问:
How to calculate $\sum_{i= 0}^{k-1}\left \{ \frac{ai+b}{k} \right \}$
How to calculate: $$F=\sum_{i= 0}^{k-1}\left \{ \frac{ai+b}{k} \right \}$$
where $a,b \in \mathbb Z$ and $(a,k)=1$; $\left \{ \frac{ai+b}{k} \right \} =$ fraction part of $\frac{ai+b}k$

Such problem so strange for me. Please help me solve it, give me solution or a hint.

Thanks

答:
Since $(a,k)=1$, the numbers $ai$ and also the numbers $ai+b$ run through all residue classes modulo $k$ exactly once. Therefore, we can simplify
$$ F=\sum_{i=0}^{k-1}\frac ik=\frac1k\frac{k(k-1)}{2}=\frac{k-1}2.$$</p>


问:
Justify whether the following equation is true
$1+2+3+...+n=\theta(n^2)$ This is the equation.I tried to find the $\theta$ value but I am not sure if it is the right way to do this.I would appreciate any tips how to solve this one

答:
$ f(n)=O{\bigl (}g(n){\bigr )}$ if there exist positive integer numbers $M$ and $n_0$ such that $f ( n )\le M g ( n )$  for all  $n \ge n_0$

In your case
$$f(n)=1+2+3+\ldots+n=\frac{n(n+1)}{2}=\frac{n^2}{2}+\frac{n}{2}$$
$$f(n)=O{\bigl (}n^2{\bigr )}$$
because for $M=1$ and $n_0=1$ we have
$$\frac{n^2}{2}+\frac{n}{2}<n^2,\forall n\ge 1$$

ContentLicense: CC BY-SA 4.0


问:
Common false/unproven "popular mathematics" claims
Motivated by the common unproven claim in memes that every finite sequence of digits appears in the digits of $\pi$ (see Does $\pi$ contain all possible number combinations?), I started wondering what other similar common claims are out there?

Criteria

A claim about math that's "commonly" stated as true by math enthusiasts who are non-mathematicians. For instance, something that one might find posted somewhere as a meme.

The claim must be either false or unproven.

答:
Let's assume playing with a fair coin.

Common claim (false)
In a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.

Contrary to this false claim we have with probability $\frac{1}{2}$ that no equalization occurs in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.
This is a consequence of the Arc sine law for last visits.


问:
The union of finite subsets $S_k$ of an infinite set $J$ with $k\in K$ with $|K|<|J|$ is strictly smaller than $J$.
In my notes I have a proof that if $B_1$ and $B_2$ are basis of a vector space, then $|B_1|=|B_2|$. The proof assumes w.l.o.g $|B_1|\ge|B_2|$ and has two parts; the second part consider the case $|B_1|$ infinite.

Let be $B_1=\{v_j\mid j\in J\}$  and $B_2=\{w_k\mid k\in K\}$.
Then $w_k=\sum_{j\in S_k}x^jv_j$, for some finite $S_k\subset J$.

The I have the following conclusion that use some set theory result:

Since $|B_1|>|B_2|$, respectively $|J|>|K|$, and the $S_k$'s are finite subsets of $J$, we have also $|J|>|\bigcup_{k\in K} S_k|$.

What is this result? (*) The union of finite subsets $S_k$ of an infinite set $J$ with $k\in K$ with $|K|<|J|$ is strictly smaller than $J$.

If I try to construct a counterexample in $\mathbb{Z}$, it is impossible, since $|K|<|\mathbb{Z}|$ means that $K$ is finite. On one site this make the result plausible, but  in $\mathbb{Z}$ there is only one type of infinity.

Does the result (*) has some name? Some reference? Some idea for the proof?

EDIT I think that the idea behind is that if the $S_k$ are finite, the cardinality of $|\bigcup_{k\in K} S_k|$ is equal to the cardinality of $K$.

答:
If $K$ is finite, then clearly $\bigcup_{k\in K}S_k$ is also finite and therefore $\left|\bigcup_{k\in K}S_k\right|<|J|$. If $K$ is infinite, then

$$|K|\le\left|\bigcup_{k\in K}S_k\right|\le|K|\cdot\aleph_0=|K|\,,$$

so $\left|\bigcup_{k\in K}S_k\right|=|K|<|J|$.


问:
How to show function is uniformly continous on certain interval
Let $ f\colon \mathbb{R} \to \mathbb{R}$ be  $$x \mapsto \begin{cases} \frac{(x-1)^2}{x^2+1}, &|x| \neq 1 \\2, &|x|=1\end{cases} $$

How can I show that $f$ is uniformly continuous on $[-2,2]\backslash \{ - 1,1\}$?
I thought it could be done with  the $ \epsilon - \delta$ criteria but I am not sure because of the specific interval.

答:
$$f(x)= 1-\frac{2x}{x^2+1}$$ for $|x|\neq 1. $
$$|f'(x)| =\left|\frac{2(1+x^2) -2x\cdot 2x }{(1+x^2)^2}\right|=\frac{2|1-x^2|}{(1+x^2)^2}\leq\frac{2(1+x^2)}{(1+x^2)^2}\leq 2 $$
Hence
$f$ is Lipschitz continuous and therefore uniform continuous for $|x|\neq 1.$


问:
Evaluate $\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx$
How to evaluate this integral or at least rewrite it as special (non-elementary) function step by step ?
$$\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx$$
Where $M$ is a positive integer that is $M=1,2,3,4,...$ , $a,b$ are two positive number and ${K_1}\left( . \right)$ is the modified Bessel function of order 1.

If you could provide some reference, I would be pretty much graceful !

Thank you for your enthusiasm !

答:
It seems useful to rearrange the integral:
\begin{align}
 I&=\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx\\
 &=\frac{2b^{-M}\sqrt{a}}{({M-1})!}\int_{0}^\infty K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-3/2}e^{-\frac{x}{b}}\,dx\\
 &=\frac{2}{({M-1})!}\left( \frac{a}{b} \right)^M\int_{0}^\infty K_1\left(\frac{2}{\sqrt s}\right)s^{M-3/2}e^{-\frac{a}{b}s}\,ds
\end{align}
which expresses that the integral is a function of $a/b$.

To obtain the Meijer function expression given by @MariuszIwaniuk in the comment, one may use the Mellin-Barnes representation for the modified Bessel function
\begin{equation}
 K_{\nu}\left(z\right)=\frac{(\frac{1}{2}z)^{\nu}}{4\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma\left(t\right)\Gamma\left(t-\nu\right)(\tfrac{1}{2}z)^{-2t}\,dt
\end{equation}
valid for $c>\max(\Re\nu,0),|\arg z|<\pi/2$. Here, with $\nu=1,z=2/\sqrt s,c>1$, we obtain, by changing the order of integration
\begin{align}
 I&=\frac{2}{({M-1})!}\left( \frac{a}{b} \right)^M\int_{0}^\infty K_1\left(\frac{2}{\sqrt s}\right)s^{M-3/2}e^{-\frac{a}{b}s}\,ds\\
 &=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)^M\frac{1}{2\pi i}
 \int_{c-i\infty}^{c+i\infty}\Gamma\left(t\right)\Gamma\left(t-1\right)\,dt\int_{0}^\infty s^{t+M-2}e^{-\frac{a}{b}s}\,ds
\end{align}
The integral over $s$ gives directly a Gamma function,
\begin{equation}
 I=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)\frac{1}{2\pi i}
 \int_{c-i\infty}^{c+i\infty}\Gamma\left(t\right)\Gamma\left(t-1\right)\Gamma(t+M-1)\left(\frac ba\right)^t\,dt
\end{equation}
From the definition of the Meijer function
\begin{equation}
{G^{m,n}_{p,q}}\left(z;\begin{array}{c}
a_1,\ldots,a_p\\
b_1,\ldots,b_q
\end{array} \right)=\frac{1}{2\pi\mathrm{i}}\int_{L%
}{\frac{\prod\limits_{\ell=1}^{m}\Gamma\left(b_{\ell}-t\right
)\prod\limits_{\ell=1}^{n}\Gamma\left(1-a_{\ell}+t\right)}{\left(\prod\limits_%
{\ell=m}^{q-1}\Gamma\left(1-b_{\ell+1}+t\right)\prod\limits_{\ell=n}^{p-1}%
\Gamma\left(a_{\ell+1}-t\right)\right)}}z^{t}\,dt
\end{equation}
we can use $z=b/a$, $a_1=1,a_2=2,a_3=2-M$, with $m=q=0, n=p=3$ and an integral along the vertical axis from $c-i\infty$ to $c+i\infty$, as the real parts of the the poles of $\Gamma\left(1-a_{\ell}+t\right)$ are all less than $1$. We deduce
\begin{equation}
 I=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)G^{0,3}_{3,0}\left(\frac{b}{a};\begin{array}{c}
1,2,2-M\\
\textrm{---}
\end{array} \right)
\end{equation}
The proposed expression can be retrieved by combining the identity
\begin{equation}
 I=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)G^{3,0}_{0,3}\left(\frac{a}{b};\begin{array}{c}
 \textrm{---}\\
0,-1,M-1
\end{array} \right)
\end{equation}
and the second identity
\begin{equation}
 I=\frac{1}{({M-1})!}G^{3,0}_{0,3}\left(\frac{a}{b};\begin{array}{c}
 \textrm{---}\\
0,1,M
\end{array} \right)
\end{equation}


问:
Find matrix $T: \mathbb{R}^2 \mapsto \mathbb{R}^2$ first rotates $2\pi/3$ rad (counterclockwise) then reflects across line $x2 = -x1$.
I know that the answer is:
$$
\begin{bmatrix}
\frac{-\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\
\frac{1}{2} & \frac{1}{2}
\end{bmatrix}
$$
but I am not sure on the steps to get to this point.

答:
So, the first thing I would note is that matrix multiplication performs operations in the order the matrices are multiplied. That is, you need to perform a rotation and a reflection on an input vector $\vec{x}$. That is
$$
M \vec{x} = Ref Rot(2\pi/3)  \vec{x}
$$

We know the rotation matrix, $Rot(\theta)$ is
$$
\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}
$$

And reflection of $x1$ about $-x2$ is simply:

$$
\begin{bmatrix}
0 & -1 \\
-1 & 0
\end{bmatrix}
$$

This can be intuited by drawing the line $y = -x$ and seeing what happens when you rotate basis vectors around it. You got it from here?


问:
Something about the proof of Bolzano-Weierstrass Theorem not making sense
So I'm trying to understand the proof of this version of the theorem

Here is the proof given

I have understood Lemma 1. and its implications.
The thing I don't get about it is:

We assume $S$ has no limit point, but then out of nowhere, because $\inf(X)\in X$ then $S$ has a limit point. Doesn't this contradict our assumption straight away?

My specific questions are:

1)Why does $X\subset S \Rightarrow \inf(X)\in X$ (and $S$ for the matter).

2)Why do we let $\inf(X)$ be a limit point? Wouldn't that go against the whole point of the proof?

答:
For any subset $A$ of $\Bbb R$ with a lower bound, if $\inf A\notin A$, then $\inf(A)$ is a limit point of $A$. That's so because, for any $n\in\Bbb N$, there is some $a_n\in\left(\inf(A),\frac1n+\inf(A)\right)\cap A$, by the definition of $\inf(A)$ and because $\inf A\notin A$. But then $\lim_{n\to\infty}a_n=\inf(A)$ and therefore $\inf(A)$ is a limit point of $A$.

So, if $\inf(X)\notin X$, then $\inf(X)$ is a limit point of $X$ and therefore a limit point of $S$. But we are assuming that $S$ has no limit points. So, $\inf(X)\in X$.


问:
Priving that the degree of an irreducible nondegenerate subvariety of $\mathbb{P^n}$ satisfies $d\ge n-k+1$
I'm reading through Dolgachev's "Classical algebraic geometry" and theorem 8.1.1 states the following

The references are not very precise, do you happen to know where, precisely, I can find the proof in any source?

答: One source is On Varieties of Minimal Degree by Eisenbud and Harris. This is available here at MSRI's website, for instance. The bibliography lists previous proofs by Bertini (1907), Harris (1981), and Xambo (1981), if you're looking for other sources.


问:
Numerical boundary-finding algorithms
I'm very familiar with numerical root-finding algorithms like Brent's method (scipy.optimize.brentq) or Chandrupatla's method. What if I am looking for a boundary numerically?

Suppose for example that I have an unknown monotonically increasing function $y = g(x)$ between $x=a$ and $x=b$ as represented by the boundary of some region $f(x,y) \ge 0$.

I cannot evaluate $g(x)$ directly; instead, I can evaluate $f(x,y)$ and compare it to zero. For any fixed value of $x=x_0$ I can use a root-finding method to determine $y$ such that $f(x_0,y)=0$.

Aside from sweeping across the interval $x\in[a,b]$ and repeating a root-finding method for each value of $x$, is there a more efficient way? (Suppose I know that $g(x)$ is continuous, for example, and that its derivative $dg/dx$ is "mostly" continuous; then the solution to $y=g(x_0+\Delta x)$ is close to $y=g(x_0)$ .)

答:
Your assumptions are very weak. If stronger assumptions are made, then I can see the some options.

If $f = f(x,y)$ is differentiable, $\frac{\partial f}{\partial y}(x,g(x)) \not =0$, and $g$ is differentiable, then $g$ satisfies the ordinary differential equation
$$ g'(x) = - \frac{\frac{\partial f}{\partial x}(x,g(x))}{\frac{\partial f}{\partial y}(x,g(x))}$$
Why? By the definition of $g$ we have $$\phi(x) = f(x,g(x)) = 0.$$ By the chain rule we have
$$0 = \phi'(x) = \frac{\partial f}{\partial x}(x,g(x)) +  \frac{\partial f}{\partial y}(x,g(x)) g'(x).$$
In principle, you can then recover $g$ using your favorite numerical method for solving ODEs. However, without some form of stabilization we will move away from the true solution. Instead, you can use the known value of $g(x)$ and the ODE to generate two different approximations of $g(x + h)$. Ex. You could use $1$ step of Euler's explicit method with stepsize $h$ to produce the $y_0 \approx g(x+h)$ and $2$ steps of Euler's explicit method with stepsize $\frac{1}{2}h$ to produce $y_1 \approx g(x+h)$. It is unlikely that $y_0 = y_1$. You can then use the secant method to solve
$$f(x+h,y) = 0$$
using $y_0$ and $y_1$ to initialize the search for the solution $y=g(x+h)$.

In the above analysis I do not see any good ways to exploit the assumption that $g$ is assumed to be increasing. It is mildly helpful to know that $g(x) \leq y = g(x+h)$, so that $g(x)$ is can be be used to bound $y$ from below, but I also need an upper bound for $y$ before I can use a safe method such as bisection to solve for $y$.

If we are only willing to assume that $g$ is differentiable, then I would slide from $g(x)$ and $g(x-h)$ to an approximation of $g(x+h)$ using linear extrapolation, i.e.
$$ g(x+h) \approx 2 g(x) - g(x-h).$$
I would use the secant method to solve $f(x,+h,y)=0$ with $y_0 = g(x)$ and $y_1 = 2g(x) - g(x-h)$. In this setting it is (mildly) useful to know that $g$ is strictly increasing. We would then be certain that $y_0 \not = y_1$ and it would be possible to do at least one secant step.

I would be interested to know to what extent you can strengthen your assumptions on $f$ and $g$.

Title: Numerical boundary-finding algorithms


问:
Problem of multivariate calculus without using cauchy mean value theorem
Suppose that $f$ is a continuous function from $\mathbb{R}$ to $\mathbb{R}^{2}$ on $[a,b]$ and differentiable over $(a,b)$. Show that $\exists c\in (a,b)$ such that $(f(b)-f(a))^{\perp}\cdot f'(c)=0$ where $(x,y)^{\perp}=(-y,x)$. Don't use Cauchy Mean Value Theorem.

I know that using Cauchy it is easy but I don't get how to prove it without using. I've been thinking and nothing comes to mind. Any help?

答: Let $g(x)=\bigl(f(b)-f(a)\bigr)^\perp.f(x)$. Then $g(b)=g(a)$. Therefore, you can apply Rolle's theorem to $g$.


问:
How does the analysis of a single perturbation problem change when $\epsilon$ changes sign?
Just for reference I am looking at the problem $\epsilon y''(x)+xy'(x)+y(x)=0$ with

$y(1)=0$ and

$y(2)=1$

for which I have found the leading-order outer and inner solutions and composite solution. There's a follow-on question which asks how the method would change if $\epsilon$ were negative and I am having trouble getting my head around this. I read somewhere (it may have been MIT lecture notes but I can't find the link anymore) that then the boundary layer is found at $x=2$ instead of $x=1$ where it is now. Is this true? Is there an intuitive explanation for it?

Thank you for your help!

答: Your inner solution are simple exponentials $C_0+C_1e^{-x_0(x-x_0)/ϵ}$. To connect them to the outer solution, you need this to be bounded (and convergent) in the connection direction. For these simple exponentials this is only possible in one direction, so the boundary layer has to be, indeed, at a boundary. With $ϵ>0$ this exponential is bounded for increasing $x$, so $x_0=1$ has to be on the left side. If $ϵ<0$, then the exponential is bounded for falling $x$, thus $x_0=2$ has to be on the right side of the interval.


问:
Intuitive understanding of a non-full functor
A functor $T : C \to B$ is full if for each $c,c' \in C$ and for each $g : T c \to T c'$,  there is at least a morphism $f : c \to c'$ such that $g = T f$.

What does this tell me about non-full functors?

My understanding is that, as soon as there's a morphism $g : T c \to T c'$ in $B$, this morphism must come from a morphism in $C$; on the other hand, if the functor is not full, <s>i.e.</s> e.g. there is no morphism $c \to c'$ in $C$, then there  must be another pair of objects $d,d' \in C$ such that

$T d = T c$,

$T d' = T c'$,

$d \neq c \vee d' \neq c'$,

$\exists h:d \to d'$ such that $T h = g$.

Is this correct?

答:
Why should there be such a pair ? It is completely possible that $h$ is simply not in the image of the functor $T$. (*)

"Most" functors aren't full, this is the "generic" situation, full functors are something special, that's it.

There's a misconception in your question "as soon as there's a morphism in $B$, it must come from $C$" : that is simply not true.

Let's look at an example, one of the most classical ones (otherwise it's easy to cook up counterexamples by hand, this one is "found in nature"): the forgetful functor $U: Grp\to Set$ from the category of group and group morphisms to the category of sets and functions between them.

This is definitely not full : given $G,H$ groups, there are tons of applications $UG\to UH$ that aren't group morphisms.

Ok, that doesn't tell me that there can't be $G',H'$ with $UG' = UG, UH' = UH$ for any application. Well, it does : here's a set theoretic condition that any group morphism must satisfy: if $f:G\to H$ is a morphims of groups then for any $x,y \in H, f^{-1}(x)$ and $f^{-1}(y)$ either have the same cardinality, or one of them is empty.

Indeed, if $x\in f(G)$ then $f^{-1}(x)$ has the cardinality of $\ker(f)$. In particular, if $h: UG\to UH$ does not satisfy this property, there is no group structure on $UG,UH$ that can "lift" $h$.

So the definition of full functors tells you nothing about non-full ones, it just tells you something about full functors.


问:
How Many Iterations for Recurrence Relation To Terminate?
I have the following recurrence, and I want to upper bound how long it takes to terminate.

$R(0) = n$

$R(t) = R(t-1) - \max\left(1, \left\lfloor\sqrt{R(t-1)}\right\rfloor\right)$

The recurrence stops when $R(\cdot)=0$. Is there a way to determine how many iterations can occur before reaching $0$ for any positive integer $n$? Experimentally it seems to end at $t\leq 2 \sqrt{n}$, but I am not sure why.

答:
Let $n_k$ be the smallest value of $n$ s.t. we will need at least $k$ iterations. We want to prove $k \leq 2 \sqrt{n_k}$, or, equivalently, $n_k \geq \frac{k^2}{4}$.

It's true for $k = 1$: $n_1 = 1 \geq \frac{1}{4}$. Assuming it's true for some $k$, lets prove it for $k + 1$.

We have $$n_{k + 1} - \sqrt{n_{k + 1}} \geq n_k \tag{*}$$ (otherwise after first iteration from $n_{k + 1}$ we will get number less than $n_k$, which will require less than $k$ more iterations for a total of at most $k$).

Assume that $n_{k + 1} < \frac{k^2 + 2k + 1}{4}$. As $n_{k + 1}$ is integer, it implies $n_{k + 1} \leq \frac{k^2 + 2k}{4}$. As function $x - \sqrt{x}$ monotonically increases when $x > 1$, this implies $$n_{k + 1} - \sqrt{n_{k + 1}} \leq \frac{k^2}{4} + \frac{k}{2} - \sqrt{\frac{k^2}{4} + \frac{k}{2}} < \frac{k^2}{4} \leq n_k$$ - contradiction with  $(*)$. So $n_{k + 1} \geq \frac{(k + 1)^2}{4}$, and our statement is proven by induction.


问:
Tensor product is unique up to unique isomorphism
Let $M,N$ be two $A$-modules. I have heard it said that then the tensor product $M\otimes_A N$ is unique up to unique isomorphism. This confuses me so I would like to check if I understand this correctly.

I know that if $(T,g)$ and $(T',g')$ both satisfy the universal property of of the tensor product then there is a a unique isomphism $j:T\rightarrow T'$ such that $g'=j\circ g$. I would to check that this does not imply that we if have $(T,g)$,$(T',g'$) and $(T'',g'')$ then we must have $(T,g)=(T'',g'')$?

For any ring $R$, $R$ is an $R$-module and $R\otimes_R R=R$. Now multiplication from $\mu:R\times R\rightarrow R $ is a bilinear map. So $(R,\mu)$ is certainly one such tensor product. Suppose $\phi,\psi:R\rightarrow R$ are two automorphisms of $R$. Then $(R,\mu)$,$(R,\phi\circ \mu)$ and $(R,\psi\circ \mu)$ are three tensor products but non are pairwise equally.

答:
From the comments:

No. In fact, you can have lots of nonisomorphic objects, no two of which are equal.

For a silly example, you can do some set theoretic trickery:

Fix one tensor product $(T,g)$. Now, for each natural $n$, look at the set $\{(t,n) ~|~ t \in T \}$ with the same algebraic structure as $T$ (we just ignore the $n$ we're carrying around).

This gives countably many isomorphic structures, all satisfying the universal property, but none of these isomorphisms are the identity.

I hope this helps ^_^


问:
How do I negate the following statements?
I need help with these homework questions. Thanks in advance!

i) $(a\land b) \lor (a\land b)$

ii) $(a\Rightarrow b) \Rightarrow  (\neg c\Rightarrow (b\Rightarrow a))$

Here's what I have so far for each problem:

i) $\neg ((a\land b) \lor (a\land b))$

$ =(\neg a\lor\neg b) \land  (\neg a\lor\neg b)$

$=(\neg a\lor\neg b)$

I'm not sure if this is right but I was thinking of using De Morgan's rules for part i.

ii) I know when negating an implication, it goes as follows:

$(a\Rightarrow b) = (a\land \neg b)$

I'm not sure if I should start on the rightmost or leftmost side but any help would be appreciated for this!

答:
Well, your first statement $(a∧b)∨(a∧b)$ is the same as $P \vee P$ if we let $P = (a \wedge b)$. We know that

$$
P \vee P \iff P
$$

The negation of $P$ is then $\neg P = \neg (a∧b) = (\neg a \vee \neg b)$ by DeMorgan.

For the second statement, $(a⇒b)⇒( c⇒(b⇒a))$, we write the contrapositve,
$$
\neg (\neg a \vee b) \vee (\neg c \vee (\neg b \vee a))
$$
Now let's negate,

$$
\neg(\neg (\neg a \vee b) \vee (\neg c \vee (\neg b \vee a))) =
(\neg a \vee b) \wedge \neg(\neg c \vee (\neg b \vee a)) =
(\neg a \vee b) \wedge (c \wedge \neg (\neg b \vee a))
$$

Can you finish distributing the negation using DeMorgan's?


问:
Operations with summatory from $ \sum_{i,j} $ to $ \sum_{i>j} $ help please.
How can I prove that

$$ \sum_{i,j} \frac{a_i-a_j}{(1+a_j)\sqrt{a_i}} = \sum_{i>j} \frac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^{2}(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} $$

Thanks.

Body: Let $A_{ij}:=\tfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}}$ so $A_{ii}=0$ and$$\sum_{ij}A_{ij}=\sum_{i\ne j}A_{ij}=\sum_{i>j}(A_{ij}+A_{ji})=\sum_{i>j}(a_i-a_j)\left(\tfrac{1}{(1+a_j)\sqrt{a_i}}-\tfrac{1}{(1+a_i)\sqrt{a_j}}\right),$$reducing the problem to$$(1+a_i)\sqrt{a_j}-(1+a_j)\sqrt{a_i}=(\sqrt{a_ia_j}-1)(\sqrt{a_i}-\sqrt{a_j}).$$


问:
Question involving disjoint open sets in $k$ topology
Let $\mathbb{R}_k$ denote the topology, whose basis elements are of the form $(a,b)$ and $(a,b)-K$,where $k=\{\frac{1}{n}|n \in \mathbb{N}\}$.

Find the closure of $\{\frac{-1}{n}| n \in \mathbb{N}\}$ and answer the question:Is it possible to find disjoint open sets $U,V$ in $\mathbb{R}_k$ such that $0 \in U$ and $K \subset V$.

Answer:the closure of  $\{\frac{-1}{n}| n \in \mathbb{N}\}$=$\{\frac{-1}{n}| n \in \mathbb{N}\} \cup \{0\}$ since any open set containing $0$, must contain points in $\{\frac{-1}{n}|n \in \mathbb{N}\}$.No it is not possible to find such sets $U$ and $V$, since $U$ would necessarily contain irrationals between $0$ and the endpoint $b$ of its open interval of the form $(a,b)$ or $(a,b)-K$, and $V$ must be an interval of the form $(a,b)$, where for the purposes of this problem it suffices to assume $V=(0,b),b>1$ which would contain the irrational points as well.

Would this be the correct approach to this problem?

答:
You are right about the closure of $\left\{-\frac1n:n\in\Bbb Z^+\right\}$.

The second part, however, needs a good bit of work. In particular, you cannot assume that $V$ has the form $(0,b)$ for some $b>0$: $V$ might, for instance, be the union of tiny pairwise disjoint open intervals around the points $\frac1n$.

Suppose that $U$ and $V$ are open nbhds of $0$ and $K$, respectively. Then there is an $a>0$ such that $0\in(-a,a)\setminus K\subseteq U$, and for each $n\in\Bbb Z^+$ there is an $\epsilon_n$ such that $0<\epsilon_n<\frac1n$ and $\left(\frac1n-\epsilon_n,\frac1n+\epsilon_n\right)\subseteq V$. There is an $n\in\Bbb Z^+$ such that $\frac1n<a$. But then

$$(-a,a)\cap\left(\frac1n-\epsilon_n,\frac1n\right)$$

is a non-empty open interval, so it contains an irrational number $x$, and clearly

$$x\in\big((-a,a)\setminus K\big)\cap\left(\frac1n-\epsilon_n,\frac1n\right)\subseteq U\cap V\,.$$

Thus, $U$ and $V$ cannot be disjoint.


问:
Confusion on pairwise disjoint and disjoint
I've just started trying to learn some set theory and topology and I've come across the definition of disjoint sets quite a lot I've seen lots of Definitions such as

A set (of sets) $\mathcal{A}$ is disjoint if $\bigcap \mathcal{A} = \emptyset$.

The set $\mathcal{A}$ is pairwise disjoint when $\forall x \in A: \forall y \in A: x \neq y \implies x \cap y = \varnothing$

I can't quite understand what the difference actually is as I saw on s.e that a pairwise disjoint set is related to a $k$-wise disjoint.

I also saw this definition

An indexed collection $\{A_i\}_{i\in I}$ of subsets of $X$ is said to be pairwise disjoint if $A_i\cap A_j=\varnothing$ whenever $i\neq j$.

This confused me even further as an indexed collection can be a surjective function so when $i=j$ it could well be that $A_i$ = $A_j$ in that case how is their intersection empty?.

What I'm really trying to understand is what is pairwise and k wise disjoint sets?
Thanks in advance

答:
To supplement the other answers and comments, let me summarize my view.

An indexed family $(A_i)_{i ∈ I}$ of sets is

pairwise disjoint if $A_i ∩ A_j = ∅$ for every $i ≠ j ∈ I$,

collectionwise disjoint if the family has empty intersection, i.e. $⋂_{i ∈ I} A_i = ∅$.

The term disjoint may serve as a shortcut either for pairwise disjoint or collectionwise disjoint depending on the used convention (but obviously not both at the same time). It seems to me that “classically” it is used for collectionwise disjoint, but at least to me (and @TonyK) it seems more useful to use it for pairwise disjoint and to call a collectionwise family just a family with empty intersection.

Note that if an indexed family of non-empty sets is pairwise disjoint, then the map $i ↦ A_i$ is one-to-one, i.e. the enumeration is faithful (as @Henno Brandsma says). But a non-empty indexed family of empty sets $(∅)_{i ∈ I ≠ ∅}$ is pairwise disjoint even though the enumeration is not faithful.

Also note that pairwise disjointness implies collectionwise disjointness unless $I = ∅$.

An unindexed family of sets $\mathcal{A}$ is whatherverwise disjoint if the indexed family $(A)_{A ∈ \mathcal{A}}$ is.


问:
A dumb question about $T_2$ space-Looking for alternative proof
Show that R is Hausdorff

I have a proof for it
I want to know is there anyway to do it

using set theory alone

For instance the distance between two sets ,
say U and V the formula is inf{||x-y||,x$\in $U,y$\in$V}

I don’t want to rely on the above too

Any hints would be nice?
Thanks

答:
To show that $\Bbb R$ is Hausdorff, you can use different approaches: it is a metric space $(X,d)$ and any metric space is Hausdorff:

Let $x \neq y \in X$. Then $r:=d(x,y)>0$ by properties of the metric. Define $s=\frac{r}{2}>0$.
Then $U=B_s(x)$ is an open neighbourhood of $x$ and $V = B_s(y)$ is an open neighbourhood of $y$ and $U \cap V = \emptyset$, for otherwise there would be some $z \in B_s(x) \cap B_s(y)$, so $d(z,x) < s$ and $d(z,y) < s$ and then

$$d(x,y) \le d(x,z) +d(z,y) < s + s =2s = r= d(x,y)$$

which is a contradiction. Hence $(X,d)$ is Hausdorff in the metric topology.

Aother approach is to use that the topology on $\Bbb R$ is induced by its linear order: and $(X,<)$ in the order topology is Hausdorff:

Let $x \neq y$ in $X$. WLOG we can assume $x < y$ (or we interchange the roles of $x$ and $y$).

Case 1:if there is some $z \in X$ such that $x < z < y$ (as will be the case in $\Bbb R$ or $\Bbb Q$, e.g.), then $x \in U:=(\leftarrow, z)$ and $y \in V:=(z,\rightarrow)$ are subbasic open in the order topology and $U \cap V = \emptyset$.

Case 2: if no such $z$ exists, we can define $U:=(\leftarrow, y), V := (x, \rightarrow)$ to be difjoint open (sub)basic open neighbourhoods of $x$ resp. $y$.

So we can prove the fact that $\Bbb R$ is Hausdorff on two possible ways to generate the topology on this set, either by metric (which is Morris' preferred way) or by order.


答:
I guess that Lemma 1 says

If $S$ is a bounded set of real numbers such that, for every nonempty subset $X$ of $S$, $\inf(X)\in X$ and $\sup(X)\in X$, then $S$ is finite.

Now your task is to show that if $S$ is a bounded set with no limit point, then $S$ is finite. Using Lemma 1 requires to consider nonempty subsets of $S$, so pick one and call it $X$.

Then we know that $\inf(X)$ exists. If $\inf(X)\notin X$, then it is a limit point of $X$. Indeed, for every $\varepsilon>0$, there exists $x\in X$ such that
$$
x<\inf(X)+\varepsilon
$$
and therefore $(\inf(X)-\varepsilon,\inf(X)+\varepsilon)$ intersects $X$ in (at least) a point different from $\inf(X)$.

Similarly, if $\sup(X)\notin X$, then $\sup(X)$ is a limit point of $X$.

On the other hand, a limit point of $X$ is also a limit point of $S$, which has none by assumption. Therefore $\inf(X)\in X$ and $\sup(X)\in X$.

Since $X$ was an arbitrary nonempty subset of $S$, Lemma 1 applies and so $S$ is finite.

Title: Something about the proof of Bolzano-Weierstrass Theorem not making sense

Body:
So I'm trying to understand the proof of this version of the theorem

Here is the proof given

I have understood Lemma 1. and its implications.
The thing I don't get about it is:

We assume $S$ has no limit point, but then out of nowhere, because $\inf(X)\in X$ then $S$ has a limit point. Doesn't this contradict our assumption straight away?

My specific questions are:

1)Why does $X\subset S \Rightarrow \inf(X)\in X$ (and $S$ for the matter).

2)Why do we let $\inf(X)$ be a limit point? Wouldn't that go against the whole point of the proof?


问:
Finding a Function That Approaches Another Function
One of my math professors gave me the following challenge. It isn't graded, it's just for fun.

Consider the function:

\begin{equation*}
f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1).
\end{equation*}

I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:

$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?

答:
Yes looks like it must be b)

Consider the following:

There is only one $x$ factor, so this eliminates choices c) and d).

The function is odd, i.e $f(-x) = -f(-x)$, so this eliminates a).

Thus b) must be the answer.

To find the answer by hand you can do the following:

Start with

$$f(x) = x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} $$

Now differentiate

$$f'(x)  = 1 + 3x^2 + 5x^4 + \cdots $$

Multiply by $x$

$$f'(x)x = x + 3x^3 + 5x^5 + \cdots $$

Differentiate again

$$(f'(x)x)' = 1 + 3^2 x^2 + 5^2 x^4 + \cdots $$

Multiply by $x$, differentiate again and then multiply by $x$ to give the answer.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 2.5


问:
Questions about epimorphisms and projectives in functor categories
Suppose $I$ is a small category, $R$ is a ring and $_R\mathrm{Mod}$ is the category of left $R$-modules. How do I show that the category $[I,~_R\mathrm{Mod}]$ of all functors from $I$ to $_R\mathrm{Mod}$ is an abelian category with enough projectives and injectives?

It's my first time verifying the axioms of an abelian category and looking at projectives/injectives in a context different than that of modules. I think I can handle the abelian part on my own. However, in order to get a feel for the projectives I would like to ask a few questions:

A natural transformation $F\stackrel{\alpha}{\Rightarrow} G$ with all $\alpha_i~(i\in I)$ epimorphisms is an epimorphism in the functor category. Are there any others?

Are the functors $P:I\to~_R\mathrm{Mod}$, with all $P_i$ projective, projective, is this condition even necessary or sufficient? Is there a description of the projectives?

Given a functor $F$, is there a simple description of a projective $P$ and an epimorphism $P\Rightarrow F\Rightarrow 0$?

Even the simplest of examples $I\equiv\bullet\rightarrow\bullet$ i don't understand.

答:
@ZhenLin This is too long to fit into a comment.

Could you please give me some details? I don't understand what you mean by "Each representable presheaf $\mathcal I(i,−)$ is free because it occurs as the image of $1$ under the left adjoint of the evaluation-at-$i$ functor." I can see that there is an "evaluation at $i$ functor" given by
$$\begin{array}{rccl}
@_i:&[\mathcal{I},\mathbf{Set}]&\to&\mathbf{Set},\\
&F&\mapsto& Fi,\\
&\lbrace F\stackrel{\alpha}{\Rightarrow}G\rbrace&\mapsto &\lbrace Fi\stackrel{\alpha_i}{\rightarrow}Gi\rbrace
\end{array},$$
and that these most likely fit together to form a functor $@:\mathcal{I}\to [[\mathcal{I},\mathbf{Set}],\mathbf{Set}]$
I also know Yoneda lemma which describes the natural transformations from a represented, set-valued functor to any other set valued functor. Given
$$\begin{array}{rcl}
h:&\mathcal I^{op}&\to&[\mathcal I,\mathbf{Set}],\\
&i&\mapsto&\mathrm{Hom}_{\mathcal I}(i,-)
\end{array},$$
there is a natural bijection
$$\mathrm{Hom}_{[\mathcal I,\mathbf{Set}]}(h(i),F)\simeq F(i)=@_i(F)$$
but I don't see how you get an adjunction (and between what categories?).


问:
How to find the angle between the line passing through two points in 3d
There are points $P(1,1,2), Q(3,-2,2), R(1,4,6), S(-1,10,30)$. How do I find the angle between the line passing through $P$ and $Q$, and the line passing through $R$ and $S$?

Am I supposed to find the distance between the points $P$ & $Q$ and $R$ & $S$? and then use that to find the angle?

答: Relative to the origin as in vectors:$$v = \arccos\left(\frac{x_1x_2 + y_1y_2 + z_1z_2}{\sqrt{(x_1^2+y_1^2+z_1^2) \cdot (x_2^2+y_2^2+z_2^2)}}\right)$$


问:
$A$ ring, $\mathfrak{m}$ maximal ideal - then $\mathfrak{m}/\mathfrak{m}^2$ is isomorphic to $\mathfrak{m}A_\mathfrak{m}/\mathfrak{m}^2A_\mathfrak{m}$
According to lemma 4.2.3 of Qing Liu's book on Algebraic Geometry, the canonical homomorphism $\mathfrak{m}/\mathfrak{m}^2 \to \mathfrak{m}A_\mathfrak{m}/ \mathfrak{m}^2 A_\mathfrak{m}$ is an isomorphism of $A/\mathfrak{m}$ vector spaces.

To show this, the book proves that $\varphi: A/\mathfrak{m} \to A_\mathfrak{m}/\mathfrak{m}A_\mathfrak{m}$ is an isomorphism and says it suffices to tensor this map by $\otimes_A \mathfrak{m}$. I don't know why this is enough, wouldn't it be required that $\mathfrak{m}$ is a flat $A$-module?

答:
$\newcommand{\m}{\mathfrak m}$
You have an exact sequence, regardless of flatness:
$\m\otimes_A \m \to \m\to \m\otimes_A A/\m\to 0$ (note that flatness would add a leftmost $0$ but we don't need it)

Moreover, the image of $\m\otimes_A\m\to \m$ is clearly $\m^2$. This is not to say that $\m\otimes_A\m \cong \m^2$ which would indeed require flatness of $\m$. It follows that $\m/\m^2\cong \m\otimes_A A/\m$

Similarly, one has an exact sequence $\m\otimes_A \m A_\m \to \m A_\m\to \m\otimes_A A_\m/\m A_\m\to 0$, and we now have to check that the image of $\m\otimes_A \m A_\m \to \m A_\m$ is $\m^2 A_\m$, but this is clear too; and so similarly $\m A_\m/\m^2A_\m\cong \m\otimes_A A_\m/\m A_\m$


问:
The limitations of the middle square method in generating pseudo-random numbers.
I have been trying to research the middle square method as a way to generate pseudo-random numbers for a school project. Everywhere I look claims that this method is limited by its short period and will either loop or repeat if used enough times. However, I don't really understand how this happens or even why it happens. I wrote a Python script that takes a seed of 6 digits and I have yet to witness any looping or repetition. Is this because my seed is too large? If so, what digit seed would be appropriate to show the limitations of this method? Lastly, why does this method apparently repeat or loop if done enough times?
Thank you!</p>

答: Since your internal state has a fixed size (it's an $n$-digit number), there are only finitely many ($10^n$) possible states, so if you iterate enough ($10^n+1$ times) you'll inevitably repeat an internal state. Since the method is deterministic, the sequence will repeat from that point onward. This is true of all PRNG algorithms, but good ones have internal states much larger than the numbers being generated. (E.g. using a 10-digit state while only outputting the last 5 digits of each state in order to generate 5-digit numbers is better than using a 5-digit state.) You should be able to explore these ideas (with sufficiently small seeds) using your Python script.


问:
Every sequence of a closed set converges in the set
I have a doubt when saying if the following assertion is true or false: "Let $F$ be a closed set of a normed vector space $X$. Then every sequence of elements of $F$ converges to an element in $F$".

One one hand, I drafted this proof which states that the assertion is true:

Let consider a sequence $(x_n)_n$ of elements of $F$ which converges to $l \in X \backslash F$.

Then, $\forall r \geq 0, \exists N_r \in \mathbb{N}, \forall n \geq N_r, x_n \in \mathcal{B}(l,r)$.

Since $x_n \in F$, and $x_n \in \mathcal{B}(l,r)$, $F \cap \mathcal{B}(l,r) \neq \emptyset$.

So $X \backslash F$ is not open. So $F$ is not closed.

So by a contraposed reasoning, $F$ is closed.

On the other, if I take $[-1,1]$ in $(\mathbb{R}, (\lvert . \rvert )$ and consider the sequence $((-1)^n)_n$ whose elements are in $[-1,1]$, then the sequence diverges, which makes the assertion false.

EDIT: I think I've found my mistake. The assertion is false. My proof assumes that the sequence is convergent. Could you confirm?

答:
The statement should read: let $F$ be a closed set of a normed vector space $X$. Then any convergent sequence of elements from $F$ converges to an element of $F$.

The last could  also be stated as

If $\{x_n\mid n \in \Bbb N\} \subseteq F$ and there is some $x \in X$ such that $x_n \to x$, then $x \in F$.

which might be clearer, and in fact holds for any closed set $F$ in any topological space.


问:
Evaluating $\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$

How to evaluate $$\int_{0}^{\infty} \frac{x^8 -1}{x^{10}  + 1} dx$$

Attempt:

$$f(x) = \frac{x^8-1}{x^{10}+1}$$

$$I = \int_{0}^{\infty} \frac{x^8 -1}{x^{10}  + 1} dx$$

Substituting $t = \frac{1}{x}$,

$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{x^{10}}  + 1} x^2dx$$

$$I = \int_{0}^{\infty} \frac{x^{4} - x^{12}}{x^{10}  + 1} dx$$

Adding the two versions of $I$,

$$2I =  \int_{0}^{\infty} \frac{x^8 - x^{12} -1 + x^4}{x^{10}  + 1} dx = \int_{0}^{\infty} \frac{(x^8-1)(1 - x^{4}) }{x^{10}  + 1} dx$$

However, that does not help much.

Putting it in WolframAlpha gives $I = 0$

I am pretty sure there is an elegant solution which I am not able to find it.

答:
with $x = e^u$ and $dx = e^u du,$ I got
$$ \int_{- \infty}^\infty \frac{e^{4u} - e^{-4u}}{e^{5u} + e^{-5u}} du =  \int_{- \infty}^\infty \; \frac{ \sinh 4u \; }{ \cosh 5u \;} du$$
which is an odd integrand over symmetric endpoints


问:
What is the least nonnegative number $a$ congruent to $3^{340}\pmod{341}$?
Find the least nonnegative number $a$ congruent to $3^{340} \pmod{341}$.

What steps should I take to get to the answer?

Body:
mod $\,11\!:\ 3^{340}\! \equiv (3^{10})^{34} \equiv 1^{34}\! \equiv 1\ $ by little Fermat. $ $ Similarly

mod $\,31\!:\ 3^{340}\! \equiv 3^{10} (3^{30})^{10}\! \equiv (3^5)^2 1^{10}\! \equiv (9(9\cdot 3)))^2\! \equiv (9(-4))^2\!\equiv (-5)^2\! \equiv \color{#0a0}{25}$

By CRT, $ $ mod $\,11\!:\ 1 \equiv 3^{340}\! = \color{#0a0}{25}+31k \equiv 3-2k,\,$ so $\, 3k\equiv 3,\,$ so $\,\color{#c00}{k\equiv 1}.\,$

Therefore $\, 3^{340}\! = 25\!+\!31\color{#c00}k = 25\!+\!31(\color{#c00}1\!+\!11n) = 56\! +\! 340n$.


答:
General note: You use $=$ for saying "Equivalent propositions", but that's not 100% correct, since $=$ is a relation, so you want to use $\equiv$ or $\iff$.

For i) you did it well!

For ii), note the outermost parentheses, they will guide you where you must apply the equivalence $\neg(p\Rightarrow q)\equiv p\land\neg q$: $$\Bigl(a\Rightarrow b\Bigr) \Rightarrow  \Bigl(\neg c\Rightarrow (b\Rightarrow a)\Bigr),$$ so negating it:
\begin{align*}
\neg[(a\Rightarrow b)\Rightarrow(\neg c\Rightarrow (b\Rightarrow a))]&\equiv(a\Rightarrow b)\land\neg[\neg c\Rightarrow (b\Rightarrow a))]\\
&\equiv(\neg a\lor b)\land\neg(c\lor(\neg b\lor a))\\
&\equiv(\neg a\lor b)\land\neg c\land b\land\neg a.
\end{align*}

Hope you can finish.

Title: How do I negate the following statements?

Body:
I need help with these homework questions. Thanks in advance!

i) $(a\land b) \lor (a\land b)$

ii) $(a\Rightarrow b) \Rightarrow  (\neg c\Rightarrow (b\Rightarrow a))$

Here's what I have so far for each problem:

i) $\neg ((a\land b) \lor (a\land b))$

$ =(\neg a\lor\neg b) \land  (\neg a\lor\neg b)$

$=(\neg a\lor\neg b)$

I'm not sure if this is right but I was thinking of using De Morgan's rules for part i.

ii) I know when negating an implication, it goes as follows:

$(a\Rightarrow b) = (a\land \neg b)$

I'm not sure if I should start on the rightmost or leftmost side but any help would be appreciated for this!


问:
Proving equality for natural numbers
Given an equality such as $15(s^2 + s) = (t^2 + t)$, if we want to solve for $(t,s) \in \mathbb{N^2}$ which satisfies this equation using only peano axioms of $\mathbb{N}$, how would we do that? I notice that if $s = t $ then the left term is gonna be 15 times bigger than the right term. So i tried a bunch of different $s$ and $t$ values and found that the only solution is $s = t = 0$. But how do i show this formally with peano axioms? I find them very difficult to work with since it doesn't use integers, subtraction, rationals and division.

答:
Note that $$15(2s+1)^2= (2t+1)^2+14.$$

This is an example of a standard Pell's equation problem.  https://en.wikipedia.org/wiki/Pell%27s_equation

This equation has infinitely many solutions some of which can be generated as follows:-

Let $X=2s+1, Y=2t+1$, then $$15X^2= Y^2+14.$$

For example, this has the obvious solution $X_1=1,Y_1=1$. From this solution we can generate further solutions from the relations $$X_{n+1}=4X_n+Y_n, Y_{n+1}=15X_n+4Y_n.$$

So $X_2=5,Y_2=19$ and $X_3=39, Y_3=151$ and so on.

Then $(s,t)=(0,0),(2,9),(19,75)$ and so on.

Title: Proving equality for natural numbers


问:
Prove that a real normed space $V$ of dimension $n$ is disconnected in two connected parts by any linear subspace of dimension $n-1$

Theorem

All norms on a finite dimensional vector space $V$ are equivalent.

Proof. Omitted.

So to follow we will use the infty norm $\lVert\cdot\lVert_\infty$ given by the equation
$$
\lVert\vec v\lVert_\infty=\max\{|v^i|:i=1,...,n\}
$$
where $v^i$ is for each $i=1,..,n$ the $i$-th componet of $\vec v$ with respect any basis.

So let be $V$ a finite dimensional real vector space and so let be $W$ the subspace spanned by the $(n-1)$ vectors $\vec v_1,...,\vec v_{n-1}$, that is $W$ is the set of linear combinations of these such vectors. Now by standards theorems of finite dimensional vector spaces it must exist a vector $\vec v_n$ that is not linear combination of the above vectors and additionally any vector of $V$ is a linear combination of this vector and of the preceding ones, that is for any $\vec v\in V$ there must exist $n$ scalars $v^1,...,v^n$ such that
$$
\vec v=v^i\vec v_i
$$
where in the above identity I have used the Einstein convention. So we define the sets $X$ and $Y$ by letting that
$$
X:=\{\vec v\in V:\lambda^n>0\}\,\,\,\text{and}\,\,\,Y:=\{\vec v\in V:\lambda^n<0\}
$$
so that $V\setminus W=X\cup Y$ and thus if we prove that these sets are open then the statement follows. So for $\vec v\in V$ we observe that if $\vec u\in B(\vec v;v^n)$ then
$$
|u^n-v^n|\le\max\{(u^i-v^i):i=1,...,n\}=\lVert\vec u-\vec v\lVert_\infty<v^n
$$
so that
$$
-|v^n|<u^n-v^n<|v^n|
$$
from which we conclude that if $\vec v\in X$ then $\vec u\in X$ and if $\vec v\in Y$ then $\vec u\in Y$ so that these two sets are open.

So I ask if what I tried to prove is effectively true and if my arguments are correct. So could someone help me, please?

答:
The proof seems correct, but it's overly complicated.

Let $H$ be a subspace of $V$ of dimension $n-1$. Pick a basis $\{v_1,v_2,\dots,v_{n-1}\}$ of $H$ and complete it to a basis $\{v_1,\dots,v_{n-1},v_n\}$ of $V$.

Define the linear map $f\colon V\to\mathbb{R}$ by
$$
f(v_1)=f(v_2)=\dots=f(v_{n-1})=0,\quad f(v_n)=1
$$
Then the map $f$ is continuous (prove it) and its kernel is $H$, which is therefore closed. Now
$$
A=\{v\in V:f(v)>0\},\quad B=\{v\in V:f(v)<0\}
$$
are nonempty disjoint open sets and $V\setminus H=A\cup B$. They are not empty because $v_n\in A$ and $-v_n\in B$.


问:
Show that if $\mathbb{E}X_n = 0$ and $\sup\limits_{n \in \mathbb{N}} \mathbb{E}|X_n|^4 \leq C$ then $\frac{1}{n} (X_1+\dots+X_n) \to 0$ almost surely
Suppose $\{X_n\}_{n=1}^{+\infty}$ is a sequence of independent variables.

I'm trying to prove that if $\forall n \in \mathbb{N}: \mathbb{E}X_n = 0$ and $\exists C\in \mathbb{R}: \sup\limits_{n \in \mathbb{N}} \mathbb{E}|X_n|^4 \leq C$, then $\frac{1}{n} (X_1+X_2+\dots+X_n) \to 0$ almost surely.

First, my question is, is that true that the second premise $\sup\limits_{n \in \mathbb{N}} \mathbb{E}|X_n|^4 \leq C$ is equivalent to this: $\forall n \in \mathbb{N}: \mathbb{E}|X_n|^4 \leq C$?

I was trying to apply Kolmogorov's strong law. For this, I have to prove that this series converges:
$$\sum\limits_{n=1}^{+\infty} \dfrac{\text{Var}(X_n)}{n^2}$$

How can I approach this problem? I would appreciate any help.

答:
Kolmogorov's strong law is overkill in this scenario. Let $S_n=\sum_{i=1}^nX_i$. Then, since $\mathbb E[X_i] = 0$ for all $i$,

$$\mathbb E[S_n^4] = \sum_{i=1}^n\mathbb E[X_i^4] + 2\sum_{1\le i<j\le n}\mathbb E[X_i^2]\mathbb E[X_j^2] \le nC + n(n-1)C = n^2C,$$

where the Cauchy-Schwarz inequality is used to bound the second sum.
Hence using Chebyshev's inequality, for any $\varepsilon>0$ one has

$$ \mathbb P\left(\left|\frac1nS_n\right|>\varepsilon\right) \le \frac{\mathbb E[S_n^4]}{n^4\epsilon^4} \le \frac{C}{n^2\epsilon^4}. $$

It follows that $\sum_{n=1}^\infty \mathbb P\left(\left|\frac1nS_n\right|>\varepsilon\right) < \infty$, and so by Borel-Cantelli we have that $\left|\frac1nS_n\right|>\varepsilon$ only finitely many times, with probability one. Since this is true for any $\varepsilon>0$, it follows that $\frac1nS_n\to0$ almost surely.


问:
Markov chain, equilibrium measure and limits
We consider a Markov Chain $\left(X_n\right)_{n \in \mathbb{N}}$ on $S=\{1,2,3,4,5\}$ with transition matrix

Q = $\begin{pmatrix}
0 & 1/2 & 1/2 & 0 & 1/3\\
1 & 0 & 1/2 & 0 & 0 \\
0 & 1/4 & 0 & 1 & 0\\
0 & 1/4 & 0 & 0 & 1/3 \\
0 & 0 & 0 & 0 & 1/3
\end{pmatrix}$

We have the two classes $\{5\}$(which is transient) and $\{1,2,3,4\}$(which is recurrent)
It is clear that under $P_1(X_n+i) = \mathbb{P}(X_n=i \lvert X_0=1)$ the chain is irreducible, since it will always move around the four states.
To compute the equilibrium measure I solved $Q\mu = \mu$ since under $P_1$ the chain is irreducible it is unique and I obtained: $\mu = \left(\frac{3}{10},\frac{2}{5},\frac{1}{5}, \frac{1}{10}\right)$

Now I am asked to computed the following limits: $\lim_{n \rightarrow \infty}P_1(X_n=3)$ and $\lim_{n \rightarrow \infty}P_5(X_n=3)$.

How can I compute these? I think $\lim_{n \rightarrow \infty}P_1(X_n=3)$ = $\mu_3 = \frac{1}{5}$

But what about $\lim_{n \rightarrow \infty}P_5(X_n=3)$?

答:
You are correct that $\lim_{n\to\infty}\mathbb P_1(X_n=3)$ is given by the equilibrium probability $1/5$.

Now observe that for $n\geq1$, we have that
$$\mathbb P_5(X_n=3)=\frac{\mathbb P_1(X_{n-1}=3)+\mathbb P_4(X_{n-1}=3)+\mathbb P_5(X_{n-1}=3)}{3}$$
So
$$\lim_{n\to\infty}\mathbb P_5(X_n=3)=\lim_{n\to\infty}\frac{\mathbb P_1(X_n=3)+\mathbb P_4(X_n=3)}{2}=\frac{1}{5}.$$


问:
Extensions of bounded linear operators of finite rank (understanding a proof)
I came across a question about how to prove that we can extend bounded linear operators (on  normed spaces) of finite rank here: Extending a bounded linear operator of finite rank.

For clarity, I will copy/paste it below:

Hint: Say $b_1,\dots,b_n$ is a basis for the finite-dimensional space
$T(W)$. Then for every $w\in W$ there is a unique expansion
$$Tw=\sum_{j=1}^n a_jb_j.$$Say $$a_j=\Lambda_jw\quad(w\in W).$$Since
every linear functional on a finite-dimensional space is bounded,
there exist $c$ so that $$|\Lambda_j w|\le c||Tw||.$$

The accepted solution seems to be relying on the claim that $\Lambda_j$ is a linear functional on a finite dimensional space. But the question does not assume that $W$ is finite dimensional, only that $T(W)$ is. Am I missing something or is this "proof" not correct?

答:
You are making a fair point, but there is no harm done: the proof is correct, but (in my humble opinion) the person that posted this has made a small mistake (a typo, actually): it should have been
$$a_j=\Lambda_j(Tw),\;\;\;w\in W$$
followed by

$|\Lambda_j(Tw)|\leq c\|Tw\|\leq c\|T\|\|w\|$, so that the functionals $\Lambda_j$ are defined on $T(W)$ which is finite dimensional (thus they are immediately continuous). The key is then to consider the functionals $\Lambda_j\circ T:W\to\mathbb{F}$ which are now immediately continuous as compositions of continuous maps and extend those on the entire domain.

I will add this as a comment to the answer in the original post too.

Edit: In the comments, OP asks what can we say about $\|T\|$ and $\|T'\|$ when $T$ is onto, where $T'$ denotes the extension. If $T$ is onto then we can assume without loss of generality that $Y=\mathbb{C}^n$ for some $n\geq1$, since $T$ is of finite rank. We equip $Y$ with the maximum norm, since all norms are equivalent. Let's look at the construction:

we begin with $T:W\to\mathbb{C}^n$ and we compose with the functional $\varepsilon_i:\mathbb{C}^n\to\mathbb{C}$ acting as $(z_1,\dots,z_n)\mapsto z_i$, so $\epsilon_i\circ T:W\to\mathbb{C}$ are bounded functionals. By Hahn-Banach we extend those to functionals $\varphi_i:X\to\mathbb{C}$ in a norm-preserving fashion, i.e. $\|\varphi_i\|=\|\varepsilon_i\circ T\|$. We now define $T':X\to\mathbb{C}^n$ as $T'(x)=(\varphi_1(x),\dots,\varphi_n(x))$. This is bounded, extends $T$ and observe that
$$\|T'\|=\sup_{x\neq0}\frac{\|T'(x)\|}{\|x\|}=\sup_{x\neq0}\sup_{1\leq j\leq n}\frac{|\varphi_j(x)|}{\|x\|}=\sup_j\sup_x\frac{|\varphi_j(x)|}{\|x\|}=\sup_{1\leq j\leq n}\|\varphi_j\|=\sup_{1\leq j\leq n}\|\varepsilon_j\circ T\|$$
But $\|\varepsilon_j\circ T\|\leq\|T\|$, so $\|T'\|\leq \|T\|$. On the other hand, $T'$ extends $T$, so obviously $\|T\|\leq\|T'\|$.


问:
Convergence in probability for two sequences of random variables
Let $\{X_n\}$ and $\{Y_n\}$ be sequences of variables and suppose that $Y_n$ converges in probability to some random variable $Y$, i.e. $Y_n\xrightarrow{p}Y$. Is it true then that:

$$\lim_{n\rightarrow\infty}\mathbb{P}[|X_n-Y_n|>\epsilon]=0 \text{ implies } X_n\xrightarrow{p}Y$$

If so, how can I show this?

答:
Assume that (where I conveniently replaced Y with Z)
$$\begin{split}X_n-Y_n&\overset p {\rightarrow} 0\\
Y_n&\overset p {\rightarrow} Z\end{split}$$

Fix $\epsilon.$ Notice that $|X_n-Y_n|\le\frac \epsilon 2$ and $|Y_n-Z|\le\frac \epsilon 2$ implies that $|X_n-Z|\le\epsilon$, by the triangle inequality. Reversing the logic, this means that $|X_n-Z|>\epsilon$ implies that $|X_n-Y_n|>\frac \epsilon 2$ (inclusive) or $|Y_n-Z|>\frac \epsilon 2$. In particular, if an event implies that at least one of two other events has occurred, this means that $A\subset B\cup C$, i.e. $P(A)\le P(B\cup C)$.

Now

$$\begin{split}P(|X_n-Z|>\epsilon)&\le P(|X_n-Y_n|>\frac \epsilon 2\cup|Y_n-Z|>\frac \epsilon 2)\text { what we just said}\\
&\le P(|X_n-Y_n|>\frac \epsilon 2)+P(|Y_n-Z|> \frac \epsilon 2)\text { definition of union}
\end{split}$$

Take the limit to get $lim_{n\rightarrow\infty}P(|X_n-Z|>\epsilon)\le0$. Since probabilities are positive, it is 0.


答:
"every linear functional on a finite-dimensional space is bounded" does not refer to the functionals $\Lambda_i$ which are indeed defined on $W$.

Rather, let $\{b_1^*, \ldots, b_n^*\}$ be the dual basis to the basis $\{b_1 ,\ldots, b_n\}$ of $T(W)$, i.e. those are functionals on $T(W)$ such that $b_j^*(b_i) = b_{ij}$.

Now $\Lambda_i$ are defined with
$$Tw = \sum_{i=1}^n \Lambda_i(w)b_i, \quad w \in W$$
and they are evidently well-defined and linear. On the other hand, using the dual basis gives
$$Tw = \sum_{i=1}^n b_i^*(Tw)b_i, \quad w \in W$$
so by comparison $\Lambda_i = b^*_i \circ T$. Now $b^*_i$ are bounded since $T(W)$ is finite-dimensional so
$$|\Lambda_i(w)| = |b_i^*(Tw)| \le \|b^*_i\|\|Tw\| \le \|b^*_i\|\|T\|\|w\|.$$

Now extend the bounded functionals $\Lambda_i$ to bounded functionals on $X$...

Title: Extensions of bounded linear operators of finite rank (understanding a proof)

Body:
I came across a question about how to prove that we can extend bounded linear operators (on  normed spaces) of finite rank here: Extending a bounded linear operator of finite rank.

For clarity, I will copy/paste it below:

Hint: Say $b_1,\dots,b_n$ is a basis for the finite-dimensional space
$T(W)$. Then for every $w\in W$ there is a unique expansion
$$Tw=\sum_{j=1}^n a_jb_j.$$Say $$a_j=\Lambda_jw\quad(w\in W).$$Since
every linear functional on a finite-dimensional space is bounded,
there exist $c$ so that $$|\Lambda_j w|\le c||Tw||.$$

The accepted solution seems to be relying on the claim that $\Lambda_j$ is a linear functional on a finite dimensional space. But the question does not assume that $W$ is finite dimensional, only that $T(W)$ is. Am I missing something or is this "proof" not correct?


Title: How to find the angle between the line passing through two points in 3d

Body:
There are points $P(1,1,2), Q(3,-2,2), R(1,4,6), S(-1,10,30)$. How do I find the angle between the line passing through $P$ and $Q$, and the line passing through $R$ and $S$?

Am I supposed to find the distance between the points $P$ & $Q$ and $R$ & $S$? and then use that to find the angle?

Body:
$P(1,1,2), Q(3,-2,2), R(1,4,6), S(-1,10,30)$

equation of line $PQ$ is $(x,y,z)=(1,1,2)+t(3-1,-2-1,2-2)=(1,1,2)+t(2,-3,0)$
$$
PQ:\begin{cases}
x=1+2t\\
y=1-3t\\
z=2\\
\end{cases}\to \vec{a}=(2,-3,0)
$$
$RS$ has equation $(x,y,z)=(1,4,6)+u(-1-1,10-4,30-6)=(1,4,6)+u(-2,6,24)$
$$
RS:\begin{cases}
x=1-2u\\
y=4+6u\\
z=6+24u\\
\end{cases}\to \vec{b}=(-2,6,24)
$$

call $\alpha$ the angle between the two lines.

$\vec{a}$ and $\vec{b}$ are the direction vector of the two lines.

$$\cos\alpha=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$$
That is

$$\cos\alpha=\frac{(2,-3,0)\cdot(-2,6,24)}{||(2,-3,0)||\cdot||(-2,6,24)||}=-\sqrt{\frac{11}{182}}$$
$$\alpha\approx 104.23°$$


问:
Condition that the faces of a graph be unique?
There is a theorem that if a planar graph is 3-vertex-connected, then it has a unique embedding up to a reflection. (See e.g. here.) This means that its faces and its dual graph are uniquely defined.

Is there a generalization of this result to graphs of higher genus (e.g. a graph that is not planar but can be embedded on the torus)?  Are there simple conditions that guarantee that the minimum-genus embedding is unique?

答:
If you're looking for generalizations of properties of planar embeddings then you definitively should have a look at results involving the face-width of a graph.

Directly answering your question, I believe no equivalent result exists. At least not for general surfaces (I vaguely remember some theorems regarding unique embeddings for the Torus and Projective Plane).

However the following result, due to Thomassen[1], may be close to what you want.

The edge-width of an embedding $\Pi$ of $G$ is the length of the shortest $\Pi$-noncontractible cycle. Large-edge-width embeddings are embedding whose edge-width are larger than the maximum length of a facial walk.

Theorem[1]: Let $G$ be a 2-connected graph that has a LEW-embedding in a surface $S$. Then a cycle $C$ in $G$ is contractible in every embedding of $G$ in $G$ if and only if $C$ is contractible in some embedding of $G$ in $S$. If, in addition, $G$ is a subdivision of a 3-connected graph, then $G$ is uniquely embeddable in $S$.

You can find this result and many other on the 5th chapter of the Graphs on Surfaces book.

[1] Thomassen, Carsten. "Embeddings of graphs with no short noncontractible cycles." Journal of Combinatorial Theory, Series B 48.2 (1990): 155-177.


问:
almost sure convergence of expected value
Let $X:\Omega \to \overline{\mathbb{R}}= \mathbb{R} \cup \{-\infty,+\infty \}$ a random variable.

If $\mathbb{E}[|X|]=0,$ how can I show that $X=0$ almost sure.

答: By Markov's inequality, $\mathbb P(|X| > \varepsilon) \le \mathbb{E}[|X|]/\varepsilon = 0$ for all $\varepsilon > 0$.  Therefore $|X| \le \varepsilon$ a.s. for all $\varepsilon > 0$ and therefore $|X| = 0$ a.s.


问:
Measure, Integration & Real Analysis Sheldon Axler SEction 2B Exercise 12
the question can be found here, but is also repeated below.

Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is a function.

(a) For $k\in\mathbb{Z}^{+}$, let

$$G_k = \{a\in\mathbb{R}: \exists\delta>0 \text{ s.t. } \lvert f(b) - f(c)\rvert < \frac{1}{k}, \forall b,c\in(a-\delta, a+\delta\}.$$

Prove that $G_k$ is an open subset of $\mathbb{R}$ for each $k\in\mathbb{Z}^{+}$.

(b) Prove the set of points where $f$ is continuous equals $\cap_{k=1}^{\infty}G_{k}$.

(c) Conclude that the set of points at which $f$ is continuous is a Borel set.

So, I'm really just looking for a help at where to start. I think that  I have an idea of where to set for (b) since the definition of $G_k$ is essentially just the definition of continuity using $\varepsilon_k = \frac{1}{k}$, but I'm fairly lost on what to do for the other sections.

Thank you!

答:
Welcome to MSE!

Hint:

For (a), what happens if we wiggle $a$ a little bit? That is, we know that $a \in G_k$ satisfies some property. Can you show that actually some open set around $a$ is also in $G_k$? Do you see why that would show that $G_k$ is open?

As for showing it, we know that if we stay within $\delta$ of $a$, something good happens. Can you show that for every point within $\delta/2$ of $a$ (say), there's a smaller radius around those points for which good things keep happening?

For (b), you're completely right!

For (c), you've shown the points of continuity is a (countable!) intersection of open sets $G_k$. Why must this be borel?

I hope this helps ^_^


问:
How many 5-bead necklaces can be made with beads of 21 colours?
How many 5-bead necklaces can be made with beads of 21 colors?

I get the answer using Polya's Enumeration Theorem as 413 049. Can someone confirm?

Does it include cases other than all five beads being distinct?

Many thanks.

答:
The answer has been found correct by comparing with
http://oeis.org/A060446/list</p>


问:
Rewriting arguments using quantifiers, variables, and predicate logic
There are two statements which sounds similar to me but their answers are different according to answer sheet.

All birds can fly

Answer: ∀ x [B(x) → F(x)]

Some men are genius

Answer: ∃ x [M(x) ^ G(x)]

In the first part implication is used but in second part AND is used. There is no mention of 'if then' or 'And' in  statements so why the answers are different for similar sounding statements?

答:
There are two ways to look at this. In the first statement
about birds that can fly, the implication, B->F, just as any
other implication can be written as F or -B, or else,
-(B and -F). In the second statement about men geniuses,
it concerns the existence of a man who is a genius. It is
the same as stating that there is a genius who is a man.
Here, being a man and being a genius have symmetric roles.
In the first statement, being a bird and flying do not
have symmetric roles and implication is not symmetric.
For example, a single flying insect would invalidate
the statement that all flying things are birds.</p>


问:
Poker cards combinatorics question
Question

Suppose you draw 7 cards from a standard deck of 52 cards. What is the probability that you draw three pairs of cards if

the three pairs must have distinct denominations?

the three pairs don't necessarily have distinct denominations? (two cases)

Note: Do not include the possibility of having exactly three cards of any denomination.

My attempt

For #1, my solution is $$probability = \frac{{{\binom{4}{2}}^3\binom{13}{3}(46)}}{\binom{52}{7}}$$
And I'm unsure what #2 could be.

I'd appreciate any help!

答:
For second,

We either choose $3$ denominations or $2$. For pairs from $3$ denominations, it is same as the first and for pairs from $2$ denominations, we first choose the denominations and then we choose which we will have $4$ cards of and which $2$ ($2$ ways do so).

So, probability $\displaystyle = \frac{\binom{13}{3} \ {{\binom{4}{2}}^3 \cdot 40} + \binom{13}{2} \cdot 2 \cdot {\binom{4}{2}} \cdot 44}{\binom{52}{7}}$

Also for first, instead of $46$ choices for the $7$th card, you should have only  $40$ choices (as $12$ cards cannot be considered - no three cards can have the same denomination).

Title: Poker cards combinatorics question


问:
If $A$ is a $n\times n$ matrix with entries from $\mathbb R$ or $\mathbb C$ (complex) and ${A^{T}}A=0$ prove $\operatorname{Rank}(A)\le \frac{n}{2}$
If $A$ is a $n\times n$ matrix with entries from $\mathbb R$ or $\mathbb C$ (complex) and ${A^{T}}A=0$ prove that $$\operatorname{Rank}(A)\le \frac{n}{2}.$$

i know how to prove it when the entries are Real numbers  but i have trouble when the entries are complex numbers.

答:
We can use Sylvester rank inequality
$$\operatorname{rank}(A) + \operatorname{rank}(B) \le \operatorname{rank}(AB) + n$$ which is valid for $n\times n$ matrices $A,B$ over any field to obtain
$$2\operatorname{rank}(A)=\operatorname{rank}(A^T) + \operatorname{rank}(A)\le\operatorname{rank}(A^TA) + n = n.$$

In case of real numbers from $A^TA = 0$ we actually get $A=0$ since $\operatorname{Tr}(A^TA)$ is the sum of squares of all elements of $A$.


问:
Expectation over a minimum
Let $X \geq 0$ a random variable.Using only the definition of the expectation (without using Fatou Lemma, BCT,MCT) show that : $$\lim_{t \to \infty}\mathbb{E}[X \wedge t] = \mathbb{E}[X],$$ where $t$ is a continous variable , i.e $t \in \mathbb{R}.$

My effort

Assume that $t > 0$ otherwise $X \wedge t = t$ which is trivial. Then $Y := X \wedge t$ is a nonnegative variable and we have

\begin{align*}
E[Y] &= \int_0^\infty P( Y > k) dt \\
       &= \int_0^\infty P( X > k, t > k) dt \\
       &= \int_0^\infty P( X > t)\mathbf{1}\{ t > k\} dt \\
       &= \int_0^t P(X > k)dt
\end{align*}

But does this closes the proof?

答:
$\newcommand{\dd}{\mathop{}\!\mathrm{d}}$
You are basically there already, modulo some typos.
$$\mathbb E[X\wedge t] = \int_0^\infty\mathbb P(X\wedge t>k)\dd k =\int_0^t\mathbb P(X>k)\dd k.$$
Now
$$\lim_{t\to\infty}\mathbb E[X\wedge t]=\lim_{t\to\infty}\int_0^t\mathbb P(X>k)\dd k=\int_0^\infty\mathbb P(X>k)\dd k=\mathbb E[X].$$
The middle equality in the above line is just the definition of what it means for one of the limits of the integral to be $\infty$.

Title: Expectation over a minimum


问:
Something about the proof of Bolzano-Weierstrass Theorem not making sense
So I'm trying to understand the proof of this version of the theorem

Here is the proof given

I have understood Lemma 1. and its implications.
The thing I don't get about it is:

We assume $S$ has no limit point, but then out of nowhere, because $\inf(X)\in X$ then $S$ has a limit point. Doesn't this contradict our assumption straight away?

My specific questions are:

1)Why does $X\subset S \Rightarrow \inf(X)\in X$ (and $S$ for the matter).

2)Why do we let $\inf(X)$ be a limit point? Wouldn't that go against the whole point of the proof?

答:
Regarding your comment "out of nowhere", what's happening here is a contradiction proof within a contradiction proof. As you say, we have assumed that $S$ has no limit point.

Next comes the following line of deduction regarding any nonempty subset $X$ of $S$: If we also assume that $\inf(X) \not\in X$, then $\inf(X)$ is a limit point of $X$ and therefore $\inf(X)$ is a limit point of $S$, and this is a contradiction. Therefore we conclude that $\inf(X) \in X$.

So we have proved: every nonempty subset of $S$ contains its own infimum.

And then, similarly, every nonempty subset of $S$ contains its own supremum.


问:
Is there a formula to calculate future value with compound interest where only part of the interest of each period is applied?
Based on FV=C0*(1+r)^n

Where

C0 = Cash flow at the initial point (Present value)

r = Rate of return

n = number of periods

Is there a way to adapt this formula to

(1) In each period, only a% of the interest gained in that period remains in the sum? (in other words: in each period, b% of the interest gained in that period is taken off the sum)

(2) In each period, N (given N is an absolute number) is taken off the sum?

In a real-world scenario, I'm looking for a formula where people will, in every period, take a certain amount (either a % or an absolute value) from the interest of their investment.

答:
For your first question
$$ FV = C_0(1+a\times r)^n $$
In this formula I supposed that the percentage remaining ($a$%) is fixed and does not depend on the period.

For your second question
$$ ((C_0(1+r)-N)(1+r)-N)\cdots(1-r) - N $$
In this case the formula is trickier and I do not think there is a tidier way to write it. This happens because the interest compounds but the amount taken off is constant. If we have had worked with simple interest (instead of compound interest) then this formula would have been
$$ C_0(1+n\times r)-n\times N $$


问:
$n$-th Derivative $\frac{d^{n}}{d x^{n}} e^{-\sqrt{x} |\omega|}$ via Recursive Product Rule
Let $g(x) = x^{-\frac{1}{2}}$ and $f(x) = e^{-\sqrt{x} |\omega|}$. I am trying to find an expression for the $M$-th derivative of their product:
\begin{align}
    \frac{d^M}{dx^M} \left[ f(x) g(x) \right] = \sum_{k=0}^{M} \binom{M}{k} f^{(M-k)}(x) g^{(k)}(x)
\end{align}
where $M$ is an even integer and $\omega \in \mathbb{R}$ is a constant parameter.

The derivative for $g(x)$ is easy:

\begin{align}
    \frac{d^{k}}{d x^{k}} g(x) = \frac{(2k-1)!!}{2} x^{-\frac{2k+1}{2}}
\end{align}

However, for $f(x)$, it is not so easy, and this is where I get stuck.

For the $M$-th derivative of $f(x)$, Mathematica gives me:

```<code>M! DifferenceRoot[
   Function[{\[FormalY], \[FormalN]}, {(2 + 6 \[FormalN] +
          4 \[FormalN]^2) \[FormalY][1 + \[FormalN]] +
       4 (1 + \[FormalN]) (2 + \[FormalN]) x \[FormalY][
         2 + \[FormalN]] - \[FormalY][\[FormalN]] Abs[w]^2 ==
      0, \[FormalY][0] ==
      E^(-Sqrt[x] Abs[w]), \[FormalY][1] == -((
       E^(-Sqrt[x] Abs[w]) Abs[w])/(2 Sqrt[x]))}]][M]
</code>```

But I have no idea what any of that means, and their site is not very helpful...

It seems that the solution will involve some kind of recursive application of the product rule, however, I can't seem to find a pattern after I do a handful of layers...

Question:

What is $\frac{d^{n}}{d x^{n}} f(x) = \frac{d^{n}}{d x^{n}} e^{-\sqrt{x} |\omega|}$ when $n$ is an even integer?

答:
It's sufficient to consider the case $\omega=1$: if we let
$$
\color{DarkBlue}{f_n(x):=e^{\sqrt{x}}\sqrt{x}\frac{d^n}{dx^n}\frac{e^{-\sqrt{x}}}{\sqrt{x}}},
$$
then the needed expression is
$$
\frac{d^n}{dx^n}\frac{e^{-|\omega|\sqrt{x}}}{\sqrt{x}}=|\omega|^n\frac{e^{-|\omega|\sqrt{x}}}{\sqrt{x}}f_n(\omega^2 x).
$$

To compute $f_n(x)$, we consider (for $x>0$ and $|t|<x$)
$$
f(x,t):=\sum_{n=0}^\infty f_n(x)\frac{(-t)^n}{n!}=e^{\sqrt{x}}\sqrt{x}\frac{e^{-\sqrt{x-t}}}{\sqrt{x-t}}=\frac{e^{\sqrt{x}-\sqrt{x-t}}}{\sqrt{1-t/x}}.
$$

Now we use the formula (for $|z|<1$) $$\frac{(1-\sqrt{1-z})^k}{\sqrt{1-z}}=2^k\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{z}{4}\right)^n.$$ To prove it, one may use the known formula
$$
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^k=\sum_{n=0}^\infty\binom{2n+k}{n}\frac{k}{2n+k}z^n;
$$
here $k/(2n+k)$ is interpreted as $1$ when $n=k=0$ (recipe: multiply this by $z^k$, replace $z$ with $z/4$ and $k$ with $k+1$, and take the derivative).

Returning to the computation, we have
$$
f(x,t)=\sum_{k=0}^\infty\frac{1}{k!}\frac{(\sqrt{x}-\sqrt{x-t})^k}{\sqrt{1-t/x}}
=\sum_{k=0}^\infty\frac{(2\sqrt{x})^k}{k!}\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{t}{4x}\right)^n,
$$
which, after (exchanging the summations and) extracting the coefficient of $t^n$, results in
$$
f_n(x)=(-1)^n\sum_{k=0}^n\frac{(2n-k)!}{k!(n-k)!}(2\sqrt{x})^{k-2n}
\color{DarkBlue}{=(-1)^n\sum_{k=0}^n\frac{(n+k)!}{k!(n-k)!}(2\sqrt{x})^{-n-k}}.
$$

Another approach is to use the following representation:
$$
\frac{e^{b(s-\sqrt{a^2+s^2})}}{\sqrt{a^2+s^2}}=\int_0^\infty e^{-st}J_0\left(a\sqrt{t^2+2bt}\right)dt,
$$
where $J_0(z)=\sum_{n=0}^\infty(-z^2/4)^n/n!^2$ is a Bessel function.

Here we put $s=\sqrt{x}$, $a=\sqrt{y}$ and $b=1$; the LHS is then $f(x,-y)/\sqrt{x}$, and the RHS may be "evaluated" by substituting the power series for $J_0$ directly, and integrating the result termwise, using the binomial formula for $(t+2)^n$. The outcome is the same.


问:
Does gradient descent for linear regression select the minimal norm solution?
I was told that Gradient Descent finds the weights of smallest norm.

This is what I understood in the linear regression setting:

$f_w(x)=w^\top x$ are the linear functions $ \mathbb{R}^n \rightarrow \mathbb{R} $.

$ \mathcal{D}=\{(x_i,y_i) \in \mathbb{R}^d\times\mathbb{R} \mid i=1,...,n\} $ is a dataset

Then I choose some loss function $L$ to get a cost function
$ C_0(w)=\frac{1}{n} \sum_i L(f_w(x_i),y_i) $

Finally, I define a regularized cost $ C_\lambda(w)=C_0(w)+\frac{\lambda}{2n}||w||^2 $

Then there must be some sort of equivalence between (and it looks it should be some classical result):

minimizing $C_\lambda(w)$ for an opportune $ \lambda \ge 0 $

minimizing $C_0(w)$ with Gradient Descent

My questions:

Can somebody clarify a bit this fact? By providing some hypothesis (e.g. on $L$), some insight, some reference...

Does this apply in a similar way to the linear classification setting ($y_i\in\{-1,+1\}$)? In that case, I would expect some equivalence bewteen:

maximizing $M(w)= \min_{i=1...n} y_i \langle w, x_i \rangle $ over $||w||=1$

minimizing $||w||$ over $ y_i \langle w, x_i \rangle \ge 1 $

maybe the previous 1. and 2. points above (those regarding regression)

答:
(2nd answer) Direct approach: the Gradient Descent (GD) update is given as

$$w_{k+1} = w_k - \tfrac{\eta}{N} X^T(Xw_k-y) = (I - \tfrac{\eta}{N}X^TX)w_k -  \tfrac{\eta}{N}X^Ty $$

By induction, we find

$$w_{k} = (I - \tfrac{\eta}{N}X^TX)^kw_0 -\Big[\sum_{j=0}^{k-1}(I - \tfrac{\eta}{N}X^TX)^j \Big]\tfrac{\eta}{N}X^Ty$$

Now, consider the Singular Value Decomposition $X=U\Sigma V^T$. In particular, $X^TX = V\Sigma^T\Sigma V^T$ is the eigendecomposition of $X^TX$ with singular values $\sigma_i$.

Claim: If $\eta<N/\sigma_{\max}^2$, then GD, starting at $w_0 = v_0+v_0^\perp$ with $v_0\in\ker(X)$, $v_0^\perp\in\ker(X)^\perp$ converges against $v_0+X^+y$

Proof:
The second term converges against $X^+y$ because

$$\begin{aligned}
\Big[\sum_{j=0}^{k-1}(I - \tfrac{\eta}{N}X^TX)^j\Big]\tfrac{\eta}{N}X^T
=V\Big[\sum_{j=0}^{k-1}(I - \tfrac{\eta}{N}\Sigma^T\Sigma)^j\Big]\tfrac{\eta}{N}\Sigma^T U^T
\overset{(*)}{\longrightarrow} V\Sigma^+ U^T=X^+
\end{aligned}$$

And regarding the first term, for any eigenvalue $v_i$ we have

$$(I - \tfrac{\eta}{N}X^TX)^kv_i = V(I-\tfrac{\eta}{N}\Sigma^T\Sigma)^k V^T v_i = (1-\tfrac{\eta}{N}\sigma_i^2)^kv_i \to v_i\cdot\begin{cases}0:&\sigma_i>0 \wedge \eta<\frac{N}{\sigma_i^2}\\ 1:& \sigma_i=0\end{cases}$$

And thus $\lim\limits_{k\to\infty}(I - \tfrac{\eta}{N}X^TX)^k(v_0+v_0^\perp) = v_0$

(*) Because

$$V\Big[\sum_{j=0}^{k-1}(I - \tfrac{\eta}{N}\Sigma^T\Sigma)^j\Big]\tfrac{\eta}{N}\Sigma^T U^T = V\tfrac{\eta}{N}\Sigma^T\Big[\sum_{j=0}^{k-1}(\Pi - \tfrac{\eta}{N}\Sigma\Sigma^T)^j\Big] U^T$$

where $\Pi_{ii}=1$ if $\sigma_i>0$ and zero else. Thus $D=\Pi - \tfrac{\eta}{N}\Sigma\Sigma^T$ is a diagonal matrix with $$(D)_{ii} = \begin{cases} (1-\tfrac{\eta}{N}\sigma_i^2)^j:& \sigma_i>0\\ \hfill 0:&\sigma_i=0\end{cases}$$

Thus, since $\lim\limits_{k\to\infty}\sum_{j=0}^k q^k = \frac{1}{1-q}$ whenever $|q|<1$, $\lim\limits_{k\to\infty}\sum_{j=0}^{k}D^j = \begin{cases} 1/(\frac{\eta}{N}\sigma^2):& \sigma_i>0\\ \hfill 0:&\sigma_i=0\end{cases}$.

Here $|\tfrac{\eta}{N}\sigma_i^2|<1$ for all $i$ if $\eta<N/\sigma_{\max}^2$.
In this case $\lim\limits_{k\to\infty}\tfrac{\eta}{N}\Sigma^T\Big[\sum_{j=0}^{k-1}(\Pi - \tfrac{\eta}{N}\Sigma\Sigma^T)^j\Big] = S$ diagonal matrix with $S_{ii}=\begin{cases} 1/\sigma_i:& \sigma_i>0\\ \hfill 0:&\sigma_i=0\end{cases}$, i.e. $S = \Sigma^+$


问:
Extensions of scalars and tangent map of morphism of schemes
Suppose I have a morphism of rings $\varphi: A \to B$, $M$ a $B$-module, $N$ an $A$-module and $f: M \to N$ a homomorphism of $A$-modules (here, $M$ is considered an $A$-module by restriction of scalars). How do I properly "extend" $f$ to a morphism of $B$-modules $F: M \to N \otimes_A B$? You may assume that $A, B$ are fields, and $M, N$ are vector spaces.

I'm asking this because I'm confused about the definition of the map $df_x: T_xX \to T_{f(x)} Y \otimes_{k(f(x))} k(x)$ for a morphism of schemes $f: X \to Y$. I know that we have a $k(f(x))$-linear map $T: \frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}} \to \frac{\mathfrak{m}_x}{\mathfrak{m}^2_x}$ induced from the map of stalks $f^\#_x: \mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ - $k(f(x)$ linearity of $T$ follows from the inclusion $k(f(x)) \to k(x)$. Taking the dual of $T$, we have a  $k(f(x))$-linear map $T^\lor: T_x X \to T_{f(x)} Y$. How is $df_x: T_xX \to T_{f(x)} Y \otimes_{k(f(x))} k(x)$ defined in terms of $T$ and $T^\lor$? I tried considering the map $T^\lor \otimes_{k(f(x))}Id_{k(x)}$ - but I wasn't sure that $T_xX \otimes_{k(f(x))} k(x) \cong T_xX$.

答: The dualization here is crucial to the existence of the map in question.  The $k(f((x))$-linear map $T:\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}} \to \frac{\mathfrak{m}_x}{\mathfrak{m}^2_x}$ induces a $k(x)$-linear map $T':\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}\otimes_{k(f(x))}k(x) \to \frac{\mathfrak{m}_x}{\mathfrak{m}^2_x}$ by $T'(v\otimes a)=aT(v)$.  The map $df_x$ is then defined as the dual of this map $T'$ over $k(x)$, identifying the $k(x)$-dual of $\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}\otimes_{k(f(x))}k(x)$ with $\left(\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}\right)^\vee\otimes_{k(f(x))}k(x)$ (which only works if either $\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}$ or $k(x)$ is finite-dimensional over $k(f(x))$).


问:
Can each edge of a Wythoffian polytope be flipped by a reflection?
A Wythoffian polytope $P\subset\Bbb R^d$ is an orbit polytope of a finite reflection group, that is,

$$P:=\mathrm{Orb}(\Gamma,x):=\mathrm{conv}\{Tx\mid T\in\Gamma\},$$

where $\Gamma$ is a finite reflection group, and $x\in\Bbb R^d$.

Question: Is it true, that if $e$ is an edge of $P$, then there is a reflection $T\in\Gamma$ that fixes $e$ set-wise but flips its orientation?

I am pretty sure that this is true, but I cannot find a concise proof from first principles. I would be most happy with a short self-contained argument (possibly using some well-known properties of reflection groups).

Partial proof

Suppose that the generator $x\in\Bbb R^d$ was chosen from the interior of a Weyl chamber of $\Gamma$ (the resulting polytope is sometimes called a $\Gamma$-permutahedron or omnitruncated uniform polytope). Since $\Gamma$ acts regularly on the Weyl chambers, every Weyl chamber contains a single vertex. In particular, if $e$ is an edge of $P$, its end vertices are in different chambers and the edge must cross a bounding reflection hyperplane of the chamber. Since reflection on this hyperplane is a symmetry of $P$, the edge must be perpendicular to the hyperplane and the reflcetion on it must flip its orientation.

My hope is that this can be extended to an argument for general placements of $x$.
Each Wythoffian polytope can be obtained as a "limit" (in some sense) of such $\Gamma$-permutahedra, and maybe this sufficec to prove the statement. However, I was not able to make a clear case for why the edges of the resulting polytope are "limits" of edges of the $\Gamma$-permutahedra.

答:
Isn't that exactly the definition of being Wythoffian?

I.e. by virtue of kaleidoscopical construction the edges occur only as convex hulls of a seed point (vertex) and its mirror image wrt. to some mirror out of the reflection group. In fact one which occurs as boundary of the fundamental domain. Thence this very mirror fixes this very edge setwise, but reverses its orientation.

--- rk


问:
explaining the derivative of $x^x$
You set the following exercise to your calculus class:

Q1. Differentiate $y(x) = x^x$.

A student submits the following solution:

Let $g(a)=a^x$ and $f(x)=x$.  Then $y(x) = g(f(x))$, so by the chain rule, $y'(x) = f'(x) g'(f(x)) = 1 \cdot x \cdot (x^{x-1}) = x^x$.

How would you explain to the student why their solution is incorrect?

To be clear, I know why this is wrong but am interested in good ways to explain it to undergraduate or high school students.

In this question someone has problems differentiating $x^x$, but they didn't take the approach of my hypothetical student.

答:
The $g(a) = a^x$ is invalid, because the argument is $a$ and $x$ appears as a free variable. If a function's body has a free variable, that has to be defined somewhere as a constant or a globally understood parameter.

A function cannot have a free variable that refers to the argument of another function.

The correct version of $g$ is $g(a) = a^a$.

The function $g$ raises the argument to itself; it does not raise its argument to $x$. $x$ is a dummy variable in the definition of $f(x)$, which is another function. And what if the dummy variable in $f(x)$ is some thing else, like $f(b) = b$?

Think about how this would work in some functional programming language in computing:

```<code>function f(x)
   return x

function g(a)
   return pow(a, x)
</code>```

Look, g is referring to x, but the only x that we see is local in another function's scope! There is no visible binding for x in the scope of g's body, so it is a free variable. Moreover, we can edit f into:

```<code>function f(b)
   return b
</code>```

without changing the meaning of the program. So now there isn't an x anywhere.

In mathematics, functions can often be regarded as macros! But they are not blind, textual macros; they are hygienic macros. Substitution in math formulas is on structure not text, and it obeys scoping rules by maintaining lexical transparency.

For instance consider this following student reasoning error. Let $f(x) = x\times x$ and let $g(y) = y + y$. Therefore, since manipulation of mathematic formulas is just dumb textual substitution like preprocessor macros in the C language, $f(g(3)) = 3 + 3 * 3 + 3$, and so the value is $f(g(3)) = 3 + 9 + 3 = 15$. And in general $f(g(z)) = z^2 + 2z$. What is wrong?

Since substitution in math formulas follows structure, $f(g(z))$ cannot be $z^2 + 2z$. $f$ multiplies whatever object comes out of $g$ without breaking apart its syntax, and so the necessary parentheses have to be shown when it is all put together: $f(g(z)) = (z + z)(z + z) = 4z^2$.

The rule of lexical transparency is this: that when we substitute a formula $\alpha$ as a replacement for some symbol in another formula $\beta$, all of the symbols in $\beta$, all of the symbols in $\alpha$ continue to have exactly the same meaning. Even if $\beta$ has symbols which have the same names, those symbols do not capture any of the references in $\alpha$.

To prevent confusion, if such a situation arises where there is a confusion between symbols in an inserted formula and in the target of insertion, we should perform a variable renaming.

Title: explaining the derivative of $x^x$

Body:
The $g(a) = a^x$ is invalid, because the argument is $a$ and $x$ appears as a free variable. If a function's body has a free variable, that has to be defined somewhere as a constant or a globally understood parameter.

A function cannot have a free variable that refers to the argument of another function.

The correct version of $g$ is $g(a) = a^a$.

The function $g$ raises the argument to itself; it does not raise its argument to $x$. $x$ is a dummy variable in the definition of $f(x)$, which is another function. And what if the dummy variable in $f(x)$ is some thing else, like $f(b) = b$?

Think about how this would work in some functional programming language in computing:

function f(x)
   return x

function g(a)
   return pow(a, x)

Look, g is referring to x, but the only x that we see is local in another function's scope! There is no visible binding for x in the scope of g's body, so it is a free variable. Moreover, we can edit f into:

function f(b)
   return b

without changing the meaning of the program. So now there isn't an x anywhere.

In mathematics, functions can often be regarded as macros! But they are not blind, textual macros; they are hygienic macros. Substitution in math formulas is on structure not text, and it obeys scoping rules by maintaining lexical transparency.

For instance consider this following student reasoning error. Let $f(x) = x\times x$ and let $g(y) = y + y$. Therefore, since manipulation of mathematic formulas is just dumb textual substitution like preprocessor macros in the C language, $f(g(3)) = 3 + 3 * 3 + 3$, and so the value is $f(g(3)) = 3 + 9 + 3 = 15$. And in general $f(g(z)) = z^2 + 2z$. What is wrong?

Since substitution in math formulas follows structure, $f(g(z))$ cannot be $z^2 + 2z$. $f$ multiplies whatever object comes out of $g$ without breaking apart its syntax, and so the necessary parentheses have to be shown when it is all put together: $f(g(z)) = (z + z)(z + z) = 4z^2$.

The rule of lexical transparency is this: that when we substitute a formula $\alpha$ as a replacement for some symbol in another formula $\beta$, all of the symbols in $\beta$, all of the symbols in $\alpha$ continue to have exactly the same meaning. Even if $\beta$ has symbols which have the same names, those symbols do not capture any of the references in $\alpha$.

To prevent confusion, if such a situation arises where there is a confusion between symbols in an inserted formula and in the target of insertion, we should perform a variable renaming.


问:
Finding side lengths of a trapezoid given the distance between its diagonal intersection and midpoint of a diagonal

The Question

My friend recently gave me a problem that I was interested in, but could not completely solve. I've already constructed a diagram that roughly represents the problem, and it's below.

Consider a trapezoid $ABCD$ with $AB \parallel CD$. Additionally, side $BC = CD = 43$ and $AD \perp BD$. Given that the length between the midpoint of diagonal $BD$ and intersectional of diagonals $AC$ and $BD$ is $11$, find the side length $AD$. Express your answer in simplified radical form.

My Understanding

So my first idea was to draw it out, and I've managed to get a good approximation of it.

So I gave some names to the other points, calling the midpoint of the diagonal $M$ and the intersection $I$. The first thing that I noted was that because $BC = CD$, I found that drawing it down to the midpoint of diagonal $BD$ created two right triangles. Additionally, I noted congruent angles within the triangle $CPO$ and triangle $ADO$, so I deduced they were similar. But I'm not sure how to proceed from here, as I'm not able to establish the side lengths of any more segments. Does anyone know how to proceed?

答:
$CM // AD$. Extend it to meet $AB$ in $E$. Alternate interior angles $\angle MBE=\angle MDC$ which is equal to $\angle MBC$. Thus $\triangle MBC \cong \triangle MBE$. So $M$ bisects $CE$.

$AECD$ is a parallelogram.  So $AD=CE$. We conclude $\triangle ADI \sim \triangle CMI$ with similarity ratio $2$.

Thus $MI=11\Rightarrow ID=22 \Rightarrow MD=33$.

From Pythagoras, you can find $CM$ and $AD=2CM$.


问:
Base and dimension in a subspace
I need help interpreting the answer to a question about the base and dimension of a subspace within linear algebra. I have a subspace W of $R^5$ that is spanned by the vectors:

$${v_1}=\begin{pmatrix} 1 \\ 2 \\ 3 \\ -1 \\ 1 \end{pmatrix} ,  {v_2}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ -1 \\ 2 \end{pmatrix} ,  {v_3}=\begin{pmatrix} -1 \\ -1 \\ -2 \\ 1 \\ 2 \end{pmatrix}
    \text{and} \,{v_4}= \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ -4 \end{pmatrix}$$

and I will determine the dimension of W and also specify the base of W.

To do this I put the vectors as columns in a matrix and simplified this using gauss:

$$\begin{pmatrix}1 & 0 & -1 & 1 \\ 2 & 1 & -1 & 0 \\ 3 & 1 & -2 & 1 \\ -1 & -1 & 1 & 0 \\ 1 & 2 & 2 & -4\end{pmatrix}\sim\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}.\\$$

Here I concluded that the dimension max can be equal to three as we have three rows of pivot elements, then I also concluded that you can express v4 = -v2-v3 (I have also tested this and it is true) and the two vectors v2 and v3 must then be linearly independent. But what I have a hard time interpreting is if this means that v2 and v3 are the basis of W (and not v1,v2 and v3 as my first thought was)? And if the dimension in that case becomes two and not three (as the maximum can be)?

Thanks in advance!

答: The dimension is exactly three since you have three linearly independent vectors $v_1, v_2, v_3$ in the generating system of $W$ and $v_4 = -v_2-v_3$.

Title: Base and dimension in a subspace


问:
How to evaluate $ \sum_{n=r}^{\infty} n^2 \binom{n-1}{r-1} p^r (1-p)^{n-r}$?
I want to evaluate the following sum

$$ \sum_{n=r}^{\infty} n^2  \binom{n-1}{r-1} p^r (1-p)^{n-r} $$

where $r \in \mathbb{N}$ and $0<p<1$.

Using WA I know that the sum evaluates to $\frac{r(r-p+1)}{p^2}$ but I don't know how to show this. I tried simplifying the sum a bit using that
$$
n  \binom{n-1}{r-1} = r  \binom{n}{r}
$$
Which gives me the new equivalent sum
$$
\sum_{n=r}^{\infty} n^2  \binom{n-1}{r-1} p^r (1-p)^{n-r} = r\sum_{n=r}^{\infty} n  \binom{n}{r} p^r (1-p)^{n-r}
$$
but I didn't know how to continue. I have also previously shown that
$$
\sum_{k=s}^{\infty} \binom{k-1}{s-1} (1-t)^{k-s} t^{s} = 1
$$
for $\lvert 1 - t \rvert <1$. I tried to apply this but I couldn't find a way to get rid of the factor of $n$ multiplying the binomial coefficients. Can anyone give me any hints as to how I can show this?

答:
$n\binom{n}{r}=(n+1)\binom{n}{r}-\binom{n}{r}=(r+1)\binom{n+1}{r+1}-\binom{n}{r}$

I assume you can finish up.


问:
L1 constraint on first-differences of sequential binary optimization
I'm working on a sequential decision making problem and am looking at a constraint that I'm not sure how to deal with. We want to maximize the sum of a finite known sequence of bounded functions $F_{t}: U \rightarrow \mathbb{R}$. The decision variables $u_{t}$ are one-hot encoding vectors, $U = \left\{ u \in \{0, 1\}^{n} | \sum_{i} u_{i} = 1 \right\}$. A solution $u^{*}$ is a sequence of these one-hot vectors, but I am constrained in the number of times the one-hot vector changes in that sequence, i.e. $u_{t+1} \neq u_{t}$. Let $b$ be the constraint value. The optimization program looks like this:

$$
\begin{align}
\text{maximize} & \,\,\, \sum_{t = 1}^{T} F_{t}(u_{t})\\
\text{subject to} & \,\,\, \sum_{t = 1}^{T-1} \frac{1}{2} || u_{t+1} - u_{t}||_{1} \leq b
\end{align}$$

where $||\cdot||_{1}$ is the L1 norm. For some $t$ the constraint is $0$ if $u_{t+1} = u_{t}$ and $1$ otherwise. Most examples from dynamic programming I'm familiar with view budget constraints on the values of u_{t} and not the first differences in sequence. I'm uncertain as to how to unroll this problem starting at T-1 and working backwards in a dynamic programming fashion.

答:
You can use Dynamic Programming, where the state is the current time, the budget used so far ($x$), and the last $u_t$. The recurrence relation is:

$$f(t, x, u_t) = F_t(u_t) + \max\{ f(t-1, x, u_t), f(t-1, x-1, 1-u_t) \}.$$


问:
Understanding viewing $M$ as both an $A$ and $B$ module
For the purpose of this discussion, all rings are commutative with a unit.

Any time we have a ring morphism $f: A\rightarrow B$, we can view $B$ as an $A$-module. Moreover, if $N$ is a $B$-module, we can give it the structure of an $A$-module by $a\cdot n:=f(a)\cdot n$. It seems that people often say in this case that we can 'view $N$ as an $A$ module'. If I understand correctly we say this because $A$-module and $B$-module structures are compatible in the sense that $a\cdot_A(b\cdot_B n)=(a\cdot_A b)\cdot_B n$ (where '$\cdot_A$' represents the multiplication from the $A$-module structure).

Its obviously not true that $M$ viewed as a $B$ module is the same as $M$ as a $M$ as an $A$-module, take $f:A\rightarrow B$ to be the zero map for example. But there are some nice things that are true:

For example exercise 2.15 in Atiyah-Macdonald gives us an associativity over tensor products

If $P$ is an ($A$-$B$)-module  and $M$ (resp. $N$) is an $A$ (resp. $B$)-module then $(M\otimes_A P) \otimes_B N \cong M\otimes_A (P \otimes_B N)$

I find this very confusing. My questions are:

Is there a notion for $M$ and an $A$-module being isomorphic to $M$ as $B$ module?

If $f:A\rightarrow B $ is a ring morphism, does this induce a functor from $B$-modules to $A$-modules.

It seems there is a lot more we could say about such things. For example, if $M$ and $N$ are $(A,B)$-modules via $f:A\rightarrow B$, is $M\otimes_A N$ the restriction of $M\otimes_B N$?

Is there a good reference where I could read more about these modules, or some good exercises to get used to this notion?

答:
I think that you should think of it as follows: Given $f:A\to B$ and a $B$-module $(M,+,\cdot_B)$, there is an induced $A$-module structure on (the underlying abelian group of) $M$, i.e. there is an $A$-module structure $(M,+,\cdot_A)$ in the way you are familiar with. I can't speak for other people but I typically think of the $B$-module structure on $M$ and the induced $A$-module structure as distinct things, i.e. I don't bother to think of it as a bi-module structure, etc., and you don't want to ask questions like whether these are isomorphic (because these are modules over different rings).

Let's go through some of your questions:

Yes, this induces a functor from $B$-modules to $A$-modules. Given a $B$-module homomorphism $\varphi:M\to M'$, it is straightforward to check that the same map $\varphi:M\to M'$ is an $A$-module homomorphism when $M$ and $M'$ are equipped with their induced $A$-module structures, and since we are using the same set-theoretic map it is trivial that this preserves the identity, composition, whatever else you need for a functor.

No, you should not expect $M\otimes_A N$ to be the restriction of $M\otimes_B N$. If you take $\mathbb R\to\mathbb C$ and $M=N=\mathbb C$ you get a counterexample, proved here.

I don't have a good idea of where you can find exercises dealing with this specific construction; I say just keep reading Atiyah-Macdonald and working with it when it comes up and soon you will find it a comfortable thing to work with.


问:
Does a $\mathrm{C}^*$-algebra generated by projections contain support projections
I think we have for a (normal?) state $\varphi$ on a von Neumann algebra $\mathcal{M}$ a projection $p_\varphi$ with some nice properties called its support. It arises as follows:

Define the following null space:

$$N_\varphi=\left\{g\in \mathcal{M}\,|\,\varphi(|g|^2)=0\right\}.$$

This set is a $\sigma$-weakly closed left ideal. Therefore there exists a projection $q_\varphi$ such that $N_\varphi=\mathcal{M}q_\varphi$. Some properties include the fact that $g\in N_\varphi$ if and only if $gq_\varphi=g$. Also, for all $f\in \mathcal{M}$ we have
$$\varphi(q_\varphi)=\varphi(fq_\varphi)=\nu(q_\varphi f)=0.$$

Also if we define the projection $p_\varphi:=1_{\mathcal{M}}-q_\varphi$. We have
$$\varphi(f)=\varphi(fp_\varphi)=\varphi(p_\varphi f)=\varphi(p_\varphi fp_\varphi),$$
and $\varphi(p_\varphi)=1$.

I understand that a $\mathrm{C}^*$-algebra generated by projections is not necessarily a von Neumann algebra... but

Question: Does a $\mathrm{C}^*$-algebra generated by projections have support projections?

答:
Let $X$ be a compact space and $\phi$ be a state on $C(X)$ induced by a probability measure  $\mu$ on $X$, whose support is the closed subset $S\subseteq X$.

Then $N_\phi$ is the ideal $C_0(X\setminus S)$, which is unital iff $S$ is clopen.

Thus $\phi$ has a support projection iff the support of the associated measure is clopen.

Now, if $X$ is totally disconnected but not discrete, such as the Cantor set, then $C(X)$ is generated (as a C*-algebra) by projections, but since $X$ contains some closed sets which are not clopen, any measure with support in such a set will fail to have a support projection in $C(X)$.


问:
symmetric decreasing rearrangement
Let $f:\mathbb{R}\rightarrow\left[0,\infty\right)$ be a continuous function. We denote by $f^{*}$ the symmetric decreasing rearrangement for $f$. Is it possible to guarantee that $f(0)\leq f^{*}(0)$?

答:
Sure, this follows from the fact that $f^\ast(0)$ is the (essential) supremum of $f$.  Indeed, we have that
$$
f(0) \le \Vert f \Vert_{L^\infty} = f^\ast(0).
$$

Title: symmetric decreasing rearrangement


问:
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$
Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$

saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.

\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}

so $x > 100$

答:
Working a little differently, you get an exact result:
$$
\begin{align}
x &= \left(\frac 21 \times \frac 22\right) \times \left(\frac 43 \times \frac 44\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{10000}\right) \\
&= \frac{1}{10000\, !} [2\times 4 \times \cdots \times 10000]^2\\
&= \frac{1}{10000\, !} [2^{5000}\times 1\times 2 \times \cdots \times 5000]^2\\
&= \frac{1}{10000\, !} 4^{5000}[5000 \, !]^2\\
&= \frac{4^{5000}}{\binom{10000}{5000}}
\end{align}
$$
Now proceed as in @Thomas Andrews's solution above.


问:
Specific counterexample involving convergence of integral
I have proven that if $X$ is a measurable space and $f, f_n: X \to [0,\infty]$ are measurable functions with $f_n \to f$ a.e. and $\lim \int f_n = \int f < \infty$, then $\int_{E}f_n \to \int_{E}f$ for all measurable $E \subseteq X$.

Is there a counterexample that breaks this statement if $\lim \int f_n = \int f = \infty$? I'm struggling to see any.

答:
$X = \mathbb R$ with Lebesgue measure; $f_n(x) = f(x) = 1$ for $x < 0$;
for $x > 0$, $f(x) = 0$ while $f_n(x) = 1$ for $n < x < n+1$, $f_n(x) = 0$ otherwise.
$E = [0,\infty)$.


问:
Evaluate $\lim_{x\to 3} \frac{h(x)}{e^x}$
Say we know that the following holds:
$$h(x) \leq 7e^x \leq \frac{3h(x)}{x}$$
Evaluate the following limit:$$\lim_{x\to 3} \frac{h(x)}{e^x}$$

I'm aware of the squeeze theorem, but I'm not sure how to "squeeze" the $\frac{h(x)}{e^x}$ algebraically. I thought about inverting the whole thing (not sure if that's even allowed) and ended up with $21 \leq \frac{h(x)}{e^x} <= 21$ but that doesn't seem to be accurate. Would appreciate some help here, thanks in advance!

答:
hint

We know that $$(\forall x>0) \;\;e^x>0$$

So

$$h(x)\le 7e^x\le\frac{3h(x)}{x}\implies$$

$$\frac{7x}{3}\le \frac{h(x)}{e^x}\le 7$$


问:
A 2nd countable metric space is separable

A 2nd countable metric space is separable

Can anyone give an idea on how to prove that, I could prove the converse quite easily but I am not sure which set to consider in order to show that it is dense.

答: The metric assumption is really useless, it works in every second countable topological space: pick an element from each open basic set (using AC) and collect them in a set. This countable set is dense in the original topological space.</p>


答:
Hint.

You have two inequalities,
$$
h(x)\le 7e^x,\quad 7e^x\le\frac{3h(x)}{x}
$$
each of which gives you an inequality for $\frac{h(x)}{e^x}$.

Title: Evaluate $\lim_{x\to 3} \frac{h(x)}{e^x}$

Body:
Say we know that the following holds:
$$h(x) \leq 7e^x \leq \frac{3h(x)}{x}$$
Evaluate the following limit:$$\lim_{x\to 3} \frac{h(x)}{e^x}$$

I'm aware of the squeeze theorem, but I'm not sure how to "squeeze" the $\frac{h(x)}{e^x}$ algebraically. I thought about inverting the whole thing (not sure if that's even allowed) and ended up with $21 \leq \frac{h(x)}{e^x} <= 21$ but that doesn't seem to be accurate. Would appreciate some help here, thanks in advance!


问:
Combinatorics - Painting the Unit Squares of a 3x3 Square
I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine).

I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here?

答:
There are $3$ colours for the central square. For each such case we have:-

The four  squares adjacent to the central square either consist of

A. $3$ squares of the same colour - $8$ possibilities.

B. $2$ squares of each colour opposite each other- $2$ possibilities.

C. $2$ squares of each colour not opposite each other- $4$ possibilities.

The corner squares can then be chosen in the following number of ways

A. $1$

B. $0$

C. $1$

Total number of arrangements is $3\times (8+4)=36$

N.B. This number will be smaller if we count arrangements which are the same under rotation as actually being the same.