问:
The matrix diagonalizable coefficients in a field of characteristic different from $2$
I read a property that I did not understand:

Let $M\in GL(2,\mathbb{F}_3)$ the matrix such that $M= I_2$ or $M^2=I_2$. As it has coefficients in a field of characteristic different from $2$, it is diagonalizable.

Is it a theorem? Can you help me prove this result

答: If $M=I_2$ then $M$ is already diagonal.  Same if $M = -I_2$.  If $M^2 = I_2$ with $M \ne \pm I_2$ (so that no linear polynomial annihilates $M$) then $p(M) = M^2 - I_2 = (M-I)(M+I) = 0$ so $p(x) = (x-1)(x+1)$ is the minimal polynomial of $M$, and in a field of characteristic different from 2, the two factors $x-1$ and $x+1$ are distinct.  This implies that $M$ is diagonalizable.


问:
Why $v_1 \in L$ and $v_2 \in L^\perp$?
Let $L$ be a one dimensional subspace of $R^2$ (we may view $L$ as a line in the plane through the origin).

Suppose $\alpha$ is the angle from the positive x axis to $L$.

Let $v_1=(\cos \alpha, \sin \alpha)$ and $v_2=(-\sin \alpha, \cos \alpha)$.

Then $v_1 \in L$ and $v_2 \in L^\perp$.

I can't understand the last line. Why $v_1 \in L$ and $v_2 \in L^\perp$?

答: The vector $ \ \overrightarrow{v_1} \ $ is a unit vector which lies in the line L , so it would be considered an element of the vector subspace $ \ L \ $ , which is the set of all vectors pointing along the line L (in either direction).  Since $ \ \overrightarrow{v_1}  \cdot   \overrightarrow{v_2} \ = \ 0 $ , the second vector is perpendicular to the first, and so must lie in the vector subspace in $ \ \mathbb{R}^2  $ which is designated as $ \ L^{\bot} \ $ ; it is also a unit vector, so   $ \ \overrightarrow{v_2} \ $ would be one element of  $ \ L^{\bot} \  . $

ContentLicense: CC BY-SA 3.0

Title: Why $v_1 \in L$ and $v_2 \in L^\perp$?


问:
Differentiation problem regarding unspecified variable or function.
I am given
$$
y = \alpha + \frac{1}{\alpha}
$$
and asked to prove that
$$
\alpha^2 y'' + \alpha y-2 = 0
$$
but when I use $\alpha$ as a variable I get $y-2 = 0$. Am I missing something? No other explanation is given regarding $\alpha$ or $y$.

答: It is wrong. $y$ is a function of $\alpha$. We have $y'=1-\frac 1 {\alpha^{2}}$ and $y''=\frac 2 {\alpha^{3}}$. Hence $\alpha^{2}y''+\alpha y-2=\frac 2 {\alpha}+\alpha^{2}+1-2$ which is not $0$.

ContentLicense: CC BY-SA 4.0


问:
$ \log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right| $ one formal way to say we won't choose all solutions in degree 2 equation

$$
\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right|
$$

After applying some properties we have:
$$
\log\left|\frac{x^2-4x}{3x-10}\right| = 0.
$$

After applying some operations, this equation becomes:
$$
x^2-7x+10 = 0\\
x=2 ;x=5.
$$

The problem is: if we substitute $x$ with "$5$" it's good, but if we substitute with "$2$" there's an obvious error, since $2$ is out of the logarithm's domain.

Are there some 'formal' ways to say why we won't choose number 2, that just say: "we won't choose number 2 because it doesn't make sense"?

EDIT:
Here is another exercise to explain that there are some solutions that doesn't work even with absolute values in the logarithms' arguments:

$$
\log(\sqrt{x+14})+\log(\sqrt{x+7})-\log(1.2)=1 \\
\cdots\\
x^2+21x-46=0\\
x_1=-23; x_2=2.$$

But again: what happened with $-23$ ? Does it just disappear ?

Note: I'm working in $\mathbb R.$

答:
Cant comment but I believe there's a special criterion you have to satisfy first and foremost when dealing with Logarithmic inequalities.

For example, in $\log|x-4|$ you have to ensure the number being logged (I call it the argument) is always a non-negative number (that is $|x-4|$ in this case).

As a generalisation in $\log_ab$,
$b>0  ,a>0$ and $a$ not equal to $0$ are the conditions your logarithmic inequality must satisfy before you do the actual cancelling of the logs in both the sides.

In our case in $\log|x-4|$, $b>0$ as it is a modulus, but don't forget from  $b>0$ you can get an equation $|x-4|>0$

I am leaving the rest for you, don't forget to do an intersection of the sets

Note: I haven't checked if the numbers are "good" or "bad", lone student seems to be saying they are good enough, but I think what I showed is the "formal proof" to reject a solution if it does not satisfy the original condition we had (i am pretty sure there's another formal name for this kind of rejection) you need to keep this in mind when you are doing inequalities

Note 2: i realised I misread the question but I wrote a lot anyway so I am going to keep  it. as for  the question

$\log\left|\frac{x^2-4x}{3x-10}\right| = 0$

you will get $log 1=0$ and then proceed to cancel the log, however, you are still left with the modulus, so you should be getting 4 answers


问:
How to get the area of the triangle in a trigonometric circumpherence when there's a negative angle?
The problem is as follows:

The figure from below represents a trigonometric circle or unit circle. Assume $PQ=|\sin\theta|$. Find the area of the region represented by $ABC$.

The choices given are as follows:

$\begin{array}{ll}
1.&\frac{\sin\theta}{2}\\
2.&1\\
3.&\frac{1}{2}\\
4.&-\frac{\sin\theta}{2}\\
\end{array}$

Well, I'm totally lost in this problem. As I don't know how to understand the negative value which it has been given in the figure.

The image has been reconstructed from the original source which isn't very clear, so I did to my best of drawing abilities to put all the information but it is not very clear if the angle starts on point which lies on $\textrm{x-axis}$ or in point $B$ so, please consider this in your answer, so it may match the answers given.

In short my biggest complain is that weird arrow how should I understand it?. Can you please address this doubt of mine in the heading of your answer?

Other than that what I could notice is that catch here is how to get the value of the base of the triangle and the height for $\triangle ABC$. The thing is that I'm confused how to use the fact that it confuses me that $PQ$ is given as an absolute value.

Since this was found in an old trigonometry practice sheet the sort of strategy which I'm looking to get is by relying in the concepts of trigonometric circle.

As an extension to this I also notice that the radius here is one unit.

Thus can someone guide me here on exactly what to do, and what should I do to get one of the answers given in the choices?.

Since I'm not very sure with this topic I appreciate that the answer given would include the necessary steps to get to that answer.

答:
The negative value is just a distraction. We only need the magnitude of the angle.
$$A_{ABC}=A_{AOC}-A_{AOB} $$
Assuming a unit circle, the base of both triangles, $OA$, is $1$. Also $OQ=1+PQ=1+|sin\theta|$
$$\therefore A_{AOC}=\frac {1\times (1+|sin\theta|)} 2=\frac {1+|sin\theta|} 2  $$
$$A_{AOB}=\frac {1\times altitudeB} 2=\frac {|sin(180-|\theta|)|} 2$$
But $sin(180-\theta)=-sin\theta$
$$\therefore A_{ABC}= \frac {1+|sin\theta|} 2-\frac {|sin\theta|} 2=\frac 1 2$$


问:
How to do modular arithmetic on power tower?
I am struggling to learn modular arithmetic for power towers (tetration). Like for example, how would I go about $$\left[{13^{13^{13^{13^{13^{13^{13^{13}}}}}}}}\right] \bmod (1000)$$ Or for any number n, what do I do?

答: The powers of $n$ modulo $k$ repeat with some period that is a factor of $\phi(k)$, which is known as Euler's theorem.  Here the powers of $13$ repeat with a period of $100$. One easy way to calculate that is to make a spreadsheet with $1$ in a cell, then put =mod(up*13,1000) in the cell below and copy down.  Just look until you see $1$. That means you only need to know the stack with one less $13$ modulo $100$.  The powers of $13 \bmod 100$ repeat with period $20$ so you need the stack with two less $13$s $\bmod 20$.  After a few steps you will have period $1$ and don't need to go any farther.  One more step gets you down to $4$


问:
For which values of $a$ and $b$ is the integral convergent?
I have to find the values $a, b > 0$ that makes the following integral convergent:

$\int_{1}^{\infty} \frac{1}{x^a(1+x^b)} \,dx$

I've tried integrating $\frac{1}{x^a(1+x^b)}$ by plugging in different values of $a$ and $b$, and found out that the value of $a$ doesn't matter (I am not too sure if this is right) and $b \leq 1$ for the integral to be convergent, but I am not sure how I can justify this. Can someone help me?

答:
Compare with $\int_1^{\infty} \frac 1 {x^{a+b}}dx$. It is convegent iff $a+b>1$.

$\frac 1 {2(x^{a+b})}\leq \frac 1 {x^{a}(1+x^{b})} \leq \frac 1 {x^{a+b}}$ for all $x \in (1,\infty)$.


问:
A basis for a subspace of $\mathbb R^3$
I have the following question:

Find the basis of the following subspace in $\mathbb R^3$:
  $$2x+4y-3z=0$$

This is what I was given. So what I have tried is to place it in to a matrix $[2,4,-3,0]$ but this was more confusing after getting the matrix $[1,2,-3/2,0]$. This was done to get a leading 1. Now I solved for $x,y,z$.

$$1x + 2y - \frac3 2z = 0$$ from the matrix. Then
$x = \frac3 2z - 2y$
so
$$\left(\frac3 2z-2y, 2y, \frac3 2z\right)$$

Now I have $$(x,y,z) = z(3/2, 0, 3/2) + y(-2, 2, 0) \implies
{(3/2,0,3/2),(-2,2,0)}$$

which has two dimensions.

But this solution that I am giving is wrong. The actual answer according to my text book is ${(3,0,2),(-2,1,0)}$. Can any one please tell me where I messed-up? Thanks.

答:
You can get a basis quickly, as follows. You have $$2x+4y-3z=0$$.

This means, $$z=\frac{2x+4y}3$$

So if $\vec x$ solves the above, $$\vec x=\left(x,y,\frac 2 3 x+\frac 4 3 y\right)$$

That is, $$\vec x =x\left(1,0,\frac 23\right)+y\left(0,1,\frac 43\right)$$

So you can choose your basis to be $\{(3,0,2),(0,3,4)\}$ upon scaling.

In general, if you're working on $\Bbb R^3$; you know $ax+by+cz=0$ will be a subspace of dimension two (a plane through the origin), so it suffices to find two linearly independent vectors that satisfy the equation. To that end, make a coordinate vanish, say $x=0$, and find what $y,z$ may be. For example, in you case we get $4y-3z=0$, so we can take $y=3,z=4$, to get the first vector to be $(0,3,4)$. Now make $y=0$. You get $2x-3z=0$, so this suggests taking $x=3,z=2$. This gives the basis $$\{(0,3,4),(3,0,2)\}$$ as above. Note that if we chose to make say $z=0$; we could put $x=-2,y=1$ to get $$\{(-2,1,0),(3,0,2)\}$$

This works nicely because we have one equation, but wouldn't work so swimmingly if we had something of the form $$\begin{cases} 2x+3y-2z=0\\-x+y+3z=0\end{cases}$$

In fact, setting any coordinate to $0$ in the above gives nothing. However, setting say $z=1$ gives $x=11/5$ and $y=-4/5$, which correctly gives the solution set is $$\langle (11,-4,5)\rangle$$

Title: A basis for a subspace of $\mathbb R^3$


问:
Is it possible to make straight line trajectory of a certain point on the unit circle rolling on the special curve?
I'm interested in cycloid and make another problem about it.

If a unit circle rolls one lap on the straight line, the point on the circle draws a cycloid trajectory. Then what if a unit circle rolls one lap on the special curve $l$ so that the corresponding trajectory is a straight (horizontal) line?

I don't even know if the curve really exists, but try to find the curve on these settings:

$l: x=f(s), y=g(s)$ where $s \in [0,2\pi]$ means 'rolling distance on $l$'

$C$: unit circle rolling on $l$ with its center point $c(s)$

$p(s)$: the point on $C$ which makes the horizontal line trajectory we want.

$p(s)=(p_1(s),p_2(s))$ can be expressed by defining three vectors:

$p(s)=q(s)+u(s)+v(s)$ .... (2)

where

$q(s)=(f(s), g(s))$ is the tangent point at which $l$ meets $C$,

$u(s)$ is the unit vector from $q(s)$ to $c(s)$, and

$v(s)$ is the unit vector from $c(s)$ to $p(s)$

and make parametric equations below:

$\int_{0}^{s}{\sqrt{f'(u)^2+g'(u)^2}}=s$ .... (1)

$p_2(s)=0$ .... (2)

Eqn(1) is from the curve length formula. If $C$ rolls distance $s$, the curve length behind becomes $s$.

Eqn(2) is from the condition where $p(s)$ is on the horizontal line.

Finding exact form of $u(s)$, $v(s)$ and simplifying them result in the system of 1st order nonlinear ODEs below:

$(f'(s))^2 + (g'(s))^2 = 1$

$-g(s)+\sin(s)g'(s)+(\cos(s)-1)f'(s)=0$

so terrible equations... even WolframAlpha doesn't solve it. I cannot go further.

Are there some well-known keywords about the initial problem?

答: The "special curve" is a circle with twice the radius. This produces the Tusi couple.


问:
Rolling four dice, what are the odds of having at least a pair of sixes?
Let's say I roll four dice. I want to know the odds of having at least two 6. There can be more too.

I did: 6C1 * 6C1, which gave me 36. I assumed the total possible combinations was 1296. I don't know if it's good, I doubt it is...

How do I know what the odds are of throwing 4 dice and at least getting two 6?

答:
HINT:

Count the odds of 6 showing up 0x and 1x time, then use complement rule.

$\to$  0x time: $\frac{5^4}{6^4}$

$\to$  1x time:  $\frac{{4 \choose 1 }5^3}{6^4}$


问:
Axler exercise 2.B.7

Prove or gibe a counterexample: If $v_1, v_2, v_3, v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1, v_2 \in U$ and $v_3 \not \in U$ and $v_4 \not \in U$, then $v_1, v_2$ is a basis of $U$.

It turns out that there is a counterexample, but I'm trying to figure out how to figure out a-priori whether a statement like this is true. The first thing I always try to do is prove it and see if I run into a problem. Provided that there are edge cases where my proof fails, I'll use them to try to find a counterexample. However, I managed to write down a proof of this and couldn't find any holes in it, though there surely is one.

Let $u \in U \subset V$, so there exist $a_1, \ldots, a_4 \in F$ such that
$$u = a_1 v_1 + \ldots + a_4 v_4,$$
since $v_1, \ldots, v_4$ is a basis of $V$ and $u \in V$ .But $u - (a_1 v_1 + a_2 v_2) \in U$ since $v_1, v_2 \in U$, which would imply that $a_3 v_3 + a_4 v_4 \in U$. Since $v_3, v_4 \not \in U$, this implies that $a_3 v_3 + a_3 v_4 = 0$. That is, $u = a_1 v_1 + a_2 v_2$, so $v_1$ and $v_2$ span $U$. Furthermore, if there exist coefficients $a_1, \ldots, a_4$ such that
$$a_1 v_1 + \ldots + a_4 v_4 = 0,$$
then $a_1 = a_2 = a_3 = a_4 = 0$ by linear independence, so $v_1$ and $v_2$ are linearly independent. So $v_1, v_2$ is a basis for $U$.

The only step of this that I can see may be a leap of logic is the conclusion "so $a_3 v_3 + a_4 v_4 = 0$. This seems like a jump, but I can't find another case where $a_3 v_3 + a_4 v_4 \in U$ but $v_3, v_4$ are not.

答: You're right.  Let $\{v_1, v_2, v_3, v_4\}$ be the standard basis.  Let $U$ have $\{v_1, v_2, v_3 + v_4\}$ for a basis. Then $v_3$ is not in $U$, and $v_4$ is not in $U$. But the subspace is 3 dimensional, not 2.

Title: Axler exercise 2.B.7


答:
I'll follow your method as closely as I can to make it clearer to you.

We get an equation $$2x+4y-3z=0 \implies x+2y-\boxed{\frac3 2z}=0 \implies x = \frac3 2z-2y$$
so that
$$(x,y,z)=\left(\frac 3 2 z-2y,y,z\right) = z\left(\frac 3 2,0,1\right) + y\left(-2,1,0\right)$$

Note that there's no reason why the basis you get and the basis they get should be the same, because there are infinitely many bases spanning the same space. For example, for any basis $\mathbf{x},\mathbf{y}$ the vectors $\mathbf{x+y},\mathbf{x-y}$ are a basis too.

Title: A basis for a subspace of $\mathbb R^3$

Body:
I have the following question:

Find the basis of the following subspace in $\mathbb R^3$:
  $$2x+4y-3z=0$$

This is what I was given. So what I have tried is to place it in to a matrix $[2,4,-3,0]$ but this was more confusing after getting the matrix $[1,2,-3/2,0]$. This was done to get a leading 1. Now I solved for $x,y,z$.

$$1x + 2y - \frac3 2z = 0$$ from the matrix. Then
$x = \frac3 2z - 2y$
so
$$\left(\frac3 2z-2y, 2y, \frac3 2z\right)$$

Now I have $$(x,y,z) = z(3/2, 0, 3/2) + y(-2, 2, 0) \implies
{(3/2,0,3/2),(-2,2,0)}$$

which has two dimensions.

But this solution that I am giving is wrong. The actual answer according to my text book is ${(3,0,2),(-2,1,0)}$. Can any one please tell me where I messed-up? Thanks.


问:
Find the minimum $n$ so that $\gamma_n$ is less than $10^{-4}$ from $\gamma$.

It is known that the sequence $\gamma_n = 1+ \frac{1}{2}+\cdots + \frac{1}{n}-\log(n)$ converges to $\gamma$ = 0.5772156649, wich is called Euler's constant. Find the minimum $n$ so that $\gamma_n$ is less than $10^{-4}$ from $\gamma$.

I am solving some exercises of numerical analysis but when I get to this I have no idea how to start. My first idea was to directly replace $ n $ but I have not made any progress. Could you give me any suggestions or what material to read in order to solve the problem? Thanks

答:
As @cavok commented, if
$$\gamma_n = -\log(n) + \sum_{k=1}^n \frac{1}{k}=-\log(n)+H_n$$ and you want to find $n$ such that
$$|\gamma_n-\gamma| \leq \epsilon$$ use the asymptotics of harmonic numbers and you need to find $n$ such that
$$\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120
   n^4}+O\left(\frac{1}{n^6}\right) \leq \epsilon$$ For sure, you could take the first and second terms and solve the quadratic equation.

I suppose that $n=\frac 1{2 \epsilon}$ will be very close.

Let us try with $\epsilon=0.001$ and use $n=500$
$$\gamma_{500}=0.578215 \implies |\gamma_n-\gamma|=0.000999667$$


问:
Check if nD point lies within nD Hypersphere defined by n+1 points
I am writing some code where I need to be able to determine if a specific n-D point lies within a n-D hypersphere defined by n+1 points. The 2D version of this is clear to me; to check if a point, p,  lies within a circle defined by a triangle abc, one can just compute the determinate of the following matrix:

[a_x, a_y, |a|, 1]

[b_x, b_y, |b|, 1]

[c_x, c_y, |c|, 1]

[p_x, p_y, |p|, 1]

(|n|= n_x^2+n_y^2)

if the determinate is greater than zero then the point lies outside of the circle, if its zero, its on the circle, else it is within the circle.

I am hoping that this can scale up to how ever many dimensions I need, but I am unsure if it will as I do not know what would go in the additional column would be, obviously the new row would represent the new point.

Any help or suggestion would be greatly appreciated! thanks!

答:
Your simple formulation has a flaw -- if I change points $A,B$ around, the determinant get multiplied by $-1$ so inside turned into outside.  It is either a feature or a bug, I don't know your intention.

To get the usual formulation,
$$
\det
\begin{pmatrix}
a_x^2+a_y^2 & a_x & a_y & 1\\
b_x^2+b_y^2 & b_x & b_y & 1\\
c_x^2+c_y^2 & c_x & c_y & 1\\
x^2+y^2 & x & y & 1
\end{pmatrix}=0
$$
is the Cartesian formula for the circle through $A,B,C$ (assuming the points are not collinear, i.e. the coefficient of $x^2+y^2$ is nonzero).  So a point is strictly outside (resp inside) the circle iff
$$
\det
\begin{pmatrix}
a_x^2+a_y^2 & a_x & a_y & 1\\
b_x^2+b_y^2 & b_x & b_y & 1\\
c_x^2+c_y^2 & c_x & c_y & 1\\
p_x^2+p_y^2 & p_x & p_y & 1
\end{pmatrix},
\det
\begin{pmatrix}
a_x & a_y & 1\\
b_x & b_y & 1\\
c_x & c_y & 1
\end{pmatrix}
$$
have the same (resp. opposite) sign.  (Remember we want the coefficient of $x^2+y^2$ to be 1 in the usual Cartesian equation of circle $\Gamma\colon C=0$ before saying $C<0$ is the inside and $C>0$ is the outside.)

This is the formulation that generalises: if the $n+1$ points $P^{(1)},\dots,P^{(n+1)}$ does not lie on a hyperplane, then point $P$ is inside/on/outside the hypersphere defined by $P^{(i)}$ if
$$
\det
\begin{pmatrix}
\lvert P^{(1)}\rvert^2 & p^{(1)}_1 & \dots & p^{(1)}_{n+1} & 1\\
\lvert P^{(2)}\rvert^2 & p^{(2)}_1 & \dots & p^{(2)}_{n+1} & 1\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
\lvert P^{(n+1)}\rvert^2 & p^{(n+1)}_1 & \dots & p^{(n+1)}_{n+1} & 1\\
\lvert P\rvert^2 & p_1 & \dots & p_{n+1} & 1
\end{pmatrix}
$$
has the opposite/zero/same sign as
$$
\det
\begin{pmatrix}
p^{(1)}_1 & \dots & p^{(1)}_{n+1} & 1\\
p^{(2)}_1 & \dots & p^{(2)}_{n+1} & 1\\
\vdots & \ddots & \vdots & \vdots\\
p^{(n+1)}_1 & \dots & p^{(n+1)}_{n+1} & 1
\end{pmatrix}.
$$


答:
Note that you should have divided $-3z$ by $2$ to obtain: $x=-2y+\dfrac{3}{2}z$, which yields:

$$\left[ \begin{array}{c}
x \\
y \\
z
\end{array} \right]
=
\left[ \begin{array}{c}
-2y+\dfrac{3}{2}z \\
y \\
z
\end{array} \right]
=
\left[ \begin{array}{c}
-2y \\
y \\
0
\end{array} \right]
+
\left[ \begin{array}{c}
\dfrac{3}{2}z \\
0 \\
z
\end{array} \right]
=
y\left[ \begin{array}{c}
-2 \\
1 \\
0
\end{array} \right]
+
\dfrac{1}{2}z\left[ \begin{array}{c}
3 \\
0 \\
2
\end{array} \right]
$$
where I factored out the $1/2$ to avoid fractions (this step wasn't necessary, but makes the answer prettier). Thus, we obtain the desired basis.

Title: A basis for a subspace of $\mathbb R^3$


问:
Proving that a set and operation form a group

$S$ is a nonempty finite set with associative operation $∗$ such that, $$∀ a, b ∈ S, ∃ c,d ∈ S\text{ s.t. }a ∗ c = b \text{ and }d ∗ a = b$$Prove that $(S, ∗)$ is a group.

I'm stuck on a difficult problem from some PSET I found online (I'm self studying Algebra).

I have been trying to show the existence of an identity in $S$ on $*$ as I think the rest will follow, but I cannot seem to make any progress.

My work so far is not very solid, but it is an attempt to show that there is at least one $e$ exists such that

$$a ∗ e = e ∗ a = b, \text{ where }a,b\in S$$

Reasoning as follows:

From the problem statement:

$$∀ (a, b) ∈ S × S\ ∃ (c, d) ∈ S × S:\ a*c = d*a = b$$

I would like to show that at least one such $(a, b)$ has $c = d$.

I try a proof by contradiction:

Assume

$$∀ (a, b) ∈ S × S\ \exists (c, d) ∈ S × S\ c ≠ d:\ a*c = d*a = b$$

Since $S$ is finite, so let $|S| = n$.

$$\therefore|S × S| = n^2$$

But there are $n(n-1)$ pairs of the form $(c, d) ∈ S × S$ such that $c ≠ d$.

From here I am trying to use the pigeonhole principle to show that there exists $(a,x)$ and $(b,y)$ such that

$$(a = b\text{ and }x ≠ y)\text{ or }(x = y\text{ and }a ≠ b)$$
that satisfy

$$a*c = c*a = x$$

$$b*c = c*b = y$$

leading to an obvious contradiction.

I think there is a way to derive this existence, since intuitively there are $n$ "excess" pairs after relating pairs $(a, b)$ one-to-one with pairs $(c, d)$.

I would greatly appreciate a hint.

Thanks!

答: The conditions can be restated that left and right multiplication by every element of your set is bijective. So, $x\to a x$ is a permutation $\pi_a$ of your set. Since every permutation has finite order, $\pi_a^k = \pi_{a^k}$ is the identity permutation for some $k,$ and that shows that $a^k$ is the (left) identity element. Similarly $a^l$ is the right identity, so your set has a left and right identity. A standard argument shows that these are the same. Since left and right multiplications are bijective, it shows that you have a left and a right inverse for every element (and and another standard argument shows that those are the same.)


问:
Computing expected number of induced subgraphs of $G(n,\frac12)$ without graph removal lemma.
Define $G(n,\frac12)$ to be a graph on $n$ vertices where each edge has a $\frac12$ probability of being formed. We may compute the number of subgraphs of $G(n,\frac12)$ isomorphic to, say, $K_3$ as follows.

$K_n$ has $\binom{n}{3}$ such subgraphs; we can enumerate them as $H_1,\cdots, H_{\binom{n}{3}}$ and define $$X_i=\begin{cases} 1 & H_i\subseteq G(n,\frac12) \\ 0 & \text{otherwise}\end{cases}$$ to obtain $\mathbb{E}[X] = \sum_i \mathbb{E}[X_i] = \binom{n}{3}\left(\frac12\right)^3$.

But how can we count the number of induced subgraphs isomorphic to $K_3$?

It appears that the graph removal lemma gives me what I want for very special cases, but it is beyond the scope of my graph theory course. Surely there has to be a simpler approach when we are interested in "simple" subgraphs like $P_n$, $C_n$, or $K_n$.

答:
I think this is right...

For any smaller graph $H$, with $k$ vertices and $e$ edges,
$$
\begin{align}
\mathbb E[\text{# subgraphs $\cong H$}] &=\binom{n}k\frac{k!}{|\operatorname{Aut}(H)|}(1/2)^{e}
\\
\mathbb E[\text{# induced subgraphs $\cong H$}] &=\binom{n}k\frac{k!}{|\operatorname{Aut}(H)|}(1/2)^{k(k-1)/2}
\end{align}
$$
$\binom{n}k\frac{k!}{|\operatorname{Aut}(H)|}$ counts the number places $H$ can appear in $G$; you choose the vertices, given them names, and note that two re-namings which result in the same $H$ should not be counted twice. The only difference is that having an induced subgraph requires the edges not present in $H$ to not be present in $G$ as well, leading to extra factors of $1/2$.

Some examples:

If $H$ is a complete graph, then $|\operatorname{Aut}(H)|=k!$, so you get $\binom{n}{k}(1/2)^{k(k-1)/2}$ in either case.

If $H$ is a cycle, then $|\operatorname{Aut}(H)|=(k-1)!$ so you get either $\binom{n}{k}\cdot k(1/2)^{n}$ subgraphs, and $\binom{n}{k}\cdot k(1/2)^{n(n-1)/2}$ induced subgraphs, on average.


问:
How to prove this by mathematical induction?
ok! so i don't have a clue as to how to prove it.

$$\sum\limits_{i=m+1}^n i=((n-m)(n+m+1))/2$$ so the base case is equal to zero because $$\sum\limits_{i=1+1}^1 i=0$$ is this correct? I don't know how to go for inductive step.

Thanks in advance

答:
I suggest you induct on $m$. Fix $n = N$, and then the base case $m=0$ is the familiar arithmetic progression
$$\sum_{i=1}^N i = \frac{N(N+1)}{2}.$$
Suppose the result is true for $k = 0, \dots, m$. Then
$$
\sum_{i=m+1}^N i = \sum_{i=m}^N i - m = \frac{(N-(m-1))(N+m)}{2} - m = \frac{N^2+N-m^2-m}{2} = \frac{(N-m)(N+m+1)}{2}.
$$


问:
Finding $\lim_{t\to 0}(\frac{\sin t}{t})$
graph
The graph posted is for this: $$\int^\infty _{-\infty} \frac{\sin t}{t}dt$$
looking at the behaviour of the graph as $\lim_{t\to 0}$, we can see that the value is $1$. I came across the proofs as to why is it so, but they relied on arc lengths of an unit circle and slight modifications to get the desired result. What confuses me about the limit is how counter-intuitive it is. Going by the standard way of finding limit at a point, if we plug in $t=0$ we get $$\frac{\sin0}{0}=\frac{0}{0}=0$$ But this is not the case, as we have proofs claiming the limit is $1$. Is there an intuitive way to explain this?

答:
This entire topic is very confusing, because of the confusion around the definition of $\sin x$ and $\cos x.$  Since the OP's query makes it clear that (from his perspective) the domain of the sine function (and presumably also the cosine function) is a dimensionless number, then there are various choices available for how the sine and cosine function will be defined.

Apostol defines these functions as 2 functions that satisfy the following axioms (if such functions might exist).

A1 : The sine and cosine functions are defined for all $x$ in $\mathbb{R}.$

A2: $\cos (0) = \sin(\pi/2) = 1, \cos(\pi) = -1.$

A3: $\cos(y - x) = \cos y \cos x + \sin y \sin x.$

A4: For $0 < x < \pi/2, ~0 < \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}.$

Apostol then makes the following points:

Examination of the traditional sine and cosine functions against the backdrop of the unit circle (centered at the origin) demonstrates that functions that satisfy the axioms do exist, as long as the domain of the functions is construed to be the dimensionless length of the corresponding arc (rather than an angle).

The sine and cosine functions can alternatively be axiomatically defined via the corresponding Taylor series.

All of the customary theorems around the cosine and sine functions are derivable, based on Apostol's four axioms.

The $\lim_{x\to 0} \frac{\sin x}{x}$ is immediately seen to be $1$, by applying the squeeze theorem to axiom 4.

The point of this rant is that the issue of how to demonstrate that $\lim_{x\to 0} \frac{\sin x}{x} = 1$ is totally dependent on how the sine and cosine functions are defined.  This means that in Apostol's world, since the derivative of the sine and cosine functions are dependent on his axioms, it would not make any sense to attempt to apply L'Hopital's rule to compute the limit.

That would constitute circular reasoning.


问:
Inclusion of $L^p$ spaces, bounded or unbounded
Problem statement:

We have show that $L^q [a,b] \subseteq L^p [a,b]$, for $1 \leq p \leq q$. Show that the
  inclusion is proper by showing that there exists a function $f \in L^p [a,b]$, but $f \notin L^q [a,b]$. For simplicity, take $[a,b]$ to be $[0,1]$. Show that the inclusion is valid if the interval $[a,b]$ is unbounded by giving an example of a function $f \in L^q [1, \infty)$,
  but $f \notin L^p [1, \infty)$, where $1 \leq p \leq q$.

The $L^p$ space is the space of all functions satisfying $(\int f^{p})^\frac{1}{p} < \infty$. Even the first statement of the problem is tricky-- is there a reference online or in a textbook that explicitly shows this? It would probably be an easier problem if I have seen the
first inclusion which the book does not show.

Also, I looked through the other questions which handled this problem in a general measure space, but we are just working with functions of one variable so we must solve from first principles. Do I proceed with Holder's inequality?

答:
For the first statement, that $L^q[a,b] \subset L^p[a,b]$, $1 \leq p < q$ (if $p = q$ it's trivial), you just need to use Holder's inequality:
Assume that $f \in L^q[a,b]$, then:

$$
\int_a^b |f|^p \leq \left[\int_a^b |f|^q\right]^{p/q}\cdot \left[\int_a^b 1\right]^{1-p/q} = (b-a)^{1-p/q}\left[\int_a^b |f|^q\right]^{p/q}
$$
Thus, $f \in L^p[a,b]$ since $f \in L^q[a,b]$. Note that this works since $p < q$, it follows that $q/p > 1$ in Holder.


问:
Symmetry in Laplace PDF
I am unable to understand the symmetry stated in my textbook, as below. Can someone please explain in detail?

We want to find the PDF of Z=X-Y, assuming X and Y are independent and exponentially distributed random variables with parameter $\lambda$.

First calculate CDF $F_Z(z)$ for z >=0. $F_Z(z)=P(X-Y<=z)=1-P(X-Y>z)$

which is calculated as $1-\frac{e^{-\lambda x}}{2}$

For the case z<0, the symmetry of the situation implies that the random variable Z=X-Y and -Z=Y-X have the same distribution. We have $$F_Z(z)=P(Z<=z)=P(-Z>=-z)=P(Z>=-z)=1-F_Z(-z)$$

I am unable to understand here, how $P(-Z>=-z)=P(Z>=-z)$, and where is the symmetry. Can someone please explain?

答:
This is the same as to say

$$X-Y\dot{=}Y-X$$

and this is due to the fact that $X,Y$ are independent exponential with THE SAME PARAMETER.

To understand this symmetry do a drawing and calculate $P[(X-Y)<z_0]$ and $P[(Y-X)<z_0]$

Example: let's fix $z=z_0<0$

the area to be integrated in the two cases are the following

It think now the symmetry of the area is self evident and the fact that the two probabilities are also the same is evident too considering that $X,Y$ have the same exponential distribution.

Title: Symmetry in Laplace PDF


问:
What test to use for antibiotics resistance?
I have 23 samples of milk and 14 samples of cheese. For each of the 7 antibiotics there is a given number of samples that were resistant to it. Note there is only a single strain of bacteria.

Which method should I use to answer these questions?

TLDR: Is (each of) the given antibiotic more effective against cheese bacteria, milk bacteria, or is there no statistical difference at p=0.05?

Another way to put it:

For each antibiotic (and at p=0.05):

Are bacteria in milk sample more resistant to given antibiotic than bacteria in cheese sample?

Are bacteria in cheese sample more resistant to given antibiotic than bacteria in milk sample?

Or is there no statisticaly significant difference in the resistance to the given antibiotic between milk and cheese samples?

I think I only need to count p-values somehow. The problem is the different number of samples. Could I use standardized mean difference (Cohen's d) to overcome  this?

答:
Seeing that 5 of the 6 listed antibiotics contain too few observations in the contingency table, one cannot use the chi-square test. When 5 or less observations are present, you have to use the Fisher's exact test instead.

https://www.statology.org/fishers-exact-test-excel/
https://youtu.be/p2c4C2D7ZZc


答:
The question seems to be about motivating the limit, and not about the definite integral that inspired your question, so I'll stick to that.

What's "intuitive" varies from person to person, but if you have a mental picture of the graph of $\sin x$ near zero (or trust a device to generate one for you) and compare it to the graph of $x$ near zero (ideally, graph both functions on the same set of axes), the fact that $(\sin x)/x \to 1$ as $x \to 0$ is reflected in the fact that the graphs of $\sin x$ and $x$ become near-indistinguishable the closer $x$ gets to $0$.  If two nonzero numbers are nearly indistinguishable, then their quotient is going to be very close to $1$.  That's what's happening here.  How you justify it depends on the formalism you have or are comfortable with.

Title: Finding $\lim_{t\to 0}(\frac{\sin t}{t})$

Body:
graph
The graph posted is for this: $$\int^\infty _{-\infty} \frac{\sin t}{t}dt$$
looking at the behaviour of the graph as $\lim_{t\to 0}$, we can see that the value is $1$. I came across the proofs as to why is it so, but they relied on arc lengths of an unit circle and slight modifications to get the desired result. What confuses me about the limit is how counter-intuitive it is. Going by the standard way of finding limit at a point, if we plug in $t=0$ we get $$\frac{\sin0}{0}=\frac{0}{0}=0$$ But this is not the case, as we have proofs claiming the limit is $1$. Is there an intuitive way to explain this?


问:
Simple divisibility problem

Suppose $a$ and $b$ are distinct integers and $n\mid a^n-b^n$. Prove that $n\mid\frac{a^n-b^n}{a-b}$.

Here is how I do it: Write $a=b+d$, then $a^n-b^n=(b+d)^n-b^n=\binom{n}{1}b^{n-1}d+\binom{n}{2}b^{n-2}d^2+ \cdots + \binom{n}{n-1}bd^{n-1}+d^n \equiv 0 \pmod{n}$. Since $n| \binom{n}{k}$ for $k=1,2,\cdots,n-1$, $n|d^n$, which implies $n|d^{n-1}$. Therefore $(a^n-b^n)/(a-b)=\binom{n}{1}b^{n-1}+\binom{n}{2}b^{n-2}d+ \cdots + \binom{n}{n-1}bd^{n-2}+d^{n-1} \equiv 0 \pmod{n}$ and we are done. I wanna ask why $n|d^n$ will implies $n|d^{n-1}$ and that if there are any other possible solutions, thanks in advance.

答:
We will repeatedly use the trick that if $n=xy$, then $a^n$ is also an $x$th power.

We will approach this by first showing it is true for prime powers.

Base case: $p$ is a prime and $p | a^p - b^p$. Then, we have $a \equiv a^p \equiv b ^p \equiv b \pmod{p}$, and hence $\frac{a^p - b^p}{a-b} \equiv p a^{p-1} \equiv 0 \pmod{p}$.

If $p^{k} | a^{p^k} - b^{p^k}$, then $p \mid \frac{ a^{p^i} - b^{p^i}} { a^{p^{i-1}}-b^{p^{i-1}}}$ for all values of $i$, and hence if you take the product of all these terms (which telescope), $p^k | \frac{a^{p^k} - b^{p^k}}{a-b}$.

Now, we will show that if $p^k || n$, and $n| a^n-b^n$, then $p^k | \frac{a^n-b^n}{a-b}$.

Set $n = p^k \alpha$. Then, $p^k | (a^\alpha) ^{p^k} - (b^\alpha)^{p^k}$, and applying the above we get $p^k | \frac{ (a^\alpha) ^{p^k} - (b^\alpha)^{p^k}} { a^\alpha - b^\alpha}$. Now, multiply this by $\frac{ a^\alpha - b^\alpha} { a-b}$ which is an integer. Hence, $p^k | \frac{a^n-b^n}{a-b}$, and we are done.

Title: Simple divisibility problem


问:
The integral of $\int \frac{x^3}{4-\sqrt{x}} dx$?
How to get the integral of :
$$\int \frac{x^3}{4-\sqrt{x}} dx$$
I have tried the substitution $u=\sqrt x$

答:
This substitution might not give wanted results.

$I=\int \frac{x^{3}}{4-\sqrt{x}} dx\\
Let\ x=( t+4)^{2}\\
dx=2( t+4) dt\\
I=\int \frac{( t+4)^{6} \cdot 2( t+4)}{4-( t+4)} =-2\int \frac{( t+4)^{7}}{t} dt\\$

Use the binomial theorem to expand out the numerator, and then use standard intgration techniques to evaluate each individual expression in the expansion.


答:
You have various options.  If I was doing this, I would use double induction as follows : outer induction on $n$ and inner induction on $m$.

First, I would manually show that it holds for $n = 2$, when $m$ is any value in $\{0,1\}.$

Then, I would assume that the conjecture holds for all $n=N$ when $m$ is any value in $\{0,1,\cdots, N-1\}.$

Then, I would show that because it holds for $n = (N)$, it also holds for $n=(N+1)$, when $m$ is any value in $\{0,1,\cdots, N-1, N\}.$

I would do this by holding the value of $n$ as fixed at $(N+1)$, showing that it holds for $m=0$, and then showing that if it holds for any $M$ in $\{0,1,2,\cdots, N-1\}$ with $n = (N+1)$, then it must also hold for $m = (M+1).$

Title: How to prove this by mathematical induction?

Body:
ok! so i don't have a clue as to how to prove it.

$$\sum\limits_{i=m+1}^n i=((n-m)(n+m+1))/2$$ so the base case is equal to zero because $$\sum\limits_{i=1+1}^1 i=0$$ is this correct? I don't know how to go for inductive step.

Thanks in advance


问:
Converting an English sentence to WFF
Lately, I have been practicing conversion to WFF and am trapped in the following example:

A person may steal something if the person is a thief and the person likes the thing and the thing is valuable.

Converting this statement to WFF could have the possible solutions:

```<code>1. ∀x ∃y person(x) ^ steal(x,y) → thief(x) ^ likes(x,y) ^ valuable(y)
2. ∀x ∃y person(x) ^ thief(x) ^ likes(x,y) ^ valuable(y) → steal(x,y)
</code>```

Now I am confused which one is correct. Since the statement has if later it shall represent the second because one may read it as "If the person is a thief and the person likes the thing and the thing is valuable then the person may steal something." but first one is a little more intuitive.

What would be the correct formula for the given statement?

答:
I think the sentence is fundamentally ambiguous. I can see two ways to read it based on the interpretation of "something". If we read "something" as just a generic predicate, then the sentence is laying some conditions under which theft is likely but not guaranteed. In that case the sentence should be a $\forall\,\forall$ sentence. For what it's worth, this intepretation seems more natural to me.

Another way to interpret it is to do exactly what you did for your first example, which is to say that every thief has at least one likely target. Based on how the sentence is structured I think this is a less plausible interpretation.

In the first reading, we are not promising the existence of at least one possible valuable item for every thief to steal, we are laying out some rules for when theft is likely. $x$ may steal $y$ can also be interpreted as "it is permissible for $x$ to steal $y$", but I will go with the "likely" interpetation for this answer.

Let's use $P(x)$ for person, $T(x)$ for thief, $L(x, y)$ for likes and $V(x)$ for valuable. Let's use $M(x, y)$ to denote $x$ is likely to steal $y$.

Slightly paraphrasing your sentence gives us something the following:

A person is likely to steal a valuable thing if they are a thief and they like
it.

We can express that as follows, using the notation $\forall x : P . \varphi(x)$ as a convenient alternative notation for $\forall x \mathop. P(x) \to \varphi(x)$.

$$ \forall x : P \mathop.\; \forall y : V \mathop.\; T(x) \land L(x, y) \to M(x, y) $$

This is equivalent to

$$ \forall x \mathop. \forall y \mathop. P(x) \land V(y) \land T(x) \land L(x, y) \to M(x, y) $$

This is similar to your second sentence, but the quantifiers are $\forall \, \forall$ rather than $\forall \, \exists$.

Another possible way to translate this sentence that captures an existential reading of "something" is the following. I think this interpretation is less likely given the sentence structure of the prompt, but it is possible.

For every thief, there exists some valuable thing that they like and are likely to steal.

$$ \forall x : P \cap T \mathop. \exists y : V \mathop. L(x, y) \land M(x, y) $$

$$ \forall x \mathop. P(x) \land T(x) \to \exists y : V \mathop. L(x, y) \land M(x, y) $$

$$ \forall x \mathop. \lnot P(x) \lor \lnot T(x) \lor \exists y : V \mathop. L(x, y) \land M(x, y) $$

$$ \forall x \mathop. \exists y \mathop. \lnot P(x) \lor \lnot T(x) \lor V(y) \land L(x, y) \land M(x, y) $$

$$ \forall x \exists y \mathop. P(x) \land T(x) \to V(y) \land L(x, y) \land M(x, y) $$

Title: Converting an English sentence to WFF


问:
Meaning of finite sets of formulas as right argument to $\models$
If a finite set of well-formed formulas appears as the right argument to $\models$, how is it usually interpreted? Is it usually interpreted with an implicit disjunction like a sequent, or an implicit conjunction?

This question references this book, which the following definition

Suppose that $\Delta$ and $\Gamma$ are $\mathcal{L}$-formulas. We will
say that $\Delta$ logically implies $\Gamma$ and write $\Delta \models \Gamma$
if for every $\mathcal{L}$-structure $\mathfrak{A}$,
if $\mathfrak{A} \models \Delta$, then $\mathfrak{A} \models \Gamma$.

I'll use the equivalent statement below for this question, which changes the notation and language slightly.

$$ \textbf{(I)}\;\;\; \Delta \models \Gamma \iff \text{for every $L$-structure $M$ where $M \models \Delta$, $M \models \Gamma$ holds} $$

So, $\Delta \vdash \Gamma$ as a sequent, is equivalent to $\bigwedge \Delta \vdash \bigvee \Gamma$. By analogy, I'll assume for a second that $\Delta \models \Gamma$ is equivalent to $\bigwedge \Delta \models \bigvee \Gamma $.

$$ \textbf{(II)} \;\small\textbf{BAD} \;\; \bigwedge \Delta \models \bigvee \Gamma \iff  \text{for every $L$-structure $M$ where $M \models \bigvee \Delta$, $M \models \bigvee \Gamma$ holds } $$

(II) clearly isn't true. We're quantifying over all structures that satisfy any statement in $\Delta$ instead of all of them.

This makes me think that $\Delta \models \Gamma$ is meant to be equivalent to $\bigwedge \Delta \models \bigwedge \Gamma$. Is this how $\models$ is normally used?

答:
I have literally never seen the expression "$\Gamma\models\Delta$" interpreted as "Every structure satisfying each sentence in $\Gamma$ satisfies some sentence in $\Delta$." Now granted, logicians (of which I am one!) are truly terrible about notation and terminology so it is totally possible that somebody somewhere did do that, but in all the logic texts I've read and presentations I've attended I've only ever seen the "conjunct-conjunct" use.

(An ill-informed historical gloss: remember that the "conjunct-disjunct" style of sequents was introduced for its technical value: it provides a very convenient framework for analyzing deductions. But this is very specific to the details of the proof apparatus; there isn't really a similar semantic counterpart. So nothing motivated a "conjunct-disjunct" version of satisfaction.)


问:
Uniform, weak and strong convergence
I need to research on the uniform, weak and strong convergence

$T_n:L_2(\Bbb{R})\rightarrow L_2(\Bbb{R})$

$T_n \;\,f(x) = f(x+n)$

I've concluded that this sequence converges to zero or diverges.

Do you have any ideas?

Thank you for your time.

答:
If you've already ruled out the possibility that $T_n$ converges to something other than 0, then it should be very easy to see that it doesn't converge to 0 strongly.  ($\|T_n f\| = \|f\|$ for all $n$.)  So it can't converge uniformly (i.e. in operator norm) either.

To show that $T_n \to 0$ weakly, try first showing that for any $f \in L^2(\mathbb{R})$ and for any $\epsilon > 0$ there exists sufficiently large $n$ so that $\int_{[-n,n]^C} f^2 < \epsilon$.  (Use dominated convergence.)  Given $f,g$ and $\epsilon$, find an $n$ that works for both $f$ and $g$.  Then estimate $(T_n f, g)$.

Title: Uniform, weak and strong convergence


问:
Are there any famous/notable mathematicians who have their own YouTube channel?
Are there any famous/notable mathematicians who have their own YouTube channel?

I found this amazing  video regarding the book.
The YouTube channel name is The Math Sorcerer,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician.

I want to know more about YouTube channels of notable mathematicians.

答:
Here are some YouTube channels of mathematicians that I really enjoy. Their levels of fame and notoriety vary, from PhD students at one end to Dr. Borcherds (a fields medalist) at the other. Presented in no particular order:

Alvaro Lozano-Robledo -- Recordings of a graduate course on elliptic curves, some videos about algebra for kids, and more!

Paul VanKoughnett -- Recordings of a currently-running seminar on stable homotopy theory.

Richard Borcherds -- Dr. Borcherds is uploading new content at an incredible pace on various undergraduate- and graduate-level topics.

Michael Penn -- Mostly videos about solving contest-style problems.

Daniel Litt -- Recordings of a graduate course on étale cohomology, among other things.

Billy Woods -- Has a lovely series of videos on algebraic number theory.

Boarbarktree -- Currently just 3 videos, on homology. I look forward to more!

Kristaps John Balodis -- Videos on number theory, geometry, set theory, and more, as well as interesting interviews with current math PhD students!

I'll also take this opportunity to plug my own YouTube channel. Currently the content there is sparse and not of the highest quality, but I aim to improve over time! Here's a 30-second video I made on computing $\frac{\mathrm{d}}{\mathrm{d}x} f^{-1}(x)$, which I'm quite happy with.


问:
Prove this identity using inequalities
For some diagonal matrix A = $(a_{ij})$ where the main diagonal are all positive numbers ($a_{ii} > 0$ for $1\le i \le n$), prove:

$$
\textbf{v}^T \textbf{A} \textbf{w} \le \|\textbf{A}^{1/2}\textbf{v}\|_2 \|\textbf{A}^{1/2}\textbf{w}\|_2
$$

We also define the matrix square root of A as $\textbf{A}^{1/2} = (a_{ij}^{1/2})$.

I tried playing around with the Cauchy-Schwarz inequality to prove this, but I think I'm just confusing myself even more. Any solution or suggestions?

答:
What is confusing is that the definition of the square root seems odd at first; but all simplifies once you remember that $A$ is assumed to be diagonal:
$$
\mathbf{v}^\top A \mathbf{w} = \sum_{i,j} a_{ij}\mathbf{v}_i \mathbf{w}_j
\stackrel{(1)}{=} \sum_{i} a_{ii}\mathbf{v}_i \mathbf{w}_i
\stackrel{(2)}{=}\sum_{i} \sqrt{a_{ii}}\mathbf{v}_i \cdot \sqrt{a_{ii}} \mathbf{w}_i
\stackrel{(3)}{\leq}\sqrt{\sum_{i} a_{ii}\mathbf{v}^2_i } \cdot \sqrt{\sum_{i} a_{ii}\mathbf{w}^2_i }
$$
where (1) uses the fact that $A$ is diagonal, (2) the fact that the entries are nonnegative, and (3) is Cauchy-Schwarz.

Now, you can easily check that
$
\| A^{1/2} \mathbf{v}\|_2^2 = \sum_{i} a_{ii}\mathbf{v}^2_i
$
and similarly for $
\| A^{1/2} \mathbf{w}\|_2^2$, so the RHS above is exactly
$\| A^{1/2} \mathbf{v}\|_2\cdot \| A^{1/2} \mathbf{w}\|_2$.

Title: Prove this identity using inequalities


问:
Eigenvalues of vectors of 2*2 matrix
A (2 x 2) matrix $\begin{bmatrix}a & c+id\\c-id & b\end{bmatrix}$ where a, b, c, d are real constants will have two different eigenvalues unless it is a multiple of the identity matrix.

i used this
|A − λI| = det(A − λI) = 0 to prove that it will have 2 eigenvalues. How do i prove that it will only have 1 when it is a multiple of the identity matrix?

答:
The characteristic polynomial is $(\lambda-a)(\lambda-b)-(c^2+d^2)$. Note that the first term already has roots $a,b$, so the discriminant is already positive; $c^2+d^2\ge0$ and increasing $c^2+d^2$ will only increase the discriminant further.

The only possibility for the matrix to have only one eigenvalue is therefore $c=d=0$ and $a=b$, i.e. the matrix is a multiple of the identity matrix.

Title: Eigenvalues of vectors of 2*2 matrix


问:
Is residue class of $x$ irreducible in $\mathbb{R}[x,y]/(x^2+y^2-1)$?
I have been trying to prove that the residue class of $x$ is irreducible in $\mathbb{R}[x,y]/(x^2+y^2-1)$.

I have tried by assuming say $X=\bar{x}$ is reducible. So $X=p(X,Y)q(X,Y)$ where $Y= y +(x^2+y^2-1)$. But calculations are getting very difficult.

How do I prove this ? Is there any other way to do this than to assume $X=p(X,Y)q(X,Y)$ and proceed ?

Please help. Thanks in advance.

答:
Use the norm: writing $\Bbb R[x,y]/(x^2+y^2-1)$ as the free $\Bbb R[x]$-module on the basis $\{1,y\}$, we can write the matrix representation of the element $p(x)+yq(x)$ as $$\begin{pmatrix} p(x) & (1-x^2)q(x) \\ q(x) & p(x) \end{pmatrix}$$ and the determinant of this map is $p(x)^2-(1-x^2)q(x)^2$. The map $\nu$ which sends $p(x)+yq(x)$ to $p(x)^2-(1-x^2)q(x)^2$ is a monoid homomorphism from the elements of $\Bbb R[x,y]/(x^2+y^2-1)$ under multiplication to the elements of $\Bbb R[x]$ under multiplication, so if $x=ab$ then $\nu(x)=\nu(a)\nu(b)$. Computing, we can see that $\nu(x)=x^2$, so if $x=ab$ with $a,b$ nonunits, then we need to be able to factor $x^2$ as a product of two non-units in the image of $\nu$.

As $\Bbb R[x]$ is a UFD, we'd have to find elements with norm $cx$ for some $c\in\Bbb R^*$, which is equivalent to solving $cx=p(x)^2-(1-x^2)q(x)^2$. But this is impossible: if $c>0$, plugging in $x=-1$ we get that $-c=p(-1)^2$, while if $c<0$, plugging in $x=1$ we get that $c=p(1)^2$.


问:
How to approximate a sharply peaked function with deltas?
I have the function

$$
f(x)=\cases{\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}} & $-1 \leq x \leq1$ \\ 0 & otherwise}
$$

Where $a \in \mathbb{R}^+$. Here is a plot of $f$:

It would be very convenient if I could justify (in any way: maybe distributionally?) that

$$
f(x)=A\left[ \delta(x-1)+\delta(x+1) \right]+B+\mathcal{O}(?) \quad ; \quad a \to \infty \tag{1}
$$

For some $x$ independent constants $A$ and $B$. I am uncertain of how to do this, or how to estimate the error.

What I've tried

My thought was to look at the Fourier series of $f$, and associate terms with the Fourier series of (2). By expanding (1) around $x= \pm1$, I can find a Laurent series in $(x \pm 1)$, and perform the Fourier integral term by term. The result is messy, and worse: does not allow easy comparison to the Fourier series of (2). This may not be a fruitful approach.

By looking at the wiki page about mollifiers, it seems I need to claim something like: $a f(x) \to \delta (x \pm1), \ a \to \infty$. Superficially, this appears to work, but I'm not sure how to make it concrete, or if there should be any other factors next to $af(x)$ (other than a factor to ensure normalization to unity).

Questions

How can I estimate the error and find $A$ and $B$ in (1)?

If $A$ and $B$ cannot be found, how can I find the ratio $A/B$?

Background

$f(x)$ will eventually appear beneath an integral with a Green's function $G$. I'd like to formulate this for arbitrary $G$. We can take some liberties about the 'niceness' of $G$, as necessary. In particular, it has a well behaved Fourier transform.

答:
Too long for comments.

Starting from @Han de Bruijn's answer, computing
$$N(a) = \int_{-1}^{+1} \left[\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}}-\frac{1}{\sinh(a)}\right] dx$$ is not so bad using elliptic integrals
$$N(a) =-\frac{2  }{a}\,\text{csch}(a)\left(a+i F\left(i
   a\left|-\text{csch}^2(a)\right.\right)\right)$$
$$N^*(a) = N(a)\,a\,e^a=-2 e^a \text{csch}(a) \left(a+i F\left(i
   a\left|-\text{csch}^2(a)\right.\right)\right)$$ which approach very fast it horizontal asymptote. For $a=100$, the value is
$$2.772588722239781237668928485832706272302\cdots$$ This number differs from $\log(16)$ by $5.5 \times 10^{-85}$.


问:
Triangle Properties application
In a triangle ABC, AD is the altitude from A. Given b>c, $\angle C = {23^o}$,$AD = \frac{{abc}}{{{b^2} - {c^2}}}$. Then  $\angle B = \_\_\_{\_^o}$

My approach is as follow

$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R$

$AD = \frac{{8{R^3}sinA\sin B\sin C}}{{4{R^2}{{\sin }^2}B - 4{R^2}{{\sin }^2}C}} = \frac{{2RsinA\sin B\sin C}}{{\sin \left( {B - C} \right)\sin \left( {B + C} \right)}} = \frac{{2R\sin B\sin C}}{{\sin \left( {B - C} \right)}}$

How do I approach from here

答:
We can use the extended law of sines and the area of a triangle in terms of sides and circumradius formulae.

$$\bigtriangleup=\frac{1}{2}a\cdot AD \implies AD= \frac{2\triangle}{a} = \frac{{abc}}{{{b^2} - {c^2}}}=\frac{4R \bigtriangleup}{{b^2} - {c^2}}$$

$$\implies 2Ra={b^2} - {c^2}$$
$$\implies 2R(2R\cdot \sin(A))=(2R\cdot \sin(B))^2 - {(2R\cdot \sin(C))}^2$$
$$\implies \sin(A)=\sin^2(B)-\sin^2(C)$$

Can you take it from here?

 $$B=C+90°$$

There is also another version of this problem here.

Title: Triangle Properties application


问:
Why $v_1 \in L$ and $v_2 \in L^\perp$?
Let $L$ be a one dimensional subspace of $R^2$ (we may view $L$ as a line in the plane through the origin).

Suppose $\alpha$ is the angle from the positive x axis to $L$.

Let $v_1=(\cos \alpha, \sin \alpha)$ and $v_2=(-\sin \alpha, \cos \alpha)$.

Then $v_1 \in L$ and $v_2 \in L^\perp$.

I can't understand the last line. Why $v_1 \in L$ and $v_2 \in L^\perp$?

答: Another way to visualizing it is noting that $$\cos(a+\frac{\pi}{2}) = -\sin(a)$$ and $$\sin(a+\frac{\pi}{2}) = \cos(a)$$ so basically you are rotating the vector by $\frac{\pi}{2}$ </p>

Title: Why $v_1 \in L$ and $v_2 \in L^\perp$?


问:
Naively adding a dedicated error value to set-builder notation
Naive set-builder notation includes expressions that don't denote a set (in ZFC), like $\{ x \mathop| \top \}$. I'm wondering if we can do something extremely silly and ad hoc and just say by fiat that such expressions refer to a dedicated error value set $E$. Do we end up with a consistent theory (assuming that ZFC is consistent)?

This question is similar to this one, but I am deliberately making an odd choice for the semantics of set builder notation. Set-builder notation is not part of the syntax of ZFC, but is a bit of notation defined on top of the core theory and justified by the axiom schema of specification, which gives us existence under certain circumstances, and axiom of extensionality, which gives us uniqueness. I'm going to treat it informally as a core bit of syntax in an extended theory. For the purposes of this question, I'm thinking of set builder notation as a core bit of syntax that creates new terms, similar to a function symbol.

Let $E$ be a propertyless constant symbol in the theory of sets. Intuitively, it is an error value, but it is a set, not an urelement or a proper class.

Let $\{ t(\vec{x}) \mathop| \varphi(\vec{x}) \}$ be an expression in set-builder notation where $t(x)$ is a term with $\vec{x}$ as a sequence of free variables and $\varphi(x)$ is a formula with $x$ as a sequence of free variables.

Some expressions in set-builder notation do not name sets in ZFC, for example.

$$ \{ x \mathop| \top \} \;\; \text{is not a set in ZFC} $$

$$ \{ x \mathop| x \not\in x \} \;\; \text{is not a set in ZFC} $$

I'm wondering whether it's possible to make set-builder notation total in a certain sense by saying that any instance of it that would fail to name a set instead refers to $E$.

$$ \{ x \mathop| \top \} = E \;\; \text{because there is no universal set} $$

I've tried to come up with an additional pair of axiom schemas based on the axiom schema of specification that captures this new property. These axiom schemas do not replace the axiom schema of specification; they are additional and possibly redundant axiom schemas.

Every predicate has at most one set equal to its extension.

$$ \forall w_1 w_n \mathop. \exists_{\le 1} B \mathop. \forall x \mathop. (x \in B \leftrightarrow \varphi(x, w_1, w_n)) $$

A term using set-builder notation is equal to $E$ if-but-not-only-if it has no extension as a set. The outermost connective can't be a biconditional because of $\{ x \mathop| x \in E \}$.

$$ \forall w_1 w_n \mathop. (\nexists B \mathop. \forall x \mathop. (x \in B \leftrightarrow \varphi(x, w_1, w_n)) \to \{z \mathop| \varphi(z, w_1, w_n) \} = E ) $$

答:
There is actually a very general version of this which works for any first-order theory $T$ with the additional property that there is some formula $\varphi$ such that $T\vdash\exists!x\varphi(x)$. (This $\varphi$ will pick out the "default value" - this is different from Mark Saving's answer, where the "default value" is considered a "new" object, but this is really superficial and which approach is more natural will depend on context.)

Specifically, we work with a variant of Hilbert's $\tau$ operator$^1$ (which I swear I've seen before but can't find at the moment). Suppose $T,\varphi$ are as above. To the usual syntax of first-order logic we add a new term-forming-symbol $\digamma$, and to $T$ we add the following additional rules:

From $\exists!x\psi(x)$ infer $\psi(\digamma(\psi(x)))$.

From $\neg(\exists!x\psi(x))$ infer $\varphi(\digamma(\psi(x)))$.

Basically, $\digamma$ picks out the unique object with a property if such an object exists, and otherwise defaults to the unique thing satisfying $\varphi$. (I'm being rather sloppy here, specifically around the handling of free variables/parameters, but there's nothing subtle going on.) This modification is conservative in the sense that models of $T$ in the usual sense corresponds exactly to models of $T$ in this enriched sense (= the behavior of $\digamma$ is completely determined by the rules we've added governing it; basically, this is a "one-level-higher" version of an expansion by definitions). In the particular case of $\mathsf{ZFC}$ or similar, in fact, something even nicer happens: we can expand the separation and replacement schemes to include formulas using the new $\digamma$ syntax without changing the power of the theory!

In the case you're interested in, we're specifically looking at formulas $\psi(x)$ of the form $$\psi(x)\equiv x=\{y: \xi(y)\},$$ or more precisely of the form $$\psi(x)\equiv\forall y(y\in x\leftrightarrow\xi(y)).$$ Extensionality says that at most one such $x$ exists, so $\digamma$ does what we want it to. (Note that $\digamma$ applies more broadly than that of course.)

Note that the uniqueness condition was in fact crucial: the full $\tau$ operator is more fragile. While it remains conservative in an appropriate sense, it is no longer the case that we can safely expand the separation/replacement schemes of $\mathsf{ZF}$ without getting any new theorems: the "$\tau$-schematic" version of $\mathsf{ZF}$ proves the axiom of choice, and indeed the (scheme version of) global choice.

$^1$Actually Hilbert used "$\epsilon$," not "$\tau$" (which was introduced for an essentially-equivalent operator by Bourbaki), but "$\epsilon$" in th


问:
Solving $\frac{\sqrt {x}}{2}=-1$, I get $x=4$. Where did I go wrong?
We have the equation
$$\frac{\sqrt {x}}{2}=-1$$ I proceed as follows
$$\sqrt {x}=-2 $$
$$x=4$$
Which does not certainly solve the equation. Where did I go wrong?

答:
By definition of the square root :

If $\sqrt x=a$, then

$$\begin{cases} x=a^2 \\ a≥0 \end {cases}$$

Now, you can check your solution.

Note that, $-2$ is the real-valued root of $4$. But, we must take only principal square root, which equals to $2.$

Small supplement:

If $\sqrt x=a$, then you get $x=a^2$. This is correct. But, if the condition $a≥0$ doesn't hold,  then the solution doesn't exist.


问:
How to prove this by mathematical induction?
ok! so i don't have a clue as to how to prove it.

$$\sum\limits_{i=m+1}^n i=((n-m)(n+m+1))/2$$ so the base case is equal to zero because $$\sum\limits_{i=1+1}^1 i=0$$ is this correct? I don't know how to go for inductive step.

Thanks in advance

答:
Bearing in mind that your claim - when written out explicitly as a formal predicate - should look as follows:
$$R \colon \equiv (\forall m, n)\left(m,n \in \mathbb{N} \wedge m \leqslant n \Rightarrow \sum_{k=m+1}^n k=\frac{(n-m)(n-m+1)}{2}\right),$$
by far the easiest approach is to consider the predicate:
$$P(m) \colon \equiv (\forall n)
\left(n \in \mathbb{N} \wedge m \leqslant n \Rightarrow \sum_{k=m+1}^n k=\frac{(n-m)(n-m+1)}{2}\right)$$
and realise that $R \Leftrightarrow (\forall m)(m \in \mathbb{N} \Rightarrow P(m))$. Therefore, in order to establish (prove) $R$ it will suffice to fix arbitrary $m \in \mathbb{N}$ and to establish the predicate $P(m)$.

In order to supply a proof for $P(m)$, the most natural choice is to perform induction on the well-ordered set $\{n \in \mathbb{N} \mid n \geqslant m\}$, since this is precisely the domain over which the universal quantifier comprised in $P(m)$ acts.

More explicitly, you would have to perform just one induction where the base case is $n=m$ and the inductive step consists of assuming the summation relation valid for a certain (arbitrary) $n \geqslant m$ in order to deduce it for $n+1$.


问:
Regarding the Tensor Product of Quotient Rings of a Euclidean Domain
Consider a field $k$ of characteristic $0$ and a positive integer $n.$ In the proof of Theorem 4.19 of Polytopes, Rings, and K-Theory by Bruns and Gubeladze, it is stated that $k[x] / (x^n - 1)$ is a tensor product of integral domains; however, I fail to see why this is true. I have made the following observations but to no avail.

Considering that $k$ has characteristic $0,$ we have that $x^n - 1 = \prod_{d \,|\, n} \Phi_d(x),$ where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. Consequently, the polynomial $x^n - 1$ splits into a product of pairwise relatively prime irreducible polynomials; however,  I don't see how $k[x] / (x^n - 1) \cong \otimes_{d \,|\, n} k[x] / (\Phi_d(x)),$ if this is indeed what the authors assert. Even more, it seems to me that the tensor product of the quotients of $k[x]$ by two relatively prime irreducible ideals would be zero, as we would have that $$\frac{k[x]}{(p(x))} \otimes_{k[x]} \frac{k[x]}{(q(x))} \cong \frac{k[x]}{(p(x), q(x))} = \frac{k[x]}{\gcd(p(x), q(x))} = \frac{k[x]}{k[x]} \cong 0.$$

I would greatly appreciate any insight, comments, or suggestions. Thank you.

答:
I think everything you say is correct. The authors are only arguing $k[x]/(x^n-1)$ is reduced, so they probably meant to observe that it is the direct product of integral domains, which is just the Chinese Remainder Theorem.

Edit: In fact, when I look at the Google Books preview version of the book, it says exactly that. So perhaps this error is fixed in later editions.


问:
Natural numbers that can be expressed as the product of two and three consecutive natural numbers
Looking for natural numbers larger than 210 which can be expressed as the product of two continuous natural numbers or the product of three continuous natural numbers at the same time.

$6 = 1 \cdot 2 \cdot 3 = 2 \cdot 3$

$210 = 5 \cdot 6 \cdot 7 = 14 \cdot 15$

I have made a program to find that numbers smaller than $1000000000$, only to find $6$ and $210$. Are there any other ones?

答:
This is equivalent to finding solutions of the elliptic curve Diophantine equation
$$y(y+1)=y^2+y=x^3-x=(x+1)x(x-1)$$
which is 37.a1 in LMFDB. Obviously $x,y>0$; there are nevertheless only a finite number of integral solutions, which are listed in LMFDB. With the restriction on $x,y$ there are only two solutions:
$$x=2,y=2\implies N=6$$
$$x=6,y=14\implies N=210$$
Hence $6$ and $210$ are the only numbers of the form you mentioned.


问:
Check the differentiability of a function from $\Bbb R^2$ to $\Bbb R$ on the boundary of unit circle.
Define the function $f:\Bbb R^2 \to \Bbb R$ by
$$f(x,y):=\begin{cases}\sqrt{x^2+y^2} &\text{ if } x^2+y^2 \ge 1\\ 1 &\text{ if } x^2+y^2<1\end{cases}$$

Checck the differentiability of the function $f$. I know the definition of differentiability at a point $(a,b)$ of several variable function. But to check for this function, we have to check the differentiability of each point on the boundary of the unit circle.

Hopw to chech this? Any hint. please..

答:
If the function were differentiable, then the function would be differentiable when restricted to the $x$ axis. On this set, it is the function
$$ f(x,0) = \max(1,|x|)$$
which is patently not differentiable at $x=1$. Also, by the rotational symmetry of $f$, the function is not differentiable anywhere on the unit circle.

Alternatively (actually the same argument) verify that at each $r\in \partial B(0,1)$, the one-sided directional derivatives in the directions $\pm r$ do not agree; so it is not (Gateaux) differentiable on the unit circle.


问:
Find the inverse of $f(x,y,z) = (\frac{x}{x^2+y^2+z^2}, \frac{y}{x^2+y^2+z^2}, \frac{z}{x^2+y^2+z^2})$
I attempted to switch to cylindrical / spherical coordinates, but I keep getting stuck. Note that $(0,0,0)$ is not in the domain or codomain.

答: Hint We can write this map in coordinate-free notation as $$f({\bf x}) = \frac{\bf x}{||{\bf x}||^2} .$$


答: $f$ maps a vector with length $r=\sqrt{x^2+y^2+z^2}$ to the same-direction vector but with length $\frac1r$. Therefore $f^{-1}=f$.

Title: Find the inverse of $f(x,y,z) = (\frac{x}{x^2+y^2+z^2}, \frac{y}{x^2+y^2+z^2}, \frac{z}{x^2+y^2+z^2})$

Body: I attempted to switch to cylindrical / spherical coordinates, but I keep getting stuck. Note that $(0,0,0)$ is not in the domain or codomain.


问:
Prove identities for matrices.
Let $A,B,C,D$ be $n \times n$ real valued matrices such that $$AC-BD=Id$$ and $$AD+BC=O$$
Prove $$CA-DB=Id$$ and $$DA+CB=0$$

One idea that I had is to consider the matrices $A,B,C,D$ as blocks of a larger matrix.
Consider the matrix $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} $$ and $$ \begin{pmatrix} C & D \\ D & -C \end{pmatrix} $$
Then $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & D \\ D & -C \end{pmatrix} =\begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$
So I want to prove $$\begin{pmatrix} C & D \\ D & -C \end{pmatrix} \begin{pmatrix} A & B \\ -B & A \end{pmatrix} = \begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$ but I haven't had any luck yet, can you help?

答:
The $n\times n$ complex matrices $A+iB$ and $C+iD$ satisfy
$$
(A+iB)(C+iD)=(AC-BD)+i(BC+AD)=I
$$
so they are inverses of each other in $GL_n(\mathbb{C})$.  Hence
$$
I=(C+iD)(A+iB)=(CA-DB)+i(DA+CB)
$$
and you have the desired result by taking real and imaginary parts.

Title: Prove identities for matrices.


答:
There were sign errors in the first version of your question. These are fixed now, but it is more convenient to write the hypothesis as

$$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & -D \\ D & C \end{pmatrix} =\begin{pmatrix} I_n & O \\ O & I_n \end{pmatrix} \, ,$$

and the desired conclusion as

$$\begin{pmatrix} C & -D \\ D & C \end{pmatrix} \begin{pmatrix} A & -B \\ B & A \end{pmatrix} = \begin{pmatrix} I_n & O \\ O & I_n \end{pmatrix} \, , $$

so that we have ($2n$-dimensional) unit matrices on the right-hand side.

These equations are equivalent because for square matrices, $XY = I$ is equivalent to $YX = I$, i.e. the left and right inverse of a matrix are the same.

Remark: The statement holds (and the proof works) not only for real-valued matrices, but more generally for complex-valued matrices, for matrices over any commutative ring, or even for elements of a finite-dimensional K-algebra. For more information, see

If $AB = I$ then $BA = I$

Right invertible and left zero divisor in matrix rings over a commutative ring

Title: Prove identities for matrices.

Body:
Let $A,B,C,D$ be $n \times n$ real valued matrices such that $$AC-BD=Id$$ and $$AD+BC=O$$
Prove $$CA-DB=Id$$ and $$DA+CB=0$$

One idea that I had is to consider the matrices $A,B,C,D$ as blocks of a larger matrix.
Consider the matrix $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} $$ and $$ \begin{pmatrix} C & D \\ D & -C \end{pmatrix} $$
Then $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & D \\ D & -C \end{pmatrix} =\begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$
So I want to prove $$\begin{pmatrix} C & D \\ D & -C \end{pmatrix} \begin{pmatrix} A & B \\ -B & A \end{pmatrix} = \begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$ but I haven't had any luck yet, can you help?


问:
Finding the intersection of the graphs of two functions algebraically.
Here is the question:

Consider the fucntions $f_1 (x) =x$, $f_2 (x) = 2 + ln(x)$. The graphs of the following functions intersect:

once in $(0,1)$ and never in $(1,\infty)$ 2.once in $(0,1)$ and once in $(e^2, \infty)$ 3. once in $(0,1)$ and once in $(e, e^2)$ 4. more than twice in $(0, \infty)$

By drawing the graphs (using a little bit of derivatives to know the shape of $f_2$) we can find the intersections intervals. I'm thinking if we can find the solution of this problem just by algebra (analysis). So, to find the intersections we shall put
$$f_1 = f_2 \\
x = 2 + ln(x) \\
ln (x) = x -2 $$

But I don't know what to do after this.

答:
$f_2$ is concave: any line that intersects its graph can only do so at most twice. Since
$$f_2(1/e^3)<f_1(1/e^3)$$
$$f_2(1)>f_1(1)$$
$$f_2(e)>f_1(e)$$
$$f_2(e^2)<f_1(e^2)$$
there are exactly two solutions, one in $(0,1)$ and one in $(e,e^2)$.


问:
If $I[x]$ is a radical ideal of $R[x]$, then show that $I$ is a radical ideal of $R.$
Show that $I$ is a radical ideal of a commutative ring $R$ iff $I[x]$ is a radical ideal of $R[x].$

The problem is in the ''only if'' part.If $I[x]$ is a radical ideal of $R[x],$ then $I[x] = \sqrt J$ for some ideal $J$ of $R[x].$

Then, how to find an ideal of $R$ whose radix will be $I?$
Will such an ideal be unique?

答: Note that $I \subset R$ is radical if and only if the quotient ring $R/I$ is reduced (has no non-trivial nilpotents). Your statement is now immediate since $R/I \cong R[x]/I[x].$


问:
Isomorphism between two finite fields
We have $k_1:= \mathbb F_7(\alpha)$ and $k_2 := \mathbb F_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F_7$. I have to show that these two are isomorphic.

Let $\phi:k_1 \rightarrow k_2$ be a homomorphism which preserves $1 \in k_1$. Then

$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$

where

$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F_7$$

Thus

$$(x+y\beta)^2 = \phi(\alpha)^2  = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$

So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F_7$. So $x = 0$ such that

$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$

Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?

答: Hint: $ $ mod $\,7\!:\  \alpha^2 \equiv 3 \equiv -4 = 4\beta^2\! = (2\beta)^2,\,$ so check  $\,\alpha \to 2\beta\,$ preserves sums and products.</p>

Title: Isomorphism between two finite fields


问:
$x_n =\sum_{i=0}^{n}\frac{1}{i!}$ is Cauchy in $\mathbb{Q}$
From Topology Without Tears:

Show that $x_n=\sum_{i=0}^{n}\frac{1}{i!}$ is a Cauchy sequence in $\mathbb{Q}$

First thing, I believe being convergent in $\mathbb{Q}$, means sequence and  the converging point must also belong to $\mathbb{Q}$. What does it mean sequence is Cauchy in $\mathbb{Q}$?Does it just mean sequence is entirely in $\mathbb{Q}?$

I was trying to show Cauchy:
let $n>m$
\begin{equation}
\begin{split}
|x_n-x_m|&=|\sum_{i=0}^{n}\frac{1}{i!}-\sum_{i=0}^{m}\frac{1}{i!}|\\
&=|\sum_{i=m+1}^{n}\frac{1}{i!}|\\
&=|\frac{1}{m+1!}+...+\frac{1}{n!}|\\
& \leq \frac{1}{m+1!}+...+\frac{1}{n!}\\
&<\frac{1}{m}+...\frac{1}{n}
\end{split}
\end{equation}
Since the number of terms are $n-m $.I was thinking to take each term $<\frac{\epsilon}{n-m}$, to get $n_0\in \mathbb{N},\ such\ that\ |x_n-x_m|<\epsilon,\forall n,m>n_0$.

But I think this will be wrong and looks odd,saying choosing $m>\frac{\epsilon}{n-m}$

How should I show this is Cauchy?

答: $(m+1)!$ is a product of $m+1$ numbers all except one of which is $2$ or more. So $(m+1)!\geq 2^{m}$. Similarly,  $(m+2)!\geq 2^{m+1}$ and so on. So $\frac 1 {m+1}!+..+\frac 1 {n!} \leq \frac 1 {2^{m}}+\frac  1 {2^{m+1}}+...$. Can you write down this geometric sum and show that it is less than $\epsilon$ if $m$ is large enough?


问:
homeomorphism between spectrum and compact set
Let $\mathcal{A}$ be a commutative unital Banach Algebra.

Suppose $a,b \in \mathcal{A}$, and $\mathcal{B}$ is a Banach subalgebra generated by $a,b$.

Let $\Delta_{\mathcal{B}}$ denote the maximal ideal space of $\mathcal{B}$. Show that there exists a homeomorphism
$ \Gamma:\Delta_{\mathcal{B}} \rightarrow \sigma(a,b)=\{ (\phi(a), \phi(b) ) : \phi \in \Delta_{\mathcal{B}}\}$.

Attempt:

I thought of using this theorem:

Suppose $X$ is a compact space and $Y $ is a Hausdorff space. If $\Gamma: X \rightarrow Y$ is bijection and continuous. Then $\Gamma$ is a homeomorphism.

I know that $\hat{a}, \hat{ b} \in C(\Delta_{\mathcal{B}}).$ Also, $\sigma(a) = Range (\hat{a})$ and $\sigma(b) = Range (\hat{b})$

And $\Delta_{\mathcal{B}}$ and $\sigma(a,b)$ are compact and Hausdorff.

From here I'm not sure how to show it's injective.

Any help will be appreciated.

Thank you.

答: To see injectivity: If $\phi$ and $\phi'$ are two elements of $\Delta_{\mathcal B}$ such that $(\phi(a), \phi(b)) = (\phi'(a), \phi'(b))$, then we must have $\phi = \phi'$ on all of $\mathcal B$ because $a$ and $b$ generate $\mathcal B$. (In particular, the elements of $\Delta_{\mathcal B}$ are compatible with Banach algebra operations.)


问:
If the set of linear transformation is a vector space, can we change the underlying field?
If we let $T$ be a linear transformation from $V$ to $W$, the set of linear transformations is a vector space. But I am curious about the underlying field for this vector space. The field for $V$ or $W$ is usually the real or complex number, but when it comes to the set of linear transformations, we only define the field to be the reals, but why can't it be other fields such as the complex number or some other field?

答:
The algebraic way is to use tensor products, called extension of scalars.

If $K$ is a field, $V$ a $K$-vector space, and $L$ a field extension of $K$ (a field containing $K$), then $V\otimes_K L$ is an $L$-vector space.


答:
Since $z$ is real, then $f(z)$ is real, and $z$ is pure imaginary, then $f(z)$ is not real or pure imaginary. So the answer is between 1. and 2.

Put $z=i \in H^+$, $f(z)\in H^+$, so the answer is 1.

Title: $f(z)={2z+1\over 5z+3}$ maps

Body:
Define

$H^{+}=\{z:y>0\}$

$H^{-}=\{z:y<0\}$

$L^{+}=\{z:x>0\}$

$L^{-}=\{z:x<0\}$

$f(z)={2z+1\over 5z+3}$  maps

$1.$ $H^+\to H^+$ and $H^-\to H^-$

$2$. $H^+\to H^-$ and $H^-\to H^+$

$3.$ $H^+\to L^-$ and $H^-\to L^+$

$4.$ $H^+\to L^+$ and $H^-\to L^-$

If I take $z\in H^{+}$ then ${3\over z}\in H^{-}\Rightarrow {3\over z}+5={5z+3\over z}\in H^{-}$ Could any one tell me how to proceed then?


问:
Can the joint entropy be expressed in terms of pairwise measures?
Assume we have $N$ random variables $x_1, x_2, \dots, x_N$. Is there a way to express the joint entropy $H(x_1, x_2, \dots, x_N)$ in terms of single-variable or pairwise measures such as the  pairwise mutual information $I(x_i; x_j)$ or the entropy $H(x_i)$ of each variable? In other words, are there functions $g_{ij}(\cdot,\cdot)$ such that

$$
H(x_1, x_2, \dots, x_N) = \sum_{i\in \{1, \dots, N\}}\sum_{j\in \{1, \dots, N\}} g_{ij}(x_i, x_j).
$$

答: This is not possible. One can have three random variables $X_1,X_2,X_3$ that are pairwise independent but not mutually independent (e.g., Example of Pairwise Independent but not Jointly Independent Random Variables?). The pairwise data therefore do not provide enough to deduce whether $X_1$, $X_2$, and $X_3$ are mutually independent. However, the joint entropy equals $H(X_1) + H(X_2) + H(X_3)$ if and only if $X_1$, $X_2$, and $X_3$ are mutually independent.


问:
How to get $-4(S+\cos\theta)$ when it depends on the area of the triangle in the unit circle?
The problem is as follows:

Assume $S$ square units $(u^2)$ is the area of the region whose color is blueberry. Consider that in this unit circle the diameter of the circle is $1$ decameter. Given what it is known in the figure from below, find $-4(S+\cos\theta)$.

The choices given are as follows:

$\begin{array}{ll}
1.&\cos 2\theta\\
2.&\sin 2\theta\\
3.&-\sin\theta\cos\theta\\
4.&-\sin\theta\cos\theta\\
\end{array}$

I'd like to note that there might be some typo in the given choices as the original source it is not very clear in this aspect. Thus I'm okay with an answer as long it is well justified. My main question here is not exactly trying to get an answer matching the choices but rather a way to understand the meaning of the arrows indicating the direction of the angle in the unit circle with that being mentioned. Please read my question from below.

For this problem I'm also stuck, and the reason for that is that the angle $\theta$ I'm not sure if it comes from the $\textrm{x-axis}$ or from the vertex on the right triagle which is contiguous to the small square looking from the bottom.

When an angle goes as this, in the clockwise direction I assume it is negative right?. But it doesn't help me much on how should I properly put this information in the solution.

I mean I always get confused when that arrow goes in the clockwise direction.

Because of this weird looking figure (at least to my understanding) there is not proper way to find that area. Does it exist something that can I do?. Because of the way how I'm stuck with this it would really help a lot a solution which would solve this sort of problem.

It seems that a drawing might be required to solve this problem thus can someone help me?. Since the part which confuses me the most is the way how it is used the arrow in the clockwise direction I kindly appreciate and beg someone could explain this to me in the right way to avoid getting confused.

答:
The clockwise arc starting at the positive $x$-axis and ending at the arrowhead is slightly thicker than the rest of the circle. So I will assume that $\theta$ is measured from the positive $x$-axis, and is rendered opposite from the usual counter-clockwise orientation. Let $\alpha$ be the angle $\angle NOM$ as in @NimaS's drawing. Traveling clockwise from the positive $x$-axis we will encounter $\frac\pi2$, then $\alpha$, then $-\theta$ to complete a full circle:
$$
\frac\pi2 + \alpha -\theta =2\pi
$$
and therefore $\sin\alpha=\sin(2\pi+\theta-\frac\pi2)=-\cos\theta$.

I agree with @NimaS that the area $S$ equals $\frac14\sin\alpha$ (since the diameter of the circle is $1$). Given all this, there is no chance that any of the choices will match $-4(S+\cos\theta)$, which seems to be a rather strange quantity to compute. Notice that if we drop the "$+$" then the quantity $-4(S\cos\theta)$ equals
$$
-4S\cos\theta=-\sin\alpha\cos\theta=\cos^2\theta
$$
which is almost one of the choices.


问:
Evaluate $\int_{0}^{[x]}\left(\int_{0}^{[t]}[s]-\left[s-\frac{1}{2}\right] \mathrm{d}s\right) \mathrm{d}t$
Evaluate
$$\int_{0}^{[x]}\left(\int_{0}^{[t]}[s]-\left[s-\frac{1}{2}\right]  \mathrm{d}s\right) \mathrm{d}t$$

Where $[x]$ denotes the greatest integer function . But I don't know how to approach this kind of problem. So I want to know how to think to solve this type of problems.

If there are more than one way to approach, I would like to gather the knowledge.

Thanks in advance.

Edit: Edited after the comments.

答:
Since $x$ only appears in a floor function in a limit of the outer integral, we can assume $x$ is an integer:
$$\int_0^x\int_0^{[t]}[s]-[s-1/2]\,ds\,dt$$
$[s]-[s-1/2]$ is $0$ when the fractional part of $s$ is greater than $\frac12$ and $1$ when it is not. Since the bounds of the inner integral are also integers, it simply evaluates to half of the distance between the bounds:
$$=\int_0^x\frac12[t]\,dt$$
Now assuming $x>0$ (the other case is similar), $t$ assumes the same value for each of $x$ intervals of length $1$: the values from $0$ to $x-1$ inclusive. Thus the integral becomes a finite sum.
$$=\frac12(0+1+\dots+(x-1))=\frac{x(x-1)}4$$
The final answer is $\frac{[x]([x]-1)}4$.


问:
Complex large exponentiation
I have the next exponentiation of complex number $$z ={(1+ i\sqrt 3)}^{2020}$$
I was using the Theorem of Moivre but I got a $$2 ^ {2020}$$
Then how can I get this exponentation without computer calculations? or how can I that power?

答: Hint: $1+\sqrt{3}i= 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)$. Note that the right answer is $2^{2020}\times \text{something}$, and this "something" is the outcome of the De Moivre's formula that you are trying to use.

Title: Complex large exponentiation


问:
Difference between $\emptyset$ and $\{\emptyset \}$
I was tasked to write the following by listing the elements between braces, $$\{X\subseteq\mathscr{P}(\{1,2,3\}): |X|\leq 1\}$$ $\{\emptyset \}$

My textbook gives the solution: $$\big\{\emptyset, \{\emptyset\}, \{\{1\}\}, \{\{2\}\}, \{\{3\}\}, \{\{1, 2\}\}, \{\{1,3\}\}, \{\{2,3\}\}, \{\{1,2,3\}\} \big\}$$

I got everything except the elment $\{\emptyset\}$. What's the difference between $\emptyset$ and $\{\emptyset\}$? Isn't $\emptyset=\{\}$, and likewise $\{\emptyset\}=\{\{\}\}$ which should be equivalent to $\{\}$?

答:
First one is a an empty set and the second one is a set containing only one element - an empty set.

If you want an analogy from the real world - imagine (a) an empty bag and then imagine (b) a bag that has another bag inside, which is empty.

The reason that you have $\emptyset$ as one of the answers is that an empty set is a subset of any set, so it qualifies too.


问:
A basis for a subspace of $\mathbb R^3$
I have the following question:

Find the basis of the following subspace in $\mathbb R^3$:
  $$2x+4y-3z=0$$

This is what I was given. So what I have tried is to place it in to a matrix $[2,4,-3,0]$ but this was more confusing after getting the matrix $[1,2,-3/2,0]$. This was done to get a leading 1. Now I solved for $x,y,z$.

$$1x + 2y - \frac3 2z = 0$$ from the matrix. Then
$x = \frac3 2z - 2y$
so
$$\left(\frac3 2z-2y, 2y, \frac3 2z\right)$$

Now I have $$(x,y,z) = z(3/2, 0, 3/2) + y(-2, 2, 0) \implies
{(3/2,0,3/2),(-2,2,0)}$$

which has two dimensions.

But this solution that I am giving is wrong. The actual answer according to my text book is ${(3,0,2),(-2,1,0)}$. Can any one please tell me where I messed-up? Thanks.

答:
First you dont need to put it in a matrix, as it is only one equation, you can solve right away.  For gettin the generators of that subspace all you need to do is put a variable in function of the other ones, be $3z-4y = 2x$ for example, then you have that $x = \frac{3}{2} z-2y$ replacing in a vector you get $(\frac{3}{2} z-2y,y,z)$ separating the sum as x and y are not in function of each other $z(\frac{3}{2} , 0, 1) + y(-2,1,0)$ and multiplying the firt by 2 as as it doesnt change the solution, they're linearly independant you get a base {${(3,0,2)(-2,1,0)}$}, and the dimension of this is 2.

Your mistake was to put 2y and  1/3 z in the second and third coordinates, as this only affect X (the first coordinate), there it should be y and z

Title: A basis for a subspace of $\mathbb R^3$


问:
Is the surface of a sphere and a crayon the same manifold?
In Schutz, Geometrical Methods of Mathematical Physics book, pg. 29, it was said that the sphere $S^2$ and the surface of a crayon has the same global structure:

I also read that two spaces are the 'same' as manifolds if they are diffeomorphic.

But I find it hard to visualise that these two surfaces have the same global structure. They clearly look very different in 3D space. What exactly does global structure mean?

答:
There is an old joke that a topologist is a mathematician who doesn't know the difference between a donut and a coffee cup.

Topology considers them the same because they are both closed surfaces that have one hole in them. Likewise, a crayon and a sphere have no holes.

The actual shape doesn't matter. topology studies the properties of an object that stay the same when the object is smoothly distorted. Tearing or poking holes is not allowed. Number of holes is such a property.

Such a smooth distortion allows you to map points from the original shape onto the final shape. Every point will be near the same points it was near in the original shape. You can reverse the distortion. Each point will return to its original location. The mapping is reversible.

OwnerDisplayName: TaeNyFan


答:
Geometry is taken to be the study of distances and angles, hence the study of the metric. If we forget the metric, we still have what is called the smooth structure, and if we forget the smooth structure, we still have the topological structure.

Whilst the surface of a crayon is not metrically equivalent to the surface of a sphere, they are smoothly equivalent, and hence also, topologically equivalent.

Smooth equivalence, roughly speaking, allows deforming the object through squeezing and stretching so long as no creases or folds are introduced and furthermore no cuts.

Imagine the crayon was made out of plasticine, then we can easily see that we can deform this crayon to a ball, simply by squeezing it and rolling it up into a ball, and hence the surfaces are diffeomorphic. Its also notable here, that the bodies of the two objects and not just their surfaces are diffeomorphic. This need not be the case, for example a hollow sphere and a ball have the same surface and so are diffeomorphic, but the hollow sphere has no interior whereas the ball does, so their bodies are not diffeomorphic.

Title: Is the surface of a sphere and a crayon the same manifold?

Body:
In Schutz, Geometrical Methods of Mathematical Physics book, pg. 29, it was said that the sphere $S^2$ and the surface of a crayon has the same global structure:

I also read that two spaces are the 'same' as manifolds if they are diffeomorphic.

But I find it hard to visualise that these two surfaces have the same global structure. They clearly look very different in 3D space. What exactly does global structure mean?

OwnerDisplayName: Mozibur Ullah


答:
The notation $\emptyset$ is the set with no elements, but it is a set by itself, it does not mean that $\emptyset$ is nothing.

For example: the zero vector $\overset{\to}{0}$ in normed space $(V,\|\cdot\|)$ is the only vector $v$ with $\|v\|=0$ (the norm of a vector is a "scale" of it "size"), $\overset{\to}{0}$ is an element in the normed space, and one can think about $\overset{\to}{0}$ as "nothing", because $\forall v\in V: v+\overset{\to}{0}=v$, but it does not mean that $\overset{\to}{0}$ is nothing.

By the same intuition, the empty set is not nothing, it is just a wallet with no money, which leads to $\{\emptyset\}\neq\emptyset$, because the set $\{\emptyset\}$ is a set with one element, and $\emptyset$ has zero elements.

Title: Difference between $\emptyset$ and $\{\emptyset \}$

Body:
I was tasked to write the following by listing the elements between braces, $$\{X\subseteq\mathscr{P}(\{1,2,3\}): |X|\leq 1\}$$ $\{\emptyset \}$

My textbook gives the solution: $$\big\{\emptyset, \{\emptyset\}, \{\{1\}\}, \{\{2\}\}, \{\{3\}\}, \{\{1, 2\}\}, \{\{1,3\}\}, \{\{2,3\}\}, \{\{1,2,3\}\} \big\}$$

I got everything except the elment $\{\emptyset\}$. What's the difference between $\emptyset$ and $\{\emptyset\}$? Isn't $\emptyset=\{\}$, and likewise $\{\emptyset\}=\{\{\}\}$ which should be equivalent to $\{\}$?


问:
If I have three points, is there an easy way to tell if they are collinear?
Points $(a,b)$, $(m,n)$, and $(x,y)$ are selected at random. What is the quickest/easiest way to tell if they are collinear? At first I thought it was a matter of comparing slopes but that doesn't appear to be enough.</p>

答:
Another way is to choose any two of the three points to construct a pair of parametric equations for the line they lie on, say,

$$X \ = \ a \ + \ (m - a) \cdot t \ , \ Y \ = \ b \ + \ (n - b) \cdot t \ , $$

with $ \ t \ $ being the parameter.  For the third point, $ \ (x, y) \ $ , find the value of a parameter $ \ s \ $ which solves $ \ x \ = \ a \ + \ (m - a) \cdot s \ $ .  If the same value $ \ s \ $ produces an equation for $ \ y \ = \ b \ + \ (n - b) \cdot s \ , $ then the three points are collinear; otherwise... they aren't!

This approach can be extended without much difficulty to deal with the reference line being horizontal or vertical (so there is no complication with things like undefined slopes).


问:
What is the definition of derivation of tensor algebra?
Derivation $d$ of a Lie algebra $L$ is defined as $d([x,y])=[d(x),y]+[x,d(y)]$ for all $x,y \in L$. Let consider $T(L)$ be the tensor algebra and $S(L)$ be the symmetric algebra of $L$. How a derivation map on $T(L)$ is defined?

答:
Let $A$ be a $K$-algebra, not necessarily associative, with $K$-bilinear product $(x,y)\mapsto x\cdot y$. A derivation of $A$ is a $K$-linear map $D\colon A\rightarrow A$ satisfying the Leibniz rule
$$
D(x\cdot y)=D(x)\cdot y+x\cdot D(y)
$$
for all $x,y\in A$. The tensor algebra $A=T(V)$ is an associative algebra, so the definition also applies here. The space of all derivations of $A$ is denoted by ${\rm Der}(A)$. It forms a Lie subalgebra of the linear Lie algebra $\mathfrak{gl}(A)$.


问:
Double summation with dependent variables
Currently i have taken part in a coding contest where i was asked this question https://atcoder.jp/contests/abc186/tasks/abc186_d. In the editorial solution they have given something like this:

Here especially in the second last step how the outer summation changes from $\sum_{i=1}^{N-1}$ to $\sum_{i=1}^N$ and the inner changes from $\sum_{j=i+1}^N$ to $\sum_{j=1}^N$?

I am not able understand this double summation variable change. Can someone please help me with it?

答:
For brevity write $x_{ij}:=|A_i-A_j|$. Work backwards. The final quantity is
$$\frac12\sum_{i=1}^N\sum_{j=1}^N x_{ij}$$
which is a sum over all pairs $(i,j)$ where $i$ and $j$ each run from $1$ to $N$. The trick is to observe that this sum can be broken into three pieces:
$$\frac12\sum_{i<j} x_{ij} + \frac12\sum_{i=j}x_{ij} + \frac12\sum_{i>j}x_{ij}.$$
By symmetry, the first and last sums are equal to each other (since $x_{ij}=x_{ji}$). Meanwhile, the middle sum equals zero (since $x_{ii}=0$). So now the sum is
$$
\sum_{i<j}x_{ij}
$$
which is the first quantity. If you plot in the $i,j$-plane all possible values for the pairs $(i,j)$ you'll get a square array of dots. The above maneuver corresponds to splitting the square into a diagonal (where $i=j$), and two triangles with the same $x_{ij}$ values.

EDIT: To go from the second-last line to the last line, the steps in detail are:
$$\begin{aligned}
&\frac12\sum_{i=1}^{N-1}\sum_{j=i+1}^N (x_{ij} + x_{ji})\\
\stackrel{(1)}=&\frac12\sum_{i=1}^{N-1}\sum_{j=i+1}^N x_{ij} + \frac12\sum_{i=1}^{N-1}\sum_{j=i+1}^Nx_{ji}\\
\stackrel{(2)}=&\frac12\sum_{i=1}^{N-1}\sum_{j=i+1}^N x_{ij} + \frac12\sum_{j=1}^{N-1}\sum_{i=j+1}^Nx_{ij}\\
\stackrel{(3)}=&\frac12\sum_{i=1}^{N-1}\sum_{j=i+1}^N x_{ij}+\frac12\sum_{i=1}^N x_{ii} + \frac12\sum_{j=1}^{N-1}\sum_{i=j+1}^Nx_{ij}\\
\stackrel{(4)}=&\frac12\sum_{i=1}^N\sum_{j=1}^N x_{ij}
\end{aligned}
$$
In step (1) we split the sum into two pieces. In step (2) we reindex the second sum by relabeling index $i$ as $j$ and index $j$ as $i$. In step (3) we are adding zero. In step (4) we observe that the three summations cover all possible pairs of $(i,j)$ from $1$ to $N$.


答: Since $f$ maps $\mathbb R \to \mathbb R$ it is one of the first two options. Since $f(i)=(1+2i)/(3+5i)=(3-5i)(1+2i)/(\cdots)^2 = (i+\cdots)/\cdots$ it's 1.</p>

Title: $f(z)={2z+1\over 5z+3}$ maps


答:
We have $1+\alpha\in \Bbb Q(\alpha)$, since $1,\alpha\in \Bbb Q(\alpha)$, and this is a field, so that also its sum is in it. Conversely, $\alpha\pm 1\in \Bbb Q(1+\alpha)$, so that $2\alpha=(\alpha+1)+(\alpha-1)\in \Bbb Q(1+\alpha)$.
Hence we have $\Bbb Q(\alpha)=\Bbb Q(1+\alpha)$.

Title: Is $\mathbb{Q}(i\sqrt[4]{2}) = \mathbb{Q}(1 + i\sqrt[4]{2})$?

Body:
Let $\alpha = i\sqrt[4]{2}$. Then, $1 + \alpha = 1 + i\sqrt[4]{2}$.

Then, $\mathbb{Q}(\alpha) = \{ c_0 + c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 \mid c_0, c_1, c_2, c_3 \in \mathbb{Q} \}$.

and $\mathbb{Q}(1 + a) = \{ b_0 + b_1 (1 + \alpha) + b_2 (1 + \alpha)^2 + b_3 (1 + \alpha)^3 \mid b_0, b_1, b_2, b_3 \in \mathbb{Q} \}$.

Since, $b_0 + b_1 (1 + \alpha) + b_2 (1 + \alpha)^2 + b_3 (1 + \alpha)^3 = b_0 + b_1 + b_1 \alpha + b_2 + 2 b_2 \alpha + b_2 \alpha^2 + b_3 + 3 b_3 \alpha + 3 b_3 \alpha^2 + b_3 \alpha^3$

$= (b_0 + b_1 + b_2 + b_3) + (b_1 + 2b_2 + 3b_3) \alpha + (b_2 + 3b_3) \alpha^2 + b_3 \alpha^3$.

Notice that $b_0,b_1,b_2,b_3 \in \mathbb{Q}$.
Then, $(b_0 + b_1 + b_2 + b_3)$, $(b_1 + 2b_2 + 3b_3)$ and $(b_2 + 3b_3)$ are all in $\mathbb{Q}$ as well.

So, let $c_0 = (b_0 + b_1 + b_2 + b_3)$, $c_1 = (b_1 + 2b_2 + 3b_3)$, $c_2 = (b_2 + 3b_3)$ and $c_3 = b_3$.

Then, $\mathbb{Q}(\alpha) = \mathbb{Q}(1 + \alpha)$.

So, $\mathbb{Q}(i\sqrt[4]{2}) = \mathbb{Q}(1 + i\sqrt[4]{2})$


问:
Monotonicity property for a simple semidefinite program (SDP)
Let $A \in S^{n}_{+}$ be a positive semi-definite matrix. Consider the following simple semi-definite program (SDP)

$$\begin{array}{ll} f(A) = &\underset{T \text{ is diagonal}}{\text{minimize}} \;\; \mbox{Tr}(T)\\ &\text{subject to} \;\; A \preceq T\end{array}$$

The dual problem is
$$\begin{array}{ll} \underset{X \in S^{n}_{+}}{\text{maximize}} & \mbox{Tr} (A X)\\ \text{subject to} & X_{ii} = 1, \quad \forall\, i \leq n\end{array}$$

Suppose $A_1$ and $A_2$ are two positive semi-definite matrices, and the eigenvalues of $A_1$ are pairwisely larger than the eigenvalues of $A_2$. In another word, let $\lambda_i(A)$ denote the $i$-th largest eigenvalue of a symmetric matrix $A$; then we have $\lambda_i(A_1) \geq \lambda_i(A_2) \geq 0$ for all $i \leq n$.

I wonder if the above implies $f(A_1) \geq f(A_2)$? This monotonicity property seems intuitive to me, and I wonder if there is a way to formally prove or disprove this. The difficulty I met is that the fact that $A_1$ has larger eigenvalues does not mean the difference $A_1 - A_2$ is positive semi-definite, so this monotonicity property is not immediately obvious.

(A related question is asked here, which is on whether the above problems have analytical solutions).

答:
I think in general we do not have the implication $f(A_1)\ge f(A_2)$. The issue is what you alluded to, namely $A_1$ and $A_2$ may not be simultaneously diagonalizable, i.e. there may not be an orthogonal matrix $V$ such that

$$A_k = V\Sigma_k V^\top, \;\; \text{for $k=1,2,$}\quad (1)$$
where $\Sigma_k$ is a diagonal matrix  with eigenvalues of $A_k$ on the diagonal.

First, if there does exist such a $V$ that $(1)$ holds. Then, since $\text{Tr}(AX)=\text{Tr}(V\Sigma V^\top X)=\text{Tr}(\Sigma (V^\top X V))$, we can rewrite the dual problem as
$$\begin{align*} f(A_k) = \max_{X\succeq 0} \quad & \text{Tr}(\Sigma_k (V^\top X V))\\ \text{s.t.} \quad & X_{ii}=1, \;\;\forall i\\
= \max_{Y\succeq 0} \quad & \text{Tr}(\Sigma_k Y)\\ \text{s.t.}\quad & (VYV^\top)_{ii}=1, \;\;\forall i\end{align*}$$
From this we can see that if $\Sigma_1 \succeq \Sigma_2$, then $\text{Tr}(\Sigma_1 Y)\ge \text{Tr}(\Sigma_2 Y)$ for all $Y\succeq 0$. Therefore, $f(A_1)\ge f(A_2)$.

Second, if (1) does not hold, then we can look at the simple case where $\Sigma_1 = \Sigma_2 = \Sigma$, but $A_1 = V_1V_1^\top$ and $A_2=V_2V_2^\top$ as their eigendecomposition and $V_1\ne V_2$. Then,
$$\begin{align*} f(A_k) = \max_{Y\succeq 0} \quad & \text{Tr}(\Sigma Y)\\ \text{s.t.}\quad & (V_kYV^\top_k)_{ii}=1, \;\;\forall i\end{align*}$$
But you see that since these two problems now have the same objective function and different linear equality constraints, there is no reason $f(A_1)$ would be equal to $f(A_2)$.
In particular, think about the 2-by-2 case, let $\Sigma=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$, and $V_1 = I$ and $V_2 = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 & 1\end{bmatrix}$, then it can be directly computed that $f(A_1) = 1$ but $f(A_2)=2$. If the computation is correct, then this example indicates that even when $A_1$ and $A_2$ share the same set of eigenvalues, $f(A_1)-f(A_2)$ can be either positive, negative, or zero.


问:
Let $a:[0,\infty) \to \mathbb R$ is continuous function.Consider the ODE $y'(x)=a(x)y(x),x>0, y(0)=y_0 \ne0.$ Then which of the following are true?

Assume that $a:[0,\infty) \to \mathbb R$ is a continuous function. Consider the ordinary differential equation, $y'(x)=a(x)y(x),x>0, y(0)=y_0 \ne0.$ Then which of the following statement(s) are true ?

$1.$ If $\int_0^ \infty |a(x)|dx< \infty$, then $y$ is bounded.

$2.$ If $\int_0^ \infty |a(x)|dx< \infty$, then $\lim_{x \to \infty} y(x)$ exists.

$3.$ If $\lim_{x \to \infty}a(x)=1$, then $\lim_{x \to \infty} y(x)= \infty$

$4.$ If $\lim_{x \to \infty}a(x)=1$, then $y$ is monotonic.

I am not getting any clue how to approach this problem. Can anyone solve it.

答:
Given ordinary differential equation is $$y'(x)=a(x)y(x)\implies \dfrac{dy(x)}{y(x)}=a(x)dx$$Integrating between the limit $x_0$ to $x$, we have
$$\int_{x_0}^x\dfrac{dy(x)}{y(x)}=\int_{x_0}^xa(x)dx\implies y(x)=y(x_0)\exp\left(\int_{x_0}^xa(x)dx\right)$$Putting $~x_0=0~,$ we have $$y(x)=y_0\exp\left(\int_{0}^xa(x)dx\right)\tag1$$
For option $\bf{(1)}$ and  $\bf{(2)}$ :

If $~\int_{0}^x|a(x)|dx<\infty~,$ then $~y(x)<\infty\implies y~$ is bounded.
$\implies \lim_{x\to\infty}y(x)~$ exists.
Hence both options $(1)$ and $(2)$ are correct.

For option $\bf{(3)}$ : As $~\lim_{x\to\infty}a(x)=1\implies\exists~ k\in\mathbb R^+~$ such that $~a(x)>1/2~\forall~x>k~,$

$$\implies \int_k^\infty a(x)dx>\int_k^1 1/2 ~dx\implies \int_k^\infty a(x)dx\to\infty$$Now $$\int^\infty_0a(x)dx=\int_0^ka(x)dx+\int_k^\infty a(x)dx$$
As $~a(x)~$ is continuous on compact set $~[0,k]~,$ the  it is also bounded i.e.,
$~\int_0^k a(x) dx\le m\in\mathbb R~.$ So
$$\int^\infty_0 a(x)dx\to\infty.$$Therefore from $(1)$,  $$\lim_{x\to\infty} |y(x)|=y_0\exp\left(\int_{0}^xa(x)dx\right)\to\infty~.$$Hence option $(3)$ is also correct.

For option $\bf{(4)}$ : Take $$a(x)=1-\dfrac{8x}{1+4x^2}\implies \lim_{x\to\infty}a(x)=1$$
$$\implies y(x)=y_0\exp\left[\int_{0}^x\left(1-\dfrac{8x}{1+4x^2}\right)dx\right]$$
$$=y_0\exp\left[x-\ln(1+4x^2)\right]=y_0\dfrac{e^x}{1+4x^2}$$which is not monotonic as the function has both increasing and decreasing portions on $[0,\infty)$.

Hence option $(4)$ is also incorrect.


答: Its not 1x+2y-1/3z=0, but 1x+2y-1/2z=0. That is your main mistake. You can also avoid fractures by looking at vector (2x,2y,2z)=(3z-4y,2y,2z)=z.(3,0,2)+y.(-4,2,0)</p>

Title: A basis for a subspace of $\mathbb R^3$