问:
In a Locally Compact Hausdorff and second countable space every compact subset is a $G_\delta$ set
Let $X$ be a second countable LCH (Locally Compact Hausdorff) then every compact subset of $X$ is a $G_{\delta}$ set.

A set $E$ is $G_{\delta}$ if is a countable intersection of open sets.

I have two questions,

How to prove this result,

I am also wondering if that would be true for the closed sets in this space, or what would be necessary to be true for closed sets.

Edit:

I posted a resolution, which I found very interesting, of question 1.

答:
Let $\mathcal{B}$ be a countable base for the topology of $X$, and let $K\subseteq X$ be a compact. We have that $X$ is Hausdorff, so for each $x\in K^c$ and $y\in K$, there is an open $U_{xy}$ such that $y\in U_{xy}$ but $x\notin U_{xy}$. Moreover, there exists $V_xy \in \mathcal{B}$ such that $y\in V_{xy}\subseteq U_{xy}$

Since $K$ is compact, for each $x\in K^c$ there is a finite set $Y_x \subseteq K$ such that,

$$K\subseteq \bigcup_{y\in Y_x} V_{xy} \qquad (*)$$

We have,

$$K=\bigcap_{x\in K^c}\left(\,\bigcup_{y\in Y_x}V_{xy} \right)$$
since $x\notin V_{xy}$ for all $x\in K^c$ and $y\in Y_x$.

Since the set,
$$\left\{ \bigcup_{U\in \mathcal{A}} U : \mathcal{A}\subseteq \mathcal{B} \text{ is finite}\right\}$$

is a countable collection of open sets, to demonstrate this, just prove that the set of all finite subsets of $\mathbb{N}$ is countable.

Therefore the intersection in (*) is countable, because, although the intersection is indexed by a possible uncontable quantity, there is only an countable number of distinct sets, by the previous result.

This shows that $K$ is $G_\delta$.


问:
Increasing sequence of a random variable
Could anyone give me an example of an increasing sequence of random variable that converges in probability to a random variable but their expectation does not? Thanks

I tried to formulate this example, but not sure it is increasing sequence.

$$X_n = \begin{cases}
n^2 \quad \ \ \text{if}\ \ 0\le \omega\le \frac{1}{n}\\
0 \quad \quad\text{if}\ \ \frac{1}{n}< \omega \le 1
\end{cases}$$

Thanks.

答:
As pointed out in the comments above your sequence is not increasing.

If $X_n$ increases to $X$ and all these random variables have finite expectation then $EX_n \to EX$. This follows by an application of Montone Convergence Theorem to the sequence $(X_n-X_1)$.

[Your sequence cannot be increasing because of this result].

ContentLicense: CC BY-SA 4.0


问:
Distribution of the negative sign to the components of the vectors in a vector space
Is there any vector space $V$ such that $\exists A \in V$ where $A=(a_1,a_2, ...,a_n)$ and $-A \ne (-a_1,-a_2, ...,-a_n)$?

In case failed my translation from English to math:

I am basically just trying to see if there exist a counter example for the proposition that you can just distribute the negative sign to the components of any vector in any vector spaces regardless of the field, the definition of addition and the definition of scalar multiplication for that vector space.

答:
No. For any $a\in\mathbb{R}^n$ where $a=(a_1,\cdots,a_n)$, by definition, $-a=(b_1,\cdots,b_n)$ is a vector such that $a+(-a)=0$. It follows by the definition of vector addition that $b_k=-a_k$ for each $k$.

In general, for an abstract vector space $V$, you can only write $v$ as an n-tuple when you have an ordered basis of $V$.


问:
A result involving dimension of inverse image (Vector spaces)
Let $V$ be a vector space, $T:V \to V$ a linear operator and consider $W \subset V$ a subspace. I was thinking about the follow inequality:

$$\dim T^{-1}(W) \leq \dim \ker T + \dim W.$$

How can I prove that? I've tried this way:

$$\dim W + \dim \ker T = \dim W + \ker T - \dim W \cap \ker T \geq \dim W+ \ker T.$$ Then if $T^{-1}(W) \subset W + \ker T$, the result follows. But I guess it is not correct.

Can you help me?

答:
Clearly, $\ker T\subset T^{-1}(W)$. Let $\{e_1,\ldots,e_k\}$ be a basis of $\ker T$ and extend it to a basis $\{e_1,\ldots,e_n\}$ of $T^{-1}(W)$. Then $\bigl\{T(e_{k+1}),\ldots,T(e_n)\bigr\}$ is a linearly independent subset of $W$, because if there were scalars $a_{k+1},\ldots,a_n$, not all of which are equal to $0$, such that $a_{k+1}T(e_{k+1})+\cdots+a_nT(e_n)=0$, then $T(a_{k+1}e_{k+1}+\cdots+a_ne_n)=0$, which means that $a_{k+1}e_{k+1}+\cdots+a_ne_n\in\ker T$, which is impossible, by the way that $e_{k+1},\ldots,e_n$ were chosen. Since $\bigl\{T(e_{k+1}),\ldots,T(e_n)\bigr\}$ is a linearly independent subset of $W$ with $n-k$ elements, $n-k\leqslant\dim W$.

Therefore\begin{align}n&=\dim T^{-1}(W)\\&=k+(n-k)\\&\leqslant\dim\ker T+\dim W.\end{align}


问:
Please help with this probalility problem.
Vince buys a box of candy that consists of six chocolate pieces, four fruit pieces and two mint pieces.  He selects three pieces of candy at random without replacement.

Calculate the probability that the first piece selected will be fruit flavored and the other two will be mint.

Calculate the probability that all three pieces selected will be the same type of candy.

答: Hint: For the first part, employ negative binomial distribution. For the second part, employ hypergeometric distribution.

ContentLicense: CC BY-SA 3.0


问:
The sum of mutually singular measures
I have been trying to make an exercise related to mutually singular measures. Namely the following:

Exercise Let $\mu$ be a positive measure and $\nu_1, \nu_2$ be arbitrary measures all defined on the same measurable space. If $\nu_1 \perp \mu$ and $\nu_2 \perp \mu$, then $\nu_1 +
\nu_2 \perp \mu$.

Here I use the following definition of mutually singular

Definition (Mutually Singular) We call $\mu$ and $\nu$ mutually singular if there exists disjoint measurable sets on $E$ and $F$ such that $\nu(A) = \nu(A \cap E)$ and $\mu(A) = \mu(A \cap F)$.

Note: This phenomena of a set $E$ having the property that $\nu(A \cap E) = \nu(A)$ for all measurable $A$'s is dubbed as the measure $\nu$ being concentrated on $E$.

Naturally, I would begin by writing out what I have:
$\exists E,F, G,H \in \Sigma$ such that
\begin{align}
\nu_1(A) = \nu_1(A \cap E)\  \text{ and }\  \mu(A) = \mu(A \cap F)\ \text{ for all $A \in \Sigma$}.\\
\nu_2(B) = \nu_2(B \cap G)\  \text{ and }\  \mu(B) = \mu(B \cap H)\ \text{ for all $B \in \Sigma$}.\\
\end{align}
I have tried writing out various combinations of $E$ and $G$, but to no avail. From here I have tried to find some hints on the internet and this StackExchange post seems to have a solution to this problem. But the answer uses another definition, so from here I was wondering two things:

Is there a way to directly prove it using this definition?

Is there a reference to see how the definition provided in the answer of the StackExchange post and this definition are the same. Maybe after seeing how those definitions are equivalent, it might much more clear how to directly prove it.

Extra Attempt:
I have been searching for a reference on point 2. During this search, I found that mathworld claimed that the notion of $\nu$ being concentrated on a certain set $E$ is equivalent to saying $\nu(A) = 0$ whenever $A \cap E = \emptyset$ (Source).
It is easy to see how $\nu(A \cap E) = \nu(A)$ implies this, but the other way around I have not been able to prove. With this other definition of concentration, I might be able to more easily prove the original question.

答:
Let me give you part hint, part thought process.  There are some details to work out here, but please give it a try.

Trying to connect $E$ and $G$ is hard, because we don't have much direct connection between $\nu_1$ and $\nu_2$. I think you also have too many sets floating around, which makes things into a symbol soup.  Let's see if we can clear it up.

In principal, the idea here is this: we know that all of the mass of the measure $\mu$ is isolated from the masses of the measures $\nu_1$ and $\nu_2$. In your existing breakdown of sets ($E$, $F$, $G$, and $H$), the real difficulty you're grappling with is the fact that these were produced by some theorem factory and it is hard to see how they can relate to eachother. So, how about this: can we prove that it is possible to find just two disjoint sets, call them $U$ and $I$, such that we can write the following?
$$
\nu_1(A)=\nu_1(A\cap U),\qquad\nu_2(A)=\nu_2(A\cap U), \qquad\mu(A)=\mu(A\cap I)\tag{1}
$$
You can think of this as a sort of simultaneous decomposition of the space into a part that hosts $\mu$ and a part that hosts both $\nu_1$ and $\nu_2$, whereas you can think of what you are given as the ability to decompose the space once for $\nu_1$/$\mu$ and once for $\nu_2$/$\mu$.

IF you can prove this, then your result is easy to come by, because
$$
(\nu_1+\nu_2)(A)=\nu_1(A)+\nu_2(A)=\nu_1(A\cap U)+\nu_2(A\cap U)=(\nu_1+\nu_2)(A\cap U).
$$

So, how do we prove (1)?

Take the sets $E$, $F$, $G$, and $H$ as you defined them above.  We know that all of the mass of $\mu$ is concentrated inside $F$; but, we also know that all the mass of $\mu$ is concentrated inside of $H$.  Can you show that this implies that for any $A$,
$$
\mu(A)=\mu(A\cap(F\cap H))?
$$
If you can prove that, then $F\cap H$ is a great candidate for $I$ above.

What, then, should $U$ be?  The masses of $\nu_1$ and $\nu_2$ are restricted to $E$ and $G$, respectively; could we take $U:=E\cup G$? You need to show that this is disjoint from $I:=F\cap H$, and you need to show that the intersection property holds for both $\nu_1$ and $\nu_2$.

Title: The sum of mutually singular measures


问:
Prove that $\Phi$ is surjective. $\Phi: \ ^C(^B A) \to\ ^{C \times B}A $ where $\Phi (f)(c,b) = \big(f(c)\big)(b)$
I would like to prove that the following function...

$\Phi: \ ^C(^B A) \to\   ^{C \times B}A $ where $\Phi (f)(c,b) = \big(f(c)\big)(b)$

...is surjective. I don't believe this is useful for the task at hand, but I have proven that $\Phi$ is injective.

Any confusion with notation can be addressed here: Notation used in the proof of $^C(^B A) \approx \ ^{C \times B}A$: $\ \ \ \Phi (f)(c,b)$ and $\big(f(c)\big)(b)$

Here is what I have so far, but I am concerned that there may be a leap in my logic...or at least an unacceptable degree of informality:

Claim to Prove - $\forall g \in \  ^{C \times B}A \Big [ \exists f \in \  ^C(^B A) \ \big [\forall c \in C, b \in B \big (g(c,b)=f(c)(b) \big) \big] \Big ]$

Consider an arbitray $g' \in \  ^{C \times B}A$.

$g' \in \  ^{C \times B}A \rightarrow \forall c \in C, b \in B \Big(
\exists ! a \in A \big(g'(c,b)=a\big) \Big)$

Consider $c' \in C$ and $b' \in B$. Then we know that $g'(c',b')=a'$.

There are two families of functions relevant for this next part:

i) functions of the form... $\quad f: C \to \ ^B A$ where $\quad f \in \ ^C(^B A)$

ii) functions of the form... $\quad h: B \to \ A$ where $\quad h \in \  ^B A$

Note that for $c\in C$, $f(c)$ necessarily maps to a function of the "ii" form.

By definition of $^B A$, there exists at least one $h \in \  ^B A$ that maps $b'$ to $a'$. In other words, $\exists h  \in \  ^B A \Big (h(b')=a' \Big)$. Call one such function $h'$.

By definition of $^C(^B A)$, there exists at least one function in $^C(^B A)$ that maps $c'$ to $h'$. In other words, $\exists f  \in \  ^C(^B A) \Big (f(c')=h' \Big)$. Call one such function $f'$.

Therefore, we can see that the following is true: $f'(c')(b')=h'(b')=a'=g'(c',b')$.

I will refer to this block as Procedure 1

Now, the issue that I believe I am running into is that I have only demonstrated that a function $f$ exists that can "handle" a single arbitrary pair of elements from $C$ and $B$. However, what I need to show is that there exists an $f$ that can handle all pairs of elements from $C$ and $B$.

Essentially, I believe I proved:

$\forall g \in \  ^{C \times B}A \Big [  \forall c \in C, b \in B \big [\exists f \in \  ^C(^B A) \Big(g(c,b)=f(c)(b) \Big ) \ \big] \Big]$

Where I am stuck is whether or not the immediately above statement implies the surjective claim because of the definition of $ ^C(^B A)$. If that implication is true, I am unsure of how to formally express it.

Added thoughts:

It seems like for a given $g'$, I could repeat Procedure 1 as many times as necessary using all of the different combinations of elements from $C$ and $B$. Necessarily, this would create subsets of functions from $ ^C(^B A)$ that satisfy the criteria used in Procedure 1.

For example, I would have the following sets:

$\{\text{functions in } ^C(^B A) \text{ that satisfy } c_1, b_1\}, \{\text{functions in } ^C(^B A) \text{ that satisfy } c_1, b_2\},...,\{\text{functions in } ^C(^B A) \text{ that satisfy } c_2, b_1\},...$

It seems like I could take the intersection of this family of sets and if I could prove that the resulting set is non-empty, then I would demonstrate that there is a function that satisfies the necessary criteria of $\exists f \in \ ^C(^B A) \Big [ \forall c \in C, b \in B \big (g'(c,b)=f(c)(b) \big) \Big ]$

答:
You started fine by looking at an arbitrary function from $C\times B$ to $A$, but after that you got bogged down in notation and details.

Let $f$ be any function from $C\times B$ to $A$; we want to find a function $g$ from $C$ to ${^BA}$ such that $\Phi(g)=f$. That is, for each $c\in C$ the function $g$ should pick out a whole function $h_c$ from $B$ to $A$, and there is a natural way to do that. For each $c\in C$ let

$$h_c:B\to A:b\mapsto f(c,b)\,,$$

and let

$$g:C\to{^BA}:c\mapsto h_c\,.$$

This is a more general instance of something that you may already have seen in a more familiar setting. If $f:\Bbb R^2\to\Bbb R$ is a real-valued function on $\Bbb R^2$, you can define a ‘slice’ function $f_x:\Bbb R\to\Bbb R:y\mapsto f(x,y)$ for each $x\in\Bbb R$. It’s these ‘slice’ functions that you look at when you consider $f_y$, the partial derivative of $f$ with respect to $y$.

Now we need only check that $\Phi(g)=f$. This is straightforward: by definition

$$\Phi(g)(c,b)=\big(g(c)\big)(b)=h_c(b)=f(c,b)$$

for each $\langle c,b\rangle\in C\times B$, so $\Phi(g)=f$.


问:
Dimension of vector space and span
I have the following problem

a) I say that the vectors are linearly independent because they are not scalar multiples of each other.

b) This one I'm struggling with. How do I find the dimension of the vector space? Is it 3 dimension due to the fact that the vectors only have 3 coordinates? Is it so simple?

c) I'm not sure what to say here. I don't think it belongs belongs to the span. If this is true, can it be explained by the same reasoning as in a)?

答:
Answering each of your questions:

a) True, they are linearly independent (it's easily seen that they're not schalar multiples of each other). You can see this from noticing that in the second component, $a_1$ has $-1$ and $a_2$ has $1$, but the other components aren't the same with opposite symbols.

b) Since they are linearly independent, then the vector space generated by both vectors (or how it's written: $\langle a_a,a_2\rangle$) is of dimension $2$ (it would have been dimension $1$ if they were linearly dependent).

c) Since $a_1$ is not a schalar multiple of $a_2$ (because they are linearly independent), you conclude that $a\notin \langle a_2\rangle$.

As you see, all of your questions have to do with the same important question: are those vectors linealy independent? The answer to that is yes, and the following statements are concluded from there.

Title: Dimension of vector space and span


答:
B) Let $W=span(a_1,a_2)$. $\dim(W)=2$. In fact the dimension of a subaspace is the cardinality of a basis. In this case, since $a_1$ and $a_2$ are generators(by definition) of the subspace, and are linearly indpendent they are a basis

C)If $\; a_1 \in span(a_2) \Rightarrow a_1=ka_2$, which is not true.

Title: Dimension of vector space and span

Body:
I have the following problem

a) I say that the vectors are linearly independent because they are not scalar multiples of each other.

b) This one I'm struggling with. How do I find the dimension of the vector space? Is it 3 dimension due to the fact that the vectors only have 3 coordinates? Is it so simple?

c) I'm not sure what to say here. I don't think it belongs belongs to the span. If this is true, can it be explained by the same reasoning as in a)?


问:
Can we claim that $A \cap B$ or $A \cup B$ is not in $\mathcal{B}(\mathbb{R}^n))$?
This question can be posed for more general sigma algebras, but let's stick to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ for now. Suppose $A \in \mathcal{B}(\mathbb{R}^n))$, and $B_n \notin \mathcal{B}(\mathbb{R}^n))$.

Can we claim that $A \cap B$ or $A \cup B$ is not in $\mathcal{B}(\mathbb{R}^n))$?

Then what about $\bigcup_{n\in \mathbb{N}} B_n$ or $\bigcap_{n\in \mathbb{N}} B_n$; can we claim that these sets are not measurable?

(I'm asking because I don't have a proof. This is a question that has crossed my mind several times, but I've never been able to satisfactorily prove/disprove.)

答:
The answer to 1 is negative, $A\cap B$ can be measurable, for example if $A$ and $B$ are disjoint (or meet in a single point if you don't want them to be disjoint), while $A\cup B$ can be measurable as well, for example if $B\subseteq A$.

For 2 the answer is also negative, countable unions of nonmeasurable sets can be measurable, in fact even finite unions can, just pick a nonmeasurable set and its complement for example.

Edit: after reading Daniel's answer I realized I probably misread the "or" in your question. My example for question 2 however does show that both $\bigcup B_n$ and $\bigcap B_n$ can be measurable when none of the $B_n$ are, so I'll leave my answer here


答:
Hints for the first question

In how many ways can Vince select 3 pieces of candy without replacement?

In how many ways can Vince select 3 pieces of candy such that 1 is fruit flavored and the other two mint?

Title: Please help with this probalility problem.

Body:
Vince buys a box of candy that consists of six chocolate pieces, four fruit pieces and two mint pieces.  He selects three pieces of candy at random without replacement.

Calculate the probability that the first piece selected will be fruit flavored and the other two will be mint.

Calculate the probability that all three pieces selected will be the same type of candy.


答: For 1: Suppose, to the contrary, that both $A\cup B$ and $A\cap B$ were measurable.  Then $A \mathop{\triangle} B = (A\cup B) \setminus (A\cap B)$ is also measurable, where $A \mathop{\triangle} B$ denotes the symmetric difference.  Therefore, $B = A \mathop{\triangle} (A \mathop{\triangle} B)$ would also be measurable, giving a contradiction.  Thus, we conclude that if $A$ is measurable but $B$ is not measurable, then at least one of $A\cup B$ or $A\cap B$ must be non-measurable.

Body:
This question can be posed for more general sigma algebras, but let's stick to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ for now. Suppose $A \in \mathcal{B}(\mathbb{R}^n))$, and $B_n \notin \mathcal{B}(\mathbb{R}^n))$.

Can we claim that $A \cap B$ or $A \cup B$ is not in $\mathcal{B}(\mathbb{R}^n))$?

Then what about $\bigcup_{n\in \mathbb{N}} B_n$ or $\bigcap_{n\in \mathbb{N}} B_n$; can we claim that these sets are not measurable?

(I'm asking because I don't have a proof. This is a question that has crossed my mind several times, but I've never been able to satisfactorily prove/disprove.)


答:
$a)$ For the vectors to be independent, the relationship $\lambda_{1}a_1+\lambda_{2}a_2=0 $

must yield $\lambda_1=\lambda_2=0$.

You can verify that this is the case here.

$b)$ You have two linearly independent vectors. This means that any $v \in V$, where $V$ is the vector space created by $a_1, a_2$ can be written as their linear combination. That is, $v=\kappa_1a_1+\kappa_2a_2$.

Now all you have to do is (re)check the definitions found on your course notes/book to see if it holds true.. Same for $c)$.

Title: Dimension of vector space and span


问:
Linear matrix inequality given the bounds
If $A\succeq A_{\min}\succ 0$, $B\succeq B_{\min}\succ 0$, will the following be true?
$$ABA^\top \succeq AB_{\min}A^\top$$
How can I prove it?

答: We can prove by using definition of positive-definiteness. For any vector $v\in\mathbb{R}^N$ and $v\neq 0$, it's obvious that from the definition $v^\top (B - B_{\min})v \geq 0$. Since $A\succeq 0$, for any vector $x\in\mathbb{R}^N$ and $x\neq 0$, we have $A^\top x\neq 0$ because $A$ has full rank. Thus $x^\top A(B - B_{\min})A^\top x>0$. Therefore $A(B - B_{\min})A^\top\succeq 0$ and $ABA^\top\succeq AB_{\min}A^\top$.


问:
Cancelling Simple Tensors?
Let $M,N$ be R-Modules with tensor product $M\otimes N$. Let $\{m_1,\dots,m_n\}$ be a collection of non zero elements of $M$ and $\{p_1, \dots,p_n\} , \{q_1,\dots ,q_n\}$ be collections of elements of $N$. If

$$m_1\otimes p_1+\dots +m_n\otimes p_n = m_1\otimes q_1+\dots +m_n\otimes q_n$$

can we conclude that $p_i=q_i$ for $1\leq i \leq n$?

答: No, and this does not even hold in the case $n=1$. For instance, let $R=\mathbb{Z}$ and $M=\mathbb{Z}\big/2\mathbb{Z}$, $N=\mathbb{Z}$. Then $\overline{1}\in M$ is non-zero, but $\overline{1}\otimes 2k=\overline{1}\otimes 0$ for any $k\in\mathbb{Z}$.

Title: Cancelling Simple Tensors?


问:
Prove that $R_n < M_n < L_n$ ,for right, mid-point and left Riemann sum
Let $a, b \in \mathbb{R}, a<b$. Suppose that $f$ is continuous and strictly decreasing over $[a, b]$. Let $L_n$ be the left Riemann sum, $R_n$ the right Riemann sum, and $M_n$ the midpoint (Riemann) sum for $f$ on $[a, b]$ with $n ∈ \mathbb{Z}^{+}$ subintervals, respectively.

(b) Prove that $R_n < M_n < L_n$

I know how we can prove this by using a graph, but I don't know how to prove this numerically.
do we need to factor out 1/n ??

答: I think this is immediate, since $f$ is decreasing so for any $\xi_i\in (x_{i-1}, x_i)$ we then have $f(x_i)<f(\xi_i)<f(x_{i-1})$ then $$ R_n=\sum_{i=1}^nf(x_i)(x_i-x_{i-1})<M_n=\sum_{i=1}^nf(\xi_i)(x_i-x_{i-1})<L_n=\sum_{i=1}^nf(x_{i-1})(x_i-x_{i-1})$$ it is clear now? Since $f$ is Riemann integrable, all of this sums converges to $\int_a^b fdx$.


问:
Finding the interval in a random variable
I need to find a in the interval $[1,a]$ with $a>1$ of a random variable with with a continuous uniform distribution, the only thing I was given is that $E(X) = 6Var(X)$, how can I approach this problem?

$$f_X(x)  = \begin{cases}
\frac 1 {(a-1)}&\text{for $1 \leq x\leq a$}\\
0&\text{otherwise}\\
\end{cases}$$

Now by properties of variance we have:

\begin{align}
6 Var (X) = Var(\sqrt6 X)
\end{align}

But I think I'm not going anywhere with this, can anyone give me a hint on how to approach this problem?

答:
Double-check your work, good practice. Here's what I'm seeing:
\begin{align}
\text{mean} &= (1 + a)/2, \\
\text{var} &= (a - 1)^2 / 12.
\end{align}
So if you happen to know $\text{mean} = 6\, \text{var}$, it
has to be the case that $(1 + a)/2 = 6 (a - 1)^2/12$. ($a = 3$.)


问:
Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$
I'm practicing for my maths term test mainly on binomial coefficients. I can't seem to find out how to prove the following identity. Any advice?

$$ \sum\limits_{k=1}^n (-1)^{k+1} k{{n}\choose k} = 0 $$

Thanks in advance.

Body:
$$\sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}=\sum_{k=1}^n(-1)^{k+1}k\frac{n}{k}\binom{n-1}{k-1}= $$
$$=n\sum_{k=1}^n(-1)^{k+1}\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}(-1)^{k+2}\binom{n-1}{k}=$$
$$n\sum_{k=0}^{n-1}(-1)^{k}\binom{n-1}{k}=n\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^{k}=$$
$$n(1+(-1))^{n-1}=n\cdot 0=0$$
From binomial theorem $$(1+x)^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}x^k$$
for $x=-1$ we get that
$$(1+(-1))^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^{k}=0$$


答:
These questions are related to the concept of a subspace.  The ambient space here is $\mathbb{R}^3$, but there are many subspaces of $\mathbb{R}^3$.

What are the subspaces of $\mathbb{R}^3$?  There are two trivial spaces.  The space consisting only of the vector $\langle 0, 0, 0\rangle$ is a zero-dimensional subspace and the whole of $\mathbb{R}^3$ is a three dimensional subspace of $\mathbb{R}^3$.  The other subspaces are one and two dimensional.

The usual way to define a subspace is to specify a spanning set, that is, a set of vectors.  If $S$ is a set of vectors, then the spanning set of $S$, usually denoted $\textrm{span}(S)$, is the set of all vectors that can be written as linear combinations of elements of $S$.

In this case, you want to know the dimension of $\textrm{span}\left(\left\{\vec{a_1}, \vec{a_2}\right\}\right)$.

It can't be zero dimensional because it contains non-zero vectors.  And it can't be three dimensional (or higher) because there is a spanning set with two elements.

Could this subspace be one dimensional?  It it were there would have to be a single vector $\vec{x}$ that spanned the whole subspace.  But that would mean that there would be (non-zero) scalars $\lambda_1$ and $\lambda_2$ such that:

$\lambda_1\vec{x}=\vec{a_1}$

$\lambda_2\vec{x}=\vec{a_2}$

But then we could write $$\frac{1}{\lambda_1}\vec{a_1} - \frac{1}{\lambda_2}\vec{a_2}=\mathbf{0}$$ which would imply that $\vec{a_1}$ and $\vec{a_2}$ are not linearly independent,  contradicting your answer to question 1.

All of this is a roundabout way to get to the conclusion that if $S$ is a finite set of linearly independent vectors, then $$\dim(\textrm{span}(S))=\vert S\vert.$$

For part 3, you already have the right idea. To show that $\vec{a_1}$ is not in $\textrm{span}(\{\vec{a_2}\})$, you just need to show that there is no $\lambda$ such that $\lambda\vec{a_2}=\vec{a_1}$, which will follow directly from the fact that $\vec{a_1}$ and $\vec{a_2}$ are linearly independent.

Title: Dimension of vector space and span


问:
What is the probability a 10 digit phone number contains at most one zero?
A telephone number consists of ten digits, of which the first digit is one of $1,2,3,4,5,6,7,8,9$ (first digit cannot be zero).  The others can be $0,1,2,3\dots 9$.
What is the probability that zero appears at most once in a telephone number if all the digits are chosen completely at random?

答:
Since $0$ can appear at most once, calculate the probability that it appears one time or zero times and sum them (considering also that the first digit cannot be $0$).

The probability of no $0$ is: $P_0= \left(\frac{9}{10}\right)^{9}$

The probability of exactly $1$ zero is: $P_1=9\frac{1}{10}\left(\frac{9}{10}\right)^{8} = \left(\frac{9}{10}\right)^{9}$

So you conclude that the probability of the phone number having at most one $0$ is:

$$\boxed{ P = \left(\frac{9}{10}\right)^{9}+\left(\frac{9}{10}\right)^{9} = 2\left(\frac{9}{10}\right)^{9} }$$


问:
Prove $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$
I want to make a combintronic proof for this. So far, I have we want to form a committee of k members with a president from n amount of people.

Thus on the left side we can select the k members from the n amount of people and then select a president from there.

How would I go about doing the Right side though?

答:
Using the factorial definition of $\binom{n}{k}$, you get that
$$\binom{n}{k}=\frac{n!}{k!\cdot(n-k)!}=\frac{n\cdot (n-1)!}{k\cdot (k-1)!\cdot(n-k)!}=\frac{n}{k}\binom{n-1}{k-1}.$$


问:
Minimizing `L2` norm with an inverse matrix.
This question is from the MIT opencourseware website. Here's how the problem is stated

We now turn to the problem of interest. Givne a real $m \times n$ matrix $A$ of full column rank, and a real $m$-vector y, we wish to approximately satisfy the equation $y = Ax$. Specifically, let us choose the vector $x$ to minimize $||y-Ax||^2 = (y-Ax)'(y-Ax)$. By invoking the above results on orhogonal matrices, show that (in the notation introduced earlier) the minimizing x is $\hat x = R^{-1} y_1$ where y_1 denotes the vector formed form the first $n$ component of $Uy$.

The final piece that is necessary to know is that

$$UA =
\begin{pmatrix}
    R \\ 0
\end{pmatrix}
$$

And $R$ is nonsingular, upper-triangular matrix.

I think what the problem is saying that even though we can't find a perfect map to take $y_1$ back to $R$ we can find a close approximation which is by using the portion of our data that we can inverse. I also know that $R^{-1}$ is analogous to $(X'X)^{-1}X'$ and I know how to derive that just fine using matrix calculus.

I think I could do most of it if I had some help wit hthe setup but I'm not sure how to get form $A$ to $UA$ for minimizing the the 2 norm.

答:
Write $A = U^{T}\tilde{R}$, where $U$ is an orthogonal $m \times m$ matrix and $\tilde{R}$ is a $m \times n$ upper triangular matrix.  It is possible to do this whenever $A$ is an $m \times n$ matrix; the procedure is called the $QR$ decomposition since typically we use the symbol $Q$ instead of $U$.

Based on the wording of your question, you (or the MIT courseware you're citing) seem to assuming that the range space of $\tilde{R}$ equals $\mathbb{R}^{n} \times \{0\}$.  This isn't a restrictive assumption --- you can always change coordinates so that it's true --- but it's worth thinking carefully about.  Given this assumption, we can write
\begin{equation*}
\tilde{R} = \left( \begin{array}{c}
R \\
0
\end{array} \right) (*).
\end{equation*}

Here $R$ is a $n \times n$ upper triangular matrix with non-zero elements on the diagonal (hence nonsingular).

Given $y \in \mathbb{R}^{n}$, we want to find the $x \in \mathbb{R}^{n}$ so that
\begin{equation*}
\|Ax - y\| = \min \left\{ \|A\tilde{x} - y\| \, \mid \, \tilde{x} \in \mathbb{R}^{n} \right\},
\end{equation*}

that is, the so-called least squares solution of the system $Ax = y$.  This is equivalent to finding the global minimum of the function $x \mapsto \|Ax-y\|^{2}$.

Using the decomposition $A = U^{T}\tilde{R}$ and the fact that $U$ is orthogonal, we have
\begin{equation*}
\|Ax - y\|^{2} = \|\tilde{R}x - U y\|^{2}.
\end{equation*}
Notice that $\tilde{R}x$ is constrained to the range space of $\tilde{R}$, no matter how we choose $x$.  Furthermore, the range space of $\tilde{R}$ equals $\mathbb{R}^{n} \times \{0\}$ by $(*)$ above.  Thus, if we let $P : \mathbb{R}^{m} \to \mathbb{R}^{n} \times \{0\}$ denote the orthogonal projection, then, by the (generalized) Pythagorean theorem,
\begin{equation*}
\|Ax - y\|^{2} = \|Rx - PUy\|^{2} + \|Uy - PUy\|^{2}.
\end{equation*}
The right-hand term doesn't depend on $x$ so we can ignore it and instead minimize $x \mapsto \|Rx - PUy\|^{2}$.  Since $PUy$ is in the range space of $R$, this is minimized precisely when it's zero, that is, when $x = R^{-1}PUy$.

Hence
\begin{equation*}
\|Uy - PUy\|^{2} = \min \left\{ \|Ax - y\|^{2} \, \mid \, x \in \mathbb{R}^{n} \right\}
\end{equation*}
and the minimum is achieved if and only if $x = R^{-1} PUy$.


答:
If $T|_{T^{-1}(W)} : T^{-1}(W) \to V$ is the restriction of $T$ on $T^{-1}(W)$, we see that
\begin{align} \ker T|_{T^{-1}(W)} &= (\ker T) \cap T^{-1}(W) \\ &= T^{-1}(\{0\}) \cap T^{-1}(W) \\ &= T^{-1}(\{0\} \cap W) \\ &= T^{-1}(\{0\}) = \ker T \end{align}
and that $$\operatorname{im} T|_{T^{-1}(W)} = T(T^{-1}(W)) = W \cap \operatorname{im} T.$$ Thus, by the rank-nullity theorem we have that $$\begin{align} \dim T^{-1}(W) &= \dim \ker T|_{T^{-1}(W)} + \dim \operatorname{im} T|_{T^{-1}(W)} \\ &= \dim \ker T + \dim(W \cap \operatorname{im} T).\end{align}$$

Title: A result involving dimension of inverse image (Vector spaces)

Body:
Let $V$ be a vector space, $T:V \to V$ a linear operator and consider $W \subset V$ a subspace. I was thinking about the follow inequality:

$$\dim T^{-1}(W) \leq \dim \ker T + \dim W.$$

How can I prove that? I've tried this way:

$$\dim W + \dim \ker T = \dim W + \ker T - \dim W \cap \ker T \geq \dim W+ \ker T.$$ Then if $T^{-1}(W) \subset W + \ker T$, the result follows. But I guess it is not correct.

Can you help me?


问:
Finding the intersection of three sets

80 students were asked if they like math, science or humanities. 24
students did not like either of the subjects, 9 liked math only, 16
liked science only, 9 liked humanities only, 12 liked math and
humanities, 7 liked math and science and 9 liked humanities and
science.

a) How many students like all three subjects?

b) How many students like math or science?

c) How many students don't like humanities?

Here's a venn diagram displaying the given information:

a) finding the intersection of sets M, S and H

|M∩S∩H|=|M∪S∪H|−(|M|+|S|+|H|)+|M∩S|+|M∩H|+|S∩H|

-2 |M∩S∩H|= (80 - 24) - (9 + 16 + 9) - (12 + 7 + 9)

-2 |M∩S∩H|= 56 - 34 - 28

-2 |M∩S∩H|= 22 - 28

-2 |M∩S∩H|= -6

|M∩S∩H|= 3

I can't do b) or c) because when I say the intersection is 3, then all the other numbers in the venn diagram change (obviously). For example, if the intersection is 3, then the number of people who like math and science = 4 (7 - 3) and the number of people who like math and humanities = 9 (12 - 3). But when I add up the newfound numbers (3 + 9 + 4), I get 16 and I can't do 9 - 16 (which is -5, A NEGATIVE NUMBER!!!) Could someone please let me know what I have done wrong and how the heck I'm supposed to figure out the intersection of three sets?! Any help would be greatly appreciated.

答:
The Venn diagram doesn't seem quite correct.  If you add up the numbers, you get 86, which is greater than the universe of 80 students. I think that the 12 who like math and humanities includes those who like math, humanities, and science (and likewise for the 7 who like math and science, and the 9 who like science and humanities).

So, when we add up the numbers, we get 86 students, but we have counted the students who like all three subjects three times each - two times more than we should have.

We know that we should have 80 students so our overcount is 86-80=6. Students who like math, science, and humanities have each been counted two times too many.  So there must be 6/2=3 such students.

Once we know this, we can conclude that there are:

12-3=9 students who like math and humanities but not science

7-3=4 students who like math and science but not humanities

9-3=6 students who like science and humanities but not math.

Fill these numbers into the Venn diagram and you should be able to complete the rest of the problems.

Title: Finding the intersection of three sets


问:
$(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$
Let $$f(x) = x\cos\frac{1}{x}$$ for x in $[1,\infty)$

Now I need to prove or disprove the difference $$f(x+2) - f(x) > 2$$ for all $x$ in the domain.

I tried a lot but I don't seem to be getting anywhere. My approach was to try and use graphs, but somehow that doesn't seem to work out that well.

Any ideas or suggestions please?

答:
For an alternative approach consider $g(y) = \cos(y)+y\sin(y)$.
$$g^\prime(y) = -\sin(y)+\sin(y)+y\cos(y) = y\cos(y) >0$$ for $y\in \left(0,\frac{\pi}{2}\right]$. Thus
$$g(y) > g(0) = \cos(0)+0\sin(0) = 1$$ for $y\in \left(0,\frac{\pi}{2}\right]$.

Now calculate $f^\prime(x) = \cos\left(\frac{1}{x}\right)+\frac{1}{x}\sin\left(\frac{1}{x}\right)$. By the previous paragraph, $f^\prime(x)>1$ for $\frac{1}{x}\in \left(0,\frac{\pi}{2}\right]$ or equivalently $x\geq\frac{2}{\pi}$.

Hence, we conclude $\displaystyle f(x+2)-f(x) = \int_x^{x+2} f^\prime(y)\,\mathrm{d}y > \int_x^{x+2} 1\,\mathrm{d}y = 2$ for all $x\geq\frac{2}{\pi}$.


问:
Applying the Poincaré Recurrence Theorem to Billiards
I am studying this application of Poincaré Recurrence Theorem and I have some questions.

1 - it states that If $S$ is a bounded space with measure $µ$
and $T : S \to S$ is a measure-preserving transformation, then for any set with
positive measure $B \subseteq S$, the subset $A \subseteq B$ of points that never recur to $B$
has measure zero.

Does $S$ must be bounded? All the other references that I've seen does not require it. Why does the author require $S$ to be bounded?

2 - Page 12:  The Poincaré recurrence theorem states that the
set of points in $B$ that will never return to $B$ no matter how many time
intervals are observed has measure zero. This means that with probability
one, the system’s initial configuration is not in this subset, and eventually
will return to $B$.

I don't get this sentence. How can we guarantee that the poll balls are not in this mesuare zero subset? And then the system's initial configuration will return to $B$?

Again, on page 13, Finally, although the recurrence theorem guarantees recurrence with probability one, it does not guarantee that the system recurs for every single point
in B. In fact, there may be infinitely many points that do not return to B.
However, the theorem does assert that the set of points in B that do not
return to B has measure zero – even though B has positive measure. Thus,
although it is possible that the billiard table’s initial configuration is one that
never recurs, it is a zero probability event; consequently, the probability of
recurrence is one.

I don't get this, is it saying that the it is possible that the billiard table's initial configuration can never recurs or it will always recur?

thanks in advance.

答:
Answering directly to your questions:

$1$. Of course that $S$ need not be bounded.

$2$. "This means that with probability one, the system’s initial configuration is not in this subset, and eventually will return to $B$."

is intended to mean

"This means that with probability one, the system’s initial configuration is not in this subset, and so it is with probability one in the complement of that subset, which means that it eventually will return to $B$."

$3$. "Thus, although it is possible that the billiard table’s initial configuration is one that never recurs, it is a zero probability event; consequently, the probability of recurrence is one."

is intended to mean

"Thus, although it is possible that the billiard table’s initial configuration is one that never recurs, in which case it is a zero probability event; consequently, the billiard table’s initial configuration is one that recurs with probability one and so the probability of recurrence is one."

I'm not trying to change the language, only to add the missing text. The language could improve substantially in this thesis.

Note that the author makes a very basic error in the text that you quote: he seems to think that the opposite of "recurring" is to "never recur". This is a nonsense since points can recur finitely often.


答:
Knowing $0$ cannot be at the first place, we have $9$ more to care about.

Logically, "Appears at most once" is the same as showing $0$ times, or showing $1$ time only. No more!

If each digit is chosen randomly, we have $\frac{1}{10}$ for each, thus making the choice of $0$ being $\frac{1}{10}$ and not zero $1 - \frac{1}{10} = \frac{9}{10}$

What is the probability we won't have a zero anywhere? it is only when we choose those $\frac{9}{10}$ (remember, this is the probability of not choosing a zero) the choices are independent of one another so we multiply:
$$ \overbrace{\frac{1}{10} \cdot \frac{1}{10} \cdot \dots \cdot \frac{1}{10}}^{\text{9 times}} = \left ( \frac{9}{10} \right )^9 $$

Now we need to deal with the $1$ zero appearing. What we need to do is first choose a place for it to be, how many places available? $9$. Then we should multiply this by the probability an actual zero would land there i.e. $\frac{1}{10}$ while multiplying the rest to assure we get another digit, not a zero i.e. $$\overbrace{\frac{1}{10}}^{\text{choosing one zero}} \cdot \overbrace{\frac{1}{10} \cdot \frac{1}{10} \cdot \dots \cdot \frac{1}{10}}^{\text{8 times}} = \left ( \frac{9}{10} \right )^9$$

Thus the whole probability is the addition of the two - because it is asking for $0$ zeros or $1$ zero.

$$ 2 \cdot \left ( \frac{9}{10} \right)^9 $$

Title: What is the probability a 10 digit phone number contains at most one zero?

Body:
A telephone number consists of ten digits, of which the first digit is one of $1,2,3,4,5,6,7,8,9$ (first digit cannot be zero).  The others can be $0,1,2,3\dots 9$.
What is the probability that zero appears at most once in a telephone number if all the digits are chosen completely at random?


问:
Proof of the theorem that there is no counterexample
If $\rho(A) $ is a spectral radius and $ \|\cdot\|$ is norm, the relation $\rho(A) \leqslant \|A\| $ is always established.
I want to prove that if $\rho(A) <1$ then $ \|A\|<1$.
Note that using the computer, I checked a large number of random matrices and there were no Counterexample. This is probably a mathematical theorem, but I can not prove it. please help. Thankful

$$ $$

A friend found a counterexample for it so I change the condition to $0<\rho(A) <1$. Can the theorem be proved now?

答:
Suppose $ V^{-1} A V = J$ where $J$ is the Jordan normal form. Write
$J= \Lambda+ T$, where $\Lambda$ is diagonal and $T$ is strictly upper triangular.

Let $\|x\|_B = \|B^{-1} x\|$, where the latter is the Euclidean norm. It is easy to check that the corresponding induced norm is $\|A\|_B = \|B^{-1}A B \|$.

Note that if $A$ is diagonalisable (that is, $T =0$) then we see that
$\|A\|_V = \|\Lambda\| = \rho(A)$.

If $T \neq 0$ we need an adjustment. The idea is straightforward but the description is a little messy.

Let $\Sigma=\operatorname{diag}(\sigma_1,...,\sigma_n)$, suppose all $\sigma_k>0$.
Note that $\Sigma \Lambda \Sigma^{-1} = \Lambda$ and
$[ \Sigma T \Sigma^{-1}]_{ij} = {\sigma_i \over \sigma_j} [T]_{ij}$.
Note that the latter term is always zero for $i \le j$.

Pick some $\epsilon>0$ and let $\sigma_1 = 1$, we will choose the remaining
$\sigma_k$ such that $|{\sigma_i \over \sigma_j} [T]_{ij}| < \epsilon$.
Suppose we have chosen $\sigma_1,...,\sigma_i$, then pick some $t >0$
such that $|{\sigma_i \over t} [T]_{ij}| < \epsilon$ for $j=i+1,...,n$ and
let $\sigma_{j+1} = t$.

Then we have
$\| A\|_{V \Sigma^{-1}} = \|\Sigma V^{-1} AV\Sigma^{-1}\| = \|\Sigma (\Lambda+T)\Sigma^{-1}\| \le \rho(A)+\| \Sigma T\Sigma^{-1} \|$
and it is straightforward to check that $\| \Sigma T\Sigma^{-1} \| \le n \epsilon$.


答:
There are $9 \times 10^9$ possible valid numbers:  $1\leftrightarrow 9$ in the first slot and $0 \leftrightarrow 9$ in each of the remaining $9$ slots.

The number of numbers that have exactly zero $0$s is:  $9 \times 9^9$, as there are only $9$ permissible digits for the first slot and $9$ permissible digits ($1\leftrightarrow 9$) in each of the $9$ remaining slots.

The number of numbers that have exactly one $0$ is $9 \times 9^8 \times 9$, as there are $9$ possible digits for the first slot and $9$ possible digits in each of $8$ remaining slots and $9$ choices for which slot must contain the $0$.

Can you put these results together?


问:
$\bigcup\limits_{1\leq r < s} l^r \neq l^s$
We had to prove for $s > 1$ :

$$\bigcup\limits_{1\leq r < s} l^r \neq l^s$$

, i. e. we have to find an $x \in l^s$ which is for all $r < s$ not in $l^r$.

The solution used the sequence $x_n := (n \log(n)\log(n))^{-1/s}$. It is said that $x \in l^s$ (using integral criterion), but this is not the case: Due to WolframAlpha, we get:
$$\int\limits_0^\infty (x \log(x)\log(x))^{-1} dx = \left[-\frac{1}{log(x)}\right]_{x \to 0}^{x\to\infty}$$

But the right side diverges according to WolframAlpha and my own calculations. Which other easy example does in which way prove the given result?

答: Take $x_n=(n log (n) log (n))^{-1/s}$ for $n \geq 2$ and $x_1=x_2=0$. Then everything works fine.   The integral is now from $2$ to $\infty$.


问:
Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?

Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?

I already did substitution for $X$ so we get the ring $R[x,x^{1/2},x^6,x^2]$ but I don't know to which ring this is isomorphic.

答:
Note that $T^3 - Y^{12}$ divisible by $T- Y^4$, so we have the equality of ideals
$$\langle T- Y^4, T^3 - Z\rangle = \langle T- Y^4,  T^3 - Y^{12}, T^3 - Z\rangle =\langle T- Y^4, Z- Y^{12}\rangle$$
and so
$$\langle X-Y^2, T- Y^4, T^3 - Z\rangle = \langle X-Y^2, T- Y^4, Z- Y^{12}\rangle$$

Now the evaluation map

$R[Y,X,Z,T]\to R[Y]$,
$X\mapsto Y^2$, $Z\mapsto Y^{12}$, $T\mapsto Y^4$ gives the isomorphism
$$R[Y,X,Z,T]/\langle X-Y^2, T- Y^4, Z- Y^{12}\rangle \simeq R[Y]$$\


问:
Just one vertex not in a cycle means there is a vertex cut?
I saw this assertion:

Any connected graph $G$ with $|V(G)| \geq 3$ and at least $2$ vertices that are not in a cycle must have a vertex cut

and realise that we just need one vertex not in any cycle. I don't know if I made a mistake or my claim is ok:

First, if $G$ does not have a cycle we are done because every vertex that is not a leaf is a vertex cut

So let $C$ be a cycle and $v_1$ a vertex such that $v_1u_1 \in E(G)$ for some $u_1 \in C$ and $v_1$ is not in any cycle. These $C$,$v_1$ exists because if we have any vertex $v$ out of a cycle, we take a path $P$ between $v$ and some cycle $C$, minimal in this sense (so $P$ does not go through any other cycle). Then $P \cap C =\{u_1\}$ with $u_1 \in C$ and $u_1v_1 \in E\,$ for some $v_1\in P$, that is not in any cycle.

For any $u_2 \in C$ there is no $(v_1,u_2)-$path that do not use $u_1$. Otherwise $u_1v_1Pu_2u_1$ would be a cycle that contains $v_1$. Then $u_1$ is a vertex cut (it separates $v_1$ from $C$)

is this ok? or am I missing something?

答:
Your claim is correct.

I suspect that the result you are thinking of is the following:

In a connected graph with $|V(G)|\ge 3$ and no cutpoints, there is a cycle containing any two given points. (Whitney 1927) .

This is actually consistent with your claim since if a connected graph with $|V(G)|\ge 3$ has no cycle through a vertex $u$ then it can have no cycle through a pair of vertices $u,v$ and therefore it is not 2-connected i.e. it has a cutpoint.


答:
So your first issue is that your Venn diagram does not display the given information, as you note at the end. This is confusing you, despite your solution being essentially correct, if somewhat oddly calculated (though I have no idea why you're trying to calculate $9 - 16$: you don't need to adjust the outer values, because those are already in your Venn diagram correctly; indeed, they're given in the question).

Let's call the number of students who like all three subjects $x$. Then your actual Venn diagram looks like this:

Now, summing all of those values, we see that $80 = 24 + 9 + 16 + 9 + 7 - x + 9 - x + 12 - x + x = 86 - 2x$, and so $2x = 6$, and $x = 3$. Thus, the full Venn diagram looks like this:

We can now solve the questions by just reading off the diagram.

Title: Finding the intersection of three sets

Body:
80 students were asked if they like math, science or humanities. 24
students did not like either of the subjects, 9 liked math only, 16
liked science only, 9 liked humanities only, 12 liked math and
humanities, 7 liked math and science and 9 liked humanities and
science.

a) How many students like all three subjects?

b) How many students like math or science?

c) How many students don't like humanities?

Here's a venn diagram displaying the given information:

a) finding the intersection of sets M, S and H

|M∩S∩H|=|M∪S∪H|−(|M|+|S|+|H|)+|M∩S|+|M∩H|+|S∩H|

-2 |M∩S∩H|= (80 - 24) - (9 + 16 + 9) - (12 + 7 + 9)

-2 |M∩S∩H|= 56 - 34 - 28

-2 |M∩S∩H|= 22 - 28

-2 |M∩S∩H|= -6

|M∩S∩H|= 3

I can't do b) or c) because when I say the intersection is 3, then all the other numbers in the venn diagram change (obviously). For example, if the intersection is 3, then the number of people who like math and science = 4 (7 - 3) and the number of people who like math and humanities = 9 (12 - 3). But when I add up the newfound numbers (3 + 9 + 4), I get 16 and I can't do 9 - 16 (which is -5, A NEGATIVE NUMBER!!!) Could someone please let me know what I have done wrong and how the heck I'm supposed to figure out the intersection of three sets?! Any help would be greatly appreciated.


答:
You are overlooking that the $7$ who like math and science, and the $12$ who like math and humanities and the $9$ who like humanities and science, OVERLAP and each of those groups include those who like all three.

You assumed those were all separate and each group excluded the $?$ that liked all three.

====

Use this image instead

When you say $12$ like math and humanities it is ambiguous as to whether is is meant there are $12$ who like math, humanities and dislike science.  Or if there are $12$ who like math and humanities and may or may not like science.

If you interpret it the first way, there are $12$ who like math and humanities but do not like science and use the drawing you drew you will get a negative number for those who like all three.

But if you interpret it the second way (which is the logical and literal and mathematical way to interpret it; if you are told they like math and humanities that means everyone who likes math and humanities regardless of what else they may or may not like) then if there are $?$ who like all three then there are $12$ who like math and humanities, and may or may not like science; and there are $12 -?$ who like math and humanities and don't like science.

Another way to view this is:

But here is should be clear then regions of $7,12,9$ overlap.

Title: Finding the intersection of three sets


问:
Necessary and sufficient conditions for commutativy of Cartesian product of sets
Let $A, B$ be sets. I know that $A\times B=B\times A$ if $A=B$ or $A=\emptyset$ or $B=\emptyset$. But, is the converse also true? or what meaningful equivalence is known?

答: The converse is indeed true.  Suppose $A$ and $B$ are both nonempty but $A\times B=B\times A$.  Pick elements $a_0\in A$ and $b_0\in B$.  Then for any $a\in A$, $(a,b_0)\in A\times B=B\times A$, which means $a\in B$ (and $b_0\in A$).  Similarly, for any $b\in B$, $(a_0,b)\in A\times B=B\times A$ so $b\in A$.  Thus $A=B$.


问:
Does $\sum_{k=0}^\infty \frac{16(2k+1)^2}{(4m^2 - (2k+1)^2)^2}=\pi^2$?
I think the following series has the value

$$\sum_{k=0}^\infty \frac{16(2k+1)^2}{(4m^2 - (2k+1)^2)^2}=\pi^2$$

for any positive integer $m$, but I am not sure how to go about computing it. Plugging the above sum into a computer for a few different values of $m$ seems to verify the result. I don't have much experience computing any infinite sums of the above form, so I am not sure where to begin. The context for computing this series is to show that $\cos(2mx)$ can be written as an infinite sum of the functions $\sin(kx)$. On a related note, I would also like to show that

$$\sum_{k=0}^\infty \frac{64m^2}{(4m^2-(2k+1)^2)^2} = \pi^2$$

where again, $m$ is a positive integer; this sum appears in showing the opposite relation, that $\sin(2mx)$ can be written as a sum of $\cos(kx)$. Again, I have verified on a computer that this is true for at least the first few values of $m$. This second sum would be reasonable to compute if it were an integral, but I'm not sure of a way I can immediately apply that idea (if we consider $f(x) = c^4/(c^2-x^2)^2$ then we could compute the integral of this function from $0$ to $\infty$ by substitution $x=c\tanh\theta$ fairly easily). I'm aware that there are easier ways to show these relations between sines and cosines, but I'm invested enough in this approach that I would like to find a way to compute these sums. Any tips would be appreciated.

Edit: I am actually not 100% sure that the first sum converges to $\pi^2$ for all $m$; in Wolfram Alpha, the sum comes out to be $\pi^2$ for $m=1,2$ but for any $m\geq 3$ it simply approximates the sum, and the approximated value is clearly less than $\pi^2$, so I may have made an error. The value of the second sum definitely seems to be $\pi^2$ though.

答: Let $$f(m) = \sum_{k=0}^\infty \frac{16(2k+1)^2}{(4m^2-(2k+1)^2)^2}, \\ g(m) = \sum_{k=0}^\infty \frac{64m^2}{(4m^2-(2k+1)^2)^2}.$$  Then $$f(m) = \pi^2 \sec^2 m\pi + \frac{\pi}{m} \tan m \pi, \\ g(m) = \pi^2 \sec^2 m \pi - \frac{\pi}{m} \tan m\pi, \\ \quad m \ne 0.$$  A proof of this might be possible using a suitable contour integral.


问:
Please help with this probalility problem.
Vince buys a box of candy that consists of six chocolate pieces, four fruit pieces and two mint pieces.  He selects three pieces of candy at random without replacement.

Calculate the probability that the first piece selected will be fruit flavored and the other two will be mint.

Calculate the probability that all three pieces selected will be the same type of candy.

答:
Since there are a total of $6+4+2=12$ candies and candies are not replaced, notice that:
$$\begin{align*}
P(\text{fruit} \to \text{mint} \to \text{mint}) &= P(\text{fruit})P(\text{mint | fruit})P(\text{mint | fruit,mint}) \\
&= \frac{4}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}\\
&= \frac{1}{3}\cdot\frac{1}{11}\cdot\frac{1}{5}\\
&= \frac{1}{165}
\end{align*}$$

Can you see why the $12$ decreased to $11$ and then to $10$? Can you see why the $2$ decreased to $1$?

Side Note: If replacement was allowed, then we would instead calculate:
  $$\frac{4}{12}\cdot\frac{2}{12}\cdot\frac{2}{12}$$

See if you can do the second one.


答: Yes, assume $A, B\neq\emptyset$ and let $x\in B$ then $< x, b>\in A\times B$ for all $b\in B$ but we have $<x,b>\in B\times A$ so $x\in B$. $A\subset B$ the other direction is the same. I hope that helps

Title: Necessary and sufficient conditions for commutativy of Cartesian product of sets

Body: Let $A, B$ be sets. I know that $A\times B=B\times A$ if $A=B$ or $A=\emptyset$ or $B=\emptyset$. But, is the converse also true? or what meaningful equivalence is known?


问:
Every infinite set has a countably infinite subset.
Prove every infinite set has a countably infinite subset.

I was wondering if this approach is completely correct for this problem, my approach was to recursively define a function $f: \mathbb{N} \rightarrow$ countably infinite subset of an infinite set.

Attempt:
Suppose $A$ is infinite. We will construct a function $f: \mathbb{N} \rightarrow B$, where $B \subset A$, and $B$ is countably infinite. Since $A \neq \emptyset$, we can choose an element $a_{1} \in A$. Set $f(1)=a_1$. Since $A$ is infinite, choose an element $a_2 \in A-\{a_1\}$.Set $f(2)=a_2.$Assume for each $m<n$, $f(m)$ has been chosen. Since $A$ is infinite there exists in element $a_{n} \in A-\{a_1,...a_{n-1}\}$. Set $f(n)=a_n$. Now that $f$ has been defined inductively, $B=\{a_1,a_2,...\} \subset A$ is countably infinite.

答:
Nerding on Wikipedia I discovered the following facts. A set that admits an injective function from natural numbers is called "Dedekind-infinite". An infinite set is one that is not in bijection with a finite set $\{0, \ldots, n\}$.

Suppose we assume the Zermelo-Frankel (ZF) axioms. Then:

We always have that a Dedekind infinite set is infinite, because there exist no injection from $\mathbb{N}$ to a finite set.

Assume the countable axiom of choice ($AC_{\omega}$). Then your proof can actually be formalized to choose all the values $f(n) $ "together". This is a very subtle point, but you should be aware that generally without the axiom of choice very pathologicsl things can happen, see for example the Banach-Tarski paradox.

Logicians would write this point as $ZF +AC_{\omega} $ proves "infinite $\Rightarrow$ Dedekind-infinite".

If we don't assume the countable axion of choice, surprisingly there exist infinite sets which are not Dedekind-infinite! As far as I understood, this proof is not constructive however, so it is kind of a monster.

In this case, logicians would say that the equivalence "Dedekind-infinite $\Leftrightarrow$ Infinite " is strictly weaker than the countable axiom of choice.


问:
Is there any meaning or basis-free description for coevaluation in FDVect?
It is a fact that the category $\textbf{FDVect}_k$ of finite-dimensional vector space is a rigid category, i.e. it is symmetric monoidal and has the notion of dual object and (co)evaluation satisfying some condition. In particular, the coevaluation of $V\in\textbf{FDVect}_k$ is given by
$$\text{coev}_V:k\to V\otimes V^*,\quad 1\mapsto \sum_{i=1}^n v_i\otimes v_i^*$$
where $\{v_1,\cdots,v_n\}$ is a basis of $V$ and $\{v_i^*,\cdots,v_n^*\}$ is the corresponding basis of $V^*$ (given by $v_i^*(v_j)=\delta_{ij}$). The coevaluation is, in fact, independent from the choice of basis. So is there any "fundamental" relationship (or meaning) between this mysterious quantity $\text{coev}_V(1)=\sum_{i=1}^n v_i\otimes v_i^*$ and $V$? Or alternatively, is there any basis-free description for $\text{coev}_V(1)$?

答:
For a finite-dimensional vector space $V$, the tensor product $V \otimes V^{\ast}$ is naturally isomorphic to $\text{End}(V)$. Its action on $V$ is given by the map

$$(V \otimes V^{\ast}) \otimes V \cong V \otimes (V^{\ast} \otimes V) \to V$$

given by applying the evaluation map $V^{\ast} \otimes V \to k$. Now, $\text{End}(V)$ has a distinguished element in it, namely the unit $\text{id}_V \in \text{End}(V)$, and if you transport this map across the isomorphism above you get exactly the coevaluation map (keeping in mind the natural identification between vectors in $W$ and linear maps $k \to W$). This is just the statement that the "coevaluation tensor" $\sum v_i \otimes v_i^{\ast}$ satisfies

$$ \left( \sum_i v_i \otimes v_i^{\ast} \right) \left( \sum_j c_j v_j \right) = \sum_{i, j} c_j v_i ( v_i^{\ast}(v_j)) = \sum_{i, j} c_j v_i \delta_{ij} = \sum_i c_i v_i$$

which is equivalent to the definition of the dual basis. This is also a simple explanation why the evaluation and coevaluation maps are sometimes called the counit and unit.


问:
Is $\operatorname{diam}(A\cap B)=\operatorname{diam}(\overline{A}\cap B)$ true when $B$ is a ball?
Let $X$ be a normed space, and let $B$ be the closed unit ball. Let $A$ be any subset of $X$ such that $A\cap B\neq\emptyset$.

Is it true that $\operatorname{diam}(A\cap B)=\operatorname{diam}(\overline{A}\cap B)$?

I proved that the equality holds if $B$ is the open unit ball. For if $x\in \overline{A}\cap B$ when $B$ is the open unit ball, then $\left\|{x}\right\|<1$ and we can find a sequence $(x_n)\subseteq A$ with $\left\|{x_n}\right\|<1$ converging to $x$. This gives an obvious upper bound for any $d(x,y)$ with $x,y\in \overline{A}\cap B$.

But the same thing can't be done when $B$ is the closed unit ball, because if $\left\|{x}\right\|\le 1$, we can't guarantee that there is a sequence contained in $B$ converging to $x$.

Is the statement false with the closed ball maybe?

答: The statement is false. You can construct a counter-example in $ \mathbb{R}^2 $ if you carefully think about when the sequence $ x_n \to x $ is not inside $ B $.


问:
Distance between fundamental domain and its translates
Let $(M,g)$ be a Riemannian manifold and $\tilde{M}$ its universal cover. Suppose that $\pi_1(M)$ is finitely generated. Let $\tilde{g}$ be the lift of $g$ to $\tilde{M}$, and let $d$ denote the induced distance function on $\tilde{M}$.

Let $L\colon\pi_1(M)\to\mathbb{N}$ be a fixed length function, and let $F\subseteq\tilde{M}$ be a fundamental domain for the action of $\pi_1(M)$ on $\tilde M$.

(By fundamental domain, I mean that $F$ is the closure of an open subset of $\tilde{M}$ such that for any $g\in\pi_1(M)$, the interiors of $g\cdot F$ and $F$ have no common points, and $\{g\cdot F\}_{g\in\pi_1(M)}$ form a cover of $\tilde M$.)

Question: Given any sequence $(g_i)_{i\in\mathbb{N}}\subseteq\pi_1(M)$ such that $\lim_{i\to\infty}l(g_i)=\infty$, is it true that
$$\lim_{i\to\infty}d(F,g_i\cdot F)=\infty,$$
where $g_i\cdot F$ is the translate of $F$ by the action of $g_i$?

Remark: If $M$ is compact, then the answer should be yes, since the action of $\pi_1(M)$ on $\tilde{M}$ is always proper. In the non-compact case, I would expect the answer to be no, but would like an example that shows this.

答:
In my answer, I will assume that
$$
d(A,B)=\inf_{a\in A, b\in B} d(a,b).
$$

Here is a simple and classical example. Consider the modular group $\Gamma=PSL(2, {\mathbb Z})$ and its standard fundamental domain $F$. Let $u_n\in \Gamma$ be any sequence corresponding to (strictly) upper-triangular matrices. Then  for each $u_n$, $d(u_nF,F)=0$.

Or, for a simpler example consider the subgroup $\Gamma< PSL(2, {\mathbb R})$ generated by the translation $u: z\mapsto z+1$ (in the upper half-plane model). As $F$ take
$$
\{z: Im(z)>0, 0\le Re(z)\le 1\}.
$$
Then for each $\gamma\in \Gamma$,
$$
d(F, \gamma F)=0.
$$


问:
How to compute: $\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$
This question was asked in a masters of mathematics exam for which I am preparing.

Compute $\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$.

I could only think of substituting $y^3 = x$ and that does not change much.

Could somebody post a solution using residues or in ways besides here in this link?:
How to compute the integral $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$

Edit : I am interested in the answers which use contour integration and residue calculus.

答:
First, let's do a substitution to reduce the number of poles to be computed. In this problem, it wouldn't be necessary, since the function is fairly easy and there aren't too many poles, however it's a good a practice since in the future you may encounter more intricated problems.

$$\int_{0}^{\infty}{\frac{x^\frac{1}{3}}{x^2+1}dx}=\frac{1}{2}\int_{0}^{\infty}{\frac{z^{-\frac{1}{3}}}{z+1}dz}$$

Now, let’s define our function and integration path. Bear in mind that since we are working with complex functions $z^{-\frac{1}{3}}=\exp{\left(-\frac{Log\left(z\right)}{3}\right)}$, wich means that we must take care of the branch point. Hence, the selection of the keyhole contour.
$$f(z)=\frac{z^{-\frac{1}{3}}}{z+1}$$

$$\oint f(z)dz=\left(\int_{0+ir}^{R+ir}+\int_{\Gamma}+\int_{R-ir}^{0-ir}+\int_\gamma \right)f(z)dz$$

1)Let's start working with the two integrals in the positive axis of the Real Line:
$$\displaystyle{\lim_{R\rightarrow \infty\\ r\rightarrow 0}}\left(\int_{0+ir}^{R+ir}\int_{R-ir}^{0-ir} \right)f(z)dz$$

Remember that:
$$Log(z)=\log|z|+i\text{Arg}(z)$$

$$\int_0^\infty \frac{\exp\left(-\frac{\log\left(z\right)+0i}{3}\right)}{z+1}dz+\int_\infty^0 \frac{\exp\left(-\frac{\log\left(z\right)+2\pi i}{3}\right)}{z+1}dz=\left(1-e^{-\frac{2\pi i}{3}}\right)\int_0^\infty\frac{z^{-\frac{1}{3}}}{z+1}dz$$

To compute the others integrals let's make use of the the Estimation Lemma:
$$\left\lvert\int_\Gamma f(z)dz\right\rvert\leq\displaystyle{\lim_{R\rightarrow \infty}}\left\lvert\int_0^{2\pi}\frac{\left(Re^{it}\right)^{-\frac{1}{3}}}{Re^{it}+1}Rie^{it}dt\right\rvert\leq\displaystyle{\lim_{R\rightarrow \infty}}2\pi R^{-\frac{1}{3}}\rightarrow 0$$

$$\left\lvert\int_\gamma f(z)dz\right\rvert\leq\displaystyle{\lim_{r\rightarrow 0}}\left\lvert\int_{2\pi}^{0}\frac{\left(re^{it}\right)^{-\frac{1}{3}}}{re^{it}+1}rie^{it}dt\right\rvert\leq\displaystyle{\lim_{r\rightarrow 0}}2\pi r^{\frac{2}{3}}\rightarrow 0$$

Finally, let's compute the residue:
$$\oint f(z)dz=2\pi i\displaystyle{\lim_{z\rightarrow -1}}\frac{z^{-\frac{1}{3}}}{z+1}(z+1)=2\pi ie^{-\frac{\pi i}{3}}$$

Gathering all results:
$$\int_0^\infty\frac{z^{-\frac{1}{3}}}{z+1}dz=\frac{2\pi ie^{-\frac{\pi i}{3}}}{1-e^{\frac{-2\pi i}{3}}}=\frac{\pi }{\frac{e^{\frac{\pi i}{3}}-e^{-\frac{\pi i}{3}}}{2i}}=\frac{\pi}{\sin\left(\frac{\pi}{3}\right)}=\frac{2\pi}{\sqrt{3}}$$

Hence:
$$\boxed{\int_{0}^{\infty}{\frac{x^\frac{1}{3}}{x^2+1}dx}=\frac{\pi}{\sqrt{3}}}$$


问:
Need help finding support set to calculate probability for joint PDF
Suppose $X, Y$ have joint density

$f(x, y) = \cfrac{1}{16}$ if $−2 ≤ x ≤ 2$ and $−2 ≤ y ≤ 2$

and $f(x,y) = 0$ otherwise.

Find $P(|X-Y|\leq 1)$

I was told that the fist step to deal with a problem like this is to find the "support set" the set of $(x,y)$ that is desirable.

So I wrote that the support set is {$(x,y): |X-Y| \leq 1$} = {$(x,y): -1\leq X-Y \leq 1$} = {$(X,Y) = X-1 \leq Y \leq X+1$}.

So $P(|X-Y|\leq 1)$ asks us to calculate the probability between $y = x-1$ and $y = x+1$

So $P(|X-Y|\leq 1) = \int_{-2}^1\int_{x-1}^{x+1}\cfrac{1}{16}dydx$ but because we have to account for the constraints $−2 ≤ x ≤ 2$ and $−2 ≤ y ≤ 2$ (which makes the desired region not symmetrical we have to stop at $1$ for the 1st integral and have another integral for $1 \leq x \leq 2$) so I think I need to add $\int_1^2\int_{x-1}^2 \cfrac{1}{16}dydx$ to account for the asymmetrical region.

So in conclusion,
$P(|X-Y|\leq 1) = \int_{-2}^1\int_{x-1}^{x+1}\cfrac{1}{16}dydx + \int_1^2\int_{x-1}^2 \cfrac{1}{16}dydx$

Do I have this correct? If not what's wrong with my line of thinking? Can someone lend me a hand? Thank You!

答:
You can check your calculation by looking at the correct answer.

A few things I found:

The notation you use for the "support set" of the density function is not quite right. You should use all lower case letters.

Looking at the picture below, your integral is not quite right.

Since the pair $(X,Y)$ is uniformly distributed, it is much easier to use the geometric meaning of double integrals.

Denote the set $A=\{(x,y):|x-y|\le 1\}$. The probability
$$
P(|X-Y|\le 1) = \iint_A f(x,y)dxdy=\frac1{16} \textrm{Area}(A\cap Q)
$$
where $Q=[-2,2]\times[-2,2]$.

It is easy to see from a picture that

$$
\textrm{Area}(A\cap Q)=16-3\cdot 3=7
$$
So the answer if $\frac{7}{16}$.


问:
For $x_{n+1}=x_n-x_n^3$, with $|x_1|>1$. What about the convergence?
I see Let $\{ x_n \}_{n=1}^{\infty}$ such that $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$ that if $|x_1| \le 1$, then $x_n\to 0$. However, if $|x_1|>1$, what can we say about this sequence? It seems hard to find the tendency of it. For $x_1$ near $1$, it converges to $0$, but for $x_1$ large, it is not the case.

答:
So if

$$|x_{n}| > \sqrt{2}$$

The sequence diverges to infinity and negative infinity with the evens and odds.

If $$|x_{n}| = \sqrt{2}$$

The sequence is periodic with period 2.

If $$|x_{n}-x^3_{n}| < 1$$

Then the next step is in the attracting basin you already know about.
Finding the roots we get that we can expand the attracting basin out to

$$|x_{n}-x^3_{n}|<(\frac{2}{3 (9 - \sqrt{69}))})^{1/3} + \frac{(\frac{1}{2} (9 - \sqrt{69}))^{1/3}}{3^{2/3}}$$

There is a small unsolved area. I will try to repeat the process, expanding the attracting basin out to the $\sqrt{2}$.

So letting the above function be called g, we have a set of nested invariant attracting basins consisting of the intervals

$$A_{n}=[-a_{n},a_{n}]$$

With $$a_{1} = 1$$ and $$a_{n+1}=g(a_{n})$$

So that if $$x_{n} \in A_{k}$$ then $$x_{n+1} \in A_{k-1}$$

$$g(\sqrt{2}) = \frac{(5\sqrt{6}-9\sqrt{2})^{1/3}}{2^{1/3}3^{2/3}} - \frac{2^{1/3}3^{-1/3}}{(5\sqrt{6}-9\sqrt{2})^{1/3}}$$

So since g(1) > 1 and $g(\sqrt{2}) < \sqrt{2}$

g has an equilibrium less than $\sqrt{2}$. We can extend the attracting basin of [-1,1] out to [-e,e] where e is the equilibrium of g, however this still leaves a small unknown region between e and $\sqrt{2}$.

To find out what happens in the region where $|x_{n}|$ is between e and $\sqrt{2}$, consider that $|x_{n+1}| < |x_{n}|$, therefore the sequence $|x_{n}|$ converges since it is bounded and monotonically decreasing. It can either converge to zero or a positive value between e and $\sqrt{2}$.

If $|x_{n}|$ converges to zero then $x_{n}$ converges to zero. Otherwise x_{n} will converge to a periodic solution of period 2 consisting of the points p and -p where $$p \in [e,\sqrt{2}).$$

p must be a root of the polynomial $$x= (x-x^{3}) - (x - x^{3})^{3}.$$

The complex roots are $x = -(-1)^{1/6}, x = (-1)^{1/6}, x = -(-1)^{5/6}, x = (-1)^{5/6}$

The real roots are $x=0, x=\sqrt{2}, x=-\sqrt{2}.$

So there is no such p periodic. Therefore if $|x_{n}|<\sqrt{2}$ then $x_{n}$ converges to zero.


问:
Possibility of duplicates in hash table
I have a problem where I am trying to find the relationship between parts and the parts that make them up. Like if a boiler is made with two side plates and one bottom plate, and the side plates have five bolts and the bottom four, than a tractor has $2*5 +1*4 = 14$ bolts. Except think 3.5 million parts and possibly ten levels deep.

Anyways, We want to prevent repeating interactions. If engines a and b require boilers we don't want to print out how many bolts are in the boiler twice.

Background out of the way, this is what I came up with. I generated the square roots of the first fifteen primes. My theory is that a linear combination of integers from one to 3.5 million multiplied by these will form an orthoginal set.

I set up the structure like a tree, repeated parts and all.

Then I go down to the deepest level. Say the connection between part 1 and part 5. My hash fuction between the two is the part number of the lower level times the first prime, so $1*\sqrt2$.

each level I go up I sum the hash functions for the children with the current lowest part number multiplied with the Nth prime, where N is the highest N of all its children plus one.

So $H(3, 7) = 7 * \sqrt5+H(7, 11)+H(7,5)+H(7, 2)$

with $\sqrt5$ being the 3rd prime, and $H(7, 2)$ having a "depth" of 2

The way that you compute the connection between parts not directly connected is the sum of the hashes of the connections that contain that part minus the sum of connections from there.

So $H(3, 11) = H(3, 7) - H(7, 11)$

And $H(12, 7) = H(12, 3) - H(3, 7) + H(12, 13) - H(13, 7)$

$= H(12, 3) - H(3, 7) + H(12, 13) - (H(13, 10) - H(10, 7))$

$= H(12, 3) - H(3, 7) + H(12, 13) - H(13, 10) + H(10, 7)$

I calculated all the first degree connections and some higher degree ones.

The question is, does a hash uniquely determine not only a relationship between the two parts, but also where on the tree they are, so that if the indicies are the same and hashes different the numbers should be summed like the total number of bolts above. And if two hash numbers are identical they must be talking about the same two indicies and suggest that they are repeated?

答:
First, a potential easier solution: you already have unique hashes, in the form of your (pairs of, though I'm not sure why you're using pairs exactly) part IDs: why not just use them? While building your output, just keep track of which part IDs you've already run your recursion on and how many bolts they needed, and if you come across a part ID that you've already calculated, just use that value (and don't print the duplicate).

Now, to answer your actual question: yes, in theory, but with implementation issues. Your hashing should never produce exact duplicates: the linear independence of the $\sqrt{p}$ over $\mathbb{Q}$ combined with the fact that a part's highest-$p$ coefficient is exactly its part ID. However you are not working in an infinite-precision setting, so it's possible that you'll have clashes in cases where the theoretical hashes are not exactly equal, but are closer than the precision that you are operating to.

Title: Possibility of duplicates in hash table


答:
You are correct, and to make it even easier:

For the graph $G$ with $|V(G)|\geq 3$ find a vertex $a$ not in any cycle.

If $a$ has degree $1$ (a leaf), its sole adjacent vertex is a cut vertex, separating $a$ from the rest of $G$

If $\text{deg}(a)>1$, $a$ is a cut vertex, since there is no other path between any two of its adjacent vertices, otherwise $a$ would be in a cycle.

Title: Just one vertex not in a cycle means there is a vertex cut?

Body:
I saw this assertion:

Any connected graph $G$ with $|V(G)| \geq 3$ and at least $2$ vertices that are not in a cycle must have a vertex cut

and realise that we just need one vertex not in any cycle. I don't know if I made a mistake or my claim is ok:

First, if $G$ does not have a cycle we are done because every vertex that is not a leaf is a vertex cut

So let $C$ be a cycle and $v_1$ a vertex such that $v_1u_1 \in E(G)$ for some $u_1 \in C$ and $v_1$ is not in any cycle. These $C$,$v_1$ exists because if we have any vertex $v$ out of a cycle, we take a path $P$ between $v$ and some cycle $C$, minimal in this sense (so $P$ does not go through any other cycle). Then $P \cap C =\{u_1\}$ with $u_1 \in C$ and $u_1v_1 \in E\,$ for some $v_1\in P$, that is not in any cycle.

For any $u_2 \in C$ there is no $(v_1,u_2)-$path that do not use $u_1$. Otherwise $u_1v_1Pu_2u_1$ would be a cycle that contains $v_1$. Then $u_1$ is a vertex cut (it separates $v_1$ from $C$)

is this ok? or am I missing something?


问:
How to prove that $A\cap B\subseteq C$ and $A^c\cap B\subseteq C$ imply that $B\subseteq C$?
How do you solve this problem??

Suppose that $A\cap B\subseteq C$, and $A^c\cap B\subseteq C$. Prove that $B$ is a subset of $C$.

I don't know where even to begin

Can anyone help?

Thank you

答:
Hint:

Pick an element $b\in B$. If it's in $A$, conclude that it's in $C$, and if it isn't in $A$, it's in the complement of $A$ -- conclude that it is in $C$.

Those are the two options, so we covered all of them.

Work closely with the definitions, and you'll be fine.


问:
Detect my mistake: $\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+4)^{n}}$
Detect my mistake:
$\displaystyle\int_{-\infty}^{\infty}
\frac{{\rm d}x}{\left(x^{2} + 4\right)^{n}}\ $.

My try:

We shall take a half circle around the upper part of the plane.

The arc will tend to zero.

Now we should calculate residue:
$\displaystyle%
\frac{\left(x - 2{\rm i}\right)^{n}}
{\left(x^{2} + 4\right)^{n}} =
\frac{1}{\left(x + 2{\rm i}\right)^{n}}$.

We shall take $\left(n - 1\right)$ derivatives:
$$
\frac{\partial^{n - 1}}{\partial x^{n - 1}}
\left[\frac{1}{\left(x + 2{\rm i}\right)^{n}}\right] =
\frac{\left(2n - 2\right)!}{n!\left(x + 2{\rm i}\right)}
= \frac{\left(2n - 2\right)!}{n!\left(4{\rm i}\right)}.
$$
Now the residue is incorrect, which leads to the integral result to be incorrect.

Does anyone see my mistake $?$.

答:
$$I=\int_{-\infty}^\infty\frac{dx}{(x^2+4)^n}$$
$x=2u\Rightarrow dx=2du$
$$I=\int_{-\infty}^\infty\frac{2du}{(4u^2+4)^n}=2^{1-2n}\int_{-\infty}^\infty\frac{du}{(u^2+1)^n}=2^{2(1-n)}\int_0^\infty\frac{du}{(u^2+1)^n}$$
now try making the substitution $u=\tan t\Rightarrow du=\sec^2t\,dt$ so:
$$I=2^{2(1-n)}\int_0^{\pi/2}\frac{\sec^2 t\,dt}{(\sec^2 t)^n}$$
now use the fact that:
$$(\sec^2t)^{-n}=(\cos^2t)^n$$
so we have:
$$I=2^{2(1-n)}\int_0^{\pi/2}\cos^{n-2}(t)dt$$
now you can use the beta function


问:
Quick Volume of a Solid of Revolution Question
So I was given the following prompt:

"What is the volume of the solid generated when the region bounded by the graph of $x=\sqrt{(y-2)}$ and the lines $x=0$ and $y=5$ is revolved about the $y$-axis?"

I understand the "washer" and "disc" method and the application of both of these, but for some reason I'm stuck on what the formula for the volume of these equations would look like here. Any help would be appreciated!

答:
Hint.

The volume is given by
$$
\int_2^5 A(y)\,dy
$$
where
$$
A(y)=\pi (\sqrt{y-2})^2
$$

Drawing a picture is helpful.


问:
How are these two graphs related (if they even are)?
I was trying to understand factoring numbers from a different perspective. Through a long convoluted route, I came up with a directed graph where if you can find the edges pointing to a given vertex you can factor the number at that vertex. This came from looking at the sequences of multiples of two numbers with a given mean, and trying to turn those sequences into a directed graph (each number points to every other number which follows it in some such sequence) - if you have one number in this sequence, the previous number in the sequence will tell you which sequence you're looking at, which corresponds to the mean of some two multiples, which is enough information to figure out those two multiples without needing to factor the number.

That graph is defined with the natural numbers as vertices, and edges directed from every $x \rightarrow cx - c^{2} + c$ for every natural number $c$ (where the right hand side is still restricted to the natural numbers). Let's call this graph A.

Another graph which has this property (the property that knowing the edges pointing to a given vertex allows you to factor the number at that vertex) is the graph where natural numbers point to their multiples: so exactly the same as the above, except using $x \rightarrow cx$ instead of $x \rightarrow cx - c^{2} + c$. Let's call this graph B.

So clearly in graph B, to go backwards from any $cx$ to an $x$ is equivalent to finding some $c$ (equivalent to factoring).

And also, in graph A, to go backwards from any $cx - c ^ {2} + c$, you also need to find some factor $c$ (since that expression factors into $c(x - c + 1)$, giving us factors $c$ and $(x - c + 1)$).

But on some level, I'm still confused, because both of these graphs look quite different when you draw them out/visualize them.

I would guess these graphs are somehow related, but I'm struggling to see how. Perhaps there is some transformation which takes one graph to the other and maintains the structure? Or perhaps we can filter out some obviously unnecessary edges from graph B (since A seems more hand-wavily "sparse") and then we can come up with some correspondence between them?

I might also just be confusing myself a lot here and there might be a much simpler explanation for what these two graphs are doing which I would also appreciate.

答:
The graphs are related but in a perhaps confusing way.

For each number $c$ the connection is straightforward. Suppose that Graph A has an edge :$$u\rightarrow v$$
This corresponds exactly with the edge in Graph B: $$u+c\rightarrow v+c$$

Simple proof

In A, $v=uc$.

In B, $u+c$ maps to $c((u+c)-c+1)=uc+c.$

So for a fixed $c$ the graphs are isomorphic.

However, these graphs are actually drawn for lots of different values of $c$. Therefore there are different sets of edges in A, each corresponding in a different way to a set of edges in B.


问:
A restricted complex vector space
Assume that $x$ is an n-dimensional vector of complex values, namely $x= x_{re} + ix_{im}$ where both $x_{re}$ and $x_{im}$ are n-dimensional real column vectors.

Assume that we restrict $x_{im}$ to be always the same vector, while $x_{re}$ can be any real vector.

What are the properties of corresponding restricted complex vector space? Does such a restricted complex vector space have significance? Is there a mathematical notion describing such a situation?

答:
I'll notate with tuples for the simplicity, but we're still talking about the same thing. At any rate, you'd have something like $\{(x_1+ai, ..., x_n+ai): a,x_1,...,x_n \in \mathbb{R})\}$. We'll call this $A$.

The most notable property of this is that it is, in fact, not a vector space. Recall our requirements (here is a nicely done list that was at the top of my Google results).

If we ever have $a$ (that is, $x_{im}$) non-zero, we run into some issues. I'll leave this to you, but will tag on that it's the same reason as to why the empty set isn't a vector space (at least, that's probably the lowest-hanging fruit; it breaks more than just one of our criteria).

 If $a\neq 0$, then $0\notin A$.

 If $a\neq 0$, it is not closed under addition nor scalar multiplication. Consider $u+v$ for $u,v \in A$. Any one coordinate would have complex part $a$, but $u+v$ would have complex part $2a$, thereby meaning that $u+v\notin A$. Scalar multiplication is similar.

 I'll drop an et cetera here. Your defined set of vectors does not only break those three axioms alone.

As to a mathematical notion for this: we'd need $a=0$ for this to be a vector space, but, at that point, we'd just have $\mathbb{R}^n$ with some extra steps. The remaining pieces of your questions were rather directly tied to this being a vector space; delving into such discussion in the terms of vector spaces isn't really bringing us much, though the notion is interesting! That being said, this math.SE post may be of interest as well. This one is probably of value to link as well. Thirdly, you may find abstract algebra interesting if you like exploring questions like these; the field deals with such structures.

Edit: a second thought that might be somewhat more insightful.

Consider constructing this proposed set of vectors as the span of a vector or two. What would those vectors be here? Definitely tough to put together (and really, you can't). Fix $n$. Pick a set of $n$ basis vectors to span our proposed vector space. ...

Where does this fall apart?

Title: A restricted complex vector space


问:
Units in $\mathbb{Z}[\omega]$
I'm considering the ring $\mathbb{Z}[\omega]$, where $\omega=\frac{1+\sqrt{-3}}{2}=e^{i \pi/3}$.

I'm trying to prove that $a+b\omega \in \mathbb{Z}[\omega]$ is invertible if and only if $|a+b\omega|=1$.

Here I'm defining $| \cdot |$ to be the usual complex modulus $\sqrt{(a+b\omega)(a+b\overline{\omega})}$.

Any tips as to how to approach this proof?

答: Hint: Use $N(z)=|z|^2= z \bar z$. Then $z \in \mathbb Z[\omega] \implies N(z) \in \mathbb Z$.

Title: Units in $\mathbb{Z}[\omega]$


问:
If $f(x)=g(a^Tx)$, where $g:\mathbb{R}\to\mathbb{R}$ and $a, x\in\mathbb{R}^n$, why is the Hessian equal to $g''(a^Tx)aa^T$?
I am trying to understand the solution to below problem:

Let $f(x)=g(a^Tx)$, where $g:\mathbb{R}\mapsto\mathbb{R}$ is continuously differentiable and $a\in\mathbb{R}^n$ is a vector. What are $\nabla f(x)$ and $\nabla^2f(x)$?

The answer is

$\nabla f(x)=g'(a^Tx)a$

$\nabla^2 f(x)=g''(a^Tx)aa^T$

I understand the answer for the gradient, but why does the answer for the hessian contain $a^T$ at the end?

答:
Since $\nabla f(x)=g'(a^Tx)a$, the $i$-th component of $\nabla f(x)$, call it $h_i(x)$, is $g'(a^Tx)a_i$.

By definition, the $i$-th row of $\nabla^2f(x)$ is the transpose of the gradient of the $i$-th component of $\nabla f(x)$. In other words, the $i$-th row of $\nabla^2f(x)$ is the transpose of $\nabla h_i(x)$.

Well, $\nabla h_i(x)=g''(a^Tx)a_ia$. The transpose of this is $g''(a^Tx)a_ia^T$ (because the first two factors are constants).

So, $\nabla^2f(x)$ has the form

$$\begin{pmatrix}g''(a^Tx)a_1a^T\\\vdots\\g''(a^Tx)a_na^T\end{pmatrix}=g''(a^Tx)\underbrace{\begin{pmatrix}a_1a^T\\\vdots\\a_na^T\end{pmatrix}}=g''(a^Tx)\underbrace{aa^T}$$

The equivalence of the two underlined expressions is just a simple consequence of the definition of matrix multiplication.


问:
finding max of quartic function $y=ax^4+bx^3+cx^2+dx+e$ when x is in range [0.0, 1.0]
How do I find the maximum value of a quartic function when x is in range between 0 and 1?
Solving $0=4ax^3+3bx^2+2cx+d$ to get possible candidate for x that gives maximum value is not elegant and messy, as that means I have to solve cubic equation which I want to avoid.
I want to know if there are better ways to do this.

答:
Taking the derivative and then applying the cubic formula is the obvious route, though I agree that the cubic formula is rather gross in practice; to my knowledge, there aren't any particularly easy ways to find zeroes to a cubic function. The other obvious routes include approximating it graphically, etc. I'm assuming that won't suffice either.

However, look into the AM-GM inequality. I wasn't able to find it for some reason, but I do know that there's a PDF floating around out there from UGA on how to use it in such a context. There's also Find maximum value by using AM-GM inequality, which may be of some assistance. Most of the techniques you might want to use are going to come from the field of optimization.

Edit:

Here is the UGA article I mentioned.

I'm not sure where you are in your education, but get used to a lot of manipulation and "things coming from nowhere." It is what it is, but the creativity and ingenuity that requires is half of what makes math such a beautiful subject. At any rate, I was able to find this video which is a roughly similar problem to what you might need to do.


问:
Measuring distances on any coordinate system
I was reading the book The ABC of Relativity from Betrand Russell, and at some point, the author mentions a method for measuring the distance between 2 points on any coordinate system. He says that the formula was discovered by Gauss and is a generalization of the pythagorean theorem. Although, it doesn't writes the explicit formula in mathematics. Since I have no more information, couldn't find anything on it. If I understand the formula correctly, it should be something like this when the coordinates can be expressed with 2 components $(x;y)$:

$$d(A,B)^2= (x_A-x_B)^2+(y_A-y_B)^2+(x_A-x_B)(y_A-y_B)$$

Is this formula correct and where it comes from?

答:
I'm unsure if my answer will cover what you want, but since you've found this statement in a book of Relativity this probably is in the context of differential geometry. Well, the idea is that you can generalize the notion of distance to any manifold you like by introducing the notion of a Riemannian Metric and the notion of a Geodesic. When I first learned General Relativity the books were a little confusing about that so I understand your worry.

Well, what are all of these and what the hell they have to do with Gauss? Well, first, manifold is a generaliztion of surface. Gauss studied surfaces in $\mathbb{R}^3$, however, mathematicians saw that the results obtained didn't depend on the structure of the background euclidean space, this leads to a generalization to a much more general object called manifold, which in simplest terms is one space on which we can do geometry as we do on surfaces. A manifold is characterized by the nice property that each of it's points has a neighbourhood equivalent in some sense to the euclidean space.

The classical way to tackle the problem of distances in surfaces is to introduce the object called the first fundamental form. The idea is that at each point $p$ of the surface $S$ we plug the tangent plane $T_pS$ and the elements of the tangent plane are the tangent vectors to $S$ at the point $p$ (imagine this, is not hard). We will be able to measure distances if we introduce on each of the tangent planes one inner product. Since the surface is a subset of $\mathbb{R}^3$ we already have one candidate to the inner product: we just restrict that of $\mathbb{R}^3$. Then if $v,w\in T_pS$ let's denote $\left\langle v, w\right\rangle_p$ the inner product of $\mathbb{R}^3$ restricted to $T_pS$. Now we introduce the first fundamental form as $I_p : T_pS \to \mathbb{R}$ by requiring that:

$$I_p(w)=\left\langle w, w\right\rangle_p$$

Now look at what we did: in $\mathbb{R}^3$ alone the inner product allows us to compute the length of vectors by taking the square root of the inner product of the vector and itself. Now the tangent plane is a subset of $\mathbb{R}^3$ so we can restric the inner product to the tangent plane and use it's inner product to define that quadratic form. That quadratic form gives us the notion of length of vectors in the tangent plane. Now if you recall what is the length of a curve and how do we compute it you'll see that introducing this quadratic form we are allowed to compute the length of curves over surfaces.

But think for a while, in three-space from all possible curves joining two points $A$ and $B$, which of them we select the length to call distance from $A$ to $B$? Well, we select that that gives us the minimum length which is the line segmenet joing the points. So in a surface something similar happens, we define a curve in the surface that is the analog of a line in a surface, this curve we call a geodesic, and when we pick a geodesic between points we call the length of the geodesic the distance between the points.

You might say: that's all cool, but I want General Relativity! It's a four dimensional space time not embedded into any other higher dimensional space! And what I tell you is: calm down, it is all related. The problem is that when you remove the background space you have to be more careful, you'll need to introduce the notion of a manifold, and since we don't have the background space we'll have to refine the notion of a tangent space and of the inner product. In practice, the inner product association to each point will become a tensor field called the metric tensor (which is the term $g_{ab}$ which appears in Einstein's Equation).

And finally how do you come to learn all of this? Well, my recommendation is: to get started study Do Carmo's Differential Geometry of Curves and Surfaces - it's a book about the differential geometry of surfaces in $\mathbb{R}^3$, good to get started. Then you'll need to move to the differential geometry of manifolds. My recommendation is that you pick Spivak's A Comprehensive Introduction to Differential Geometry. Spivak however uses the notion of metric spaces, if you don't know anything about them you can go after a book about metric spaces, and to study this my recommendation is to get Rudin's Principles of Mathematical Analysis and read the chapter of metric spaces and after it get some topology book and read the chapter of metric spaces.

I hope this helps you somehow. Good luck!


问:
Does it hold that $f(x)=O(x^{k+1}) \implies f(x)=o(x^k)$?
Does it hold that

$$f(x)=O(x^{k+1}) \implies f(x)=o(x^k)$$

as $x\to 0$?

答:
Yes, as $\frac{x^{k+1}}{x^k} = x$ and $x \rightarrow 0$ as $x\rightarrow 0$.

Use this to fill in the details yourself: If $f$ satisfies $f(x) \le Cx^{k+1}$ around a neighborhood of 0 then what about $\lim_{x \rightarrow 0} \frac{f(x)}{x^k}$. And if the limit of this fraction is 0 [it is], then what is the definition of $o(x^k)$ again.


问:
Image of compact set is bounded in $\mathbb{R}^n$
$\newcommand{\R}{\mathbb{R}^n}$
Consider a function $f:\R\rightarrow \R$, not necessarily continuous, and a compact subset $W$ of $\R$. Then, can we say that $f$ is bounded on $W$? If not, what is the counterexample?

答: I don't know if i'm missing something but i think this is false. For example the function piecewise defined: Fix a point $p\neq 0$. $f\colon \mathbb{R}^n \to \mathbb{R}^n$ given by $f(x) = \frac{p}{\|x\|}$ if $x \neq 0$ and $f(0)=0$ the unit closed ball $B = \{ x \in \mathbb{R}^n : \|x\|\leq 1 \}$ is compact but near $0$ (and hence on $B$) $f$ is not bounded. Hope this helps.


问:
Solve $\eta_{xx}-w\eta_x=0$ where $\eta(x,y)$
I am solving a system of equations.
These are the two equations I have solved:

\begin{align}
\eta_{xx}-w\eta_x&=0\\
\eta(x,y) &= a(y) + b(y)e^{wx}\\
\end{align}

\begin{align}
\xi_{yy}-\frac{1}{y}\xi_y&= 0\\
y^2\xi_{yy} - y\xi_y &= 0\\
\xi(x,y)&= d(x) + f(x)y^2\\
\end{align}

Is this correct?

Body:
For the first PDE I got this:
$$\eta_{xx}-w\eta_x=0$$
$$\eta_{x}-w\eta=C_1$$
$$(e^{-wx}\eta )'=C_1e^{-wx}$$
$$e^{-wx}\eta =-\dfrac {C_1}we^{-wx}+C_2$$
$$\eta =e^{wx}(-\dfrac {C_1}we^{-wx}+C_2)$$
$$\eta (x,y) ={C(y)}+C_2(y)e^{wx}$$
For the second PDE:
$$\xi_{yy}-\frac{1}{y}\xi_y= 0$$
$$y\xi_{yy}-\xi_y= 0$$
$$\left(\dfrac { \xi_y} y \right)'=0$$
$$ \xi_y =C_1y$$
$$ \xi =Cy^2+C_2$$
$$ \xi (x,y) =C(x)y^2+C_2(x)$$
So everything is correct.


问:
number of all inconstant maps f from A to A
Let $ A=\{1, 2, 3,..., n\}$. Find the number of all nonconstant maps $f: A \rightarrow A$  for which $f(k) \le f(k + 1)$ and $f(k) = f(f(k + 1))$ for $k = 1, \dots, n-1$..</p>

答: Hint: If $f(k+1)>k$ for some $k$, then $f(k)\le f(k+1)\le \cdots\le f(f(k+1))$, but since the last term equals the first all inequalities are equalities, so $f(k)=f(k+1)$.</p>


问:
Find the vector given the dot product and cross product of a set of vectors.
This question comes from Ted Shifrin's Multivariable Mathematics. The question states:

Given the nonzero vector $\textbf{a} \in \mathbb{R}^{3}$, $\textbf{a} \cdot \textbf{x} = b \in \mathbb{R}$, and $\textbf{a} \times \textbf{x} = \textbf{c} \in \mathbb{R}^{3}$, can you determine the vector $\textbf{x} \in \mathbb{R}^{3}$? If so give a geometric construction of $\textbf{x}$.

I'm trying to visualize what the vector $\textbf{x}$'s would look like in this scenario, but I think I'm getting confused by everything happening. So $\textbf{a} \cdot \textbf{x} = b$ is giving me the equation of a plane, in particular an affine plane from the one at the origin. As well $\textbf{a} \times \textbf{x} = \textbf{c}$ is giving me vector orthogonal to $\textbf{a}$ and $\textbf{x}$ respectively. The norm of this cross product also gives me the area of the parallelogram spanned by $\textbf{a}$ and $\textbf{x}$. Even with all these properties I'm still having trouble figuring out how to solve for $\textbf{x}$.

Would I be able to get some assistance? If I'm lucky maybe the man himself actually might pop in to provide guidance.

答: BIG HINT: Note that the equation $\mathbf a\times\mathbf x = \mathbf c$ tells us that $\mathbf a$ and $\mathbf c$ are orthogonal. Draw a picture of an example. Say $\mathbf c$ is a vector along the positive $\mathbf e_3$-axis. Then $\mathbf a$ must be a vector in the $xy$-plane. Draw it in a convenient location. Last, relative to it, where must $\mathbf x$ be located in order for the equation $\mathbf a\times\mathbf x = \mathbf c$ to hold? You mentioned the area of a parallelogram. Draw the possible locations (and there are infinitely many) of $\mathbf x$ to arrive at that.


问:
Classifying tensor fields thought of as multilinear maps
This is the definition of a tensor field I'm familiar with:

A tensor field $T$ of type $(k,l)$ is is a $C^{\infty}(M)$-multilinear map $T: \Omega^1(M) \times ...\Omega^1(M) \times \mathfrak{X}(M) \times... \times\mathfrak{X}(M)\rightarrow C^{\infty}(M)$. It eats $k$ covector fields and $l$ vector fields and spits out a smooth real function defined on a manifold $M$.

Now I came across the following claim:

The torsion $T$ of an affine connection $\nabla$ is a (1,2) tensor field. To check this it's sufficient to show that that the map defined by $T$ is $C^{\infty}(M)$-bilinear.

However, this doesn't make sense to me. A torsion tensor takes two vector fields so why is it a (1,2) tensor field? Where is the third argument? Also to check if $T$ is indeed a (1,2) tensor field, shouldn't we show that it is trilinear?

答:
Associated to $T : \mathfrak{X}(M)\times\mathfrak{X}(M) \to \mathfrak{X}(M)$ is a map $T' : \Omega^1(M)\times\mathfrak{X}(M)\times\mathfrak{X}(M) \to C^{\infty}(M)$ given by

$$T'(\alpha, X, Y) = \alpha(T(X, Y)).$$

When people say that $T$ is a $(1, 2)$-tensor field, they really mean the associated map $T'$ is a $(1, 2)$-tensor field. To check that $T'$ is a tensor field, you need to check $C^{\infty}(M)$-linearity in all three arguments, but note that it is linear in the first argument as

\begin{align*}
T'(f\alpha + g\beta, X, Y) &= (f\alpha + g\beta)(T(X, Y))\\
&= f\alpha(T(X, Y)) + g\beta(T(X, Y))\\
&= fT'(\alpha, X, Y) + gT'(\beta, X, Y).
\end{align*}

So we only need to check that $T'$ is $C^{\infty}(M)$-linear in the second and third arguments, but this is the case if and only if $T$ is $C^{\infty}(M)$-linear in its two arguments.


问:
Proving that the existence of a section implies the direct sum.
Here is the question I want to prove so far:

Let $R$ be a commutative ring. For $R-$modules $L,M,N$ show that the following conditions are equivalent.(all functions are $R-$ module homomorphisms.)

a- $M \cong_{R} L \oplus N.$

c- There exists a right-split short exact sequence $0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$

My trial is as follows:

$a \implies c.$

Also, using the fact that the direct sum of a finite number of modules coincides with their direct product we will prove the required.

Assume that $M \cong_{R} L \oplus N.$ We want to show that there exists a right-split short exact sequence $$0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$$ i.e., $\exists$ a section $s: N \rightarrow{M}$ s.t.$$p \circ s = id_N \quad (2)$$ Where $p: M \rightarrow N $ and it is onto. So we need to find the functions $s,p$ that satisfies $(2)$ above.

So, since $M \cong_R L \oplus N,$ we can say that we have $0 \rightarrow L \rightarrow L \oplus N  \rightarrow N \rightarrow 0,$ with $i:  L \rightarrow L \oplus N$ the embedding $i(l) = (l,0_N)$ and  $p:  L \oplus N \rightarrow N$ the projection $p(l,n) = n.$

Also, if we define $s: N \rightarrow L \oplus N  $ with $s(n) = (0_L,n)$ then we have $(p\circ s)(n) = p ((0_L,n)) = n $ as required.

Now, we want to verify that $s$ is an $R-$module homomorphism:

Let $n,n_1, n_2 \in N$ and $r\in R,$ then we have the following:

1- $s((n_{1} + n_{2})) = (0_L, n_{1} + n_{2})) = (0_L, n_{1}) + (0_L, n_{2}) = s(h_{1})s(h_{2}).$

Where the second equality is by definition of multiplication in the direct product.

2- $s(rn) = (0_L, rn) = r(0_L, n) = rs(n).$

And hence $s$ is an $R-$module homomorphism as required.

$c \implies a$

I do not know exactly how to define the isomorphism in that case, I am just guessing that it should be $\varphi :=(i,s): L \oplus N \rightarrow M$ for the following ses:

$$0 \longrightarrow L \stackrel{i}{\longrightarrow} M \stackrel{p}{\longrightarrow} N \longrightarrow 0$$

Where $i$ is injective and $p$ is surjective. Is my isomorphism function $\varphi$ defined correctly?

If so, I know that $\varphi $ is an $R-$module homo. (I can easily prove that), but how can I show that it is injective and bijective?

Could anyone help me in answering those questions please?

答:
Define the $R$-module homomorphism $\varphi : L \oplus N \to M$ by $\varphi(l,n) = i(l) + s(n)$. Note that if $\varphi(l,n)=0$, then
$$\begin{align} 0 &= p(\varphi(l,n)) \\ &= p(i(l))+p(s(n)) \\ &= p(s(n)) = n \end{align}$$
($p \circ i = 0$ by exactness) and then $0=\varphi(l,n)=i(l)$ implies $l=0$ (since $i$ is injective). Hence $$\varphi(l,n)=0 \implies (l,n)=(0,0)$$ which means that $\varphi$ is injective. Can you now figure out how to prove the surjectivity of $\varphi$?


问:
Definition of Matroid in two cases
For matroids M = (S, U), the following property applies to U:

If A, B ∈ U and |B| = |A| + 1 , then there must exist an x ∈ {B \ A} such that A∪ {x} ∈ U

I want to prove this property for the following two matroids

a)

S = set of edges of a connected graph G

U = all subsets of S that do not contain circles

My approach: If | B | = | A | +1, then there is at least one edge x in B, which can be added to A. A∪ {x} is a forest and A is a forest, but how do I know that A∪ {x} does not create a circle in A?

b)

S = set of vectors from $
\mathbb{R}^{n}$

U = all sets of linearly independent vectors from S

My approach: If there are |B| linearly independent vectors, and A contains |B|−1 vectors, then there must be a vector that is linearly independent of the vectors in A. Otherwise A would already be a basis of $
\mathbb{R}^{n}$, and B could not have more elements than A.

Would be glad for a scrutinizing look

答:
Your approach to (b) is on the right track, but you need to show that there is a vector in $B$ that is linearly independent of the vectors in $A$; you can do this be showing that if $B\subseteq\operatorname{span}A$, then $B$ is not linearly independent.

For (a), suppose that $A$ and $B$ are the edge sets of forests $G_A$ and $G_B$, respectively, and that $|B|=|A|+1$. Let $V_A$ and $V_B$ be the vertex sets of $G_A$ and $G_B$, respectively. Suppose further that for each $e\in B$, $G_A+e$ contains a circuit. This can happen only if the vertices adjacent to $e$ are already in $V_A$, so $V_B\subseteq V_A$. Moreover, the vertices adjacent to $e$ must lie in the same component of $G_A$, so $G_A$ and $G_B$ have the same number $k$ of components.

Since $G_A$ and $G_B$ are forests, $|B|=|V_B|-k$ and $|A|=|V_A|-k$, and therefore

$$|V_B|-k=|V_A|-k+1\,.$$

But then

$$0\le|V_A|-|V_B|=-1\,,$$

which is absurd.

Title: Definition of Matroid in two cases