问:
How to evaluate $\int x\cdot\sin x\cdot\text{cosec} \,3x \,\mathrm dx$
Evaluvate integral $\int x\sin (x)\csc(3x) ~\mathrm{d}x$

My Try

Using a trigonometric identity I was able to simplify,

$$\int \frac{x\sin x}{3\sin x-4\sin^3x}\,\mathrm{d}x$$

$$=\int \frac{x}{3-4\sin^2x}~\mathrm{d}x$$

How can I proceed? Should I use a substitute for this? Any hint would be appreciated.  Thank you.

答:
$$I=\int x\sin (x)\csc(3x) \,dx$$ One integration by parts
$$dv=\sin (x)\csc(3x) \,dx \implies v=\frac 1 {\sqrt 3}\tanh ^{-1}\left(\frac{\tan (x)}{\sqrt{3}}\right)$$ which could be transformed with logarithms.

So, what is left is to compute
$$J=\int \tanh ^{-1}\left(\frac{\tan (x)}{\sqrt{3}}\right)$$ which has a closed form. Have a look at the monster.

Using @Ishraaq Parvez's answer
$$\int \log(\cos(x))\,dx=\frac{i}{2}  \text{Li}_2\left(-e^{2 i x}\right)+\frac{i x^2}{2}-x \log
   \left(1+e^{2 i x}\right)+x \log (\cos (x))$$
Not much more funny !


问:
Intuition behind stationary points in a multivariable function
Why should the local optima for a differentiable function $f(x,y)$ be at the stationary point(s)? What is the intuition behind setting the first partials equal to zero? Is it because if one of the partials is positive (negative), then we can move in (away from) that variable's direction and increase the function value?

答:
From geometry, $f(x,y)$ is a 2D surface, $df$ is the vector field on this 2D surface, while its dual $\star df$ tangent to the contours on the surface.

From the aspect of vector field $df$. The stationary points of $f(x,y)$ are those points, the vectors at which are of zero length. In other words, all the vectors either converge to those point, or emit from them.

From the aspect of dual vector field $\star df$. The stationary points of $f(x,y)$ are those points, around which all the dual vectors are running.

ContentLicense: CC BY-SA 4.0


问:
Explanation of the statement of the fundamental theorem of arithmetic.
It's from the book Mathematics Made Difficult.

If $L^+(P,N_0)$ is the set of functions $f:P\rightarrow N_0$ with a property such that
$$\exists\; n_0 \in N_0 \; \forall \; p \in P \;$$

$$ p\ge n_0 \implies f(p) = 0 $$
then there exists a bijection $N_1 \rightarrow L^+(P,N_0) $ such that if $n \rightarrow f$ then
$$n = \prod_{p\in P} p^{f(p)}$$
Here $P$ is the prime numbers and $N_1 = N_0 - \{0\} $

I know this is <s>related to</s> the fundamental theorem of arithmetic and the book mentioned above made this statement a bit complicated, intentionally. Out of curiosity, I am interested in what this statement is actually saying?

答:
Here $L^+$ is the set of functions which takes a prime number $p$ and outputs some natural number $n$ with the bonus property that $f(p) = 0$ for all but finitely many primes.

For instance, we may consider the function

$$
\begin{cases}
g(2) = 3 \\
g(3) = 1 \\
g(5) = 0 \\
g(7) = 3 \\
g(p) = 0 & \text{for all other primes}
\end{cases}.
$$

The theorem is saying that the set of all such functions canonically corresponds to the natural numbers. How? Perhaps a definition by examples is the best option:

To $g$ as defined above, we can associate this function with the number

$$
8232 =
2^3 \cdot 3^1 \cdot 5^0 \cdot 7^3 =
2^{g(2)} \cdot 3^{g(3)} \cdot 5^{g(5)} \cdot 7^{g(7)}
$$

Here we think of $g(p)$ as telling us what the exponent on $p$ should be. The fact that all but finitely many $g(p)$ are $0$ means that we get a finite number when we multiply them all together (do you see why?).

Conversely, say we're given a number like $80864$. Then we can factor it as
$2^5 \cdot 7 \cdot 19^2$, and so we associate it to the function

$$
\begin{cases}
f(2) = 5 \\
f(7) = 1 \\
f(19) = 2 \\
f(p) = 0 & \text{for all other primes}
\end{cases}
$$

It's easy to see that these two maneuvers undo each other, and so they define a bijection.

This is a fancy way of saying that we can identify a natural number with the exponents of its prime factors. But this is a fancy way to say that every natural number admits a unique prime factorization.
I.e. the fundamental theorem of arithmetic.

I hope this helps ^_^


问:
what is the meaning of this sequence function $f_n=\mathbb 1_{[n, \infty)}?$
what is the meaning of this sequence of functions $f_n=\mathbb 1_{[n, \infty)}?$

Actually after reading this solution, I got confused and didn't understand the given answer

答:
For a subset $S$ of some domain $X$, $\mathbf{1}_S$ usually denotes the indicator function of $S$, i.e., the function
$$
\mathbf{1}_S\colon X \to \{0,1\}
$$
such that $\mathbf{1}_S(x)=1$ if $x\in S$, and $0$ otherwise.

In your case, the sets considered are intervals of the form $[n,\infty)$: so, for any integer $n$, $f_n$ is the function that takes value $0$ on $(-\infty, n)$ and $1$ on $[n,\infty)$.


问:
$f: \mathbb{N} \rightarrow [1,2,3, ..., n] $
Is there a function that is of the form $f: \mathbb{N} \rightarrow [1,2,3, ..., n] $ with $ n \in \mathbb{N} $?

I am looking for a way to assign a natural number to a natural number, but this second it falls in an interval $ [1, n] $
Is this possible?

答:
A function can be any mapping so just map them anyway you want.

For example I can assign $f(x) = 7$ always so that $f(x) \in \{1,2,3,4,5,6,7\}$ because $f(x) = 7$ always.

Or I can assign $f(1) =3$ and $f(2) = 5$ and $f(3) = 1$ and $f(4) = 3$ and $f(5)=5$ and $f(6) =1$ and just repeat over and over again.  Then $f(x) \in \{1,2,3,4,5\}$ because $f(x) = 1,3,$ or $5$ always.

Or you could use the remainder theorem.  For any integer $x$ there is a unique quotient and remainder so that $x = qn + r$ (and $0\le r < n$).  So jest let $f(x) = \begin{cases} r&\text{if }r\ne 0\\n&\text{if }r = 0\end{cases}$.

That way if $n=7$ we'd have $f(1)=1; f(2) =2.....,f(6) = 6; f(7) = 7; f(8) = 1; f(9)=2$.....etc.


问:
Proof easy version of Burnside's Lemma
I'm trying to understand why a group $G$ of order $pq$ is solvable, for $p,q \in \mathbb{N}. p\neq q$ prime.

Proof:
Let $G$ act on its subsets by conjugation:
$$\cdot: G \times \mathbb{P}(G)\longrightarrow \mathbb{P}(G)$$
$$(g,A) \longmapsto g \cdot A=\{gag^{-1}:g\in G \}$$
Then clearly $N_G(H)=Stab_G(H)$ for all subgroups $H<G$. Let $H_p$ be a $p$ Sylow-subgroup. If $N(H_p)=G$, then we are done. Else $N(H_p)=H_p$ has index $q$ in $G$. Now the part that confuses me. By the orbit stabilizer theorem: $G/Stab_G(H_p)\cong G \cdot H_p$. Since $Stab_G(H_p)=N_G(H_p)=H_p$ this means $|G \cdot H_p|=|G/H_p|=q$. Hence the amount of $p$ Sylow-subgroups is $|Syl_p(G)|=q $ Further are these subgroups conjugated to each other and have trivial intersection (since $H_p$ is cyclic).

On the other hand, there is the third Sylow-theorem. I guess its proof uses a similar argument so I wish to fully understand it. Applying this theorem to $G$ we get: $|Syl_p(G)| \equiv 1 \, mod \, p$ which to me looks like a contradiction.

I'd be really thankful if someone could enlighten me.

答:
Actually in this easy case, no Sylow theorems are necessary. Say $p < q$, and let $H$ be the $q$-Sylow. Since $p$ is the smallest prime dividing the order of $G$, and $[G:H]=p$, it is now a standard argument that $H$ is normal.

To spell it out, $G$ acts on $G/H$ on the left, and so there is a homomorphism $\phi : G\to S_p$, where $S_p$ is the symmetric group, and $\ker \phi\le \text{Stab}(H) = H$. Then $\ker \phi$ is either trivial or equal to $H$. But $\phi$ is not injective since $pq$ does not divide $p!$, so $\ker \phi = H$. This means that $H$ acts trivially on $G/H$, so for all $g\in G$ we have $HgH = gH \implies Hg=gH$. This shows that $H$ is normal, so you're done.

As far as your Sylow theorems argument goes, it's basically correct as far as I can tell. One way to finish would be to WLOG assume $q < p$. Then $q\equiv 1\mod p$ is indeed a contradiction. I was a bit unclear what you were asking for here; if you want to understand the content/proof of the third Sylow theorem, any book on abstract algebra would have that information (e.g. Dummit and Foote).

Another way to finish would be to observe that if there are $q$ of the $p$-Sylow subgroups, these account for $pq-q+1$ elements of $G$, meaning there is a unique $q$-Sylow accounting for the other $q-1$ elements. This $q$-Sylow must be normal in $G$.

It is important to realize that there is only necessarily a normal subgroup of the larger prime order. For example in a dihedral group of order $2q$, a subgroup of order $2$ generated by a reflection is not normal.

Title: Proof easy version of Burnside's Lemma


问:
Derive the inequality $x<\arcsin \left(x\right)<\frac{x}{\sqrt{1-x^2}}$
I found that I could do this with the Mean Value theorem and also found a similar question here in the stackexchange and Socratic but couldn't solve the right hand side of the inequality.

I figured that by specifically choosing the function and the interval (combinations of the given question) we could prove that. But  I am lost tin selecting the appropriate function.

Can you guys please suggest some ideas and also some example sums where I could check these kind of sums.

Thanks

答:
Let $f(x)=x-\sin^{-1} x \implies f'(x)=1-\frac{1}{\sqrt{1-x^2}}=\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}}<0$
So $f(x)$ is a decreasing function.
$$x\ge 0 \implies f(x)\le f(0) \implies f(x)\le 0$$
Next, we take $g(x)=\sin^{-1} x-\frac{x}{\sqrt{1-x^2}} \implies g'(x)=\frac{-x^2}{\sqrt{1-x^2}}<0.$
So $g(x)$ is decreasing function, then
$$ x \ge 0 \implies g(x)\le g(0)=0 \implies
\sin^{-1}x \le \frac{x}{\sqrt{1-x^2}}.$$


问:
Calculate $\int_{\mu}^{2t} \frac{1}{E+2\mu -2\epsilon}\frac{1}{\sqrt{4t^2-\epsilon^2}} \mathrm dx$
I stumbled upon this integral while doing Solid-state physics homework:
$$\int_\mu^{2t} \frac{1}{E+2\mu -2\epsilon}\frac{1}{\sqrt{4t^2-\epsilon^2}} \mathrm d\epsilon$$
where $E<0$, $\mu \in (-2t,2t)$ and $t>0$.
I tried plugging it into Mathematica:

```<code>Integrate[1/(e + 2 \[Mu] - 2 x) 1/Sqrt[4 t^2 - x^2], {x, \[Mu], 2 t},
Assumptions -> -2 t < \[Mu] < 2 t && t > 0 && e < 0]
</code>```

and got the following answer:
$$-\frac{1}{2} \pi  \sqrt{\frac{1}{E^2+4 E \mu +4 \mu ^2-16 t^2}}-\frac{\arctan\left(\frac{\mu  (E+2 \mu )-8 t^2}{\sqrt{\left(4 t^2-\mu ^2\right) \left(E^2+4 E \mu +4 \left(\mu ^2-4 t^2\right)\right)}}\right)}{\sqrt{E^2+4 E \mu +4 \left(\mu ^2-4 t^2\right)}}$$

This result gives correct "physical" interpretation, so now I would like to know how this integral was
calculated.
My first guess was to use substitution $\epsilon = 2t\sin{x}$ and transform this integral to
$$\int_{\arcsin{\frac{\mu}{2t}}}^{\frac{\pi}{2}} \frac{\mathrm dx}{E+2\mu - 4t\sin{x}}
$$
Here I thought about trying to evaluate indefinite integral by doing the standard substitution $u = \tan{\frac{x}{2}}$, however I hesitated after reading this comment.
Is there any other way to obtain the result in closed form like the one obtained from Mathematica?

答:
If you just ask Mathematica to compute the antiderivative without any assumption, you should receive
$$I=\int \frac{dx}{\sqrt{4 t^2-x^2} (e+2 \mu -2 x)}=$$
$$\sqrt{-e^2-4 e \mu -4 \mu ^2+16 t^2}\,I=-\log (e+2 \mu -2 x)+$$ $$\log \left(\sqrt{4 t^2-x^2} \sqrt{-e^2-4 e \mu -4 \mu ^2+16 t^2}-e x+8 t^2-2 \mu
   x\right)$$  and I suppose that the arctangent comes from the combination of the logarithms.


问:
Is there a category-theoretic generalisation of function application?
(I'll be using functional programming jargon, because my category theory is not too good).

Is the notion of function application, i.e. the application f a of a function f :: a -> b to a value a :: a to yield a value b :: b, peculiar to functions, or is there a more general category-theoretic concept to generalise it?

答: Functions are morphisms in the category of sets. A morphism  $$e:S\rightarrow X$$ from any singleton S set to another set X (object in category Set), "picks out" an element of X. The composition of a morphism $$f:X\rightarrow Y$$ with the morphism e, $$f\circ e$$ "picks out" the value of f applied to the "value" of e.


答:
Here's one way to "translate" it.

If $L^+(P,N_0)$ is the set of functions $f:P\rightarrow N_0$ with a property such that
$$\exists\; n_0 \in N_0 \; \forall \; p \in P \;$$

$$ p\ge n_0 \implies f(p) = 0 $$

Define $L^+(P,N_0)$ as the set of all expressions of the form

$$2^{x_2} \ 3^{x_3} \ 5^{x_5} \ \cdots,$$

where the bases are the prime numbers, and $x_2, x_3, x_5, \ldots$ are non-negative integers, only finitely many of which are nonzero.

In other words, $L^+(P,N_0)$ is the set of all possible prime factorizations.

then there exists a bijection $N_1 \rightarrow L^+(P,N_0) $ such that if $n \rightarrow f$ then
$$n = \prod_{p\in P} p^{f(p)}$$

Then there is a function which maps positive integers to prime factorizations, such that the value of the prime factorization is (as we would hope) the number which produced it. Furthermore, this function is a bijection.

In other words, each positive integer has exactly one prime factorization, and furthermore, each possible prime factorization is in fact the prime factorization of exactly one positive integer.

Title: Explanation of the statement of the fundamental theorem of arithmetic.

Body:
It's from the book Mathematics Made Difficult.

If $L^+(P,N_0)$ is the set of functions $f:P\rightarrow N_0$ with a property such that
$$\exists\; n_0 \in N_0 \; \forall \; p \in P \;$$

$$ p\ge n_0 \implies f(p) = 0 $$
then there exists a bijection $N_1 \rightarrow L^+(P,N_0) $ such that if $n \rightarrow f$ then
$$n = \prod_{p\in P} p^{f(p)}$$
Here $P$ is the prime numbers and $N_1 = N_0 - \{0\} $

I know this is <s>related to</s> the fundamental theorem of arithmetic and the book mentioned above made this statement a bit complicated, intentionally. Out of curiosity, I am interested in what this statement is actually saying?


问:
Hodge decomposition seems to say co-exact component is zero
Hodge Decomposition Theorem says smooth  $k$-forms on on compact oriented Riemannian (smooth) $m$−manifold $(M,g)$ (I think M need not be connected, but assume connected if need be or you want) decompose into exact, co-exact and harmonic.

About this powerpoint,

If a smooth $k$-form $\omega = \omega_d \oplus \omega_\delta \oplus \omega_\Delta$ is closed, then is its co-exact component $\omega_\delta$ zero?

What exactly is going on in slides 28-41?

2.1. It seems to be proving (1), but actually it appears we have the stronger assumption that $\omega$ is harmonic. In this case, for a smooth $k$-form $\omega$, I think $\omega$ is harmonic if and only if the exact and co-exact components, resp $\omega_d$ and $\omega_\delta$, are both zero. However, I think for a smooth closed $k$-form $\omega$, we have only that $\omega_\delta = 0$ (and $\omega_d]$, all we have is $[\omega_d]=[0]$, but I believe this doesn't even require $\omega$ to be closed).

Update: Based on this, I think: (1) is true, and I think the name for this fact is called 'short Hodge decomposition (theorem)'. As for (2), I think author forgot to explicitly exclude the assumption of harmonic from the claim, which is meant assume closed (instead of harmonic).

答: (copying 'Update' into an answer): Based on this, I think: (1) is true, and I think the name for this fact is called 'short Hodge decomposition (theorem)'. As for (2), I think author forgot to explicitly exclude the assumption of harmonic from the claim, which is meant assume closed (instead of harmonic).


问:
Why does the multiplicative order divide $|\mathbb{F}|-1$?
This source describes Lagrange's theorem for finite fields (page 6). It generalizes Fermat's little theorem. In particular, it states that if $\mathbb{F}$ is a field with finitely many elements, with $|\mathbb{F}|=m$, then
$$a^{m-1}=1$$
for every $a\in\mathbb{F}\setminus\{0\}$.

To prove this, they note that $\mathbb{F}\setminus\{0\}$ has $m-1$ elements. But then, they suddenly conclude that this implies $\text{ord}_{\mathbb{F}}(a)$ divides $m-1$. The result then follows.

How is the divisibility claim reached?

答:
In a field, the nonzero elements always form a group (under multiplication).  The order of that group is of course $|\Bbb F|-1$.  The conclusion is then reached by Lagrange's theorem.

Indeed Fermat's little theorem then follows by application to the field $\Bbb Z_p$, for any prime $p$.


问:
Can a walk be equivalently viewed as a graph?
https://en.wikipedia.org/wiki/Path_(graph_theory)#Walk,_trail,_path says

A walk is a finite or infinite sequence of edges which joins a sequence of vertices. Let G = (V, E, ϕ) be a graph. A finite walk is a
sequence of edges (e1, e2, …, en − 1) for which there is a sequence of
vertices (v1, v2, …, vn) such that ϕ(ei) = {vi, vi + 1} for i = 1, 2,
…, n − 1. (v1, v2, …, vn) is the vertex sequence of the walk. This
walk is closed if v1 = vn, and open else. An infinite walk is a
sequence of edges of the same type described here, but with no first
or last vertex, and a semi-infinite walk (or ray) has a first vertex
but no last vertex.

A trail is a walk in which all edges are distinct.

A path is a trail in which all vertices (and therefore also all edges) are distinct.

Is it correct that

A walk is by definition a sequence of edges, so a walk is not a graph?

A path, which is defined as a special walk, is not a graph by definition? A lot of graph theory books say that a path is a graph. Are these two different meanings of a path equivalent?

Can a walk be equivalently viewed as a graph? ("equivalently" means without loss of information. Since a walk may repeat an edge, while a graph's edge set doesn't contain an edge more than once, I doubt a walk can be equivalently viewed as a graph)

Can a trail be equivalently viewed as a graph? (I guess so, because a trail is defined to not have repeated edges, which makes it equivalent to a graph)

Thanks.

答:
The two notions of a path are as follows:

A "path graph": a graph isomorphic to one with $n$ vertices $v_1, v_2, \dots, v_n$ and $n-1$ edges $v_1 v_2, v_2 v_3, \dots, v_{n-1} v_n$. Commonly denoted $P_n$.

A path inside a graph $G$. Sometimes this is defined as a type of walk, as you did; less commonly, this is defined as a subgraph of $G$ isomorphic to $P_n$ (for some $n$).

The most careful way to define a walk is a sequence of vertices and edges: $v_0, e_1, v_1, e_2, v_2, \dots, e_n, v_n$, where each edge $e_i$ joins $v_{i-1}$ to $v_i$. But of course just the sequence of edges is enough to infer the vertices. In a simple graph, the sequence of vertices is also enough to infer the edges, so we could use either definition.

For any walk in a graph $G$, we could instead consider the subgraph of $G$ consisting of all the edges included in the walk. But we always lose some information by doing this.

In the case of a walk, as you've noticed, we lose the most information. An edge could be part of the walk multiple times, and the subgraph idea forgets all about that.

Even for a trail, we forget some information by passing to the subgraph. Suppose we have a complete graph on vertices $1,2,3,4,5$. The closed trail that goes $1 \to 2 \to 3 \to 1 \to 4 \to 5 \to 1$ is different in many ways from the trail that goes $5 \to 4 \to 1 \to 2 \to 3 \to 1 \to 5$, but they have the same edge set. Or, take any Eulerian graph: there could be many different Eulerian trails, but they would all give the same subgraph, which is the whole graph.

We lose almost no information for a path, but we still lose one thing: orientation. We might want to distinguish the paths $1 \to 2 \to 3 \to 4 \to 5$ and $5 \to 4 \to 3 \to 2 \to 1$, but they definitely give the same subgraph.

Sometimes, though, we want to lose the information, and consider the subgraph instead of the sequence. This is especially common for cycles: closed walks in which no vertex is repeated except for the first and the last vertex being the same. If we take an $n$-vertex cycle, there are $2n$ closed walks total with that set of edges: we can start at any of $n$ vertices, and go in either of $2$ directions. But we don't really think of those as different cycles.


问:
Finite group is generated by a set of representatives of conjugacy classes.
Could you tell me how to prove that a finite group is generated by a set of representatives of conjugacy classes?

I've read this https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class but I don't know

why if $H$ is a proper  subgroup of a finite group $G$  intersecting every conjugacy class of $G$, then $G = \bigcup_{g \in G} g H g^{-1}$

and why the fact that we can't cover $G$ with sets in $\bigcup_{g \in G/H} g H g^{-1}$ implies that the group is generated by a set of representatives of conjugacy classes?

Maybe you know a simpler proof?

Thank you.

答:
Let $x \in G$. $x$ is in a conjugacy class, $x^G = \{ g^{-1} x g : g \in G \}$, and $H \cap x^G \neq \varnothing$ by assumption, so let $h \in H \cap x^G$. Since $h \in x^G$, there is some $g \in G$ with $g^{-1} x g = h$ and so $x = ghg^{-1} \in gHg^{-1} \subseteq \cup_{g \in G} gHg^{-1}$.

Since $G$ is not in fact such a union (by the Lemma below), there is no such subgroup $H$. Every subgroup that intersects all conjugacy classes must not be proper, it must be all of $G$.

If $H$ is generated a set of representatives of the conjugacy classes, then it intersects every conjugacy class (for instance, in that chosen generator), and so it cannot be proper; instead $H=G$.

Lemma: No finite group is the union of conjugates of a proper subgroup.

Proof: [Standard and included in Hagen's answer] Let $G = \cup_{g \in G} gHg^{-1}$ for some subgroup $H \leq G$. Since $gh H = gH$ and $H(gh)^{-1} = Hh^{-1} g^{-1} = Hg^{-1}$, we don't need to union over all $g \in G$, since $g$ and $gh$ give the same subgroup $gHg^{-1} = ghH(gh)^{-1}$. We only need $[G:H]$ different $g$. Now consider the non-identity elements of $G$. Each one has to be in at least one of the $gHg^{-1}$, but it won't be the identity element of that subgroup. Hence the $|G|-1$ different non-identity elements of $G$ lie in one of the subsets $gHg^{-1} \setminus \{1\}$ of size $|H|-1$. Since there are at most $[G:H]$ of those subsets we get an inequality, $|G| - 1 \leq [G:H](|H|-1)$, but the right hand side is just $|G|- [G:H]$. Simplifying $|G|-1 \leq |G|-[G:H]$ gives $[G:H] \leq 1$, but since $[G:H]$ is a positive integer, this means $[G:H]=1$ and $H=G$. $\square$

ContentLicense: CC BY-SA 3.0


问:
Is the complexification of $\mathfrak{sl}(n, \mathbb{C})$ itself?
More precisely if I complexify the algebra, is the following isomorphism true? $$\mathfrak{sl}(n, \mathbb{C})_\mathbb{C} \cong \mathfrak{sl}(n, \mathbb{C})$$
Given that the complex dimension is already $n^2-1$, and each element of the matrix can already be complex, do we gain anything by complexifying here?

答: For any real finite-dimensional vector space $V$ with dimension $n$, $\dim V_{\Bbb C}$ is $2n$ over $\Bbb R$ (and $n$ over $\Bbb C$). So, that isomorphism is not true: $\mathfrak{sl}(n,\Bbb C)_{\Bbb C}\not\simeq\mathfrak{sl}(n,\Bbb C)$, since$$\dim\mathfrak{sl}(n,\Bbb C)_{\Bbb C}=2\dim\mathfrak{sl}(n,\Bbb C)\neq\dim\mathfrak{sl}(n,\Bbb C).$$


问:
Permutations in lexicographic order
A permutation is a bijection from a set to itself.

$$\pi: \{1,\ldots , n\} \mapsto \{1,\ldots , n\}$$

One way to get permutations in lexicographic order is based on the algorithm successor which finds each time the next permutation.

This procedure works as follows:

We start from the end of the list and we find the first number that is smaller from its next one, say $x$.

Then we start again from the end of the list and we find the first number that is bigger than $x$, say $y$.

We replace $x$ and $y$.

We reverse the sublist that starts from the next position of (the new position) of $y$ till the end.

The algorithm is :

I haven't really understood the idea of that algorithm. Could you explain that further to me? Do we change the order of the elements at the end of the list to get the next permutation in lexicographic order?

$$$$

Other methods to get get permutations in lexicographic order is with rank and unrank.

These algorithms are:

Is the idea of these ones that we write all the permutations in lexicograhic order and enumerate them to get the rank and the unrank is, given a number (rank), that we can give the respective permutation?

Is that correct? Or is the idea of rank or unrank related to permutations elsewise?

$$$$

EDIT :

I am trying to understand the formula that is used in permLexRank(n,L).

The formula is $$\text{rank}(\pi,n)=(\pi(1)-1)(n-1)!+\text{rank}(\pi',n-1) \ \text{ with } \ \text{rank}([1],1)=0$$ where $\pi'$ is the permutation that we get if at $\pi$ we delete the first element and subtract all elements that are greater than that by $1$.

The idea for this formula is :  At the lexicographic order of permutations of $S=\{1,\ldots  n\}$ first there are $(n-1)!$ permutations that start by $1$ then there are $(n-1)!$ permutations that start by $2$, etc.

So it holds that $$(\pi(1)-1)(n-1)!\leq \text{rank}(\pi)\leq \pi(1)(n-1)!-1$$
If $r'$ is the rank of $\pi$ in the set of $(n-1)!$ permutations that start with $\pi(1)$, then $r'$ is the rank of $[\pi(2), \ldots , \pi(n)]$ if we conside it as a permutation of $\{1, \ldots , n\}\setminus \{\pi(1)\}$.

If we subtract each element of $[\pi(2), \ldots , \pi(n)]$ by $1$ we get a permutation $\pi'$ of $\{1, \ldots , n-1\}$ that has again rank $r'$.

I haven't really understood that. Could you explain that to me?

答:
As I understood, the algorithm PermLexSuccessor(n,L), given a permutation $L$ of numbers $1,\dots,n-1$ constructs a permutation which is lexicographically next to $L$. To feel how it works we can take the identity permutation for some small $n$ (say $4$ or $5$), and consecutively applying the algorithm, to obtains the list of all permutations of numbers $1,\dots,n-1$ in the lexicographic order.

Is the idea of these ones that we write all the permutations in lexicograhic order and enumerate them to get the rank and the unrank is, given a number (rank), that we can give the respective permutation?

As I understood, given $n$, the algorithms permLexRank(n,L) and permLexRank(n,r) provide an enumeration $e:S\to \{0,\dots, (n+1)!-1\}$ of a set $S$ of all bijections between sets $\{0,1,\dots, n\}$ and $\{1,\dots, n+1\}$. Namely, given $L\in S$, the algorithm permLexRank(n,L) calculates $e(S)$, and given an integer $r\in \{0,\dots, (n+1)!-1\}$, the algorithm permLexRank(n,r) constructs $e^{-1}(r)$.

Title: Permutations in lexicographic order


答:
The complexification of $\mathfrak{sl}_n(\mathbb{C})$ is $\mathfrak{sl}_n(\mathbb{C}) \times \mathfrak{sl}_n(\mathbb{C})$.

This follows from the observation that $\mathfrak{sl}_n(\mathbb{C})$ is itself the complexification of $\mathfrak{sl}_n(\mathbb{R})$, so its complexification can be computed as

$$\left( \mathfrak{sl}_n(\mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \right) \otimes_{\mathbb{R}} \mathbb{C} \cong \mathfrak{sl}_n(\mathbb{R}) \otimes_{\mathbb{R}} \left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right)$$

and then computing that

$$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[x]/(x^2 + 1) \cong \mathbb{C}[x]/((x + i)(x - i)) \cong \mathbb{C} \times \mathbb{C}.$$

So we get two copies of the complexification of $\mathfrak{sl}_n(\mathbb{R})$.

This generalizes to an arbitrary complex Lie algebra $\mathfrak{g}$ but it's a bit more subtle. The general statement is that the complexification $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$ of the underlying real Lie algebra of $\mathfrak{g}$ is $\mathfrak{g} \times \overline{\mathfrak{g}}$ where $\overline{\mathfrak{g}}$ is the conjugate - the same real Lie algebra as $\mathfrak{g}$ but with the conjugate complex multiplication. This follows from repeating the above calculation but using $\mathfrak{g} \otimes_{\mathbb{C}} (\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C})$ instead and being more careful about complex structures.

If $\mathfrak{g}$ is already the complexification of a real Lie algebra (equivalently, has a real form) then it's isomorphic to its conjugate. In particular every semisimple (even reductive) complex Lie algebra has this property. Probably there are solvable complex Lie algebras that don't but I don't know any off the top of my head.

This also isn't specific to Lie algebras and also applies to commutative algebras, associative algebras, vector bundles, group representations, all sorts of stuff.

Given that the complex dimension is already $n^2-1$, and each element of the matrix can already be complex, do we gain anything by complexifying here?

When we complexify we ignore the existing complex structure and freely add a new one. Concretely you can think about this as follows: replace each complex number $a + bi$ in a matrix entry of an element of $\mathfrak{sl}_n(\mathbb{C})$ with the $2 \times 2$ real matrix $\left[ \begin{array}{cc} a & b \\ -b & a \end{array} \right]$. This is how you can think about $\mathfrak{sl}_n(\mathbb{C})$ as a real Lie algebra (in fact as a real Lie subalgebra of $\mathfrak{sl}_{2n}(\mathbb{R})$). When we complexify a second time we allow each of these real numbers to also be complex.

Title: Is the complexification of $\mathfrak{sl}(n, \mathbb{C})$ itself?

Body:
More precisely if I complexify the algebra, is the following isomorphism true? $$\mathfrak{sl}(n, \mathbb{C})_\mathbb{C} \cong \mathfrak{sl}(n, \mathbb{C})$$
Given that the complex dimension is already $n^2-1$, and each element of the matrix can already be complex, do we gain anything by complexifying here?


问:
Infinite Rational numbers add up to give irrational numbers.
$$e=1+\frac{1}{1!}+\frac{1}{2!}+...$$

The rational numbers are closed under addition that means RHS should be rational number but LHS is an irrational number. So, I think that in RHS there are infinitely many terms that's why this is happening.

More such examples:

$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\dfrac{1}{3^2}+...$$

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-...$$

How can you show systematically that such infinite rational terms add up to give an irrational number?

I'm not sure even how to start.

Sorry if this question is too trivial for MSE.

答:
A finite sum of rational numbers is rational as $\mathbb Q$ is a field. So the fact that you won't get an irrational number by summing up a finite number of rational numbers is always the case.

An infinite sum of rational numbers can be either rational or irrational.

Your second question is about how to prove that a number is not rational. And this is a difficult topic. It usually goes through proceeding by contradiction.


问:
Is this a correct way of proving that $Y$ is homeomorphic to a one-point compactification of a Hausdorff space $X$?
I need to prove that $Y$ is homeomorphic to a one-point compactification of a Hausdorff space $X$. I want do the following:

Show that $Y$ is compact.

Show that $Y \setminus \{a\}$ is homeomorphic to $X$ for some particular $a \in Y$.

I believe that the uniqueness of one-point compactification of Hausdorff spaces allows me to do this. Is this correct?

答: Yes, that is correct, assuming that $X$ is not compact. Otherwise it fails (but I suppose that if $X$ was not compact, then you would not be asking this question).


问:
A semicontinuous diagonal function?
Let $X$ be a compact Hausdorff space and let $(x_\alpha)_\alpha$ be a net in $X$ on the directed set $(A, >)$ that converges to $x$.

Let $f: X \times X \to [0, \infty]$. Suppose $f$ has the following properties:

(1) $f(x,x) \in \mathbb R$ for all $x \in X$.

(2) $f(y,y) \leq f(y',y)$ for all $y,y' \in X$.

(3) For all $\beta \in A$, the net $(f(x_\alpha,x_\beta))_\alpha$ converges to $f(x, x_\beta)$.

Does it follow that $\limsup_\alpha f(x_\alpha, x_\alpha) \leq f(x,x)$?

I've tried to start by using (2) to extract a subnet from $(f(x_\alpha, x_\alpha))_\alpha$ that converges to $f(x,x)$, but I've been unable to see that this can be done, and even if it can I'm not sure whether it will imply the result.

答:
Here is a counterexample.

Let $X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual topology; $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ is a net converging to $0$. For $n,m\in\Bbb Z^+$ let

$$f\left(\frac1n,\frac1m\right)=\begin{cases}
1,&\text{if }n=m\\
2,&\text{otherwise.}
\end{cases}$$

For $n\in\Bbb Z^+$ let $f\left(\frac1n,0\right)=f\left(0,\frac1n\right)=2$, and let $f(0,0)=0$. Then

$f$ takes non-negative real values on $X\times X$;

$f(x,x)\le f(y,x)$ for all $x,y\in X$; and

for each $m\in\Bbb Z^+$ the sequence $\left\langle f\left(\frac1n,\frac1m\right):n\in\Bbb Z^+\right\rangle$ converges to $2=f\left(0,\frac1m\right)$,

so the conditions are satisfied. But

$$\limsup\limits_{n\in\Bbb Z^+}f\left(\frac1n,\frac1n\right)=1>0=f(0,0)\,.$$

Title: A semicontinuous diagonal function?


答:
Let $S$ be a set containing an element from each conjugacy class, and let $H=\langle S\rangle$.
By assumption, for any $a\in G$ there is a conjugate of $a$ in $S\subseteq H$, that is $g^{-1}ag\in H$ for some $g$ or equivalently $a\in gHg^{-1}$ for some $g$.

The rest of the argument on the linked question goes like this:
$\bigcup_{g\in G/H}gHg^{-1}\setminus\{1\}$ has at most $|G/H|\cdot (|H|-1)=|G|-|G/H|$ elements. But if the union is all of $G$ (because $H$ contains elements from all conjugacy classes), we must have $|G|-1$ elements after removing $\{1\}$. Hence $|G/H|=1$, i.e. $G=H=\langle S\rangle$.

Title: Finite group is generated by a set of representatives of conjugacy classes.

Body:
Could you tell me how to prove that a finite group is generated by a set of representatives of conjugacy classes?

I've read this https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class but I don't know

why if $H$ is a proper  subgroup of a finite group $G$  intersecting every conjugacy class of $G$, then $G = \bigcup_{g \in G} g H g^{-1}$

and why the fact that we can't cover $G$ with sets in $\bigcup_{g \in G/H} g H g^{-1}$ implies that the group is generated by a set of representatives of conjugacy classes?

Maybe you know a simpler proof?

Thank you.


答:
The rational numbers are closed under addition

True.

that means RHS should be rational number

False.

The right hand side is not an addition operation. It is a limit operation, i.e.

$$\sum_{n=0}^\infty a_n$$ is defined as $$\sum_{n=0}^\infty a_n = \lim_{N\to\infty} \sum_{n=0}^N a_n.$$

Since rational numbers are not closed under limits, there is no contradiction.

In fact, any real number $x\in\mathbb R$ can be the result of an infinite sum. In fact, that's what decimal expansions are. They are a way of writing any real number as a sum of powers of $10$.

How can you show systematically that such infinite rational terms add up to give an irrational number?

What do you mean by "such" in the sentence above? In general, an infinite sum can have either a rational sum (see $\sum_{n=0}^\infty 2^{-n}$) or an irrational sum (the ones above). In fact, for any real number $x\in\mathbb R$, you can find an infinite sum that sums up to $x$. So, since there is no way to "systematically" prove an arbitrary number is irrational, we can't expect to have a systematic proof that an arbitrary sum is irrational.

Title: Infinite Rational numbers add up to give irrational numbers.

Body:
$$e=1+\frac{1}{1!}+\frac{1}{2!}+...$$

The rational numbers are closed under addition that means RHS should be rational number but LHS is an irrational number. So, I think that in RHS there are infinitely many terms that's why this is happening.

More such examples:

$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\dfrac{1}{3^2}+...$$

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-...$$

How can you show systematically that such infinite rational terms add up to give an irrational number?

I'm not sure even how to start.

Sorry if this question is too trivial for MSE.


问:
Prove that a set $F \subseteq \mathbf{R}$ is closed $\Longleftrightarrow$ every Cauchy sequence in $F$ has a limit in $F$
Exercise 3.2.5 in Stephen Abbott's Understanding Analysis asks to prove the below theorem. I would like to ask, if my proof is sound.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Theorem. A set $F \subseteq \mathbf{R}$ is closed if and only if every Cauchy sequence contained in $F$ has a limit that is also an element of $F$.

My Attempt.

($\Longrightarrow$) Assume $F \subseteq \mathbf{R}$ is closed. By definition, a set is closed, if and only if it contains all its limit points. Let $x \in F$ be an arbitrary limit point of $x$. Thus, $V_\epsilon(x)$ intersects $F$ in some point other than $x$. To produce a Cauchy sequence in $F$, we let $\epsilon = 1/n$. Then, there exists a point $x_n \in F$, where
\begin{align*}
 x_n \in V_\epsilon(x) \cap F
\end{align*}

with the stipulation that $x_n \ne x$.

It is easy to see, that $(x_n) \to x$. To see this, choose $N > 1/\epsilon$. Then, for all $n \ge N$, we have,
\begin{align*}
 \absval{x_n - x} < \epsilon
\end{align*}

Convergent sequences are Cauchy and Cauchy sequences are convergent. Convergent Sequence $\Longleftrightarrow$ Cauchy sequence.

Since, $F$ contains all its limit points, all Cauchy sequences in $F$ have their limiting value in $F$.

($\Longleftarrow$) Assume that every Cauchy sequence in $F$ has a limit that is also an element of $F$. Therefore, $\lim x_n = x$, $x_n \ne x$. By the definition of convergence, given any $\epsilon > 0$m there exists a term $x_N$ in the sequence satisfying $\absval{x_N - x} < \epsilon$. So, $V_\epsilon(x) \cap F$ contains elements other than $x$.

答:
Your proofs in both directions are wrong. For $\implies$ assume that $F$ is closed and start with any Cauchy sequence  $(x_n)$ in $F$. [This is imporatnt]. Since any Cauchy sequence of real numbers converges we see that $x =\lim x_n$ exists as a real number. Since each $x_n \in F$ and $F$ is closed it follows that $x \in F$.
For the converse let $(x_n)$ be  as sequence in $F$ converging to some $x$. We have to prove that $x \in F$. Now $(x_n)$ is a Cauchy sequence in $F$. By assumption the limit $x$ is in $F$.


问:
Prove the roots of $z^7+7z^4+4z+1=0$ lie inside a circle of radius $2$.
Problem: Prove that all the roots of the equation $z^7+7z^4+4z+1=0$ lie inside the circle of radius $2$ centered at the origin.

This question came in the roots of unity section of a summer course. My attempts have been to factor and hope something nice pops out, which ended up in the following: $$z(z^3+2)^2+3z^4+1=0$$ $$\frac{z^8-1}{z-1}+3z^4=(z^4+z)(z^2+z-3).$$ They all look horrendous. Other thoughts would be to use some inequality to prove $|z|<2$ (am-gm, cauchy, etc. seem to fail me). Not entirely sure what to do from here.

答:
Use Rouché's theorem with the choice $f(z) = z^7 + 1$ and $g(z) = 7z^4 + 4z$.  Then you need to show that for $z = 2e^{i\theta}$; i.e., on the boundary of the disk of radius $2$ at the origin, $|g(z)| < |f(z)|$.  Then $f+g$ and $f$ have the same number of zeroes in the disk.  Then because $f$ has all seven zeroes on the unit circle, the result follows.

(I edited the previous version to make it a bit easier to show $g$ satisfies the condition.)


答:
Assume that $z$ is a root of the equation with $|z| \ge 2$. Then
$$
 1 = \left|\frac{7z^4+4z+1}{z^7} \right|
\le \left|\frac{7}{z^3}\right| + \left|\frac{4}{z^6}\right| +\left| \frac{1}{z^7}
\right| \\
\le \frac{7}{8} + \frac{4}{64} + \frac{1}{128} = \frac{121}{128} < 1
$$
gives a contradiction.

The idea is to show that if $|z| \ge 2$ then the modulus of $z^7$ is larger than the modulus of $7z^4+4z+1$, so that the sum of these terms can not be zero.

Remark: According to WolframAlpha, the equation $z^7 + 7z^4+4z+1 = 0$ has one solution $z \approx -1.86283$, so that the estimate $|z| < 2$ for all solutions is not bad.

Title: Prove the roots of $z^7+7z^4+4z+1=0$ lie inside a circle of radius $2$.

Body:
Problem: Prove that all the roots of the equation $z^7+7z^4+4z+1=0$ lie inside the circle of radius $2$ centered at the origin.

This question came in the roots of unity section of a summer course. My attempts have been to factor and hope something nice pops out, which ended up in the following: $$z(z^3+2)^2+3z^4+1=0$$ $$\frac{z^8-1}{z-1}+3z^4=(z^4+z)(z^2+z-3).$$ They all look horrendous. Other thoughts would be to use some inequality to prove $|z|<2$ (am-gm, cauchy, etc. seem to fail me). Not entirely sure what to do from here.


答:
This question can be solved using Roché's Theorem. Consider the holomorphic function

\begin{align}
h(z) := f(z) + g(z) = z^7 + 7z^4 + 4z + 1
\end{align}
with $f(z) = z^7$ and $g(z) = 7z^4+4z+1$. Consider $K := \{z \in \mathbb{C}: |z| < 2\}$. Then $\partial K$ is a circle of radius $2$ centered at the origin. On $\partial K$, that means for $|z| = 2$, we have

\begin{align}
|g(z)| = |7z^4+4z+1| \leq 7|z|^4 + 4|z| + 1 = 121 < 128 = |z|^7 = |f(z)|.
\end{align}

Then, by Rouché's Theorem, we have that $f$ and $f+g$ have the same number of zeros inside $K$. Since $f$ has all his seven zeros inside $K$ so does $f+g$ and we are done.


问:
Natural topology on a group acting on a topological space
I'm wondering if there is a general way to give a topological structure to a group acting on a topological space. For example, the rotation group $SO(3)$ acts on the space $\mathbb{R}^3$, and it is normally given the subspace topology when you think of it as a a subspace of the collection of $3\times 3$ matrices. Is there a more natural way of arriving at this topology that is coordinate independent?

Here's what I am considering:
Suppose there is a group action by a group $G$ on a topological space $X$. Suppose that for $x\in X,$ $f_x:G\to X$ is a map so that $f_x(g)=g\cdot x.$ Define a topology on $G$ by the minimal topology such that for all $x\in X$, the map $f_x$ is continuous. Explicitly this would involve taking pre-images of open subsets of $X$ under $f_x$ for each $x\in X$ as a subbasis.

Basically I want the action $\cdot:G\times X \to X$ to be continuous. For this to even be possible, the action must already be continuous w.r.t. change in the $X$ component. Does anyone know a good way to do this?

答:
When $X, Y$ are two topological spaces there is a commonly used topology on the set of continuous functions $X \to Y$ called the compact-open topology. Under mild hypotheses ($X$ locally compact, which is true here) this describes the exponential object $Y^X$ in the category of topological spaces and so has a very pleasant universal property, namely that there is a natural bijection

$$\text{Hom}_{\text{Top}}(X, Z^Y) \cong \text{Hom}_{\text{Top}}(X \times Y, Z).$$

For example, a homotopy between two continuous functions $X \to Y$ is a map $[0, 1] \to Y^X$, the evaluation map $X \times Y^X \to Y$ is continuous, etc. So for $G$ a topological group and $X$ a locally compact space we have

$$\text{Hom}_{\text{Top}}(G, X^X) \cong \text{Hom}_{\text{Top}}(G \times X, X).$$

This implies in particular that if $G$ acts faithfully on $X$ and carries the subspace topology with respect to the inclusion $G \to X^X$, then the action map $G \times X \to X$ is continuous (although it doesn't imply that $G$ is a topological group with respect to this topology!). On the other hand, your proposal only asks for continuity in $G$ for fixed $x \in X$ and doesn't seem to me to get joint continuity.

So, what topology do we get when $SO(3)$ is topologized as a subspace of the space of maps $\mathbb{R}^3 \to \mathbb{R}^3$ with the compact-open topology? We can actually prove a more general fact about the compact-open topology on the space of linear maps between two finite-dimensional real vector spaces, as follows. First, recall that a finite-dimensional real vector space has a unique topology with respect to which addition and scalar multiplication are continuous (which agrees with the product / Euclidean topology if we pick a basis and an isomorphism to some $\mathbb{R}^d$). This is also the topology induced by any norm, and below I'll call it the Euclidean topology.

Proposition: Let $V, W$ be finite-dimensional real vector spaces (with Euclidean topologies). Consider the vector space $[V, W]$ of linear maps $V \to W$, topologized as a subspace of the space $\text{Hom}_{\text{Top}}(V, W)$ of continuous maps $V \to W$, with the compact-open topology. This topology agrees with the Euclidean topology.

Proof. We'll use the fact that when $Y$ is a metric space the compact-open topology on $Y^X$ is the topology of compact convergence. Pick norms $\| \cdot  \|_V, \| \cdot \|_W$ on $V, W$ (which you can take to be Euclidean if you prefer). The Euclidean topology on $[V, W]$ is induced by the operator norm

$$\| T \| = \sup_{\| v \|_V \le 1} \| T(v) \|_W.$$

We'll show that the topology of compact convergence, and hence the compact-open topology, is also the topology induced by the operator norm. The topology of compact convergence is, by definition, the topology in which a sequence $T_i$ of linear maps $V \to W$ converges iff it converges uniformly on compact subspaces of $V$. Since every compact subspace of $V$ is contained in a closed ball centered at the origin (which conversely is compact by the Heine-Borel theorem), equivalently $T_i$ converges compactly iff it converges uniformly on closed balls centered at the origin. And since linear maps are homogeneous, by scaling, $T_i$ converges compactly iff it converges uniformly on the closed ball of radius $1$ centered at the origin. But this is the same as converging with respect to the operator norm. $\Box$

(This argument is actually new to me and suggests the nice intuition that the compact-open topology is like a nonlinear generalization of the operator norm, at least if $X$ is locally compact.)

Corollary: The compact-open topology on $SO(3)$, thought of as a subspace of $\text{Hom}_{\text{Top}}(\mathbb{R}^3, \mathbb{R}^3)$, agrees with the Euclidean topology.


问:
Inequality between two limits
Suppose $a_n \rightarrow a$, $b_n \rightarrow b$ and $ a < b$.
Is it true that there is $ N \in \mathbb{N} $ such that for any $ m > N $, $a_m < b_m $?

I think it's trivial but couldn't prove it.

答: Hint: There exists some $\epsilon>0$ such that $a+\epsilon < b - \epsilon$.

Title: Inequality between two limits


答:
Hint:

You may recast as

$$a_n-b_n\to a-b<0$$ and apply the $\delta/\epsilon$ definition of a limit to $a_n-b_n$.

Note that this approach assumes that the theorem on the limit of a sum is known.

Title: Inequality between two limits

Body:
Suppose $a_n \rightarrow a$, $b_n \rightarrow b$ and $ a < b$.
Is it true that there is $ N \in \mathbb{N} $ such that for any $ m > N $, $a_m < b_m $?

I think it's trivial but couldn't prove it.


问:
Linear control theory and transition matrix problem
So I want to prove the following statement:

Consider an $n$-dimensional time-varying system $\dot{x} = A(t)x$, where $A(t)$ is continuous. If $A^{\intercal}(t)=-A(t)$ $\forall t\in\mathbb{R}$, then $\Phi^{\intercal}(t,s)=\Phi^{-1}(t,s)$, where $\Phi(t,s)$ is the state transition matrix.

So first of all, I know the following:

\begin{align}
\frac{\partial}{\partial s}\Phi(t,s) = -\Phi(t,s)A(s) \iff \frac{\partial}{\partial s}\Phi^{\intercal}(t,s) = -A^{\intercal}(s)\Phi^{\intercal}(t,s) = A(s)\Phi^{\intercal}(t,s).
\end{align}

So how do I go from here to show $\Phi^{\intercal}(t,s)=\Phi^{-1}(t,s)$? It feels like I am on the track, but I am stuck.

Edit

If you are unfamiliar with the transition matrix, this appears in the solution to system of differential equations:

$x(t)=\Phi(t,s)x(s)$

where $x(s)$ is the initial state.

答:
Consider

$$\dot{x} = Ax, \; x(t_0) = x_0 \tag{1}\label{1}$$
$$\dot{z} = -A^\top z, \; z(t_0) = z_0 \tag{2}\label{2}$$

Let $\Phi_x(t,s)$ be the state transition matrix for $(1)$, and $\Phi_z(t,s)$ be the transition matrix for $(2)$. These satisfy the differential equations

\begin{cases}
    \frac{\partial}{\partial t} \Phi_x(t,s) = A \Phi_x(t,s),& \Phi_x(s,s) = I\\
    \frac{\partial}{\partial t} \Phi_z(t,s) = -A^\top \Phi_z(t,s),& \Phi_z(s,s) = I
\end{cases}

Now consider

$$\frac{\partial}{\partial t}\Phi_z^\top(t,s) \Phi_x(t,s) = (\frac{\partial}{\partial t}\Phi_z^\top(t,s)) \Phi_x(t,s) + \Phi_z^\top(t,s) (\frac{\partial}{\partial t} \Phi_x(t,s)) = -\Phi_z^\top(t,s) A \Phi_x(t,s) + \Phi_z^\top(t,s) A \Phi_x(t,s) = 0$$

First, we conclude that $\Phi_z^\top(t,s) \Phi_x(t,s)$ is constant. Second, note that $\Phi_z^\top(s,s) \Phi_x(s,s) = I$. These statements imply that $\Phi_z^\top(t,s) \Phi_x(t,s) = I$ for all $t$. We can then rearrange the equation to obtain $\Phi_z^\top(t,s) = \Phi_x^{-1}(t,s)$. In the special case when $A = -A^\top$, we can see that

$$\Phi^\top (t,s) = \Phi^{-1}(t,s).$$


答:
Being closed under addition means that two rational numbers added together is a rational.

And by induction that means any finite number of rational numbers added together is a rational.  (We add up the first two and that makes one number, then we add the third and that is one number, than the fourth).

But an infinite sum is .... not actually a sum.  The "sum"  $\sum_{k=1}^\infty a_k$ is actually defined as a limit $\lim_{n\to \infty} \sum_{k=1}^n a_k$ (assuming that such a limit actually exists)

Now it is true that if all $a_k$ are rational, and if we write $q_n = \sum_{k=1}^n a-k$ then all $q_n$s are rational.  But then $\sum_{k=1}^{\infty} a_k = \lim_{n\to \infty} q_n$.  So there is no reason at all, that a converging sequence of rational numbers will converge to rational number!

Indeed that is the fundamental property of real analysis!  Every real number, irrationals included, is actually a limit of a converging sequence of rational numbers.

And if we think about it that is not a surprise.  Consider the decimal expansion of an irrational number.  We've been dealing with those for years!

$3.1415.....  = 3 + \frac 1{10} + \frac 4{100} + \frac 1{1000} + \frac 5{10000}+.....$


问:
Show that $f^*:B \to A$ is a function iff $f$ is bijective as follows.
Let $f: A \to B$. Let $f^*$ be the inverse relation, i.e.
\begin{equation*}
f^* = \{(y,x) \in B \times A \mid f(x)=y \}.
\end{equation*}
Show that $f^*:B \to $ is a function iff $f$ is bijective.

Attempt:
Right now, I'm only have a problem for the right direction, but only on the injectivity proof.

Let $f^*$ is a function from $B$ to $A$.
Let $x,z \in A$ and take some $y \in B$ such that $f(x) = f(z) = y$. Then,
$(y,x),(y,z) \in f^*$. Since $f^*$ is a function, then we must have $x=z$. Hence, $f$ is injective.

Is the injective proof correct?

答:
That looks fine, but you could also do this, which is a tad more elegant in my humble opinion:

$\implies$direction: Suppose that the inverse relation $f^* \subset  B  \times A$ is in fact a function $f^*: B \to A$. By definition, that means $f$ is invertible. But $f$ is invertible iff $f$ is bijective. Therefore $f$ is bijective. (See Theorem 1 in these notes from Northwestern: https://sites.math.northwestern.edu/~scanez/courses/300/notes/functions.pdf)


问:
Image of a union of collection of sets as union of the images
I am having problems with establishing the following basic result. Actually, I found a previous post that is close in nature (it is about inverse image), but I was interested in this specific one, with the following notation, because it is the one I found in the book I am self-studying and I guess that most of my problems are actually related with the notation itself. Morevoer I would like to take it as an opportunity to find out how to write proof with equalities, because I tend in those cases to write them in a cumbersome way, with necessary and sufficient conditions.

[I assume that this is partly due to the influence of "How to prove it: a structured approach", a book that I loved, but that left me the tendency to be quite mechanical in proving any sort of result].

Theorem:

Let $X$ and $Y$ be nonempty sets and $f \in Y^X$. Prove that, for any (nonempty) classes $\mathcal{A} \subseteq 2^X$, we have
$ f(\cup \mathcal{A}) = \cup \{f(A):A \in \mathcal{A} \} $

Here, it's how I approach the problem.

Tentative Proof:

First of all, by definition of (direct) image of a function, we have
$$f(\cup \mathcal{A}) := \{f(x):x\in \cup \mathcal{A} \}. \hspace{1cm} (*)$$
Let $X$, $Y$ be arbitrary (nonempty) sets. Let $f$ be an arbitrary function that maps from $X$ to $Y$ and let $\mathcal{A}$ be a family of sets, subset of the powerset of $X$. By $(*)$, the result we have to prove becomes
$$ \{f(x):x \in \cup \mathcal{A} \} = \cup \{ f(A):A \in \mathcal{A}\} \hspace{1cm} (1)$$
In order to prove it, rephrase $(1)$ as
$$ \forall y ( y \in \{f(x):x \in \cup \mathcal{A} \} \leftrightarrow y \in \cup \{ f(A):A \in \mathcal{A}\}). \hspace{1cm} (2)$$
We start by proving the necessary condition. Let $y$ be an arbitrary element and assume that $y$ is a member of $\{f(x):x \in \cup \mathcal{A} \}$. This means that $\exists x(x \in \cup \mathcal{A} \land y=x)$. At the same time, we have to prove that $\exists A(A \in \mathcal{A} \land y\in f(A))$.

And here I got stuck...

I really don't see how from my premises I can get the desired result. Indeed, what $A$ should be?

I assume there is a problem with the way in which I translate the problem in logical terms.

答: The following are equivalent: $$y\in f\left(\bigcup\mathcal{A}\right)\\\exists x\in\bigcup\mathcal{A}:y=f(x)\\\exists A\in\mathcal A:\exists x\in A:y=f(x)\\\exists A\in\mathcal{A}:y\in f(A)\\y\in\bigcup\{f(A):A\in\mathcal{A}\}$$ You should be able to justify each step by definition of union or image.</p>


问:
Surface area (cylindrical coordinates)
I'm having troubles when calculating the area of part of surface bonded with:$$x^2+y^2+z^2=a^2, x^2+y^2=ax, x^2+y^2=-ax.$$ The other two cylinders should give me the bounds for integral, so after I set cylindrical coordinates $x=r\cos\phi, y=r\sin\phi$ for first cylinder I get $-\frac{\pi}{2}\le \phi\le\frac{\pi}{2}$ and for the other cylinder $\frac{\pi}{2}\le \phi\le \pi$. Now not sure what of those two to use when calculating the surface area.

答:
It is not very clear which surface area the question seeks you to find but given it says bounded by all three of them, it is more likely than not that it is seeking surface area of the sphere bounded between both cylinders. If not, we subtract it from $4 \pi a^2$ and that should give us the sum of surface area bound between each cylinder and sphere.

As you would notice both cylinders are of radius $\frac{a}{2}$ with centers at $(-\frac{a}{2}, 0, z)$ and $(\frac{a}{2}, 0, z)$ and are tangent to each other on $z-$axis.

It is much easier to do this in spherical coordinates. The surface of the sphere is defined by

$x = a \cos \theta \sin \phi$, $y = a \sin \theta \sin \phi$, $z = a \cos \phi$ $(0 \leq \phi \leq \pi, 0 \leq \theta \leq 2 \pi) \,$ where $\phi$ is the polar angle and $\theta$ is azimuthal angle.

Now we know that both cylinders intersect the surface of the sphere such that,

At $z = 0 \,(or \, \phi = \frac{\pi}{2}): x = a, y = 0$ for one of the cylinders and and $x = a, y = 0$ for the other. Similarly at $z = a \, (or \, \phi = 0)$, they both intersect sphere at $(0, 0, \pm a)$

From equation of the first cylinder and sphere, $\,x^2 + y^2 = ax \implies a^2 \sin^2 \phi = a^2 \cos \theta \sin \phi$

At intersection $\theta = \cos^{-1} (\sin \phi)$. For the second cylinder, $\theta = \cos^{-1} (-\sin \phi)$

So if we consider the surface only bound between $0 \leq \theta \leq \pi$ and $0 \leq \phi \leq \frac{\pi}{2}$, we can write

$\theta = \frac{\pi}{2} - \phi$ at the intersection of sphere and the first cylinder.

Similarly, $\theta = \frac{\pi}{2} + \phi$ at the intersection of sphere and the second cylinder.

So surface area of the surface bounded between three of them is given by

$ S = 4a^2 \displaystyle \int_0^{\pi/2} \int_{\pi/2 - \phi}^{\pi/2 + \phi } \sin \phi \, d\theta \, d\phi = 8a^2$

If you are doing it in cylindrical coordinates, equations of the surfaces are $r^2 + z^2 = a^2, r = \pm a \cos \theta$. Based on one of the previous questions I saw from you, you already know how to find surface area in cylindrical coordinates. So leaving those details aside, your integral will be,

$4 \displaystyle \int_{0}^{\pi} \int_{a |\cos \theta|}^a \frac{a \, r}{\sqrt{a^2-r^2}} dr \, d\theta$

Title: Surface area (cylindrical coordinates)


问:
Limit involving log

I was asking why this limit is $-\infty$ and not $+\infty\ ?$.

I thought it was $+\infty$, but Wolfram Alpha says $-\infty$.

My guess is that $\log\left(\log\left(x\right)\right) -\log\left(\log\left(x - 1\right)\right)$ goes to zero faster than $-0.5x$ goes to $-\infty$.

$$
\lim_{x \to \infty}\left\{\left(\frac{x^{2}}{2} - x\right) \left[\log\left(\log\left(x\right)\right) -
\log\left(\log\left(x - 1\right)\right)\right] -
\frac{x}{2} -\frac{x}{\log\left(x\right)}\right\}
$$

答:
The first thing I should work in order to get rid of is
$$\log (\log (x))-\log (\log (x-1))=\log \left(\frac{\log (x)}{\log (x-1)}\right)$$
$$\frac{\log (x)}{\log (x-1)}=1+\frac{1}{x \log (x)}+\frac{\log (x)+2}{2 x^2 \log^2(x)}+O\left(\frac{1}{x^3}\right)$$
$$\log \left(\frac{\log (x)}{\log (x-1)}\right)=\frac{1}{x \log (x)}+\frac{\log (x)+1}{2 x^2 \log
   ^2(x)}+O\left(\frac{1}{x^3}\right)$$
All of that gives for the expression
$$-\frac{x (\log (x)+1)}{2\log \left(x\right)}+\frac{1-3 \log (x)}{4 \log
   ^2(x)}+O\left(\frac{1}{x}\right)$$

For $x=10$, the above truncated expression gives $-7.45004$ to be compared to the exact value $-7.46945$.

For $x=100$, the above truncated expression gives $-61.0084$ to be compared to the exact value $-61.0093$.

We are quite far away from $\infty$ and it seems to be working quite well.

Title: Limit involving log


问:
How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital?
$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$

$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:

$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$

 I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?

答:
Always try to simplify algebraic calculations in a manner which reduces typing effort and visual clutter.

Clearly we can take $n$ common from both terms in denominator and hence denominator can be written as $n(a-b) $ where both $a, b$ tend to $1$. Further we can see that $a^3,b^2$ are radical free and hence we have $$n(a-b) =n(a-1-(b-1))=n\left((a^3-1)\cdot\frac{a-1}{a^3-1}-(b^2-1)\cdot\frac{b-1}{b^2-1}\right)\tag{1}$$ Just note that $$n(a^3-1)=n\left(\frac{1}{n^2}+\frac {1}{n^3}\right)\to 0$$ and $$n(b^2-1)=n\left(-\frac{1}{n}+\frac{2}{n^2}\right)\to - 1$$ It now follows from equation $(1)$ that denominator $n(a-b) $ tends to $$0\cdot\frac{1}{3}-(-1)\cdot\frac{1}{2}=\frac{1}{2}$$ and the expression under limit thus tends to $2$.


问:
Derive the inequality $x<\arcsin \left(x\right)<\frac{x}{\sqrt{1-x^2}}$
I found that I could do this with the Mean Value theorem and also found a similar question here in the stackexchange and Socratic but couldn't solve the right hand side of the inequality.

I figured that by specifically choosing the function and the interval (combinations of the given question) we could prove that. But  I am lost tin selecting the appropriate function.

Can you guys please suggest some ideas and also some example sums where I could check these kind of sums.

Thanks

答:
For $|x|<1$,

$$1<\frac1{\sqrt{1-x^2}}<\frac1{(1-x^2)^{3/2}}$$ seems obvious. Then by integration between $0$ and $x$ you get the claimed inequalities.


答:
Hoping that you are aware of the digamma function and harmonic numbers, we can even approximate the partial sums before going to the limit.

Let

$$a_k= \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6}$$
$$S_p=\sum_{k=1}^p a_k=\frac{1}{6} \psi \left(p+\frac{7}{6}\right)+\frac{1}{6} \psi
   \left(p+\frac{3}{2}\right)+\frac{1}{6} \psi
   \left(p+\frac{11}{6}\right)-$$ $$\frac{1}{2} \psi (4 p+4)-\frac{\gamma
   }{2}+\frac{11}{12}-\frac{\psi \left(\frac{11}{6}\right)}{6}-\frac{\psi
   \left(\frac{3}{2}\right)}{6}-\frac{\psi \left(\frac{7}{6}\right)}{6}$$ Switching to harmonic numbers
$$S_p=\frac{1}{2} H_{3 p+\frac{5}{2}}-\frac{1}{2}H_{4 p+3}-\frac{37}{60}+\log (2)$$ Using the asymptotics of harmonic numbers
$$S_p=\left(\frac{\log (3)}{2}-\frac{37}{60}\right)+\frac{1}{16 p}-\frac{199}{3456 p^2}+O\left(\frac{1}{p^3}\right)$$

Using this truncated expansion, we would hve $S_{10}=-0.06169$ while the exact value would be $-0.06164$.

Title: Evaluate $ \sum_{k=1}^\infty ( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6})$

Body:
I need to evaluate the sum given by:

$$\displaystyle \sum_{k=1}^\infty \left( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6} \right)$$

I know that:

for $k = 1$ I get:
$\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}\mathbf{-\frac{1}{14}}$

for $k = 2$ I get:
$\frac{1}{13}+\frac{1}{15}+\frac{1}{17}-\frac{1}{16}-\frac{1}{18} \mathbf{-\frac{1}{20}-\frac{1}{22}}$

for $k = 3$ I get:
$\frac{1}{19}+\frac{1}{21}+\frac{1}{23}-\frac{1}{24} \mathbf{-\frac{1}{26}-\frac{1}{28}-\frac{1}{30}}$

for $k = 4$ I get:
$\frac{1}{25}+\frac{1}{27}+\frac{1}{29} \mathbf{ -\frac{1}{32}-\frac{1}{34}-\frac{1}{36}-\frac{1}{38}}$

I can write that sum for $k = 4$ as:
$$ \frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15}-\frac{1}{16}+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}+\frac{1}{21}-\frac{1}{22}+\frac{1}{23}-\frac{1}{24}+\frac{1}{25}-\frac{1}{26}+\frac{1}{27}-\frac{1}{28}+\frac{1}{29} -\frac{1}{30} \mathbf{ -\frac{1}{32}-\frac{1}{34}-\frac{1}{36}-\frac{1}{38}}$$

And I can write that sum for $k = 8$ as:
$$ \frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15}-\frac{1}{16}+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}+\frac{1}{21}-\frac{1}{22}+\frac{1}{23}-\frac{1}{24}+\frac{1}{25}-\frac{1}{26}+\frac{1}{27}-\frac{1}{28}+\frac{1}{29} -\frac{1}{30} + \frac{1}{31}-\frac{1}{32}+\frac{1}{33}-\frac{1}{43}+\frac{1}{35}-\frac{1}{36}+\frac{1}{37} -\frac{1}{38}+\frac{1}{39}-\frac{1}{40}+\frac{1}{41}-\frac{1}{42}+\frac{1}{43}-\frac{1}{44}+\frac{1}{45}-\frac{1}{46}+\frac{1}{47}-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}+\frac{1}{51}-\frac{1}{52}+\frac{1}{53} -\frac{1}{54} \mathbf{ -\frac{1}{56}-\frac{1}{58} -\frac{1}{60}-\frac{1}{62}-\frac{1}{64}-\frac{1}{66}-\frac{1}{68} -\frac{1}{70}}$$

I see that for every k I get $1$ extra negative element at the end of the sum. I had an idea to rewrite it that way:
$$\displaystyle \sum_{k=1}^\infty \left( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6} \right) = $$
$$\displaystyle \sum_{k=1}^\infty \left( (-1)^{n+1}\frac{1}{n} \right) -1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6} - \displaystyle \sum_{k=1}^\infty X$$

That way I know that the first sum converges (Dirichlet), but still I don't know how to evaluate that expression. I don't know how to evaluate $\displaystyle \sum_{k=1}^\infty ( (-1)^{n+1}\frac{1}{n}$ ) and I don't know how to include those negative elements in my sum (those are marked as X).


Title: How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital?

Body:
$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$

$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:

$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$

 I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?

Body: For large $n$, $(1+n^{-2}+n^{-3})^{1/3}\in 1+\tfrac13n^{-2}+O(n^{-2})\subseteq 1+o(n^{-1})$, so$$\begin{align}\frac{n^{-1}}{(1+n^{-2}+n^{-3})^{1/3}-(1-n^{-1}+2n^{-2})^{1/2}}&\in\frac{n^{-1}}{1+o(n^{-1})-1+\tfrac12n^{-1}+o(n^{-1})}\\&=\frac{n^{-1}}{\tfrac12n^{-1}+o(n^{-1})}\\&\stackrel{n\to\infty}{\sim}2.\end{align}$$


问:
Inverse Laplace Transform. Computing the integral.
This question is related to this one, but I'm hereby taking a different approach.

Problem: Solve
$\ddot x+\delta\dot x+\omega_0^2x=\gamma\cos\omega t$.

Find the stationary points and examine their stability.

Attempted Solution: Let us use the Laplace transform to solve the equation as it is. Let us denote the Laplace transform of $x(t)$ by $X(s)$. Then
$$s^2X(s)-sx(0)-sX(0)+x(0)+\delta(sX(s)-x(0))+\omega_0^2X(s)=\frac{\gamma s}{s^2+\omega^2}\Leftrightarrow$$
$$(s^2+\delta s+\omega_0^2)X(s)=\frac{\gamma s}{s^2+\omega^2}+x(0)(s-1+\delta)+sX(0)\Leftrightarrow$$
$$X(s)=\frac{\frac{\gamma s}{s^2+\omega^2}+x(0)(s-1+\delta)+sX(0)}{s^2+\delta s+\omega_0^2}$$
We now need to apply the inverse transform. So
$$x(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\xi-iT}^{\xi+iT}e^{st}\frac{\frac{\gamma s}{s^2+\omega^2}+x(0)(s-1+\delta)+sX(0)}{s^2+\delta s+\omega_0^2}ds$$.

Can I get some help with this nasty integral? How are we supposed to find $X(0)$ in the first place?

答:
You made a mistake there somewhere: the value of $X(0)$ makes no sense.  Here's what I get for the LT:

$$X(s) = \frac{\dot{x}_0  + (s+\delta) x_0}{s^2+\delta s+\omega_0^2} + \frac{\gamma s}{(s^2+\delta s+\omega_0^2)(s^2+\omega^2)} $$

where $x_0 = x(0)$ and $\dot{x}_0 = \dot{x}(0)$.  As you point out, the ILT is given by

$$\frac{1}{i 2 \pi} \lim_{R \to \infty} \int_{c-i R}^{c+i R} ds \, e^{s t} \, X(s)$$

I think you are under the impression that you actually have to evaluate this integral directly.  I'm sure it could be done, but that's not the best way to do it.  The way to evaluate this integral is to use the Residue theorem.  If you are not familiar with it, then I cannot give a entire course on it here, but recommend that you read up in some introductory text in Complex Analysis such as, e.g., Churchill & Brown, so you can follow what I do below.

For $t>0$, close a contour $C$ to the left of $\Re{s}=c$ in the complex-$s$ plane with a semicircle of radius $R$; the integral about the circular arc approaches zero as $R \to \infty$, so the integral about the closed loop $C$ is equal to $i 2 \pi$ times the ILT.  The integral about this closed loop $C$, on the other hand, is $i 2 \pi$ times the sum of the residues of the poles inside $C$; therefore, the ILT is the sum of these residues.

The poles of course are given by the zeroes of the denominator:

$$s^2+\omega^2=0 \implies s=\pm i \omega$$

$$s^2+ \delta s+\omega_0^2 = 0 \implies s_{\pm} = \frac12 \left ( -\delta \pm \sqrt{\delta^2-4 \omega_0^2}\right)$$

Now, for simple poles like these, the residue of a pole $s_0$ of a function of the form $p(s)/q(s)$ is $p(s_0)/q'(s_0)$.  So finding the ILT is just a matter of handling the algebra behind evaluating the residues of the LT times $e^{s t}$ at the four poles.  The sum of the residues due to the first two poles at $s=\pm i \omega$ above is straightforward:

$$x_1(t) = \frac{\gamma}{2} \frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2+4 \delta^2 \omega^2} \cos{\omega t} + \frac{2 \gamma \delta \omega}{(\omega_0^2-\omega^2)^2+4 \delta^2 \omega^2} \sin{\omega t} $$

This first part corresponds to the oscillations induced by the forcing term.  The other contribution to the full solution determined by the other poles are more tricky, because you have to examine 3 cases: 1) $\delta > 2 \omega_0$, 2) $\delta < 2 \omega_0$, and 3) $\delta = 2 \omega_0$ exactly.  Let's examine the first case.  The algebra involved in evaluating these residues is obviously tedious, but there are clear ways to simplify a little.  For example, $2 s_{\pm}+\delta = \pm \sqrt{\delta^2-4 \omega^2}$.  Otherwise, there is no good way to illustrate the algebra other than to just show the result:

$$x_2(t)=\frac{e^{-\delta t/2}}{\sqrt{\delta^2-4 \omega_0^2}} \left \{ \left [ -\dot{x}_0  + \frac12 x_0 (\delta+\sqrt{\delta^2-4 \omega_0^2})\right] e^{\sqrt{\delta^2-4 \omega_0^2} t/2} +  \left [ \dot{x}_0 +\frac12 (\delta-\sqrt{\delta^2-4 \omega_0^2}) \right] e^{-\sqrt{\delta^2-4 \omega_0^2} t/2} \right \}$$

$$x_3(t) =\frac{\gamma}{2} \frac{e^{-\delta t/2}}{\sqrt{\delta^2-4 \omega_0^2}} \left \{ \frac{-\delta + \sqrt{\delta^2-4 \omega_0^2}}{\omega^2-\omega_0^2+\frac12 \delta^2 -\frac12 \delta \sqrt{\delta^2-4 \omega_0^2}} e^{\sqrt{\delta^2-4 \omega_0^2} t/2} + \frac{\delta + \sqrt{\delta^2-4 \omega_0^2}}{\omega^2-\omega_0^2+\frac12 \delta^2 +\frac12 \delta \sqrt{\delta^2-4 \omega_0^2}} e^{-\sqrt{\delta^2-4 \omega_0^2} t/2}\right \}$$

Then the solution to the original equation is

$$x(t) = x_1(t) + x_2(t) + x_3(t)$$

This corresponds to the case of pure damping even though the $x_1$ solution is oscillatory.  Note that this solution is valid for the case $\delta<2 \omega_0$, although you will have some algebra involving complex numbers to do.  There, the solution has both a damping and oscillatory component.

I leave the case $\delta = 2 \omega_0$ to the reader.

For $t<0$, close the contour to the right of $\Re{s}=c$; by design, there are no poles here, so the integral is zero as you'd expect.


问:
Showing something is an open two cell (Lee page 134)
In Lee's Introduction to Topological Manifold on page 134, it states that the set $\mathbb{B}^2\setminus \{(x,0): x\in[0,1)\}$ is an open 2-cell. However, I'm not quite sure as to how to define a homeomorphism from $\mathbb{B}^2$ onto the set. ($\mathbb{B}^2$ is defined to be the open disc in $\mathbb{R}^2$).

答:
There is probably a much easier way to do it but here we go:

Let $X=\{z\in\mathbb{C}\mid z\notin[0,1]\subset\mathbb{R}\}$. By applying the square-root, i.e. "halfing the angle", we get a biholomorphic map from $X$ to the upper half disk, i.e. $E=\{z\in\mathbb{C}:\left\rvert z\right\rvert<1,\operatorname{Im} z>0\}$. Now look at the Möbius-transform $T:E\to\mathbb{C}$, $T(z)=\frac{z+1}{-z+1}$. If I didn't mess up the calculations this should map $E$ to the first quadrant. Now by squaring we get a biholomorphism with the upper half complex plane $\mathbb{H}$. Now define a similar Möbius-transform that gives the isomorphism between $\mathbb{H}$ and the open disc, e.g. the Cayley-transform.

Composing all those maps we get a biholomorphic map from $X$ to the open unit disc.


问:
show that two rational functions do not have a root in common
Let $$f(x)=8-\binom{4}{0}\frac{1}{x+5}-\binom{4}{1}\frac{1}{x+3}-\binom{4}{2}\frac{1}{x+1}-\binom{4}{3}\frac{1}{x-1}-\binom{4}{4}\frac{1}{x-3}$$ and $$g(x)=8+\binom{4}{0}\frac{1}{x-5}+\binom{4}{1}\frac{1}{x-3}+\binom{4}{2}\frac{1}{x-1}+\binom{4}{3}\frac{1}{x+1}+\binom{4}{4}\frac{1}{x+3}.$$
Show that there are no common roots for $f(x)=0$ and $g(x)=0$.

My approach is to use the intermediate value theorem. By looking at the graph of $f(x)$, we can find $5$ distinct intervals each of which contains exactly one root for $f(x)$. Similarly, we can find them for $g(x)$. Finally, I was able to finish the proof by showing that any two intervals among the $10$ distinct intervals do not overlap.

I wonder if there are any other ways to prove this.

答:
Here's a mechanical way that doesn't require any cleverness, although it does require one tool that you might not have seen yet.

It suffices to show that the polynomials
\begin{align*}
p(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) f(x) \\
&= 8 x^6-16 x^5-264 x^4+480 x^3+1720 x^2-2384 x-1080 \\
q(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) g(x) \\
&= 8 x^6+16 x^5-264 x^4-480 x^3+1720 x^2+2384 x-1080
\end{align*}
have no common roots. One can then compute the resultant of the two polynomials, which turns out to be
$$
\mathop{\rm res}(p,q) = -65{,}355{,}707{,}448{,}508{,}518{,}802{,}391{,}040;
$$
since the resultant is nonzero, the two polynomials have no common roots.


问:
Why $ \lim_{ n \to \infty} \sqrt[n]{n!}= \infty $ and not $1$?
I found on some online sites that

$$  \lim_{ n \to \infty}  \sqrt[n]{n!}= \infty $$

But I did not undertand ow they get it.

My attempt:
$$  \lim_{ n \to \infty}  \sqrt[n]{n!}= \lim_{ n \to \infty}  e^{\ln[ (n!)^{\frac1n}]} = \lim_{ n \to \infty}   e^{\frac1n \ln(n!)} = \lim_{ n \to \infty}   e^{ \frac{\ln(n)}{n} +\frac{\ln(n-1)}{n}+ \cdots } =   e^{ 0 +0+ \cdots }=   e^0 =1 $$

Body:
$n!  > (n/2)^{(n/2) };$

$(n!)^{(1/n)} >(n/2)^{(1/2)};$

Hence?


答:
As mentioned above, each term of the sequence
$$
\frac{\ln n}{n} + \frac{\ln(n-1)}{n} + \cdots + \frac{\ln 1}{n}
$$
tends to zero, but you have $n$ of them. What is the sum of many small things? Well, it depends how many, and how small.

To solve this, you can indeed use Stirling's formula. Without it, a simple way is to see that, for $m_n = \lfloor n/2 \rfloor$
$$
n! = n \times (n-1) \times \cdots \times 1 \geq n \times (n-1) \times \cdots \times m_n \geq m_n^{m_n},
$$
and thus
$$
\frac{\ln (n!)}{n} \geq \frac{\ln(m_n^{m_n})}{n} = \frac{m_n}{n} \ln m_n \to + \infty,
$$
since $m_n/n \to 2$ and $\ln m_n \to + \infty$.

Title: Why $ \lim_{ n \to \infty} \sqrt[n]{n!}= \infty $ and not $1$?

Body:
I found on some online sites that

$$  \lim_{ n \to \infty}  \sqrt[n]{n!}= \infty $$

But I did not undertand ow they get it.

My attempt:
$$  \lim_{ n \to \infty}  \sqrt[n]{n!}= \lim_{ n \to \infty}  e^{\ln[ (n!)^{\frac1n}]} = \lim_{ n \to \infty}   e^{\frac1n \ln(n!)} = \lim_{ n \to \infty}   e^{ \frac{\ln(n)}{n} +\frac{\ln(n-1)}{n}+ \cdots } =   e^{ 0 +0+ \cdots }=   e^0 =1 $$


问:
Is this a correct way of proving that $Y$ is homeomorphic to a one-point compactification of a Hausdorff space $X$?
I need to prove that $Y$ is homeomorphic to a one-point compactification of a Hausdorff space $X$. I want do the following:

Show that $Y$ is compact.

Show that $Y \setminus \{a\}$ is homeomorphic to $X$ for some particular $a \in Y$.

I believe that the uniqueness of one-point compactification of Hausdorff spaces allows me to do this. Is this correct?

答:
No. This is not (completely) correct.

It depends on definition of compactness. Nowadays it is common to NOT include Hausdorff requirement in compactness (while it was so some years ago).

So, if you don't check that Y is Hausdorff, then the facts that Y is compact and that $Y\setminus \{a\}$ is homeomorphic to $X$, are not sufficient to conclude that $Y$ is the one point compactification of $X$ (the one-point compactification is unique in the realm of Hausdorff spaces, not in general).

Here an explicit example:

Let $X=\mathbb R$ with the usual topology (or any other space).
Let $Y=X\cup \{\infty\}$ with the following topology:
$A\subseteq Y$ is open if and only if it is either an open set of $X$ or if $A=Y$.

The space $Y$, with that topology, is compact. Indeed if $\{U_i\}$ is an open cover of $Y$, then there must be an open $U_{i_0}$ containing $\infty$, but the unique open set with that property is $Y$. So the sub-cover consisting of the sole $U_{i_0}$ is a finite sub-cover of $\{U_i\}$. Therefore any open cover has a finite sub cover and therefore $Y$ is compact.

Also, $Y\setminus\{\infty\}$ is clearly homeomorphic to $X$.

But $Y$ is not Hausdorff, so it is not the one-point compactification of $X=\mathbb R$ (which is $S^1$).


问:
What kind of functions can be moment-generating functions for a random variable?
Given an infinitely differentiable function $ g: \mathbb{R} \rightarrow \mathbb{R}$, can we always find a distribution function $f_X$ of some random variable $X$ so that
$g(t) = \int_{-\infty}^\infty e^{tx}f_X(x) dx$?

If my question is too vague or ill-posed, can anyone recommend any literature on the characterization of moment-generating functions?

答:
No, there are a few necessary conditions, I don't know if there's a sufficient one.

If $g(t) = \mathbb E(e^{tX})$ we must have $g(0) = 1$ and $\lim_{t\to -\infty}g(t) = \mathbb P(X=0) \in [0,1]$.

we also have $\mathbb E(X) = f'(0)$ and Var$(X) = f''(0) - f'(0)^2 >0$.
We also have by Jensen's inequality

$$\begin{array}{rl}g(t) &=\mathbb E(e^{tX}) \\&\geq e^{t\mathbb E(X)} \\ &= e^{tf'(0)}\end{array}.$$

So there are lots of probabilistic statements you can make that must be satisfied by anything that's a probability generating function. It's quite easy to find counterexamples.


问:
Prove that $e$ is irrational
Could you please verify whether my attempt is fine or contains logical gaps/errors? Thank you so much!

Lemma: $0<e-\sum_{k=0}^{n} \frac{1}{k !}<\frac{1}{n n !}$ for all $n \in \mathbb N ^+$.

Assume the contrary that $e$ is rational, i.e. $e = p/q$ for some $p,q \in \mathbb Z^+$. Let $x=q!\left(e -\sum_{k=0}^{q} \frac{1}{k !}\right)$. It's easy to verify that $x \in \mathbb Z^+$. On the other hand, $x < q!/(qq!) = 1/q <1$ by Lemma, which contradicts the fact that $1 < e <3$. As such, $e$ is irrational.

Update: As suggested by @lulu, I added a proof of the lemma.

We have $$\begin{aligned}e-\sum_{k=0}^{n} \frac{1}{k !} &= \sum_{k=0}^{\infty} \frac{1}{k !} -\sum_{k=0}^{n} \frac{1}{k !} &&=\sum_{k=n+1}^{\infty} \frac{1}{k !} \\&= \sum_{k=n+1}^\infty \frac{1}{n! \prod_{j=n+1}^k j} &&= \frac{1}{n!} \sum_{k=n+1}^\infty \frac{1}{\prod_{j=n+1}^k j} \\& < \frac{1}{n!} \sum_{k=n+1}^\infty \frac{1}{(n+1)^{k-(n+1)+1}} &&= \frac{1}{n!} \sum_{k=n+1}^\infty  \left (\frac{1}{n+1} \right)^{k-n} \\&= \frac{1}{n!} \frac{\frac{1}{n+1}}{1-\frac{1}{n+1}} && = \frac{1}{nn!} \end{aligned}$$

答:
I am trying to adapt an elementary way .
$$e=1+1+{1 \over 2!}+...+{1\over n!}+...$$ $e-1-1-{1\over 2!}-...-{1\over n!}={1\over (n+1)!}+{1\over (n+2)!}+...>0 \,\forall n\in \mathbb N$.

Let $e$ be rational and $e=\frac pq:p,q\in \mathbb N$. Choose a large $n$ s.t. $n>q$ and consider the number $a=n![e-1-1-{1\over 2!}-...-{1\over n!}]$. Since $q|n!, a\in \mathbb N$. But $a=n![{1\over (n+1)!}+{1\over (n+2)!}+...]={1\over n+1}+{1\over (n+1)(n+2)}+...<{1\over n+1}+{1\over (n+1)^2}+...=\frac{1\over n+1}{1-{1\over n+1}}=\frac 1n$.

Now $a\in \mathbb N \land a<\frac 1n$ (there is no positive integer $<\frac 1n$) is a contradiction.  Hence $e$ is irrational.

Title: Prove that $e$ is irrational


问:
Total number of ways in which 7 different prizes can be given to 5 students such that each student gets at least 1 prize
Find the total number of ways in which 7 different prizes can be given to 5 students such that each student gets at least 1 prize

答:
I'm considering the given prizes as well as the students as distinguishable.

First we make from the given prizes $5$ nonempty packages. Their sizes can be $(3,1,1,1,1)$ or $(2,2,1,1,1)$. In the first case we can form the $3$-pack in ${7\choose3}=35$ ways, in the second case we can form two $2$-packs in ${1\over2}{7\choose2}{5\choose2}=105$ ways.

We now have $5$ different packages that can be dealt  to the students in $5!$ ways. The total number of choices then is
$$N=(35+105)\cdot5!=16\,800\ .$$
(In certain cases $N$ could be smaller: The given prizes could be six balls of  colors blue, red, white, yellow, black, and orange, and a package containing three balls of colors blue, red, white.)


问:
How to measure magnitude of chaotic behavior?
Not necessarily Lyapunov time. In a higher dimensional model, suppose you have many rooms in a house with various heating sources. This is then modeled by a large matrix of coefficients of heat transfer equations, but such a large system of nonlinear equations is prone to chaotic behavior. Realistically though, the temperatures of many rooms don't fluctuate too rapidly after the house attains some kind of equilibrium but retains some due some from the culmination of many turbulent flows.

How do you quantify this notion of "not too chaotic"? Because the manifolds in the solution space could be very turbulent, but their overall magnitude in proportion to the equilibrium temperature may not be that great.

答: This sounds like you should just go for the amplitude. You do not seem to actually care about chaoticity (whatever that may be). It doesn’t matter for your application whether the oscillation is periodic, quasiperiodic or chaotic.</p>


答:
You’re confusing yourself by using more notation than is necessary. Suppose that $y\in f\left[\bigcup\mathscr{A}\right]$; then by definition there is an $x\in\bigcup\mathscr{A}$ such that $y=f(x)$. By the definition of union we know that $x\in A_x$ for some $A_x\in\mathscr{A}$, so $y=f(x)\in f[A_x]\subseteq\bigcup\left\{f[A]:A\in\mathscr{A}\right\}$. Since $y$ was arbitrary, this shows that

$$f\left[\bigcup\mathscr{A}\right]\subseteq\bigcup\left\{f[A]:A\in\mathscr{A}\right\}\;.$$

That’s all you have to say, and it’s far more readable than something cluttered with quantifiers and other formal logical notation.

The argument to show the reverse inclusion is essentially just running the same steps in reverse.

Title: Image of a union of collection of sets as union of the images

Body:
I am having problems with establishing the following basic result. Actually, I found a previous post that is close in nature (it is about inverse image), but I was interested in this specific one, with the following notation, because it is the one I found in the book I am self-studying and I guess that most of my problems are actually related with the notation itself. Morevoer I would like to take it as an opportunity to find out how to write proof with equalities, because I tend in those cases to write them in a cumbersome way, with necessary and sufficient conditions.

[I assume that this is partly due to the influence of "How to prove it: a structured approach", a book that I loved, but that left me the tendency to be quite mechanical in proving any sort of result].

Theorem:

Let $X$ and $Y$ be nonempty sets and $f \in Y^X$. Prove that, for any (nonempty) classes $\mathcal{A} \subseteq 2^X$, we have
$ f(\cup \mathcal{A}) = \cup \{f(A):A \in \mathcal{A} \} $

Here, it's how I approach the problem.

Tentative Proof:

First of all, by definition of (direct) image of a function, we have
$$f(\cup \mathcal{A}) := \{f(x):x\in \cup \mathcal{A} \}. \hspace{1cm} (*)$$
Let $X$, $Y$ be arbitrary (nonempty) sets. Let $f$ be an arbitrary function that maps from $X$ to $Y$ and let $\mathcal{A}$ be a family of sets, subset of the powerset of $X$. By $(*)$, the result we have to prove becomes
$$ \{f(x):x \in \cup \mathcal{A} \} = \cup \{ f(A):A \in \mathcal{A}\} \hspace{1cm} (1)$$
In order to prove it, rephrase $(1)$ as
$$ \forall y ( y \in \{f(x):x \in \cup \mathcal{A} \} \leftrightarrow y \in \cup \{ f(A):A \in \mathcal{A}\}). \hspace{1cm} (2)$$
We start by proving the necessary condition. Let $y$ be an arbitrary element and assume that $y$ is a member of $\{f(x):x \in \cup \mathcal{A} \}$. This means that $\exists x(x \in \cup \mathcal{A} \land y=x)$. At the same time, we have to prove that $\exists A(A \in \mathcal{A} \land y\in f(A))$.

And here I got stuck...

I really don't see how from my premises I can get the desired result. Indeed, what $A$ should be?

I assume there is a problem with the way in which I translate the problem in logical terms.


问:
We have a germ that contains finite, infinite, oscillating and infinitesimal parts. How can we separate them?
Suppose a germ of a function at a point (from one direction) or at infinity has finite, infinite, oscillating and infinitesimal parts.

Is it possible to separate it into all four, so that

The infinite part was monotonously growing (by absolute value and without changing sign), with all its derivatives also were growing monotonously.

The finite part was a constant equal to the regularized value of the integral of the derivative of the function at the point.

The oscillating part was neither growing, nor decreasing by absolute value, but have no limit and changing sign, also having convergent Fourier series.

The infinitesimal part heaving limit $0$?

Is there an algorithm?

答:
This is not an anwser, but the OP might find it interesting.

I don't think there is an algorithmic way to do this. For now I am not even sure that such a decomposition exists even for analytic functions. However this is difficult to tell because I don't understand the third category (I take the second one to be "constant function"), because for me, Fourier series regard periodic functions. But perhaps you meant convergent (or existing) Fourier transform?

But if one is willing to consider only certain types of functions as explained below, one may have a simpler (but more restrictive) decomposition.

Let $\mathcal{R}$ denote the set of germs $[f]$ at $+\infty$ of real-valued partial functions $f$ such that for all $n \in \mathbb{N}$, there is $a_n \in \mathbb{R}$ such that $f$ is defined and $n$-times differentiable on $(a_n,+\infty)$. Under pointwise sum and product, this is a ring. Under pointwise derivation, it is a differential ring. It is equipped with the partial order $[f]<[g]$ iff $f(x)<g(x)$ for sufficiently large $x \in \mathbb{R}$.

A Hardy field is a differential subfield $\mathcal{H}$ of $\mathcal{R}$ which contains the set of germs of constant functions. The structure $(\mathcal{H},<)$ is then an ordered field. So in $\mathcal{H}$, any element is either in category $1$, or a sum of one unique element of category $2$ and one unique element of category $4$.

Hardy fields include the fields generated by real constants and the identity function, the exponential function, the logarithm, all functions on neighborhoods of $+\infty$ which are first order definable in the language of exponential fields, or in any o-minimal structure on $\mathbb{R}$. It is conjectured that every Hardy field embeds into a Hardy field where differential polynomials have the intermediate value property. So one can get quite big while satisfying an even stronger sort of separation than that you propose.

By a theorem of Aschenbrenner, van den Dries and van der Hoeven, every Hardy field embeds as a differential field in a field of generalized transseries (surreal numbers). Every such transseries $f$ can be written as $f=f_{\succ}+f_=+f_{\prec}$ where $f_{\succ},f_=$ and $f_{\prec}$ correspond to categories $1$, $2$ and $4$ respectively. The difference here with the general case of Hardy fields is that the function $f\mapsto f_{\succ}$ is an additive morphism (so are the other two).

However the embedding is not unique and not at all constructive in general. In fact, assuming a subfield $\mathcal{H}'$ of $\mathcal{H}$ is embedded as a differential field of transseries, and $\varphi \in \mathcal{H} \setminus \mathcal{H}'$ is d-transcendant over $\mathcal{H}'$, meaning that any differential equation $g_n {\varphi}^{(n)}+...+g_1 \varphi'+ g_0 \varphi=0$ where $g_0,...,g_n \in \mathcal{H}'$ must satisfy $g_0=...=g_n=0$, then there are many transseries $f$ that $\varphi$ could be identified with, and in some cases one can arrange that $f_=$ for instance be any real number.


问:
Expanding a quadratic term
In a derivation involving the normal distribution, I saw this used:

$$ \sum (X_i - \mu)^2 = \sum(X_i - \bar{X} + \bar{X} - \mu)^2 = \sum(X_i - \bar{X})^2 + \sum(\bar{X} - \mu)^2$$.

I don't follow the last step. Is this a general trick/formula that one can use, with any value in the place of $\bar{X}$?

答:
Its same as
$(A+B)^2=A^2+B^2+2AB$

So there is an additional term but it vanishes.

$$\begin{align*}\sum_i (X_i - \mu)^2 &= \sum_i(X_i - \bar{X} + \bar{X} - \mu)^2 \\&= \sum_i(X_i - \bar{X})^2 + \sum_i(\bar{X} - \mu)^2+2(\bar X - \mu )\sum_i(X_i - \bar{X})\end{align*}$$

The last term vanishes because of which property?

 \begin{align} \text{The sum of deviations of the items from their arithmetic mean is always zero} \end{align}

Title: Expanding a quadratic term


问:
Finding the area of a circle with changing radius?
Hi guys I need some help with a math question I have. It is about a tethered dog and calculating the accessible area of the dog on a given leash. Let us say the pole has dimensions radius=0.1 meters and lets say the leash is legnth=2 meters. The leash of the dog is tied at the very top of the pole (facing north) and it is free to walk around on its leash. You would see that the rope would interact with the rope as the dog walks around the pole in such a way that the leash gets shorter which changes the radius of accessible circle of the dog. How would I go about solving this accessible area for the dog? I know its a circle but the radius (leash) of the circle gets shorter as he walks around the pole as the leash gets tied around it. If you have further questions, always ask me! I would really appreciate some help!!! Thanks everyone!!!  - Skillzz9</p>

答:
$\underline{\mathrm{Introduction}}$

$\mathrm{Fig.\space 0}$ shows the general scenario of tethering a dog to a pole of diameter $2r$ using a rope of length $s$. However, in this answer, we consider a special case of this scenario, in which the rope lies horizontal (i.e. $b = h$), whenever the dog pulls it taut. The rope is tied to the pole at $D$. Dog’s total accessible area is designated as $A$. It is easy to comprehend that the part of this area that lies on the left hand side of the line $DF$ has the shape of a semicircle of radius $s$, which has its center at $D$. Please note that this area is not shown in our diagrams. We denote it as $A_0$, which can be expressed as
$$A_0 = \frac{1}{2}\pi s^2.$$

The rest of the accessible area, which is equal to $A-A_0$, takes the shape of an involute and lies on the right hand side of the line $DF$. When we set on to find a method to calculate this area, we realized that we need to differentiate between two different scenarios. One of them can be called short-rope scenario and the other long-rope scenario. To give the words “short” and “long” mathematically precise meanings, it is necessary to define a dimensionless parameter $\lambda$ as
$$\lambda = \frac{s}{r}, \mathrm{\quad where\quad \infty \ge \lambda \gt 0}.$$

For short-rope cases, we have $\space\pi \ge \lambda \gt 0\space$ and, for long-rope cases, $\space\infty \ge \lambda \ge \pi.\space$ Note that the case $\lambda = \pi$ can be considered as either a short or a long-rope case.

$\underline{Note}$: Values of all angles mentioned in this answer are given in radians. Furthermore, we assume that the reader has gone through the answer given at the link mentioned by @Travis in his comment.

$\mathrm{\underline{Case\space of\space Short\space Rope} \space\left(\pi \ge \lambda \gt 0\right)}$

$\mathrm{Fig.\space 1\space to\space 3}$ show three short-rope cases. The solution of the case depicted in $\mathrm{Fig.\space 3}$ is described in the answer given at the mentioned link.
As shown in our diagrams, we denote the end point of the involute as $E$ and it is the decisive factor to the size of the sought area. For all short-rope cases, this point lies on the circumference of the circle depicting the pole.
We denote half of the area that lies on the right hand side of the line $DF$ as $A_1$. As shown in $\mathrm{Fig.\space 2}$, we need to know the angle $\phi$ in order to determine $A_1$. Since the arc $DE$ is equal to the length of the rope, we shall write,
$$\phi = \frac{s}{r} = \lambda.\tag{1}$$

The area $dA$ of the honeycombed sector-like region $PMNQ$ (see $\mathrm{Fig.\space 3}$) can be expressed after  approximating it with an isosceles triangle.
$$dA=PMNQ = \frac{1}{2} \left(s-r\theta\right)^2 d\theta$$

When we integrate this, we have,
$$A_1 = \int^\phi_0dA =   \int^\phi_0 \frac{1}{2} \left(s-r\theta\right)^2 d\theta=\frac{1}{6}\phi\left[3s\left(s-r\phi\right)+r^2\phi^2\right], \tag{2}$$

which, can be simplified by replacing $\phi$ using (1) to obtain
$$A_1 = \frac{1}{6}\frac{s^3}{r}.$$

$$\therefore\quad A=A_0 + 2A_1 = \frac{1}{2}\pi s^2 + \frac{1}{3}\frac{s^3}{r} = \frac{1}{6}\frac{s^2}{r}\left(3\pi r+2s\right).\tag{3}$$

For example, if $\lambda = 2$, the dog has the freedom to traverse an area equal to $\space 2.237463 s^2\space$ or $\space 8.94985 r^2$. If the reader need self-convincing, then try his/her hand at solving the case of $\lambda=\pi$, the answer of which is $\frac{5}{6}\pi^3 r^2$.

$\mathrm{\underline{Case\space of\space Long\space Rope} \space\left(\infty \ge \lambda \ge \pi\right)}$

Several long-rope scenarios are portrayed in $\mathrm{Fig.\space 4\space to\space 6}$. As shown in the diagrams, we define a special point $E$ as the first point of intersection between the involute and the extended $DO$.

There is a special case (of which no sketch is given) with $\lambda = \infty$ (i.e. $r=0$ and $s\gt 0$), in which the involute takes the shape of a semicircle of radius $s$ with its center at $D$. In this instance, the point $E$ is to be found exactly a distance $s$ from $D$ (i.e. $DE=s$). For all the other long-rope cases, we have $DE \lt s$. Because of this, as shown in $\mathrm{Fig.\space 4}$, the part of each involute lying beyond $E$ (see the hatched area) do not contribute to expand the accessible zone already demarcated by the parts of the two involutes up to $E$.

Now, pay your attention to $\mathrm{Fig.\space 7}$. Here too, we are looking for an angle $\phi$. However, this has nothing to do with the angle $\phi$ that came up in short-rope scenarios. Its size depends on the position of the point $E$. You can obtain this angle by first constructing the tangent from point $E$ to the circle depicting the pole. Then, draw the radius of this circle to pass through the tangential point. The angle between this radius and the line $OD$ is equal to $\phi$, which can have values only in the range
$$\space\pi\ge\phi\gt\frac{\pi}{2}. \tag{4}$$

As shown below, we can derive an expression for $\phi$ by considering the right-angled triangle $EPO$.
$$s-r\phi=r\tan\left(\pi-\phi\right)=-r\tan\left(\phi\right)\quad\rightarrow\quad\frac{s}{r}-\phi+\tan\left(\phi\right)=0 \quad\rightarrow\quad\lambda-\phi+ \tan\left(\phi\right)=0$$

Since this is a nonlinear equation, one has to resort to numerical methods, such as Newton-Raphson method, to obtain the value of $\phi$ within the range specified by (4). An iteration formula based on the said method is given below.
$$\phi_{n+1}=\phi_n-\frac{\lambda-\phi_n}{\tan^2\left(\phi_n\right)}- \frac{1}{\tan\left(\phi_n\right)}, \quad \mathrm{with}\quad \phi_0=1.57079633 \tag{5}$$

To find the area bounded by the involute from $F$ to $E$, the circumference of the pole, and the line segment $BE$, we have to determine areas of two adjacent regions denoted as $A_1$ and $A_2$ separately. As shown below, equation (2) can be used to find the area $A_1$.
$$A_1 = \frac{1}{6}\phi\Bigl[3s\left(s-r\phi\right)+r^2\phi^2\Bigr]$$

As illustrated in $\mathrm{Fig.\space 8}$, the area of $A_2$ can be calculated using a similar method. First, an expression for the area $dA$ of the honeycombed sector-like region $HMNG$ is derived after approximating it with an isosceles triangle. Then, we integrate that expression to obtain $A_2$ as,

$$A_2=\frac{1}{2}\int^{\pi-\phi}_0 r^2\tan^2\left(\pi-\phi-\omega\right)d\omega=\frac{1}{2}r^2\Bigl[\phi-\tan\left(\phi\right) -\pi\Bigr].$$

$$\therefore A=A_0+2A_1+2A_2=\frac{1}{2}\pi s^2  +\frac{1}{3}\phi\Bigl[3s\left(s-r\phi\right)+r^2\phi^2\Bigr] + r^2\Bigl[\phi-\tan\left(\phi\right) -\pi\Bigr]\tag{6}$$

For example, if $\lambda = 4$, the angle $\phi$ is equal to $2.04325\space rad$. As a consequence, the dog has the freedom to traverse an area equal to $\space 2.80169 s^2\space$ or $\space 44.82708 r^2$.

It is up to the reader to check the formula (6) by applying it to the two cases $\lambda=\pi$ and $\lambda=\infty$. The latter case, where $r=0$, is bit tricky and, if the reader is able to show that the accessible area is equal to $\pi s^2$, then he/she has mastered the method.

$\mathrm{\underline{Additional\space of\space Information}}$

It is possible to express the total accessible area as $A=\mu s^2$, where $\mu$ is a function of $\lambda$. $\mathrm{Fig.\space 9}$ shows how $\mu$ varies with $\lambda$ for a given length of a rope. It is easy to see that, as $\lambda$ goes from $0$ to $\infty$, the total accessible area $A$ monotonically increases from $\frac{1}{2}\pi s^2$ to $\pi s^2$.


问:
Is "Convolution operator" well-defined and injective?
Problem: Let $g_t(x)=\frac{1}{\sqrt{2\pi}} e^{ \frac{- (x-t)^2}{2}}$ be the Gaussian function centered at $t\in \mathbb R$. Let $I$ be a subset of $\mathbb R$. Consider the "convolution operator"
\begin{align}
\mathcal A: L^1(I) & \longrightarrow L^2(I)\\
f & \longmapsto \left(x\longmapsto  \int_{I} g_t(x)f(t) dt \right).
\end{align}
How can I prove (or disprove) that the operator is well-defined and injective?

My attempts: Here are some of my attempts to prove the well-defined property.

I have tried to use Cauchy inequality but with no success:
\begin{align}
\int_{I} \left( \int_{I} g_t(x)f(t)dt\right)^2 dx \leq \int_{I} \int_{I} g_t^2(x)dx dt \int_{I}f^2(t)dt.
\end{align}
as the right-hand-side may be infinite as $f$ does not belong to $L^2(I)$.

I have tried to use the boundedness of Gaussian function but nothing better. We have
\begin{align}
\int_{I} \left( \int_{I} g_t(x)f(t)dt\right)^2 dx \leq
M \left( \int_{I} f(t)dt \right)^2 \int_I dx.
\end{align}
and the right-hand-side may be infinite as $I$ is unbounded.

答:
\begin{align*}
\left|\int_I \left(\int_I e^{-(x-t)^2}f(t)dt\right)^2dx\right| &= \left|\int_I \int_I \int_I e^{-(x-t)^2}f(t)e^{-(x-s)^2}f(s)dsdtdx\right| \\ &= \left|\int_I \int_I f(t)f(s)\left(\int_I e^{-(x-t)^2}e^{-(x-s)^2}dx\right)dsdt\right| \\ &\le \int_I\int_I |f(t)|\hspace{1mm}|f(s)| \left(\int_\mathbb{R} e^{-(x-t)^2}e^{-(x-s)^2}dx\right)dsdt \\ &\le C\int_I \int_I |f(t)|\hspace{1mm} |f(s)|dsdt \\ &= C\left(\int_I |f(t)|dt\right)^2
\end{align*}