问:
Examples of measures that assign positive measure to countable sets of $\mathbb{R}^n$?
Let $A\subset\mathbb{R}^n$ be a countable (and finite) set of Real-valued $N$-dimensional vectors. The Lebesgue measure of this set $A$, denoted $\lambda(A)$, is just $\lambda(A)=0$ because the Lebesgue measure assigns measure zero to all countable subsets. The Dirac measure of this set $A$, denoted $\delta(A)$, may or may not be zero, since:
\begin{gather}
\delta_x(A)=
\begin{cases}
1&\text{if }x\in A\\
0&\text{if }x\notin A
\end{cases}
\end{gather}
for some $x\in\mathbb{R}^n$. My first question is this: in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$?

BONUS QUESTION: consider some (possibly uncountable and infinite) set $B\subset\mathbb{R}^N$ and consider this construction:
\begin{gather}
\delta_B'(A)=
\begin{cases}
1&\text{if }A\cap B\neq\emptyset\\
0&\text{if }A\cap B=\emptyset
\end{cases}
\end{gather}
My bonus question is as follows: is the construction $\delta_B'(A)$ a measure? Why or why not?

EDIT: While $\delta_B'$ satisfies the properties of 'non-negativity' and 'null empty set', I think it fails to satisfy 'countable additivity'. Consider two sets $A,A'\subset\mathbb{R}^N$ such that $A\cap A'=\emptyset$ yet $A\cap B\neq\emptyset$ and $A'\cap B\neq\emptyset$. Then, $\delta_B'(A)=1$, $\delta_B'(A')=1$ and $\delta_B'(A\cup A')=1$. However, $\delta_B'(A)+\delta_B'(A')=2$, thus implying $\delta_B'(A\cup A')\neq\delta_B'(A)+\delta_B'(A')$ and therefore countable additivity fails to be satisfied. Is this reasoning correct?

答:
Your analysis of $\delta_B'$ is accurate.

Seperately, measures of the sort you seek do not really exist.  Suppose $C$ is a countable set with $\mu(C)>0$.  Suppose moreover that, for each $x\in C$, the set $\{x\}$ is measurable.  Then, by countable additivity, we must have $\mu=\sum_{x\in C}{\mu(\{x\})\delta_x}$ (as measures on $C$).  So any countably-supported measure on $\mathbb{R}^n$ with the Lebesgue $\sigma$-algebra is expressible in terms of Dirac masses.

But that caveat about $\sigma$-algebras cannot be removed!  For example, consider the set $\mathbb{Z}^+\times\{0,1\}$, with the $\sigma$-algebra $\{S\times\{0,1\}:S\subseteq\mathbb{Z}^+\}$.  (That is, any measurable set cannot determine the second coordinate.)  Dirac masses aren't measures with respect to this $\sigma$-algebra, so (for example) the measure $$\mu(S\times\{0,1\})=\sum_{s\in S}{2^{-s}}$$ cannot be written in terms of them.


问:
Function with compact support has uniformly continuous integral over $(-\infty,x]$?
If $f$ is a bounded Lebesgue measurable function that has compact support, then is $\int_{-\infty}^x f(t)dt$ uniformly continuous?

I have just learned of compact support and do not see how compact support can lead to uniform continuity since the integral here is not over a compact set? I believe this is a condition for a distribution function in probability and would like to understand why. Also, what if the function $f$ was just integrable, then would this integral still be uniformly continuous?

答:
Define $F(x)=\int_{-\infty}^{x}f(t)dt$. It is well-known that $F$
is absolutely continuous, which implies that $F$ is uniformly continuous.
Note that the only condition is "$f$ is Lebesgue integrable".
We do not require that $f$ is bounded nor having compact support.

Proof: That proposition is well-known and the proof can be found in
every real analysis textbook. For the sake of completeness, I include
it.

Claim 1: If $f$ is measurable and bounded, then for each $\varepsilon>0$,
there exists $\delta>0$ such that $\int_{A}|f|<\varepsilon$ whenever
$m(A)<\delta$.

Proof of Claim 1: Choose $M>0$ such that $|f(t)|\leq M$ for all
$t\in\mathbb{R}$. Let $\varepsilon>0$ be given. Choose $\delta=\frac{\varepsilon}{2M}.$
For any measurable set $A$ with $m(A)<\delta$, we have that $\int_{A}|f|\leq M\cdot m(A)<\varepsilon.$

Claim 2: If $f$ is integrable, then for each $\varepsilon>0$, there
exists $\delta>0$ such that $\int_{A}|f|<\varepsilon$ whenever $m(A)<\delta$.

Proof of Claim 2: For each $n\in\mathbb{N},$ define $f_{n}=\min(|f|,n)$,
then $|f_{n}|\leq|f|$ and $f_{n}\rightarrow|f|$. By Lebesgue Dominated
Convergence Theorem, we have that $\int\left|f_{n}-|f|\right|\rightarrow0$.
Let $\varepsilon>0$ be given. Choose $n$ such that $\int\left|f_{n}-|f|\right|<\varepsilon.$
Note that $f_{n}$ is bounded and integrable, so we may choose $\delta>0$
such that $\int_{A}|f_{n}|<\frac{\varepsilon}{2}$ whenever $m(A)<\delta$.
Now, let $A$ be an arbitrary measurable set with $m(A)<\delta$,
we have that
\begin{eqnarray*}
 &  & \int_{A}|f|\\
 & = & \int_{A}\left(|f|-f_{n}\right)+\int_{A}f_{n}\\
 & \leq & \int\left|f_{n}-|f|\right|+\int_{A}|f_{n}|\\
 & < & \varepsilon.
\end{eqnarray*}

Now, the uniform continuity of $F$ follows immediately. For, let
$\varepsilon>0$ be given. Choose $\delta>0$ such that $\int_{A}|f|<\varepsilon$
whenever $A$ is measurable and $m(A)<\delta$. For any $x_{1}<x_{2}$
with $x_{2}-x_{1}<\delta$, we have that
\begin{eqnarray*}
|F(x_{2})-F(x_{1})| & = & \left|\int_{(x_{1},x_{2}]}f\right|\\
 & \leq & \int_{(x_{1},x_{2}]}|f|\\
 & < & \varepsilon.
\end{eqnarray*}

ContentLicense: CC BY-SA 4.0


问:
I roll a fair die 4 times. Let X be the number of different outcomes that I see. Find $\mathbb{E}[X]$
I roll a fair die 4 times. Let X be the number of different outcomes that I see. Find $\mathbb{E}[X]$.

My attempt:

I know that I can write X as a sum of indicator random variables and then I can use the fact that $\mathbb{E}[1_A]=\mathbb{P}(A)$. Thus,

\begin{equation}
I_{A_1}=
\begin{cases}
1 & \text{if only one kind}\\
0 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
I_{A_2}=
\begin{cases}
1 & \text{if two kinds}\\
0 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
I_{A_3}=
\begin{cases}
1 & \text{if three kinds}\\
0 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
I_{A_4}=
\begin{cases}
1 & \text{all different}\\
0 & \text{otherwise}
\end{cases}
\end{equation}

Then the probabilities of event A happening for each is
$$\mathbb{P}(I_{A_1})=\frac{6\cdot1\cdot1\cdot1}{6^4}$$
$$\mathbb{P}(I_{A_2})=\frac{6\cdot5\cdot2\cdot2}{6^4}$$
$$\mathbb{P}(I_{A_3})=\frac{6\cdot5\cdot4\cdot3}{6^4}$$
$$\mathbb{P}(I_{A_4})=\frac{6\cdot5\cdot4\cdot3}{6^4}$$

The sum of these should be my desired expectation.

My question is whether or not I have found the probabilities correctly. This is my thought process for how I counted the number of choices for the numerator, using $I_{A_2}$ as an example:

There are 6 choices for choosing the first number, then we don't want to get that number again so then there are 5 choices. After that, we only want to get the first or second number again, thus there are 2 choices for the third roll and 2 choices for the fourth roll. Hence, $6\cdot5\cdot2\cdot2$.

Is this the correct way of thinking about it? Thanks

答:
Let $(X_i)$ be an indicator r.v. that equals $1$ if face $i$ appears in $4$ rolls, and $0$ otherwise

Let $X$ be the number of distinct faces that have appeared, then

$ X = X_1 + X_2 + ... +X_6$

and $\Bbb E(X) = \Bbb E(X_1) + \Bbb E(X_2) + ... + \Bbb E(X_6) =6\Bbb E(X_i)$ by symmetry

Now the expectation of an indicator variable is just the probability of the event it indicates, so

$\Bbb E(X_i) = \Bbb P(X_i) = 1 - \left(\frac 5 6\right)^4$

and $\Bbb E(X) = 6\Bbb E(X_i) = 6\left(1- \left(\frac 5 6\right)^4\right) = \dfrac{671}{216}, \approx 3.1065$


问:
How to prove the rep of SU(2) on homogeneous polynomials in 2 variables is irreducible?
The group $\mathrm{SU}(2)$ has a tautologous representation on the space $\mathbb{C}^2$ and thus a representation on the $d$th symmetric power $S^d (\mathbb{C}^2)$.  What's the easiest way to prove that this representation is irreducible?

I want to explain this in a course I'm teaching.  I want to prove this "from scratch", not using powerful tools like Schur-Weyl duality or the general classification of representations of compact simple Lie groups or simple Lie algebras.   I have an elementary method but it involves some mildly annoying calculations.  Ideally there would be some slick, quick, elementary and insightful proof.

If it helps, I'm happy to think of $S^d(\mathbb{C}^2)$ as the space of homogeneous degree-d polynomial functions on $\mathbb{C}^2$ and describe the representation in this way.  I'm happy to consider it either as a representation of $\mathrm{SU}(2)$, $\mathrm{SL}(2,\mathbb{C})$, or the Lie algebra $\mathfrak{su}(2)$ or $\mathfrak{sl}(2,\mathbb{C})$: it's pretty painless to prove irreducibility of the representation of any one of these groups or Lie algebras on $S^d(\mathbb{C}^2)$ implies irreducibility of all the others.

I'd be equally happy to prove indecomposability, since I've proved an indecomposable representation of a compact Lie group is irreducible.

答:
Depending on the background of your audience the following may be simple enough.

Let $x,y$ be the indeterminates, so $S^d(\Bbb{C}^2)$ is the linear span of $x^jy^{d-j},0\le j\le d$.
Assume contrariwise that there exists a non-trivial submodule $V\subset S^d(\Bbb{C}^2)$.

Claim 1. $V$ contains a monomial. That is, a weight vector.

Proof. Let $P=\sum_{j=0}^dc_j x^{d-j}y^j$ be the element of $V$ with the lowest possible number $m$ of non-zero terms. Assume contrariwise that $m>1$. Then there
exist indices $\ell\neq k$ such that $c_\ell\neq0\neq c_k$.

Action by $g(\theta)=\operatorname{diag}(e^{i\theta},e^{-i\theta})$ will map $x\mapsto e^{i\theta}x$ and $y\mapsto e^{-i\theta}y$. Therefore
$$
g(\theta)\cdot P=\sum_{j=0}^dc_je^{i(d-2j)\theta}x^jy^{d-j}.
$$
By selecting $\theta$ to be small but positive we can ensure that $e^{i(d-2\ell)\theta}$
and $e^{i(d-2k)\theta}$ are distinct. It follows that
$$
g(\theta)\cdot P-e^{i(d-2\ell)\theta}P
$$
is an element of $V$ that contains no monomials other than those of $P$, is missing the $x^{d-\ell} y^{\ell}$-term, and still had the $x^{d-k}y^k$-term. This contradicts the choice of $P$, and proves the claim.

Claim 2. $V$ must be all of $S^d(\Bbb{C}^2)$

Proof. If you have explained the ladder operators (here acting as $x\partial_y$ and
$y\partial_x$) this is easy starting from the monomial created in the first Claim. If you have $SL_2(\Bbb{C})$, then an option is to
use elements of a root subgroup (together with a Vandermonde matrix) to show that all the monomials must be there as well. You probably know the drill.


问:
Prove that $V$ is the direct sum of $W_1, W_2 ,\dots , W_k$ if and only if $\dim(V) = \sum_{i=1}^k \dim W_i$
Let $W_1,\dots,W_k$ be a subspace of a finite dimensional vector space $V$ such that the $\sum_{i=1}^k W_i = V$.

Prove: $V$ is the direct sum of $W_1, W_2 , \dots, W_k$ if and only $\dim(V) = \sum_{i=1}^k \dim W_i$.

答:
Hint 1: Finite dimensional vector spaces are isomorphic if and only if they have the same dimension.

Hint 2: What is the dimension of $W_{1}\oplus W_{2}$?

ContentLicense: CC BY-SA 3.0


问:
How to find the solution set for $\tan ^2 x= (\sqrt{2}-1)^2$?
The problem is as follows:

Find the set solution for the equation from below:

$\sqrt{2}-\tan (2x)=\cot \left(\frac{\pi}{4}+x\right)$

Well after doing all the necessary algebra I'm getting the following expression from below.

$\tan ^2 x= (\sqrt{2}-1)^2$

Well my intuition tells me that I have to take the inverse function for both sides of the equation not before taking the square rooth in both sides which will yield two solutions:

$\tan x= \pm \sqrt{2}-1$

But taking the inverse function will yield:

$x=\tan^{-1}\left(\sqrt{2}-1\right)$

But what it is in the brackets it is not an arc which I can remember. Could it be that I got something wrong here or what?.

Can someone help me here please?. I will really appreciate to know how exactly can I get the set solution for this.

答:
$$\sqrt2-1=\csc\pi/4-\cot\pi/4=\cdots=\tan\pi/8$$

Now if $\tan^2x=\tan^2y\implies\sin^2x=\cdots=\sin^2y$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

$$x=n\pi\pm y$$


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
If you have a quadratic polynomial $f(x)$ and an exponential function $b(x) = b^x$ where $b>1$, you can show that $b(x)$ surpasses the polynomial by showing that eventually the growth rate of $b(x)$ exceeds the polynomials growth rate.

Since the polynomial's leading term has the biggest effect when $x$ grows very big, that means that the other terms don't matter, so let $f(x) = ax^2$ for an $a>0$. Now, compute the ratio between $f(x+1)$ and $f(x)$: $$
\frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, .
$$
As you can see, as $x$ grows very big, that ratio between $x+1$ and $x$ grows close to $1$, thus $f(x+1)$ barely increases from $f(x)$ because it is being multiplied by a number close to $1$. Now analyze the ratio between $b(x+1)$ and $b(x)$. By definition, the exponential function multiplies by its base $b$ evey time you increase by $1$, so the ratio between $b(x+1)$ and $b(x)$ is always $b$. However, we stated already that $b>1$. We also found out that $f(x)$ ratio approaches $1$ as $x$ gets really big. Thus, there is a point when $b(x)$ ratio exceeds $f(x)$ ratio, which means that $b(x)$ will start growing faster than $f(x)$ and will eventually outgrow $f(x)$.

Note that I just used terminology like approach and really big because a $7$th grader would not know of limits, so don't nitpick that.

Secondary note: I said $a>0$ and $b>1$ because I assumed that both of the functions would be traveling upwards as you moved right along the $x$-axis.

If you want a downward-facing parabola with $a<0$ and a downward-facing exponential with $0<b<1$, then you can just note that the exponential will just tend towards $0$ when $x$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.


问:
Determinant of a matrix having a row or columns of 1s
So I read this property of determinants that

If in |A| ( i.e determinant of matrix A), if any row/column is proportional to any other corresponding row/column then its determinant is zero.

OR if,

Ri   =    kRj

then |A|=0 (zero)

AN example would look like the following in a matrix format:

a11
b12
c13

d21
e22
f23

2a31
2b32
2c33

Here, R3=2R1

Hence, determinant here would be 0.

My problem is what if one of the rows is a all 1s?

i.e.

a11
b12
c13

d21
e22
f23

131
132
133

In this case, why aren't all rows multiple of third row and determinant not 0?

答:
Your claim $R_3=2R_1$ holds only if $(a_{31},a_{32},a_{33}) =(a_{11},a_{12},a_{13}) $. Otherwise
the replacing in the matrix the row $(a_{31},a_{32},a_{33}) $ with $(2a_{31},2a_{32},2a_{33}) $ does not make the determinant of the matrix 0, this doubles it instead. The reason is of course the fact that multiplying  a row with a non-zero constant does not make the raw to become linearly dependent on the others. Particularly a matrix containing $(1,1,\dots,1) $ row can have any deteminant.


问:
How can the components of the metric in one chart be used to construct an endomorphism with "the same" components?
For a chart $x:U\to\mathbb{R}^n$ on a pseudo-Riemannian manifold $(M,g)$, the volume form can be expressed as
$$
\mathrm{vol}_g = \sqrt{|\det(g_{\cdot\cdot})|}\ \ \mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n
$$
where $\det(g_{\cdot\cdot})$ is the determinant of the matrix representation of $g$ in the coordinates $x^i$. Calculation-wise, this works splendidly. However, formalism tells us that we can only take the determinant of endomorphisms (i.e. linear transformations) which are $(1,1)$-tensors. What I wonder is this: what is the endomorphism $\phi_{g,x}$ that is truly being determined in
$$
\det(\phi_{g,x}) = \det(g_{\cdot\cdot})
$$
I know that $\phi_{g,x}$ depends on $x$ because the matrix representation of $g$ under coordinates $x^i$ produces the extra factor needed to keep the volume form invariant, but I do not know how this map can be defined.

Question: What is the endomorphism $\phi_{g,x}$ that satisfies $\phi^a_b=g_{ab}$ and thus $\det(\phi) = \det(g_{\cdot\cdot})$ for coordinates $x^i:U\to\mathbb{R}$?

答:
Let $V$ be an $n$-dimensional vector space over $\Bbb{R}$, let $g:V\times V\to\Bbb{R}$ be a non-degenerate symmetric bilinear form and let $\beta$ be a basis of $V$. Then, relative to this basis, we can calculate the matrix representation $[g]_{\beta}$, and this has a certain determinant. To do this "abstractly", what we need to do is consider the basis induced isomorphism $\psi_{\beta}:V\to V^*$ (i.e the one which maps the basis $\beta$ onto the dual basis $\beta^*$), and consider the "flat map" or the "musical isomorphism" $g^{\flat}:V\to V^*$ defined by $g^{\flat}(v):=g(v,\cdot)$.

Then, consider the composition $\phi_{g,\beta}:= \psi_{\beta}^{-1}\circ g^{\flat}:V\to V$. This is the right thing to consider as we easily see that the matrix representation of this endomorphism relative to the basis $\beta$ (we use this basis on the domain and target) is exactly what we'd like: $[\phi_{g,\beta}]_{\beta}^{\beta}= [g]_{\beta}$. Therefore,
\begin{align}
\det(\phi_{g,\beta})=\det\left([g]_{\beta}\right),
\end{align}
where the determinant on the left is the "abstract" one defined for any endomorphism.

In the manifold setting, given a chart $(U,x)$, we consider the chart-induced basis $\left\{\frac{\partial}{\partial x^i}\right\}_{i=1}^n$ and apply the above construction pointwise at every tangent space to get the desired $\phi_{g,x}$.


问:
Jacobi $sd(u,m)$ hyperbolic approximation for $m \to 1$
I am following a paper (Fink 1976); the details are a mathematical problem.

We are given a solution (paraphrased from the paper for clarity)

$$y^2 = \alpha m_{1n} sd^2(u,\sqrt{m_n}),$$

where $sd$ is a Jacobi Elliptic Function, and $m_n + m_{1n} = 1$.

In a particular limit, where $m_n \to 1$, we are told that we have an approximation for this equation:

$$y^2 = \alpha m_{1n} \left[\text{cosh}^2(u) \left(1 - \frac{1}{4}m_{1n}\text{sinh}(2u)\right) -1\right].$$

To my eye, it appears that this is a combination of the hyperbolic expansions as given for example in this Wikipedia page or this page on JE functions from NIST:

$$\operatorname{sn}\left(z,k\right)=\tanh z-\frac{{k^{\prime}}^{2}}{4}(z-\sinh z%
\cosh z){\operatorname{sech}}^{2}z+O\left({k^{\prime}}^{4}\right),$$
$$\operatorname{dn}\left(z,k\right)=\operatorname{sech}z+\frac{{k^{\prime}}^{2}}
{4}(z+\sinh z\cosh z)\tanh z\operatorname{sech}z+O\left({k^{\prime}}^{4}\right).$$

No such expression is given as an approximation for $sd$.

What I cannot understand is how the factors of $ \pm z$ in the terms can possibly be removed from these expressions. Any advice or guidance would be appreciated - I have tried using other approximations since $m_{1n}$ is small such as $1/(1-x)$ expansions to deal with the denominators etc.

EDIT:
An additional check is given by solving this resulting expression for $m_{1n}$ and the limit that $u$ is large:

$$m_{1n} \approx 4 e^{-2u}[ 1 - \sqrt{1-y^2}].$$

答:
The definition of Jacobi elliptic functions is given in terms of amplitude $\phi$ by the relation \begin{align}u&=\int_0^{\phi}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{1}\\
\operatorname {sn} (u, k) &=\sin\phi\tag{2}\\
\operatorname {cn} (u, k) & =\cos\phi\tag{3}\\
\operatorname {dn} (u, k) & =\sqrt{1-k^2\sin^2\phi}\tag{4}
\end{align}
The function $\operatorname {sd} $ in your question is defined by $$\operatorname {sd} (u, k) =\frac{\operatorname {sn} (u, k)} {\operatorname {dn} (u, k)} \tag{5}$$ When $k=1$ then we have $$u=\log(\sec\phi+\tan\phi), \sec\phi=\cosh u, \tan\phi=\sinh u\tag{6}$$ And then using $(5)$ we have $$\operatorname {sd}^2 (u, k) =\tan^2\phi=\sinh^2 u$$ However this does not resemble the expression in question.

Perhaps there are some more details which are not available in the question.


问:
Given an semi-ellipse inscribed about a square, how do I find the equation of the ellipse?
Given the following diagram:

Where:

W = (-1, 0)

X = (-1, 2)

Y = (1, 2)

Z = (1, 0)

How can I find M?

The ellipse can be assumed to be a semi-ellipse with one of the foci on $\bar{XY}$. I'm guessing that this means that one focus is at (0, 2), with the other focus at (0, -2). Now, by the definition of an ellipse, I know that the sum of the distances from those two points to any other point on the ellipse is a constant. But, having reasoned that far, I've hit a dead end. Where do I go from here?

答:
Recall the equation of an ellipse: $\boxed{\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1}$

Furthermore, recall that if $c$ is the distance from the focus to the vertex, then: $\boxed{c^2=a^2-b^2}$

Since the focus is on $XY$, we know that $c=2$ so that $2^2=a^2-b^2$ or $\boxed{b^2=a^2-4}$.

Since $Y(1,2)$ is on the ellipse, we may substitute it into its equation, which yields:
$$ \begin{align*}
\dfrac{1^2}{a^2} + \dfrac{2^2}{b^2} &= 1 \\
\dfrac{1}{a^2} + \dfrac{4}{a^2-4} &= 1 \\
1(a^2-4)+4(a^2) &=a^2(a^2-4) \\
5a^2-4 &=a^4-4a^2 \\
0 &=a^4-9a^2+4 \\
a^2 &= \dfrac{9 \pm \sqrt{65}}{2} \\
a &= \sqrt{\dfrac{9 \pm \sqrt{65}}{2}} \quad (\approx 0.6847 \text{ or } 2.9208)
\end{align*} $$

However, notice that from the diagram, $M(a,0)$ lies to the right of $Z(1,0)$, implying that $a>1$. Hence, we reject the negative version and conclude that:
$$
\boxed{a = \sqrt{\dfrac{9 + \sqrt{65}}{2}} = 2.9208...}
$$


问:
How to evaluate from the number of solutions for $\sin 2x +\cos 3x - \sin 4x = 0$?
The problem is as follows:

First find the number of solutions for the equation from below:

Assume $x \in [0,2\pi]$

$\sin 2x +\cos 3x - \sin 4x = 0$

Let $n$ be the number of solutions. Using this $n$ find the sum for:

$5\tan^2\left(\frac{n\pi}{18}\right)+1.5n^2$

I'm not sure how to solve this, what it came to my mind was this:

$\sin 2x +\cos 3x - \sin 4x = 0$

$\sin 4x - \sin 2x - \cos 3x = 0$

Using Prosthaphaeresis formulas then I'm getting to:

$\cos 3x \cdot (2\sin x - 1)=0$

From this it can be inferred that there will be six solutions for

$\cos 3x=0$

In the interval mentioned. Or at least I think so,

these will be:

$\frac{\pi}{6}$, $\frac{3\pi}{6}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$, $\frac{9\pi}{6}$, $\frac{11\pi}{6}$.

While for the other guy it will be this:

$\sin x = \frac{1}{2}$

There will be two solutions in that interval, namely:

$\frac{\pi}{6}$, $\frac{5\pi}{6}$

Thus the number of solutions between the two would be eight. But when replacing in what it is being asked it doesn't seem to yield something reasonable.

Therefore can someone help me here?. Which part went wrong?.

答: The solutions you got for $\sin x =\frac 1 2$ are already included in the other six solutions you got. So $n=6$ and the answer is $5\tan^{2}(\frac {\pi} 3)+(1.5)(36)=69$.


问:
How to show AC=BD..

If we have the picture above, and $AB$ is parallel   to $CD$, also, $\angle{ACB}=\angle {ADB}$, then how to show $AC=BD$?

答:
There are many ways to see this, but I wanted to use area.  $(A(\bigtriangleup))$ denotes area of a triangle.

Let the center of the trapezoid be $O$.

Observe:
$$\bigtriangleup COA \sim \; \bigtriangleup DOB$$

Also observe that (from equal height and base):

$$A(\bigtriangleup CBA)=A(\bigtriangleup DAB)$$
Hence:
$$A(\bigtriangleup COA)=A(\bigtriangleup DOB)$$

Similar triangles with equal areas are congruent:
$$AC=BD$$

Title: How to show AC=BD..


问:
Definition of limit and finding a solution
As creating a link of limits like
$$\lim_{a\rightarrow2b} \lim_{b\rightarrow2c} \lim_{c\rightarrow3}abc$$
is this possible? If yes, what is the formal solution steps of this question?

答:
As I said in a comment to another answer, we have a notation problem here, rather than calculation problem. Chains like that are normally evaluated right-to-left, i.e. your formula is effectively:

$$\lim_{a\to 2b}\left(\lim_{b\to 2c}\left(\lim_{c\to 3}abc\right)\right)$$

Now the innermost expression is $3ab$. Note that $c$, as a bound variable, has disappeared:

$$\lim_{a\to 2b}\left(\lim_{b\to 2c}3ab\right)$$

This means that $c$ in the "middle" limit is "some other $c$" and the result is $6ac$ with $b$ disappearing:

$$\lim_{a\to 2b}6ac$$

Now, the "outer" limit has introduced a "new" $b$ and so its value is $12bc$ with two free variables $b$ and $c$...

So, the bottom line is: the limit as you stated it is possible to calculate, but is extremely confusing. Normally, for clarity, one should not use the same letter to mean two different things anywhere near to each other, let alone in the same formula!

Note that the limit in which the order is reversed:

$$\lim_{c\to 3}\left(\lim_{b\to 2c}\left(\lim_{a\to 2b}abc\right)\right)$$

is perfectly fine and does not introduce double labelling. The innermost limit is $2b^2c$ (with $a$ disappearing), the middle limit is $8c^3$ (with $b$ disappearing) and the overall limit is then $8\cdot 3^3=216$.


答: Call the intersection point $O$. Note that $\triangle AOC$ is similar to $\triangle BOD$, since all angles are same. Also $\triangle OCD$ is similar to $\triangle OBA$ by parallelity. Now call $OC = a$, $OD = b$, $OA = c$ and $OB = d$. By similarity, we have $$\frac{a}{d} = \frac{b}{c}$$ and, by the other similarity, we have, $$\frac{a}{c} = \frac{b}{d}$$ Write $a = \frac{bd}{c}$ in the last equation to get $$\frac{bd}{c^2} = \frac{b}{d}$$ and so $c = d$. Therefore, $\triangle AOC$ and $\triangle BOD$ are the same triangle and the desired line segments are equal.

Title: How to show AC=BD..

Body: If we have the picture above, and $AB$ is parallel   to $CD$, also, $\angle{ACB}=\angle {ADB}$, then how to show $AC=BD$?


问:
A summation of a series based on the Fibonacci sequence.
A sequence is defined as follows: $$a_n=a_{n-1}+a_{n-2}\  \forall \ n\geq3\  ,\ n\in Z. $$

If $a_1=a$ and $a_2=b$, find $$S=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{a_{2i-1}a_{2i+1}}$$ in terms of a and b.

My Approach:

I first found the n-th term as $a_n=F_{n-1}a_2+F_{n-2}a_1$ where $F_i$ is the i-th term of the well-known Fibonacci sequence $1,1,2,3,5,8,13...$

I also noticed that by setting $a=b$ the resulting sum comes to $\frac{1}{a^{2}}\left(\frac{1}{F_1F_3}+\frac{1}{F_3F_5}+\frac{1}{F_5F_7}+...\right)$. And by this theorem,(https://link.springer.com/article/10.1186/s13662-016-0860-0)

The final sum should be, $\frac{1}{a^{2}}*\frac{1}{F_2}=\frac{1}{a^{2}}$.

I also tried expanding the i-th term of $S$ but it did not simplify well.

Another result which may be of use is $\frac{1}{a_{2i-1}a_{2i+1}}=\frac{1}{a_{2i}}\left(\frac{1}{a_{2i-1}}-\frac{1}{a_{2i+1}}\right)$

A solution or any hint/insight on how to solve this problem will be much appreciated.

Thank you

答:
Let's try again to reduce the induction proof to a statement about a telescoping sum.

Working backwards, if $b_n=\frac{F_{2n}}{a_1a_{2n+1}}$ then
$$
b_{n}-b_{n-1} = \frac{1}{a_1} \frac{F_{2n}a_{2n-1}-F_{2n-2}a_{2n+1}}{a_{2n-1}a_{2n+1}}.
$$
It remains to simplify the denominator. From three successive recursion equations one can eliminate the even-index terms as
\begin{align}
a_{2n-1}-a_{2n-3}&=a_{2n-2}\\
a_{2n-1}&=a_{2n}-a_{2n-2}\\
a_{2n+1}-a_{2n-1}&=a_{2n}\\
\hline
a_{2n+1}-3a_{2n-1}+a_{2n-3}&=0
\end{align}
This holds equally well for the Fibonacci sequence
$$\begin{align}
\pmatrix{F_{2n}&a_{2n+1}\\F_{2n-2}&a_{2n-1}}
&=\pmatrix{3&-1\\1&0}^{n-1}\pmatrix{F_{2}&a_3\\F_{0}&a_1}
\\[.8em]
F_{2n}a_{2n-1}-F_{2n-2}a_{2n+1}&=\det\pmatrix{1&a_3\\0&a_1}=a_1
\end{align}$$
In total this confirms the term of the series as a telescoping difference, and the thus the circle can be closed,
$$
b_n-b_{n-1}=\frac{1}{a_{2n-1}a_{2n+1}}
\implies
\sum_{k=1}^n\frac{1}{a_{2k-1}a_{2k+1}}=b_n-b_0=\frac{F_{2n}}{a_1a_{2n+1}}.
$$


问:
What is the relation between order in a number field and order in a quaternion algebra?
The study of number field theory and quaternion algebra theory confuses me a lot:
there are always defined symmetrical quantities (for example the trace, the norm, the orders).
But I can't understand the underlying link.

Is there a correspondence of some kind between the norm of an element
in a number field and the norm in a quaternion algebra?

Is there a correspondence of some kind between the trace of an
element in a number field and the trace in a quaternion algebra?

Is there a correspondence of some kind between an order in a number
field and an order in a quaternion algebra?

Thanks for the help.

答:
The (reduced) norm, (reduced) trace, (reduced) characteristic polynomial and orders have generalizations to general finite-dimensional (unitary, associative) algebras over a field.

The principal invariants. Take a finite-dimensional algebra $A$ over a field $K$, and choose a basis $(e_i)$.
The principal polynomial of $A$ (w.r.t. the basis) is the minimal polynomial of the "generic" element $\sum x_i e_i \in A \otimes_K K(x_1, \ldots, x_n)$. You can specialize it by plugging in values $\lambda_i \in K$ for every $x_i$, and you then call this the principal polynomial of the element $a = \sum \lambda_i e_i$. Denote it by $P_{A, a}$. (It does not depend on the choice of basis.) When $A$ is a field extension of $K$, the principal polynomial is just the usual characteristic polynomial. When $A$ is a quaternion algebra (or more generally, a central simple algebra) the principal polynomial is the reduced characteristic polynomial. One can show that
$$P_{A, a}(T) = \prod_{ (\rho, V) \in \operatorname{Irr}(A \otimes_K \overline K)} \chi_{\rho(a)}(T) \,,$$
where the product runs over the isomorphism classes of simple left $A \otimes_{K} \overline K$-modules, and $\chi_b(T)$ denotes the characteristic polynomial of $b \in \operatorname{End}(V)$. (See Jacobson's The Theory of Rings, Chapter 5 §18.)

Now define the principal trace of $a \in A$ to be minus the coefficient of $T^{\deg P_A - 1}$ in $P_{A, a}$, and define the principal norm of $a \in A$ to be $(-1)^{\deg P_A}$ times the constant term of $P_{A, a}$. The above expression for $P_{A, a}$ implies that the principal trace is a linear map and that the principal norm is multiplicative. For field extensions and central simple algebras, they coincide with  the usual trace and norm resp. reduced trace and reduced norm.

Integrality and orders. Suppose $A$ is a finite-dimensional $K$-algebra with $K$ the field of fractions of some Noetherian and integrally closed ring $R$. You call an element $a \in A$ integral (over $R$) when the following equivalent conditions are satisfied:

$a$ is the root of a monic polynomial in $R[X]$.

The minimal polynomial of $a$ is in $R[X]$.

The characteristic polynomial of $a$ is in $R[X]$.

The principal polynomial of $a$ is in $R[X]$.

The $R$-algebra generated by $a$ is finite (=finitely generated as an $R$-module).

$a$ is contained in a finite $R$-algebra.

Define an order in $A$ to be a finite $R$-subalgebra that generates $A$ over $K$. When either

$A$ is separable (think $A$ = field extension of $K$ with $\operatorname{char}(K) = 0$ or $A$ = quaternion algebra over $K$)

$A$ is only semi-simple, but $K$ is a global field and $R$ is the localization of $\mathcal O_K$ at finitely many places (think $K = \mathbb F_q(T)$, $R = \mathcal O_K$, $A$ = field extension that can be inseparable)

one can show that an $R$-algebra $\mathcal O \subset A$ is an order iff it consists of integral elements and generates $A$ over $K$. (The implication $\implies$ is immediate, $\Longleftarrow$ is harder.)


问:
Uniform convergence of a function defined as an integral
I have to prove that this function :
$$f_n(x) = \int\limits_0^n e^{-\left(1+x^2\right)t^2}\,\mathrm{d}t$$ converges uniformly.

In the solution of this exercise, they use a sequence $a_n = \int_{n-1}^n e^{-t^2}dt $ and they prove that $|f_n(x)-f_{n-1}(x)|<a_n $. Then they use that to conclude on the convergence of $f_n$.

I don't really understand the reasoning and why it works.
Could you please help me?

答: $\sum a_n=\int_0^{\infty} e^{-t^{2}} dt <\infty$. Apply M-test to conclude that $\sum [f_n(x)-f_{n-1}(x)]$ converges uniformly. But the $n-$th partial sum of this series is nothing but $f_n(x)$ so this proves that $(f_n(x))$ converges unifromly. [Here I define $f_0$ as $0$].


问:
Convergence of series and disk or convergence

I applied the Ratio Test and I got

$$x + 2y < 1$$

Shouldn't this give me a half plane? The answer says it is (D). The only reason why I think it could be (D) is because $y$ could be positive or negative?

答: What you want is $|x+2y|<1$ or $x+2y=-1$: this last option gives the alternated harmonic series. Note that $$ax+by+c<0$$ is the half "plane" determined by a line, and $ax+bx+c=0$ is a line.</p>


问:
Why $\gcd(12,30)$ equals the number of homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$?
The following is a snippet from Gallian's Contemporary Abstract Algebra:

Chapter 10-Example 10:

We determine all homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$.

By

$\phi(g^n) = (\phi(g))^n$ for all $n\in\Bbb Z$",

such a homomorphism is completely specified by the image of $1$.

That is, if $1$ maps to $a$, then $x$ maps to $xa$.

By Lagrange’s Theorem and by " If $|g|$ is finite, then $|f(g)|$ divides $|g|$"

we require that $|a|$ divide both $12$ and $30.$

So, $|a|$ can be  $1, 2, 3$, or $6$.

Thus, $a = 0, 15, 10, 20, 5,$ or $25.$

This gives us a list of candidates for the homomorphisms.

That each of these six possibilities yields an operation-preserving, well defined function can now be verified by direct calculations.
[Note that $\gcd(12, 30) = 6.$ This is not a coincidence!]

Can someone please explain why $\gcd(12,30)=6$ isn't a coincidence? I mean to ask why $\gcd(12,30)$ equals the number of homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$.

答:
GCD(12,30)=6,the factor of 6 are {1,2,3,6), i.e., you can have smaller subgroup of codomain Group ${Z_{30}}$  only of above order. That is what the property you mentioned above.

In Homomorphism if we sacrifice bijection requirement of Isomorphism, then we can have onto relation with codomain group of a domain Group.

GCD confirms that any of divisor of two group below its value, is a candidate for homomorphism, considering a important order related property described above.

Same Order of mapped elements is a required property, but not a sufficient property. It still needs to obey the  ${\phi(ab)=\phi(a)\phi(b)}$


问:
Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $
Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $

this isn't really a computable integral, But the only idea I have in mind is that:

derivative from $x=0$ to $x=1$ of:  $\frac{d}{dx}\left({e^x}-{x})\right) = e -1 $, but I would still have to compute $\frac{e^x}{x^{2}+1}$,

because I'm not sure the derivatives will be sufficient in order to prove the inequality.

答:
Because of $\frac{1}{x^2+1}\leq 1$ for all $x$, we have $$\int_0^1 \frac{e^x}{x^2+1}dx\leq \int_0^1 e^x dx= e-1.$$

The inequality $\frac{1}{x^2+1}\leq 1$ follows from $x^2+1\geq 1$. This inequality is not very sharp though: The integrals equals, roughly, 1.271 whereas the upper bound is $e-1=1.718$.


问:
Mirror property of angle bisector

In $\Delta ABC, A\equiv (1,2)$. If the angle bisector of $\angle B$ is $y=2x+10$ and perpendicular bisector of $AC$ is $y-x=0$, find the equation of $BC$

The solution given says that:

The image of point $A$ about the angle bisector of $C$ lies on $BC$.

The problem becomes easy after this since we now know $2$ points on the line $BC$(Another being $C$ itself)

Is the statement given in the answer a well known theorem/property? Im not exactly sure how I can prove its validity.

答:
Maybe the solution should say '...about the angle bisector of $∠B$ lies on $BC'$.  This is true, if we reflect $A$ in the angle bisector it will land on $BC$ (not necessarily at point $C$, maybe $BC$ produced), it will form an isosceles triangle with $A,B$ and the image of $A$, ($A'$), so will have a line of symmetry that is the angle bisector, try a quick sketch of any triangle.

The point $M(-3,4)$ is where  $AA'$ crosses and is perpendicular to $y=2x+10$ so the other point on $BC$ is $(-7,6)$, (AM is 4 left 2 up, then repeat).

However the original question doesn't seem realistic, as can be seen by drawing diagrams, i.e is it really possible to have an angle bisector at $B$ of the equation given?  There seems to be a mistake somewhere in the solution or the question you were set, but hope this helps.

Title: Mirror property of angle bisector


答:
A better bound can be achieved by applying the Integral Chebyshev inequality to $f(x) = \frac{1}{1+x^2}$ and $g(x) = e^{x}$. Since $f$ and $g$ are of opposite monotonicity on $[0, 1]$,
$$
 \int_0^1 \frac{e^x}{1+x^2} \, dx \le \int_0^1 \frac{1}{1+x^2} \, dx \cdot \int_0^1 e^x \, dx = \frac{\pi}{4} (e-1) \approx 1.3495 \, .
$$

Body:
Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $

this isn't really a computable integral, But the only idea I have in mind is that:

derivative from $x=0$ to $x=1$ of:  $\frac{d}{dx}\left({e^x}-{x})\right) = e -1 $, but I would still have to compute $\frac{e^x}{x^{2}+1}$,

because I'm not sure the derivatives will be sufficient in order to prove the inequality.


答:
I Think $n=6$.

Algebraically:

$$\sin 2x +\cos 3x -\sin 4x = \\ 2\sin x \cos x +\cos 2x \cos x-2\sin x \cos x \sin x -4 \sin x
 \cos x \cos 2x= \\ 2 \sin x \cos x +(1-2 \sin^2 x)\cos x - 2 \sin^2 x \cos x -2 \sin x \cos x(2-4 \sin^2 x) = \\ 2 \sin x \cos x (1-2+4 \sin^2 x) + \cos x (1- 2 \sin^2 x - 2 \sin^2 x)= \\ 2\sin x \cos x(-1+4\sin^2 x) + \cos x (1-4 \sin^2 x)=  \\ 2\sin x \cos x(4\sin^2 x -1) - \cos x (4 \sin^2 x -1)= \\ (4 \sin^2 x -1)(\cos x)(2\sin x -1)=0 $$

so we have as solutions: $\frac{7\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6}$ for the first term, $\frac{\pi}{2},\frac{3\pi}{2}$ for the second, and $\frac{\pi}{6},\frac{5\pi}{6}$ for the third. Six solutions in total.

Title: How to evaluate from the number of solutions for $\sin 2x +\cos 3x - \sin 4x = 0$?

Body:
The problem is as follows:

First find the number of solutions for the equation from below:

Assume $x \in [0,2\pi]$

$\sin 2x +\cos 3x - \sin 4x = 0$

Let $n$ be the number of solutions. Using this $n$ find the sum for:

$5\tan^2\left(\frac{n\pi}{18}\right)+1.5n^2$

I'm not sure how to solve this, what it came to my mind was this:

$\sin 2x +\cos 3x - \sin 4x = 0$

$\sin 4x - \sin 2x - \cos 3x = 0$

Using Prosthaphaeresis formulas then I'm getting to:

$\cos 3x \cdot (2\sin x - 1)=0$

From this it can be inferred that there will be six solutions for

$\cos 3x=0$

In the interval mentioned. Or at least I think so,

these will be:

$\frac{\pi}{6}$, $\frac{3\pi}{6}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$, $\frac{9\pi}{6}$, $\frac{11\pi}{6}$.

While for the other guy it will be this:

$\sin x = \frac{1}{2}$

There will be two solutions in that interval, namely:

$\frac{\pi}{6}$, $\frac{5\pi}{6}$

Thus the number of solutions between the two would be eight. But when replacing in what it is being asked it doesn't seem to yield something reasonable.

Therefore can someone help me here?. Which part went wrong?.


问:
How to find the number of solutions in $2\cot 4x=3-3\cot 2x$?
The problem is as follows:

Find the number of solutions from the equation from below in the given
range.

$2\cot 4x=3-3\cot 2x$

Assume $x\in \left[0,\frac{3\pi}{4}\right]$

I am not sure what to do here. All what I was able to do was to find the values of the cotangent function as follows:

For brevity purposes I'm posting the simplification which I arrived after algebraic manipulation:

$4\cot^2 2x -3\cot 2x -1 =0$

Then this gets me to:

$(4\cot 2x+1)(\cot 2x-1)=0$

which means:

$\cot 2x=-\frac{1}{4}$

and

$\cot 2x = 1$

But now what?. This is exactly where I'm stuck. It would help me a lot if someone could gently guide me on what should be done next to find the number of solutions in the required interval.

答: Let $f(x) = 2\cot 4x-3+3\cot 2x$. Observe that $f'(x) <0$. Now find discontinuities of $f(x)$ to happen at multiples of $\frac{\pi}4$. Hence, there only $3$ roots in $[0,3\pi/4]$.


答:
I prove a general result here (of which yours is a special case):

Theorem:
There are $\gcd(m,n)$ homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$.

Proof:
As you've already noted, we just need to find the number of possible images of $1$ in $\mathbb Z_n$. Note that order of the image of an element is a factor of the order of the element (this is because, if $s\in\mathbb Z_m$ has order $k$, then $0=\varphi(sk) = \varphi(s)k$, therefore, $k$ divides order of $\varphi(s)$). Since the order of $1$ in $\mathbb Z_m$ is $m$, therefore, order of the image of $1$ has to be a factor of $m$. But at the same time, it also is a factor of $n$, by Lagrange's theorem. Therefore, the image of $1$ in $\mathbb Z_n$ has to be a common factor of both $m$ and $n$, which are ultimately the factors of $\gcd(m,n)$. Since there are $\phi(k)$ elements of order $k$ in $\mathbb Z_n$ (where $\phi(\cdot)$ is the Euler's totient function), therefore, the number of possible images of $1$, and hence the number of homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$ is given by $$\sum_{d|\gcd(m,n)}\phi(d) = \boxed{\gcd(m,n)}$$

Title: Why $\gcd(12,30)$ equals the number of homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$?

Body:
The following is a snippet from Gallian's Contemporary Abstract Algebra:

Chapter 10-Example 10:

We determine all homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$.

By

$\phi(g^n) = (\phi(g))^n$ for all $n\in\Bbb Z$",

such a homomorphism is completely specified by the image of $1$.

That is, if $1$ maps to $a$, then $x$ maps to $xa$.

By Lagrange’s Theorem and by " If $|g|$ is finite, then $|f(g)|$ divides $|g|$"

we require that $|a|$ divide both $12$ and $30.$

So, $|a|$ can be  $1, 2, 3$, or $6$.

Thus, $a = 0, 15, 10, 20, 5,$ or $25.$

This gives us a list of candidates for the homomorphisms.

That each of these six possibilities yields an operation-preserving, well defined function can now be verified by direct calculations.
[Note that $\gcd(12, 30) = 6.$ This is not a coincidence!]

Can someone please explain why $\gcd(12,30)=6$ isn't a coincidence? I mean to ask why $\gcd(12,30)$ equals the number of homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$.


问:
Show that $A$ is dense in $X$ if and only if every nonempty open set of $X$ contains a point of $A$.
Show that $A$ is dense in $X$ if and only if every nonempty open set of $X$ contains a point of $A$.

I did not think there was much to prove for this, if using the theorem:$x \in \bar A$ if for each open set $U$ containing $x$, $U \cap A \neq \varnothing$.

Attempt:

$(\Rightarrow)$
If A is dense, then $\bar A=X$. So for each $x \in X$, there is an open set $U$ containing $x$ with $U \cap A \neq \varnothing$. Since $x$ was arbitrary this holds for each $x \in X$, so each open set in $X$ contains points in $A$.

$(\Leftarrow)$If each open set contains points in $A$, for any $x \in X$ and open set $U$ containing $x$, $U \cap A \neq \varnothing$. So $\bar A=X$.

I do not know if this proof is actually okay. I have seen a different proof in "topology without tears", that is much more involved, using a proof by contradiction in one direction. Am I over simplifying this, and is my proof unsuccessful? Any help will or criticism be appreciated.

答:
It's indeed that simple: if $\overline{A} = X$ and $U$ is non-empty open, then we can argue as you did: let $x \in U$ then $x \in \overline{A}$ , so any open neighbourhood of $x$ intersects $A$ (in particular $U$ too) so $U \cap A \neq \emptyset$.

Conversely, if $A$ intersects every non-empty open set, then $\overline{A}=X$ or else $U = X\setminus \overline{A}$ would be a non-empty open set disjoint from $A$, or: let $x \in X$, $U$ any open neighbourhood of $x$, by assumption $A$ intersects $U$ so by the criterion you mentioned $x \in \overline{A}$, so $X \subseteq \overline{A}$ and so $X=\overline{A}$.


答:
Hint

$$\tan2x=-4\implies2x=n\pi+\arctan(-4)=n\pi-\arctan4$$

Now $\arctan4>\arctan\sqrt3=\dfrac\pi3$

$$\tan2x=1=\tan\dfrac\pi4\implies 2x=?$$

Title: How to find the number of solutions in $2\cot 4x=3-3\cot 2x$?

Body:
The problem is as follows:

Find the number of solutions from the equation from below in the given
range.

$2\cot 4x=3-3\cot 2x$

Assume $x\in \left[0,\frac{3\pi}{4}\right]$

I am not sure what to do here. All what I was able to do was to find the values of the cotangent function as follows:

For brevity purposes I'm posting the simplification which I arrived after algebraic manipulation:

$4\cot^2 2x -3\cot 2x -1 =0$

Then this gets me to:

$(4\cot 2x+1)(\cot 2x-1)=0$

which means:

$\cot 2x=-\frac{1}{4}$

and

$\cot 2x = 1$

But now what?. This is exactly where I'm stuck. It would help me a lot if someone could gently guide me on what should be done next to find the number of solutions in the required interval.


答: When you do the ratio test you evalute the limit of the absolute value hence you must consider this inequality $ |x+2y|<1$</p>

Title: Convergence of series and disk or convergence

Body:
I applied the Ratio Test and I got

$$x + 2y < 1$$

Shouldn't this give me a half plane? The answer says it is (D). The only reason why I think it could be (D) is because $y$ could be positive or negative?


问:
Proof that ${2p\choose p}\equiv 2\pmod p$
While trying to prove that ${2p\choose p}\equiv 2\pmod p$, I saw a few successful methods, many of which seemed rather complicated or relied on obscure theorems (at least from the point of view of someone with my limited knowledge). Since I am quite new with number theory, I would appreciate it if someone could tell me if a method I came up with works, or, if not, where it goes wrong. My idea is that by expanding and reducing  ${2p\choose p}$, we get $$2\prod\limits_{i=1}^{p-1} \frac{2p-i}{p-i} \equiv 2\prod\limits_{i=1}^{p-1} 1 \equiv 2 \pmod {p}$$ because $2p-i \equiv p-i \pmod{p}$.

Is this valid? I know division$\pmod{p}$ is slightly weird, especially when the denominator is $\equiv 0\pmod{p}$, so I'm not sure if all of the statements I made are true. Let me know if I need to clarify anything, and thank you in advance for any help you can provide.

P.S. If there are any other [more accurate] proofs of this congruence that aren't too complicated, I'd love to see them.

答:
An elementary method to prove this is to note that;
$$p+1\equiv 1\pmod{p}$$
$$p+2\equiv 2\pmod{p}$$
$$\vdots$$
$$2p-1\equiv p-1\pmod{p}.$$

Multiplying all these Equivalences gives us;
\begin{align*}
(p+1)(p+2)\cdots(2p-1)
&\equiv 1\cdot2\cdots(p-1)\pmod{p}\\
&\equiv (p-1)!\pmod{p}.
\end{align*}

Hence, we can say that;
$$p\mid (p+1)(p+2)\cdots(2p-1)-(p-1)!$$
$$\implies p\mid \frac{(p+1)(p+2)\cdots(2p-1)-(p-1)!}{(p-1)!}$$
$$\implies p\mid \frac{(p+1)(p+2)\cdots(2p-1)}{(p-1)!}-1$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)}{(p-1)!}\equiv 1\pmod{p}$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)(2p)}{(p-1)!(2p)}\equiv 1\pmod{p}$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)(2p)}{(p-1)!(p)}\equiv 2\pmod{p}$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)(2p)}{(p)!}\equiv 2\pmod{p}$$
$$\implies \frac{(2p)!}{(p)!^2}\equiv 2\pmod{p}$$
$$\implies \binom{2p}{p}\equiv 2\pmod{p}.$$

Hence, we know that $p$ divides $\binom{2p}{p}-2,$ as required.


问:
Fourier transform of a triangular function
How do I calculate Fourier Transform? I want to use the formula to calculate the Fourier transform, but I can not come to the right answer,anything wrong?

答:
The function in your plot is
$$f (t) = \left\{
\begin{array}{ll}
0 & t\le-1\\
1+t & -1<t\leq 0 \\
1-t & 0< t \leq 1 \\
0 & 1  <t. \\
\end{array}
\right. $$

Note that you can split the FT up into $4$ pieces, where two of them are zero. So
$$\mathcal{F}\{f\}(\omega) = \int\limits_{-\infty}^{\infty}f(t)e^{-i\omega t}\,\mathrm{d}t = \int\limits_{-1}^{0}(1+t)e^{-i\omega t}\,\mathrm{d}t+\int\limits_{0}^{1}(1-t)e^{-i\omega t}\,\mathrm{d}t.$$
From now on this should be easy to solve. Multiply out, use linearity of the integral and solve four easy integrals of exponential functions.


问:
How to find the number of times a weather station registered a certain temperature?
The problem is as follows:

The temperature in the city of Daegu on March 1st, 2020 is given by:

$15+5\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)$ in celcius, where
$t\in [0,24]$. Assume $t$ is the time elapsed in hours from midnight.
Using this information. Find how many times the temperature of the
city was measured to be $17.5$ celcius.

What I've attempted to do in order to solve this problem isn't much. What I arrived was this:

$$15+5\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=17.5$$

Then this is reduced to:

$$\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=\frac{1}{2}$$

But now what? I could use the inverse sine function as:

$$\frac{\pi t}{12}=\sin ^{-1}\left(\frac{1}{2}\right)-\frac{\pi}{2}$$

But from looking at this effort, I don't seem this would help me to get the value of $t$, needless to say. Assuming that I could get that value. How can I get how many times that temperature was attained?. Therefore I need assistance in this problem.

答:
from $$\sin\left(\frac{\pi t}{12} + \frac{\pi} 2\right)=\frac12$$

We have $$\frac{\pi t}{12} + \frac{\pi}2 = k\pi + (-1)^k\cdot \frac{\pi}6, k \in \mathbb{Z}$$

Simplifying, we have
$$t + 6 = 12k + 2(-1)^k, k \in \mathbb{Z}$$

$$t  = 12k - 6 + 2(-1)^k, k \in \mathbb{Z}$$

Now, we have the restriction of $t$ should be in $[0, 24]$.

When $k=0,$ we have $t<0$

When $k=1$, we have $t=12-6-2=4.$

When $k=2$, we have $t=24-6+2=20$

When $k=3$, we have $t = 36-6-2 > 24$.

Hence the answer is two times.


问:
meaning of topology on a finite set
I have just started learning topology on my own and it's been quite intuitive and well-motivated while it was defined on the set of real numbers. However, in many books authors, for the sake of simplicity,  usually begin demonstrating a topology on some small finite sets such as $\{1,2,3\}$. For example, they say that $\{\{\}, \{1,2\}, \{1,2,3\}\}$ is a valid topology, since it satisfies three axioms. And i didnt get it. I cant grasp idea of topology on finite sets. Ideas of open sets, or open balls (when a metric defined) are clear on real numbers, but for me they lose the meaning on finite sets, however it should be vice versa.

That's why i have a question:

What does we obtain conceptually when we define a topology on a finite set? Sense of nearness, connectedness? May be by defining a topology $\{\{\}, \{1,2\}, \{1,2,3\}\},$ we show continuity within the $\{1,2\}$ and in a sense isolate $3$?

答: The main interest of defining a topology $\tau$ on a finite set $F$ is a pedagogical one. Since there are one finitely many subsets, given $A\subset F$ there are only finitely many choices for the closure and for the interior of $A$. Besides, $\mathring A$ must be an element of $\tau$ (which is, again, a finite set), and the same thing applies to $F\setminus\overline A$. So, you can find $\mathring A$ and $\overline A$ by simply picking the right element from $\tau$, and this will help you to grasp the idea that $\mathring A$ is the largest open subset of $A$ and that $\overline A$ is the smallest closed set containing $A$.

Title: meaning of topology on a finite set


问:
Does $\sum{\frac{\sin (n+1/n)}{ \ln n}} \sim \sum{\frac{\sin n}{\ln n}}$?
As was mentioned in the title, can we say that both the series
$$
\sum{\frac{\sin (n+1/n)}{\ln n}} \quad \text{ and } \quad \sum{\frac{\sin n}{\ln n}}
$$
converges or diverges simultaneously? I mean, it looks very intuitive since
$$
\sin(n+1/n) \approx \sin(n)
$$
for sufficiently large $n$. But I didn't find any rigorous Test or Rule to show that.

答:
Since $\sin(a+b) = \sin a \cos b + \cos a \sin b$, by a Taylor expansion around $0$ we get
$$\begin{align*}
\sin\!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \\
&= \sin n\left(1-\frac{1}{n^2} + o\!\left(\frac{1}{n^2}\right) \right)
+ \cos n\left(\frac{1}{n} + o\!\left(\frac{1}{n^2}\right) \right)
\end{align*}$$
which means that
$$\begin{align*}
\frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n}
&= \frac{\sin n}{\ln n}
+ \frac{\cos n}{n\ln n}+ o\!\left(\frac{1}{n^2}\right)
\end{align*}$$
since $|\sin|,|\cos|\leq 1$. This shows that, if you take for granted that
$$
\sum_{n=2}^\infty \frac{\cos n}{n\ln n}
$$
converges, then indeed
$$
\sum_{n=2}^\infty \frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n},\quad\sum_{n=2}^\infty \frac{\sin n}{\ln n}
$$
have same nature (since the remainder, "$\sum_n o\!\left(\frac{1}{n^2}\right)$", is absolutely convergent by comparison). The question then becomes: can one easily show that
$$
\sum_{n=2}^\infty \frac{\cos n}{n\ln n}
$$
converges? (One way: Dirichlet's test.)


问:
Infimum problem lemma to prove $\mathbb{R}^2$ is regular
I am seeing this infimum problem which is serving as a helper proof to prove $\mathbb{R}^2$ is regular, and having difficulty finding a way to use the infimum property.

Problem:Consider $\mathbb{R}^2$ with the standard topology. Let $p \in \mathbb{R}^2$ be a point not in a closed set $A$. Show that $\text{inf}\{d(a,p)|a \in A\}>0$.

Attempt$(1)$:Assume $p \in \mathbb{R}^2$ is not in $A$, but $d(a,p)=0$. Then $a=p$ and $a \in A$ contradiction to $p \notin A$.So each $d(a,p)>0$ for each $p \in \mathbb{R}^2-A$. Now I am having difficulty seeing the contradiction if the infimum is potentially $0$. So I do not think this proof is sufficient.And I am nowhere using the infimum property anywhere.

Attempt$(2)$:Assume $\text{inf}\{d(a,p)|a \in A\}=0$. Then for any $\epsilon >0$, $\epsilon$ is not a lower bound of $\text{inf}\{d(a,p)|a \in A\}$.So there's a sequence $\{\epsilon_i\}$ of small numbers with $\epsilon>\epsilon_2>\dots$ converging to $0$ and $\{p_i\}$ not in $A$ with $d(a,p_i)<\epsilon_i$. Then $a=\text{lim}p_i$ and so $\{p_{i}\}$ is sequence in $\mathbb{R}^2-A$ converging to $a \in A$. Again I am stuck.

How should I approach this problem differently?

答:
Hint: use that $p \in \overline{A}$ iff for every ball $B(p,r)$ with $r>0$ we have that $B(p,r) \cap A \neq \emptyset$. This holds in any metric space, not just $\Bbb R^2$.

It follows that if $p \notin A$ where $A$ is closed, we have some $r>0$ so that $B(p,r) \cap A = \emptyset$.


答:
My topology professor used to say that topology is the "study of continuous functions".
It is clear to most that continuous functions mapping $\mathbb{R}\rightarrow\mathbb{R}$ are very useful.
Later, this is generalized to metric spaces.
As you say continuous functions can also be defined there.
Topology goes even further than this.
The idea is that one can define continuity, without necessarily having a metric defined over your space.

This is the case in your example: There is no metric over $\{1,2,3\}$, for which the sets $\{\},\{1,2\},\{1,2,3\}$ are the only open sets.
Your question is justified: What's the point of this?
This topology does not have any intuitively appealing properties.
As José already pointed out, the point of this example is a pedagogical one.

As you study topology, you will learn definitions to classify topological spaces (connectedness, compactness, etc.).
These help to describe the behavior of continuous functions, defined over those spaces.
In my opinion, a good argument to study topology at this level of generality is that there are topological spaces with such useful properties, for which there is no metric (not metrizable).
An example is the weak topology over an infinite-dimensional normed vector space.

Title: meaning of topology on a finite set

Body:
I have just started learning topology on my own and it's been quite intuitive and well-motivated while it was defined on the set of real numbers. However, in many books authors, for the sake of simplicity,  usually begin demonstrating a topology on some small finite sets such as $\{1,2,3\}$. For example, they say that $\{\{\}, \{1,2\}, \{1,2,3\}\}$ is a valid topology, since it satisfies three axioms. And i didnt get it. I cant grasp idea of topology on finite sets. Ideas of open sets, or open balls (when a metric defined) are clear on real numbers, but for me they lose the meaning on finite sets, however it should be vice versa.

That's why i have a question:

What does we obtain conceptually when we define a topology on a finite set? Sense of nearness, connectedness? May be by defining a topology $\{\{\}, \{1,2\}, \{1,2,3\}\},$ we show continuity within the $\{1,2\}$ and in a sense isolate $3$?


问:
Is it possible to eliminate a variable from these two differential equations?
I have two second order differential equations in three variables that I need to use to establish a relation between two of the variables.

The equations are

$\alpha$ $ x $ + $\beta$$y$ = $A$$(cx + x'')$$ $ + $\gamma$r

$\alpha$ $ x $ + $\beta$$y$ = $B$$(cx + x'')$$ $ + $\delta$r

where $x$, $y$, $r$ are functions of $z$, and $\alpha$, $\beta$, $\gamma$, $\delta$, $A$, $B$ are constants, and $x''$ = $\frac{\partial^2x}{\partial^2z}$.

I need to get a relation between $y$ and $r$ by eliminating $x$ from the equations, but cannot manage to do this. Is it actually possible to eliminate $x$ from these equations?

Any help would be greatly appreciated.

答:
$$
\cases{B\alpha x  + B\beta y = B A(cx + x'') + B\gamma r \\
A\alpha x  + A\beta y = AB(cx + x'')  + A\delta r
}$$

subtracting the second from the first ode we have

$$
\alpha(B-A)x+\beta(B-A)y = (B\gamma-A\delta)r
$$

and now if $A\ne  B$

$$
x = \left(\frac{B\gamma-A\delta}{B-A}\right)r-\beta y
$$

after that we can substitute the found $x$ into the odes.


问:
When does a rectangular matrix A $\in$ $\mathbb C^{\times }$ have the property such that $\|Ax\|_2=\|x\|_2$, where $x \in \mathbb C^ $
This is one of my homework questions which I am trying to solve I started with the claim that $\|Ax\|_2$ will be less than equals to $\|A\|_2$ $\|x\|_2$ and from here I can only get the condition for $\|Ax\|_2 = \|x\|_2$, that $\|A\|_2 \ge 1$ should hold.

pls suggest me approach for solving this question

答:
Below follows an analysis that hinges on the polarization identity:
$$ \forall x, y \in \mathbb{C}^n \: : \: \langle x, y \rangle_n = \frac{1}{4} \left( \|x + y\|_n^2 + \|x - y\|_n^2 - i \|x + iy\|_n^2 + i \|x - iy\|_n^2\right).$$
Here $\langle x, y \rangle_n$ denotes the inner product between $x, y \in \mathbb{C}^n$ and $\|x\|_n$ denotes the norm given by $$\|x\|_n^2 = \langle x, x \rangle_n.$$ The proof of the polarization identity will be omitted. I use the subscript $n$ to emphasize the dimension of the vector space.

Returning to the problem at hand. We have $A \in \mathbb{C}^{m \times n}$ such that
$$\forall x \in \mathbb{C}^n \: : \: \|Ax\|_m = \|x\|_n.$$
We say that $A$ preserves norms. We mention in passing $A$ is necessarily one-to-one. Specifically if $Ax=0$, then $$\|x\|_n = \|Ax\|_m = 0,$$ so $x=0$ and we must have $m \ge n$. Now from the polarization identity we have
$$ \forall x, y \in \mathbb{C}^n \: : \: \langle Ax,Ay \rangle_m = \langle x,y \rangle_n.$$
We see that $A$ also preserves inner products.
It follows that $$ \forall x, y \in \mathbb{C}^n \: : \: \langle x,A^*Ay \rangle_n = \langle Ax, Ay \rangle_m =\langle x, y \rangle_n.$$
Now let $B = [b_{ij}] \in \mathbb{C}^{n \times n}$ be given by $B = A^*A$ and let $e_k$ denote the $k$ column of the identity matrix $I_n$ of dimension $n$. Then
$$ b_{ij} = \langle e_i, Be_j \rangle_n = \langle e_i, e_j \rangle_n = \begin{cases} 1 & i=j \\ 0 & i \not =j \end{cases}.$$
We conclude that $B = A^*A = I_n$. It follows that $A$ is an orthogonal matrix.


问:
Solve this integral $\int_0^9x^9\sqrt{x^3 + 2}dx$
$I = \displaystyle\int\limits_0^9x^9\sqrt{x^3 + 2}dx$

I let $u = \sqrt{x^3 +2}$

Then $I = \dfrac23\displaystyle\int\limits_{\sqrt2}^{\sqrt{731}}u^2(u^2 -2)^2\sqrt[3]{u^2 -2}du$

After that, I let $t = u^2 -2$

And I got:

$I =\dfrac13\displaystyle\int\limits_0^{729}t^2\sqrt{t +2}\sqrt[3]{t}dt$

And I can't go furthur.

And idea or better way to solve it, please help!??

答:
This is a very tricky approach to the solution.

Consider that
$$\sqrt{x^3+\epsilon}=x^{3/2}+\frac{\epsilon }{2 x^{3/2}}-\frac{\epsilon ^2}{8 x^{9/2}}+\frac{\epsilon
   ^3}{16 x^{15/2}}+O\left(\epsilon ^4\right)$$ I do not introduce the next term because muliplied by $x^9$, it will contain a negative power of $x$ (what we cannot afford.

Make $\epsilon=2$ and muliply by $x^9$. So we compute
$$\int_0^9 \Big[x^{21/2}+x^{15/2}-\frac{x^{9/2}}{2}+\frac{x^{3/2}}{2}  \Big]\,dx=\frac{176374123741638}{21505}$$ which is $\sim 8.20154028094\times 10^9$ while the exact solution (obtained using the hypergeometric function is
$$\frac{176132680620646 }{34155}\sqrt{\frac{43}{17}}-\frac{7168 \sqrt{2} }{580635}\,
   _2F_1\left(-\frac{2}{3},\frac{1}{2};\frac{1}{3};-\frac{729}{2}\right)$$ that is to say $8.20154028011\times 10^9$


问:
Assertion on Compactness
I'm trying to prove the following:

"Let $f :X\rightarrow f(X) $ be a continuous, 2-to-1 function, $X$ metric space.

If $F\subset X$ is a closed, non compact subspace, then $f(F)$ is non compact$.

At first, I tried decomposing $X$ into the disjoint union $X=A \cup B$, such that $f |_A$ and $f |_B$ are injective.

Then the subset $F$ could be written as $F= (F\cap A)\cup(F \cap B)$.

The thing is, I'm trying to use that "$f:X \rightarrow f(X)$ is injective, continuous function, and $F \subset X$ is closed, non compact set, then $f(F)$ is non- compact"

But, in order to do that, I need $A$ or $B$ to be closed.

Do you have any ideas to fulfill the attempt? Or am I in the wrong direction?

Thanks in advance.

答:
Here's a counterexample . . .

Let metric spaces $X,Y$ be given by
\begin{align*}
X&=\mathbb{Z}\,{\large{\setminus}}\{0\}\\[4pt]
Y&=\{0\}\cup\Bigl\{{\small{\frac{1}{n}}}{\;{\Large{\mid}}\;}n=2,3,4,...\Bigr\}\\[4pt]
\end{align*}
with metrics inherited from the standard metric on $\mathbb{R}$.

Note that the topology on $X$ is discrete, so every subset of $X$ is both open and closed.

Regarding compactness,

Every infinite subset of $X$ is non-compact.$\\[4pt]$

Every closed subset of $Y$ is compact.

Let $f:X\to Y$ be defined by
$$
f(x)=
\begin{cases}
0&\text{if}\;|x|=1\\[4pt]
{\Large{\frac{1}{|x|}}}&\text{otherwise}\\
\end{cases}
$$
Then $f$ is a continuous $2$-to-$1$ function from $X$ onto $Y$.

Now let $F$ be any infinite subset of $X$ such that $F\cap\{-1,1\}\ne{\large{\varnothing}}$.

Then $F$ is closed but not compact, whereas $f(F)$ is compact.

Title: Assertion on Compactness


问:
Why is $\operatorname{argmax}_{\pi} P(\pi \mid x) = \operatorname{argmax}_{\pi} P(x, \pi)$?
Why is $\operatorname{argmax}_{\pi} P(\pi \mid x) = \operatorname{argmax}_{\pi} P(x, \pi)$?

Could it be because of Bayes Theorem

$$P(\pi \mid x) = \frac{P(x, \pi)}{P(\pi)}$$

and since $\operatorname{argmax}_{\pi}P(\pi \mid x)$ gives possible values of $\pi$ for which $P(\pi \mid x)$ is maximum and since $P(\pi \mid x) = \frac{P(x, \pi)}{P(\pi)}$ so when $P(\pi \mid x)$ is maximum $P(x, \pi)$ will be maximum too. So $\operatorname{argmax}$ will be the same for both.

Am I right here?

答: Your reasoning is right but your formula is wrong. Bayes rule says that $P(\pi | x) = \frac{P(x, \pi)}{P(x)}$, where the denominator is clearly independent of $\pi$ so the $\mbox{argmax}_\pi$ of either $P(\pi | x)$ or $P(x,\pi)$ is the same.</p>


问:
What is the name for a magma which is neither a quasigroup nor a semigroup yet has both an identity and inverses?
Is there a name which is more specific than `unital magma' for a magma whose only requirements are that it should have both an identity and (L/R symmetric) inverses for all elements?

The following Cayley Table is the simplest non-trivial object of the type I am interested in, though of course you could make much more complicated examples:

```<code>  | 0 1 2
--+------
0 | 0 1 2
1 | 1 0 1
2 | 2 1 0
</code>```

This object it is not a quasigroup (or a loop) as the table is not a Latin Square.

This object is not a semigroup as it is not associative.  For example: (1*1)*2 = 0*2 = 2 while 1*(1*2) = 1*1 = 0.

It does, however, have an identity, 0.  Furthermore, every element has a unique inverse (which is itself).

For what it's worth the multiplication rule is a*b = |a-b| on the set {0,1,2}.

A part of an older question ( Magma with inverse and identity yet not a quasigroup ) could be interpreted as requesting an answer to the same question I am asking today. However, the one answer which that question received did not answer the bit which I would like to see answered. I have therefore judged that it's helpful to ask the missing part here more succinctly to maximise the chance that it gets an answer.

答: "Cancellative unital magma" is more-specific than "unital magma".  This phrase would technically be a valid answer to the question, as invertibility implies cancellation, but I would prefer something more specific.  I'd suggest "Invertible unital magma" except that I cannot find any evidence that this is a term that is in use outside of this post, whereas at least "Cancellative magma" has appears from time to time, such as here.</p>


答:
In addition to what others have already said (especially the point raised in one of the answers - that their use is mainly "pedagogical"), those finite examples are good for two more things:

An endless source of examples and counterexamples. Want to show that being a "Hausdorff" space is a thing? Here you go: your finite example is one that is not Hausdorff. If you have a sequence $a_n$ in your space converging to $1$, it will at the same time converge to $2$.

Some topological constructions already work on finite spaces, and may be the easiest to understand in that context. I am thinking of e.g. discrete topology (every subset is an open set) and indiscrete/trivial topology (the only open sets are empty set and the whole set). Note that the former is even metrizable - you can take the distance between any two distinct elements to be $1$ and this metric yields the discrete topology.

Title: meaning of topology on a finite set


问:
Property of a continuous convex function with additional properties
Let $f:[0,1]\to\mathbb{R}$ be a continuous and convex function such that $f(0)=0$ and $\int\limits_0^1 f(x) ~dx=0$. Prove that
$$\int\limits_0^1 |f(x)| ~dx=2\cdot \max\limits_{x\in [0,1]}\left|\int\limits_0^x f(t) ~dt\right|.$$

For now I considered just the simpler case when $f(1)\leq 0$. Since a convex function defined on a closed interval attains its maximum on one end of the interval and since $f(0)=0$, it follows that $f(x)\leq 0$ on $[0,1]$; therefore, the function $\int\limits_0^x f(t)~dt$ is decreasing on $[0,1]$, but since its values in $0$ and $1$ are both $0$ this function is constant ($0$) on the whole interval and the conclusion follows easily.

答:
$$
a = \max \{ x \in [0, 1] \mid f(x) \le 0 \}
$$
is well-defined since $f(0) = 0$ and $f$ is continuous. Then $f(x) \le 0$ on $[0, a]$ (since $f$ is convex), and $f(x) > 0$ on $(a, 1]$.

The function $F(x) = \int_0^x f(t) \, dt$ is decreasing on $[0, a]$ and increasing on $[a, 1]$ with $F(0) = F(1) = 0$. Now we have
$$
\begin{align}
\int_0^1 |f(x)| \, dx &= \int_0^a (-f(x)) \, dx + \int_a^1 f(x) \, dx \\
&= -(F(a)-F(0) + F(1)-F(a) \\
&= -2 F(a)
\end{align}
$$
and
$$
 \max_{x \in [0, 1]} |F(x)| = \max_{x \in [0, 1]} (-F(x)) = -F(a) \, .
$$


问:
Catalogue of propositional logic laws
I have searched for it in Wikipedia and this site and I haven't found it yet. What I have found isn't complete enough -maybe a short list of ten or so equivalence laws, perhaps forgetting all together laws related to other connectors different from the disjunction or conjunction. I am interested in all logic laws, inference rules too and in general all relevant tautologies which I can use in my deductions in the framework of propositional logic.

I'm looking for an organised, logical and neat catalogue of derivation rules. Biconditional tautologies or equivalence laws, conditional tautologies or inference rules, maybe the law of excluded middle should have a special place in this assortment since it is a disjunction tautology, and the law of non contradiction because it is a negation tautology, I'm not sure. Another criteria which can be taken into account to put order in the catalogue could be the number of propositions to which the logic law is applied. For example, the double negation law is only applied to one proposition but Modus Ponens to two of them.

Could anybody point me to a good resource where I can find this kind of list? My textbook has got a list of this kind with about 40 or so of these laws but I would like a different resource to check it and improve it before memorising it. I don't trust it that much!

答:
You might want to check:

193 propositional logic test cases from Principia

https://gist.github.com/jburse/4ab909be4406b0c960494c692fc7a37f#file-principia-pl

They can be proved by various methods:

Prolog Tautology Checking, who does it better?

https://qiita.com/j4n_bur53/items/28d667cf3412fa950bc7

Title: Catalogue of propositional logic laws


答:
short way
$$
\|Ax\|_2^2 = (Ax)^\dagger(Ax) = x^\dagger A^\dagger A x,\qquad \text{and} \qquad \|x\|_2^2 =  x^\dagger x
$$
where $\dagger$ denotes the conjugate transpose. Now, since $\forall x \in \mathbb C^n : \|Ax\|_2 = \|x\|_2$ we get
$$
x^\dagger (A^\dagger A) x = x^\dagger x \iff  x^\dagger (A^\dagger A -E_n) x = 0 \\
$$
Consider a quadratic form $Q_A:\mathbb C^n \to \mathbb C$ such that $ x \mapsto x^\dagger(A^\dagger A -E_n)x$, now if this is the zero quadratic form, i.e. $\forall x \in \mathbb C^n :Q_A(x) = x^\dagger(A^\dagger A -E_n)x = 0$ then its matrix is zero $A^\dagger A - E_n = 0$, and consequently
$$
A^\dagger A = E_n \in \mathbb C^{n\times n}
$$

More explicitly:

$$
y_i = (Ax)_i = \sum_{j=1}^n A_{ij} x_j \\\implies \|y\|_2^2 = \sum_{i=1}^m (\sum_{j=1}^n A_{ij} x_j)^*(\sum_{j=1}^n A_{ij} x_j) = \sum_{i=1}^m \sum_{j,k=1}^n A_{ij}^* A_{ik} x_j^* x_k
$$
which should be equal to
$$
\|x\|_2^2 = \sum_{i=1}^n \sum_{j,k=1}^n \delta_{ij} \delta_{ik} x_j^* x_k
$$
So we get
$$
\sum_{i=1}^n \sum_{j,k=1}^n \delta_{ij} \delta_{ik} x_j^* x_k = \sum_{i=1}^m \sum_{j,k=1}^n A_{ij}^* A_{ik} x_j^* x_k \\
\sum_{i=1}^n \sum_{j,k=1}^n \delta_{ij} \delta_{ik} x_j^* x_k - \sum_{i=1}^m \sum_{j,k=1}^n A_{ij}^* A_{ik} x_j^* x_k = 0 \\
\sum_{j,k=1}^n \left(\sum_{i=1}^n \delta_{ij} \delta_{ik} - \sum_{i=1}^m A_{ij}^* A_{ik}\right) x_j^* x_k = 0 \\
\overset{\forall x \in \mathbb C^n}{\implies} \sum_{i=1}^n \delta_{ij} \delta_{ik} = \sum_{i=1}^m A_{ij}^* A_{ik}
$$
from the last equation
$$
\delta_{jk} = \sum_{i=1}^m A_{ij}^*A_{ik}, \quad j,k \in \{1,\cdots,n\}
$$

Title: When does a rectangular matrix A $\in$ $\mathbb C^{\times }$ have the property such that $\|Ax\|_2=\|x\|_2$, where $x \in \mathbb C^ $

Body:
This is one of my homework questions which I am trying to solve I started with the claim that $\|Ax\|_2$ will be less than equals to $\|A\|_2$ $\|x\|_2$ and from here I can only get the condition for $\|Ax\|_2 = \|x\|_2$, that $\|A\|_2 \ge 1$ should hold.

pls suggest me approach for solving this question


问:
Hausdorff compactly generated is locally compact?
I was trying to prove the "theorem of exponential correspondence" and "exponential law" with $X,Y,Z$ compactly generated and Hausdorff.

According to Hedwin H.Spanier-Algebraic Topology and Tammo tom Dieck-Algebraic topology we have that locally compact seems necessary in order to prove the statement.

The true hyphotesis given on the spaces are Hausdorff locally compact, but as far as I can see, compactly generated and Hausdorff doesn't imply locally compact, am I right ? Are my hypothesis wrong ? Are there any counterexample.

Thanks in advance

答: You can take any non-locally compact metric space, for some familiar examples. E.g., the Baire space $\mathbb{N}^\mathbb{N}$.


答:
I would like to add that there are topologies defined on finite sets that arise naturally in other mathematical constructions.

An example is a scheme from algebraic geometry. See here for basics on the spectrum of a ring.

In short terms, if $A$ is a commutative ring, then the set $\operatorname{Spec}(A)$ of all prime ideals of $A$ has a natural topology, called the Zariski topology. By definition, a subset $X \subseteq \operatorname{Spec}(A)$ is closed if there is an ideal $I$ of $A$ such that $X = \{P \in \operatorname{Spec}(A): I \subseteq P\}$.

Now consider for example the spectrum of a discrete valuation ring $A$ with maximal ideal $m$. The spectrum $\operatorname{Spec}(A)$ then consists of only two elements $\{0, m\}$.

The Zariski topology on $\operatorname{Spec}(A)$ has the following open sets: $\emptyset, \{0\}, \{0, m\}$.

This is a typical example in algebraic geometry, and one indeed has the notions of connectedness, continuity, etc. which have geometric interpretations.

For more information, a classical reference is the GTM book Algebraic Geometry by Robin Hartshorne.

Title: meaning of topology on a finite set


问:
Let $A, B,C$ be subsets of $\{1,2,...n\}$. What is $P((A\cap B)\subseteq C)$?
My reckoning is that there are totally $2^{n} \cdot 2^{n} \cdot2^{n}$ options for $3$ subsets and for $(A\cap B)\subseteq C$ ,  for each of $\sum_{k=0}^{n}\binom{n}{k}$ options for C there are $\sum_{i=0}^{k}\binom{k}{i}$ for $A$ and $\sum_{i=0}^{k}\binom{k}{i}$ for $B$ and so in total we have:

$$P((A\cap B)\subseteq C)=\frac{\sum_{k=0}^{n}\binom{n}{k}\cdot 2^{k}\cdot 2^{k}}{2^{n}\cdot 2^{n}\cdot 2^{n}}=\frac{\sum_{k=0}^{n}\binom{n}{k}\cdot 4^{k}}{8^{n}}=\frac{5^{n}}{8^{n}}=(\frac{5}{8})^{n}$$

Is this reckoning right?

答:
We want to calculate $P(A\cap B\subseteq C)=P((A\cap B)\setminus C=\emptyset)$.

For each $i$ in $\{1,2,\dots,n\}$ let $E_i$ be the event $i\notin(A\cap B)\setminus C$.

As the events $E_1,E_2,\dots,E_n$ are independent, we have
$$P((A\cap B)\setminus C=\emptyset)=P(E_1\cap E_2\cap\cdots\cap E_n)=P(E_1)P(E_2)\cdots P(E_n)=\left(\frac78\right)^n$$
since
$$P(E_i)=1-P(i\in(A\cap B)\setminus C)=1-P(i\in A,\ i\in B,\ i\notin C)=1-\frac18=\frac78.$$


问:
Proving homomorphism from functions
I am really struggling to grasp the concept of homomorphism as it has been poorly explained to me in courses, and I have not been able to find any clear examples of how to actually classify a function correctly. For instance, I have the function:

substract, that takes a list of integers $[x_1,...,x_n]$ as input and computes the result as a function of: $$x_1 - \sum_{i=2}^n x_i,\quad n > 0.$$

So with an example this would be:

$\operatorname{substract}([19,5,8]) = 19 - (5 + 8) = 6$.

But how should I approach showing whether this is a $homomorphism$ or not? From my understanding a $homomorphism$ is suc that I can find an operator $\oplus$ so that:

$f(x_1 * x_2) \mapsto f(x_1) \oplus f(x_2)$

For which $*$ resembles the concatenation operator on list structures.

But what would be the best case to prove an example as the one provided above? I have to classify a series of such functions with various signatures, but I want to understand the HOW I would go about doing this, as I clearly (and hornestly) cannot grasp it.

答: This is a very confusing question. A homomorphism is a concept in algebra describing mappings which preserves the algebraic structure. To talk about homomorphism, you need an algebraic structure, (e.g. groups, rings, modules, vector space), and a homomorphism is something which preserves the operation of that structure.</p>

Title: Proving homomorphism from functions


答: A simple example would be $\Bbb Q$, endowed with its usual topology. It is metrizable, and therefore compactly generated. But it is not locally compact.

Title: Hausdorff compactly generated is locally compact?

Body:
I was trying to prove the "theorem of exponential correspondence" and "exponential law" with $X,Y,Z$ compactly generated and Hausdorff.

According to Hedwin H.Spanier-Algebraic Topology and Tammo tom Dieck-Algebraic topology we have that locally compact seems necessary in order to prove the statement.

The true hyphotesis given on the spaces are Hausdorff locally compact, but as far as I can see, compactly generated and Hausdorff doesn't imply locally compact, am I right ? Are my hypothesis wrong ? Are there any counterexample.

Thanks in advance


问:
Left coset and right coset of subgroup $H=\langle(234)\rangle$ in alternating group $A_4$
My homework question is:

Partition $G=A_4$ into left cosets of the subgroup $H=\langle (234)\rangle$

but I am not sure how to start with.

I know that $$A_4= \{(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)\},$$ am I right?

Also, the left coset of H is $xH=\{xh:h\in H\}$.

答: A 3-cycle has order 3, and $A_4$ has order 12, so there will be four left cosets by Lagrange's theorem. Recall the Klein four group is a subgroup of $A_4$, and it intersects trivially with $C_3$ since their elements have different orders, so each element of $K_4$ induces a left coset of $C_3$.


答:
Let $.^H$ be the Hermetian (conjugate) transpose, and let $\langle \cdot,\cdot\rangle$ be the standard inner product on $\mathbb{C}^n$.

In the forward direction, assume that $\lVert Ax \rVert = \lVert x\rVert$ for $x$ nonzero. Then this implies that:
$$
\langle Ax, Ax \rangle = \langle x,A^H Ax \rangle = \langle x,x\rangle \implies \langle x, A^H Ax - x\rangle = 0.
$$

Note that:
$$
\lVert A^H x\rVert^2 = \langle x,AA^H x\rangle\leq \lVert x\rVert\lVert AA^H x\rVert = \lVert x\rVert\lVert A^H x\rVert \implies \lVert A^Hx\rVert\leq \lVert x\rVert.
$$

But using Cauchy-Schwarz:
$$
\begin{split}
0 = \langle x,A^H Ax - x\rangle^2  \leq \lVert x\rVert^2\lVert A^H Ax - x\rVert^2 =\\ \lVert x\rVert^2(\lVert A^H Ax\rVert^2 - \langle A^H Ax,x\rangle - \langle x,A^H Ax\rangle + \lVert x\rVert^2) \leq\\
 \lVert x\rVert^2(\lVert Ax\rVert^2 - \langle A^H Ax,x\rangle- \langle x,A^H Ax\rangle +\lVert x\rVert^2) =\\ -\lVert x\rVert^2(\langle x,A^H Ax - x\rangle - \overline{\langle x,A^H Ax - x\rangle }) = 0,
\end{split}
$$
which implies that $\lVert A^H Ax - x\rVert = 0$, which in turn implies that $A^H A x = x\implies A^H A = I_n$

In the backward direction, assume $A^H A = I_n$. But then we trivially have
$$
\lVert x\rVert^2= \langle x,x\rangle = \langle x,A^H A x\rangle = \langle Ax,Ax\rangle= \lVert Ax\rVert^2.
$$


问:
Continued fraction $[0;0,0,0,\ldots] = \pm 1$
I would like to represent each $x\in\mathbb{R}_{>0}$ as a unique continued fraction. We represent each $x$ with $[a_{0};a_{1},a_{2},\ldots]$ as usual for the continued fraction. for irrational numbers, we know the sequence is infinite. For rational numbers we have a finite sequence. To have a unique representation in this notation I would like to extend the sequence with zero's. For example, $\frac{415}{93} = [4;2,6,7]$ would be extended to $[4;2,6,6,0,0,\ldots]$. We compute $$0+\frac{1}{0+\frac{1}{0+\ddots}},$$

$x_{n+1}=0+\frac{1}{x_{n}}\Rightarrow x^{2}=1\Rightarrow x=\pm 1.$

Would this be a unique representation for $x\in\mathbb{R}_{>0}$, or would this give a problem since the continued fraction of $[0;0,0,\ldots]$ is also $-1$. Possible way to work around this problem?

答:
My gut feeling is that you should not proceed in this direction at all, unless this new notation buys you something really big (some result that is impossible or very hard to get otherwise) - which is very unlikely, as it would be a result, essentially, about rational numbers.

I will elaborate.

This is because, with ordinary continued fractions, even if they are infinite, the resulting irrational number is the limit (in the ordinary calculus sense) of the sequence of the truncated (finite) continued fractions. However, once you start the tail of zeros, truncation is impossible. You will not be able to obtain the value of $0+\frac{1}{0+\frac{1}{0+\ldots}}$ using standard analytic limit. Your technique ("$x=0+\frac{1}{x}$") is useful to calculate limits that already exist, but here you don't even have a sequence that should converge to anything, because any finite truncation results in division by zero!

Of course, you may neglect all that and say "let me introduce some new notation, in which $0+\frac{1}{0+\frac{1}{0+\ldots}}$ has the value (say, $1$), and let me develop the mathematics that proves some rules about that new notation. It might be possible (although my guess is that it is very unlikely you will get to something that you cannot prove without using this notation).

A few last remarks:

Introducing zeros will make the continued fractions ambiguous. Say, $[1;1,1,1,1,1,\ldots]=\varphi$ but if you introduce zeros, $[1;0,0,1,1,0,0,1,1,0,0,1,1,\ldots]$ would be $\varphi$ too.

Also, what is $[1;0,1,0,1,0,1,\ldots]$? (It's got to be a number $x$ such that $x=x+1$.)

Altogether, I think by adding the possibility that some numbers in the continued fraction expansion are zeros, you lose some important properties of continued fractions - do you gain enough that would justify that loss?


问:
Challenge: find angle x in the figure via logical approach
Background: this problem was from social app, but I cannot find original source. This is an interesting problem that I have not seen similar ones. It is a hard problem, cannot be solved with angle chasing.  I worked out via analytic geometry programming, but cannot figure out a logical approach. Is there a logical approach to solve it? May need help from some theorems that I am not familiar with.

Problem: There are 5 known angles in the figure. Find the angle x. Is there an approach to solve it through deduction?

答:
Label the bottom two points (starting from left) with $A$, $B$, and the top points (starting from left) with $C$, $D$, $E$. Let $F$ be the intersection of $AD$ and $BC$. Let $\omega$ be the circumcircle of $ABF$. Let $G$ be the second common point of $\omega$ and $AE$. Let $H$ be the second common point of $BD$ and $\omega$. Finally, let $I$ be the point on the ray $BA^\to$ such that $AG=AI$.

Since $AI=AG$, we have $\angle AIG = \frac 12 \angle BAG = \frac 12 \cdot 44^\circ = 22^\circ$. Then we find that $\angle GBI = \angle FBI + \angle GBF = 49^\circ + \angle GAF = 49^\circ + 30^\circ = 79^\circ$, and $\angle IGB = 180^\circ - \angle BIG - \angle GBI = 180^\circ - 22^\circ - 79^\circ = 79^\circ$. Hence $IBG$ is isosceles and $IB=IG$.

Let $\Omega$ be the circle with center $I$ and radius $IB=IG$. Let $C'\in\Omega$ be such that the chord $BC'$ passes through $F$. Note that $\angle GIC' = 2 \angle GBC' = 2\angle GBF = 2\cdot 30^\circ = 60^\circ$, hence triangle $GIC'$ is equilateral. In particular, $C'I=C'G$. Since $AI=AG$, it follows that $AC'$ bisects the angle $GAI$. Hence $\angle GAC' = \frac 12 \angle GAI = \frac 12 \cdot (180^\circ - \angle BAG) = 90^\circ - \frac 12 \cdot 44^\circ = 90^\circ - 22^\circ = 68^\circ = \angle GAC$, and therefore $C'$ lies on the line $AC$.

Moreover, $\angle C'BI = 90^\circ - \frac 12 \angle BIC' = 90^\circ - \frac 12(\angle BIG + \angle GIC') = 90^\circ - \frac 12 \cdot (22^\circ + 60^\circ) = 49^\circ = \angle CBI$, and therefore $C'$ lies on the line $BC$.

Hence $C'$ is the intersection of $AC$ and $BC$, hence $C'$ coincides with $C$.

Next, we shall prove that $C, G, H$ are collinear. Indeed, $\angle CGH = \angle CGA + \angle AGH = \angle AIC + (180^\circ - \angle HBA) = \angle GIC + \angle BIG + 180^\circ - 82^\circ = 60^\circ + 22^\circ + 180^\circ - 82^\circ = 180^\circ$.

Applying Pascal theorem for (degenerated) hexagon $FBBHGA$ we see that $BE$ is tangent to $\omega$. Therefore $\angle EBD = \angle BAH = \angle BAF - \angle HAF = 74^\circ - \angle HBF = 74^\circ - 33^\circ = 41^\circ$.

Remark: This is a special case of the following phenomenon:

The proof presented above can be adapted to the general case.


问:
What is the relationship between $(u\times v)\times w$ and $u\times(v\times w)$?
Given three vectors $u$, $v$, and $w$, $(u\times v)\times w\neq u\times(v\times w)$. This has been a stated fact in my recent class. But what is the ultimate relationship between them? I would presume that one is a scalar multiple of the other?</p>

答:
What you see here by the inequality of the left-hand side and the right-hand side is that the cross product (in this case, the vector triple product) is not associative $(\dagger)$:

$$(\vec u\times \vec v)\times \vec w\neq \vec u\times(\vec v\times \vec w)$$

Using boldface to represent the vectors:

$$ \mathbf{u}\times (\mathbf{v}\times \mathbf{w}) = (\mathbf{u}\cdot\mathbf{w})\mathbf{v} - (\mathbf{u}\cdot\mathbf{v})\mathbf{w}\tag{1}$$

$$(\mathbf{u}\times \mathbf{v})\times \mathbf{w} = -\mathbf{w}\times(\mathbf{u}\times \mathbf{v}) = -(\mathbf{w}\cdot\mathbf{v})\mathbf{u} + (\mathbf{w}\cdot\mathbf{u})\mathbf{v}\tag{2}$$

See Wikipedia for more about triple products of vectors

$(\dagger)$ "In mathematics the Jacobi identity is a property that a binary operation [like the triple vector cross product] can satisfy which determines how the order of evaluation behaves for the given operation. Unlike for associative operations, order of evaluation is significant for operations satisfying Jacobi identity. It is named after the German mathematician Carl Gustav Jakob Jacobi." (Wikipedia).


问:
Why is $M(X)=C_c(X)^*=C_0(X)^*$?
Let $X$ be a a locally compact, separable and complete metric space.and $M(X)=\{$finite signed measures on $X\}$. Now I learned that $M(X)=C_c(X)^*$ i.e. it is the dual of continuous functions compactly supported in $X$. Now it seems $M(X)=C_0(X)^*$ as well (dual of continuous functions vanishing at $\infty$ as well. But why is that? Does it have something to do with the fact that $C_0(X)$ is the set of uniform limits of elements of $C_c(X)$?

答: If $X$ is a normed linear space and $M$ is dense subspace of $X$ then $M$ and $X$ have the same dual. This is because any continuous linear functional on $M$ has a unique extension to a continuous linear functional on $X$.

Title: Why is $M(X)=C_c(X)^*=C_0(X)^*$?


问:
A question regarding exponential distribution
Charlie and Bella and their friends Mark and Leonard each have a toy. Each toy breaks at a time that is exponentially distributed with expectation $24$ hours. Assume the toys are independent of each other.

What is the cumulative distribution function of the time until Charlie's toy fails?

Let $X$ be the time until Charlie's toy fails. $E[X]=\frac{1}{\lambda}=24$. Then $\lambda=\frac{1}{24}$. And so the cumulative distribution function is

$$F(x) = \begin{cases}
1-e^{-x/24}& \text{ if } x \ge 0, \\
0 & \text{ if } x \lt 0.
\end{cases}$$

Let $T$ be the time until all the toys have failed. Compute the cumulative distribution function of $T$.

The toys are independent of each other, so $E[T]=96$. Then $\lambda=\frac{1}{96}$. Hence the cumulative distribution function of $T$ is

$$F(t) = \begin{cases}
1-e^{-x/96}& \text{ if } t \ge 0, \\
0 & \text{ if } t \lt 0.
\end{cases}$$

Let $S$ be the time until the first toy has failed. Compute the cumulative distribution function of $S$.

Are the attempts above correct? How do I do the last one? Thank you.

答:
The first part is solved correctly.

Let $T$ be the time until all the toys have failed. The probability that $T\le t$ is the probability that all the toys have failed by time $t$. By independence this is the fourth power of the cdf you calculated in the first part. In symbols,
$$F_T(t)=\left(1-e^{-t/24}\right)^4.$$
The time $T$ is usually called $\max(X_1, X_2,X_3,X_4)$, where the $X_i$ are the lifetimes of the toys of the various people.

For the last, let $S$ be the time until the first failure. The probability this is $\gt s$ is the probability all the toys are alive at time $s$. This is $(e^{-s/24})^4$. So
$$F_S(s)=1-\left(e^{-s/24}\right)^4.$$
The time $s$ is usually called $\min(X_1, X_2,X_3,X_4)$.

Remark: In your solution of the second problem, you asserted that the mean of $T$ is the sum of the means of the $X_i$. There is no reason to think that. The mean of the sum of the $X_i$ is the sum of the means, but $T$ is not the sum of the $X_i$. You also assumed that $T$ has exponential distribution. It doesn't. However, $S$ does.


问:
Show that solution of a Cauchy problem is not globally defined
Hi everyone: I need a check on the following exercise

\begin{cases} u'(t)=\sin(t)^2 - u(t)^2 \\ u(0)=-1\end{cases}

Show that the maximal solution $u:[\alpha,\beta] \rightarrow \mathbb{R}$ is such that $\beta < \infty$

First of all, the problem is well posed: the r.h.s. of the ODE is smooth and therefore we can infer there exists a unique local solution. Then, I note immediately that $$\sin(t)^2-u^2(t) > - u^2(t)$$ for all the $t$'s for which the solution is defined

Hence $$u'>-u^2$$ and integrating from $0$ to $t$ I obtain $$\frac{1}{u(t)} + 1 < t$$ and then $$u(t) > \frac{1}{t-1}$$ again for all $t$'s such that the solution is defined. But the r.h.s explodes as $t \rightarrow 1^{-}$ and therefore we can say that the $$\beta<1$$ is necessary for the solution norm not to explode

This seems to be confirmed also numerically. Is everything okay?

答:
At $t=0$ the right side is negative. It is easy to observe, perhaps harder to formulate, that this implies that at no point the right side can become positive, $u$ is continuously falling and thus always smaller than $-1$. Then for $t\in[0,1]$
$$
u'(t)=\sin^2t-u(t)^2\le t^2-1\implies u(t)\le -1-t+\frac{t^3}3=v(t).
$$
so the solution moves decidedly away from the level $-1$ downward and
$$
u(1)\le-\frac53.
$$

Now use $\sin^2t\le 1\le -u(t)$, $u(t)^{-1}+1\ge0$ to derive a simple upper bound for the solution for $t\ge 1$,
$$
u'\le -u-u^2\implies (u^{-1}+1)'=-u^{-2}u'\ge u^{-1}+1
\\~\\
\implies
u(t)^{-1}+1\ge (u(h)^{-1}+1)e^{t-h}
$$
So in the numerically simple case $h=1$ one gets for $t\ge1$
$$
u(t)^{-1}+1\ge\frac25e^{t-1}
\iff
u(t)\le-\frac{5}{5-2e^{t-1}}
$$
This upper bound diverges to $-\infty$ at $t=1+\ln\frac52$, thus forces the solution to diverge in this direction at some time before that.

This is a rather large bound, as the numerical solution diverges shortly after $t=1$, closely following the lower bound $-\frac1{1-t}$ discussed in the question.


问:
What is the greatest integer that divides $p^4-1$ for every prime number $p$ greater than $5$?
What is the greatest integer that divides $p^4-1$ for every prime number $p > 5$?  This was on a practice math GRE so it's probably really easy.</p>

答:
I suppose this is more of a brute force method. By FlT, you know $p^4\equiv 1\pmod{5}$, so $5$ is a factor. Also, factoring, you get

$$p^4-1=(p-1)(p+1)(p^2+1),$$

and since $p$ is odd, each factor is divisible by $2$. Moreover, one of the factors $p-1$ or $p+1$ is divisible by $4$, so $4\cdot2\cdot 2=16$ is another factor. Finally, of the three consecutive integers, $p-1$, $p$, and $p+1$, $3$ is a factor of one. Since $p$ is prime and larger than $5$, $3\nmid p$, so $3|p^4-1$. Hence $16\cdot 3\cdot 5=240$ is a factor of $p^4-1$ for any prime $p\gt 5$.

It still remains to see that $240$ is the greatest such divisor. This can be confirmed by looking at $7^4-1=2400$ and $11^4-1=14640$.


问:
Prove that $V$ is the direct sum of $W_1, W_2 ,\dots , W_k$ if and only if $\dim(V) = \sum_{i=1}^k \dim W_i$
Let $W_1,\dots,W_k$ be a subspace of a finite dimensional vector space $V$ such that the $\sum_{i=1}^k W_i = V$.

Prove: $V$ is the direct sum of $W_1, W_2 , \dots, W_k$ if and only $\dim(V) = \sum_{i=1}^k \dim W_i$.

答:
Define $\ \ B_i$ a basis of $ \ \ W_i\ \ $  for $\ \ i \in {1,..,k } $. Since sum( $W_i$)$= V$ we know that  $\bigcup_{i=1}^k B_i$ is a spanning list of $V$.

Now we add the condition : sum dim($B_i$)$=$sum dim($W_i$)$=$dim($V$).

This means that $\bigcup_{i=1}^k B_i$ is a basis. (Spanning list of $V$ with same dimension).

Can you conclude from here the direct sum property?

If we define Basises as before and we know that $W_i$ is a direct sum can you conclude ( knowing that sum( $W_i$)$= V$ , wich is given)  that $\ \ \bigcup_{i=1}^k B_i$ is a Basis of $V$?


问:
Minimize $\quad z=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}$ s.t $\quad x_{1}+x_{2} \leq 2 \ ,x_{1}, x_{2} \geq-1$ without KKT conditions
The problem is to minimize $\quad z=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}$
Subject to the following constraints.
$$\quad x_{1}+x_{2} \leq 2$$$$ x_{1}, x_{2} \geq-1$$
I don't want to solve this problem with the KKT conditions. I am looking for a more intuitive way, maybe using geometry or inequalities. a simple approach but I can't seem to think of anything outside KKT which I can't use.

Any suggestions??

答:
Make a figure!

The feasible domain $\Omega$ is a right triangle with hypotenuse $h$ connecting the points $(-1,3)$ and $(3,-1)$. The objective function is $f(x_1,x_2):=4(x_1-1)^2+(x_2-2)^2$. Its level lines are ellipses centered at the point $(1,2)$ slightly above $h$. It is then obvious that the smallest ellipse intersecting $\Omega$ is the ellipse touching $h$ at some point. It follows that we have to find the point on $h$ having smallest value of $f$.

To this end we parametrize $h$ by
$$h:\qquad t\mapsto\bigl(x_1(t),x_2(t)\bigr):=(t,2-t)\quad(-1\leq t\leq 3)\ ,$$
and consider the pullback of $f$ to $h$:
$$\phi(t):=f\bigl(x_1(t),x_2(t)\bigr)=4(t-1)^2+\bigl((2-t)-2\bigr)^2=5t^2-8t+4\ .$$
The minimum is taken at $t={4\over5}\in[-1,3]$ and leads to the minimum value $z_*=f\bigl({4\over5},{6\over5}\bigr)={4\over5}$.


问:
Find $\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$
Find the limit of $$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$$ without l'Hospital.

I tried expanding $1-\cos(x)$ to get something this kind $$\lim_{x\to 0}\frac{\sin x}{x}$$ but couldn't solve it. The answer is $1/p$.

答:
A little bit of trigonometry does the job:
$$\begin{align}
\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)} &= \lim_{x\to 0}\frac{1+\sqrt 2\sin(x-\pi/4)}{1+\sqrt 2 \sin(px-\pi/4)} \\[1mm]
&= \lim_{x\to 0}\frac{\frac{1}{\sqrt 2} + \sin(x-\pi/4)}{\frac{1}{\sqrt 2} + \sin(px - \pi/4)} \\[1mm]
&= \lim_{x\to 0}\frac{\sin(\pi/4) + \sin(x-\pi/4)}{\sin(\pi/4) + \sin(px - \pi/4)} \\[1mm]
&= \lim_{x\to0} \frac{\sin(x/2)\cos(\pi/2)}{\sin(px/2)\cos(\pi/2)} \\[1mm]
&= \lim_{x\to0} \underbrace{\frac{\sin(x/2)}{\color{red}{x/2}}}_{\to 1}\underbrace{\frac{\color{blue}p\color{red}{x/2}}{\sin(px/2)}}_{\to 1}\frac {1}{\color{blue}p} \\[1mm]
&= \frac 1p\end{align}$$

See MSE discussions about $\lim_{x\to0}\frac{\sin x}{x}=1$


答:
$$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}=\lim_{x\to 0} \frac{\sin x+2\sin^2 x/2}{\sin(px)+2\sin px/2}.$$
Make use of the properties of the limit:
$$ \frac{\lim_{x\to 0}\sin x+\lim_{x\to 0}2\sin^2 x/2}{\lim_{x\to 0}\sin(px)+\lim_{x\to 0}2\sin px/2} = \lim_{x\to 0}\frac{\sin x}{\sin px}.$$
Algebraic manipulation:
$$\lim_{x\to 0}\frac{x\sin x}{x\sin px}=\lim_{x\to 0}\frac{\sin x}{x}\lim_{x\to 0}\frac{x}{\sin px}=1\cdot\lim_{x\to 0/p}\frac{x/p}{\sin x}=\frac{1}{p}.$$

Here is a geometric approach (without l'hopital) that proves $\lim_{x\to 0}\frac{\sin x}{x}=1.$

Title: Find $\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$

Body:
Find the limit of $$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$$ without l'Hospital.

I tried expanding $1-\cos(x)$ to get something this kind $$\lim_{x\to 0}\frac{\sin x}{x}$$ but couldn't solve it. The answer is $1/p$.


答: Let $f(x) = \sin{x} - \cos{x}$. Then, $$1 = f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{\sin{h} - \cos{h} + 1}{h}$$ Make the substitution $h\mapsto ph$ and this limit becomes $$1 = f'(0) = \lim_{h \to 0} \frac{\sin{ph} - \cos{ph} + 1}{ph}$$ Finally, note that: $$ 1 = \frac{1}{1} =  \frac{f'(0)}{f'(0)} =  \lim_{h \to 0} \left ( \frac{\frac{\sin{h} - \cos{h} + 1}{h}}{\frac{\sin{ph} - \cos{ph} + 1}{ph}} \right )$$ So, $$\lim_{h\to 0} \frac{1+\sin h-\cos h}{1+\sin{ph}-\cos{ph}} = \frac{1}{p}$$


问:
CDF of $1-X-Y$ where $X\sim U[0,1]$ and $Y\sim U[0, 1-X]$
I am trying to prove that the random variable:

$Z = 1-X-Y$

where $X\sim U[0,1]$ and $Y\sim U[0, 1-X]$

is $Z \sim U[0,1]$. I'm not sure this is even true, but it feels correct. I'm having trouble with the formalities of how to go about attempting to prove this, especially since $Y$ depends on $X$. Any guidance would be much appreciated.

Body: $Z$ is not uniform. For $0<t<1$ we have $$P(Z \leq t)$$ $$=P(1-X-Y \leq t)=E(P(Y \geq 1-X-t|X)$$ $$=E\frac {(1-X)-(1-X-t)} {1-X} I_{0\leq 1-X-t \leq 1}+P(X>1-t)$$ $$=E\frac t {1-X} I_{X\leq 1-t}=\int_0^{1-t} \frac t {1-x} dx+t=t-t \ln t$$.


问:
Prove that $AG+GB+GH+DH+HE\ge CF.$
In the diagram, $ABCDEF$ is a hexagon with $AB=BC=CD$ and $DE=EF=FA$. Angles $BCD$ and $EFA$ both equal $60°$. $G$ and $H$ are two points taken from inside the hexagon such that angles $AGB$ and $DHE$ both equal $120°$. Prove that $AG+GB+GH+DH+HE\ge CF.$

I tried to construct equilateral triangles by using the angle bisector of the two $120°$ angles, but still could not link anything to the conclusion. I also know that the conclusion is equivalent to $AB+GH+DE\ge CF$ by Triangle Inequality. Please advise.

答:
$|AB| = |BD|,\ |DE|=|AE|$ Hence we have $$ \Delta ABG \sim \Delta
 DBG',\ \Delta DHE \sim \Delta AH'E$$ s.t. $|GH|=|G'H'|$

And we have $G'',\ H''$ s.t. $$
 \Delta BDG' \sim \Delta CDG'',\ \Delta
    AEH' \sim\Delta AFH''$$

Hence clearly $$|CG''| + |G''G'|+|G'H'| + |H'H''| + |H'' F|\geq |CF|$$

Title: Prove that $AG+GB+GH+DH+HE\ge CF.$


问:
Differential equation based on initial values
I have a differential equation that looks like:

$$y'=y(4-y)$$
I have identified the critical points to be $0$ and $4$.
I need to identify the behavior of the equation as $t \rightarrow \infty$ and this behavior it seems depends on the initial value.

Now depending on the differential equation I can say that

If $y > 4$, then $y$ is decreasing

If $0<y<4$, then $y$ is increasing

If $y<0$, then $y$ is decreasing

And I infer this based on the slope values calculated from the $y'$ equation.

So based on this I was able to conclude that as $t \rightarrow \infty$, $y$ converges to $4$ and and diverges from $0$. But how do I identify this behavior from initial values? I feel like I'm missing something here.

答:
There are two equilibrium points (constant solutions) $y = 0$ and $y = 4$. By the uniqueness theorem other solutions cannot intersect with those two. The stable point $y = 4$ is a hyperbolic attractor and has a basin of attraction $(0, +\infty)$. The unstable point $y = 0$ is a hyperbolic repeller with basin $(-\infty, 0)$. For example, as you have figured out, $y$ is increasing on phase curves in $(0, 4)$ with some bounded below speed in $(\delta, 4-\delta)$ for any positive small enough $\delta$. But any trajectory with initial values in $(0,4)$ cannot leave the region. So it will converge to $4$ asymptotically.

Actually, it is possible to just find a general solution, so you can use it for anything you what to prove.