问:
Recurrence sequences
So i got a recurrence sequence of $$a_n=2a_{n-1}+4a_{n-2}-8a_{n-3}$$
With                    $$b_n=a_n-2a_{n-1}$$
I was able to show that      $$b_n=A2^n+B(-2)^n$$
Was the general solution for the sequence $b_n$.

Now I need to use induction on k to prove that

$$a_n=kA2^n+B(-2)^{n-1} ((-1)^k-1)+2^k a_{n-k}\tag 1$$      for $k\ge 1$
And deduce that
$$a_n=nA2^n-B2^{n-1}-B(-2)^{n-1}+2^n a_0$$

No matter how i shifted the numbers and indicied i cant get my inductive step to work. Also confused about the deduce part, Does the sequence b play a part in this?

Title: Recurrence sequences

Body:
We want to prove $a_n=kA2^n+B(-2)^{n-1} ((-1)^k-1)+2^k a_{n-k}\tag 1$

When $k=1$ it's true. If it's true for $k$, we have

$$a_n=kA2^n+B(-2)^{n-1} ((-1)^k-1)+2^k a_{n-k} \\
=kA2^n+B(-2)^{n-1} ((-1)^k-1)+2^k (b_{n-k}+2a_{n-k-1}) \\
=kA2^n+B(-2)^{n-1} ((-1)^k-1)+2^k (A2^{n-k}+B(-2)^{n-k}+2a_{n-k-1})\\
=kA2^n+A2^n+B(-2)^{n-1}((-1)^k-1-2(-1)^k)+2^{k+1}a_{n-k-1}\\
=(k+1)A2^n+B(-2)^{n-1}((-1)^{k+1}-1)+2^{k+1}a_{n-k-1}
$$

Note that $(-1)^k-1-2(-1)^k=(1-2)(-1)^k-1=(-1)^{k+1}-1$.


问:
Group homomorphism between quotient groups.
Let $G$ and $H$ be two groups and $M, N$ be normal subgroups of $G$ and $H$ respectively. Consider a relation $\phi: G \to H$ be a homomorphism.
Now, my question is

what is the necessary condition on $\phi$ such that it induces a homomorphism between $\frac{G}{M}$ and $\frac{H}{N}.$

答:
The map $\pi\circ \phi\colon G\to H/N$ factors through $G/M$ if and only if $M\subseteq \mathrm{ker}(\pi\circ\phi)$. The kernel of $\pi\circ\phi$ is $\phi^{-1}(\ker(\pi)) = \phi^{-1}(N)$.

So a necessary and sufficient condition is $M\subseteq \phi^{-1}(N)$.

ContentLicense: CC BY-SA 4.0


问:
Localization is finitely generated implies, that multiplicative group only contains invertible elements
I am currently stuck on this problem:

"Let $R$ be an integral domain, let $0 \notin S \subseteq R$ be a multiplicative group of $R$. If $S^{-1}R$ is finitely generated as a $R$-module, then every element of $S$ is invertible (in $R$)"

I just started doing Algebra last month and I have some difficulties grasping these abstract concepts. So let's assume, that $S^{-1}R$ is finally generated by $\frac{1}{s_1}, ... , \frac{1}{s_n}$ with $s_1, ... , s_n \in S$. What can I do now? My first guess, that $S$ has to be finite as well, because if I understood it right, the localization $S^{-1}R$ describes a Ring, that is a little bit bigger than $R$ that also contains the multiplicative inverses of all Elements of $S$. So for example, if $R = \mathbb{Z}$ and $S = \mathbb{N}$, then $S^{-1}R = \mathbb{Q}$.
Is that so far correct? But in this case, $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$-module, so it's not a very good example, I tried it on $Z_{11}$, and as a field it obviously contains multiplicative inverses, so every multiplicative group is a unit. Also, it looks like if all such Rings $R$ would already satisfy $R = S^{-1}R$, is that true?

Nevertheless, I still don't know how to formally prove it, can someone maybe give me a small tip, because right now I don't even know where to start.

答: Hint for a direct argument: suppose that $S^{-1}R$ is generated as an $R$-module by finitely many $r_i/s_i$.  Let $s = \prod s_i$.  Rewrite $r_i / s_i = r_i'/ s$ where $r_i' = r_i \prod_{j \not= i} s_j$.  Now for any $t \in S$, consider the element $1 /ts \in S^{-1}R$.  Write $1/ts$ as an $R$-linear combination of the $r_i'/s$ and mess around with the expression to deduce that $t$ is a unit of $R$.


问:
Consecutive Integers Pythagorean Triplets
I have a question for which I was not able to find an answer online. I was wondering how many Pythagorean Triplets we have found till now which consists of three consecutive integers like $(3,4,5)$.

答:
$$x^2 +(x+1)^2=(x+2)^2\\\implies x^2 +(x+1)^2-(x+2)^2=0\\\implies x^2 - 2 x - 3 = 0$$

Using the quadratic equation, we find
$$x\in\{3,-1\}\implies \text{the set of triples}\quad S=\{(3,4,5),(-1,0,1)\}$$
There are no other solutions.


问:
Compactness preserved in subspace
Suppose $A,B$ are topological spaces, $B $ is a subspace of $A $ and  $X\subseteq B $.

If $X $ is compact in $ A$ then is $X $ compact in B?

If an open covering of $X$ in $B$ is in fact an open covering in $A$ then by compactness of $A$ we would have a finite sub covering.

答:
Suppose $X$ is compact relative to $A$  and let $ \{O\}_i$ be collection of open sets relative to $B$, such that $X\subset \bigcup_i O_i$.
By definition of subspace topology,
there exists $N_i$ open in $A$, such that $O_i=B\cap N_i$, for all $i$. Since $X$ is compact in $A$.
Therefore,
$$X\subset N_{i_1}\cup\dots \cup N_{i_k}$$ for some choice of finitely many indices $i_1,\dots,i_k$.

Since $X\subset B$, $\Longrightarrow$
$X\subset O_{i_1}\cup\dots \cup O_{i_k}$.

Title: Compactness preserved in subspace


问:
Check for stability and asymptotic stability of the zero solution for a system
I am trying to show that the zero solution for a system is unstable even though both eigenvalues are negative.
This is the system:
$$ \vec{y'} = \begin{pmatrix}
-1 & e^{2t} \\
0 & -1
\end{pmatrix} \vec{y}.$$
I know that I have to find the fundamental matrix, but I am stuck.
This is what I have:
$$|A-\lambda I| = \begin{vmatrix}
-1- \lambda & e^{2t} \\
0 & -1-\lambda
\end{vmatrix} = (-1-\lambda)^2,$$ which means that both eigenvalues are equal to -1.
The only eigenvector is: $\begin{bmatrix}
1  \\
0
\end{bmatrix}$. Is this the fundamental matrix:
$$\begin{pmatrix} e^{-t} & e^{-t} \\ 0 &0
\end{pmatrix}$$
Do the zeros in the second imply that the zero solution is unstable, because there would be infinitely many solutions?
I want to know if I am at least on the right track; I would really appreciate the help.

答:
Your system is unusual in that the coefficients of the matrix $A$ in the equation $y' = Ay$ are not constant.

I think it is best to solve this "from scratch" as a system of equations. In terms of the separate components of $\vec y$, the system is
$$
\frac{dy_1}{dt} = -y_1 + e^{2t}y_2\\
\frac{dy_2}{dt} = -y_2.
$$
Solving the second equation yields $y_2 = C_1 e^{-t}$. Plugging this solution into the first system gives us
$$
\frac{dy_1}{dt} = -y_1 + C_1 e^t \implies\\
\frac{dy_1}{dt} + y_1 = C_1 e^t \implies\\
e^t\frac{dy_1}{dt} + e^ty_1 = C_1 e^{2t} \implies\\
\frac d{dt}[e^t y_1] = C_1 e^{2t} \implies\\
e^t y_1 = \frac {C_1}{2} e^{2t} + C_2 \implies\\
y_1 = \frac{C_1}{2} e^t + C_2 e^{-t}.
$$
In other words, every solution to this differential equation can be written in the form
$$
\vec y = \pmatrix{e^{-t} & \frac 12 e^t \\0 & e^{-t}}\pmatrix{C_2\\C_1}.
$$


问:
Condition under which a ring is isomorphic to a right ideal.
Defintion: an element $a\in R$ is said to be right regular if, there is no nonzero element $b\in R$ such that $ab=0$.

My Questuion: Let $R$ be a ring with identity. Then $R$ is $R$-isomorphic to a right ideal $I$ of $R$ (as a right $R$-module) if and only if $I=aR$ for some right regular element of $R$.

答:
Now I have find a solution of my problem which is as follows:

First suppose that there exists an $R$-isomorphism $f:R\rightarrow I$. Let $a=f(1)\in R$.
Then $ab=0\implies f(1)b=0\implies f(b)=0\implies b=0$ for any $b\in R$, as $f$ is one-one.  Let $s\in I$ then there exist $r\in R$ such that $s=f(r)=f(1)r=ar$, as $f$ is onto.
Thus $a$ is a right regular element such that $I=aR$.

Conversely suppose that there exists a right regular element $a\in R$ such that $I=aR$. Define a map $f:R\rightarrow I$ by $f(r)=ar,\forall r\in R$. Then $f$ is an $R$-isomorphism.


问:
Arithmetic progression $(a,b,c)$ with $a^2+b^2=c^2$ and $\gcd(a,b,c) = 1$
I was wondering whether or not primitive Pythagorean triples could form an arithmetic sequence with their terms. I know that $(3,4,5)$ satisfies this condition, but I was wondering whether or not there are other ones, and how many there could be. Thanks!

答:
Among primitives
$$x^2 +(x+1)^2=(x+2)^2\\\implies x^2 +(x+1)^2-(x+2)^2=0\\\implies x^2 - 2 x - 3 = 0$$
Using the quadratice equation, we find
$$x\in\{3,-1\}\implies \text{the set of triples}\quad S=\{(3,4,5),(-1,0,1)\}$$
There are no other primitive solutions and, for non-primitives there are only mulitples of these like

$$(6,8,10),(9,12,15),(12,16,20)\cdots\quad (-2,0,2),(-3,0,3),(-4,0,4)\cdots$$


问:
How to find the eigenvalues of this matrix in a simple way?
Can anyone help how to find the eigenvalues of the following matrix in a simple way? I expand the characteristic polynomial being,
$$
    \lambda(\lambda-3)(\lambda - 2k) = 0
$$
and get the answer but intuition is that there must be a simple way to find it.
$$
    \begin{bmatrix} 1 + k & 1 & 1 - k \\ 1 & 1 & 1 \\ 1 - k & 1 & 1 + k \\ \end{bmatrix}
$$

答:
Glancing at the matrix might suggest cancelling $k$:

Multiply your matrix by $\begin{bmatrix}1 & y & 1\end{bmatrix}^T$ to get $\begin{bmatrix}2 & 2 & 2\end{bmatrix}^T + y\begin{bmatrix}1 & 1 & 1\end{bmatrix}^T$.

Setting $y = -2$ gives $0$ as an eigenvalue and $y = 1$ gives $3$ as an eigenvalue. (Find these values for $y$ either by inspection or solving $(2+y)y = 2 + y$.)

Now the trace constraint gives the last eigenvalue.


问:
A integral involving hyperbolic and trigonometric functions
So I have been pondering this question for a while, and nothing is coming through. I have tried multiple $u$-substitutions, and been attempting many other tricks I have learned, but it simply didn't come out. Also, Wolfram Alpha can't find a solution. This is the problem:
$$\int_0^1 \tanh(\cos(x)) dx$$

I understand there may be no closed form solution, but at least it is not an indefinite integral. At the least, a series representation would be okay.

答:
I think that the faster would be to use the series expansion of $\tanh(t)$ which gives
$$ \tanh(\cos(x))=\sum_{n=1}^\infty \frac{4^n \left(4^n-1\right) B_{2 n} }{(2 n)!}\cos ^{2 n-1}(x)$$ which leads to
$$I=\int_0^1\tanh(\cos(x))\,dx=$$ $$I=\frac{1}{2}\sum_{n=1}^\infty  \left(4^n-1\right) B_{2 n} \left(\frac{\pi }{n \Gamma
   \left(n+\frac{1}{2}\right)^2}-\frac{4^n B_{\cos
   ^2(1)}\left(n,\frac{1}{2}\right)}{(2n)!}\right)$$

Computing the partial sums from $n=1$ to $n=p$, some results
$$\left(
\begin{array}{cc}
p & \text{result} \\
 1 & 0.841471 \\
 2 & 0.627183 \\
 3 & 0.697668 \\
 4 & 0.673052 \\
 5 & 0.681934 \\
 6 & 0.678660 \\
 7 & 0.679885 \\
 8 & 0.679422 \\
 9 & 0.679598 \\
 10 & 0.679531 \\
 11 & 0.679557 \\
 12 & 0.679547 \\
 13 & 0.679551 \\
 14 & 0.679549 \\
 15 & 0.679550 \\
 16 & 0.679549
\end{array}
\right)$$

Composing the Taylor series in terms of $x$ would be a pure nightmare.

However, for a quick and dirty approximation, we could use the $[2n,2]$Padé approximant of the integrand. For example
$$\tanh(\cos(x))\sim \frac {t+a x^2+b x^4}{1+c x^2}$$ with
$$a=\frac{15 t^2-61 t+15}{30 (6 t-1)} \qquad b=\frac{21 \left(t^2-1\right)}{40 (6 t-1)}\qquad c=\frac{-90 t^2+30 t+29}{30 (6 t-1)}$$ and $t=\tanh(1)$. This would give as a result
$$I \sim  \frac {a+b}c-\frac{b}{c^2}+\frac{ c (c t-a)+b}{c^{5/2}}\tan ^{-1}\left(\sqrt{c}\right)\sim 0.679076$$


问:
Validity of proving identities by showing LHS-RHS =0 or using reversible steps?
When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:

Method 1:
Showing that $\mathrm{LHS}-\mathrm{RHS}=0$

For instance, if we are required to prove

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

We instead show that $\mathrm{LHS}-\mathrm{RHS}=0$ as follows:

$\mathrm{LHS} - \mathrm{RHS} = x^2\cos^2(x)+\sin^2(x) - x^2 + x^2\sin^2(x) - 1 + \cos^2(x)$

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

$= x^2 - x^2 + 1 - 1 = 0$

Method 2: Showing that $\mathrm{LHS}=\mathrm{RHS}$ is equivalent with another equation which is true (taking care that we can always reverse the steps and showing it by putting $\iff$):

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

$ \iff  x^2\cos^2(x)+x^2\sin^2(x) - x^2 = 1 - \sin^2(x) - \cos^2(x)$

$ \iff x^2 - x^2 = 1 - 1$

$ \iff 0 = 0$

Are these two methods of proving valid? Are there any cases where we can make fallacious argument by using these methods? Are any one of them better than the other?

答: This question works both from LHS $\rightarrow$ RHS and RHS $\rightarrow$ LHS to prove it.

ContentLicense: CC BY-SA 3.0


问:
Is there a smooth curve connecting $\left(\frac{1}{2},2\right)$ and $\left(2,\frac{1}{2}\right)$ that maximizes $\int_{C}\textbf{F}\cdot d\textbf{r}$?
IMPORTANT: This is a problem I created for fun, so if it is ill-posed in any way, please let me know in the comments. Also, I think my title is a bit vague, so please suggest any improvements that come to mind.

Let's consider the vector field
$\textbf{F}(x,y)=x\textbf{i}-y\textbf{j}$. Out of all the smooth
curves lying in $(0,\infty)^2$, is there one connecting the points
$\left(\frac{1}{2},2\right)$, $\left(2,\frac{1}{2}\right)$ that
maximizes the value of $\int_{C}\textbf{F}\cdot d\textbf{r}$? Here, a
curve is said to be smooth if it admits a parameterization
$\textbf{r}$ with a continuous, non-vanishing derivative.

I've spent some time thinking about this problem, and am beginning to suspect that the curve, if it exists, is the hyperbola $y=\frac{1}{x}$, where $\frac{1}{2}\leq x\leq 2$. Here's how I arrived at this conclusion:

If we interpret the line integral $\int_{C}\textbf{F}\cdot d\textbf{r}$ as the work done by $\textbf{F}$ on a particle moving along the "track" drawn out by $\textbf{r}(t)=x(t)\textbf{i}+y(t)\textbf{j}$, then my intuition of work suggests that it will be maximized when the track "flows" with $\textbf{F}$ (if it didn't, then $\textbf{F}$ would act against the particle at certain times). More precisely, I think that $\textbf{r}'(t)=\textbf{F}\left(\textbf{r}(t)\right)$ should be true for all $t$ in the domain of $\textbf{r}$, as this condition will ensure that the particle points in the direction of the force field at all times.

Using the fact that $\textbf{r}'(t)=x'(t)\textbf{i}+y'(t)\textbf{j}$ and $\textbf{F}\left(\textbf{r}(t)\right)=\textbf{F}\left(x(t),y(t)\right)=x(t)\textbf{i}-y(t)\textbf{j}$, we can transform the equation $\textbf{r}'(t)=\textbf{F}\left(\textbf{r}(t)\right)$ into the system

\begin{align*}
x'(t) &= x(t)\\
y'(t) &= -y(t)
\end{align*}

This gives $x(t)=Ce^t$ and $y(t)=Ke^{-t}=\frac{K}{e^t}$ (clearly $K,C\neq 0$). We deduce that the Cartesian equation of the curve is

$$y=\frac{K}{\frac{x}{C}}$$

or

$$y=\frac{CK}{x}$$

This curve passes through $\left(\frac{1}{2},2\right)$ and $\left(2,\frac{1}{2}\right)$, so we end up with the following system

$$2=\frac{CK}{\frac{1}{2}}\iff CK=1$$
$$\frac{1}{2}=\frac{CK}{2}\iff CK=1$$

They're both the same equation, so we solve for one of the constants, say $K$.

$$K=\frac{1}{C}$$

Substituting this into the equation of the curve gives

$$y=\frac{C\cdot\frac{1}{C}}{x}=\frac{1}{x}$$

which is the section of the hyperbola I talked about (I found the domain of $x$ with a graph).

This seems reasonable, but I have no idea how to prove that the hyperbola actually maximizes the value of the line integral $\int_{C}\textbf{F}\cdot d\textbf{r}$. Heck, I don't even know if there exists a curve that maximizes the integral, let alone how to prove its existence. Any help is appreciated.

Note: the line integral of $\textbf{F}$ over this hyperbola is easily determined to be $\frac{15}{4}$, so if you find a curve where the line integral is greater than this amount, please share it with me.

答:
Consider $$\phi=\frac12(x^2-y^2)$$
Then $$\vec F=\nabla \phi$$
Therefore the force is conservative, and your integral depends only on the initial and final points. See the gradient theorem. No matter the path, your integral is the same.


问:
Probability of a deviation when Jensen’s inequality is almost tight
Let $X>0$ be a random variable. Suppose that we knew that for some $\epsilon \geq 0$,
\begin{eqnarray}
\log(E[X]) \leq E[\log(X)] + \epsilon
\tag{1} \label{eq:primary}
\end{eqnarray}
The question is: if $\epsilon$ is small, can we find a good bound for
\begin{eqnarray*}
P\left( \log(X) > E[\log(X)] + \eta \right)
\end{eqnarray*}
for a given $\eta > 0$. One bound can be obtained this way:
\begin{eqnarray*}
P\left( \log(X) > E[\log(X)] + \eta \right) &=&  P\left( X > \exp(E[\log(X)] + \eta) \right) \\
& \leq & E[X] / \exp(E[\log(X)] + \eta)  \\
& = & \exp( \log E[X] - E[\log(X)] - \eta ) \\
& \leq & \exp(\epsilon - \eta)
\end{eqnarray*}
where the first inequality follows from Markov’s inequality. This seems like a good bound due to the exponential decay with $\eta$, but upon closer examination it appears that it can be significantly improved. If we have $\epsilon = 0$, then this bounds gives
\begin{eqnarray}
P\left( \log(X) > E[\log(X)] + \eta \right) & \leq & \exp(-\eta)
\tag{2} \label{eq:good_but_not_best}
\end{eqnarray}
However, from Jensen's inequality applied to (\ref{eq:primary}) with $\epsilon = 0$ we obtain $\log(E[X]) = E[\log(X)]$ and therefore $X$ is a constant almost everywhere. As a consequence, for any $\eta>0$,
\begin{eqnarray*}
P\left( \log(X) > E[\log(X)] + \eta \right) = 0.
\end{eqnarray*}
which is (of course) infinitely better than (\ref{eq:good_but_not_best}).

It would appear that a better bound should decay to zero as $\epsilon$ decays, and ideally preserve the exponential decay with $\eta$. Any suggestions?

答: I cross-posted on Math Overflow and quickly obtained a very nice answer. See https://mathoverflow.net/questions/377913/probability-of-a-deviation-when-jensen-s-inequality-is-almost-tight</p>


问:
Proving $\frac{1}{2}(5x+4),\;2 < x,,\;\text{isPrime}(n)\Rightarrow n = 10k+7$
How is it possible to establish proof for the following statement?

$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$

Where $n,x,k$ are $\text{integers}$.

To be more verbose:

I conjecture that;

If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$

How could one prove this?

答:
Since $n=5\,x/2\ +2$ is integer, we have that $x$ is even, and that $n$ gives remainder $2$ modulo $5$ (written $n\equiv 2\pmod{5}$). So, modulo $10$ there are only two possibilities: $2$ and $7=5+2$. As all primes $>2$ are odd, but $10k+2$ is even, it can't be prime.

(So, instead of 'prime' and $x>2$ you could have also said simply 'odd'.)


问:
How does one determine what group homomorphisms are injective and which ones are surjective?
I have been tasked to solve both parts of this question.

(a): How many group homomorphisms $\phi :\mathbb{Z}_{45} \to \mathbb{Z}_{27}$ are there?

(b): How many of these $\phi$ are injective and how many are surjective?

For part (a): I recall, that in general, the # of group homomorphisms from $Z_m\to Z_n$ can be found by the $\gcd(m,n)$. So here, I got:
$\gcd(45,27) = 9$.

For part (b) though, I'm actually entirely lost on how to determine which of these $\phi$ are injective and surjective. Is there a general way to find these? and if so, how would one do that with this example?

Thanks in advance. Any help, advice, or direction would be greatly appreciated.

答:
There are no surjective ones, because $27\not|45$.  Here I've used Lagrange's theorem.

There are no injective ones, because $45\gt27$.


问:
How to graph a zoomed in circular segment
First, apologies for having forgotten my high school math and terminology. Shame! Shame!

The following equation ... https://www.desmos.com/calculator/ejzmw8pefu ... does a fine job of graphing a perfect top-left quadrant of a circle:

With the above formula, I can change the amplitude via a, and move it vertically up and down with v ... but I still need to move the circle horizontally left/right. My first question is, how do you do that?

My goal is to flatten out that curve to varying degrees without it becoming parabolic (parabolas and quadratic equations will begin or end "faster", rather than being perfectly symmetrical along the diagonal). If I can horizontally adjust it, I can sort of mess with the variables to graph the red and yellow (photoshopped) circular curve segments:

My second question would be, is there a better way of creating these symmetrical curves?

NOTE: All of this is just to solve for y given x on a nice curve.

答:
I am interpreting your question as a request for the family of circular arcs that pass through $(0,0)$ and $(b,b)$ for some $b > 0$.  The center of the family of such circles lies on the perpendicular bisector of the line joining these two points, namely $x + y = b/2$.  If we parametrize the center as $(t, b/2 - t)$, then the radius of the circle satisfies the Pythagorean relationship $$r^2 = t^2 + (b/2 - t)^2,$$ hence the circle has equation $$(x - t)^2 + (y - b/2 + t)^2 = t^2 + (b/2 - t)^2.$$  Equivalently, $$x^2 - 2tx + y^2 + (2t-b)y = 0.$$  Solving for $y$ in terms of $x$ and $t$ yields
$$y = \frac{b - 2t \pm \sqrt{(b-2t)^2 + 4x(2t-x)}}{2}.$$  The choice of sign depends on which portion of circle to plot.  If $t \le 0$, we plot the negative sign corresponding to the lower half of the circle.  If $t \ge b/2$, we plot the positive sign corresponding to the upper half.  If $0 < t < b/2$, we must plot a portion of both halves.


问:
For what value of $a>1$ is $\int_a^{a^2}\frac{1}{x}\ln(\frac{x-1}{32})$ minimum?

For what value of $a>1$ is $\int_a^{a^2}\frac{1}{x}\ln(\frac{x-1}{32})$ minimum?

What is the relation between minimum value and integral?
Appreciate for any suggestions.

答:
$$f(a) = \int_a^{a^2}\frac{1}{x}\ln\left(\frac{x-1}{32}\right)dx$$

$$f'(a) = \frac{2}{a}\ln\left(\frac{a^2-1}{32}\right)-\frac{1}{a}\ln\left(\frac{a-1}{32}\right)$$

Now when $f'(a) = 0$, we will have minimum value

Can you proceed?

EDIT

Using Leibniz rule

$$\frac{d}{da}\int_{f_1(a)} ^ {f_2(a)} g(a,x)dx = g(a, f_2(a)) \cdot \frac{d}{da}(f_2(a)) - g(a, f_1(a)) \cdot \frac{d}{da}(f_1(a)) + \int_{f_1(a)} ^{f_2(a)} \frac{\partial}{\partial a} g(a,x)dx$$

Since $g(a,x)$ is a pure function of $x$, the third term will vanish and give you the equation I have directly used


问:
Showing set in subspace of $\mathbb{R}$ is not open
How do you show that there is no open set $U$ in  $\mathbb{R}$ such that $U\cap [2,4] = [3,4] $?

I can see why it can’t be true if $U$ is an open interval since the intersection won’t have $3$ as an endpoint. But for a general open set I’m stuck and not sure what argument I’m missing.

答: Since $3\in U\cap [2,4]$, we conclude that $3\in U$. Since $U$ is open, it contains a small interval around $3$ - $I=(3-\epsilon,3+\epsilon)\subset U$. But for $\epsilon$ small enough, $I\subset [2,4]$ as well, so $I\subset U\cap [2,4]=[3,4]$. This is a contradiction.


问:
how to determine transient and recurrent state from transition matrix
I wonder how can I determine the transient and recurrent state from transition matrix ? I mean if I have 10 states It would be very hard to draw diagram for them so how to analyse the matrix?
     For example consider :
$$
\begin{bmatrix}
0.1 & 0.2 & 0.7 & 0 \\
0.7 & 0 & 0.2 & 0.1 \\
0.6 & 0.4 & 0 & 0\\
0 & 0.5 & 0.5 & 0
\end{bmatrix}
$$
Can anyone tell me how to approach this problem I only want to know how to look at transition matrix and identify transient and recurrent states?

Thanks.

答:
The formal way to do this and as defined in the book Introduction to Probability Models by Sheldon Ross is:

A state $i$ is recurrent if $\sum_{n=1}^{\infty}p_{ii}^n = \infty$

A state $i$ is transient if $\sum_{n=1}^{\infty}p_{ii}^n < \infty$

You can also define this as:

A state $i$ where $i \in S$ is said to be recurrent if $f_i$ = $P$(ever returns to $i$| starts in $i$) = 1

A state $j$ where $j \in S$ is said to be recurrent if $f_j$ = $P$(ever returns to $j$| starts in $j$) < 1

You would need to do this for all states in your matrix and solve it.


问:
Quotient rings of Gaussian integers $\mathbb{Z}[i]/(2)$, $\mathbb{Z}[i]/(3)$, $\mathbb{Z}[i]/(5)$,
I am studying for an algebra qualifying exam and came across the following problem.

Let $R$ be the ring of Gaussian Integers. Of the three quotient rings
  $$R/(2),\quad R/(3),\quad R/(5),$$
  one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?

I know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.

I hope that this can be answered easily and quickly. Thanks.

答:
I find quotients of polynomial rings easier to reason with algebraically than subfields of $\mathbb{C}$.

Because $\mathbb{Z}[i] \cong \mathbb{Z}[x] / (x^2 + 1)$, we can do arithmetic with rings and ideals. Calculating in more detail than is usually necessary:

$$\begin{align} \mathbb{Z}[i] / (3) &\cong \left( \mathbb{Z}[x] / (x^2 + 1) \right) / (3)
\\&\cong \mathbb{Z}[x] / (3, x^2 + 1)
\\&\cong \left( \mathbb{Z}[x] / (3) \right) / (x^2 + 1)
\\&\cong \left(\mathbb{Z} / (3)\right)[x] / (x^2 + 1)
\\&\cong \mathbb{F}_3[x] / (x^2 + 1)
\end{align}$$

and so we've reduced the problem to one of quotients of polynomial rings over the finite field $\mathbb{F}_3$.


问:
Why v²/2 comes equal to (-k/m)*(x²/2) while doing indefinite integration to express v as a function of x in case of spring-mass system.
We know that, using indefinite integration over f'(x) we can get f(x) whose derivative is f'(x). That's why integration is also called antiderivative. But when we do definite Integration, we get ∆f(x) for ∆x.
But in the above mentioned case, indefinite integration gives an absurd result.

The case is as follows,
 A body of mass m connected to a spring displaces from its equilibrium position..

On applying force on the body for a while  the body displaces  to x=x(i.e. position w.r.t. equilibrium position is a certain value x) and at that position it has velocity equals to V. But due to its inertia it moves to further to x=a. Durin this displacement from x to a, the body experiences varying spring force and this causes the body to retardate and at x=a, velocity becomes 'zero'.

According to *Hook's law', F(spring force)= -kx, where k is the spring constant.
Now using this formula and doing a bit of indefinite integration to express V in terms of x, we get the following equation,
Expressing V as a function of x in case of spring-mass system.

Since Velocity can't be imaginary,hence this equation is quite absurd. However applying the upper and lower limits 0 and V & a and x respectively for both sides, we get **V = √{(k/m)*(a²-x²)}, which seems quite correct. But the question is why the equation from where we have derived this one seems to be so 'incorrect' or 'invalid' ?

答:
$v \frac {dv}{dx} = -\frac {k}{m} x\\
\int v\ dv = \int -\frac {k}{m} x\\
\frac 12 v^2 = -\frac 12 \frac {k}{m} x^2 + C\\
v^2 = -\frac {k}{m} x^2 + C\\
v = \sqrt {C - \frac {k}{m} x^2}$

Now, what you would normally do...

Since $C$ is an arbitrary constant we can replace $C$ with $\frac {k}{m}C.$ And since C is necessarily positive we can replace $C$ with $\frac {k}{m} C^2$

$\frac {dx}{dt} = \sqrt {\frac {k}{m}C^2 - \frac {k}{m} x^2}\\
\int \frac {dx}{\sqrt {C^2 - x^2}} = \int \sqrt {\frac km}dt\\
\arcsin \frac {x}{C}  = \sqrt {\frac {k}{m}}t + \phi\\
x = C\sin (\sqrt{\frac km} t + \phi)$

$C$ is the maximum amplitude of the oscillation.


问:
Is every numerical or algebraic expression a relation?
This may be a simpleton question since I am trying to understand functions and relations. However, every arithmetic operation: addition, subtraction, multiplication and division are defined as 'functions." Well, functions are relations. Does this imply that even in a simple numerical expression such as 2+2 that this is simply a relation between two quantities(function form of a relation)? That is,the input x, being the quantity 2, going through the operation or  function of x, namely adding another quantity of 2? In other words F(x)= x+2? Simply put, this would mean that every expression, numerical, algebraic and polynomial would be a relation (not necessarily a function)? It just seems to be true especially when I realized that a rational number is a ratio of two integers, but a ratio is defined as a quantitative relationship between two amounts and shows how many times one value contains or is contained by the other. Thoughts please.</p>

答:
let $f(x)$ be a function, that means for each unique input $x$ we will get unique value of $f(x)$.

But, let $g(x)$ be a relation, then for for each unique input $x$ we may or may not get unique values of $g(x)$.

Conclusion: Every function is a relation but every relation is not necessarily a function.


问:
Equivalence of wedge and tensor product with Levi-Civita symbol
In this answer the following is stated in the 2-dim case:

\begin{eqnarray}
v\land w & = & \frac{1}{2!}(v\land w-w\land v) \\
& = & \frac{1}{2!}\epsilon_{\mu\nu}v^{\mu}\land w^{\nu} \\
& = & \frac{1}{2!}\epsilon_{\mu\nu}(v^{\mu}\otimes w^{\nu}-w^{\nu}\otimes v^{\mu}) \\
& = & \epsilon_{\mu\nu}v^{\mu}\otimes w^{\nu}.
\end{eqnarray}

NB:

I wanted to see the index mechanics at play replicating what was done in this answer, building a 2-vector from two vectors in $v, w \in\mathbb R^3, $ such as $v=1e_1+3e_2-2e_3$ and $w=5e_1+2e_2+8e_3:$

$$v\wedge w=(1\cdot e_1 + 3 \cdot e_2 - 2 \cdot e_3) \wedge (5\cdot e_1 + 2 \cdot e_2 + 8 \cdot e_3) = \\[2ex]
1\cdot 5 \cdot e_1 \wedge e_1 + 1\cdot 2 \cdot e_1 \wedge e_2 + 1\cdot 8 \cdot e_1 \wedge e_3 \\
+3\cdot 5 \cdot e_2\wedge e_1 +3\cdot 2 \cdot e_2\wedge e_2 +3\cdot 8 \cdot e_2\wedge e_3  \\
-2\cdot 5 \cdot e_3\wedge e_1 -2\cdot 2 \cdot e_3\wedge e_2 -2\cdot 8 \cdot e_3\wedge e_3 = \\[2ex]
5 \cdot \mathbb O + 2 \cdot e_1 \wedge e_2 - 8 \cdot e_3 \wedge e_1 \\
-15 \cdot e_1\wedge e_2 +6 \cdot \mathbb O +24 \cdot e_2\wedge e_3  \\
-10 \cdot e_3\wedge e_1 +4 \cdot e_2\wedge e_3 -16 \cdot \mathbb O = \\[2ex]
\bbox[5px,border:2px solid red]
{
28 \cdot e_2\wedge e_3-18 \cdot e_3\wedge e_1 - 13 \cdot e_1\wedge e_2}$$

starting off at the end, and trying to calculate $\epsilon_{\mu\nu}v^\mu\otimes w^\nu:$

$$\epsilon_{\mu\nu}v^\mu\otimes w^\nu= \\[2ex]
\color{blue}{\epsilon_{11}} 1\cdot 5 \cdot e_1 \otimes e_1 + \color{blue}{\epsilon_{12}} 1\cdot 2 \cdot e_1 \otimes e_2 + \color{blue}{\epsilon_{13}} 1\cdot 8 \cdot e_1 \otimes e_3 + \\
\color{blue}{\epsilon_{21}} 3\cdot 5 \cdot e_2\otimes e_1 + \color{blue}{\epsilon_{22}} 3\cdot 2 \cdot e_2\otimes e_2 + \color{blue}{\epsilon_{23}} 3\cdot 8 \cdot e_2\otimes e_3 + \\
\color{blue}{\epsilon_{31}} (-2)\cdot 5 \cdot e_3\otimes e_1 +\color{blue}{\epsilon_{32}}(-2)\cdot 2 \cdot e_3\otimes e_2 +\color{blue}{\epsilon_{33}}(-2)\cdot 8 \cdot e_3\otimes e_3 = \\[2ex]
\color{blue}0\cdot 1\cdot 5 \cdot e_1 \otimes e_1 + \color{blue}1\cdot 1\cdot 2 e_1 \otimes e_2 + \color{blue}1 \cdot 1\cdot 8  e_1 \otimes e_3 + \\
\color{blue}{(-1)}\cdot 3\cdot 5  e_2\otimes e_1 + \color{blue}0 \cdot 3\cdot 2  e_2\otimes e_2 + \color{blue}1 \cdot 3\cdot 8  e_2\otimes e_3 + \\
\color{blue}{(-1)}\cdot (-2)\cdot 5  e_3\otimes e_1 +\color{blue}{(-1)}\cdot(-2)\cdot 2  e_3\otimes e_2 +\color{blue}0\cdot (-2)\cdot 8 \cdot e_3\otimes e_3 = \\[2ex]
\bbox[5px,border:2px solid red]
{
2 e_1 \otimes e_2 + 8  e_1 \otimes e_3 - 15  e_2\otimes e_1 + 24 e_2\otimes e_3 + 10  e_3\otimes e_1 + 4  e_3\otimes e_2}
$$

How do I reconcile these two results?

NB: This is impossible to reconcile as per the comments: A change of signs cannot relate $v\otimes w$ to $w\otimes v$ - the initial equations are not correct.

The second issue is reflected on my extended comment / "answer" below, and makes reference to the use of the Levi-Civita symbols seemingly undoing the usual change of signs when permuting wedge products...

The essential issue is that the LeviCivita symbols don't seem to naturally "handle" the basis of the wedge product, as well as they do, say, in the case of the cross product. How should the LC symbols be applied in the wedge product?

答:
I feel that the answer you copied, in order to point out some problems with it, on physics stack exchange, is not as bad as you think.  That answerer simply got mixed up between the notation of the O.P., $x^0$ and $x^1$ (passing to two Euclidean dimensions, as he did, which is a good pedagogic strategy), and his own, which was $v$ and $w$.  So, in context, it seems to me he means $x^0$ = v and $x^1$ =w.  This is another good pedagogic strategy since the superscripts on $dx^0$ do not indicate components at all.  But then he fell back into the superscripts, putting them on the v and w, which was just carelessness.  Since $w^1$ really means $v$, and $w^2$ means w, and $v^1$ means v, and $v^2$ means w, due to carelessness, the manipulation is justified by the usual splitting the sum into two sums and making a "change of variable" in one of them, which is what I think he intended, instead of "commuting", which as you and Prof. Shifrin point out, is invalid.

Indeed, you must see that he did not mean components, since you can't take the tensor product of components, so he couldn't have possibly really meant what he carelessly wrote.

I do feel that such carelessness is made less avoidable by the usual sloppy use of indices for two purposes: one, to enumerate the elements of a set, like the four elements of a basis set, and two, to enumerate the components of a vector, like the four components of a four-vector.  But this ambiguity is common, and one must get used to it, at least in Physics.


答:
If $n$ is prime , clearly $x$ must be even and moreover $x$ has only one power of 2, eg $x=2y$ where $y$ is odd, $y=2z+1$. Thus we have </p>

<p>$$n\equiv 5y+2 \pmod{10}\equiv 5+2\equiv 7$$ </p>

Title: Proving $\frac{1}{2}(5x+4),\;2 < x,,\;\text{isPrime}(n)\Rightarrow n = 10k+7$

Body:
How is it possible to establish proof for the following statement?

$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$

Where $n,x,k$ are $\text{integers}$.

To be more verbose:

I conjecture that;

If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$

How could one prove this?

Body:
If $n$ is prime , clearly $x$ must be even and moreover $x$ has only one power of 2, eg $x=2y$ where $y$ is odd, $y=2z+1$. Thus we have

$$n\equiv 5y+2 \pmod{10}\equiv 5+2\equiv 7$$


问:
What is Lie Theory/ a Lie Group, simply?
I'm studying physics, and I continually come across mentions of "Lie Theory" and "Lie Groups" as they relate to such topics as particle physics and String Theory, as well as vague mentions of "symmetry".

I've attempted to read some texts on the topic and, while I feel I could probably sift through it in due time, it is very terse and esoteric. I've had group theory, calculus up to university calc II, and a teensy bit of analysis. Until I'm able to study this proper, what is Lie Theory/ a Lie Group, put simply? What is with these vague mentions of "symmetry"? What can be understood in terms of what I know now?

答:
This was going to be a comment, but it got too long.

I don't know how "simple" this is, but the 5 word summary is "a group with manifold structure". Or perhaps if you're a topologist, "a manifold with group structure". Now that the snarky answer is out of the way, I can try to be a bit more helpful.

Remember that groups measure the symmetries of other objects. The first examples that you see are often the symmetries of discrete objects. I.e. the symmetries of a pentagon correspond to $D_{10}$, and more generally, you get a dihedral group from looking symmetries of regular polygons. If you have a polygon with $n$ sides, then you can rotate by an angle of $\frac{2\pi}{n}$ or reflect through any of a number of axes.

What happens when the object you're studying is smooth in some sense, though? For instance, instead of looking at the symmetries of a polygon, let's look at the symmetries of a circle. Now there's no "smallest angle" to rotate through. You have a continuous parameter of group elements. For each $\theta \in [0,2\pi)$ you can rotate through that angle $\theta$. This (to me) is the defining feature of a lie group. Let's forget the reflections going forward and focus on the rotations.

How do we make the idea of a "continuous parameter" of group elements precise? It turns out the "right approach" is to give your group the structure of a smooth manifold. Remember a manifold is (roughly) a thing that locally looks like $\mathbb{R}^n$. So in the case of the symmetries of a circle (for instance), every rotation $\theta$ has a neighborhood of "nearby" rotations $(\theta - \epsilon, \theta + \epsilon)$, and this neighborhood looks like a neighborhood of $\mathbb{R}$. This is what formalizes the idea that the group elements "vary continuously". You also want to be smart about how the group structure and the manifold structure interact: The multiplication/inversion operations $m : G \times G \to G$ and $i : G \to G$ should both be differentiable. There's a lot more to say, but in the interest of keeping the answer short and relatively elementary I'll leave it there.

If you're looking for a good first reference on lie groups, and you haven't at least skimmed Stillwell's "Naive Lie Theory", you're in for a treat. Like all of his books, it's a very polite read, and it covers a lot of ground with almost no prerequisites at all. He doesn't go into the nitty gritty of manifold theory, which can bog down a lot of the discussion. Instead, he focuses on groups of matrices (whose manifold structure is obvious: after all, you can see the smooth parameters in the entries of the matrix!). This brings the entire text down to a very concrete level, and makes the subject very approachable.

I hope this helps ^_^


问:
Residue theorem applied to $\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}$
I am trying to use residue theorem applied to $\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}$ over an indented contour. I am also told to consider integrating $\frac{e^{\pm2iz}-1}{z^2}$. I chose a bridge shape contour where we have $\Gamma_1:R+iy;0\leq y; \Gamma_2:x+iR.\Gamma_3=-\Gamma_1. \Gamma_{\epsilon}:\epsilon e^{2i\theta}$. I was able to show that $\Gamma_1,\Gamma_2,\Gamma_3$ go to $0$ as $R \to \infty$. For the $\Gamma_{\epsilon}$ I have tried this so far: $$\int_{\Gamma_{\epsilon}} \frac{e^{2iz}}{z^2}dz = \int_{0}^{\pi}\frac{e^{2i\epsilon e^{2i\theta}}}{\epsilon^2e^{4i\theta}} \epsilon e^{2i\theta}d\theta=2i\int_{0}^{\pi}\frac{e^{2i\epsilon e^{2i\theta}}}{\epsilon  e^{2i\theta}}d\theta$$. But from here I cant send $\epsilon \to 0$, what mistake did I make, or is this approach completely wrong?

答:
You don't need any $\epsilon$,
$$\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}dx = \int_C \frac{\sin^2(z)}{z^2}dz =  \int_C \frac{e^{2i z}-1}{-4z^2}dz+ \int_C \frac{e^{-2i z}-1}{-4z^2}dz$$
where $C$ is a piecewise linear curve $-\infty\to -1\to i \to 1\to +\infty$.

Those two integrals are found easily from the residue theorem adding to $C$ the 3 edges of an infinite square in the upper or lower half-plane.


问:
Probability of hitting a quad when the flop has a three of a kind in Texas Hold'em
Let's assume that I am playing Texas Hold'em with 5 of my friends (total 6 players). I have pockets Kings (KK). Now the flop has 3 A (AAA). I want to figure out if anyone on the board has beaten me after the flop. What would be the probability of me getting beaten?

I think the only way to beat me would be to have the final A. What would be the probability of anyone else in the game have that A?

答:
To find the probability of one of your friends holding the last ace, we need to find the number of ways that one of them can have the ace and divide this by the total number of ways the remaining $47$ cards can be arranged. A simple way to do this would to consider those $47$ cards to be in a single deck, where the bottom $5$ pairs ($10$ cards) represent your friends' hands. The total number of ways the cards can be arranged in this deck is $47!$, and the number of arrangements with the ace in the bottom $10$ cards is $10*46!$ ($10$ possible positions for the ace and $46!$ possible arrangements of the other $46$ cards). Therefore, the probability of one of your friends holding the ace is
$$\frac{10*46!}{47!} = \frac{10*46!}{47*46!} = \frac{10}{47}$$


问:
How to plot sinusoidal waves of color frequencies?
I'm looking for a function that I can use to plot a sinusoidal wave for various color frequencies. For example, the frequency of red light is 4.6 THz (4.6*10^14 Hz). The frequency of blue light is 6.3 THz (6.3*10^14 Hz). I'm not sure how I can translate these frequencies into sine functions, however, despite a fair amount of searching online. I'd appreciate any guidance toward the right direction. Here's an example:

答:
Converting colormegone's comment into an answer so it can be accepted:

If you want to make sine functions with wavelengths in units of microns (your picture uses cosine functions), and you're not fussy about things like phase, you can use $$y=\sin\left(2\pi\left[\frac{f}{3.00}\right]x\right),$$ with frequency $f$ in units of $10^{14}$ Hz. For $f=4.6$ (your red), you should get a cycle of 0.652 units on your graph, which will be microns. For $f=6.3$ (your blue), you should get 0.476 units. I'm using an amplitude of 1 for the sine function; you can multiply the function by some number to vary the "height" of the graphs.


答: $x$ must be even or$n$ is not a natural. So let $x=2y$ and your conjecture is that if $5y+2$ is prime, it is$10k+7$. As $5$ divides into $10$, $5y+2 \equiv 2,7 \pmod {10}$. Any number $2\pmod {10}$ is even and (if $\gt 2)$ not prime.</p>


问:
Generalized lexicographic order
In wikipedia I have found this about functions and the lexicographic order, it is basically the statement and I think it is a generalization of the lexicographic order.

Let's define the relationship $\preceq$ in $\mathbb{N}^\mathbb{N}$ as $f\preceq g$ if $f(n)= g(n)$ or if given $n$ the minimum natural number such that $f(n)\neq g(n)$ you have to $f(n)\leq g(n)$.
Prove that $\preceq$ is a linear order in $\mathbb{N}^\mathbb{N}$

How can you prove that result? for the case of the lexicographic order it is easy to verify that it is a linear order, but for this case of the more abstract space I hope you can help me.
Thanks in advance

Edit:

I have tried the following:

Reflexivity: Let $f\in \mathbb{N}^\mathbb{N}$, we have that $f(n) = f(n)$, then $f \preceq f$.

Antisymmetry: Let $f,g\in \mathbb{N}^\mathbb{N}$, if $f\preceq g$ and $g\preceq f$ it is trivial that $f = g$

Transitivity: Let $f,g,h\in \mathbb{N}^\mathbb{N}$ if $f\preceq g$ and $g\preceq h$, then $f=g$ and $g=h$, then $f=h$ or if given $n$ the minimum natural s.t $f(n)\neq g(n)$ and $g(n)\neq h(n)$ then $f\leq g$ and $g\leq h$ from this we can deduce that $f=h$ or if given $n$ the minimum natural number s.t. $f(n)\neq h(n)$ have $f\leq h$, then $f\preceq h$

It is right? And how can I see that the trichotomy property is fulfilled?

答:
The given proof for transitivety has two mistakes.

Assume f <= g and g <= h.

Case f = g.  Trivally f <= h.

Case g = h.  Trivally f <= h.

Case f < g and g < h.

Exists smallest n with not f(n) = g(n)

and f(n) < g(n).

Exists smallest m with not g(m) = h(m)

and g(m) < h(m).

Case n <= m.  Clearly f < h.

Case m < n.  Show f < h.

Title: Generalized lexicographic order


问:
Theoretical Probability of 3 spinners
There are 3 spinners that are identical, with 6 different colors of equal size on a spinner. What is the theoretical probability that they:

There are no matching colors

There are two matching colors

All three colors match

I'm currently confused about the 2nd criteria because I'm not sure whether or not I should count the order.

Thanks

答:
The easiest way to compute is to find P(all match) and P(none match) first, and get P(two match) as the complement.

P(all match) $= 1\cdot\frac1 6\cdot \frac 1 6 = \frac 1{36}$

P(none match) $= 1\cdot\frac5 6\cdot \frac 4 6 = \frac {20}{36}$

Thus P(two match) $= 1 - \frac{21}{36} = \frac{15}{36}$

But it is best to confirm by direct computation also.
There are three possible pairs, and we need the pair to match and the third to not match, thus Pr $= 3\cdot (1\cdot\frac1 6)\cdot\frac5 6 = \frac{15}{36}$ which confirms our shortcut.

Title: Theoretical Probability of 3 spinners


问:
Expected sum of $m$ tickets.
If $m$ tickets are drawn at random out of $n$ tickets which are numbered from $1$ to $n$, find the expected sum.

How to approach this problem?

答:
hint

number of tickets drawn $\times$ expected average value of each ticket

Title: Expected sum of $m$ tickets.


问:
Hypothesis Tests: Finding size and power for various decision rules
I have a homework question as follows, so would appreciate hints and suggestions.

Given two uniform i.i.d. random variables, $X_1$ and $X_2$, with support $[\theta ;\theta +1]$. The hypotheses are as follows - $H_0: \theta = 0$ ; $H_A: \theta \neq 0$.

Part of the assignment is to determine the size and power of a test which always rejects $H_0$ regardless of observed data, and a test which always accepts $H_0$ regardless of observed data.

How do we think about the power of a test in cases like this, where we're not thinking about Z scores or any clearly defined probabilities?

答:
Here is an analogy from a completely different every day scenario that you might want to think about. Consider two decision rules:

A) Tag someone as having a fever regardless of their temperature.

B) Tag someone as not having a fever regardless of their temperature.

In the scenario of a person truly having a fever, decision rule A will be absolutely perfect.

In the scenario of a person truly not having a fever, decision rule B will be absolutely perfect.


问:
Valid Proof that the Irrationals are Uncountable?
So I originally wanted to prove that the reals are uncountable, but the best solution I came up with was to prove the irrationals are uncountable so therefore the reals must be as well. I suppose my first question is, is this valid logic?

Next take any countable subset of $\mathbb{R} \backslash \mathbb{Q}$, put it as $S$. Since it's countable $\exists \; a : S \to \mathbb{N}$ whereas $a$ is bijective. Because it's bijective, $\exists \; b : \mathbb{N} \to S$ similarly bijective whereas $b$ is the inverse of $a$. Now let's form an irrational, put it as $s$. Start with the number defined as '$0.$'. The $n$th place after the decimal place the $n$th digit of $b_n$ with one added to it (if the $n$th digit of $b_n$ is $9$ put it to be $0$). It's clear that this number $s$ is not in the set $S$ so it follows that any countable subset of $\mathbb{R} \backslash \mathbb{Q}$ is a strict subset, so $\mathbb{R} \backslash \mathbb{Q}$ must be uncountable (otherwise it would apparently be a strict subset of itself...).

Now since $\mathbb{R} \backslash \mathbb{Q}$ is uncountable it follows that $\mathbb{R}$ is uncountable.

P.S. I tag this as topology because it's introduced in "basic topology" in Rudin

EDIT: as William and Thomas pointed out this isn't exactly valid, there is no guarantee that $s \in \mathbb{R} \backslash \mathbb{Q}$. So we can say let $S$ be a countable subset of $\mathbb{R}$ so $\exists \; a : S \to \mathbb{N}$ and similarly the inverse of $a$, $b : \mathbb{N} \to S$ and so form $s$ as the number $0.$ with the $n$th digit after the decimal point to be either the $n$th digit of $b_n$ (extending the number with as many $0$'s as needed) plus one or if that number turns out to be $10$, place $0$ there. Then it's clear that $s \not\in S$ so every countable subset is a strict subset so $\mathbb{R}$ must be uncountable. It then must follow that $\mathbb{R} \backslash \mathbb{Q}$ is uncountable since $\mathbb{Q}$ is countable, and taking a countable number of things from an uncountable number of things will clearly result in an uncountable number of things.

答:
See Theorem 2.14 of Rudin's PMA, the idea for which Rudin credits Cantor, who provided the canonical argument for the uncountability of the reals: See Cantor's Diagonalization Process.

How is your proof distinct from that proof?

Indeed, if you thought of this on your own, KUDOS!

See also Rudin's Theorem 2.43 (p. 41).


问:
Could someone explain to me why the range of the graph is R\(0,5]
I am trying to help my brother with his exam preparations, and we came across this question and found the answer a little confusing. Here is the graph and the range, taken from the answers page of the practice questions.

Shouldn't the correct range be R\[0,10)?

答:
Summarizing the contents of the comments so this question can be removed from the "unanswered" queue:

First of all, the book is wrong. The range of this graph is the entire real line except the region between $y=0$ and $y=10$. Since the function is defined at $y=0$ but not at $y=10$, the range should be $R\setminus(0,10]$ (or, more accurately, $\mathbb{R}\setminus(0,10]$).


问:
any base of the canonical topology in $\mathbb R$ can be decreased
Any base of the canonical topology in $\mathbb R$ can be decreased  .

I have found this question in Elementary Topology book. I am not quite sure what the term "decreased" mean here.

Does it mean that for a given basis B of canonical topology, there exits another basis B' such that B' $\subset$ B.

Can anyone help me with this ?

答:
Take a non-void element $U$ of $B$. $U$ is open. Let's show that $U$ can be written as a union of two open subsets $U=D_1 \cup D_2$, with both $D_i\ne U$. Indeed, take $x\in U$, $D_1 = U\backslash \{x\}$, and $D_2$ an open interval inside $U$ containing $x$, but not equal to $U$.
Now both $D_1$, $D_2$ are unions of some elements of $B$, and those are sests $\ne U$. Therefore, $U$ is a union of some elements in $B\backslash \{U\}$. We conclude that $B\backslash \{U\}$ is also a basis.


问:
When is a laurent series meromorphic?

When is a Laurent series meromorphic?

Suppose $f$ is holomorphic on $D_r(a)\setminus\{a\}$ and $f(z) = \sum\limits_{n\in Z}c_n (z-a)^n$ be its unique Laurent series on $D_r(a)\setminus\{a\}$ Then, I want to show that $f$ is meromorphic if and only if $\exists N_0 \in Z$ such that $c_n = 0\; \forall n < N_0$

I am struggling in arriving at this result.

Definition of meromorphic that I am using:

Let $f$ be holomorphic on $\Omega \setminus \{z_1, \ldots z_n\}$. Then
$f$ is meromorphic on $\Omega$, if for each $z_i$, $\exists D_p(z_i)$
inside $\Omega$ such that $g, h$ are holomorphic on $D_p(z_i)$ and $h$
is never zero and $\frac{g}{h} = f$ on $ D_p(z_i)\setminus\{z_i\}$

答:
Suppose $f$ is meromorphic in $D_r(a)$. Then there are holomorphic functions $g,h$ defined on $D_r(a)$ with $h$ not identically $0$ and $f=g/h$ in $D_r(a)\setminus\{a\}$. Since $f$ is well-defined and finite at each point in $D_r(a)$ other than $a$ we may assume $h$ is only $0$ at $z=a$. This zero must also be of finite order (otherwise $h$ would be identically $0$, say, by Taylor expansion), so we may write $h(z) = (z-a)^mk(z)$ for some holomorphic $k(z)$ satisfying $k(z)\neq 0$ in some neighborhood $D_{r'}(a)$ of $a$. In this neighborhood, we have
$$(z-a)^mf(z) = \frac{g(z)}{k(z)},$$
and the right-hand side is analytic since $k\neq 0$ here, meaning we get a power series expansion
$$(z-a)^mf(z) = \sum_{k\geq 0} d_k(z-a)^k.$$
Dividing both sides by $(z-a)^m$ gives a Laurent series expansion valid in $D_{r'}(a)\setminus\{a\}\subset D_r(a)\setminus\{a\}$. Yet the Laurent series is unique; since the coefficients of $(z-a)^n$ for $n<-m$ are zero in our new one, we must have $c_n = 0$ for $n<-m$ in our original Laurent series.

The converse is easier. If $c_n = 0$ for $n<N_0$, then in $D_r(a)\setminus\{a\}$ we have
$$(z-a)^{N_0}f(z) = c_{N_0}+c_{N_0+1}(z-a)+c_{N_0+2}(z-a)^2+\dotsm.$$
Letting $g(z)$ be the power series on the right-hand side, we see $g$ and $h(z):=(z-a)^{N_0}$ are analytic on the disk, $h(z)$ is not identically $0$, and $f=g/h$ on the punctured disk. Thus $f$ is meromorphic, as desired.

Title: When is a laurent series meromorphic?


问:
Examples of Cauchy sequences
In $\mathbb{R}$, it is true that every Cauchy sequence is convergent and vice-versa. After introducing the Cauchy sequence, usually, the explicit examples stated in almost all the books (and notes) are

constant sequence, $(\frac{1}{n})$, and any convergent sequence

The sequence $x_1=1$ and $x_{n+1}=1+\frac{1}{x_n}$.

Here the limits of the sequences are easy to compute.

Q.0 Are there any other typical examples of Cauchy sequences, which, from their expression, do not look convergent (or Cauchy)?

Q.1 Are there examples of Cauchy sequences, whose limits are not easy to find, or we can only say that it is Cauchy, without telling its limit?

答:
Q0. There is a whole class of metric spaces in which Cauchy sequences may not converge to some element inside the metric space. We call those metric spaces "Incomplete metric spaces".

As an example, consider the metric space $(\mathbb{Q} , \rho)$ where $\mathbb{Q}$ is the set of rationals, and the metric $d := |x - y|~$ for all $x , y \in \mathbb{Q}$ . It can be easily verified that $d$ is indeed a metric.

Now, consider the sequence $\{a_n\}$ in $\mathbb{Q}$ defined by $a_0 = 1$ , and $a_{n+1} = 2 \cdot \dfrac{a_n + 1}{a_n + 2}$ .

Observe that $\{a_n\}$ is strictly increasing, and $a_n \leqslant 2~$ for all $~n \in \mathbb{N}$ . So, it converges to some point, say $L$ . But, taking $n \to \infty$ , we get that $\lim_{n\to \infty} a_n = \sqrt{2}~ \not\in \mathbb{Q}$ . Now, $\{a_n\}$ is a Cauchy Sequence in $\mathbb{Q}$ which doesn't converge in $\mathbb{Q}$ .

Q1. As @Bumblebee has mentioned in the comment section, take $\{a_n\}$ in $\mathbb{R}$ such that for all $n \in \mathbb{N}$ , $$a_n = 1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n - \log n$$ This is a Cauchy sequence which converges to the limit $\gamma$ , popularly known as the Euler - Mascheroni Constant. One can easily prove that this sequence converges, i.e. $\lim_{n \to \infty} a_n$ exists and is finite. But it's very difficult to show that the limit is precisely $\gamma$ .

Title: Examples of Cauchy sequences


问:
how can I get the fourier transform of this generalized function?
when function f(x) of the generalized function is $f(x) = x+2$, the fourier transform of the generalized function will be like   ($F$ is the Fourier transform)
$$\int_{-∞}^{∞}(x+2)(F[k(x)])dx$$
when $f(x) = \sin x$, it will be
$$\int_{-∞}^{∞}\sin x(F[k(x)])dx$$
Is there a way to simplify these?
For example, I know when $f(x) = 1$, It will be
$$\int_{-∞}^{∞}1(F[k(x)])dx = \int_{-∞}^{∞}e^{i0x}(F[k(x)])dx = 2πk(0)$$
Can anyone help me?

答:
With $g$ a Schwartz function and $F(g)$ its Fourier transform then the Fourier transform of $x+2$ in the sense of distributions is the unique distribution $T$ such that for all $g$ Schwartz $$\langle T,g\rangle = \langle x+2,F(g)\rangle$$
where $\langle x+2,F(g)\rangle$ means the Riemann integral $ \int_{-\infty}^\infty (x+2) F(g)(x)dx$ because $x+2$ is a continuous function.

The Fourier inversion theorem for Schwartz functions yields
$$ \int_{-\infty}^\infty 2 F(g)(x)dx= 4\pi g(0)$$ and
$$ \int_{-\infty}^\infty x F(g)(x)dx= \int_{-\infty}^\infty (-i) F(g')(x)dx = -2i\pi g'(0)$$
thus $$T = 4\pi\delta+2i\pi \delta'$$


问:
Pointwise and Pointwise almost everywhere convergence of monic simple functions
I am doing a course on measure theory in which we have defined monic simple functions on a measure space $\left( X, \mathcal{B}, \mu \right)$ as follows:

Definition: A monic simple function is of the form $f = \alpha \cdot \chi_E$, where E is a measurable set.

Now, we consider a sequence of monic simple functions $f_n = \alpha_n \cdot \chi_{E_n}$ such that for all $n \in \mathbb{N}$, we have $\alpha_n > 0$ and $\mu \left( E_n \right) > 0$. I want to look at its pointwise and pointwise almost everywhere convergence to the zero function. We have claimed that

$f_n \to 0$ pointwise if and only if $\alpha_n \to 0$ or $\limsup\limits_{n \to \infty} E_n = \emptyset$.

$f_n \to 0$ pointwise a.e. if and only if $\alpha_n \to 0$ or $\mu \left( \limsup\limits_{n \to \infty} E_n \right) = 0$.

Here,we have defined $\limsup\limits_{n \to \infty} E_n = \bigcap\limits_{n \in \mathbb{N}} \bigcup\limits_{m \geq n} E_m$.

My ideas so far:

To prove (1), if we consider $f_n \to 0$ pointwise, we may either have $\limsup\limits_{n \to \infty} E_n = \emptyset$ or it is non-empty. In case it is empty, there is nothing to prove. Therefore, we consider the case when, $\limsup\limits_{n \to \infty} E_n \neq \emptyset$. Then, there is some $x \in \limsup\limits_{n \to \infty} E_n$. This would however, mean that for each $n \in \mathbb{N}$, there is some $m \geq n_0$ such that $x \in E_m$. Also, pointwise convergence gives for each $\epsilon > 0$ some $n_0 \left( x \right) \in \mathbb{N}$ such that $\left| f_n \left( x \right) \right| < \epsilon$.

These two information together tell us that there is some $m \geq n_0 \left( x \right)$ such that $\left| \alpha_m \right| < \epsilon$. How should we conclude that for all $m \geq n_0 \left( x \right)$, we have $\left| \alpha_m \right| < \epsilon$?

The converse is pretty simple, and I am through it.

Now, to prove (2), I have assumed that (1) is indeed true. So, if we assume that $f_n \to 0$ pointwise a.e., then again, we have either $\mu \left( \limsup\limits_{n \to \infty} E_n \right) = 0$ or $\mu \left( \limsup\limits_{n \to \infty} E_n \right) \neq 0$. In the first case, there is nothing to prove. In the second case, we conclude that $\limsup\limits_{n \to \infty} E_n \neq \emptyset$ (since it is a non-null set), and therefore $f_n \to 0$ pointwise (everywhere) on this set. Then, from the above result, because the set is non-empty, we conclude that $\alpha_n \to 0$. Is this reasoning correct? I feel that there is some problem towards the end of the reasoning, although I cannot see it clearly.

Also, in the converse, if one assumes that $\mu \left( \limsup\limits_{n \to \infty} E_n \right) = 0$, how should we claim that $f_n \to 0$ pointwise a.e? If we start with a set $E := \left\lbrace x \in X | f_n \left( x \right) \not\to 0 \right\rbrace$, we should prove that $\mu \left( E \right) = 0$. Till now, I am able to prove that $E \subseteq \limsup\limits_{n \to \infty} E_n$. However, does this imply (using the monotonocity of the measure) that $\mu \left( E \right) = 0$? If not, then how do we prove that $E = \limsup\limits_{n \to \infty} E_n$?

Thank you for your help and suggestions!

答:
The first part is actually false. Let $E$ be a  set of positive measure, $E_n=\emptyset $ for $n$ odd and $E_n=E$ for $n$ even. Let $\alpha_n=n$ for $n$ odd and $\alpha_n=\frac 1 n$ for $n$ even. Then $\alpha_n$ does not tend to $0$ and $\lim \sup E_n$ is not empty even though $f_n(x) \to 0$ for every $x$.

In the second part, yes, $E \subseteq F$ and $\mu (F)=0$ does imply $\mu(E)=0$.


答:
This is essentially Cantor's diagonalization argument.

If you came up with this by yourself, that is really impressive. It is widely regarded to be one of the most beautiful proofs in mathematics.

Title: Valid Proof that the Irrationals are Uncountable?

Body:
So I originally wanted to prove that the reals are uncountable, but the best solution I came up with was to prove the irrationals are uncountable so therefore the reals must be as well. I suppose my first question is, is this valid logic?

Next take any countable subset of $\mathbb{R} \backslash \mathbb{Q}$, put it as $S$. Since it's countable $\exists \; a : S \to \mathbb{N}$ whereas $a$ is bijective. Because it's bijective, $\exists \; b : \mathbb{N} \to S$ similarly bijective whereas $b$ is the inverse of $a$. Now let's form an irrational, put it as $s$. Start with the number defined as '$0.$'. The $n$th place after the decimal place the $n$th digit of $b_n$ with one added to it (if the $n$th digit of $b_n$ is $9$ put it to be $0$). It's clear that this number $s$ is not in the set $S$ so it follows that any countable subset of $\mathbb{R} \backslash \mathbb{Q}$ is a strict subset, so $\mathbb{R} \backslash \mathbb{Q}$ must be uncountable (otherwise it would apparently be a strict subset of itself...).

Now since $\mathbb{R} \backslash \mathbb{Q}$ is uncountable it follows that $\mathbb{R}$ is uncountable.

P.S. I tag this as topology because it's introduced in "basic topology" in Rudin

EDIT: as William and Thomas pointed out this isn't exactly valid, there is no guarantee that $s \in \mathbb{R} \backslash \mathbb{Q}$. So we can say let $S$ be a countable subset of $\mathbb{R}$ so $\exists \; a : S \to \mathbb{N}$ and similarly the inverse of $a$, $b : \mathbb{N} \to S$ and so form $s$ as the number $0.$ with the $n$th digit after the decimal point to be either the $n$th digit of $b_n$ (extending the number with as many $0$'s as needed) plus one or if that number turns out to be $10$, place $0$ there. Then it's clear that $s \not\in S$ so every countable subset is a strict subset so $\mathbb{R}$ must be uncountable. It then must follow that $\mathbb{R} \backslash \mathbb{Q}$ is uncountable since $\mathbb{Q}$ is countable, and taking a countable number of things from an uncountable number of things will clearly result in an uncountable number of things.


问:
Showing set in subspace of $\mathbb{R}$ is not open
How do you show that there is no open set $U$ in  $\mathbb{R}$ such that $U\cap [2,4] = [3,4] $?

I can see why it can’t be true if $U$ is an open interval since the intersection won’t have $3$ as an endpoint. But for a general open set I’m stuck and not sure what argument I’m missing.

答: Assume there is an open set $U=(a,b)$ such that $(a,b) \cap [2,4]=[3,4]$.  So $3$, and $4$ belongs to $U'$. Now, if $a=3$ then it must be that $U=[3,4)$ which is not open. And if $a \not =3)$ then the $(a,b) \cap [2,4]\not=[3,4]$. So there is no such open set.


问:
proof of propositional logic
$((A \land (A \rightarrow B))  \rightarrow B$

$((A \land (A \rightarrow B))$ is premise
and $B$ is goal

$((A \land (\neg A \lor B))$

$(((A \land \neg A)  \lor ( A \land B))$

How to prove this is true? Whether using contradiction or anything else?

答: Hint: replace $A \to B$ with its equivalent: $B \lor \lnot A$ .

Title: proof of propositional logic


问:
Is it true that $ - \frac{\delta'(|\mathbf{x} - \mathbf{y}|)}{ 4 \pi | \mathbf{x} - \mathbf{y}|} \ \to \ \delta^{(3)}(\mathbf{x} - \mathbf{y})$?
Is there a way in which one can assign the meaning
$$
- \frac{\delta'(|\mathbf{x} - \mathbf{y}|)}{ 4 \pi | \mathbf{x}  - \mathbf{y}|} \ \to \ \delta^{(3)}(\mathbf{x} - \mathbf{y}) \
$$
for $\mathbf{x}, \mathbf{y} \in \mathbb{R}^3$? It seems to me that this should be true. Notice that for some test function $f(\mathbf{x})$ we have (using spherical coordinates)
$$
- \int_{\mathbb{R}^3} d^3\mathbf{x} \; \frac{\delta'(|\mathbf{x}|)}{ 4 \pi | \mathbf{x}|} f(\mathbf{x}) = - \frac{1}{4\pi} \int_{\mathbb{R}_3} dX \; d\theta \; d\phi \; X^2 \sin\theta \; \frac{\delta'(X)}{X} f(X,\theta, \phi)
$$
and using integration by parts on the $\delta'$ yields
$$
\cdots = \frac{1}{4 \pi} \int_{\mathbb{R}_3} dX \; d\theta \; d\phi \; \sin\theta \; \delta(X) \left[ f(X,\theta, \phi) + X \; \partial_X f(X,\theta, \phi) \right]
$$
which evaluates to
$$
\cdots = \frac{1}{4 \pi} f(0,0,0) \int_{S^2} d\theta \; d\phi \; \sin\theta = f(\mathbf{0})\ .
$$
This seems to have the same effect as using $\delta^{(3)}(\mathbf{x})$. Is

答:
Answering my own question, but the identity holds true: I found the following formula in Kanwal's book
$$
\delta(x)\delta(y)\delta(z) = \frac{\delta(r)}{4\pi r^2}
$$
which is equation (3.1.30), where $r = \sqrt{x^2 + y^2 + z^2}$. Noting the identity $r \delta'(r) = - \delta(r)$ yields the result I quoted above.


答:
\begin{align}
((A \land (A \rightarrow B))  \rightarrow B &\equiv (( A \land (\neg A \lor B))  \rightarrow B\\
&\equiv (( A \land \neg A) \lor (A \land B))  \rightarrow B\\
&\equiv (F \lor (A \land B))  \rightarrow B\\
&\equiv (A \land B)  \rightarrow B\\
&\equiv \neg(A \land B)  \lor B\\
&\equiv (\neg A \lor \neg B) \lor B\\
&\equiv \neg A \lor (\neg B \lor B)\\
&\equiv \neg A \lor T\\
&\equiv T\\
\end{align}

As @Michael Hoppe's comment, we can use the contradiction to prove that is true.

Assuming $\neg B$. Given $A\land (A\to B)$, which we can conclude is $\neg A$. This is contradiction. So, the propositional logic $(A\land (A\to B))\to B$ is true.

Title: proof of propositional logic

Body:
$((A \land (A \rightarrow B))  \rightarrow B$

$((A \land (A \rightarrow B))$ is premise
and $B$ is goal

$((A \land (\neg A \lor B))$

$(((A \land \neg A)  \lor ( A \land B))$

How to prove this is true? Whether using contradiction or anything else?


问:
Show that $B \setminus C$ is connected.
Let A and B be connected sets and $ A\subset B$. If C is a component of $B \setminus A$, prove that $B \setminus C$ is connected.

I have tried to solve this problem but I am really having a hard time proving it.
I hope someone can help me. Thanks a lot.

答:
If $A = B$ then $C = \emptyset$ and $B \setminus C = B$ is connected. If $A = \emptyset$, then $C = B$ and $B \setminus C = \emptyset$ is connected. Assume that $A$ is a proper subset of $B$.

Let $C_i$ denote the connected components of $B \setminus A$ with $i \in I$ in some index set $I$. By definition, each connected component $C_i$ is a connected subset of $B \setminus A$ which is maximal in the inclusion order. Therefore, each $C_i$ is closed in $B \setminus A$.

Fix $k$ and assume $B \setminus C_k$ is not connected, i.e.

$$
B \setminus C_k = D_1 \cup D_2
$$

for $D_1$ and $D_2$ disjoint non-empty sets that are open in $B \setminus C_k$. We will derive a contradiction by showing that $B$ is not connected.

First, note that every $X \subset B \setminus C_k$ can be written as

$$
X = (X \cap D_1) \cup (X \cap D_2)
$$

where $X \cap D_i$ is open in $X$ for $i = 1, 2$. Therefore, every connected subset $X \subset B \setminus C_k$ is either contained in $D_1$ or in $D_2$. In particular, $A$ is contained in $D_1$ or in $D_2$ and for each $i \in I, i \ne k$, $C_i$ is contained in $D_1$ or in $D_2$.

Assume, without loss of generality, that $A \subset D_1$. Define

$$
E_1 = D_1 \cup C_k \\
E_2 = D_2
$$

and note that $E_1$ and $E_2$ are disjoint non-empty sets and $E_1 \cup E_2 = B$. We will show that both are open in $B$ by demonstrating that $E_2$ is clopen in $B$.

Recall that $E_2$ is open in $B \setminus C_k$, so there is an open set $G$ in $B$ such that $E_2 = G \cap (B \setminus C_k)$. Let $X'$ denote the closure of $X$ in $B$. Noting that $C_k' \subset C_k \cup A$ and $G \cap A = \emptyset$, we can write $E_2$ as the intersection of two sets open in $B$

$$
E_2 = G \cap (B \setminus C_k')
$$

so $E_2$ is open in $B$.

Similarly, recall that $E_2$ is closed in $B \setminus C_k$ so there is a closed set $F$ in $B$ such that $E_2 = F \cap (B \setminus C_k)$. Noting that $E_2' \subset E_2 \cup A$ and $F \cap A = \emptyset$, we can write $E_2$ as the intersection of two sets closed in $B$

$$
E_2 = F \cap E_2'
$$

so $E_2$ is closed in $B$.

Thus, $E_2$ is both open and closed in $B$ and consequently $B = E_1 \cup E_2$ is the union of two disjoint non-empty open sets contradicting the fact that $B$ is connected. Therefore, the assumption  that $B \setminus C_k$ can be written as the union of two disjoint non-empty open sets is false and thus $B \setminus C_k$ is connected.

Title: Show that $B \setminus C$ is connected.


问:
Find the number of quadratic polynomials ax² + bx + c

Find the number of quadratic
  polynomials ax² + bx + c such that:

a)
  a, b, c are distinct.

b) a, b, c ε {1,
   2, 3, …2008}

c) x + 1 divides ax² + bx
  + c

Reasoning->b=a+c  b ranges from 3 to
  2008 x1+x2=n has n-1C1 solns. Using th
  elogic we get 2+3+..2007 solutions. A
  total of 2015027
  solutions.here,condition 1 comes into
  picture telling us that a,b,c are
  distinct. Therefore the even values of
  b ie.4,6,8,..2008 will have one
  solution less ie.from the given
  solution subtract 1003 thus giving us
  20104024

答:
Ross's answer was my first thought, but as a slightly-different alternative, use the remainder theorem:

The remainder when a polynomial $P(x)$ is divided by $x-k$ is $P(k)$.

So if $x+1=x-(-1)$ divides $P(x)=ax^2+bx+c$ (leaving remainder $0$), $P(-1)=0$, which should give the same condition on $a$, $b$, and $c$ that would result from long division.


问:
What is the shape of the shadow of the rim of my lamp shade?
My lamp has a cylindrical shade and a light bulb sits in the center of it. It projectes the pictured shadow on the wall. That U-shaped curve is the shadow of the rim(s). What shape is it? A parabola?

I know I can "compute" it with some work, but is there a quick geometric answer?

答: If the bulb is more or less a point source, then two cones of light are being emitted from the upper and lower rims of the shade. When the axis of the cylindrical shade is parallel to the plane of the wall, the wall cuts both cones and you get the two opposite sections of an hyperbola.  (See Apollonius, Conics I, 12 & 14). If you incline the top of the lamp toward the wall until light is reaching the wall only from the top of the shade, you have a parabola. Tilting it farther gives an ellipse, and finally a circle.</p>


答: Regarding Q1, you can easily encode open problems into the precise value of a sum which always converges, no matter the resolution of the open problem. E.g. let $A(n)=1$ if $n$ is a Fermat prime, and $A(n)=0$ otherwise. Then $L=\sum_{n=0}^\infty \frac{A(n)}{2^n}$ converges. By replacing $A(n)$ with $1$s, we see $L$ is trivially at most $2$. Actually it's much smaller...but how much smaller? If you could compute $L$ exactly, then writing $L$ in binary essentially gives you a list of all Fermat primes (of which we only know 5 exist, but it could in principle be an infinite list).

Title: Examples of Cauchy sequences

Body:
In $\mathbb{R}$, it is true that every Cauchy sequence is convergent and vice-versa. After introducing the Cauchy sequence, usually, the explicit examples stated in almost all the books (and notes) are

constant sequence, $(\frac{1}{n})$, and any convergent sequence

The sequence $x_1=1$ and $x_{n+1}=1+\frac{1}{x_n}$.

Here the limits of the sequences are easy to compute.

Q.0 Are there any other typical examples of Cauchy sequences, which, from their expression, do not look convergent (or Cauchy)?

Q.1 Are there examples of Cauchy sequences, whose limits are not easy to find, or we can only say that it is Cauchy, without telling its limit?


问:
Does a given positive measure set in $\mathbb{R}^2$ contains product of two 1 dimensional positive measure set?
In $\mathbb{R}^2$ consider the 2 dimensional positive Lebesgue measure set,
$$D=\{re^{it}:t\in[0,2\pi),r\in[0,1]\setminus\mathbb{Q}\}  .   $$
Does there exist sets $A,B\subset\mathbb{R}$ of 1 dimensional positive  Lebesgue measure such that $$A\times B\subset D?$$

Note: The above is not true if $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin\mathbb{Q}\}.$$

答: No, it there were two such sets $A$, $B$,  then with the map $(x,y) \mapsto (x^2, y^2)$ ($t \mapsto t^2$ bi-Lipschitz on compacts inside $(0, \infty)$) , it would transform into $A_1$, $B_1$ of measure non-zero inside $\{(x,y)\ | \ x+y \not \in \mathbb{Q}^2\}$, which is not possible.


问:
$\sum_{n=1}^\infty \frac{1}{x^2 + n^2}$ is an uniform convergent series.
I was required to prove  that $\sum_{n=1}^\infty \frac{1}{x^2 + n^2}$  is an uniform convergent series.

My attempt : $ f(x) = \sum_{n=1}^\infty \frac{1}{x^2 + n^2}$

Now $\frac{1}{x^2 + n^2} \leq \frac{1}{ n^2}$ for all $x \in \Bbb R$.

Now  we know $ \sum_{n=1}^\infty \frac{1}{ n^2}$  is convergent. So  for every $\epsilon > 0$ , we will get a natural number $K$ such that $ \sum_{n=k}^\infty \frac{1}{ n^2} < \epsilon$ .

Thus ,  for every  $\epsilon > 0$ , we will get a natural number $K$ such that  $ \sum_{n=k}^\infty \frac{1}{x^2 + n^2} < \epsilon$  for all $x \in \mathbb R$.
Therefore , $ \sum_{n=1}^\infty \frac{1}{x^2 + n^2}$ is an uniform convergent series.

Can anyone please check my attempt? Have I gone wrong anywhere?

答:
Your solution is correct. However, I would use the M-test theorem for simplicity.

Let $f_n(x) = \frac{1}{x^2 + n^2}$, It is clear that

$f_n(x) \leq M_n =\frac{1}{ n^2}$  $\forall x \in \mathbb{R}$.

Since $\sum_{n=0}^{\infty} M_n$ is convergent ,($\sum_{n=0}^{\infty} M_n =\frac{\pi^2}{6}$), then the M-test (https://en.wikipedia.org/wiki/Uniform_convergence) asserts that the series $\sum_0^\infty f_n(x) = \sum_0^\infty \frac{1}{x^2 +n^2}$  is uniformly convergent.


问:
Justifying $\int_0^\infty e^{-ax^2}\ \mathrm{d}x$ for complex values
I have searched and failed to find a rigorous proof showing that
$$\int_{0}^\infty e^{-ax^2}\ \mathrm{d}x = \frac{\sqrt{\pi}}{2\sqrt{a}}$$
is true for  $\Re(a)=0$ and $\Im(a)\neq0$. For example, why does
$$
\int_{-\infty}^\infty e^{-ix^2}\ \mathrm{d}x = \sqrt{\frac{\pi}{i}}
$$
and does one need to invoke contour integration to show this? Or is the following technique valid:

Suppose $\Re (b)>0$ and $a$ is complex; then
$$
\int_{-\infty}^\infty e^{-ax^2}\ \mathrm{d}x \stackrel{?}{=} \lim_{b\to0}\int_0^\infty e^{-(a+ b)x^2}\ \mathrm{d}x = \lim_{b\to0}\frac{\sqrt{\pi}}{2\sqrt{a+b}}.
$$

I have an idea for contour integration but is there any way to justify this with real analysis? If not I would very much appreciate a hint for the complex analytic approach. Thanks!

答:
If $a$ is a pure imaginary number, using contour integration $$\int_{0}^\infty e^{-ax^2}\, dx = e^{i\frac \pi 4} \sqrt{\frac \pi {4a}}$$ (have a look here for the proof).

Where the problem starts to be more difficult is to establish that
$$\int_{0}^\infty e^{-(a+ib)x^2}\, dx =\frac{\sqrt{\pi }}{2 \sqrt[4]{a^2+b^2}} e^{-\frac{i}{2} \arg (a+i b)}$$ where $a$ and $b$ are real numbers


问:
Quotient rings of Gaussian integers $\mathbb{Z}[i]/(2)$, $\mathbb{Z}[i]/(3)$, $\mathbb{Z}[i]/(5)$,
I am studying for an algebra qualifying exam and came across the following problem.

Let $R$ be the ring of Gaussian Integers. Of the three quotient rings
  $$R/(2),\quad R/(3),\quad R/(5),$$
  one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?

I know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.

I hope that this can be answered easily and quickly. Thanks.

答: Why don't you first think through the case of $\mathbb Z/n\mathbb Z$.  When is this a field? When is it a product of fields?  When is it neither?  Try to relate your answer to the factorization of $n$ and the Chinese Remainder Theorem.  Now see if you can generalize it to $\mathbb Z[i]$ (or to any PID).</p>


答:
Let $A$ and $B$ be defined by
\begin{align*}
  A & =\big\{a\in\mathbb{R}: a^2=\sum_{i=1}^\infty a_i10^{-2i},
a_i\in\{1,2\,\dots,9\}, a^2\not\in \mathbb{Q}\big\} \\
B & =\big\{b\in\mathbb{R}: b^2=\sum_{i=1}^\infty b_i10^{-2i+1},
b_i\in\{1,2,\dots,9\}, b^2<\frac12, b^2\not\in \mathbb{Q}\big\}.
\end{align*}
Note that if $r^2=a^2+b^2\in \mathbb{Q}$ then the decimal expansion of
$r^2$ will eventually have a finite length repeating sequence. Any
such sequence would have to come from a unique repeating sequence of
even place decimals from $a^2$ and repeating odd place decimals from
$b^2$. Since $a^2,b^2\not\in \mathbb{Q}$, such repeating sequences do
not exist, thus $r^2\not\in\mathbb{Q}$.  Not entirely sure about the
measure of $A$ and $B$, but for their squares, I think the measures
are positive.

Title: Does a given positive measure set in $\mathbb{R}^2$ contains product of two 1 dimensional positive measure set?

Body:
In $\mathbb{R}^2$ consider the 2 dimensional positive Lebesgue measure set,
$$D=\{re^{it}:t\in[0,2\pi),r\in[0,1]\setminus\mathbb{Q}\}  .   $$
Does there exist sets $A,B\subset\mathbb{R}$ of 1 dimensional positive  Lebesgue measure such that $$A\times B\subset D?$$

Note: The above is not true if $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin\mathbb{Q}\}.$$


问:
Using your slope intercept form equation, find the amount of donations needed
You need to create a valentines package. 50 valentine packages cost $4300. You've started your own charity donation program. The three scenarios listed below show how much donations(y) you've collected in(x) days.

Items are price times 50.

So,

Flowers: $20 * 50 = 1000

Candy: $20 * 50 = $1000

Card: $12 * 50 = $600

Gift: $18 * 50 = $900

Dinner: $16 * 50 = $800

Total Expense for 50 valentines packages = $4300

Given three ordered pairs:

Scenario 1: (6, 305); 6 days, you've collected $305.00

Scenario 2: (14, 585); 14 days, you've collected $585.00

Scenario 3: (0, 125); at the start of the donation program, you had $125.00 saved up

Based on the three scenarios, write the equation in slope intercept form.

Using the amount of money you need to create 50 valentine packages, how many donations do you need?

I found the slope using points (0, 125) since this is when the donation program started and (14, 585) since this the two week period from when donations started. I arrived at y=230/7x + 125. Now, using this, if I solve for y = 4300, this just gives me the days it would take to earn $4300 in donations.

答:
Money obtained (y) :
$y = 280/8 x + 125$ where x is the number of days and you solve for y = 4300


问:
f Riemann integrable on [0,1]
I was trying to show that if $f$ is Riemann integrable on $[0,1]$ then
$$\int_{0}^{1} f dx  = \lim \left( \frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n}) \right)$$

I know that since $f$ is Riemann integrable, $\forall \varepsilon > 0$ $\exists P_\varepsilon$ partition of $[0,1]$ such that if $P \le P_\varepsilon$ then
$$\left| S(P;f,x)-\int_{0}^{1} f dx\right| < \varepsilon$$

Then I defined $P \le P_\varepsilon$ as $P=\left\{ x_0=0<x_1<...<x_n=1 \right\}$ and $c_k=\frac{k}{n}$ and I got that
$$S(P;f,x)=\sum_{k=1}^{n} f(c_k)[x_k - x_{k-1}] = \sum_{k=1}^{n} f\left(\frac{k}{n}\right)[x_k - x_{k-1}]$$

I think that if the intervals of the partition are equidistant I would get the desired result, but is there anything that restricts me from creating a partition $P$ as such? Or did I made any mistakes in the reasoning explained before?

Thanks in advance!

答:
Following Michael Spivak - Calculus, page 282, 2008 - if $f$ is integrable on $[a,b]$, then for every $\varepsilon > 0$, there is some $\delta>0$ such that, if $P=\{t_0,\cdots,t_n\}$ is any partition of  $[a,b]$, with all lengths $t_i-t_{i-1}<\delta$, then
$$\left|\sum\limits_{i=1}^{n}f(x_i)(t_i-t_{i-1}) -\int\limits_{a}^{b}f(x)dx \right |<\varepsilon$$
for any Riemman sum formed by choosing $x_i$ in $[t_i,t_{i-1}]$.

As you see nothing prevents you from considering  equidistant partition and choosing one border of it as value for function. So, your getting desired result is correct.

Title: f Riemann integrable on [0,1]


问:
Finding the probability of students learning material within a certain timeframe
I am trying to determine the probability of students learning a set of instructions within a given period of time. The length of time it takes for the students to learn has a normal distribution with a mean of 24min and a standard deviation of 8min. There is only one copy of the instructions, so as soon as one person has completed the learning process, the next person begins, with no intervening gap. What is the probability that the total time for 25 learners to complete the process will be less than 9 hours (i.e. 540 minutes)?

I first wrote the probability for a single learner:

540/25=21.6min

p(x<21.6) = p((x-μ)/(σ/√n)<(21.6-24)/(8/√25)

Using normal approximation:

p(z<-1.5)

=p(z>1.5)

=0.066807

Does this check out?

答:
If $X_i,$ for  $i =1,2, \dots, 25,$ are a random sample from
$\mathsf{Norm}(\mu = 24, \sigma = 8),$ then
$$S = \sum_{n=1}^{25} X_i \sim
\mathsf{Norm}(25\mu = 600, \sqrt{25\sigma^2}= 40),$$
by independence.

You seek $P(S < 540) = 0.1587.$ Computation in R below or by standardizing and using printed normal CDF tables.

```<code>pnorm(540, 600, 40)
[1] 0.0668072
</code>```

A simulation gives an approximate verification and a figure. [Using 100,000 iterations gives about two or three
significant digits of accuracy.]

```<code>set.seed(2020)
s = replicate(10^5, sum(rnorm(25, 24, 8)))
mean(s); sd(s)
[1] 599.9223   # aprx E(S) = 600
[1] 39.87919   # aprx SD(S) = 40
mean(s < 540)
[1] 0.06765    # aprx P(S < 540) = 0.0668

hist(s, prob=T, br=30, col="skyblue2",
     main="Simulated Totals")
 curve(dnorm(x,600,40), add=T, col="orange", lwd=2)
 abline(v = 540, lty="dotted", lwd=3, col="darkgreen")
</code>```

Body:
If $X_i,$ for  $i =1,2, \dots, 25,$ are a random sample from
$\mathsf{Norm}(\mu = 24, \sigma = 8),$ then
$$S = \sum_{n=1}^{25} X_i \sim
\mathsf{Norm}(25\mu = 600, \sqrt{25\sigma^2}= 40),$$
by independence.

You seek $P(S < 540) = 0.1587.$ Computation in R below or by standardizing and using printed normal CDF tables.

pnorm(540, 600, 40)
[1] 0.0668072

A simulation gives an approximate verification and a figure. [Using 100,000 iterations gives about two or three
significant digits of accuracy.]

set.seed(2020)
s = replicate(10^5, sum(rnorm(25, 24, 8)))
mean(s); sd(s)
[1] 599.9223   # aprx E(S) = 600
[1] 39.87919   # aprx SD(S) = 40
mean(s < 540)
[1] 0.06765    # aprx P(S < 540) = 0.0668

hist(s, prob=T, br=30, col="skyblue2",
     main="Simulated Totals")
 curve(dnorm(x,600,40), add=T, col="orange", lwd=2)
 abline(v = 540, lty="dotted", lwd=3, col="darkgreen")


问:
For $n$ points $P_1$, $\ldots$, $P_n$ on the unit circle show that $\sum_{i<j} d(P_i, P_j )^2 ≤ n^2$

Suppose that $P_1$, $P_2$, . . . , $P_n$ ($n \in \mathbb{N}$) are points on a circle of radius 1. For any two points $X$, $Y$ on the plane, let $d(X, Y)$ denote the standard Euclidean distance.

Prove
that $$\sum_{1≤i≤j≤n}
d(P_i, P_j)^2 ≤ n^2$$

and find when equality holds.

For this question initially my motivation was to search using the Pigeon Hole Principle, but then I realised I was on the wrong track. How do I approach this problem?

答:
HINT:

Check that
$$n(\sum_{i=1}^n \|v_i\|^2) - \|\sum_{i=1}^n v_i\|^2 = \sum_{i< j} \|v_i -v_j\|^2$$
for any vectors $v_1$, $\ldots$, $v_n$.

$\bf{Added:}$ One checks the identity in dimension $1$ ( Lagrange indentity), and then use the fact $\|w\|^2= \sum w_s^2$, where $w_s$' are the the components of $w$.

We have equality if and only if the sum of the unit vectors $v_i$ is $0$ :  $\sum v_i = 0$.


问:
Find position on surface of a lens
If I have a lens with coordinates UV on the lens surface where U, V are [-1, 1] and I want to find the real-world (x,y,z) coordinates of the UV point, how would I do that if I have the following information:

-The lens is centered on the z-axis.

-You know the lens' z-intercept value z.

-The diameter of the lens is d.

-It has a spherical radius of curvature value 'c' where c>0, c<0 and c=0 are convex, concave and flat respectively.

答:
The equation for the sag $z$ of a spherical lens of curvature $c$ is

$$z = \frac{1}{c} \left [ 1-\sqrt{1-c^2 \rho^2}\right ] = \frac{c \rho^2}{1+\sqrt{1-c^2 \rho^2}}$$

$\rho^2=x^2+y^2$ may be written in terms of your $(u,v)$ coordinates as

$$\rho^2=\frac14 d^2 (u^2+v^2)$$

Title: Find position on surface of a lens


问:
Localization is finitely generated implies, that multiplicative group only contains invertible elements
I am currently stuck on this problem:

"Let $R$ be an integral domain, let $0 \notin S \subseteq R$ be a multiplicative group of $R$. If $S^{-1}R$ is finitely generated as a $R$-module, then every element of $S$ is invertible (in $R$)"

I just started doing Algebra last month and I have some difficulties grasping these abstract concepts. So let's assume, that $S^{-1}R$ is finally generated by $\frac{1}{s_1}, ... , \frac{1}{s_n}$ with $s_1, ... , s_n \in S$. What can I do now? My first guess, that $S$ has to be finite as well, because if I understood it right, the localization $S^{-1}R$ describes a Ring, that is a little bit bigger than $R$ that also contains the multiplicative inverses of all Elements of $S$. So for example, if $R = \mathbb{Z}$ and $S = \mathbb{N}$, then $S^{-1}R = \mathbb{Q}$.
Is that so far correct? But in this case, $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$-module, so it's not a very good example, I tried it on $Z_{11}$, and as a field it obviously contains multiplicative inverses, so every multiplicative group is a unit. Also, it looks like if all such Rings $R$ would already satisfy $R = S^{-1}R$, is that true?

Nevertheless, I still don't know how to formally prove it, can someone maybe give me a small tip, because right now I don't even know where to start.

答:
I apologise for the somewhat misleading comments I've been posting above, for I have only now realised you include in the problem context the hypothesis that your original ring is an integral domain.

The problem statement remains true in a slightly more general context, which I would like to present. Assume that $A$ is a commutative ring and $S \leqslant_{\mathrm{Mon}}A$ (the notation $\leqslant_{\mathrm{Mon}}$ means submonoid of) is a multiplicative system consisting only of cancellable elements, which is equivalent to claiming that the localisation map $\iota \colon A \to S^{-1}A$ is injective. Then if $S^{-1}A$ is finitely generated as an $A$-module than necessarily $S \subseteq \mathrm{U}(A)$. For arbitrary $t \in A$ and $x \in S^{-1}A$ I will use the notation $t.x\colon=\iota(t)x$.

Let $Y \subseteq S^{-1}(A)$ be a finite generating system for $S^{-1}(A)$ as an $A$-module. By virtue of this finiteness, considering the product of all the (finitely many) denominators occurring in some fixed representations of the elements of $Y$ as fractions, one easily infers the existence of $a \in S$ such that $a.Y \subseteq \mathrm{Im}\iota$ and subsequently the existence of $X \subseteq A$ such that $a.Y=\iota[X]$ (we can argue that a finite such $X$ exists, but it is not relevant for the rest of the argument).

Since $A$ and its localisation are both commutative, the map of multiplication by $\iota(a)$ is an endomorphism of $S^{-1}(A)$ as a module over itself and hence in particular as an $A$-module. As this map is surjective (it is actually an automorphism, since $\iota(a)$ is invertible as $a \in S$), it follows that it maps the generating system $Y$ onto yet another generating system:
$$S^{-1}(A)=\iota(a)S^{-1}(A)=\iota(a)\langle Y \rangle_A=\langle \iota(a)Y \rangle_A=\langle a.Y \rangle_A=\langle \iota[X] \rangle_A=\iota[(X)],$$
where we remark that the last equality makes use of the fact that $\iota$ is in particular $A$-linear.

At this point we have ascertained that $\iota$ is surjective ($S^{-1}A=\iota[(X)]\subseteq \iota[A] \subseteq S^{-1}(A)$) and we know by hypothesis that it is also injective. As I remarked in one of the comments above, it is an elementary fact of localisation theory that the localisation map is bijective if and only if the multiplicative system consists only of units, which is in particular the case here.


问:
Issue with distance point-to-line
I am having a small issue with the distance calculation of a point-to-line and I cannot find the mistake.

I have the line:

$$x + 2y + 3 = 0$$

And I am trying to compute its distance to point $(x_0, y_0) = (0,0)$. According to the well-known formula:

$$ d = \frac{\left|ax_0+by_0+c\right|}{\sqrt{a^2 + b^2}} = \frac{\left|3\right|}{\sqrt{5}} = 1.3416$$

Now, I try to compute the same distance on a different manner. The line intersects the $y$ axes at $p=(0, -1.5)$. I am trying to solve the following triangle (sorry for the cheap plotting):

The distance computed before should correspond to the length of the red segment (perpendicular to the line that goes through the desired point), which could also be computed as:
$$ d = 1.5 \cdot cos(30^{\circ}) = 1.299$$

Both results for the length of the red segment do not match. What am I doing wrong?

Thank you in advance!

答: The angles are wrong - instead of $30^0$ it should be $\arctan(0.5) = 26.56^0$

Title: Issue with distance point-to-line


问:
Proving $\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+1}{n+1})}k>\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+3}{n+1})}k$
A few days ago I asked a question about an interesting property of the partial sums of the series $\sum\sin(nx)/n$.
Here's the link:
Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$

The proof given there left some details for me, in particular I have to prove that
$$
\sum_{k = 1}^{n}{1 \over k}
\sin\left(k\pi\frac{2m+1}{n+1}\right) >
\sum_{k=1}^{n}{1 \over k}\sin\left(k\pi\frac{2m+3}{n+1}\right)
$$
for every natural $n$ and $m$ such that $m = 0, 1, \ldots, \left\lfloor\left(n-1\right)/2\right\rfloor$. I don't think this is relatively simple; anyway I tried a few expansions with sum-to-product and viceversa formulas, but everything I tried leads to $$\sum_{k=1}^{n}\frac{\cos(2(m+1)x_k)\sin(x_k)}{k} < 0$$
where $x_k = k\pi/(n+1)$. From here I don't see any further semplification. Also I noted that $\sin(x_k)$ is always positive since $k$ is at most $n$, but we can't say much about $\cos$.
Then I tried to pair terms since for $k' = n+1-k$ we have $\sin(x_k) = \sin(x_k')$, which gives $$\sum_{k=1}^{n/2}\left[\frac{\cos(2(m+1)x_k)}{k}+\frac{\cos(2(m+1)x_{k'})}{n+1-k}\right]\sin(x_k) < 0$$ for $n$ even. Then I hoped that the term between square brackets were always negative, but it's not always the case... How can I do?

答:
Remark: One year ago, I analyzed the monotonicity of the function values at all local minimizers,
when I tried to answer Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson).
Here, I give the analysis of the monotonicity of the function values at all local maximizers.

Problem: Given positive integer $n\ge 3$, all the local maximizer of $f(x) = \sum_{k=1}^n \frac{\sin kx}{k}$ on $(0, \pi)$ are given by $x_m = \frac{2m+1}{n+1}\pi, m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor$.
Prove that $f(x_m) > f(x_{m+1})$ for all $m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor - 1$.

Proof: Since $f(x) = \int_0^x f'(t) \mathrm{d} t + f(0)$, we have
\begin{align}
f(x_{m+1}) - f(x_m) &= \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+3}{n+1}\pi} f'(t) \mathrm{d} t
= \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+2}{n+1}\pi} \Big( f'(t) + f'(t + \tfrac{\pi}{n+1})\Big) \mathrm{d} t. \tag{1}
\end{align}
Since $f'(x) = \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2} \cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}$, we have,
for all $t\in [\frac{2m+1}{n+1}\pi, \frac{2m+2}{n+1}\pi]$,
\begin{align*}
&f'(t) + f'(t + \tfrac{\pi}{n+1})\\
=\ & \frac{\sin \frac{nt}{2} \cos \frac{(n+1)t}{2}}{\sin \frac{t}{2}} + \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi)
\cos (\frac{(n+1)t}{2} + \frac{\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\
=\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\
& - \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi)\sin (\frac{(n+1)t}{2} + \frac{\pi}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{2}\\
=\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\
& - \frac{\cos (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2})\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\
\le\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\
& - \frac{\cos (\frac{nt}{2} - m\pi )\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}  \tag{3}\\
\le\ & - \frac{\cos \frac{\pi - t}{2}}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\
\le\ & 0
\end{align*}
where in (2) we have used
$\sin(\alpha - m\pi) \sin (\beta - \frac{(2m+1)\pi}{2}) = -\sin \alpha \cos \beta$,
and in (3) we have used
\begin{align}
&0 < \frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2} < \frac{nt}{2} - m\pi < \pi, \\
&0 \le \frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2} \le \frac{\pi}{2},\\
&0 < \frac{t}{2} < \frac{t}{2} + \frac{\pi}{2n+2} < \frac{\pi}{2}.
\end{align}
Also, note that $f'(\frac{2m+1}{n+1}\pi) + f'(\frac{2m+1}{n+1}\pi + \tfrac{\pi}{n+1}) = -1 < 0$.
From (1), we have $f(x_{m+1}) < f(x_m)$. We are done.


问:
solutions of $az^2 + bz + c = 0$ have the same module $r > 0$

Prove that  the solutions of the equation $az^2 + bz + c = 0$, with $a, b, c \in \mathbb{C}$ have the same module $r > 0$ if and only if $b\overline{c} = r ^ 2 a\overline {b}$.

How can I deal with it? I have tried a rather "brute" solution, but it did not work and I am out of ideas...

$r^2 = \frac{b^2\overline{ac}}{ab\overline{ab}}$ and $r = |-b - \sqrt{b^2 -4ac}| = |-b+\sqrt{b^2-4ac}$ and I tried proving it, but it gets quite complicated.

答: let $p,q$ be the roots of equation. then $$\bar p=\frac{r^2}{p},\bar q=\frac{r^2}{q}$$ Now by vieta $$\color{blue}{p+q=\frac{-b}{a}},pq=\frac{c}{a}$$ hence taking conjugate of blue colored equation $$\color{blue}{\frac{-\overline{ b}}{\bar a}=\bar p+\bar q}=r^2(\frac{p+q}{pq})=\frac{-r^2b}{c}$$ $$\iff b\bar c=r^2a \bar b$$


问:
proof that sets [4, 7) and (2,12) are equivalent
prove that [4, 7) and (2,12) are equivalent:

Cantor-Bernstein theorem

constructing bijection between these sets.

A=[4, 7) and B=(2,12)
injection from A to B, $f:A \to B$
y=x, $x \in [4,7). $
$f([4,7))=(4,7) \subset (2,12)$

injection from B to A, $g:B \to A$

$y=4+ \frac{1}{4}x$

$g((2,12))=(4.5,7) \subset [4,7)$

but what about the second proof?

答:
We seek to construct a bijection between the sets $[4,7)$ and $(2,12)$. We will do so by first creating a bijection $h:(2,12) \to (4,7)$, and then creating a bijection $g:(4,7) \to [4,7)$. Since the composition of bijections is a bijection, the bijection $g \circ h:(2,12) \to [4,7)$ will be a desired one.

For the bijection $h:(2,12) \to (4,7)$, define $h$ as the function of the line passing through $(2,4)$ and $(12,7)$. (This trick works to define a bijection between any two open intervals.) This line is

$$ h(x)= \frac{3}{10}(x-2)+4$$

Where $2<x<12$. You may graph $h(x)$ to convince yourself that this is a correct bijection.

For the bijection $g:(4,7) \to [4,7)$, I will use an approach very similar to Ben's answer to the post linked by Eric Towers in the comments.

Let $A_1= \{7-\frac{3}{2},7-\frac{3}{3},7-\frac{3}{4},\ldots\}$. Note that $A_1\subseteq (4,7)$.

Let $A_2= \{4\}\cup A_1 = \{7-\frac{3}{1}, 7-\frac{3}{2},7-\frac{3}{3},\ldots\}$.

Define the function $f: A_1 \to A_2$ by $f(7-\frac{3}{n})=7-\frac{3}{n-1}$. $f$ is a bijection.

Now define $g:(4,7) \to [4,7)$ as $g(y)=y$ if $y\notin A_1$, otherwise $g(y)=f(y)$. $g$ is a bijection.

We are now ready to create a composition $g \circ h:(2,12) \to [4,7)$.

$(g \circ h)(x) = g(h(x))=h(x)$ if $h(x)\notin A_1$, otherwise $g(h(x))=f(h(x))$.

$g \circ h$ is the desired bijection. My apologies for inaccuracies and/or incorrect etiquette; this is one of my first answers on this site. I would appreciate any feedback. I hope this is helpful!


答:
The polar form of the straight line is

$$ x\cos \alpha + y \sin \alpha = p =1.3496$$

where $\alpha = \cos^{-1}\dfrac{-1}{\sqrt 5}$ is counter-clockwise rotation of the red normal reckoned from x-axis lying in the third quadrant.

Title: Issue with distance point-to-line

Body:
I am having a small issue with the distance calculation of a point-to-line and I cannot find the mistake.

I have the line:

$$x + 2y + 3 = 0$$

And I am trying to compute its distance to point $(x_0, y_0) = (0,0)$. According to the well-known formula:

$$ d = \frac{\left|ax_0+by_0+c\right|}{\sqrt{a^2 + b^2}} = \frac{\left|3\right|}{\sqrt{5}} = 1.3416$$

Now, I try to compute the same distance on a different manner. The line intersects the $y$ axes at $p=(0, -1.5)$. I am trying to solve the following triangle (sorry for the cheap plotting):

The distance computed before should correspond to the length of the red segment (perpendicular to the line that goes through the desired point), which could also be computed as:
$$ d = 1.5 \cdot cos(30^{\circ}) = 1.299$$

Both results for the length of the red segment do not match. What am I doing wrong?

Thank you in advance!


答:
Let's go through the process step by step instead of using a calculator/grapher, and (ultimately) verify Dhanvi's claim.

So, to find the distance from the line (call it $\ell$) to $(0,0)$, we will want the line perpendicular to $\ell$ through the origin (call this one $p$).

The equation of $p$ will have negative reciprocal slope to $\ell$ and $y$-intercept $0$. Then we have that

$$y_p (x) = 2 x$$

Meanwhile, $\ell$ has the equation

$$y_\ell(x) = -\frac 1 2 x - \frac 3 2$$

So your triangle has two points already: $(0,0)$ and $(0,-3/2)$. The third is where $\ell$ and $p$ intersect. Set the two equations equal and solve for $x$:

$$2x = - \frac 1 2 x - \frac 3 2 \implies x = - \frac 3 5$$

Substitution can get us that, then, the corresponding $y$ is at $y = -6/5$.

We can then find the distances between these points, and we get that

\begin{align*}
\Big( \text{distance from } (0,0) \text{ to } (0,-1.5) \Big) &= 1.5 \\
\Big( \text{distance from } (-3/5,-6/5) \text{ to } (0,-1.5) \Big) &= \frac{3}{2 \sqrt 5} \\
\Big( \text{distance from } (-3/5,-6/5) \text{ to } (0,0) \Big) &= \frac{3}{\sqrt 5}
\end{align*}

Obviously, one angle is formed by the first and last sides. It has angle $\theta$, and it will be given by

$$\theta = \arccos \left( \frac{3 / \sqrt 5}{1.5} \right) \approx \boxed{26.57^\circ}$$

This is where your problem lies. That top angle of your triangle is not $30^\circ$.

Title: Issue with distance point-to-line