问:
What probability measure whould I choose? Calculation of probability
I am a little bit confused about the following math problem I asked myself.
Let's say there are 10 exams in a year, and each exam contains exactly three questions. A question's probability to be answered correctly by a student is given by 1/4. (If a question is not answered at all, we say it is not answered correctly.)
Now assume that after each exam the teacher chooses one out of the three tasks to be corrected, so the other two will be ignored. Each task has the same probability to be choosen by the teacher. If a student gets the chosen question right, he gets one point.

My question is: What is the probability of a student to have a least 5 points in total at the end of the year?

I thought about using Binomial distribution, but the problem is that my question is a "choose with replacement" and you do a choose without replacement when you use Binomial dsitribution...

Any help is appreciated :)
Thanks!

答:
"Choose with replacement" refers to modeling a process where the outcome of one trial does not influence the outcomes of any other trials.  So for instance, when we flip a coin, the outcome of subsequent flips does not depend on the outcome of previous flips.  Or, if we draw balls from an urn with balls of two different colors, if we replace the ball after each draw, then the outcomes of future draws does not depend on the outcomes of previous draws--unlike the case where we do not replace the ball.

As this notion applies to your specific question about exams, the intent is that each exam is an independent trial with its own outcome, so we are dealing with a binomial model because the performance on future exams does not depend on the performance on previous exams.  Whether that is a realistic assumption is a separate matter--this is what we are told to assume when it is stipulated that the student as a $1/4$ chance to get any single question correct.

Another way to think about it is that the exams are not reused--each one is different, thus in a sense the "replacement" that is going on is the replacement of one exam with another.

Note that, for all intents and purposes, the professor could simply choose to grade the first of the three questions in each exam, because the outcome for each question has the same probability distribution as any of the other questions.  And then, the second and third questions become irrelevant information, so our model for the number of correct questions is simply binomial with $n = 10$ and $p = 1/4$.

However, things get more interesting if, for example, within a single test, the questions have increasing difficulty, so although they remain independent, they are not identically distributed:  say, the chance of the first question being answered correctly is $3/4$, the second $1/2$, and the third and most difficult has only $1/4$ chance of being answered correctly.  Then the professor's random choice of which question to grade becomes relevant.


问:
Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?
Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?

I believe it is since $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[b]{a}}^\frac{1}{c}=\frac{1}{\sqrt[bc]{a}}$.

Can someone please confirm my suspicions as I just came across a question that states in the answer that $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\sqrt[bc]{a}$ and I believe that this is a typo.

答: For $a, b, c > 0$, yes, you are correct.  This is easy to check:  choose $a = 729 = 3^6$, $b = 1/2$, $c = 1/3$.  We must obviously have $((1/a)^{1/b})^{1/c} < 1$, but $\sqrt[6]{729} > 1$.

ContentLicense: CC BY-SA 4.0


问:
Prove that $a_n$ is eventually zero?
Let $(a_n)$ be a nonnegative sequence such that $\sum_{n=1}^{\infty}a_n < \infty$. Suppose we have the following

$a_{n+1}^2-a_n^2+\alpha a_n \le 0 \quad \forall n\in \mathbb{N}$ for some $\alpha >0$  $\quad (*)$

I can prove by contradiction that under the above assumptions $a_n$ is eventually zero, which means there exists $N \in \mathbb{N}$ such that $a_n=0 \quad \forall n \geq N$.

However I'm pretty curious about whether the result still holds when we substitute $\alpha a_n$ in $(*)$ by $\alpha a_{n+1}$ (all other assumptions remained). I've been working on this but haven't had an answer so far.

Any idea/answer would be appreciated.

答:
$a_{n+1}(a_{n+1} +\alpha) \le a_n^2$

$(a_{n+1} + \alpha) \le \frac {a_n^2}{a_{n+1}}$ for the $a_{n+1} > 0$.

If we say $a_{n+1} \ge \beta a_n^2$ then $\frac {a_n^2}{a_{n+1}} \ge \frac 1\beta$ and this is certainly possible.

Let $a_{n+1} = \frac 12 a_n^2; a_1 < 1$ then we want

$\frac 12a_n^4 + \alpha \frac 12a_n^2 \le a_n^2$ which can be true for $\alpha = 1$

Title: Prove that $a_n$ is eventually zero?


问:
Regular value theorem and sard's theorem for banach manifolds
I have been reading about floer homology , and alot of banach (infinite) dimensional manifolds tend to appear. To try and deal with them results such as the inifinite dimensional analogue of the regular value theorem and the sard-smale theorem that applies to fredholm appear.

Now I think to prove the regular value theorem we can just try an use the analogue of the implicit function theorem for banach spaces, but then when we are dealing with fredholm operators I belive we can determine the dimension of the submanifold by the index of the operator.

So my question is if anyone knows a reference that deals with this "classical" manifolds theorem but in infinite dimensions and dealing with tools from Functional analsysis ?

Thanks in advance.

答:
The keyword is "Sard-Smale theorem". The implicit function theorem for Banach manifolds is covered in Lang's book. It is the same statement, all one has to ask is that for all $x$ in $f^{-1}(y)$, we have that $df_x$ is surjective and that the kernel of $df_x: T_x M \to T_{f(x)} N$ be complemented (that is, there is another closed subspace $V$ so that $T_x M = \text{ker}(df_x) \oplus V$).

This assumption is automatically true for Fredholm maps (prove this: if you are going to succeed in this topic you will need to be able to prove small facts like this with ease), and this is also automatically true for arbitrary smooth maps between Hilbert manifolds (you should be able to prove this as well).


问:
How to show two elements in a group are (not) conjugate?
While stuying group theory, I was wondering how to proof two elements in a group are not conjugate?

For example, say we're in the symmetric group $S_8$, and I have to show that the two elements $(1354672)$ and $(18967543)$ are not conjugate.
I can see that the orders of both elements are not equal. And order should be conserved under the conjugation map $τ:S_8 → S_8$, since this mapping is an isomorphism. So I think I can conclude that these two elements are not conjugate.

But how do can you prove two elements are not conjugate when they have the same order? Or does same order imply conjugation?

Thanks in advance!

答:
In $S_n$, two elements are conjugate if and only if they have the same cycle type: https://groupprops.subwiki.org/wiki/Cycle_type_of_a_permutation

So, if you have two elements of $S_n$ and you want to know if they are conjugate, you have to determine their cycle type. If they have the same cycle type, they are conjugate, and if they don't have the same cycle type, they are not conjugate.


问:
$\sigma$-field of sets of measure $0$ and $1$ - independence
$(\Omega, \mathcal{F}, P)$ - probability space and $\mathcal{N}=\sigma\{A:P(A)=0\vee P(A)=1\}$. Where I can find theorem with proof which states that $\mathcal{N}$ is independent from any other $\sigma$-field on this probability space?

答: Let $\mathcal{G}$ be any sub-$\sigma$-field on $\mathcal{F}$. Let $A \in \mathcal{N}$ and $B \in \mathcal{G}$. What can you say about the relationship between $P(A \cap B)$ and $P(A)P(B)$?


答:
${ \left( {\frac{1}{a}}^{\frac{1}{b}} \right)}^{\frac{1}{c}}= { \frac{1}{a}}^{ \frac{1}{b} \cdot \frac{1}{c}} =   { \frac{1}{a} }^{ \frac{1}{bc}} = \frac{1}{a^{\frac{1}{bc}}} = \frac{1}{\sqrt[bc]{a}}$

Thus, yes (assuming $a,b,c$ are all positive reals).

Body:
Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?

I believe it is since $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[b]{a}}^\frac{1}{c}=\frac{1}{\sqrt[bc]{a}}$.

Can someone please confirm my suspicions as I just came across a question that states in the answer that $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\sqrt[bc]{a}$ and I believe that this is a typo.


问:
Hill climbing extremist
Imagine you are on a trip in mountains, but you are a bit nutter and you walk only straight uphill or downhill. If you are on the top of a mountain or in a valley you can walk in any direction.

The question is: Can you get from any point to any another point?

In more mathematical language. Imagine you have a smooth function $\phi:\mathbb{R}^2\rightarrow\mathbb{R}$. In order to dismiss simple counter examples we require that $\phi$ coercive or is zero at infinity. Now be given two points $A,B \in \mathbb{R}^2$, can you find piece-wise smooth curve $\gamma(t)$ such that $\gamma'$ has the same or opposite direction as $\nabla \phi(\gamma(t))$, except at the points where $\nabla \phi = 0$? This can be stated, after reparametrization of $\gamma$, as $|\gamma' \cdot \nabla \phi| = \|\gamma'\|\|\nabla \phi\|$.

Background: I was thinking about this differential equation for $u$
$$
f = g_i \partial_i u.
$$
you can find local solution of by integrating the function $f$ along integral curves of $g_i$. But when can you glue these solutions together and get a globally continuous solution? This should be possible when you can get from point $A$ to point $B$ by different paths but you integrate the same amount of $f$. So natural question is, can you get from any point $A$ to any point $B$?

I found this question interesting but surprisingly I was unable to find a (simple) argument why it should be true.

答:
I haven't thought this through, but at the same time it would take too much space as comment, so let me describe an idea I have for this problem.

$\newcommand{\R}{\mathbb{R}}\newcommand{\vp}{\varphi}$
Idea. The relation $x \sim y$ if there is a path joining $x$ and $y$ is an equivalence relation in $\R^2$. If you can only prove that each equivalence class is open, then your claim follows from connectedness of $\R^2$.

Other observations:

I would expect that one needs some assumptions on behavior of $\vp$ in infinity, so it could be easier to work on a torus or a sphere first, and then see what is needed in the case of $\R^2$.

It is surely important that our vector field is a gradient field. On a torus any nonzero constant vector field yields a counterexample, so one probably needs to use the growth of $\vp$ in some way.

To make things clearer, I would add an additional assumption that $\vp$ is a Morse function. This way we can be sure what happens at points where $\nabla \vp = 0$.

ContentLicense: CC BY-SA 3.0

Title: Hill climbing extremist


问:
Looking for rotation rule - 3D vectors
here is my problem... to visualize... you are standing at the origin 0,0,0 .. facing Z(forward) Y(up) X(left/right).. there are two points in space anywhere... the goal is for you to grab one point in one hand and the other point in the other hand. so basically move to a point in between the two points and rotate to grab both points with your hands aligning you perfectly in between.

the middle point is calculated from both vectors in space...

and if i calculate a lookat to one vector, you may rotate and align properly.... but if i calculate using the second vector, you wont align properly. (or it doesn't in my results at the moment)

my problem is i don't know how to determine which vector to use to get the good alignment each and every time... is there a rule to pick to correct point/vector and my lookat will work flawlessly?

Now, i am not really looking for the "math" part... but more the logical/rules part.

i have done alot of looking around but cant seem to find anything pertaining to this.

thanks in advance for assisting. I have never really posted here so apologize if I am being naive about how this forum works.

cheers

答:
In this answer I consider a more specific example offered in the comments.
Namely, the points are vertices of a polygon in the $x,z$ plane,
and it is assumed that the only rotation you will perform is a rotation around the $y$ axis, that is, whichever direction was "straight up" relative to your body will remain "straight up" relative to your body.

Let's list the vertices in sequence around the perimeter of the polygon,
starting with an arbitrary vertex and stopping when we return to that vertex.
That is, if there are $n$ vertices we can list them like this:
$P_0,P_1,P_2,\ldots,P_{n-1},P_0.$
For convenience, let $P_n$ be another name for $P_0.$
And let the coordinates of $P_k$ be $(x_k,0,z_k)$ for $0 \leq k \leq n.$

Let's also assume the polygon is a simple polygon, that is, it does not cross over itself. This way there is a clearly defined interior of the polygon
and a clearly defined exterior.

Now you can stand midway between two vertices, $P_k$ and $P_{k+1}$, and turn so that $P_k$ is directly to your left and $P_{k+1}$ is directly to your right.
You will be facing perpendicular to the edge of the polygon,
facing either the interior or exterior of the polygon.
If you follow this same procedure for $k=0,$ then $k=1,$ $k=2,$
etc. up to and including $k=n-1,$ you will visit each edge of the polygon.
Either you will be facing outward on every edge, or you will be facing inward on every edge.

So the orientations and locations you want are well defined.

Getting to the midpoint of a side is easy.
The coordinates of the midpoint of the edge from $P_k$ to $P_{k+1}$ are
$$ \left(\frac{x_k + x_{k+1}}2, 0, \frac{z_k + z_{k+1}}2\right).$$
You should be able to obtain a translation matrix $T$ that moves you from the origin to those coordinates.

To define the orientations in a mathematical way, we can use the direction of each side.

The vector from $P_k$ to $P_{k+1}$ is $(x_{k+1} - x_k, 0, z_{k+1} - z_k).$
This vector points in the same direction as the direction from
the origin to the point $(x_{k+1} - x_k, 0, z_{k+1} - z_k).$
So one way to get a matrix to orient you in that direction might be a lookat function if you have one where you can (in effect) put the camera at the origin and turn the camera to face the point $(x_{k+1} - x_k, 0, z_{k+1} - z_k).$

Another option is to find the angle that the vector
$(x_{k+1} - x_k, 0, z_{k+1} - z_k)$ makes with the positive $z$ axis in the $x,z$ plane, looking down from the positive $y$ axis.
The angle is $\operatorname{atan2}(x_{k+1} - x_k, z_{k+1} - z_k)$ radians counterclockwise.
So the rotation matrix to turn you from facing the positive $z$ direction to facing the direction $(x_{k+1} - x_k, 0, z_{k+1} - z_k)$
is the standard matrix for a rotation of
$\operatorname{atan2}(x_{k+1} - x_k, z_{k+1} - z_k)$ radians around the $y$ axis.

Assuming your transformation matrices are for transformations of world coordinates of your body, you would first apply the rotation matrix and then the translation matrix.
Alternatively, in terms of matrix multiplication, if the rotation matrix is $R$ and the translation matrix is $T$ you want the matrix $TR$ (first transformation goes on the right).

This procedure has one flaw: it leaves you facing along the edge of the polygon rather than perpendicular to it.
You need an extra $90$ degrees ($\frac\pi2$ radians) of rotation to face perpendicular to the edge. So just apply that rotation first.
If $R_{\pi/2}$ is the matrix to rotate $90$ degrees around the $y$ axis,
then the complete transformation is $T R R_{\pi/2}.$

This procedure should have you facing inward from the midpoint of each edge of the polygon, or outward from the midpoint of each edge of the polygon.
It depends on whether you originally listed the vertices of the polygon in counterclockwise sequence or clockwise sequence.

To consider how you might decide which way is which:

If you know how to measure the signed angle between two vectors in the $x,z$ plane -- that is, you can measure the angle so that a positive angle is a left turn and a negative angle is a right turn -- then add up all the angles as you go around the polygon once (so you add $n$ angles).
The sum should be either $2\pi$ radians or $-2\pi$ radians (approximately, allowing for roundoffs during the calculation).
If it's $2\pi$ radians you are going counterclockwise and the exterior of the polygon is on the right; in the other case the exterior is on the left.

Another method is to choose a number $C$ between the $z$ coordinates of vertices, not equal to the $z$ coordinate of any vertex. (You can just list all the $z$ coordinates in ascending order, choose a pair of $z$ values, and set $C$ halfway between them.)

Now consider the line $z = C.$ Your polygon must cross that line at least twice, or possibly more times (as in the example below).

Each crossing crosses the line $z = C$ at a different $x$ value.
Find the maximum $x$ value among all the crossings, and consider the edge that crosses at this $x$ value.
In the example, this edge is the one from $P_3$ to $P_4.$
We see in this example that the $z$ coordinate increases as you go from $P_3$ to $P_4,$ which means the exterior of the polygon is on the left.
If the $z$ coordinate had decreased, the exterior of the polygon is on the right.

These particular rules make a number of assumptions about the coordinate system. If one of these assumptions is wrong -- for example, the coordinate system is left-handed instead of right-handed, or the $y$ axis points downward instead of upward, or the sense or the rotation matrix is left-handed instead of right-handed -- then the procedure could say to turn the wrong way. But if it does so, it will be wrong every single time, so you just need to do the opposite and then it will always work.

Title: Looking for rotation rule - 3D vectors


问:
Does a stationary point (derivative equals zero) have to be local max/min/inflection?
I was wondering whether a function can have a point in which its derivative is zero but the point is neither local min, local max or inflection point. I was considering the function $$f(x)=\begin{cases}x^2\sin\frac1x,\ x\neq 0;\\
0,\ x=0.\end{cases}$$
The derivative at $x=0$ is zero (can be proven using limit definition of derivative). But for every $h>0$ there are positive and negative numbers for $f$ in $(-h,h)$, and therefore this isn't a local maximum or minimum. Is the point $x=0$ an inflection point or not? How can this be proven?
$$$$
Thank you

答:
Good question!

The given function $f'(x)$ is zero but discontinuous† at $0.$

So, for the same reason you cited as why it isn't a turning point, the stationary point $0$ isn't an inflection point either. (Also, neither the sign test nor second-derivative test are applicable.)

Another example of a non-inflection (curvature doesn't change sign), non-turning stationary point is the point $0$ of the function $g(x)=3.$

On the other hand, if a function has a continuous first derivative at, and is neither constant left of nor right of, its stationary point, then the stationary point is surely either a turning or inflection point.

Incidentally, the function $h(x)=|x|$ has a non-stationary turning point at $0.$

† $$f'(x)=\begin{cases}2x\sin\frac1x-\cos\frac1x,\ x\neq 0;\\
0,\ x=0.\end{cases}$$
Let $\left\langle x_n:n\in\mathbb N\right\rangle=\left\langle \frac1{2n\pi}:n\in\mathbb N\right\rangle$ so that $x_n\rightarrow0$ and $n\neq0.$
But $f'(x_n)=-1\nrightarrow0=f'(0).$


问:
What is the symmetry between the definitions of the bounded universal/existential quantifiers?
What is the symmetry between the definitions of the bounded universal/existential quantifiers?

$\forall x \in A, B(x)$ means $\forall x (x \in A \rightarrow B(x))$

$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$

These make intuitive sense, but I would expect there to be some kind of symmetry between how the definitions of the bounded quantifiers work, and I can't see one. $A \rightarrow B$ means $\lnot A \lor B$ which doesn't seem to have a direct relationship with $A \land B$. What am I missing?

答:
Consider $\forall x\in A, B(x)$. That is if $x$ is a member of $A$ then $B(x)$ is true.

This translates, as you well put to $\forall x(x\in A\rightarrow B(x))$, which again translates to $\forall x(x\notin A\lor B(x))$.

Denote $B=\{x\mid B(x)\}$, that is all the elements which satisfy $B(x)$. Now you have that $x\in B \iff B(x)$ is true.

We re-evaluate the previous sentence: $\forall x(x\in A\rightarrow x\in B)$, that is to say $A\subseteq B$, or to say $\forall x(x\notin A\lor x\in B)$.

Now consider the existential version: $\exists x(x\in A\land B(x))$ that is $\exists x(x\in A\land x\in B)$, or in simpler words $A\cap B\neq\emptyset$.

Lastly we recall that $\lnot\forall x\varphi(x) \iff \exists x\lnot\varphi(x)$. This translates well over bounded quantifiers:

$$\begin{align}
\lnot(\forall x\in A, B(x)) &\iff \lnot\forall x(x\in A\rightarrow x\in B)\\
&\iff \exists x\lnot(x\in A\rightarrow x\in B)\\
&\iff \exists x(x\in A\land x\notin B)\\
&\iff \exists x\in A,\lnot B(x)
\end{align}$$


问:
Early-exercise point of American Put
Could you please help me?
In book Rüdiger Seydel "Tools for Computational Finance" in Chapter 4.5 "American Options as Free Boundary Problem" it is provided the following explanation for case $S>S_f$ , where $S_f$ is contact point:

If $S>S_f$, then early-exercise causes immediate loss, because
$-V+K-S<0$.
Well, we have options so in case of early exercise we should borrow $1$ risky asset and seal it with price $K$. So our current payoff will be $K$.
Why here is $-V+K-S$ ?

Thank you.

答:
By the definition of $S_f$ (eqn 4.22), we have $V^{am}_P(S, t) > (K - S)^+$ whenever $S > S_f(t)$.  Rearranging, $-V^{am}_P + (K - S)^+ < 0$, so we have

$-V^{am}_P + K - S \leq -V^{am}_P + (K - S)^+ < 0$

Title: Early-exercise point of American Put


问:
Addition and multiplication of Spectrums
I know that if $A$ is a unital abelian Banach algebra, then we have $\sigma(x+y)\subset \sigma(x)+ \sigma(y)$ and $\sigma(xy)\subset \sigma(x)+\sigma(y)$.

So we have the notation $\sigma(x)+\sigma(y)$ and $\sigma(x) \sigma(y)$. So I wonder how the addition and multiplication are defined.

For example, if we consider the set of $2\times 2$ matrices.

Let $a=\pmatrix{-1&0\\0&1}$, $b=\pmatrix{1&1\\0&0}$.

Then $\sigma(a)= \{ 1, -1\}$ , $\sigma(b)= \{1, 0  \}$

What are $\sigma(x)+\sigma(y)$ and $\sigma(x) \sigma(y)$ in this case?

Thank you!

答:
For any sets $A$ and $B$, $A + B = \{a+b: a \in A,\; b \in B\}$ and similarly
$AB = \{ab: a \in A,\; b \in B\}$.

Note that in your example $a$ and $b$ don't commute, so they are not relevant to the theorem.


问:
How do you determine the coordinates of a point on a vector if you are given one point?

Vector AB - the tail is at (-4,2) and the head is at (-1,3). Determine the coordinates of point D on vector CD, if C(-6,0) and vector CD = vector AB.

I tried to find the slope of vector CD knowing that it is the same as vector AB.

Determining the slope of vector AB:

m = (y2-y1)/(x2-x1) = (3-2)/((-1)-(-4)) = 1/3

Setting the slope of vector AB to the slope of vector CD:

1/3 = (y2-0)/(x2-(-6))

I stopped here because I realized that I cannot do anything if I do not know x or y.

How would you approach this problem?

答:
Hint :

Think of $A,B,C,D$ as position vectors in plane. So $A=(-4,2)$ etc.

Vector $AB=B-A$, where $B$ is head, $A$ is its tail. Similarly $CD=D-C$.

Problem is asking for $AB=CD \Rightarrow D=B-A+C$ from above.


Title: How do you determine the coordinates of a point on a vector if you are given one point?

Body:
Vector AB - the tail is at (-4,2) and the head is at (-1,3). Determine the coordinates of point D on vector CD, if C(-6,0) and vector CD = vector AB.

I tried to find the slope of vector CD knowing that it is the same as vector AB.

Determining the slope of vector AB:

m = (y2-y1)/(x2-x1) = (3-2)/((-1)-(-4)) = 1/3

Setting the slope of vector AB to the slope of vector CD:

1/3 = (y2-0)/(x2-(-6))

I stopped here because I realized that I cannot do anything if I do not know x or y.

How would you approach this problem?

Body: So $A = (-4,2), B = (-1,3), C = (-6,0), D = (x,y)$. Thus $\vec{AB} = \vec{CD}\implies <-1 - (-4), 3 - 2> = <x - (-6), y - 0>\implies x+6=3, y = 1\implies x = -3,y=1$. So $D = (-3,1)$. Hope it helps.


问:
Dense linear span implies injectivity of a Gelfand Transform
Let $A$ be a unital commutative Banach Algebra.

Let $\Omega(A)$ be the set of non-zero homomorphisms from $A$ to $\Bbb C$.

Suppose the set of idempotents $N$ in $A$ have dense linear span.

Show the map $f:\Omega(A)\rightarrow \{0,1\}^N$ defined by $f(\phi)(e) = \phi(e)$ is injective.

In The spectrum of a commutative unital Banach algbra having linearly dense idempotents is totally disconnected, it says that the injectivity comes from the dense linear span. But I don't really see how this goes.

So if $f(\phi)(e) =f (\psi)(e)$ , we have $\phi(e)=\psi(e)$. Why does the dense linear span tell us $\phi=\psi$?

Thank you for your help!

答:
If $\phi(e)=\psi (e)$,  for every idempotent $e$,  then  $\phi(a)=\psi (a)$,  for every element $a$ which can be written as
a linear combination of idempotents (because $\phi$ and $\psi $ are linear).

The set of all such $a$ is suposedly dense,  so $\phi$ and $\psi $ coincide on a dense set,  and hence everywhere because
they are continuous.


问:
How do you prove $I_n - A$ is invertible when A is symmetric?
I've been struggling to prove that if $A$ is a $nxn$ symmetric matrix with every eigenvalue $|\lambda_i|<1$ then $I_n-A$ is invertible.

Any help is welcome.

答:
Use theorem that eigenvalue  for polynomial of matrix is this polynomial of original eigenvalue of the matrix.
Symmetric matrix has only real eigenvalues then $-A$ has only positive eigenvalues. If you additionally add $I$ to $-A$ you have all eigenvalues greater than 1. Therefore $I-A$ is invertible.


问:
Induced map on $K_{1}$-group
Let $A$ be a unital $C^{\ast}$-algebra. Any automorphism $\alpha$ of $A$ induces a map on $K_{1}(A)$ by $\alpha_{\ast}[u]=[\alpha(u)]$.

Let the automorphism $\alpha$ be inner, does it follow that $\alpha_{\ast}$ is the identity map?

I guess the answer is negative because unitary equivalence is far from homotopy of unitaries. The answer may also depend on the type of the $C^{\ast}$-algebra. Any suggestions?

答:
In short: this is essentially because $K_1(A)$ is Abelian, and it is true for every inner automorphism of every $C^*$-algebra. But if one wants to be careful, there are a few things to check and recall. I will assume like you that $A$ is unitary, but that's true in general with the same proof, replacing $A$ by $\widetilde{A}$.

1) Preliminary facts: let us denote $(K_1(A),+)$. This is an Abelian group for any $C^*$-algebra $A$ and by the standard picture, we have $K_1(A)=\{[u]\,;\,u\in U_\infty(A)\}$.

For every $v,w$ in the same $U_n(A)$, we have
  $$
[vw]=[wv]=[v]+[w].
$$

That's a direct consequence of Whitehead's lemma. Precisely
$$
\pmatrix{u&0\\0&v}=\pmatrix{u&0\\0&1}\pmatrix{0&1\\ 1&0}\pmatrix{v&0\\0&1}\pmatrix{0&1\\ 1&0}\sim_h \pmatrix{uv&0\\0&1}
$$
where the rhs homotopy follows from the continuous $2\times 2$ unitary path
$$
\pmatrix{\cos t&\sin t\\ \sin t&-\cos t }.
$$
And conjugation by the same path on the lhs allows to exchange $u$ and $v$, whence the result.

In particular, for every unitary $u_n$ in $U_n(A)$, we have
$$
0=[1_n]=[u_nu_n^*]=[u_n]+[u_n^*].
$$

Note: with the fact above, it is easy to deduce that $K_1(A)$ is indeed Abelian. With $u\in U_n(A)$ and $v\in U_m(A)$, we can assume $m=n$ up to adding some $1$'s down the diagonal. Then $[u]+[v]=[u\oplus v]=[(uv)\oplus 1]=[(vu)\oplus 1]=[v\oplus u]=[v]+[u]$.

2) Proof: now assume $\alpha$ is an inner automorphism of $A$, that is there exists $u$ unitary in $A$ such that $\alpha(x)=uxu^*$ for every $x\in A$. Let us compute $\alpha_\ast=K_1(\alpha)$ on $K_1(A)$, that is on $[v]$ for every $v\in U_n(A)$, for every $n\geq 1$.

The key point is that for $v=(v_{ij})\in U_n(A)$, still denoting by $\alpha$ the $*$-homomorphism induced by $\alpha$ on $M_n(A)$, we have
  $$
\alpha(v)=\pmatrix{\alpha(v_{ij})}=\pmatrix{uv_{ij}u^*}=\mbox{diag}(u,\ldots,u)\,v\,\mbox{diag}(u^*,\ldots,u^*)=u_nvu_n^*
$$
  where $u_n$ is the unitary of $U_n(A)$ made of a constant diagonal of $u$, and zero elsewhere.

Therefore
$$
\alpha_\ast([v])=[\alpha(v)]=[u_nvu_n^*]=[u_n]+[v]+[u_n^*]=[v]+[u_n]+[u_n^*]=[v].
$$
Thus $\alpha_\ast=K_1(\alpha)$ is the identity on $K_1(A)$.

Of course, it suffices to use the property $[vw]=[wv]$ for $v,w\in U_n(A)$ to get $[u_nvu_n^*]=[u_n^*u_nv]=[v]$. But it was a good opportunity to recall that for a unitary $u_n\in U_n(A)$, $-[u_n]=[u_n^*]$.

3) Extension: more generally, for every approximately inner automorphism of $A$, we have $K_0(\alpha)=\mbox{id}$ and $K_1(\alpha)=\mbox{id}$.

Title: Induced map on $K_{1}$-group


问:
Variance of sum of products
Let $\{X_i\}$ be iid random variables where $X_i \sim N(\mu_x,\sigma^2)$ and $\{Y_i\}$-iid random variables  where $Y_i \sim N(\mu_y,\sigma^2)$. Every $X$ is also independent from $Y$. I need to find such variance:

$$Var\Big(\sum_{\substack{i=1,...,n\\ j=1,...,n}} X_i Y_j \Big)$$

I tried to transform it into sums of variances and covariances but then i couldn't manage to calculate the covariance. I did something like this but i feel like its wrong.

$$\sum_{\substack{1\leq i < k \leq n\\ 1\leq j < l \leq n \\ i \neq k \lor j\neq l}} Cov(X_iY_j,X_k Y_l) = \sum_{\substack{i=1,...,n\\ 1\leq j < l \leq n}} Cov(X_iY_j,X_i Y_l) + \sum_{\substack{1\leq i < k \leq n\\ j=1,..., n}}Cov(X_iY_j,X_k Y_j)$$
$$+ \sum_{i\neq j \neq k \neq l} Cov(X_iY_j,X_k Y_l)+ \sum_{\substack{ i \neq j \neq k\\ j=1,..., n}}Cov(X_iY_j,X_k Y_k) + \sum_{\substack{ i \neq k \neq l\\ j=1,..., n}}Cov(X_iY_i,X_k Y_j)$$

I would appreciate any advice on how to calculate this.

答:
Notice that $$S = \sum_{1 \le i, j \le n} X_i Y_j = \sum_{i=1}^n X_i \sum_{j=1}^n Y_j.$$  Then since $$S_X = \sum_{i=1}^n X_i \sim \operatorname{Normal}(n \mu_x, n \sigma^2), \\ S_Y = \sum_{j=1}^n Y_j \sim \operatorname{Normal}(n \mu_y, n \sigma^2),$$ the variance is computed as $$\begin{align}
\operatorname{Var}[S] &= \operatorname{Var}[S_X S_Y] \\
&= \operatorname{E}[(S_X S_Y)^2] - (n^2 \mu_x \mu_y)^2 \\
&\overset{\text{ind}}{=} \operatorname{E}[S_X^2]\operatorname{E}[S_Y^2] - n^4 \mu_x^2 \mu_y^2 \\
&= (\operatorname{Var}[S_X] + n^2 \mu_x^2) (\operatorname{Var}[S_Y] + n^2 \mu_y^2) - n^4 \mu_x^2 \mu_y^2 \\
&= n^2 (n (\mu_x^2 + \mu_y^2) + \sigma^2 )\sigma^2.
\end{align}$$

Title: Variance of sum of products


问:
Let $\zeta=\{E \in \mathcal{P}(X): card(E)=2 \}$ proof that $\sigma(\zeta)=\mu_{c}$
Consider $X$ an uncountable set and $\mu_{c}:=\{A\in \mathcal{P}(X): A\quad\text{is countable or} \quad X-{A}\quad \text{is countable} \}$

Let $\zeta=\{E \in \mathcal{P}(X): card(E)=2 \}$

proof that $\sigma(\zeta)=\mu_{c}$

it is clear that $\mu_{c}$ is a sigma-algebra that contains $\zeta$, indeed, given $E\in \zeta$ then $Card(E)$=2 so, $E$ is countable and therefore $E \in \mu_{c}$  and we know that $\sigma (\zeta)$ is the smallest sigma algebra that contains $\zeta$,

therefore, $\sigma (\zeta) \subset \mu_{c} $

for the other contention I expressed $\sigma(\zeta)$ as follows:

$\sigma(\zeta)= \bigcap_{g \in G} g$, $G$ is the set of all sigma algebras that contain $\zeta$

then, to prove that $\mu_{c} \subset \sigma (\zeta)$ take a $s\in \mu_{c}$ and $g\in G$ But I have not been able to find a good reasoning that leads me to show that $s \in g$

答:
We actually have $\sigma(\zeta) = \mu_c$ whenever $|X| \neq 2$. It's just less interesting when $X$ isn't uncountable.  The case of $|X|=1$ is trivial (there's only one $\sigma$-algebra), so I'll assume $|X|\geq 3$ here.

First note, as you've done, that $\zeta \subseteq \mu_c,$ so as $\mu_c$ is a $\sigma$-algebra, $\sigma(\zeta) \subseteq \mu_c.$

For the other direction, let $x \in X$.  Since $|X| \geq 3$, there are at least two other distinct points $y,z \in X,$ so that $\{x,y\}, \{x,z\} \in \zeta$.  Therefore, $$\{x\} = \{x,y\} \cap \{x,z\} \in \sigma(\zeta),$$ so for any $x\in X$, we have $\{x\} \in \sigma(\zeta)$.

Now let $A\in \mu_c$, so by cases, either

$A \subseteq X$ is countable: Then $A = \bigcup_{x\in A} \{x\} \in \sigma(\zeta),$ as it's a countable union of elements of $\sigma(\zeta).$

$X-A \subseteq X$ is countable: Then $X-A \in \sigma(\zeta)$ as in the previous case and then $A = X-(X-A) \in \sigma(\zeta)$.

It follows that $\mu_c \subseteq \sigma(\zeta)$.


问:
Direct integral, reference
I am looking for a basic reference for the definition of a direct integral and a basic discussion. I know that there is a discussion on this in "Methods of modern mathematical physics, volume 4", but it does not feel very 'self-contained' or simple. Is anyone aware on possibly better references?</p>

答: This is in page 999 of Kadison, Richard V.; Ringrose, John R., Fundamentals of the theory of operator algebras. Volume II: Advanced theory, Pure and Applied Mathematics, 100-2. Orlando etc.: Academic Press (Harcourt Brace Jovanovich, Publishers). XIV, pp. 399-1074. (1986). ZBL0601.46054.

Title: Direct integral, reference


问:
Parabolic equations without compatibility conditions
Consider
$$
\left\{\begin{array}{rrrr}
u_t-Lu&=&f\\
u|_{\partial\Omega}&=&0\\
u(0)&=&g
\end{array}\right.
$$
where $L$ is an elliptic operator; for convenience assume $L$ has smooth coefficients and $f, g$ are all smooth. Now (for example following the PDE book of Evans) by Galerkin method one gets a weak solution $u\in L^2(0, T; H_0^1(\Omega))$ with $u'\in L^2(0, T; H^{-1}(\Omega))$. If $f, g$ satisfy compatibility conditions of all orders, one can prove $u$ is smooth on $\overline{\Omega}\times[0, T]$, as done in the book of Evans. Without the compatibility conditions, I can prove that $u$ is smooth on $\overline{\Omega}\times (0, T]$, namely the time $0$ is excluded.

MY QUESTION: is there an easy way (say along the line of the Evans book) to prove that, without compatibility conditions, $u$ is smooth on $\Omega'\times [0, T]$ with $\Omega'\subset\subset\Omega$? Note time $0$ is included.

答:
Yes, and the idea is fairly simple: multiply $u$ by a cutoff function that vanishes near $\partial\Omega$ and apply the global regularity result, noting the compatibility conditions are satisfied for this modified equation. The rest is just keeping track of the details.

Given $\Omega' \Subset \Omega,$  choose an intermediate subdomain $\Omega''$ such that $\Omega' \Subset \Omega'' \Subset \Omega$ and let $\eta,\tilde \eta \in C^{\infty}(\Omega)$ such that $1_{\Omega''} \leq \eta \leq 1_{\Omega}$ and $1_{\Omega'} \leq \tilde\eta \leq 1_{\Omega''}.$ Then $v(x,t) = \eta(x) u(x,t)$ solves the modified equation
$$ \begin{cases}
  v_t - Lv = f - [L,\eta]u & \text{ in } \Omega \times (0,T), \\
  v(x,t) = 0 & \text{ on } \partial\Omega \times (0,T), \\
  v(x,0) = \eta(x)g(x) & \text{ on } \Omega \times \{0\},
\end{cases}$$
where $[L,\eta]u = L(\eta u) - \eta Lu$ is the commutator. Note that since $\eta g$ is compactly supported in $\Omega,$ this equation satisfies compatibility conditions up to all orders and so the global higher regularity theorems apply.

We know from the improved regularity result Chapter 7 Theorem 5(i) that if $g \in H^1_0(\Omega)$ and $f \in L^2(0,T;L^2(\Omega)),$ then $u \in L^2(0,T;H^2(\Omega))$ and $\partial_tu \in L^2(0,T;L^2(\Omega)).$ Then $v = \eta u$ lies in the same regularity class (noting $\partial_tv = \eta \partial_tu$) so we deduce that $[L,\eta]u \in L^2(0,T;H^1_0(\Omega)).$

Hence* $\nabla v$ satisfies the equation
$$ \begin{cases}
  (\partial_t-L)\nabla v = \nabla \tilde f + [L,\nabla]v & \text{ in } \Omega \times (0,T), \\
  \nabla v(x,t) = 0 & \text{ on } \partial\Omega \times (0,T), \\
  \nabla v(x,0) = \nabla(\eta(x)g(x)) & \text{ on } \Omega \times \{0\},
\end{cases}$$
and note the commutator term also lies in $L^2(0,T;L^2(\Omega))$ so we can apply the improved regularity result one again to deduce that $v \in L^2(0,T;H^3(\Omega))$ and $\partial_tv \in L^2(0,T;H^1(\Omega)).$ Since $v \equiv u$ on $\Omega' \times (0,T),$ the same holds for $u$ on $\Omega'' \times (0,T).$ Now if we let $\tilde v = \tilde\eta \nabla u$ then the associated commutator term satisfies $\partial_t[L,\tilde\eta]\nabla u = [L,\tilde\eta]\partial_t \nabla u \in L^2(0,T;L^2(\Omega)$ and similarly $\partial_t[L,\nabla]u \in L^2(0,T;L^2(\Omega))$ so assuming $g,f$ are sufficiently regular by applying (ii) of the same theorem to $\tilde v$ we get
\begin{align*}
  u &\in L^2(0,T;H^3(\Omega')) \\
  \partial_tu &\in L^2(0,T;H^2(\Omega'), \\
  \partial_t^2u &\in L^2(0,T;L^2(\Omega').
\end{align*}
Now the idea is to iterate this and argue by induction to show that $\partial_t^ku \in L^2(0,T;H^{2m+2-2k}(\Omega'))$ for all $m \geq 0$ and each $\Omega'\Subset \Omega.$ The argument is essentially the same as the proof of Theorem 6, notably the compatibility conditions are vacuously satisfied thanks to the cutoff argument above.

*Added later: To see this note we have the weak formulation
$$ \langle \partial_tv(t),\phi \rangle + B[v,\phi;t] = \langle f(t),\phi \rangle $$
for almost every $t \in (0,T),$ with $\phi \in C^{\infty}_c(\Omega).$ Then taking $\phi = \nabla\varphi$ we can integrate by parts to get
$$ \langle \partial_tv(t),\nabla\varphi\rangle + B[\nabla v,\phi;t] = \langle \nabla \tilde f(t),\phi \rangle + \left( B[\nabla v, \phi ;t] - B[v,\nabla\phi;t]\right). $$
Now we can extend by density to take $\phi \in H^1_0(\Omega);$ the only term that needs justification is the $\partial_t\nabla v(t)$ term; for this we know that $\partial_tv \in L^2(0,T;L^2(\Omega))$ and $\partial_t\nabla v = \nabla \partial_t v$ by definition of weak deriavtives and Fubini, noting that
$$  \int_{\Omega} \left(\int_0^T\eta'(t) \nabla v(t,x)\,\mathrm{d}t\right) \varphi(x) \,\mathrm{d} x  = \int_0^T \eta(t) \left(\int_{\Omega} \partial_tv(t,x) \nabla \varphi(x) \,\mathrm{d}x\right) \,\mathrm{d}t.$$
This verifies the weak formulation, as required.


问:
Conditional distributions for a multinomial random variable
I'm struggling to understand how to solve the following problem.

We have (1,2,3,4) is a multinomial distributed r.vs with parameters  =10 and p =(1 =0.3,2=0.3,3=0.3,4=0.1).

Now I need to find P(1=3,2=3,3=3|4=1). I would know how to approach this if we had for example P(1=3|4=1) using:

And then just treating it as a binomial distribution, but what to do if we need to find the probability of more than 1 variable occurring, given that 1 other has occurred?

答: The formula you included tells you that the conditional distribution of the other $X_i$, given $X_4$, is also multinomial, not just binomial.  In other words, $$((X_1, X_2, X_3)^\top \mid X_4 = 1) \sim \operatorname{Multinomial}(n = 9; p_1 = p_2 = p_3 = 1/3).$$  So the desired probability is simply $$\Pr[X_1 = X_2 = X_3 = 3 \mid X_4 = 1] = \frac{9!}{3!3!3!} (1/3)^3 (1/3)^3 (1/3)^3.$$


问:
I have to calculate the following integral: $\int \lvert\sin(k\cdot x)\rvert\, \mathrm{d}x,\quad k\in\mathbb{Z}$
I have to calculate the following integral: $$\int_0^\pi \lvert\sin(k\cdot x)\rvert\, \mathrm{d}x,\quad k\in\mathbb{Z}.$$
What I've tried, but is undoubtedly wrong:

答:
$$\int_0^\pi |\sin k x| dx = k \int_0^{\frac{\pi}{k}} \sin k x \, dx = k \cdot \frac{2}{k} =2, \quad (k \in \mathbb{Z}_{>0}).$$

Similarly for $k \in \mathbb{Z}_{<0}$,$$\int_0^\pi |\sin k x| dx =2.$$

Of course $k=0$ requires special treatment:

$$\int_0^\pi |\sin k x| dx =0, \quad \text{when } k=0.$$

UPDATE

To evaluate the integral:

$$\int_0^{\frac{\pi}{k}} \sin k x dx = \left.\frac{1}{k} \cos kx\right|_{\frac{\pi}{k}}^0=\frac{1}{k}(1-(-1))=\frac{2}{k}.$$


问:
Why is the formula for the area of a cardioid $ \int_a^b \frac{1}{2} r^2 d \theta$
I've seen this expression in many places :$\int_a^b \frac{1}{2} r^2 d \theta$ and was wondering if someone can explain where this came from? I've noticed that it's sometimes explained in conjunction with double integrals but I'm not very familiar with double integrals (as in what it means using a picture). </p>

答:
There are at least two way to look at this formula. The simplest, in my opinion, is approximating the area inside the arc into small circular arcs.

$\hspace{3.5cm}$

The area inside each arc of radius $r$ and angle $\Delta\theta$ is $\frac12r^2\Delta\theta$ (this is proportional to $\pi r^2$ for the entire circle). Summing these approximations as $\Delta\theta\to0$ yields
$$
\int\frac12r^2\mathrm{d}\theta
$$

Another approach is to look at the Jacobian of the conversion from rectangular to polar coordinates.
$$
\mbox{
$\begin{align}
x&=r\cos(\theta)\\
y&=r\sin(\theta)
\end{align}
$}
\quad\text{and}\quad
\mbox{$
\begin{align}
\mathrm{d}x&=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\
\mathrm{d}y&=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta
\end{align}
$}
$$
That is
$$
\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix}
=\begin{bmatrix}\cos(\theta)&-r\sin(\theta)\\\sin(\theta)&r\cos(\theta)\end{bmatrix}
\begin{bmatrix}\mathrm{d}r\\\mathrm{d}\theta\end{bmatrix}
$$
The determinant of the Jacobian yields
$$
\begin{align}
\mathrm{d}x\,\mathrm{d}y
&=r(\cos^2(\theta)+\sin^2(\theta))\,\mathrm{d}r\,\mathrm{d}\theta\\
&=r\,\mathrm{d}r\,\mathrm{d}\theta
\end{align}
$$
if we integrate in $r$ from the origin first, this yields the area to be
$$
\int\frac12r^2\,\mathrm{d}\theta
$$


问:
Identity of Fibonacci sequence
Let $F_n$ be a Fibonacci sequence with initial terms $F_0=0, F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\geqslant 1$.

Prove that $F_n^2+F_{n+1}^2=F_{2n+1}$ for $n\geqslant 0$ (with mathematical induction).

My efforts: For $n=0$ it is true.

Suppose that our statement holds for $0\leqslant k \leqslant n$ i.e. $F_k^2+F_{k+1}^2=F_{2k+1}$

Let's try to prove it for $k=n+1$. $$F_{2n+3}=F_{2n+1}+F_{2n+2}=F_{n+1}^2+(F_{n}^2+F_{2n+2})= ?$$

Here I'm stuck and I have applied different methods but none of them brings a positive result.

Can anyone help to complete this?

答:
We can actually prove two different identities (1) and (2), of which you were asked to prove (1):
$$F_{n+1}^2+F_{n+2}^2=F_{2n+3} \quad \text{(1)}$$
$$F_{n+2}^2-F_{n}^2=F_{2n+2} \quad \text{(2)}$$
Note that the Fibonacci numbers are numbered starting from $F_1$.

We can proceed by induction. Suppose that we have the two equations
$$F_n^2+F_{n+1}^2=F_{2n+1} \quad \text{(3)}$$
$$F_{n+1}^2-F_{n-1}^2=F_{2n} \quad \text{(4)}$$
The base cases are trivial.

First Step: Prove (2).

Now, adding (3) and (4), we obtain :
\begin{align}
2F_{n+1}^2+F_n^2-F_{n-1}^2 &= F_{2n+1}+F_{2n}\\\\
(F_{n+1}-F_n)^2+F_{n+1}^2 + 2F_{n+1}F_n - F_{n-1}^2&= F_{2n+2}\\\\
(F_{n-1})^2+F_{n+1}^2 + 2F_{n+1}F_n - F_{n-1}^2&= F_{2n+2}\\\\
(F_{n+1}+F_{n})^2-F_{n}^2&= F_{2n+2}\\\\
F_{n+2}^2-F_{n}^2&= F_{2n+2} \quad \text{(2)}
\end{align}
As desired. $\square$

Second Step: Prove (1). Now that we have (2), we are home free!

Adding (2) and (3), we obtain :

\begin{align}
F_{n+2}^2-F_n^2+F_{n}^2+F_{n+1}^2 &= F_{2n+2}+F_{2n+1}\\\\
F_{n+2}^2+F_{n+1}^2 &= F_{2n+3} \quad \text{(1)}
\end{align}
And we are done. $\square$

Title: Identity of Fibonacci sequence


问:
Given the basis for an orthogonal complement, how would you find the matrix that projects all vectors onto this basis?
Let's say $A = \text{projection onto orthogonal complement}, V^{\perp}$. Are the columns of $A$ just made up of the basis vectors of $V^{\perp}$ or do we have to do some extra work (and in which case, what must be done)? We just covered projections in class and I'm slightly confused by how they work, so any input is appreciated!

答:
If your vector space $U$ is finite-dimensional and $V$ is a subspace, then it decomposes as $U=V\oplus V^\perp$. This means that giving a vector in $U$ is the same thing as giving a vector in $V$ and a vector in $V^\perp$. Consequently, defining a linear transformation in $U$ is really the same thing as defining how it acts on $V$ and how it acts on $V^\perp$.

The transformation $A$ that you define as projection onto the subspace $V^\perp$ is then clear from this point of view: it acts as the identity in $V^\perp$ and trivially on $V$ (i.e. maps everything to $0$).

So, if you have a basis of $n$ vectors of $U$ that is made up of a basis of $n-k$ vectors of $V$ and a basis of $k$ vectors of $V^\perp$, the matrix representation of $A$ is just
\begin{bmatrix}
0 & 0\\
0 & I
\end{bmatrix}

where $I$ is the $k\times k$ identity matrix.


问:
Let $T: P_2 \rightarrow P_2$ be the map $f(x) \mapsto xf'(x)$. Find the matrix $[T_B]$ associated to the basis $B = \{1, x, x^2\}$
I'm confused because $f(x^2) \mapsto x^2(2x) = 2x^3$, which is not in $P_2$ and I'm not sure how to represent it in the associated matrix. Did I do the calculation incorrectly?

答:
If $f\in P_2$ is non-constant, then $\text{deg }T\left(f\right) = \text{deg }f$. If $f\in P_2$ is a constant, then $T(f) = 0$. In particular, $T:P_2\to P_2$ is indeed a well-defined map. (The reason this is true is because the degree of $f'$ is one less than the degree of $f$ and the degree of $xf'$ is one more than the degree of $f'$, so long as $f$ is non-constant.)

We have to determine the polynomials $T(1)$, $T(x)$, and $T(x^2)$. The (three) coefficients of the polynomial $T(x^i)$ will constitute the three entries in the $(i+1)$st column of $[T_{\cal B}]$.

Hope this helps!


问:
$a_n$ and its alternating series $(-1)^n a_n$
I have 2 questions regarding series. 1. If $a_n >0$, and $\sum (-1)^n a_n$ converges, then does $\sum a_n$ converge? 2. (This one is unrelated to 1.) If $\sum a_n$ converges, then does $\sum (-1)^n a_n$ converge?

For 1., I doubt it. A counter example is $\sum (-1)^n \frac{1}{n}$, which converges, but $\sum \frac{1}{n}$ doesnt. 2. I'm not sure. If I make $a_n = \frac{sinx}{x} -2$, then does the series with $(-1)^n$ converge?

答:
For the first, take $ a_n=\frac 1n$.

$$\sum (-1)^na_n \text{ converges but}$$
$$\sum a_n \text{ diverges }$$

For the second, take $ b_n=\frac{(-1)^n}{n}$.
$$\sum b_n \text{ converges but}$$
$$\sum (-1)^nb_n \text{ diverges }$$


问:
Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

答:
Another approach. Since

$$\left(1+\frac1n\right)^n\xrightarrow [n\to\infty]{}e>2.5$$

there exists $\,M\in\Bbb N\,$ s.t.

$$n>M\;\implies\; \left(1+\frac1n\right)^n>2$$

and thus, for $\,n>M$ :

$$\left(1+\frac1n\right)^{n^2}=\left[\left(1+\frac1n\right)^n\right]^n>2^n\xrightarrow [n\to\infty]{}\infty$$


问:
Bounded Subsets of a Well-Ordered Set
A proof I'm reading relies on the following fact: Let $(J,\leq)$ be a well-ordered set, $A \subseteq J$ a subset, and suppose that there exists $i \in J$ such that $a < i$ for all $a \in A$. Then there exists $k < i$ such that $a \leq k$ for all $a \in A$.

I can't figure out how to prove this. $A$ must have a least upper bound because $J$ is well-ordered, but what stops this least upper bound from being $i$ itself?

答: This is indeed false without additional assumptions.  For instance, $A$ could be empty and $i$ could be the least element of $J$.  Or, $J$ could be $\mathbb{N}\cup\{\infty\}$, $A$ could be $\mathbb{N}$, and $i$ could be $\infty$.

Title: Bounded Subsets of a Well-Ordered Set


问:
Optimization Problem: Minimizing Total Cost of All Pairs
This is related to a real-world problem I'm solving but I'm keeping it abstract so I don't bog you down with details. An efficient answer to this feels like it should be obvious but I'm not seeing it. Let me know if there's a better place to ask this.

Given some letters and some numbers, I need to generate the most efficient letter-number pairing.

letters: a-z

numbers: 0-99

result: {a:46, b:27, c:13, ...}

For each of the C(26,2)=325 letter pairs, there exists a frequency of how likely it is for that letter pair to occur. Ex: {'th': 0.03}

For each of the C(100,2)=4950 number pairs, there exists a cost for how expensive it is to transition between those two numbers. Ex: {(32,17): 53}

Pair each number with a letter, such that the total cost of all letter combinations (frequency*cost) is as low as possible.

Hopefully this makes sense. Is there some elegant way to do this that doesn't involve brute-forcing the solution?

答: This is an instance of the quadratic assignment problem, which is a generalization of the traveling salesman problem.  It is difficult to solve large instances optimally, but using an integer linear programming solver will generally perform much better than brute force.</p>


问:
contraction in five dimensional Euclidean space
$T(u,v,w,x,y)=(v,0.5w,x,y,u)$

$S(u,v,w,x,y)=(y,u,v,0.5w,x)$

(a) Could anyone tell me whether S and T are contraction maps on any matrix norm/induced vector norm?

(b) Is their composition both way i.e ST and TS, contraction?

(c) What can we say about the Contraction property of (SSSS)(TTTT) i.e $S^4T^4$ and the other-way $T^4S^4$? In other words even if $T$, $S$ are not a contraction, but does there exists $k_1, k_2\in\mathbb N$ such that $T^{k_1}$ is a contraction and $S^{k_2}$ is a contraction?

Thank you for your response.

答:
First, $T$ and $S$ cannot be strict contraction mappings with respect to the same norm. To see this, suppose for contradiction that there exists a norm $\lVert \cdot \rVert$ on $\mathbb{R}^5$ (or $\mathbb{C}^5$, doesn't matter) with respect to which $T$ and $S$ are both strict contractions. This means that $\lVert T(x) \rVert < \lVert x \rVert$ and $\rVert S(x) \rVert < \lVert x \rVert$ for all $x \in \mathbb{R}^5 \setminus \{0\}$.

Now
$$\lVert (0,1,0,0,0) \rVert = \lVert S(1,0,0,0,0) \rVert < \lVert (1,0,0,0,0) \rVert = \lVert T(0,1,0,0,0) \rVert < \lVert (0,1,0,0,0) \rVert,$$
which is a contradiction.

On the other hand, $T$ and $S$ are both non-strict contractions with respect to the $\sup$ norm $\lVert (a,b,c,d,e) \rVert := \sup \{\lvert a \rvert, \dots, \lvert e \rvert\}$. I'll give the proof for $T$; the proof for $S$ is completely analogous.

Let $(a,b,c,d,e) \in \mathbb{R}^5$ be arbitrary. Then

$$\lVert T(a,b,c,d,e) \rVert = \lVert (b,0.5c,d,e,a) \rVert = \sup \{\lvert b \rvert, \lvert 0.5c \rvert, \lvert d \rvert, \lvert e \rvert, \lvert a \rvert\} = \sup \{\lvert a \rvert, \lvert b \rvert, 0.5 \lvert c \rvert, \lvert d \rvert, \lvert e \rvert\} \leq \sup \{\lvert a \rvert, \lvert b \rvert, \lvert c \rvert, \lvert d \rvert, \lvert e \rvert\} = \lVert (a,b,c,d,e) \rVert,$$

as desired.

In all of the above, I'm using the fact that $d(Tx, Ty) := \lVert Tx - Ty \rVert = \lVert T(x-y) \rVert$, so $d(Tx, Ty) \leq d(x,y)$ for all $x,y$ is equivalent to $\lVert Tx \rVert \leq \lVert x \rVert$ for all $x$ (in one direction, take $y = 0$).

Edit (b) and (c) seem to contain many different questions with many different answers. I'll try answer as many of them as I can identify.

Neither $ST$ nor $TS$ can be a strict contraction with respect to any norm. This is because $ST(1,0,0,0,0) = (1,0,0,0,0) = TS(1,0,0,0,0)$, and strict linear contractions cannot have nonzero fixed points.

$S^4 T^4$ is given by $$(a,b,c,d,e) \mapsto (\frac{1}{4}a, \frac{1}{2}b, \frac{1}{2}c, \frac{1}{4}d, \frac{1}{4}e).$$ This a strict contraction with respect to any $\ell^p$ norm (including the $\sup$ norm). Likewise, $$T^4 S^4 = (a,b,c,d,e) \mapsto (\frac{1}{4}a, \frac{1}{4}b, \frac{1}{2}c, \frac{1}{2}d, \frac{1}{4}e)$$ is a strict contraction with respect to any $\ell^p$ norm.

$S^5$ and $T^5$ both equal $\frac{1}{2} \operatorname{id}$, so both are strict contractions with respect to any norm whatsoever. Of course, this implies that $S^{5k} T^{5k} = T^{5k} S^{5k} = \frac{1}{2^{k+1}} \operatorname{id}$ is a strict contraction for every positive integer $k$.

For $k \geq 4$, $S^k T^k$ and $T^k S^k$ will be strict contractions with respect to any $\ell^p$ norm. For $1 \leq k \leq 3$, $S^k T^k$ and $T^k S^k$ have nonzero fixed points, so are not strict contractions with respect to any norm.


问:
Solution to $y' + \frac xy = 1$
$y' + \frac xy = 1$

I've identified it as a homogeneous differential equation and transformed it to $xv' + v = 1-\frac 1v$ (where $v = \frac yx$) but am stuck here.

答:
Rewrite as $x = (1- 1/v - v)/v^\prime,$ so
$$\frac{dx}x = \frac{dv}{1-1/v - v}.$$

Integrate both sides, to get

$$\log x + C = -\frac{1}{2} \log \left(v^2-v+1\right)-\frac{\tan ^{-1}\left(\frac{2
   v-1}{\sqrt{3}}\right)}{\sqrt{3}}.$$

Title: Solution to $y' + \frac xy = 1$


问:
Corollary 1.9 in Eisenbud's Commutative Algebra
(See statement and proof attached below.) In his proof of Corollary 1.9, he states that $\mathfrak{m}_p \supset I(X)$ iff $p\in X$. I can see one direction, namely if $p\in X$ then $\mathfrak{m}_p \supset I(X)$. Why is the other direction true? I think it has to do with the fact that $I(X)$ is a radical ideal, but I'm not sure how to leverage this fact. Thanks!

Corollary 1.9. Let $k$ be a field. For each $p=(a_1,\ldots,a_r)\in\mathbf{A}^r(k)$ the ideal $\mathfrak{m}_p:=(x_1-a_1,\ldots,x_n-a_n)\subset k[a_1,\ldots,x_r]$ is a maximal ideal. If $k$ is algebraically closed and $X\subset\mathbf{A}^r(k)$ is an algebraic set, then every maximal ideal of $A(X)$ is of the form $\mathfrak{m}_p/I(X)$ for some $p\in X$. In particular, the points of $X$ are in one-to-one correspondence with the maximal ideals of the ring $A(X)$.

Proof. It is clear that $k[x_1,\ldots,x_r]/\mathfrak{m}_p$, so $\mathfrak{m}_p$, is a maximal ideal. The natural map $k[x_1,\ldots,x_r]\rightarrow k[x_1,\ldots,x_r]/\mathfrak{m}_p=k$ may be described as evaluation at $p$. Thus $\mathfrak{m}_p\supset I(X)$ iff $p\in X$. Since the maximal ideals of $A(X)$ are the maximal ideals of $S:=[x_1,\ldots,x_r]$ containing $I(X)$, taken modulo $I(X)$, it only remains to show that every maximal ideal of $S$ has the form $\mathfrak{m}_p$ for some $p$.

Suppose $\mathfrak{n}$ is a maximal ideal of $S$. By Theorem 4.19 applied with $R=k$, $S/\mathfrak{n}$ is algebraic over $k/(\mathfrak{n}\cap k)=k$. Since $k$ is algebraically closed, $S/\mathfrak{n}=k$. Let $a_i$ be the image of $x_i$ under the map $S\rightarrow S/\mathfrak{n}=k$, and let $p=(a_1,\ldots,a_r)$. It follows that $\mathfrak{m}_p$ is contained in $\mathfrak{n}$. Since $\mathfrak{m}_p$ is maximal, $\mathfrak{m}_p=\mathfrak{n}$. $\qquad\square$

答:
The ideal being radical is essentially irrelevant for the implication you are looking for.

If $J\supseteq I$, then $\mathcal V(J)\subseteq \mathcal V(I)$. Recall also that $\mathcal{IVI}(A)=\mathcal I(A)$ and $\mathcal {VIV}(I)=\mathcal V(I)$. With these premises, $\mathfrak m_p\supseteq \mathcal I(X)$ implies $p\in\mathcal V(\mathfrak m_p)\subseteq \mathcal {VI}(X)$. Since $X$ is algebraic, there is some $I$ such that $X=\mathcal V(I)$. Therefore, $p\in \mathcal{VIV}(I)=\mathcal V(I)=X$.


问:
Why can't the Augmenting Path Algorithm be used for matching in general graphs?
I'm told that the problem lies behind the possible odd cylces in general graphs. However, I can't find any counterexamples and get confused.

For example, one says that if we do DFS in the graph below starting from 6 in the order of 4-5-1-2-3, we'll miss the augmenting path. But I think starting a search from every unsaturated vertex would solve the problem, since a DFS from 1 in the order of 6-5-1-2-3-4 results in an augmenting path.

For another example, one says that if we do DFS in an odd cycle, we'll get stuck in an infinite loop. But should we visit the vertex that has been visited before? I think that's unnecessary.

I would be extremely appreciative of any assistance!

答:
It's possible for the same issue to arise starting from every vertex.

Consider the following diagram:

Here, $1$ and $10$ are the two vertices unsaturated by the matching $\{23, 45, 67, 89\}$, and $(1,2,3,4,5,6,7,8,9,10)$ is an augmenting path. But we won't find it starting from vertex $1$ if we happen to explore from $1$ to $5$ first: we'll walk $(1,5,4,3,2)$ and get stuck. Similarly, we won't find it starting from vertex $10$ if we happen to explore from $10$ to $6$ first: we'll walk $(10,6,7,8,9)$ and get stuck.


问:
How to establish the chart-independence of this equivalence relation on curves on a manifold
Let $\mathscr{M}$ be a differentiable manifold of dim m. We define the following equivalence relation on curves on $\mathscr{M}$.

Two curves $\sigma_1$ and $\sigma_2$ are tangent at a point $p$ in $\mathscr{M}$ iff

$\sigma_1(0) = \sigma_2(0) = p$

In some local coordinate system $(x^1, \dots, x^m)$ around the point $p$ the two curves satisfy
$$\frac{dx^i}{dt}(\sigma_1(t))\Bigr|_{t=0} = \frac{dx^i}{dt}(\sigma_2(t))\Bigr|_{t=0}$$
for $i=1, \dots, m$.

How would check that if $\sigma_1$ and $\sigma_2$ are tangent in one coordinate system, then they're tangent    in any other coordinate system that covers the point $p$?

答:
Assume one has a chart $(U, \phi)$ with $p \in U$ and $x^i$ is $u^i \circ \phi$, where $u^i$ is the $i$-th projection from $\mathbb{R}^m$ to $\mathbb{R}$. Your assumption $2.$ means that

$$ D(\phi \circ \sigma_1)(0) = D(\phi \circ \sigma_2)(0) \, , $$

where $D(\phi \circ \sigma_1)(0)$ (resp. $D(\phi \circ \sigma_2)(0)$) denotes the differential of $\phi \circ \sigma_1$ (resp. $\phi \circ \sigma_2$) at $0$. Now, assume that $(V, \psi)$ is another chart with $p \in V$ and $y^i$ is $u^i \circ \psi$.

One has to prove
$$\frac{dy^i}{dt}(\sigma_1(t))\Bigr|_{t=0} = \frac{dy^i}{dt}(\sigma_2(t))\Bigr|_{t=0}$$
for $i=1, \dots, m$.

One has the following sequence of equalities:
\begin{equation*}
\begin{split}
D (y^i \circ \sigma_1)(0) & = D(u^i \circ \psi \circ \sigma_1)(0) \\
& = D(u^i \circ \psi \circ \phi^{-1} \circ \phi \circ \sigma_1)(0) \\
& = D(u^i \circ \psi \circ \phi^{-1})(\phi (p)) \circ D(\phi \circ \sigma_1 )(0) \\
& = D(u^i \circ \psi \circ \phi^{-1})(\phi (p)) \circ D(\phi \circ \sigma_2 )(0) \\
& = D(u^i \circ \psi \circ \sigma_2)(0) \\
& = D(y^i \circ \sigma_2)(0) \, ,
\end{split}
\end{equation*}
where to go from line 2 to line 3 one uses the chain rule. This is the required equality.


问:
The graph of Borel measurable function whose range is a separable metrisable space

If $Y$ is separable and $f : X \to Y$ is Borel measurable, then the graph of
  $f$ is Borel.

On page 14, Lemma 2.3, (iii) of this online note, given, $\{U_n\}_{n \in \Bbb N}$, a basis for the topology of  a metrisable space $Y$, the graph of $f$ is:

I can't see why this formula represents the graph of $f$. It seems to me it's an empty set.

答:
You're right; it's empty, as $U_n$ cover $Y$.

It's a typo, most likely meant to be
$$\bigcap_{n=0}^\infty \left( \{(x,y)\mid y\notin U_n\} \cup \{(x,y) \mid x\in f^{-1}[U_n]\} \right)$$


问:
Can we define the set of reals as the set containing all the convergences of the sequences?
My analysis teacher defined the set of reals as, and I quote, a collection of finite numbers $\{A_0, A_1, ..., A_n\}$ and a collection of finite or infinite numbers $\{d_1, d_2, ..., d_m, ...\}$ all in the range $0$ to $9$; we match this collection with

$x_+=+A_n...A_1A_0.d_1d_2...d_m$ a positive real

$x_-=-A_n...A_1A_0.d_1d_2...d_m$ a negative real

For example, $\pi=3.141...$ is defined such as $A_0=3$, $d_1=1$, $d_2=4$, $d_3=1$, ...

I find this definition very strange, as it seems to be based on syntax alone (or almost). In the rest of the course, we went on to discuss Cauchy sequences and the Cauchy criterion and what it implies:

The sequence $(U_n)_{n\in\mathbb{N}}$ converges if and only if it is a Cauchy one. This equivalence gives $\mathbb{R}$ the property of being a complete set, i.e. any Cauchy sequence of real numbers converges in $\mathbb{R}$.

My intuition then led me to think that the set of reals could be defined as the set containing all the limits of the real Cauchy suites. However, this is only an intuition and I don't know how to show this, I haven't found any result on the internet.

I wonder therefore if this intuition is good, if not (or if it is, because it is always interesting), I would like to know in what other ways we could define the reals, other than by the definition I gave above.

PS: obviously, we could define $\mathbb{R}$ in ensemblistic terms, and say that it contains all naturals, integers, rationals and irrationals, but I was looking for another definition than in terms of sets.

答:
In fact, your teacher's definition doesn't seem to be right per Arthur's comment. I suspect that what they're trying to do is make the construction of the reals as simple as possible, but there is some unavoidable complexity there and - assuming you've copied what they've said correctly - they've stumbled on some of it.

Yes (with a bit of care), this is in fact one of$^1$ the standard ways of constructing the real numbers from the rational numbers.

We start with the notion of a Cauchy sequence of rational numbers only. Intuitively, these are the sequences which should converge but might not ... within $\mathbb{Q}$, anyways! We'd like to add a number corresponding to each such sequence. However, we have to be careful: different Cauchy sequences need not get different numbers! For example, consider $$(3,3.14,3.1415,3.141592,...)\quad\mbox{versus}\quad(3.1, 3.141, 3.14159, 3.1415926,...).$$ These each "point to $\pi$" (ignoring the minor fact that $\pi$ doesn't exist for us just yet!), but are different sequences.

So instead, we work with equivalence classes of Cauchy sequences of rationals. Specifically, there is a natural way - without "looking ahead" to $\mathbb{R}$! - to tell when two Cauchy sequences $(a_i)_{i\in\mathbb{N}},(b_i)_{i\in\mathbb{N}}$ "point to the same thing:"

 $\lim_{i\rightarrow\infty}\vert a_i-b_i\vert=0$.

Writing "$\approx$" for the corresponding equivalence relation, we define $\mathbb{R}$ to be the set of $\approx$-classes of Cauchy sequences. Addition, multiplication, and so on of such classes can then be straightforwardly, if somewhat tediously, defined.

$^1$There are actually many different ways to "construct the reals." My personal favorite is via Dedekind cuts, FWIW. Given the plethora of options, this raises an interesting methodological concern: what exactly do we mean by "construct the real numbers," or even "the real numbers" for that matter? If I use a different construction than you, do we "disagree about $\mathbb{R}$?"

Addressing this concern satisfactorily turns out to be rather involved. I'm mentioning it here, however, since I think it is one which can reasonably occur early on and will only cause confusion if swept under the rug. So, even though it uses jargon which presumably you have not seen yet, I think it's worth stating the key theorem if only so that you know that such a thing exists:

There is exactly one complete real closed field up to isomorphism.

While the precise meaning of the above probably isn't clear, the general idea is quite simple: even if you and I think of $\mathbb{R}$ in terms of different constructions (e.g. Cauchy sequences vs. Dedekind cuts), "your version of $\mathbb{R}$" and "my version of $\mathbb{R}$" will be basically identical since they'll share a couple key mathematical properties. It turns out that there is a lot of subtlety here - see e.g. the discussion here - but ultimately the point is sound.


问:
Prove that a two periodic function is bounded
I've given the following assignment: Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a continous function such that $\forall x\in \mathbb{R},  \ f(x+2)= f(x)$.(two periodic function) I have to prove that:

a) The function is bounded and has a minimum and maximum value

b)$\exists x_{0}  \ such  \ that  ,  \ f(x_{0}+ \pi)= f(x_{0})$

There is a function to which applies $h(x)= f(x+\pi)-f(x)$ but can you explain how to construct the formal proof?

答:
Part A

Consider $D = f([0,2])$

As $f$ is c.t.s and $[0,2]$ is bounded we have that $D$ is a bounded interval. That is $\forall y \in D$ $B_1 \leq y \leq B_2$

However from the periodicity we know that $f(x) \in D $

Hence $B_1 \leq f(x) \leq B_2$

Part B

Let $x_{min}$ and $x_{max}$ respectively denote the points where $f$ attains its minimum and maximum bounds.

Then $h(x_{min}) = f(x_{min} + \pi) - f(x_{min} ) $ and $h(x_{max}) = f(x_{max} + \pi) - f(x_{max} ) $

However we know that $f(x_{max}) \geq f(x)$ $ \forall x$ so most certainly $f(x_{max}) \geq f(x_{max}+\pi)$

Hence $h(x_{max}) = f(x_{max} + \pi) - f(x_{max} ) \leq 0 $

Similarly for  $h(x_{min}) = f(x_{min} + \pi) - f(x_{min} ) \geq 0  $

If either of these inequalities are not strict, that is $h(x_{max})$ or $  h(x_{min}) = 0 $ then we are done as we have a root to $h(x)$ which implies the existence of a point such that $f(x+\pi) = f(x)$

So let us assume they are both strict. That is $h(x_{max}) < 0 < h(x_{min})$

Hence Intermediate Value Theorem states $\exists x_0 $ such that $h(x_0) = 0 $ and we are done


问:
A "simple" coupled system of first order PDEs
I'm trying to solve/ understand the the following system of PDEs:
$$\begin{cases}
\partial_xf(x,y)+\partial_yg(x,y) &= 0,\\
\partial_xg(x,y)-\partial_yf(x,y) &= 0,
\end{cases}$$
for the functions $f$ and $g$. Here's what I saw so far:

by deriving each line by $x$ or $y$ and substituing in the other equations, we can see that de system becomes
$$\begin{cases}
(\partial^2_x+\partial^2_y)f(x,y) &= 0,\\
(\partial^2_x+\partial^2_y)g(x,y) &= 0,
\end{cases}$$
so each function must satisfy the 2-dimensional Laplace equation, for which I know the solutions. But, if I'm not mistaken, this manipulation results in finding a particular case form of solutions, i.e. a subset of solutions in the bigger set of solutions that the initial system admits. Is that right ?

these equations are the Cauchy-Riemann equations, thus this system is exactly equivalent to asking that the complexe map
$$F(x+iy)=g(x,y)+if(x,y)$$
is holomorphic (or $\mathbb{C}$-differentiable) but, up to my knowledge, there exist no generic explicit expression for such maps. Does this means that this system of PDEs is just a constraint on the form of $f$ and $g$ but that there is no generic explicit solution ?

In general, how to solve this system ?

I'm also very interested in a way (coordinate change for example) to obtain two decoupled equations.

答:
The answer to the first question is no. It is a theorem that if $g$ and $f$ have the indicated first partial derivatives in an open connected set, then $g+if$ is analytic. In particular $g$ and $f$ have second derivatives. Therefore every solution to CR is harmonic, not just a subset of them.

For the second question I suppose you could say that a convergent power series is a generic explicit solution.


问:
What does the comma mean in the graphoid axioms?
I have been trying to understand the graphoid axioms, one of which--Decomposition--claims that

$$ (X \bot Y,W |Z) \implies (X \bot Y|Z).$$

In Pearl's book Causality, he writes $\bot$ with two vertical lines, and omits the comma.  The list is just after equation 1.29 in that text.
Axiom C4 in this lecture by Steffen Lauritzen, uses the comma. Wikipedia seems to claim that it is a logical and uses a different notation:

$$I(X,Z,Y)\ \text{ means that }\ (X \bot Y |Z).$$

There are several answers to questions like this, but as far as I can tell none of them actually give a definition of the comma.  Neither do any writings by Pearl or any other online sources I have read, except where they contradict each other.  I can't prove the statement if I can't tell what it is saying!

Now before you just say, "It's an and" or some such, I will point out that multiple people have claimed that it is a set intersection, a set union, a logical and--meaning that it isn't one statement but rather two to begin with? and I don't know what else. I tried my best guess and found a counterexample, so clearly I don't know what it is trying to say.

I want to know what is the definition of the claim.  Exactly.  No hand waving.  If you can carefully distinguish sets from random variables from probabilities that would also be a big help.

If I ever figure this out I promise to share my understanding as clearly as possible with others.

答:
Concatenation means and in Pearl's independence axioms.  So Pearl's Decomposition $(X\bot YW \;| Z)$ should be read as "$X$ is independent of $(Y\&W)$ given $Z$."

Pearl uses comma for and in probability statements.  This seems clear given (1.28)and (1.29), which are pretty standard defintions of conditional independence and regular probabilistic independence.

With this understanding, you can derive Decomposition from the definition of conditional independence in (1.28) [p. 11 in 2nd ed.].

Intersection (the set-theoretic operation, not the graph axiom) is the set-theoretic version of and.  They're equivalent because you can convert statements to sets in which they hold, and vice versa.

Comma is and in the probability statements in Wikipedia.  In $I(X,Y,Z)$ the commas distinguish arguments to the predicate $I$, which means "$X$ is independent of $Y$ given $Z$" as stated on the page.

Lauritzen seems to use comma for and, though I haven't checked carefully.  That looks right, though, and he uses $\cup$ (union, or) in other places.


答:
You are quite close. The problem is that before you define real numbers, a Cauchy sequence that ought to converge to the squar root of 2, for example, has nothing yet to converge to.

However, even without having real numbers, you can define whether two Cauchy sequences converge to the same number, and that’s when the differences between the elements of two Cauchy sequences get closer and closer. So you can split the Cauchy sequences into equivalent classes of Cauchy sequences that converge to the same limit. And you can use these equivalence classes to define the real numbers.

(That’s one of the two most common methods to define real numbers actually, the other is to define a real number as a non empty set of rational numbers with an upper bound that is not an element of the set, with the property that if x is an element of the set and y < x then x is an element of the set as well.

Title: Can we define the set of reals as the set containing all the convergences of the sequences?

Body:
My analysis teacher defined the set of reals as, and I quote, a collection of finite numbers $\{A_0, A_1, ..., A_n\}$ and a collection of finite or infinite numbers $\{d_1, d_2, ..., d_m, ...\}$ all in the range $0$ to $9$; we match this collection with

$x_+=+A_n...A_1A_0.d_1d_2...d_m$ a positive real

$x_-=-A_n...A_1A_0.d_1d_2...d_m$ a negative real

For example, $\pi=3.141...$ is defined such as $A_0=3$, $d_1=1$, $d_2=4$, $d_3=1$, ...

I find this definition very strange, as it seems to be based on syntax alone (or almost). In the rest of the course, we went on to discuss Cauchy sequences and the Cauchy criterion and what it implies:

The sequence $(U_n)_{n\in\mathbb{N}}$ converges if and only if it is a Cauchy one. This equivalence gives $\mathbb{R}$ the property of being a complete set, i.e. any Cauchy sequence of real numbers converges in $\mathbb{R}$.

My intuition then led me to think that the set of reals could be defined as the set containing all the limits of the real Cauchy suites. However, this is only an intuition and I don't know how to show this, I haven't found any result on the internet.

I wonder therefore if this intuition is good, if not (or if it is, because it is always interesting), I would like to know in what other ways we could define the reals, other than by the definition I gave above.

PS: obviously, we could define $\mathbb{R}$ in ensemblistic terms, and say that it contains all naturals, integers, rationals and irrationals, but I was looking for another definition than in terms of sets.


问:
Stuck on proof and need help to solve it

This is not homework

I do it for fun.

I am using the SD system.

Derive $L\supset T$

$L\supset\left ( C\vee T \right ).$ Assume

$\left ( \sim L\vee B \right )\wedge\left ( \sim B\vee\sim C \right ).$ Assume

As we see from the image it is valid in SL truth tree form

There does not seem to be an easy way to derive it. I figure the the easiest method is to derive a contradiction, but how to do it after the initial set up ?

$\sim\left ( L\supset T \right ).$ Assume

For "$Q$" and "$\sim Q$" from #2, we could use $B$ and $\sim B,$ but how to separate them so they could be used ?

Any help would greatly appreaciated.

答: Hint: Start with your given assumptions. Assume L. Use $P\to Q ~\equiv \neg P \lor Q$ to make a chain of implications to obtain T from L.


问:
Meaning of "defined"
What are the precise meanings of terms "defined", "well defined" and "undefined", etc.? We can't define what "defined" means since then we would run into circular definitions. (If definitiveness is established in terms of something else like "existence", then again we face the same problem.)

答:
Mathematics is not about what "define" means in English or a natural language - that is a subject for philosophy or for the study of language. But we can use natural language to explain what it means to define something in mathematics.

The most common type of definition in mathematics says that any object with a certain collection of properties is given a certain name. For example a polygon with three corners is named a triangle. These definitions are trivial in the sense that you could replace the defined word with its definition, and although it might be inconvenient there would be no formal loss of meaning.

The term well defined is used in some settings where we want to make a definition, but there is a worry that the "definition" is faulty. It is used most often when we define a function on an entire equivalence class by looking at a representative of the equivalence class; in this case we have to argue that the function gives the same value no matter which representative is chosen, and we then say that the function is "well defined".

Title: Meaning of "defined"


问:
Why scipy.stats.chi2_contingency computes incorrect Pearson's chi-squared statistic?
I computed Pearson's chi-squared statistics for the contingency table bellow.

Here are the expected values.

Method 1 (using SciKit)

```<code>from scipy.stats import chi2_contingency
from scipy.stats import chi2

table = [[21,2], [14, 0]]
stat, p, dof, expected = chi2_contingency(table)
print(stat)
</code>```

Returns

```<code>0.14814773735581194
</code>```

Method 2 (directly)

```<code>G3 = 35*23/37
H3 = 2*23/37
G4 = 35*14/37
H4 = 2*14/37
(21-G3)**2/G3+(2-H3)**2/H3+(14-G4)**2/G4+(0-H4)**2/H4
</code>```

Returns

```<code>1.2869565217391306
</code>```

I expected the returned values to be equal. Why do they differ?

答:
There are many slightly different implementations of chi-squared tests within and among various kinds of software?

Here are four tests in R, using your table:

```<code>TBL = matrix(c(21,2,14,0), byrow=T, nrow=2)
TBL
      [,1] [,2]
 [1,]   21    2
 [2,]   14    0
</code>```

Default chi-squared test in R, using Yates' continuity
correction. It gives an error message because counts
in some cells are too small for the chi-squared statistic
to have approximately a chi-squared distribution.

```<code> chisq.test(TBL)

        Pearson's Chi-squared test
        with Yates' continuity correction

 data:  TBL
 X-squared = 0.14815, df = 1, p-value = 0.7003

 Warning message:
 In chisq.test(TBL) :
  Chi-squared approximation may be incorrect

chisq.test(TBL, cor=F)
</code>```

Without the continuity correction: Another error message
because of low counts.

```<code>        Pearson's Chi-squared test

data:  TBL
X-squared = 1.287, df = 1, p-value = 0.2566

Warning message:
In chisq.test(TBL, cor = F) :
  Chi-squared approximation may be incorrect
</code>```

Simulation to give a more reliable P-value:

```<code>chisq.test(TBL, sim=T)

        Pearson's Chi-squared test
        with simulated p-value
        (based on 2000 replicates)

data:  TBL
X-squared = 1.287, df = NA, p-value = 0.5042
</code>```

Fisher exact test uses a hypergeometric distribution:

```<code>fisher.test(TBL)

    Fisher's Exact Test for Count Data

data:  TBL
p-value = 0.5165
alternative hypothesis:
  true odds ratio is not equal to 1
95 percent confidence interval:
 0.000000 8.792391
sample estimates:
odds ratio
     0
</code>```

None of the tests gives a P-value significant at the 5% level.
With only two subjects in 'Senzitvne` outcome, you have
no chance for a significant between groups.

Body:
There are many slightly different implementations of chi-squared tests within and among various kinds of software?

Here are four tests in R, using your table:

TBL = matrix(c(21,2,14,0), byrow=T, nrow=2)
TBL
      [,1] [,2]
 [1,]   21    2
 [2,]   14    0

Default chi-squared test in R, using Yates' continuity
correction. It gives an error message because counts
in some cells are too small for the chi-squared statistic
to have approximately a chi-squared distribution.

 chisq.test(TBL)

        Pearson's Chi-squared test
        with Yates' continuity correction

 data:  TBL
 X-squared = 0.14815, df = 1, p-value = 0.7003

 Warning message:
 In chisq.test(TBL) :
  Chi-squared approximation may be incorrect

chisq.test(TBL, cor=F)

Without the continuity correction: Another error message
because of low counts.

        Pearson's Chi-squared test

data:  TBL
X-squared = 1.287, df = 1, p-value = 0.2566

Warning message:
In chisq.test(TBL, cor = F) :
  Chi-squared approximation may be incorrect

Simulation to give a more reliable P-value:

chisq.test(TBL, sim=T)

        Pearson's Chi-squared test
        with simulated p-value
        (based on 2000 replicates)

data:  TBL
X-squared = 1.287, df = NA, p-value = 0.5042

Fisher exact test uses a hypergeometric distribution:

fisher.test(TBL)

    Fisher's Exact Test for Count Data

data:  TBL
p-value = 0.5165
alternative hypothesis:
  true odds ratio is not equal to 1
95 percent confidence interval:
 0.000000 8.792391
sample estimates:
odds ratio
     0

None of the tests gives a P-value significant at the 5% level.
With only two subjects in 'Senzitvne` outcome, you have
no chance for a significant between groups.


问:
Cauchy product of alternating geometric series
I want to prove that for  $ -1\lt q_1 \le q_2 \lt 1 $ then the following holds:

$$\sum_{n=0}^{\infty}(-1)^n \sum_{k=0}^n q_1^k \cdot q_2^{n-k} = \frac1{(1+q_1)(1+q_2)}$$

By the first inequality, $ -1\lt q_1 \le q_2 \lt 1 \Rightarrow \vert q_1 \vert \lt 1, \vert q_2 \vert \lt 1$ then $\sum_{n=0}^{\infty}(-q_1)^n = \frac1{1+q_1}$ and $\sum_{n=0}^{\infty}(-q_2)^n = \frac1{1+q_2}$

Now for the Cauchy product, by definition, $$\sum_{n=0}^{\infty}(-q_1)^n \sum_{n=0}^{\infty}(-q_2)^n = \sum_{n=0}^{\infty} \sum_{k=0}^n (-q_1)^k \cdot (-q_2)^{n-k} = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n (q_1)^k \cdot (q_2)^{n-k}$$

Now, by Merten's, since both series $\sum_{n=0}^{\infty}(-q_1)^n$ and $\sum_{n=0}^{\infty}(-q_2)^n$ are absolutely convergent to $\frac1{1+q_1}$ and $\frac1{1+q_2}$ respectively, we have that their Cauchy product will converge to $\frac1{1+q_1} \cdot \frac1{1+q_2}$

Ergo,

$$\sum_{n=0}^{\infty}(-1)^n \sum_{k=0}^n q_1^k \cdot q_2^{n-k} =\frac1{1+q_1} \cdot \frac1{1+q_2}= \frac1{(1+q_1)(1+q_2)}$$

Is there anything missing on my proof?

答:
Two of your $q_1^n$ should instead be $q_1^k$.  Also,
$$\sum_{n=0}^{\infty} \sum_{k=0}^n (-q_1)^k \cdot (-q_2)^{n-k} = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n q_1^k \cdot q_2^{n-k}$$
because you have already factored out $(-1)^n$.


问:
What can you say about groups if every homomorphism is trivial?
Suppose we have two groups $G,H$, and the premise is that every homomorphism from $G$ to $H$ is trivial.

This certainly can happen when either of $G,H$ is trivial, but is it possible that neither of $G,H$ are trivial?

答:
Yes.

Try $\Bbb Z_2$ and $\Bbb Z_3$.

Body:
Yes.

Try $\Bbb Z_2$ and $\Bbb Z_3$.


答:
You have very good instincts.  There are a couple of points to note.  First, your teacher's definition is not in fact well-defined, because two different sequences may end up defining the same number; e.g., $1=0.999 \ldots$.

Second, it is indeed possible to use Cauchy sequences to define the reals.  In fact, it's a standard method of doing so.  However, it's a little more intricate than you may realize.

First you define the natural numbers.  Then you use natural numbers to define the integers -- an integer is an equivalence class of ordered pairs of natural numbers under the equivalence relation $\forall a, b, c, d \in \Bbb N~((a, b) \sim (c, d) \iff a+d = b+c)$  (Think of the ordered pair $(a, b)$ as $a-b$.)

Then you use the integers to define the rational numbers:  $\forall, x, z \in \Bbb Z, y, w \in \Bbb Z \setminus \{ 0 \}~((x, y) \approx (z, w) \iff xw=yz)$.  You can think of the ordered pair $(x, y)$ as $\frac xy$.  However, you need to confirm this proposed definition is in fact well defined:  In other words, if $x_1, x_2$ are different ordered pairs of natural numbers such that $x_1 \sim x_2$, and $y_1, y_2$ are different ordered pairs of natural numbers such that $y_1 \sim y_2$, then you need to confirm $(x_1, y_1) \approx (x_2, y_2)$.

Once you have the rational numbers, you can define Cauchy sequences of rational numbers.  There are lots of different Cauchy sequences that converge to the same number, so you need to define an equivalence relation among Cauchy sequences.  One that works is that two Cauchy sequences are equivalent if the "interleaved" sequence also is Cauchy.

The set of equivalence classes under this equivalence relation is the set of real numbers.


问:
0-manifold - final step
$f$ is a Lefschetz map on a compact manifold X. And I need to show the Lefschetz fixed point is isolated.

I proved that the graph of f is transversal to the diagonal inside $X \times X$, then I don't know how to proceed from here. Thank you very much for your help.

Suppose $x_0$ is a Lefschetz fixed point. Then we know that given 1 is not an eigenvalue of $df$ at $x_0$, take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood, think of $f$ as a map from open ball in $\mathbb{R}^n$ (say $B$), to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$ such that $f(0)=0.$ Then $det(df - id)(0)\neq 0$. Hence the preimage of $f-id$ is a local diffeomorphism.

But how can I show that it is 0-manifold? (Then 0-manifold has discrete topology, so $x_0$ is isolated.) I feel there is some trivial fact leads me to the conclusion, but I couldn't get it. Would someone be able to help point out? Thank you!

答: If $g$ is a diffeomorphism then $g$ is bijective in that small neighborhood. If $f$ has another fixed point a in that neighborhood then $g(a)=0$. This will contradict the bijcetivity of $g$. Therefore the Lefschetz fixed point is a 0-manifold. In other word, isolated.</p>

Title: 0-manifold - final step


问:
Why are those statements about ideals true?

(a) If $f\::\:R\:\rightarrow \:T$ is an epimorphism, and $I\subset R$ is an ideal, then $f\left(I\right)$ is an ideal in T

(b) If I take an ideal $J\supset I$, and $\pi :\:R\:\rightarrow \:R/I$ is an epimorphism, then $\pi \left(J\right)$ is also an ideal

(c) The function $\pi$ is a bijection which retains the inclusion relation

(d) $\left\{J\supset I\::\:J\:\in Max\:R\right\}\ni L\:\rightarrow \:\pi \left(L\right)\in \:\left\{M\:\subset R/I\::\:M\:\in \:Max\:R/I\right\}$ is also a bijection which retains the inclusion relation (the same goes for Spec R and Spec R/I)

(e) $I \in Max\:R\:\Leftrightarrow \:\pi \left(I\right)=\left(0\right)\in Max R/I \Leftrightarrow  R/I\text{ is a field}$

My approach to that is:

(a) and (b) essentially describe the same problem because we are dealing with an epimorphism and this can be proved by just writing out the properties of ideals and using the homomorphism of f and $\pi$.

What I don't understand though is (c), why should $\pi$ be a bijection if we defined it as an epimorphism which is a surjection, but no injection? Also what does it mean to "retain the inclusion relation" (that's translated from my mother tongue so maybe it's different in English)

I'd really like to see how we should prove (d), I am fairly confident that we should use the homomorphism properties of $\pi$ but how?

I kind of understand the proof of (e), because you use the fact that if I is a maximal ideal then R/I is a field, but why is $\pi \left(I\right)=\left(0\right)\in Max R/I$? That's something I don't quite see

答:
For the purposes of part (a), we need to be clear that we're referring to a surjective ring homomorphism. The term "epimorphism" generally does not mean "surjective ring homomorphism" - this is not standard terminology. But it's the OP's definition, so I'll go with it.

Just to make things clearer, I'm going to write $Im_\pi(S)$ for the image of $S$ under $\pi$ (what you're calling $\pi(S)$).

Part (c) says that $Im_\pi$ is a bijection between ideals of $R / I$ and ideals of $R$ which contain $I$. It does not say that $\pi$ itself is a bijection.

"Maintaining the inclusion relation" means that $Im_\pi(J) \subseteq Im_\pi(K)$ iff $J \subseteq K$ for all ideals $J, K$ such that $I \subseteq J$ and $I \subseteq K$.

Regarding (d), we prove it as follows:

Suppose that $L$ is a maximal ideal of $R$ and $I \subseteq L$. Then $Im_\pi(L)$ is an ideal of $R / I$. Suppose we have an ideal $Z \subseteq R / I$ such that $Im_\pi(L) \subseteq Z \subsetneq R/I$. Then we can write $Z = Im_\pi(Y)$ for some ideal $Y$ such that $I \subseteq Y$. Then we have $Im_\pi(L) \subseteq Im_\pi(Y) \subsetneq Im_\pi(R)$, and since $Im_\pi$ preserves the $\subseteq$ relation in both directions, we have $L \subseteq Y \subsetneq R$. Then $Y = L$. Then $Z = Im_\pi(L)$. Thus, we see that $Im_\pi(L)$ is a maximal ideal.

Now suppose that $L$ is an ideal of $R$ such that $I \subseteq L$, and suppose that $Im_\pi(L)$ is a maximal ideal. Suppose we have $L \subseteq K \subsetneq R$ for some ideal $K$. Then we have $Im_\pi(L) \subseteq Im_\pi(K) \subsetneq Im_\pi(R) = R / I$, since $Im_\pi$ preserves the $\subseteq$ relation and is injective. Then $Im_\pi(K) = Im_\pi(L)$, since $Im_\pi(L)$ is a maximal ideal. Then $K = L$, since $Im_\pi$ is injective.

Thus, we see that $Im_\pi$ restricts to a bijection $\{L$ a maximal ideal of $R$: $I \subseteq L\} \to \{K$ a maximal ideal of $R / I\}$. Since $Im_\pi$ preserves $\subseteq$ in both directions, so does the restriction.

For proving (d) with respect to prime ideals:

Suppose $I \subseteq L$, $L$ is a prime ideal. Now suppose we have $xy \in Im_\pi(L)$. Then we write $x = \pi(a)$, $y = \pi(b)$. Then we have $\pi(ab) \in Im_\pi(L)$. Then we can write $ab = \ell + i$ where $\ell \in L$ and $i \in I$. But we see that since $I \subseteq L$, we have $\ell + i \in L$. Then $ab \in L$. Then $a \in L$ or $b \in L$. Then $\pi(a) \in Im_\pi(L)$ or $\pi(b) \in Im_\pi(L)$. Then $Im_\pi(L)$ is prime.

Now suppose $I \subseteq L$, $L$ is an ideal, and $Im_\pi(L)$ is prime. Consider $a, b$ such that $ab \in L$. Then $\pi(a) \pi(b) \in Im_\pi(L)$. Then $\pi(a) \in Im_\pi(L)$ or $\pi(b) \in Im_\pi(L)$. WLOG, take $\pi(a) \in Im_\pi(L)$. Then we can write $\pi(a) = \pi(\ell)$ for some $\ell \in L$. Then $a = \ell + i$ for some $i \in I$. And $i \in I \subseteq L$, so $\ell + i \in L$. Then $a \in L$. Thus, $L$ is prime.

(e) follows immediately from (d). By part (d), $I$ is a maximal ideal iff $Im_\pi(I)$ is. And it's easy to show that $(0)$ is a maximal ideal of a ring exactly when the ring is a field.


问:
construction of a a markov chain from two markov chain, doubt
Could anyone help me to understand Remark 6.2 on page 190 by some example?

http://janroman.dhis.org/finance/Books%20Notes%20Thesises%20etc/Probability%20Theory/PROB.ch6.pdf

One way to generate markov processes with multiple invariant measures is to start with two markov processes with transition probabilities $ \pi_1$,  $ \pi_2$,

$$\pi(x,A)= \begin{cases}
          \pi_1(x,A\cap X_1) & ;x\in X_1,\\
          \pi_2(x,A\cap X_2) & ;x\in X_2. \\
       \end{cases}$$

$X=X_1\cup X_2$, A is some Borel set in X, So what is $X_1, X_2$ anyway? Can $X_1=X_2=(0,1)$

Why the new chain $\pi(x, A)$ has several invariant measures, I didn't understand, could anyone be of little more explicit with some example, please? Thanks.

Can I say from this example that if I switch randomly two Markov chains with a unique invariant probability measure, the resulting chain may not possess a unique invariant measure?

答:
$X_1$ and $X_2$ are the state spaces of the two Markov processes we're starting with; they should be disjoint for this example to work.

Let's take a discrete example: $X_1 = \{1,2\}$ and $X_2 = \{3,4\}$ where $\pi_1(1,1) = \pi_1(1,2) = \pi_1(2,1) = \pi_1(2,2) = \frac12$ and $\pi_2(3,3) = \pi_2(3,4) = \pi_2(4,3) = \pi_2(4,4) = \frac12$.

Then $\pi(x,y)$ is equal to $\pi_1(x,y)$ if $x,y \in X_1$, it is equal to $\pi_2(x,y)$ if $x,y \in X_2$, and is equal to $0$ otherwise. If we start in $X_1$, the combined random process is going to stay in $X_1$; if we start in $X_2$, it will stay in $X_2$.

The two processes have invariant measures $\mu_1(1) = \mu_1(2) = \frac12$ and $\mu_2(3) = \mu_2(4) = \frac12$, which we can extend to measures on $X_1 \cup X_2$ by setting $\mu_1(3) = \mu_1(4) = 0$ and $\mu_2(1) = \mu_2(2) = 0$. These are still invariant measures in the combined process; in general, if we have $\mu(1) = \mu(2) = a$ and $\mu(3) = \mu(4) = \frac12 - a$, then $\mu$ is an invariant measure.


问:
Change of variables - 1D integration
I have the following integral
$$
\int dr_2 \frac{\partial}{\partial r_1} f(|r_1-r_2|)
$$
and I change variables by
$$
r = r_1-r_2, \\ dr = -dr_2, \\ \frac{\partial}{\partial r_1}=\frac{\partial r}{\partial r_1}\frac{\partial}{\partial r}
= 1 \frac{\partial}{\partial r}
$$
to get
$$
-\int dr \frac{\partial}{\partial r} f(|r|)\,.
$$
Is this correct or did I transform $\partial/\partial r_1$ incorrectly?  If it's incorrect, how can I do the change of variables correctly? Thanks

I think the correct result  should be
$$
\int dr \frac{\partial}{\partial r} f(r),
$$ but what I have is incorrect by a minus sign and absolute value sign, so perhaps I am making the change of variables incorrectly due to the absolute value sign in $f$, or due to how I'm transforming $\partial/\partial r_1$.

I don’t think the answer posted below is correct so a correct solution would be greatly appreciated.

答:
Lemma
$$
\begin{align}
\frac{\mathrm{d}h(x)}{\mathrm{d}g(x)} &= \frac{\mathrm{d}h(x)}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}g(x)}\\\\
&= \frac{\mathrm{d}h(x)}{\mathrm{d}x}:\frac{\mathrm{d}g(x)}{\mathrm{d}x}\\\\
&= \frac{\frac{\mathrm{d}h(x)}{\mathrm{d}x}}{\frac{\mathrm{d}g(x)}{\mathrm{d}x}}
\end{align}
$$
How you solved it

You set  $g(r)=r+r_2$ and $h(r)=f(\lvert r \rvert)$.
So by the Lemma:
$$\frac{\partial}{\partial(r+r_2)}f(\lvert r\rvert)=\frac{\frac{\partial f(\lvert r\rvert)}{\partial r}}{\frac{\partial(r+r_2)}{\partial r}}=\frac{\partial f(\lvert r\rvert)}{\partial r}.$$
The negative sign comes from the new differential.

How the book solved it

They set $r=\lvert r_1-r_2\rvert$. We have to solve for $r_1$:

Case 1: $r_1\ge r_2$. That means $r_1=r+r_2$.

Case 2: $r_1 < r_2$. That means $r_1=r_2-r$.

Case 1 is the exact same case we solved with your method. Case 2 is very similar. By our Lemma we have
$$\frac{\partial}{\partial(r_2-r)}f(r)=\frac{\frac{\partial f( r)}{\partial r}}{\frac{\partial(r_2-r)}{\partial r}}=-\frac{\partial f( r)}{\partial r}.$$

Here we have a negative expression and in Case 1 we have a positive expression. These two will evaluate to positve values if multiplying by the differential.

Title: Change of variables - 1D integration


问:
The quotient group of a ring and the intersection of an ideal is isomorphic to a subring of the product of the quotient ring and each ideal
Is it true that:

Let $R$ be a ring and let $I, J$ be ideals of $R$. Then for some subring of $S \subseteq R/I \times R/J$ we have
$$
R/(I \cap J) \cong S.
$$

This would be equivalent to the Chinese Remainder Theorem for non-coprime ideals. I've tried constructing a function $\phi: R/(I \cap J) \to R/I \times R/J$ and showing its injective.

答:
Proof

We construct the natural homomorphism
$
\varphi: R \to R/I \times R/J
$
such that
$$
\varphi(x) = (x+I,x+J).
$$
We know that
$$
\ker \varphi = \{x\in R: x\in I, x\in J\} = I \cap J.
$$
Since the $\text{im } \varphi$ is a subset $S$ of $R/I \times R/J$ so by the first isomorphism theorem
$$
R/(I\cap J) \cong S.
$$


问:
If $R$ is a ring and $a,b \in R$. Show that $a(-b)+2a\cdot2b=3ab$

Assume that $R$ is a ring and $a,b \in R$. Show that $$a(-b)+2a\cdot2b=3ab.$$ Keep an eye for what definitions of a ring you use in each step.

I'm trying to learn about rings from a book on algebraic structures and there's a few definitions on the operations. I have that $a(-b)=-(ab)$ so $$a(-b)+2a\cdot2b=-(ab)+2a\cdot2b.$$

It's tempting to group the last two terms as $4ab$, but I feel like there is some catch here that I'm not seeing? This would result in $$-(ab)+4ab=3ab$$ that seems to be working according to the definitions of  a ring?

答:
Going purely by the definition of a ring we get the following:
$$\begin{align}
a(-b)+2a\cdot2b &= a(-b)+(a+a)(b+b) \tag{definition}\\
&=a(-b)+a(b+b)+a(b+b) \tag{right distributivity}\\
&=a(-b)+ab+ab+ab+ab \tag{left distributivity}\\
&=a((-b)+b)+ab+ab+ab \tag{left distributivity}\\
&=a(0)+ab+ab+ab \tag{additive inverse}\\
&=a(0+b)+ab+ab \tag{left distributivity}\\
&=ab+ab+ab \tag{additive identity}\\
&=3ab \tag{definition}
\end{align}$$
It's a bit excessive, but does purely rely on the basic axioms and definitions.


问:
Differentiability Using Little Oh Notation
I'm trying to get a deeper understanding of the derivative of a function. I have been reading from the following page:

I have been thinking about why this is an equivalent statement to the original, but I've been having trouble with it.

What I've tried doing so far is by assuming $h \in \mathbb{R}$, and then $\frac{f(x+h) - f(x)}{h}$ is the slope of the secant line from $(x, f(x))$ and $(x+h, f(x+h))$, which could be a decent approximation based on how close $h$ is to zero. But after that I'm having trouble seeing how the little oh gets involved.

Can someone prove to me why the definition of differentiability and the one involving the little-oh notation are equivalent?

答:
By definition of differentiable function at $x$ you get
$$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} $$

Now if you move the number $f'(x)$ on the other side and inside the limit ( why you can do that?) obtaining
$$\lim_{h \to 0}\frac{f(x+h)-f(x)-hf'(x)}{h}= 0 $$

but this is the definition of a function being $o(h)$ so there exists a number $f'(x)$ such that
$$o(h) = f(x+h)-f(x)-hf'(x)$$

Rearranging you get the desired expression.  (Note that all passages are biconditional)


问:
Solution to $y' + \frac xy = 1$
$y' + \frac xy = 1$

I've identified it as a homogeneous differential equation and transformed it to $xv' + v = 1-\frac 1v$ (where $v = \frac yx$) but am stuck here.

答:
$$
\begin{aligned}
y' + \frac{x}{y} = 1 &\Leftrightarrow
\left|
\begin{aligned}
y(x) &= z(x)\cdot x \\
y'(x) &= z'(x)x + z(x)
\end{aligned}
\right| \Leftrightarrow \\
&\Leftrightarrow z'x + z + \frac{x}{zx} = 1 \Leftrightarrow \\
&\Leftrightarrow z'x + z + \frac{1}{z} = 1 \Leftrightarrow \\
&\Leftrightarrow z' = -\frac{1}{x}\frac{z^2-z+1}{z}\Leftrightarrow \\
&\Leftrightarrow z' = -\frac{1}{x}\frac{z^2-z+1}{z}\Leftrightarrow \\
&\Leftrightarrow \frac{zdz}{z^2-z+1} = -\frac{dx}{x} \Leftrightarrow \\
&\Leftrightarrow \int\frac{zdz}{z^2-z+1} = -\int\frac{dx}{x}
\end{aligned}
$$

Left-hand side integral:

$$
\begin{aligned}
\int\frac{zdz}{z^2-z+1} &= \frac{1}{2}\int\left(\frac{2z-1}{z^2-z+1} + \frac{1}{z^2-z+1}\right)dz = \\
&= \frac{1}{2}\left(\int\frac{d(z^2-z+1)}{z^2-z+1}+\int\frac{dz}{\left(z-\frac{1}{2}\right)^2 +\left(\frac{\sqrt{3}}{2}\right)^2}\right) = \\
&= \frac{1}{2}\left(\text{ln}\left(z^2-z+1\right) + \frac{2}{\sqrt{3}}\arctan\left(\frac{2\left(z-\frac{1}{2}\right)^2}{\sqrt{3}}\right)\right).
\end{aligned}
$$

Right-hand side integral:

$$
-\int\frac{dx}{x} = -\text{ln}(x) + C.
$$

Finaly ($z = \frac{y}{x}$),

$$
\frac{1}{2}\text{ln}\left(\frac{y^2}{x^2}-\frac{y}{x}+1\right) + \frac{1}{\sqrt{3}}\arctan\left(\frac{2\left(\frac{y}{x}-\frac{1}{2}\right)^2}{\sqrt{3}}\right) = -\text{ln}(x) + C.
$$

Title: Solution to $y' + \frac xy = 1$


问:
Proving a result about a graph with $p$ connected components
I am trying to prove the following result: let $G=(V,A)$ be a graph with $p$ connected components. Prove that

$$|V|-p\leq |A|\leq \binom{|V|-p+1}{2}$$

I have shown that if $G$ is a connected graph with $n$ vertices and $m$ edges, then $m\geq n-1$.

What I have tried to do is: as we have shown above, $|A|\geq |V|-1$, as there is at least one connected component (the graph $G$ itself), $|V|-p$ will also be true!

$$\binom{|V|-p+1}{2}=\textrm{ number of edges of the graph } K_{|V|-p+1}$$

$$|V|-p\leq |A|$$

It is clear that at most, $\binom{|V|-p+1}{2}$ can be equal to $\binom{n}{2}=m(K_n)$ since there is at least one connected component, the graph $G$ itself.

I don't know how to continue.

答:
HINT: You know that $|A|\ge|V|-1$ for a connected graph, but you cannot apply that result to $G$ here unless $p=1$, so that $G$ is connected. You can, however, prove the first inequality by applying that result to each of the $p$ components of $G$.

You know that $\binom{|V|-p+1}2$ is the number of edges in $K_{|V|-p+1}$, but that graph is not $G$ (unless $p-1$). In order to apply that result here, you will have to prove that $G$ cannot have any more edges than the graph $K_{|V|-p+1}$ has. If you add $p-1$ isolated vertices to $K_{|V|-p+1}$, you get a graph $G_0$ with $|V|$ vertices and $p$ connected components. You can prove the second inequality by showing that among all graphs with $|V|$ vertices and $p$ connected components, $G_0$ is the one with the largest number of edges; I’ll sketch one way to do.

Let $G$ be any graph with $|V|$ vertices and $p\ge 2$ connected components. Let $C_1$ and $C_2$ be two of the components of $G$, with $n_1$ and $n_2$ vertices, respectively; $C_1$ and $C_2$ have altogether at most $\binom{n_1}2+\binom{n_2}2$ edges. (Why?) Replace $C_1$ by an isolated vertex and $C_2$ by $K_{n_1+n_2-1}$, and show that the resulting graph has at least as many edges as $G$ and still has $|V|$ vertices and $p$ components. Show that by performing this operation a finite number of times you can convert $G$ to $G_0$, and conclude that $G$ has at most as many edges as $G_0$.


答: First, remember that there are $(m,n)$ number of homomorphisms between $\mathbb{Z}_n$ and $\mathbb{Z}_m$, see Find the number of homomorphisms between $\mathbb{Z}_m$ and $\mathbb{Z}_n$. Hence if $n$ amd $m$ are coprime, there is only one homomorphism. So it is not necessary for both groups to be trivial.

Title: What can you say about groups if every homomorphism is trivial?

Body:
Suppose we have two groups $G,H$, and the premise is that every homomorphism from $G$ to $H$ is trivial.

This certainly can happen when either of $G,H$ is trivial, but is it possible that neither of $G,H$ are trivial?