问:
Definition of a stalk why must we be careful about the equivalence realtion
We define a stalk as follows. Let $\mathcal{F}$ be a sheaf of abelian groups on $X$ and let $p$ be a point in $X$.

The stalk $\mathcal{F}_p:\{(U,s)\mid p\in U \subset X \text{ an open set and } s\in \mathcal{F} \}/\sim$. Where $(U,s)\sim (U',s')$ if $s\mid_V=s'\mid_V$ for some $V\subset U\cap U'$ open.

Seemingly we cannot always replace $V$ with $U\cap U'$ as then this would not be an equivalence relation, but I cannot come up with an example on how this fails to be an equivalence relation.

答: Consider what happens if we have a $U''$ properly contained in $U\cap U'$. We could have pairs $(U,s)$, $(U', s')$, and $(U'', s'')$ such that $s$ agrees with $s''$ on $U\cap U''=U''$ and $s'$ agrees with $s''$ on $U'\cap U''=U''$, but $s$ and $s'$ don't agree on the larger set $U\cap U'$. So transitivity could fail in general if we don't "cut down" appropriately.


问:
Find a basis of a vector space where F(R) is the set of all functions from R to R
F(R) is the set of all functions from R to R.
(F(R),R,+,•) is a vector space.
What is a basis of given vector space?

What if we restrict the set of functions to functions with finite domain?

答:
The theorem that states that every vector space has a basis uses the axiom of choice (particularly for infinite dimensional vector spaces such as this one). The fact that $F(R)$ has a basis follows from the axiom of choice, but that doesn't really allow us to construct a basis for it (i.e, the proof of the existence of the basis is "non constructive").

So if you use the axiom of choice, set theory will bestow upon you a nice basis, but if asked to explicitly construct or describe a basis, you will be unable to do so. That is, you need axiom of choice in a strong sense, and without choice, there are models of set theory where there is no such basis.

An intuitive way of seeing why you will never be able to describe such a basis is: given some set of functions, we may always use those functions to obtain something "stranger" (for example, if given a bunch of continuous functions, we may bring up a non continuous). In that sense any infinite family of functions $B$ that we supply will be insufficient since we will be able to construct some function $f$ sufficiently pathalogical so that $f \not \in span(B)$.

Additional links: basis for space of continuous functions, basis for $\mathbb{R}^{\mathbb{R}}$, which is the space of functions $\mathbb{R} \rightarrow \mathbb{R}$

What if we restrict the set of functions to functions with finite domain?

If $A \subset \mathbb{R}$ is finite, and $F(A, \mathbb{R}) = \{f : A \rightarrow \mathbb{R}\}$, then this is in fact a finite dimensional vector space. Suppose $A = \{a_1, ... a_n\}$ for some $n \in \mathbb{N}$. Then the basis is of the functions $f_i : A \rightarrow \mathbb{R}$ where $f_i(a_i) = 1$ and $f_i(a_j) = 0$ if $j \not = i$.

ContentLicense: CC BY-SA 4.0


答:
Let $X=\{a,b,p\}$ with the discrete topology, and let $\mathcal{F}$ be the constant sheaf with value $\Bbb Z$. Consider the following global sections, both of which assign $0$ to $p$:

$\alpha$, which assigns $1$ to $a$ and $0$ to $b$,

$\beta$, which assigns $0$ to $a$ and $1$ to $b$.

Using the relation of "agreeing where defined", these are both equivalent to the local section defined on $\{p\}$ which assigns $0$ to $p$, but they're not equivalent to each other. So the relation is not transitive.

Title: Definition of a stalk why must we be careful about the equivalence realtion

Body:
We define a stalk as follows. Let $\mathcal{F}$ be a sheaf of abelian groups on $X$ and let $p$ be a point in $X$.

The stalk $\mathcal{F}_p:\{(U,s)\mid p\in U \subset X \text{ an open set and } s\in \mathcal{F} \}/\sim$. Where $(U,s)\sim (U',s')$ if $s\mid_V=s'\mid_V$ for some $V\subset U\cap U'$ open.

Seemingly we cannot always replace $V$ with $U\cap U'$ as then this would not be an equivalence relation, but I cannot come up with an example on how this fails to be an equivalence relation.


问:
How is material implication the same as subset?
I've been told and read a number of times that implies, $\implies$, is the same as subset $\subseteq$. For example, in the wikipedia page on iff it states that "P only if Q", "if P then Q", and "P→Q" all mean that P is a subset, either proper or improper, of Q

I'm currently thinking in terms of propositional logic, and I don't understand how it makes sense to say that $P \implies Q$ means that $P$ is  a subset of $Q$. To my current understanding, $P$ and $Q$ are both propositions, which means that they are in the set $\{T,F\}$. This set contains only two elements and neither of them are sets, so how can $P$ be a subset of $Q$?

I know that the two element Boolean algebra can be realised using  $T = \{\emptyset\}$ and $F = \emptyset$, along with intersection union and complement. Then this would mean that I can represent $\implies$ using $\subseteq$, because $\emptyset \subseteq \{\emptyset\}$ etc. to give the correct truth table, but I don't see how that could relate $P \implies Q$ to $P \subseteq Q$.

I think when I'm told that $P \implies Q$ means that $P \subseteq Q$, this must be somehow taking the space of all propositions (perhaps for some specific set of axioms?) and then $P \subseteq Q$ in that space would mean that $P \implies Q$, is this correct? Can anyone elaborate on how this works and direct me to any related reading?

答:
As should be clear by now, the $\to$ is of course not the same as the subset relationship, as you yourself and others have noted.  The $\to$ is a truth-functional operator, while the $\subseteq$ is a set-theoretic predicate.

But, there is a close relationship between them. Indeed, there is a kind of isomorphism between what is going on in propositional logic and set-theory.

Consider the following axiom defining the set-theoretic operation of intersection:

$\forall x \forall y \forall z (z \in x \cap y \leftrightarrow (z \in x \land z \in y))$

So here we see a close connection between the set-theoretic $\cap$ and the logical $\land$.  Of course, the fact that these two symbols look alike is not at all a coincidence! Indeed, we also have a close conections between the logical $\lor$ and the set-theoretic $\cup$

$\forall x \forall y \forall z (z \in x \cup y \leftrightarrow (z \in x \lor z \in y))$

And between the set-theoretic complement and logical negation:

$\forall x \forall z (z \in x^C  \leftrightarrow \neg z \in x )$

And here is a connection between $\to$ and $\subseteq$:

$\forall x \forall y (x \subseteq y \leftrightarrow \forall z (z \in x \to z \in y))$


问:
$u^3 + u^2-id=0\text{ and } tr(u) \in \mathbb{Q}$ show that the dimension of the vector space is a multiple of 3
$\text{let E a vector space} \:dimE=n \\ \text{let}\:\: u \in L(E,E) \text{such as} \: u^3 + u^2-id=0\text{ and } tr(u) \in \mathbb{Q}\\ \text{show that n is a multiple of 3}$

答:
$X^3+X^2-1$ has 3 distinct roots $a_1,a_2,a_3$ in $\overline{\Bbb{Q}}$.

$E$ is a finite dimensional $K$-vector space where $K$ contains $\Bbb{Q}$, let $E' = E\otimes_K \overline{K}$ and $u':E'\to E'$. Then

$$E' = \ker(u'-a_1)\oplus \ker(u'-a_2)\oplus \ker(u'-a_3)$$

$$tr(u') = \sum_{j=1}^3 a_j \dim(\ker(u'-a_1))$$
(dimension as $\overline{K}$ vector space)

That $tr(u')\in \Bbb{Q}$ implies that $$\dim(\ker(u'-a_1))=\dim(\ker(u'-a_2))=\dim(\ker(u'-a_3))$$
ie. $$n=\dim(E')=3 \dim(\ker(u'-a_1))$$


问:
Prove that $(\log(n))^k=O(n)$ for any $k$
I'm having problem proving that $(\log(n))^k=O(n)$. I tried googling it but can't find an answer.

答:
$\log n = O(n^\epsilon)$ for any $\epsilon > 0$, since
$$\log n = \tfrac1\epsilon \log(n^\epsilon) = \tfrac1\epsilon\textstyle\int_1^{n^\epsilon}\frac{dt}{t} <\tfrac1\epsilon \int_1^{n^\epsilon}dt= \tfrac1\epsilon n^\epsilon-\tfrac1\epsilon < \tfrac1\epsilon n^\epsilon,$$

Thus $\log^kn= O(n^{k\epsilon})$ for any $\epsilon>0$. Let $\epsilon=1/k$ and you're done.


问:
Geometric probabilistic problem on a plane
Let $T$ be a triangle with vertices belonging to a given Cartesian plane $P$ and with side lengths $a$, $b$ and $c$, where $a\ge b\ge c\ge 0$. Let $L_T$ be a straight line selected uniformly at random from the set of all straight lines on $P$ intersecting $T$, which we define as follows:

$L_T$ is selecting uniformly at random from the set of all straight lines intersecting $T$ and making an angle $\theta$ with a given axis of $P$, where $\theta$ is selected uniformly at random from $[0,\pi)$.

Question: What are the probabilities $\Pr(a)$, $\Pr(b)$ and $\Pr(c)$ that each side of $T$, with length $a$, $b$ and $c$ respectively, is not intersected by $L_T$?

答:
The answer I propose is:

$$Pr(a)=\dfrac{b+c-a}{a+b+c}, \ \ \ \ Pr(b)=\dfrac{c+a-b}{a+b+c}, \ \ \ \  Pr(c)=\dfrac{a+b-c}{a+b+c} \tag{0}$$

Why that ? This question deserves to be formulated at first in terms of "measure theory",  probabilities being considered in a second step, in the framework of the domain called "Integral Geometry" or "Geometric Probability" (see Historical note below).

As all plane lines can be given a unique standardized form:

$x \cos \theta+y \sin \theta=p, \tag{1}$

Therefore, $(p,\theta)\in \mathbb{R^+} \times [0,2\pi)$ is a parameterization of the set of plane lines.

With this parametrization, a natural measure (invariant for the group of direct isometries) is

$$\mu(S)=\int_S dpd\theta$$

(see first reference at the bottom of this text).

For example the measure of the set of straight lines intersecting the disk centered in $0$ with radius $R$ is:

$$  \int_{(p,\theta)\in [0,R] \times [0,2\pi)}dpd\theta=\int_{p=0}^R\int_{\theta=0}^{2 \pi}dpd\theta=2 \pi R$$

(i.e., its perimeter).

Let $S_{A|B,C}$ be the set of lines that separate vertex $A$ from vertices $B,C$, otherwise said, that intersect $AB$ (length $c$) and $AC$ (length $b$) but not $BC$ (length $a$).

One can compute the measure of set $S_{A|B,C}$ to be:

$$ \mu(S_{A|B,C}) \ = \ \int_{S_{A|B,C}} dpd\theta \ = \ b+c-a \tag{2}$$

and similar formulas for the other cases.

It is remarkable that the positivity of measure (2) is equivalent to triangle inequality...

One has to work with conditional probabilities, i.e., by assuming that we limit our attention to lines $(L)$ intersecting triangle $ABC$.

Due to the fact that a line that intersects a triangle intersects two and only two of its sides (setting apart lines passing through one of the vertices, which have measure $0$, therefore are neglectable from the point of view of measure theory):

$$(L) \in S_{A|B,C}, \ \ (L) \in S_{B|C,A}, \ \ (L) \in S_{C|A,B}$$

are disjoint and their union is the whole set of lines (complete set of events in probabilistic terms). Therefore, we have a total measure:

$$(b+c-a) \ + \ (c+a-b) \ + \ (a+b-c) \ = \ a+b+c$$

i.e, the perimeter of the triangle (as was found for the disk above) !

As a consequence, the (conditional) probabilities you are looking for can "logicaly" been taken as given by (0).

Historical corner: "Integral Geometry" (aka "Geometric probability") was initiated by Buffon in the mid-18th century, then by Crofton in the $1880$s, followed by Poincaré (cinematic formulas, see reference below) around 1900, but especially developed by Blaschke and his students circa $1930-1950$ (see Santalo's book "Integral Geometry and Geometric Probability" for a rather clear exposition).

Remark: A cousin domain of Integral Geometry is "Stereology"

A nice presentation of Integral Geometry can be found here.

A rather accessible reference here.

For a generalisation of the perimeter result, see Cauchy-Crofton formula

For cinematic formulas see this (more difficult!) reference.

Another (very theoretical) reference.

Edit: For 3D, use Crofton formula (one of the many formulas of stereology dealing with a number of intercepts $N(...)$ of a variable plane $P$ with a fixed triangle $T$):

$$\int N(P \cap T) d\mu  \ = \ \pi L(T)$$

where $L(T)$ is the length (= perimeter) of triangle $T$.

($N=0,2$ in the case of a triangle, $N=1$ being a case with measure $0$).

Of course, practically speaking, the integral in the LHS will be approximated using a mean value...


问:
Distance between two skew lines in a cube
Let ABCD.EFGH be a cube with length 2. X is the midpoint of EF. Find distance between AX and BH.

I have no idea how to do this.

答:
HINT:

With $(i,j,k)$ unit vectors along the axes consider vector cross product

$$(i + 2 k) \times (-2i+2j+2k)$$

which lies parallel to minimum distance vector between the given skew line vectors. Identify the points on the two skew lines. There are no parallel lines given here.


问:
Do two matrices commute only if one matrix is the inverse of the other?
Is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? The only way I could believe this was the 2 commutative type properties I know of:
$$AA^{-1}=A^{-1}A=I$$
and
$$adj A*A=A*adjA = |A|I$$
where $adj A$ is the adjoint of matrix A.

The first equation when expanded with $A^{-1}=\frac{adjA}{|A|}$ you end up getting the second equation. So both are the same equation. These are the only two properties that explains commutativity as far as I'm aware of(are there any more?) and my conclusion was based on these only. Also multiplication with identity matrix is also commutative but any other counter cases? So is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? Thanks!!

答:
Another commuting matrix examples come from:

Lemma: Any square matrix $A$ conmutes with a linear combination of itself and the
identity, that is:

$A\cdot (\alpha A-\beta I)= A \cdot(\alpha A-\beta I)$

Proof:

$$A\cdot (\alpha A-\beta I) = \alpha A^2 -\beta A = A\cdot(\alpha A - \beta I)$$


问:
If set of vectors in $\mathbb{R}^n$ is a basis, there are $n$ vectors in it.
I am trying to prove this. For me the definition of basis was presented without the condition that each vector in $\mathbb{R}^n$ is a linear combination of the basis vectors. That result was derived later using the result, that a basis of $\mathbb{R}^n$ includes exactly $n$ vectors. This however isn't clear to me. Here's my attempt to prove it:

I assume, that there are m vectors in $\mathbb{R}^n$ in this linearly independent set:

$$ \sum_{k=1}^m\xi_kx_k=0, \ \ x_k=\left(\begin{matrix}
x_{k1}\\
x_{k2}\\
\vdots\\
x_{kn}
\end{matrix}\right) $$
This can be written as a matrix equation:
$$ \Leftrightarrow\ \underbrace{\left(\begin{matrix}
x_{11}&x_{21}&\dots&x_{m1}\\
x_{12}&x_{22}&\cdots&x_{m2}\\
\vdots&\vdots&\ddots&\vdots\\
x_{1n}&x_{2n}&\dots&x_{mn}
\end{matrix}\right)}_{=A} \underbrace{\left(\begin{matrix}
\xi_1\\
\xi_2\\
\vdots\\
\xi_m
\end{matrix}\right)}_{=x}=0 $$
Now the question is, when is the solution vector $x$ unique? When this is answered, it should follow that $m = n$.

答:
It's false.  One vector is linearly independent in any $\Bbb R^n$, for $n\gt1$.

Pretty clear that you meant in a basis.

For the solution to exist and be unique, in your setup, you need the matrix to be invertible.  In which case it's square, and $m=n$.


问:
Request for Clarification : Differentiability vs Analytic
Consider the function $f(x)=e^{-1/x}$ if $x>0$ and $f(x)=0$ otherwise, now previously I managed to show that $f$ is infinitely differentiable in $\mathbb{R}$ so that includes differentiable at $x>0$, $x<0$, and the most important one $x=0$. The problem lies with showing that $f$ is not analytic at $x=0$ this confuses me since I have shown that $f$ is differentiable at $x=0$. How could this be interpreted?

答:
Look back at the definition of analyticity. It doesn't just say that all of the derivatives of $f$ exist, it says that "Taylor series work well" for $f$:

$f$ is analytic iff for each $a$, the Taylor series $T^f_a$ agrees with $f$ around $a$ - that is, there is some $\epsilon>0$ such that $T_a^f$ is defined and equals $f$ on the interval $(a-\epsilon, a+\epsilon).$

Intuitively, analyticity doesn't just say that all the derivatives of $f$ exist at each point, but that the various derivatives contain "all the information about $f$:" there's no point where knowing the derivatives of $f$ won't let you figure out what $f$ is, at least "locally."

So to show that your $f$ is not analytic, what you want to do is find some point where the Taylor series for $f$ "goes wrong" somehow. Now your $f$ consists of three regions: negative $x$, positive $x$, and $x=0$ where the first two regions meet. On each of the first two regions $f$ is very "tame:" $f$ is equal to some function we already know to be analytic. So the only possible "weird point" for $f$ is $a=0$. You know that $f$ is infinitely differentiable here too, but again analyticity is more than just infinite differentiability:

What is $T_0^f$? Does this "look like $f$" even in a small neighborhood of $0$?

What we'll see is that, while the infinite differentiability of $f$ at $0$ does allow us to build the series $T_0^f$, somehow information is lost when we do so.


答:
the eigenvalues are distinct, two are complex conjugates. The minimal polynomial is the given $x^3 + x^2 -1.$ The degree of each eigenvalue term $(x- \lambda)$ in the minimal polynomial is the size of the largest Jordan block. Thus the Jordan form is diagonal; the trace is $n_1 \lambda + n_2 \mu + n_3 \bar{\mu}$
where  the real eigenvalue is $\lambda.$ For the trace to be real, we know $n_2 = n_3$

We also know that the sum of the eigenvalues is $-1.$  If $n_1 \neq n_2,$ write $t = n_1 - n_2$  so that the trace is now
$$  t \lambda  + n_2 \lambda +  n_2 \mu + n_2 \bar{\mu}= t \lambda - n_2 $$
If $t \neq 0$ this is irrational.

So, $n_1 = n_2 = n_3$ and the dimension is $3 n_1$

Title: $u^3 + u^2-id=0\text{ and } tr(u) \in \mathbb{Q}$ show that the dimension of the vector space is a multiple of 3

Body: $\text{let E a vector space} \:dimE=n \\ \text{let}\:\: u \in L(E,E) \text{such as} \: u^3 + u^2-id=0\text{ and } tr(u) \in \mathbb{Q}\\ \text{show that n is a multiple of 3}$


问:
Proving property of Euclidean projection minimization problem
Let $X \subseteq \mathbb{R}^n$ be a convex set and $y \not\in X$. Suppose $x^*$ is the Euclidean projection of $y$ onto $X$. That is, $x^*$ is the optimal solution to
$$ \min_{x\in X}\frac{1}{2}\|x-y\|_2^2. $$
I want to prove that
$$\frac{1}{2}\|x-y\|_2^2 - \frac{1}{2}\|x^* - y\|_2^2 \geq \frac{1}{2}\|x-x^*\|_2^2.$$
What I've tried:
I tried breaking up the L2-norm into a summation to prove the statement directly. Something like this:
\begin{align*}
\frac{1}{2}\|x-x^*\|_2^2 & = \frac{1}{2}\sum_{i=1}^{n}(x-x^*)_i^2\\
& \hspace{2cm} \vdots\\
& \le \frac{1}{2}\sum_{i=1}^{n} (x_i-y_i)^2 - (x^*_i - y_i)^2
\end{align*}
But I got stuck in the process. I think I am going about this wrong, as it doesn't look like I can prove a statement such as $(a-b)^2 \le (a-c)^2 - (b-c)^2.$ This looks like something akin to the triangle inequality. Any help would be appreciated!

答:
The necessary and sufficient condition for a point $x^*\in X$ to be a global minimum of a differentiable convex function $f$ over the convex set $X$ is
\begin{equation}
\langle \nabla f(x^*), x-x^* \rangle \ge 0\quad \forall x\in X.\tag{*}
\end{equation}

This can be proved as follows.

If $x^*$ satisfies $(*)$ then $f(x) \ge f(x^*) + \langle \nabla f(x^*), x-x^* \rangle \ge f(x^*) \ \forall x\in X$, which means $x^*$ is a global minimum of $f$ over $X$. To prove the converse, let $x^*$ be a global optimum and assume that there exists $x\in X$ such that $\langle \nabla f(x^*), x-x^* \rangle < 0$. Define $h(t) = f(x^* + t(x-x^*))$ for $t\in [0,1]$, we have
$$h'(t) = \langle \nabla f(x^* + t(x-x^*)), x-x^* \rangle \implies h'(0) = \langle \nabla f(x^*), x-x^* \rangle < 0.$$
Therefore, for $t > 0$ small is enough, we have $h(t)< h(0)$, i.e. $f(x^* + t(x-x^*)) < f(x^*)$, which is a contradiction of $x^*$ being a global optimum.

Now back to the main question. Applying $(*)$ with $f(x) = \frac{1}{2}\|x-y\|_2^2$ we obtain $$\langle x^*-y, x-x^* \rangle \ge 0\quad \forall x\in X.$$
Adding $\frac{1}{2}\|x^*-y\|_2^2 + \frac{1}{2}\|x-x^*\|_2^2$ to both sides of the above, we obtain the desired inequality.

Note that you can obtain a direct proof, instead of using $(*)$, by replacing $f$ by $\frac{1}{2}\|x-y\|_2^2$ in the above proof of $(*)$. The resulting proof is simple and elementary enough. Try that yourself! (Actually I was hesitating between the direct proof and the above proof for the answer, but finally decided that the latter would be better for the readers to see the picture.)


问:
Discrete Definition of a Limit
My professor defines a limit the following way:

Let $f: \mathbb N \rightarrow \mathbb R$ be a sequence and $L\in\mathbb R$. $f$ converges to $L$ if $$\forall \epsilon \in \mathbb R^+ \exists N\in\mathbb N\forall n\in \mathbb N(n>N \rightarrow |f_n-L|<\epsilon)$$

Which is the dramatic way of saying $$\lim_{n\rightarrow \infty}f_n=L$$

Can someone help me unpack this definition and how I can apply it like one applies the typical epsilon-delta definition of a limit?

答:
For instance, $f(n)=\frac 1n$ converges to zero.

This means that given any $\epsilon>0$, eventually $f(n)$ will always be within $\epsilon$ distance away from $0$. By "eventually", we mean for all $n$ bigger than some $N$. Here's a picture illustrating what it means for $a_n\to a$.

This is what your definition is saying. Indeed, by the Archimedean property, given any $\epsilon>0$, we can find an $N_\epsilon$ big enough such that $\frac 1{N_\epsilon}<\epsilon$. For all $n$ bigger than this $N_\epsilon$, we get that
$$\Big|\frac1n - 0\Big| = \frac1n \leqslant \frac{1}{N_\epsilon} < \epsilon,$$
and this proves that $\frac1n\to 0$.

Title: Discrete Definition of a Limit


问:
Prove or give a counterexample of limsup and liminf
Lets $a_n$ and $b_n$  be two sequence that $0\leqslant a_n \leqslant 1$ and $0\leqslant b_n\leqslant 1$ for each $n$.

Prove or give a counterexample that:

a) if $\displaystyle\liminf_{n\to\infty} a_nb_n = 1$  then $\displaystyle\lim_{n\to\infty} a_n = 1$

b) if $\displaystyle\limsup_{n\to\infty} a_nb_n = 0$  then $\displaystyle\lim_{n\to\infty} a_n = 0$

I was trying to find a counterexample for hours and it's not working, I'd like to get a tip or anything...

答:
(a) Clearly, we have : $0 \leqslant a_nb_n \leqslant a_n \leqslant 1$. Therefore, $$1 = \displaystyle\liminf_{n\to\infty} a_nb_n \leqslant \displaystyle\liminf_{n\to\infty} a_n \leqslant 1$$ implying : $\displaystyle\liminf_{n\to\infty} a_n= 1$

Again, $$1 \geqslant \displaystyle\limsup_{n\to\infty} a_n \geqslant \liminf_{n\to\infty} a_n = 1$$ implying : $\displaystyle\limsup_{n\to\infty} a_n =  \liminf_{n\to\infty} a_n = 1$

Hence, we must have $\displaystyle \lim_{n\to\infty} a_n = 1$ .

(b) As per J. W. Tanner's comment above, take $\{a_n\} = (1 , 0 , 1 , 0 , \cdots)$ and take $\{b_n\} = (0 , 1 , 0 , 1 , \cdots)$ .

Then, $\displaystyle\limsup_{n\to\infty} a_nb_n = 0$ , but $\displaystyle\lim_{n\to\infty} a_n$ does NOT exist.


问:
Devising equation using exponential
So, this is a bit hard to explain. I want to devise an equation that turns the numbers on the left into the numbers on the right.

8 = 20

20 = 40

32 = 80

44 = 160

56 = 320

As well as all the other values in between 8-56. As in, the numbers between 8-20 have a proportional value between 20-40. Numbers between 32-44 have a proportional value between 80-160 ect.

The values on the left go up linearly and the values on the right go up exponentially.

I hope this makes sense. It's quite hard to explain. But would appreciate any help. Thanks

答:
Increasing by 12 doubles the result so
something like
$20*2^{(x-8)/12}$.

Title: Devising equation using exponential


答:
Let $D$ be the origin of the Cartesian coordinates, with $A(2,0,0)$, $B(2,2,0)$, $H(0,0,2)$ and $X(2,1,2)$. Then, the line equations for $AX$ and $BH$ are
$$AX: \>\>\>\vec {a}_1+ t\vec{b}_1=( 2,0,0)+t(0,1,2)$$
$$BH: \>\>\>\vec {a}_2 + t\vec{b}_2=( 0,0,2)+t(2,2,-2)$$
Note that the init vector perpendicular to both lines is
$$\vec n = \frac{\vec{b}_1\times \vec{b}_2 }
{|\vec{b}_1 \times \vec{b}_2|}= \frac1{\sqrt{56}}(-6,4,-2)$$
Then, the distance between the two lines is
$$d=|(\vec{a}_1-\vec{a}_2)\cdot \vec{n}
|=\frac{2\sqrt{14}}7
$$

Title: Distance between two skew lines in a cube

Body:
Let ABCD.EFGH be a cube with length 2. X is the midpoint of EF. Find distance between AX and BH.

I have no idea how to do this.


问:
Any compact, convex subset of $\mathbb{R}^n$ is a deformation retract of $\mathbb{R}^n$
I am looking at a question from Chapter 32 of Willard's General Topology:

Any compact, convex subset of $\mathbb{R}^n$ is a deformation retract of $\mathbb{R}^n$

I'm struggling to even know where to start with this exercise, as I don't have any intuition as to why compactness is related to (deformation) retracts.

I think I will need to use the following facts / definitions:

Let $X := \mathbb{R}^n$. If $Y \subseteq X$ is convex, then any $f,g \in C(X,Y)$ are homotopic.

A subset $A \subseteq X$ is a retract of $X$ if $\exists r \in C(X,A)$ such that $r(a) = a,$ for each $a \in A$. It is a deformation retract of $X$ if $r$ is homotopic (as a map into $X$) to $1_X$ (the identity function on $X$).

A subset $E \subseteq X$ is compact if every cover of $E$ by open sets of $X$ has a finite subcover. (Edit: according to the comments, I probably want to use the fact that any compact subset of $\mathbb{R}^n$ is closed and bounded, instead).

Any help, including hints on how to begin approaching this question, is appreciated.

Edit: Based on a comment, I tried to show that the closed unit disk $D = \{ (x,y) \in \mathbb{R}^2: x^2 + y^2 \leq 1\}$ is a deformation retract of $\mathbb{R}^2$. I think I've shown that below, but I did not find this particularly illuminating, so I'm wondering if I've skipped over something or perhaps my "proof" of this is not correct.

$D$ is a compact, convex subset of $\mathbb{R}^2$ and the function $r: \mathbb{R}^2 \to D$ given by:

$r((x,y)) = \left\{
        \begin{array}{ll}
            (x,y) & \quad x^2+y^2 < 1 \\
            \big(\frac{x}{x^2 + y^2},\frac{y}{x^2+y^2}\big) & \quad x^2+y^2 \geq 1
        \end{array}
    \right.$

is a retraction, as $r(d) = d$ for each $d \in D$ and it is continuous. Indeed, $D$ is a deformation retract; if we view $r$ as a map from $\mathbb{R}^2 \to \mathbb{R}^2$, we see that it is continuous function from $\mathbb{R}^2$ into a convex subset (clearly $\mathbb{R}^2$ is a convex subset of itself) and so by Fact 1 above, any continuous function, including the identity function $1_{\mathbb{R}^2}$, is homotopic to $r$.

Any feedback on this first attempt would be appreciated, as I really don't see how this clarifies the more general situation.

答:
We shall generalize this by proving that

Any closed convex subset of $\mathbb R^n$ is a strong deformation retract of $\mathbb R^n$.

Let $C$ be a closed convex subset of $\mathbb R^n$. For each $x \notin C$ we have
$$d(x,C) = \inf\{\lVert x - y \rVert : y \in C \} > 0 ,$$
because otherwise we would find a sequence $(y_n)$ in $C$ such that $y_n \to x$. But then we would have $x \in C$ since $C$ is closed.

Note that the definition of $d(x,C)$ is based on the Euclidean norm. This will be essential for our proof. See the remark below.

There exists $y \in C$ such that $\lVert x - y \rVert = d(x,C)$. In fact, let $y_n \in C$ such that $\lVert x - y_n \rVert < d(x,C) + 1/n$. This sequence is bounded by $\lVert x \rVert + d(x,C) + 1$, thus has a convergent subsequence, so we may assume w.l.o.g. that $(y_n)$ converges to some $y \in \mathbb R^n$. Since $C$ is closed, we have $y \in C$ and $\lVert x - y \rVert = d(x,C)$.

We claim that $y$ is unique because $C$ is convex. So  assume that $y' \in C$ is point $y' \ne y$ such that $\lVert x - y \rVert = \lVert x - y' \rVert$. The points $x, y, y'$ span an affine Euclidean plane $E^2 \subset \mathbb R^n$ and form an isosceles triangle. The midpoint $y'' = 1/2 y + 1/2y'$ of the line segment between $y, y'$ is contained in $C$. The points $x,y, y''$ form a right-angled triangle, thus $\lVert x - y \rVert^2 = \lVert x - y'' \rVert^2 + \lVert y - y'' \rVert^2$ which gives  $\lVert x - y \rVert > \lVert x - y'' \rVert$, a contradiction.

Remark: As pointed out in a comment by copper.hat, we use a special property of the Euclidean norm $\lVert - \rVert$: It is strictly convex, which means that any closed ball $B$ is a strictly convex set in the sense that every point on the line segment connecting two points $x, y \in B$ other than the endpoints is inside the interior of $B$. I proved a special case of this (for the midpoint of the line segment) using the Pythagorean theorem. Note that other norms may not have this property.

Define
$$r : \mathbb R^n \to C, r(x) = \begin{cases} x  & x \in C \\ \text{unique } y \in C \text{ such that } \lVert x - y \rVert = d(x,C) & x \notin C  \end{cases}$$

Let us prove that $r$ is continuous (i.e. that $r$ is a retraction).
Continuity is obvious in all interior points of $C$.

Let us now consider a boundary point $\xi$ of $C$. Let $\epsilon > 0$ and $x \in \mathbb R^n$ such that $\lVert x - \xi \rVert < \epsilon/2$. We claim that $\lVert r(x) - r(\xi) \rVert = \lVert r(x) - \xi \rVert < \epsilon$. This is trivial for $x \in C$. For $x \notin C$ we have $\lVert r(x) - \xi \rVert \le \lVert r(x) - x \rVert + \lVert x - \xi \rVert = d(x,C) + \lVert x - \xi \rVert \le 2 \lVert x - \xi \rVert < \epsilon$.

Let us finally consider a point $\xi \notin C$. In the sequel it will be useful to draw pictures to understand geometrically what is going on.

We begin with a preparation. Let $P^{n-1}(x)$ denote the affine hyperplane which contains $r(x)$ and is orthogonal to the line through $x$ and $r(x)$ (i.e. $P^{n-1}(x) = \{r(x) + y : \langle y, x - r(x) \rangle = 0\}$ , where $\langle -, - \rangle$ denotes the standard inner product). This is the tangent hyperplane of the sphere $S^{n-1}(x;d(x,C))$ with center $x$ and radius $d(x,C)$ at the point $r(x)$. $P^{n-1}(x)$ divides $\mathbb R^n$ in two open half-spaces. Let $H^n(x)$ denote the open half-space containing $x$ (i.e. $H^n(x) = \{r(x) + y : \langle y, x - r(x) \rangle > 0\}$). We claim that $H^n(x) \cap C = \emptyset$. Assume there exists $y \in H^n(x) \cap C$. The points $x, r(x), y$ are contained in an affine Euclidean plane $E^2 \subset \mathbb R^n$ (if $y$ lies on the line through $x$ and $r(x)$, then $E^2$ is not unique, but that does not matter). The set $S' = E^2 \cap S^{n-1}(x;d(x,C))$ is a circle in $E^2$, and $L = E^2 \cap P(x)$ is the tangent line to $S'$ at $r(x)$. The circle $S'$ bounds the open disk $D^2(x,d(x,C)) \subset E^2$ with center $x$ and radius $d(x,C)$. Clearly $y \notin D^2(x,d(x,C))$ because otherwise $d(x,C) \le \lVert y - x \rVert < d(x,C)$. The line $L(y)$ through $y$ and $r(x)$ is different from $L$, thus $D^2(x,d(x,C)) \cap L(y)$ is non-empty. Let  $y' \in D^2(x,d(x,C)) \cap L(y)$. Since $y \notin D^2(x,d(x,C))$, the point $y'$ lies between $y$ and $r(x)$, thus $y' \in C$ because $C$ is convex. Therefore $d(x,C) \le d(x,y') < d(x,C)$, a contradiction.

Now let $ 0 < \epsilon \le d(x,C)$ and $x \in \mathbb R^n$ such that $\lVert x - \xi \rVert < \epsilon/2$. Note that this assures $x \in H^n(\xi)$. We claim that $\lVert r(x) - r(\xi) \rVert < \epsilon$. Let $\rho(x) \in P^{n-1}(\xi)$ be the unique point such that line $L_x$ through $x$ and $\rho(x)$ is orthogonal to $P^{n-1}(\xi)$. We have $\lVert \rho(x) - r(\xi) \rVert < \epsilon/2$: Note that in the quadrilateral with vertices $\xi, x, r(\xi), \rho(x)$ (which span an affine Euclidean plane $E^2 \subset \mathbb R^n$) the edges $\overline{\xi r(\xi)}$ and $\overline{x \rho(x)}$ are parallel with distance $\lVert \rho(x) - r(\xi) \rVert$, thus  $\lVert \rho(x) - r(\xi) \rVert \le$ length of the edge $\overline{x \xi}$ which is $\lVert x - \xi \rVert < \epsilon/2$. We have $d(x,C) \le d(x,r(\xi))$, thus $r(x)$ is contained in the closed ball $\bar D^n(x,d(x,r(\xi))) \subset \mathbb R^n$ with center $x$ and radius $d(x,r(\xi))$. Since $H^n(\xi) \cap C = \emptyset$, we must have $r(x) \in D' = \bar D^n(x,d(x,r(\xi))) \cap G^n(\xi)$, where $G^n(\xi) = \mathbb R^n \setminus H^n(\xi)$ is the closed half-space bounded by $H^{n-1}(\xi)$ and not containing $\xi$. The intersection $D'' = \bar D^n(x,d(x,r(\xi))) \cap P^{n-1}(\xi)$ is a closed ball in $P^{n-1}(\xi)$ with center $\rho(x)$ and radius $R = \lVert \rho(x) - r(\xi) \rVert < \epsilon$. Thus $D'$ is a spherical dome of $\bar D^n(x,d(x,r(\xi)))$ with base $D''$. The diameter of $D'$ equals the diameter of $D''$ which is $2R$. Thus $\lVert r(x) - r(\xi) \rVert \le 2R < \epsilon$.

$r$ is in fact a strong deformation retraction. Look at
$$H: \mathbb R^n \times I \to \mathbb R^n, H(x,t) = (1-t)x + tr(x) .$$


问:
Optimal solution to non-linear program
Find an optimal solution to the following problem:

$$\min_{x\in \mathbb{R}^n} \sum_{i=1}^{n}\frac{1}{2}(x_i-1)^2$$
$$\text{s.t. } x\ge0 \text{ and } \sum_{i=1}^{n}x_i=1.$$

答:
First, to answer your question regarding $\lambda_i = 0 \ \forall i$, I would suggest a simpler argument, as follows.

Assume that $\lambda_i > 0$ for some $i$, then
$$x_i=0\implies\mu-1=-\lambda_1< 0\implies x_j-\lambda_j<0\forall j \implies \lambda_j > 0\ \forall j \implies x_j=0\ \forall j,$$
which is of course not feasible.

Second, there's a simple and elementary proof for the problem in the question. Notice that $(x_i-1/n)^2 \ge 0$, we have that $x_i^2 \ge ax_i+b$ for some $a,b$. Sum up this inequality for all $i$ we see that the objective function in the equation is no smaller than a constant, with equality iff $x_i=1/n\ \forall i$.

Title: Optimal solution to non-linear program


问:
Prove or disprove that Y = AX-C
Let $A$ be an $m \times n$ matrix such that $\mathrm{rank}(A) = n \le m$.  Prove, or disprove using a counter example:

Every $m\times n$ matrix $Y$ has a decomposition $Y = AX-C$, where $X$ and $A^TC$ are nonnegative semidefinite symmetric matrices.

I have a feeling this is false, but I can't come up with a good counter example to show this.
I've been trying to find a $Y$ such that $X$ would have to be non semidefinite to construct this $Y$.

Is it possible that the statement is true?

This is a problem from a book on semidefinite optimization.

Any insights would be greatly appreciated.

Thanks

答:
Your proposition is false as stated.

Let $A$ be the $n\times n$ identity matrix. Then the decomposition reduces to $X-C$, where both $X$ and $C$ are symmetric. Clearly no asymmetric $Y$ has such a decomposition.

ContentLicense: CC BY-SA 3.0

Title: Prove or disprove that Y = AX-C


问:
How to solve $\arg z^3 =0$ where $z\in\mathbb{C}$?
Can someone show me how to solve $$\arg z^3 = 0$$
I don't know even how to start because of the zero.

答:
Hint:

First prove that given non-zero complex numbers $z,w$ that Arg$(zw)$ = Arg$(z) +~$ Arg$(w),~$ within a modulus of $(2\pi)$.

You will need formulas for $\cos(a + b)$ and $\sin(a + b).$

Then prove that for non-zero complex $z,~$ Arg$(z^3)$ = $3 \times $ Arg$(z),~$ within a modulus of $(2\pi)$.


问:
Finding a vector equation orthogonal to a plane
Find the vector equation of the line containing the point $(1,2,3)$ and orthogonal to the plane $x-y+2z=4$.

From what I understand $x-y+2z=4$ is written in point normal form and I should be able to take out the vector $\boldsymbol n=(1,-1,2)$. From here I should be able to get a vector $\boldsymbol v$ orthogonal to $\boldsymbol n$ by doing $\boldsymbol v·\boldsymbol n=0$.

However, I was told the correct answer is $(x,y,z) = (1,2,3) + t(1,-1,2)$. Isn't the vector in this vector equation parallel to the plane $x-y+2z=4$? I would really appreciate it if someone could explain to me why this is the right answer. Thank you!

答:
"Orthogonal" relates to perpendicularity. So you're basically finding a line perpendicular to the plane, which means a single line that's simultaneously and mutually perpendicular to every line on the plane.

What you have in the first step is a normal vector to the plane, which means a vector which is already at right angles to the plane. To get the required line perpendicular to the plane, the normal vector will simply be the direction vector of that line. In other words, you want a line in the direction of (or exactly opposite to) the normal vector of the plane, but not one that's at any other angle (including perpendicular) to the normal vector. That would defeat the purpose - if you calculated a line direction vector perpendicular to that normal vector, you're getting a line that's parallel to a line on the plane. This is not what's required.

So all you need to do is use the normal vector of the plane as the direction vector of the line in the formula (and the given fixed point to "anchor" it in space) and you're done.


答:
Think it geometrically: if a line's guiding vector is parallel to a plane's normal vector' then it is normal (thus is, orthogonal) to the plane. This is quite intuitive: a plane's normal vector defines the direction orthogonal to it. If a line is parallel to that vector, then it has its direction and, therefore, is orthogonal to the plane.

The solution they provided you is correct, for it goes through the desired point in a direction orthogonal to the plane, this is, the direction of the normal vector.

You can also think geometrically why, if a line's guiding vector is ortohogonal to a plane's normal vector, then it is one of the infinitely many lines that are parallel to the plane

Title: Finding a vector equation orthogonal to a plane

Body:
Find the vector equation of the line containing the point $(1,2,3)$ and orthogonal to the plane $x-y+2z=4$.

From what I understand $x-y+2z=4$ is written in point normal form and I should be able to take out the vector $\boldsymbol n=(1,-1,2)$. From here I should be able to get a vector $\boldsymbol v$ orthogonal to $\boldsymbol n$ by doing $\boldsymbol v·\boldsymbol n=0$.

However, I was told the correct answer is $(x,y,z) = (1,2,3) + t(1,-1,2)$. Isn't the vector in this vector equation parallel to the plane $x-y+2z=4$? I would really appreciate it if someone could explain to me why this is the right answer. Thank you!


问:
Number theory proof mod p
For a given prime $p$ and integer $k$,  $k∈<1, p-1>$ $S(k)$ is defined as a sum of remainders of $k, k^2, k^3, ..., k^{p-1}$ when divided by $p$. Prove that $S(k)=\frac{1}{2}p(p-1)$ is true for an odd number of different k.

I managed to prove that $S(p-1)=\frac{1}{2}p(p-1)$ but can't see how to do it further.

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain what stumps you.)

Hint: Prove that $S(k) = \frac{1}{2} p (p-1)$ iff there exists an $a$ such that $ k^a \equiv -1 \pmod{p}$.

Corollary: Hence, if $p -1 = 2^m \times n, $ where $n$ is odd, then the number of solutions is $ (2^m-1) \times n $, which is odd.

Proof of hint without using primitive elements.

If there exists $ k^ a \equiv -1 \pmod{p}$:

(Proof by algebraic manipulation)  Show that $$ 2 \sum_{i=1}^{p-1} ( k^i \pmod{p})   = \sum_{i=1}^{p-1} (k^i + k^{i+a} \pmod{p} )  = \sum_{j=1}^{p-1} p = (p-1)p. $$

Hence $ \sum_{i=1}^{p-1} (k^i \pmod{p}) = \frac{p-1}{2} \times p$.

If no such $a$ exists, then consider $ v = ord_p (k)$.

(Proof by contradiction) Show that $ v$ is odd, and a factor of $ p-1$.

(Proof by algebraic manipulation) Show that
$$ \sum_{i=1}^{p-1} ( k^i \pmod{p} = \sum_{j=0}^{\frac{p-1}{v}-1} \sum_{i=1}^v (k^{jv + i} \pmod{p} = \frac{p-1}{v} \times \sum_{i=1}^v (x^i \pmod{p}).$$

Observe that if $ p-1 = 2^m \times n$ with $n$ odd, then $ 2^m \mid \frac{p-1}{v} \mid \sum_{i=1}^{p-1} (k^i  \pmod{p})$, but $ 2^m \not \mid \frac{1}{2} (p-1)p$, so these 2 values can never be equal.

Hint using primitive elements. This was originally the easier way to approach. However, now that I've written out the steps which could be followed via algebraic manipulation, you should just review the above.

Consider a primitive element $g$.

Let $ k = g^b$.

Let $ p-1 = 2^m \times n$, where $n$ is odd.

Then $k$ satisfies the conditions iff $ 2^m \not \mid b$.

If $2^m \not \mid b$, then there exists $a$ such that $ ba = \frac{p-1}{2} \times o$, where $o$ is odd. This $a$ satisfies $k^a \equiv -1 \pmod{p}$, and the above pairing argument works.

If $ 2^m \mid b$, show that $2^m \mid S(k)$ while $ 2^m \not \mid \frac{1}{2} p (p-1)$, in a similar manner to the above. Hence these terms are not equal.

The description of the solution set as powers of $g$ that were not multiples of $2^m$, felt hard to motivate directly. (E.g. how can we tell that if $g^b$ doesn't work, then neither does $g^{2b}, g^{3b}, \ldots$?)

Whereas finding the $k^a \equiv -1 \pmod{p}$ condition was quite direct especially after considering small cases. Hence, that was my initial hint.

Title: Number theory proof mod p


问:
How much choice is needed to show that $|A\times \omega|=|A|$?
A recent question here (which I believe was deleted, unfortunately) came down to showing the title question, that $|A\times\omega|=|A|$ for all infinite sets $A$. In that situation, the questioner could assume AC so the question became easy, but it got me thinking about the situation without choice. I'm familiar with Tarski's theorem that $|A\times A|=|A|$ for all $A$ implies AC, and I'm pretty confident that it takes at least some choice to prove $|A\times\omega|=|A|$. For instance, it would seem to clearly fail if $A$ is Dedekind-finite: given the bijection $\beta: A\times\omega\mapsto A$, the set $B=\{\beta(\langle a,0\rangle): a\in A\}$ is a proper subset of $A$ equinumerous with it. But how much choice does it take? The natural assumption would be Countable Choice, but I don't see any sensible way of modifying the proof of Tarski's theorem I've seen to get CC from the bijection between $A\times\omega$ and $A$.

答:
We can show, without choice whatsoever that $$|A|+|A|=|A|\implies|A|\times\aleph_0=|A|.$$

So the exact choice you need is "for every infinite $A$, $|A|+|A|=|A|$". This principle implies that there are no infinite Dedekind-finite sets, but it does not follow from $\sf DC_\kappa$, for any $\kappa$, nor it implies even $\sf AC_\omega$. It is, indeed, part of the "strange principles of choice" family that is just... out there.


问:
Analytic continuation of geometric series using Taylor expansion
I am currently trying to get a better grasp of the concept of analytic continuation and so I am working through this example:

Let $$f:(-1,1)\to\mathbb{R}~,~x\mapsto\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$

Then we can use Taylor expansion to get:

$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)(x-x_0)^n}{n!} =\sum_{n=0}^\infty \frac{n!(x-x_0)^n}{n!(1-x_0)^{n+1}} =\frac{1}{1-x_0}\sum_{n=0}^\infty \left(\frac{x-x_0}{1-x_0}\right)^n$$

Now, what we have is another geometric series, which converges for
$$\left|\frac{x-x_0}{1-x_0}\right|<1$$

and in fact, the whole thing even converges to
$$\frac{1}{1-x}$$
so all is nice here. The question is, for which $x$ does it converge? We get:

$$\left|\frac{x-x_0}{1-x_0}\right|<1\Leftrightarrow |x-x_0|<1-x_0~,$$
since $1-x_0 > 0$. To maximize the RHS we let $x_0\to-1$ and we get

$$|x+1|<2\Leftrightarrow -3<x<1~.$$

Now we can repeat this procedure as many times as we like and always push the negative boundary but never the positive so that we get

$$x\in(-\infty,1)~.$$

Finally to get to my question: Can you continue $f$ in the positive direction, using Taylor expansion or do you need another method for that? Does anyone know a reasonably simple way how to do this?

Thanks in advance,

Alex

答:
The analytic continuation is performed in the complex plane, where the right side of
$$\sum_{n\in\mathbb{N}} z^n = \frac{1}{1-z}$$
shows its unique pole at $z=1$.
By the Uniqueness of Analytic Continuation for complex functions,
$$\forall z\in\mathbb{C}\setminus\{1\} \\ f(z)=\frac{1}{1-z}$$
must be the analytic continuation of the geometric series, since it defines the same correspondence rule than $\sum_{n\in\mathbb{N}} z^n$ at the open unit $z$-complex disk.

This power series expansion turn into a Fourier series expansion if we simply set some $t\in\mathbb{R}$ such that
$$z\to e^{it}.$$


问:
residue class rings - value calculations with modulo
computer science studies - residue class rings.

I have to calculate the following expression in $ \mathbb{Z_9}$

and the other one in $ \mathbb{Z_{16}}$

my question is how one should handle $-1, 23, 10$ in the "powers"?

are they really powers or do they have another meaning perhaps?

Thank you for your help

答:
Here we show the OP some techniques (tricks?) for finding the inverse elements in   $\mathbb{Z_{16}}$.

Since $15 \equiv -1 \pmod{16}$ we can factor $15$ and write

$\quad 3\cdot 5 \equiv -1 \pmod{16} \; \text{ iff } \; (-3) \cdot 5 \equiv 1 \pmod{16} \; \text{ iff } \; 5 \cdot 13 \equiv 1 \pmod{16}$

and

$\quad 3\cdot 5 \equiv -1 \pmod{16} \; \text{ iff } \; 3 \cdot (-5) \equiv 1 \pmod{16} \; \text{ iff } \; 3 \cdot 11 \equiv 1 \pmod{16}$

A good start - we've found the inverse of four residue primes. By working with our 'material', we'll find $[7]^{-1}$ by multiplying $[7]$ with known quantities and see if we transform it to $[1]$.

$\quad 7 \cdot 3 \equiv 21 \pmod{16}$

$\quad 7 \cdot 3 \equiv 5 \pmod{16}$

$\quad 7 \cdot 3 \cdot 13 \equiv 5 \cdot 13 \pmod{16}$

$\quad 7 \cdot (39) \equiv 1 \pmod{16}$

$\quad 7 \cdot 7 \equiv 1 \pmod{16}$

So $[7]^{-1} = 7$.

Here are the remaining invertible elements,

$\quad \large [9]^{-1} = \bigr([3]^{2}\bigr)^{-1}  = \bigr([3]^{-1}\bigr)^2 = [11^2] = [121] = [9]$

$\quad \large [15]^{-1} = [3\cdot5]^{-1} = [3^{-1} \cdot 5^{-1}] = [11 \cdot 13] = [15]$

(there is an easier way to find $[15]^{-1}$)

$\quad \large [1]^{-1} = [1]$

So $\mathbb{Z_{16}}$ has $8 = \phi(16)$ invertible elements.


问:
What model do you get from PA without addition and multiplication?
I have the feeling that this question is trivial, but I cannot figure the answer by myself nor from the stuff I have read. So the question is if addition (and multiplication) can be shown as a theorem of PA (without the arithmetical axioms but with successor and induction). If not they are independent and  first order PA+¬A has a model (A=addition + multiplication axioms). What does a model of such system looks like?</p>

答:
I assume you mean a standard first-order PA setup.

So we have a language  $L$ with constant symbol $0$, unary function symbol $S$, and binary function symbols $f$ and $g$ (their common names, when they are written in the middle, are $+$ and $\times$).

Make an $L$-structure by using the non-negative integers as the underlying set, and interpreting $0$ and $S$ in the usual way. We want to interpret $f$ and $g$ so that the usual axioms about addition and multiplication are false. It is awfully easy to make things false. For example, interpret $f(x,y)$ as $x^2+2y^2$, and $g(x,y)$ as $2x^2+y^2$. Almost anything will work.


答:
A complex number $z \neq 0$ can be always rewritten as:

$$z = M e^{i\theta},$$

with $M > 0$ and $\theta \in [0, 2\pi]$.

Recall that $\theta \in [0, 2\pi]$, since $e^{i\theta + 2k\pi} = e^{i\theta}e^{i2k\pi} = e^{i\theta}\cdot 1 = e^{i\theta}$ for any integer $k$.

Moreover, $\text{Arg}(z) = \theta$ by definition.

Now, observe that:

$$z^3 = M^3 e^{i3\theta} = M^3 e^{i \phi},$$

since you can always find an integer $k$ such that $3 \theta = 2k \pi + \phi$ and $\phi \in [0, 2\pi].$ Then:

$$\text{Arg}(z^3) = \phi \equiv 3 \theta ~\text{mod}~2\pi.$$

What does $\phi \equiv 3 \theta ~\text{mod}~2\pi$ mean?

It means that you can always find an integer $k \in \{\ldots, -2, -1, 0, 1, 2 \ldots\}$ such that:

$$3\theta = 2k\pi + \phi.$$

Example.

If $\theta = \frac{5}{6}\pi$, then:

$$3 \theta = \frac{5}{2}\pi = \left(2 + \frac{1}{2}\right)\pi =
2\pi + \frac{1}{2}\pi = 2k\pi + \phi,$$

where $k=1$ and $\phi = \frac{1}{2}\pi.$

Title: How to solve $\arg z^3 =0$ where $z\in\mathbb{C}$?

Body:
Can someone show me how to solve $$\arg z^3 = 0$$
I don't know even how to start because of the zero.


问:
If ${x_n}$ is a Cauchy sequence in a bounded set $S$, then ${f(x_n)}$ is a Cauchy sequence implies that $f$ is uniformly continuous on $S$.
Suppose $f$ is a function and $S$ is a bounded subset of $\mathbb{R}$.  I want to prove that

if ${x_n}$ is a Cauchy sequence in $S$, then ${f(x_n)}$ is a Cauchy sequence

implies that

$f$ is uniformly continuous on $S$.

答:
Let's prove it by contradiction.

If $f$ were not uniformly continuous, then we would have a number $\delta > 0$ and two sequences $x_n$ and $y_n$ such that $|x_n-y_n| \leq \frac{1}{n}$ and $|f(x_n)-f(y_n)| \geq \delta$.

Since $S$ is bounded, we can find a subsequence $x_{n_k}$ with a limit $l$. It's easy to see that also $y_{n_k}$ goes to $l$.

Here comes the tricky part, both $(x_{n_k})_k$ and $(y_{n_k})_k$ are Cauchy sequence, but this does not help us. Is there a "new" sequence that can do that?

Yes, there is. Consider the sequence $x_{n_1},y_{n_1},x_{n_2},y_{n_2},\ldots$

Can you finish from here?


问:
transformation of differential equations
If I have this D.E:

$$y^{''}+P(x)y^{'}+Q(x)y=0$$
which transformation $y \to u(x)$, shall I take to transform the D.E to: $u^{''}+q(x)u=0$?

答:
The answer is yes. Let $y=\rho(x)u(x)$ and then put it in the equation to get
$$ \rho''(x)u(x)+2\rho'(x)u'(x)+\rho(x)u''(x)+P(x)(\rho'(x)u(x)+\rho(x)u'(x))+Q(x)\rho(x)u(x)=0. \tag1 $$
Write (1) as an equation of $u$:
$$ \rho(x)u''(x)+(2\rho'(x)+P(x)\rho(x))u'(x)+(\rho''(x)u(x)+P(x)\rho'(x)+Q(x)\rho(x))u(x)=0. \tag2$$
Letting
$$ 2\rho'(x)+P(x)\rho(x)=0 $$
which implies
$$ \rho(x)=\exp\bigg(-\frac12\int P(x)dx\bigg) $$
and then dividing both sides of (2) by $\rho(x)$, one has
$$ u''(x)+q(x)u(x)=0 $$
for some $q(x)$ depending on $P(x),Q(x)$.


问:
Simplication from e to sin
In a lecture, I am seeing this simplification:

$$\frac{-2 \pi i}{4} \cdot (e^{\pi i/4} - e^{-\pi i/4})$$

$$= - \frac{2 \pi i }{4} \cdot 2 i \sin{\frac{\pi}{4}}$$

Why is this?

Things I know:

$e^{\pi i} = -1$ via this proof

Can someone show me how this is done?

Title: Simplication from e to sin

Body:
Using the fact that $e^{ix}=\cos(x)+i\sin(x)$:
$$\frac{-2 \pi i}{4} \cdot (e^{\pi i/4} - e^{-\pi i/4})=$$
$$= \frac{-2 \pi i}{4} \cdot \biggl(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})-\cos(-\frac{\pi}{4})-i\sin(-\frac{\pi}{4})\biggr)=$$
$$=\frac{-2\pi i}{4}\biggl(i\sin(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\biggr)=$$
$$= - \frac{2 \pi i }{4} \cdot 2 i \sin{\frac{\pi}{4}}$$


答:
From $e^{ix}$ and $e^{-ix}$,

$$e^{ix} = \cos x + i \sin x;\quad e^{-ix} = \cos x - i \sin x$$

Rearrange them to make $\cos x$ and $\sin x$ the left hand side: (a bit like solving linear equations)

$$\cos x = \frac{e^{ix} + e^{-ix}}{2};\quad \sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

The result $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ is what's being used here.

Title: Simplication from e to sin

Body:
In a lecture, I am seeing this simplification:

$$\frac{-2 \pi i}{4} \cdot (e^{\pi i/4} - e^{-\pi i/4})$$

$$= - \frac{2 \pi i }{4} \cdot 2 i \sin{\frac{\pi}{4}}$$

Why is this?

Things I know:

$e^{\pi i} = -1$ via this proof

Can someone show me how this is done?


问:
Result of multiplying a scaling matrix with a rotation matrix
I don't understand why if you multiply a scaling matrix with rotation matrix that the resulting matrix, when applied to a shape like an ellipse, only gets scaled and does not get rotated.

$$\begin{bmatrix}
3.7 & 0 \\
0 & 2.1
\end{bmatrix}
\times
\begin{bmatrix}
0.9510 & -.3090 \\
0.3090 & 0.9510
\end{bmatrix}$$

If you do the matrix multiplication the opposite way it does do both when the resulting matrix is applied to an ellipse.

答: The only way you could be "right" is for a rotation with  $\theta=n\pi/2$ ($n \in \mathbb{Z})$. If $n$ is odd you need a scaling that "reverses" the shape of the ellipse, in such a way that the major axis becomes the minor axis, and the minor axis becomes the major axis, with the same proportions as the original ellipse.</p>


问:
The Product of Power Series
I understand the Cauchy product for power series, but we have slightly different notation here. Suppose we have the following power series:

$$f(x) = \sum_{n=0}^\infty a_nx^n$$
$$g(x) = \sum_{m=0}^\infty b_mx^m$$

Their product is written as follows:
$$g(x) \times f(x) = \sum_{n=0}^\infty \sum_{m=0}^\infty  a_n b_{m}x^{n+m}$$
$$= \sum_{k=0}^\infty x^k (\sum_{n=0}^k a_n b_{k-n})$$

where, $k=n+m, m=k-n, 0\le n \le k$.

Could you please tell me how we got $$g(x) \times f(x) = \sum_{k=0}^\infty x^k (\sum_{n=0}^k a_n b_{k-n})$$

答:
If you have
$$g(x) \times f(x) = \sum_{n=0}^\infty \sum_{m=0}^{\infty}  a_n b_{m}x^{n+m}$$
then if you set $k=n+m$, the coefficient for the $k$ degree term is
$\sum_{n=0}^ka_nb_{k-n}$ becouse you have to go trough all elements of the sums that combined give $k$.

So you have that with one sum it is the seires that has that as coefficient to the $k$ term, which means:
$$g(x) \times f(x) = \sum_{k=0}^\infty(\sum_{n=0}^k a_n b_{k-n}) x^k $$
which is what you wrote as the answer.

Title: The Product of Power Series


问:
show by induction that $4^{2n+1}+3^{n+2}$ is divisible by $13$
i'm stuck in this problem, anyone can give me a hint, please?

"Show by induction that $4^{2n+1} + 3^{n+2}$ is divisible by $13$

In my attempt i got the expression

$$4^{2}\cdot 4^{2n+1} + 3\cdot 3^{n+2}$$

but i can't go more, any suggestion is thankful.

Body:
\begin{eqnarray*}
4^{2}\cdot 4^{2n+1} + 3\cdot 3^{n+2}= 16(4^{2n+1}+3^{n+2}) -13 \times 3^{n+2}.
\end{eqnarray*}


问:
Logarithmic Function With Range [0,1] for domain between [0,n]
I'm looking for a function that has logarithmic like behavior for a set of input ranging from [0,n]. The range of the output should be [0,1].

$\lim_{x\to 0} = 0$

$\lim_{x\to n} = 1$

Basically, I'm trying to have values near n be very close to 1, and slowly fall off towards zero as input gets closer to 0. Value for $n/2$ for example would be greater than $1/2$.

答: The function $f(x) = \displaystyle\frac{\log(x+1)}{\log(n+1)}$ should do the trick.


问:
Inverse Laplace transformation of $\frac{s^2}{(s^2+1)^2}$
I am a bit lost how to compute the inverse Laplace transformation of $$ \frac {s^2}{(s^2+1)^2}$$ I think it will be some combination of sine and cosine (oscillation-like), however I ran into some weird expressions when I tried to expand the expression as $$ \frac {s}{(s^2+1)}  \frac {s}{(s^2+1)}$$

Could someone please give some suggestions? Thanks in advance!

答:
The original problem was for $\frac{s^2}{(s^2 + 1)^2}$

Consider the Laplace transforms of $\sin(a t)$ and $t \, \cos(a t)$ which are
\begin{align}
\sin(a t) &\doteqdot \frac{a}{s^2 + a^2} = \frac{a \, s^2 + a^3}{(s^2 + a^2)^{2}} \\
t \, \cos(a t) &\doteqdot \frac{s^2 - a^2}{(s^2 + a^2)^{2}}.
\end{align}
Now one can notice that
$$ \frac{1}{2 a} \, \left( \sin(a t) + a \, t \, \cos(a t) \right) \doteqdot \frac{s^2}{(s^2 + a^2)^{2}}. $$

A second method is to use the convolution theorem,
$$ \bar{f}(s) \, \bar{g}(s) \doteqdot \int_{0}^{t} f(t-u) \, g(u) \, du $$
where $\bar{f}(s)$ is the transform of $f(t)$. Since
$$ \cos(a t) \doteqdot \frac{s}{s^2 + a^2}$$
then
\begin{align}
\frac{s^2}{(s^2 + a^2)^2} &\doteqdot \int_{0}^{t} \cos(a u) \, \cos(at - au) \, du \\
&\doteqdot \left[\frac{2 a t \, \cos(a t) - \sin(a (t - 2 u))}{4 a} \right]_{0}^{t} \\
&\doteqdot \frac{1}{2 a} \, \left( \sin(a t) + a \, t \, \cos(a t) \right).
\end{align}

Modified form $\frac{s^3}{(s^2+1)^2}$

By using :
\begin{align}
\cos(a t) - a t \, \sin(a t) &\doteqdot \frac{s \, (s^2 - a^2)}{(s^2 + a^2)^2} \\
\cos(a t) + a t \, \sin(a t) &\doteqdot \frac{s \, (s^2 + 3 a^2)}{(s^2 + a^2)^2}
\end{align}
then it can be shown that
$$ \frac{1}{2} \, ( 2 \, \cos(a t) - a \, t \, \sin(a t) ) \doteqdot \frac{s^3}{(s^2 + a^2)^2}. $$


问:
Projection topology
Let $(\prod_{i\in \mathcal{I}}X_{i},\tau)$ a topological space with some topology $\tau$. Consider the map $\pi_{i}:\prod_{i\in\mathcal{I}}X_{i}\longrightarrow X_{i}$. Can we consider a topology $\tau_{i}$ for the space $X_{i}$ induced by the topology $\tau$ and the projection? Or in other word, can we construct that $\tau_{i}$?

答:
You need to be somewhat careful with your indices. Let $X:= \prod_{i\in I}X_i$ and

$$\pi_j\colon X \mapsto X_j,\ \text{for}\ j\in I$$

You can always induce the final topology $\tau_j$ on $X_j$ via

$$\tau_j = \{U\subset X_j \mid \pi^{-1}(U) \in \tau\}$$

this is the coarsest topology on $X_j$ for which $\pi$ is continuous.

Title: Projection topology


问:
If $p$, $q$ are two distinct prime numbers in $\Bbb N$ then $\Bbb Q(\sqrt p, \sqrt q)) = \Bbb Q(\sqrt p + \sqrt q)$.
If $p$, $q$ are two distinct prime numbers in $\Bbb N$ then

$$\Bbb Q(\sqrt p, \sqrt q) = \Bbb Q(\sqrt p + \sqrt q)$$

How to prove that the first is included in the second, and why is it ligical to assume the equality

答:
Let $a=\sqrt p$ and $b=\sqrt q$.

Let $c=(a^2+3b^2)a+(3a^2+b^2)b.$ Then $c=a^3+3a^2b+3ab^2+b^3=(a+b)^3\in \Bbb Q(a+b).$

Let $d=(a^2+3b^2)(a+b).$ Then $d=(p+3q)(a+b)\in \Bbb Q(a+b).$

So $2(p-q)b=2(a^2-b^2)b=c-d\in \Bbb Q(a+b).$

So $\sqrt q=b\in \Bbb Q(a+b)=\Bbb Q(\sqrt p +\sqrt q).$

The rest is up to you.


问:
How to show that $\det(A+I)\ne 0$
How to show that for any skew symmetric real matrix $A$, we have $\det(A+I)\ne 0?$

Where to begin? I'm looking for some clue only.

答:
When I learned how to do proofs (I am still learning :) ) one thing that felt true was to pay very close attentions to definitions...
so my hints would be...

1). what is the definition of non-zero skew symmetric real martix A.

2)  what does adding identity matrix do to A.

3). what does $det(A + I) \neq 0$ really mean....think invertibility, which will bring us full circle back to Joseph G's help.

Happy Hunting

Title: How to show that $\det(A+I)\ne 0$


问:
Markov chain (six-sided die)
For the following question asked by someone before, I just wonder why $\mathbb {P}\left(X_{n+1}=j\vert X_{n}=i\right)=\frac {1}{6}$ when $j>i$. I think the answer should depend on $i$, which I think the answer should be $1-\frac {i}{6}$. Thanks.

is the largest number $X_n$ shown up to the nth roll a Markov chain?

答:
Suppose for example that the largest roll after $n$ attempts is $X_n=4$

If the next roll is any of $1,2,3,4$, so with probability $\frac46$, you get $X_{n+1}=4$ as the largest roll after $n+1$ attempts

If the next roll is $5$, so with probability $\frac16$, you get $X_{n+1}=5$ as the largest roll after $n+1$ attempts

If the next roll is $6$, so with probability $\frac16$, you get $X_{n+1}=6$ as the largest roll after $n+1$ attempts

and this confirms the assertion for $i=4$ and $j > i$.

You get similar results for other values of $i$.

Title: Markov chain (six-sided die)


问:
Why is it enough to test only the words up to length n ( number of state) in the given algorithm for a decidability problem
My question is related to the problem below, basically it's a decidability problem and the algorithm prooves it's decidable.

My question :

After reading the step 2 of the algorithm below, why is it enough to only test the words up to n (number of states) in the given algorithm.

It seems to have a link with the DFA being minimal but i can't put my hand on the reason why.

I drew the automaton, since there's two states, why would we only need to test the word 01  and not 000001111 ? What is the reason behind this.

Thanks a lot for your help.

The context of the problem

Given the language $K$ = $\{<M>: M$ is a finite automaton on the alphabet {0,1}) and $L(M)$ contains at least one word of the form $0^k1^l$ with $k,l\geq 0$}. In other words, describe an algorithm with the entry $M$ ( a finite automaton)  and that decide after a finite number of steps if $M$ accept at least one word of the form described above.

The algorithm for the question

Input : < M >

Convert the given FA to a minimal DFA

If minimal DFA contains n states, then generate all strings of language from 01 of length up to n.

3.Run the strings on a simulated DFA, if any one of the string is accepted then FA ∈ L. Stop and accept.

Else, stop and reject.

答:
If a finite-state automaton accepts some input, then it accepts some input without entering the same state twice. (Because you can just delete the part of the string between two entries to the same state and the FA will still accept what's left.)

Consequently, a FA which accepts some string also accepts some string whose length is at most the number of states in the FA.


问:
Problem in limits formula for sequences
Consider the following formulae applicable to limits of sequences:-

$$\lim_{n\to\infty}{a_n}^{p} = [\lim_{n\to\infty}{a_n}]^p \qquad $$

The above is given across all texts in the above format.This is also given https://en.wikipedia.org/wiki/Limit_of_a_sequence#Properties here as it is .

Just wanted to ask the community that in the above limit , is there a assumed notion that the limit of $a_n$ is positive ?

It can be possible that $a_n$ < 0 and p = $\frac{1}{2}$ , in which case it is invalid.

答: What's actually going on here is that limits commute with continuous functions: in other words, $$\lim_{n\to\infty}f(a_n) = f(\lim_{n\to\infty}a_n).$$ This implicitly assumes that $f$ is continuous at the limit; the function in question here is $f(x) = x^p$. In your final sentence, this is not the case.

Title: Problem in limits formula for sequences


问:
A "Dilation" with a Complex Number
I have a mapping $w(z) = 2iz$, where $z \in \{z\in\mathbb{C} : x > 0, |y| < 1 \}$. My goal is to find the image set under this mapping. At first, I thought that this was simply a dilation, but then I realized that under the definition of a dilation that I'm working with, the constant must be a positive real number (which it is clearly not in this case, since $2i$ is a complex number). So this mapping is not a dilation. Can I simply consider $z = x + yi$ and then do $2iz = 2i(x + yi) = 2ix - 2y$ to determine my image set, or there something else I need to do instead?

Title: A "Dilation" with a Complex Number

Body:
${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

$$w = a+ib = 2iz = -2y+2ix$$

$$|y|<1 \Rightarrow |a|<2, \quad x>0 \Rightarrow b>0.$$


问:
Difference between "there exists at most one" and "there exists only one"
What is the difference in proof of this type?
I know that for "there exists only one" existence
and uniqueness must be demonstrated but for
"there exists at most one" I don't understand very well.</p>

答: "there exists only one"  means "there exists (exactly) one". Whereas "there exists at most one" means either "there exists (exactly) one" or "there exists (exactly) none". So the statement, "there exists at most one integer between $0.5$ and $0.6$" is True, whereas the statement, "there exists only one integer between $0.5$ and $0.6$" is False.


问:
Questions on Bolzano-Weierstrass theorem
Here is how it is worded in Apostol:

Theorem. If a bounded set $S$ in $\mathbb{R}^n$ contains infinitely many points, then there is at least one point in $\mathbb{R}^n$ which is an accumulation point of $S$.

Proof. (for $\mathbb{R}^1$) Since $S$ is bounded, it lies in some interval $[-a, a]$. At least one of the subintervals $[-a, 0]$ or $[0, a]$ contains an infinite subset of $S$. Call one such subinterval $[a_1, b_1]$. Bisect $[a_1, b_1]$ and obtain a subinterval $[a_2, b_2]$ containing an infinite subset of $S$, and continue this process. In this way a countable collection of intervals is obtained, the $n$th interval $[a_n, b_n]$ being of length $b_n -a_n = a/2^{n-1}$. Clearly the sup of the left endpoints $a_n$ and the inf of the right endpoints $b_n$ must be equal, say to $x$. The point $x$ will be an accumulation point of $S$ because, if $r$ is any positive number, the interval $[a_n, b_n]$ will be contained in $B(x;r)$ as soon as $n$ is large enough so that $b_n -a_n < r/2$. The interval $B(x;r)$ contains a point of $S$ distinct from $x$ and hence $x$ is an accumulation point of $S$.

My questions:

1) Why bother with the first 3 sentences? What's wrong with just saying "Since $S$ is bounded, there is an interval $[a_1, b_1]$ containing an infinite subset of $S$". Actually, now I'm not entirely sure why it has to be bounded. Couldn't you always find some infinite subset to do this with?

2) Why does $b_n -a_n < r/2$? I would think it just has to be less than $r$. For example, these intervals could just "converge" to one end point ($x = a_1$ or $x = b_1$), in which case the length just has to be less than the radius of the ball (neighborhood).

3) I'm not entirely sure why we kept bisecting the intervals. I'm confused, but I don't really have a good concrete question.

答:
We need to start somewhere. So we start by noting $S\subset [-a,a]$ for some $a$. After we do this, we keep bisecting the interval ad infinitum. $[a_1,b_1]$ is just the result of the first bisection. You'll see why the interval needs to be bounded soon.

$r$ here is just any positive number. Every time you bisect an interval, its length halves. $[a_n,b_n]$ stands for the interval you get after bisecting $n$ times, so its length is $\frac{2a}{2^n}$. Now, the point is that by increasing $n$, you can get this number to be as small as you want, since it converges to $0$. In particular, for large enough $n$, this length is less than $r/2$. The author then uses this to show that $[a_n,b_n]$ is contained in $B(x;r)$.
(this also explains why the interval $[-a,a]$ we start out with should be bounded. Because we want its size to converge to $0$ when we bisect it; this wouldn't happen if we started with an infinite interval).

The idea is: if we keep cutting the interval in half, each time choosing where we cut it so that it contains infinitely many points of $S$, then, as mentioned in 2., the interval's size will get smaller and smaller, so it will "zero in" on a single point. We call this point $x$, and this point is the accumulation point whose existence we set out to prove. It's intuitively clear why it's an accumulation point of $S$: because the intervals $[a_n,b_n]$ are smaller and smaller sets containing $x$, but also containing infinitely many points of $S$.


Title: Inverse Laplace transformation of $\frac{s^2}{(s^2+1)^2}$

Body:
I am a bit lost how to compute the inverse Laplace transformation of $$ \frac {s^2}{(s^2+1)^2}$$ I think it will be some combination of sine and cosine (oscillation-like), however I ran into some weird expressions when I tried to expand the expression as $$ \frac {s}{(s^2+1)}  \frac {s}{(s^2+1)}$$

Could someone please give some suggestions? Thanks in advance!

Body:
$$
F(s)=\frac{s^2}{(s^2 + 1)^2} = \frac{s^2 + 1}{(s^2 + 1)^2} - \frac{1}{(s^2 + 1)^2}$$

$$
F(s)= \frac 1{s^2+1} +\dfrac {1}{2s}\dfrac {d}{ds}\frac 1{(s^2 + 1)}.
$$
Apply inverse Laplace Transform:
$$f(t)=\sin t -\dfrac 12 \int_0^t 1  \times \tau \sin \tau \ d\tau$$
$$f(t)=\dfrac 12 (\sin t +t \cos t)$$


问:
Discrete Definition of a Limit
My professor defines a limit the following way:

Let $f: \mathbb N \rightarrow \mathbb R$ be a sequence and $L\in\mathbb R$. $f$ converges to $L$ if $$\forall \epsilon \in \mathbb R^+ \exists N\in\mathbb N\forall n\in \mathbb N(n>N \rightarrow |f_n-L|<\epsilon)$$

Which is the dramatic way of saying $$\lim_{n\rightarrow \infty}f_n=L$$

Can someone help me unpack this definition and how I can apply it like one applies the typical epsilon-delta definition of a limit?

答:
This is the "epsilon-delta" style defniition for limits at infinity; the only difference being that the bound variable (i.e. $n$) ranges over the natural numbers instead of over the real numbers. It is, other than that one change, the same definition as for the value of

$$\lim_{x \rightarrow \infty} f(x)$$

for $f: \mathbb{R} \rightarrow D$, $D \subseteq \mathbb{R}$, and so $x \in \mathbb{R}$.

Intuitively, the terms involved have the same meaning: $\epsilon$ is the tolerance, and $N$ is the lower bound on the input on $f$ needed to approximate $L$ within that desired tolerance.

Insofar as how you use it, you should use it in the same way that you use the other. If you want to prove that, say,

$$\lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

you should ask yourself: "if I want to approximate $0$ to within $\epsilon$ - say, +/-0.001 - by plugging something in for $n$, how big must I make that $n$ to cause that to happen?" And you can easily see that if $N = \frac{1}{\epsilon}$, that that will provide the necessary tolerance, because if

$$n > \frac{1}{\epsilon} > 0$$

then

$$0 < \frac{1}{n} < \epsilon$$

by virtue of the elementary fact that if

$$x > y$$

then

$$\frac{1}{x} < \frac{1}{y}$$

for positive reals $x$ and $y$, and also that $\frac{1}{x}$ has the same sign as $x$ for any real $x \ne 0$.

Title: Discrete Definition of a Limit


问:
$\mathbb{Q}(\pi^2-\pi) \subset\mathbb{Q}({\pi})??$
I have a "simple" question about field extension, let me say, $\mathbb{Q}(\pi^2-\pi) \subsetneq\mathbb{Q}({\pi})??$

I know that $\pi^2-\pi \in \mathbb{Q}(\pi)$ so we have $\mathbb{Q}(\pi^2-\pi) \subset\mathbb{Q}({\pi})$, but why $\pi \notin \mathbb{Q}(\pi^2-\pi)$?

I've tried this way: If $\pi \in \mathbb{Q}(\pi^2-\pi)$ then $\mathbb{Q}(\pi^2) \subset \mathbb{Q}(\pi^2-\pi),  $ but this is ok, since $\mathbb{Q}(\pi^2) \subset \mathbb{Q}(\pi)$.

Another way is: If $\pi \in \mathbb{Q}(\pi^2-\pi) $ then $\pi = \frac {p(\pi^2 - \pi)}{q(\pi^2-\pi)}$, what is the contradiction?

I cannot see the contradiction, but my professor told me that $[\mathbb{Q}(\pi): \mathbb{Q}(\pi^2-\pi)]=2$ (I have no ideia how to prove that).

答:
Assume by contradiction that $\pi = \frac {p(\pi^2 - \pi)}{q(\pi^2-\pi)}$

Then $\pi$ is a root of $R(X)=XQ(X^2-X)-P(X^2-X) \mathbb Q[X]$. Since $\pi$ is transcendental over $\mathbb Q$ you get that $R=0$. Therefore
$$
XQ(X^2-X)=P(X^2-X)
$$
But this is not possible since the left-hand side has odd degree while the right-hand side has even degree.


答:
Indeed, a certain assumption has to be made to ensure that this property (as well as the other properties listed on that Wikipedia page) is true. And this assumption is stated there immediately before the list of properties:

Some other important properties of limits of real sequences include the following (provided, in each equation below, that the limits on the right exist).

I added the boldface to emphasize the important part.

In other words, what the property says: if both limits exist, then they are equal to each other. But existence of both must be established first. In general, existence of one of them does NOT imply existence of the other.

Also, I intentionally said "in general" above. To make perfectly rigorous statements, we can look more closely at a property and consider and state various cases that make existence implications one way or the other true. For example, look at another property from that list:
$$\lim_{n\to\infty}(a_n\cdot b_n)=\left(\lim_{n\to\infty}a_n\right)\cdot\left(\lim_{n\to\infty}b_n\right).$$
If both limits on the right exist, then the limit on the left exists and the equality holds. But if know only that the limit on the left exists, we can NOT deduce whether the two limits on the right exist.

Title: Problem in limits formula for sequences

Body:
Consider the following formulae applicable to limits of sequences:-

$$\lim_{n\to\infty}{a_n}^{p} = [\lim_{n\to\infty}{a_n}]^p \qquad $$

The above is given across all texts in the above format.This is also given https://en.wikipedia.org/wiki/Limit_of_a_sequence#Properties here as it is .

Just wanted to ask the community that in the above limit , is there a assumed notion that the limit of $a_n$ is positive ?

It can be possible that $a_n$ < 0 and p = $\frac{1}{2}$ , in which case it is invalid.


问:
Dual cone and polar cone
Dual cone and polar cone http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone are defined only on $\mathbb R^n$. Has anyone seen the extension to $\mathbb C^n$? Any references for these?</p>

答: As I see it, the definition in Wikipedia is for any Linear Space. The book by Aliprantis and Border defines polars for any Dual Set, which is even more general. These general definitions are very useful in Convex Optimization, but I've never seen it used on $\mathbb{C}^n$.</p>

Title: Dual cone and polar cone


问:
When exactly is the character space of a Banach algebra empty?
It is well-known that the character space, (i.e. the set of multiplicative characters) of a commutative, unital Banach algebra is non-empty.

But is there a complete characterization of when exactly the character space of Banach algebra is empty? Or even an incomplete characterization (ie certain classes of Banach algebras which we know have empty character space)?

Edit: I’d appreciate any responses, especially those with links for further reading!

答:
Given a Banach algebra $B$ let $[B, B]$ (the commutator ideal) be the intersection of all kernels of morphisms of Banach algebras $B \to A$ where $A$ is commutative. This is a closed two-sided ideal of $B$, and the quotient $B/[B, B]$ is the abelianization of $B$; the universal commutative Banach algebra to which $B$ maps.

(Probably $[B, B]$ is the closure of the ideal generated by commutators $[b_1, b_2]$ but this construction of it manifestly has the appropriate universal property so it doesn't matter too much either way.)

Any character of $B$ must be a character of its abelianization, so the problem for arbitrary Banach algebras reduces immediately to the commutative case. A nonzero commutative Banach algebra always admits a character (by the Gelfand-Mazur theorem), so we get:

Proposition: $B$ admits a character iff $B/[B, B]$ is nonzero.

It's worth noting that $B/[B, B]$ can in fact be zero, for example if $B$ is simple and noncommutative (e.g. $B = M_n(\mathbb{C})$) as noted in the comments.


问:
Sequences with two other sequences
Let $(z_n) \subset \mathbb{R}$ be a sequence defined, for each $n \in \mathbb{N}$, by:

$$z_n=min\{x_n,y_n\}$$

With $(x_n)$ and $(y_n)$ being two convergent sequences in $\mathbb{R}$ with $\lim{x_n}=a$ and $\lim{y_n}=b$.

Show that $(z_n)$ is also convergent and calculate the limit.

First I've separated in two cases: $a=b$ and $a \not= b$.
For the first one, we have:

$\forall \varepsilon > 0, \exists n_0 \in \mathbb{N} ; n > n_0 \Rightarrow |x_n - a|<\varepsilon$

$\forall \varepsilon > 0, \exists n_1 \in \mathbb{N} ; n > n_1 \Rightarrow |y_n - a|<\varepsilon$

So, if $n_\delta = max\{n_0,n_1\}$, we have:
$$n > n_\delta \Rightarrow |z_n-a|<\varepsilon$$
By the defnition of $z_n$ sequence. (Right?)

But what about the second case? I can't come up with anything. Leads?

答: Hint. Taking $\lim{x_n}=a $ and  $\lim{y_n}=b$ for second case let assume, that $a<b$. This means that $\exists N \in \mathbb{N}$ such that for $n>N$ $z_n=\min\{x_n,y_n\} = x_n$.

Title: Sequences with two other sequences


问:
Why is a gradient not a linear functional?
I'm reading this book and on page 104 they define:

Afterwards they said that if $m=1$, the function $f$ is real-valued and $T$ is the gradient which has to be a linear functional according to the definition.

The problem is this is not true in general, see for example:

$f:\mathbb R^2\to \mathbb R$ defined as $f(x,y)=\sin x$. The gradient is $\nabla f=(\cos x,0)$ which is not a linear functional.

答:
Technically, the gradient at each point should be a linear functional for just this reason. What that means is that, if you think about it from a linear algebra perspective, is that the gradient is a covector, or dual vector, or row vector, i.e. it inhabits $(\mathbb{R}^n)^{*}$, because such a linear functional is the definition of a covector.

You can "check" this by thinking about how the gradient transforms when the input to $f$ is subject to a coordinate transformation. Namely, because of the chain rule, the gradient's components transform covariantly, instead of contravariantly; thus it must be a covector. In a sense, your elementary Calc text lied to you when it called this a vector (but not entirely - covectors are vectors, they just live in a different vector space than the things that are commonly called "vectors" at that stage. That's the "lie". However, the two spaces are isomorphic, so you can "ignore" this in a sense ... until you can't, as here.).


答: Hint: If $a \neq b$, without loss of generality, assume $a < b$. So, as usual, there are $n_1, n_2 \in \Bbb N$ such that $\Bbb N \ni n > n_1$ implies $|x_n - a| < \frac{b - a}{2} \implies x_n - a < \frac{b - a}{2}$ and $\Bbb N \ni n > n_2$ implies $|y_n - b| < \frac{b - a}{2} \implies -\frac{b - a}{2} < y_n - b$. So what can you say about $x_n$ and $y_n$ for all $\Bbb N\ \ni n > \max\{n_1, n_2\}$, after some rearrangement?

Title: Sequences with two other sequences

Body:
Let $(z_n) \subset \mathbb{R}$ be a sequence defined, for each $n \in \mathbb{N}$, by:

$$z_n=min\{x_n,y_n\}$$

With $(x_n)$ and $(y_n)$ being two convergent sequences in $\mathbb{R}$ with $\lim{x_n}=a$ and $\lim{y_n}=b$.

Show that $(z_n)$ is also convergent and calculate the limit.

First I've separated in two cases: $a=b$ and $a \not= b$.
For the first one, we have:

$\forall \varepsilon > 0, \exists n_0 \in \mathbb{N} ; n > n_0 \Rightarrow |x_n - a|<\varepsilon$

$\forall \varepsilon > 0, \exists n_1 \in \mathbb{N} ; n > n_1 \Rightarrow |y_n - a|<\varepsilon$

So, if $n_\delta = max\{n_0,n_1\}$, we have:
$$n > n_\delta \Rightarrow |z_n-a|<\varepsilon$$
By the defnition of $z_n$ sequence. (Right?)

But what about the second case? I can't come up with anything. Leads?


问:
How to find the SVD of a matrix with one zero eigenvalue?
I am trying to find the SVD of the matrix \begin{pmatrix}1&2\\3&6\end{pmatrix}

$A^TA=\begin{pmatrix}1&3\\2&6\end{pmatrix}\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}10&20\\20&40\end{pmatrix}$

$\lambda_1=50$

$\lambda_2=0$

$\lambda_1>\lambda_2>=0$

$\Sigma=\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}$

When $\lambda_1=50$:

$v_1=\begin{pmatrix}1\\2\end{pmatrix}$

Normalize:

$u_1=\begin{pmatrix}\frac{1}{\sqrt{5}}\\\frac{2}{\sqrt{5}}\end{pmatrix}$

When $\lambda_1=0$:

$v_2=\begin{pmatrix}-2\\1\end{pmatrix}$

Normalize:

$u_2=\begin{pmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{pmatrix}$

$V=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{-2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$V^T=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$A=U\Sigma V^T$

$\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}u_1&u_2\\u_3&u_4\end{pmatrix}\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}5\sqrt{2}u_1&0\\5\sqrt{2}u_3&0\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}\sqrt{10}u_1&2\sqrt{10}u_1\\\sqrt{10}u_3&2\sqrt{10}u_3\end{pmatrix}$

$u_1=\frac{1}{\sqrt{10}}$

$u_3=\frac{3}{\sqrt{10}}$

How can I find $u_2$ and $u_4$?

答: You have correctly noticed that the second column of $U$ doesn't really matter in terms of satisfying $A=U\Sigma V^\top$. In order to be a proper SVD, you just need to find a vector orthonormal to the first column of $U$ to make $U$ an orthogonal matrix (but again, this is not necessary for the equality $A = U \Sigma V^\top$ to hold).


问:
Partial Derivative Calculation Help
Question: Use the $\newcommand{\p}{\partial}$chain rule to find $\p z/\p s$ when $(s, t) = (1, 2)$ for the equation:
$$z = \arctan(x^2 + y^2)$$
provided that $x = st$ and $y = t^2$.

My attempted Solution: For the first step, I converted $\p z/\p s$ into:
$$\frac{{\p z}}{ds} = \frac{{\p z}}{\p x} \cdot\frac{{\p x}}{\p s} + \frac{{\p z}}{\p y} \cdot \frac{{\p y}}{\p s}$$
This would allow me to calculate the partial derivative of each section and put them together. However, I'm not very sure how the calculations are done in this case.

I would be much appreciated if someone could show me how this is done.

答:
First of all, all derivatives here are partial derivatives, since we're working with functions of several variables. What you set up for the multivariate Chain Rule is essentially correct, except for the notation. Instead, it must be:
$$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial s}.$$
Then you just need to calculate the four partial derivative that appear in this expression and put them in there for the final answer. Here are a couple of them:

$\displaystyle \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}\left[\arctan(x^2+y^2)\right]=\frac{1}{1+(x^2+y^2)^2}\cdot2x=\frac{2x}{1+(x^2+y^2)^2}$, using the "usual" Chain Rule;

$\displaystyle \frac{\partial x}{\partial s}=\frac{\partial}{\partial s}\left[st\right]=t$;

and similarly you can find the rest.

Title: Partial Derivative Calculation Help


问:
Evaluating limits involving absolute values: $\lim_{x\to\pm6} \frac{2x+12}{|x+6|}$
I have two questions regarding limits involving absolute values. How do I evaluate the following:

$\displaystyle\lim_{x\to -6} \frac{2x+12}{|x+6|}$

$ \displaystyle \lim_{x\to 6} \frac{2x+12}{|x+6|}$.

To handle the first problem, I considered two cases: $x\gt -6$ and $x\lt -6$ and thus get $2$ and $-2$ respectively for the right and left hand limits.

However, I am bit confused regarding the second problem. How do I handle that?

Body:
As $\ x \to 6^+, \ |x+6| \to 12^+ > 0$, so we can replace $|x + 6|$ with $ x+ 6$

As $\ x\to 6^-,\  |x+6| \to 12^- > 0,$ so we can replace $|x+ 6|$ with $x + 6$:

$$ \lim_{\large x\to 6^+} \ \frac{2x+12}{|x+6|} \quad =\quad \lim_{\large x \to 6^-} \ \frac{2x+12}{|x + 6|} \quad= \quad \frac{2(x+6)}{x+ 6} \quad= \quad2$$


问:
Interpretation of complex trilinear coordinates
The point $X_{5374}$ in the Encyclopedia of Triangle Centres has trilinear coordinates
$$\sqrt{\cot A}:\sqrt{\cot B}:\sqrt{\cot C}$$
If the reference triangle is obtuse, one (and only one) of these coordinates becomes a non-real complex number by virtue of taking the square root of a negative number. There are a few other ETC points that may have complex coordinates, like $X_{5000}$; the new triangle centre I asked about with trilinears $\frac1{\sqrt{a\cos A}}:\frac1{\sqrt{b\cos B}}:\frac1{\sqrt{c\cos C}}$ falls in the same category.

Trilinear coordinates, by definition, are the ratio of directed distances from a point to the triangle edges, but this interpretation only works if all coordinates are real. If a point has complex trilinear coordinates, how can I interpret it? A geometric interpretation would be preferred.

答:
This answer will be a brief and superficial survey of references with interpretations of "imaginary" entities in geometry. When doing Euclidean plane geometry you are implicitly working in the complex Euclidean plane.  In the real plane, a line and circle may not intersect, but in the complex plane they will. You can work purely algebraically, but often there will be geometric interpretations.  For example, the intersection of a disjoint circle and line will yield a conjugate pair of imaginary points, but the line through them will be a "real" line that is in fact the original line.  The two imaginary points of intersection of two disjoint circles will define a real line that is the radical axis of the two circles (see this answer).  And the four imaginary points of intersection of two conics will define two lines that generalize the radical axis and also correspond to a degenerate member of the pencil defined by the two conics.

Back to trilinears, Chapter IX of Whitworth's Trilinear Coordinates, 1866 discusses imaginary points and lines.  The URL is set up with a search term that lets you browse through references to imaginary entities.

Synthetic geometers back in the 19th century also had ways of interpreting the imaginary.  Some of this goes back to Poncelet's "Principle of Continuity" which argued that, for example, the intersection of a conic and line didn't abruptly disappear when they became disjoint. For example, a conic a determines on every straight line an involution by means of pairs of collinear conjugate points and when the double points of this involution are real, they are the points of intersection of the line and conic.  By analogy, involutions where the double points are off the line represent pairs of imaginary points.

It seems that von Staudt pushed these ideas the furthest. See Hatton's Theory of the Imaginary in Geometry for much more detail.  Also the last part of Coolidge's Geometry Of The Complex Domain.

Hamilton and Kettle's Graphs and imaginaries may be worth a glance.

I suspect that this may not directly answer your question of how to interpret a complex trilinear coordinate, but these texts from the past at least give an idea of how geometers have tried to work with imaginaries.  Working out how to do it in synthetic geometry was an impressive achievement but probably less fruitful than analytic methods in forging ahead, so this is perhaps an all but forgotten branch of mathematics.


问:
Can't find n (algebraic manipulation)
I‘m trying to understand a proof on my textbook that says $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational, where $n \in \mathbb{N} $. However I can't seem to show that
$$n^2 - 1 = \frac{(\frac{p^2}{q^2} - 2n)^2}{4} \to n =  \frac{(p^4+4q^4)}{4p^2q^2}$$,
where $p,q \in \mathbb{Z}$，and the gcd of $p,q$ is 1. I feel really stupid, is there something I don't see?

The furthest I got is
$$n = \frac{\sqrt{2p^4+5p^4-2p^2q^2}-(p^2-q^2)}{2q^2}$$

Title: Can't find n (algebraic manipulation)

Body:
$$n^2 - 1 = \frac{\left(\dfrac{p^2}{q^2} - 2n\right)^2}{4}= \frac{\dfrac{p^4}{q^4} + 4n^2-4n\dfrac{p^2}{q^2} }{4}=\dfrac {p^4}{4q^4}+n^2-n\dfrac{p^2}{q^2}. $$

Subtract $n^2$ from both sides to get $$-1=\dfrac {p^4}{4q^4} -n\dfrac{p^2}{q^2} ,$$

which is $$n\dfrac{p^2}{q^2}=1+\dfrac {p^4}{4q^4}.$$

Can  you take it from here (solve for $n$)?


问:
Uniform convergence of $\sum_{n=1}^{\infty} (-1)^n \frac{\sin nx}{n^ {\ln n} } $, $ x\in \Bbb{R}$,
$\sum_{n=1}^{\infty} (-1)^n \frac{\sin nx}{n^ {\ln n} } $, $ x\in \Bbb{R}$, converges

A) only for $x=0$

B) uniform only for $x\in [-\pi, \pi] $

C) uniform for {$x \in \Bbb{R} $- {$n\pi:n\in\Bbb{Z}$}}

D) uniform for all $x \in \Bbb{R} $

$\implies$ let $u_n(x) =\sin nx$ . and $v_n(x) = \frac{1}{{n}^{\ln n}} $ then $s_n(x) =\sin x +\sin 2x +... + \sin nx$ = $\frac{{\sin \frac{nx} {2}} \sin \frac{(n+1)x}{2}}{\sin \frac{x} {2}}$
Hence $|s_n(x)|=|{\frac{1}{\sin \frac{x} {2}} |} $. Now if $x=0$ then $|s_n(x)|$ will be unbounded. Hence option A, B, D are false. If we consider C then $|s_n(x) |$ will be bounded there and$v_n(x)$ is a monotone decreasing sequence. So by Dirichlet test U.C.follows.
I request experts to check the above and help me refine my arguments if I am on the right track. Also any independent new ideas are cordially welcome. Thanks in advance.

答:
We can use the Weierstrass M-test to prove the result. For $n\geq 3$ we have that $1 < \ln(3)\leq \ln(n)$ and so $\frac{1}{n^{\ln(n)}} \leq \frac{1}{n^{\ln(3)}}$. Then by $p-$test the series $\sum_{n=3}^\infty \frac{1}{n^{\ln(3)}}$ converges. Since $|\sin(n x)|\leq 1$. Then
$$\sum_{n=1}^\infty\Big|\frac{(-1)^n\sin(nx)}{n^{\ln(n)}}\Big|\leq \sum_{n=1}^\infty \frac{1}{n^{\ln(n)}}\leq 1 + \frac{1}{2^{\ln(2)}}+\sum_{n=1}^\infty \frac{1}{n^{\ln(3)}} < \infty$$
Thus we have uniform convergence for all $x\in\mathbb{R}$.


Title: Can't find n (algebraic manipulation)

Body:
I‘m trying to understand a proof on my textbook that says $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational, where $n \in \mathbb{N} $. However I can't seem to show that
$$n^2 - 1 = \frac{(\frac{p^2}{q^2} - 2n)^2}{4} \to n =  \frac{(p^4+4q^4)}{4p^2q^2}$$,
where $p,q \in \mathbb{Z}$，and the gcd of $p,q$ is 1. I feel really stupid, is there something I don't see?

The furthest I got is
$$n = \frac{\sqrt{2p^4+5p^4-2p^2q^2}-(p^2-q^2)}{2q^2}$$

Body:
If
$$n^2 - 1 = \frac{(\frac{p^2}{q^2} - 2n)^2}{4}$$
then
$$n^2 = \frac{(\frac{p^2}{q^2} - 2n)^2+4q^2}{4}$$
then
$$n^2 = \frac{(p^2 - 2q^2n)^2+4q^2}{4q^2}$$
then
$$4q^2n^2 = (p^2 - 2q^2n)^2+4q^2$$
then
$$4q^2n^2 = 4q^4n^2-4p^2q^2n+p^4+4q^2$$
so
$$0=-4p^2q^2n+p^4+4q^2$$
and finally
$$n=\frac{p^4+4q^4}{4p^2q^2}$$


问:
Error bounds for $e$
Prove that for all $n\in\mathbb{N}_+$, we have $$(1+\frac{1}{n})^n>\sum_{k=0}^n \frac{1}{k!}-\frac{e}{2n}.$$</p>

Title: Error bounds for $e$

Body:
Note that when $x>0$, we have:
$$\log(1+x)>x-\frac{x^2}{2},\quad e^{-x}>1-x.$$
It follows that
$$(1+\frac{1}{n})^n=e^{n\log(1+\frac{1}{n})}>e^{n(\frac{1}{n}-\frac{1}{2n^2})}=e\cdot e^{-\frac{1}{2n}}>e(1-\frac{1}{2n})>\sum_{k=0}^n\frac{1}{k!}-\frac{e}{2n}.$$


问:
$\{f_n\}$ be a sequence of continuous functions converging pointwise to $0$, show $\lim \int_0^1 f_n dx = 0$
This problem comes from Rudin's Real and Complex Analysis and is Exercise 2.10. Let $\{f_n\}$ be a sequence of continuous functions such that $0\leq f_n\leq 1$ and $f_n(x)\to0$ for all $x\in [0,1]$. We need to show that $$\lim_{n\to\infty}\int_0^1 f_n(x)dx = 0$$
In particular I am trying to show this without using Measure Theory or Lebesgue Integration.

What I have tried so far is to consider the sequence of functions $\{g_n\}$ defined by $$g_n(x)= \min\{f_1(x),\ldots f_n(x)\}$$
Then $\{g_n\}$ is a continuous, decreasing sequence of functions which converge pointwise to $0$. Since the sequence of continuous functions decreases to a continuous function, then we may apply Dini's Theorem to know that $g_n \to 0$ uniformly. Hence $$ \lim_{n\to 0}\int_0^1 g_n(x)dx = \int_0^1\lim_{n\to0}g_n(x)dx = 0$$
The only thing left to show is that $$ \lim_{n\to 0}\int_0^1 f_n(x)dx= \lim_{n\to 0}\int_0^1 g_n(x)dx $$
But this is where I am stuck. Again I am avoiding measure theory proofs.

答:
For $\epsilon> 0$ and $n\ge 1$ define
$$A_{n,\epsilon} \colon = \bigcap_{k\ge n} f_k^{-1}([0, \epsilon])$$
then
$$A_{1,\epsilon} \subset A_{2,\epsilon} \subset \ldots$$
and
$$\bigcup_{n} A_{n, \epsilon} = [0,1]$$

Consider a set $A_{n,\epsilon}$. For every $k \ge n$, the function $f_k$ takes values $\le \epsilon$ on $A_{n, \epsilon}$. By continuity, and the compactness of $A_{n,\epsilon}$, there exists a finite union of open intervals $U= U_{k,n, \epsilon}$ containing $A_{n, \epsilon}$ such that $f_k$ takes values $< 2 \epsilon$ on $U$. The complement of $U$ in $[0,1]$ is a finite union $E$ of closed intervals. We have
$$\int_{[0,1]} f_k = \int_{\bar U}f_k + \int_{E}f_k \le 2 \epsilon + m(E)$$

Note that $E$ is an elementary subset (finite union of closed intervals) of $A_{n, \epsilon}^c$, an open subset of $[0,1]$.

Basic fact, proved below: If $U_n$ is a decreasing sequence of open subsets of $[0,1]$ with void intersection, and $E_n$ are elementary subsets of $U_n$ then $m(E_n) \to 0$. Once we prove this, we  have the result.

...................................................

Define for an open subset $U$ of $[0,1]$, $m(U)\colon= \sup m(E)$, where $E$ is an elementary subset of $U$  ( so $m(U)$ is the interior Jordan measure of $U$).

One sees easily that $m$ is monotone: $m(U)\le m(V)$ if $U \subset V$.

For every $U$,  and $\epsilon>0$, there exists $E_{\epsilon}\subset U$ elementary sucht that $m(E_{\epsilon}) > m(U) - \epsilon$. It's clear then that $m(U\backslash E_{\epsilon}) < \epsilon$.

Any elementary subset $E$ of $U\cup V$ is the union of elementary subsets $F$, $G$ of $U$, $V$, with intersections only at the boundary (Lebesgue covering lemma). Therefore, $m(U\cup V) \le m(U) + m(V)$.

Basic statement: If $U_n$ is a decreasing sequence of open subsets of $[0,1]$ with void intersection then $m(U_n) \to 0$.

Indeed, let $\epsilon > 0$. For every $n$ consider $E_n$ elementary such that $m(U_n \backslash E_n) < \epsilon/2^{n+1}$.
Let $E'_n = E_1 \cap \ldots \cap E_n$.  We have
$$m(U_n \backslash E'_n) \le \sum_{k=1}^n m(U_n \backslash E_k) \le \sum_{k=1}^n m(U_k \backslash E_k)< \epsilon$$

Now $E'_n$ are elementary and form a decreasing sequence with empty intersection. By compactness,
there exists $n$ such that $E'_n=\emptyset$. For that $n$ we have $m(U_n)< \epsilon$


问:
intersection and union of two independent sets is independent in matroid
As we know that, a matroid $M$ is a pair $(U,\mathcal{I})$ where $U$ is a finite set and $\mathcal{I}\subseteq \mathcal{P}(U)$ satisfying

$\varnothing \in \mathcal{I}$,

if $Y \in \mathcal{I}$ and $X\subseteq Y$, then $X\in \mathcal{I}$,

$X,Y\in \mathcal{I}$ with $|X|<|Y|$, there exists $y\in Y-X$ such that $X\cup \{y\} \in \mathcal{I}$.

The set $U=U(M)$ is called ground set of $M$ and $\mathcal{I}=\mathcal{I}(M)$ is called the collection of independent sets of $M$.

How to show that intersection and union of two independent set is independent or not? Anybody can help?

答:
Recall the definition of matroid:

The first axiom shows that any matroid is non-empty.

The second axiom is called hereditary, it means that all the subsets of an independent set is also independent.

The third axiom is called independence augmention axiom. Note that there is a important condition that $|X| < |Y|$.

The intersection of two independent sets is independent.

Let $A, B$ are independent set, and $A \cap B$ is the subset of $A$ (or $B$). According the hereditaty axiom, $A \cap B$ is also independent.

The union of two independent sets is not independent in general.

Suppose $(U, \mathcal{I})$ is a maroid.
Let $$U' = \bigcup_{A \in I}A$$
Thus, every element in $U'$ is at least in one independent set.
And $(U', \mathcal{I})$ is also a maroid.

If the union of two independent sets is independent, then the union of all independent sets is independent, too.
It means that $U'$ is an independent set.
Also, we can conclude a strange result:

For $(U', \mathcal{I})$, we have $\mathcal{I} = \mathcal{P}(U')$.

For $(U, \mathcal{I})$, $$X \in \mathcal{I} \Leftrightarrow X \subseteq U'$$
which leads that $(U, \mathcal{I})$ and $(U', \mathcal{I})$ become trivial.

One can easily to construct a countexample, such as $U = \{a, b\}$, $\mathcal{I} = \{ \varnothing, \{a\}, \{b\} \}$ and $U \not\in \mathcal{I}$. Note that since $|\{a\}| = |\{b\}|$, the independence augmention axiom is also holds in this case.


问:
Solve:$x^x=5$ and $x^{(1/x)}=1/5$
The first equation comes from standard log equation:
$\log(x)= \frac{1}{x}$

When base of log is $5$ then it will be reduced to
$x^x=5$

The second equation comes from standard log equation:
$\log(x)=-x$

When base of log is $5$ then it will be reduced to
$x^{1/x}= \frac{1}{5}$

答:
The first equation is
$$x^x=5$$
First, take the natural log of both sides:
$$x\log x=\log5$$
$$\implies e^{\log x} \log x = \log 5$$
Now, take the product log (Lambert W function) of both sides:
$$W(e^{\log x} \log x)=W(\log 5)$$
Which gives
$$\log x=W(\log 5)$$
i.e.,
$$x=e^{W(\log 5)}$$
Similarly, the second can be solved. The solution of the second is $e^{-W(\log5)}$.

Title: Solve:$x^x=5$ and $x^{(1/x)}=1/5$


问:
Finding the vector equation of a plane in 3 space
I'm working on a question that asks me to find the vector equation of the plane in $\mathbb R^3$ that contains the point P(-1,6,0) and is parallel to the plane x+2y-z=5. I'm trying to make sure my answer is right versus the answer that I was provided with, which I think is wrong.

I pulled out the normal vector n = (1,2,-1). I proceeded to find two vectors v and u that are orthogonal to n but are not parallel to each other. I followed the theorem that states that when the dot product of 2 vectors equals zero, those vectors are orthogonal. I came up with v = (1,1,3) and u = (4,1,6). So I ended with the vector equation (x,y,z) = (-1,6,0) + t(1,1,3) + s(4,1,6).

It's my understanding that there are a number of different answers for this problem as there are a number of different vectors, besides the ones I found, that are orthogonal to the normal vector. The answer I was provided for this question stated that the vector equation is (x,y,z) = (-1,6,0) + t(3,-4,1) + s(1,-4,-1). I'm fairly certain that this answer is incorrect as the vectors in the vector equation are not orthogonal to the normal equation. Is my thinking correct?

答:
The plane parallel to $x+2y-z=5$ containing $(-1,6,0)$ is $x+2y-z=-1+2(6)-0=11$.

Your parametrization satisfies that:  $(-1+t+4s)+2(6+t+s)-(3t+6s)=11$.

The other answer does not:  $(-1+3t+s)+2(6-4t-4s)-(t-s)\not=11$ generally.


问:
Consequences of a class bijection $V\times V\to V$
A famous theorem of Tarski (proved in ZF) says that, if $\mathfrak{m}^2=\mathfrak{m}$ for every infinite cardinal $\mathfrak{m}$, then the Axiom of Choice holds.

Consider the language $\{\in,F\}$, where $F$ is a binary function symbol, and let $T$ be the theory ZF+"$F$ is a bijection". Informally, $V\times V\approx V$ (where $X\approx Y$ means that there is a bijection $X\to Y$).

By standard arguments, $L$ is a model of $T$, so $T$ is consistent.

My questions are:

Does $T\vdash$ AC?

Does $V\times V\approx V$ reflect to any initial segment of the universe? Or to any cardinal?

Does $T\vdash$ Global Choice? (there's an issue here regarding how to state Global Choice in a first order fashion, but still)

答:
ZF can prove that such a bijection exists (that is, there is a specific class function it can prove is a bijection).  There are obvious injections $V\times V\to V$ (the inclusion) and $V\to V\times V$ ($a\mapsto(a,\emptyset)$, say).  Schroder-Bernstein can then combine these two injections to get a bijection (a little care is needed to carry out the proof of Schroder-Bernstein for classes; see this answer for one way to do it).

Since you asked about initial segments of the universe, I would add that the same argument gives that there is a bijection between $V_\alpha\times V_\alpha$ and $V_\alpha$ for all limit ordinals $\alpha$ (and that in fact there is such a bijection that is uniformly definable in $\alpha$).  It follows that such a bijection also exists when $\alpha$ is an infinite successor, since if $X$ is infinite then $X\times X\approx X$ implies $X\times 2\approx X$ by Schroder-Bernstein and so $2^X\times 2^X\approx 2^{X\times 2}\approx 2^X$.


问:
Proving that this sum $\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$
How does one prove that for a prime $p \geq 5$ the sum :  $$\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.

Since each term of $\displaystyle \sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 < k < \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.

How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$

答:
From your last equation one gets
$$\frac1p{p\choose k}\equiv(-1)^k\frac 1k\pmod p.$$
So the problem is equivalent to
$$\sum_{0 < k < 2p/3}(-1)^k\frac1k\equiv0\pmod{p}.$$

The case $p=1979$ was set as the first problem at the 1979 IMO.
The method used for this extends to all primes, and no doubt
you can find solutions for the IMO problem out on the interweb.

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