问:
Which is the correct application of Chebyshev's inequality in the below case?
Starting from a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$, a standard normal random variable $Z\sim\mathsf{N}\left(0,1\right)$ and a positive constant quantity $\varepsilon$, if I have to say something about the following probability:
$$\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)\tag{1}$$
By applying Chebyshev's inequality, doesn't one have that (since $Z$ is a standard normal random variable, its second moment equals $1$):
$$\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)\leq 1-\dfrac{1}{\dfrac{\varepsilon}{\Delta t^2}\left(1+\Delta t-\dfrac{\Delta t}{\varepsilon}\right)^2}\tag{2}$$
?

If so, how can $(2)$ be furtherly expanded/simplified? My goal is to be able to say how $\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)$ behaves as $\varepsilon$ increases/decreases and as $\Delta t$ increases/decreases.

If my reasoning is wrong, which would be the correct application of Chebyshev's inequality to $(1)$?

答:
Chebyshev bounds are "loose" bounds and useful when you don't know the distribution. Here you know the distribution of $Z$. Why not directly work with the distribution function?

For example, let $y = -\frac{\sqrt{\epsilon}}{\Delta t}(1 + \frac{\epsilon - 1}{\epsilon}\Delta t)$. Then,
$$
N(y) = P(Z < y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{y}e^{-\frac{x^2}{2}}\;dx
$$
$$\frac{\partial N(y)}{\partial \epsilon} = N'(y) \frac{\partial y}{\partial \epsilon}$$
$$\frac{\partial N(y)}{\partial \Delta t} = N'(y) \frac{\partial y}{\partial \Delta t}$$


问:
How does Cantor's diagonal argument work?
I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the wikipedia article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):

\begin{align}
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.

My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:

\begin{align}
s_0 &= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
s_4 &= (0,1,1,\mathbf{1},\dots)\\
s_4 &= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\
s_5 &= (1,0,0,0,\dots)\\
s_6 &= (1,0,0,1,\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:

For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,

$$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$

...snip...

Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.

Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.

I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:

$$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 & \mathrm{if\ } s_{m,m} = 1\\1 & \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$

This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because... it's turtles all the way down!?  Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.

Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".

What am I missing?

答:
First, let me give you a proof of the following:

Let $\mathbb{N}$ be the natural numbers, $\mathbb{N}=\{1,2,3,4,5,\ldots\}$, and let $2^{\mathbb{N}}$ be the set of all binary sequences (functions from $\mathbb{N}$ to $\{0,1\}$, which can be viewed as "infinite tuples" where each entry is either $0$ or $1$).

If $f\colon\mathbb{N}\to 2^{\mathbb{N}}$ is a function, then $f$ is not surjective. That is, there exists some binary sequence $s_f$, which depends on $f$, such that $f(n)\neq s_f$ for all natural numbers $n$.

What I denote $2^{\mathbb{N}}$ is what Wikipedia calls $T$.

I will represent elements of $2^{\mathbb{N}}$ as tuples,
$$(a_1,a_2,a_3,\ldots,a_n,\ldots)$$
where each $a_i$ is either $0$ or $1$; these tuples are infinite; we think of the tuple as defining a function whose value at $n$ is $a_n$, so it really corresponds to a function $\mathbb{N}\to\{0,1\}$. Two tuples are equal if and only if they are identical: that is,
$$(a_1,a_2,a_3,\ldots,a_n,\ldots) = (b_1,b_2,b_3,\ldots,b_n,\ldots)\text{ if and only if } a_k=b_k\text{ for all }k.$$

Now, suppose that $f\colon\mathbb{N}\to 2^{\mathbb{N}}$ is a given function.
For each natural number $n$, $f(n)$ is a tuple. Denote this tuple by
$$f(n) = (a_{1n}, a_{2n}, a_{3n},\ldots,a_{kn},\ldots).$$
That is, $a_{ij}$ is the $i$th entry in $f(j)$.

I want to show that this function is not surjective. To that end, I will construct an element of $2^{\mathbb{N}}$ that is not in the image of $f$. Call this tuple $s_f = (s_1,s_2,s_3,\ldots,s_n,\ldots)$. I will now say what $s_k$ is. Define
$$s_k = \left\{\begin{array}{ll}
1 &\mbox{if $a_{nn}=0$;}\\
0 &\mbox{if $a_{nn}=1$.}
\end{array}\right.$$

This defines an element of $2^{\mathbb{N}}$, because it defines an infinite tuple of $0$s and $1$s; this element depends on the $f$ we start with: if we change the $f$, the resulting $s_f$ may change; that's fine.  (This is the "diagonal element").

Now, the question is whether $s_f = f(n)$ for some $n$. The answer is "no." To see this, let $n\in\mathbb{N}$ be any natural number. Then
$$f(n) = (a_{1n},a_{2n},a_{3n},\ldots,a_{nn},\ldots)$$
so the $n$th entry of $f(n)$ is $a_{nn}$. If the $n$th entry of $f(n)$ is $0$, then by construction the $n$th entry of $s_f$, $s_n$ is $1$, so $f(n)\neq s_f$. If the $n$th entry of $f(n)$ is $1$, then by construction the $n$th entry of $s_f$, $s_n$, is $0$. Then $f(n)\neq s_f$ again, because they don't agree on the $n$th entry.

This means that for every $n\in\mathbb{N}$, $s_f$ cannot equal $f(n)$, because they differ in the $n$th entry. So $s_f$ is not in the image of $f$.

What we have shown is that given a function $f\colon\mathbb{N}\to 2^{\mathbb{N}}$, there is some element of $2^{\mathbb{N}}$ that is not in the image of $f$. The element depends on what $f$ is, of course; different functions will have possibly different "witnesses" to the fact that they are not surjective.

Think of the function $f$ being hauled before a judged and accused of Being Surjective; to prove its innocence, $f$ produces a witness to verify its alibi that it's not surjective; this witness is $s_f$, who can swear to the fact that $f$ is not surjective because $s_f$ demonstrates that $f$ is not surjective: $s_f$ is not in $\mathrm{Im}(f)$; if the police hauls in some other function $g$ and accuse that function of being surjective, $g$ will also have to produce a witness to verify its alibi that it isn't surjective; but that witness does not have to be the same witness that $f$ produced. The "witness" we produce will depend on who the "accused" is.

The reason this is called the "diagonal argument" or the sequence $s_f$ the "diagonal element" is that just like one can represent a function $\mathbb{N}\to \{0,1\}$ as an infinite "tuple", so one can represent a function $\mathbb{N}\to 2^{\mathbb{N}}$ as an "infinite list", by listing the image of $1$, then the image of $2$, then the image of $3$, etc:
$$\begin{align*}
f(1) &= (a_{11}, a_{21}, a_{31}, \ldots, a_{k1},\ldots )\\
f(2) &= (a_{12}, a_{22}, a_{32}, \ldots, a_{k2},\ldots)\\
&\vdots\\
f(m) &= (a_{1m}, a_{2m}, a_{3m},\ldots, a_{km},\ldots)
\end{align*}$$
and if one imagines the function this way, then the way we construct $s_f$ is by "going down the main diagonal", looking at $a_{11}$, $a_{22}$, $a_{33}$, etc.

Now, remember the definition of "countable":

Definition. A set $X$ is said to be countable if and only if there exists a function $f\colon\mathbb{N}\to X$ that is surjective. If no such function exists, then $X$ is said to be uncountable.

That means that the theorem we proved above shows that:

Theorem. The set of all binary sequences, $2^{\mathbb{N}}$, is not countable.

Why? Because we showed that there are no surjective functions $\mathbb{N}\to 2^{\mathbb{N}}$, so it is not countable.

How does this relate to the real numbers? The real numbers are bijectable with the set $2^{\mathbb{N}}$. That is, there is a function $H\colon 2^{\mathbb{N}}$ to $\mathbb{R}$ that is both one-to-one and onto. If we had a surjection $\mathbb{N}\to\mathbb{R}$, then composing this surjection with $H$ we would get a surjection from $\mathbb{N}$ to $2^{\mathbb{N}}$, and no such surjection exists. So there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$, so $\mathbb{R}$ is not countable (that is, it is uncountable).

Bijecting $\mathbb{R}$ with $2^{\mathbb{N}}$ is a bit tricky; you can first biject $\mathbb{R}$ with $[0,1]$; then you would want to use the binary representation (as in wikipedia's article), so that each sequence corresponds to a binary expansion, and each number in $[0,1]$ corresponds to a binary sequence (its digits when written in binary); the problem is that just like some numbers in decimal have two representations ($1$ and $0.999\ldots$ are equal), so do some numbers have two representations in binary (for example, $0.01$ and $0.0011111\ldots$ are equal). There is a way of fixing this problem, but it is a bit technical and may obscure the issue, so I would rather not get into it.

Instead, let me note that the set $2^{\mathbb{N}}$ can be mapped in a one-to-one way into $(0,1)$: simply take a binary sequence
$$(a_1,a_2,a_3,\ldots,a_n,\ldots)$$
and map it to the decimal number that has a $5$ in the $k$th decimal position if $a_k=0$, and has a $6$ in the $k$th decimal position if $a_k=1$. Using $5$ and $6$ ensures that each number has only one decimal representation, so the map is one-to-one. Call this map $h$. Define $H\colon\mathbb{R}\to 2^{\mathbb{N}}$ as follows: given a real number $x$, if $x$ is in the image of $h$, then define $H(x)$ to be the unique sequence $s$ such that $h(s)=x$. If $x$ is not in the image of $h$, then define $H(x)$ to be the sequence $(0,0,0,\ldots,0,\ldots)$. Notice that $H$ is surjective, because $h$ is defined on all of $2^{\mathbb{N}}$.

This is enough to show that there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$: suppose that $f\colon\mathbb{N}\to\mathbb{R}$ is any function. Then the function $H\circ f\colon \mathbb{N}\stackrel{f}{\to}\mathbb{R}\stackrel{H}{\to}2^{\mathbb{N}}$ is a function from $\mathbb{N}$ to $2^{\mathbb{N}}$. Since any function from $\mathbb{N}$ to $2^{\mathbb{N}}$ is not surjective, there is some $s\in 2^{\mathbb{N}}$ that is not in the image of $H\circ f$. Since $s$ is in the image of $H$, there exists some $x\in\mathbb{R}$ such that $H(x)=s$. That means that $f(n)\neq x$ for all $n$ (since $H\circ f(n)\neq s$).

Since there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$, that means that $\mathbb{R}$ is uncountable.

So, as to your questions. First, you should understand that the diagonal argument is applied to a given list. You already have all of $s_1$, $s_2$, $s_3$, etc., in front of you. Nobody is allowed to change them. You construct the "diagonal number" (my $s_f$ above) on the basis of that list. Yes, if you change the list then you can put the diagonal number $s_f$ in the new list; but $s_f$ is only a witness to the fact that the original list was not a list of all sequences. If you change to a different list, then I will have to produce a different witness. The witnesses depend on the given list. You know that $s_4$ is not equal to $s_f$ because $s_f$ is constructed precisely so that it disagrees with $s_4$ in the $4$th position, and one disagreement is enough to guarantee inequality.

Wikipedia's presentation seems to argue by contradiction; I don't like to introduce that into these discussions because the argument is difficult enough to "grok" without the added complication.  (The "Otherwise..." part is an argument by contradiction, arguing that if you could 'list' the elements of $T$, then you would apply the argument to show that this 'complete list' is not 'complete', etc). There's no need. Simply, there is no surjection from $\mathbb{N}$ to $T$, as discussed above.

Now, there is a common "first reaction" that this argument would apply "just as well" to the natural numbers: take a list of natural numbers listed in binary, and engineer an argument like the diagonal argument (say, by "reflecting them about the decimal point", so they go off with a tail of zeros to the right; or by writing them from left to right, with least significant digit first, instead of last) to produce a "number" not on the list. You can do that, but the problem is that natural numbers only corresponds to sequences that end with a tail of $0$s, and trying to do the diagonal argument will necessarily product a number that does not have a tail of $0$s, so that it cannot represent a natural number. The reason the diagonal argument works with binary sequences is that $s_f$ is certainly a binary sequence, as there are no restrictions on the binary sequences we are considering.

I hope this helps.

ContentLicense: CC BY-SA 4.0

ContentLicense: CC BY-SA 3.0


问:
Differences between “for each” and “for all” according to Pugh's book
On page 6 of C. C. Pugh's Real Mathematical Analysis (2017)：

Avoid reading $\forall$ as "for all," which in English has a more inclusive connotation.

Instead, one should read $\forall$ always as for each, as the writer suggests.

Could anyone please point out for me what the more inclusive connotation Pugh is referring to?

答:
I'll take a crack at it.

In English, to a non-mathematician, the statement "All green crows pay taxes" might be regarded as false because it suggests that there is at least one green crow that is paying taxes.

Alternatively, even to a non-mathematician, the statement "For each crow that is green, the crow pays taxes" might be regarded as true, because the wording suggests that in order to dispute the statement, the non-mathematician must be concerned about finding a tax-dodging green crow.

So the alternative wording seems to emphasize that in order to dispute the statement, you must find a tax-dodging green crow.  Then the non-mathematician will ask himself: How can I do that if there are no green crows in the first place?


问:
Why are all linear maps that sends one basis to another basis or to 0 still linear?
I notice that in a lot of linear transformation proofs involves making assumptions about mapping a basis vector to another basis vector or to 0. I was wondering why this is even allowed and is this true for all linear maps?

So for example , take an arbitrary map L(U,W). I often see proofs that says, "assume (u1, u2, u3...um) are basis vectors for U and take a linear map where T(u1) is mapped to w1 where w1 is a basis vector for W and T(v2) is mapped to 0 in W."

I was wondering why you are can freely allowed to do this.

答:
A linear mapping $f:U\rightarrow W$ is usually defined by assigning images to a basis $u_1,\ldots,u_n$ of $U$.
Then the mapping is well-defined, i.e., each vector $u=\sum_i a_iu_i$, with $a_i$ scalars, is then assigned an image vector $f(u) = \sum_i a_i f(u_i)$.

The image of $f$ is generally a subspace of $W$ generated by the vectors $f(u_1),\ldots,f(u_n)$.


问:
How is, by its nature, a class differ from a set?
A class is a collection of sets in which the sets are members of the class.

So, may I ask what are the differences between a class and a set (or more precisely a set of sets) by their nature? What motivates mathematicians to give it a separate name?

答:
A set is an element of some class.

We typically identify a set with the class that has the same elements, and say that class is a set.

A class that isn't a set is not an element of any class or thereby identified with an element of any class, is larger than any set (according to the axiom schema of replacement), and is called a proper class.

While arbitrary classes of sets exist, a class is often proven not to be identifiable with any set by showing such a set's existence would imply a contradiction.

A theory with only sets has classes in its metalanguage, but discussing proper classes can be convenient. A first-order theory with classes has an in-a-set predicate, typically defined as being an element of a class. There are also second-order theories, where classes don't have the same quantifiers as sets.


问:
convex function in open interval almost everywhere differentiable
Let $I \subset \mathbb{R}$ be an open interval and $f: I\to \mathbb{R} $ is convex. (That is, whenever $a, b \in I$ and $0\leq t\leq 1$ then $$f((1-t)a + tb) \leq (1-t)f(a) +tf(b).$$ How can I prove that $f$ is differentiable almost everywhere? The proof of Alexandrov theorem or the contents in Rockafellar's "Convex analysis", Theorem 25.5 mentioned in Convex function almost surely differentiable. are too difficult for me, who is studying elementary real analysis. Are there any other proof of this? (I hope there is because it is the simplest case in which $ n =1$.)

答:
This is only a broad outline and not a detailed answer.

$f$ has right-hand and left-hand derivatives at every point. These are both monotone functions and hence they are continuous except at countable number of points. At any point where one the derivatives is continuous the function is actually differentiable. Hence $f$ is differentiable at all but countably many points, in particular almost everywhere.


问:
Simplify summation with factorial and binomial coefficients
I would like to know how to simplify the following summation:

$$\sum_{p=0}^n\quad n!\frac{(2p)!}{(p!)^2}\frac{(2(n-p))!}{((n-p)!)^2}$$

Which rules should I use to simplify it?
Thanks!

答:
The sum is just:
$$
n! \sum_{0 \le p \le n} \binom{2 p}{p} \binom{2 (n - p)}{n - p}
$$
Vandermonde's convolution says:
$$
\sum_{0 \le k \le r} \binom{m}{k} \binom{n}{r - k} = \binom{m + n}{r}
$$
so your mistery sum is just:
$$
n! \binom{2 p + 2 (n - p)}{n} = n! \binom{2 n}{n!} = (2 n)^{\underline{n}}
$$
(here $x^{\underline{n}} = x (x - 1) \ldots (x - n + 1)$).</p>


问:
Prove that if for a number x ∈ R x + x ^(-1) ∈ Z ..... I can't get started it would be very important. please
** Prove that if for a number x ∈ R    x + x ^(-1) ∈ Z ,  then for all n ∈ Z cases x^n + x^(−n) ∈ Z.strong text**

答:
Hint

If $x + \frac{1}{x} = a$, then solve for $x$ in terms of $a$.

Then, prove the following lemma:

If $a$ is an integer and $b,n$ are a positive integers, then

$$(a + \sqrt{b})^n + (a - \sqrt{b})^n ~\text{is an integer}.$$

Hint-2

$$\left[a \times \frac{a \pm \sqrt{a^2 - 4}}{2}\right] ~-~ 1 ~=~ \left[\frac{a \pm \sqrt{a^2 - 4}}{2}\right]^2.$$


问:
How to define piecewise function for joint CDF
Suppose I have $X,Y$, which are continuous random variables, with a joint PDF:
$$
f_{X,Y}(x,y)=
\begin{cases}
    24xy & 0 \leq x \text{ and } 0 \leq y \text{ and } x+y \leq 1 \\
    0 & \text{otherwise} \\
\end{cases}
$$

I want to find the joint CDF, which I have calculated to be $6x^2y^2 $. However, I'm not sure how to define the piecewise function for the joint CDF.

I wrote out the CDF below, but I'm not sure if the cases of $x$ and $y$ are correct for all $x,y \in \mathbb{R}$.
$$F_{X,Y}(x,y)=
\begin{cases}
0 & x < 0 \text { or } y < 0\\
6x^2y^2 & 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \text{ and } x+y \leq 1\\
1 & x > 0 \text{ and } y > 0 \text{ and } x+y > 1\\
\end{cases}$$

答:
Look at the following drawing:

Thus the CDF is the following:

$$F_{XY}(x,y)=6x^2y^2\mathbb{1}_{(0;1)}(x)\mathbb{1}_{(0;1-x)}(y)+\mathbb{1}_{(0;1)}(x)\mathbb{1}_{(1-x;\infty)}(y)+\mathbb{1}_{(1;\infty)}(x)\mathbb{1}_{(0;\infty)}(y)$$


问:
how do I calculate the ismorphism group of a six-nodes-tree?
How do I calculate the ismorphism group of a connected six-nodes-tree? The tree has a node centred and the other 5 nodes are leaves of the graph. I already know the answer is 6, which is the quotient between 720/120. But I do not know where 120 comes from. </p>

答:
You appear to be seeking the number of labelled graphs isomorphic to the star graph $K_{1,5}$, i.e., the size of its isomorphism class.

In general, the size of the isomorphism class a graph $G$ belongs to is given by $$\frac{|V(G)|!}{|\mathrm{Aut}(G)|},$$ where $V(G)$ is the set of vertices of the graph, and $\mathrm{Aut}(G)$ is the automorphism group of the graph.

For $K_{1,5}$, the automorphism group has size $5!=120$: we can permute the $5$ degree-one vertices arbitrarily and the degree-5 vertex must be fixed.  Hence the size of the isomorphism class is $$\frac{6!}{5!}=6.$$

The six labelled graphs in this isomorphism class are:

(Recall that two labelled graphs on the same vertex set are equal if they have the same edge set.)


答:
We can in general have that a linear map in $\mathcal{L}(U,W)$ sends some vectors to $0$. But as long as we assume that $T$ is not the trivial map (ie we are assuming that $T(u) \neq 0$ for at least one vector in $U$), then we can call one such vector $u_{1}$ and extend it to a basis $u_{1}, \ldots, u_{n}$ for $U$ such that $T(u_{1}) \neq 0$; then since $T(u_{1}) \in W \setminus \{0\}$, we can define $w_{1} = T(u_{1})$ and extend $\{w_{1}\}$ to a basis $\{w_{1}, \ldots, w_{m}\}$ for $W$.

This is why it is common to 1: Assume that $T$ is not the zero map, 2: take bases for $U$ and $W$ for which $T(u_{1}) = w_{1}$. If you start assuming that, for instance $T(u_{1}) = w_{1}$, $T(u_{2}) = w_{2}$, etc are linearly independent vectors in $W$, then you are making more complicated assumptions about the rank of $T$; but assuming $T$ is nontrivial is usually reasonable (because the trivial map can be rather easily considered separately).

Title: Why are all linear maps that sends one basis to another basis or to 0 still linear?

Body:
I notice that in a lot of linear transformation proofs involves making assumptions about mapping a basis vector to another basis vector or to 0. I was wondering why this is even allowed and is this true for all linear maps?

So for example , take an arbitrary map L(U,W). I often see proofs that says, "assume (u1, u2, u3...um) are basis vectors for U and take a linear map where T(u1) is mapped to w1 where w1 is a basis vector for W and T(v2) is mapped to 0 in W."

I was wondering why you are can freely allowed to do this.


问:
Joint Probability Density Function P(Y < 1/2) integration
Let X and Y be continuous random variables that have the following joint probability density function:
$$
f(x,y) =
\begin{cases}
3y\ ,\ 0\le x\le y \le1 \\
0 \ \ \  ,\ elsewhere
\end{cases}
$$

I'm trying to find $P(Y < 1/2)$ I'm having difficulty understanding the precise steps, and because 1/2 is a constant, I'm wondering if I'm supposed to find this probability via $\int_0^{1/2}3y \ dy$, $\int_x^{1/2}\int_0^y 3y \ dy$, or some other limits and ordering here. Every resource, including online, textbook, and accompanying exercises includes another variable in the parentheses.

Body: $f_Y(y)=\int f_{X,Y} (x,y)dx=3y^{2}$ for $0\leq y \leq 1$. Hence $P(Y<\frac 1  2)=\int_0^{\frac 1  2} 3y^{2}dy=\frac 1 8$.


问:
Convergence of a geometric series but the sum takes square integers
I want to know if there is a closed form for the convergence of a series of this kind:

$$\sum_{j=1}^\infty r^{j(j-1)/2}$$  given that $0<r<1$. As it is possible to see the terms of the sum are $r^{(j^2-j)/2}$, which gives problems if trying to solve by the usual way of a geometric series.

And also I have the same problem with a series which is similar to the last one:

$$\sum_{j=1}^\infty \gamma^{j}r^{j(j-1)/2}$$  I want to know the closed form for this one also, given that $0<r<1$ and $0<\gamma<1$.

I will really appreciate any idea to solve these series.

答:
$$\sum_{j=1}^\infty r^{j(j-1)/2}=\sum_{j=1}^\infty {\sqrt r}^{\,j(j-1)}=\frac{\vartheta _2\left(0,\sqrt{r}\right)}{2 \sqrt[8]{r}}$$ where appears Jacobi theta function.

$$\sum_{j=1}^\infty \gamma^{j}r^{j(j-1)/2}= ???$$


问:
Converting set of inequalities to LP
I have a problem where I have a bunch of inequalities in the form:

$a_{1,1}b_{1,1} + a_{1,2}b_{1,2} + ... + a_{1,n}b_{1,n} >
   a_{2,1}b_{2,1} + a_{2,2}b_{2,2} + ... + a_{2,n}b_{2,n}$

$a_{2,1}b_{2,1} + a_{2,2}b_{2,2} + ... + a_{2,n}b_{2,n} >
   a_{3,1}b_{3,1} + a_{3,2}b_{3,2} + ... + a_{3,n}b_{3,n}$

$a_{1,1}b_{1,1} + a_{1,2}b_{1,2} + ... + a_{1,n}b_{1,n} <
   a_{3,1}b_{3,1} + a_{3,2}b_{3,2} + ... + a_{3,n}b_{3,n}$

etc.

I've randomly distributed the < and > signs in the inequalities for the example; they can be any of ≤, ≥, <, or >. All the b variables are dummy variables (either 0 or 1).

I want to find values of all the a variables, assuming possible given the data. This seems like a linear programming problem, but I'm struggling to convert it to one.

One further problem is that I may come to a point where there is not solution by classic linear programming (ex. the example above). In this case, I would want to find the point that is "closest" to being a valid point (think OLS error reduction). I know how to do this for linear programming (another reason I jumped to lp), but am unsure on how to do it for the problem statement above. As a side note, I plan to do this mainly computationally.

What method would I use to solve the above?

答: Accepting Micheal Adamaszek's answer that he commented. It's already in LP form, just move the terms to one side to get to standard form.</p>

Title: Converting set of inequalities to LP


答:
Let $V$ and $W$ be two vector spaces.

For the sake of simplicity assume that $\dim(V)=n$ and let $\{v_1,...,v_n\}$ be a basis.

Also, let $w_1,...,w_n$ just any choice of $n$ vectors in $W$, possibly with repetitions.

Then there is a unique linear map $T:V\rightarrow W$ such that $T(v_i)=w_i$ for all $i=1,...,n$. Since any vector $v\in V$ can be expressed in a unique way as a linear combination $v=a_1v_1+\cdots+a_nv_n$ the map $T$ can be defined as
$$
T(v)=a_1w_1+\cdots+a_nw_n.
$$
It is a standard exercise to check that $T$ is well defined, linear and unique in the sense that if $T^\prime$ is another linear map such that $T^\prime(v_i)=w_i$ for all $i$ then $T=T^\prime$.


问:
Problem Finding the limit with 2 parameters where $I(n,m) = \lim\limits_{x \to \pi} \frac{\sin (nx)}{\sin (mx)}$
Here i have a problem since x approaches pi

So here i have to calculate the Limit and then add it

I (2019,2020) + I (2018,2020) + I (2019,2021) rounded upto 4 decimal places

I got a answer of 2.9975 which is wrong i need  to know how to approach this problem

Body:
Hospital:

$\lim_{x \rightarrow π}\dfrac{n\cos (nx)}{m\cos(mx)}=$

$(n/m)(-1)^n(-1)^m=$

$(n/m)(-1)^{n+m}.$


问:
Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$
Since we have $\frac00$,I Applied L'Hopital rule :

$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$

But as you can see it is getting very ugly.

Body:
$$y=\dfrac{(1+x)^{\tfrac1x}-e}{x}$$
$$z=(1+x)^{\tfrac1x}\implies \log(z)=\tfrac1x\log(1+x)$$ Now Taylor
$$\log(z)=\tfrac1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$
$$z=e^{\log(z)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$
$$y=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$


问:
Proving $ab^T$ is diagonalizable $\iff a^T b \neq 0$
Assuming $a$ and $b$ as two $n$-element vectors, how can we show that $$a b^T$$ is diagonalizable, if and only if their dot product is nonzero ( = they are not orthogonal)?

I've tried to prove $ab^T$ has $n$ linearly independent eigenvectors and then, automatically as a result, the beginning statement would be true, but I've not been successful in proving it.

Any help would be appreciated.

答:
(It's understood that $a$ and $b$ are nonzero, right?)

Here are some ideas to get you started: Nonzero multiples of $a$ are the only possible eigenvectors with $\lambda \neq 0$. The eigenspace belonging to $\lambda=0$ is the orthogonal complement of $b$.


答:
The matrix $ab^T$ has a very specific structure. If the coordinates of $a$ and $b$ are $a_i$ and $b_i$ then
$$ab^T=\begin{pmatrix}
a_1b_1 & a_1b_2 & \cdots & a_1b_n \\
a_2b_1 & a_2b_2 & \cdots & a_2b_n \\
\vdots & \vdots & \vdots & \vdots \\
a_nb_1 & a_nb_2 & \cdots & a_nb_n
\end{pmatrix}$$
Since the rows are all multiple of one another, it shows that $0$ is an eigenvalue with high multiplicity, try finding the corresponding eigenspace. Then find the remaining eigenvectors.

Title: Proving $ab^T$ is diagonalizable $\iff a^T b \neq 0$

Body:
Assuming $a$ and $b$ as two $n$-element vectors, how can we show that $$a b^T$$ is diagonalizable, if and only if their dot product is nonzero ( = they are not orthogonal)?

I've tried to prove $ab^T$ has $n$ linearly independent eigenvectors and then, automatically as a result, the beginning statement would be true, but I've not been successful in proving it.

Any help would be appreciated.


问:
Projection transormation, proof of existence and uniqueness
Question:
A Projection Transformation is defined to be a linear transformation from V to V that satisfies $T^2=T$.

Let $V=U \oplus W$ . Show that there exists only one linear projection $T:V \to V$ s.t. $KerT=W, ImT=U$

What I did

I managed to prove that if $KerT=V, ImT=${$0$} then $T^2=T$. I didn't succeed in proving that other cases are impossible (or are they possible)?

答:
Nontrivial projections on vector spaces certainly exist: take $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $[x,y]^T \mapsto [x-y, x-y]^T.$ Then $\ker T $ is precisely the line $y=x,$ a one-dimensional subspace of $\mathbb{R}^2.$

For $v\in V,$ write $v= u + w.$ This representation is unique by the definition of the direct sum operation. Since $T(w) =0,$ we have $T(v) = T(u).$ Moreover, since $T^2 (u) = T(u),$ considering the restriction map $T' = T_{|U} : U \rightarrow U  = T(V)$ we have that $T'$ is a bijection - applying inverses shows that $T'(u) = u$ for all $u\in U.$ Hence $T(u+w)=u$ for all $v = u+w\in V.$ Linearity and well-definedness are easy to check.


问:
Help with a Riemann-Stieltjes integral
Given the function g:[0,1]->R: g(x)=0 if 0$\le$x$\le$$\frac{1}{2}$,g(x)=1 if $\frac{1}{2}$$\lt$x$\le$1. Prove that f es g- integrable in [0,1] if and only if  f is continuos from the right in $\frac{1}{2}$. In this case the value of the integral is f($\frac{1}{2}$).

Any insight is helpful :)

答:
If I'm not mistaking, this is wrong.

Take:

$$f(x)=\begin{cases}
\sin \left(\frac{1}{1/2-x}\right) & \text{if } 0 \le x \lt 1/2\\
0 & \text{if } 1/2 \le x \le 1
\end{cases}$$

$f$ is bounded and right continuous at $1/2$. However, you can find a sequence of partitions $\{P_n\}$ whose norms approach zero and for which $S(P_n,f,g) = (-1)^n$.

Title: Help with a Riemann-Stieltjes integral


问:
Understanding nonstandard Peano arithmetic
I've had the idea of nonstandard Peano arithmetic introduced to me in the comments of this question. The concept that we could write down the axioms which produce the natural numbers and also produce something else from them is totally crazy to me! I did not expect this.

I've been doing some reading here, to try and understand what these non-standard models of the Peano axioms are. It's difficult for me to follow the details of this paper, but I've understood that the most simple non-standard model of the Peano axioms is
$$
M = \{\mathbb{N}, +, ×, ≤, 0, 1, c\}
$$

I understand this to be the natural numbers along with an additional constant $c$, is that correct?

So from this understanding, it makes sense to me to ask the questions,

What is $n+c$ for $n$ a natural number?

What is $n\cdot c$ for $n$ a natural number?

What is $c+c$?

What is $c\cdot c$?

Is $c≤n$  for $n$ a natural number?

Are these sensible questions? If not, is there some similar concrete way to understand this model?

答:
The appearance of nonstandard models of Peano arithmetic can be understood intuitively as an entropic effect. You see, one of the Peano axioms is called the induction axiom. It gives us sufficient conditions to determine if a set of natural numbers encompasses all of them - namely:

Any set of natural numbers $A$, which has the two properties that $0 \in A$, and for every $n \in A$, its successor $S(n)$ is also in $A$, contains all the natural numbers.

In effect, this process is precisely to "weed" such non-standard numbers out by taking each way you can form, in some candidate for a model of the natural numbers, such a set containing 0, some natural numbers and all their successors, and then throwing away whatever numbers aren't included in it, and repeating this over an over (technically, a "supertask"), until all the unwanted numbers are flushed.

Non-standard models appear when we weaken this statement from "any set" to "any describable set" in some description language. (Typically, this description language is first-order logic.) In this case, there are "indescribable" sets, i.e. random sets (hence "entropic", from before) in some regard, that have no better way to describe them than to list out every element, or do some other equivalently infinitely detailed process. But "describability" implies a finite description, hence those sets are not covered. Some nonstandard numbers make it through.

That also answers your question regarding what such a model "looks like". We can't really write one down explicitly, for this very reason: the sets containing the nonstandard numbers, while also meeting the hypotheses of the above principle, have to be sneaky enough in what they do and don't include in order to "evade" every single statement we could pose in our description language that might suss out the nonstandard numbers in them. Worse, we can't just switch to another, because it can be shown that answering this kind of question about languages is one of those "undecidable" problems, in the same sense as the Halting Problem. No physically possible computing machine can perform the necessary computation to generate such a model.

However, all is not entirely hopeless. Consider the case of a countable nonstandard Peano model - it has "as many" (in the sense of infinite cardinality) naturals as the standard one, but it also includes nonstandard numbers. One can prove that the basic order structure of the model is that beyond all the natural numbers, there is a "furry infinityey nebulaline" of nonstandard, "infinitely big" natural numbers looming out there that consists of doubly-unended chains of such that are themselves ordered densely in the same way the rational numbers are. In particular, you might be tempted to say that

a standard natural number is of the form $n$, for $n$ a natural number,

a nonstandard natural number is of the form $0_q + m$, where $q$ is a rational number and $m$ is an integer.

The "origin" constants $0_q$ are ordered so that if $q_1 < q_2$ as rationals, then $0_{q_1} < 0_{q_2}$, and every $0_q$ also is beyond every natural $n$, i.e. $n < 0_q$ for every standard $n$.

What is indescribable is what the subscripts for sums $0_{q_1} + 0_{q_2}$ and products $0_{q_1}0_{q_2}$ have to be, to ensure that the weakened induction axiom holds.

(This also works for uncountable models, too - only here now the index set of constants becomes some larger, uncountable set, and the "nebula" becomes denser, hazhyer, and worse. Only now, it's even worse to try and "describe" the sums and products.)


问:
How to solve this indefinite integral of a sine curve?
I am carrying out through my report but I got into a part with integration that I don’t know how to solve because I am still a beginner on integration.
$$
\int-0.807 \sin \left(\frac{π}{12}  t\right) e^{0.269} dt
$$

答:
There is a small trick to be used in this case. First, we have to review a bit of differentiation.

If we have a function $f(x)$ and we know its derivative $f'(x)$, then a "scaled" version of $f(x)$ is differentiated like so
$$ \tag{1}
\frac{d}{dx}f(kx) = kf'(kx)
$$
This is an important result to learn and know, before learning integration.

So back to the question itself. We have the following integral (I'll remove the constants):
$$
\int \sin \left( \frac{\pi}{12}t \right)\, dt
$$
How would we solve this? We certainly know that $\int \sin t \, dt = -\cos t +C$, but how do we deal with the factor $\frac{\pi}{12}$ ? Well, we can modify the expression a bit:
$$
\int \sin \left( \frac{\pi}{12}t \right)\, dt = \frac{12}{\pi} \int \frac{\pi}{12} \sin \left( \frac{\pi}{12} t \right) \, dt
$$
I have done nothing else than to multiply and divide by the same constant. Please convince yourself that this is true. But how does this help us? Well, looking at the integral now, we see that
$$
\int \underbrace{ \frac{\pi}{12} \cdot \sin \left( \frac{\pi}{12} t \right)}_{\text{this}} \, dt
$$
reminds of the thing that we learned from differentiation! It looks like the right-hand side of Equation (1). Perhaps it could be that $k = \frac{\pi}{12}$ and $f'(kt) = \sin \left( \frac{\pi}{12} t \right)$... Since integration is the reverse operation of differentiation, we only need to figure out what function $f(x)$ would have this feature. In other words
$$
\int k f'(kt) \, dt = f(kt) + C
$$
Can you continue from here?


问:
Proving a limit with epsilon delta
For the sequence
$$s_n = \frac{n}{n+\sqrt n},$$
I am trying to prove that the limit as $n\to+\infty$  is $1$ with $\varepsilon-\delta$ type proof.

So i currently have that:
$|f(x)-L| < \varepsilon$ if $x>\delta$

$|x/(x+\sqrt x)-1|<\varepsilon$

I dont know how to manipulate the LHS or what epsilon to choose that will help me with the proof any help is appreciated, thanks.

答:
Observe that
\begin{align*}
\biggl|\frac{n}{n+\sqrt{n}} -1\biggr| = \biggl|\frac{\sqrt{n}}{n+\sqrt{n}}\biggr|\leq \frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}
\end{align*}
Hence you have to choose an $n$ such that $\epsilon >\frac{1}{\sqrt{n}}$ happens.

Title: Proving a limit with epsilon delta


答:
For the case of sequences, the $\varepsilon-\delta$ type proof is as follows:

Given $\varepsilon>0$, you need to find $N\in\Bbb N$ (depending on $\varepsilon$) such that for all $n\ge N$, it holds that $|s_n-L|<\varepsilon$ (in your case, $|s_n-1|$).

For that purpose, observe that $$|s_n-1|=\left|\frac{n}{n+\sqrt n}-1\right|=\left|\frac  {n-(n+\sqrt n)}{n+\sqrt n}\right|=\frac{\sqrt n}{n+\sqrt n}=\frac{1}{\sqrt n+1}.$$

So, given $\varepsilon>0$, let $N\in\Bbb N$ such that $1+\sqrt N>\varepsilon$ (for example, $N:=\lfloor(\varepsilon-1)^2\rfloor$).

I think you can finish from here.

Title: Proving a limit with epsilon delta

Body:
For the sequence
$$s_n = \frac{n}{n+\sqrt n},$$
I am trying to prove that the limit as $n\to+\infty$  is $1$ with $\varepsilon-\delta$ type proof.

So i currently have that:
$|f(x)-L| < \varepsilon$ if $x>\delta$

$|x/(x+\sqrt x)-1|<\varepsilon$

I dont know how to manipulate the LHS or what epsilon to choose that will help me with the proof any help is appreciated, thanks.


问:
Why is this derivative incorrect?
We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$

I would like to know why my approach is incorrect:

$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)}  - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$

$$ =  \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$$

$$ = \dfrac {2 \sin(x) - \sin(2x) \cdot \cos(x)}{\cos^3(2x) \cdot \sin^2(x)}$$

p.s. - To avoid confusion ; I wanted to get rid of the $\tan$. I'm sure there is a shorter method than this but I don't want it; I just want to know why this is wrong.

答:
Third line is: $ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)\cdot\cos^2(2x)}{\cos^2(2x) \cdot \sin^2(x)}$
instead of $\\\\ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$</p>

Title: Why is this derivative incorrect?


问:
How to solve a minimization problem with infinite norm objective?
Recently, I came up with an idea on my research about beamforming. It can be formulated as,

$$\begin{array}{ll} \underset{w \in \mathbb{C}^{1\times N}}{\text{minimize}} & \| H w^H \|_{\infty} \\ \text{subject to} & a w^H = 1\end{array}$$

where matrix $H \in \mathbb{C}^{M \times N}$ and vector $a \in \mathbb{C}^{1\times N}$ are given.

I haven't found any paper or method to solve it. My question is how to solve this kind of infinite norm objective function. Is there any fixed way to solve this kind of problem?

答:
If the problem was set over real numbers the you could do the following.
You can introduce an auxiliary variable $t\ge0$, that encodes the $\infty$-norm:
$$
-t \le (Hw^H)_i \le t.
$$
Then minimize $t$ subject to these linear constraints. The resulting problem is now a linear programming problem.

For this complex case, one can do similar:
$$
|(Hw^H)_i| \le t.
$$
You end up with more constraints ...


问:
How Can I calculate the standard deviation of a yes/no question?
Is it possible to calculate the standard deviation of a single yes/no question? For example, 100 people took the survey and 60 of them chose "yes", 40 of them chose "no".

I'm thinking using binomial distribution sd formula:

$$
\sqrt{np(1-p)}
$$

so it would be
$\sqrt{100*(60/100)*(1-60/100)}$

is this correct?

答:
Comment: The previous comment from @JMoravitz refers to the variance of a binomial random variable. Perhaps your question is about a sample. So here is an example with a sample of $n=100$ Yes/No observations (coded as 1=Yes, 0=No).

Suppose you roll a fair die 100 times, counting 1s as Yes and any other number as No. Then you might assign the value $1$ to 1 (Yes) and the value $0$ to other outcomes (No). We might
generate such a sample of $0$s and $1$s in R as follows:

```<code>set.seed(2020)
x = rbinom(100, 1, 1/6)
x
 [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1
[26] 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0
[51] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0
[76] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
table(x)
x
 0  1
86 14
</code>```

In this particular sample you have 14 Yes's and 86 No's.

You could use the usual formal
$S = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2$ to find the
sample standard deviation. The computation from R is as follows:

```<code>var(x)
[1] 0.1216162
</code>```

Note: This sample might be considered as 100 observations from
the binomial (Bernoulli) distribution $\mathsf{Binom}(n=1, p= 1/6),$ which has mean $\mu = np = 1/6 = 0.6667$ and variance $\sigma^2 =
np(1-p) = 5/36 =  0.1389.$  Notice that the sample variance $S = 0.1216$ is a fairly close estimate of $\sigma^2$---about as good
an estimate as one can expect from a sample of size $n=100.$

In four additional samples of size $n = 100,$ I got sample variances
$0.1616, 0.1789, 0.1067,$ and $0.1216,$ as shown below.

```<code>var(rbinom(100,1,1/6))
[1] 0.1616162
var(rbinom(100,1,1/6))
[1] 0.1788889
var(rbinom(100,1,1/6))
[1] 0.1066667
var(rbinom(100,1,1/6))
[1] 0.1216162
</code>```

Body:
Comment: The previous comment from @JMoravitz refers to the variance of a binomial random variable. Perhaps your question is about a sample. So here is an example with a sample of $n=100$ Yes/No observations (coded as 1=Yes, 0=No).

Suppose you roll a fair die 100 times, counting 1s as Yes and any other number as No. Then you might assign the value $1$ to 1 (Yes) and the value $0$ to other outcomes (No). We might
generate such a sample of $0$s and $1$s in R as follows:

set.seed(2020)
x = rbinom(100, 1, 1/6)
x
 [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1
[26] 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0
[51] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0
[76] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
table(x)
x
 0  1
86 14

In this particular sample you have 14 Yes's and 86 No's.

You could use the usual formal
$S = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2$ to find the
sample standard deviation. The computation from R is as follows:

var(x)
[1] 0.1216162

Note: This sample might be considered as 100 observations from
the binomial (Bernoulli) distribution $\mathsf{Binom}(n=1, p= 1/6),$ which has mean $\mu = np = 1/6 = 0.6667$ and variance $\sigma^2 =
np(1-p) = 5/36 =  0.1389.$  Notice that the sample variance $S = 0.1216$ is a fairly close estimate of $\sigma^2$---about as good
an estimate as one can expect from a sample of size $n=100.$

In four additional samples of size $n = 100,$ I got sample variances
$0.1616, 0.1789, 0.1067,$ and $0.1216,$ as shown below.

var(rbinom(100,1,1/6))
[1] 0.1616162
var(rbinom(100,1,1/6))
[1] 0.1788889
var(rbinom(100,1,1/6))
[1] 0.1066667
var(rbinom(100,1,1/6))
[1] 0.1216162


答:
The claim is not true. If $a=0$ or $b=0$ then $ab^T$ is trivially diagonalizable.

If $a$ and $b$ are non-zero but $a^Tb=0$ then the matrix $ab^T$ is a non-zero nilpotent matrix, hence not diagonalizable.

Now assume that $a$ and $b$ are both non-zero, $a^Tb\ne0$, and $n\ge2$. Then the matrix has rank one but is not invertible. So $\lambda=0$ is an eigenvalue. Since the matrix has rank one, the eigenspace of $\lambda=0$ is $n-1$ dimensional. In order that $ab^T$ is diagonalizable there must be a non-zero eigenvalue. Its corresponding eigenvector has to be an element of the column space of the matrix. Hence it has to be a multiple of $a$. It is now easy to verify that $a$ is an eigenvector to the eigenvalue $a^Tb\ne0$. This gives us a basis of eigenvectors, and the matrix is diagonalizable.


问:
Numerical Analysis or Numerical Linear Algebra Book with Available Solutions?
I am a beginning graduate student in Mathematics soon and I am planning to self-study Numerical Analysis and Numerical Linear Algebra. I know there are already reference-request questions about this, but I am looking for some more specific books.

I am looking for books in the following categories:

Books with difficult problems, and solutions in the back (ideal)

Books with difficult problems

Books with solutions in the back

Books with separate solution manuals

Problem books (ideally with solutions in the back, or solution manuals)

Could you please recommend me some textbooks, and tell me in which category they are? Thank you very much, any recommendations are immensely appreciated!

答:
For a more proof-heavy treatment, I'd suggest

Isaacson, E., & Keller, H. B. Analysis of numerical methods.

A comprehensive treatment of several methods, with emphasis on proofs and theory. No solutions in the back, though.


问:
Simplify summation with factorial and binomial coefficients
I would like to know how to simplify the following summation:

$$\sum_{p=0}^n\quad n!\frac{(2p)!}{(p!)^2}\frac{(2(n-p))!}{((n-p)!)^2}$$

Which rules should I use to simplify it?
Thanks!

答:
You may start with the generating function of $\displaystyle \frac{(2k)!}{(k!)^2}$ which is :
$$f(x):=\sum_{k=0}^\infty\ \frac{(2k)!}{(k!)^2}x^k=\sum_{k=0}^\infty\ \binom{2k}{k}x^k$$
but this is the well known central binomial generating function :
$$f(x)=\frac 1{\sqrt{1-4x}}$$
and compute the square of $f(x)$ so that the coefficients of the 'diagonal terms' $\;\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{(n-k)}x^kx^{n-k}\,$ will be your sums for increasing values of $n$

(ignoring the 'constant' factor $n!$) :
$$f(x)^2=\frac 1{1-4x}=\sum_{n=0}^\infty\ 4^n\;x^n$$
From this we deduce simply :
$$\boxed{\displaystyle n!\sum_{p=0}^n\ \frac{(2p)!}{(p!)^2}\frac{(2(n-p))!}{((n-p)!)^2}=\ n!\,4^n}$$


问:
If $A$ is a subgroup of $B$, just it hold that $\text{Aut}(A)$ is a subgroup of $\text{Aut}(B)$?
Let $\text{Aut}(G)$ denote the group of automorphisms of $G$ and let $A\subseteq B$ denote $A$ is a subgroup of $B$. Does the following hold:

$$A\subseteq B\implies \text{Aut}(A)\subseteq \text{Aut}(B)$$

If not, is there a necessary and sufficient condition for this to hold?

答:
This question is hard, and there's not a simple answer. In general, there's no reason a function preserving only the $A$ structure should extend to one preserving all the extra $B$ structure. You can find a discussion about which automorphisms can be extended here, but the answer may not satisfy you.

So I suppose the answer is "no", and the answer to the follow-up question about a necessary and sufficient condition is "not a simple one".

There is also some good discussion in the comments of this question

I hope this helps ^_^


问:
What is the surface equation of a spiral strip?
I have a strip in the $xy$ plane, whose inner side is a spiral. The strip has a finite width. The image of the strip is shown below. Since, the inner side is spiral (shown in red), the polar equation of it is:  $r_1=a.e^{b\theta}$ and the other side's equation is, $r_2=r_1+d$  ,where $d$ is the width of the strip. for both $r_1$ and $r_2$ , $0\leq\theta\leq2\pi$. Now my question is how can we express the surface equation? My attempt is simple, the equation (in cylindrical co-ordinate) is: $$z=0, r_2\leq \rho \leq r_1 , 0\leq\theta\leq2\pi$$
Is there any better way to write down the strip equation?

答:
If $d$ is intended to represent the minimum distance between the two edges of the spiral strip, then your parametrization $$r_1 = a e^{b \theta}, \quad r_2 = r_1 + d,$$ is not correct.  However, it is true that the curves $r_1$, $r_2$ are separated by a constant width; it just isn't $d$, as your own figure shows.  In your figure, $d$ is the length of the black line segments.  However, the width of the strip is measured along a line perpendicular to a given point on either curve.

You have in essence stumbled upon a unique property of the logarithmic spiral $$r = a e^{b \theta},$$  namely any ray from the origin intersecting the spiral does so at a constant angle.  This is a consequence of the spiral's self-similarity.  The composite transformation $$(r, \theta) \mapsto (r e^{b \varphi}, \theta + \varphi)$$comprising an anticlockwise rotation of $\varphi$ and dilation $e^{b \varphi}$ will map the spiral to itself for any angle $\varphi$.

Unfortunately, the relationship between the strip width and the value of $d$ does not have an elementary closed form.


问:
Lebesgue measure of $\partial A$ given $m(A^o) = m(\overline{A})$
Suppose that $A \subseteq \mathbb{R}$, and consider the Lebesgue outer measure $m: \mathcal{P}(\mathbb{R}) \to [0,\infty]$ on $\mathbb{R}$. Denote by $A^o$, $\overline{A}$ and $\partial A$ the interior, closure and boundary of $A$, respectively.

Suppose that $m(A^o) = m(\overline{A})$. Is it then true that $m(\partial A) = 0$?

This is certainly true if $m(A) < \infty$. Since open sets and closed sets are Lebesgue measurable, we have that $A^o$ and $\overline{A}$ are Lebesgue measurable, and hence so is $\partial A = \overline{A} \setminus A^o$. Therefore, we can write $m(\overline{A}) = m(A^o) + m(\partial A)$. By the monotonicity of the Lebesgue outer measure, we also have that $m(A^o) < \infty$. Hence, we can cancel to conclude. But what about the case $m(A) = \infty$?

If we take for example $A = \bigcup_{n \geq 1} (n - 1, n)$, then $m(A^0) = m(\overline{A}) = \infty$, but $m(\partial A) = m(\mathbb{N}) = 0$. Are there any counter examples?

答: Let $A$ be the set you have constructed and let $C$ be  a fat Cantor set, i.e. a Cantor like set of positive measure. The $C \cup A$ would satisfy your requirements.


问:
Convergence or divergence of the gaussian hypergeometric function $_{2}F_1(a,b;c;z)$ where $a,b$ and $c$ are complex.
I'm studying the convergence of the hypergeometric function and I have some questions. First, I define the hypergeometrical functions as
\begin{equation}
_{2}F_{1} (a,b;c;z)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(a)_n(b)_n}{(c)_n}\dfrac{1}{n!}z^n,
\end{equation}
where $(w)_n$ is the rising factorial and it is defined by
\begin{equation}
(w)_n=\left\lbrace \begin{array}{cc}
1 & \text{for}~n=0\\
w(w+1)\cdots (w-n+1) & \text{for}~n\neq 0
\end{array}\right.
\end{equation}

and the parameters $a$, $b$ and $c$ are complex.

Then, I know that:

For $|z|=1,z\neq 1$, we have that it is not absolutely convergent, but convergent for $0\leq \Re(a+b-c)<1$. But I don't know how to prove that it is convergent there. Moreover, in this case, if $\Re(a+b-c)>1$ we obtain that it is divergent and if $\Re(a+b-c)=1$ and $\Re(a+b)\leq\Re(ab)$ it is also divergent. But I don't know again how to see it.

Thanks a lot.

答:
Assume that $a$, $b$, and $c$ are complex parameters.

For large $n$, we have $$ \begin{align}\frac{(a)_n(b)_n}{(c)_n}\frac{1}{n!} &= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \frac{\Gamma(a+n) \Gamma(b+n)}{\Gamma(c+n)} \frac{1}{\Gamma(1+n)} \frac{\Gamma(n)}{\Gamma(n)} \\ &= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \frac{\Gamma(a+n)}{\Gamma(n)} \frac{\Gamma(b+n)}{\Gamma(n)} \frac{\Gamma(n)}{\Gamma(c+n)} \frac{1}{n} \\ & \sim  \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} n^{a}n^{b}n^{-c}n^{-1} \tag{1} \\&=  \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} n^{a+b-c-1}. \end{align} $$

Therefore, by the p-series test, the series converges absolutely for $|z|=1$ and $\Re(a + b -c) <0$.

$(1)$ $\Gamma(\beta+n)  \sim \Gamma(n)n^{\beta} $ as $n \to \infty$

To show that the series converges conditionally for $|z|=1$, $z \ne 1$, and $0 \le \Re(a+b-c) <1$,  we can use the proof of Dirichlet's test .

Saying that $|z|=1$, $z \ne1$, is equivalent to saying that $z^n= e^{i n \theta} = \cos(n \theta) + i \sin(n\theta)$ for some $\theta$ between $0$ and $2 \pi$.

And the partial sums of $\cos(n \theta)$ and $\sin (n \theta)$ are bounded.  (See here, for example.)

Looking at the proof of Dirichlet's test, we need to show that $$ \sum_{n=1}^{\infty} \left| n^{a+b+c-1} - (n+1)^{a+b+c-1}\right| \tag{2}$$ converges for $\Re(a+b-c) <  1$.

To show that it converges, let $\alpha = a+b-c$.

Then for large $n$, we have $$ \begin{align} n^{\alpha-1}-(n+1)^{\alpha-1} &= n^{\alpha-1}-n^{\alpha-1} \left(1+ \frac{1}{n} \right)^{\alpha-1} \\ &=n^{\alpha-1} -n^{\alpha-1} \left(1+ \frac{\alpha-1}{n} + \mathcal{O}(n^{-2})  \right) \\ &= (1- \alpha) n^{\alpha-2} + \mathcal{O}(n^{\alpha-3}). \end{align}$$

Therefore, as $n \to \infty$, $$|n^{\alpha-1} - (n+1)^{\alpha-1}| \sim |1-\alpha|n^{\Re(\alpha)-2},$$ which, by the p-series test, shows that $(2)$ converges for $\Re(\alpha) = \Re(a +b -c) < 1$.


问:
Solve linear algebra expression $- \nabla . G$
Let's consider the vector $G$

\begin{equation}
G = \begin{pmatrix}
-x(\alpha + \beta) + y \beta \\
x \beta - y (\alpha + \beta)
\end{pmatrix}
\end{equation}

with constants $\alpha$ and  $\beta$. I would like to solve $$- \nabla . G$$

答: The vector function $G=(G_1,G_2)$ (not a matrix) has derivatives $$\frac{\partial G}{\partial x}=\begin{pmatrix}-\alpha-\beta\\\beta\end{pmatrix},\qquad\frac{\partial G}{\partial y}=\begin{pmatrix}\beta\\-\alpha-\beta\end{pmatrix}$$ Hence $$\nabla\cdot G=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}=-2(\alpha+\beta)$$


答:
Consider $q_\bullet:\Bbb N\setminus\{0\}\to\Bbb Q\cap (0,1)$ a bijective map, and call $$B=\bigcup_{n=1}^\infty \left(q_n-\min\left\{\frac123^{-k}, \lvert q_n\rvert\right\};q_n+\min\left\{\frac123^{-k}, \lvert 1-q_n\rvert\right\}\right)$$

Then, $B=B^\circ$, $\overline B=[0,1]$ and $\partial B=[0,1]\setminus B$. It is clear that $$0<m(B^\circ)\le\sum_{k=1}^\infty 3^{-k}=\frac12\\ m(\partial B)\ge\frac12$$ Now, call $A=\bigcup\limits_{n\in\Bbb Z}n+B$ and you'll see that $A^\circ =A$, $\overline A=\Bbb R$ and $\partial A=\Bbb R\setminus A=\bigcup\limits_{n\in\Bbb Z}n+\partial B$. Notice that $A$ is a disjoint union, and that $\mu((n+\partial B)\cap(m+\partial B))=0$ for all $m\ne n$. Therefore $m(A)=m(A^\circ)=\infty\times m(B)=\infty$, $m(\overline A)=m(\Bbb R)=\infty$ and $m(\partial A)=\infty\times m(\partial B)=\infty$.

Title: Lebesgue measure of $\partial A$ given $m(A^o) = m(\overline{A})$

Body:
Suppose that $A \subseteq \mathbb{R}$, and consider the Lebesgue outer measure $m: \mathcal{P}(\mathbb{R}) \to [0,\infty]$ on $\mathbb{R}$. Denote by $A^o$, $\overline{A}$ and $\partial A$ the interior, closure and boundary of $A$, respectively.

Suppose that $m(A^o) = m(\overline{A})$. Is it then true that $m(\partial A) = 0$?

This is certainly true if $m(A) < \infty$. Since open sets and closed sets are Lebesgue measurable, we have that $A^o$ and $\overline{A}$ are Lebesgue measurable, and hence so is $\partial A = \overline{A} \setminus A^o$. Therefore, we can write $m(\overline{A}) = m(A^o) + m(\partial A)$. By the monotonicity of the Lebesgue outer measure, we also have that $m(A^o) < \infty$. Hence, we can cancel to conclude. But what about the case $m(A) = \infty$?

If we take for example $A = \bigcup_{n \geq 1} (n - 1, n)$, then $m(A^0) = m(\overline{A}) = \infty$, but $m(\partial A) = m(\mathbb{N}) = 0$. Are there any counter examples?


问:
Question about the thought process at multivariable limit calculation
Is the following calculation correct, using the substitution method?

$$\lim_{(x,y)\to(0,0)}\frac{{x^2 - y^2}}{{2x + y}} = \lim_{(x,y)\to(0,0)}\frac{{x(1 - y/x)}}{{2 + y/x}} = \lim_{(x,y) \to(0,0)}\frac{{1 - (y/x)^2}}{{2 + y/x}} \cdot \lim_{x\to0} x,$$ where $u = y/x \to 1/2 \cdot 0 = 0$.

I'd like to know, firstly if this is correct, and after that if there is other way to solve problems like this, thank you.

答:
You have no control over $\frac y x$ so your answer is wrong.

The expression is not defined along that lime $y=-2x$ so the limit is not even well defined.

Also along the  sequence $(\frac 1  n, \frac 1 {n^{2}}-\frac 2 n)$ you get a non-zero limit.


问:
What is the deepest / most interesting known connection between Trigonometry and Statistics?
I'm teaching both at the same time to different classes in high school, so I just wondered about this.

Added by OP on 16.May.2011 (Beijing time)

I mean Statistics only, without Probability. In other words, Descriptive Statistics only. This rules out Buffon’s Needle Problem.

The occurrence of π is counted only as a connection to geometry. By Trigonometry is meant explicit, non-gratuitous, occurrence of the sine, cosine, tangent, or their reciprocals.

Yes, the Law of Cosines fits the bill, but it is on the surface: everyone knows about it. It would be a hugely interesting meta theorem that this were the “deepest” connection between Trigonometry and Descriptive Statistics. My suspicion/hope is that there are deeper connections, somewhat along the line of the surprising use of trig in solving the cubic in closed form, or the use of trigonometric substitutions in evaluating certain integrals. The comment below about the arcsine transformation at first blush seems to be something along this line, but when you follow the link you see that someone is bringing it up only to say how bad it is.

So, I hope the intent of my question is now much clearer.

答:
No doubt it's the Law of Cosines. The correlation between two data sets follows the generalized $n$-dimensional Law of Cosines.

EDIT: Maybe I'll make this a little more explicit. Take two data sets $A = (5,7,2...)$ and $B = (12, 4, 9...)$ and ask if they are correlated. One way is to treat them as vectors and look at the data set $C = A+B = (17, 11, 11...)$ where the sum of the data sets (vectors) is taken pointwise. Okay...it's not the length of the vectors that works like the Law of Cosines, but the standard deviation. If the two data sets are randomly correlated then you should expect the standard deviations to add like the Law of Pythagoras, so that if $\text{StDev}(A) = 3$ and $\text{StDev}(B)=4$ then $\text{StDev}(C)$ should equal $5$. For $100\%$ correlations, the standard deviation of $C$ would have to be $7$ (or $1$ for negative correlation). It's the Law of Cosines where the correlation is the cosine of the angle between two vectors of length $3$ and $4$.


问:
Outer measure question (disjoint set st $\lambda^*(A \cup B) < \lambda^*(A) + \lambda^*(B)$)
Let $\lambda^*$ be the outer measure on $\mathcal{P}(\mathbb{R})$. Is it true that there exist $A,B \in \mathcal{P}(\mathbb{R})$ such that
$$A \cap B = \emptyset$$ and $$\lambda^*(A \cup B) < \lambda^*(A) + \lambda^*(B)?$$
Do you have any idea if such $A$ and $B$ sets exist?
Thanks in advice!

答:
This statement is equivalent to the existence of non-Lebesgue-measurable sets.

According to Caratheodoy's Criterion (see also here), a set $E\subset\Bbb R$ is measurable iff $$\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^c)$$ for every $A\subset\Bbb R$.

If you assume the Axiom of Choice, then you can construct the Vitali Set, wich is the classical example of non-Lebesgue-measurable set on $\Bbb R$.

Now, if you denote $\Bbb V$ this non-measurable set, by Caratheodory's Criterion, there is some set $A_0\subset \Bbb R$ such that
$$\lambda^*(A_0)\ne \lambda^*(A_0\cap \Bbb V)+\lambda^*(A_0\cap \Bbb V^c), $$
wich (thanks to subadditivity of $\lambda^*$) is equivalent to
$$\lambda^*(A_0)< \lambda^*(A_0\cap \Bbb V)+\lambda^*(A_0\cap \Bbb V^c).$$

And here are your counterexample: $A:=A_0\cap \Bbb V$ and $B:=A_0\cap\Bbb V^c$.


问:
Randomly take $51$ numbers from the set 1, 2, ..., 159. Find variance of their sum.
We randomly take $51$ numbers from 159 natural numbers $1,...,159$ without replacement. Let $\alpha$ be a random variable equal to the sum of the selected numbers. Find the variance of $\alpha$.

Firstly I need to understand something about $\alpha$ destribution. There are totally $$C^{51}_{159} = \frac{159!}{51!108!}$$ kinds of sums. A lot of them are equal, because $$\sum_{i=1}^{51}i = 1326\leq\alpha\leq\sum_{i=109}^{159}i=6834$$
Consequenlty, I want to know how many subsets of $51$ numbers have the sum equal to $N$, where  $1362\leq N\leq6834$. I'm stuck here because I don't know how to do it.

答:
Comment: You can get a reasonable approximation to $Var(\alpha)$ by simulation. In the simulation, I assume
the 51 numbers are selected without replacement.

```<code>set.seed(2020)
alpha = replicate(10^5, sum(sample(1:159, 51)))
summary(alpha)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   2915    3897    4081    4081    4266    5275
</code>```

Notice that among the 100,000 samples I summed, all of the totals
are between the two numbers you mention in your question.

```<code>var(alpha)
[1] 74069.39
sd(alpha)
[1] 272.1569
</code>```

A histogram of the simulated values of $\alpha$ looks approximately normal, so I show the best-fitting normal
density along wit the histogram.

```<code>hist(alpha, prob=T, col="skyblue2")
 curve(dnorm(x, mean(alpha), sd(alpha)), add=T, col="red")
</code>```

With replacement, the variance is somewhat larger.
(Again here the distribution of $\alpha$ seems approximately normal; histogram not shown.)

```<code>set.seed(1130)
alpha = replicate(10^6, sum(sample(1:159, 51, rep=T)))
summary(alpha)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   2593    3859    4080    4080    4302    5590
var(alpha)
[1] 107274.7
</code>```

Possible solution: If you consider the population to be numbers 1 through 159, then the population has variance 2120, and the sum of a random sample with replacement should have variance 51 times as large, which is 108,120, which seems to agree with the simulated result within the margin of simulation error.

```<code>var(1:159)
[1] 2120
51*var(1:159)
[1] 108120
</code>```

Body:
Comment: You can get a reasonable approximation to $Var(\alpha)$ by simulation. In the simulation, I assume
the 51 numbers are selected without replacement.

set.seed(2020)
alpha = replicate(10^5, sum(sample(1:159, 51)))
summary(alpha)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   2915    3897    4081    4081    4266    5275

Notice that among the 100,000 samples I summed, all of the totals
are between the two numbers you mention in your question.

var(alpha)
[1] 74069.39
sd(alpha)
[1] 272.1569

A histogram of the simulated values of $\alpha$ looks approximately normal, so I show the best-fitting normal
density along wit the histogram.

hist(alpha, prob=T, col="skyblue2")
 curve(dnorm(x, mean(alpha), sd(alpha)), add=T, col="red")

With replacement, the variance is somewhat larger.
(Again here the distribution of $\alpha$ seems approximately normal; histogram not shown.)

set.seed(1130)
alpha = replicate(10^6, sum(sample(1:159, 51, rep=T)))
summary(alpha)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   2593    3859    4080    4080    4302    5590
var(alpha)
[1] 107274.7

Possible solution: If you consider the population to be numbers 1 through 159, then the population has variance 2120, and the sum of a random sample with replacement should have variance 51 times as large, which is 108,120, which seems to agree with the simulated result within the margin of simulation error.

var(1:159)
[1] 2120
51*var(1:159)
[1] 108120


问:
Calculate the number of order $p$ elements in a group (using Sylow)
I'm having trouble understanding how to calculate the number of elements of order $p$ (being $p$ prime number) in a group $G$ so that
$$|G|=p^a\cdot m,$$
being $a\in\mathbb{N}$, $a\geq 1$, and $p\nmid m$. I need this for problems of the form: "Prove that there aren't simple groups of order $x$" and so. For these, I first calculate the minimum number of Sylow p-subgroups of $G$ to verify that $G$ is simple (that's the smallest number possible greater than $1$). I do that for every prime factor in $G$, and calculate the number of elements of each prime order (This is the operation I don't know), and if I've done everything right, the sum of all them plus $1$ (the neutral element) must be greater that $|G|$, thus a contradiction.

What is the "formula" for calculating these number of elements of each prime order $p$ in $G$ (knowing |G| and the number of Sylow $p$-subgroups)?

Any help will be appreciated, thanks in advance.

Example:

Prove that groups of order $132$ aren't simple.

For this problem, I know that $132=2^2\cdot 3\cdot 11$, and, if it's not simple, then the minimum nomber of Sylow subgroups are $n_11=12$, $n_3=4$ and $n_2=3$. So I know there are $12\cdot (11-1)=120$ elements of order $11$ and $4\cdot (3-1)=8$ elements of order 3. My book then states that there are at least $4$ elements of order $4$ (but I don't know why is this, the exponent of $2$ is $2$ and I don't know how to calculate the minimum number of elements of order $2$). Finally he sees that
$$120+8+4+1=133>132$$
so $G$ is not simple.

My question is: why a minimum of $4$ order $2$ elements?

答:
This technique is possible only when the exponent $a=1$. In which case the $p$-Sylow subgroups will be cyclic of order $p$ and there can't be any overlap between the $p$-Sylow subgroups other than the identity element i.e. $P_i \cap P_j =\{e\}$ because the cardinality of the intersection of those two subgroups $|P_i \cap P_j|$ should divide $p=|P_i|=|P_j|$. In general if $a>1$ there can be overlap between the subgroups.

A good exercise for the above technique the following problem

Let $G$ be a finite group of order $pqr$ where $p,q,r$ are distint primes. Show that $G$ cannot be simple

 $\mathbf{Hint}$: Show that there is always a normal $p$ or $q$ or $r$ Sylow subgroup


问:
Set that is not well-ordered
I want to find a set that is transitive, every non-empty subset of it has $\in$-maximum and is not $\in$-well ordered.

Any hint?

答:
Let $S$ be a non-empty set such that every non-empty subset of it has an $\in$-maximum.

Then for $x,y \in S$ with $x \ne y$, we have either $x \in y$ or $y \in x$ because by hypothesis, $\{x,y\}$ has an $\in$-maximum. Therefore, $(S, \in)$ is a linear order. In the degenerate case that $S$ has only one element, clearly $(S, \in)$ is also a linear order.

Now suppose that a non-empty subset $T \subseteq S$ had no $\in$-smallest element. By the Axiom of Foundation, $T$ does have an $\in$-minimal element, i.e. an $x \in T$ with $x \cap T = \varnothing$.

But this then means that for all $y \in T$, we have $y \notin x$. Since $\in$ is a linear ordering on $S$, we conclude that $x \in y$ (if $x \ne y$). But then $x$ is an $\in$-smallest element; hence $S$ is well-ordered by $\in$.

Therefore, no set $S$ matching your conditions can exist. Note that the assumption that $S$ is transitive was not necessary to arrive at this conclusion.

Title: Set that is not well-ordered


问:
Counting intersection points of circles and lines

Find maximum number of points of intersection of 7 straight lines and 5 circles when 3 lines are parallel and 2 circles are concentric.

My attempt: Total intersection points = Total intersection points of lines with lines ($T_{ll}$) + Total intersection points of circles with circles ($T_{cc})$ + Total intersection points of lines with circles $(T_{lc})$.

$T_{ll}$ = ${7}\choose{2} $ - ${3}\choose{2}$ = 18. Each pair of lines intersect once except for the three parallel lines.

$T_{cc}$ = $2\cdot {4\choose2}$ + $2\cdot {4\choose 2}$ = 24. The three non concentric circles intersect with each of the two concentric circles at 2 points each.

$T_{lc}$ = $ 5 \cdot 7 \cdot 2$ = 70 . Each line can intersect w/ each circle twice at max.

Total = 112

But the answer in the book is given 106.

I think i might have gone wrong in calculating $T_{lc}$ or  $T_{cc}$ but I am not sure.

I am not very good at combinatorics (as you can probably see) so I would really appreciate it if someone could tell me not just the solution but the thought process behind the solution too and also where i went wrong. Also if you have some tips on how to tackle these kinds of problems and combinatorics in general, do let me know.

Thanks!

答:
Your $T_{cc}$ is overcounting because in first $2\cdot {4 \choose 2}$ you already counted the intersection points of non-concentric circles. In second $2\cdot {4 \choose 2}$, you're still counting these. Correction would be $$2 \cdot {4 \choose 2} + 2\cdot 3=18$$

You can also find $T_{cc}$ in following, slightly more illustrating manner. Two circles can intersect in maximum two points. Draw two concentric circles. Now introduce third circle. This gives $4$ intersection points, two with each of previous ones. Fourth circle gives $6$ more points. Fifth gives $8$ more points. So $$T_{cc}=4+6+8=18$$


Title: Most efficient method in simplifying quotient determinant(Cramer's rule) with binomial coefficients

Body:
The matrix entries are defined as:

$$ c_{n,i} = \binom{n}{i} \binom{3}{i} i!$$

For $ i , n \in [0,3]$

As a matrix it looks like:

$$ \begin{bmatrix} \binom{1}{0} \binom{3}{0} 0! &  \binom{1}{1} \binom{3}{1} 1! & 0& 0\\
\binom{2}{0} \binom{3}{0} 0! &  \binom{2}{1} \binom{3}{1} 1! & \binom{2}{2} \binom{3}{2} 2!& 0\\
\binom{3}{0} \binom{3}{0} 0! &  \binom{3}{1} \binom{3}{1} 1! & \binom{3}{2} \binom{3}{2} 2!&  \binom{3}{3} \binom{3}{3} 3! \\
\binom{4}{0} \binom{3}{0} 0! &  \binom{4}{1} \binom{3}{1} 1! & \binom{4}{2} \binom{3}{2} 2!&  \binom{4}{3} \binom{3}{3} 3!\end{bmatrix} = A$$

I have a linear system of the form:

$$ A \begin{bmatrix} q\\w\\e\\r \end{bmatrix} = \begin{bmatrix} a\\b\\c\\d  \end{bmatrix}$$

And, I need to find the quantity '$q$' that is by cramer's rule,

$$ q=  \frac{\begin{vmatrix} a&  \binom{1}{1} \binom{3}{1} 1! & 0& 0\\
b &  \binom{2}{1} \binom{3}{1} 1! & \binom{2}{2} \binom{3}{2} 2!& 0\\
c &  \binom{3}{1} \binom{3}{1} 1! & \binom{3}{2} \binom{3}{2} 2!&  \binom{3}{3} \binom{3}{3} 3! \\
d &  \binom{4}{1} \binom{3}{1} 1! & \binom{4}{2} \binom{3}{2} 2!&  \binom{4}{3} \binom{3}{3} 3!\end{vmatrix}}{\begin{vmatrix} \binom{1}{0} \binom{3}{0} 0! &  \binom{1}{1} \binom{3}{1} 1! & 0& 0\\
\binom{2}{0} \binom{3}{0} 0! &  \binom{2}{1} \binom{3}{1} 1! & \binom{2}{2} \binom{3}{2} 2!& 0\\
\binom{3}{0} \binom{3}{0} 0! &  \binom{3}{1} \binom{3}{1} 1! & \binom{3}{2} \binom{3}{2} 2!&  \binom{3}{3} \binom{3}{3} 3! \\
\binom{4}{0} \binom{3}{0} 0! &  \binom{4}{1} \binom{3}{1} 1! & \binom{4}{2} \binom{3}{2} 2!&  \binom{4}{3} \binom{3}{3} 3!\end{vmatrix}} $$

Now, we can simplify this to get:

$$ q=  \frac{\begin{vmatrix} a&  \binom{1}{1}  & 0& 0\\
b &  \binom{2}{1} & \binom{2}{2} & 0\\
c &  \binom{3}{1} & \binom{3}{2} &  \binom{3}{3}  \\
d &  \binom{4}{1} & \binom{4}{2}&  \binom{4}{3} \end{vmatrix}}{\begin{vmatrix} \binom{1}{0}  &  \binom{1}{1} & 0& 0\\
\binom{2}{0} &  \binom{2}{1} & \binom{2}{2} & 0\\
\binom{3}{0}  &  \binom{3}{1}& \binom{3}{2}&  \binom{3}{3}  \\
\binom{4}{0}  &  \binom{4}{1} & \binom{4}{2} &  \binom{4}{3}\end{vmatrix}} $$

Progress: Turns out the denominator is always equal one for any size of the matrice, now only need to work out num

Any advice on how to simplify these determinants further? Prefer general method for similar matrices

Body:
This may not be the best method out there, but for the denominator, you could do this:

$$
\begin{vmatrix} \binom{1}{0}  &  \binom{1}{1} & 0& 0\\
\binom{2}{0} &  \binom{2}{1} & \binom{2}{2} & 0\\
\binom{3}{0}  &  \binom{3}{1}& \binom{3}{2}&  \binom{3}{3}  \\
\binom{4}{0}  &  \binom{4}{1} & \binom{4}{2} &  \binom{4}{3}\end{vmatrix}\\[8ex]
= \begin{vmatrix} 1  &  1 & 0& 0\\
1 &  2 & 1 & 0\\
1 &  3& 3&  1  \\
1  &  4 & 6 &  4\end{vmatrix} \;\;\;\;\; \text{(Simplifying the binomial coeffs.)}\\
=\begin{vmatrix}
1 & 0 & 0 &0 \\
1 & 1 & 1 & 0 \\
1 & 2 & 3 & 1 \\
1 & 3 & 6 & 4
\end{vmatrix}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ \ \  \text{($\mathrm{C}_2 \to \mathrm{C}_2-\mathrm{C}_1$)} \\[2ex]
=1\cdot
\begin{vmatrix}
1 & 1 & 0\\
2 & 3 &1\\
3 & 6 & 4
\end{vmatrix}
=
\begin{vmatrix}
1 & 0 & 0\\
2 & 1 &1\\
3 & 3 & 4
\end{vmatrix}
=1
$$

Now for the numerator:

$$\begin{vmatrix}
a&  \binom{1}{1}  & 0& 0\\
b &  \binom{2}{1} & \binom{2}{2} & 0\\
c &  \binom{3}{1} & \binom{3}{2} &  \binom{3}{3}  \\
d &  \binom{4}{1} & \binom{4}{2}&  \binom{4}{3} \end{vmatrix}
\\[8ex]
=\begin{vmatrix}
a&  1  & 0& 0\\
b &  2 & 1 & 0\\
c &  3 & 3 & 1 \\
d &  4 & 6 & 4
\end{vmatrix}
\\[4ex]
=\begin{vmatrix}
a & 1 & 0 & 0\\
b & 2 & 1 & 0\\
c & 3 & 0 & 1\\
d & 4 & -6& 4\\
\end{vmatrix} \;\;\;\;\;\;\;\;\;\text{($\mathrm{C_3}\to\mathrm{C_3}-3\mathrm{C}_4$)}\\[8ex]
=a
\begin{vmatrix}
2 & 1 & 0\\
3 & 0 & 1\\
4 & -6& 4\\
\end{vmatrix}
-
\begin{vmatrix}
b & 1 & 0\\
c & 0 & 1\\
d &-6 & 4\\
\end{vmatrix}\\[8ex]
=4a-6b+4c-d
$$
Finally, we have :
$$\boxed{q=4a-6b+4c-d}$$


问:
Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$
While solving a question based on Binomial series, I encountered a definite integral as : $$\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$$ where $m,n$ are natural numbers.
My approach : wrote the above expression in two ways (1)- $$\int_0^1\sum_{r=0}^n \binom{n}{r} \left( {x}\right)^{m+r}dx $$ and also as (2)- $$\int_0^1\sum_{r=0}^m \binom{m}{r} \left(-1)^{r}( {1+x}\right)^{m+n-r}dx$$ and tried to solve using both the above mentioned expressions. Did some other simplification of expression (1) , But was unable to proceed further. Can anyone guide me to a "general answer based on any m,n" ? Any help would be highly appreciated. Thanks in advance.!

答:
Let $ n $, $ m $ be positive integers. Denoting $ u_{m}':x\mapsto x^{m} $, and $ v_{n}:x\mapsto \left(1+x\right)^{n} $, we have : \begin{aligned}\int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}&=\int_{0}^{1}{u_{m}'\left(x\right)v_{n}\left(x\right)\mathrm{d}x}\\ &=\left[u_{m}\left(x\right)v_{n}\left(x\right)\right]_{0}^{1}-\int_{0}^{1}{u_{m}\left(x\right)v_{n}'\left(x\right)\mathrm{d}x}\\ &=\frac{2^{n}}{m+1}-\frac{n}{m+1}\int_{0}^{1}{x^{m+1}\left(1+x\right)^{n-1}\,\mathrm{d}x}\end{aligned}

Now choosing $ k \in\mathbb{N} $, we'll have : $$ \int_{0}^{1}{x^{m+k}\left(1+x\right)^{n-k}}=\frac{2^{n-k}}{m+k+1}-\frac{n-k}{m+k+1}\int_{0}^{1}{x^{m+k+1}\left(1+x\right)^{n-k-1}\,\mathrm{d}x} $$

Multiplying both sides by $ \left(-1\right)^{k}\prod\limits_{j=1}^{k}{\frac{n-j+1}{m+j}} $, we'll get : $$ \scriptsize\left(-1\right)^{k}\prod_{j=1}^{k}{\frac{n-j+1}{m+j}}\int_{0}^{1}{x^{m+k}\left(1+x\right)^{n-k}\,\mathrm{d}x}=\left(-1\right)^{k}\frac{2^{n-k}}{m+k+1}\prod_{j=1}^{k}{\frac{n-j+1}{m+j}}+\left(-1\right)^{k+1}\prod_{j=1}^{k+1}{\frac{n-j+1}{m+j}}\int_{0}^{1}{x^{m+k+1}\left(1+x\right)^{n-k-1}\,\mathrm{d}x} $$

Denoting $ \left(\forall p\in\mathbb{N}\right),\ u_{n,m}\left(p\right)=\left(-1\right)^{p}\prod\limits_{j=1}^{p}{\frac{n-j+1}{m+j}}\int_{0}^{1}{x^{m+p}\left(1+x\right)^{n-p}\,\mathrm{d}x} $.

Summing up everything : $$ \sum_{k=0}^{n-1}{\left(u_{n,m}\left(k\right)-u_{n,m}\left(k+1\right)\right)}=\sum_{k=0}^{n-1}{\left(-1\right)^{k}\frac{2^{n-k}}{m+k+1}\prod_{j=1}^{k}{\frac{n-j+1}{m+j}}} $$

Since : \begin{aligned} \sum_{k=0}^{n-1}{\left(u_{n,m}\left(k\right)-u_{n,m}\left(k+1\right)\right)}&=u_{n,m}\left(0\right)-u_{n,m}\left(n\right)\\ &=\int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}-\left(-1\right)^{n}\prod_{j=1}^{n}{\frac{n-j+1}{m+j}}\int_{0}^{1}{x^{m+n}\,\mathrm{d}x}\\ &=\int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}-\frac{\left(-1\right)^{n}}{m+n+1}\prod_{j=1}^{n}{\frac{n-j+1}{m+j}} \end{aligned}

And, for any $ k\in\mathbb{N}^{*} $ : \begin{aligned}\prod_{j=1}^{k}{\frac{n-j+1}{m+j}}&=\frac{n!m!}{\left(n-k\right)!\left(m+k\right)!}\\ &=\frac{\binom{n}{k}}{\binom{m+k}{m}}\end{aligned}

We have : $$ \fbox{$\begin{array}{rcl}\displaystyle\int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}=\sum_{k=0}^{n}{\left(-1\right)^{k}\frac{2^{n-k}\binom{n}{k}}{\left(m+k+1\right)\binom{m+k}{m}}} \end{array}$}$$


问:
Prove that if for a number x ∈ R x + x ^(-1) ∈ Z ..... I can't get started it would be very important. please
** Prove that if for a number x ∈ R    x + x ^(-1) ∈ Z ,  then for all n ∈ Z cases x^n + x^(−n) ∈ Z.strong text**

答:
Induction hint: The cases $n=0$ and $n=1$ are trivial.

Suppose $x^{n-1} + x^{1-n}$ and $x^{n-2} + x^{2-n}$ are integers.

Consider the product $(x + x^{-1})(x^{n-1} + x^{1-n})$, and show that $x^n + x^{-n}$ must be an integer.


答:
$7$ lines can intersect in $\dfrac{7\cdot6}2=21$ points. But if $3$ are parallel, you deduce $\dfrac{3.2}2=3$.

$7$ lines and $5$ circles can intersect in $2\cdot3\cdot5=70$ points.

$5$ circles can intersect in $2\dfrac{5\cdot4}2=20$ points but if $2$ are concentric, you deduce $2$.

Anyway, this is an upper bound, there is no guarantee that all are possible simultaneously.

Title: Counting intersection points of circles and lines

Body:
Find maximum number of points of intersection of 7 straight lines and 5 circles when 3 lines are parallel and 2 circles are concentric.

My attempt: Total intersection points = Total intersection points of lines with lines ($T_{ll}$) + Total intersection points of circles with circles ($T_{cc})$ + Total intersection points of lines with circles $(T_{lc})$.

$T_{ll}$ = ${7}\choose{2} $ - ${3}\choose{2}$ = 18. Each pair of lines intersect once except for the three parallel lines.

$T_{cc}$ = $2\cdot {4\choose2}$ + $2\cdot {4\choose 2}$ = 24. The three non concentric circles intersect with each of the two concentric circles at 2 points each.

$T_{lc}$ = $ 5 \cdot 7 \cdot 2$ = 70 . Each line can intersect w/ each circle twice at max.

Total = 112

But the answer in the book is given 106.

I think i might have gone wrong in calculating $T_{lc}$ or  $T_{cc}$ but I am not sure.

I am not very good at combinatorics (as you can probably see) so I would really appreciate it if someone could tell me not just the solution but the thought process behind the solution too and also where i went wrong. Also if you have some tips on how to tackle these kinds of problems and combinatorics in general, do let me know.

Thanks!


问:
Determine if integer can be factored as $ab(a+b)$
For $N,a,b \in \mathbb{Z}$ and distinct $a,b>1$, how can I determine if it is possible to factor $N$ as any $a \times b \times (a+b)$? (I don't necessarily need all $a,b$ if that helps...)

My first thought was to factor $N$, then enumerate the factors into three buckets, $a$, $b$, and \$leftover, then check if \$leftover equals $a+b$. But this grows quite badly, for n total factors ~ $\mathcal{O}(n!)$ (OEIS A200978), so this approach seems infeasible.

答:
If you fix the value of $a$, you have a simple quadratic equation $b(b+a) = \tfrac{N}{a}$ with $b$ the unknown, you can solve it to get two possible values, then you check if any of those are natural numbers and divisors of $N$. So to find out if such $a$ and $b$ exist, you need to check all values of $a$ which are divisors of $N$, and for every such $a$ find $b$.

For a number with $k$ divisors this algorithm's complexity is $O(k) + O(factorize(N))$ where $factorize(N)$ is complexity of factorizing a number.


答: I'm not an expert in Set Theory, but I'd say that if you're working in $ZF$, i.e, in particular, if you assume the Axiom of Foundation, then you can't find an example of what you're looking for. Indeed, if a set $x$ is transitive and each of its non empty subsets has $\in -$ maximum element, then trivially $x$ is totally (linearly) ordered by the membership relation. But in $ZF$ we can prove that a set $x$ is an ordinal (defined as a transitive set which is well-ordered by membership) if and only if it is transitive and totally ordered by $\in$ (you might try to show this as an exercise), and then there's nothing more to say.</p>

Title: Set that is not well-ordered

Body:
I want to find a set that is transitive, every non-empty subset of it has $\in$-maximum and is not $\in$-well ordered.

Any hint?


问:
Infinite sum of semisimple Ideals of a Lie algebra
Following the paper by Kubok (Infinite-dimensional Lie algebras with null Jacobson radical), we have the following definitions. Let $L$ be a Lie algebra. If every finite subset of $L$ is contained in a finite-dimensional subalgebra (resp. a solvable subalgebra) of $L$ then $L$ is called locally finite (resp. locally solvable). For a locally finite Lie algebra $L$, $\sigma(L)$ is the maximal locally
solvable ideal of $L$ and $L$ is said to be semisimple if $\sigma(L)=0$.

Suppose that $L$ is an arbitrary Lie algebra over a field of characteristic zero and that $L=\sum_{i\in\mathcal I}L_i$, where each $L_i$ is a finite-dimensional semisimple ideal of $L$.

If $\mathcal I$ is finite then $L$ is semisimple. In order to show this, it is enough to verify the case $L=L_1+L_2$ and apply induction. As $L_1$ is semisimple the Killing form $\kappa(\cdot,\cdot)$ of $L$ restrict $L_1$ is non-degenerated and thus $L=L_1\oplus L_1^\perp$, where $L_1^\perp=\{x\in L\mid \kappa(x,y)=0, \mbox{ for } y\in L_1 \}$. Now, since $L=L_1+L_2$ it follows that
$$L_1^\perp \cong L/L_1=(L_1+L_2)/L_1\cong L_2/(L_1\cap L_2)$$is semisimple. Hence $L=L_1\oplus L_1^\perp$ is semisimple.

But, what happens if $\mathcal I$ is infinite? Is $L$ semisimple? I believe that the answer is no, but I can't find a counterexample. Is there some condition on $L$ or $L_i$ that guarantees the semisimplicity of $L$?

I appreciate any help.

答:
If $L=L_1\oplus L_2$ is a direct sum of simple Lie algebras, then $L$ is semisimple. However, for the sum $L_1+L_2$, an arbitrary sum, this need not be true in general - see our Example $4.10$ here:
$$
L=\mathfrak{sl}_n(\Bbb C)\rtimes V(n)=\mathfrak{sl}_n(\Bbb C)+\mathfrak{sl}_n(\Bbb C)
$$
is not semisimple. So $\sum_i L_i$ need not be semisimple, even for finitely many summands.

Edit: An infinite direct sum $L$ of simple Lie algebras will be infinite-dimensional. What is your definition of "semisimple" for infinite-dimensional Lie algebras?


问:
Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$
I have tried to wrap my head around this for some time now, and quite frankly I am stuck.

Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$

Then what is the value of $a/b$?

I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?

Body:
Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$,

$$\tan(2n+1)x=\dfrac{\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}}{\binom{2n+1}0-\binom{2n+1}2t^2+\cdots+(-1)^n\binom{2n+1}{2n}t^{2n}}$$

where $t=\tan x$

So, the roots of $$\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}=0$$  i.e., $$t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(-1)^n(2n+1)t=0$$ are $t_r=\tan\dfrac{r\pi}{2n+1}; r=0,\pm1,\pm2,\cdots,\pm n$

As $r=0\implies \tan\dfrac{r\pi}{2n+1}=0$, the roots of $$t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(-1)^n(2n+1)=0$$  are $t_r=\tan\dfrac{r\pi}{2n+1}; r=\pm1,\pm2,\cdots,\pm n$

Writing $t_r^2=p_r,$

the roots of $$p^n-\binom{2n+1}{2n-1}p^{n-1}+\cdots+(-1)^n(2n+1)=0$$  are $p_r=\tan^2\dfrac{r\pi}{2n+1}; r=1,2,\cdots, n$

Use Vieta's formula


问:
Let $f:A \to B$ be a surjective homomorphism between rings. If $B$ and ker$f$ has ascending chain condition, then $A$ has it too.
Let $f:A \to B$ be a surjective homomorphism between rings, where $A$
is abelian. Let $B$ and $\text{ker} f$ fulfill the ascending chain
condition. I want to show that $A$ fulfills the ascending chain
condition, namely that given
$$A_{1} \subseteq A_{2}\subseteq ...\subseteq A,$$
there is an integer $N$ such that $A_{M}=A_{N}$ for $M \geq N$. Here's
my attempt. Let $$A_{1} \subseteq A_{2}\subseteq ...\subseteq A.$$

Since $f$ preserves ideals, we have that $$f(A_{1}) \subseteq f(A_{2})
\subseteq ... \subseteq f(A).$$ Since $B$ satisfies the ascending chain
condition, there is an integer $N$ such that $$f(A_{M})=f(A_{N})$$
for $M \geq N.$ Is it valid to say that $$f ^{-1} (f(A_{M}))=A_{M}=f ^{-1}( f(A_{N}))=A_{N}$$
for $M \geq N,$ and thus $A$ also satisfies the ascending chain
condition? I'm not sure where to go from here and how to use the fact that ker $f$ is acc. I also thought about proving the contrapositive or getting
a contradiction. I take this statement to logically be

$$f \text{ surjective homomorphism } \land B \text{ acc} \land
\text{ker} f \text{ acc} \implies A \text{ acc} .  $$
Therefore the contrapositive would be
$$A \text{ not acc} \implies f \text{ not surjective homomorphism }
\lor B \text{ not acc } \lor \text{ker} f \text{ not acc.} $$
But if we assume that $B$ isn't acc, then given any ideals
$$A_{1} \subseteq A_{2}\subseteq ...\subseteq A$$
they never stop increasing. But this contraducts our assumption that ker
$f$ is acc. Is this reasoning correct?

答:
Recall that an $R$-module $M$ is defined to be Noetherian if the following equivalent conditions hold:

$M$ has the ascending chain condition.

Every submodule of $M$ is finitely generated.

Every set of submodules of $M$ has a maximal element (we won't use this condition).

The ring $R$ is Noetherian if it is Noetherian as a module over itself, which (by 2.) is equivalent to all of its ideals being finitely generated.

Let $I = \operatorname{ker}f$. Then $I$ is an ideal of $A$ and $B \cong A/I$ by the First Isomorphism Theorem. So we are assuming that $I$ and $A/I$ have the ascending chain condition. That is, $I$ is a noetherian $A$-module and $A/I$ is a Noetherian ring.

Let $J$ be an ideal of $A$. Then by the Second Isomorphism Theorem,
$$
\frac{J}{I\cap J} \cong \frac{I+J}{I}.
$$
Since $I\cap J$ is a submodule of $I$, which is Noetherian, so $I\cap J$ is finitely generated as an $A$-module (since submodules of Noetherian modules are finitely generated). Suppose that $I \cap J$ is generated by $f_1,\ldots, f_m \in I\cap J$ as an $A$-module.

Now, $\frac{I+J}{I}$ is an ideal of $A/I$, so since $A/I$ is Noetherian it is finitely generated. By our isomorphism above, $J/(I\cap J)$ is also finitely generated, say by $\bar{g}_1,\ldots, \bar{g}_n$. For each $i$, there is an element $g_i\in J$ such that $\bar{g}_i = g_i + I\cap J$.

Now let $x \in J$. Then $x + I\cap J \in J/(I\cap J)$, so there exist $a_1,\ldots, a_n \in A$ such that
$$
x + I\cap J = a_1g_1 + \ldots + a_ng_n + I\cap J.
$$
Then $x - a_1g_1 - \ldots a_ng_n \in I\cap J$, so there are elements $b_1,\ldots, b_m \in A$ such that
$$
x - a_1g_1 - \ldots - a_ng_n = b_1f_1 + \ldots + b_mf_m.
$$
It follows that $x \in M$, where $M$ is the $A$-module generated by the $f_i$ and the $g_i$. Since $x \in J$ was arbitrary, we have $J\subseteq M$. But each $f_i \in I\cap J \subseteq J$ and $g_i \in J$, so $M\subseteq J$. Therefore $J = M$ is finitely generated.

So every ideal of $A$ is finitely generated, which means that $A$ is Noetherian, so $A$ has the ascending chain condition.


问:
Drawing 3D Surface
I would like to know to plot this kind of 3D surface as depicted below. Anybody help plot it, give me suggestions to choose an appropriate pc program for this matter???https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/15%3A_Multiple_Integration/15.3%3A_Double_Integrals_in_Polar_Coordinates

答:
You may try searching "desmos 3d", and you would find an array of 3D graphs represented on a 2D plane.

Here's my graph
originally from another graph.

In the example, input your explicit function in terms of $x$ and $y$, followed by a constraint enclosed with braces.  In your example, that would be $\{(x-1)^2+y^2 \ge 1\}$.

Title: Drawing 3D Surface


问:
I don't think W/O replacement events are dependent.

Let say there're cards numbered 1 to 10. P(A) is the probability you pick a card whose number is the multiple of 2. P(B) is the probability you pick a card whose number is the multiple of 5. So now you draw two cards without replacement. You don't place the first card you just drew back in the original deck. Then, you draw again from that 9 remaining cards.

In the picture, P(A) is shown as the ordered pairs colored red, and P(B) is shown as the ordered pairs shaded with light blue.

Then I counted them all.

P(A) = (9+9+9+9+9) / (100-10) = 45/90 = 1/2

P(B) = (9+9) / (100-10) = 18/90 = 1/5

P(A∩B) = 9 / 90 = 1/10

So here, I found out that P(A∩B) is actually the multiple of P(A) and P(B), which shouldn't be if they are dependent, meaning they are independent. I've also found out, in a different model, that the probabilities could be dependent each other in the case of W/O replacement such as this one.

I know events with replacements are independent. But it feels like events without replacements doesn't mean dependency. It's more like the case of contraposition of the first one. Events with replacements mean independent, which means, by the contraposition rule, dependent events are the ones done without replacements.

But all the other internet pages say like "if they are done without replacement, then they are dependent". This claim is so prevalent. But one of us, they or I, must be wrong. What am I missing here?

答:
In the first version of your question, $P(A)$ only applied to the first card and $P(B)$ only applied to the second card. In that case, your workings are correct: $P(A)=\frac12,P(B)=\frac15,P(A\cap B)=\frac1{10}=P(A)P(B)$.

However, you then made this comment:

Let say you do the case A first, and then case B. If you draw 5 or 10 in your first drawing, P(B) is 2/10 times 1/9. And if you didn't draw 5 or 10 in your first drawing, P(B) is 8/10 times 2/9.

This suggests an alternate interpretation: $P(A)$ is the probability of picking at least one even card from the two draws, $P(B)$ is the probability of picking at least one multiple-of-$5$ card from the two draws.

Then $P(A)$ is the complement of the probability that you don't pick an even card in either draw:
$$1-\frac5{10}\cdot\frac49=\frac79$$
$P(B)$ is similar, the complement of the probability of not picking a multiple-of-$5$ card in either draw:
$$1-\frac8{10}\cdot\frac79=\frac{17}{45}$$
$P(A\cap B)$ is the probability of picking one multiple-of-$5$ card and one even card:
$$\frac{2(2\cdot5-1)}{90}=\frac15$$
(There are $2$ ways to pick the multiple-of-$5$ card and $5$ ways to pick the even card, but then subtract $1$ for the impossible $(10,10)$ case. Then there are $2$ ways in which those cards can be drawn out, out of $90$ possible draws.)

Clearly, with this interpretation, $P(A\cap B)\ne P(A)P(B)$. Thus you cannot say straight away that "events that arise from sampling without replacement are dependent" – usually this is true, but your original formulation is a counterexample.


问:
Differentiation of $x^{\sqrt{x}}$, how?
The answer is (I think)
$x^{\sqrt{x}-0.5} (1+0.5\ln(x))$, but how?</p>

答:
I'll probably get stones thrown at me for suggesting this, but let's view $x^{f(x)}$ as a hybrid of a power function $x^k$ and a composite exponential function $b^{f(x)}$. It's actually legitimate to take derivatives of these two simpler functions and add them together to get the derivative. $$\begin{align}\frac{d}{dx}x^{f(x)}&=\overbrace{f(x)x^{f(x)-1}}^{\text{power rule}}+\overbrace{x^{f(x)}\ln(x)}^{\text{exponential rule}}\cdot\overbrace{f'(x)}^{\text{chain rule}}\\
&=x^{f(x)} \left(\frac{f(x)}{x}+\ln(x)f'(x)\right)\end{align}$$ which is completely consistent with the "correct" approach of rewriting the function as $e^{f(x)\ln(x)}$ and using the chain and product rules. I don't recommend that you think this way, I just find this curious. I'd love to see a rigorous explanation for this happy coincidence that does not essentially appeal to rewriting as $e^{f(x)\ln(x)}$.

I think I found one. Let $F(y,z)=y^z$ with $y=g(x)$ and $z=f(x)$, then actually $F(y(x),z(x))=g(x)^{f(x)}$. Now using the chain rule for multivariate functions, $$\begin{align}\frac{d}{dx}g(x)^{f(x)}=\frac{dF}{dx}&=\frac{\partial F}{\partial y}\frac{dy}{dx}+\frac{\partial F}{\partial z}\frac{dz}{dx}\\&=zy^{z-1}\frac{dy}{dx}+y^z\ln(y)\frac{dz}{dx}\\&=f(x)(g(x))^{f(x)-1}g'(x)+g(x)^{f(x)}\ln(g(x))f'(x)\\&=g(x)^{f(x)}\left(\frac{f(x)g'(x)}{g(x)}+\ln(g(x))f'(x)\right)\end{align}$$

Title: Differentiation of $x^{\sqrt{x}}$, how?


问:
Calculate the angle x
In the figure, AB is diameter, PM = MH and PN = NB.
If PEB arc = 108 °, calculate "x"

Drawing to obtain the right triangle in P: $\Delta APB $.  $\hat{A}$ as $\hat{B}$  are angles inscribed on the circumference, they are equal to half the arc you see. Therefore,  $\hat{A} = 54^o$ , $\hat{B} = 36^o$ and  note also that $\overline {MN}$ t is  average base $\Delta HPB $.

If $\hat{B} = 36^o$. $\hat{H} = 90^o$ then  $H\hat{P}B = 54^o$

In the right triangle, the middle base is parallel to the base of the original triangle. Therefore, the triangle $\Delta PMN$ is $\hat{M}$ rectangle is $P \hat{N} M = 36^o$ and In addition $R \hat{Q}N$ and $O \hat{Q}M$ are opposed by the vertex. From this, it is easy to conclude that $\Delta OMQ$ $\simeq$ $\Delta NRQ$
Therefore,  in the triangle $\Delta OMQ$ we obtain that $ x + 90^o + 36^o = 180^o \Rightarrow x = 54^o $

how to demonstrate the similarity of $\Delta OMQ$ $ and \Delta NRQ$ ?

答:
The sketch shown in $\mathrm{Fig.\space 1}$ is an extended version of the diagrams you have included in your problem statement. We denote the sought angle $\measuredangle RQN$ as $\alpha$.  Let $F$ be the center of the semicircle $BPA$ and $\measuredangle NRO=\phi$.

As you have mentioned in your incomplete answer, $\measuredangle PAB$ and $\measuredangle ABP$ can be easily determined as $54^o $ and $36^o$ respectively, because $\measuredangle PFB =108^o$. It is also given that $M$ and $N$ are the midpoints of the segments $PH$ and $PB$ respectively. Therefore $MN$ is parallel to $AB$, which makes $\measuredangle MNR = 36^o$.

To solve this seemingly difficult problem we have to bring Ceva’s theorem in to play. That is why we have extended the line $NM$ to intersect $PA$ at $D$. This makes $D$ the midpoint of $PA$. As you see below, solving of the problem now becomes as easy as eating a piece of cake. Consider the triangle $PAN$, in which three cevians $ND$, $PO$, and $AR$ are concurrent at $M$. Applying Ceva’s theorem to $\triangle PAN$, we obtain,
$$\frac{PD}{DA}\frac{AO}{ON}\frac{NR}{RP}=1. \tag{1}$$

Since $PD=DA$, equation (1) can be simplified to get,
$$\frac{AO}{ON}=\frac{RP}{NR}. \tag{2}$$

Equation (2) implies that $PA$ is parallel to $RO$. Therefore, we have $\phi = \measuredangle NRO = \measuredangle BPA = 90^o$. Using triangle $RQN$, we obtain,
$$\alpha = \measuredangle RQN =180^o-\measuredangle QNR\space\space –\space\phi =180^o – 36^o – 90^o = 54^o.$$

Title: Calculate the angle x