答: The dimension is defined as the number of elements of a basis. So in order to know that the dimension is what you think it is it is sufficient to prove that the proposed set of matrices is a basis. One useful criterion for that is that any matrix in the pertinent vector space should be expressible as a linear combination of to-be-basis vectors with uniquely determined coefficients. An argument that this is indeed true has been explained in the answer of Eric Towers.</p>

Title: Basics Question on Matrix Spaces: Selecting Basis

Body:
I am trying to understand the concept of a new vector space where vectors are instead matrices. In this example, I am looking at space M made of all 3 by 3 matrices. The text is as follows:

The dimension of M is 9; we must choose 9 numbers to specify an element of M. The space M is very similar to $\mathbb{R}^9$. A good choice of basis is:
$$
\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right], \ldots\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]
$$

The text goes on to specify: the subspace of symmetric 3 x 3 matrices S has dimension 6. When choosing an element of S we pick three numbers on the diagonal and three in the upper right, which tell us what must appear in the lower left of the matrix. One basis is the collection:
$$
\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]
$$

The text goes on to do the same for the subspace of all upper triangular matrices (also of dimension 6).  Lastly, it describes the intersection of symmetric and upper triangular to be the subspace of all 3 x 3 diagonal matrices.

Question: The basis and dimension of M, all 3 x 3 matrices, it makes sense that it would require 9 different matrices to span that space. It makes sense each one of the elements in the basis would be a matrix with 1 in a different position of the matrix.  However for the symmetric example, I don't see an easy way to come about the dimension or basis.  I am not clear how/why it's quickly apparent that the dimension of all symmetric matrices is 6. Is this just by intuition?


问:
Prove that an integral is real and ignore complex parts
(This question comes from this video)

Imagine we are trying to solve the integral :

$$I=\int_{0}^{\pi/2} \ln(2\cos x)\,dx$$
This integral is defined on the real numbers. However, if I decide to use eulers complex cosine formula ($\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$), I can reduce the integral to this (this is shown in the video) :

$$ I = \frac{i\pi^2}{8} - i( 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{9^2}+\cdots)$$

As you can the the result found this way is a pure imaginary number. In the video, it is then explained that since the integral is defined on the real numbers, therefor the complex result is simply a "residue" of the calculations. My question is, is this reasoning valid ? If not, what could bring us to the same result, that $I = 0$ (I doubt the youtube video would assume something wrong)

I turns out that the result is correct and saying that the result is equal to zero
means that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{9^2}+\cdots = \frac{\pi^2}{8}$. We can the continue and use this result to show that $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$.

答:
One can prove the integration result by pure real methods.

First using the substitution $t=\frac\pi2-x$ observe:
$$
I=\int_0^{\pi/2}\log(2\cos x)dx=\int_0^{\pi/2}\log(2\sin x)dx.
$$

Next add both above integrals to obtain
$$\begin{align}
2I&=\int_0^{\pi/2}\log(4\cos x\sin x)dx\tag1\\
&=\int_0^{\pi/2}\log(2\sin 2x)dx\tag2\\
&=2\int_0^{\pi/4}\log(2\sin 2x)dx\tag3\\
&=\int_0^{\pi/2}\log(2\sin t)dt\tag4\\
&=I
\end{align}$$
where we used:

$\log a+\log b=\log(ab)$

$2\cos x\sin x=\sin 2x$

$\sin(x)=\sin(\pi-x)$

$x\mapsto \frac t2$

Finally:
$$2I=I\implies I=0.
$$

ContentLicense: CC BY-SA 4.0


问:
Computing $\int \cos^3(x)\,dx$
So I'm trying to figure out how the reduction formula can help me solve integrands like $\int \cos^3(x)\,dx$.
I've figured that if I'd use integration by parts, I would get to the following:
$$\int u\,dv = uv - \int v\,du$$
$$u = \cos^m(x)$$
$$du = -m\sin(x)\cos^{m-1}\,dx$$
$$dv = dx$$
$$v = x$$

$$\int \cos^m(x)\,dx = x\cos^m  + \int x \cdot m\cdot \sin(x)\cos^{m-1}\, dx$$

Which 1. doesn't seem to really help me and 2. doesn't seem to be true.

Would really appreciate some help, thanks in advance!

答:
Integrating by parts, take

$$u=\cos^2x\implies \mathrm du=-2\cos x\sin x\,\mathrm dx\\
\mathrm dv=\cos x\,\mathrm dx\implies v=\sin x$$

(Picking $\mathrm dv=\mathrm dx$, as you've done, indeed isn't of much help.)

Then

$$\begin{align}
\int\cos^3x\,\mathrm dx&=uv-\int v\,\mathrm du\\[1ex]
&=\cos^2x\sin x+2\int\cos x\sin^2x\,\mathrm dx
\end{align}$$

In the remaining integral, substitute $t=\sin x$ and $\mathrm dt=\cos x\,\mathrm dx$ to get

$$\begin{align}
\int\cos^3x\,\mathrm dx&=\cos^2x\sin x+2\int t^2\,\mathrm dt\\[1ex]
&=\cos^2x\sin x+\frac23t^3+C\\[1ex]
&=\cos^2x\sin x+\frac23\sin^3x+C
\end{align}$$

Differentiate to confirm:

$$\frac{\mathrm d}{\mathrm dx}\left[\cos^2x\sin x+\frac23\sin^3x+C\right]=(-2\cos x\sin^2x+\cos^3x)+2\sin^2x\cos x\stackrel{\checkmark}{=}\cos^3x$$

Title: Computing $\int \cos^3(x)\,dx$


问:
Solve $\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $
Solve
$$\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $$
under degrees (not radians).

There is a well known identity

$$4 \sin x \sin(60 ^{\circ} −x) \sin(60 ^{\circ} +x) = \sin3x$$

but it doesn't quite help here. I can guess a solution $x = 6^{\circ} $, but I would like a more satisfactory solution other than guessing ...

答:
Take five points equally spaced around the circle.  Their centre is at the origin, so
$$\cos0^o+\cos72^o+\cos144^o+\cos216^o+\cos288^o=0\\
\cos36^o-\cos72^o=1/2\\
2\sin54^o\sin18^o=1/2\\
\sin6^o=4\sin54^o\sin6^o\sin18^o$$


问:
GCD proof of (m,2n)=(m+2n,n)
(m,2n)=(m+2n,n). I was testing cases before believing that this was true, and to me it only seems to be true for odd m, n, is this correct?</p>

答:
Say $d$ is the GCD of $m+2n$ and $n$; i.e., $d$ divides both $n$ and $m + 2n$.  Then it must divide their difference, $d \mid m + 2n - n = m+n$.  Again by the same logic, $d$ must divide $m + n - n = m$.  So the GCD of $m+2n$ and $n$ is equal to the GCD of $m$ and $n$.

But if that is true, then there are obvious cases where the GCD of $m$ and $n$ is not the same as the GCD of $m$ and $2n$; specifically, if $m$ is even and $n$ is odd, then the GCD of $m$ and $n$ cannot be even, because $n$ is not divisible by $2$; but the GCD of $m$ and $2n$ will be even, since $2$ divides both.

Title: GCD proof of (m,2n)=(m+2n,n)


问:
Evaluate the following integral : $\int_0^\infty \frac{1}{2 \sqrt{x}(x+1)} dx$
I am unsure how to do this. I have been told that I must separate it into 2 limits since it is discontinuous at 0 and infinity. Could anyone help?

Update: I integrated by substitution by setting $u= \sqrt{x}$ and this was equal to $arctan{x}$. It is just the limits which I am unsure of.

答: Hint: $I = \int_{0}^1 + \int_{1}^\infty$. For the first one, put $u = \sqrt{x}$, can you take it from here?


问:
Determine the number of integer solutions of the equation $ x_{1}+x_{2}+x_{3}+x_{4}=20 $
Determine the number of integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=20
$$
under the restrictions:
$$
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{array}
$$
I did $x_1=y_1+2$ then $0\leq y_1 \leq4$

$x_2=y_2$ then $0\leq y_2 \leq5$

$x_3=y_3+2$ then $0\leq y_3 \leq6$

$x_4=y_4+2$ then $0\leq y_4 \leq4$

So $y_1+y_2+y_3+y_4 \leq 19$
So the number of solutions is

$\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
k
\end{array}\right)=\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
3
\end{array}\right)$
is right?

答:
Actually there is a trick to do this, to remove inequality introduce a new dummy variable :

$$ y_1 + y_2 + y_3 + y_4 + c =19$$

Now, this represent same as inequaliy if we let $ c$ vary from $[0,19]$, can you think why? [Well I couldn't haha :P, see my question about it here]

After this use generating functions as shown in this post

Comment on your method:

Stars and bars method of finding solutions assumes constraints on each variable is same, in your question each variable has different upper limit , so it doesn't work. See the derivation for formula for understanding why. A modified version however works see here( Thanks to  user2661923's comment)


问:
How do I split 16 people into 4 groups, 5 different times with each group being unique (no repeats)?
I have a group of 16 people that are working together on a committee. We want to get to know each other better by splitting into smaller groups of 4 for icebreakers.

I believe if we did this 5 times, every member would be able to meet every other member, with nobody meeting the same person twice. I can't figure out the formula for this. Can you help?

答:
Here is a solution. Arrange the $16$ people in a square. First break them up by rows, then by columns, and then as follows:

```<code>Round 3:      Round 4:      Round 5:
A B C D       A C D B       A D B C
B A D C       B D C A       B C A D
C D A B       C A B D       C B D A
D C B A       D B A C       D A C B
</code>```

Body:
Here is a solution. Arrange the $16$ people in a square. First break them up by rows, then by columns, and then as follows:

Round 3:      Round 4:      Round 5:
A B C D       A C D B       A D B C
B A D C       B D C A       B C A D
C D A B       C A B D       C B D A
D C B A       D B A C       D A C B


答:
Hint: Use the substitution $\;t=\sqrt x$, $\:\mathrm dx=2t\,\mathrm dt$.

The integral becomes
$$\int_0^\infty\frac{2t\,\mathrm dt}{2t(t^2+1)}=\int_0^\infty\frac{\mathrm dt}{t^2+1},$$
which shouldn't be too hard.

Title: Evaluate the following integral : $\int_0^\infty \frac{1}{2 \sqrt{x}(x+1)} dx$

Body:
I am unsure how to do this. I have been told that I must separate it into 2 limits since it is discontinuous at 0 and infinity. Could anyone help?

Update: I integrated by substitution by setting $u= \sqrt{x}$ and this was equal to $arctan{x}$. It is just the limits which I am unsure of.


答:
HINT

To find the indefinite integral, make the substitution $u=\sqrt x$, and the integrand is reduced to a nicer one.

About splitting it up, there's no need to split up the integral once you've performed the substitution as the new integral has only one limit that makes it improper.


问:
Which is the best way define Lie derivative?
I read 2 main definitions of the Lie derivative. One is done using the concept of flow and the pullback-pushfoward while the other one uses the algebraic with axioms. Both of these can be found on wikipedia. I know that these ways are equivalent but I sort of need to find a principal one. Since I am studying this alone and I don't have a unique definition given by a professor I started wondering which one is the best/most correct way to define the Lie derivative for vector and tensor fields so here I am in trouble also in understanding it. So, in which way should I define this in differential geometry ?</p>

答: I can write my opinion. Consider that I am just a student. The ideal variant is to define something axiomatically, then to give an example of such object, a realization of it. You ask why? I give an explonation. I think you know that real numbers can be defined with at least three ways. The first is a set of classes of equivalence of fundamental sequences of rational numbers. The second is a set of Dedecind's sections of rational numbers. The third is a set of sequences of numbers from zero to nine. All these constructions can be used as a definition. But from set theory point of view these sets are not the same, they are different as sets. "Geometrically" they are about one thing, but formally they are different. So where are "real" real numbers? To avoid this problem it is better as I think to define real numbers with a list of axioms. I hope you heard about these axioms. Then you prove two things. The first is the existence, for it you may give one of constructions I mentioned. Then you prove uniqness up to an "isomorphism". I hope you get my idea. This way is ideal for me and as I think it is good to try to correspond with defining something.</p>


问:
Intuition behind order isomorphisms
So I've tried to teach myself some discrete math and topology and I've recently come to a grinding halt on the topic of order relations and specifically order isomorphisms.
I'm finding it difficult to grasp how isomorphic sets are equal and the definition of the function itself.

Is there any intuition and basic examples of $2$  isomorphic sets as I can't quite understand what exactly it means for 2 sets to be order isomorphic and what an isomorphism is.
Looking on Wikipedia I came across this definition

Formally, given two posets

Formally, given two posets $(S,\leq _{S})$ and $(T,\leq _{T})$, an order isomorphism from ${\displaystyle (S,\leq _{S})}$ to ${\displaystyle (T,\leq _{T})}$ is a bijective function ${\displaystyle f}$ from ${\displaystyle S}$ to ${\displaystyle T}$ with the property that, for every ${\displaystyle x}$ and ${\displaystyle y}$ in ${\displaystyle S}$, $x\leq y $ if and only if  $f(x)\leq f(y)$. That is, it is a bijective order-embedding.

So my confusion is what exactly does it mean when they are isomorphic how is it that they can be essentially said to be the same is there any simple example of such a function that obeys this rule that would give some intuition?.

Furthermore I can't understand why the function need be bijective doesn't the definition of the sets being partially ordered imply infectivity? If not are  order isomorphisms not applicable to strict total orders?.

Thanks in advance.

答:
Look at the set $S = \{0, 1, 2\}$ with partial order given by $0 \le 0$, $1 \le 1$, $2 \le 2$, $0 \le 1$, $0 \le 2$, and nothing else. You might represent this partially ordered set with a V, where the bottom of the V is $0$, and the two heads of the V are $1$ and $2$.

Now look at another set $T = \{3, 4, 5\}$ with partial order given by $3 \le 3$, $4 \le 4$, $5 \le 5$, $3 \le 4$, $3 \le 5$, and nothing else. Again, you can represent this with a V, where the bottom is $3$, and the heads are $4$ and $5$.

You may complain that these two partially ordered sets are really "the same", just that the names of the elements are different. In particular, if we map $x \in S$ to $x+3 \in T$, then we seem to be able to identify (= make identical) the two partially ordered sets.

How do we formalize this notion of being "the same up to renaming of elements"?

Well, renaming elements is just specifying a bijection, because a bijection gives you a one-to-one correspondence between elements of one set and elements of the other set.

But it cannot be just any bijection. It has to preserve the partial order. For example, you wouldn't consider the function mapping $x \in S$ to $5-x \in T$ to be an appropriate renaming, because $0$ and $1$ are mapped to $5$ and $4$ respectively, and $0 \le 1$, but $5 \not\le 4$.

So, "preserving the partial order" means that $x \le y$ if and only if $f(x) \le f(y)$, where $f$ is our bijection.

Here's a little exercise if you made it to the end: can you show that there are exactly $2$ order-isomorphisms from $S$ to $T$?

Title: Intuition behind order isomorphisms


问:
How to find the area of this triangle? I tried to find the area of other little triangles and substract to the entire area but can't solve.
$MPQF$ trapezoid has MF and PQ bases,  $MF$=$24$,  $PQ$=$4$, the height of this trapezoid equals to $5$.

we have a point N on the side $MP$, such that $MN=3NP$.

find the area of the triangle $NQF$

for example $S$=$70$ (trapezoid), then I am trying to do this:
$S-S(NMF)-S(NPQ)$,  which gives me the answer, but I am struggling to find the area of $NMF$ and $NPQ$ because I don't know the heights of these triangles and can't connect those heights to trapezoid height

答:
Drop a perpendicular from $P$ onto base $MF$ meeting it at the point $X.$ Then triangle $MXP$ is a right triangle with right angle at $X.$ We know $XP=5.$

Now let $\theta$ be angle $XMP$ so that $MP\sin \theta=5.$ Then since $MN=3NP$ the height of triangle $MNF$ is $15/4$ and height of triangle $NPQ$ is $5/4.$ This gives a way to finish the area calculation as you outlined.

Note I think you didn't want the letter $S$ at the start of the two smaller removed triangles.

Added. I did it as if point $X$ lies between $M$ and $C$ but think the same heights result if that is not the case, if $X$ is on the extension of $MF.$


问:
$\int_0^a \mid f(t) \mid dt \rightarrow 0 \Rightarrow f = 0 $?
I was reading a proof and I stucked at some point that I don't understand. Let's consider norm : $$||f||_1 = \int_0^1 |f(t)| dt$$

Now I will cite a part of the proof that I don't get

I don't understand this reasoning marked in yellow rectangle. The inequalities above says that:
$$0 \le \int_0^a|f(t)|dt \le something$$ where $something \rightarrow 0 $. So how can say by this argument that $\int_0^a|f(t)|dt = 0$ ?

If something is not understandble you can find full proof here: https://www.math.kit.edu/iana3/~schnaubelt/media/fa14-skript.pdf (on page 5)

答:
Set $something = k(n)$ and suppose
$$  0 < \int_0^a |f(x)| \,\mathrm{d}x = \varepsilon \leq k(n) \text{.}  $$

Since $k(n) \rightarrow 0$, there exists an $N$ such that for $n > N$, $k(n) < \varepsilon/2$, violating the inequality.  Therefore, $\varepsilon \not > 0$ and the integral is zero.

In different words, the integral is squeezed between $0$ below and an infinite set of values of $k(n)$ above and the infimum of these upper bounds is also $0$, so the integral has nowhere to be other than $0$.


问:
Disjoint closed sets in subspace means closure is disjoint from set
Let $Y$ be a subspace of a topological space $X$, and let $A$ and $B$ be two disjoint closed subsets of $Y$ in the subspace topology. Show that $\bar A \cap B=\varnothing$ and $A \cap \bar B=\varnothing$.

Attempt:Let $A$ and $B$ be two disjoint closed sets in $Y$. Then $A=\bar A \cap Y$,where $\bar A$ is the closure of $A$ in $X$. $B=\bar B \cap Y$, where $\bar B$ is the closure of $B$ in $X$.Then if $y \in \bar A,y \notin B$ since $X-\bar A$ is an open set containing $B$ disjoint from $A$. Similarly if $y \in \bar B$, $y \notin A$. Thus $\bar A \cap B=\varnothing$ and $A \cap \bar B=\varnothing$.

Comment: Unfortunately I think it is ok to assume if $A$ is closed in $Y$ $A=\bar A \cap Y$ for $\bar A$ the closure of $A$ in X, similarly for $B$, but I cannot understand why. Is this justified, if so what is the reason this is justified? Also, I am unsure whether it is correct to say that $X-\bar A$ contains $B$ and similarly for $X-\bar B$. Is this ok? If so why is this justified?

答:
Yes, it’s correct.

Let $A$ be a relatively closed subset of $Y$. Then $A=Y\cap F$ for some $F$ that is closed in $X$, and $\operatorname{cl}_XA$ is the intersection of all closed subsets of $X$ that contain $A$, so $\operatorname{cl}_XA\subseteq F$, and therefore

$$A\subseteq Y\cap\operatorname{cl}_XA\subseteq Y\cap F=A\,,$$

which immediately implies that $A=Y\cap\operatorname{cl}_XA$.


答: Since $\frac12-\frac1n\geqslant a$, you have that $f_n(t)=0$ on $[0,a]$, and therefore$$\int_0^a|f(t)|\,\mathrm dt=\int_0^a|f(t)-f_n(t)|\,\mathrm dt.$$So, actually $\int_0^a|f(t)-f_n(t)|\,\mathrm dt$ does not depend upon the choice of $n$. But, on the other hand,$$\lim_{n\to\infty}\int_0^a|f(t)-f_n(t)|\,\mathrm dt=0.$$So, $\int_0^a|f(t)|\,\mathrm dt=0$.

Title: $\int_0^a \mid f(t) \mid dt \rightarrow 0 \Rightarrow f = 0 $?

Body:
I was reading a proof and I stucked at some point that I don't understand. Let's consider norm : $$||f||_1 = \int_0^1 |f(t)| dt$$

Now I will cite a part of the proof that I don't get

I don't understand this reasoning marked in yellow rectangle. The inequalities above says that:
$$0 \le \int_0^a|f(t)|dt \le something$$ where $something \rightarrow 0 $. So how can say by this argument that $\int_0^a|f(t)|dt = 0$ ?

If something is not understandble you can find full proof here: https://www.math.kit.edu/iana3/~schnaubelt/media/fa14-skript.pdf (on page 5)


答: You have a squeeze theorem from a theory of limits. If you have three sequences $\{x_n\},\{y_n\},\{z_n\}\subseteq \mathbb{R}$ such that $x_n\leq y_n \leq z_n$ for all $n$ and if $x_n$ and $z_n$ converge to $A\in \overline{\mathbb{R}}$ then $y_n$ also coverges to $A$.


问:
Number of solutions
I have a polynomial in integers $\psi (x)$ of degree $k$.
Consider the number of solutions
$$
\psi(z) \equiv u (\mod p^r)
$$

with
$$
(\psi'(z),p)=1.
$$

I was wondering how can I show that the number of solutions is $O(1)$?

Thank you!

答:
There are at most $\max(k,p)\leq p$ solutions to $\psi (z) \equiv u \pmod{p}$.

Consider lifting solutions from $p^r$ to $p^{r+1}$.

Hensel's lifting lemma tells us that if $z$ is a solution $\pmod{p^r}$, then there exists a unique solution $z^*$ in the higher modulus power, such that $z^* \equiv z \pmod{p^r}$ if $\psi'(z) \not \equiv 0 \pmod{p^r} $.

Hence, there are at most $p$ solutions where $\psi'(z) \neq 0$, and thus there are at most $p$ solutions to $(\psi'(z), p) = 1$.

ContentLicense: CC BY-SA 3.0

Title: Number of solutions


问:
Looking for a counterexample regarding operator norm and spectra
I have proven that if $\mathcal{A}$ is a unital Banach algebra and  $r>0$, $x,y$  commuting elements in $\mathcal{A}$ such that $\Vert x - y \Vert < r$ then  $\sigma_\mathcal{A}(y) \subseteq B_r (\sigma_\mathcal{A}(x)):=\bigcup_{t\in \sigma_\mathcal{A}(x)}B_r(t).$ What I'm now looking for is a counterexample to show that this does not hold in general for non-commuting elements. However, despite expecting I could find nice and easy counterexamples in $M_n(\mathbb{C})$, I have not been able to find any. Do they exist, or do I need to look at a nastier Banach algebra?

答:
Let $x=\begin{bmatrix} 0 & \frac{1}{\epsilon} \\ 0 & 0 \end{bmatrix}$, $y=x +\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. Then $x$ has spectrum $\{ 0 \}$, $y-x$ has norm of order $1$ and $y$ has an eigenvalue of order $\frac{1}{\epsilon}$. So for sufficiently small $\epsilon$ the desired inclusion is violated.

Remark: this behaviour is impossible if $x$ is a normal matrix (or even a normal element of a $C^*$ algebra). This is because the resolvent of $y$ admits a geometric series expansion in powers of $(x-z)^{-1}(y-x)$ and for $x$ normal, $\| (x-z)^{-1} \|$ is bounded by the inverse of distance between the scalar $z$ and the spectrum of $x$. Thus for $\| y-x \| $ smaller than this distance the series converges, and hence $z$ is in the resolvent set.


问:
Disjoint union for monotonicity

Let $\mu$ be a measure on $(\Omega,\mathcal{F})$. Show that if $A\subset B$, then $\mu(A) \le \mu(B)$.

Proof

Wikipedia defines

I don't see how that "disjoint union" works here. Is this the wrong definition of "disjoint union", or am I missing something?

答: They just mean a union of two disjoint (i.e. with empty intersection) sets. Ignore the definition in Wikipedia for this case.

Title: Disjoint union for monotonicity


答:
If there is an order isomorphism between two posets, a good picture to have in mind is that they are the same poset, just with differently labelled vertices. If $S$ and $T$ are order isomorphic with isomorphism $f:S\to T$, this means that for every $s\in S$, we an "relabel it" with $f(s)$. What we get when we do this is exactly a copy of $T$. What this means is that $s_1 \leq s_2 \iff f(s_1)\leq f(s_2)$, so all the order relationships that exist in $S$ also exist in $T$. Picture $S$ and $T$ as being directed graphs. Them being order isomorphic means that the graphs are the same shape.

To illustrate why the map needs to be bijective, consider for example the posets $\mathbb{Z}\to \mathbb{Z} \amalg \mathbb{Z}$ (where $\amalg$ is the disjoint union. The order on $Z$ is the standard order, and the order on $\mathbb{Z} \amalg \mathbb{Z}$ is such that elements are comparable only when they come from the same copy of $\mathbb{Z}$. This is essentially just taking two copies of the poset $\mathbb{Z}$ and putting them next to each other. Then the map $f:\mathbb{Z}\to \mathbb{Z} \amalg \mathbb{Z}$ which just sends $\mathbb{Z}$ to the first factor. Then $f$ satisfies $a \leq b \iff f(a)\leq f(b)$ for all $a$ and $b$, but clearly we don't want to consider these two to be the same poset.

Title: Intuition behind order isomorphisms

Body:
So I've tried to teach myself some discrete math and topology and I've recently come to a grinding halt on the topic of order relations and specifically order isomorphisms.
I'm finding it difficult to grasp how isomorphic sets are equal and the definition of the function itself.

Is there any intuition and basic examples of $2$  isomorphic sets as I can't quite understand what exactly it means for 2 sets to be order isomorphic and what an isomorphism is.
Looking on Wikipedia I came across this definition

Formally, given two posets

Formally, given two posets $(S,\leq _{S})$ and $(T,\leq _{T})$, an order isomorphism from ${\displaystyle (S,\leq _{S})}$ to ${\displaystyle (T,\leq _{T})}$ is a bijective function ${\displaystyle f}$ from ${\displaystyle S}$ to ${\displaystyle T}$ with the property that, for every ${\displaystyle x}$ and ${\displaystyle y}$ in ${\displaystyle S}$, $x\leq y $ if and only if  $f(x)\leq f(y)$. That is, it is a bijective order-embedding.

So my confusion is what exactly does it mean when they are isomorphic how is it that they can be essentially said to be the same is there any simple example of such a function that obeys this rule that would give some intuition?.

Furthermore I can't understand why the function need be bijective doesn't the definition of the sets being partially ordered imply infectivity? If not are  order isomorphisms not applicable to strict total orders?.

Thanks in advance.


问:
$a=\frac{2}{3}b+\frac{1}{3}c$ as Average Value
$f(x)$ is a continuous function and a is the average value of f on the interval [1, 4]. b is the average value of f on the interval [1,3]. c is the average value of f on [3, 4]. Is it true that $a=\frac{2}{3}b+\frac{1}{3}c$? I hypothesize this to be false. For a continuous function, say sinx, the average value over these ranges does not form this equation here. But I wanted a formal mathematical proof for this. Where do I proceed?

答: We have $$a = \frac{1}{4-1} \int_{x=1}^4 f(x) \, dx, \\ b = \frac{1}{1-3} \int_{x=1}^3 f(x) \, dx, \\ c = \frac{1}{4-3} \int_{x=3}^4 f(x) \, dx.$$  Consequently $$3a = \int_{x=1}^4 f(x) \, dx = \int_{x=1}^3 f(x) \, dx + \int_{x=3}^4 f(x) \, dx = 2b + c,$$ or $$a = \frac{2}{3} b + \frac{1}{3} c.$$  This applies as long as each of the individual integrals is well-defined.


问:
How to obtain conditional cumulative probability
Two random variables $x$, and $y$, with PDF $f_x(x)$ and $f_y(y)$.

Then a third variable $z = x + y + c$, with $c$ being a constant.

What is the probability of $x > x'$, given $z$? ($x'$ is a cutoff value of $x$.)

That is, what is $P(x > x' | z)$. Thank you so much.

答:
Let me rewrite this:

Consider some independent random variables $X$ and $Y$, with density $f_X$ and $f_Y$ respectively, and $Z=X+Y+c$ for some constant $c$. Compute $P[X\gt x\mid Z]$, for every $x$.

There is no general answer to this question, but, when $X$ and $Y$ are both normal, one can go further.

Thus, assume without loss of generality that $X$ is centered normal vith variance $a^2$ and that $Y$ is centered normal vith variance $b^2$, then $(X,Z)$ is a centered gaussian vector with covariance matrix $\begin{pmatrix}a^2&a^2\\ a^2&a^2+b^2\end{pmatrix}$. In particular, $X=cZ+dU$, where $U$ is standard normal and independent of $Z$, for some suitable $(c,d)$. One can check that $c=a^2/(a^2+b^2)$ and $d=ab/\sqrt{a^2+b^2}$. Thus, the conditional distribution of $X$ conditionally on $Z$ is gaussian with mean $cZ$ and variance $d^2$, in particular,
$$
P[X\gt x\mid Z]=P\left[U\gt\frac{x-cZ}d{\Large\mid} Z\right]=1-\Phi\left(\frac{x-cZ}d\right).
$$


答:
My opinion is that the definition using flows is the clearest, because it shows in what sense this is a derivative. Indeed, for $\theta$ the flow of $X\in \mathfrak{X}(M)$, and $W$ another vector field we get
$$
(\mathcal{L}_XW)_p=\lim_{t\to 0}\frac{d(\theta_{-t})_{\theta_t(p)}W_{\theta_t(p)}-W_p}{t}
$$
which shows that the Lie derivative is measuring how $W$ changes as we flow along $X$. Ideally, you would then prove the algebraic characterization.

Title: Which is the best way define Lie derivative?

Body: I read 2 main definitions of the Lie derivative. One is done using the concept of flow and the pullback-pushfoward while the other one uses the algebraic with axioms. Both of these can be found on wikipedia. I know that these ways are equivalent but I sort of need to find a principal one. Since I am studying this alone and I don't have a unique definition given by a professor I started wondering which one is the best/most correct way to define the Lie derivative for vector and tensor fields so here I am in trouble also in understanding it. So, in which way should I define this in differential geometry ?</p>


问:
Proving norm of coset is same as norm of some vector in that coset
Let $V$ be a nontrivial subspace of a finite dimensional vector space $U$ with a positive definite scalar product. I want to prove that for each $[u]\in U / V$ there is a $w\in[u]$ such that $\|[u]\|=\|w\|$.

I have the definition of a norm over a quotient space as $\|[u]\|=\inf\{\|x\|\mid x\in[u]\}$ for $[u]\in U/V$. Doesn't this definition have built into it that the there is a vector $w\in U$ such that $\|[u]\|=\|w\|$? In this case doesn't my definition directly imply what I am trying to prove?

答:
Hint: in any normed space, finitely dimensional subspaces are closed. On the other hand, in finitely dimensional normed spaces a subset is comapct if and only if it is closed and bounded.

Fix $x \in V$ and consider $F = x+U$. Since the translation $T_x(y) = y+x$ is a homeomorphism, the set $F$ is closed. Note that

$$
\|[x]\| = \inf \{\|y\| : y \in x+U\} = d(0,F).
$$

However, for some arbitrary $y_0 \in F$ we moreover have

$$
\|[x]\| = \inf\{\|y\| : y \in F\} = \inf\{\|y\| : y \in F, \|y\| \leq\|y_0\| \}.
$$

Can you take it from here?


答:
To prove the monotonicity property, the author has made use of the finite additivity property and the fact that measures are non-negative.

More precisely, given a collection of pairwise disjoint sets $A_{k}\in\mathcal{F}$, where $1\leq k\leq n$, one has that
\begin{align*}
\mu\left(\bigcup_{k=1}^{n}A_{k}\right) = \sum_{k=1}^{n}\mu(A_{k})
\end{align*}

At your case, one has that $A_{1} = A$ and $A_{2} = B\cap A^{c}$. Consequently, given that $A\subseteq B$, one has that
\begin{align*}
\mu(B) = \mu(A\cup(B\cap A^{c})) = \mu(A) + \mu(B\cap A^{c}) \geq \mu(A)
\end{align*}

and we are done.

Hopefully this helps!

Title: Disjoint union for monotonicity

Body:
Let $\mu$ be a measure on $(\Omega,\mathcal{F})$. Show that if $A\subset B$, then $\mu(A) \le \mu(B)$.

Proof

Wikipedia defines

I don't see how that "disjoint union" works here. Is this the wrong definition of "disjoint union", or am I missing something?


Title: $a=\frac{2}{3}b+\frac{1}{3}c$ as Average Value

Body: $f(x)$ is a continuous function and a is the average value of f on the interval [1, 4]. b is the average value of f on the interval [1,3]. c is the average value of f on [3, 4]. Is it true that $a=\frac{2}{3}b+\frac{1}{3}c$? I hypothesize this to be false. For a continuous function, say sinx, the average value over these ranges does not form this equation here. But I wanted a formal mathematical proof for this. Where do I proceed?

Body:
$\displaystyle 3a = \int_{1}^4 f(x) \text{d}(x)$, $\;\;\displaystyle 2b = \int_{1}^3 f(x) \text{d}(x) $, $\;\;\displaystyle c = \int_{3}^4 f(x) \text{d}(x) $

$\implies 3a = 2b + c \iff a = \frac{2}{3}b + \frac{1}{3}c$

For $f(x) = \sin(x)$ we get to $5$s.f

$a = 0.39798$,  $b = 0.76515$,  $c = -0.33635$ and indeed

$3(0.39798) = 2(0.76515) + (-0.33635)$ (at least to 4.s.f or so)


问:
Given that $x^3=x^2+x+1$ find $x^4$ in terms of $x^2$ and a constant.
I don't know is this can be done,  but I'm just interested in it, and I'm hoping that it is possible. Here's the question:

Given that $x^3=x^2+x+1$ find $x^4$ in terms of $x^2$ and a constant.

I simply cannot seem to do it (and no, I don't mean something like $x^4=(x^2)^2$, I mean in terms of $x^2$ with some constant coeficcient :-) ) . The best I can do is show that $x^4=2x^3-1$, but nothing more.

Thank you for your help.

答: $x^3=x^2+x+1$. Multiply by $x$: $x^4=x^3+x^2+x$. Replace $x^3$ by $x^2+x+1$: $x^4=2x^2+2x+1$. And I think that's as good as it gets, I don't see anyway to eliminate the linear term, without some further information.


问:
From geometric sequence to function
I have this question:

Find the functions which equal the sums:
$$
x + x^3 + x^5 + ..
$$

Now, I can see in my result list, that its supposed to give

$$
\frac{x}{(1-x^2)}
$$

I can see why the numerator should be x, but I fail to see why the denominator should be
$$
(1-x^2)
$$
Can someone please explain this to me?

David

答:
This is a infinite geometric series which has first term $a=x$ and common ratio $r=\dfrac {x^3}{x}\implies r=x^2 $

sum of infinite geometric series is $s_{\infty}=\dfrac {a}{1-r} when |r|<1 $
$$s_{\infty}=\dfrac {x}{1-x^2}$$

Title: From geometric sequence to function


问:
Solve $\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $
Solve
$$\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $$
under degrees (not radians).

There is a well known identity

$$4 \sin x \sin(60 ^{\circ} −x) \sin(60 ^{\circ} +x) = \sin3x$$

but it doesn't quite help here. I can guess a solution $x = 6^{\circ} $, but I would like a more satisfactory solution other than guessing ...

答:
Apply $\sin(60-x)=\frac{\sqrt3}2\cos x -\frac12\sin x$ and
rearrange the equation to isolate $x$

\begin{align}
\cot x &=\frac{1+2 \sin6\sin18} {2\sqrt3\sin 6\sin 18}
\end{align}
Then, substitute $ \sin 18 =\frac{\sin36}{2\cos18}
= \frac{\sin72}{4\cos18\cos36}
= \frac1{4\cos36}$
to obtain
\begin{align}
\cot x &=\frac{2\cos36+\sin6} {\sqrt3\sin 6}
=\frac{\cos36+(\sin54 +\sin6)} {\sqrt3\sin 6}\\
&=\frac{\cos36+\cos24} {\sqrt3\sin 6}
= \frac{2\cos30\cos6} {\sqrt3\sin 6}= \cot 6
\end{align}
Thus, $x = 6^\circ+ 180^\circ n$.


问:
How many length 10 words can we make that start with "TRY" or end with "TRY"?
With replacement and repetition is allowed, if we can choose from the 26 letters of the alphabets, how many length 10 strings/ words can we make that either start with the substring "try" or end with "try". I know this is about the inclusion exclusion principle, so it would be words that start with TRY + words that end with TRY - the duplicates between them. This is how I started it:

Words that start with try: 26^7 (because first 3 letters are not changeable).

Words that end with try: 26^7 (last 3 letters are not changeable)

But how do I calculate the numbers that both begin and end with try (the duplicates) ?
Thanks.

答:
There are $N=26^4$ string which both start and end with TRY. The next action depends on the fact if such words are allowed (this is not quite clear from the question).

If they are allowed you should subtract $N$ once, and if not allowed - twice.


问:
Show that the substitution $t=\tan\theta$ transforms the integral ${\int}\frac{d\theta}{9\cos^2\theta+\sin^2\theta}$, into ${\int}\frac{dt}{9+t^2}$
To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:

$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\theta$$

I tried working backwards

$$\frac{d}{d\theta}\tan\theta=\sec^2\theta\,\,\,\,{\Rightarrow}\,\,\,\,d\,\tan\theta=\sec^2\theta\,d\theta$$
$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{9+\tan^2\theta}$$

$$\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\,\,{\Rightarrow}\,\,\,\,\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$

$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{\left(9+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}$$

$$9=\frac{9\cos^2\theta}{\cos^2\theta}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta+\sin^2\theta}{\cos^2\theta}\right)}$$

$${\Rightarrow}\,\,\,\,{\int}\frac{\color{red}{\cos^2\theta\,\sec^2\theta}\,d\theta}{9\cos^2\theta+\sin^2\theta}$$

Now I have to prove that $$\cos^2\theta\,\sec^2\theta=1$$
but I don't think it is... What have I done wrong? Regards Tom

答:
An alternative method. Let
$$I={\int}\frac{1}{9\cos^2x+\sin^2x}\,dx$$
Dividing the denominator and numerator by $\cos^2x$ we have
$$I=\int\frac{\sec^2x}{9+\tan^2x}dx$$
Make the substitution $t=\tan x$, so that $\sec^2x~dx=dt,$ and we are done.


问:
Linear transformation of a matrix by given bases
I am trying to find out the matrix for the following linear transformation with respect to the given bases:
$$
T:\mathbb{R}[x]_{\le 2} \rightarrow \mathbb{R}[x]_{\le 3} \space \text{given by}\space  T(f)=xf \space \text{in the bases}\space  {1,x,x^2} \space \text{for} \space \mathbb{R}[x]_{\le 2} \space \text{and} \space {1,x,x^2,x^3} \space \text{for} \space \mathbb{R}[x]_{\le 3}
$$

I know that the matrix should have $3$ columns and $4$ rows.

Here is an example of a similar question that I got given in a multiple-choice quiz:

The matrix for the linear transformation $T:\mathbb{R}[x]_{\le3} \rightarrow \mathbb{R}$ given by $T(f)=f(3)$ in the basis {${1,x,x^2,x^3}$} for $\mathbb{R}[x]_{\le3}$ and the basis {${1}$} for $\mathbb{R}$ is $(1,3,9,27)$.

This one I found to be pretty straightforward as it's $4$ columns and $1$ row, so you just substitute $3$ in and you get the answer $(1,3,9,27)$.

How can apply what I know from this question to the one that I'm trying to answer?

答: The first element of $\{1,x,x^2\}$ is $1$, and $T(1)=x$, which is the second element of $\{1,x,x^2,x^3\}$. Also, the second element of $\{1,x,x^2\}$ is $x$, and $T(x)=x^2$, which is the third element of $\{1,x,x^2,x^3\}$. Finally, the third element of $\{1,x,x^2\}$ is $x^2$, and $T(x^2)=x^3$, which is the fourth element of $\{1,x,x^2,x^3\}$. So, the matrix that you are after is$$\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$


问:
What does the "-" in front of the arccos do to its result, when wanting to compute the polar coordinates of a complex number?
so I'm still training for my calculus exam but I stumbled upon a confusing thing again :( I think I kind of feel certain with computing polar coordinates now, but I just don't know if my train of thought with this one is right.

Soooo I have these three exercises, where I'm trying to come up with the respective $\arccos$ part that needs to be computed for $\varphi$ in $z=re^{i\varphi}$:

$$(-\frac{\sqrt{3}}{2}-\frac{1}{2}i) \Rightarrow r=1 \Rightarrow -\arccos{(-\frac{\sqrt{3}}{2})} = -(\frac{5\pi}{6}) = \frac{7\pi}{6}.$$

my understanding of the whole thing is that, for every negative minus sign, we move one quadrant to the left in the polar grid, to the corresponding value of $\cos$. I'm really really bad at explaining so I'll just draw what I mean:

the blue arrow is for the minus in front of the argument within $\arccos$, the red arrow for the $-$ in front of $\arccos$.

and for another exercise, it's pretty much the same:

$$(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i) \Rightarrow r=1 \Rightarrow -\arccos{(-\frac{\sqrt{2}}{2})} = -(\frac{3\pi}{4}) = \frac{5\pi}{4}.$$

with the same argumentation as above.

HOWEVER,, now, when I try to use this trick for the following exercise

$$(\frac{\sqrt{3}}{2}-\frac{1}{2}i) \Rightarrow r=1 \Rightarrow -\arccos{(\frac{\sqrt{3}}{2})} = -(\frac{\pi}{6}) = \frac{11\pi}{6}.$$

that is the sample solution. BUT, according to what I've gotten accustomed to, the arrow should only be moving from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$, not to $\frac{11\pi}{6}$.

like demonstrated here.

could someone please explain to me in simple terms what in particular is wrong with my way of thinking?... I would appreciate it so much ;-;

thank you guys in advance <3

答:
Why are you not using
\begin{align*}
|-\sqrt{3}/2 - \mathrm{i}/2| &= 1  \text{,}  \\
\Re(-\sqrt{3}/2 - \mathrm{i}/2) &= -\sqrt{3}/2  \text{, and }  \\
\Im(-\sqrt{3}/2 - \mathrm{i}/2) &= -1/2  \text{, so}  \\
\theta &= 7\pi/6  \text{.}
\end{align*}

The first guarantees that we have extracted $r$, so have a point on the unit circle.  If hte magnitude is not $1$ extract $r$, then proceed.  The second and third pull out the coordinates of the point on that circle, $(x,y) = (-\sqrt{3}/2, -1/2)$ using the real part ($\Re$) and imaginary part ($\Im$) functions.  The last is the angle in your unit circle diagram for the point that has those coordinates.


问:
Three possible positions of a vertex of a parallelogram
The coordinates of the point A ,B and C are $(5 ,-1 ,0) , (2,4 ,10) , (6 ,-1 ,4) $ respectively

The point D is such that ABCD is a parallelogram .

Find three possible positions of D

My turn :

The midpoint of ACis $(11/2 , -1 ,2) $
Which the same as the midpoint of BD

Then we can find one position of D

But i do not know how can i get the other two positions ?

答:
The three positions that $D$ may take--call them $D_1, D_2, D_3$--form a triangle such that the midpoints of $D_1 D_2$, $D_2 D_3$, and $D_3 D_1$, are $A$, $B$, and $C$ in some order.

So if you want to extend your process to find the other two points, you would find the midpoint of $BC$, then find the point such that $AD$ shares the same midpoint; and then do the same for the midpoint of $AB$.


问:
Average Value of a function $f_{avg}$
Let $f_{avg}$ be the average value of a continuous function f(x) on the interval [a,b]. Assume that there's only one c $\in$ [a , b] for which $f(c)=f_{avg}$. The question is, if this is the case, then must the function be either increasing or decreasing on [a, b]? This questions seems to jump from the assumption that $f(c)=f_{avg}$ for 1 value of c to the conclusion above, and I'm having trouble making sense of the question - obviously this is about the average value of a function, which is denoted by $\frac{1}{b-a}\int_{a}^{b}f(x)dx$, but how does it relate to whether the function is increasing or decreasing? One way I know of finding out is by taking the derivative of f(x), but we don't know what f is in the first place. So...plz help :D

答: I think it is just a wrong claim. Let $a=0, b=2$ and define $f$ with that way: $f(x)=x$ if $x\in[0,1]$ and $f(x)=-10x+11 $ if $x\in[1,2]$. Then $f$ is continuous and not monotonic on $[0,2]$. But $f_{avg}=\frac{1}{2}(\frac{1}{2}-15+11)=-\frac{7}{4}$. But there is only one $x\in [0,2]$ with $f(x)=f_{avg}$.

Title: Average Value of a function $f_{avg}$


问:
Negative factors of a number
Can a factor of a number be negative?

Is $-5$ a factor of $25$ or $-25$?

A number is said to be prime if it has two factors : $1$ and the number itself. So if $-5$ can be a factor of $5$, how to define the prime numbers?

Thanks.

答: The condition you have given does not suffice to make a number prime. It is also generally required that it be greater than $1$, and that the divisors considered be positive.</p>

Title: Negative factors of a number


问:
Determine the number of integer solutions of the equation $ x_{1}+x_{2}+x_{3}+x_{4}=20 $
Determine the number of integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=20
$$
under the restrictions:
$$
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{array}
$$
I did $x_1=y_1+2$ then $0\leq y_1 \leq4$

$x_2=y_2$ then $0\leq y_2 \leq5$

$x_3=y_3+2$ then $0\leq y_3 \leq6$

$x_4=y_4+2$ then $0\leq y_4 \leq4$

So $y_1+y_2+y_3+y_4 \leq 19$
So the number of solutions is

$\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
k
\end{array}\right)=\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
3
\end{array}\right)$
is right?

答:
With $x_{1}+x_{2}+x_{3}+x_{4}=20$ and the the restrictions:
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{array}

You have a good idea to switch to secondary variables to simplify the limits; there are various ways to do this but I would move to zero lower limits with:
\begin{array}{l}
y_1=x_1-2 & & 0 \leq y_{1} \leq 4 \\
y_2=x_2   & & 0 \leq y_{2} \leq 5 \\
y_3=x_3-2 & & 0 \leq y_{3} \leq 6 \\
y_4=x_3-2 & & 0 \leq y_{4} \leq 4
\end{array}
and $x_{1}+x_{2}+x_{3}+x_{4}=20$ then means that $y_{1}+y_{2}+y_{3}+y_{4}=14$.

Now without the upper limits we would have the standard method to determine options at $\binom{14+3}{3}$. However this counts a number of cases when one or more of the variables
exceed their limits. So we need to count those cases too by "pre-loading" the smallest illegal value into the each variable in turn in order to remove these. So to formalize this we'll consider a new set of variables:
\begin{array}{l}
z_1=y_1-5  \\
z_2=y_2-6  \\
z_3=y_3-7 \\
z_4=y_4-5
\end{array}
which then means that any $z_i\geq 0$ would imply that the corresponding $y_i$ (and $x_i$) has broken its upper constraint.

So then we consider the cases (with options shown):
\begin{array}{l}
\color{red}{z_{1}}+y_{2}+y_{3}+y_{4}=14-5  & & \binom{9+3}{3} \\
y_{1}+\color{red}{z_{2}}+y_{3}+y_{4}=14-6  & & \binom{8+3}{3} \\
y_{1}+y_{2}+\color{red}{z_{3}}+y_{4}=14-7  & & \binom{7+3}{3} \\
y_{1}+y_{2}+y_{3}+\color{red}{z_{4}}=14-5  & & \binom{9+3}{3}
\end{array}
with a total of $C_1 = 2\binom{12}{3}+\binom{11}{3}+\binom{10}{3} = 725$

However then we are into an inclusion-exclusion cycle to avoid double counting the cases when more than one constraint is broken, which will need to be added back in. There are six viable cases of two constraints broken:
\begin{array}{l}
\color{red}{z_{1}}+\color{red}{z_{2}}+y_{3}+y_{4}=14-5-6  & & \binom{3+3}{3} \\
\color{red}{z_{1}}+y_{2}+\color{red}{z_{3}}+y_{4}=14-5-7  & & \binom{2+3}{3} \\
\color{red}{z_{1}}+y_{2}+y_{3}+\color{red}{z_{4}}=14-5-5  & & \binom{4+3}{3} \\
y_{1}+\color{red}{z_{2}}+\color{red}{z_{3}}+y_{4}=14-6-7  & & \binom{1+3}{3} \\
y_{1}+\color{red}{z_{2}}+y_{3}+\color{red}{z_{4}}=14-6-5  & & \binom{3+3}{3} \\
y_{1}+y_{2}+\color{red}{z_{3}}+\color{red}{z_{4}}=14-7-5  & & \binom{2+3}{3}
\end{array}
with a total of $C_2 = \binom{7}{3}+2\binom{6}{3}+2\binom{5}{3}+\binom{4}{3}=99$

There is no combination that allows three constraints to be broken simultaneously broken so we can stop there; the total combinations is then
$$\binom{17}{3}-C_1+C_2$$

The trick of looking at the inverse space in the answer by @quantus14 provides a very nice shortcut with the same basic approach.

Defining $w_i$ by how far off each $x_i$ is from its top limits, that is
\begin{array}{l}
w_1=6-x_1 & & w_1 \leq 4 \\
w_2=5-x_2 & & w_2 \leq 5   \\
w_3=8-x_3 & & w_3 \leq 6  \\
w_4=6-x_4 & & w_4 \leq 4
\end{array}
we can see that $w_1+w_2+w_3+w_4 = (6+5+8+6) - (x_1+x_2+x_3+x_4) = 25-20 = 5$. Now unconstrained the options for dividing the $5$ amongst the $w_i$ looks like $\binom{5+3}{3} = 56$ but we have two cases that break the value range available, since $w_1$ and $w_4$ have a maximum value of $4$. So the result is $\binom{8}{3}-2$, matching the previous result.


问:
Every isotropic subspace is contained in a lagrangian subspace.
Let $(V,\Omega)$ be a symplectic vector space. A subspace $L$ is called Lagrangian if $L^\perp=L$. A subspace $W$ is called isotropic if $W\subset W^\perp$.

For any isotropic subspace $W$ there is a Lagrangian subspace $W$ such that $W\subset L$

This is easy to see if we let $e_1,...,e_k$ be a basis of $W$ and extend this to a symplectic basis of $V$. Then we can take $L=span\{e_1,...,e_n\}$. I would like to prove this without resorting to a basis to get used to working with the machinery.

答:
I think it is important here to assume that the pertinent vector spaces are finite dimensional; otherwise the statement is false. I proceed to a proof.

Let $W$ be isotropic. If there exists an isotropic subspace $W'$ containing $W$ properly ($W \subsetneq W'$), we replace $W$ by $W'$. We repeat inductively after there exists no such $W'$, i.e. until we obtain a maximal isotropic subspace. This always happens after finitely many steps, because the dimensions of consecutive subspaces grows and is bounded by $\dim(V) < \infty$. This proves that every isotropic subspace is contained in a maximal isotropic subspace.

The proof will be completed by showing that every maximal isotropic subspace is Lagrangian (the converse is also true; I suggest showing that as an exercise). Suppose that $W$ is maximal isotropic. We have to show that $W^{\perp} \subset W$. For the sake of contradiction assume that $x$ is an element of $W^{\perp}$ which is not in $W$. Then the linear span of $x$ and $W$ is an isotropic subspace properly containing $W$, which is absurd because $W$ is maximal.


问:
Prove There is a set for which {1}∈X and {1}⊆X?
I am new to proof I found this question in worksheet I have no idea how to do it? Can anyone willing explain it to it would much appreciated. this question Prove There is a set for which {1}∈X and {1}⊆X?

This is what I have tired:
suppose {1}EX by definition it means 1 is an element of x hence {1} is subset of X therefore  {1}∈X and {1}⊆X is true

答: Hint: Sets can contain other sets! (It would be more helpful if you could tell us what you tried.)</p>


问:
To show that a sequence is bounded above by $3$.
Let $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2 x_n}$.

I would like to know if there is a way to see that this sequence is bounded above by 3 without induction.

I know how to use induction to show that it is in fact bounded by 2. I would like to not use induction though. Thanks for your time.

答:
Induction is pretty ... pervasive.  Even a simple comment like for $2^x$ is increasing because $2^{x+1} = 2\times 2^x > 2^x$ is using induction.

I could quibble and say let $x_n \ge 3$.  Then $x_n = \sqrt{2 x_{n-1}}$ and

$x_{n-1} = (\frac {x_n}2)^2 \ge (\frac 32)^2 > 3$.

So the is no least $n$ where $x_n \ge 3$ violating the well ordering principal.  So that's a contradiction and there is no $x_n \ge 3$.

What? that's not induction that's the well-ordering principal!

.... but the well-ordering principal is equivalent to induction so it is induction.

I'm not saying it's not provable with induction but I bet no-one here will be able to.

I trust Lee Mosher will attempt and succeed in finding where induction is hiding under the table.


答:
I think for the most part in primary and secondary school, we stick to factoring positive integers to sidestep this issue. We deal with it in higher mathematics though!

Technically, when factoring integers, factors that differ only by $\pm 1$ are considered to be the same factor.

So really, $25=5\cdot5=(-5)\cdot(-5)$ are both factorizations of 25, but we consider them to be "the same" because the factors only differ from each other by $-1$. By convention, we usually just talk about the positive versions as being "the factors."

In ring theory, the significance of prime numbers in $\Bbb Z$ is that the ideal $(a)\lhd\Bbb Z$ is a prime ideal iff $a$ is a prime number. Since $(5)=(-5)$ as ideals, then it seems fair that both 5 and -5 are considered as "prime."

For principal ideal domains more complicated than $\Bbb Z$, it gets even more interesting. If $p$ is a prime in a domain $R$ and $u$ is a unit of the ring, then $pu$ and $p$ are considered them same prime, since $(pu)=(p)$ is a prime ideal. This applies to the integers because in $\Bbb Z$, the only units are $\pm 1$.

Title: Negative factors of a number

Body:
Can a factor of a number be negative?

Is $-5$ a factor of $25$ or $-25$?

A number is said to be prime if it has two factors : $1$ and the number itself. So if $-5$ can be a factor of $5$, how to define the prime numbers?

Thanks.


问:
An example of a non-constant function with the following properties
Let $f: \mathbb{R} \to \mathbb{R}$ be a non-constant function with the following property: Given $\epsilon > 0$, there exists a positive real number $r > 0$ and a set $D \subset \mathbb{R}$ such that

Any interval $I$ with length $r >0$, is such that $D \cap I \ne \emptyset$

$|f(x+t) - f(x)| < \epsilon$ for any $t \in D$ and $x\in \mathbb{R}$

I'm having a hard time picturing a function that is non-constant that satisfies both properties. If we take $D = \mathbb{Q}$ then the first property is easily satisfied, but I can't figure out what kind of function satisfies the second property. Can someone give me an example?

答: Let $f(x)=\sin x$ and (no matter what $\epsilon$ is), let $r=2\pi$, $D=2\pi\Bbb Z$.


答: Consider the equivalence relation $\sim$ on $\Bbb{R}$ defined by $x \sim y$ iff $x - y \in \Bbb{Q}$ and let $P$ be the set of equivalence classes of $\sim$. Choose any non-constant function $g :P \to \Bbb{R}$ and define $f : \Bbb{R} \to \Bbb{R}$, by $f(x) = g([x])$ where $[x]$ denotes the $\sim$-equivalence class of $x$. Then with $D = \Bbb{Q}$ any interval $I$ of positive length meets $D$ and $|f(x + t) - f(x)| = 0$ for any $t \in D$.

Title: An example of a non-constant function with the following properties

Body:
Let $f: \mathbb{R} \to \mathbb{R}$ be a non-constant function with the following property: Given $\epsilon > 0$, there exists a positive real number $r > 0$ and a set $D \subset \mathbb{R}$ such that

Any interval $I$ with length $r >0$, is such that $D \cap I \ne \emptyset$

$|f(x+t) - f(x)| < \epsilon$ for any $t \in D$ and $x\in \mathbb{R}$

I'm having a hard time picturing a function that is non-constant that satisfies both properties. If we take $D = \mathbb{Q}$ then the first property is easily satisfied, but I can't figure out what kind of function satisfies the second property. Can someone give me an example?


问:
Solving unknown complex number w
I have a complex number $w$= a + bi such that $w$$\bar{w}$ = 25  and $w$ + $\bar{w}$ = 8. I assume from this that the complex number in question hast two values. I've already solved the $w$ + $\bar{w}$ = 8 equation, but I'm having problems with $w$$\bar{w}$ = 25.

I've narrowed the equation down to $a^2$+$b^2$=25 but I don't know how to solve the rest. Or if I'm at least on right tracks.

I  suck at this stuff and I've tried over and over again but just can't make it.

答: $w+\bar w=8$ means $a=4$. Then we can find $b$ using $a^2+b^2=25$.

Title: Solving unknown complex number w


问:
How do you find the domain and range of a rational function(or any other function) using Calculus
I know that the only ways to find the range of a rational function is using an inverse, graphing, or using calculus. How would you use calculus to find the range of a rational function (or any other type of function) using Calculus?</p>

答: You can find its limiting behaviour as $x \to \infty$ and $x \to -\infty$, and also find local maxima and local minima. Don't forget to account for vertical asymptotes that arise when the denominator is $0$.


问:
Binomial formula in $GF(2^m)$
there is a binomial formula:

$$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$$

When operations are done in $GF(2^m)$ then all positive integers are reduced $\bmod2$, so binomial formula for $n=2^i$ in $GF(2^m)$ is:

$$(x+y)^{2^i}=x^{2^i} + y^{2^i} $$

So now the question. If there are reduced all binomial coefficients $\binom{n}{k}$, then why exponents like $2^i$ are not reduced?

答:
It's because multiplication by a coefficient is periodic in the coefficient with the right period $P$:
$$ (a+P)\cdot b = a\cdot b + P\cdot b \equiv a\cdot b \quad {\rm mod} \quad P$$
because the equivalence modulo $P$ is defined so that it allows me to subtract multiples of $P$ such as $P\cdot b$ above - while powers are not periodic functions of the exponent with the period $P$:
$$ a^{b+P} \neq a^b \quad {\rm mod}\quad P$$
For a simple example, look at powers of $2$ computed modulo $2$:
$$2^0 = 1, \quad 2^1 = 2 = 0, \quad 2^2 = 4 = 0, \dots $$
All the following powers are equal to $0$ modulo $2$ but the initial one, the zeroth power, is not. (Pick more complicated numbers to get the point - to see how complicated the powers may be especially if $P$ is large.) So the results of the exponentiation are not periodic as a function of the exponent, and you are never allowed to simplify the exponents by considering them modulo $P$.</p>

Title: Binomial formula in $GF(2^m)$


问:
Local coefficient System and universal cover
We work with a topological space $B$ which is path-connected and locally path-connected.

I have troubles writing down a formal proof for the following proposition:

Prop: Any local coeeficient system $A\hookrightarrow E \to B$ is of the form  $$ A \hookrightarrow \tilde{B}\times_{\pi_1B} A \to B $$ i.e. is associated
  to the principal $\pi_1B$-bundle given by the universal cover
  $\tilde{B}$ of $B$ where the action is given by a homomorphism $\pi_1B \to $Aut$(A)$.

I'm able to show that the natural monodromy action of the fundamental group on $A$ gives a group homomorphism $\pi_1B \to $Aut$(A)$. But I'm not able to show that this implies that $E \cong \tilde{B}\times_{\pi_1B} A $. Can you help me on that?

Note: The author of the book I'm reading says this should be easy to show using a standard covering space argument.

Here are some definitions:

Def: A local coefficient system is a fiber bundle $p:E\to B$ such that

The fiber is a discrete abelian group $A$

The structure group $G$ is a subset of Aut$(A)$

Def: Let $p:P\to B$ be a principal $G$-bundle, where $G$ acts on an abelian group $A$. The Borel contruction is the quotient
$$ P\times_G A = P\times A \;/ \sim $$
where the equivalence relation $\sim$ is defined by $(p,a) \sim (pg, g^{-1}a) $, for $g\in G$.

The map $q:P\times_G A \to B$ given by $q([p,a]) = p(p)$ gives a fiber bundle with fiber $A$.

答:
I think one can proceed as follows :

Let us call $p : E \to B$ the local coeffiscient system, and let $x$ be a point in $B$ Let us also call $\varphi : \pi_B \to \mbox{Aut}(A)$ the holonomy homomorphism. Every choice of a point $a\in p^{-1}(x)$ determines a sub-covering in $E$, so that $a$ and $b$ live in the same sub-covering if and only if they are related by an element of $\pi_1B$. So each of these sub-coverings has fibre $\mbox{Im} (\varphi) \cong \pi_1B/\ker \varphi$, with $\pi_1B$ acting on the global covering as one would expect, so they are all isomorphic (to $\tilde{B}/\ker(\varphi)$).
Furthermore there are obviously $A/\pi_1B$ ($=A/\mbox{Im}(\varphi)$) copies of this bundle in $E$, so that $E\cong \tilde{B}/\ker(\varphi) \times A/\mbox{Im}(\varphi)$.

Now
\begin{align*}\tilde{B}\times_{\pi_1B}A = (\tilde{B}\times A)/\pi_1B &=((\tilde{B}\times A)/\ker(\varphi))/(\pi_1B/\ker(\varphi))\\ &=((\tilde{B}/\ker(\varphi))\times A)/(\pi_1B/\ker(\varphi))\end{align*} because the action of $\ker(\varphi)$ on $A$ is trivial (by definition). For the last step, choose a $y\in p^{-1}(x)$, and define the following map :
$$
((\tilde{B}/\ker(\varphi))\times A)/(\pi_1B/\ker(\varphi)) \to \tilde{B}/\ker(\varphi) \times A/\mbox{Im}(\varphi) \\
[([z],a)] \mapsto ([\tilde{z}],[a])
$$
where $[ - ]$ stands for `class of $-$' and  $\tilde{z}$ denotes any element of the $\pi_1G$-orbit of $z$ that is on the same sub-bundle as $y$. The $\tilde{z}$ are all related by an element of $\ker(\varphi)$, so this map is well defined, one can compute an inverse, so it is bijective, and one can check that it is a fibre bundle morphism.

I hope this helped, and wasn't too messy.


问:
Kronecker's delta complex sum form
I hope you're all doing well.

I'm trying to prove the following identity:
$$\frac{1}{N} \sum_{j=1}^{N} e^ {\frac{2i\pi(n-n')j}{N}} = \delta_{nn'}$$

but I'm having some troubles.

This is what I tried:

We know that for a finite geometric series, the entire sum is equal to
$$\sum_{j=1}^{N} ar^{bj} = \frac{ar^b(r^{bN} - 1)}{r^b - 1}$$

In your case we have:
$$r = e$$
$$a = \frac{1}{N}$$
$$b = \frac{2i\pi(n-n')}{N}$$

Using this I found

$$\frac{1}{N} \sum_{j=1}^{N} e^ {\frac{2i\pi(n-n')j}{N}} = \frac{1}{N} \frac{e^{\frac{2i\pi(n-n')}{N}}(e^{{2i\pi(n-n')}} - 1)}{e^{\frac{2i\pi(n-n')}{N}} - 1}$$

but after this I'm not able to see how these exponentials are gonna to result in the Kronecker's delta.

答:
Be careful with $\delta_{nn'}$: your formula is true if $n, n'$ are either regarded as integers modulo $n$ or restricted to set $\{ 1,...,N \}$. Can you see why the value of the sum is unchanged if $n$ or $n'$ is shifted by an integer multiple of $N$?

Take the formula you derived. In the numerator you have $e^{2 \pi i (n-n')}-1=1-1=0$. This is not quite right if $n-n'$ is divisible by $N$ because then also denominator vanishes. In fact in this special case your formula is wrong, because geometric series formula is valid only if the ratio of consecutive terms of the series is not equal $1$. So in this special case you have to compute the sum in a different way.

Title: Kronecker's delta complex sum form


问:
probability of extracting consecutively an even numbered ball and the following numbered one
An urn contains 2n numbered balls that are extracted one by one until no one remains. Which is the probability that for some integer $k \geq 1$ the $(2k-1)$-numbered ball is extracted immediately before the $(2k)$-numbered one?

I tried using the inclusion-exclusion principle. For one single consecutive couple the probability is $\frac{1}{(2n)!} \cdot (2n-1) \cdot (2n-2)!=\frac{1}{2n}$. For two couples I can use the conditional probability and calculate that:

if the first ball of the first consecutive couple is extracted in one of the n possible odd places, then the chance of extracting a second consecutive couple is $ \frac{ (2n-2) \cdot (2n-4)!}{(2n-2)!}=\frac{1}{2n-3} $;

if the first ball of the first consecutive couple is extracted in one of the $n-1$ possible even places, then the chance of extracting a second consecutive couple is $ \frac{ (2n-3) \cdot (2n-4)!}{(2n-2)!}=\frac{1}{2n-2} $;

therefore the chance of extracting two couples of consecutive (odd then even) balls is $\frac{1}
{2n} \cdot \left(\frac{n}{2n-3}+\frac{n-1}{2n-2} \right)$.

Then I tried generalizing to the extraction of more consecutive couples and using the inclusion-exclusion principle, but I got stuck trying to write down the formula that looks quite long with multiple summations and products. I'm convinced that there's a simpler way to solve the problem but this looked the most promising way to me, considering the complementary seemed worse.

答:
The inclusion-exclusion setup itself isn’t too bad if you think in terms of the $(2n)!$ permutations of the balls (i.e., the possible sequences of draws).

For $k\in[n]$ let $A_k$ be the set of permutations in which $2k-1$ immediately precedes $2k$. Then the inclusion-exclusion principle says that

$$\begin{align*}
\left|\bigcup_{k\in[n]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}A_k\right|\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{n}k(2n-k)!\,.
\end{align*}$$

To see that last step, note that there are $\binom{n}k$ $k$-element subsets $I$ of $[n]$, and if we treat each pair $\{2k-1,2k\}$ for $k\in I$ as a single element, there are $2n-k$ elements to be permuted: the pairs $\{2k-1,2k\}$ for $k\in I$, and the remaining $2n-k$ single elements. The desired probability is then

$$\frac1{(2n)!}\sum_{k=1}^n(-1)^{k+1}\binom{n}k(2n-k)!\,,$$

which can be rewritten

$$\sum_{k=1}^n(-1)^{k+1}\frac{\binom{n}k}{k!\binom{2n}k}\,,$$

for what it’s worth. Unfortunately, I do not at the moment see how to evaluate this in a closed form.


答: Let $t= 2 \pi (n-n')$ and $r=e^{\frac {it } N}$.  Then the given expression is $\frac  1N \sum\limits_{j=1}^{N} r^{j}$. This is $\frac  1N  \frac {r^{N+1}-r} {1-r}$. And $r^{N+1}=r$ if $n \neq n'$ (because $r^{N}=1$).  The case $n=n'$ is trivial.

Title: Kronecker's delta complex sum form

Body:
I hope you're all doing well.

I'm trying to prove the following identity:
$$\frac{1}{N} \sum_{j=1}^{N} e^ {\frac{2i\pi(n-n')j}{N}} = \delta_{nn'}$$

but I'm having some troubles.

This is what I tried:

We know that for a finite geometric series, the entire sum is equal to
$$\sum_{j=1}^{N} ar^{bj} = \frac{ar^b(r^{bN} - 1)}{r^b - 1}$$

In your case we have:
$$r = e$$
$$a = \frac{1}{N}$$
$$b = \frac{2i\pi(n-n')}{N}$$

Using this I found

$$\frac{1}{N} \sum_{j=1}^{N} e^ {\frac{2i\pi(n-n')j}{N}} = \frac{1}{N} \frac{e^{\frac{2i\pi(n-n')}{N}}(e^{{2i\pi(n-n')}} - 1)}{e^{\frac{2i\pi(n-n')}{N}} - 1}$$

but after this I'm not able to see how these exponentials are gonna to result in the Kronecker's delta.


答:
Hint: Try using the Fourier transform. The discrete Fourier transform of $\delta_{nn'}$ is given by:

$$ \mathcal F(\delta_{nn'}) = \sum_{n=1}^N \delta_{nn'} e^{-2 i\pi j n/N} = e^{-2 i\pi j n'/N} $$

We can simply invert $\mathcal F$ to obtain an expression for $\delta_{nn'}$. The inverse Fourier transform of $e^{-2 i\pi j n'/N}$ is given by,

$$ \delta_{nn'} = \mathcal F^{-1}(e^{-2 i\pi j n'/N}) = \frac{1}{N} \sum_{j=1}^N e^{-i 2\pi j n'/N} e^{2 i\pi j n/N} = \frac{1}{N} \sum_{j=1}^N e^{2 i \pi j (n-n')/N} $$

As desired!

Title: Kronecker's delta complex sum form


问:
Consider functions f:{1,2,3,4,5}→{a,b,c,d,e,f}. How many functions have the property that f(1)≠c or f(2)≠f, or both?
So I'm working through this problem, and I think my error is that I'm over counting, but I'm not sure.

Question:

Consider functions f:{1,2,3,4,5}→{a,b,c,d,e,f}. How many functions have the property that f(1)≠a or f(2)≠b, or both?

And I'm trying this.

6^5 - C(5,4)1! + C(5,4)1!

Am I thinking about this the right way? Any help to get me started would be great.

答: HINT: The functions $f$ that you don’t want are those such that $f(1)=a$ and $f(2)=b$; if $B$ is the set of such functions, there are $6^5-|B|$ functions that you do want. If $f\in B$, $f(3)$, $f(4)$, and $f(5)$ can be any members of $\{a,b,c,d,e,f\}$, so what is $|B|$?


问:
Disjoint closed sets in subspace means closure is disjoint from set
Let $Y$ be a subspace of a topological space $X$, and let $A$ and $B$ be two disjoint closed subsets of $Y$ in the subspace topology. Show that $\bar A \cap B=\varnothing$ and $A \cap \bar B=\varnothing$.

Attempt:Let $A$ and $B$ be two disjoint closed sets in $Y$. Then $A=\bar A \cap Y$,where $\bar A$ is the closure of $A$ in $X$. $B=\bar B \cap Y$, where $\bar B$ is the closure of $B$ in $X$.Then if $y \in \bar A,y \notin B$ since $X-\bar A$ is an open set containing $B$ disjoint from $A$. Similarly if $y \in \bar B$, $y \notin A$. Thus $\bar A \cap B=\varnothing$ and $A \cap \bar B=\varnothing$.

Comment: Unfortunately I think it is ok to assume if $A$ is closed in $Y$ $A=\bar A \cap Y$ for $\bar A$ the closure of $A$ in X, similarly for $B$, but I cannot understand why. Is this justified, if so what is the reason this is justified? Also, I am unsure whether it is correct to say that $X-\bar A$ contains $B$ and similarly for $X-\bar B$. Is this ok? If so why is this justified?

答:
It's OK, and you don't actually need that $\overline{A} \cap Y = A$, even though it's true:

$A$ closed in $Y$ just means there is some closed $C$ in $X$ so that $C \cap Y= A$. But then $B \subseteq X-C$ (or else any point in $B \cap C \subseteq Y \cap C=A$ contradicts the disjointness) and so every point of $B$ has an open neighbourhood (namely $X-C$) that is disjoint from $A$, so $x \in B \to x \notin \overline{A}$ showing in effect that $\overline{A} \cap B = \emptyset$. Doing the same argument for the closed $B$ shows $\overline{B} \cap A= \emptyset$ too.


问:
If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$?
Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?

I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.

How can I proper formalize this intuition?

答:
The correct statement that corresponds to the intuition that you had is as follows: if $\omega$ is a nowhere-vanishing top degree form on an orientable closed manifold $M$, then the integral of $\omega$ over $M$ is nonzero.

To put it differently, if $\int_M \omega =0$, then $\omega$ vanishes at some point of $M$.


问:
is $\hat{\theta}$ unbiased

consider a random sample of size n from a distribution with pdf $f(x;\theta)=\frac{1}{\theta}$  $0<x\leq \theta$ and zero otherwise. $0< \theta$

Now the first question was to find the MLE of $\hat{\theta}$ which I found to $X_{n:n}$ , now they want to find out if it is unbiased. My work so far:
$$
\begin{align}
E[\hat{\theta}] &=E[X_{n:n}] \\
&= E[n\frac{1}{\theta}[ln(\theta)]^{n-1}]
\end{align}
$$
now this is probably where i went wrong. isnt the cdf of $X_{n:n}$:
$$
nf(x)[F(x)]^{n-1}
$$
?

答:
The cdf of the maximum is given by $F(x)^n$. Thus, the pdf is $n f(x) F(x)^{n-1}$, In your case, we have for $0 < x \le \theta$:

$$f(x) = \frac{1}{\theta}$$

$$F(x) = \frac{x}{\theta}$$

Thus, the expected value of $\hat{\theta}$ is given by:

$$\int_0^{\theta}\frac{nx^{n-1}}{\theta^n}dx$$

That should hopefully help.

Title: is $\hat{\theta}$ unbiased


答: For each $n \in \mathbb Z^+$, let $$b_n = -1 + \log_2 x_n.$$  Then $b_1 = -\frac{1}{2}$ and $$b_{n+1} = \frac{b_n}{2}.$$  So if $b_n < 0$, it follows that $b_{n+1} < 0$.  And here is the "induction under the table step"--therefore, $b_n < 0$ for all positive integers $n$.  Consequently, $\log_2 x_n < 1$, or $x_n < 2 < 3$ for all $n \in \mathbb Z^+$.  This is about as naked as we can make the issue, since the Peano axioms have induction built in.

Title: To show that a sequence is bounded above by $3$.

Body:
Let $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2 x_n}$.

I would like to know if there is a way to see that this sequence is bounded above by 3 without induction.

I know how to use induction to show that it is in fact bounded by 2. I would like to not use induction though. Thanks for your time.


答:
Your answer isn't correct.  I'm not sure where you're getting the terms you're subtracting from $6^5.$

There are $6^5$ functions $f: \{1, 2, 3, 4, 5 \} \to \{a, b, c, d, e, f \}$.  Of those, you have to subtract the functions where $f(1)=a$ and $f(2)=b$.  Any function that doesn't satisfy both of those constraints is fine.  There are $6^3$ "bad" functions, so your answer should be $6^5-6^3 = 35 \cdot 216 =7560.$

Title: Consider functions f:{1,2,3,4,5}→{a,b,c,d,e,f}. How many functions have the property that f(1)≠c or f(2)≠f, or both?

Body:
So I'm working through this problem, and I think my error is that I'm over counting, but I'm not sure.

Question:

Consider functions f:{1,2,3,4,5}→{a,b,c,d,e,f}. How many functions have the property that f(1)≠a or f(2)≠b, or both?

And I'm trying this.

6^5 - C(5,4)1! + C(5,4)1!

Am I thinking about this the right way? Any help to get me started would be great.


问:
Evaluate $\int \cos(x)\sin^2(x)e^{\sin(x)} \, \text{d} x$.
Evaluate $\int \cos(x)\sin^2(x)e^{\sin(x)} \, \text{d} x$. When I first see it, I think it's use integration by parts. Because there's $e^{\sin(x)}$. But when I use integration by parts, then I use $u=e^{\sin(x)}$, it's a bad idea. Next step I use $u$ sustitution $u= \sin^2(x)$ and since it's derivative and it cancels out. How to do it next?

答: Hint: $u=\sin x$ gives $\int u^{2}e^{u}du$. Integrate by parts twice to finish.


答:
The combined values of the maximum value for the $x$'s is 25. We can then take four part ordered partitions of $5$ (including zeros) and subtract each of those from the maximum.

The first set of partitions is $(5,0,0,0)$ with the $5$ being in four potential locations.
You cannot subtract $5$ from $x_1$ or $x_4$ so there are $2$ ways of doing this.

The second set of partitions is
$(4,1,0,0)$. There are $12$ ways of ordering these numbers.

The third set of partitions is $(3,2,0,0)$.There are $12$ ways of ordering these numbers.

The fourth set is $(3,1,1,0)$.There are $12$ ways of ordering these numbers.

The fifth set is $(2,2,1,0)$.There are $12$ ways of ordering these numbers.

Finally the sixth and last set is $(2,1,1,1)$. There are four locations for the $2$. Therefore there are $4$ ways of ordering these numbers.

$2+12+12+12+12+4=54$

There are $54$ solutions to the equation $x_1+x_2+x_3+x_4=20$.

Title: Determine the number of integer solutions of the equation $ x_{1}+x_{2}+x_{3}+x_{4}=20 $

Body:
Determine the number of integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=20
$$
under the restrictions:
$$
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{array}
$$
I did $x_1=y_1+2$ then $0\leq y_1 \leq4$

$x_2=y_2$ then $0\leq y_2 \leq5$

$x_3=y_3+2$ then $0\leq y_3 \leq6$

$x_4=y_4+2$ then $0\leq y_4 \leq4$

So $y_1+y_2+y_3+y_4 \leq 19$
So the number of solutions is

$\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
k
\end{array}\right)=\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
3
\end{array}\right)$
is right?


答: Substitute $u = \sin x$, $du = \cos x$. Then the integral becomes $\int u^2 e^u du$. This can be computed by integrating by parts (twice).

Body: Evaluate $\int \cos(x)\sin^2(x)e^{\sin(x)} \, \text{d} x$. When I first see it, I think it's use integration by parts. Because there's $e^{\sin(x)}$. But when I use integration by parts, then I use $u=e^{\sin(x)}$, it's a bad idea. Next step I use $u$ sustitution $u= \sin^2(x)$ and since it's derivative and it cancels out. How to do it next?


问:
Exactness for special quadrature formula
Given the quadrature formula $Q(f):= w_0f(-a) + w_1f(a)$ and $I(f):= \int_{-1}^1 f(x) dx$ and $a\in [0,1]$. I need to find $a$ so that $Q(f)$ is

a) exact for polynomials of degree $1$.

b) exact for polynomials of degree $2$.

Exactness for degree $n$ means that $R(p):= Q(p)-I(p)$ for all polynomials $p$ of degree $\leq n$. I try to understand the idea behind this.

Is the right way to take a polynomial $p(t)=m_0+m_1t$ of degree $1$ and then just put it into the formulas above? That would mean finding $a$ would be done by solving:
$w_0p(-a) + w_1p(a) = \int_{-1}^1 p(x) dx$? Another idea was that I know (or have shown in the lecture) that $Q(f)$ is exact for polynomials of max. degree $n$ if and only if $Q(f)=\sum_{j=0}^n w_j f(t_j)$ with $w_j$ being Lagrange polynomials. So if that is the case, isn't $Q(f):= w_0f(-a) + w_1f(a)$ exact for polynomials of degree max. $1$ if $w_0,w_1$ are the Lagrange polynomials and can't $a$ in that case be $a$ a value from the whole interval $[0,1]$?

答:
The formula is linear, so you just need to check that $Q(x^i) = I(x^i), i=0,1, \cdots$.So,

a)
$$
\begin{cases} Q(1) = I(1)\\ Q(x) = I(x)\end{cases} \Leftrightarrow
\begin{cases} w_0 + w_1 = 2\\ w_0 (-a) + w_1 a = 0\end{cases}
$$

The solutions are: i. $a=0$, with any $w_0,w_1$ such that $w_0+w_1=2$; ii. $w_0=w_1=1$, with any $a$.

b)
$$
\begin{cases} Q(1) = I(1)\\ Q(x) = I(x)\\Q(x^2=I(x^2)\end{cases} \Leftrightarrow
\begin{cases} w_0 + w_1 = 2\\ w_0 (-a) + w_1 a = 0 \\ w_0 a^2 + w_1 a^2 = \frac 23\end{cases}
$$

From last equation you get $a= \frac{\pm 1}{\sqrt{3}}$ and from the first two equations you then deduce that $w_0=w_1=1$.


答:
Hint:

Use the substitution $\:u=\sin x$, $\:\mathrm d u=\cos x\,\mathrm dx$. You obtain the standard integral $\;\int u^2\mathrm e^u\,\mathrm du$.

Title: Evaluate $\int \cos(x)\sin^2(x)e^{\sin(x)} \, \text{d} x$.


问:
Minimax criterion in decision theory
I have the following risk functions and I want to apply minimax criterion to both of them.

$inf_c sup_x$ ($c^2+(c-1)^2x^2$)

$inf_c sup_x$ ($xc^2+(c-1)^2x^2$)

For the first case I think that the minimax rule is for $c=1$, and for the second case I think that the minimax rule does not exist.

However I am not sure about my calculations, and so I need some help. Also, I need some guidelines how to work on such problems.

答:
If by minimax rule you mean an optimal solution for the outer inf, then yes, for 1, when c=1, the objective function $f(c):=\sup_x c^2 + (c-1)^2 x^2$ takes the value $f(1)=1$, and $f(c)=+\infty$ for $c\ne 1$, so $\inf_c f(c)=1$ at $c=1$.

For 2, $g(c):=\sup_x c^2 x + (c-1)^2 x^2$ always has $g(c)=+\infty$ for all $c\in\mathbb{R}$. So $\inf_c g(c)=+\infty$. $g$ in this case is not a proper convex function.

Title: Minimax criterion in decision theory


问:
Let $G$ be solvable. Let $g\in G$. What can we say about $g=[g^{x_1}, g^{y_1}]\dots [g^{x_n}, g^{y_n}]$ with $x_i,y_i\in G,i=1,...,n$?
We have $G$ a solvable group. Let g be an element of G. What can we say about

$$g = [g^{x_{1}}, g^{y_{1}}]\cdot\cdot\cdot[g^{x_{n}}, g^{y_{n}}]$$

with $x_{i}, y_{i} \in G, i = 1,..., n$?

答:
If $G$ is finite, then by an absurd-sounding result of John Wilson, $n\leq 55$. This appears in J. London Math. Soc., entitled Finite Axiomatization of Finite Soluble Groups. The theorem is as follows:

A finite group $G$ is soluble if and only if no non-trivial element $g$ can be expressed as a product of 56 commutators of conjugates of $g$.

It is likely that the number 56 can be reduced.


问:
Transformation between a hypercube and a polygon?
Is there a straightforward way to generalize the answer here to how to transform the points within a square to a polygon to n-dimensional cubes and polygons? Specifically, given a convex region $\mathcal{C}\in\mathbb{R}^n$, bounded by hyperplanes on all sides, is there a one-to-one mapping between this region and the hypercube $[0,1]^n$?

A region I'm thinking of would be something like
$$
\mathcal{C}:\{(x,y,z)\in\mathbb{R}^3\big|\; 0\leq x\leq2,
-1\leq y \leq 3, -3\leq z \leq -1, \\
-1\leq 2x-y+z, 0\leq x+y\leq 3 \}.
$$

答: Any convex body, $C$ in $\Bbb{R}^n$ is homeomorphic to $[0, 1]^n$. To see this, pick a point $p$ in the interior of $C$ and choose $\alpha > 0$ small enough such that $D = p + \alpha[0, 1]^n$ is contained in the interior of $C$. Then design a radial projection $f : C \to D$ that maps a point $c$ on the boundary of $C$ to the point $c_0$ where the line segment $[p, c]$ meets the boundary of $D$ and maps $p + \lambda (c - p)$ to $p + \lambda (c_0 - p)$ for $0 \le \lambda < 1$. Composing $f$ with the obvious affine homeomorphism from $D$ to $[0, 1]^n$ gives the homeomorphism you are asking for.


问:
Countable algebraically closed field inside an incountable algebraically closed of characteristic zero
I would like  to know if the following claim is true. I found this claim in a paper without proof :(.

Let $k$ an uncountable algebraically closed field of characteristic $0$.

Let $S=\operatorname{Spec}(k[X_1,..., X_m]/I(S))=\operatorname{Spec}(k[X_1,..., X_m]/\langle f_1,...,f_n\rangle)$ be an integral affine scheme of finite type over $k$.

Claim: We can choose a countable algebraically closed subfield $k_0\subset k$ such that there is an irreducible quasi projective scheme $S_0$ over $k_0$ with $S=S_0\times_{\operatorname{Spec}(k_0)}\operatorname{Spec}(k)$.

答: Let $T$ be the finite set of coefficients appearing in the $f_i$. Then $k_0=\overline{\Bbb Q(T)}$ and $S_0=\operatorname{Spec} k_0[X_1,\cdots,X_m]/(f_1,\cdots,f_m)$ suffice. The verification that this has all the properties you request is straightforward; if you are stuck on any part, please leave a comment.


问:
Prove $\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx = \int_{0}^{\frac{\pi}{2}}f(\sin(x)dx$
Assume $f$ is continuous, prove $\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx = \int_{0}^{\frac{\pi}{2}}f(\sin(x))dx$. Of course this immediately brings to mind of the relation between $\cos(x)$ and $\sin(x)$. I know that $\cos(x)=-\sin(x+\pi)$. So should I rewrite the sine function like this? I've tried this and it does not seem to help. Any answers?

答: Hint: $\sin(x) = \cos\left(\dfrac\pi2-x\right)$.


问:
Prove There is a set for which {1}∈X and {1}⊆X?
I am new to proof I found this question in worksheet I have no idea how to do it? Can anyone willing explain it to it would much appreciated. this question Prove There is a set for which {1}∈X and {1}⊆X?

This is what I have tired:
suppose {1}EX by definition it means 1 is an element of x hence {1} is subset of X therefore  {1}∈X and {1}⊆X is true

答:
I think you are misunderstanding what is being asked.  It isn't asking that one statement implies the other.  That would be false.  You can certainly have $1$ in a set without having $\{1\}$ in the set (and vice versa).

(You statement $\{1\} \in X$ implies $1$ is an element of $X$ is just plain wrong.  The set $\{1\}$ is not the same thing as the number $1$ and the set $\{1\}$ being an element of $X$ will not mean the number $1$ will be an element.)

$\{1\} \in X$ means that $\{1\}$ (the set, not $1$ itself) is in the set $X$.  SO if we were to list out all the elements of $X$ as $X$ contains: $......., \{1\}........$ then $\{1\}$ will be one of them.

And $\{1\}\subset X$ means $\{1\}$ is as subset of $X$.  That means if we list the elements of $\{1\}$, that is $\{1\}$ contains the following elements:  $1$; that is all.... then all the  element of $\{1\}$ will also be an element of $X$.  So if we list out all the elements of $X$ as: $......, 1,.....$ then $1$ will be in there because $1$ is in $\{1\}$.

All you are being asked is that there is a set with both the elements $\color{blue}{\{1\}}$ and $\color{red}1$ in it.

So if we listed out all the elements of $X$ it would have: $......., \color{blue}{\{1\}},.........,\color{red}1,.....$ in it.

And that should be trivial: just put them in a set.  $X= \{  \color{blue}{\{1\}} ,\color{red}1\}$

That's it.

Of course you need axioms that sets of elements exist for the elements you want, I'm not up an my set theory axioms but I believe there is one that states just that.... goes off to google.... Ah, yes... that is the Axiom of Specification.

If we want to get formal:

$\emptyset$ exists.

So by Axiom of Specification. $1 := \{\emptyset\}$ exists.

And as $1$ exists, by Axiom of Specification $\{1\}$ exists.

So we have $1$ and $\{1\}$ exists.  So for the third time by Axiom of Specification $X:=\{1, \{1\}\}$ exists.

And $1 \in X$. So $\{1\} \subseteq X$.  And $\{1\} \in X$.

So the statements hold for $X$.  That proves it.


问:
CFG for a Language
I am given the language:

$L = $ {${a^kb^k | k > 0 }$}*

And i am asked to design a Context Free Grammar for this language. Can anyone help me in this one? Like if there is a step by step guide that i should follow? I'm kind of confused with the kleene star in the end as i don't know how is the information supposed to be saved in order to be repeated. I am new and i don't know many things about CFG's. Please if you think that my question is too general please don't down-vote, please tell me where to go, like find a tutorial somewhere because i have seen a few in YouTube and i didn't find anything that would help me in this kind of CFG.  Thanks.

答: You don’t have to save anything. If $L_0=\{a^kb^k:k>0\}$, your $L$ is simply $L_0^*$, the language of all finite strings of words in $L_0$. Thus, besides the empty word you get all words in $L_0$, all words in $L_0^2$, and so on. For instance, you get $aabb$, $abaaabbb$, and $abaabbabaaabbb$. If you can design a CFG for $L_0$, it takes only a very minor addition to convert it to a CFG for $L$, so I would begin by working on a CFG for $L_0$; there is one with only two productions.

Title: CFG for a Language


问:
How to show surjectivity in this task?
Let $R$ be a commutative ring with unity and $e$ an idempotent, $e\neq 1, e\neq0$. I need to show that $Re$ and $R(1-e)$ are subrings with unity and $R \cong Re \times R(1-e)$.

Notice that $(1-e)$ is also idempotent.

So, first show that $Re=\{re:r\in R\}$ is subring with unity:

$0_R\in Re$ because $0_Re=0_R$

$r_1e-r_2e=(r_1-r_2)e \in Re$ (those are elements of $R$, so distributivity holds)

$(r_1e)(r_2e)=r_1e^2r_2=r_1er_2=r_1r_2e\in R$

$1\in R \Rightarrow 1e=e\in Re$.    $ree=re^2=re, ere=ree=re^2=re$, so $e$ is unity.

Similar for $R(1-e)$.

Now show $R \cong Re \times R(1-e)$. Define $\phi:R\rightarrow Re \times R(1-e)$.

injectivity: Let $(r_1e,r_1(1-e))=(r_2e,r_2(1-e))\Rightarrow...\Rightarrow r_1=r_2$

$\phi(r_1+r_2)=((r_1+r_2)e,(r_1+r_2)(1-e))=...=\phi(r_1) + \phi(r_2)$

$\phi(r_1r_2)=(r_1r_2e,r_1r_2(1-e))=(r_1r_2e^2,r_1r_2(1-e)^2)=...=\phi(r_1) \phi(r_2)$

And I'm stuck with surjectivity. Any help?

答: You haven't stated your proposed definition of $\phi$. But I think it's implicit that $\phi(x) = (xe, x(1-e))$.  In that case, if $(x, y) \in Re \times R(1-e)$, say $(x, y) = (x'e, y'(1-e))$, you will have $\phi(x' + y') = (x, y)$, so $\phi$ is a surjection of $R$ onto $Re \times R(1-e)$.

Title: How to show surjectivity in this task?


问:
$m(X) > 0$ Show that $Y \subset X$ where $Y$ is measurable can be such that $m(Y) = \delta$ for any $0 < \delta < m(X)$
Let $X$ be a Lebesgue Measurable subset of $\mathbb{R}$ where $m(X) > 0$.

Prove that $\forall 0 < \delta < m(X)$ that $\exists Y \subset X$, measurable such that $m(Y) = \delta$

Hint: consider $f(x) = m(X \cap [-x,x]) \ \forall x > 0$

I'm having trouble figuring out how to use the hint. I am not sure what I should do with it. Maybe attempt to show it's continuous and maybe do something from there? Exactly what to do though, I'm not sure. Any advice would be appreciated!

答: Note that $f(0)$ is also well-defined and $f(0)=0$. Also, $f$ takes value in $[0,\infty)$. By continuity of measure, we can prove that $f$ is continuous and $f(x)\rightarrow m(X)$ as $x\rightarrow \infty$. Since $\lim_{x\rightarrow \infty}f(x) = m(X)>\delta$, there exists $x_2$ such that $f(x_2)>\delta$. Now, $0= f(0) < \delta < f(x_2)$. Apply Intermediate Value Theorem, the result follows.


问:
$0\leq \int_{\mathbb{R}}fdm < \infty \implies m(E) = 0$?
I am currently reading my textbook on measure theory and started reading the author's proof for the following lemma regarding the Lebesgue integral.

Let $f$ be a Lebesgue measurable function on $\mathbb{R}$. Then $f$ is finite almost everywhere.

I understand all of the first bit, which he writes as (in my own words):

It suffices to consider the case $f \geq 0$. Let $n \in \mathbb{N}$, and define $E_n:=\{x \in \mathbb{R}: f(x) \geq n\}$. Therefore take $E:= \bigcap_{n=1}^{\infty}E_n$ such that $$\int_{\mathbb{R}}fdm \geq \int_{E_n}fdm \geq n\cdot m(E_n) \geq n \cdot m(E),$$ and It follows that $m(E) \leq \frac{1}{n}\int_{\mathbb{R}}fdm \ \forall n \in \mathbb{N}$ . However, following this sentence is where I start to get lost. He states that "since $0\leq \int_{\mathbb{R}}fdm < \infty$ we can conclude that $m(E) = 0$". What is the reasoning behind this? For some reason, I can't seem to see why $0\leq \int_{\mathbb{R}}fdm < \infty$ implies that $E$ is null. I am pretty new to Lebesgue integration so if this is due to some basic fact, forgive me.

Any feedback is welcome.

答:
In line 3 you should say 'Let $f$ be a Lebegue integrable function  on $\mathbb R$'.

The inequality holds for all $n$. If $m(E)>0$ you get a contradiction by chosing an integer $n >\frac {\int f dm} {m(E)}$.


问:
Closure of a subset
Show that the closure $\overline E$ of a subset $E \subset \mathbb R^n$ is the smallest closed set containing $E$.

I am new to topology so I was wondering how to establish a solid proof of this statement that seems basic.

Here is my attempt:
First we can show that $\overline E$ is a closed set containing $E$ :

$\overline E = E \cup \partial E $ so it surely contains $E$.

$\overline E^C = E^C \cap (\partial E)^C$. I need to show that $\forall \vec{x} \in \overline E^C, \exists \delta_x > 0 \text{ such that } B(\vec{x},\delta_x) \subset \overline E^C$ but I struggle to show it properly.

Then I need to show that $\overline E$ is the smallest set with the previous conditions. I wanted to do it by contradiction by supposing that $\exists F \subset \overline E$ (so smaller than $\overline E$) and then find a contradiction but it seems too strange to prove.

答:
For your attempt toward contradiction, you assume that $F$ contains $E$, is closed, and is strictly contained in $\overline{E}$. If such a set could exist, then $\overline{E}$ would not be the smallest closed set containing $E$. There are several equivalent definitions of the closure, this being one of them, but I will proceed with the fact that $\overline{E}$ is the set $E$ together with all its limit points.

Assuming $E \subset F \subset \overline{E}$ with $F$ closed, then every sequence in $E$ is also a sequence in $F$. Using the fact that closed sets contain their limit points and the aforementioned definition of the closure, can you arrive at a contradiction from here?

Title: Closure of a subset


问:
How to distribute 10 distinct envelopes into 3 (identical) mailboxes?
As in the title, I would like to distribute 10 distinct envelopes into 3 mailboxes. Let us also assume that some mailboxes can remain empty.

If they were identical envelopes, I know that we can use the stars and bars method, which would let us arrive at the answer of $\binom {12} {2}$.

However, here is where a friend and I started an extended discussion as we tweaked the question slightly.

What if, the envelopes were distinct and not identical?

My logic

Similar to the stars and bars method, 3 mailboxes is equivalent to 2 bars, so the question is equivalent to arranging 12 objects (10 distinct, 2 identical) in a line. Thus, my answer is $\frac {12!} {2!}$.

My friend's logic

My friend says that each envelope has 3 mailboxes to go into and since we have 10 envelopes, so the answer should be $3^{10}$.

Clearly, both of our answers are very different and I would obviously like to know who is correct and more importantly, why. We are currently at a standstill because neither of us can find the flaw in the other's logic. Any intuitive explanations will be greatly appreciated!

答:
Stars and bars works fine when you have to distribute 10 identical envelopes into 3 distinct mailboxes, but not when the envelopes are distinct. Your argument gives the right answer ($12!/2!$) to the following question:

How many ways are there to arrange 10 distinct envelopes and two identical bars in a line?

If you regard the two bars as the dividers between three mailboxes, then under your approach the envelopes to the left of the first bar are sent to the first mailbox, the envelopes between the first and second bar are sent to the second mailbox, and the envelopes to the right of the second bar are sent to the third mailbox. The flaw in your approach is that you are overcounting what are equivalent assignments: once you've placed the bars, rearranging the items between the bars will lead to the same distribution of envelopes into the three mailboxes.

For grins, let's modify your reasoning to arrive at the correct answer ($3^{10}$):

For each of the possible ways to place the two identical bars, you are currently counting $10!$ permutations. This leads to a total of
${12\choose 2}10!$, which equals $12!/2!$. To repair this, replace $10!$ with a number that recognizes the overcounting going on. Specifically, suppose you place the bars at positions $a<b$. Then there are $a-1$, $b-1-a$, and $12-b$ envelopes going into the respective mailboxes. To fix the overcount, divide $10!$ by the number of ways to permute the envelopes between the bars. This leads to the total
$$
\sum_{1\le a<b\le 12}\frac{10!}{(a-1)!(b-1-a)!(12-b)!},
$$
which equals $3^{10}$, by tedious calculation, or by the following

Claim:
$$\sum_{1\le a<b\le N+2}\frac{N!}{(a-1)!(b-1-a)!(N+2-b)!}=3^N.$$
Proof:
Write the sum in the form
$$\sum_{a=1}^{N+1}\frac {N!}{(a-1)!}\sum_{b=a+1}^{N+2}\frac1{(b-1-a)!(N+2-b)!}
\stackrel{(*)}=\sum_{a=1}^{N+1}\frac {N!}{(a-1)!}\sum_{c=0}^{N+1-a}\frac1{c!(N+1-a-c)!}
$$
where in ($*$) we apply the change of index $c:=b-a-1$. The inner sum on the RHS equals $2^{N+1-a}/(N+1-a)!$, by the identity
$${k\choose 0}+{k\choose 1}+\cdots+{k\choose k}=2^k,$$ which can be proved using the binomial theorem. To finish the proof, use the binomial theorem again to find the identity
$$
3^N=(1+2)^N=\sum_{i=0}^N{N\choose i}1^i2^{N-i}=\sum_{a=1}^{N+1}{N\choose a-1}2^{N+1-a}.$$


问:
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

答:
Here are some remarks complementing lab bhattacharjee's answer.

The general problem is as follows : find the “first component” of
a third root of $z=a+b\sqrt{D}$, where $a,b,D$ are integers and $D$
is a non-square (when $D<0$, this “first component” will simply be the real
part).

Now, the generic heavy algebraic machinery tells us that this “first component”
(call it $x$) is annulated by the polynomial

$$
P=64x^9 - 48ax^6 + (-15a^2 + 27b^2D)x^3 - a^3
$$

and that $P$ is irreducible in general, for “almost all” values of $(a,b,D)$.

The OP’s example is special in this regard because the norm $a^2-Db^2$ is a perfect
cube. But this condition alone cannot guarantee that $P$ can be factorized ; it will
again be irreducible for  “almost all” values of $(a,b,D)$ satisfying this additional condition.

So the question now becomes : what is it about the values in the OP that allows
us to find a rational value in the end ? (this corresponds to the “lucky” factorization by $x-7$ in lab bhattacharjee’s second solution, which seems to come out of nowhere). Is there just a finite set of exceptional
points $(a,b,D)$ having this property, or do they form a curve ?


问:
Proof of Lagrange theorem - Order of a subgroup divides order of the group
The Lagrange theorem states:

If $G$ is a finite group, and $H$ a subgroup of $G$, then the order of $H$ will divide the order of $G$. More precisely, $|G| = |H| \cdot (\text{number of left cosets of }H)$:

$$|G| = |H| \cdot (G:H)$$

The proof I have in my notes says:

$G$ consists of $\{G:H\}$ cosets, each of them consists of $|H|$ elements, the cosets are disjoint.

That's it. How does this prove the theorem?

答:
[ Here is one way we could proceed ]

Let $ G $ be a finite group, and $ H \subseteq G $ a subgroup.

Consider the map $ f : G \longrightarrow f(G) $ sending $ g \mapsto gH $, where $ gH := \{ gh : h \in H \} $. In trying to understand fibres of $ f $, we find $ f^{-1} ( \{ gH \} ) = gH $.

[ For a surjective map $ f : X \longrightarrow Y $, the sets $ f^{-1} ( \{y\} ) := \{ x \in X : f(x) = y \} $ where $ y \in Y $ are called "fibres" of $ f $. These fibres partition the set $ X $ ]

Also each fibre has the same size as $ H $ (because $ H \longrightarrow gH $ sending $ h \mapsto gh $ is a bijection), and hence $ |H| $ divides $ |G| $.


问:
rank of matrix created from 2 vectors
Take two vectors $a,b\in \mathbb{R}^{12}$ and create a 12x12 matrix $C$ so that $c_{ij}=(a_i-b_j)^2$. What are the possible values of rank(C)?

Not sure how to proceed. One idea I had was to express C as a product of two other matrices and then I might put together an estimate of some sort using $rank(AB)\leq min \{ rank(A),rank(B) \}$. Though having tested it out on a couple of concrete 3x3 matrices I found they always had full rank. So maybe that's the answer?

Either way, appreciate any help.

答:
could be any value from number 0 to 12

because any 12×12 matrix C ，cᵢⱼ≥0
is equivalent to diagonal matrix P whose Rank(P)
could be 0 to 12.

Title: rank of matrix created from 2 vectors


问:
The sum of a product of two random variables: how do I know if each summand is independent?
So I have a question that involves using SLLN and almost sure convergence:

I'm given $Y_{1},Y_{2},Y_{3}\ldots$ as iid r.v.'s with a finite mean $k$.

I'm also given this sum
$$
\frac{1}{n}\sum_{i=1}^{n}Y_{i}Y_{i+1}\
\overset{\rm a.s.}{\rightarrow}\ k^{2}
$$
and asked if the SLLN applies here.

My question is:

Is there a  surefire way to tell if each summand $\left(~Y_{1}Y_{2},Y_{2}Y_{3}, \mbox{etc.}~\right)$ is iid with finite mean $?$.

What would be the case for a different sum, say,  $$
\frac{1}{n}\sum_{i = 1}^{n}Y_{2i - 01}\ Y_{2i}\ ?
$$

答:
In your case, the summands are not i.i.d. in general (rather, they are identically distributed, but not independent). An easy way to see it is to consider the following: letting $Z_i := Y_i Y_{i+1}$ for every $i$, you have
$$
\mathbb{E}[Z_i Z_{i+1}]
= \mathbb{E}[Y_i Y_{i+1}\cdot Y_{i+1}Y_{i+2}]
= \mathbb{E}[Y_i Y_{i+1}^2Y_{i+2}]
= \mathbb{E}[Y_i] \mathbb{E}[Y_{i+1}^2]\mathbb{E}[Y_{i+2}]
= \mathbb{E}[Y_1]^2\mathbb{E}[Y_1^2]
$$
(using independence for the second-to-last equality, and the i.d. assumption for the last).
On the other hand,
$$
\mathbb{E}[Z_i]\mathbb{E}[Z_{i+1}]
= \mathbb{E}[Y_i Y_{i+1}]\mathbb{E}[Y_{i+1}Y_{i+2}]
= \mathbb{E}[Y_1]^4
$$
so if the expectation is non-zero$^{\dagger}$ the two differ unless $\mathbb{E}[Y_1^2]=\mathbb{E}[Y_1]^2$, i.e., unless $\mathrm{Var}[Y_1]=0$ (that is, unless $Y_1$ is constant a.s.).

For the second case where the summands are $Y_{2i} Y_{2i+1}$, however, you would have independence as the indices of different summands do not overlap.

$(\dagger$) Overall, one can prove the following result (based on this, and the comment below for the case $\mathbb{E}[Y_1]=0$, in which case you apply the same argument to $\mathbb{E}[Z_i^2Z_{i+1}^2]$ and conclude that $|Y_1|$ is constant a.s.):

Theorem. Let $(Y_n)_{1\leq n\leq \infty}$ be a sequence of real-valued i.i.d. random variables, and set $Z_n := Y_{n}Y_{n+1}$ for all $n\geq 1$. Then, the $Z_n$'s are not pairwise independent unless either

$Y_1$ is constant a.s.; or

$Y_1$ is uniformly distributed on $\{-a,a\}$ for some $a>0$.

Conversely, in both cases (1) and (2) the $Z_n$'s are mutually independent.


问:
How can I write this as a recurrence relation?
Can someone please help me...how can I write this in the form of a recurrence relation?

$$ \sqrt{5 + \sqrt{6 + \sqrt{7 + ... }}}$$

Thank you

答:
Let $a_n = \sqrt{ n+4 + \sqrt{n+5 + \sqrt{ n+ 6 + \sqrt{ \ldots } } } } $

Hint: Consider $a_n^2 - a_{n+1}$.