n.$$

STEP 6: Plugging in the first integral to get the final answer.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$

$$= \int_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n+\int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \; \mathrm{d}\theta$$
$$+\int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

I am unsure of how to simplify this further or if this is even a decent approach. Does anybody have a good resource for this?

My goal is to write this derivative as multiple contour integrals, bar any $\gamma$-parameterized functions.

(Disclaimer: This same question has been posted by myself to MathOverflow)

答:
I think what we need is to notice is that while $\partial_x \gamma_{\theta}(\theta,x)=\gamma_{\theta x}$ what we really have is $\partial_x \gamma_{\theta}(\gamma(\theta, x))=\partial_x (\gamma_{\theta}\circ\gamma(\theta, x))=\gamma_{\theta n}\gamma_x$ now the second expression can also be written as $\partial_x \gamma_{\theta}(\gamma(\theta, x))=\partial_x \partial_{\theta}\gamma(\theta, x)=\partial_{\theta}\partial_{x}\gamma(\theta, x)=\partial_{\theta} \gamma_x(\gamma(\theta, x))=\partial_{\theta}(\gamma_x\circ\gamma(\theta, x))=\gamma_{xn }\gamma_{\theta}$ (@) so using this last expression we have that

$
\frac{d}{dx}\oint_{\partial \Omega(x)} fdn \\
=
\frac{d}{dx}\int_0^{2pi} f\gamma_{\theta}d\theta\\
=
\int_0^{2pi} \frac{\partial}{\partial_x}(f\gamma_{\theta})d\theta\\
=
\int_0^{2pi} (\frac{\partial}{\partial_x}(f)\gamma_{\theta}+f\frac{\partial}{\partial_x}(\gamma_{\theta}))d\theta \\
=_{@}
\int_0^{2pi} (\frac{\partial}{\partial_x} (f)\gamma_{\theta}+f\gamma_{xn}\gamma_{\theta})d\theta\\
=
\int_0^{2pi} (\frac{\partial}{\partial_x} (f)+f\gamma_{xn})\gamma_{\theta}d\theta\\
=
\oint_{\partial \Omega(x)}(\frac{\partial}{\partial_x} (f)+f\gamma_{xn})dn\\
=
\oint_{\partial \Omega(x)}(f_x+f_n\gamma_x+f\gamma_{xn})dn\\
=
\oint_{\partial \Omega(x)}(f_x+\frac{\partial}{\partial_n}(f\gamma_x))dn\\
=
\oint_{\partial \Omega(x)}f_x dn+\oint_{\partial \Omega(x)}\frac{\partial}{\partial_n}(f\gamma_x)dn\\
=
\oint_{\partial \Omega(x)}f_x dn+\oint_{\partial \Omega(x)}d(f\gamma_x)\\
=
\oint_{\partial \Omega(x)}f_x dn\\
$

OP says this proof is wrong and another one has been posted after which has been accepted by him so I'm only leaving this in case someone can point what is wrong as OP has already tried but I have failed to understand and I wish to clear my misconceptions.

What I had in mind was not the Leibniz integral rule/differentiation under the integral sign but the Reynolds transport theorem which is a generalization, the proof can be found in the link but I have put the steps here for comparison

$
\frac{d}{dt}\int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)\,dV\\
=
\frac{\partial}{\partial t}\int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)\, J(\mathbf{X},t)\,dV_0\\
=
\int_{\Omega_0} \frac{\partial
}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t)\, J(\mathbf{X},t))\,dV_0\\
=
\int_{\Omega_0} [\frac{\partial}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t))\, J(\mathbf{X},t)+\hat{\mathbf{f}}(\mathbf{X},t)\, \frac{\partial}{\partial t}(J(\mathbf{X},t))]\,dV_0\\
=
\int_{\Omega_0} [\frac{\partial}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t))\, J(\mathbf{X},t)+\hat{\mathbf{f}}(\mathbf{X},t)\,  J(\mathbf{X},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,dV_0\\
=
\int_{\Omega_0} [\frac{\partial}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t))+\hat{\mathbf{f}}(\mathbf{X},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,J(\mathbf{X},t)dV_0\\
=
\int_{\Omega(t)} [\frac{\partial}{\partial t}(\mathbf{f}(\mathbf{x},t))+\mathbf{f}(\mathbf{x},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,dV\\
=
\int_{\Omega(t)} [\frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}+(\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t))\cdot \mathbf{v}(\mathbf{x},t)+\mathbf{f}(\mathbf{x},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,dV\\
=
\int_{\Omega(t)} [\frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}+\boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})]\,dV\\
=
\int_{\Omega(t)} \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}\,dV+\int_{\Omega(t)}\boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\,dV\\
=
\int_{\Omega(t)} \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}\,dV+\int_{\Omega(t)}(\mathbf{f}\otimes\mathbf{v})\cdot n\,dA\\
=
\int_{\Omega(t)} \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}\,dV+\int_{\partial\Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}\,dA\\
$


问:
Determine whether $f(x, y) = \frac{1}{x(xy)^{1/2}}$ is integrable over some open sets
I am studying analysis, using Munkres' book "Analysis on Manifolds". I am in the chapter about improper integrals and I want to solve the following question:

Let $f(x, y) =  \frac{1}{x(xy)^{1/2}}$ for $x > 0$ and $y > 0$. Let:
\begin{align}
A_0 &= \{(x, y) \ | \ 0 < x < 1 \text{ and } x < y < 2x \} \\
B_0 &= \{(x, y) \ | \ 0 < x < 1 \text{ and } x^2 < y < 2x^2 \}
   \end{align}
Determine whether $\int_{A_0} f$ and $\int_{B_0} f$ exist; if so, calculate.

My idea is to use Theorem 15.2:

Theorem 15.2 Let $A$ be an open set in $\mathbb{R}^n$; let $f: A \rightarrow \mathbb{R}$ be continuous. Choose a sequence $C_n$ of compact rectfiable subsets of $A$ whose union is $A$ such that $C_N \subset Int \ C_{N+1}$ for each $N$. Then $f$ is integrable over $A$ if and only if the sequence $\int_{C_N} |f|$ is bounded. In this case, $\int_{A} f = \lim_{N \to \infty} \int_{C_N} |f|$.

I plan to choose $C_N = \{(x, y) \ | \ \frac{1}{n} \leq x \leq 1 - \frac{1}{n} \text{ and } x  \leq y \leq 2x \}$ for $A_0$.

I plan to choose $C_N = \{(x, y) \ | \ \frac{1}{n} \leq x \leq 1 - \frac{1}{n} \text{ and } x^2  \leq y \leq 2x^2 \}$ for $B_0$.

Is this the right way to proceed? Thank you in advance.

ContentLicense: CC BY-SA 4.0

Body:
Yes. First, notice that in both cases we have $f$ = $|f|$.

For $A_0$ you will obtain that $\int_{A_0} f$ does not exist:

Note that:
$$ \int_{C_N} |f| = \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \int_{x}^{2x} \frac{1}{x \sqrt{xy}} dy \ dx =  \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \frac{1}{x \sqrt{x}} \ (2 \sqrt{y})|_{x}^{2x} = \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \frac{1}{x \sqrt{x}} \ 2 \sqrt{x} (\sqrt{2} - 1) = \\ 2 (\sqrt{2} - 1) \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \frac{1}{x} = 2 (\sqrt{2} - 1) \ [\ln(1 - \frac{1}{n}) - \ln(1/n)].$$
Since $\lim_{N \to \infty} \ln(1 - 1/n) = 1$ and $\lim_{N \to \infty} \ln(1/n) = - \infty$, the sequence $\int_{C_N} |f|$ is not bounded and therefore $\int_{A_0} f$ does not exist.

For $B_0$ you will obtain that $\int_{B_0} f$ does exist and it's value is: $4(\sqrt{2} - 1)$.

Note that:
$$ \int_{C_N} |f| = \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \int_{x^2}^{2x^2} \frac{1}{x \sqrt{xy}} dy \ dx =  \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \frac{1}{x \sqrt{x}} \ (2 \sqrt{y})|_{x^2}^{2x^2} = \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \frac{1}{x \sqrt{x}} \ 2 x (\sqrt{2} - 1) = \\ 2 (\sqrt{2} - 1) \int_{\frac{1}{n}}^{1 - \frac{1}{n}} \frac{1}{\sqrt{x}} = 4 (\sqrt{2} - 1) \ [\sqrt{1 - \frac{1}{n}} - \sqrt{\frac{1}{n}}].$$
We have that $\int_{C_N} |f|$ is bounded and $\int_{B_0} f = \lim_{N \to \infty} \int_{C_N} |f| = 4(\sqrt{2} - 1) $.


问:
Factorial of Infinite Cardinal
I have been thinking about the following problem:

Let $A$ be a set of cardinality $k$ and denote $\sum_A$ the set of all bijection from $A$ to $A$.

Also denote $k! = \mathrm{card}\left(\sum_A \right)$. Prove that $k!=2^k$.

My proof consists of finding a bijection $F:\sum_A\to P(A)$ which associates each bijection from the left to the set of its fixed points. Then the result would follow.
($P(A)$=the power set of $A$).

Since this proof seems quite easy I am afraid it is wrong. Can someone enlighten me?
Thank you very much!

答:
First of all, unless you limit yourself to infinite sets this statement is wrong. It is wrong because for $k=3$ we have $3!=6<8=2^3$.

For infinite sets, assuming the axiom of choice, this is true. To see why note that $f\colon A\to A$ means that $f\subseteq A\times A$, so $\Sigma_A\subseteq\mathcal P(A\times A)$.

Assuming the axiom of choice $|A|=|A\times A|$ and therefore $|\Sigma_A|\leq 2^k$. You still have to show the other direction holds as well. In order to show that, you need to find $2^k$ distinct bijections from $A$ to itself.

Hint: There are $2^k$ different pairs $A_1,A_2\subseteq A$ such that $\{A_1,A_2\}$ is a partition of $A$ and $|A_1|=|A_2|$. For every such pair we define a unique $f\colon A\to A$ which sends $A_1$ to $A_2$ and vice versa, conclude the wanted equality of cardinals.

ContentLicense: CC BY-SA 3.0

Title: Factorial of Infinite Cardinal


问:
For Brownian motion $B_t,$ we have $\lim_{t\to \infty}\frac{B_t}{t^{\alpha}}=0$ almost surely.
This might be a silly problem. I am trying to show that $\lim_{t\to \infty}\frac{B_t}{t^{\alpha}}=0$ almost surely if $\alpha>\frac{1}{2}.$ I can apply Chebyshev's inequality to show that $P(|B_{t}/t^{\alpha}|>\epsilon)\to 0$ as $t\to \infty.$ This tells me that $\frac{B_t}{t^{\alpha}}$ converges to $0$ in probability. But I am not sure how to argue the almost sure limit. I found at the very beginning of a book I am reading, so I am looking forward to something which does not use $0-1$ laws if that's possible. I am wondering if this can be done cheaply using the Borel-Cantelli lemma.

答:
Let $\beta=(\alpha-1/2)/2$. By the law of iterated logarithm,
$$
\limsup_{t\to\infty}\frac{|B_t|}{t^\alpha}\le  \limsup_{t\to\infty}\frac{|B_t|}{t^{
\beta}\sqrt{t\ln\ln t}}=\lim_{t\to\infty}\frac{1}{t^{\beta}}\times\limsup_{t\to\infty}\frac{|B_t|}{\sqrt{t\ln\ln t}}=0\quad\text{a.s.}
$$


问:
Showing a metric space is not complete by showing the set is neither open nor closed
I have to determine whether a metric space $(X,d)$ where $X=\{(x,y,z) \in \mathbb R^3 : x^2+y^2+z^2<1, x\geq 0\}$ and $d$ is the induced metric from $\mathbb R^3$ (which I believe is $d_2 = (\sum_{j=1}^k |x_j-y_j|^2)^{\tfrac{1}{2}}$) is complete or not.

The approach I am trying to take is to show that $X \subset \mathbb R^3$ is neither open nor closed in $\mathbb R^3$, which is complete with respect to the $d_2$ metric and as such, it fails the completeness characterization for subsets theorem which states that if a metric space $(X,d)$ is complete, then a subset $A \subset X$ is complete with respect to the induced metric if and only if $A$ is closed.

I am still not sure that even if I do manage to show that it is neither open nor closed, I can actually apply the theorem since I have no clue whether it applies to sets which are neither (I haven't been able to find an answer for that either, but I digress). The lemma we were shown to prove such a set is closed in another is to show that for every sequence $x_n \in A$ converging in $X$, one also has that the $lim_{n \to \infty} (x_n) \in A$. My "proof" does not make use of this lemma. It is as follows:

Consider some point $x_1=(0,\tfrac{1}{2},0)$. Then, $\forall \epsilon>0, B_{\epsilon}(x_1)$ (ball with radius $\epsilon$ centered on the point $x_1$) contains the point $(0-\tfrac{1}{2}\epsilon,\tfrac{1} {2},0) \notin X$ showing that $B_{\epsilon}(x_1) \not\subset X \implies X$ is not open. A similar argument is used to show that the complement $\mathbb R^3 \setminus X$ is not open $\implies$ $X$ is not closed. As such, it fails the aforementioned theorem for the completeness characterization for subsets (I think?) and as a result, $(X,d)$ is not complete.

I also know that I did not make use of the metric space definition of open balls. Intuitively, I am aware how that definition applies, but I am not able to formalize it in this context. Definition is as follows; For any metric space $(X,d_X)$, the open ball with centre $x \in X$ and radius $r$ is the subset $B_r(x) = \{y \in X : d_X(y,x) < r\}$.

I am not entirely confident with my approach and any hints are much appreciated. Thank you in advance!

答:
The theorem you are trying to apply is:

if a metric space $(X,d)$ is complete, then a subset $A \subset X$ is complete with respect to the induced metric if and only if $A$ is closed.

Now comes a somewhat confusing part: In your case the "$X$" of the theorem is $\Bbb R^3$, while the "$A$" of the theorem is your set "$X$". Rather than maintain this mix up of the meaning of "$X$", let's just relabel that variable in the theorem:

if a metric space $(S,d)$ is complete, then a subset $A \subset S$ is complete with respect to the induced metric if and only if $A$ is closed.

The implications here are clear: $A$ is complete if and only if $A$ is closed.

If $A$ is complete, then $A$ is closed. ("only if" clause)

If $A$ is closed, then $A$ is complete. ("if" clause)

If $A$ is not complete, then $A$ is not closed. (contrapositive of "if" clause)

IF $A$ is not closed, then $A$ is not complete. (contrapositive of "only if" clause)

Now in your application, $S = \Bbb R^3$, and $A = X$.

I am still not sure that even if I do manage to show that it is neither open nor closed, I can actually apply the theorem since I have no clue whether it applies to sets which are neither (I haven't been able to find an answer for that either, but I digress).

The answer is  right there in the theorem. If $X$ is not closed, then it is not complete.

"Open" has nothing to do with it. It doesn't matter whether $X$ is open or not. All you need to do is show that $X$ is not closed in $\Bbb R^3$, and the theorem implies that $X$ is not complete.


问:
Reduction from 'combination with repetition' to 'permutation with repetition'
The combination with repetition can be written as follows:

$$\left(\!\!\binom{n}{r}\!\!\right) = \binom{n+r-1}{r}$$

This formula can be deduced by considering a system that gives an equivalent number of cases of mixing - a system in which there are $n-1$ identical dividers and $r$ identical objects.

I thought, what if we say that the objects are not identical (where dividers are still identical)? Then can we not reduce this to the permutation with repetition, $n^r$?

I actually used an analogy to come up with the above idea - $n$ menus in a restaurant and food for $r$ people is to be served and we are finding the number of combinations of menus, without considering who gets to eat what. So, if we don't think of the objects as being identical, then it is same as caring who eats what.

But the following formula, $$\binom{n+r-1}{r} \times r! = \frac{(n+r-1)!}{r! \times (n-1)!} \times r!$$ $$= \frac{(n+r-1)!}{(n-1)!} = (n+r-1) \times (n+r-2)\times ... \times (n+1) \times n$$ does not seem to lead to the expression $n^r$.

I think I considered who gets what well, but what is wrong here?

答:
You are using $r!$ as the factor to pass from combinations to permutations, however not every combination has $r!$ permutations you can create from it.

The problem is that there can be repetitions on the menus  : compare the case where all the people get different menus and the case where there are two people receiving the same menus.


问:
Clarification of solution set of a system of linear equations in three variables.
I am reviewing Jordan Canonical form:

I have a situation where I have two linearly independent eigenvectors for a 3x3 matrix and I am trying to find the third.

In the example it is saying that by solving the following:

$$ \begin{bmatrix}
    0 & 1 & 1  \\
    0 & 0 & 0 \\
    0 &0 & 0
  \end{bmatrix}     \cdot  \begin{bmatrix} x  \\y \\ z\end{bmatrix} = \begin{bmatrix} 1  \\0 \\ 0\end{bmatrix} $$

we will get the solution set $\,\,x \begin{bmatrix} 1  \\0 \\ 0\end{bmatrix} + y \begin{bmatrix} 0  \\1 \\ -1\end{bmatrix} + \begin{bmatrix} 0  \\0 \\ 1\end{bmatrix}$

But for whatever reason I am not understanding why that is the solution set. Looking for some explanation here.

答:
We get that the solutions are $\{(x,y,-y)+(0,0,1)|x,y\in\Bbb R\}$.  Just back substitute to get the solution to the associated homogeneous system; and note that $(0,0,1)$ is a particular solution.

That is, it's correct.  The general form for the solution of a linear system is $\text{general solution}=\{\text{particular solution} + \text{elements of the kernel}\}$.


问:
Find the maximum of $2xy^2 - 2x^2$
I got the constraint $g(x,y)=x^2+y^2$ (since the domain is $x^2+y^2\ge 1$ and $x\ge 0$) (it is a half circle domain) and the function $f(x,y)=2xy^2-2x^2$. I got the critical point to be $(0,0)$. The Hesse gave me $0$ which is of course no conclusion. I have a problem expressing $x$ and $y$ in terms of $\lambda$ but here is where my problem occurs. I get $\lambda=y^2-2$ and $\lambda=-y^2-2$. What am I doing wrong, is it the start?

答:
$f_x=2y^2-4x,f_y=4xy$ so $2y^2-4x=2\lambda x$ and $4xy=2\lambda y$. The Lagrange condition in 2D is usually handled most easily by dividing one equation by the other to eliminate $\lambda$. In this case you then obtain $\frac{y}{2x}-\frac{1}{y}=\frac{x}{y}$. Clearing denominators you get $y^2/2 - x = x^2$. That in combination with the constraint $x^2+y^2=1$ is enough to find some points of interest, I think.

However this division is not in general legitimate; you needed $4xy \neq 0$ and $2 \lambda y \neq 0$, so you needed none of $x$ or $y$ or $\lambda$ to be zero.

So you need to go back and check those cases. For $\lambda=0$ you get either $x$ or $y$ being zero. In the case $x=0$ you get $y=0$, which violates the constraint. In the case $y=0$ you get $-4x=2\lambda x$ so either $\lambda=-2$ (contradiction) or $x=0$ (violates the constraint). So $\lambda \neq 0$.

If $x=0$ then immediately $2y^2=0$ so $y=0$, again violating the constraint.

If $y=0$ then $-4x=2\lambda x$ so either $x=0$ (violating the constraint) or $\lambda=-2$. In the latter case $x$ can be anything and still satisfy the second equation, so by using the constraint you conclude that $(1,0)$ is another point of interest.

Title: Find the maximum of $2xy^2 - 2x^2$


问:
How to find a set of vectors given its span?
We're learning about what vector spaces and subspaces are in linear algebra, and this question is really confusing me:

Let W be the set of all vectors of the form shown, where a, b, and c represent arbitrary real numbers. In each case, either find a set S of vectors that spans W or give an example to show that W is not a vector space.

$\begin{bmatrix}a - b\\b - c\\c-a\\b\end{bmatrix}$

I know that this can be written as the span of a set of vectors. I also know that this span would be a 1D subspace (line) in R4, since it has 1 pivot and 4 rows. I just don't know how I would go about finding the set of vectors that this is the span for. Any help is appreciated!

答:
Hint: write the given general vector in the form
$$\begin{bmatrix}a-b\\b-c\\c-a\\b\end{bmatrix} = a \begin{bmatrix}?\\?\\?\\?\end{bmatrix} + b \begin{bmatrix}?\\?\\?\\?\end{bmatrix} + c \begin{bmatrix}?\\?\\?\\?\end{bmatrix}$$


问:
Integer programming problem, 5 pens and 10 bags
I am trying to solve an Integer Programming question (trying to come up with a model). The question goes like this: There are 5 pens a person has. He wants to have all 5 of them at the same time. And he has 10 different weighted bags. He wants to cary minimum weight as he can while having all five of the pens. Different bags carry different pens, the distribution is like this:

The weights of the bags are like this:
bag 1 --> 4
bag 2 --> 6
bag 3 --> 4
bag 4 --> 2
bag 5 --> 3
bag 6 --> 2
bag 7 --> 4
bag 8 --> 3
bag 9 --> 6
bag 10 --> 3

What I thought is this:
Let xi = 1 be that he brings the ith bag and xi = 0 be that he does not bring the ith bag. Then the obj. function is the minimum of sum of all of the x variables multiplied with their weight. And I could think of only one constraint: The sum of all the x variables i = 1..10 should be at most 10 and at least 3.

I could not think of anything else. Dont even know if this is the right path to the solution.

答:
This is a weighted set cover problem.  Let $a_{b,p}$ indicate whether bag $b$ contains pen $p$.  You want to minimize $\sum_b w_b x_b$ subject to
\begin{align}
\sum_b a_{b,p} x_b &\ge 1 &&\text{for all $p$}\\
x_b &\in \{0,1\} &&\text{for all $b$}
\end{align}
For your instance, you want to minimize
$$4x_1 + 6x_2 + 4x_3 + 2x_4 + 3x_5 + 2x_6 + 4x_7 + 3x_8 + 6x_9 + 3x_{10}$$
subject to
\begin{align}
x_1 + x_4 + x_7 &\ge 1 \\
x_2 + x_4 + x_6 + x_9 &\ge 1 \\
x_3 + x_5 + x_8 + x_{10} &\ge 1 \\
x_1 + x_2 + x_7 + x_{10} &\ge 1 \\
x_3 + x_6 + x_9 &\ge 1 \\
x_b &\in \{0,1\} &&\text{for all $b$}
\end{align}
The unique optimal solution is $x=(0,0,0,1,0,1,0,0,0,1)$, with objective value $2+2+3=7$.  For a short certificate of optimality, use the optimal linear programming dual variables $(2, 0, 2, 1, 2)$ for the five constraints and $(1, 5, 0, 0, 1, 0, 1, 1, 4, 0)$ for the lower bounds $x_b \ge 0$.


问:
Show $\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n) = 0$?
I am trying to show that
$$\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n)=0.$$
I have tried expanding $e^{-nx}$ with its Taylor series and $(1-x)^n$ with Newton's binomial theorem. The resulting Taylor series is $\sum_{k=2}^{\infty}  (-1)^k \left(\frac{n^k}{k!} - {n \choose k}\right)x^k.$

I also tried taking the first derivative with respect to $x$ to find the point in $(0,1)$ where $e^{-nx} - (1-x)^n$ is maximized (in $x$ as a function of $n$), but I was only able to find the hard-to-solve $\frac{n}{1-n} = \frac{\ln(1-x)}{x}.$

答: To get rid of the $e^{-n x}$ term, use the formula that you just found: $\frac{n}{1-n} = \frac{\log(1-x)}{x}$ (just move the $x$ in the denominator of the RHS to the LHS). That is going to give you a formula of the form $(1-x)^{n-1} - (1-x)^n$. From here, the result should be almost immediately.


问:
When do I use N for the variance of x-bar instead of n-1
if the $X_i$'s are all normally distributed

always

if the $X_i$'s are independent of each other

if the $X_i$'s are identical distributed

答:
In the simulation below (using R), we have $X_1 = \sqrt{12}\,U,$ where $U\sim\mathsf{Unif}(0,1);$ $X_2 \sim \mathsf{Norm}(0,1);$ $X_3\sim\mathsf{Exp}(1),$ the exponential distribution with rate $1.$ That is, the $X_i$ are independent and $Var(X_1) = Var(X_2) = Var(X_3) = \sigma^2 = 1.$ The simulation illustrates that
$$Var(\bar X) = Var\left[\frac 13(X_1+X_2+X_3)\right] = \frac 19 Var[X_1+X_2+X_3]\\ = \frac 19 [\sigma^2+\sigma^2+\sigma^2] =\frac{\sigma^2}{3} = \frac 13,$$
where independence is used to get from the first line above to the second.
[Note that $Var(X+Y) = Var(X)+Var(Y),$ if $X$ and $Y$ are independent.]

The $X_i$ are not all normally distributed, are not identically distributed, and are independent. Also, the equation in your Question is true for my random variables.
So what do you think?

```<code>set.seed(2020)
m = 10^6
x1 = runif(m)*sqrt(12)
x2 = rnorm(m)
x3 = rexp(m)
a = (x1 + x2 + x3)/3  # sample mean
var(a)
[1] 0.3328322         # aprx Var(A) = 1/3
</code>```

Body:
In the simulation below (using R), we have $X_1 = \sqrt{12}\,U,$ where $U\sim\mathsf{Unif}(0,1);$ $X_2 \sim \mathsf{Norm}(0,1);$ $X_3\sim\mathsf{Exp}(1),$ the exponential distribution with rate $1.$ That is, the $X_i$ are independent and $Var(X_1) = Var(X_2) = Var(X_3) = \sigma^2 = 1.$ The simulation illustrates that
$$Var(\bar X) = Var\left[\frac 13(X_1+X_2+X_3)\right] = \frac 19 Var[X_1+X_2+X_3]\\ = \frac 19 [\sigma^2+\sigma^2+\sigma^2] =\frac{\sigma^2}{3} = \frac 13,$$
where independence is used to get from the first line above to the second.
[Note that $Var(X+Y) = Var(X)+Var(Y),$ if $X$ and $Y$ are independent.]

The $X_i$ are not all normally distributed, are not identically distributed, and are independent. Also, the equation in your Question is true for my random variables.
So what do you think?

set.seed(2020)
m = 10^6
x1 = runif(m)*sqrt(12)
x2 = rnorm(m)
x3 = rexp(m)
a = (x1 + x2 + x3)/3  # sample mean
var(a)
[1] 0.3328322         # aprx Var(A) = 1/3


问:
Link between probability space and random variable
Suppose $\Omega=\{A,B\}$, $\mathcal{F}$ consists of all possible subsets, $P\{A\}=P\{B\}=\frac{1}{2}$, $X(A)=-1$ and $X(B)=1$. Write the expression of $\mu_X(\mathcal{B})$, where $\mathcal{B}$ is any Borel set in $\mathbb{R}$.

Is it correct that when $-1 \in \mathcal{B}$ and $1 \in \mathcal{B}$, $\mu_X(\mathcal{B})=1$? If not, I wonder where I made a mistake. Thank you so much for your help!

答:
$\mu_X(\mathcal{B})=P(\{\omega\in\Omega: X(\omega)\in \mathcal{B}\})=\begin{cases}0,&-1,1\not\in\mathcal{B} \\
P(B)=1/2,&-1\not\in\mathcal{B}, 1\in\mathcal{B}\\
P(A)=1/2,&1\not\in\mathcal{B},-1\in\mathcal{B}\\
P(A\cup B)=1,&-1,1\in\mathcal{B}\end{cases}$


问:
There is enough of gold to...
I read that there is enough of gold layering the surface of planet Earth with 45cm layer of gold.

My question is: Suppose that our planet is a sphere(without mountains and stuff), then how much kilograms of gold is on "Earth"?

(I now that the volume of gold is: $V=V_l-V_E$, where: $V_l=\frac{4}{3}\pi (r+45)^3$ and $V_E=\frac{4}{3}\pi r^3$ )

答: Hint: Gold has a density of about $19\frac{\mathrm g}{\mathrm{cm}^3}$. Multiply this with the number of $\mathrm{cm}^3$ you find.

Title: There is enough of gold to...


问:
Finding constants so that inequality holds
Find $b > 0$ such that $b^x \geqslant x + 1$ holds for all $x$ values of $x$.

This is apparently related to differentiation but since you can’t differentiate inequalities, I don’t know how this works.

答:
Here is a picture of $x+1$ and $5^x$  and $\left( \frac{5}{4} \right)^x$

Note how both curves are sometimes below the straight line.

What needs to happen so that $b^x$  is always above or touching the line? It  touches when $x=0$


问:
Why -(A&B)=-A|-B
Let me repeat that.

Why not (A and B) means not A or not B

I use | to mean or and & to mean and.

It seems that if the not (or - sign) enter the bracket it changes or into and and via versa.

How true is it?

Also what's about the extension of it?

Is -(A|B&C|D) = -A & -B | -C & -D

Basically I want the strongest theorem and if possible, the most elegant proof.

答:
Look up de Morgan's Law.  But, basically:

When it is not the case that both are true, then it must be the case that at least one among them is false.$$\lnot(A\land B)\models(\lnot A)\lor(\lnot B)$$

Likewise, when it is the case that at least one among them is false, then it must not be the case that both are true.$$(\lnot A)\lor(\lnot B)\models\lnot(A\land B)$$

Title: Why -(A&B)=-A|-B


问:
Hyperboloid Equation
canonical hyperbola equation:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $$
or
$$ d_{kl}=\sqrt{(x-x_{k})^2+(y-y_{k})^2}-\sqrt{(x-x_{l})^2+(y-y_{l})^2} $$
where the focus of hyperbola are $(x_{k},y_{k})$ and $(x_{l},y_{l})$

canonical hyperboloid equation is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 $$
can I use
$$ d_{kl}=\sqrt{(x-x_{k})^2+(y-y_{k})^2+(z-z_{k})^2}-\sqrt{(x-x_{l})^2+(y-y_{l})^2+(z-z_{l})^2} $$
as hyperboloid equation? If can, which one is it: one sheet or two sheet?
Or, is it another shape equation?
Thank you.

答:
Note that the equation
$$ d_{kl} = \left\lvert\sqrt{(x-x_k)^2+(y-y_k)^2}
             - \sqrt{(x-x_l)^2+(y-y_l)^2}\right\rvert $$
is simply
$d_{kl} = d(\mathbf x, F_1) - d(\mathbf x, F_2)$
where $F_1$ and $F_2$ are the foci of the hyperbola,
$\mathbf x = (x,y)$ is any point on the hyperbola,
and $d(P,Q)$ is the function giving the distance between points $P$ and $Q.$
(Note the absolute value on the right-hand side, which enables the equation to product both branches of the hyperbola.)

The equation
$$ d_{kl} = \left\lvert\sqrt{(x-x_k)^2+(y-y_k)^2+(z-z_k)^2}
                         - \sqrt{(x-x_l)^2+(y-y_l)^2+(z-z_l)^2}\right\rvert \tag1 $$
also is simply $d_{kl} = d(\mathbf x, F_1) - d(\mathbf x, F_2)$,
except that now $\mathbf x = (x,y,z)$ is a point in three dimensions instead of two.
This gives the equation of a hyperboloid produced by taking a hyperbola with foci $F_1$ and $F_2$ in some plane through the two foci and rotating that hyperbola about its transverse axis (the axis through the foci).
The result is a hyperboloid of two sheets contained within a double cone.

The equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1, \tag2$$
on the other hand, is a hyperboloid of one sheet.
If $\lvert a\rvert = \lvert b\rvert$ the hyperboloid can be produced by rotating a hyperbola around its conjugate axis, which we can see by writing its equation in cylindrical coordinates (with $z$ as the cylindrical axis) as
$$\frac{r^2}{a^2}-\frac{z^2}{c^2}=1.$$
If $\lvert a\rvert \neq \lvert b\rvert,$ however,
the hyperboloid is not a surface of revolution at all,
although it still has only one sheet.

So I don't think you'll have much luck finding the equivalence between Equations $(1)$ and $(2).$ But you might try
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{b^2}=1,$$
or in cylindrical coordinates (with the $x$ axis as the cylindrical axis),
$$\frac{x^2}{a^2}-\frac{r^2}{b^2}=1.$$

For a more general hyperboloid of two sheets (not equivalent to a surface of revolution), try
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1.$$

Title: Hyperboloid Equation


问:
Show that $\alpha$ is bijective and find $\beta: \mathbb{R} \rightarrow \mathbb{R}$ such that $(\beta \alpha)(a) = a$ for all $a \in \mathbb{R}$
Define $\alpha: \mathbb{R} \rightarrow  \mathbb{R}$ via

$\alpha (a)=\begin{cases}
 4a-3& \text{ if } a\leq 1 \\
 a^2 & \text{ if } a>1
\end{cases}$

Show that $\alpha$ is bijective and find $\beta: \mathbb{R} \rightarrow  \mathbb{R}$ such that $(\beta \alpha)(a) = a$ for all $a \in \mathbb{R}$.

My attempt:

If $a,b\leq1$, we have $4a-3=4b-3$, so $a=b$

If $a,b>1$, then $a^2=b^2$, so $a=b$

now if I consider the cases $a\leq1, b>1$ and $a>1,b\leq1$, I have that  $b=\sqrt{4a-3}$ and $a=\sqrt{4b-3}$, which only happens if $a,b=1$

How can I interpret this?

How can I conclude that $\alpha$ is indeed injective, surjective?

Who would $\beta$ ?

答:
The function is injective: In any case $(a,b\leq1), (a,b>1), (a\leq 1 \; \text{and}\;  b>1)$ and $(a>1 \; \text{and}\;b\leq1)$ we have that $\alpha(a)=\alpha(b)$ implies that $a=b$, so $\alpha$ is injective.

The function is surjective: If $c\in \mathbb{R}$,Then

For $a \leq1, a=\frac{c+3}{4}$

$\alpha(\frac{c+3}{4})=c$

And for $a>1, a=\sqrt{c}$

$\alpha(\sqrt{c})=c$

That is, for any $c\in \mathbb{R}$ there exists $a\in \mathbb{R}$ such that $\alpha(a)=c$. So $\alpha$ is surjective and thus, $\alpha$ is bijective.

In fact, we will use $\beta: \mathbb{R}\rightarrow \mathbb{R}$ given by

$\beta(c)=\begin{cases}
 \frac{c+3}{4}& \text{ if } c\leq 1 \\
 \sqrt{c} & \text{ if } c>1
\end{cases} $

We have that if $a\leq 1$

$ \beta(\alpha(a))=\frac{(4a-3)+3}{4}=a$

And if $a>1$

$ \beta(\alpha(a))=\sqrt{a^2}=a$


问:
Prove by double inclusion $(A \Delta B) - C = (A - C) \Delta (B - C)$
$\subseteq$ Let A, B, C be sets.
Must first prove that:
$$
(A \Delta B) - C \subseteq (A - C) \Delta (B - C)
$$

Let $x \in (A \Delta B) - C$. Note:

$$
(A \Delta B) = (B - A) \cup (A - B)
$$

Therefore, $x \in ((B - A) \cup (A - B)) - C$.

If $x \in A$, then $x \notin B$. If $x \in B$, then $x \notin A$.

I'm a bit confused as to how to continue. I know that when subtracting C, you are left with the x's that are not in C and that are not shared by A and B. However, I am confused as to how to put that into notation. Should I use separate x and y?

Also, how would I go about proving that the right side of the equation is a subset of the left side as to prove by double inclusion?

答:
Since $x \in ((B-A)\cup (A-B))-C$ we have $x \in (B-A)-C$ or $x \in (A-B)-C$. Hence $x \in B, x \notin A, x \notin C$ or $x \in A, x \notin B, x \notin C$. So $x \in B-C, x\notin A-C$ or $x \in A-C, x\notin B-C$. Which implies that $x \in (B-C)-(A-C)$ or $x \in (A-C)-(B-C)$. And we have $x \in ((B-C)-(A-C))\cup ((A-C)-(B-C))=(A-C)\Delta(B-C)$.

For the other inclusion suppose $x \in (A-C)\Delta(B-C)=((B-C)-(A-C))\cup ((A-C)-(B-C))$. Then $x \in (B-C)-(A-C)$ or $x \in (A-C)-(B-C)$. Suppose $x \in (B-C)-(A-C)$. Then $x \in B, x \notin C, x \notin A-C$. Since $x \notin C$ and $x \notin A-C$ we have that $x \notin A$. So $x \in B-A$ and $x \notin C$. Then $x \in (B-A)-C$. If $x \in (A-C)-(B-C)$ show by a similar reasoning that $x \in (A-B)-C$. So $x \in ((B-A)-C) \cup (((A-B)-C)=(B-A) \cup (A-B))-C $ which by definition is $A\Delta B-C$.


答:
In fact
\begin{eqnarray}
(A - C) \Delta (B - C)&=&(A\cap C^c)\Delta(B\cap C^c)\\
&=&[(A\cap C^c)\cap(B\cap C^c)^c]\cup [(B\cap C^c)\cap(A\cap C^c)^c]\\
&=&[(A\cap C^c)\cap(B^c\cup C)]\cup [(B\cap C^c)\cap(A^c\cup C)]\\
&=&[A\cap C^c\cap B^c]\cup[B\cap C^c \cap A^c]\\
&=&[A\cap B^c]\cup[B\cap A^c]\cap C^c \\
&=&(A\Delta B)-C.
\end{eqnarray}

Title: Prove by double inclusion $(A \Delta B) - C = (A - C) \Delta (B - C)$

Body:
$\subseteq$ Let A, B, C be sets.
Must first prove that:
$$
(A \Delta B) - C \subseteq (A - C) \Delta (B - C)
$$

Let $x \in (A \Delta B) - C$. Note:

$$
(A \Delta B) = (B - A) \cup (A - B)
$$

Therefore, $x \in ((B - A) \cup (A - B)) - C$.

If $x \in A$, then $x \notin B$. If $x \in B$, then $x \notin A$.

I'm a bit confused as to how to continue. I know that when subtracting C, you are left with the x's that are not in C and that are not shared by A and B. However, I am confused as to how to put that into notation. Should I use separate x and y?

Also, how would I go about proving that the right side of the equation is a subset of the left side as to prove by double inclusion?


问:
Is there a way to find the center point of this circle?
I am facing a problem that I have difficulty to solve.

I have a line that originates from the origin (point $(0,0)$) and has a known angle to the $x$ axis (angle $\theta$)

Somewhere between the $x$ axis and this line is a circle. This circle touches both lines, but the dimension and center of this circle are not known.

On the right side of the circle is a point $P$ that lies on the circumference of the circle and of which the coordinates are known.

How can I find the coordinates of the center of the circle?

I can't seem to add an image so I have used an external website to make the drawing.

========================================================

I now realize that I have over-simplified the problem because it is hard to explain in words. However, now I know that I can send a drawing in another way I would like to post the original problem:

You can see the drawing over here

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known.
A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines.
On the arc of the circle is a point P of which the coordinates are known.
The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?

答: A drawing of my understanding is below.  Please make sure it matches your understanding.  We know $\theta$ and the coordinates of $P$.  Because the circle is tangent to the two lines, $D$ is on the bisector of the angle.  That gives you the slope of the line from the origin through $D$, which gives a relationship between the $x$ and $y$ coordinates of $D$.  You can now write the law of cosines for the triangle $ADP$ in terms of the $x$ coordinate of $D$ and solve the equation.


问:
Why is a relative forward error enough for computations?
I am quite new to Numerical Analysis and I was wondering if someone could guide me into understanding this concept.</p>

答:
Let $T$ denote the target value for our computation and let $A$ denote the computed approximation.

Our objective should always be to produce accurate approximations $A$ as well as an upper bound for the error $E = T- A$ or a reliable error estimate $E \approx E_{\text{est}}$.

We say that $A$ approximates $T$ with at least $k$ significant figures if the relative error $$R = \frac{E}{T} = \frac{T - A}{A}$$ is bounded by $\tau = \frac{1}{2} \times 10^{-k}$, i.e., if
$$ |R| \leq \tau.$$

If we have established the relative error bound
$$|R| \leq B$$
then we do not need to know $B$ very accurately before we can determine the largest value of $k$ that can be justified.

As an example, let us consider three special cases:

Suppose that we have determined $B \approx 4 \times 10^{-7}$. The true value of $B$ must necessarily be in the interval $(3.5,4.5)\times 10^{-7}$. In all cases we can conclude $k=6$, because $B \in (0.35,0.45) \times 10^{-6}$.

Suppose that we have determined $B \approx 6 \times 10^{-7}$. The true value of $B$ must necessarily be in the interval $(5.5,6.5)\times 10^{-7}$. We can only conclude that $k \ge 5$, because $B \in (0.055,0.065) \times 10^{-5}$.

Suppose that we have determined $B \approx 5 \times 10^{-7}$. The true value of $B$ must necessarily be in the interval $(4.5,5.5)\times 10^{-7}$. As in the previous case we can only say that $k\ge 5$.


问:
Coordinate transform
Can anyone see what transformation $$r\to f(r)$$ transforms $$\exp(2\phi(r))(dr^2+r^2d\theta^2)$$ to
$$df^2+\sinh^2(f)d\theta^2$$?
I there a systematic way to attack such a problem -- rather than just hoping that  I somehow spot it?

Thank you!

答:
Well, $\phi$ can't be very creative. Since the curvature of this surface must be invariant under the change of coordinates, with a bit of computation we'll get (from the latter) $K=-1$ and (from the former) $K=-e^{-2\phi(r)}\big(\phi''(r)+\frac1r\phi'(r)\big)$. Thus, $\phi$ must satisfy the ODE
$$\phi''(r)+\frac1r\phi'(r)=e^{2\phi(r)}\,.\tag{$\star$}$$
(This should look somewhat familiar if you've computed curvature of a conformally flat metric before.)
It also makes us think we're looking at some hyperbolic metric, of course.

Starting at the other end of the problem, you need
\begin{align*} f'(r) &= e^{\phi(r)} \\
\sinh f(r) &= re^{\phi(r)}\,,
\end{align*}
and we can actually integrate patiently to obtain
$$\cosh f(r) = \frac{Cr^2+1}{Cr^2-1}\,,$$
so $$f(r) = \log\left(\frac{(Cr+1)^2}{Cr^2-1}\right)\,.$$
Amazingly, if we now set $\phi(r) = \log f'(r)$, we do in fact (!) get a solution of the ODE ($\star$) above.

Title: Coordinate transform


问:
Interpreting $e^x+e^y<e^z$
Suppose that $(x,y,z)\in[0,1]^3$ for simplicity. It is easy for me to interpret "$x+y<z$": this just means "$z$ is larger than the sum of x and y." Obviously "$e^{x+y}<e^z$" has the same interpretation. I'm having trouble interpreting "$e^x+e^y<e^z.$" Any suggestions for an intuitive interpretation of this inequality would be greatly appreciated!

An observation:
Plotting $\left\{(x,y,z)\in[0,1]^3:e^x+e^y<e^z\right\}$ in MATLAB seems to suggest that in order to be in this set, $z$ must be larger than $x+y$ plus some additional buffer. Intuition behind why this happens?

答:
the whole thing with equality is a surface. Here is a picture of level curves of $e^x + e^y$

Note that the level curves are identical, take the one that goes through the origin and shift  by  vector $(t, t).$  In the picture use  $(\frac{1}{2}, \frac{1}{2}).$

Here it is with some diagonal lines drawn. If you tilt your head $45^\circ$ so that the diagonal lines appear to be horizontal, the idea that the curves are identical becomes more believable. I do not know how to rotate a jpeg or png by $45^\circ$

Title: Interpreting $e^x+e^y<e^z$


问:
Show that: $F$ is a ring with addition and multiplication given by matrix multiplication in $Mat(2;\mathbb{F_5})$
Let $$F:=\left \{ \begin{pmatrix}
a & b\\
2b & a
\end{pmatrix}\left. \right | a,b \in \mathbb{F}_5 \right \}$$
Show that: $F$ is a ring with addition and multiplication given by matrix multiplication in $Mat(2;\mathbb{F_5})$.

This question seems relatively simple, but I'm quite new to fields and rings and there are some things regarding notation and terminology that I'm unsure about.

In the equation for $F$, the $:=$ symbol is used, would I be correct to assume that it has no real special meaning and is equivalent to the $=$ symbol?

Since all fields are rings, it seems strange that the questions would ask to show that $F$ is a ring.

As far as I understand, $Mat(2;\mathbb{F}_5)$ is the map of all $2 \times 2$ matrices to the field $\mathbb{F}_5$, so is the question asking me to take $x,y,z \in F$ and show that they satisfy the ring axioms? This is the part where I'm really struggling to understand what the question is asking.

答: For the first part, typically the := means “is defined as”. That notation just represents that it is being defined this way. For the second part, are you assuming that it is a field because it is labeled F? You would definitely need to show it’s a ring before it’s a field which I’m sure that you know. For the third part, I interpret $M(2;\mathbb{F}_5)$ as the space of $2\times2$ matrices with coefficients in the finite field with 5 elements. Addition and Multiplication are defined by matrix addition and multiplication but you need to be careful since you’re working in $\mathbb{Z}/5\mathbb{Z}$. So really the last part is just asking you to show that matrices of the given form form an abelian group with standard matrix addition and satisfy the ring multiplication axioms with standard matrix multiplication.


问:
existence of a derivative $x\cdot f(x)$
Lets say we have a function $f(x)$ that has a derivative at point $a$.
Can we prove that the function $x\cdot f(x)$ has also a derivative at point $a$?
If this is not true, can anybody give an example that shows this ...</p>

答:
To avoid using anything but the definition:

Let $g(x)=xf(x)$
$$\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h\to 0} \frac{(x+h)f(x+h)-xf(x)}{h} = \lim_{h\to 0} \frac{xf(x+h)-xf(x)+hf(x+h)}{h} = x\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} +\lim_{h\to 0}f(x+h) = xf^\prime(x)+f(x)$$

Title: existence of a derivative $x\cdot f(x)$


答:
actually if you focus on the limit
$$\lim_{n \to \infty}\displaystyle e^{\displaystyle-nc} -(1-c)^n$$
where c is a real number between $0$ and $1$then if $c \neq 0$ then the exponential term goes to zero and the other bracket also and if $c = 0$ the value of the limit is zero so this can be proved directly without evaluating the maximum

Title: Show $\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n) = 0$?

Body:
I am trying to show that
$$\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n)=0.$$
I have tried expanding $e^{-nx}$ with its Taylor series and $(1-x)^n$ with Newton's binomial theorem. The resulting Taylor series is $\sum_{k=2}^{\infty}  (-1)^k \left(\frac{n^k}{k!} - {n \choose k}\right)x^k.$

I also tried taking the first derivative with respect to $x$ to find the point in $(0,1)$ where $e^{-nx} - (1-x)^n$ is maximized (in $x$ as a function of $n$), but I was only able to find the hard-to-solve $\frac{n}{1-n} = \frac{\ln(1-x)}{x}.$


答:
$\alpha(1) = 1.$

Take any $c < 1.$

Set $a = \frac{c + 3}{4} \implies a < 1 \implies$

$\alpha(a) = \left(4 \times \frac{c + 3}{4}\right) - 3 = c.$

Take any $c > 1.$

Set $a = \sqrt{c} \implies a > 1 \implies \alpha(a) = a^2 = c.$

Thus, for all values of $c$, there exists a value for $a$ such that
$\alpha(a) = c.$

Therefore $\alpha$ is a surjective function.

If $a < 1,~$ then $~\alpha(a) < 1. \tag1$

If $a > 1,~$ then $~\alpha(a) > 1. \tag2$

Therefore, the only value for $a$ such that $\alpha(a) = 1$ is $a = 1. \tag3$

To show that $\alpha$ is an injective function, I need to show that
$[\alpha(a) = \alpha(b)] \implies [a = b].$

$\underline{\text{case 1}}$

$\alpha(a) = \alpha(b) = 1.$

By (3), $a = 1 = b.$

$\underline{\text{case 2}}$

$\alpha(a) = \alpha(b) < 1.$

By (1) and (2), $a < 1$ and $b < 1.$
Therefore, $4a - 3 = 4b - 3 \implies a = b.$

$\underline{\text{case 3}}$

$\alpha(a) = \alpha(b) > 1.$

By (1) and (2), $a > 1$ and $b > 1.$
Therefore, $\sqrt{a} = \sqrt{b} \implies a = b.$

Thus in all three cases, $[\alpha(a) = \alpha(b)]
\implies [a = b].$

Therefore, $\alpha$ is an injective function.

Body:
Define $\alpha: \mathbb{R} \rightarrow  \mathbb{R}$ via

$\alpha (a)=\begin{cases}
 4a-3& \text{ if } a\leq 1 \\
 a^2 & \text{ if } a>1
\end{cases}$

Show that $\alpha$ is bijective and find $\beta: \mathbb{R} \rightarrow  \mathbb{R}$ such that $(\beta \alpha)(a) = a$ for all $a \in \mathbb{R}$.

My attempt:

If $a,b\leq1$, we have $4a-3=4b-3$, so $a=b$

If $a,b>1$, then $a^2=b^2$, so $a=b$

now if I consider the cases $a\leq1, b>1$ and $a>1,b\leq1$, I have that  $b=\sqrt{4a-3}$ and $a=\sqrt{4b-3}$, which only happens if $a,b=1$

How can I interpret this?

How can I conclude that $\alpha$ is indeed injective, surjective?

Who would $\beta$ ?


问:
Partial Sum of $\sum^{n-1}_{i=0} \frac{n-i-1}{n-i}$
I am looking to find the partial sum of the following sum for an algorithm I am analyzing: $\sum^{n-1}_{i=0} \frac{n-i-1}{n-i}$.

I can see it can be simplified to:
$\sum^{n-1}_{i=0} \frac{n-i-1}{n-i}$  $= \sum^{n-1}_{i=0} (\frac{n-i}{n-i}-\frac{1}{n-i})$ $= \sum^{n-1}_{i=0} (1-\frac{1}{n-i})$

But the $\frac{1}{n-i}$ looks like the harmonic series to me. Which doesn't have a closed form. Have I done my analysis of the summation right or am I missing something?

答:
You are correct. We have:

$$\sum_{i=0}^{n-1}\frac{n-i-1}{n-i}$$

$$\sum_{i=0}^{n-1}(1-\frac{1}{n-i})$$

$$n - \sum_{i=0}^{n-1}\frac{1}{n-i}$$

$$n-\sum_{k=1}^{n}\frac{1}{k}$$

Here, we make the substitution $k = n-i$. Thus:

$$\sum_{i=0}^{n-1}\frac{n-i-1}{n-i} = \boxed{n-H_{n}}\approx n - \ln(n) - \gamma$$

The last approximation is $H_{n}\approx \ln(n) + \gamma$, where the Euler-Mascheroni Constant $\gamma \approx 0.577$. This approximation is very good for larger values of $n$.


问:
Could anyone prove this property of orthonormal basis?
Suppose $a_{1},a_{2},\cdots,a_{n}$ form an orthonormal basis of $\mathbb{R}^{n}$. Want to prove

$$\sum_{i}^{n}\sum_{j,k}a_{ij}a_{ik}=n$$

After some numerical experiments, this is true. Also, this is a sufficient condition for another theorem.

答: An orthonormal basis can be written as the image of an orthogonal matrix by writing the vectors as column vectors in a single matrix. Your inner expression is this matrix times its transpose. But for an orthogonal matrix $A$, we have $A A^T = 1$, i.e. $A^T = A^{-1}$. Then the latter sum is just taking the trace. The trace of the identity matrix is $n$, so this is your numerical observation.


问:
Countable comeager set in an (uncountable) non perfect polish space
Is it possible to have an uncountable and non perfect Polish space with a countable comeager set? Furthermore, is it possible for this space to have a comeager collection of isolated points?

I've been banging my head against the wall to construct such a space for a while now, but coming up with nothing. Maybe I'm missing something obvious.

答: Let $C$ be the usual Cantor set, obtained from the interval $[0,1]$ by repeatedly deleting middle thirds. Let $M$ be the set of midpoints of the deleted intervals.  Then $C\cup M$ is an uncountable, compact subset of $[0,1]$, and the countable set $M$ of isolated points is open and dense in it, hence comeager.


问:
Show $\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n) = 0$?
I am trying to show that
$$\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n)=0.$$
I have tried expanding $e^{-nx}$ with its Taylor series and $(1-x)^n$ with Newton's binomial theorem. The resulting Taylor series is $\sum_{k=2}^{\infty}  (-1)^k \left(\frac{n^k}{k!} - {n \choose k}\right)x^k.$

I also tried taking the first derivative with respect to $x$ to find the point in $(0,1)$ where $e^{-nx} - (1-x)^n$ is maximized (in $x$ as a function of $n$), but I was only able to find the hard-to-solve $\frac{n}{1-n} = \frac{\ln(1-x)}{x}.$

Body:
Let
$f_n(x)
=e^{-nx} - (1-x)^n
$.

I will show that
for $0 < x < 1$,
$f_n(x) \ge 0$
and
if $x > \dfrac1{\sqrt{n}}$
then
$f_n(x)
\lt e^{-\sqrt{n}}
$.
and if
$0 < x < c_0
\approx 0.683803
$
(the root of
$\ln(1-x)+x+x^2
=0$)
then
$f_n(x) < x
$.

Therefore
$0 < f_n(x)
\le\dfrac1{\sqrt{n}}
$.

For $0 < x < 1$,
$e^x < \dfrac1{1-x}
$
(compare the power series)
so
$e^{-x} > 1-x
$
so
$e^{-nx} > (1-x)^n
$.

Therefore
$f_n(x) > 0$.

If $x > \dfrac1{\sqrt{n}}$ then
$e^{-nx}
\lt e^{-\sqrt{n}}
$
so
$f_n(x)
\lt e^{-\sqrt{n}}
$.

$\begin{array}\\
f_n(x)
&=e^{-nx}-(1-x)^n\\
&=e^{-nx}-e^{n\ln(1-x)}\\
&=e^{-nx}(1-e^{nx+n\ln(1-x)})\\
&<e^{-nx}(1-e^{nx+n(-x-x^2)})
\qquad 0 < x < c_0\ (*)\\
&=e^{-nx}(1-e^{-nx^2})\\
&<e^{-nx}(1-(1-nx^2))\\
&=nx^2e^{-nx}\\
&\le x
\qquad\text{since } z e^{-z} \le 1\\
\end{array}
$

$(*)
\ \ln(1-x)
\gt -x-x^2
$
for
$0 < x < c_0
\approx 0.683803
$.


问:
Find values for which the integral converges
Find all the values of $\alpha$ and $\beta$ for which the following integral converges:

$$ \int_{0}^{\infty} \bigg(|1-x|^{-\alpha} - (1+x)^{-\alpha}\bigg)x^{-\beta}dx $$

Of course integral converges when $0 < \alpha <1$ and $1 - \alpha < \beta < 1$. Integral also converges when (for example) $\beta = 1.5$ which I checked in Wolfram Alpha.

I would appreciate any tips or hints.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}\pars{\verts{1 - x}^{-\alpha} - \pars{1 + x}^{-\alpha}}x^{-\beta}\,\dd x
\label{1}\tag{1}
\\[5mm] = &\
\int_{0}^{1}\bracks{\pars{1 - x}^{-\alpha} - \pars{1 + x}^{-\alpha}}x^{-\beta}\,\dd x
\label{2}\tag{2}
\\[2mm] + &
\int_{1}^{\infty}\bracks{\pars{x - 1}^{-\alpha} - \pars{1 + x}^{-\alpha}}x^{-\beta}\,\dd x
\label{3}\tag{3}
\end{align}

When $\ds{x \sim 1}$ $\ds{\pars{~\mbox{see} (\ref{2})\ \mbox{and}\ (\ref{3})~}}$, the convergence is enforced by
\begin{equation}
\Re\pars{-\alpha} > -1 \implies \bbx{\Re\pars{\alpha} < 1} \label{4}\tag{4}\\
\end{equation}

When $\ds{x \gtrsim 0}$, the integrand is $\ds{\propto x^{1 - \beta}.\,\,\,}$ See (\ref{1}).
$$
\mbox{Then,}\quad
\Re\pars{1 - \beta} > - 1 \implies \bbx{\Re\pars{\beta} < 2} \\
$$

When $\ds{x \to \infty}$, the integrand is $\ds{\propto x^{-\alpha -1 - \beta}\,.\,\,}$
See (\ref{3}). Then, $\ds{\Re\pars{-\alpha - \beta} < 0 \implies \bbx{\Re\pars{\beta} > -\Re\pars{\alpha}}}$

Summarizing,
$\ds{-\Re\pars{\alpha} < \Re\pars{\beta} < 2}$
which obviously yields $\ds{\Re\pars{\alpha} > -2}$.

With (\ref{4}):
$$
\left\{\begin{array}{rcccl}
\ds{-\Re\pars{\alpha}} & \ds{<} & \ds{\Re\pars{\beta}} & \ds{<} & \ds{2}
\\[2mm]
\ds{-2} & \ds{<} & \ds{\Re\pars{\alpha}} & \ds{<} & \ds{1}
\end{array}\right.
$$


答: For each $i$, {$a_{ik}$} is an orthonormal basis, so by definition $\sum _{jk}a_{ij}a_{ik}=1$.  Therefore the sum of $n$ such bases $=n$.

Title: Could anyone prove this property of orthonormal basis?

Body:
Suppose $a_{1},a_{2},\cdots,a_{n}$ form an orthonormal basis of $\mathbb{R}^{n}$. Want to prove

$$\sum_{i}^{n}\sum_{j,k}a_{ij}a_{ik}=n$$

After some numerical experiments, this is true. Also, this is a sufficient condition for another theorem.


问:
Show that $\sum_{i=1}^k|a_ib_i|c_i\leq \left(\sum_{i=1}^k|a_i|^pc_i\right)^{1/p}\left(\sum_{i=1}^k|b_i|^qc_i\right)^{1/q}$, where $c_1,...c_k >0$
I know that this immediately follows from Holder's inequality if all the $c_i=1$, and I take $\mu$ as the lebesgue measure and define $f(x)=\sum a_i I_{[i-1,i)}(x)$, $g(x)=\sum b_i I_{[i-1,i)}(x)$. But I'm not sure how to handle the $c_i$'s.

Is there perhaps some way to show that $$\int|fg|h\leq\left(\int|f|^ph\right)^{1/p}\left(\int|g|^qh\right)^{1/q}$$

if $h>0$, and $f\in L^p,g\in L^q$?

Maybe some well known identity/inequality that would help push the initial problem further, or perhaps some intuition that would help me think of it problem more clearly?

Thanks in advance.

答:
For $p>0$, $q>0$ such that $\frac{1}{p}+\frac{1}{q}=1$ by Holder we obtain:
$$ \left(\sum_{i=1}^k|a_i|^pc_i\right)^{1/p}\left(\sum_{i=1}^k|b_i|^qc_i\right)^{1/q}\geq\left(\sum_{i=1}^k\left(|a_i|\cdot c_i^{\frac{1}{p}}\cdot|b_i|\cdot c_i^{\frac{1}{q}}\right)^{\frac{1}{\frac{1}{p}+\frac{1}{q}}}\right)^{\frac{1}{p}+\frac{1}{q}}=\sum_{i=1}^k|a_ib_i|.$$
Actually. The Holder's inequality it's the following.

let $a_i$, $b_i$, $\alpha$ and $\beta$ are positive numbers, $n\geq1$ be an integer number. Prove that:
$$\left(\sum_{i=1}^na_i\right)^{\alpha}\left(\sum_{i=1}^nb_i\right)^{\beta}\geq\left(\sum_{i=1}^n\left(a_i^{\alpha}b_i^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$


问:
Computing last two digits of $27^{2018}$
For abstract algebra I have to find the last two digits of $27^{2018}$, without the use of a calculator, and as a hint it says you should work in $\mathbb{Z}/100\mathbb{Z}$.

I thought breaking up the problem into $\mod(100)$ arguments. Thus:
$27^{2}=729\equiv 29 \mod (100)$, and

$27^{4}=(27^{2})^{2} \equiv 29^{2}=861\equiv 61 \mod (100)$ and

$27^{8}=(27^{4})^{2} \equiv 61^{2}=3421 \equiv 21 \mod(100)$ and so on until I would find something that repeated itself. But I've done quite some terms no and I've not seen any iteration yet. So I'm thinking this is the wrong way.

Any suggestions?

答:
Working in the multiplicative group of integers modulo $100$, $(\Bbb Z /100 \Bbb Z)^\times$, you can discover all that is needed to solve the problem with elementary algebra.

If $n$ is an integer denote by $[n]$ its $\text{modulo-}100$ residue class in  ${\displaystyle \mathbb {Z} /100 \mathbb {Z}}$.

Exercise 1: The subset $\mathcal V = \bigr\{[20n+d]\mid n \in \Bbb Z \land d \in \{1,3,7,9\}\bigr\}$ of ${\displaystyle \mathbb {Z} /100 \mathbb {Z}}$ is actually a subset of $(\Bbb Z /100 \Bbb Z)^\times$, forming a subgroup with $20$ elements.

Since $[27] \in \mathcal V$, $\,27^{20} \equiv 1 \pmod{100}$.

To solve $27x \equiv 1 \pmod{100}$, observe that $x$ has the form $20n +3$. Multiplying,

$\tag 1 (20 + 7)(20n+3) \equiv 60 + 40n + 21 \equiv 81 + 40n \pmod{100}$

Examining $\text{(1)}$ we determine that $x \equiv 63 \pmod{100}$.

So

$\; 27^{2018} \equiv 27^{2020} \cdot 63^{2} \equiv 63 \cdot 63 \equiv 80 + 80 + 9 \equiv 69 \pmod{100}$.


问:
Finding the Expected Value of T=($\sum X_i^2$, $\sum X_i$)
Suppose we have $X_1,…, X_n$ as a sample of $iid$ $N(0, )$ random variables with $ \in (0, ∞)$. And I'm being asked to determine the following:

a) A sufficient statistic $T$ for $$.

b) $E(T)$

c) An unbiased estimator of $$ that is a function of $T$.

$\textbf{My Solution:}$

Suppose $s = x_1, x_2,....,x_n$

For part (a) I got $T = (T_1(s), T_2(s)) = (\sum_{i=1}^{n}x_i^2, \sum_{i=1}^{n}x_i)$ but I'm not sure if that is correct. Can someone confirm if this is correct or not?

Assuming it is correct, how do we start part(b)? I know how to do in the case of single variable (e.g. if $T = \sum_{i=1}^{n}x_i^2$ only), but T is multi-variable in this case. I don't know how to solve that. Please help.

答:
The given density is not a general Gaussian Family but it is a particular one of Gaussian, Centered in ZERO. Thus the solution is very simple

(a) A sufficient statistic is $T=\sum_{i=1}^{n}X_i^2$

In fact the likelihood is the following

$$L(\theta)=\Bigg(\frac{1}{2\pi\theta}\Bigg)^{\frac{n}{2}}e^{-\frac{1}{2\theta}\sum_i X_i^2}$$

(b)

$$\mathbb{E}[T]=\mathbb{E}\Bigg[\sum_{i=1}^{n}X_i^2\Bigg]=n\mathbb{E}[X_1^2]=n\theta$$

(c)

$$T^*=\frac{T}{n}$$

...and you are done!


问:
How to solve a system of three linear equations?
When presented with $3$ equations, to solve for $a$, $b$ and $c$ for me becomes a guessing game. If I have just two equations for example $3a + 2b = 5$ and $6a - 7b = 8$, you could easily solve for $a$ and $b$ through elimination or substitution. However, when I get three equations, e.g.,

$$\begin{aligned} 4a + 2b + c &= -3 \\ a + b + c &= -1 \\ a + b + c &= -1 \end{aligned}$$

it becomes a guessing game.

I usually start with elimination for a system of three equations but my biggest problem is substituting or elimination to get the same equation hence going around in circles. I then do blind/random substitutions to try and eliminate one variable, but I have no idea what I am doing or if it will work although I do get there in the end. Is this normal? Is there a set of steps or method which will be able to solve 3 for three unknowns without guessing randomly?

答:
As Will Jagy comments, Gaussian Elimination is the way to go, especially to account for situations where there are no or infinitely many solutions.

In the scenario where there is a unique solution, it can be useful to think geometrically. The solutions of a linear equation in three variables form a plans in 3D space, and as long as two equations do not define parallel planes, the solution of two equations here is a line.

What we can do is find the equation of the line of intersection of the first two planes, then of the second and third planes, then find the intersection of these two lines, the intersection of the planes.

In practise, solve all three equations for one of the variables in terms of the other two.

Say
$$\begin{aligned}
c&=f_1(a,b)
\\ c&= f_2(a,b)
\\ c&=f_3(a,b)
\end{aligned}$$

Set $f_1=f_2$ for the first line (aka equation in $a$ and $b$) and $f_2=f_3$ for the second.

Then find the intersection of these two lines in terms of $a$ and $b$ and substitute into (any) one of the $f_i$ to find the third coordinate of intersection.

Draw some pictures to understand.


答:
If you don't know the prodouct rule (yet), you can calculate explitly for $g(x)=xf(x)$
$$\begin{align}g'(a)&=\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}\\&=\lim_{h\to 0}\frac{(a+h)f(a+h)-af(a)}{h}\\&=\lim_{h\to 0}\frac{af(a+h)-af(a)+hf(a+h)}{h}\\&=\lim_{h\to 0}\left(a\cdot\frac{f(a+h)-f(a)}{h}+f(a+h)\right)\\
&=a\lim_{h\to0}\frac{f(a+h)-f(a)}{h}+\lim_{h\to0}f(a+h)\\&=af'(a)+f(a).\end{align}$$
Make yourself clear that each step is justified.</p>

Title: existence of a derivative $x\cdot f(x)$

Body:
Lets say we have a function $f(x)$ that has a derivative at point $a$.
Can we prove that the function $x\cdot f(x)$ has also a derivative at point $a$?
If this is not true, can anybody give an example that shows this ...</p>


问:
Proof verification: ideal of $\mathbb{C}[x, y]$ generated by x, y is not a free module
Let $R$ be $\mathbb{C}[x, y]$ and let $M$ be the ideal of $R$ generated by the two elements $x$ and $y$. Prove or disprove: $M$ is a free R-module.

This problem comes from Artin First Edition. It is problem 12.2.1. I am working on some problems near it 12.2.(4-7) as part of an informal study group, but this problem itself is excluded from the problem set.

This problem is similar to this one, but is not a strict duplicate because the answer to that question assumes a result that is proven later on in the problems. I wrote the solution to this problem before reading that answer.

Is my attempted solution to this problem correct? Is there a way to make the same argument that's simpler without appealing to more general results about $K[x, y]$?

I think this the statement is false and here is my attempt to prove it.

First a word on notation:

I'm going to use the following highly nonstandard notation.

Let $\mathop{\mu} \begin{bmatrix} S \cr R \end{bmatrix} $ denote the free R-module on S. Equivalently, $\mathop{\mu} \begin{bmatrix} S \cr R \end{bmatrix} $ is the set of formal linear combinations of $S$ with coefficients taken from $R$ and the symbols $\{0, +, -, *, =\}$ defined in appropriate way for formal linear combinations. It is a priori a free module.

Let $\mathop{\mu^e} \begin{bmatrix} S \cr R \end{bmatrix} $ denote the module derived from the free R-module on S by evaluating the formal linear combinations and identifying elements that are equal.

In prose, $X$ contains $Y$ means $X \in Y$. $X$ contains $Y$ by inclusion means $X \subset Y$.

Lemma: No element of $\mathop{\mu^e} \begin{bmatrix} \{x, y\} \cr \mathbb{C}[x, y] \end{bmatrix}$ has a non-zero constant term.

Proof: For any formal linear combination of elements of $\{x, y\}$ with coefficients in $\mathbb{C}[x, y]$, each of the terms has a zero constant subterm. Therefore, there is no way of producing an element of $\mathbb{C}[x, y]$ with a non-zero constant term in $\mathop{\mu ^ e}\begin{bmatrix} \{x, y\} \cr \mathbb{C}[x,y] \end{bmatrix}$.

We are asking ourselves, is $\mathop{\mu ^ e}\begin{bmatrix} \{x, y\} \cr \mathbb{C}[x,y] \end{bmatrix}$ free?

Equivalently, does there exist some subset $S$ of $\mathbb{C}[x,y]$ such that

$$ \mathop{\mu ^ e}\begin{bmatrix} \{x, y\} \cr \mathbb{C}[x,y] \end{bmatrix} \;\;\text{is isomorphic to}\;\; \mathop{\mu}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix} $$

My claim is no.

Suppose $0$ is an element of $S$, then consider $S'$ with $0$ removed.

Suppose $S$ has exactly two elements, $\alpha$ and $\beta$.

$ \mathop{\mu^e}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix} $ is not free because $0$ has a nontrivial representation before the evaluation step is applied.

$$ (-\beta) \cdot \alpha + (\alpha) \cdot \beta = 0 $$

Since $0$ has a nontrivial representation, that means that $ \mathop{\mu}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix} $ has stray equalities when evaluation is considered.

Suppose $S$ has more than two elements. Pick any two elements and apply the argument for the 2-element case.

Suppose $S$ has one element. Let $\cup S$ denote the sole element of $S$, borrowing notation from set theory.

$\cup S$ cannot be zero. $\cup S$ also cannot be a non-zero constant. If $\cup S$ were a non-zero constant, then we would be able to generate $2$ as an element of $\mathop{\mu^e}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix}$, which contradicts the lemma.

$\cup S$ cannot have degree 2 or greater because then it would be unable to generate $x$.

$\cup S$ cannot have degree 1 because it must be able to generate both $x$ and $y$ and they have no common factors of degree 1.

$S$ cannot be the empty set because then the resulting module would not contain $x$.

答:
Since you show yourself quite open to simpler alternative approaches, may I try and present you with one, the inspiration for which comes from one of the exercises of the "Algèbre" of Bourbaki.

Consider a nonzero commutative ring $A$, the two indeterminate polynomial ring $B=A[X, Y]$ and the ideal $I\colon=BX+BY$. We aim to show that the $B$-module $I$ is not free, and we first claim that in order to prove this it will suffice to establish that in the tensor product (square) $E\colon=I \otimes_B I$ the element $X \otimes Y-Y \otimes X \neq 0_E$ is not null, however $XY(X \otimes Y-Y \otimes X)=(YX)\otimes(XY)-(XY)\otimes(YX)=0_E$.

Indeed, having established the above claim regarding the mentioned element of the tensor product let us assume by contradiction that $I$ were free, in other words $I \approx B^{(T)} \hspace{3pt} (\operatorname{\mathit{B}-\mathbf{Mod}})$ for a certain set $T$. Since $B$ is commutative, it would follow that the tensor product $E=I \otimes_B I \approx B^{(T)} \otimes_B B^{(T)} \approx B^{(T \times T)} \hspace{3pt} (\operatorname{\mathit{B}-\mathbf{Mod}})$ is also free. In general, for an arbitrary ring $C$ and a free (more generally, projective) left $C$-module $L$, it can be easily shown that given element $x \in L$ and operator $\lambda \in C$ which is not a left zero-divisor (this means that $\lambda$ is left simplifiable, or that $\lambda \mu=0_C \Rightarrow \mu=0_C$ for any $\mu \in C$), the condition $\lambda \in \mathrm{Ann}_C(x)$ entails $x=0_L$. In our particular instance, we have that $XY \in \mathrm{Ann}_B(X \otimes Y-Y \otimes X)$ and $XY$ is indeed a simplifiable element of $B$, however $X \otimes Y-Y \otimes X \neq 0_E$, which constitutes a contradiction.

Now on to justifying the claim concerning the tensor product. Consider the unique surjective unitary $A$-algebra morphism $\sigma \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Alg_1}}}(B, A)$ given by $\sigma(X)=\sigma(Y)=0_A$ -- this is nothing else than the morphism of substitution, taking any polynomial $f \in B$ to $f(0_A, 0_A)$ -- and consider ${}_{[\sigma]}A$, the $B$-module structure on $A$ obtained by scalar restriction via $\sigma$ from the canonical $A$-module structure on $A$. Explicitly, we are considering the $B$-module structure on $A$ for which $f.\lambda=f(0_A, 0_A)\lambda$ for any $f \in B$ and $\lambda \in A$.

Further consider the $B$-module $B^2$, with the canonical basis constituted by $e\colon=(1_A, 0_A)$ and $e'\colon=(0_A, 1_A)$ (we keep in mind that $0_B=0_A$ and $1_B=1_A$, since by construction $A$ is a subring of any ring of polynomials over it). We go on to define the unique $B$-bilinear map $\varphi \colon B^2 \times B^2 \to A$ such that $\varphi(e,e)=\varphi(e,e')=\varphi(e',e')=0_A$ and $\varphi(e',e)=1_A$. The general behaviour of this map is described by $\varphi\left((f,g), \left(f', g'\right)\right)=\left(gf'\right).1_A=\left(gf'\right)(0_A, 0_A)$.

Consider the $B$-module surjection $\rho \in \mathrm{Hom}_{\operatorname{\mathit{B}-\mathbf{Mod}}}\left(B^2, I\right)$ given by $\rho(e)=X$ and $\rho(e')=Y$. As Daniel Schepler pointed out in his above comment, it is easy to show that $\mathrm{Ker}\rho=B(Y, -X)$. We wish to remark here that the above bilinear map $\varphi$ is compatible with the surjection $\rho \times \rho \colon B^2 \times B^2 \to I \times I$, in the precise sense that $\rho(u)=\rho(u')$ and $\rho(v)=\rho(v')$ entail $\varphi(u, v)=\varphi(u',v')$ for any elements $u, u', v, v' \in B^2$. This is easily ascertained by remarking that $\varphi((Y, -X), u)=\varphi(u, (Y, -X))=0_A$ for any $u \in B^2$, relation which in turn stems from the fact that $X, Y \in \mathrm{Ann}_B\left({}_{[\sigma]}A\right)$. By virtue of a fundamental proposition in elementary set theory, the map $\varphi$ induces a unique quotient map $\psi \colon I \times I \to A$ such that $\varphi=\psi \circ (\rho \times \rho)$. This relation allows us to immediately ascertain the fact that $\psi$ is also $B$-bilinear and that $\psi(X, Y)=0_A$ respectively $\psi(Y, X)=1_A$.

By virtue of the universal property of tensor products, $\psi$ induces a unique $B$-linear map $\psi' \in \mathrm{Hom}_{\operatorname{\mathit{B}-\mathbf{Mod}}}(E, A)$ such that $\psi'(f \otimes g)=\psi(f, g)$ for any $f, g \in I$. It follows that $\psi'(X \otimes Y-Y \otimes X)=-1_A \neq 0_A$, whence necessarily $X \otimes Y-Y \otimes X \neq 0_E$.

The same argument can be used to show that $I$ is neither projective nor flat as a $B$-module (both these statements being stronger than the claim that $I$ is not free).

Before concluding, I would like to remark that the answer you included a link of in your posting is by far the most simple and efficient in the particular case when $A$ is a domain. The fact that $(X, Y)$ is not principal in $A[X, Y]$ can be immediately established.


问:
Is there a better way to show $a^3 -4ab +8c = 0$ for this polynomial
$x^3 +ax^2 +bx +c = 0 $ has roots $\alpha$, $\beta$ and  $\gamma$, where $\gamma=\alpha+\beta $. Show that $a^3 -4ab +8c = 0$ . I have done it through relations of roots ($\alpha$ + $\beta$ + $\gamma$ = $\frac{-b}{a} = -a$ and etc.) and substituing $\gamma$ into every single equation, then substituing them into what is required to be shown (sub in $ a, b$ and $c$). Is there a faster method to show this, maybe a smarter substitution?

Body:
Maybe so:
$$a^3-4ab+8c=-8(\alpha+\beta)^3-4(-2\alpha-2\beta)\cdot(\alpha\beta+(\alpha+\beta)^2)-8\alpha\beta(\alpha+\beta)=0.$$


问:
$a^n-1$ and $b^n-1$ have the same set of prime factors for each $n\in\Bbb{Z}^+$, show that $a=b$.
Let $a,b$ be two positive integers, if $a^n-1$ and $b^n-1$ have the same set of prime factors for each $n\in\Bbb{Z}^+$, then $a=b$.

It seems to be not hard, but I have no idea. Any hint is welcome.

Edit: As the comment points out, it is probably not easy. I thought it is easy because it is kind of intuitively correct. I tried to study the Zsigomondy primes as $n$ increases, but it is not successful.

答:
The condition that $a^n-1$ and $b^n-1$ have the same prime factors for every $n$ is equivalent to $a$ and $b$ having the same order mod $p$ for every $p$. Equivalently, for every positive integer $k$ and every prime $p$, $a$ is a $k$-th power mod $p$ if and only if $b$ is a $k$-th power mod $p$. I will argue that for fixed $a\neq b$ and every sufficiently large prime value of $k$, there is a positive proportion of primes $p$ for which $a$ is a $k$-th power mod $p$ but $b$ is not.

Suppose $2\leq a< b$. Let $q>b$ be a prime, and consider the field $K=\mathbb{Q}(\sqrt[q]{a},\sqrt[q]{b},\zeta_q)$, where $\zeta_q$ is a primitive $q$-th root of unity. Then
$$
G:=\operatorname{Gal}(L/\mathbb{Q})\cong \mathbb{F}_q^\times \ltimes\left(\mathbb{F}_q\times\mathbb{F}_q\right).
$$
Then $\mathbb{Q}(\sqrt[q]{a})$ is the fixed field of $H_1:=\mathbb{F}_q^\times \ltimes\left(\{0\}\times\mathbb{F}_q\right)$, and $\mathbb{Q}(\sqrt[q]{b})$ is the fixed field of $H_2:=\mathbb{F}_q^\times \ltimes\left(\mathbb{F}_q\times \{0\}\right)$. If $p$ is a rational prime coprime to $a$, $b$, and $q$, then $a$ is a $q$-th power mod $p$ if and only if the Frobenius conjugacy class of $p$ in $G$ intersects $H_1$, and $b$ is a $q$-th power mod $p$ if and only if the Frobenius conjugacy class of $p$ in $G$ intersects $H_2$. Now, by the Chebotarev density theory, there exist a positive proportion of primes $p$ whose Frobenius conjugacy class contains $(1,0,1)\in G$. Then the conjugacy class intersects $H_1$ but not $H_2$, so for such $p$, $a$ is a $q$-th power mod $p$ but $b$ is not.


问:
Algebraic geometry: an exercise
Let X be the Riemann sphere with local coordinate $z$ in one chart and $w=1/z$ in the other chart. Let $\omega$ be a meromorphic 1-form on X. Show that if $\omega=f(z)dz$ in the coordinate z, then f must be a rational function of z.

EDIT: I'm sorry but I didn't know the rules of this site and I'm sorry for my awful English. However, I have to solve the following exercise: "Miranda - Algebraic Curves and Riemann Surfaces" pag. 111 ex. IV.1 A, that's to say

Let X be the Riemann sphere with local coordinate $z$ in one chart and $w=1/z$ in the other chart. Let $\omega$ be a meromorphic 1-form on X. Show that if $\omega=f(z)dz$ in the coordinate $z$, then f must be a rational function of z.

Now, my idea is: let $\omega=\{\omega_{i}\}_{i=1,2}$ such that $\omega_{1}=f(z)dz$ and $\omega_{2}=g(w)dw$ where $z$ and $w$ are the local coordinate with respect to the chart $(U_1,h_1)$ and $(U_2,h_2)$ [where $U_1=\mathbb{C}$ and $U_2=\mathbb{C}_{\infty} \setminus\{0\}$]. We know that $\omega_1$ transforms under $h_2 h_1 ^{-1}$ into $\omega_2$, therefore $g(w)=f(1/w)(-1/w^{2})$ is a meromorphic function.

But now I can't conclude the exercise. Can you help me, please, using Miranda's notations?

答:
In the context of the book, the answer of Jyrki Lahtonen is kind of reinventing the wheel.

For $\omega$ meromorphic,  we have for $\omega|_{\mathbb C} = f(z) dz$ and $\omega|_{\mathbb C_{\infty} \setminus 0} = g(w) dw$, where I guess $g(w) = f(\frac 1 w) \frac{-1}{w^2}$ on $\mathbb C \setminus 0 (= \mathbb C \cap (\mathbb C_{\infty} \setminus 0$)), that the functions $f: \mathbb C \to \mathbb C$ and $g: \mathbb C_{\infty} \setminus 0 \to \mathbb C$ are meromorphic.

We can show that for $f: \mathbb C \to \mathbb C$ meromorphic, $f$ extends to $\tilde f: \mathbb C_{\infty} \to \mathbb C$ meromorphic.

It's proved earlier in Theorem II.2.1 and Example II.1.18 that any rational function $h: \mathbb C \setminus \{\text{zeroes of denom}\} \to \mathbb C$ is equivalent to meromorphic $h_m: \mathbb C_{\infty} \to \mathbb C$.

Actually, $h_m = \tilde f$ iff $h=f$ such that for $f: \mathbb C  \to \mathbb C$ meromorphic, we have $f: \mathbb C \setminus \{\text{poles(f)}\}  \to \mathbb C$ holomorphic with $\{\text{poles(f)}\} = \{\text{zeroes of denom}\}$

Title: Algebraic geometry: an exercise


答:
Hint: you  dont need to sub in every equation

$$\alpha+\beta+\gamma=-a$$
$$\gamma=-a/2$$

As this is a root ....substituiting in cubic $x^3+ax^2+bx+c=0$ we have $$a^3-4ab+8c=0$$

Title: Is there a better way to show $a^3 -4ab +8c = 0$ for this polynomial

Body: $x^3 +ax^2 +bx +c = 0 $ has roots $\alpha$, $\beta$ and  $\gamma$, where $\gamma=\alpha+\beta $. Show that $a^3 -4ab +8c = 0$ . I have done it through relations of roots ($\alpha$ + $\beta$ + $\gamma$ = $\frac{-b}{a} = -a$ and etc.) and substituing $\gamma$ into every single equation, then substituing them into what is required to be shown (sub in $ a, b$ and $c$). Is there a faster method to show this, maybe a smarter substitution?


问:
Is the ring of germs of $C^\infty$ functions at $0$ Noetherian?
I'm considering the property of the ring $R:=C^\infty(\mathbb R)/I$, where $I$ is the ideal of all smooth functions that vanish at a neighborhood of $0$. I find that $R$ is a local ring of which the maximal ideal is exactly $(x)$. I also want to determine if $R$ is a Noetherian ring, but I have no idea about how to tackle with it. Can anyone help me? Thanks in advance.</p>

答:
I think this should work:

Consider the function $f_n$ defined by
$$f_n(x) = e^{-1/x^2}\sin(2^n/x).$$
Intuitively, the exponential vanishes so incredibly quickly that the craziness of the sine term causes no problems.  More formally, the $n$th derivative (away from zero) of $f_1$ is, by induction, of the form
$$e^{1/x^2}[P(1/x)\sin(1/x)+Q(1/x)\cos(1/x)]$$
which is easily seen to be continuous everywhere by the fact that $e^{1/x^2}/x^n$ goes to zero as $x$ goes to zero, for any given $n$, and the trig terms are bound between $-1$ and $1$. $f_n$ should be similar. But the vanishing locus of $f_{n}$ strictly contains that of $f_{n+1}$ (and intersects $(-\varepsilon, \varepsilon)$ infinitely many times), so
$$(0) \leq (f_1) \leq (f_1, f_2) \leq \cdots$$
is an infinite ascending chain.


问:
Nontrivial examples of ring homomorphisms which restrict to the identity on an ideal and induce the identity on the quotient
Suppose that $A$ is an abelian group and $B$ is a subgroup of $A$. Then there are usually plenty of homomorphisms (actually they must be automorphisms) $\theta:A \to A$ which restrict to the identity on $B$ and induce the identity on $A/B$. You can think of them as skew transformations, acting along the cosets of $B$. They can all be uniquely expressed in the form $\theta = \mathrm{id}_A + T\circ \pi$ where $\pi$ is the projection $A \to A/B$ and $T : A/B \to B$ is an arbitrary homomorphism.

But what if $R$ is a ring and $I$ is an ideal in $R$? Then you are hard-pressed to find ring homomorphisms $\theta$ which restrict to the identity on $I$ and induce the identity on $R/I$. Of course you can take the zero multiplication on $R$ (assuming your worldview is accepting of nonunital rings) which brings us back to the setting of abelian groups, so at least there are some examples, but I'm not sure if there are any more compelling ones.

My recollection is that there are no such maps, apart from the identity, in the context of C*-algebras and (closed, $*$-invariant) ideals, but I can't recall at the momement whether this is an analytic fact (say using the uniqueness of the norm on the quotient or something) or if it just follows from some purely algebraic considerations which I am overlooking.

答:
Here's a simple example.  Let $R=\mathbb{Z}[x,y]/(xy,y^2)$ and let $f:R\to R$ be given by $f(x)=x+y$ and $f(y)=y$.  This restricts to the identity on the ideal $I=(y)$ and induces the identity $R/I\to R/I$, but is not the identity.

One simple obstruction is that if a (left) ideal $I\subseteq R$ contains an element $a$ that is not a (right) zero divisor, then any homomorphism $f:R\to R$ that restricts to the identity on $I$ must be the identity.  Indeed, for each $r\in R$, $ra\in I$ so $ra=f(ra)=f(r)f(a)=f(r)a$, which implies $r=f(r)$ since $a$ is not a zero divisor.  A bit more generally, the same conclusion holds by a similar argument if $I$ is a faithful $R$-module.

More strongly, suppose $f:R\to R$ restricts to the identity on $I$ and induces the identity $R/I\to R/I$.  Then for each $r\in R$, $f(r)-r\in I$ (since $f$ induces the identity $R/I\to R/I$), but the calculation above shows that $(f(r)-r)I=0$.  So, if $f$ is not the identity, there must be a nonzero element of $I$ that annihilates $I$.  This is impossible for a $*$-ideal $I$ in a $C^*$-algebra, since if $a\in I$ then $a^*\in I$ but $aa^*=0$ implies $a=0$.


问:
Polyhedra with identical faces
The isohedra have identical faces. They have symmetries acting transitively on their faces -- any face can be mapped to any other face to give the same figure.

There are also polyhedra where all faces are the same, but the faces are not transitive.  For example, take an antiprism and make caps with the same triangles.

I just found that this net seems to work, with all faces identical.  The long edges all have length 1, with angles of 60 and 90 degrees.

Have polyhedra like this been explored?  Is there a name for non-isohedra where all faces are the same?

答: I just posted a question about these polyhedra, which are monohedral but not isohedral. I've included your shape in the question, which seems to have been unknown at least prior to 1996. I've included every example I know of there, though I very much doubt it encompasses all known non-isohedral convex monohedra.

Title: Polyhedra with identical faces


问:
Similar matrices involving the flow of a vector field X
Question: Let $\phi$ be the flow of $X$ a vector field of class $C^1$ defined on an open set of $\mathbb{R}^n$ and $\gamma=\{\phi(t,p);0 \leq t \leq T\}$ a periodic orbit. Show that the matrix $D_2(T,p)$ is similar to $D_2(T,q)$ for every $q \in \gamma$.

Attempt: Given $q \in \gamma$ there is some $t_0 \in [0,T]$ such that $q=\phi(t_0,p)$. Notice that using group property of the flow

$$
\phi(-t_0,\phi(T,\phi(t_0,x)))=\phi(-t_0,\phi(T+t_0,x))=\phi(-t_0+(T+t_0),x))=\phi(T,x).
$$
Differentiating the expression above on the point $p$ using the chain rule
$$
D_2 \phi(T,p)=D_2\phi(-t_0,\phi(T,\phi(t_0,p)))D_2\phi(T,\phi(t_0,p))D_2\phi(t_0,p)
$$
$$
D_2 \phi(T,p)=D_2\phi(-t_0,\phi(T,q))D_2\phi(T,q)D_2\phi(t_0,p).
$$
Then I tried to show that $D_2\phi(-t_0,\phi(T,q))=(D_2\phi(t_0,p))^{-1}$ and got stuck. Any hints on how to proceed from here are appreciated. If there is a mistake or some other way of proving it, please let me know. Thanks in advance ;)

答:
Since no answer was given, I will post how I ended up answering the question and leave it open for others to find here.

First notice that using group property of the flow and that $\gamma$ is a periodic orbit
$$
\phi(T,q)=\phi(T,\phi(t_0,p))=\phi(T+t_0,p)=\phi(t_0,p)=q.
$$
So to prove that $D_2\phi(-t_0,\phi(T,q))=(D_2\phi(t_0,p))^{-1}$ is equivalent to prove that $D_2\phi(-t_0,q)=(D_2\phi(t_0,p))^{-1}$. But again using group property of the flow
$$
x=\phi(0,x)=\phi(-t_0+t_0,x)=\phi(-t_0,\phi(t_0,x)).
$$
Then differentiating the expression above on the point p using the chain rule
$$
E=D_2\phi(-t_0,\phi(t_0,p))D_2\phi(t_0,p)=D_2\phi(-t_0,q)D_2\phi(t_0,p)
$$
where $E$ is the identity matrix. Hence $D_2\phi(-t_0,q)=(D_2\phi(t_0,p))^{-1}$.


问:
Let $G$ be a group generated by two involutions $x$ and $y$. Show that there is some $N\unlhd G$ with $[G:N]=2$.
This is Exercise 3.35 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search, it is new to MSE.

The Details:

Presentations are not covered in the book so far, so, presumably, there is a way to answer it without thinking of $G$ given by (some quotient of) the free product

$$\Bbb Z_2\ast\Bbb Z_2\cong \langle x,y\mid x^2,y^2\rangle.$$

The combinatorial-group-theory tag does not apply.

Since I sometimes get involutions and idempotents mixed up, here is a

Definition: An element $f$ of a group $G$ with identity $e$ is an involution if $f^2=e$.

Since there is a plethora of definitions for a normal subgroup, the following is from the book cited above.

Definition 2: A subgroup $H$ of a group $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.

The Question:

Let $G$ be a group generated by two involutions $x$ and $y$. Show that $G$ has a normal subgroup of index two.

Thoughts:

It won't help to consider, without loss of generality, that the normal subgroup - let's call it $N$ - is $\langle x\rangle$, say, since, although $x\langle x\rangle=\langle x\rangle x$, trivially, the same cannot be said in general for $y\langle x\rangle$ and $\langle x\rangle y$.

It does give me the idea, though, that, to test the normality of $N$ in $G$, we need only show that $xN=Nx$ and $yN=Ny$.

Exactly where the $[G:N]=2$ comes from is a mystery to me. My hope is that some pithy choice of generators for $N$ would work, but, well, that's using techniques not yet covered in the book.

I feel like this is a question I ought to be able to answer myself. I've given it a few days and this is all I have.

Please help :)

答:
Consider $\langle xy\rangle$. Note that $(xy)^{-1}=yx$. Therefore, $x(xy)x^{-1} = xxyx = yx = (xy)^{-1}\in\langle xy\rangle$, and $y(xy)y^{-1} = y(xy)y = yx = (xy)^{-1}\in\langle xy\rangle$. Thus, $\langle xy\rangle$ is normal. Call it $N$.

Now note that $G/N$ is abelian, since $[x,y]=xyxy\in N$ and $G/N$ is generated by $xN$ and $yN$. On the other hand, if $x\in N$ then there exists $n\in\mathbb{Z}$ such that $x=(xy)^n$, and since $x$ has order $2$ we may assume $n\gt 0$. Pick $n$ minimal with this property. Then $x=x(yx)^{n-1}y$, so $(yx)^{n-1}y=e$. Therefore $(yx)^{n-1}=y$, so $y(xy)^{n-2}x = y$. Therefore, $(xy)^{n-2} = x$. By the minimality of $n$, either $n=1$ or $n=2$. If $n=1$, then $y=(yx)^0=e$, which contradicts the assumption that $y$ has order $2$. Therefore, $n=2$. But then $yx
=y$, so $x=e$, again a contradiction. Thus $x\notin N$. Symmetrically, $y\notin N$. Thus, $G/N$ is abelian, nontrivial, generated by two elements of order $2$. But since $xN=yN$ (as $y=xxy\in xN$), $G/N$ is cyclic of order $2$. So $N$ will work.


问:
Defining multiples of principal-region angles in inverse trig terms
Consider $tan^{-1}x = \theta$. Then apparently $3\theta=tan^{-1}{}\frac{3x-x^{3}}{1-3x^{2}}$.

How does this work out if $3\theta$ doesn't end up in the first or third quadrant, i.e., if it's not in the range of arctan?

答: The arctangent is a multivalued function since $\tan$ is periodic with period $\pi$. Thus, to get the correct value of $3\theta$, we have to add or subtract $\pi$ repeatedly.


问:
Power of sums of matrices
What is the theory behind the following expression, where A + B represents square matrices nxn

I know how it is done in algebra but how to arrive at this expression? Obviously binomial theorem does not hold..

答:
Matrices are non-commutative in general with respect to multiplication. Thus the order of multiplication must be preserved when expanding powers of sums of matrices.

$(A+B)^3$ is really $(A+B)(A+B)(A+B)$, and the expansion is determined by picking one term from each factor and combining them in order.

Title: Power of sums of matrices


问:
There is enough of gold to...
I read that there is enough of gold layering the surface of planet Earth with 45cm layer of gold.

My question is: Suppose that our planet is a sphere(without mountains and stuff), then how much kilograms of gold is on "Earth"?

(I now that the volume of gold is: $V=V_l-V_E$, where: $V_l=\frac{4}{3}\pi (r+45)^3$ and $V_E=\frac{4}{3}\pi r^3$ )

答:
Where did you read that? The estimated gold mined till now is about 150 000 tons that is a cube with edges of length 20 meters. This will be 8000 $m^3$.

If we use the calculation in your exercise with the earth equtorial radius ($6.3781370\cdot 10^6$ m) the volume would be about $2.3 \cdot 10^{14} \, m^3$ which is about $3\cdot 10^{10}$ times more than it should be (that factor is higher than world population). That would make about $4\cdot 10^{17}$ kilograms of gold. That is about 1 percent of the amount of salt in the oceans.

Title: There is enough of gold to...


答:
It's the distributive law, applied a whole bunch of times. For raising to the second power, and writing out all the steps:

$$(A+B)^2 = (A+B)(A+B) = A(A+B) + B(A+B) = AA + AB + BA + BB$$

Since you can't replace $BA$ with $AB$ for matrices (in general), that's as far as you can go with simplifying, except for collecting runs of the same matrix as a power (e.g. $AA = A^2$). Can you see how this would work when you multiply by $(A+B)$ one more time?

Title: Power of sums of matrices

Body:
What is the theory behind the following expression, where A + B represents square matrices nxn

I know how it is done in algebra but how to arrive at this expression? Obviously binomial theorem does not hold..


问:
Is there a way to find the center point of this circle?
I am facing a problem that I have difficulty to solve.

I have a line that originates from the origin (point $(0,0)$) and has a known angle to the $x$ axis (angle $\theta$)

Somewhere between the $x$ axis and this line is a circle. This circle touches both lines, but the dimension and center of this circle are not known.

On the right side of the circle is a point $P$ that lies on the circumference of the circle and of which the coordinates are known.

How can I find the coordinates of the center of the circle?

I can't seem to add an image so I have used an external website to make the drawing.

========================================================

I now realize that I have over-simplified the problem because it is hard to explain in words. However, now I know that I can send a drawing in another way I would like to post the original problem:

You can see the drawing over here

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known.
A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines.
On the arc of the circle is a point P of which the coordinates are known.
The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?

答:
The equation of the line at angle $\theta$ is $$y=x\tan\theta$$
Let's call the coordinates of the center of the circle $(x_c,y_c)$. Then you know that the distance from the center to the line is equal to the distance from the center to point $P$ (note that this is on a parabola with focus $P$, and the directrix is the tilted line).
We write this as $$\frac{(x_c\tan\theta-y_c)^2}{\tan^2\theta+1}=(x_c-x_P)^2+(y_c-y_P)^2$$
I've used the formula on this link, but you can just search for it. Since the circle touches the line and the $x$ axis, it means that the center is on the bisector of the angle. Therefore $$\tan\frac\theta 2=\frac{y_c}{x_c}$$
Take $y_c$ from the second equation, plug it into the first one, and you just get a quadratic. One of the solution corresponds to $P$ on the right side of the circle, one on the left. In your case, you want the smaller number.


问:
Quick Integral Approximation Question
So, I was studying for a test and I came across this problem:

Sorry for the disorganized format, but it seemed like a lot of info to throw into text. I'm a bit confused on how I'd go about finding the answer here, I understand how to integrate and derive, but I'm a bit lost on how I'd go about accomplishing this problem.

答:
Let $u=x^2-1$ then $du=2xdx$. Evaluating u on the limit points the integral becomes
$$
\frac{1}{2}\int_3^8 f'(u)du.
$$
What can you say now using the graph given?

Title: Quick Integral Approximation Question


问:
How do I sum this series $\sum_{r=3}^{\infty}(r-1)(r-2)\left(qt\right)^{r-3}$?
An archer shoots at a target. It may be assumed that each shot is independent of all other shots and
that, on average, she hits the bull’s-eye with probability $p$. The total number of shots when she hits the bull’s-eye for the third time is denoted by $Y$.

Suppose I want to derive the probability generating function $G_Y(t)$ of $Y$ in order to determine $\mathrm{E}(Y)$ and $\mathrm{Var}(Y)$ and other related results:

I know that for $r\geqslant 3$, $\mathrm{P}(Y=r)=\frac{1}{2}(r-1)(r-2)p^3q^{r-3}$, where $q=1-p$, thus
$$G_Y(t)=\mathrm{E}(t^Y)=\sum_{r=3}^{\infty}t^r\times\frac{1}{2}(r-1)(r-2)p^3q^{r-3}$$

$$\Rightarrow G_Y(t)=\frac{1}{2}p^3t^3\sum_{r=3}^{\infty}(r-1)(r-2)\left(qt\right)^{r-3}$$

How would one proceed to sum this series to obtain an expression for the PGF in a simplified form?

答: The formula for the sum of a geometric series is $$\sum_{n = 1}^{\infty}x^{n-1} = \dfrac{1}{1-x} \quad \text{for} \quad |x| < 1.$$ Differentiating w.r.t. $x$ twice gives us $$\sum_{n = 3}^{\infty}(n-1)(n-2)x^{n-2} = \dfrac{2}{(1-x)^3} \quad \text{for} \quad |x| < 1.$$ Assuming $|qt| < 1$ in your problem, you can simply apply the above formula for $x = qt$.


答:
Reindex the sum:
$$\sum_{k=0}^\infty(k+1)(k+2)(qt)^k=\frac2{(1-qt)^3}$$
This follows by differentiating the geometric series. Thus
$$G_Y(t)=\left(\frac{pt}{1-qt}\right)^3$$

Title: How do I sum this series $\sum_{r=3}^{\infty}(r-1)(r-2)\left(qt\right)^{r-3}$?

Body:
An archer shoots at a target. It may be assumed that each shot is independent of all other shots and
that, on average, she hits the bull’s-eye with probability $p$. The total number of shots when she hits the bull’s-eye for the third time is denoted by $Y$.

Suppose I want to derive the probability generating function $G_Y(t)$ of $Y$ in order to determine $\mathrm{E}(Y)$ and $\mathrm{Var}(Y)$ and other related results:

I know that for $r\geqslant 3$, $\mathrm{P}(Y=r)=\frac{1}{2}(r-1)(r-2)p^3q^{r-3}$, where $q=1-p$, thus
$$G_Y(t)=\mathrm{E}(t^Y)=\sum_{r=3}^{\infty}t^r\times\frac{1}{2}(r-1)(r-2)p^3q^{r-3}$$

$$\Rightarrow G_Y(t)=\frac{1}{2}p^3t^3\sum_{r=3}^{\infty}(r-1)(r-2)\left(qt\right)^{r-3}$$

How would one proceed to sum this series to obtain an expression for the PGF in a simplified form?


问:
Subtracting two infinities
I am Curious if the following is mathematically correct:

Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.

I take from $a$ some of its elements, say integers $10$, $11$, and $12$ only.

So now we have a new set $a'$ that is the infinite set of all nonnegative integers $0,1,2,3...$ except for $10$, $11$, and $12$.

If I subtract: $a-a'$ the result is a set comprised of $10$, $11$, $12$ only. Is this correct? Can one subtract infinities like this?

If yes, does this mean that $a \gt a'$(despite that both are infinite)?

答: The set of positive integers is denoted $\mathbb{Z}^+=\{1,2,3,\dots\}$ and has cardinality $\aleph_0$. The set of positive integers excluding $10,11,12$ also has cardinality $\aleph_0$. The difference between the sets is indeed $\{10,11,12\}$ and its cardinality is $3$. To answer your question, you should read more about cardinal number arithmetic.</p>

Title: Subtracting two infinities


问:
Why substitution in irrational equation doesn't give equivalent equation?
I have two examples of irrational equations:

The first example:  $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$

In solution, they take cube of both sides and do following:

\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x}) +6+x=27\\
&\iff&
3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27   \iff \sqrt[3]{(3-x)(6+x)} = 2\\
&\iff&
x^2+3x-10=0\\
&\iff&   x=2\quad  \text{ or }\quad x= -5
\end{eqnarray*}
They conclude that both values are solutions, they satisfy the original equation.

The second example:  $\sqrt[3]{x+1} +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}$

Here they do the following:
\begin{eqnarray*}
&\sqrt[3]{x+1}& +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}\\
&\iff&
x+1 + 3 \sqrt[3]{(x+1)(3x+1)} (\sqrt[3]{x+1} +  \sqrt[3]{3x+1} ) + 3x+1 = x-1\\
&\implies&
3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3\\
&\iff&
(x^2-1)(3x+1)= -(x+1)^3\\
&\iff&
x= 0 \quad\text{ or }\quad x = -1
\end{eqnarray*}

but the only solutions is $x= -1$, because $0$ doesn't satisfy the equation.

What is difference between these examples, why do we have all equivalence signs in the first, and implication in the secondnd example?

The explanation in solution is that we have substituted  $\sqrt[3]{3-x} + \sqrt[3]{6+x}$ by a number in the 1st example, and in the 2nd example we substituted by another expression which depends on x. I don't understand that, when does substitution  give equivalent equation? Do we always have to verify if solutions satisfy the original equation in the end (is that the answer)?

Please help with this explanation, I need to understand better the irrational equations. Thanks a lot in advance.

答:
These problems don't happen if you'll use the following identity.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Since, $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we see that $\sum\limits_{cyc}(a-b)^2=0$ for $a=b=c$ only and it can give a extraneous root of the equation.

Now, we can solve your equations by using this idea.

$$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$$
Here $a=\sqrt[3]{3-x},$ $b=\sqrt[3]{6+x}$ and $c=-3.$

Since $$\sqrt[3]{3-x}=\sqrt[3]{6+x}=-3$$ is impossible, our equation is equivalent to:
$$3-x+6+x-27-3\sqrt[3]{(3-x)(6+x)}(-3)=0$$ or
$$\sqrt[3]{(3-x)(6+x)}=-2,$$ which gives the answer:
$$\{2,-5\}$$
2. $$\sqrt[3]{x+1} +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}.$$
Since $$x+1=3x+1=-(x+1)$$ is possible for $x=0$ and $0$ is not a root of our equation, we need to remove this root before we'll write the answer.

Id est, we obtain:
$$x+1+3x+1-(x-1)+3\sqrt[3]{(x+1)(3x+1)(x-1)}=0$$ or
$$\sqrt[3]{(x+1)(3x+1)(x-1)}=-x-1$$ or
$$x^2(x+1)=0,$$ which gives the answer:
$$\{-1\}$$


问:
Probability and Statistics : Joint and Marginal PDF
The question is from Probability and Statistics:

Q. A point is selected at random inside the triangle $T=\{(x,y) : 0\leq y\leq x\leq 1\}$. Assume the point is equally likely to fall anywhere in the triangle. Find the joint and marginal pdf of X and Y.

I tried solving this but was not able to come up to the answer.
Please someone help me come to the right answer.
Thanks!
Have a nice day!

答:
Do a drawing of your problem and realize that $(X,Y)$ domain is the lower triangle of the unit square (the triangle under the line $X=Y$)

As per the fact that the point is equally likely to fall in any poin of the triangle, the bivariate density is uniform, thus its density is the reciprocal of Triangle area

$$f_{XY}(x,y)=2\cdot\mathbb{1}_{0\leq y\leq x\leq1}$$

Now to find the marginals is enough to integrate the other rv

$$f_X(x)=\int_0^x 2dy=2x\cdot\mathbb{1}_{[0;1]}(x)$$

$$f_Y(y)=\int_y^1 2dx=2(1-y)\cdot\mathbb{1}_{[0;1]}(y)$$

To understand well the integration extremes just have focus on the given domain

$$\underbrace{0<y<x}_{\text{y-integration interval}}<1$$

$$0<\underbrace{y<x<1}_{\text{x-integration interval}}$$


问:
If two dice are thrown 5 times what is the probability that sum of the numbers coming up will be 10 in exactly 5 out of these 5 trials.
I know there can be 3 possible pairs, so we have 1/12 possibilities to toss one of them.  But I'm not sure how to prove it for 5 out of 5 times.</p>

答: You want a ten on all five tosses, which are independent, so the probabilities multiply. The answer is $\frac1{12^5}$.


问:
Understanding stable and unstable manifolds
I want to understand better the subject of stable and unstable manifold, I want to know what is the motivation to discuss this, and if there is way to imagine this or think about this?
Because I see the definition and some theorems and I understand some proofs but not what comes behind this.</p>

答:
Consider the fixed point of a dynamical system—which is just a point where the system's state doesn't change over time.

The stable manifold of that fixed point is the collection of all states that will asymptotically approach that fixed point as time goes to infinity.

The unstable manifold of that fixed point is the collection of all states that would asymptotically approach that fixed point as time goes to infinity in the inverted system—so if you "ran the system backwards", the unstable manifold would constitute all the states that would eventually approach the fixed point asymptotically.

Knowing what the stable and unstable manifolds of a fixed point are can be useful for simple purposes—for example, knowing which initial conditions of a system of ODE's lead to which asymptotic behavior. It is also very useful for spotting "exotic" behavior in dynamical systems—for example, homoclinic orbits occur when the stable and unstable manifolds of a fixed point are connected. If the stable and unstable manifolds of a fixed point intersect non-tangentially, a structure called a homoclinic tangle forms, which is indicative of transient chaotic dynamics within the dynamical system.


问:
Find values of $a$ for which $|x-a|=(x-2)^2$
Find all values of $a$ for which $|x-a|=(x-2)^2$.

This is a calculus question and I'm told I'm suppose to use the derivative. But the thing is that the absolute value function on the left side looks nothing alike to the parabola function on the right, so how could there be any real number $a$ where this is true?

答: I agree with you, a linear can't be a parabola, though it can have 1 or 2 solutions for various values of "a"!! I have tried to show it through the image!!