问:
What does $\mathbb{R}^{\mathbb{N}}$ mean?
Can someone explain me what $\mathbb{R}^{\mathbb{N}}$ actually means?
I know the definition, but I'm not really understanding in practice what it means.

答:
Formally, $X^Y$ is the set of all functions from $Y$ to $X$. (You can keep straight which is the base and which is the exponent by thinking about finite combinatorics: how many functions from a $3$-element set to a $2$-element set are there, $2^3$ or $3^2$?)

However, this can be difficult to visualize at first, especially if $Y$ is infinite; some infinite sets, including $\mathbb{N}$, allow more intuitive descriptions. In particular, you can think of $X^\mathbb{N}$ as the set of infinite sequences of elements of $X$. To see the connection, given a sequence $S$ of elements of a set $X$ think of the function $f_S:\mathbb{N}\rightarrow X$ sending $i$ to the $i$th term of the sequence $S$.

So $\mathbb{R}^\mathbb{N}$ in particular is the set of infinite sequences of real numbers. These are things you've already played with, e.g. when constructing the reals via Cauchy sequences or when asking questions about convergence of sequences.


问:
Show that if $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and $f(x)=0$ for all $x\in \mathbb{Q}$, then $f(x)=0$ for all $x\in \mathbb{R}$.
I was proving the statement "if $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and $f(x)=0$ for all $x\in \mathbb{Q}$, then $f(x)=0$ for all $x\in \mathbb{R}$."

My idea was that if $f(x)=a$ for any $a\neq 0$ and for any $x\notin \mathbb{Q}$, then the  theorem of continuity won't work for all $\varepsilon $.

Precisely,

Let $a$ be any real number such that $a\neq 0$.

Suppose that $f(x)=0\; \forall x\in \mathbb{Q}$, and $f(x)=a\;  \forall x\notin \mathbb{Q}$.

Since $f$ is continuous on $\mathbb{R}$, it is continuous at $x_{0}\in \mathbb{Q}\subseteq \mathbb{R}$.

Thus, $\forall\varepsilon> 0$, $\exists \delta > 0$ s.t

$x\in \mathbb{R},\left | x-x_{0} \right |< \delta$ implies $\left | f(x)-f(x_{0}) \right |<\varepsilon  $.

If $x\notin \mathbb{Q}$, then $\left | f(x)-f(x_{0}) \right |=\left | a \right |$.

This implies that for $x\notin \mathbb{Q}$, we can't pick any $\varepsilon$ s.t $0<\varepsilon\leq \left | a \right |$.

But, by the theorem, we must be able to pick any $\varepsilon>0$.

Therefore, this is a contradiction.

Thus, $f(x)\neq a\;  \forall x\notin \mathbb{Q}\Leftrightarrow f(x)=0 \; \forall x\notin \mathbb{Q}$

Hence, $f(x)=0 \; \forall x\in \mathbb{R}$ $$\blacksquare $$

However, my proof seems wrong, but I can't tell why. "If there is something to fix," which part should I fix, and which way is better to prove this statement?

答:
On the contrary, your approach is good. What I will say is that the presentation needs some tidying, which I suspect is the source of your doubt. Let's try putting it in a slightly different order:

Suppose $f(x_0) = a \neq 0$. Since $f$ is continuous, it is continuous at $x_0$. Considering $\varepsilon = |a| \neq 0$, there exists some $\delta > 0$ such that
$$|x - x_0| < \delta \implies |f(x) - a| < |a| \implies |a| - |f(x)| < |a| \implies |f(x)| > 0 \implies f(x) \neq 0.$$
Thus, every point $x \in (x_0 - \delta, x_0 + \delta)$ satisfies $f(x) \neq 0$. But, since $\Bbb{Q}$ is dense in $\Bbb{R}$, there must exist a rational point $q \in (x_0 - \delta, x_0 + \delta)$, which satisfies $f(q) = 0$. This is a contradiction.

ContentLicense: CC BY-SA 4.0


问:
Let $a_n = \frac{n^p}{e^n}$ Show that $a_n$ converges for all $p>0$
In my work on this problem I've set up a limit:
$$\lim _{x\to \infty }\left(\frac{x^p}{e^x}\right)$$

I can see that for any number of p, successive applications of L'Hopital's Rule will eventually bring the numerator to a finite number, if p is an integer. If p is a fraction successive applications of L'Hopital's rule will eventually bring the $x$ term from the numerator to the denominator. Therefore in each case the limit will equal zero because the denominator's $e^x$ term will still be the same.

How do I express this logic in math form?

答:
Very simple:

If $p\in\mathbb Z^{+}$

$$\begin{align}\lim _{x\to \infty }\left(\frac{x^p}{e^x}\right)=\lim _{x\to \infty }\frac{p!}{e^x}=0.\end{align}$$

If $p>0, p\in\mathbb R^{+}$

$$0≤\left(\frac{x^p}{e^x}\right)≤\left(\frac{x^{\lceil p \rceil}}{e^x}\right)$$

Then apply the Squeeze theorem.


答:
It is enough to know this well fact . If $f,g\colon \Bbb R \to \Bbb R$ continuous functions and $$f(x)=g(x)$$ for all $x\in D$ where $D$ is dense subset of $\Bbb R$ then $f(x)=g(x)$ for every $x\in \Bbb R$ chekc?

Now, $f(x)=O(x)$ for every $x\in \Bbb Q$ where $O$ is a zero function. By using the fact above $f(x)=0$ for every $x\in\Bbb Q$

Title: Show that if $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and $f(x)=0$ for all $x\in \mathbb{Q}$, then $f(x)=0$ for all $x\in \mathbb{R}$.

Body:
I was proving the statement "if $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and $f(x)=0$ for all $x\in \mathbb{Q}$, then $f(x)=0$ for all $x\in \mathbb{R}$."

My idea was that if $f(x)=a$ for any $a\neq 0$ and for any $x\notin \mathbb{Q}$, then the  theorem of continuity won't work for all $\varepsilon $.

Precisely,

Let $a$ be any real number such that $a\neq 0$.

Suppose that $f(x)=0\; \forall x\in \mathbb{Q}$, and $f(x)=a\;  \forall x\notin \mathbb{Q}$.

Since $f$ is continuous on $\mathbb{R}$, it is continuous at $x_{0}\in \mathbb{Q}\subseteq \mathbb{R}$.

Thus, $\forall\varepsilon> 0$, $\exists \delta > 0$ s.t

$x\in \mathbb{R},\left | x-x_{0} \right |< \delta$ implies $\left | f(x)-f(x_{0}) \right |<\varepsilon  $.

If $x\notin \mathbb{Q}$, then $\left | f(x)-f(x_{0}) \right |=\left | a \right |$.

This implies that for $x\notin \mathbb{Q}$, we can't pick any $\varepsilon$ s.t $0<\varepsilon\leq \left | a \right |$.

But, by the theorem, we must be able to pick any $\varepsilon>0$.

Therefore, this is a contradiction.

Thus, $f(x)\neq a\;  \forall x\notin \mathbb{Q}\Leftrightarrow f(x)=0 \; \forall x\notin \mathbb{Q}$

Hence, $f(x)=0 \; \forall x\in \mathbb{R}$ $$\blacksquare $$

However, my proof seems wrong, but I can't tell why. "If there is something to fix," which part should I fix, and which way is better to prove this statement?


答:
Just to prove that for any $\epsilon > 0 $ have the following,
$$
|f(x)|<\epsilon
$$
this implies that $f(x) \equiv 0$

what we know is $f(x) = 0$ for $x\in\mathbb{Q}$,
for $x \not\in \mathbb{Q}$ , we can find a sequence $x_n \in \mathbb{Q}$ such that $x_n \to x $ as $ n \to \infty $, because $f$ is continuous , that means for any $\epsilon > 0$ ,we can find the $\delta > 0$ such that only $|x_n-x| < \delta $ for some $n>N$ , there have the following
$$
|f(x_n)-f(x)| < \epsilon
$$
note that $f(x_n) = 0$ , then we get what we want to prove.


问:
Show that the norms $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are both continuous.

Show that the norms  $\|\cdot\|_1 : (\mathbb{R}^n,d_2) \to (\mathbb{R}, |\cdot|)$ and $\|\cdot\|_\infty : (\mathbb{R}^n,d_2) \to (\mathbb{R}, |\cdot|)$ are both continuous.

For the first one if $x \in \mathbb{R}^n$, then $\|x\|_1=|x_1|+|x_2|+\dots+ |x_n|$ this seems just to be a sum of multiple projections e.g $\operatorname{pr}_j: \mathbb{R}^n \to \mathbb{R}$, where $j  \in \{0, \dots n\}$. So as the sum of continuous functions this is continuous? For the second one I'm not sure how to approach. I know that if I have for example $\max\{f,g\}$ then this equals $\frac12(f+g)+\frac12|f-g|$ so do I have that $$\max\{|x_1|,|x_2|, \dots|x_n|\}= \frac1n(|x_1|+\dots|x_n|)+\frac1n||x_1|-|x_2|-\dots-|x_n||$$ and I could show that this is continuous by the fact that all $x_i$ are continous since the previous result?

答:
It suffices to show that there are constants $C,K > 0$ such that
$$
\|x\|_1 \le C \|x\|_2  \quad \text{and} \quad \|x\|_{\infty} \le K \|x\|_2.
$$
This is because if $(x_n)$ is a sequence with $\|x_n - x\|_2 \to 0$ then $\|x_n -x\|_1 \le C \|x_n - x\|_2 \to 0$, and similarly for the other norm. Hence, the sequential characterisation of continuity implies that $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are continuous with respect to the Euclidean distance on $\mathbb{R}^n$.

For the first inequality we can use the Cauchy-Schwarz inequality to get
$$
\|x\|_1 = \sum_{i=1}^n |x_i| =  \sum_{i=1}^n 1 \cdot |x_i| \le \Big(\sum_{i=1}^n 1^2\Big)^{1/2}\Big(\sum_{i=1}^n |x_i|^2\Big)^{1/2} = \sqrt{n} \, \|x\|_2.
$$
For the second inequality
$$
\|x\|_{\infty} = \max_{1 \le i \le n} |x_i| = \sqrt{\max_{1 \le i \le n} |x_i|^2} \le \sqrt{\sum_{i=1}^n |x_i|^2} = \|x\|_2.$$


问:
When $G/H$ is cyclic?
Clearly $G/H$ is cyclic if $G$ is cyclic. But sometimes $G/H$ is cyclic without $G$ necessarily cyclic, for example: $(\mathbb{Z}_6\times \mathbb{Z}_6)/\left \langle (2,3) \right \rangle \cong \mathbb{Z}_6$

In cases like the latter, how can one determine if the group is cyclic? I mean, is there something in general that lets me know?

For example

Write $(\mathbb{Z}_{20}\times \mathbb{Z}_6)/\left \langle (10,2) \right \rangle $ as an external direct product of cyclic groups of prime power order.

Is $(\mathbb{Z}_{20}\times \mathbb{Z}_6)/\left \langle (10,2) \right \rangle $ isomorphic to $\mathbb{Z}_{4}\times\mathbb{Z}_{5}$ or isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_2\times \mathbb{Z}_5$?

答:
If $|G/H|$ is prime,  then of course it's cyclic.

In general,  the Smith normal form can be used to compute the structure of quotients of $\mathbb Z^n $.  But every finitely generated abelian group is such a quotient. After determining $G/H$ as a quotient of the free abelian group on  $n$ generators, this gives an approach.

Title: When $G/H$ is cyclic?


问:
Let $A⊆B⊆C$ and $X<Y$ denote $X$ is an elementary substructure of $Y.$ Does $A<B$ and $A<C$ imply $B<C$?
Let $A \subseteq B \subseteq C $ and $X<$Y denote $X$ is an elementary substructure of $Y$. Does $A < B$ and $A < C$ $\implies B<C$? My inkling is that this is not true. Here is why:

$A<B$ means for any sentence $\phi(\overline{x})$, $A \models \phi(\overline{a})$ iff $B \models \phi(\overline{a})$ for any tuple $\overline{a} \in A$.

Similarly,

$A<C$ means for any sentence $\psi(\overline{x})$, $A \models \psi(\overline{\alpha})$ iff $C \models \psi(\overline{\alpha})$ for any tuple $\overline{\alpha} \in A$.

We want to show that for any $\theta(\overline{x})$, $B \models \theta(\overline{b})$ iff $C \models \theta(\overline{b})$ for any tuple $\overline{b} \in B$.

From the first two we can only conclude that for any $\theta(\overline{x})$, $B \models \theta(\overline{\text{a}})$ iff $C \models \theta(\overline{\text{a}})$ for any tuple $\overline{\text{a}} \in A$, but not for any tuple $\overline{b} \in B$.

However, I cannot think of a counter example.

答:
Here's another example. The structure in question is a bit simpler in my opinion. The verification is definitely harder, but it's a good exercise.

Consider a language with a single binary relation symbol $E$. Let $T$ be the first-order theory in this language whose models are the structures in this language in which:

$E$ is an equivalence relation,

every $E$-class has either $1$ or $2$ elements, and

each of the two possible types of equivalence class occurs infinitely often.

(Whipping up such a $T$ is a good exercise.)

For the rest of this answer, restrict attention to models of $T$.

Let $A$ be some infinite model of $T$, and let $B$ be an extension of $A$ gotten by adding a new element $v$ not $E$-related to anything in $A$ (so $B$ adds a class to $A$). Now let $C$ be the extension of $B$ gotten by adding a new element $w$ and $E$-relating $w$ to $v$ (so $C$ adds an element to $B$ but does not add a new equivalence class). Clearly $B\not\preccurlyeq C$:

 The sentence-with-parameters $\forall x(vEx\leftrightarrow v=x)$ is true in $B$ but not in $C$.

On the other hand, it's a good exercise to show that $A\preccurlyeq B$ and $A\preccurlyeq C$. One way to do this is via automorphisms, via (proving and checking the applicability of) the following:

Lemma. Let $X, Y$ be isomorphic structures with $X\subseteq Y$ (remember that it's perfectly possible for a structure to be isomorphic to a proper substructure of itself - take e.g. $(\mathbb{R},<)$ vs. $((0,1), <)$). Suppose that for each finite tuple $a_1,...,a_n\in X$ there is an isomorphism $f:Y\rightarrow X$ such that $f(a_i)=a_i$ for each $1\le i\le n$. Then $X\preccurlyeq Y$.

HINT: isomorphisms preserve satisfaction of formulas ...


问:
Solving $|2x-5| + |7-2x| =2 $
$$|2x-5| + |7-2x| =2 $$

I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-

1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.

2nd case, where $2.5<x\le3.5$ :-
$$2x-5+2x-7 = 2 \Rightarrow x=3.5$$
We can say that one solution exists in this range.

3rd case, where $x>3.5$ :-
$$2x-5+7-2x = 2 \Rightarrow 2=2$$
We can say that infinite solutions exist in this case.

So, as per above calculations, what I can conclude that for all $x\geq2.5 $ solution exists.

But the solution provided in the book says that $|2x-5| + |7-2x| =2 $ will only be valid when $2.5 \leq x \leq 3.5$.

Can someone please help me out on how to decide the solutions and solution ranges for these type of questions?

Thanks in advance !

答:
The short cut way: let $y=2x$, what are the values that $y$ can take such that the distance from $5$ and the distance from $7$ are exactly $2$. If $y$ is less than $5$, then the distance from $7$ is beyond $2$. SImilarly if $y$ is more than $7$, then the distance from $5$ is beyond $2$. We can easily check that any values of  $y$ between $5$ and $7$ satisfies the condition and hence $x$ is in between $2.5$ and $3.5$.

Now, as a practice, let's work it out.

If $x \le 2.5$, then $|2x-5|=5-2x$, $|7-2x|=7-2x$.

Hence $$5-2x+7-2x=2$$
$$10=4x$$

and hence the only solution is $x=2.5$.

If $2.5 < x <3.5$, then $|2x-5|=2x-5$ and $|7-2x|=7-2x$.

$$2x-5+7-2x=2$$

which is always true.

If $x \ge 3.5$, $|2x-5|=2x-5$ and $|7-2x|=2x-7$

$$4x-12=2$$

$$2x=7$$ the only solution is $3.5$.

Hence taking union: $\{2.5\} \cup (2.5, 3.5) \cup \{3.5\}=[2.5,3.5]$ is the desired result.

Title: Solving $|2x-5| + |7-2x| =2 $


问:
Solving a system
I have the following system:

$a b x_0 y - \sqrt{a^2 - x_0^2} (a^2 x - a^2 x_0 + b^2 x_0) = 0$

$ x_0^3 ( a^2-b^2) = a^4 x$

I need to solve it and get $(ax)^{2/3} + (by)^{2/3} = (a^2- b^2) ^{2/3}$

This is the only part I have left on a exercice about computing the evolute of the ellipse using the normal rects (I have to do it that way)

ContentLicense: CC BY-SA 3.0

Title: Solving a system

Body:
$x_0(a^2-b^2)=\dfrac{a^4x}{x_0^2}, abx_0y=\sqrt{a^2-x_0^2}(a^2x-x_0(a^2-b^2))=\sqrt{a^2-x_0^2}\left(a^2x-\dfrac{a^4x}{x_0^2}\right) \to $

$(by)^2=(ax)^2\left(\dfrac{a^2}{x_0^2}-1\right)^3 \to (by)^{\frac{2}{3}}=(ax)^{\frac{2}{3}}\left(\dfrac{a^2}{x_0^2}-1\right)  \to (ax)^{2/3} + (by)^{2/3}=\dfrac{a^{8/3}x^{2/3}}{x_0^2}=\left(\dfrac{a^8x^2}{x_0^6} \right)^{1/3}=\left(\dfrac{a^4x}{x_0^3} \right)^{2/3}=(a^2-b^2)^{2/3}$


问:
Covariance of two normal distributions
Why does the covariance of the following two normal distributions, $X \thicksim N(67,20), Y \thicksim N(9,1),$ be $0?$ My simulations show that it should always be $0$ in these conditions, but I'm looking for some examples, proof or references.

Thank you

I got the question from here.

答: Since $X$ is the random variable that represents people's height and $Y$ is the random variable that represents the time people woke up. Intuitively speaking, two things should be independent because your height will not effect the time you wake up and the time you wake up will not effect your height. Since they are independent, we could say their covariance is zero. But we must be careful because there is no rigorous proof. It's more like an assumption.

Title: Covariance of two normal distributions


问:
Showing Position Operator is Unbounded
For example 1, on the Unbounded Operators section in Reed's "Functional Analysis", they mention that the position operator $T: D(T) \to L^2(\mathbb{R})$, with $(T\varphi)(x) = x \varphi(x)$ is unbounded. The domain is defined as

$$D(T) = \left\{\varphi \in L^2(\mathbb{R}) : \int_{\mathbb{R}} x^2|\varphi(x)|^2dx < \infty\right\}.$$

To show it is unbounded in general, can we just quickly apply the definition of boundedness here?

For any $\varphi \in D(T)$,

$$
\begin{align}
||T|| &= \sup_{||\varphi||_2 = 1} ||T\varphi||_2 \\
&= \sup_{||\varphi||_2 = 1} \left(\int_{\mathbb{R}} x^2|\varphi(x)|^2dx\right)^{1/2} \\
&= \sup_{||\varphi||_2 = 1} \left(\int_{\mathbb{R}} x^2dx\right)^{1/2}\left(\int_{\mathbb{R}}|\varphi(x)|^2dx\right)^{1/2} \tag{$\varphi$ constant in $L^2$-norm}\\
&= \left(\int_{\mathbb{R}} x^2dx\right)^{1/2} \\
&= \infty.
\end{align}
$$

EDIT: For completeness, using the accepted we have

$$||T|| = \sup_{||\varphi||_2 = 1} ||T\varphi||_2 \geq \sup_{n \in \mathbb{N}} ||T\chi_{(n, n+1)}||_2 = \sup_{n \in \mathbb{N}} \sqrt{\frac{2n + 1}{2}} = \infty.$$

Title: Showing Position Operator is Unbounded

Body: $\|\chi_{(n,n+1)} \|^{2}=1$ for each $n$ and $\|T\chi_{(n,n+1)} \|^{2}\to \infty$. Hence, $T$ is unbounded.


问:
Longest-ever time between problem posing and solving?
Guiness World Record claims that Goldbach's conjecture is the oldest unsolved problem. A natural related question is what solved problem went unsolved for the longest time. In other words, of all the now-solved problems, which was open for the longest?

Perhaps it depends how we define the moment when a problem becomes "posed," but we could define this moment as the first time the problem is referenced in any record we have.

答:
Some geometric construction problems - e.g. doubling the cube - were "posed" in antiquity but only solved in the $19$th century. This gets us a gap of about $\sim 2000$ years at least.

Re: the scare quotes above, there is a slight issue here about what exactly constitutes posing a problem. The texts I'm aware of all phrase it as a positive imperative, "figure out how to double the cube using only compass and straightedge." Since the punchline is that that can't be done, there's a disconnect between the way the problem was posed and the way it should have been posed. I personally think this still counts though.


问:
How to show that this ideal is not primary?
I am stuck at the following exercise:

Let $R$ be the subring of $\mathbb{Z}[X]$ consisting of the polynomials whose
coefficient of $x$ is even. Let $P$ the ideal of those polynomials in $R$ whose constant
coefficient is zero. Then $P$ is prime, and $P^2$ is not primary.

I understand that $P$ is a prime ideal, but I do not see how I can show that $P^2$ is not be primary. I think one way to do this might be to show that $\sqrt{P^2}$ is not prime, but again I do not see how. Could you help me?

答:
Your proof approach is not quite right; in fact, for an arbitrary commutative ring $R$ and an arbitrary prime ideal $P<R$, we have $\sqrt{P^n}=P$ for any $n\in\mathbb{N}$. Indeed, certainly $P\subseteq\sqrt{P^n}$, since $p^n\in P^n$ for any $p\in P$. On the other hand, suppose $q\in \sqrt{P^n}$. Then $q^k\in P^n$ for some $k\in\mathbb{N}$, and hence, since $P^n\subseteq P$, in particular $q^k\in P$. Since $P$ is prime, this forces $q\in P$.

Now let's restrict to the $R$ and $P$ of your question. Consider the element $4x^2=(2x)^2\in P^2$, and decompose this element as a product $ab$, where $a=2x^2$ and $b=2$. We have $a\notin P^2$, since the coefficient of the $x^2$ term in any element of $P^2$ must be divisible by $4$. (Indeed, an arbitrary element of $P$ looks like $2m_1x+m_2x^2+m_3x^3+\dots$, where the $m_i$ lie in $\mathbb{Z}$. So, the product of two arbitrary elements of $P$ is of the form
\begin{align*}
&(2m_1x+m_2x^2+m_3x^3+\dots)(2n_1x+n_2x^2+n_3x^3+\dots) \\
=\text{ }&4m_1n_1x^2+2(m_1n_2+m_2n_1)x^3+(2m_1n_3+m_2n_2+2m_3n_1)x^3+\dots.
\end{align*} In particular, the coefficient of the $x^2$ term of such an element is divisible by $4$.) But also $b\notin\sqrt{P^2}=P$, since $b$ has non-zero constant term. Thus $P^2$ is not primary, as desired.


问:
Finding a paper
Sometimes I have trouble finding particular papers.  Does anyone have any general  suggestions?

I find myself in this situation frequently:  An author references a paper.  I find it on MathSciNet and read the abstract.  I then realize I want to read the paper.  Then I cannot find the paper.

Is there any better approach besides just using Google Scholar, Google, the author's webpage, JStor and MathSciNet?  I am at a University, and have some access to catalogues.

Specifically I wanted to read this but I can't seem to do it online.

Thanks a lot!

Update:  I ended up finding the paper by talking to a Professor I know who had a PDF on hand.  None of the links below lead to the correct paper, which is why I have not accepted an answer.

答:
UPDATE

Actually, I just took a stab at searching again, and came up with this PDF, by Moree, with same title...hope it helps...(I love researching stuff!)
Here's what might be the paper you're looking for (same author, title as the article you're looking for): arXiv: Moree, pdf.

Of course, you may still want a copy of the published article, and if not available at your university, requesting an inter-library loan (photocopy of published article) would be your best bet.  In the meantime, you can peruse the arXiv article?

You can find the article at SpringerLink here. There, the article of interest is available to individuals for purchase, though it would be worth checking if your Uni has institutional access if you can wait!

Also, I've had good luck with Google, Google Scholar, authors webpages...emailing author (yes it works, I've had wonderful encounters that way!).

UPDATE 2: Please be sure to scroll down to read J.M.'s "process" post; it's worth the read in terms of approaching a search for academic publications!

Title: Finding a paper


问:
Eigenvectors of matrices which commute with a projection
Just a quick question. Cant seem to prove it or find any relevant references! Maybe it's really simple :\ Is the following statement true (for square matrices of the same finite dimension)?

If there exists no simultaneous right and left eigenvectors of M, then there exists no (besides $P=0$ or $P=I$) projection matrix $P=P^2$ such that $[P,M]=0$.

Thanks for the help/potential counter example!

答:
A simultaneous right/left eigenvector for $A$ is $x\neq 0$ such that $Ax=\lambda $ and $x^TA=\mu x^T$, i.e. $A^Tx=\mu x$. That is a common eigenvector of $A$ and $A^T$. I assume the coefficients belong to a field $K$.

1) Here is a $2\times 2$ counterexample:
$$
A=P=\pmatrix{1&1\\0&0}.
$$
Clearly, $A$ and $P$ commute, and $P$ is idempotent ($P^2=P$). The two eigenvalues of $A$ and $A^T$ are $0$ and $1$. And we have, denoting the canonical basis of $K^2$ by $\{e_1,e_2\}$:
$$
\ker A=K(e_1-e_2)\quad\ker(A-I_2)=Ke_1\quad \ker A^T=Ke_2\quad\ker (A^T-I_2)=K(e_1+e_2).
$$
So $A$ and $A^T$ don't share any nonzero eigenvector.

2) Here is a $4\times 4 $ counterexample with $P$ self-adjoint + idempotent:
$$
A=\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0}\qquad P=\pmatrix{1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0}.
$$
Clearly, $P$ does commute with $A$. Yet the only eigenvalue of $A$ and $A^T$ is $0$. And the only solution to $Ax=A^Tx=0$ is $x=0$.


Title: Solving $|2x-5| + |7-2x| =2 $

Body:
$$|2x-5| + |7-2x| =2 $$

I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-

1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.

2nd case, where $2.5<x\le3.5$ :-
$$2x-5+2x-7 = 2 \Rightarrow x=3.5$$
We can say that one solution exists in this range.

3rd case, where $x>3.5$ :-
$$2x-5+7-2x = 2 \Rightarrow 2=2$$
We can say that infinite solutions exist in this case.

So, as per above calculations, what I can conclude that for all $x\geq2.5 $ solution exists.

But the solution provided in the book says that $|2x-5| + |7-2x| =2 $ will only be valid when $2.5 \leq x \leq 3.5$.

Can someone please help me out on how to decide the solutions and solution ranges for these type of questions?

Thanks in advance !

Body:
WLOG for real $x,$

$$|2x-5|=2\sin^2t,|7-2x|=2\cos^2t$$

$$(2x-5)^2-(7-2x)^2=4(\cos^2t-\sin^2t)$$

$$\implies2x=6-\cos2t\implies-1\le2x-6\le1\iff 5\le2x\le7$$

In that case $|2x-5|=+(2x-5)$ and $|7-2x|=+(7-2x)$


问:
Cofactor-Trace Identity
I have a question regarding a lemma in the paper Total Positivity and Toda Flow. $b_1, b_2, ..., b_p$ are column vectors of $B_p$. For a matrix $C$, the product $Cb_i$ is well defined. Prove that the following equality holds.

$$\sum_{j=1}^p det(b_1, ..., b_{j-1}, Cb_j, b_{j+1}, ..., b_p) = tr(C)detB_p$$

The author states that a proof will rely on the notion that $\sum_{i=1}^p a_{ij} A_{ik} = \delta_{jk} det(A)$ with $\delta$ being a Kronecker delta.

I feel that I already have an intuition for the special case when $b_j$ is an eigenvector of $C$; however, I fail to see the method required for a general proof. Could one successively factor out "pieces" of $C$, which will yield a sum or product of $tr(C)$?

Thank you all.

答:
By a continuity argument, we may assume that $B$ is invertible. Let $X=B_p^{-1}CB_p$. Denote the $j$-th column of $X$ by $x_j$. Then $B_px_j=Cb_j$. Therefore, by Cramer's rule, the $j$-th entry of $x_j$ is given by
$$
\frac{\det(b_1,\ldots,b_j,Cb_j,b_{j+1},\ldots,b_p)}{\det(B_p)}.
$$
Since the $j$-th entry of $x_j$ is the $j$-th diagonal element $x_{jj}$ of $X$, we have
$$
\operatorname{tr}(C)
=\operatorname{tr}(B_p^{-1}CB_p)
=\operatorname{tr}(X)
=\sum_{j=1}^px_{jj}
=\sum_{j=1}^p\frac{\det(b_1,\ldots,b_j,Cb_j,b_{j+1},\ldots,b_p)}{\det(B_p)}
$$
and the result follows.

Title: Cofactor-Trace Identity


答:
A quite quick way to solve this inequality is using the triangle inequality:

$$|2x-5| + |7-2x|\geq |2x-5+7-2x| = 2$$

The triangle inequality $|a| + |b| \geq |a+b|$ turns in an equality $|a|+|b| =|a+b|$ if and only if $ab=|ab|$. (Just square $|a| + |b| \geq |a+b|$ and rearrange).

Hence, you only need to find those $x$ for which $2x-5$ and $7-2x$ have the same sign:

Case $\geq 0$:

$2x-5 \geq 0 \Leftrightarrow x\geq \frac 52$.

$7-2x \geq 0 \Leftrightarrow x\leq \frac 72$.

So, this gives the solution $\boxed{x\in \left[\frac 52,\frac72\right]}$.

Case $\leq 0$:

$2x-5 \leq 0 \Leftrightarrow x\leq \frac 52$.

$7-2x \leq 0 \Leftrightarrow x\geq \frac 72$.

Hence, no further solution.

Title: Solving $|2x-5| + |7-2x| =2 $


问:
Prove that the function $f(x) = \frac{2\sin(x) + \sqrt x}{x}$ is not uniformly continuous on the domain $(0, \infty)$
Consider$$f(x) = \frac{2\sin(x) + \sqrt x}{x}$$

When $x$ approaches zero,  $\frac{2sinx}{x}$ goes to $2$, but $\frac{\sqrt x}{x}$ goes to infinity, whose graph indicates that its derivative when $x$ approaches zero goes to negative infinity. It suggests that no single $\delta$ should be able to respond to all $\epsilon$ s' challenges, thus $f(x)$ is not uniformly continuous.

To prove rigorously, my intention was to choose two series $x_n$ and $y_n$ so that |$x_n - y_n$|$\to 0$  and |$f(x_n) - f(y_n)$|$\ge \epsilon_0$, where $\epsilon_0 $ is chosen and $\gt 0$. But it turned out that it could be very hard to prove $|f(x_n)-f(y_n)|\ge$ any fixed number.

However, I have no idea how to apply another strategy, i.e. choose $\epsilon = \epsilon_0$, given any $\delta \gt0$, choose particular $x$ and $y$ s.t. $|x-y|\lt\delta$ and $|f(x)-f(y)|\ge\epsilon_0$. I tried to set $x = min\{1,\delta\}$ and $y = \frac x2$, it doesn't work either.

Could you guys help me out of this? I am fresh to analysis and this question has stung me for a while.

答:
It is not uniformly continuous on $(0,a)$ for any $a >0$. If it  is uniformly continuous then it would be be bounded as $x \to 0+$. But $\frac {\sin x} x \to 1$ as $ x\to 0+$ and $\frac 1 {\sqrt x} \to \infty$ as $x \to 0+$.

If you don't want to use the boundeness criterion you can argue as follows: Supppose there exists $\delta >0$ such that $|f(x)-f(y)| <1$ whenever $|x-y| <\delta $. Then there exists $n_0$ such that  $|f(\frac 1 n)-f(\frac  1 m)| <1$ whenveer $n,m \geq n_0$. Put $m=n_0$ and take limit as $ n \to \infty$ to get a contradiction.


答:
Here's a neater proof:

Let $x\in \mathbb{R}$. It suffices to show that $|f(x)|=0$. Let $\epsilon>0$. As $f$ is continuous on $\mathbb{R}$, there exists some $\delta>0$ such that whenever $y\in (x-\delta,x+\delta)$ then $|f(x)-f(y)|<\epsilon$. So in particular choosing a rational $q$ in $(x-\delta,x+\delta)$, we have $|f(x)-f(q)|=|f(x)|<\epsilon$.  $\square$

Additionally, let us adapt the technique in the proof above to the following result:

Proposition:

Let $D\subseteq \mathbb{R}$ be a dense subset. Suppose $f,g:\mathbb{R}\rightarrow \mathbb{R}$ are continuous and they agree on $D$. Then $f=g$ on $\mathbb{R}$.

Proof:

It suffices to show that if $h: \mathbb{R}\rightarrow \mathbb{R}$ is a continuous function such that $h=0$ on $D$ then $h=0$ on $\mathbb{R}$.

Let $x\in \mathbb{R}$ and $\epsilon>0$. By continuity of $h$ on $\mathbb{R}$, there exists some $\delta>0$ such that whenever $y\in (x-\delta,x+\delta)$ we have $|h(x)-h(y)|<\epsilon$.
As $D$ is dense in $\mathbb{R}$, $D\cap (x-\delta,x+\delta) \neq \varnothing$. Thus picking one element, $d$, from the intersection we obtain $|h(x)-h(d)|=|h(x)|<\epsilon$. $\square$


问:
Graphing an integrated max equation
Evaluate the integral and sketch the region in $x-y$ plane,

$$\int^1_0\int^1_0 x\ \operatorname{max}(x^2,y)\ \mathrm{d}y\ \mathrm{d}x.$$

I have already integrated the equation with when $x^2 > y$ and when $y > x$. Both also have the result of $\frac14$. I just wanna know how do I graph this to show which region is when $x^2$ is larger than $y$ and which region is when $y$ is larger than $x^2$?

My attempt

答:
After following @lisyarus advice I have concluded this graph
Attempt 2</p>

Title: Graphing an integrated max equation


问:
$H$ and $K$ are supposed to be closed in $X$, Why?
Let $X$ be a topological space and $A$ a closed subspace of $X$. Let $H$ and $K$ be closed in $A$. Now $H$ and $K$ are supposed to be closed in $X$, Why?</p>

答:
If A is closed in X then X\A must be open in X.
If H (or K) is is closed in A then A\H must be open in A. Hence there exists some open set B such that A\H $\subset$ B $\subset$ X and H $\not \subset$ B.
Now X\H = X\A $\cup$ B which is a union of open sets so this must be open.
Hence H is closed in X.</p>


问:
How to factorise $x^2 + 1$
Keep in mind that I am not asking how to find the solutions to the equation $x^2 + 1 = 0$. I know that there will be imaginary answers In this case. I just want to know how to factor this using a logical method and obtaining 2 factors. Thanks for answering.

Title: How to factorise $x^2 + 1$

Body:
HINT:

Remember $a^2-b^2=(a+b)(a-b)$, $i^{2}=-1$.

So we can write, $a^2+b^2=a^2-(ib)^2=(a+ib)(a-ib)$

Putting $a=x$ and $b=1$ we get,

$x^2+1=(x+i)(x-i)$


问:
Recurrence General formulas
In this post I will share general formulas for recurrences I have worked out, just for fun.

And you are more than welcome to share yours.

I will also include a table for a quick check.

答:
$$u_{n} = au_{n-1} + b$$
$$\Downarrow$$
i: $$u_n = u_0a^n + \frac {b(a^n-1)} {a-1} $$
or ii: $$u_n = u_0a^n + Σ_{m=1}^{n} ba^m$$

Proof:

Introduce another recurrence $v$, define $v_n = u_n + \frac b {a-1}$

$\therefore u_n = v_n - \frac b {a-1}$

$v_n = u_n + \frac b {a-1} = (au_{n-1} + b) + \frac b {a-1} = au_{n-1} + \frac {ab}{a-1} = a(u_{n-1} + \frac b {a-1}) = av_{n-1}$

$\therefore v_n = av_{n-1}$ (definition of $v$ in terms of $v$)

$v_n = a^rv_{n-r} = a^nv_0$ (general formula for $v$)

$\because u_n = v_n - \frac b {a-1}$ (definition of $v$ in terms of $u$)

$\therefore u_n = a^nv_0 - \frac b {a-1} = a^n(u_0 + \frac b {a-1}) - \frac b{a-1}$ (substitute $v$ s with $u$ s)

$u_n = a^nu_0 + \frac {a^nb} {a-1} - \frac b {a-1} = a^nu_0 + \frac {a^nb-b} {a-1} = a^nu_0 + \frac {b(a^n-1)} {a-1}$

$\therefore$
$$u_n = a^nu_0 + \frac {b(a^n-1)} {a-1}$$

Notice that the second term is the formula for a geometric series (Kadry, 2014) $ba^{n-1}$

Therefore the expression above equals to:
$$u_n = u_0a^n + Σ^n_{m=1} ba^m$$

Reference:

Seifedine Kadry,
Mathematical Formulas for Industrial and Mechanical Engineering,
Elsevier,
2014,
Pages 65-111,
ISBN 9780124201316,
https://doi.org/10.1016/B978-0-12-420131-6.00005-1.
(https://www.sciencedirect.com/science/article/pii/B9780124201316000051)

Title: Recurrence General formulas


问:
Nullspace of a block matrix
Suppose I have $M \in \mathbb{R}^{2n\times 2n}$ such that
$$M = \begin{bmatrix} A &B\\0  & C\end{bmatrix}$$
for some $A,B,C \in \mathbb{R}^{n \times n}$.

I wish to:

(i) Find the nullspace of $M$

(ii) Prove that rank$(M) \geq \text{rank}(A) + \text{rank}(C)$

For (i) it is clear that for some vector $[x,  y]^T$ will be in the null space if and only if $y \in \text{null}(C)$ and that $Ax + By = 0$, but I am unsure how to proceed from here. For (ii) I can see why this is true, but am not sure how to approach proving it.

Thanks.

答:
Let's set $\operatorname{rank}(A) = r$ and $\operatorname{rank}(C) = s$. Choose $i_1, \dots, i_r$ such that $Ae_{i_1},\dots,Ae_{i_r}$ (which are just the $i_1,\dots,i_r$-th columns of $A$) are linearly independent and similarly choose $j_1, \dots, j_s$ such that $Ce_{j_1},\dots,Ce_{j_s}$ are linearly independent. It is enough to show that the $r + s$ columns
$$ M \begin{bmatrix} e_{i_1} \\ 0 \end{bmatrix}, \dots, M \begin{bmatrix} e_{i_r} \\ 0 \end{bmatrix}, M \begin{bmatrix} 0 \\ e_{j_1} \end{bmatrix}, \dots, M \begin{bmatrix} 0 \\ e_{j_s} \end{bmatrix} $$
of $M$ are linearly independent. And indeed, assume that
$$ \lambda_1 M \begin{bmatrix} e_{i_1} \\ 0 \end{bmatrix} + \dots + \lambda_r M \begin{bmatrix} e_{i_r} \\ 0 \end{bmatrix} + \mu_1 M \begin{bmatrix} 0 \\ e_{j_1} \end{bmatrix} + \dots + \mu_s M \begin{bmatrix} 0 \\ e_{j_s} \end{bmatrix} = \\
\lambda_1 \begin{bmatrix} Ae_{i_1} \\ 0 \end{bmatrix} + \dots + \lambda_r \begin{bmatrix} Ae_{i_r} \\ 0 \end{bmatrix} + \mu_1 \begin{bmatrix} Be_{k_1} \\ Ce_{k_1} \end{bmatrix} + \dots + \mu_s \begin{bmatrix} Be_{k_s} \\ Ce_{k_s} \end{bmatrix} = \\
\begin{bmatrix} \lambda_1 Ae_{i_1} + \dots + \lambda_r Ae_{i_r} + \mu_1 Be_{j_1} + \dots + \mu_s Be_{j_s} \\ \mu_1 Ce_{j_1} + \dots \mu_s Ce_{j_s} \end{bmatrix} =  \begin{bmatrix} 0 \\ 0 \end{bmatrix}. $$

Since $Ce_{j_1},\dots,Ce_{j_s}$ are linearly independent this implies that $\mu_1 = \dots = \mu_s = 0$ but then we get the equation $\lambda_1 Ae_{i_1} + \dots + \lambda_r Ae_{i_r} = 0$ and since $Ae_{i_1},\dots,Ae_{i_r}$ are also linearly independent we get that $\lambda_1 = \dots = \lambda_r = 0$.

Title: Nullspace of a block matrix


问:
Confusion on Proof of Gram-Schmidt Conjugation Process in context of Conjugate Gradient Method
First of all, I'd like to make clear that I'm talking about the Gram-Schmidt Conjugation process (in the context of conjugate gradient method for optimization), which instead of making a set of n independent vectors orthonormal, makes a set of n independent vectors conjugates - that is: expressable in the form

(where vector u_i and vector u_j are conjugates, or sometimes, "A-orthogonal").

Understandably, this formula is going to be very similar to Gram-Schmidt Orthogonalization, since the process is the same - but instead of projecting (and subtracting) some vector v onto all the previous orthogonal vectors and subtracting, we're projecting on to the all the previous conjugate vectors and subtracting.

So, the question is what this "projection matrix" is. In the paper/explanation I'm working through to understand Conjugate Gradient, there is two pages dedicated to the explanation of Gram-Schmidt Conjugation Process (pages 31-32 in pdf) and there is a proof of what the "projection" matrix is, but I don't understand it. I recommend you look at it in the paper but here is a screenshot:

I get equation (36), but kind of get lost after that. I get why they're premultiplying, since the whole definition of conjugate is that they abide to that property of u_i(T) * A*u_j = 0, but I really don't get the new subscript "j" which I haven't seen before and the removal of the summation.

If someone could explain it step by step that would be really good,

Thank you!

答:
The goal is to find $\beta_{ij}$ such that $d_{(i)}^TAd_{(j)}=0$.

We transpose equation $(36)$ and premultiply it to $Ad_{(j)}$

In the very next line, the summation seems to vanish, that is because if $k \ne j$, then $d_{(k)}^TAd_{(j)}=0$. Hence only one term is left.

After which, we can solve for $\beta_{ij}$.


问:
Proof related to a connected graph of diameter 2
Let G be a connected graph of diameter 2. So if κ(G) represents vertex connectivity and λ(G) represents the edge connectivity, then prove that κ(G) = λ(G) for
such G.

What I have tried: Since the diameter is 2, the maximum distance between two vertices should be 2. So I am considering x and y vertices which are connected to another set of vertices. So there is no direct connection between x and y making the diameter 2 as the path to reach from x to y is through another vertex from the set of vertices to which both x and y are connected. The diagram for the same is as below:

Since it's a connected graph and it should have a diameter of 2 therefore the degree of vertices inside the set should be 2 or else the diameter may increase which is not desired. And the degree of any vertices should not be 1 or else the diameter will also increase as shown below:

So far looks good to me but I am not sure if this is correct. Moreover, I do not know what to do next. Could you please help me to solve this? Thanks in advance.

答: The statement you are trying to prove is false; the butterfly graph is a counterexample, with $\operatorname{diam}(G)=2$, $\kappa(G)=1$, $\lambda(G)=2$. The question you linked to is about graphs satisfying a different condition, namely, graphs with maximum degree $\Delta(G)\le3$.


问:
If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x) $ exist?
I want to prove or disprove this problem:
If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L<\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.

When I assume problem below:

If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?

I can use mean-value theorem to show that.

So my question is:

If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?

答:
When $f'$ is continuous, you can avoid l'Hôspital's rule by saying:

Let $\epsilon >0$. There exists $x_0 \in \mathbb{R}$ such that $x \geq x_0$ implies $$L-\epsilon < f'(x)+f(x) <L+\epsilon$$

Hence $$(L-\epsilon)e^x < (f'(x)+f(x))e^x <(L+\epsilon)e^x$$
$$\int_{x_0}^t(L-\epsilon)e^xdx < \int_{x_0}^t (f'(x)+f(x))e^xdx <\int_{x_0}^t (L+\epsilon)e^xdx$$
$$(L-\epsilon)e^t-(L-\epsilon)e^{x_0} < e^tf(t)-f(0)<(L+\epsilon)e^t-(L+ \epsilon)e^{x_0}$$
$$(L-\epsilon) - (L-\epsilon)e^{x_0-t} < f(t)-f(0)e^{-t} < (L+ \epsilon)-(L+\epsilon)e^{x_0-t}$$

for $t \geq x_0$. When $t$ is large enough, we get $$L-2\epsilon < f(t) < L+2\epsilon$$

Therefore, $\lim\limits_{x \to + \infty} f(x)=L$.


问:
Solving $|2x-5| + |7-2x| =2 $
$$|2x-5| + |7-2x| =2 $$

I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-

1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.

2nd case, where $2.5<x\le3.5$ :-
$$2x-5+2x-7 = 2 \Rightarrow x=3.5$$
We can say that one solution exists in this range.

3rd case, where $x>3.5$ :-
$$2x-5+7-2x = 2 \Rightarrow 2=2$$
We can say that infinite solutions exist in this case.

So, as per above calculations, what I can conclude that for all $x\geq2.5 $ solution exists.

But the solution provided in the book says that $|2x-5| + |7-2x| =2 $ will only be valid when $2.5 \leq x \leq 3.5$.

Can someone please help me out on how to decide the solutions and solution ranges for these type of questions?

Thanks in advance !

答:
You have the info about $7-2x$ completely backwards.  I $x \le 3.5$ then $7-2x \ge 0$ and $|7-2x| = 7-2x$.

And if $x >3.5$ then $7-2x < 0$ and $|7-2x| = 2x - 7$.

So case 1: $x <2.5$ then $|2x - 5|+|7-2x| = (5-2x)+(7-2x) = 2$ and $12-4x = 2$ and $x =2.5$ but $x < 2.5$ so that isn't a solution.

Case 2: $2.5 \le x \le 3.5$ and $|2x-5|+|7-2x|=(2x-5)+(7-2x)= 2$ and $2=2$ has infinite solutions.

Case 3: $x > 3.5$ then $|2x-5|-|7-2x| = (5-2x) -(2x-7)= 12 - 4x = 2$ so $x=-2.5$ but $x > 3.5$ so that isn't a solution.

....

Note $|7-2x| = |2x-7|$ so the critical points don't have to mean $x < critical\implies (central stuff) < 0$ and $x > critical \implies (central stuff) > 0$.

It could be the exact opposite.  One key thing to note $2x -7$ has a positive coefficient so it has $<> critical \implies <> 0$.  But a negative coefficient would imply $<> critical \implies >< 0$.

Title: Solving $|2x-5| + |7-2x| =2 $


问:
Solving IVPs with one-sided initial condition using Laplace Transform
Consider an initial value problem of first order linear ODE.
$$\frac{\mathrm{d}}{\mathrm{d}t}y(t)+2y(t)=e^{-t}H(t),\lim_{t \to 0^-}y(t)=2$$
where $H(t)$ is the Heaviside function. Let $L[y: t \to s]=Y(s)$ and applying Laplace Transform to both side of the equation.
$$sY(s)-y(0)+2Y(s)=\frac{1}{s+1}.$$
Then I find $y(t)$ to be
$$Y(s)=\frac{1}{(s+1)(s+2)}+\frac{1}{s+2}y(0) \Rightarrow y(t)=y(0)e^{-2t}+(e^{-t}-e^{-2t}).$$
My main question is whether it is appropriate to use $y(0^-)$ in this case as the value of $y(0)$, which in this case only the value of left sided limit is given.

答:
Note that the solution $y(t)=e^{-t}+e^{-2t}$ you obtained is only valid for $t>0$. For $t<0$ the solution of the now homogeneous equation is $y(t)=2e^{-2t}$, so that in total
$$
y(t)=2e^{-2t}+H(t)(e^{-t}-e^{-2t}).
$$


问:
Bound of the sum $\sum_{p\le n}\frac{1}{\log(p)}$
While doing a sum I came to the sum $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}$.
Where the $\log$ is the natural logarithm.

It was easy to prove that $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}<\pi(n)+\text{const.}$.

Thus $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}=o\left(\frac{n}{\log(n)}\right)$.
Using my computer I can see that $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}=o\left(\log(n)^{3/2}\right)$.

I can go further and see that it is $o\left(\log(n)^{1.4}\right)$. It may be true that $o\left(log(n)^{1.35}\right)$.

(I have checked $n$ till $10^{19}$. My C++ code can't be trusted anymore because $64$ bit integers can't take in more values and precision of long double is up to like $16$ digits.)

My conjecture is that it is $O(\log(n))$. Is this true? Are there any other bounds? I don't want advanced proofs as I am a novice in Maths. If there is any bounds just post them here without the proofs and with some comments, if necessary.

答:
By partial summation, we have

$$
\sum_{p\le x}{1\over\log p}={\pi(x)\over\log x}+\int_2^x{\pi(t)\over t\log^2t}\mathrm dt
$$

Due to the prime number theorem that

$$
\pi(x)={x\over\log x}+\mathcal O\left(x\over\log^2x\right)
$$

we have

$$
\sum_{p\le x}{1\over\log p}={x\over\log^2x}+\mathcal O\left(x\over\log^3x\right)+\mathcal O\left(\int_2^x{\mathrm dt\over\log^3t}\right)
$$

By partitioning the remaining integral into

$$
\int_2^{\sqrt x}+\int_{\sqrt x}^x
$$

it easily follows that

$$
\int_2^x{\mathrm dt\over\log^3t}\asymp{x\over\log^3x}
$$

Consequently, we have

$$
\sum_{p\le x}{1\over\log p}={x\over\log^2x}+\mathcal O\left(x\over\log^3x\right)
$$


问:
True or False? If the limit of $[f(x)-g(x)]$ as x approaches $a = 0$, then the limit of $f(x), g(x)$, as $x$ approaches $a$, are equal.
TRUE OF FALSE?
$$\lim_{x\to\alpha}[f(x)-g(x)]=0\implies\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha} g(x)$$

I'm in senior high school and just recently took a test with this as one of the questions. I initially answered true — I mean, it obviously looks too true. Simple distribution and manipulation of terms would easily show $\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha}g(x)$ — but just before submitting, I changed it to false. I got a perfect score, but I do feel uneasy not knowing why my teacher remarked this as false.

Just for context, maybe the counterexample to make this statement false might involve infinities.  When the limit is $+\infty$ or $-\infty$, we simply write is as "does not exist."
Also we only tackle real number. And our limit definition specifies that $\lim_{x\to\alpha}f(x)$ only exists when $\lim_{x\to\alpha^{+}}f(x)=\lim_{x\to\alpha^{-}}f(x)$.

Qn: The actual question looked like this Here. Link

答:
What about $\alpha =0$ and
$$f(x)=1+\sin\left(\frac{1}{x}\right) \text{ and } g(x)= \sin\left(\frac{1}{x}\right)$$

as a counterexample? If you want $f,g$ to be defined at $0$ just set $f(0)=g(0)=0$.

If you suppose either $f$ or $g$ to have a limit at $\alpha$, then the result becomes true.


问:
Finding the determinant of a Cauchy matrix
Found this nice matrix:

$$\begin{pmatrix} \dfrac{1}{a_1+b_1} & \dfrac{1}{a_2+b_1} & \cdots & \dfrac{1}{a_n+b_1} \\ \dfrac{1}{a_2+b_1} & \dfrac{1}{a_2+b_2} & \cdots &  \dfrac{1}{a_n+b_2}\\ \vdots & \vdots &\ddots & \vdots \\ \dfrac{1}{a_n + b_1} & \dfrac{1}{a_n + b_2} & \cdots & \dfrac{1}{a_n+b_n}\end{pmatrix}$$

What is its determinant?  Is there a way to simplify it into a more compact form?

I was able to find the determinant for $2\times 2$ case:

$$\dfrac{(a_1-a_2)(b_1-b_2)}{(a_1+b_1)(a_2+b_2)(a_1+b_2)(a_2+b_1)}$$

Is there a nice form for the $n\times n$ case?

答: This is a Cauchy matrix. By applying the formula on Wikipedia to your specific Cauchy matrix, we get that the determinant is $$\dfrac{\prod_{i = 2}^{n}\prod_{j = 1}^{i-1}(a_i-a_j)(b_i-b_j)}{\prod_{i = 1}^{n}\prod_{j = 1}^{n}(a_i+b_j)}.$$


问:
Manageable project to learn some arithmetic geometry
OK, this may be an unusual question, since I'm asking you for advice, the way I would ask a supervisor were I a grad student.

I am a thirty-something maths teacher with an unresolved attraction to arithmetic geometry, and would like to know more about themes I find fascinating when I read texts about arithmetic geometry, such as:

the local-global principle

the way geometry influences the arithmetic behaviour of Diophantine equations, like in Faltings's theorem

the way scheme theory helps number theory by making a lot of geometric constructions possible, starting from base change

One of the reasons this attraction is unresolved is that I never managed to seriously learn enough algebraic number theory, algebraic geometry, etc. in a "bottom-to-top" way: I have often started to read for instance R. Vakil's The rising sea, D. Eisenbud & J. Harris The geometry of schemes or Q. Liu's Algebraic geometry and arithmetic curves  and find them to be very good texts, but I don't have the stamina to ingest hundreds of pages without a more specific goal.

Because of this there are a number of subjects I'm vaguely familiar with, but don't know in any depth. For instance, I know what a scheme is, I even think I understand in part their raison d'être, but am probably unable to pass an scheme theory exam (and the same goes for elliptic curves, derived categories, class field theory...).

To change this, I would like to try and understand a specific result and working my way from the top down to acquire the needed skills. The result has to be interesting enough to preserve my motivation in the long run, but "small" enough so that I could realistically understand it in a few months. I would love it to really mix number theory and geometry.

So the question is:

what theorem would constitute a reasonable-size project?

Personal facts that might be relevant: my PhD is in low-dimensional topology, so I'm quite comfortable with things like cohomology and Riemann surfaces, and I like it when geometric intuition is useful. Large doses of homological algebra or category theory don't frighten me.

答:
This is complementary to hm2020's answer, in that learning about the Hurwitz theorem and Riemann Roch are fantastic places to see some beautiful geometry. I think one more arithmetic direction this can go is to prove the (easier parts of) the Weil conjectures for smooth curves over finite fields. These conjectures motivated massive amounts of algebraic geometry, and are very down to earth, its all just counting solutions to equations over finite fields!

This is all done very well in Dino Lorenzini's book An Invitation to Arithmetic Geometry, and he goes out of his way to emphasise the arithmetic=geometry dictionary.

In the last chapter he even proves the hard part of the Weil conjectures, the Riemann Hypothesis for curves, which gives a nice perspective of why the integer version might be true (that the primes are as well distributed as possible).


答: Take $f=g$ to be any function that does not have a limit at $\alpha$, for example $\tan x$ at $\alpha=\frac\pi2$.

Title: True or False? If the limit of $[f(x)-g(x)]$ as x approaches $a = 0$, then the limit of $f(x), g(x)$, as $x$ approaches $a$, are equal.

Body:
TRUE OF FALSE?
$$\lim_{x\to\alpha}[f(x)-g(x)]=0\implies\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha} g(x)$$

I'm in senior high school and just recently took a test with this as one of the questions. I initially answered true — I mean, it obviously looks too true. Simple distribution and manipulation of terms would easily show $\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha}g(x)$ — but just before submitting, I changed it to false. I got a perfect score, but I do feel uneasy not knowing why my teacher remarked this as false.

Just for context, maybe the counterexample to make this statement false might involve infinities.  When the limit is $+\infty$ or $-\infty$, we simply write is as "does not exist."
Also we only tackle real number. And our limit definition specifies that $\lim_{x\to\alpha}f(x)$ only exists when $\lim_{x\to\alpha^{+}}f(x)=\lim_{x\to\alpha^{-}}f(x)$.

Qn: The actual question looked like this Here. Link


问:
Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$
I am trying to show that  $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:

\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.

答:
$$y=\frac{x^{2x}}{(x+1)^{x+1}}\implies \log(y)=2x \log(x)-(x+1)\log(x+1)$$
$$\log(y)=2x \log(x)-(x+1)\left(\log(x)+\log \left(1+\frac{1}{x}\right)\right)$$
$$\log(y)=(x-1) \log (x)-(x+1)\log \left(1+\frac{1}{x}\right)$$
Now, by Taylor series
$$(x+1)\log \left(1+\frac{1}{x}\right)=(x+1)\Bigg[\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right) \Bigg]$$
$$(x+1)\log \left(1+\frac{1}{x}\right)=1+\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ Putting all together
$$\log(y)=(x-1) \log (x)-1-\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ which already tends to $\infty$. If you want to continue, use
$y=e^{\log(x)}$


问:
proof reading and a little question and refine english about college-algebra

Welcome to edit my post to revise any mistakes, both English and Math, thanks.

Theorem 12

polyIC=integer polynomial

polyRC=rational polynomial

Suppose $f(x)=a_nx^n+a_{n-1}x^{n-1}+\text{...}+a_0$ is a polyIC, and $\frac{r}{s}$ is a rational root, where $(r,s)=1$.

Then we must have $s\left|a_n\right.,r\left|a_0\right.$. Specially, If the leading coefficient of $f(x)$ satisfies $a_n=1$,

then the rational roots of $f(x)$ are all integer roots, and they are factors of $a_0$.

Proof$\unicode{ff1a}$

Because $\frac{r}{s}$ is an rational root of $f(x)$. On rational number field $\left.\left(x-\frac{r}{s}\right)\right|f(x)$.

Thus $[\text{Mark1}]$ $(\text{sx}-r)|f(x)$.  For $(r,s)=1$, $s x-r$ is an primitive polynomial.

Conclude from above, $f(x)=(s x-r)\left(b_{n-1}x^{n-1}+\text{...}+b_0\right)$,where $b_{n-1},\text{...},b_0$ are all integers.

Compare the coefficients of two sides, that is $a_n=s b_{n-1},a_0=-r b_0$.

Hence $s\left|a_n\right.,r\left|a_0\right.$.

one little question

Does $[\text{Mark1}]$ so obvious?

$\left.\left(x-\frac{r}{s}\right)\right|f(x)$ $\Longrightarrow $ $s|f(x)$

of course, this is obvious in this example: f below, has one rational root-4/5$\unicode{ff0c}$so 5(x-4/5)$|$f(x)

$\pmb{f=25\left(x-\frac{4}{5}\right)(x-2)\left(x+\frac{4}{5}\right)\text{//}\text{Expand}}$

$32-16 x-50 x^2+25 x^3$

$\pmb{\text{Solve}[f==0,x]}$

$\left\{\left\{x\to -\frac{4}{5}\right\},\left\{x\to \frac{4}{5}\right\},\{x\to 2\}\right\}$

$\pmb{\text{Factor}[f]}$

$(-2+x) (-4+5 x) (4+5 x)$

$\pmb{5(x-4/5)\text{//}\text{Expand}}$

$-4+5 x$

答: When you divide a polynomial $f$ with rational coefficients by $x-t$ with $t$ rational, you get a quotient polynomial with rational coefficients and a rational number as remainder; $$f(x)=(x-t)q(x)+r$$ Now let $x=t$, and you get $f(t)=r$. So if $f(t)=0$, then $r=0$, so $f(x)=(x-t)q(x)$, so $x-t$ divides $f(x)$. </p>


问:
What does double vertical bars notation mean in probability?
I'm studying on generative adversarial networks and I've come across the following formula for inception score :

$$\text{IS}(G) \approx \exp \left(\frac{1}{N}\sum_{i=1}^ND_{KL} \left(p \left(y\vert x^{(i)} \Vert \hat{p} \left(y \right) \right) \right) \right).$$
I just want to know what does those $\Vert$ (Double vertical bars) mean in that formula?

答: I don't know what just bumped this question to front page, but the correct answer to this question is given in the comments; namely, there is a paranthesis missing in the original question, and the double bars are traditionally used to indicate the two inputs to KL divergence (presumably so that you can put conditional distributions in there.)</p>


答:
The key is that you can only say two limits are equal if they exist, and the two limits on the RHS cannot be guaranteed to exist with the given information. You are right in suggesting that the answer can involve infinity, but there are finite examples that are badly behaved as well.

Consider the function $f(x)$ where $f(x)=1$ for $x$ rational and $f(x)=0$ for $x$ irrational. It is a well defined function, but has no limit anywhere. If we let $g(x)=f(x)$, then $f(x)-g(x)=0$ everywhere so the limit is zero for every $\alpha$, but $f(x)$ and $g(x)$ have no limit no matter what value of $\alpha$ is chosen.


问:
Coordinate systems and Vector spaces
Coordinate systems are used to uniquely define different points in space(sometimes not so as in the case of origin described by spherical or cylindrical systems).Whereas, a vector space is a set containing objects(vectors), which satisfy certain axioms.A basis vector set is any subset of the Space, whose linear combination of its elements describes the entire space.

When combining these ideas to describe space in cartesian system, the basis is fixed, while in other orthogonal systems they are position dependent. This sense of direction makes sense when  using bases to describe vector fields, but when they are position dependent, does it mean our basis vector is a different set at every position in space ,as the "direction is changing",or is this notion of no significance when talking in terms of sets?

Also , I want to have some intuition regarding the relation between coefficients of the basis vectors we choose and the coordinates.

答:
A basis $B$ of a vector space $V$ is a linearly independent subset of $V$ that spans $V$. By definition, you can write every vector $v \in V$ as a linear combination of the basis vectors. To define a coordinate system, you must have an ordering on the basis vectors, and the coordinates are defined by the coefficients.

For example, consider the basis $B = \{b_1, \cdots, b_n\}$. We can write any vector $v \in V$ as
$$v = \alpha_1 v_1 + \cdots + \alpha_n b_n$$
where $\alpha_1, \cdots, \alpha_n$ are the referred to as coordinates. Keep in mind, the ordering of the basis is important. So, by definition, an ordered basis defines a coordinate system.

Now, if a coordinate system A is translated with respect to another coordinates system B, then the basis alone is not adequate for capturing this information. In such cases, you must define the origin of coordinate system A with respect to coordinates system B (or the other way around). In most applications (in my experience), we describe such relations using transformation matrices. I suggest looking at the Special Orthogonal Group $SO(n)$ (comprises of rotations) and Special Euclidean Group $SE(n)$ (comprises of rotations and translations).

Title: Coordinate systems and Vector spaces


答:
Start from a simple case in which this is true:
$$ f=x, \quad g=0 $$ when $ \alpha \to 0 $.

We do have the limit $$ \lim_{\alpha \to 0} f(x)-g(x)=0 $$

For any situation similar to this, I can simply plus a so-called "pathological" term to both $f$ and $g$ to make their limits at $0$ do not exist

$$ f=x+\sin \frac{1}{x}, \quad g=\sin \frac{1}{x} $$

or

$$ f=x+e^{\frac{1}{x}}\sin \frac{1}{x^2}, \quad g=e^{\frac{1}{x}}\sin \frac{1}{x^2} $$
Thus our counterexamples do not have to be infinity cases.
Hope this would help.


问:
What can we conclude about $P(X_1+X_2+\dots+X_n\le na)$?
If for a sequence of random variables $X_1, X_2, \dots, X_n$, we know
$$P(X_i\le a)\le e^{-ca}, \quad i=1,2,3,\dots, n$$

What can we conclude about $P(X_1+X_2+\dots+X_n\le na)$?

Can we get
$$P(X_1+X_2+\dots+X_n\le na)=P(X_i\le a, \mbox{ for any $i=1,2,\dots,n$})=(e^{-ca})^n$$
if we know that $X_i$ is independent?

答: $P(X_1+X_2+\cdots+X_n \leq na) \leq P(\bigcup_i (X_i \leq a)) \leq ne^{-ca}$ even without independence.


问:
find the roots $(2z+3)^3=\frac{1}{64}$
There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$  so $z=\frac{-11}{8}$  however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you

答:
There is a simpler method. Just use

$$\begin{align}a^3=b^3 \Longleftrightarrow a^3-b^3=0\Longleftrightarrow (a-b)(a^2+ab+b^2)=0\end{align}$$

To find the complex roots, just  solve the quadratic:

$$a^2+ab+b^2=0.$$

In your case you can take $$a=2z+3, b=\dfrac 14.$$

Title: find the roots $(2z+3)^3=\frac{1}{64}$


问:
Finding maximum volume
A box has corner (0,0,0) and all edges parallel to the axes. If the opposite corner (x,y,z) is on the plane $$ 3x+2y+z=1 $$, what position gives maximum volume? Show first that the problem maximizes
$$
xy-3x^2y-2xy^2
$$

Can somebody please explain how this should be solved?

答:
From the description of the problem, we have each edge parallel to an axis, one corner at $(0,0,0)$ and one corner at $(x,y,z)$. This means that we can think about the sides of the box each having length $x$, $y$, and $z$. This gives us an area function $A(x,y,z)=xyz$.

We know that the corner $(x,y,z)$ is on the plane defined by $3x+2y+z=1$. If we solve this equation for $z$, we have $z=1-3x-2y$. We can the substitute this into our area function and have $$A(x,y,z)=xy(1-3x-2y)=xy-3x^2y-2xy^2.$$

Now, we need to maximize this function. To get started, we need to take partial derivatives of $A$ with respect to both $x$ and $y$. Please ask if you need help with this part.

Title: Finding maximum volume


Title: find the roots $(2z+3)^3=\frac{1}{64}$

Body:
There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$  so $z=\frac{-11}{8}$  however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you

Body:
$$(2z+3)^3=\frac{1}{64} \implies 2z+3=\frac{1}{4}, \frac{\omega}{4}, \frac{\omega^2}{4}$$
$$\implies z=\frac{-11}{8}, \frac{\omega -12}{8}, \frac{\omega^2-12}{8}.$$ Here $\omega$ is cube-root of unity.


问:
Automorphisms of Klein group
I have a problem I can't solve I need to show all the automorphism in Klein group and their order and after that, I need to find that what is the group that  these automorphisms are isomorph with.

I have half of it I have the automorphisms in the group I used representatives like $Id=e$   $(12)(34)=x$  $\;\;(13)(24)=y$ $\;\;(14)(23)=z$ and I found $6$ automorphism

$xyz\rightarrow xyz$

$xyz\rightarrow xzy$

$xyz\rightarrow yxz$

$xyz\rightarrow yzx$

$xyz\rightarrow zxy$

$xyz\rightarrow zyx$

And my question is about the second part. How do I get the order each of these automorphisms and with what group this automorphismgroup isomorph with?

Thank you in advance.

答:
Hint: let $V=\{1,a_1(=a_2a_3),a_2(=a_1a_3),a_3(=a_1a_2)\}$ be Klein's $4$-group. Prove that the map $\varphi\colon S_3\to\operatorname{Aut}(V)$, defined by:

\begin{alignat}{1}
&\varphi(\sigma)(1)=1 \\
&\varphi(\sigma)(a_i)=a_{\sigma(i)}
\end{alignat}

is an isomorphism (good-definiteness, injectivity, surjectivity, operation-preservation [homomorphism]).

Title: Automorphisms of Klein group


问:
GRE topology definition
So on a certain GRE subject problem they define for a given a subset $Y$ of a topological space $X$

$$Y' = \{x \in X : x \in \text{cl}(Y \setminus \{x\})\}$$

Where $\text{cl}$ is closure. Is this just a fancy definition for the set of limit points of $Y$

Thanks!

答: Yes, this is clear after you know that $\operatorname{cl}(Y)= \{x \in X: \forall O \text{ open }: (x \in O) \to (O \cap Y \neq \emptyset) \}$ for all $Y \subseteq X$ the closure of $Y$ is the set of points $x$ such that each (open) neighbourhood of $x$ intersects $Y$. A limit point of $Y$ is the same except that each neighbourhood of $x$ must contain a point of $Y$ different from $x$ itself, which in fact says that $x$ is in the closure of $Y\setminus \{x\}$.

Title: GRE topology definition


问:
Pullback, 2-form invariant
Let $\mathbb H^2 = \{(x, y) \in \mathbb{R}^2 : y> 0\}$ and consider the $2$-form in $\mathbb H^2$ defined by
$$\varphi = \dfrac{dx \wedge dy}{y^2}.$$

Show that $\varphi$ is invariant $(T^*(\varphi)= \varphi)$ under the transformation $T$ from $\mathbb H^2$ to $\mathbb H^2$ given by
$$T(z) = \dfrac{az + b}{cz + d}$$

where $z \in \mathbb H^2 \subset \mathbb{C}$ and $a, b, c, d \in \mathbb{R}$ with $ad - bc \neq 0$.

My attempt was to try to express $T(x, y)$ and calculate $dx $ and $ dy $ to make $ dx \wedge dy $, and calculate the pullback. I couldn't, the expression for $T$ was strange.

答:
Write it in terms of $z = x + iy$ and $\overline{z} = x - iy$.

$$\frac{dx \wedge dy}{y^2} = -2i \cdot \frac{dz \wedge d\overline{z}}{(\overline{z} - z)^2}$$

Then plug in and simplify. You will find that

$$\frac{d\left(\frac{az + b}{cz+d}\right) \wedge d\left(\frac{a\overline{z} + b}{c\overline{z} + d}\right)}{\left( \left(\frac{a\overline{z} + b}{c\overline{z} + d}\right) - \left(\frac{a
{z} + b}{c
{z} + d}\right)\right)^2} =  \frac{dz \wedge d\overline{z}}{(\overline{z} - z)^2}$$

Title: Pullback, 2-form invariant


问:
Method to find if a polynomial has irrational roots?

Question

Let's say I define a polynomial $P(x)$ whose roots $\alpha_i$ and degree $> 1$.
We also add constraint that the coefficient of the highest power of $P(x)$ is $1$ and all other coefficients are integers.

when $P(0) \neq 0$. Then:

$$ \frac{P(|P(0)|)}{P(0)} = \prod_{i=1}^{\deg P(x)} (1 - \frac{P(0)}{\alpha_i}) $$

where |x| is the modulus of x. Let, the number of prime factors of ${P(|P(0)|)}/{P(0)}$ be $z$. Then if,

$$ \deg P(x) > z$$

and $P(2) \neq 0$ this implies $P(x)$ has some irrational roots.

Example

$$P(x) = x^4 -x^3  + 3x^2 + 3$$

thus,

$$\frac{P(|P(0)|)}{P(0)} = 7 \times 2 \times 2$$

Hence, the number of primes are $3$ but $\deg P(x) = 4 > 3$. We conclude, $P(x)$ has irrational roots.

Question

Is it possible to quantify how often a polynomial with some irrational roots of the form $P(x)$ satisfies  $ \deg P(x) > z$?

答:
If $P$ of degree $n$ and with $P(0)\ne 0$ has only rational roots, then these must in fact be (non-zero) integers and $P(0)$ is (up to sign) their product. Thus
$$\tag1 \frac{P(|P(0)|)}{P(0)}=\frac{\prod_{i=1}^n(|P(0)|-\alpha_i)}{\prod_{i=1}^n(0-\alpha_i)}=\prod_{i=1}^n\left(1-\frac{|P(0)|}{\alpha_i}\right)$$
is the product of $n$ integers. As $P(0)\ne 0$, none of the factors is $=1$. If additionally $P(|P(0)|)\ne 0$, none of the factors is $=0$.

Case 1. If none of the factors in $(1)$ is $=-1$, then all $n$ factors are integers $\notin\{-1,0,1\}$ and hence each contributes at least one prime factor. So in this case $z\ge n$, as desired.

Case 2. Now suppose that at least one of the factors in $(1)$ is $=-1$, which happens precisely when $|P(0)|=2\alpha_i$ for some $i$.
Note that in this case, the product of the other $n-1$ roots is $\pm2$, i.e., exactly one of them is $\pm2$ (but by the additional condition that $P(2)\ne 0$, it must be $-2$) and the rest are $\pm1$. In other words,
$$ P(X)=(X-\alpha)(X+ 2)(X-1)^{n_1}(X+1)^{n_2}$$
for some positive integer $\alpha$ and with $n_1+n_2=n-2$.
Explicitly, this makes $|P(0)|=2\alpha$ and $$\tag2\frac{P(|P(0)|)}{P(0)}=\frac{P(2\alpha)}{P(0)}=(-1)\cdot (\alpha+1)\cdot (1+2\alpha)^{n_1}(1-2\alpha)^{n_2}$$

If $\alpha=1$, this becomes $(-1)^{n_2+1}\cdot 2\cdot 3^{n_1}$ and has $z=n_1+1$ prime factors. As this makes $z<n$, we obtain counterexamples to the claim!
For a concrete counterexample, consider
$$ P(X)=(X-1)(X+2)=X^2+X-2,$$
for which $\frac{P(|P(0)|}{P(0)}=\frac{P(2)}{-2}=-2$ has only one prime factor, but has only rational roots.

We cannot have $\alpha=2$ as that would make $P(2)=0$.

If $\alpha\ge 3$, each factor in $(2)$ except the first is $\notin\{-1,0,1\}$, hence the only way to have $z<n$ is when all of these are $\pm$ a prime. In particular $(2\alpha-1,2\alpha+1)$ must be a pair of twin primes. As $\alpha=2$ is excluded, these twin primes must be of the form $6k\pm1$, i.e., we must have $\alpha=3k$ and $\alpha+1=3k+1$ also prime. This happens when $a\in\{6, 30,36,96,156,210,330,546,576,660,726,810,936,966,\ldots\}$ and so gives rise to many more counterexamples. Here's a concrete les trivial one:
$$ P(X)= X^4 - 94X^3 - 193X^2 + 94X + 192$$
has $P(0)=192$, $P(192)=686536512$, $\frac{P(192)}{P(0)}= 97\cdot 191\cdot 193$, so $z=3<n=4$.


问:
Proof for using the formula of cross product for area and volume of a shape
For two vectors $2i + 3j - 4k$ and $i + 2k$. The way I understood crops product is that we can find these two vectors magnitude. Let us say (Assumption ) $= 10$ and $7$. Now , to find area . We do $\frac 12 \times 10 \times  7 \times \sin \theta$. Probably because $10$ and $7$ are parallel to each other. And if we draw something that is not parallel to $10$ units. Then , we can find an area.
This is one way.

Second way is that we write it in discriminant form.

Now , my confusion is that solving it in discriminant form , It gives area. Now , how did we take care of sin theta here and why do we write it in discriminant form.

Just a picture of how I view these Values.

答: Area of a triangle formed by two adjacent vectors $\vec A$ and $\vec B$ is given by $\frac{1}{2} |\vec A \times \vec B|= \frac{1}{2} |\vec A| |\vec B| \sin \theta$. If you find area by LHS in above $\sin \theta $ as taken care of. Other thing is the cros product is represented by the determinant which gives you a vector, them find its modulus.


问:
names for "sub" discrete topology
Given a finite discrete set $X$. Its powerset $P(X)$ is often referred to as discrete topology.

I'm wondering what are the names for the topology of $X$: $S(X)$ ? Which is only a proper subset of $P(X)$.

$S(X) \subset P(X)$.

I'm not interested in a general subset of (), but the ones are the topology of $X$.

And $S(X) \neq \emptyset$, so it's not the other popular topology trivial topology.

答:
In fact, $\mathcal{T}=P(X)$ is always knwon as the discrete topology (for both finite and infinite sets). It's a topology you can put on any set.

Any proper subset of $P(X)$ that is a topology is just called a non-discrete topology, if it's given a name at all.

The trivial, or indiscrete, topology s just $\{\emptyset, X\}$, the minimal topology on any set $X$.

Title: names for "sub" discrete topology


问:
How to factorise $x^2 + 1$
Keep in mind that I am not asking how to find the solutions to the equation $x^2 + 1 = 0$. I know that there will be imaginary answers In this case. I just want to know how to factor this using a logical method and obtaining 2 factors. Thanks for answering.

答:
I will suggest you $3$ different ways.

If you want to factorise $x^2+1$, remember that

$$x^2+1=(x-a)(x-b)$$

For general factorization we will find  $a,b\in\mathbb C$ such that $x^2+1=(x-a)(x-b)$.

We have,

$$x^2+1=x^2-x(a+b)+ab$$

$$ab=1, ~a+b=0$$

$$-a^2=1,~ -b^2=1$$

$$a^2=-1,~ b^2=-1$$

$$a=\pm \sqrt {-1}, ~b=-a=\mp \sqrt {-1}$$

$$a=\pm i, ~ b=\mp i$$

$$x^2+1=(x-i)(x+i)\\~~~~~~~~~~~~ =(x+i)(x-i).$$

But you can do faster,

$$\begin{align}x^2+1&=x^2-(-1)
\\&=x^2-i^2
\\&= (x-i)(x+i).\end{align}$$

Alternatively, the most common way for factorization you need to find the roots of polynomial. If, $r_1,r_2$ are the roots of $x^2+1$, then

$$x^2+1=(x-r_1)(x-r_2)$$

Thus, you need just to solve

$$x^2+1=0 \implies x_{1,2}=±i.$$

It is enough to take $r_1=i, r_2=-i.$

You get again the same result:

$$x^2+1=(x-r_1)(x-r_2)\\
~~~~~~~\thinspace=(x-i)(x+i).$$

Title: How to factorise $x^2 + 1$


答:
For any $z = a + ib \in \mathbb{C}$,

$$
z = \rho\left(\cos{\varphi} + i\sin{\varphi}\right),
$$

where

$$
\rho = \sqrt{a^2 + b^2}, \text{ and }\varphi = \arctan\left(\frac{b}{a}\right).
$$

Then,

$$
\sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos{\left(\frac{\varphi + 2\pi k}{n}\right)} + i\sin{\left(\frac{\varphi + 2\pi k}{n}\right)}\right), \quad k = 0, 1, \ldots, n-1.
$$

In our example,

$$
\frac{1}{64} = \frac{1}{64} + i0, \text{ i. e., }a = \frac{1}{64}\text{ and }b = 0 \Rightarrow
$$

$$
\frac{1}{64} = \frac{1}{64}\left(\cos(0) + i\sin(0)\right) \Rightarrow
$$

$$
\sqrt[3]{\frac{1}{64}} = \sqrt[3]{\frac{1}{64}}\left(\cos\left(\frac{2\pi k}{3}\right) + i\sin\left(\frac{2\pi k}{3}\right)\right), \quad k = 0, 1, 2.
$$

$$
\begin{array}{rl}
k = 0 \Rightarrow & z_1 = \frac{1}{4} \\
k = 1 \Rightarrow & z_2 = \frac{1}{4}\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\frac{1 - i\sqrt{3}}{8}\\
k = 2 \Rightarrow & z_3 = \frac{1}{4}\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1 + i\sqrt{3}}{8}
\end{array}
$$

Title: find the roots $(2z+3)^3=\frac{1}{64}$


问:
Weak convergence in $L^2$ of powers of a sequence which is weakly convergent in $W^{1,2}$
Let $\Omega\subset \mathbb{R}^3$ be a bounded domain. Assume that $(f_n)_{n\in \mathbb{N}}\subset W^{1,2}(\Omega)$ is weakly convergent to $f\in W^{1,2}(\Omega)$.
Consider now $(f_n^3)_{n\in \mathbb{N}}$ which is bounded in $L^2(\Omega)$ by the embedding $W^{1,2}(\Omega)\hookrightarrow L^6(\Omega)$, so that in converges weakly in $L^2(\Omega)$. Is it possible to show that the weak limit is acutally $f^3$?

答:
By compact embeddings of $W^{1,2}$ into $L^p$, $p<6$, we have $f_{n_k}\to f$
weakly in $H^1$, strongly in $L^p$ for $p<6$, and pointwise almost everywhere.
This implies $f_{n_k}^3 \to f^3$ pointwise almost everywhere.

Since $(f_n^3)$ is bounded in $L^2$ it has a weakly converging subsequnec, so we could have chosen $f_{n_k}$ from the beginning such that $f_{n_k}^3 \rightharpoonup g$ in $L^2$ for some $g$. Pointwise a.e. and weak limits coincide, so $f^3=g$.

By a subsequence-subsequence argument, the whole sequence $(f_n^3)$ converges weakly to $f^3$.


问:
Numerical Approximation for 2D Curvature
I have a list of points (x, y) that are taken from an unknown 2D parametric curve $\vec{f}(t)$. These points are monotonically increasing in t (ie: they're a "connect the dots" version of $\vec{f}(t)$). I would like to numerically approximate the curvature at all of these points.

I know that

$\kappa = |\frac{dT}{ds}|
        =  \frac{|\vec{T}'(t)|}{|\vec{r}'(t)|}
        =  \frac{|\vec{r}'(t) \times \vec{r}'(t)|}{|\vec{r}'(t)|^3}$

and I was planning on using the last formula, approximating the derivative as:

$\frac{dx}{dt} \approx \frac{x(t + \Delta t) - x(t - \Delta t)}{2\Delta t}$

Is there a better way of doing this? (Maybe by fitting a curve to the points?)

答:
Have a look at chapter "10.4 The Curvature of a Planar Digital Curve" in KLETTE, Reinhard; ROSENFELD, Azriel. Digital geometry: Geometric methods for digital picture analysis. Morgan Kaufmann, 2004.

See also
Adina CIOMAGA, Pascal MONASSE, and Jean-Michel MOREL, The Image Curvature Microscope: Accurate Curvature Computation at Subpixel Resolution, Image Processing On Line, 7 (2017), pp. 197–217. https://doi.org/10.5201/ipol.2017.212


问:
Second pair of matching birthdays
The "birthday problem" is well-known and well-studied.  There are many versions of it and many questions one might ask.  For example, "how many people do we need in a room to obtain at least a 50% chance that some pair shares a birthday?"  (Answer: 23)

Another is this: "Given $M$ bins, what is the expected number of balls I must toss uniformly at random into bins before some bin will contain 2 balls?"  (Answer: $\sqrt{M \pi/2} +2/3$)

Here is my question: what is the expected number of balls I must toss into $M$ bins to get two  collisions?  More precisely, how many expected balls must I toss to obtain the event "ball lands in occupied bin" twice?

I need an answer for very large $M$, so solutions including summations are not helpful.

Silly Observation:

The birthday problem predicts we need about 25 US Presidents for them to share a birthday.  It actually took 28 presidents to happen (Harding and Polk were both born on Nov 2).  We see from the answers below that after about 37 US Presidents we should have a 2nd collision.  However Obama is the 43rd and it still hasn't happened (nor would it have happened if McCain had won or Romney had won; nor will it happen if H. Clinton wins in 2016).

答:
We will change the problem slightly and make some approximations.  Instead of the expected value of $n$ to get two collisions, we will think about the value of $n$ to get a $50\%$ chance of two collisions.  They should be very close, as the probability of two collisions will rise quickly from near $0$ to near $1$.  In the single collision case, it is the difference between a factor of $\sqrt{2 \log 2}\approx 1.177$ and $\sqrt {\frac \pi 2}\approx 1.253$.  The approximation is that each pair of chosen elements has the same probability to match, $\frac 1M$.  This ignores the correlations between the pairs.

In this case, the distribution of number of collisions is Poisson, with $\lambda$ being the expected number of collisions.  The chance that we will not have two collisions is $(1+\lambda)\exp(-\lambda)$ with solution $\lambda \approx 1.67835$ so we want $\frac {n^2}{2M}=1.67835$ or $n=\sqrt{2M1.67835}$ or about $55\%$ more than the number to get the first collision.

Title: Second pair of matching birthdays


答:
Rewrite $x^2 + 1$ as $x^2 - (-1)$. This is a difference of two squares, which is easy to factorise.
$x^2 - (-1)\\$
$= x^2 - i^2\\$
$= (x + i)(x - i)$

Title: How to factorise $x^2 + 1$

Body: Keep in mind that I am not asking how to find the solutions to the equation $x^2 + 1 = 0$. I know that there will be imaginary answers In this case. I just want to know how to factor this using a logical method and obtaining 2 factors. Thanks for answering.


问:
True or False? If the limit of $[f(x)-g(x)]$ as x approaches $a = 0$, then the limit of $f(x), g(x)$, as $x$ approaches $a$, are equal.
TRUE OF FALSE?
$$\lim_{x\to\alpha}[f(x)-g(x)]=0\implies\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha} g(x)$$

I'm in senior high school and just recently took a test with this as one of the questions. I initially answered true — I mean, it obviously looks too true. Simple distribution and manipulation of terms would easily show $\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha}g(x)$ — but just before submitting, I changed it to false. I got a perfect score, but I do feel uneasy not knowing why my teacher remarked this as false.

Just for context, maybe the counterexample to make this statement false might involve infinities.  When the limit is $+\infty$ or $-\infty$, we simply write is as "does not exist."
Also we only tackle real number. And our limit definition specifies that $\lim_{x\to\alpha}f(x)$ only exists when $\lim_{x\to\alpha^{+}}f(x)=\lim_{x\to\alpha^{-}}f(x)$.

Qn: The actual question looked like this Here. Link

答:
If atleast one of $\lim f(x)$ or $\lim g(x)$ exists on $\mathbb R$ i.e. exists finitely then the given statement is true.

WLOG let $\lim_{x\to a}g(x)=l$, where $l\in \mathbb R$

Given that: $\lim_{x\to a}(f(x)-g(x))=0$

So $f(x)=(f(x)-g(x))+g(x)\implies \lim_{x\to a}f(x)=0+l=l=\lim_{x\to 0}g(x)$

However, if you consider $f(x)=[x]$, where $[.]$ is floor function and $g(x)=-\{x\}$, where $\{.\}$ is fractional part function. Indeed we have $\lim _{x\to 0} (f(x)-g(x))=0$ but neither $\lim_{x\to 0}f(x)$ exists nor does $\lim_{x\to 0} g(x)$.

Note: It is to be noted that even if either of the limits $\lim_{x\to a} f(x)$ or $\lim_{x\to a} g(x)$ exists on extended reals that is on the set $\mathbb R\cup \{-\infty, \infty\}$, still if $\lim_{x\to a}(f(x)-g(x))=0$
then $f(x)$ and $g(x)$ have the same limiting behaviour as $x\to a$.

For example: Consider $f(x)=x+\frac 1{|x|}, g(x)=\frac 1{|x|}$, then $f(x)-g(x)=x\to 0$ as $x\to 0$. Also, $\lim_{x\to 0}f(x)=\infty$ and $\lim_{x\to 0}g(x)=\infty$


问:
Show that $A = \lambda I_n$ if $\lambda$ is an eigenvalue of $A$ with multiplicity $n$.
Let $A$ be a diagonalizable $n \times n$ matrix, and suppose that $\lambda$ is an eigenvalue of $A$ with multiplicity $n$. I want to show that $A= \lambda I_n$.

I started by saying that the determinant $\det (A- \lambda I_n)$ yields a characteristic polynomial with root $\lambda$. Thus, the eigenvalues of A are just $\lambda$ with multiplicity $n$. To find the eigenvectors now, we need to find the null space of $A- \lambda I_n$.

This is where I get confused. In order for $A = \lambda I_n$ to be true, $A - \lambda I_n = 0$ must also be true. Moreover, let $\vec v$ be an eigenvector of $A$ belonging to an eigenvalue $\lambda$. Then $v ∈ N(A − λI)$. But I'm not sure how to tie these concepts together or solve for the eigenvectors of A.

Any help is greatly appreciated!

答:
Once you understand what the hypotheses mean, there is almost nothing left to prove here. A linear operator (and by extension a square matrix) is diagonalisable if the sum of the associated eigenspaces (which is always a direct sum) fills the whole space. The multiplicity of an eigenvalue can have different meanings, but in the diagonalisable case they are all the same, and the multiplicity of $\lambda$ is the dimension$~d(\lambda)$ of the eigenspace $E_\lambda$ for $\lambda$. (In general this is called the geometric multiplicity. Since you mention the characteristic polynomial $\chi$ you seem to be thinking of the algebraic multiplicity, which is the multiplicity of$~\lambda$ as root of$~\chi$; in the diagonalisable case each eigenspace$~E_\lambda$ contributes a factor $(X-\lambda)^{d(\lambda)}$ to$~\chi$ and these contributions give all of$~\chi$, so the algebraic multiplicity of$~\lambda$ is $d(\lambda)$ as well.)

Now given that $A$ of size $n\times n$ is diagonalisable with an eigenvalue$~\lambda$ of multiplicity$~n$ means that $\dim(E_\lambda)=n$ which implies $E_\lambda$ is the whole space $K^n$ on which $A$ acts. But if $A$ acts on the whole space as scalar multiplication by$~\lambda$ (which is how it acts on $E_\lambda$ by definition) this means that $A=\lambda I$.


问:
Let $a_n = \frac{n^p}{e^n}$ Show that $a_n$ converges for all $p>0$
In my work on this problem I've set up a limit:
$$\lim _{x\to \infty }\left(\frac{x^p}{e^x}\right)$$

I can see that for any number of p, successive applications of L'Hopital's Rule will eventually bring the numerator to a finite number, if p is an integer. If p is a fraction successive applications of L'Hopital's rule will eventually bring the $x$ term from the numerator to the denominator. Therefore in each case the limit will equal zero because the denominator's $e^x$ term will still be the same.

How do I express this logic in math form?

答:
Let $ N \in \mathbb N$ such that $N \ge p.$ For $x>1$ we get with the Taylor series of $e^x$:

$$e^x > \frac{x^{N+1}}{(N+1)!} \ge \frac{x^{p+1}}{(N+1)!} .$$

Thus

$$\frac{e^x}{x^p} \ge \frac{x}{(N+1)!},$$

hence

$$0 \le \frac{x^p}{e^x} \le \frac{(N+1)!}{x}.$$

This gives

$$ \lim_{x \to \infty}\frac{x^p}{e^x}=0.$$


答:
Hope you got it now, it's a property of the polynomial, the equation $z^2-1$ is satisfied by $z =1, z= -1$, because a quadratic has two roots

While $z^3-1$ is satisfied by $z=1$ as the obvious real solution, but a cubic has three roots, so the other two are unreal, but if we can factor it or substitute $z=a+bi$, to get the remaining roots
$$ z^2+z+1$$
$$ z_1 = -\frac{1}{2}  + \frac{\sqrt{ -3}}{2}$$
$$ z_2 = -\frac{1}{2}  - \frac{\sqrt{ -3}}{2}$$
They are conjugate of eachother, If you cube it using the property of complex number the result is $z^3 = 1$, so $z_0 , z_1, z_2$ all from the cube root of unity

Anytime you take the cube root of a number $n$, the solutions are $\sqrt[3]{n} , \sqrt[3]{n} \cdot z_1, \sqrt[3]{n} \cdot z_2$

Title: find the roots $(2z+3)^3=\frac{1}{64}$


问:
If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x) $ exist?
I want to prove or disprove this problem:
If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L<\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.

When I assume problem below:

If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?

I can use mean-value theorem to show that.

So my question is:

If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?

答:
Consider the function $$g(x)=e^{x} f(x).$$ Then $$Dg(x)=e^{x}f(x)+e^{x}Df(x)=e^{x} \left( f(x)+Df(x) \right).$$
Now,
$$
\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{g(x)}{e^x} = \lim_{x \to +\infty} \frac{Dg(x)}{e^x} = \lim_{x \to +\infty} Df(x)+f(x)
$$
by De l'Hospital's theorem.

N.B. I think this exercise was solved by G. Hardy in one of his books.


问:
Probability of rolling 2 dice 10,000 times and getting snake eyes between 280-300 times using probability theory.
You roll 2 dice 10,000 times.  You are interested in the probability of getting snake eyes(two ones) between 280 and 300 times.  Do two things:  using software, find the exact probability.Second, approximate the probability using methods we have seen in class.  Compare the two results.

I think I might be using the wrong method to find out what the answer is, I used sterlings approximation with gaussian distribution to find the answer and got .00015 as my answer but I think that is incorrect. Any help would be greatly appreciated.

Thanks.

答:
I read the question to mean between 280 and 300 inclusive.

In R: Exact binomial computation,

$$P(280 \le X \le 300) = 0.3698562.$$

```<code>diff(pbinom(c(279,300), 10000, 1/36))
[1] 0.3698562
</code>```

Normal approximation,

$$X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu = 277.7778, \sigma=16.66667).$$

```<code>mu = 10000*1/36;  mu
[1] 277.7778
vr = mu*36/36;  vr
[1] 277.7778
sd = sqrt(vr);  sd
[1] 16.66667
</code>```

Using continuity correction:
$$P(279.5 < X< 300.5) = 0.3724605.$$

[You can get close
to this answer by standardizing and using printed
standard normal CDF tables.]

```<code>mu = 10000*1/36;  mu
[1] 277.7778
vr = mu*36/36;  vr
[1] 277.7778
sd = sqrt(vr);  sd
[1] 16.66667
diff(pnorm(c(279.5,300.5), mu, sd))
[1] 0.3724605
</code>```

Figure:

```<code>x = 275:315;  pdf = dbinom(x, 10000, 1/36)
hdr = "PDF of BINOM(10000, 1/36) with Approx. Normal Density"
plot(x, pdf, type="h", ylim=c(0,.025), xaxs="i", lwd = 2, main=hdr)
 curve(dnorm(x, mu, sd), add=T, col="orange", lwd=2)
 abline(h=0, col="green2")
  abline(v=c(279.5, 300.5), col="blue", lwd=2, lty="dotted")
</code>```

Body:
I read the question to mean between 280 and 300 inclusive.

In R: Exact binomial computation,

$$P(280 \le X \le 300) = 0.3698562.$$

diff(pbinom(c(279,300), 10000, 1/36))
[1] 0.3698562

Normal approximation,

$$X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu = 277.7778, \sigma=16.66667).$$

mu = 10000*1/36;  mu
[1] 277.7778
vr = mu*36/36;  vr
[1] 277.7778
sd = sqrt(vr);  sd
[1] 16.66667

Using continuity correction:
$$P(279.5 < X< 300.5) = 0.3724605.$$

[You can get close
to this answer by standardizing and using printed
standard normal CDF tables.]

mu = 10000*1/36;  mu
[1] 277.7778
vr = mu*36/36;  vr
[1] 277.7778
sd = sqrt(vr);  sd
[1] 16.66667
diff(pnorm(c(279.5,300.5), mu, sd))
[1] 0.3724605

Figure:

x = 275:315;  pdf = dbinom(x, 10000, 1/36)
hdr = "PDF of BINOM(10000, 1/36) with Approx. Normal Density"
plot(x, pdf, type="h", ylim=c(0,.025), xaxs="i", lwd = 2, main=hdr)
 curve(dnorm(x, mu, sd), add=T, col="orange", lwd=2)
 abline(h=0, col="green2")
  abline(v=c(279.5, 300.5), col="blue", lwd=2, lty="dotted")


问:
Why can’t prove two vectors to be collinear if the vectors have difference signs using guess?
Vectors are $A = xi + 2j + zk$ and $B = 2i + yj + 1k$

Question is to check if they ($A$ and $B$) are parallel or collinear.

So, in my book it says that $x= 2 , y = 2$ and $z = 1$. Then they are parallel. That looks pretty obvious too.

Now, for another $Q$ with negative values.
$A = 2i + 3j -4k$ and $B = 3i-aj +bk$.

Now, here they didn’t assume $a = -3$ and $b = 4$. But, solved by another method i.e. that their ratio of coefficients should be equal.

$\frac{2}{ 3} $= $\frac {3}{-a}$= $\frac
{-4}{b}$. So,

$b \times 2 = -4 \times 3$ , $b = -6$. Now, here the values were solved in a different way and have different answers. Why is that ? I think due to negative sign but for the 1st $Q$. I am not sure.

$\frac{x}{2} = \frac{2}{y} =\frac{z}{1}$ . Here, we can’t solve it the way we did 2nd $Q$. Since there are 3 variables . Then, how to solve this

答:
For the first case, we could also have written that $A$ and $B$ are parallel if $\frac{x}{2}=\frac{2}{y}=\frac{z}{1}$. By taking the given particular vectors, we can make $A$ a scalar multiple of $B$. It has $3$ variables, so cannot be directly solved. But, you may use trial and error to find appropriate values.

In the second case, if we write $\frac{2}{3}=\frac{3}{-a}=\frac{-4}{b}$, and compare $2$nd and $3$rd term with the first one, we get $a=-\frac{9}{2}$ and $b=-6$. The first case was direct, probably that's why it was not explicitly stated.


答:
You are not requested to check that the vectors are parallel. You are asked to find values of the unknowns that make them so.

$$\frac{x}{2}=\frac{2}{y}=\frac{z}{1}.$$ For convenience, you can assume that the common ratio is $1$.

$$\frac{2}{3}=\frac{3}{-a}=\frac{-4}{b}.$$ Here the common ratio is $\dfrac23$, giving $a=-\dfrac92,b=-6$.

Title: Why can’t prove two vectors to be collinear if the vectors have difference signs using guess?

Body:
Vectors are $A = xi + 2j + zk$ and $B = 2i + yj + 1k$

Question is to check if they ($A$ and $B$) are parallel or collinear.

So, in my book it says that $x= 2 , y = 2$ and $z = 1$. Then they are parallel. That looks pretty obvious too.

Now, for another $Q$ with negative values.
$A = 2i + 3j -4k$ and $B = 3i-aj +bk$.

Now, here they didn’t assume $a = -3$ and $b = 4$. But, solved by another method i.e. that their ratio of coefficients should be equal.

$\frac{2}{ 3} $= $\frac {3}{-a}$= $\frac
{-4}{b}$. So,

$b \times 2 = -4 \times 3$ , $b = -6$. Now, here the values were solved in a different way and have different answers. Why is that ? I think due to negative sign but for the 1st $Q$. I am not sure.

$\frac{x}{2} = \frac{2}{y} =\frac{z}{1}$ . Here, we can’t solve it the way we did 2nd $Q$. Since there are 3 variables . Then, how to solve this


问:
How to transform pitch and roll to a rotated reference frame?
My problem is the same as that in this unanswered question, as well as this one and this one.

I would like to calculate the pitch and roll of an object with respect to a rotated reference frame. For instance say we have an aircraft fitted with an attitude sensor which is rotated 90 degrees clockwise to the aircraft's reference frame. When the aircraft pitches down 10 degrees, the sensor will roll 10 degrees to the left.

The approach I have taken so far is to use rotation matrices. I create a rotation matrix representing the object orientation and another for the transformation to the rotated reference frame. Then I multiply the matrices and extract pitch and roll from the result. However, when tested in Excel* I do not get exactly the expected result from this method (...although it is close).

Example :-

X positive right, Y positive forwards, Z positive upwards.

I am testing the case where the sensor is mounted on the aircraft with a rotation of 45 degrees around the z axis i.e. yaw. Here I would expect any pitch by the aircraft to be measured as simultaneous pitch and roll by the sensor, with the measured pitch and roll being identical.

For 10 degrees of pitch I construct the following rotation matrix. Numbers are in degrees here for clarity ; in the formulas they are of course radians:

matrixPitch = $\begin{bmatrix} cos(10) & 0 & sin(10) \\ 0 & 1 & 0 \\ -sin(10) & 0 & cos(10) \end{bmatrix}$

And for 45 degrees of yaw:

matrixYaw = $\begin{bmatrix} cos(45) & -sin(45) & 0 \\ sin(45) & cos(45) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

I then multiply the two using the MMULT() worksheet function i.e.

=MMULT( matrixPitch , matrixYaw )

I get the following result, with numbers shown to six decimal places:

$\begin{bmatrix} 0.696364 & -0.696364 & 0.173648 \\ 0.707107 & 0.707107 & 0 \\ -0.122787 & 0.122788 & 0.984808 \end{bmatrix}$

I then extract the pitch and roll from this matrix, using the following formulas (from here):

$R = \begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix}$

$\tan \phi = \frac{R_{32}}{R_{33}}$

$\tan \theta = \frac{-R_{31}}{\sqrt{R_{32}^2 + R_{33}^2}}$

Where $\phi$ is roll and $\theta$ is pitch.

I get the following (in degrees and, again, to six decimals):

$\phi$ = 7.107076

$\theta$ = 7.053022

I would expect these numbers to be essentially identical, and I don't think that there is much opportunity for significant rounding errors in the calculations. I have also performed the same calculations using C# (using double precision) and get the same results.

Am I doing something wrong? To validate the method I would have liked the resulting pitch and roll to agree to at least two decimal places.

* Other spreadsheet applications are available

答:
The calculations are correct, but I am unsure of what you are expecting e.g., what are you trying to visualize? In general, you need to define a rotation matrix from the sensor to the aircraft, then you need to define the rotations from one step to another step, and so on...

Let $I$ represent the frame of the IMU, and let $B$ represent the body of the aircraft. The rotation matrix from the body to the IMU can then be denoted as ${}^I \mathbf{R}_{B}$ (e.g., 45 degrees yaw in your example).

Now, the aircraft has some motion (e.g., 10 degrees pitch as in your example), then you define the rotation matrix from $B_0$ to $B_1$, which is the rotation from the body at step $t_0$ to the body at step $t_1$. This can be denoted as ${}^{B_1} \mathbf{R}_{B_0}$.

If you want the orientation of the sensor at step $t_1$, then you need to calculate the rotation matrix as follows:
${}^{B_1}R_{I} = {}^{B_1} \mathbf{R}_{B_0}  {}^B \mathbf{R}_I$

Now, you can extract the roll, pitch, and yaw of your sensor. Perhaps, you want to calculate a different quantity. If so, the procedure is similar. Note, in general, ${}^B\mathbf{R}_A = {}^A \mathbf{R}_B^{-1}$ for reference frames $A$ and $B$. Be careful with signs because ${}^A \mathbf{R}_B$ represents the orientation of $B$ with respect to $A$.


问:
Solving $\;x+y+z =8 ; \;\;\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10 $
Solve the problem

\begin{cases}x+y+z =8 \\ \\
\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10
\end{cases}

with $(x,y,z) \in \mathbb R^3$

I have already solved it, but I'd like to see others creative solutions and before all, share this funny problem with the community.

答:
Think Length of path on $\mathbb{R}^2$ $(0,0)-(x,1)-(x+y,3)-(x+y+z,6)$.
Minimum length of path is 10. When those points are on one line.
$x=8/6,y=16/6,z=24/6$.</p>


问:
Bracket $[X,Y]$, where $X$ and $Y$ are differentiable vector fields.
Let $X$ and $Y$ be linearly independent vector fields in $\mathbb{R}^3$
and let $\omega$ be a 1-form
whose kernel (for each point q of $\mathbb{R}^3$
) is the plane generated by $X (q)$ and $Y (q)$. Show that
the condition $d\omega\wedge\omega = 0 $ is equivalent to the fact that, for each point of $\mathbb{R}^3$
, $[X, Y](q)$ is contained in the plane generated by $X(q)$ and $Y(q)$.

This problem is not going away at all. Anyone who could explain to me, please?

答:
Hint:

Recall the formula:
$$
\mathrm{d}\omega(X,Y) = X\cdot \omega(Y) - Y\cdot \omega(X) - \omega\left([X,Y]\right).
$$
This tells us that, if $X$ and $Y$ are in the kernel of $\omega$ (that is, if $\omega(X) = \omega(Y) = 0$):
$$
\mathrm{d}\omega(X,Y) = - \omega([X,Y]).
$$

Can you go on from there ?

Edit:

Suppose $(X,Y,Z)$ are -locally- linearly independant. They form a basis of $\mathbb{R}^3$. If $X$ and $Y$ are in $\ker \omega$, it follows that:
$$
\omega \wedge \mathrm{d}\omega \left( X,Y,Z\right) = \pm \omega(Z) \mathrm{d}\omega(X,Y) = \mp \omega(Z) \omega([X,Y])
$$
where the $\pm$ and $\mp$ are here because I don't really want (and care about) the exact sign. As $\omega$ is supposed non-zero (its kernel $\ker \omega = \mathrm{span}(X,Y)$ is two dimensional) and as $X,Y,Z$ are linearly independant, $\omega(Z) \neq 0$. Therefore:
\begin{align}
\omega\wedge\mathrm{d}\omega = 0 &\iff \omega\wedge \mathrm{d}\omega(X,Y,Z)= 0 \\
& \iff \omega(Z)\omega([X,Y]) = 0 \\
& \underset{\omega(Z)\neq 0}{\iff} \omega([X,Y]) = 0 \\
& \iff [X,Y] \in \ker \omega \\
& \iff [X,Y] \in \mathrm{span}(X,Y).
\end{align}
This is a particular case of the Frobenius integrability theorem (in dimension 3).


问:
What is the difference between formal group and formal group law?
I heard a formal group law is a formal group with chosen coordinate.
But I cannot grasp this meaning.

My understainding; （1-dimmensional）formal group and formal group law are both element of formal power series with two variables, which satisfies some conditions（like group axiom）.
For example, $X＋Y∈K[X,Y]$ is formal additive group
law.

Could you tell me the difference between formal group and formal group law with examples?

Thank you so much for your help.

答:
For Silverman's purposes, I think it's best if you just thought of this as a notational choice and moved on quickly to something with more content.

In any case, Silverman is clearly thinking of formal groups as objects in some category and formal group laws as a structure internal to those objects, so he uses $\mathcal{F}$ for the formal group (object in a category) and $F(x, y)$ for the formal group law (a power series). However, for his purposes in Section IV.2 there is no additional structure to $\mathcal{F}$ beyond $F(x, y)$, so you're asking why there are two symbols.

Well, you could just use $F(x, y)$, but if you did things would get awkward. For instance, a homomorphism between formal groups would then be a function $f \colon F(x, y) \to G(x, y)$ such that $G(f(x), f(y)) = f(F(x, y))$--the notation specifying the domain and codomain of $f$ is pretty misleading.


问:
Why are combinations used in the expansion of $(a + b)^4$?
The difference between Permutation and Combination is that in Permutation order matters, Combination order doesn't matter.

So in $ABC, ACB$, the permutation is $2$, and combination is $1$.

But, to expand $(a+b)^4$, the $(a^3)b$ term's coefficient is $4\choose {1}$$  = 4$ ways of getting three $a$'s and $1$ $b$ multiplied together,

namely:

$aaab,  aaba,  abaa,   baaa$. But that's the same combination, shouldn't it (combination) equal to 1? Why 4?

$4\choose 1$$ = \frac{4!}{(1!) \cdot (3!)} = 4$

That's permutation isn't it?

I understand that adding the $(a^3)b$s there are $4$ altogether, but why that's combination?

I'm really confused.

答:
When you expand
$$(a + b)^4 = (a + b)(a + b)(a + b)(a + b)$$
each term is determined by selecting either an $a$ or a $b$ from each of the four factors.  For instance, if you choose an $a$ from the first three factors and a $b$ from the fourth factor, you obtain $aaab = a^3b$.  The coefficient of $\binom{4}{1}$ for $a^3b$ in the binomial expansion
$$(a + b)^4 = \binom{4}{0}a^4b^0 + \binom{4}{1}a^3b^1 + \binom{4}{2}a^2b^2 + \binom{4}{3}a^1b^3 + \binom{4}{4}a^0b^4$$
represents the number of ways a $b$ can be selected from exactly one of the four factors.  We use combinations in the binomial expansion since we are selecting the subset of factors from which a $b$ is drawn.


问:
Inverse of an algebraic function is irrational under a condition
I've recently been looking to Function Theory/Real Analysis and I've encountered a bit of a problem I'm not entirely sure how to approach. This isn't from a textbook, rather something I have noticed that may be the case or may not be. Here's the problem:

Let $f: \mathbb{R} \to \mathbb{R}$ be an algebraic function with an existent inverse over $\mathbb{R}$. If $\alpha \in \mathbb{R}$ is such that $f(\alpha) \in \mathbb{Q}$, is $f^{-1}(\alpha) \not\in \mathbb{Q}$? Is the converse, inverse and contrapositive of this statement true?

EDIT: As pointed out, $f(x) = x$ is a counterexample to this, but since this is obvious, I'll elect to exclude this as a function to be considered.

I've been tempted to generalise everything and brute force from there, although I highly doubt that is the correct approach, let alone a reasonable way to approach this. I feel like I could be missing something really obvious here however I have not been able to even get a proper foothold. Any help or guidance would be greatly appreciated!

EDIT 2: Found a simple counterexample again, so this question isn't really that troubling anymore.

答:
Focus on examples. Take the counterexample of $f(x) = x = f^{-1}(x)$.

For the forward direction, try $\alpha = 1$, which is rational.

For the converse, try $\alpha=\pi$, which is irrational.

For the inverse, try $\alpha=\pi$.

For the contrapositive, try $\alpha=1$.

Each of these give a counterexample to the relevant statement. Remember that, for the statement $P \implies Q$, we have

the converse being $Q \implies P$

the inverse being $\neg P \implies \neg Q$

the contrapositive being $\neg Q \implies \neg P$