问:
How to find a $k$ and $B$ such that $\forall n\in\mathbb N, n\ge 1,|u_n-\sqrt{a}|\le Bk^{2^n}$?
Let $a\in\mathbb R_+^*$ and $(u_n)_{n\in\mathbb N}$ a sequence such that : $u_0=1$ and
$\forall n \in \mathbb N, u_{n+1}=\dfrac{1}{2}\left( u_n+\dfrac{a}{u_n}\right)$

Here the goal is to find $k$ and $B$ such that :
$$\forall n\in\mathbb N, n\ge 1,|u_n-\sqrt{a}|\le Bk^{2^n} \text{ with }k\in[0,1[\text{ and }B\in\mathbb R_+^*$$

After some work, i found out that : $$\forall n \ge 1, u_{n}\ge\sqrt{a} \text{ and } (u_n)_{n\in\mathbb N^*}\text{ is decreasing}$$

So : $$B\ge \dfrac{u_1-\sqrt{a}}{k^2} = \dfrac{\dfrac{1+a}{2}-\sqrt{a}}{k^2}=\dfrac{(\sqrt{a}-1)^2}{2k^2} = \left(\dfrac{\sqrt{a}-1}{\sqrt{2}k}\right)^2$$

After that i don't know what i can do to just find something.

Yet even if i find something, i don't know how i would prove the statement.

Thanks for your future help !

EDIT : I notice that : $$\forall n\in\mathbb N, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=\left(\dfrac{u_{n}-\sqrt{a}}{u_{n}+\sqrt{a}}\right)^2$$

$$\lim_{n\to \infty}u_n=\sqrt{a} \text{ so } \forall \varepsilon\in\mathbb R^+, \exists n_0\in \mathbb N,\forall n\ge n_0,|u_n-\sqrt{a}|\le\varepsilon$$
$$\text{So }, \forall k\in[0,1[, \forall l\in\mathbb N^*,\exists n_0\in \mathbb N,\forall n\ge n_0,|u_n-\sqrt{a}|\le k^{2^l}$$

I don't know but maybe it could be useful ?

答:
Thanks to Neat Math, i found a $k$ and a $B$. Because $\forall n\in\mathbb N, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=\left(\dfrac{u_{n}-\sqrt{a}}{u_{n}+\sqrt{a}}\right)^2$, we can say :$\forall n\in\mathbb N, \dfrac{u_{n}-\sqrt{a}}{u_{n}+\sqrt{a}}=\left(\dfrac{u_{0}-\sqrt{a}}{u_{0}+\sqrt{a}}\right)^{2^n} \text{ or } \forall n\in\mathbb N^*, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=\left(\dfrac{u_{1}-\sqrt{a}}{u_{1}+\sqrt{a}}\right)^{2^{n}}$

But $\dfrac{u_{1}-\sqrt{a}}{u_{1}+\sqrt{a}}=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)^2$ so $1\gt\dfrac{u_{1}-\sqrt{a}}{u_{1}+\sqrt{a}}\ge0$

Therefore, we can let $k = \left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)^2$ and we would have :
$\forall n\in\mathbb N^*, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=k^{2^{n}}$.

Then, ${u_{n+1}-\sqrt{a}}=k^{2^{n}}(u_{n+1}+\sqrt{a})$ and because $(u_n)_{n\in \mathbb N^*}\text{ is decreasing}$, we can deduce that :
$${u_{n+1}-\sqrt{a}}=k^{2^{n}}(u_{n+1}+\sqrt{a})\le k^{2^{n}}(u_{1}+\sqrt{a}) \Longleftrightarrow k^{2^{n}}(u_{1}+\sqrt{a})\ge \dfrac{1}{2}\left( u_n+\dfrac{a}{u_n}\right)-\sqrt{a}$$
$$\Longleftrightarrow 2k^{2^{n}}(u_{1}+\sqrt{a})\ge u_n+\dfrac{a}{u_n}-\sqrt{a}\ge u_n-\sqrt{a}=|u_n-\sqrt{a}| \text{ because } \forall n \in \mathbb N^*, u_n\ge\sqrt{a}$$
$2(u_1+\sqrt{a})=(\sqrt{a}+1)^2\gt0$ so let $B = (\sqrt{a}+1)^2$

In final, we have : $\forall n \in \mathbb N^*, Bk^{2^n}\ge |u_n-\sqrt{a}|$

And that's what we wanted !


问:
Prove using the definition of limit $\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 14}{7 - xy}=2$
I've been trying for a while to prove $$\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 14}{7 - xy}=2$$, but I don't even know how to continue after $|\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 2xy}{7 - xy}|$. Any hints or solutions?

答:
The function is continuous at the point (0,0) so you can literally just plug in the values (0,0)

$ \begin{array}{l}
f( x,y) =\frac{4x^{3} y-x^{4} y^{5} +14}{7-xy}\\
\lim _{(x,y)\rightarrow ( 0,0)}\left[\frac{4x^{3} y-x^{4} y^{5} +14}{7-xy}\right] =f( 0,0) =\frac{4( 0)^{3} \cdot 0-( 0)^{4}( 0)^{5} +14}{7-( 0)( 0)} =\frac{14}{7} =2
\end{array}$

I'm not familiar with this kind of a limit, so maybe I'm wrong, but I believe regular rules apply to this limit.

Someone tell me if my method is incorrect though.

ContentLicense: CC BY-SA 4.0


问:
How to prove that the points $E, F, G$ are collinear

Given $\triangle ABC$ and $\odot(ABC) $, $AG$ and $AD$ are the angle bisectors of $\angle A$ .  $CD$ and $AB$ intersects at $E$. $DB$ and $AC$ intersects at $F$. Prove that $E, F, G$ are collinear.

I found that $DC=DB$. By Menelaus theorem I need to prove $$\frac{BG} {CG} \cdot \frac{CF} {AF} \cdot \frac{EA} {EB} =1$$ and do I need to use the angle bisector theorem also?

答:
Although you already have a solution using projective geometry, we would like to give you a proof that can be worked out using a train of thoughts similar to yours, i.e. applying Menelaus theorem. However, we are going to use Ceva’s theorem in our proof.

Consider $\triangle ABC$. For brevity, let $AB=c$ and $CA=b$. It is given that the segment $AG$ is the angle bisector of $\measuredangle CAB$. We shall write,
$$\frac{BC}{GC}=\frac{AB}{AC}=\frac{c}{b}. \tag{1}$$

In a similar vein, consider $\triangle AFB$, wheret the segment $AD$ is the angle bisector of $\measuredangle BAF$. Here too we shall write,
$$\frac{FD}{DB}=\frac{AF}{AB}=\frac{AF}{c}. \tag{2}$$

Let us hypothesize that $F$, $E$, and $G$ are collinear. Now, if the three lines $FG$, $BA$, and $CD$, which are concurrent at $E$, are cevians of $\triangle FBC$, then, according to Ceva’s theorem, the following condition should be satisfied.
$$ \frac{CA}{AF}\frac{FD}{DB}\frac{BG}{GC}=1 \tag{3}$$

To check, whether this actually holds, let us substitute (1) and (2) in (3).
$$ \frac{CA}{AF}\frac{FD}{DB}\frac{BG}{GC}=\frac{b}{AF}\frac{AF}{c}\frac{c}{b}=1$$
This proves that our hypothesis is true.


问:
Characterizing the solutions of the functional equation $ f ( 3 x ) - f ( 2 x ) = f ( 2 x ) - f ( x ) $

Problem: Find all continuous and strictly increasing functions $ f : ( 0 , + \infty ) \to \mathbb R $ with
$$ f ( 3 x ) - f ( 2 x ) = f ( 2 x ) - f ( x ) $$
for all $ x > 0 $.

A class of solutions is given by $ f ( x ) = a x + b $, where $ a > 0 $ and $ b $ is a constant.

Do we know whether or not these are the only solutions?

答:
In this answer, we prove that linear polynomials are indeed the only possible solutions:

Claim. Let $f : (0, \infty) \to \mathbb{R}$ be a continuous, non-decreasing function such that
$$ f(x) = \frac{f(\frac{1}{2}x) + f(\frac{3}{2}x)}{2} \tag{FE} $$
holds for all $x > 0$. Then $f(x) = mx + c$ for some $m \geq 0$ and $c \in \mathbb{R}$.

Proof of the claim modulo key lemmas. We will use the following lemmas.

Lemma 1. Let $f : (0, \infty) \to \mathbb{R}$ be as in the above claim. Then for any $0 < a < 1 < b$ and $x > 0$, we have
$$ f(x) \geq f(\tfrac{1}{2}ax) \biggl( \frac{2b-2}{2b-a} \biggr) + f(bx) \biggl( \frac{2-2a}{3b-2a} \biggr) \tag{1} $$
and
$$ f(x) \leq f(ax) \biggl( \frac{3b-2}{3b-2a} \biggr) + f(\tfrac{3}{2}bx) \biggl( \frac{2-a}{2b-a} \biggr). \tag{2} $$

Intuitively, this lemma follows by repeatedly applying the functional equation $\text{(FE)}$ only to the terms $f(\square)$ satisfying $ax < \square < bx$. This argument can be conveniently implemented by using martingale. Since the proof is a bit lengthy, we will postpone it until the end.

Lemma 2. Let $g : \mathbb{R} \to \mathbb{R}$ be a bounded, continuous function such that
$$ g(u) = \frac{1}{4}g(u-\log 2) + \frac{3}{4}g(u+\log(3/2)). $$
Then $g$ is constant.

Now let us check that this lemma indeed proves the desired conclusion. First, fix $b$ and set $x = 1$. Then $\text{(2)}$ shows that

$$ \biggl( \frac{3b-2a}{3b-2} \biggr) \biggl[ f(1) - f(\tfrac{3}{2}b) \biggl( \frac{2-a}{2b-a} \biggr) \biggr] \leq f(a), $$

hence $f(a)$ is bounded from below as $a \to 0^+$. Together with the montonicity of $f$, it follows that

$$c = \lim_{a \to 0^+} f(a)$$

exists. Now define $\tilde{f}(x) = f(x) - c$. Then $\tilde{f}$ also satisfies the condition of OP's problem, and in addition we have $\tilde{f}(0^+) = 0$. Then letting $ a \to 0^+$ to both $\text{(1)}$ and $\text{(2)}$ yields

$$\frac{2}{3b} \tilde{f}(bx) \leq f(x) \leq \frac{1}{b} \tilde{f}(\tfrac{3}{2}bx).$$

In particular, plugging $x=1$ shows that $ \tilde{f}(b)/b \leq 3\tilde{f}(1)/2 $, hence $\ell = \limsup_{b\to\infty} \tilde{f}(b)/b$ is finite. Then taking $\limsup$ as $b\to\infty$ to the above inequality, we obtain

$$ \frac{2\ell}{3} \leq \frac{\tilde{f}(x)}{x} \leq \frac{3\ell}{2}. $$

Now define $g(u) = e^{-u}\tilde{f}(e^u)$. The above inequality tells that $g$ is bounded. Also, $\text{(FE)}$ for $\tilde{f}$ tells that $g$ satisfies the condition of Lemma 2. Therefore $g$ is constant, proving the main claim.

Proof of Lemma 1. Let $X_1, X_2, \dots$ be i.i.d. random variables such that

$$ \mathbb{P}(X_i = \tfrac{3}{2}) = \mathbb{P}(X_i = \tfrac{1}{2}) = \frac{1}{2}. $$

Then the discrete-time process $M = (M_n)_{n\geq 0}$ defined by

$$ M_0 = 1, \qquad M_{n+1} = X_{n+1} M_{n} $$

is a positive martingale with respect to the natural filtration $\mathcal{F}_n = \sigma(X_1,\dots,X_n)$. Now this martingale is related to the original problem in such way that, if $f : (0, \infty) \to \mathbb{R}$ is as in OP's problem, then for each $x > 0$, the functional equation $\text{(*)}$ shows that

$$ \mathbb{E}[ f(M_{n+1} x) \mid \mathcal{F}_n]
= \mathbb{E}[ f(X_{n+1}M_n x) \mid \mathcal{F}_n]
= \frac{f(\frac{3}{2}M_n x) + f(\frac{1}{2}M_n x)}{2} = f(M_n x), $$

and so, $(f(M_n x))_{n\geq 0}$ is also a martingale. Now fix $0 < a < 1 < b$ and define

\begin{gather*}
S = \inf\{ n \geq 0 : M_n \leq a \}, \\
T = \inf\{ n \geq 0 : M_n \geq b\}.
\end{gather*}

Then by the SLLN or by other means, we can easily check that $\mathbb{P}(S < \infty) = 1$. Moreover, $\frac{a}{2} \leq M_{n\wedge S\wedge T}$ for all $n$, and so, both $(M_{n\wedge S\wedge T})_{n\geq 0}$ and $(f(M_{n\wedge S\wedge T} x))_{n\geq0}$ are bounded martingales. So by the optional stopping theorem,

$$ 1 = \mathbb{E}[M_0] = \mathbb{E}[M_{S\wedge T}] = \mathbb{E}[M_{S}\mathbf{1}_{\{S \leq T\}}] + \mathbb{E}[M_{T}\mathbf{1}_{\{S \geq T\}}] $$

as well as

$$ f(x) = \mathbb{E}[f(M_0 x)] = \mathbb{E}[f(M_{S\wedge T} x)] = \mathbb{E}[f(M_{S} x)\mathbf{1}_{\{S \leq T\}}] + \mathbb{E}[f(M_{T} x)\mathbf{1}_{\{S \geq T\}}]. $$

Now if we write $p_a = \mathbb{P}(S \leq T)$ and $p_b = \mathbb{P}(S \geq T)$, then $\frac{a}{2} \leq M_S \leq a$ and $b \leq M_T \leq \frac{3}{2}M_T$, together with the monotonicity of $f$, shows that

$$ \tfrac{1}{2}ap_a + bp_b \leq 1 \leq a p_a + \tfrac{3}{2}b p_b $$

and

$$ f(\tfrac{1}{2}a x)p_a + f(b x)p_b \leq f(x) \leq f(a x) p_a + f(\tfrac{3}{2}b x) p_b . $$

The first inequality, together with $p_a + p_b = 1$, implies that

$$ \frac{2b-2}{2b-a} \leq p_a \leq \frac{3b-2}{3b-2a},
\qquad\qquad \frac{2-2a}{3b-2a} \leq p_b \leq \frac{2-a}{2b-a}. $$

Then applying this to the second inequality proves the desired lemma.

Proof of Lemma 2. Let us consider the random walk $S_n = Z_1 + Z_2 + \dots + Z_n$ on the lattice $\mathbb{L} = \{a + ib \in \mathbb{C} : a, b \in \{0,1,2,\dots\} \}$, where $Z_k$'s are i.i.d. and

$$ \mathbb{P}(Z_k = 1) = \frac{1}{4} \qquad\text{and}\qquad \mathbb{P}(Z_k = i) = \frac{3}{4}. $$

Let $g$ be as in the assumption, and define $h : \mathbb{L} \to \mathbb{R}$ by

$$ h(a + ib) = g(- a \log 2 + b \log(3/2)). $$

Then $h$ is bounded and satisfies

$$ h(z) = \frac{1}{4}h(z+1) + \frac{3}{4}h(z+i) = \mathbb{E}[h(z+Z_1)] $$

for all $z \in \mathbb{L}$. In particular, this tells that $h(S_n)$ is a bounded martingale and hence converges a.s. by the martingale convergence theorem. We will show that $h$ is constant.

To see this, fix $z = a + ib \in \mathbb{L}$ and write $m = a + b$. Also, let $(Z'_k)_{k\geq 1}$ be an independent copy of $(Z_k)_{k\geq 1}$ and define $S'_n = z + Z'_1 + \dots + Z'_n$. Then

$$ Y_n := \operatorname{Re}(S_{m + n} - S'_n) $$

is a random walk on $\mathbb{Z}$ which is irreducible and recurrent. In particular,

$$T := \inf \{ n \geq 0 : Y_n = 0 \}$$

is a.s. finite. Now by noting that $S_{m+n} = S'_n$ if andy only if $Y_n = 0$, we find that the process $(\tilde{S}_n)$ defined by

$$ \tilde{S}_n := \begin{cases}
S'_n, & n \leq T \\
S_{m+n}, & n > T
\end{cases} $$

has the same law as $(S'_n)$. So

$$ h(z)
= \mathbb{E}[h(S'_n)]
= \mathbb{E}[h(S_{m+n})\mathbb{1}_{\{n > T\}}] + \mathcal{O}(\mathbb{P}(n \leq T)) $$

as $n \to \infty$. Now by the dominated convergence theorem,

$$ h(z)
= \lim_{n\to\infty} \mathbb{E}[h(S'_n)]
= \mathbb{E}\Bigl[\lim_{n\to\infty} h(S_{m+n})\mathbb{1}_{\{n > T\}}\Bigr]
= \mathbb{E}\Bigl[\lim_{n\to\infty} h(S_{n}) \Bigr]
= h(0). $$

So $h$ is constant as claimed. However, since $\frac{\log(3/2)}{\log 2}$ is irrational, the set

$$\{ - a \log 2 + b \log(3/2) : a, b \in \{0,1,2,\dots\}\}$$

is dense in $\mathbb{R}$. Since $g$ is continuous and is constant on a dense subset of $\mathbb{R}$, it follows that $g$ is also constant, completing the proof.


问:
Figuring out big O'notation of a naive depth-first-search scheduling algorithm
(I'm not sure if this question belongs here, but stack overflow doesn't work with LaTeX so here it is)

An inefficient solution to a scheduling problem similar to the 2018 Google Hashcode problem (unless you already know of this problem I wouldn't recommend reading it for this, it will likely overcomplicated this).

Given $n$ tasks and $m$ agents and that 1 agent can do 1 task at a time, you want to assign agents to tasks such that all tasks are completed in the minimum amount of time.

An extremely expensive algorithm is an application of depth-first-search, constructing a tree of every possible assignment and from that finding the shortest path (minimum time required).

Considering a problem with 2 agents and 3 tasks, for each node in each layer you have so many outgoing edges (considering at layer $i$ there are $n+1-i$ tasks remaining):

$mn = 2 \cdot 3 = 6$

$m(n-1)=2 \cdot 2 = 4$

$m(n-2)=2 \cdot 1 = 2$

As such the number of outgoing edges you have in each layer are:

$mn=\frac{n!}{(n-1)!}m^1=2 \cdot 3 = 6$

$mn\cdot m(n-1)=\frac{n!}{(n-2)!}m^2=(2 \cdot 3)(2 \cdot 2) = 24$

$mn\cdot m(n-1)\cdot m(n-2)=\frac{n!}{(n-3)!}m^3=(2 \cdot 3)(2 \cdot 2)(2 \cdot 1) = 48$

Therefore one could give the total number of edges in the network as $t$ where:

$t=\sum\limits_{i=1}^{n} \frac{n!}{(n-i)!}m^i$

At this point I am stuck and do not know how I could simplify it further.

Body:
$$t_{m,n}=\sum\limits_{i=1}^{n} \frac{n!}{(n-i)!}m^i=e^{\frac{1}{m}}\, m^n\, \Gamma \left(n+1,\frac{1}{m}\right)-1$$ where appears the incompleta gamma function.

$$\left(
\begin{array}{ccc}
m & n & t_{m,n} \\
 1 & 1 & 1 \\
 1 & 2 & 4 \\
 1 & 3 & 15 \\
 1 & 4 & 64 \\
 1 & 5 & 325\\
 2 & 1 & 2 \\
 2 & 2 & 12 \\
 2 & 3 & 78 \\
 2 & 4 & 632 \\
 2 & 5 & 6330\\
 3 & 1 & 3 \\
 3 & 2 & 24 \\
 3 & 3 & 225 \\
 3 & 4 & 2712 \\
 3 & 5 & 40695 \\
 4 & 1 & 4 \\
 4 & 2 & 40 \\
 4 & 3 & 492 \\
 4 & 4 & 7888 \\
 4 & 5 & 157780 \\
 5 & 1 & 5 \\
 5 & 2 & 60 \\
 5 & 3 & 915 \\
 5 & 4 & 18320 \\
 5 & 5 & 458025
\end{array}
\right)$$


问:
Prove that if (a,b,c) is a Pythagorean triple, then 3 divides either $a$ or $b$ without remainder classes
This number theory book I have proved that if $(a, b, c)$ are a primitive Pythagorean triple (i.e., $a, b, c$ have no common factors and $a^2 +b^2 = c^2$), then either $a$ or $b$ is odd. The book did that by assuming that both $a$ and $b$ are even (or odd) and derived a contradiction each time. (If $a$ and $b$ are both even, it showed $c$ would be too; if they're both odd, it ends up with an equation where the left-hand side is odd and the other even.)

As an exercise, the book asks to use a similar reasoning to prove that either $a$ or $b$ must be a multiple of 3. The book didn't get to remainder classes or modular algebra yet, so I guess this "similar reasoning" wouldn't involve them. I don't think the book has even mentioned the word "remainder" yet.

Keeping in mind that $a$ is even, $b$ is odd, and $c$ is odd, I tried to either assume that neither $a$ or $b$ are divisible by 3 and tried to derive a contradiction, or that if either wasn't divisible by 3 then the other was. Neither approach led anywhere, except calculations so long and tedious that you end up forgetting why you're even doing them.

The book comes with no answers to problems, and I should note I suck at figuring out divisibility. Any hint would be greatly appreciated.

答:
If neither a nor b
is divisible by 3
(which could be the start of a poem)
then they are of the form
$3n\pm 1$
so their squares are of the form
$3n+1$
so the sum of their squares
is of the form $3n+2$
and no square is of this form.


问:
Lefschetz-Hopf fixed point theorem
According to Wikipedia, the Lefschetz-Hopf fixed point theorem states that if $X$ is a compact triangulable space and $f:X\rightarrow X$ is continuous and has a finite number of fixed points, then
$$ \sum_{x\in X,f(x)=x}i(f,x)=\sum_{k\geqslant 0}(-1)^k{\rm tr}\left(f_*:H_k(X,\mathbb{Q})\rightarrow H_k(X,\mathbb{Q})\right) $$
where $i(f,x)$ denotes the index of the fixed point $x$. I can't find out how $i(f,x)$ is defined. I am also looking for a proof of this theorem. Thanks in advance.

答:
$i(f,x)$ is what is called the index, you may find the definition and several related things about index in the beautiful book on Differential Topology by V. Guillemin and A. Pollack.

Just to point, a variant of the above theorem is also true for compact simplicial complex for which you can find a proof in the Algebraic Topology book of A.Hatcher.

Title: Lefschetz-Hopf fixed point theorem


问:
Expected Value of a Random Variable M
Question: After your complaint about their service, a representative of an insurance company promised to call you "between $7$ and $9$ this evening". Assume that this means that the time T of the call is uniformly distributed in the specified interval.
Assume that you know in advance that the call will last exactly $1$ hour. From $9$ to $9:30$, there is a game show on $TV$ that you wanted to watch. Let $M$ be the amount of time of the show that you miss because of the call. Compute the expected value of $M$.

What I have understood is $P(M | X < 8:00) = 0$, i.e. probability that show will be missed is $0$ when call is received before $8:00$.
If I consider time b/w $8:00$ to $8:30$, then expected value is $(8:30-8:00)/2 = 15.$ Is it the right way to proceed. I don't know what is actual answer.

答:
If the call arrives before 8, M=0

If call arrives in the interval $[8;8:30]$ your M is uniform in $[0;30]$. This happens with probability $p=\frac{1}{4}$

If the call arrives after 8:30 you miss all your TV show. This happens with probability $p=\frac{1}{4}$

Thus

$$E(M)=15\frac{1}{4}+30\frac{1}{4}=11.25$$

You can expect to miss about 11 minutes of your show

Title: Expected Value of a Random Variable M


问:
Calculating the Limit . $\lim_{x \rightarrow 0}\space \left[ \int_0^{1}\ [ by+a(1-y)]^{x}dy\right]^{1/x}$
Find the value of  $$\lim_{x \rightarrow 0}\quad
 \left[  \int_0^{1}\ [ by+a(1-y)]^{x}dy\right]^{1/x}\\ \quad where\space (a,b) \in \begin{align*}\
\mathbb{R}
\end{align*}
$$
My approach : since it is of the form $ ( \rightarrow {1})^{\rightarrow\infty}$
so wrote it in the form of exponential i.e $e^ {(f(x)-1)(g(x))}\quad where,\ f(x)=\left[  \int_0^{1}\ [ by+a(1-y)]^{x}dy\right] \quad and\quad g(x)= \frac{1}{x}$ .

But was unable to solve from from here onwards .
Any help would be appreciated.!

Thanks !

Body:
By L'Hopital's Rule, one has
\begin{eqnarray}
&&\lim_{x\to0^+}\ln\bigg\{\left[  \int_0^{1}\ [ by+a(1-y)]^{x}dy\right]^{1/x}\bigg\}\\
&=&\lim_{x\to0^+}\frac{\ln\bigg\{\int_0^{1}\ [ by+a(1-y)]^{x}dy\bigg\}}{x}\\
&=&\frac{1}{\int_0^{1}\ [ by+a(1-y)]^{x}dy}\int_0^{1}\frac{d}{dx} [ by+a(1-y)]^{x}dy\bigg|_{x=0}\\
&=&\int_0^1(b-a)\ln[(b-a)y+a]dy\\
&=&\int_a^b\ln tdt\\
&=&b\ln b-a\ln a-(b-a).
\end{eqnarray}


问:
Defining a subset
The question I have to answer is as following in Swedish:

```<code>Hur många mängder X uppfyller {a, b, c} ⊆ X ⊆ {a, b, c, d, e}?
</code>```

Loosely translated (I do not know mathematical terminology well in English):

```<code>How many sets X satisfy {a, b, c} ⊆ X ⊆ {a, b, c, d, e}?
</code>```

I do not know how to solve this question. For instance, would {{a,b}, a} be two valid subsets of {a, b, c, d, e}?

Any insight into the main question is greatly appreciated.

答: Since you want $\{a,b,c\}\subseteq X$ all you have to do is to choose the other possible elements to expand the smaller set $\{a,b,c\}$. You can do this choosing none or $d$ or $e$ or both $d,e$.</p>

ContentLicense: CC BY-SA 3.0

Title: Defining a subset


问:
Efficiency of the prime generating constant $2.920050977316 \dots$ for the purpose of compressing a list of primes.
The constant $c \approx 2.920050977316 \dots$ has the property that if $\{a_n\}$ is the sequence defined by $a_1 = c, a_{n+1} = \lfloor a_n \rfloor (a_n - \lfloor a_n \rfloor + 1),$ then $\lfloor a_n \rfloor$ is the $n$th prime.

We need to calculate primes in order to calculate the constant rather than the other way around, so this does not yield a method of generating new primes. However, we might be able to use the constant to compress an already known list of primes. The question then becomes, how many digits of $c$ are needed to generate the first $n$ primes? Note that it takes $\sim n \log n$ digits to store the first $n$ primes naively and $\sim n \log \log n$ digits by storing the differences between consecutive primes.

答:
We can refine the result of the paper as follows, where we will use $\lfloor x\rfloor$ for the floor function and $\{x\}=x-\lfloor x\rfloor$ to be the fractional part function.

Let $s_0,s_1,s_2,\ldots,s_n$ be a finite sequence of integers such that $s_k \leq s_{k+1} < 2s_k$ for every $k$. Let $I$ be the set of $c$ for which the sequence inductively defined by $a_1=c$ and $a_{n+1}=\lfloor a_n\rfloor(1+\{a_n\})$ has the property that $\lfloor a_k\rfloor = s_k$ for each $k$. There is some $z$ such that
$$I=\left[z, z + \frac{1}{\prod_{i=0}^{n-1}s_i}\right).$$

This tells us that we need to choose the starting value in an interval of length $\frac{1}{\prod_{i=0}^{n-1}s_i}$, which generically takes $\log_{10}\left(\prod_{i=0}^{n-1}s_i\right)$ digits of precision. However, just writing out all the terms in the sequence would require writing about the same number of digits since $\log_{10}(\prod s_i) = \sum \log_{10}(s_i)$. Representing sequences this way doesn't have any advantage at least asymptotically - and this is not surprising, since this method is a fairly literal transcription of sequences satisfying the given inequalities - and this inequality alone is not a very helpful way to single out the sequence of primes.

Let's define $f(x)=\lfloor x\rfloor (1+\{x\})$. Note that $f$ is just a linear function of slope $n$ on each interval $[n,n+1)$. Define its iterates as $f^0(x)=x$ and $f^{n+1}(x)=f(f^{n}(x))$. Let's say that a given constant $c$ represents a sequence $s_0,\ldots,s_n$ if $\lfloor f^k(x) \rfloor = s_k$ for each $k$. Let me give two inductive proofs of the claimed theorem using this terminology.

Proof 1: We prove the statement directly by induction on $n$. The statement is trivial if $n=0$. Suppose that we wish to represent $s_0,s_1,s_2,\ldots,s_{n+1}$. By the inductive hypothesis, the set of representations of $s_1,s_2,\ldots,s_{n+1}$ is an interval $I$ of the prescribed length. A representation of $s_0,s_1,s_2,\ldots,s_{n+1}$ is just some $x$ such that $\lfloor x\rfloor = s_0$ and $f(x) \in I$. The second part of this condition simplifies to saying $s_0(x-s_0+1) \in I$, since we know that floor of $x$. Note that, since $I$ is a subset of $[s_1,s_1+1) \subseteq [s_0,2s_0)$, the condition that $s_0(x-s_0+1) \in I$ implies $\lfloor x\rfloor = s_0$. The set of representations of $s_0,s_1,s_2,\ldots, s_{n+1}$ therefore is the preimage of $I$ under a linear function of slope $s_0$, hence has length $\frac{1}{s_0}$ times the length of $I$. This immediately proves the desired statement.

Proof 2: For this proof, we use an inductive argument that builds up the sequence from the end rather than from the start, which enables somewhat clearer computation of the interval itself. We need to prove an additional statement as part of the inductive hypothesis, however:

If $I$ is the interval of representations for $s_0,s_1,s_2,\ldots,s_n$, then $f^n$ is linear when restricted to $I$ and has image $[s_n,s_n+1)$.

Essentially, we need that $f$ behaves well when we know what it represents and that it hits every possible terminating value. Clearly, if $n=0$, the sequence has only one term and is represented exactly on the interval $[s_0,s_0+1)$, on which $f^0$ is the identity function.

Next, let $I$ be the interval of $x$ representing the sequence $s_0,s_1,\ldots,s_{n-1}$. Note that $f^{n-1}$ restricted to $I$ is, by hypothesis, some linear function taking $I$ to $[s_{n-1},s_{n-1}+1)$. However, $f$ restricted to the interval $[s_{n-1},s_{n-1}+1)$ is also linear, satisfying $f(x)=s_{n-1}(x-s_{n-1}+1)$. The composition $f^n$ is therefore also linear on the interval $I$ - and, by calculation, takes $I$ to the interval $[s_{n-1},2s_{n-1})$. Note that, by hypothesis, the interval $[s_n,s_n+1)$ is contained within $[s_{n-1},2s_{n-1})$ and, in fact, is a proportion of exactly $\frac{1}{s_{n-1}}$ of that interval, hence its preimage under $f^n$ within $I$ - which is the set of sequences representing $s_0,s_1,\ldots,s_{n-1},s_n$ is the same proportion of $I$. In particular, since $I$ had length $\frac{1}{\prod_{i=0}^{n-2}s_i}$, it must be that this preimage has length $\frac{1}{\prod_{i=0}^{n-1}s_i}$, as was desired. Moreover, observe that we have already established the refinement we used for the inductive hypothesis.

If we work out the details of the calculation implied here, we end up determining that
$$z=s_0+\sum_{k=0}^{n-1} \frac{s_{k+1}-s_k}{\prod_{i=0}^k s_k}$$
which also extends to finding the (unique) $c$ that generates an infinite sequences satisfying the condition that $s_k \leq s_{k+1}<2s_k$ for every $k$ and the additional condition that $s_{k+1}<2s_k-1$ for infinitely many $k$.


问:
Mathematica vs Wolfram Alpha integration results
When I insert the following integration command in wolframalpha:

```<code>int(pi^2*0.05*Exp[-0.1]*14142135/2*x^-1*
  Exp[-x]*y*Exp[-pi*0.1*y^2/2]*
  Hypergeometric2F1[1, 2/3, 5/3, -2*y^-4*(2*10^14)^(3/4)]*
  BesselI[2, x*y*Sqrt[2*pi]],{x,0,0.25},{y,0,Infinity})
</code>```

I get the following numerical result 6.68805

However, inserting the same command in Mathematica as follows:

```<code>NIntegrate[pi^2*0.05*Exp[-0.1]*14142135/2*x^-1*
  Exp[-x]*y*Exp[-pi*0.1*y^2/2]*
  Hypergeometric2F1[1,2/3,5/3,-2*y^-4*(2*10^14)^(3/4)]*
  BesselI[2,x*y*Sqrt[2*pi]],{x,0,0.25},{y,0,Infinity}]
</code>```

I get the following output (no result)

Am I doing something wrong in Mathematica? Does the wolframalpha site has extended functionality compared to Mathematica (in the above concept).

Thanks in advance

PS: I suppose that wolfralpha translates int() as NIntegrate in the specific case.

答:
I think the apparent answer to your problem is that you entered in Mathematica:

```<code>NIntegrate[pi^2 ... blah ]
</code>```

... and in Mathematica, Pi is the exact real number $\pi$ = 3.1415 ... but pi is just a name like cat or mouse. Capital letters matter. Wolfram Alpha is more flexible about such things, perhaps because it needs to be. This explains why you would get the message about your expression not being numerical, because it contains a pi (or a cat or a dog).

Body:
I think the apparent answer to your problem is that you entered in Mathematica:

NIntegrate[pi^2 ... blah ]

... and in Mathematica, Pi is the exact real number $\pi$ = 3.1415 ... but pi is just a name like cat or mouse. Capital letters matter. Wolfram Alpha is more flexible about such things, perhaps because it needs to be. This explains why you would get the message about your expression not being numerical, because it contains a pi (or a cat or a dog).


问:
Given a $2\times 2$ matrix $A$, does two unique eigenvalues guarantee that $A$ is diagonizable?
This question relates to a seminar I've been working on, so I do not wish to disclose the whole question but simply ask how this would be handeled in theory.

$A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}$
$Q=(\lambda I - A) = \begin{bmatrix}\lambda - a & -b \\ -c & \lambda - d \end{bmatrix}$

We then want to find the eigenvalues of $A$, which responds to solving $\det(Q)=0$. My question is, given that we would get two unique eigenvalues, does that mean we are guaranteed that matrix $A$ is diagonalizable? I have only found a theorem that says if we have two distinct eigenvectors for a $2\times 2$ matrix $A$, then $A$ is diagonalizable...

Help would be much appreciated!

答:
If you have $2$ unique eigenvalues, it means your characteristic polynomial will look like this: $(λ-a)(λ-b)$, where $a$ and $b$ are your eigenvalues.

Now a matrix is diagonizable if for each of its eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.

In our case you have an algebraic multiplicity of $1$ (for each eigenvalue), and thus their geometric multiplicity is also equal to $1$ ($0<\text{geometric multiplicity} \leq \text{algebraic multiplicity}$) thus the matrix is diagonizable.

So for a general $n\times n$ matrix if you have $n$ unique eigenvalues, it's diagonizable.


问:
Composition of orientation preserving and reversing homeomorphisms
Let $S^1$ be the unit circle and $f:S^1 \to S^1$ is a homeomorphism.

We say $f$ is a orientation preserving homeomorphism if any lifting of $f$ to the covering space $\mathbb{R}$ is
strictly increasing and It is called orientation reversing if any lifting of $f$ to the covering space $\mathbb{R}$ is strictly decreasing.

Now I have some question about them :

1- I want to prove that any homeomorphism $f:S^1 \to S^1$ is either orientation preserving or else orientation reversing.

2- The composition of an orientation preserving homeomorphism and an orientation reversing is orientation reversing.

3- If $a,b \in S^1$ then $f$ is orientation preserving if $f(a,b)=(f(a),f(b))$.

And also why the degree of a homeomorphism is either $1$ or $-1$ and does it related to the question 2?
because we know $deg(fog)=deg(f)deg(g)$

答:
Let us first show

Each continuous injection $\phi : J \to \mathbb R$ defined on an interval $J \subset \mathbb R$ is either strictly increasing or strictly decreasing. (Note that intervals may be open, half-open, closed / bounded, unbounded.)

Proof. Let $H = \{(x,y) \in J \times J \mid x < y \}$. It is easy to verify that this is a convex subset of $\mathbb R^2$, thus it is path connected and a fortiori connected.

Let $A = \{(x,y) \in H \mid \phi(x) < \phi(y) \}$ and  $B= \{(x,y) \in H \mid \phi(x) > \phi(y) \}$. Clearly $A \cap B = \emptyset$ and $A \cup B  = H$ (note $\phi(x) = \phi(y)$ implies $x = y$ because $\phi$ is injective). Since $\phi$ is continuous, both $A$ and $B$ are open in $J \times J$, thus also open in $H$. Since $H$ is connected, one of $A$ or $B$ must be $= H$ and the other $= \emptyset$. This means
that $\phi$ is either strictly increasing or strictly decreasing.

Next let us show

If $H : \mathbb R \to  \mathbb R$ is a lift of a homeomorphism $h : S^1 \to S^1$, then $H$ is a homeomorphism.

Proof. Let $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$, be the standard covering map. As a lift of $f : S^1 \to S^1$ we denote any map $F : \mathbb R \to \mathbb R$ such that $p \circ F = f \circ p$.

You certainly know that if $F,F'$ are lifts of $f$, then
$$(*) \phantom x F'(t) = F(t) + k \text{ for all } t \text{ with a } \textbf{fixed } k \in \mathbb Z. $$
In fact, we have $e^{2\pi i(F(t) -F'(t))} = e^{2\pi iF(t)}/e^{2\pi i F'(t)} = (p \circ F)(t)/ (p \circ F')(t) = (f \circ p)(t)/ (f \circ p)(t) = 1$, thus $(F - F')(t) = F(t) -F'(t) \in \mathbb Z$ and by continuity of $F - F'$ we see that $(F - F')(t) = k$ for some fixed $k \in \mathbb Z$. This means that $F' = \tau_k \circ F$ with the translation homeomorphism $\tau_k : \mathbb R \to \mathbb R, \tau_k(t) = t + k$.

If $F, G$ are lifts of $f, g$, then $p \circ G \circ F = g \circ p \circ  F = g \circ f \circ  p$, thus $G\circ F$ is a lift of $g \circ f$.

Let $h$ be a homeomorphism with inverse homeomorphism $h^{-1}$ and let $H, \bar H$ be lifts of $h, h^{-1}$. Then $\bar H \circ H$ is a lift of $h^{-1} \circ h = id$. Since also $id : \mathbb R \to \mathbb R$ is a lift of $id : S^1 \to S^1$, we get $(\tau_k \circ \bar H ) \circ H = \tau_k \circ (\bar H  \circ H) = id$ for some $k \in \mathbb Z$. Similarly we get $H \circ \bar H = \tau_r \circ id = \tau_r$ for some $r \in \mathbb Z$. The latter implies $H \circ (\bar H \circ \tau_r^{-1}) = id$. Thus $H$ has a left inverse $H' = \tau_k \circ H$ and a right inverse $H'' = \bar H \circ \tau_r^{-1}$. But now $H'' = id \circ H'' = H' \circ H \circ H'' = H' \circ id = H'$, thus $H$ is a homeomorphism with inverse $H^{-1} = H' = H''$.

Your question 1 is answered by the above two theorems.

By the degree formula $\deg(f \circ g) = \deg(g)\deg(f)$ we see that any homeomorphism $h$ has degree $\pm 1$ (since $\deg(id) = 1)$). In fact, $\pm 1$ are the only elements of $\mathbb Z$ which have a multiplicative inverse. Let $H$ be a lift of $h$. It is a homeomorphism, thus $H$ is either strictly increasing or strictly decreasing. In the first case it must have positive degree, in the second case negative degree. Thus

A homeomorphism is orientation preserving iff it has degree 1; it is orientation reversing iff it has degree -1.

Thus the degree formula also answers your question 2.

Concerning 3. : It is not really precise how you define "open intervals" $(a,b) \subset S^1$. It seems that if $a, b \in S^1$ are two distinct points, then you move counterclockwise from $a$ to $b$ and all points strictly between $a$ and $b$ constitute $(a,b)$. I think my answer to Open sets on the unit circle $S^1$ explains it more precisely. The open intervals $(a,b) \subset S^1$ are precisely the images $p((s,t))$ of open interval $(s,t) \subset \mathbb R$ such that $0 < t - s < 1$, where we have $a = p(s)$ and $b = p(t)$. Let us show that an orientation preserving homeomorphism $h$ maps $(a,b)$ onto $(h(a),h(b))$.

Clearly $h(a) \ne h(b)$. Let $H$ be a lift of $h$. Then $H(s) < H(t)$ and $H$ maps $(s,t)$ homeomorphically onto $(H(s),H(t))$. We have $p(H(s)) = h(a), p(H(t)) = h(b)$ and $0 < H(t) - H(s) < 1$. Concerning the last inequality: If $H(t) - H(s) = 1$, then $h(b) = h(a)$ which is impossible. If $H(t) - H(s) > 1$, then $p \mid_{(H(s),H(t))}$ is not injective, thus $p \circ H \mid_{(s,t)}$ is not injective which is a contradiction since $p \circ H \mid_{(s,t)} = h \circ p \mid_{(s,t)} = h \mid_{(a,b)}$.


答:
$$\lim_{(x, y)\to (0, 0)}|\frac{4x^3y - x^4y^5 + 2xy}{7 - xy}|$$

$$|\frac{4x^3y - x^4y^5 + 2xy}{7 - xy}|=|xy||\frac{4x^2 - x^3y^4 + 2}{7 - xy}| < \epsilon$$

So $\sqrt{x^2+y^2} < \delta$, so $|x|<\delta$ and $|y|<\delta$

Choose a $\delta \le 1$. So $|x|,|y|<1$. So what's an upper bound M for $|4x^2-x^3y^4+2|$ and what's an upper bound N for $|\frac{1}{7-xy}|$?

You want $|xy|MN < \epsilon$. so $|xy|< \frac{\epsilon}{MN}$

Now you can do your epsilon delta proof using $\delta = \min\{1,\sqrt{\frac{\epsilon}{MN}}\}$

Title: Prove using the definition of limit $\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 14}{7 - xy}=2$

Body: I've been trying for a while to prove $$\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 14}{7 - xy}=2$$, but I don't even know how to continue after $|\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 2xy}{7 - xy}|$. Any hints or solutions?


问:
Construct $f$ such that $\partial X=f^{-1}(0)$

Given a (smooth) manifold $X$ with boundary, show that there is a smooth map $f:X \to \mathbb R$ with $0$ as a regular value and $\partial X=f^{-1}(0)$.

I haven't been able to make any progress. Hints are welcome.

答:
Well, smooth manifolds with boundary admit partitions of unity, so what if you assigned a nonnegative real valued function to every element of a partition of unity and then added them up?

Then you essentially only have to define such a function $f$ on whatever space your manifolds with boundary are modeled on (by this I mean what the codomain of a chart is defined to be). And I think you should find that considerably less difficult. :)

Elaboration on the hint: First suppose that we can solve your problem when $X = \mathbb{H}^k$ (i.e. by finding such a nonnegative $f:X \to \mathbb{R}$)---then I claim we are done. Why? Because if $\{ g_i \}$ is a partition of unity of $X$ (necessarily with each $g_i$ nonzero on the interior of some chart $\psi_i : U_i \to \mathbb{H}^k$), then we can multiply each $g_i$ by the pullback $f \circ \psi_i$ of the solution $f$ to each chart domain $U_i$. We are then allowed to take the sum $\widehat{f} = \sum_i g_i \cdot (f \circ \psi_i)$, and this sum vanishes at a point $x \in X$ exactly if each $g_i  \cdot (f \circ \psi_i)$ vanishes there (since we chose $f$ to be nonnegative).

So now assume that $\widehat{f}(x) = \sum_i g_i(x) \cdot (f \circ \psi_i)(x) = 0$. By the definition of a partition of unity there must be some $i$ for which $g_i(x)$ is nonzero, and we still must have $g_i(x) \cdot (f \circ \psi_i)(x) = 0$, so the only possibility is $f(\psi_i(x)) = 0$. But now by the definition of $f$ we must have that $\psi_i(x)$ lies on the boundary of $\mathbb{H}^k$, and since $\psi_i$ is a diffeomorphism $x$ must lie in the boundary $\partial X$. Conversely if $x \in \partial X$ then each pullback $f \circ \psi_i$ is zero at $x$, so each product $g_i \cdot (f \circ \psi_i)$ is also zero there, and thus $\widehat{f}(x) = 0$.

It only remains to solve your problem on $\mathbb{H}^k$ with a nonnegative $f : \mathbb{H}^k \to \mathbb{R}$. How about you just let $f$ be the product of all of the coordinate projections?


问:
Show that $\ell _2 ^2$ is not isometric to $\ell _2 ^\infty$

Show that $\ell _2 ^2$ is not isometric to $\ell _2 ^\infty$.

I have seen similar questions but none of them were between those two spaces. The definition of the two isometric spaces is as follows:

A metric space $X$ is isometric to a metric space $Y$ if there is a bijection $f$ between $X$ and $Y$ such that $d(x,y)=d(f(x),f(y))$.

I know that for every $x \in \ell_2^2$ the norm is $$||x||_2=\left(\sum_{i=0}^2 |x_i|^2 \right)^\frac{1}{2}$$ and for every $x \in \ell_\infty^2$ the norm is $$||x||_\infty=\max_i |x_i|.$$

Those two definitions don't really help me to prove $\ell _2 ^2$ is not isometric to $\ell _2 ^\infty$.

On the other hand, I have found a theorem for this:

Every $n$-dimentional normed space $X$ is isomorphic to the Euclidean space $\ell_2^n$. Consequently, all $n$-dimentional normed spaces are isomorphic to each other.

Maybe there is a way to prove those two finite dimensional normed spaces are not isomorphic and then it will follow they are not isometric?

答:
Since $\ell_2^2$ is a norm induced from an inner product it must satisfy the parallelogram law. On the other hand, this law doesn't hold in $\ell_\infty^2$. For example, take $e=(1,0)$ and $f=(1,1)$. Then:

$||e||_\infty^2+||f||_\infty^2=1+1=2$

$\frac{1}{2}(||e+f||_\infty^2+||e-f||_\infty^2)=\frac{1}{2}(2^2+1^2)=\frac{5}{2}$

It follows that there can't be a norm-preserving isomorphism between these two normed spaces.


问:
"If x is a difference of squares, prove that 3x is a difference of squares as well."
I'm banging my head against the wall with this task:

Prove that if $x$ is a difference of integer squares, then $3x$ is a difference of integer squares as well.

What strategies could I utilise in order to prove this?

Body: Write $x=u^2-v^2$ with $u,v\in \Bbb Z$, then $3x=3(u^2-v^2)=3(u+v)(u-v)=(3u+3v)(u-v)=((2u+v)+(u+2v))((2u+v)-(u+2v))=(2u+v)^2-(u+2v)^2$


问:
Special case of the Eilenberg-Watts theorem for the base ring
Let $A$ and $R$ be commutative rings and let $F$ be a functor $\text{Mod}(A) \to \text{Mod}(R)$ from $A$-modules to $R$-modules. The Eilenberg-Watts theorem says that if $F$ is right exact and preserves small coproducts then $F$ is isomorphic to the functor $- \mapsto M \otimes_A - $ given by tensoring with an $(R, A)$-bimodule $M$.

My question is: Is anyone aware of a (hopefully abelian category flavored) condition on such $F$ which characterizes when the underlying $(R, A)$-bimodule is just $R$ equipped with a ring morphism $f : A \to R$?

I think one answer is when $F$ is a strong monoidal functor, but this is a little bit clumsy because a monoidal structure on $F$ is extra data. Can anyone see a better answer?

答:
First, let me point out that you shouldn't really expect there to be such a characterization that does not involve some extra data, because the $R$-module isomorphism from $M$ to $R$ is not unique.  So, you are going to need some sort of extra data to encode the choice of that isomorphism.  A related issue is that if you allow noncommutative rings, the property you are asking about is not Morita-invariant: if you identify $\text{Mod}(R)$ with $\text{Mod}(M_2(R))$ in the canonical way, then a functor $F:\text{Mod}(A)\to\text{Mod}(R)$ with your property would not have that property as a functor $\text{Mod}(A)\to\text{Mod}(M_2(R))$.  So you are going to need to use something about the commutativity of your rings, which basically means using the monoidal structure.

You don't need to go all the way to giving $F$ a strong monoidal structure, though: it suffices for it just to preserve the unit.  That is, the data of an isomorphism from $F$ to a functor $R\otimes_A -$ for some $A$-algebra structure on $R$ is equivalent to the data of an isomorphism $F(1_{\text{Mod}(A)})\to 1_{\text{Mod}(R)}$.  Indeed, identifying $F$ with $M\otimes_A-$ for some bimodule $M$, an isomorphism $F(1_{\text{Mod}(A)})\to 1_{\text{Mod}(R)}$ is just an isomorphism $M\to R$ as an $R$-module.  It is then automatic that the $A$-module structure on $M$ comes from a homomorphism $A\to R$, since the $A$-module structure must commute with the $R$-module structure and every $R$-module endomorphism of $M$ is multiplication by an element of $R$.


问:
Confusing symbol in papers on hybrid logic
In literature about hybrid logic I'm reading for my thesis I've come across the following symbol:

```<code>::=
</code>```

Now, I've never seen this notation before. I can also not find it explained anywhere (for instance here). Mostly I've worked under the assumption it is equal to $:=$, which its usage seems to suggest. Am I correct in assuming that $::= \ \equiv  \ :=$ (talk about a nicely confusing formulation...).

Since the concepts of hybrid logic are not the main focus of the thesis, but its explanation and correctness of fundamental importance I wan't to make sure I'm understanding everything correctly.

Also note that it is NOT a typo, as I've seen it on more than one occasion, in different papers on the subject.

答: Yes, $:=$ and $::=$ have similar meanings. A lot of times that notation is used to mean "equal by definition or by stipulation". So for instance, you might say, "Define the formula $\varphi ::= \forall x \exists y R(x,y)$," or you might say, "Let "$\varphi \vee \psi$" be an abbreviation for "$\neg (\neg\varphi \wedge \neg\psi)$", i.e. let $\varphi \vee \psi ::= \neg(\neg\varphi \wedge \neg\psi)$." You wouldn't say "Therefore, $(x^2 + 1) ::= y^4$", since if there is an equality involved in this case, it's not really equality by stipulation or definition, but rather some derived equality. You can basically read it as "is equal to" but it's more emphatic, in that it's equality by definition or stipulation, rather than some derived equality.


问:
Show that a set is open iff its complement is closed
I know that there are a lot of questions like this here on StackExchange, but they all use limit points.

My question says that if $(X, d)$ is a metric space and $U \subseteq X$ is open, then the set $U^c = X \backslash U$ is closed, and vice versa.

 A set $U \subseteq X$ is open if $\forall u \in U, \exists \epsilon \in (0, \infty) \ni B_\epsilon(u) \subseteq U$

 A set $Y \subseteq X$ is closed if $\exists (a_n) \in Y \ni (a_n) \rightarrow y \ni y \in Y$

 Plus I can't use limit points, just the definition of open and closed sets.

答:
Suppose $U$ is open.    Let $(a_n)$ be any converging sequence of points where $a_n \not \in U$ (i.e. $a_n\in U^c$).  Suppose $a_n \to y$.

Your Task: Prove $y \not \in U$. (i.e. $y \in U^c$).

Then, by your definition, $U^c$ is closed.

(Hint:  For any $\epsilon > 0$.  As $a_n \to y$ there is a $N$ where $n > N\implies d(a_n,y) < \epsilon$.  So $n > N$ implies $a_n \in B_{\epsilon} y$.)

.......

Now suppose $U$ is a set and $U^c$ is closed.  Do a proof by contradiction.  Suppose $U$ is not open.  That means there is a point $a \in U$ where for every $\epsilon > 0$ we have $B_{\epsilon}(a)\not \subset U$.  That is to say there must be a $b\in B_{\epsilon}(a)$ where $b\not \in U$.

Your Task:  Create a $(b_n); b_n \in U^c; b_n \to a$.

This means that because $U^c$ is closed that $a\in U^c$ and that's a contradiction.

(Hint:  Consider $B_{\frac 1n}(a)$.  Pick a $b_n \in B_{\frac 1n}(a)$ where $b_n \not \in U$.  Prove that $b_n \to a$.)


问:
Prove that sequence of elements in (Q,<) is indiscernible iff the sequence is either constant or strictly increasing/decreasing
I want to prove that a sequence of elements in ($Q, <$) is indiscernible if and only if it is either strictly increasing or strictly decreasing or constant.

For the reverse direction: I note that the ($Q, <$) is a model of dense linear order(DLO). Then by quantifier elimination in DLO, the type of a tuple ($x_1, x_2, ..., x_n$) over $\varnothing$ is completely determined by the relative ordering of $x_i$s. So, take a strictly increasing sequence $a_1<a_2 ...,$, if $i_1, ..., i_n$ and $j_1, ...,j_n $ then, type($a_{i_1}, ..., a_{i_n} $/$\varnothing$) = type($a_{j_1}, ..., a_{j_n} $/$\varnothing$) since both the tuples (type($a_{i_1}, ..., a_{i_n} $) and type($a_{j_1}, ..., a_{j_n}$) are strictly increasing. So, the sequence is indiscernible.

I think the same (exact) argument can be used for strictly decreasing and constant sequences.

However, I don't know how to do the forward direction? I have to show that $\phi(a_{i_1}, ..., a_{i_n})$ $\leftrightarrow$ $\phi(a_{j_1}, ..., a_{j_n} )$ holds only when the sequence is strictly increasing/decreasing or constant. Am I supposed to prove this by contradiction?

答: The forward direction is actually very simple: given a nontrivial indiscernible sequence $\bar a = (a_i)_{i\in I}$, if you pick any $i_1<i_2\in I$, then, since DLO implies that $<$ is a total ordering, you have either $a_{i_1}<a_{i_2}$, $a_{i_1}=a_{i_2}$, or $a_{i_1}>a_{i_2}$, and indiscerniblity trivially implies that in each case, $\bar a$ is strictly increasing, constant or strictly decreasing (respectively).


问:
Calculate number of nodes in a tree from degrees
If I have a tree with the first $N$ nodes having degree $\{2, 3, 4,\dots, N+1\}$ and the remaining nodes having degrees of $1$, is it possible to calculate the total number of nodes?

I've no idea how to go about starting this.

I first thought I could try:

$ \sum_{k=1}^{N} k -1 = \frac{N(N+1)}{2} - 1$

for part of it before realising that degrees are the total number of edges that are incident to a vertex and not necessarily the number of edges that connect to its child nodes. So, there could be overlapping edges for the same nodes.

Edit: realised that I confused how you connect nodes in a tree with how you can connect them in a standard undirected graph. Ignore the final sentence.

答:
My bad (again).  You can't for generic graphs, but you can for trees.  My graph that disproved it had a loop, so it wasn't a tree.

The formula is:

$$Nodes = \frac{N*(N+1)}{2} + 2$$


问:
Importance of Neatness / Organization / Speed in Math?
Pretty simple question here but it does relate to math. I ask this as my writing is quite messy, possibly a cause of silly mistakes.

How important is neatness in math?

Does having messy writing put you at a disadvantage?

What is the tradeoff of speed vs. neatness?

Should I look to increase the neatness of my writing?

What are some simple tips to prevent mistakes that can be solved by simple organization / tidyness/structure?

In timed settings (exams), neat vs. speed?

答:
My handwriting is pretty bad, I love $\LaTeX$, I've lectured on chalkboards/blackboards a couple of times, and given computerized presentations.

My general feeling is that you should make the general direction of everything you write in exams crystal clear, and keep letters/symbols distinct, but otherwise don't waste too much time on lovely handwriting. Structure and layout is far more important. Brief, simple sentences saying what you're about to do, bullet points, headings are all sensible measures.

Lecturing and teaching, however, are different. People should never even think about your handwriting. It should be completely out of their minds - completely normal, legible, consistent, boring.

The magic is always in the maths, and its flow through structured discussion, nowhere else.


问:
Count the number of word pairs with different characters
Suppose you have an alphabet of fixed size $n$. How many pairs of words of equal length $m$ can you compose such that words in a pair do not have a character in common?

Examples:

Let's call such word pairs unique.

$n=2, m=2$: if we only have 2 characters (say A and B) and our words are of length two, then only (AA, BB) and (BB, AA) are unique from 16 possibilities in total.

$n=3, m=1$: if the alphabet has 3 characters and the words are singletons, then 6 of them are unique ((A, B), (B, A), etc.).

$n=3, m=2$: if the alphabet consists of 3 characters and we are composing words of length 2, there are 18 unique pairs.

One approach would be to count the ways of allocating characters to the left and right words, then counting the number of sequences that can be composed from the allocation. At the moment I cannot wrap my head around taking care of the double counting.

Curious about your suggestions on a better way to look at the problem!

答:
Given your example, both elements of the pair are distinct. I am going by that.

Choose $1$ from $n$ alphabets and make word of $m$ letters. Make other word of the pair from $(n-1)$ remaining letters.

Number of such word pairs $ = \, {n \choose 1} \times 1 \times (n-1)^m$.

Now, choose $2$ from $n$ alphabets and make words of $m$ letters but now we need to be careful as all words just with one letter are already made. So we need to subtract that. We make second word of the pair from remaining (n-2) letters.

Number of such word pairs $ = {n \choose 2} \times (2^m - 2) \times (n-2)^m$.

Similarly when we choose $3$ letters, we would have to exclude all words with just two letters and with $1$ letter. As we try and exclude words with $2$ letters, we would subtract words with $1$ letter more times than we should so we have to keep applying P.I.E.

Now choose $2$ from $n$ alphabets and make words of $m$ letters but now we need to be careful as all words just with one letter are already made. So we need to subtract that.

Number of such word pairs $ = {n \choose 3} \times (3^m - {3 \choose 2} 2^m + {3 \choose 1}) \times (n-3)^m$.

We can convert this into using Stirling Number of the second kind and multiply by number of ways to arrange the letters.

So total number of possible word pairs

$ = \displaystyle \sum \limits_{i = 1}^m {n \choose i} \times (i)! \times S2(m,i) \times (n-i)^m$


问:
Evaluation of a nested sum
Given

$\sum_{i=1}^5 \sum_{j=i}^{12} {(2j-7)}$

My questions are :

(1)when $(j=i)$ what that actually means ?

(2) I tackled the inner sum

$\sum_{j=i}^{12}{(2j-7)}$ = 2$\sum_{j=i}^{12} {(j)}$ - $\sum_{j=i}^{12} {(7)} $

I know how to evaluate $\sum_{j=i}^{12} {(7)} $=$-7(13-i)$

I tried to use same logic with 2$\sum_{j=i}^{12}{(j)}$ it didn’t work and the solution of the textbook says that the sum of this, is the sum of the first 12 integers minus the sum of the first i-1 integers.

I have no idea what that means and how so.

Any clarifications please ?

Thanks in advance.

Title: Evaluation of a nested sum

Body:
$$S=\sum_{i=1}^{5} \sum_{j=i}^{12}(2j-7)=\sum_{i=1}^{5} \left[\sum_{j=1}^{12}(2j-7)-\sum_{j=1}^{i-1}(2j-7)\right].$$
$$S=\sum_{i=1}^{5}\left[ (12.13-7.12)-(i(i-1)-7(i-1)\right]=72.5-\sum_{i=1}^{5}(i^2-8i+7)$$ $$=360-55+8.5.3-7.5=390$$
We have used $\sum_{k=1}^{n}=\frac{1}{2}n(n+1), \sum_{k=1}^n k^2=\frac{1}{6}n(n+1)(2n+1)$


问:
Prove that in a triangle $ABC$ $\forall P, P\neq M_{a},M_{b} ,M_{c}:min\{ \frac{|PA|}{|PM_{a}|},\frac{|PB|}{|PM_{b}|},\frac{|PC|}{|PM_{c}} \}\le 2.$
Suppose a triangle $ABC$. Define $M_{a},M_{b} ,M_{c}$ as the midpoints of the sides $a,b,c$ respectively. Then prove that for any point $P$ different from $M_{a},M_{b} ,M_{c}$, the smallest ratio of magnitudes of lines from $P$ to $M_{x}$ and from $P$ to $x$ is $\le 2$.

I have had certain success in prooving it for special cases, such as triangle $ABP$ being isosceles, but I cannot see a way to generalize this properly. I would think this is a pretty important (or at least very interesting result) but I have not seen it anywhere before (or its proof) so I'm intrigued to hear your ideas.

答:
It seems to me (I may be wrong) that the two geometric solutions given to this problem so far have not been very clear to beginners so I want to give
an algebraic solution.

The triangle $\triangle{ABC}$ can be given by its vertices $A (0,0), B (a, 0), C (b, c)$ with the condition of $c\ne0$. The locus of the points $P(x,y)$ such that $\dfrac{\overline{PA}}{\overline{PM_A}}=2$ is a circle and the equations of the three circles involved, noted $\Gamma_A,\Gamma_B,\Gamma_C$  are
$$\Gamma_A:\hspace 0.5cm 3x^2-4(a+b)x+3y^2-4cy+(a+b)^2+c^2=0..........(1)\\\Gamma_B:\hspace 0.5cm 3x^2-2(2b-a)x+3y^2-4cy+b^2+c^2-a^2=0..........(2)\\\Gamma_C:\hspace 0.5cm3x^2-2(2a-b)x+3y^2+2cy+a^2-b^2-c^2=0..........(3)$$
The set union of intersections $X=(\Gamma_A\cap\Gamma_ B)\cup(\Gamma_A\cap\Gamma_ C)\cup(\Gamma_B\cap\Gamma_ C)$ turns out to be a kind of irregular trifolium and we prove algebraically that the only point belonging to the three foliums is the centroid $G$ of the triangle (which is graphically evident) so that the only solution of the system of the three equations above is given by the known formula $G=\left(\dfrac{a+b}{3}, \dfrac{c}{3}\right)$.

This is straightforward because $(1)-(2)$ gives a single possible value $x=\left(\dfrac{a+b}{3}\right)$ and $(1)-(3)$ and $(2)-(3)$ give a linear system in $x$ and $y$ whose solution is $(x,y)=\left(\dfrac{a+b}{3}, \dfrac{c}{3}\right)$.

Now, noting $\overset{\circ}\Gamma_A$ the interior of the circle $\Gamma_A$,  by construction of $\Gamma_A$ we have $\dfrac{\overline{PA}}{\overline{PM_A}}\gt2\iff P\in\overset{\circ}\Gamma_A$ and this is so for the other two circles. It follows that for two of $\left\{ \dfrac{|PA|}{|PM_{a}|},\dfrac{|PB|}{|PM_{b}|},\dfrac{|PC|}{|PM_{c}} \right\}$ be greater than $2$ it is necessaire that the point $P$ belongs to the interior of the intersection of two circles.

For example, if $P(x,y)\in\overset{\circ}\Gamma_A\cap\overset{\circ}\Gamma_ B$ then $\dfrac{\overline{PA}}{\overline{PM_A}}\gt2$ and $\dfrac{\overline{PB}}{\overline{PM_B}}\gt2$ but in this case $P$ must be outside the interior of $\Gamma_C$ since $\{G\}=\Gamma_A\cap\Gamma_ B\cap\Gamma_ C$ and $P\ne G$ (because $\dfrac{\overline{GA}}{\overline{GM_A}}=2)$. This means that  $\dfrac{\overline{PC}}{\overline{PM_C}}\lt2$.

We are done.


问:
Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ .

Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1}-2}$ for $n \in \mathbb{N}$. Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ .

I divided this problem into $3$ cases but I don't know whether it is correct or not if there is anything wrong please tell me?
I will explain what I did briefly?

$1^{st}$ case

if $x_0=2$ or $x_0=3$ it constant sequence so that either   $\lim_{n \to \infty}x_n=2$ or $3$ .

$2^{nd}$case

if $2<x_0<3$ then I showed $x_n>x_{n-1}$ (increasing sequence) and $x_{n-1}<3$ using induction

$3^{rd}$case

if $x_0>3$ then I showed $x_n<x_{n-1}$ (decreasing sequence) and $x_{n-1}>3$ using induction

then we can say $\lim_{n \to \infty}x_n=2$ or $3$ (Monotone convergence theorem)
since in either case $\{x_n\}$ is increasing or decreasing and bounded below or bounded above

In other word in either case,
$$L=\lim_{n\to \infty}x_{n+1}=2+\sqrt{L-2}\iff L=2\text{ or }L=3$$

答:
I don't think you can use monotone convergence directly to show that the limit is $2$ or $3$. As you have done though, you can show that it means for any $x_0 \geq 2$ the limit exists. Once you have this, you can write (using continuity of $\sqrt{.}$) $L = 2 + \sqrt{L-2}$ as you have done, and solve for $L$. So just to summarize:

Use monotone convegence to show the limit exists and is nonnegative for any $x_0$

Use continuity of $\sqrt{.}$ to get $L = 2 + \sqrt{L-2}$.

Conclude $L = 2$ or $L = 3$.


问:
Why do we divide by standard deviation when standardizing a normal distribution?
We have this random variable $Y= \frac{x - μ}{\sigma}$ to convert a normal distribution $N(\mu, \sigma)$ to a $N(0, 1)$. It is quite intuitive to subtract $\mu$, since you move all the values in the $x$-axis, and thus move the mean $\mu$ to the origin of coordinates. But it does not seem intuitive to divide by the standard deviation.

This answer is basically this, however, I have not understood the answers, specifically this:

$$E[Y] = \frac{E[X] - \mu}{\sigma} = \frac{\mu-\mu}{\sigma} = 0.$$

$$\text{Var}(Y) = \frac{1}{\sigma^2}\text{Var}(X) = \frac{1}{\sigma^2}\sigma^2 = 1.$$

And I would also like to get an intuitive answer.

答:
The way I see it, the Normal Variable Z with which mathematicians have built the Z-table isn't very special at all. It's only a matter of conveniency that it has parameters Z~N(0,1).

Now, to answer your question. Consider a normal distribution X with mean μ and σ=10. On the x-axis, every point can be represented as the number of σ from the mean as it usually is. We now want to CONVERT the values of X to the normal distribution Z. To do that we have to translate the graph by µ units to the left hence the new function becomes:
$$X - \mu$$

Now, I assume you can intuitively see how this can be. Next, you have to change the x-axis in such a way that σ should equal 1. This is done by dividing by the value of X's σ as it changes the values from -30,-20,-10,0,10,20,30 to -30/10,-20/20,-10,10,0,10/10 etc In other words, the values have all been converted by a constant multiplier. And that multiplier is σ because we want our values to match the Z random variable in order to use the table.

Hence becoming:

```<code>                    (X-μ)/σ
</code>```

It could have been any value, really. Z could've had a σ of 5 and we would have converted the values accordingly by the constant multiplier of 1/2 in this example.

I hope this helps. I'm no mathematician but this is how I visualise it.

Body:
The way I see it, the Normal Variable Z with which mathematicians have built the Z-table isn't very special at all. It's only a matter of conveniency that it has parameters Z~N(0,1).

Now, to answer your question. Consider a normal distribution X with mean μ and σ=10. On the x-axis, every point can be represented as the number of σ from the mean as it usually is. We now want to CONVERT the values of X to the normal distribution Z. To do that we have to translate the graph by µ units to the left hence the new function becomes:
$$X - \mu$$

Now, I assume you can intuitively see how this can be. Next, you have to change the x-axis in such a way that σ should equal 1. This is done by dividing by the value of X's σ as it changes the values from -30,-20,-10,0,10,20,30 to -30/10,-20/20,-10,10,0,10/10 etc In other words, the values have all been converted by a constant multiplier. And that multiplier is σ because we want our values to match the Z random variable in order to use the table.

Hence becoming:

                    (X-μ)/σ

It could have been any value, really. Z could've had a σ of 5 and we would have converted the values accordingly by the constant multiplier of 1/2 in this example.

I hope this helps. I'm no mathematician but this is how I visualise it.


问:
Prove that if $f:U\to V$ is holomorphic and injective, then $f'(z)\neq 0$ for all $z\in U$.
This is from Stein's Complex Analysis page 207 on the chapter about conformal equivalence.

Proof:

We argue by contradiction and supposed that $f'(z_0)=0$ for some
$z_0\in U$. Then

$f(z)-f(z_0)=a(z-z_0)^k+G(z)$ for all $z$ near $z_0$,

with $a\neq 0,k\geq 2$ and $G$ vanishing to order $k+1$ at $z_0$. For
sufficiently small $w$, we write

$f(z)-f(z_0)-w=F(z)+G(z)$, where $F(z)=a(z-z_0)^k-w$.

Since $|G(z)|<|F(z)|$ on a small circle centered at $z_0$, and $F$ has
at least two zeros inside that circle, Rouche's theorem implies that
$f(z)-f(z_0)-w$ has at least two zeros there. Since $f'(z)\neq 0$ for
all $z\neq z_0$ but sufficiently close to $z_0$ it follows that the
roots of $f(z)-f(z_0)-w$ are distinct, hence $f$ is not injective, a
contradiction.

I saw many similar questions on MSE, but I couldn't find a satisfactory answer unfortunately.
I understand we can do the taylor expansion $f$ about $z_0$ and simplify as the above equation. But I don't know why we need sufficiently small $w$. I know Rouche's theorem and that it is used here, but I don't really understand why the above inequality holds and all roots are distinct. I think pictorial explanation will help, and I can't seem to think of any proper one. Can you please explain what's going on in the proof?

答:
But I don't know why we need sufficiently small $w$.

The larger $w$ is, the larger the circle needs to be to ensure that $F$ has roots inside of it. If $G$ grows quickly enough, this may push the circle into regions where $|F| \ge |G|$ becomes false. We know exactly how $F$ behaves, but we have very little information about the behavior of $G$.

For simplicity, I'll assume that $z_0 = 0$ and $f(0) = 0$. The general case follows by translation.

There are two things that must be satified. For the $w$ we use, there has to be an $r$ such that

on the circle of radius $r, |G| \le |F|$, and

The magnitude $\left|\frac wa\right|^{1/k}$ of the zeros of $F$ must be less than $r$.

Now $G(z) = z^{k+1}g(z)$ for some holomorphic $g(z)$. Because $g$ is continuous at $z = 0$, it is bounded on a neighborhood. So there is some $M > 0$ and some $R > 0$ such that for all $z \le R, |g(z)| < M$, and therefore $|G(z)| \le |z|^{k+1}M$. To assure both conditions are met, we need to choose $r$ such that
$$r <\min\left\{\frac{|a|}M, R\right\}$$ Then choose a $w$ such that
$$|w| < \min\left\{|a|r^k, r^k(|a| - rM)\right\}$$

Since $|w| < |a|r^k$, we have $\left|\frac wa\right|^{1/k} < r$, so the roots of $F$ are inside the circle, and since $|w| < r^k|a| - r^{k+1}M$ we have for $|z| = r < R$,

$$|G(z)| \le Mr^{k+1} < |a|r^k - |w| = |a||z^k| - |w| \le |az^k - w| = |F(z)|$$

Since $k \ge 2$, $F$ has at least two roots, which are all within the circle. By Rouche's theorem, so does $f(z) - w = F(z) + G(z)$. Which means there are at least two $z$ for which $f(z) = w$, though this counts multiplicities.

But this works for any $0 < |w| < r$ and as $|w| \to 0$, so do the roots of $f(z) = w$. Since $f'(0) \ne 0$, and $f'$ is continuous, there is some neighborhood of $0$ in which $f' \ne 0$ anywhere. For small enough $w$, the roots $z$ must be in the neighborhood, and since $f'(z) \ne 0$, it cannot be a multiple root. Hence the roots of $f(z) = w$ must be distinct.


答:
It may just be my fanatical opinion, but I think that clarity is one of the most important attributes of performing mathematics, regardless of level.

For your purposes, let me present an example: Regardless of what course I am teaching, whether it be first year calculus or fourth year topology, I always have students who submit messy, poorly structured, incomprehensible nonsense on their assignments. They typically receive a very poor mark as a result, even if their answer/solution is entirely correct. They of course come and complain that because their answer is correct, they should receive full marks. This is my response:

"A mathematical solution is not just a correct number at the end of a computation: it is a logical sequence of events which is clearly motivated, explained, and justified at every step. If you present non-sense that happens to give you the right number, then while your answer may be correct, your solution is wrong."

Of course, this idea should transcend undergraduate mathematics and manifest in research mathematics as well. A proof is not a proof unless the community can verify your result.

Title: Importance of Neatness / Organization / Speed in Math?

Body:
Pretty simple question here but it does relate to math. I ask this as my writing is quite messy, possibly a cause of silly mistakes.

How important is neatness in math?

Does having messy writing put you at a disadvantage?

What is the tradeoff of speed vs. neatness?

Should I look to increase the neatness of my writing?

What are some simple tips to prevent mistakes that can be solved by simple organization / tidyness/structure?

In timed settings (exams), neat vs. speed?


问:
Finding the basis for a kernel of a polynomial linear transformation?
Say given the function:
$$ T: \mathbb R_{2}[t] \rightarrow \mathbb R_{1}[t]$$
$$ at^{2} + bt + c \rightarrow at + 2c - b $$

We were tasked to find the basis for the kernel of $T$ and $T(\mathbb R_2[t])$.

Where do I start? Could somebody point me in the right direction?

答:
I'll show you a general method to solve this kind of problems.

Take the canonical basis for $\mathbb R_2[x]$, which is $\mathcal C=\{x^2,x,1\}$ and find the images of its vectors.

I suggest you to consider the isomorphism $\varphi:\mathbb R_n[x]\longrightarrow\mathbb R^{n+1}$ to write polynomials as column vectors.

Consider the polynomial $p(x)=ax^2+bx+c$, since we have an isomorphism $\varphi: \mathbb R_n[x]\longrightarrow \mathbb R^{n+1}$ we can map polynomials to column vectors.

$$ax^2+bx+c \mapsto \begin{pmatrix}a\\b\\c \end{pmatrix}$$
Now we can find the images of the polynomials of any basis through $f$:

$$p(x)=ax^2+bx+c \underset{\varphi}{\mapsto} \begin{pmatrix}a\\b\\c \end{pmatrix} \underset{f}{\mapsto} \begin{pmatrix}a\\2c-b \end{pmatrix}  $$
Rmk: You work with column vectors to solve the exercise, but when you have to show a basis for your space of polynomials you have to change column vectors with polynomials.

$$\begin{pmatrix}1\\0\\0 \end{pmatrix}\underset{f}{\mapsto} \begin{pmatrix}1\\0 \end{pmatrix}, \begin{pmatrix}0\\1\\0  \end{pmatrix}\underset{f}{\mapsto} \begin{pmatrix}0\\-1 \end{pmatrix},\begin{pmatrix}0\\0\\1 \end{pmatrix} \underset{f}{\mapsto} \begin{pmatrix}0\\2\end{pmatrix}$$
Now we can write the matrix associated to the map $T$:
$$A:=\mathcal M_{\mathcal C}^{\mathcal C}(T)=\begin{pmatrix}1 && 0 &&0\\0&&-1 && 2 \end{pmatrix}$$
Studying the rank of $A$ you can find the linearly independent columns of the matrix which span $Im(T)=T(\mathbb R_2[x])$ and solving the system $(A|\underline 0)$ you get the vectors that generate $Ker(T)$. Remember that $3=dim(\mathbb R_2[x])=dim(Im(T))+dim(Ker(T))$.


问:
On the proof of limsup is the largest cluster point
I'm studing the following proof:

the proof for the case $x^* \in \mathbb{R}$ is as follows. Define $y_n = \sup_{n \geq k} x_n$ so $x^* = \inf_{n} y_n$:

My problem is at the last paragraph: "the interval $(x^* - \epsilon, x^* + \epsilon)$ must contain infinitely many terms of the sequence $(x_n)$". I know that for each $n$ we can find $x_k$ such that $x_k > x^* - \epsilon$, but why is it true that $x_k < x^* + \epsilon$ ?

答: As pointed out in my previous comment, you can justify $x_k < x^{*} + \varepsilon$ by noting that $x^{*}$ is by definition the infimum $\inf_n y_n$, so for each $\varepsilon > 0$ there exists an $n$, such that $y_n < x^{*} + \varepsilon$. $y_n$ is itself a supremum over $x_k$, so we have for any $k \geq n$ the desired inequality: $x_k \leq y_n < x^{*} + \varepsilon$.


问:
Simplify $\sin(1 + \tan{x} * \tan{\frac{x}{2}})$
I have to simplify $$\sin(1 + \tan{x} * \tan{\frac{x}{2}})$$ and I have got this $$\sin(\frac{1}{\cos{x}})$$
Can I even simplify more or not?

答: That's as simple as it gets. The composition of trigonometric functions doesn't yield anything with another name. Personally, though, I'd write it as $\sin\sec x$.


问:
Equivalence between rep$(Q)$ and modules over $KQ$. Does it preserve indecomposability?
I am studying representation theory at the moment and I was wondering the following (Since I found no source discussing it):

Let $Q$ be a quiver and let $KQ$ be its path algebra. Let Rep($Q$) be the category of representations of $Q$ and Mod($KQ$) be the category of modules over $KQ$.

Now in representation theory one finds that there is an equivalence between Rep($Q$) and Mod($KQ$).

Now my question is: Given a indecomposable representation of $Q$, is the corresponding module over $KQ$ also indecomposable? It would make sense to me but the following example shows otherwise.

Here is a contradictory example:
Let $K \stackrel{id_K}{\rightarrow} K \rightarrow 0$ be a indecomposable representation of the quiver $1 \rightarrow 2 \rightarrow3$. By the equivalence theorem the corresponding module is $K^2$ which is decomposable.

答:
Actually, $K^2$ is an indecomposable module over $KQ$. What is your proposed direct sum decomposition? If you are proposing $K \oplus K$ (with the factors chosen the "natural" way) this doesn't work, because the first factor is taken to the second factor by multiplication by the element $1 \to 2$ of the path algebra. I think you were just thinking that $K^2$ is decomposable as a $K$-module, which of course is true.

The difference between a module over $KQ$ and an element of your $\text{Rep}(Q)$ is really just expanding some definitions; you'll find that any proof that $K \xrightarrow{\text{id}_K} K \to 0$ is indecomposable directly translates to a proof that the module $K^2$ you've defined is indecompsable, and vice versa. This correspondence will extend to any more complicated (sensibly-defined) property that quiver representations and $KQ$-modules can have.


问:
Calculating $\int_0^{2\pi} (e^{it} - a)^k e^{int} dt$
I'm trying to calculate

$$ \int_0^{2\pi} (e^{it} - a)^k e^{int} dt$$

where $a \in \mathbb{C}$ and $k, n \in \mathbb{Z}$.

My first thought was the residue theorem but I end up with permutations of different cases depending on if $|a| > 1$, $k > n$, $k, n < 0$.  Is there a way to approach the integral that reduces the number of cases I need to consider?

答:
Substituting $e^{it}=z$ simplifies the integral to
$$-i\oint_{\text{unit circle}}(z-a)^kz^{n-1}\,dz$$
By the residue theorem this evaluates as
$$\newcommand{Res}{\operatorname{Res}}2\pi(\Res((z-a)^kz^{n-1},0)[n\le0]+\Res((z-a)^kz^{n-1},a)[|a|<1\land k\le-1])$$
where $[\cdot]$ is the Iverson bracket ($1$ if the predicate inside is true, $0$ otherwise). To evaluate the residues, the following two facts can be used:
$$\Res(z^{-n}(z-a)^m,0)=\begin{cases}
0&n\le0\\
\binom{n-m-2}{n-1}a^{m-n+1}(-1)^m&n>0,m<0\\
\binom m{m-n+1}(-a)^{m-n+1}&n>0,m\ge0
\end{cases}$$
$$\Res(z^n(z-a)^m,a)=\Res((z+a)^nz^m,0)$$
The first fact can be obtained by performing a partial fraction decomposition of $z^{-n}(z-a)^m$; the residue is the coefficient of $\frac1z$. The second fact is obvious by substituting $z+a$ for $z$.

```<code>#!/usr/bin/env python3
from mpmath import *

def q1(a, n, k):
    return quad(lambda t: (expj(t)-a)**k * expj(n*t), [0, 2*pi])

def bproduct_res(a, n, m):
    # Res(z**-n (z-a)**m, 0)
    if n <= 0:
        return 0
    if m < 0:
        return binomial(n-m-2, n-1) * power(a, m-n+1) * (-1)**m
    return binomial(m, m-n+1) * power(-a, m-n+1)

def q2(a, n, k):
    rsum = 0
    if n <= 0:
        rsum += bproduct_res(a, 1-n, k)
    if abs(a) < 1 and k <= -1:
        rsum += bproduct_res(-a, -k, n-1)
    return 2*pi*rsum

test_tuples = ((1.4, 3, 4), (1.4, 3, -4), (1.4, -3, 4), (1.4, -3, -4),
               (0.4, 3, 4), (0.4, 3, -4), (0.4, -3, 4), (0.4, -3, -4))
for tup in test_tuples:
    print(chop(q1(*tup)))
    print(chop(q2(*tup)))
    print()
</code>```

Body:
Substituting $e^{it}=z$ simplifies the integral to
$$-i\oint_{\text{unit circle}}(z-a)^kz^{n-1}\,dz$$
By the residue theorem this evaluates as
$$\newcommand{Res}{\operatorname{Res}}2\pi(\Res((z-a)^kz^{n-1},0)[n\le0]+\Res((z-a)^kz^{n-1},a)[|a|<1\land k\le-1])$$
where $[\cdot]$ is the Iverson bracket ($1$ if the predicate inside is true, $0$ otherwise). To evaluate the residues, the following two facts can be used:
$$\Res(z^{-n}(z-a)^m,0)=\begin{cases}
0&n\le0\\
\binom{n-m-2}{n-1}a^{m-n+1}(-1)^m&n>0,m<0\\
\binom m{m-n+1}(-a)^{m-n+1}&n>0,m\ge0
\end{cases}$$
$$\Res(z^n(z-a)^m,a)=\Res((z+a)^nz^m,0)$$
The first fact can be obtained by performing a partial fraction decomposition of $z^{-n}(z-a)^m$; the residue is the coefficient of $\frac1z$. The second fact is obvious by substituting $z+a$ for $z$.

#!/usr/bin/env python3
from mpmath import *

def q1(a, n, k):
    return quad(lambda t: (expj(t)-a)**k * expj(n*t), [0, 2*pi])

def bproduct_res(a, n, m):
    # Res(z**-n (z-a)**m, 0)
    if n <= 0:
        return 0
    if m < 0:
        return binomial(n-m-2, n-1) * power(a, m-n+1) * (-1)**m
    return binomial(m, m-n+1) * power(-a, m-n+1)

def q2(a, n, k):
    rsum = 0
    if n <= 0:
        rsum += bproduct_res(a, 1-n, k)
    if abs(a) < 1 and k <= -1:
        rsum += bproduct_res(-a, -k, n-1)
    return 2*pi*rsum

test_tuples = ((1.4, 3, 4), (1.4, 3, -4), (1.4, -3, 4), (1.4, -3, -4),
               (0.4, 3, 4), (0.4, 3, -4), (0.4, -3, 4), (0.4, -3, -4))
for tup in test_tuples:
    print(chop(q1(*tup)))
    print(chop(q2(*tup)))
    print()


问:
Uniqueness of the solution of $Ax=b$
Given a matrix equation
$$ {A}_{n\times m} x_{m\times 1}=b_{n\times 1},$$

how to find out if the solution vector x is unique?

答: The matrix $A$ that you get is a linear transformation from $\mathbb{R}^m$ to $\mathbb{R}^n$. For the solution if it exists, is unique if the null space of the matrix is trivial. That means that rank($A$)$=m$. But for all this to make sense $m\leq n$.

Title: Uniqueness of the solution of $Ax=b$


问:
Prove that $\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0$.

Let $f$ be a continuous function on the interval $[a,b]$. Prove that $$\displaystyle\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0.$$

Attempt:

Want to show that for a given $\epsilon > 0$, there exists $N \in \mathbb N$ s.t. if $n \geq N$, then $$\left|\int_a^b f(x) \sin^n (x) dx\right| < \epsilon.$$

Since $f$ is continuous on a closed interval, it is bounded, i.e., $|f(x)| \leq M, \forall x \in [a,b]$.

For a given $\epsilon > 0$, choose $N$ s.t. $|\sin^N(x)| < \dfrac{\epsilon}{M(b-a)}$. [Is this allowed? I am thinking of the case where $\sin(x) = 1$, then this doesn't hold.]

Hence, $$\left|\int_a^b f(x) \sin^n (x) dx\right| \leq \int_a^b |f(x)|| \sin^n(x) | dx < \epsilon$$ for $n \geq N$.

答: Some points have $\sin x=1$, so you cannot take $|\sin^Nx|<\epsilon/M(b-a)$. The solution is to remove small neighborhoods around those points (bounding the integral by the length of those small neighborhoods, times a constant). For the remaining parts you have $|\sin x|<1-\delta<1$, so you can bound the integral by $(1-\delta)^N$ times a constant.


问:
A question about $\prod_{x\in \mathbb{R}^{*}}{x}$
When I was an elementary school student, I encountered some puzzles like "What's product of numbers of hair of each people in the world?", which answer was zero, because there's people with no hair(bald), and 0*any number = 0. There was similar question like "What's the product of every number in a telephone?".

After know that '0' makes the whole answer 0, I made three questions and asked first two questions to my friends (to induce the last question):

"What's the product of every numbers?" (or $\prod_{x\in \mathbb{R}} x = ?$)

"What's the sum of every numbers?" (or $\sum_{x\in \mathbb{R}} x = ?$)

They answered '0' in the first question easily, and they also answered '0' in the second question, though they spent several seconds to think (that $x+(-x) = 0$).

After they answered the second question, I asked the last question:

"The answer of the first question was zero, because 'every numbers' includes zero. Then, what's the product of every numbers, except zero? (or $\prod_{x\in \mathbb{R}^{*}} x = ?$)"

My answer was -1, because except 1 and -1, every number $x$ has it's inverse, $\frac{1}{x}$, and the product of two numbers is 1. Therefore, the answer = 1*-1*1*1*.... = -1.

It was quite interesting, because some answered 'infinite?', some answered '1', and some answered '-1', which was the correct answer I thought, before one of my friend says that

"Because the set of real numbers are uncountable, your question doesn't have an answer."

I don't remember whether my friend said "doesn't have an answer", "'s answer is infinite", or other things...

Anyway, the question is, what is the answer of $\prod_{x\in \mathbb{R}^{*}} x$? is it -1? or undecidable?

If it's not -1 (undecidable or infinite or others), then what's the answer of above two questions? or $\prod_{x\in \mathbb{Q}^{*}} x$?

If it's 1, then what's the answer in hyperreal number?

答:
First of all, it is almost impossible to have zero hairs. Even bald people presumably have some amount of hair on their chest, their arms, or their legs. I shudder to think what kind of chemicals you'd need to douse on yourself to get rid of all of your hair. A much more reasonable answer is something like $(3 \times 10^5)^{7 \times 10^9}$: I happen to know that the average human head has about $1.5 \times 10^5$ hairs, so counting all of the hair on a human body should get you about twice that on average, and there are about $7$ billion people in the world. This is a very, very rough estimate, though (the logarithm of it should be within an order of magnitude). Edit: Apparently it is possible to have zero hairs; see the comments.

Second, while these are interesting answers, mathematicians have a formal definition of infinite sums and infinite products, and in those definitions

you need to specify an order to take the sum or product in, which you haven't, and

for the vast majority of orders, none of these expressions are well-defined.

The reason for the first is the Riemann rearrangement theorem, which says that under certain conditions it is possible to rearrange a series of numbers to have any sum, and the reason for the second is that a sum of uncountably many positive numbers always diverges. This is because for some positive integer $n$ there must be uncountably many numbers in the sum greater than $\frac{1}{n}$. Thus, mathematicians generally do not consider sums or products over an uncountable set at all.

Nevertheless, the real numbers are an example of a field, and it is possible to apply the reasoning that you've described to finite fields. In a finite field of characteristic not equal to $2$, it is true for precisely the reasons you describe that the sum of every element is $0$ and that the product of all of the nonzero elements is $-1$. More generally, in any finite abelian group, the product of all the elements is equal to the product of all the elements of order $2$, which is equal to $1$ unless there is exactly one non-identity such element $-1$, in which case it is equal to $-1$. This is a nice exercise, and the proof is very similar to the one you've already described.

Edit: It occurs to me that I don't need to go quite so far as the Riemann rearrangement theorem to convince you that you can't naively apply symmetry arguments to infinite sums and products. You claim that the sum of all real numbers is zero because a number cancels with its inverse. Similar reasoning would tell you that

$$1 - 1 + 1 - 1 \pm ... = 0$$

because you can write it as

$$(1 - 1) + (1 - 1) + (1 - 1) + ... = 0.$$

However, you can also write it as

$$1 + (-1 + 1) + (-1 + 1) + ... = 1.$$

So by applying symmetry in a different way, you get a different answer. This is a basic example known as Grandi's series which highlights the problem with thinking about infinite sums this way, although in the particular case of Grandi's series there's a funny answer: in certain senses, the "right value" of the above sum is $\frac{1}{2}$, the average of the two values above!


答:
What you did is correct... except that you noticed that it doesn't work when $\vert \sin x\vert = 1$, i.e. when $x \in A= \pi \mathbb Z + \pi/2$.

To take care of this notice that $B =[a,b] \cap A$ is finite. Let says that $B$ has $m$ elements $\{b_1, \dots, b_m\}$. You can write $$[a,b] = C \cup ([a,b] \setminus C)$$
where $C=\displaystyle\bigcup_{1 \le i \le m}\left(b_i- \frac{\epsilon}{2M m}, b_i+\frac{\epsilon}{2M m}\right)$.

Then on $[a,b] \setminus C$, you can proceed as you did. And on $C$ you have
$$\left\vert \int_C f(x) \ dx \right\vert \le M m  \frac{\epsilon}{M m} = \epsilon.$$

You then get the desired conclusion.

Title: Prove that $\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0$.

Body:
Let $f$ be a continuous function on the interval $[a,b]$. Prove that $$\displaystyle\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0.$$

Attempt:

Want to show that for a given $\epsilon > 0$, there exists $N \in \mathbb N$ s.t. if $n \geq N$, then $$\left|\int_a^b f(x) \sin^n (x) dx\right| < \epsilon.$$

Since $f$ is continuous on a closed interval, it is bounded, i.e., $|f(x)| \leq M, \forall x \in [a,b]$.

For a given $\epsilon > 0$, choose $N$ s.t. $|\sin^N(x)| < \dfrac{\epsilon}{M(b-a)}$. [Is this allowed? I am thinking of the case where $\sin(x) = 1$, then this doesn't hold.]

Hence, $$\left|\int_a^b f(x) \sin^n (x) dx\right| \leq \int_a^b |f(x)|| \sin^n(x) | dx < \epsilon$$ for $n \geq N$.


问:
Can we check if limit exists using $\epsilon$-$\delta$ method.
We can prove that limit exists using $\epsilon$-$\delta$ method. But, can we conclude a limit's existence using this method? As I saw some other questions where limit does not exist, it felt we can form a relation between $\epsilon$ & $\delta$ using inequalities. At this moment i am really confused. Some help will be appreciated.
for example I can't prove that limit does not exist for $\frac{x^2y^2}{x^4+y^4}$ using $\epsilon$-$\delta$ method.

Thanks

答:
We say that the limit of $f(x, y)$ at a point $(a,b)$ exists if and only if there is $c \in \mathbb R$ such that for all $\varepsilon > 0$ there is $\delta_\epsilon > 0$ with
$$\|(x, y) - (a, b)\|_2 \le \delta_\epsilon \Rightarrow |f(x, y) - c|\le  \varepsilon.$$
In this case we write $\lim_{(x, y) \to (a, b)} f(x, y) = c$.

Now I will show that there is no $c \in \mathbb R$ that satisfies the above formula for $(a, b) = (0, 0)$ and $f(x, y) = x^2y^2/x^4 + y^4$. We do that by contradiction by assuming that there exists $c \in \mathbb R$ with $\lim_{(x, y) \to (0, 0)} f(x, y) = c$ (we assume that $c > 0$, the other cases can be done using the same argument).

Let us fix $\varepsilon > 0$ and the associate $\delta_\epsilon > 0$. We should have that for all $(x, y)$ smaller than $\delta_\epsilon$, $|f(x, y) - c|$ is smaller than $\varepsilon$, or, in other words, $c - \varepsilon \le f(x, y) \le c + \varepsilon$. However, if you take
$$(x, y) = (\min\{\delta_\epsilon, 1/\sqrt{c + \varepsilon}\}, 0),$$
you find that $\|(x, y)\|_2 = \sqrt{x^2 + y^2} = \min\{\delta_\epsilon, 1/\sqrt{c + \varepsilon}\} \le \delta_\varepsilon$ and
$$f(x, y) = \frac{x^2y^2}{x^4 + y^4} = \frac{1}{\min\{\delta_\epsilon, 1/\sqrt{c + \varepsilon}\}^2} > c + \varepsilon$$
which is a contradition to our assumption. Therefore $f(x, y)$ diverges.


答:
Hint:

Divide $[a,b]=A\cup B$, where $A=\bigcup_\limits{(2k+1)\pi/2\in[a,b]}[(2k+1)\pi/2-\delta, (2k+1)\pi/2+\delta]$.

It follows by
$$
\left| \int_A f(x) \sin^n (x) dx\right|\leqslant (1-\epsilon)^n\left| \int_A f(x)dx\right|\to 0\quad\text{as}\:\:n\to\infty
$$
And $$
\left| \int_B f(x) \sin^n (x) dx\right|\leqslant \left| \int_B f(x)dx\right|\leqslant M m(B)<MN\delta
$$


问:
How to prove that $\log(x)<x$ when $x>1$?
It's very basic but I'm having trouble to find a way to prove this inequality

$\log(x)<x$

when $x>1$

($\log(x)$ is the natural logarithm)

I can think about the two graphs but I can't find another way to prove it, and, besides that, I don't understand why should it not hold if $x<1$

Can anyone help me?

Thanks in advance.

答:
This is equivalent to prove that $e^x>x$.

It is obvious that $e^x>x$ if $x<0$ since the LHS is positive and the RHS is negative.

Suppose that for some $a\ge 0$, the inequality $e^a\le a$ holds.

Then $a\ge e^a\ge 1$ since $e^a\ge e^0$ because $a\ge 0$. But now we can see that $a\ge 1$ and again, $a\ge e^a\ge e^1$ and so $a\ge e$. We continue applying the same observation and conclude that $a\ge e^{^{e}}$ and so on, which means that $a$ is unbounded which is a contradiction.


问:
Show that the function satisfies $x^{2}f''(x)+xf'(x)+x^{2}f(x)=0$
If $f(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{4^{n}(n!)^{2}}$, show that $x^{2}f''(x)+xf'(x)+x^{2}f(x)=0$ $\forall x$.

I developed the function's first and second derivatives:

$$f'(x)=\sum_{n=0}^{\infty}\frac{(n+1)(-1)^{n+1}x^{2n}}{4^{n+1}((n+1)!)^{2}}$$

$$f''(x)=\sum_{n=0}^{\infty}\frac{(n+2)(n+1)(-1)^{n+2}x^{2n}}{4^{n+2}((n+2)!)^{2}}$$

And then when I plug them in the differential equation, I don't get 0, no matter how much I try to simplify it. Should I look at the partial sums? Proof by induction? Any help would be appreciated! It's for an analysis course.

Body:
$$f(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{4^{n}(n!)^{2}}$$
The derivative of $f(x)$ is not correct. You should have:
$$f'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}(2n)x^{2n-1}}{4^{n}(n!)^{2}}$$
$$f''(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}(2n(2n-1))x^{2n-2}}{4^{n}(n!)^{2}}$$
And:
$$xf'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}(2n)x^{2n}}{4^{n}(n!)^{2}}$$
$$x^2f''(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}(2n(2n-1))x^{2n}}{4^{n}(n!)^{2}}$$
So that;
$$x^2f''(x)+xf'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}(4n^2)x^{2n}}{4^{n}(n!)^{2}}$$
Start the indice of the sum at $0$:
$$x^2f''(x)+xf'(x)=-\sum_{m=0}^{\infty}\frac{(-1)^{m}x^{2m+2}}{4^{m}(m!)^{2}}$$
$$x^2f''(x)+xf'(x)=-x^2f(x)$$
Since you have this property:
$$(x^n)'=nx^{n-1}$$


问:
Show that $\lim_{n \to \infty}\mu(A_n) = 0$
Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X) < \infty$, and let $f \in L^1(X)$, be such that $|f(x)| > 0$ a.e. Let $A_n \in \mathcal{M}$, $n \in \mathbb{N}$ and suppose that
$$\lim_{n \to \infty}\int_{A_n}|f|d\mu = 0$$

Now, I would like to show that $\lim_{n \to \infty}\mu(A_n) = 0$. Any hints would be appreciated!

答:
Hint: Write $X=\bigcup_m B_m$, where $B_m=\{x\in X\mid \lvert f(x)\rvert\geq 1/m\}$ and observe that $\lim_{m\to \infty}\mu(X\setminus B_m)= 0$ and for each $m$, $\lim_{n\to \infty} \mu(A_n\cap B_m)=0$. Finish by noticing that $$\mu(A_n)\leq\mu(A_n\cap B_m)+\mu(X\setminus B_m).$$

(By the way, if $\mu(X)$ is infinite, then $\lim_{m\to \infty}\mu(X\setminus B_m)=0$ might fail, which is why you can have examples like $f(x)=\frac{1}{x^2}$, $A_n=(n,n+1)$.)


问:
interior and closure strictly from definitions.
Let $X$ be a topological space and $A \subset X$.

Let int$A$ (interior of $A$ in $X$) be the union of all open sets contained in $A$. That is,

int$A= \bigcup_{B \in \tau_A} B$.

Let cl$A$ (closure of $A$ in $X$) be the intersection of every closed set containing $A$.

I need to show that the boundary of $A$, which is the set difference of the two defined above, contains every limit point of $A$.

Attempt:

Let $x$ be a limit point of $A$, then every neighborhood of $x$, say $U$ intersects $A$ nontrivially, that is, there exists a $y \neq x$ such that $y \in U \cap A$.

I need to show $x \in$ every closed set containing $A$ and that it is not in any open set contained in $A$. really stuck.

答: You cannot prove it, since it is false. If, say, $X=\Bbb R$, endowed with its usual topology, and if $A=(-1,1)$, then the boundary of $A$ is $\{1,-1\}$.  But the set of limit points of $A$ is $[-1,1]$, and $[-1,1]\not\subset\{1,-1\}$.


问:
$\det(N) = \det(Q^{-1} B Q) = \det(Q^{-1}) \det(B) \det(Q) = \det(B) = -5 $
Let the matrix $B$ of order $3$ with $\det(B) = -5$.

Let $Q$ the orthogonal matrix  of order $3$ and $N$ a matrix of order $3$ such that $B = Q N Q^t$. What is the determinant of $N$.

I am not sure I am on the right track. $N = Q^t B Q = Q^{-1} B Q$. $\det(N) = \det(Q^{-1} B Q) = \det(Q^{-1}) \det(B) \det(Q) = \det(B) = -5 $. Am I right to say that?

答:
You're on the right track, and you're right,

but perhaps you should show more details to be convincing.

Matrix multiplication is not commutative, but multiplication of determinants is commutative.

Thus, $\det (N)=\det(Q^{-1}BQ)=\det (Q^{-1})\det( B)\det (Q)=\det(Q^{-1})\det (Q)\det( B)$

$=\det(Q^{-1}Q)\det (B)=\det(I)\det (B)=1\times\det( B)=\det(B).$


问:
No. of 7 bit strings that are equidistant
I need the total number of 7-bit strings such that any pair disagree with each other on exactly 4 bits.</p>

答:
This can be formulated as a maximum clique/maximum independent set problem as follows:

Construct a graph $G=(V,E)$ with 128 vertices corresponding to the strings of 7 bits.  For each pair of vertices, add an edge if the two bitstrings disagree in exactly four positions.

Any clique in this graph corresponds to a set of bitstrings that are pairwise different in exactly four positions.  Any maximum clique for this graph (there will be many of them due to symmetry under permutation of the bit positions) is a solution to your problem.

The maximum clique problem is also equivalent to a maximum independent set problem on the complement of $G$.

You can formulate this problem as a 0-1 integer linear programming problem and solve it with widely available software.  I'm not aware of any way to solve the problem by hand.


问:
The matrix M with complex entries

Let $B$ be a matrix of order $3$ with $\det(B) = -5$

Does there exist an invertible matrix $M^{-1}$ with real entries where $M = 4 B M^{-1}$?

I am very confused because I thought I had to verify that $\det(M) \not= 0$. So I computed $\det(M) =  det(4 B M^{-1}) = 4^3 \det(B) \frac{1}{\det(M)}$. I got $\det(M) = \sqrt{-320}$. So the answer might be simply NO because $\det(M)$ is not a real value, so we might find complex value entries, right?

答:
The equation $M = 4 B M^{-1}$ is equivalent to $M^2 = 4B$. As you noticed, it is not possible to find a real matrix $M$ satisfying such equation as it implies

$$(\det M)^2 = -320.$$

However, the equation has at least one solution in term of complex matrix using Schur decomposition. See this wikipedia entry for more details.

Title: The matrix M with complex entries


答:
Here is an important result called Rouchè-Capelli Theorem:
$$Ax=b \ \ \text{admits solution(s)} \iff \text{rank}(A|b)=\text{rank}(A)$$
Now if this condition is verified then:
$$Ax=b \ \ \text{admits an unique solution} \iff  \text{rank}(A)=m$$
So withouth any premise:
$$Ax=b \ \ \text{admits an unique solution} \iff  \text{rank}(A|b)=\text{rank}(A)=m$$

Title: Uniqueness of the solution of $Ax=b$

Body:
Given a matrix equation
$$ {A}_{n\times m} x_{m\times 1}=b_{n\times 1},$$

how to find out if the solution vector x is unique?


问:
Help in a proof of a result in Hungerford's book
I need help to proof the last part of this corollary:

I didn't understand the part (IV) because the author proves just the canonical projection case and the statement says "every nonzero homomorphism of rings $R\to S$ is a monomorphism".

Thanks in advance.

答: If every non-zero homomorphism of rings $R\to S$ is a monomorphism, then certainly the canonical projections are. On the other hand, if $h:R\to S$ is a surjective homomorphism, then $S\cong R/\ker h$, and you’re essentially looking at the canonical projection $\pi:R\to R/\ker h$ anyway.</p>


问:
If $m$ is a positive integer, show that $3m+2$ and $5m+3$ are relatively prime
I tried proving it by assuming the opposite.
So (3m+2 , 5m+3)= k , k>1
3m+2=ka ; 5m+3=kb;

5m+3=3m+2+2m+1;
5m+3=ka + 2m+1;
kb=ka +2m+1;
2m+1=kb-ka;
2m+1= 5m+3-3m+2;
2m+1=2m+1;
Which means that they arent relatively prime , but if you test this with numbers you can clearly see that they are.
What am i doing wrong ?

Body:
You just proved that $2m+1=2m+1$.

Try this (Euclidean algorithm) to show the gcd is $1$:

$$5m+3=1(3m+2)+(2m+1)$$

$$3m+2=1(2m+1)+(m+1)$$

$$2m+1=1(m+1)+m$$

$$m+1=1(m)+1$$


Title: If $m$ is a positive integer, show that $3m+2$ and $5m+3$ are relatively prime

Body:
I tried proving it by assuming the opposite.
So (3m+2 , 5m+3)= k , k>1
3m+2=ka ; 5m+3=kb;

5m+3=3m+2+2m+1;
5m+3=ka + 2m+1;
kb=ka +2m+1;
2m+1=kb-ka;
2m+1= 5m+3-3m+2;
2m+1=2m+1;
Which means that they arent relatively prime , but if you test this with numbers you can clearly see that they are.
What am i doing wrong ?

Body: $$(5m+3;3m+2)=(2m+1;3m+2)=(2m+1;m+1)=(m;m+1)=1$$


答:
If $\lim_{(x,y)\to (0,0)} \frac {x^2y^2}{x^4 + y^4}$ exist there would be an $L$ so that for any $\epsilon > 0$ we could find a $\delta_\epsilon$ so that whenever $0 < d((x,y),(0,0)) < \delta$ we must have $|\frac {x^2y^2}{x^4+y^4} - L| < \epsilon$.

If that is so then for any two $(x_0,y_0),(x_1,y_1)$ where $0 < d((x_0,y_0),(0,0)) < \delta$ and $0 < d((x_1,y_1),(0,0)) < \delta$ we must have $|\frac {x_0^2y_0^2}{x_0^4+y_0^4} -\frac {x_1^2y_1^2}{x_1^4+y_1^4}|   \le d|\frac {x_0^2y_0^2}{x_0^4+y_0^4}-0| +  |\frac {x_1^2y_1^2}{x_1^4+y_1^4}-0| < 2\epsilon$.

So Contrpositively, if we can show for any $\delta > 0$ there will always be $(x_0,y_0),(x_1,y_1)$ where $|\frac {x_0^2y_0^2}{x_0^4+y_0^4} -\frac {x_1^2y_1^2}{x_1^4+y_1^4}| \ge C$ for some positive constant that will prove no limit can exist.  (Because it contradicts the paragraph above.)

So for any $\delta > 0$ we have $(0,\frac 12\delta)$ is within $\delta$ of $(0,0)$.  And $\frac {0^2\cdot (\frac 12\delta)^2}{0^4 + (\frac 12\delta)^4}=0$.

But $(\frac 12 \delta, \frac 12\delta)$ is also withing $\delta$ of $(0,0)$.  And $\frac {(\frac 12\delta)^2\cdot (\frac 12\delta)^2}{(\frac 12\delta)^4+ (\frac 12\delta)^4}= \frac {\frac 1{16}\delta^4}{\frac 18\delta^4} = \frac 12$.

So $|\frac {0^2\cdot (\frac 12\delta)^2}{0^4 + (\frac 12\delta)^4}-\frac {(\frac 12\delta)^2\cdot (\frac 12\delta)^2}{(\frac 12\delta)^4+ (\frac 12\delta)^4}|=\frac 12$ which is a positive constant.

So for any $\frac 14 > \epsilon > 0$ we can't find any $\delta$ that does what we want for any $L$.

So no limit exists.

Title: Can we check if limit exists using $\epsilon$-$\delta$ method.

Body:
We can prove that limit exists using $\epsilon$-$\delta$ method. But, can we conclude a limit's existence using this method? As I saw some other questions where limit does not exist, it felt we can form a relation between $\epsilon$ & $\delta$ using inequalities. At this moment i am really confused. Some help will be appreciated.
for example I can't prove that limit does not exist for $\frac{x^2y^2}{x^4+y^4}$ using $\epsilon$-$\delta$ method.

Thanks


问:
Jacobian linearization and equilibrium points for any inputs
I have 3 nonlinear differential equations and implement Jacobian procedure to create state space model. As you know, Jacobian matrices should be calculated for equilibrium points.

My model contains 3 state variables and 2 inputs. So, i have to obtain totaly 5 equilibrium points for my system. My inputs are like ramp input, so they are not constant. Magnitude of inputs are different in each step of time. At this point, i have confusion:

Should i change state space model in every step because of input?

If first is true, how can i obtain new equilibrium points in every step?

May new equilibrium points make my system unstable locally? If i face this kind of problem, what should i do?

Thanks for your support!

https://www.mathworks.com/help/slcontrol/ug/exact-linearization-algorithm.html

答:
There exist linearization around a point or along a trajectory. When you stud the linearization around a point, you obtain a linear time-invariant system. But it should be linearization around a point, i.e., constant values of the state and the input signal.
On the other hand, you can also linearize along a trajectory. Then you have some nominal trajectory, e.g., for a ramp input and the state trajectory driven by this input. You study the dynamics of deviations of your trajectory from the nominal one and obtain a linear time-varying system.

But the key question is: are you interested in the behavior around a point or along a trajectory?


答: If $d$ divides both $3m+2$ and $5m+3$, it must also divide $5(3m+2)-3(5m+3)=1$.


问:
Finding recursive formula for repeated term on sequence
I have a sequence

$$3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,...$$

The pattern is so difficult. The best i can do is knowing that after jumping to the "next number" it has length $2+2n$. I mean i have $3$ as initial value. And first repeated term has lenght $2+2(1)=4$.  ($4$ repeated term) The second number is $4$ and its length is $2+2(2)=6$. ($6$ repeated term). The third number is $5$ and its length is $2+2(3)=8$. ($8$ repeated term)

How to find the recursive formula? I'm really stuck. Please help me.

答:
$$a_0=3;\;a_n=a_{n-1}+\left\lfloor -\text{frac}\left(\frac{1}{2} \left(\sqrt{4 n+9}-3\right)\right)\right\rfloor +1$$
where $\text{frac}(x)$ is the fractional part of $x$.

From $0$ to $20$ we have
$$\{3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6\}$$

This works because $\frac{1}{2} \left(\sqrt{4 n+9}-3\right)$ is integer when $n=k^2+3k$ which give exactly the sequence $4,10,18,\ldots$ when the number of the given sequence must increase by one.

I mean $3,3,3,3,\mathbf{4},4,4,4,4,4,\mathbf{5},5,5,5,\ldots$


问:
Find $x$ when $x^2\equiv y\bmod p$
Given $p$ is prime and $y$ is a constant. What's the fastest possible way to find $x$ where $x^2\equiv y\bmod p$?

Example: $x^2\equiv97\bmod101$ would give us $x=81$ as one of the solutions. What's the fastest way to compute any one of the solutions of $x$?

Constraints:

$0\le p\le10^9$

$0\le y\le p$

$0\le x<p$

答: The standard approach to compute square roots modulo primes is the Tonelli–Shanks algorithm.</p>

Title: Find $x$ when $x^2\equiv y\bmod p$


问:
why is every element of the disk algebra A generated by the function $z \mapsto z$
The disk algebra is the set of all functions on the unit disc $D$ which are analytic on the interior of the disc and continuous on the boundary. Addition and multiplication are defined obviously. Why is the disk algebra generated by 1 and the map defined above?

答:
Let $f\in A$ so that $f\in C(D) \cap H(\operatorname{int}(D))$.

Since $f$ is continuous on the compact set $D$, it is also uniformly continuous here.
Let $\varepsilon >0$. There is an $0<r<1$ such that $|f(z)-f(rz)| < \varepsilon$ for all $z\in D$.

Since $f(r\cdot)$ is holomorphic in a neighbourhood of $D$, we can find a sequence of polynomials $p_N$ converging uniformly to $f(r\cdot)$ on $D$.

Using this uniform convergence we see that for $N$ sufficiently large, we have for each $z\in D$,
$$
|f(z)-p_N(z)| \leq |f(z)-f(rz)| + |f(rz) - p_N(z)| \leq 2\varepsilon.
$$
Hence the algebra generated by the functions $z\mapsto 1, z\mapsto z$ is dense in $A$.