问:
Derangement, gift exchange
Five people have each bought their own Christmas gift. They put the gifts in a sack which then everyone can pull their gift from. In how many ways can the gifts be distributed so that no one gets the Christmas gift they have bought themselves?

So I have figured $93$ combinations in which only $1$ and then $2, 3$ or $4$ and $5$ get their gift but I haven't figured out how to calculate that other people don't get their gift. For example

If $2$ people get their gifts how do I count that the other $3$ don't get their own gifts etc? I have read about derangement and seen different formulas but I haven't quite understood the logic.

答:
You can use the concept of derangement. It is derived from Principle of exclusion and inclusion, For example if there are two tasks- $A$ and $B$, let $A$: Getting right gift for person $1$ and $B$ be getting right gift for person $2$. Then you want:
$U-(A \cup B)$

Which can also be expressed as total ways of $U - (A + B - A \cap B)$ i.e. total ways - number of ways in which Event $A$ happens- Event $B$ + Ways where $A$ and $B$ occur simultaneously. This whole process for a general scenario becomes $n!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\cdots+ \frac{(-1)^n}{n!}\right)$.

Title: Derangement, gift exchange


问:
Pointwise Convergence In Integral
This question's my profesors. I know $x^n$ is a pointwise convergent sequence. But I don't know how can I show that this limit. Can you show me solution of this question.

$f:[0, 1]\rightarrow \mathbb{R}$ continuous function.

$$\lim_{n\rightarrow \infty }\int_0^{1}x^nf(x) \, dx$$

Find the limit.

答:
Since $f$ is continuous, it is bounded on $[0,1]$, say by $M >0$. Then,

$$\left\lvert \int_0^1 x^n f(x) \,dx \right\rvert \leq \int_0^1 x^n |f(x)| \,dx \leq M \int_0^1 x^n \,dx = \frac{M}{n+1} \to 0$$

as $n \to \infty$.

ContentLicense: CC BY-SA 4.0

Title: Pointwise Convergence In Integral


问:
How can we show that there is a zero of $f$
Let $$f(x)= x^2+2020x+1$$ a polynomial.
We define $F_n(x)$ to be $f(f(f...f(x)))$ applied $n$ times .
Show that for all positive integer $n\ge1$ there exists a real $x$ such that $F_n(x)=0$

Please help me, I've tried induction but it doesn't seem to work. $n=1$ is trivial but then it's hard. I don't know much about polynomials. Any ideas?

答:
First, note that for each $n$, $F_n(x)$ is a polynomial of degree $2n$ (with leading coefficient $1$). This is easy to show by induction as

$$F_1(x)=f(x)=x^2+2020x+1$$

Then if $r(x)$ is the remaining polynomial such that $F_{n-1}(x)=x^{2(n-1)}+r(x)$ (note then that the degree of $r(x)$ is less than $2(n-1)$) we have

$$F_n(x)=f(F_{n-1}(x))=(x^{2(n-1)}+r(x))^2+2020 (x^{2(n-1)}+r(x))+1$$

$$=x^{2n}+2x^{2(n-1)}r(x)+r(x)^2+2020x^{2(n-1)}+2020r(x)+1$$

Since this clearly has degree $2n$ the proposition is proved. This implies that

$$\lim_{x\to\infty}F_n(x)=\infty$$

or that there exists $X\in\mathbb{R}$ such that $x\geq X$ implies $F_n(x)\geq 1$. Of course, this implies that $F_n(X)=1$. Now, let $x_0$ be the lesser fixed point of $f(x)$. That is

$$f(x_0)=x_0$$

$$x_0^2+2020x_0+1=x_0$$

$$\Rightarrow x_0=\frac{-2019\pm\sqrt{4076357}}{2}$$

Since we want the lesser fixed point, we may conclude that

$$x_0=\frac{-2019-\sqrt{4076357}}{2}$$

But for this $x_0$ we have

$$F_1(x_0)=f(x_0)=x_0$$

$$F_2(x_0)=f(F_1(x_0))=f(x_0)=x_0$$

$$\vdots$$

$$F_n(x_0)=x_0=\frac{-2019-\sqrt{4076357}}{2}<0$$

Thus, $F_n(x_0)<0$ and $F_n(X)=1>0$. By the mean value theorem, there exists some $x\in (x_0,X)$ such that $F_n(x)=0$.


答:
the logic behind the formula for number of derangement is based on recurrence relations.
assume Dn is the number of derangements for n objects.
Dn=En+Fn where En= number of derangements in which the nth object swaps with one of the other n-1 objects and Fn= number of derangements in which the nth object doesn't swap with other objects.
for example on 4 elements: 2 1 4 3 in this case the fourth element has swapped with the 3rd one hence it is counted as En. but in 4 3 2 1 a swap has not happened and it is counted as Fn. now it is time to calculate En and Fn. for En we have n-1 choices for the object we want to swap with the nth object and after we have chosen that, n-2 objects will be left which can freely have their own derangement,
therefore: En=(n-1).Dn-2
now for Fn: since n doesn't swap with anything, some r goes to the nth place and there are obviously n-1 choices for r, now you are left with n-1 objects than can have their own derangement, so: Fn=n-1.Dn-2
and therefore, Dn=(n-1).(Dn-2 + Dn-1)
what it left now is finding the closed form you have probably seen somewhere based on this recurrence relation.
by doing some operations on the relation you can simply get
Dn - n.Dn-1 = -(Dn-1 - (n-1).Dn-2)
a crucial point you have to notice in here is that the expression on the left is the negative of the expression on the right (check for yourself) and by iterating on that we will get Dn - n.Dn-1 = (-1)^n.
divide all terms by n! :

source: a first course in discrete mathematics, Ian Anderson

Title: Derangement, gift exchange

Body:
Five people have each bought their own Christmas gift. They put the gifts in a sack which then everyone can pull their gift from. In how many ways can the gifts be distributed so that no one gets the Christmas gift they have bought themselves?

So I have figured $93$ combinations in which only $1$ and then $2, 3$ or $4$ and $5$ get their gift but I haven't figured out how to calculate that other people don't get their gift. For example

If $2$ people get their gifts how do I count that the other $3$ don't get their own gifts etc? I have read about derangement and seen different formulas but I haven't quite understood the logic.


问:
Why velocity and accelleration must be linearly independent?
I am struggling to understand the definition of biregular curve.

Definition

A curve $I \ni t \mapsto P(t) \in \mathbb{R}^3$ is said to be biregular when the velocity and accelleration vectors are linearly independent, that is when their cross product is different from zero $$dP/dt \land d^2P/d^2t \neq 0$$

Q.

I read that this is equivalent to the fact that $$d \underline{t}/ds \neq 0$$ and this is what I am having trouble understanding. Why is so?

Recall that $$s= \int_0^t \|dP/dt\| \, dt,$$ is the curvilinean coordinate, and $$\underline{t}(s)=dP/ds$$ so that $$dP/dt= dP/ds \  \dot{s}= \underline{t} \ \dot{s}$$ where $$\dot{s}=ds/dt.$$

Moreover $$d^2P/d^2t= d\underline{t}/ds \ \dot{s}^2 + \underline{t} \ \ddot{s}.$$ Hence we have

$$dP/dt \land d^2P/d^2t = \underline{t} \ \dot{s} \land (d\underline{t}/ds \ \dot{s}^2 + \underline{t} \ \ddot{s})$$$$ = \dot{s}^3(\underline{t}\land d\underline{t}/ds)+\ddot{s}(\underline{t}\land \underline{t})=\dot{s}^3(\underline{t}\land d\underline{t}/ds).$$

So, the curve is biregular $\iff$ $\dot{s}=\|dP/dt\|\neq 0$ and $\underline{t}\land d\underline{t}/ds \neq 0.$

How is this equivalent to the fact that $d \underline{t}/ds \neq 0$?

Why is it not possible that $d\underline{t}/ds$ and $\underline{t}$ are both different from zero and yet linearly dependent?

答:
I can't bring myself to use little letters for these vectors when we already have the time parameter $t$.

Because $\|T(s)\| = 1$ for all $s$, it follows that $dT/ds$ is orthogonal to $T$, so they will always be linearly independent provided $dT/ds\ne 0$.


问:
Why is it important to have a discrepancy between image and codomain?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.

What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?

I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?

答:
The codomain tells us what possible values a function may take and the image tells us what actual values it takes. We can have many functions to the same codomain and we may distinguish them by their images.

For example, the codomain may represent the total set of names for all objects in some limited context while an individual map may be an assignment of names to a set of particular objects.

ContentLicense: CC BY-SA 3.0


问:
Multiset identity.
How does one get to the following combinatorial identity: $$\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{i}=\binom{n+k-1}{k}$$

I'm well aware of the definition of a multiset, as well as of the derivation of cardinalities of such. The closest one I came across is a special case of the Chu-Vandermonde identity, namely for any integers $i,k$ and $n$ satisfying $0\leq i\leq k \leq n$ the following holds true $$\sum_{m=0}^n\binom{m}{i}\binom{n-m}{k-i}=\binom{n+1}{k+1}$$

Any help is appreciated. Thanks in advance.

答:
The left-hand side can be rewritten as $\sum_{i=0}^{k-1} \binom{n}{i+1}\binom{k-1}{k-1-i}$.

If the sum were $\sum_{i=-1}^{k-1}$, then  applying the Vandermonde identity would yield $\binom{n+k-1}{k}$. Is there a typo?

Title: Multiset identity.


问:
What’s a good (fairly advanced) book on probability, cumulants and CLT please?
Can anyone recommend a book covering probability, cumulants and a proof of the CLT (ideally including a proof of Levy’s continuity theorem).

I studied Math undergrad and PhD many years ago, but don’t work in the field.

I’ve read Yuval Filmus‘s note on the CLT and the Wiki pages.  Cumulants make intuitive sense, but e.g. I can’t see a proof of Levy’s continuity theorem online.

As a bonus, if there’s any version of the CLT for correlated random variable size , that would be interesting.

Thanks

答: Thanks all.  In the end I went for Probability with Martingales by David Williams.  Seems like tough going but hopefully will be worthwhile in the end.</p>


问:
Finding a portion of the volume of a 3D object
I was learning about volumes of revolution and I have a question. Here is a curve, $y=5^\left(x-2\right)-\frac{1}{25}$, which is bounded by, $y=5$ and $x=0$, graph.

Finding the volume of the shape as it's rotated about the y-axis. First, I solved for x, because I'm rotating about the y-axis. And thus, $\textrm{radius}=\log_5\left(y+\frac{1}{25}\right)+2$, we know that the area of a given cross-section at any height of the object y is, $\pi\left(\log_5\left(y+\frac{1}{25}\right)+2\right)^2$. The total volume of the solid is, $\int^5_0{\pi\left(\log_5\left(y+\frac{1}{25}\right)+2\right)^2}\delta y$, which yield's, $\approx95.976$. If this is the total volume of the solid, how could I find a specific portion of the volume? If I wanted to know at exactly what height half of the volume lies I could use, $\int^x_0{\pi\left(\log_5\left(y+\frac{1}{25}\right)+2\right)^2}\delta y=\frac{\textrm{Total Volume}}{a}$, where in this case a would equal 2, (for half of the total volume), and x would equal the height at which the solid would need to be cut in-order for the volume from 0 to x would equal half of the total volume.

How could I solve for x? I'm sure there is a way to do it with the integral there, or I'm making an assumption that there must be a way to get that height after doing some operation with the original 2D curve.

I tried guessing and checking the height to, $\approx3.15193$, with that height for x the total volume is, $\approx47.98819$, so I assume there must be a way to calculate the value of x without guessing and checking or iteration (sums are fine).

An explanation including every step would be nice.

Body:
$$V=\pi  \int_0^5 \left(\log _5\left(y+\frac{1}{25}\right)+2\right){}^2 \, dy=^*\frac{2 \pi  \left(125+63 \log ^2(126)-126 \log (126)\right)}{25 \log ^2(5)}\approx 95.9764$$
Therefore to know where is the $y^*$ which divides the solid in half I set (RHS is half $V$):
$$\pi  \int_0^t \left(\log _5\left(y+\frac{1}{25}\right)+2\right){}^2 \, dy=\frac{\pi  \left(125+63 \log ^2(126)-126 \log (126)\right)}{25 \log ^2(5)}$$
which gives the non-algebraic equation
$$\frac{\pi  \left(2 t+\frac{1}{25} (25 t+1) (\log (25 t+1)-2) \log (25 t+1)\right)}{\log ^2(5)}=\frac{\pi  \left(125+63 \log ^2(126)-126 \log (126)\right)}{25 \log ^2(5)}$$
Setting $25t+1=z$ and simplifying
$$2 z+z \log ^2(z)-2 z \log (z)=127+63 \log ^2(126)-126 \log (126)$$
Mathematica gives
$$z\approx 79.7981551 \to t\approx 3.1519262$$

$^*$

$$ \int \left(\log _5\left(y+\frac{1}{25}\right)+2\right){}^2 \, dy=\int \left(\frac{\log \left(y+\frac{1}{25}\right)}{\log (5)}+2\right)^2 \, dy=$$
$$=\frac{1}{\log ^2(5)}\int \left(\log ^2\left(y+\frac{1}{25}\right)+4 \log (5) \log \left(y+\frac{1}{25}\right)+4 \log ^2(5)\right)\,dy=$$
$$F(y)=\frac{1}{25 \log ^2(5)} \left(50 y+(25 y+1) \log ^2(25 y+1)-2 (25 y+1) \log (25 y+1)\right)+C$$
Then $V=F(5)-F(0)$


答:
We have two bags. First bag has $n$ balls, second bag has $k-1$ balls.

We pick $i+1$ balls from first bag and reject $i$ balls from second bag. So we are drawing $i+1+(k-1-i)=k$ balls in total for each $i$.

Number of ways is $$\binom{n+k-1}{k}$$

Title: Multiset identity.

Body:
How does one get to the following combinatorial identity: $$\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{i}=\binom{n+k-1}{k}$$

I'm well aware of the definition of a multiset, as well as of the derivation of cardinalities of such. The closest one I came across is a special case of the Chu-Vandermonde identity, namely for any integers $i,k$ and $n$ satisfying $0\leq i\leq k \leq n$ the following holds true $$\sum_{m=0}^n\binom{m}{i}\binom{n-m}{k-i}=\binom{n+1}{k+1}$$

Any help is appreciated. Thanks in advance.


问:
Understanding integration by substitution
The integration by substitution technique is dervied from the following statement:
$$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,du$$
Now almost all the explanations I saw about the actual technique start playing around with the Leibniz notation and $dx$, $du$ to get to the new form, e.g stating $du=\varphi '(x)\,dx$ from ${\frac {du}{dx}}=\varphi '(x)$.

Now the problem is that $dx$ and $du$ are said to have no actual meaning, at least in elementary calculus, and yet we're playing around with them as if they were actual variables forming actual quotients. As far as I am aware this wouldn't be a problem in nonstandard analysis, but how should one make sense of it in the elementary settings? How can we apply this technique without "heuristic" methods (which actually seems to be used almost all the time for different techniques)?

Note: My problem is actually applying the method, not the proof, as there are many proofs available not using the Leibniz's notation.

答:
It's quite simple once the notion of differential of a function has been defined. For a differentiable function $\varphi$; its differential $\mathrm d\varphi$ at a point $x$ is its best linear approximation at $x$, which is shown to be the map of the variable $h$:
$\enspace h\longmapsto f'(x)h$. All linear maps ar clearly their own differentials.

In particular, the differential of the identity map is itself, since it's a linear map, and, as by abuse of language, this identity map is often denoted $x$, its differential is denoted $\mathrm dx$.

In this context, if in the integral $\;\int_a^b f\bigl(\varphi(x)\bigr)\varphi'(x)\,\mathrm dx$, you set $u=\varphi (x)$, its differential is precisely $\mathrm du=\varphi'(x)\,\mathrm dx$,  and the corresponding bounds for $u$ are clearly $\varphi(a)$ and $\varphi(b)$, so that the integral can quite naturally be written as
$$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,\mathrm dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,\mathrm du.$$


问:
A question about linear bounded functionals

Let $X$ be a banach space. $(\phi_n)_n \subset X^*$ a sequence of linear, bounded functionals that satisfies for all $x\in X$, $\sum_n \phi_n(x)$ converges.
Show that $\sum_n \frac{||\phi_n||}{n^2}$ converges.

I am trying to see this using the uniform bounded theorem.

$\sum_n \phi_n(x)$ converges, then the sequence $T_k(\phi_n)=(\sum_{n=1}^{k} \phi_n(x))_{k\geq 1}$ converges so $T_k$ is bounded.And $sup T (\phi_n(x)) < \infty$.Or we can say that there is $N$ such that for every for every $n>N$ $|\phi_n(x)| \leq 1$ so $||\phi_n||=sup {|\phi_n(x)|  , x\in X , \in N} < \infty$ .Thus by the uniform bounded theorem $sup ||\phi_n|| <\infty$.
But how to finish it.

答:
Since for all $x \in X$, the series $\sum_n \phi_n(x)$ converges, we necessarily have $\sup_n |\phi_n(x)| < \infty$ for all $x \in X$. By the uniform boundedness principle, it follows that $M:=\sup_n \|\phi_n \| < \infty$. Consequently
$$\sum_n \frac{\|\phi_n \|}{n^2} \leq M \sum_n \frac{1}{n^2} < \infty$$
as desired.


问:
What qualities of an algebraic torus make it like a torus?
Why do we call it an algebraic torus?

What qualities of an algebraic torus make it like a torus?

D_S gives the following classical definition of an algeraic torus here Formal Definition of an Algebraic Torus:

"Clasical definition: Let $H \subseteq \operatorname{GL}_n(\overline{k})$ be a connected linear algebraic group over $k$.  So $H$ is a Zariski closed subgroup of $\operatorname{GL}_n(\overline{k})$ which is irreducible, and the polynomials of the radical ideal $I(H)$ corresponding to $H$ can be chosen to have coefficients in $k$, rather than just $\overline{k}$.   We say that $H$ is an algebraic torus if there exists a $g \in \operatorname{GL}_n(\overline{k})$ such that $gHg^{-1}$ consists of diagonal matrices."

答:
This was a comment on Qiaochu’s answer, but it got too long so I thought I’d made it into an answer.

As Qiaochu said, one can define tori in several ways, but from a representation theoretic standpoint perhaps the most natural definition is the following:

Definition: A compact torus is a compact Lie group $\mathcal{T}$ isomorphic to $(S^1)^n$ for some $n$.

The real usage of such objects is in the theory of compact Lie groups where for a compact Lie group $\mathcal{G}$ one has a unique conjugacy class of maximal compact tori (i.e. there exists a compact torus $\mathcal{T}\subseteq\mathcal{G}$ which is maximal amongst compact tori contained in $\mathcal{G}$ and that such a maximal compact torus is unique up to conjugacy). The reason they are so useful is that representation theory of $\mathcal{G}$ is encapsulated entirely in terms of $\mathcal{T}$.

To make this precise, it’s useful to note denote by $\mathrm{Rep}(\mathcal{G})$ the representation ring of a compact Lie group $\mathcal{G}$. We then have the following:

Theorem (Highest weight theory, summarized): Let $\mathcal{G}$ be a connected compact Lie group and $\mathcal{T}$ a maximal torus. Then, the restriction map map
$$\mathrm{Rep}(\mathcal{G})\to \mathrm{Rep}(\mathcal{T})$$
is an isomorphism of $\mathrm{Rep}(\mathcal{G})$ onto the subring $\mathrm{Rep}(\mathcal{T})^W$ of Weyl group invariant elements of $\mathrm{Rep}(\mathcal{T})$.

A very similar thing happens for tori in a reductive algebraic group over $k$ an algebraically closed field (one can extend this to the non-algebraically closed situation, but it’s not worth the effort here).

Definition: Let $k$ be a field. Then, an algebraic torus over $k$ is an algebraic group $T$ over $k$ such that $T_{\overline{k}}$ is isomorphic to $\mathbb{G}_{m,\overline{k}}^n$ for some $n$.

Again, it’s true that every (connected) reductive group $G$ over a field $k$ (reductive means, at least in characteristic $0$, that its category of algebraic representations is semisimple) has a maximal torus (defined in the same way) which is unique up to conjugacy if $k$ is algebraically closed. One also has a highest weight theory:

Theorem (Highest weight theory, summarized, [2, §22.b]): Let $k$ be an algebraically closed field and $G$ a (connected) reductive group over $k$. Then,
the restriction map

$$\mathrm{Rep}(G)\to\mathrm{Rep}(T)$$

is an isomorphism of $\mathrm{Rep}(G)$ onto the subring
$\mathrm{Rep}(T)^W$ of Weyl group invariant elements of
$\mathrm{Rep}(T)$.

Thus, one sees that there is a strong analogy between compact tori and algebraic tori from the perspective of representation theory. But, in fact, one can make the connection more literal.

Namely, we have the following beautful theorem:

Theorem (Chevalley—Tanaka, [1, Theorem D.2.4]): The association $G\mapsto G(\mathbb{R})$ is an equivalence of categories
$$\left\{\begin{matrix}\text{Connected reductive anisotropic groups}\\ \text{over }\mathbb{R}\end{matrix}\right\}\to\left\{\begin{matrix}\text{Connected compact Lie groups}\\ \text{over }\mathbb{R}\end{matrix}\right\}$$
Moreover, a connected reductive group $G$ is anisotropic if and only if $G(\mathbb{R})$ is compact.

Let us say briefly what the terminology `anisotropic’ means. Namely, let us call a torus $T$ over a field $k$ split if $T\cong \mathbb{G}_{m,k}^n$ (i.e. it’s isomorphic to powers of the multiplicative group not just over $\overline{k}$, but over $k$). A reductive group $G$ over $k$ is then called anisotropic it it contains no positive dimensional split tori. If $k=\mathbb{R}$ then this is equivalent to the claim that $G(\mathbb{R})$ is compact.

Let us obtain the following corollary of this result:

Corollary: Let $T$ be a reductive group over $\mathbb{R}$. Then, $T$ is an anisotropic torus if and only if $T(\mathbb{R})$ is a compact torus. Moreover, for an anisotropic (connected) reductive group $G$ over $\mathbb{R}$ the association $T\mapsto T(\mathbb{R})$ forms a bijection
$$\left\{\begin{matrix}\text{Maximal algebraic tori}\\ T\subseteq G\end{matrix}\right\}\to \left\{\begin{matrix}\text{Maximal compact tori}\\ \mathcal{T}\subseteq G(\mathbb{R})\end{matrix}\right\}$$

Proof: It suffices to prove the first statement, since the second then follows immediately from the Tannaka—Chevalley theorem. Suppose first that $T$ is an anisotropic torus. Then, by the Tannaka—Chevalley theorem we have that $T(\mathbb{R})$ is a compact connected Lie group. But, since $T$ is abelian so is $T(\mathbb{R})$. Then, by standard Lie group theory this implies that $T$ is a compact torus. Conversely, suppose that $T(\mathbb{R})$ is a compact torus. By Tanaka—Chevalley we have that $T$ is a connected reductive anisotropic group over $\mathbb{R}$. We need to show that $T$ is abelian, since then $T$ is automatically a torus (e.g. see [2. Proposition 21.7]). But, since $T(\mathbb{R})$ is dense in $T$ (e.g. see [2, Theorem 17.93]) we see that the map $T\times T\to T$ given by $(t_1,t_2)\mapsto t_1t_2t_1^{-1}t_2^{-1}$ is trivial on a dense subset, so is trivial (since all schemes involved are affine). $\blacksquare$

From this, we see that compact tori can, in a very literal sense, be thought of as a special case of algebraic tori over $\mathbb{R}$.

References:

[1] Conrad, B., 2014. Reductive group schemes. Autour des schémas en groupes, 1, pp.93-444.

[2] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.


问:
Sample $x$ from $g(x)$
I got confused with all this randomness and probability functions. I was trying to implement the rejection sampling method which (apparently) is really simple. I was reading from Rejection Sampling in Wikipedia and the first step says sample $x$ from $g(x)$ and $u$ from $U(0,1)$ what does this mean?

If I have a uniform distribution, when someone says "Generate/Sample/Draw a sample from $U(0,1)$" means that I have to take a random value from the uniform distribution right? In matlab terms, using the function rand will generate a sample from $U(0,1)$ right?

Now, for any pdf $g(x)$ defined over $(a, b)$. If I want to sample from $g(x)$ I have to pick a value of $x$ between $a$ and $b$ and evaluate the function $g(x)$ ? Is this correct ?
Then to pick the value of $x$ I can use a uniform distribution $U(a,b)$ to sample from, is this true?

I got confused easily. Thanks in advance.

UPDATE

I appreciate your help. But this is still not clear to me. Say I have a probability density function $g(x)=3x+1$. What would be the result of draw a sample from g(x) or pick any $x_0$ distributed according to $g(x)$?

UPDATE

I think I got it. I was confused with the graph of the PDF. When someone asks "draw a sample from a PDF, say $g(x)$" it should return a value for $x$ and this value has to be distributed as the function, that means that more samples will be drawn where the value of $g(x)$ is bigger. Does this sounds about right?

Finally, if I want to use the function I said before $g(x)=3x+1$ I have to use for instance the inverse method to draw samples form that distribution properly.

I think the confusion arose because I thought of the value of $g(x)$ as the probability but this was wrong, the value of $g(x)$ is some kind of density (is this correct?). Hope you can correct me or confirm my hypotheses.

答:
In its simplest form, rejection goes like this: Assume $g\colon [a,b]\to [0,M]$ is the pdf of the desired distribution.
Then generate a uniformly distributed $x\in[a,b]$ (that is $x=a+b\cdot{\mathbf {rand}}()$) and independently $y\in[0,M]$ (that is $y=M\cdot{\mathbf {rand}}()$). Accept $x$ as result if $y\le g(x)$, otherwise repeat.

That is: We pick a random point in $[a,b]\times[0,M]$ and accept its $x$-value if it is below the graph of $g$. Thus $x$ with $g(x)$ small are less likely to be picked. Unfortunately, the probability of success is $\frac1{(b-a)M}$, which may be small, i.e. lots of resampling takes place. To remedy this, one starts with a different distribution for $x$ (of course a distribution that can be simulated faster than by rejection) and replaces the constant $M$ by a suitable multiple $Mf$ of the pdf $f$ for this $x$ (we need the scaling to ensure $MF\ge g$). The easiest way to grasp may be if you use a staircase function.

Title: Sample $x$ from $g(x)$


问:
Equation using $f(x)=\frac{\sin\pi x}{x^2} $
If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$

Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.

Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.

Then which of the options is(are) correct.

(A) $x_1<y_1$

(B) $x_{n+1}-x_n>2$ for all $n$

(C) $x_n \in (2n,2n+\frac{1}{2})$ for every $n$

(D) $|x_n-y_n|>1$ for every $n$
The official answer is $B,C,D$

My approach is as follow

$$y' = \frac{x^2 \times \pi  \times \cos \pi x - 2x\sin \pi x}{x^4}$$

$$y' = \frac{x^2 \times \pi  \times \cos \pi x - 2x\sin \pi x}{x^4} = \frac{x \times \pi  \times \cos \pi x - 2\sin \pi x}{x^3} = \frac{2\cos \pi x \left( \frac{\pi x}{2} - \tan \pi x \right)}{x^3}$$

We should consider $x\in\frac{1}{2},\frac{3}{2},\ldots.$ as inflection point even though $\tan\pi x$ is not defined because we need to consider the numerator part $\pi x\cos \pi x - 2\sin \pi x$

答:
Only the option $A)$ is correct.
$$x_1=\frac 12\;\;,\; y_1=\frac 32$$

Let $$f(x)=\frac{\sin(\pi x)}{x^2}$$

it is clear that
$$(\forall x>0)\;\; -\frac{1}{x^2}\le f(x)\le \frac{1}{x^2}$$

The local maximums are such that
$$f(x_i)=\frac{1}{x_i^2}$$
or
$$\sin(\pi x_i)=1$$
thus
$$x_i=\frac 12+2n$$
with $n\ge 0$.

So, $$x_i\in\left\{\frac 12,\frac 52,\frac 92,\ldots\right\}$$
by the same,
$$y_i\in\left\{\frac 32,\frac 72,\frac{11}{2},\ldots\right\}$$


问:
Minimal Continuous Maps in Topological Space
Definition: A nonempty open set U of X is said to be a minimal open set if
and only if any open set which is contained in U is ∅ or U.

Definition: A nonempty open set U of X is said to be a
maximal open set if any open set which contains U is X or U.

Definition: A proper nonempty closed subset F of X is said to be a minimal closed set if any closed
set which is contained in F is ∅ or F.

Definition: A proper nonempty closed subset F of X is said to
be a maximal closed set if any closed set which contains F is X or F.

Note: Complement of a minimal open set is maximal closed set.

Definition: Let X and Y be topological spaces. A map $f:X\rightarrow Y$ is called minimal continuous if $f^{-1}(M)$ is an open set in X for every minimal open set M in Y.

Question: Let X and Y be the
topological spaces. A map $f:X\rightarrow Y$ is
minimal continuous if and only if the inverse
image of each maximal closed set in Y is a
closed set in X.

My Attempt: $(\implies)$ Let $f:X\rightarrow Y$ be a minimal continuous map. That is, $f^{-1}(M)\in\tau_{X}$ for every minimal open set $M\in Y$. We have that $Y-M$ is maximal closed set because the complement of a minimal open set is maximal closed set.. We need to show that $f^{-1}(Y-M)\in\kappa_{X}$ or $f^{-1}(Y-M)=X-f^{-1}(M)$ which is closed in $X$ since $f^{-1}(M)\in\tau_{X}$. Bu how?

答: That $f^{-1}[Y\setminus M]=X\setminus f^{-1}[M]$ is basic elementary set theory. Just show that each side is a subset of the other. If $x\in X\setminus f^{-1}[M]$, for instance, then $f(x)\notin M$, so $f(x)\in Y\setminus M$, and therefore $x\in f^{-1}[Y\setminus M]$. Now just prove the opposite inclusion.


问:
1st order nonlinear homogeneous ODE and integral equations
I'm currently studying ODEs, I've been using a book about the topic, and for now I've seen only first order ODEs and a few methods to solve them like the integrating factor, the Bernoulli, and the two other methods that concerns exact and separable equations.

Then my book gave this exercice under the chapter Miscellaneous Methods of Solving a First Order Differential Equation.

$x^2dy-(xy+y\sqrt{x^2+y^2})dx=0$

I basically used the substitution method for homogeneous equation with $u$ as $u=\frac{y}{x}$ and $dy=udx+xdu$.

I ended up with this: $\int{\frac{\frac{du}{dx}}{u\sqrt{u^2+1}}}dx=ln(\left| x \right|) + C$

My book seemed to assume that I knew how to solve this kind of integral with an unknown function of $x$ or its derivative, I just learnt that it was named an integral equation.

My questions are:

Where can I find informations about how to solve this kind of integrals ? I can't find anything on the Internet about it (is it something we learn later in ODE courses under a different name or something?)

Maybe I should have known how to do this before studying ODEs ? (I'm self thaught, I've only seen Calculus 1 to 3 and integral equation seems to be really physics-related).

答:
$$I=\int{\frac{\frac{du}{dx}}{u\sqrt{u^2+1}}}dx$$
$$I=\int{\frac{{du}}{u\sqrt{u^2+1}}}$$
$$I=\int{\frac{{udu}}{u^2\sqrt{u^2+1}}}$$
Substitute $u^2+1=v^2 \implies udu=vdv$ then use fraction decomposition.
$$I=\int \frac{{dv}}{v^2-1}$$


问:
three-uniform hypergraph on $n$ vertices with at least $n/3$ edges contains an independent set of size at least $\frac{2n^{3/2}}{3\sqrt{3m}}$
Here is question 3 from chapter 3 Part 1 of The Probabilistic Method, 4th edition.

Prove that every three-uniform hypergraph with $n$ vertices and $m \ge n∕3$ edges contains an independent set (i.e., a set of vertices containing no edges) of size at
least $\frac{2n^{3/2}}{3\sqrt{3m}}$.

I'm trying to solve this using a probabilistic approach. I've only been able to get $\frac{n^{3/2}-mn^{1/2}}{\sqrt{3m}}$ as a bound by defining $S\subset [n]$ by randomly including a vertex in $S$ with probability $p$. We call the resulting induced graph $G':=G[S]$ and denote its vertex set and edge set by $V'$ and $E'$. Then $\mathbb E[|V'|]-\mathbb E[|E'|]=np-mp^3$ because for each edge to be inside $S$ we need all three of its vertices to be chosen. Also, removing one vertex per edge in $G'$ results in an edge-less graph, so $\alpha(G)\ge\alpha(G')\ge np-mp^3$, where $\alpha$ is the size of the largest independent set. The inequality comes from taking the expectation on $\alpha(G')\ge \mathbb E[|V'|]-\mathbb E[|E'|]$.

Then the idea is to optimize over $p$, so make the 1st derivative w.r.t. $p$ equal to zero and get $p=\frac{n^{1/2}}{\sqrt{3m}}$ and the optimized RHS of the equality is $\frac{n^{3/2}-mn^{1/2}}{\sqrt{3m}}$.

Now the problem is that there is a minus there and the hypothesis is that $m\ge n/3$ so $-mn^{1/2}\le-\frac{-n^{3/2}}{3}$ and so $\alpha(G)\ge f(m,n)$ with $f(m,n)\le \frac{2n^{3/2}}{3\sqrt{3m}}=:B$, which does't prove anything. I think I must have made a mistake somewhere...

I only showed that the largest independent set is bounded from below by something which is below $B$. If I had showed that it is bounded by $B$ then that would probably be enough to say that there exists an independent set of size at least $B$.

答:
Your algebra is fine. I think you just misunderstand what optimizing over $p$ does.

The inequality
$$
   \alpha(G) \ge np - mp^3
$$
holds for any $m, n$ and any $p \in [0,1]$, because the random process you've described works for any $m, n$ and any $p \in [0,1]$. It will produce a random independent set $S$, with $\mathbb E[|S|] \ge np - mp^3$; therefore, with positive probability, it will produce an $S$ with $|S| \ge np - mp^3$; therefore, the largest independent set has size at least $np - mp^3$.

At this point, if all you want to do is to write a proof, you can just write:

For no particular reason, we are inspired to take $p = \sqrt{\frac{n}{3m}}$, which satisfies $0 \le p \le 1$ provided that $m \ge n/3$. With this value of $p$, we conclude that $\alpha(G) \ge \frac{2n^{3/2}}{3\sqrt{3m}}$, which was what we wanted.

If we want to actually find a good value of $p$ to pick, so that we can come up with the proof, then optimizing over $p$ is a good idea. When $m \ge n/3$, we conclude that $p = \sqrt{\frac{n}{3m}}$ is the best probability to use. In other words, we deduce the inequality $$np - mp^3 \le \frac{2n^{3/2}}{3\sqrt{3m}}.$$ You're right that sticking this inequality into our proof would produce nonsense! But this inequality doesn't belong in our proof to begin with. It is part of our outside-the-proof reasoning for picking $p$.

All we actually need to know is that the right-hand side $\frac{2n^{3/2}}{3\sqrt{3m}}$ is achievable by some choice of $p$. We don't, strictly speaking, need to know that it's the largest possible value of $np - mp^3$. (But if it weren't, we could have written a better proof.)

In more complicated problems, it is common to skip taking the derivative, because we wouldn't get a closed form for the answer. Even here, we could be lazy and think "hmm... we want $np$ and $mp^3$ to be about the same size. Let's try setting $mp^3 = \frac12 np$. This gives us $p^2 = \frac{n}{2m}$, which gives some answer of the form $O(n^{3/2}/m^{1/2})$. Good enough!" Then, we could write a proof using $p = \sqrt{\frac{n}{2m}}$. That proof would work equally well, even though it no longer proves the best lower bound that this method can prove.


问:
Understanding Leibniz rule in the framework of Banach space.

Let $E,F_1,F_2,G$ be Banach spaces and $U \subseteq E$ be an open subset. Let $f_j : U \longrightarrow F_j$ be $C^r$ maps and $B : F_1 \times F_2 \longrightarrow G$ be a bounded bilinear map. Then $g : x \mapsto B(f_1(x),f_2(x))$ is $C^r.$ Also for $k=1,2,\cdots,r$ we have $$D^kg(x) (h_1,h_2,\cdots,h_k) = \sum\limits_{j=0}^{k} \binom {k} {j} B\left (D^jf_1(x) (h_1,h_2,\cdots,h_j) ,D^{k-j}f_2(x) (h_{j+1},h_{j+2}, \cdots, h_k) \right ).$$

I have managed to prove that $g$ is $C^1$ and for any $h \in E$ we have $$Dg(x) (h) = B \left (f_1(x),Df_2(x)(h) \right ) + B \left (Df_1(x)(h),f_2(x) \right ).$$

Now how do I prove that $g$ is $C^k$ for $k \gt 1\ $? How do I prove first that $Dg \in C^1\ $? Here $Dg : U \longrightarrow \mathcal L(E,G).$ Can $Dg$ be written as a composition of Frechet differentiable functions? Can anybody please give me some suggestion? Then I will try to prove it for any $k \gt 1.$

Thanks for your time.

EDIT $:$ What I get is that $$D^2 g(x) (h_1,h_2) = B \left (f_1(x),D^2f_2 (x) (h_1,h_2) \right ) + B \left (Df_1(x)(h_1), Df_2 (x) (h_2) \right ) + B \left (Df_1(x) (h_2), Df_2 (x) (h_1) \right ) + B\left (D^2f_1(x) (h_1,h_2),f_2(x) \right).$$ But I don't know why it has been claimed in the proposition that $$B \left (Df_1(x)(h_1), Df_2 (x) (h_2) \right ) = B \left (Df_1(x) (h_2), Df_2 (x) (h_1) \right ).$$

答:
I won't prove the $k^{th}$ product rule formula, because it is already tedious enough to do in one-dimension (though if you keep your cool, it is just a matter of induction and carefully playing around with some binomial coefficient identities... try to mimic the one-dimensional proof as much as possible).

As a general remark, whenever you want to prove certain maps are of class $C^k$, very useful are the canonical injections and projections $\iota_i:V_i\to V_1\times V_2$, and the canonical projections $\pi_i:V_1\times V_2\to V_i$ (for example $\iota_1(x):=(x,0)$ and $\pi_1(x,y):=x$). These are bounded linear maps hence $C^{\infty}$ (their second derivative vanishes).

For your particular question, there are a few ways of arguing things, depending on how much stuff you've already proven. For example, if you've already proven that the composition of $C^k$ functions is $C^k$, more precisely:

Let $k\geq 1$ be an integer, $E,F,G$ be Banach spaces (all over the same field), $U\subset E$, $V\subset F$ be non-empty open sets, $f:U\to V$, $g:V\to G$ be $C^k$ functions. Then, $g\circ f:U\to G$ is also $C^k$.

then this is quite simple. Here's how we apply in your specific case: we can write
\begin{align}
g= B\circ \left(\iota_1\circ f_1 + \iota_2\circ f_2\right)
\end{align}
Just to see that everything makes sense, observe that $f_1:U\to F_1$, $\iota_1:F_1\to F_1\times F_2$ so the composition is $\iota_1\circ f_1:U\to F_1\times F_2$. Similarly, $\iota_2\circ f_2:U\to F_1\times F_2$. Hence, their sum is also a function $U\to F_1\times F_2$. So, finally by composing with $B:F_1\times F_2\to G$, we get the function $g:U\to G$.

Now, $f_1,f_2$ are $C^r$ by assumption, $\iota_1,\iota_2$ are bounded and linear (which implies $C^{\infty}$ and hence trivially also $C^r$), so by the remark above, $\iota_1\circ f_1$ and $\iota_2\circ f_2$ are $C^r$ functions. Therefore, their sum is also $C^r$ (the fact that sum of $C^r$ functions is $C^r$ is an easy exercise in induction and the simple rule that $D(\alpha+\beta)_x= D\alpha_x+D\beta_x$). Finally $B$ is a bounded bilinear mapping, it is also $C^{\infty}$ hence $C^r$ (more generally, for any bounded $p$-multilinear mapping, if you take its $(p+1)^{th}$ derivative you get $0$... again an exercise in induction... therefore it is easily seen to be $C^{\infty}$). Therefore, $g=B\circ \left(\iota_1\circ f_1 + \iota_2\circ f_2\right)$ is the composition of $C^r$ functions hence $C^r$. In short,

\begin{align}
g= B\circ \left(\iota_1\circ f_1 + \iota_2\circ f_2\right) ``=" C^{\infty}\circ \left(C^{\infty}\circ C^r + C^{\infty}\circ C^r\right) = C^r
\end{align}

Edit:

Your calculation is right, the given formula is wrong, because the formula on the right is not symmetric in $h_1,\dots, h_k$. For example, when $k=2$, the given formula says that
\begin{align}D^2g(x)(h_1,h_2) &= B(f_1(x), D^2f_2(x)(h_1,h_2)) + 2B(Df_1(x)(h_1), Df_2(x)(h_2)) \\
&+ B(D^2f_1(x)(h_1,h_2),f_2(x)).
\end{align}
The LHS $D^2g(x)$ being a second derivative must be symmetric in $h_1,h_2$, but the RHS is clearly not symmetric, so the given formula is wrong. If I'm not mistaken, the correct formula is obtained by symmetrizing your given formula.

So, if we let $\lambda_k(x):\underbrace{E\times\cdots \times E}_{\text{$k$ times}}\to G$ be the multilinear map
\begin{align}
&\lambda_k(x)(h_1,\dots, h_k)\\
&:=
\sum_{j=0}^k \binom{k}{j} B\left (D^jf_1(x) (h_1,h_2,\cdots,h_j) ,D^{k-j}f_2(x) (h_{j+1},h_{j+2}, \cdots, h_k) \right ),
\end{align}
then $D^kg_x = \text{Sym}_k(\lambda_k(x))$, or more explicitly,
\begin{align}
D^kg(x)(h_1,\dots, h_k)&=\dfrac{1}{k!}\sum_{\sigma\in S_k}\lambda_{k}(x)\left(h_{\sigma(1)}, \dots, h_{\sigma(k)}\right),
\end{align}
where $S_k$ is the set of all permutations on $k$ objects (i.e the set of all bijections $\sigma:\{1,\dots, k\}\to \{1,\dots, k\}$).


问:
Evaluating $\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$
I am trying to evaluate the following integral:

$$\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$$

I thought about using the substitution $\beta \doteq \tfrac{(t-a)}{C}$ so the integral would become:

$$\int_{0}^{\infty} \beta^2 \cdot \exp (- \beta^2 ) \cdot \exp(i w (\beta C - a)) \ C \ d\beta$$

However, I don't know how to evaluate this either.

Any idea on how to do this?

EDIT: As Caran-d'Ache pointed out, I did in fact lose a minus! I'm still just learning Mathematica, how did you enter the integral?

答: I think you may be able to integrate by parts twice if we have $e^{-(...)}$ instead of $e^{(...)}$, the integral in this current case does not converge  </p>


问:
Maximize the variance of sum of Bernoulli random variables
Let $Z_1, \cdots, Z_n \in \{0,1\}$ be $n$ Bernoulli random variables, with known joint probability distribution. I am considering to solve the following problem in general:
$$
\mathrm{maximize}_{x_1,\cdots,x_n \in \{0,1\}} {\mathbb{V}\mathrm{ar}}\Big[ \sum_{i=1}^{n}{x_i Z_i} \Big]
$$
i.e., select a subset of random variables such that their summation has the largest variance. I feel the intuition is to select a subset of variables that are positively correlated. I wonder if this is a known problem and I am really looking forward to getting some thoughts here.

答:
If you know the joint distribution, then you know the covariance matrix $\Sigma$ (whose $i,j$ entry is $\text{Cov}(Z_i, Z_j)$).
Then the variance of $x^\top Z = \sum_{i=1}^n x_i Z_i$ is
$$x^\top \Sigma x.$$
If the maximization problem were constrained to be in the unit ball $\|x\|=1$, then the maximizing $x$ would be the eigenvector corresponding to the largest eigenvalue of $\Sigma$.

However, your optimization problem is combinatorial; I believe it can be cast as a binary optimization problem. The linked PDF has some algorithms to solve them.


答:
Another way:

$$\int \frac{du}{u\sqrt{u^2+1}} $$

Substitute $u=\tan t \implies du=\sec^2 t dt $: $$\int \frac{\sec^2 t}{\sec t \tan t} dt \\ =\int \csc t \ dt \\ = \ln |\csc t -\cot t| + c \\ = \ln \left | \frac{\sqrt{1+u^2}}{u} -\frac 1u \right| + c$$ And now after replacing $u$ by $\frac yx$, you’ll get your solution.

Title: 1st order nonlinear homogeneous ODE and integral equations

Body:
I'm currently studying ODEs, I've been using a book about the topic, and for now I've seen only first order ODEs and a few methods to solve them like the integrating factor, the Bernoulli, and the two other methods that concerns exact and separable equations.

Then my book gave this exercice under the chapter Miscellaneous Methods of Solving a First Order Differential Equation.

$x^2dy-(xy+y\sqrt{x^2+y^2})dx=0$

I basically used the substitution method for homogeneous equation with $u$ as $u=\frac{y}{x}$ and $dy=udx+xdu$.

I ended up with this: $\int{\frac{\frac{du}{dx}}{u\sqrt{u^2+1}}}dx=ln(\left| x \right|) + C$

My book seemed to assume that I knew how to solve this kind of integral with an unknown function of $x$ or its derivative, I just learnt that it was named an integral equation.

My questions are:

Where can I find informations about how to solve this kind of integrals ? I can't find anything on the Internet about it (is it something we learn later in ODE courses under a different name or something?)

Maybe I should have known how to do this before studying ODEs ? (I'm self thaught, I've only seen Calculus 1 to 3 and integral equation seems to be really physics-related).


问:
What is the expected time to A reach C, given the probabilities and travel time?
A space ship walks randomly between 3 planets: A. B and C. It does countless trips among then, always going from planet X to planet Y. The probabilities (are constants) and travel times are on the table below. Example: leaving from planet B, there is a 50% chance that it arrives on planet A, and 50% at planet B (yes, it can go back to the same place it left). ORIGIN (row) DESTINY (columns)

The problem: there is a space ship leving from A that will only stop the journey when it reaches C. What will be the expected mean time (in months) to reach C?

PROBABILITY MATRIX
DESTINATION

ORIGIN
A
B
C

A
0.05
0.90
0.05

B
0.50
0.50
0

C
0
0
1

TRAVEL TIME MATRIX (in moths)
DESTINATION

ORIGIN
A
B
C

A
5
2
10

B
3
3
0

C
0
0
0

[markov chain sketch with the probabilities and travel time][1]
[1]: https://i.stack.imgur.com/XTKz7.png

What I have so far:

let "a" be the number of times it will bounce on a journey from A to C;
"b", the number of times it will bounce on a journey from B to C; and
"c", the number of times it will bounce on a journey from C to C.

a = 1 + 0.05a + 0.9b + 0.05c (c is the absorbing state, so c=0)

thus:  a = 1 + 0.05a + 0.9b

b = 1+ 0.5a + 0.5b

I came up with a = 56 and b = 58.

BUT, these numbers represents the number of times it moves from one planet to another. It does not give me the times in months that I need. (I'm stuck here)

答:
Can't you do essentially the same thing by taking the travel time into account?  Let $m_A$ be the expected number of months of the journey if the ship is at $A$, and let $m_b$ be the expected number of months if the ship is at $B$.

Then we have $$m_A=.05(5+m_A)+.9(2+m_B)+.05\cdot10$$  We get a similar equation for $m_B$, and solve for $m_A$ and $m_B$.  Then $m_A$ is the answer to the question.


问:
Finding the steady state error in the Laplace domain
I have the following block diagram:

Now I like to find the steady state error for theta_ref being a step input and for several values of n, Td, K1 and K2.

For the moment we can assume all gains are simple scalars, so no vectors and matrices involved.

I like to find an expression for e(s) so that I can apply Laplace's final value theorem.

I came up with:

$e(s) = \theta_{ref}(s) - \theta(s)$

$\theta(s) = (e(s)K_2-n\cdot M)N^{-1}....$

But I don't know how to handle the inner loop.

I have Matlab / Simulink on my computer, so if I can obtain steady state error information using this software it is find with me.

答:
If you're looking for tracking error, you can use Mason's formula to find the transfer function between $\Theta_{ref}$ and $\Theta$. First find the forward path gains. There is only one forward path:

$$F_1 = K_{2}*\frac{1}{N}*K_1\frac{1}{s}*N*\frac{1}{s}=\frac{K_1K_2}{s^2}$$

Now find the loop gains. there are two loops, the inner loop ($L_1$) and the outer loop($L_2$):

$$L_1=-1*K_1*\frac{1}{s}=-\frac{K_1}{s}$$
$$L_2=-1*K_2*\frac{1}{N}*K_1*\frac{1}{s}*N*\frac{1}{s}=-\frac{K_1K_2}{s^2}$$

Finding the determinant and plugging into Mason's formula, you will obtain:
$$\frac{\Theta}{\Theta_{ref}}=\frac{K_1K_2}{s^2+K_1s+K_1K_2}$$
$$T_e(s)=1-\frac{\Theta}{\Theta_{ref}}$$

You can do steady state analysis on this transfer function to find the tracking error, and you could use the same process to find the transfer function for the noise and disturbances and find the steady state error for those inputs as well.


问:
Separability and Continuity
I've always had a little bit of an issue with the general definition of continuous function for a topological space. We say a function $f : X \rightarrow Y$ is continuous if for ever open set $U$ of $Y$, $f^{-1} (U)$ is open in $X$. I can understand this definition when it comes to metric spaces, but have always had trouble establishing intuition for general topological spaces.

Then I came across the concept of separated sets. Two sets are $A$ and $B$ are "separated by neighborhoods" if there are disjoint open sets U and V such that U contains A and V contains B. Using this definition it's not difficult to show that if two sets $A$ and $B$ are inseparable and f is continuous then the images $f(A)$ and $f(B)$ will also be inseparable.

This seems intuitive to me. A function separating something that was inseparable is "tearing something apart" and so discontinuous. It'd be nice if I could think of this as an "alternative" definition of continuous. But it's not clear to me whether the converse holds. That is, if a function satisfies this property does it mean that the function must be continuous? If not, are there any additional reasonable axioms I can add to make them equivalent?

答:
The converse need not hold.

Let $Y$ be an infinite set with the cofinite topology; then any two non-empty open sets in $Y$ have non-empty intersection. Thus, for any non-empty $A,B\subseteq Y$, $A$ and $B$ are inseparable. Let $X$ denote the same set with the indiscrete topology, and let $f:X\to Y$ be the identity map. Then $f$ is not continuous, but $f$ does preserve inseparability, since there are no separable pairs of non-empty sets in $Y$.

Title: Separability and Continuity


问:
Application of Lang' theorem about finite groups
Let $G$ be a connected algebraic group defined over the finite field $\mathbb{F}_q$, and let $F: G \to G$ be the Frobenius morphism.

Show that $G^F = \{g \in G\mid g = F(g)\}$ is a finite group;

For $x\in G^F$, let $Z_G(x) = \{g\in G\mid gx = xg\}$. Show that $Z_G(x)$ is $F$-stable, i.e., $g\in Z_G(x)$ implies $F(g) \in Z_G(x)$;  and if $Z_G(x)$ is connected, $y\in G^F$ is conjugate to $x$ in $G$, then $y$ is conjugate to $x$ in $G^F$.

I have proved 1. and the first half of 2. About the 2nd half

Let $h \in G$, s.t. $hyh^{-1} = x$, hence I want to prove $\{F^n(h)\}_{n \geq 0} \cap G^F \neq \emptyset$.

But I have no idea using the connectivity of $Z_G(x)$ to prove this.

Any help will be appreciate.

答:
I’m not quite sure what the blockquoted statement is about.

Let

$$L_G:G\to G:x\mapsto x^{-1}F(x)$$

be the Lang isogeny. Lang’s theorem then says that if $G$ is smooth and connected (which I assume is what you’re assuming) then $L$ is surjective. Note that $G^F=\ker(L)$.

So, for 2. we want to show that $Z_G(x)$ is $F$-stable if $x$ is in $G^F$. But, this is simple since if

$$gx=xg$$

then applying $F$ we get

$$F(g)F(x)=F(x)F(g)$$

but $F(x)=x$ and so this says

$$F(g)x=xF(g)$$

so $F(g)$ is in $Z_G(x)$. We now want to show that if $Z_G(x)$ is connected and if $y$ in $G^F$ is conjugate to $x$ in $G$ then it’s conjugate to $x$ in $G^F$. But, note that since $y$ is conjugate to $x$ in $G$ we can write

$$x=z^{-1}yz$$

for some $z$ in $G$. But, note that then that $L_G(z)$ is in $Z_G(x)$. Indeed,

$$(z^{-1}F(z))x(F(z)^{-1}z)=z^{-1}F(zxz^{-1})z=z^{-1}F(y)z=z^{-1}yz=x$$

But, since $L_{Z_G(x)}$ is surjective this implies that we can write $L_G(z)=L_G(w)$ for some $w$ in $Z_G(x)$. Let us then note that since

$$z^{-1}F(z)=w^{-1}F(w)$$

that

$$wz^{-1}=F(wz^{-1})$$

so $wz^{-1}\in G^F$. But, note that

$$(wz^{-1})y(wz^{-1})^{-1}=wz^{-1}yzw^{-1}=wxw^{-1}=x$$

where the last equality holds since $w$ is in $Z_G(x)$. Thus, $wz^{-1}$ is in $G^F$ and conjugates $y$ to $x$ as desired.


问:
Inequality and sequence of functions
I am trying to prove that the sequence of functions $f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$ does not converge for any $x\in\mathbb{R}$.

My idea was to fix $x\in\mathbb{R}$ and then obtain a bound like $|f_n(x)|\geq $ constant $\times$ (function of $n$ which goes to infinity as $n\to\infty$) so $|f_n(x)|=|\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}|\geq ||\sqrt{n}\cos(nx)|-|\frac{\sin(nx)}{2n\sqrt{n}}||\geq \sqrt{n}||\cos(nx)|-\frac{1}{2n^2}|$ which is almost what I wanted but I don't see how to go further so I would appreciate an hint. Thank you.

答:
$f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$

$\frac{\sin(nx)}{2n\sqrt{n}} \rightarrow 0$ when $n\rightarrow +\infty$.

And $\sqrt{n} \cos (nx)$ diverges (we can easily extract a subsequence that diverges to $+\infty$)

Title: Inequality and sequence of functions


问:
Integration of product of exponential function and natural logarithm
I am trying to calculate the following integral:

$$\int_{-1}^{1}\ln(1+\gamma x)\exp\left(-\frac{(z-x)^2}{2}\right)dx$$

with $\gamma\in[-1,1]$, $z\in\mathbb{R}$ and ln being the natural logarithm. I tried a lot of thinks like substitution, integration by parts, used the series expansion of the natural logarithm resp of the exponential function. I also searched for some help in Literatur but couldn't find anything helpful. My question is: Can we solve this integral?

答:
Considering first the restricted problem with $z=0$.

We can find an approximate series for $\gamma\rightarrow 0$. First, expand the logarithm

$$\ln(1+\gamma x)=\gamma x - \frac{(\gamma x)^2}{2}+ ... $$

Now notice that on the integration domain, all terms $x^{odd}$ will integrate to zero. We are left with

$$ I(\gamma)=-\int_{-1}^{1}dx\ \left(  \frac{(\gamma x)^2 }{2}+    \frac{(\gamma x)^4 }{4}+...    \right)e^{-x^2/2}$$

Term by term integration gives

$$ I(\gamma)\sim\sum_{n=1}^\infty A_n \gamma^{2n}, \ \gamma\rightarrow 0$$

With the coefficients

$$ A_n =-\frac{2^{n-1/2}}{n}\left( \Gamma(n+1/2)-\Gamma(n+1/2,1/2) \right)$$

The degree to which this is an improvement depends on how you feel about the incomplete gamma function. By repeated use of the identities

$$\Gamma(s+1,x)=s \Gamma(s,x)+x^s e^{-x} $$

$$ \Gamma(1/2,x)=\sqrt{\pi}  \operatorname{erfc}(\sqrt{x})$$

Every term in the series may be cast into the form of $\operatorname{elementary function} \times \operatorname{erfc}(2^{-1/2}) $. This is an improvement because now all those $\operatorname{erfc}$s appear as constants in the series. The first term alone is a fine approximation

$$ I(\gamma)\sim -\gamma^2 \left(\sqrt{\pi/2}  \left( 1-\operatorname{erfc}(2^{-1/2})\right)-e^{-1/2}     \right),\gamma\rightarrow0$$

More importantly we learn that for small $\gamma$, the integral scales as $-\gamma^2$.

Here is a plot of the first term together with the exact (numerically integrated) result

And the same plot with just three terms

We can also (somewhat) extend this to the case $z\rightarrow 0$.

To avoid issues of non-uniform convergence, I'll consider the integral

$$I(\epsilon)= \int_{-1}^1 \ dx \ln(1+\gamma \epsilon x) \exp \left(-\frac{(x-\epsilon z)^2}{2}\right) $$

This of course is still restricted compared to the original. Expanding the exponential around $\epsilon=0$ will produce a series of 'corrections' to the $\ln$ series. The first nonzero term still occurs at $\epsilon^2$. I find

$$ I(\epsilon)\sim -\epsilon^2 \frac{\gamma(\gamma-2z)}{2\sqrt{e}}\left( \sqrt{2e\pi} \operatorname{erf}(2^{-1/2})-2\right),\epsilon\rightarrow 0$$

We learn the integral scales with $-\gamma(\gamma-2z)$ for small $\epsilon$. Here's a plot with $\gamma=1/2$

As expected, when $z>1$ the approximation is bad: this is when the peak of the Gaussian passes outside of the integral bounds.

For large $z$, I think we should be able to write the exponential in such a way as to use Laplace's method, but I don't see how right now.


问:
Cutting space by hyperplanes: a direct proof
The number of regions formed by cutting $k$ generic hyperplanes out of $\mathbb R^n$ is:
$$C_n(k)=\binom{k}{0}+\binom{k}{1}+\dots+\binom{k}{n}$$
The usual proof is by induction, counting how many new regions are formed by adding the $k^\text{th}$ plane and utilizing the formula $C_n(k)=C_n(k-1)+C_{n-1}(k-1)$. This formula resembles Pascal's rule $\binom{n}k=\binom{n}{k-1}+\binom{n-1}{k-1}$ and indeed, it can be proven by rearranging sums and applying Pascal's rule.

But we can also view $C_n(k)$ as the number of subsets of $\{1,2,\dots,k\}$ of size $\leq n$. Then the formula $C_n(k)=C_n(k-1)+C_{n-1}(k-1)$ admits essentially the same combinatorial proof as Pascal's rule, where we break up $C_n(k)$ depending on whether or not the subset contains $k$.

To better understand the relationship between $C_n(k)$ and hyperplane cutting, I wonder if there is a direct proof (i.e. not using induction) of the above-mentioned result? Since there are $k$ hyperplanes, it seems reasonable that $C_n(k)$ might count subsets of these hyperplanes of size $\leq n$. But the question that currently stumps me is: how does one then biject these subsets of the planes with the regions?

答:
Choose coordinates for $\mathbb R^n$ so none of the $k$ hyperplanes are parallel to any coordinate planes, and so that all of the hyperplane intersections occur in the orthant where all coordinates are positive.

First, we account for the $\binom{k}n$ term. Any $n$ of the $k$ hyperplanes will intersect in a point, and that point will be the "lowest" point of one of the regions, where "lowest" means having the least $x_1$ coordinate. Conversely, any region with a lowest point determines a set of $n$ hyperplanes intersecting at that point.

The only regions which have not been accounted for are those which do not have a lowest point. To count these, let $E_1$ be the coordinate plane perpendicular to the $x_1$ axis, so by assumption all the intersection points are above $E_1$. Then the original regions with no lowest point correspond to the regions $E_1$ is divided into by the $k$ planes. Since the dimension of $E_{1}$ is one less than $\mathbb R^n$, you can recursively give a bijection of subsets of size at most $n-1$ of the $k$ planes with the regions of $E_{1}$.

All in all, the bijection between regions of $\mathbb R^n$ and subsets of the $k$ planes of size at most $n$ is as follows:

If the size of the subset is $i$, then find the intersection $P$ of that subset of planes with the first $n-i$ coordinate planes, $E_1,E_2,\dots,E_{n-i}$, where $E_j=\{(x_1,\dots,x_n):x_j=0\}$. There is a unique region of $\mathbb R^n \cap E_1\cap \dots \cap E_{n-i}$ which has $P$ as its furthest point in the direction of the negative $x_{n-i}$ axis, and this is contained in a unique region of $\mathbb R^n$.


问:
What is the probability of a person winning a dice game on the first roll, given that he/she won the game?
Let Jill and Jack be two people. They play a game with a dice. The first person to roll $6$ wins the game. Jill goes first.

I want to compute the probability that Jack won on his first roll given that he won the game.

It can be shown that the probability of Jill winning is $6/11$ and the probability of Jack winning is then $5/11$. Obviously the probability of Jill winning on her first roll is $1/6$, but since Jack rolls second, the probability is different, although I am not sure what. Suppose it is $x$, then this would be a conditional probability question: i.e., computing the probability of Jack winning first intersected with the probability of him winning the game, so $x+5/11$, divided by $5/11$.

答:
The probability of Jack winning on his first roll includes the probability that Jill doesn't win on her first roll ($\frac56$). Hence $\frac56×\frac16=\frac5{36}$. The conditional probability is then
$$\frac{5/36}{5/11}=\frac{11}{36}$$


答:
As the sequence of functions $\{g_n\}$ with
$$g_n(x)=\frac{\sin(nx)}{2n\sqrt{n}}$$ converges to $0$ for all $x \in \mathbb R$, it is enough to prove that $\{h_n\}$ does not converge for any $x \in \mathbb R$ with

$$h_n(x) = \sqrt{n}\cos(nx).$$

And this is clear as $\cos nx$ never converges to zero which would be required for $\{h_n(x)\}$ to converge as $\lim\limits_{n \to \infty} \sqrt n = \infty$.

Proof that $\{\cos nx\}$ never converges to zero:

As $$\cos(n+1)x = \cos nx \cos x - \sin nx \sin x,$$ if $\{ \cos nx \}$was converging to zero, we would have

$$\lim\limits_{n \to \infty} \sin x \sin nx = 0$$
which implies either $\sin x = 0$, i.e. $x \in {\pi \mathbb Z}$ and a contradiction as $\{\cos np\pi\}$ doesn't converge to zero. Or $\lim\limits_{n \to \infty} \sin nx = 0$ and another contradiction as
$$\cos^2 nx + \sin^2 nx = 1.$$

Title: Inequality and sequence of functions

Body:
I am trying to prove that the sequence of functions $f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$ does not converge for any $x\in\mathbb{R}$.

My idea was to fix $x\in\mathbb{R}$ and then obtain a bound like $|f_n(x)|\geq $ constant $\times$ (function of $n$ which goes to infinity as $n\to\infty$) so $|f_n(x)|=|\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}|\geq ||\sqrt{n}\cos(nx)|-|\frac{\sin(nx)}{2n\sqrt{n}}||\geq \sqrt{n}||\cos(nx)|-\frac{1}{2n^2}|$ which is almost what I wanted but I don't see how to go further so I would appreciate an hint. Thank you.


问:
Optimal sampling for an arbitratry area
I have a closed area (2D) of arbitrary shape and some number of sample points to take from the area.

Now I would like to find optimal distribution of the sample points, so that distances from sample to sample are roughly the same and the sample points cover the area as uniformly as possible:

I am doing this in a context of image processing, so the area can be defined as a set of (gray) points or by its boundary.

There may be some clever algorithm (probably iterative) to compute the positions of sample points for a given count.

Thanks for any ideas.

答:
Here is a very simple proposal :) Note that this can totally converge to a local minimum, so if you want to be confident that you're as close as possible to the best solution, you should run the whole procedure several times, and select the one with the smallest energy... Not great, but it's not an easy problem :/

To answer your last comment, this has nothing to do with uniform sampling either: uniform sampling is randomly distributed with uniform distribution accross the shape-space. This might be suitable to initialize the current model at best, but what you're trying to achieve here is not random sampling, it's equirepartite sampling. Very much different, and extremely hard for arbitrary shapes; in fact, analytical solutions do not exist in the general case.

Dynamic Model

Let $N$ particles be defined by their 2D coordinates at time t, $p_i(t) = \big( x_i(t), y_i(t) \big)$. We define the interaction between particles $i$ and $j$ ($i\neq j$) at time $t$ as:
$$
I_{i\to j}(t) = \frac{ \vec{u}_{i\to j}(t) }{ \epsilon + \| p_i(t) - p_j(t) \|_2^2 }
$$

For the sake of simplicity, given a timestep $\delta t$, we suppose that changes on position and its derivatives occur only at discrete times $t_k,\ k\geq 0$.
We find the following update rules for the coordinate $x$ (similarly for $y$), arbitrarily taking a unitary mass:
$$
\frac{\partial x_i}{\partial t}( t_{n+1} ) = \frac{\partial x}{\partial t}( t_n ) + \delta t \sum_{j\neq i} \frac{ u_{j\to i}^x (t_n) }{ \epsilon + \| p_i(t_n) - p_j(t_n) \|_2^2 } \\
x_i(t_{n+1}) = x_i(t_n) + \delta t \frac{\partial x}{\partial t}( t_n ) + \frac{\delta t^2}{2} \sum_{j\neq i} \frac{ u_{j\to i}^x (t_n) }{ \epsilon + \| p_i(t_n) - p_j(t_n) \|_2^2 }
$$

Quantitative Choices

First, the dimensions of the shape being studied should be normalized, for instance, by setting the largest dimension of the bounding box to a fixed constant $L$.
$\delta t$ can be modeled as a variable of time too, recomputed at each timestep to limit the displacement of each particle to a fraction of $L$.
This way, we can limit the numerical errors caused by our simplistic model.

Say we chose to limit the maximum displacement to a constant $\frac{d}{\sqrt{2}}$; from the second update equation in $x$, we find a second order polynomial in $\delta t$, which roots can be computed for each point $p_i$.
Taking the smallest positive timestep found for $x$ and $y$ coordinates, we can then assign:
$$
\delta t = \min( \delta t_x, \delta t_y )
$$
and finalize the update of position and velocity.

Running the Sampling

There are two additional things to consider before running the sampling:

We need a function $S$ which takes 2D coordinates and returns either; the same coordinates if they are inside the shape, OR the orthogonal projection of the input coordinates on the border of the shape otherwise. This ensures that the particles won't leave the shape. (Note: the velocity should be nulled if the coordinates are projected).

The particles should be initialized with random positions inside the shape (rejection sampling in the bounding box with the previous function $S$ should do), and random velocities (direction and magnitude). That's 5 random values per particle.

Repeat the update until the maximum displacement between two timesteps goes below a certain fixed threshold, that's you convergence criterion.

Final note: I'm absolutely not sure whether this yields accurate results for fancy shapes. If you get unsatisfactory/weird results, a straightforward extension would be to cancel the interactions between particles that are not "facing" each-other (ie, $[p_i,p_j]$ crosses the boundary).

Title: Optimal sampling for an arbitratry area


答:
According to Bayes' formula,
\begin{align*}
\mathbb{P}(\text{Jack wins on first roll} \, | \, \text{Jack wins})
&=
\mathbb{P}(\text{Jack wins} \, | \, \text{Jack wins on first roll})\frac{\mathbb{P}(\text{Jack wins on first roll})}{\mathbb{P}(\text{Jack wins})} \\
&=
\frac{\mathbb{P}(\text{Jack wins on first roll})}{5/11}.
\end{align*}
The probability that Jack wins on first roll is
\begin{align*}
\mathbb{P}(\text{Jack wins on first roll})
&=
\mathbb{P}(\text{Jack wins on first roll} \, | \, \text{Jill doesn't win on first roll})\mathbb{P}(\text{Jill doesn't win on first roll}) \\
&=1/6 \cdot 5/6 = 5/36,
\end{align*}
since the event that Jill doesn't win on first roll is contained in the event that Jack wins on first roll, and since the dice are independent.

All in all,
\begin{align*}
\mathbb{P}(\text{Jack wins on first roll} \, | \, \text{Jack wins})
=
\frac{5/36}{5/11} = \frac{11}{36}.
\end{align*}

Title: What is the probability of a person winning a dice game on the first roll, given that he/she won the game?

Body:
Let Jill and Jack be two people. They play a game with a dice. The first person to roll $6$ wins the game. Jill goes first.

I want to compute the probability that Jack won on his first roll given that he won the game.

It can be shown that the probability of Jill winning is $6/11$ and the probability of Jack winning is then $5/11$. Obviously the probability of Jill winning on her first roll is $1/6$, but since Jack rolls second, the probability is different, although I am not sure what. Suppose it is $x$, then this would be a conditional probability question: i.e., computing the probability of Jack winning first intersected with the probability of him winning the game, so $x+5/11$, divided by $5/11$.


问:
Derangement, gift exchange
Five people have each bought their own Christmas gift. They put the gifts in a sack which then everyone can pull their gift from. In how many ways can the gifts be distributed so that no one gets the Christmas gift they have bought themselves?

So I have figured $93$ combinations in which only $1$ and then $2, 3$ or $4$ and $5$ get their gift but I haven't figured out how to calculate that other people don't get their gift. For example

If $2$ people get their gifts how do I count that the other $3$ don't get their own gifts etc? I have read about derangement and seen different formulas but I haven't quite understood the logic.

答:
It sounds like you're reading an inclusion-exclusion argument, but are misunderstanding it. We number these people $1$ through $5$, and let $[5]$ denote the set $\{1,2,3,4,5\}$. For a subset $X$ of $[5]$, let $f(X)$ count the number of permutations such that every person in $X$ gets their own gift, and every person outside of $X$ gets someone else's gift. In other words, $f(X)$ counts the number of derangements of $[5]\setminus X$. In particular, $f(\varnothing)$ counts the number of derangements of $[5]$, which is what we want to count. The idea of this argument is that $f(X)$ is in general hard to compute, so we'll count it using a function that's easier to compute.

For $X\subseteq [5]$, let $g(X)$ count the number of permutations such that every person in $X$ gets their own gift, but people outside of $X$ may also get their own gift. While $f(X)$ counts the permutations where $X$ is exactly the set of people who get their own gift (in other words, permutations where $X$ is the set of fixed points), while $g(X)$ counts the permutations where at least everyone in $X$ gets their own gift (permutations such that the set of fixed points contains $X$).

While $f(X)$ is difficult to compute, $g(X)$ is easy to compute- we let everyone in $X$ get their own gift, and let everyone else in $[5]\setminus X$ get any of the remaining gifts, so $g(X)=(5-|X|)!$. Now we use the principle of inclusion-exclusion. To use inclusion-exclusion, we think of counting objects with certain properties. In this example, the objects are permutations of $[5]$, and the properties are that $1$ is a fixed point, $2$ is a fixed point, etc, so we have $5$ properties to consider. One phrasing of inclusion-exclusion is that if $f(X)$ counts objects with exactly the properties in $X$, and $g(X)$ counts objects with at least the properties in $X$, then we can relate $f$ and $g$ through the equation $$f(X)=\sum_{Y\supseteq X} (-1)^{|Y|-|X|}g(Y).$$

I won't go into detail on how this works, but it sounds like this is the type of proof that you're reading. We'll use the formula above, setting $X=\varnothing$ to count the number of derangements. We already know that $g(Y)=(5-|Y|)!$, so we can use that to solve for $f(\varnothing)$. Furthermore, since $g(Y)$ only depends on the size of $Y$, we can write $g(i)=g(Y)$ where $|Y|=i$. Then instead of iterating over all $Y\subseteq [5]$, we can just consider all $i$ from $0$ to $5$. For each $i$, we have $g(i)=(5-i)!$, and there are ${5\choose i}$ subsets of $[5]$ with size $i$. Thus, we have $$f(\varnothing)=\sum_{Y} (-1)^{|Y|}g(Y)=\sum_{i=0}^5 (-1)^i{5\choose i}(5-i)!=\sum_{i=0}^5 (-1)^i \frac{5!}{i!}.$$

The last equality uses the definition of ${5\choose i}$.

Title: Derangement, gift exchange


问:
How can I prove $\int_{0}^{+\infty }\frac{x-sinx}{\sqrt{x^{7}}}dx$ converges/diverges?
I need to determine if the following integral converges/diverges:

$$\int_{0}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx$$

We can write the integral as:

$$\int_{0}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx=\int_{0}^{a }\frac{x-\sin x}{\sqrt{x^{7}}}dx+\int_{a}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx=I_{1}+I_{2}$$

+For $I_{1}=\int_{0}^{a}\frac{x-\sin x}{\sqrt{x^{7}}}dx$.

Since $\sin x= x-\frac{x^{3}}{3!} +O(x^{3})$, I know that $I_{1}$ is convergent.

+For $I_{2}=\int_{a}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx$.

I don't know how to use $x-\sin x$ for other ideas such as the comparison test. The only thing I can exploit is $x - \sin x>0 , x>0$.

Can you give me some hints? Thank you very much!

答:
Since $ \frac{x-\sin{x}}{x^{\frac{7}{2}}}\underset{x\to 0}{\sim}\frac{1}{6\sqrt{x}} $, there is no problem at $ 0 $. And since $ \frac{x-\sin{x}}{x^{\frac{7}{2}}}\underset{x\to +\infty}{\sim}\frac{1}{x^{\frac{5}{2}}} $ there is no problem at $ +\infty $ either. Thus our integral converges.

Let's try to find its value using Fubini theorem and the two following identities : \begin{aligned}\left(\forall x\in\mathbb{R}^{*}\right),\ \frac{x-\sin{x}}{x^{3}}=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\left(1\right)\\ \int_{0}^{+\infty}{\cos{\left(x^{2}\right)}\,\mathrm{d}y}=\frac{\sqrt{\pi}}{2\sqrt{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\left(2\right)\end{aligned}

The first equality can be easily proven, doing some integrations by parts, the second one (Fresnel Integral) is a well known identity that I'm sure has been proven a lot of times in this website.

\begin{aligned}\int_{0}^{+\infty}{\frac{x-\sin{x}}{x^{\frac{7}{2}}}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{x^{2}-\sin{\left(x^{2}\right)}}{x^{6}}\,\mathrm{d}x}\\ &=\int_{0}^{+\infty}{\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(x^{2}y\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\left(1-y\right)^{2}\int_{0}^{+\infty}{\cos{\left(x^{2}y\right)}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=\int_{0}^{1}{\frac{\left(1-y\right)^{2}}{\sqrt{y}}\int_{0}^{+\infty}{\cos{\left(\mu^{2}\right)}\,\mathrm{d}\mu}\,\mathrm{d}y}\\ &=\sqrt{\frac{\pi}{2}}\int_{0}^{1}{\frac{\left(1-y\right)^{2}}{2\sqrt{y}}\,\mathrm{d}y}\\ &=\sqrt{\frac{\pi}{2}}\int_{0}^{1}{\left(1-\nu^{2}\right)^{2}\,\mathrm{d}\nu}\\ \int_{0}^{+\infty}{\frac{x-\sin{x}}{x^{\frac{7}{2}}}\,\mathrm{d}x}&=\frac{8}{15}\sqrt{\frac{\pi}{2}}\end{aligned}


问:
Is there a special way to finding a nonzero vector v that belongs to both the null space and the column space of a square matrix?
In my textbook is it written that for some special cases the column space and the nullspace of a square matrix can have nonzero vectors (v) in common. I wonder if there is any special type of matrix for which this statement is always true, or how one would go about to finding such vectors v in the general case?

答:
Such a vector $\mathbf v$ satisfies

$\mathbf v = A \mathbf w$ for some $\mathbf w$.

$A \mathbf v = \mathbf 0$.

Putting these two facts together tells us that $A^2 \mathbf w = \mathbf 0$, which is the first step to finding such a $\mathbf v$.

We can find all the vectors $\mathbf w$ that will work by finding a basis for the null space of $A^2$. Then, we can multiply each of the vectors in the basis by $A$. If any of the results are nonzero, they give a possible value of $\mathbf v$. If not, then there is no intersection.

This generalizes to finding the intersection between the null space of a matrix and the column space of another matrix.


问:
Don't follow proof that Hadamard differentiable implies compactly differentiable
A function $f:X \to Y$ between Banach spaces is said to be compactly differentiable if there is a function $f'_x:X\rightarrow Y$ such that
$$\lim_{t \to 0} \frac{f(x+th) - f(x) }{t} -f'_x(h) =0$$
where the limit holds uniformly in $h \in K \subset X$, where $K$ is a compact set.

We say $f$ has a Hadamard derivative if
$$\lim_{n \to \infty} \frac{f(x+t_nh_n) - f(x)}{t_n} - f'_x(h)=0$$
holds where $h_n \to h$ and $t_n \to 0$ are any sequences.

There is a result (Prop. 3.3 of this
) that states Hadamard diff. implies compact diff. The proof starts as follows.

Let $f$ be Hadamard diff., then in order to show it's also compact diff. it is enough to show that for any compact set $S$ and sequences $h_n \in S$ and $t_n \to 0$, we have
$$\lim_{n \to \infty} \frac{f(x+t_nh_n) - f(x)}{t_n} -f'_x(h_n) = 0.$$
The rest is omitted.

Question: how is this limit enough to show it? I don't see where the uniformity comes in at all. Could someone explain it please?

答:
This is an old problem but since the source of the OP states a more general result that the stated in the OP,  I think it is worthwhile to present the complete result and a proof as presented by the source:

In what follows, $X$ and $Y$ are linear topological spaces (linear spaces equipped with a topology in which addition and scalar product are continuous operations, and singletons are closed), and $f$ is a function from $x+U$ to $Y$, where $U$ is some neighborhood of $0\in X$. For definitions of Gateaux derivative and Hadamard derivative see the comment section below.

Theorem: Hadamard directional differentiability implies compact directional differentiability. Conversely, if $f$ admits a compactly directional derivative $f'_x$ at $x$, and if $f_x':X\rightarrow Y$ is sequentially continuous on $X$, then $f$ is directional differentiable in the sense of Hadamard.

Proof:
The direct part (which is the one the OP is concerned) is based on the fact (see the comment section) that if $f$ has  Hadamard derivative $f'_x$, the $f'_x:X\rightarrow Y$ is sequentially continuous. Assume $f$ is not compactly differentiable at $x$. This means that there exists a compact set $K\subset X$ and an open neighborhood $V$ of $0\in Y$ such that for any $n\in\mathbb{N}$, there is $(t_n,h_n)\in (0,\tfrac1n)\times K$ such that
$$
\begin{align}
\frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h_n)\notin V\tag{0}\label{zero}
\end{align}
$$
Let $W$ be a symmetric neighborhood of $0\in Y$ such that $W+W\subset V$. By the compactness of $K$, we may assume that $h_n\xrightarrow{n\rightarrow\infty} h$ for some $h\in K$ (the fact that $X$ is $T_1$ separable linear space -hence Hausdorff- implies that every sequence in a compact subset of $X$ admits a convergent subsequence). As $f$ is Hadamard differentiable, we have that there is $n_0\in\mathbb{N}$ such that
$$
\begin{align}
\frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h)&\in W \quad n\geq n_0\tag{1}\label{one}\\
f'_x(h_n) -f'_n(h)&\in W\quad n\geq n_0\tag{2}\label{two}
\end{align}
$$
It follows from \eqref{one} and \eqref{two} that
\begin{align}
\frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h_n)&=\Big(\frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h)\Big) +\\
&\quad\quad  \Big(f'_x(h) -f'_n(h_n)\Big)\in W+W\subset V
\end{align}
for all $n\geq n_0$ in contradiction to \eqref{zero}.

Conversely, assume that $f$ admits a compactly directional derivative $f'_x:X\rightarrow Y$ which is sequentially continuous. Let $(t_n,h_n)\in (0,\infty)\times X$ a sequence such that $t_n\xrightarrow{n\rightarrow\infty}0$ and $h_n\xrightarrow{n\rightarrow\infty}h$. The set $H=\{h_n,h:n\in\mathbb{N}\}$ is a compact subset of $X$. For any open neighborhood $V$ of $0\in Y$ choose a symmetric  neighborhood $W$ of $0\in Y$ such that $W+W\subset V$. By assumption, there are $\delta>0$ and $n_0\in\mathbb{N}$ such that for all $0<t<\delta$
$$
\begin{align}
\frac{f(x+th)-f(x)}{t}-f'_x(h)\in W,&\qquad (t,h)\in(0,\delta)\times H\tag{4}\label{four}\\
 f'_x(h)-f'_x(h_n)\in W,&\qquad n\geq n_0\tag{5}\label{five}\\
t_n\in(0,\delta),&\qquad n\geq n_0\tag{6}\label{six}
\end{align}
$$
Combining \eqref{four},\eqref{five}, and \eqref{six}, we obtain that for all $n\geq n_0$
$$\begin{align}
\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)&=\Big(\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h_n)\Big)+\\
&\quad\quad \Big(f'_x(h_n)-f'_x(h)\Big)\in W+W\subset V
\end{align}
$$
This shows that $f$ has Hadamard directional derivative $f'_x$.

Comments:

Definition 0: $f$ is directionally differentiable at $x$ in the sense of Gâteaux if there is a function $f'_x:X\rightarrow Y$ such that for any $h\in X$
\begin{align}
\lim_{t\rightarrow0+}\frac{f(x+th)-f(x)}{t}-f'_x(h)=0
\end{align}

Definition 1: $f$ is directionally differentiable at $x$ in the sense of Hadamard, if there is a function $f'_x:X\rightarrow Y$ such that for any $h\in X$ and any $h_t\xrightarrow{t\rightarrow0+}h$

\begin{align}
\lim_{t\rightarrow0+}\frac{f(x+th_t)-f(x)}{t}-f'_x(h)=0
\end{align}
(Notice that $x+th_t\in x+U$  for all sufficiency small $t$.)

Remark 1a: It is a simple exercise to show that this definition is equivalent to saying that for any sequence $(t_n,h_n)\in (0,\infty)\times X$ such that $(t_n,h_n)\xrightarrow{n\rightarrow\infty}(0,h)$,
\begin{align}
\lim_{n\rightarrow\infty}\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)=0
\end{align}

Remark 1b:
When $X$ is a normed space, it is an easy exercise to show that differentiability at $x$ in the sense of Hadamard is equivalent to
\begin{align}
\lim_{(t,k)\rightarrow(0+,h)}\frac{f(x+tk)-f(x)}{t}-f'_x(h)=0
\end{align}
for all $h\in X$.

Remark 1c: Clearly differentiability in the sense of Hadamard implies differentiability in the sense of Gateaux. Furthermore, the function $f'_x$ is positive homogeneous, that is $f'_x(\alpha h)=\alpha f'_x(h)$ for all $\alpha\geq0$ and $h\in X$. In most applications, it is also required in  the definitions above that $f'_x$ be a linear bounded operator from $X$ and $Y$.

Definition 2: $f$ is compactly differentiable at $x$ if there is a function $f'+x:X\rightarrow Y$ such that for any compact $K\subset X$
\begin{align}
\lim_{t\rightarrow0+}\frac{f(x+th)-f(x)}{t}-f'_x(h)=0
\end{align}
uniformly on $K$. (Notice that since $K$ is compact, it is bounded and so, $x+tK\subset x +U$ for all $t>0$ small enough.)

The main result used on the proof above is the following

Lemma: If $f$ has directional derivative $f'_x$ at $x$ in the sense of Hadamard, then $f'_x:X\rightarrow Y$ is sequentially continuous.

Proof:
Let $\{h_n:n\in\mathbb{N}\}$ be a sequence in $X$ that converges to some $h\in X$. For any open neohborhood $V$ of $0\in Y$, let $W$ be a symmetric open neighborhood of $0\in Y$ such that $W+W\subset V$. Since Differentiability in the sense of Hadamard implies differentiability in the sense of Gâteaux differentiability, we have that for any $n$ there is $0<t_n$ such that $t_n\xrightarrow{n\rightarrow\infty}0$ and
\begin{align}
\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h_n)\in W\tag{7}\label{seven}
\end{align}
Differentiability in the sense of Hadamard on its own implies that there is $n_0\in\mathbb{N}$ such that
\begin{align}
\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)\in W,\qquad n\geq n_0,\tag{8}\label{eight}
\end{align}
Combining \eqref{seven} and \eqref{eight} we obtain for all $n\geq n_0$
\begin{align}
f'_x(h_n)-f'_x(h)&=\Big(\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)\Big) -\\ &\qquad\quad\Big(\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h_n)\Big)\in W+W\subset V
\end{align}
This proves that $f'_x$ sequentially differentiable.


问:
Degree of $\mathbb{Q}(2^{\frac{1}{3}},\omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$
Consider $f(x) = x^3 - 2 \in \mathbb{Q}[x]$.

The roots of $f(x)$ are $2^{\frac{1}{3}}, 2^{\frac{1}{3}}\omega, 2^{\frac{1}{3}}{\omega}^2$.

$\mathbb{Q}(2^{\frac{1}{3}},\omega)$ is the minimal splitting field of $f(x)$ over $\mathbb{Q}$.

What is degree of $\mathbb{Q}(2^{\frac{1}{3}},\omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$?

I have been told that the degree is $3$ because it satisfies degree $3$ irreducible polynomial. But how were we able to come to that conclusion?

答:
We know that

$$|\mathbb{Q}(2^{1/3},w):\mathbb{Q}(2^{1/3})| \leq |\mathbb{Q}(w):\mathbb{Q}| = 2.$$

So $|\mathbb{Q}(2^{1/3},w):\mathbb{Q}(2^{1/3})| \in \{1,2\}.$ But if $|\mathbb{Q}(2^{1/3},w):\mathbb{Q}(2^{1/3})| = 1$ then $w \in \mathbb{Q}(2^{1/3})$ which of course isn't true.

Explicitly finding a minimal polynomial for $w$ over $\mathbb{Q}(2^{1/3})$ would be doable, but often this approach is relatively painful compared to the above approach. The idea is the use the inequality stated in the first line. (It holds since any polynomial over $\mathbb{Q}$ with $w$ has a root is also a polynomial over $\mathbb{Q}(2^{1/3})$ with $w$ as a root).


问:
Applications of the concept of homomorphism
What are some interesting applications of the concept of homomorphism?

Example: If there is a homorphism from a ring $R$ to a ring $r$ then a solution to a polynomial equation in $R$ gives rise to a solution in $r$. e.g. if $f:R \rightarrow r$ and $X^2+Y^2=0$ then $f(X^2+Y^2)=f(0), f(X^2)+f(Y^2)=0, f(X)^2+f(Y)^2=0, x^2+y^2=0$

答:
A professor of mine, Francisco Raggi, once said in class:

There are two ways to study a ring. You can stare at it, and stare at it, and stare a it, until you can spot interesting features of the ring, and then you can start describing those interesting features. Or you can study its images and the way it acts on other structures, by considering homomorphic images and modules over the ring. The latter is usually easier and more fruitful.

When you are studying a particular structure, you can study it in isolation: you can consider a vector space, and its subsets and subspaces, and say a lot of interesting things about that vector space; but in the end, it doesn't get you all that far. The same is true when you study a particular ring (and you can learn a lot just by staring at that particular ring; think of classical Number Theory as the result of staring intently at the integers looking for interesting features), or groups, or topological spaces, etc. But it turns out that studying the functions between similar objects that preserve the structure (linear transformations for vector spaces, ring and group homomorphisms, continuous maps, etc) leads to a much richer palette, with a lot more information and a lot more structure to work with. In some cases, like vector spaces, the new objects have themselves the same structure that you were studying (the set of all linear transformations $T\colon\mathbf{V}\to\mathbf{W}$ can be given a natural structure of a vector space, so everything you know about vector spaces applies) or some other structure (if $A$ and $B$ are abelian groups, then the set of all group homomorphism $A\to B$ can be given a ring structure by pointwise addition and composition, so you can study it as you would study a ring) which in turn gives you a wealth of information about the structures themselves. And of course, things like the Isomorphism Theorems can give you a lot of information about the structure you are looking at by looking at its images, which will often be "simpler" or "smaller" than the one you were originally concerned with.

Even in the case were we have found a lot of good information by simply staring (as in classical number theory), just look at how big a leap forward was achieved with the introduction of congruences (homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$), or with algebraic number theory (mapping to larger rings and then trying to import the information back into $\mathbb{Z}$).

ContentLicense: CC BY-SA 2.5


问:
1st order nonlinear homogeneous ODE and integral equations
I'm currently studying ODEs, I've been using a book about the topic, and for now I've seen only first order ODEs and a few methods to solve them like the integrating factor, the Bernoulli, and the two other methods that concerns exact and separable equations.

Then my book gave this exercice under the chapter Miscellaneous Methods of Solving a First Order Differential Equation.

$x^2dy-(xy+y\sqrt{x^2+y^2})dx=0$

I basically used the substitution method for homogeneous equation with $u$ as $u=\frac{y}{x}$ and $dy=udx+xdu$.

I ended up with this: $\int{\frac{\frac{du}{dx}}{u\sqrt{u^2+1}}}dx=ln(\left| x \right|) + C$

My book seemed to assume that I knew how to solve this kind of integral with an unknown function of $x$ or its derivative, I just learnt that it was named an integral equation.

My questions are:

Where can I find informations about how to solve this kind of integrals ? I can't find anything on the Internet about it (is it something we learn later in ODE courses under a different name or something?)

Maybe I should have known how to do this before studying ODEs ? (I'm self thaught, I've only seen Calculus 1 to 3 and integral equation seems to be really physics-related).

答:
Note that in this ODE we have

$$
A(x,y)dy-B(x,y)dx=0
$$

and making $y = \lambda x$

$$
\cases{
A(x,y) = x^2A(1,\lambda)\\
B(x,y) = x^2B(1,\lambda)\\
dy = \lambda dx + x d\lambda}
$$

$(A,B)$ are same order homogeneous functions: then

$$
\frac{dy}{dx} = \lambda+x\frac{d\lambda}{dx} = \frac{A(1,\lambda)}{B(1,\lambda)}
$$

and finally

$$
\frac{dx}{x}=\frac{d\lambda}{\frac{A(1,\lambda)}{B(1,\lambda)}-\lambda}
$$


问:
How to prove similarity of a constructed triangle?
There is a triangle $ABC$ and a point $X$ inside it. The lines $AX$, $BX$, and $CX$ intersect the circumcircle of the triangle $ABC$ also at points $P$, $Q$, and $R$. Let $U$ be a point on the line segment $XP$ so that it is in between the points $X$ and $P$. Let's assume a line, that goes through $U$ and is parallel to $AB$, intersects the line segment $XQ$ at point $V$. Let's also assume that there's a line going through $U$, that is parallel to $CA$, and intersects the line segment $XR$ at point $W$.

Prove that triangles $ABC$ and $UWV$ are similar.

A picture of the situation

Now, here's some things I have noticed/tried:

First of all, it's fairly obvious that the angle $\angle U$ in $\triangle UWV$ is congruent to angle $\angle A$ from the "same" parallel lines/the parallelogram formed. So, the similarity would be proved if we can find one of the other angles in $\triangle UWV$ or if we can find the side lengths of that triangle in terms of the original one, but the latter seems almost impossible to do, so I decided to go angle chasing.

I'm sure that this is possible by chasing angles but everytime I was close to proving $\angle XWV=\angle WCB$ (as the other part of $\angle UWX$ is congruent to $\angle ACW$) I found out that some angle was missing and couldn't find that one in terms of the parts of the angles of the original triangle.

So, I gave up on angle chasing and thought of a different way to prove this. (Well, what follows is essentially angle chasing) If we construct the line segment $QR$, we get the quadrilateral $RWVQ$.

Here's a picture of that situation:

If now the two triangles are actually similar, it follows that $RWVQ$ is a cyclic quadrilateral, as $\angle RQV=\angle RCB$ and as $AB\parallel VU$ and $AC\parallel WU$ then $BC\parallel WV$, which means that $\angle CWV=\angle WCB$ and thus $\angle VWR=\pi-\angle CWV=\pi-\angle WCB$ and so $\angle VWR+\angle QVW=\pi-\angle WCB+\angle WCB=\pi$. And thus if $RWVQ$ were a cyclic quadrilateral then $\angle CWV=\angle WCB$ and as also $\angle UWX=\angle ACW$ then the triangles would be similar.

So, if we could prove that the point $R$, $W$, $V$, and $Q$ were on the same circle, then the triangles would be similar. But I feel this is more difficult than our original questions but I thought it would be wisest to put everything I've thought so far to this.

So, if you have any ideas on how to do this, I would appreciate any hints as I would like to solve it on my own, but right now I'm honestly stuck so I need a nudge in the right direction. (Probably I'm just overthinking the whole question.)

答:
Similarity of $XUV$ and $XAB$ gives $XU/XA=UV/AB$.

Similarity of $XUW$ and $XAC$ gives $XU/XA=UW/AC$.

Therefore $UV/UW = AB/AC$.

I'm not sure what the points $P,Q,R$ and the circumscribed circle are for.


答:
I have been told that the degree is 3 because it satisfies degree 3 irreducible polynomial.

The number $\omega$ is a root of the cubic $x^3-1$. But that is not irreducible. One factor is $x^2+x+1$, which $\omega$ is a root of.

The minimal polynomial of $\omega$ (be it over $\mathbb{Q}$ or $\mathbb{Q(2^{1/3})}$) has degree $2$, not $3$.

Title: Degree of $\mathbb{Q}(2^{\frac{1}{3}},\omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$

Body:
Consider $f(x) = x^3 - 2 \in \mathbb{Q}[x]$.

The roots of $f(x)$ are $2^{\frac{1}{3}}, 2^{\frac{1}{3}}\omega, 2^{\frac{1}{3}}{\omega}^2$.

$\mathbb{Q}(2^{\frac{1}{3}},\omega)$ is the minimal splitting field of $f(x)$ over $\mathbb{Q}$.

What is degree of $\mathbb{Q}(2^{\frac{1}{3}},\omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$?

I have been told that the degree is $3$ because it satisfies degree $3$ irreducible polynomial. But how were we able to come to that conclusion?


问:
Method to solve this indefinite integral $ \int \frac{x^{4}}{\sqrt{x^{2}+9}}dx $
I'm not able to solve this integral. Can anybody help me?
I tried doing it by parts noticing that
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .
$$
where we know that $$ D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9} $$

so that we can view the integral as

$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx  .
$$
which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it

答:
hint

We know that
$$(\forall t\in \Bbb R)\;\; \sqrt{1+\sinh^2(t)}=\cosh(t)$$

and

$$\sqrt{9+(3\sinh(t))^2}=3\cosh(t)$$
So, with $t=3\sinh(t) $,

the integral becomes

$$81\int \sinh^4(t)dt$$

now, use linearization by expanding
$$(\frac{e^x-e^{-x}}{2})^4$$


答:
Well, I will give you some hints:

Substitute $x=3\tan\left(\text{u}\right)$;

For the integrand you will end up with, use the trigonometric identity:

$$\tan^2\left(x\right)=\sec^2\left(x\right)-1\tag1$$

Prove the reduction formula (using integration by parts);

$$\int\sec^\text{n}\left(x\right)\space\text{d}x=\frac{\sin\left(x\right)\sec^{\text{n}-1}\left(x\right)}{\text{n}-1}+\frac{\text{n}-2}{\text{n}-1}\int\sec^{\text{n}-2}\left(x\right)\space\text{d}x\tag2$$

Apply the reduction formula;

Use the fact that:

$$\int\sec\left(x\right)\space\text{d}x=\ln\left|\tan\left(x\right)+\sec\left(x\right)\right|+\text{C}\tag3$$

Use the fact that:

$$\sec\left(\arctan\left(x\right)\right)=\sqrt{1+x^2}\tag4$$

Title: Method to solve this indefinite integral $ \int \frac{x^{4}}{\sqrt{x^{2}+9}}dx $

Body:
I'm not able to solve this integral. Can anybody help me?
I tried doing it by parts noticing that
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .
$$
where we know that $$ D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9} $$

so that we can view the integral as

$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx  .
$$
which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it


问:
Good introductory book for self-studying quasigroups?
I'm looking for an undergraduate or beginning graduate level text from which to self-learn quasigroup theory. An emphasis on using quasigroups to understand the structure of groups would be appreciated. I prefer texts that are leisurely and systematic. I like having only a few examples, but those examples worked out in great detail.</p>

答:
Bruck R.H.: A Survey of Binary Systems. Springer-Verlag, Berlin-Gottingen-Heidelberg, 1966.

Moreover, there is a good book

Belousov V.D.: Foundations of the theory of quasigroups and loops, Moscow, 1967,

but as you see, it is only in Russian


问:
What is a good complex-analysis textbook for someone without too much real analysis background?
I'm a sophomore in HS and I've been self-studying pure maths for about 6 months now. I really enjoy complex analysis and have had a lot of fun tackling dogbones, keyholes, and even evaluating the basel problem and fresnel integrals, as well as understanding how important results such as Jordan's Lemma, Cauchy's Integral Theorem, and residues at infinity are derived. I only have basic Real Analysis background, but I want to have a solid textbook for self-study. It should have a decent amount of exercises (and solutions), stay semi-rigorous, and not leave anything important out. I don't mind a challenge as long as it doesn't assume I have crazy prior knowledge about Real Analysis and Number Theory. I know most of univariate and a good chunk of multivariate Calc. Thanks!!</p>

答:
I would say that two standard references are Lars Ahlfors Complex analysis, an introduction to the theory of analytic functions of one complex variable and Serge Lang Complex Analysis. They are for a first (but rigorous) introduction to complex analysis. Note that since they are both well-known books, you can find nice exercises and also some of them solved here or online. Ahlfors' book also has a chapter that helps you bridge the possible gap you may have from introductory real analysis to analysis in metric spaces and some multivariable analysis (implicit function theorem...), with a bit of topology of metric spaces that you may need. Another book may be also John Conway Functions of one complex variable, it is introductory, it starts from the very basics of complex numbers and has the needed material for topology of $\mathbb{C}$ and metric spaces. These 3 are basically equivalent, it's a matter of tastes, you may try to have a quick view at some libraries and then pick, also based on the solutions of some exercises that you may want to solve and then check the (suggested) solutions. Sometimes Gamelin's Complex Analysis is suggested since very intuitive and with many applications, but I honestly have never seen it.

Another suggestion may be to follow the syllabi and the websites of some professors that on their website publish the notes of their classes. Sometimes they are well-written and may be useful. You can basically type the name of the university, the name "complex analysis" and see if you find some stuff.


问:
Let $G$ be a group. What is ${\rm Hom}(\mathbb{Z}_n \times \mathbb{Z}_m, G)$?
$\DeclareMathOperator{\Hom}{Hom}$

Let $G$ be a group. What is $\Hom(\mathbb{Z}_n \times \mathbb{Z}_m, G)$?

I think that $$\Hom(\mathbb{Z}_n \times \mathbb{Z}_m, G) = \{\phi_{a,b} : a, b \in G, a^n = e,b^m = e\},$$ where $\phi_{a,b}(k,l ) = a^k b^l$ for $(k,l)\in\mathbb{Z}_n\times \mathbb{Z}_m.$

By definition, $\Hom(\mathbb{Z}_n \times \mathbb{Z}_m, G)$ is the set of homomorphisms from $\mathbb{Z}_n \times \mathbb{Z}_m$ to $G.$ Let $\phi\in \Hom(\mathbb{Z}_n \times \mathbb{Z}_m, G)$ and $a = \phi((1,0)).$ Let $b = \phi((0,1)).$ Then for all $(k,l)\in \mathbb{Z}_n \times \mathbb{Z}_m, \phi((k,l)) = \phi(k(1,0) + l(0,1)) = \phi(k(1,0))\phi(l(0,1)) = a^k b^l$ so that $\phi = \phi_{a,b}$ and $a^n = \phi((n,0)) = \phi((0,0)) = e$.

Similarly, $b^m = e.$ The map $\phi_{a,b}$ is also well-defined because if $k\equiv l \mod n$ and $p\equiv q \mod m,$ then $a^k = a^l$ and $b^p = b^q.$ Also, $\phi_{a,b}$ is a homomorphism because $a^{k_1 + k_2} b^{l_1 + l_2} = (a^{k_1} b^{l_1})(a^{k_2}b^{l_2})$. Thus, $\Hom(\mathbb{Z}_n \times \mathbb{Z}_m, G) = \{\phi_{a,b} : a, b \in G, a^n = e,b^m = e\},$ where $\phi_{a,b}(k,l ) = a^k b^l$ for $(k,l)\in\mathbb{Z}_n\times \mathbb{Z}_m.$

Is this correct?

答:
Here is the general result:

Theorem. Let $H$ and $K$ be groups, and let $G$ be a group. Then the morphism $H\times K\to G$ are in one-to-one correspondence with pairs $(f,g)$, where $f\colon H\to G$ and $g\colon K\to G$ are group homomorphisms, and $f(H)\subseteq C_G(g(K))$ (equivalently, $g(K)\subseteq C_G(f(H))$.

Proof. First, let us note that if $f(H)\subseteq C_G(g(K))$, then each $f(h)$ commutes with each $g(k)$, and hence each $g(k)$ lies in $C_G(f(H))$. Thus, the conditions $f(G)\subseteq C_G(g(K))$ and $g(K)\subseteq C_G(f(H))$ are indeed equivalent. (Alternatively, $f(H)\subseteq C_G(g(K))$ implies $C_G(f(H))\supseteq C_G(C_G(g(K)))\supseteq C_G(g(K))$.)

Let $\varphi\colon H\times K\to G$ be a group morphism. Identify $H$ with $H\times\{e\}$ and $K$ with $\{e\}\times K$ in $H\times K$. We define $f=\varphi|_{H\times\{e\}}$ and $g=\varphi|_{\{e\}\times K}$, identifying $H$. Note that since $H\times\{e\}$ and $\{e\}\times K$ generate $H\times K$, the two restrictions uniquely determine $\varphi$, so the association is one-to-one from maps on $H\times K$ to pairs.

To verify that $f(H)\subseteq C_G(g(K))$, let $h\in H$ and $k\in K$. Then $$f(h)g(k) = \varphi(h,e)\varphi(e,k) = \varphi(h,k) = \varphi( (e,k)(h,e)) = \varphi(e,k)\varphi(h,e) = g(k)f(h).$$ Thus, every element of $f(H)$ commutes with every element of $g(K)$, as required.

Conversely, given $f\colon H\to G$ and $g\colon K\to G$ with $f(H)\subseteq C_G(g(K))$, define $\varphi\colon H\times K\to G$ by $\varphi(h,k) = f(h)g(k)$. To verify this is a group homomorphism, we have
$$\begin{align*}
\varphi(h_1,g_1)\varphi(h_2,g_2) &=f(h_1)g(k_1)f(h_2)g(k_2)\\
&=f(h_1)f(h_2)g(k_1)g(k_2) &\text{(since }f(h_2)\text{ commutes with }g(k_1)\text{)}\\
&= f(h_1h_2) g(k_1k_2)\\
&= \varphi(h_1h_2,k_1k_2)\\
&= \varphi\Bigl( (h_1,k_1)(h_2,k_2)\Bigr).
\end{align*}$$
Now note that the pair associated to this map are the original $f$ and $g$, showing that the association is also surjective and giving the correspondence. $\Box$

Now, to complete what you want, we have:

Theorem. Let $G$ be a group, $n\gt 0$. The morphism $\mathbb{Z}/n\mathbb{Z}\to G$ are in one-to-one correspondence with elements $g\in G$ such that $g^n=e$.

Proof. If $f\colon\mathbb{Z}/n\mathbb{Z}\to G$ is a morphism, then $f(1)=g$ satisfies $g^n = f(n) = f(0) = e$. Conversely, if $g\in G$ has exponent $n$, then the map $f\colon\mathbb{Z}/n\mathbb{Z}\to G$ given by $f(\overline{a}) = g^a$ is easily verified to be a well-defined group homomorphisms. The two associations are inverses of each other. $\Box$

Putting the two together we see that your result is only missing one detail, as indicated by Lukas Heger in the comments: that your elements $a$ and $b$ should centralize each other, i.e., commute.