问:
Problem with Veronese (or d-uple) embedding
In Hartshorne, ex I.2.12 we see that the image of the d-uple embedding, $\varphi_d$ is exactly $Z(Ker \theta)$ where $\theta: K[y_0,\dots,y_N]\rightarrow K[x_0,\dots,x_n]$ is defined by sendind $y_i$ to $M_i$ where $M_0,\dots, M_N$ are all the monomials of degree d in $x_0,\dots x_n$).

I was wondering: what is exactly $Ker\theta$?

Reading online I found this:

Let $w_I$ where $I=(i_0,\dots,i_n)$ such that $i_0+\dots+i_n=d$, be the homogeneous coordinates of $\mathbb{P}^N$

the image of $\varphi_d$ is $Z(w_Iw_J-w_Kw_L | I+J=K+L)$ here $I+J=(i_0+j_0,\dots,i_n+j_n)$.

So if I rewrite $\theta: K[\{w_I\}]\rightarrow k[x_0,\dots,x_n]$ sending $w_I$ to $x_0^{i_o}\dots x_n^{i_n}$

Is $Ker\theta=(w_Iw_J-w_Kw_L | I+J=K+L)$?

Of course $(w_Iw_J-w_Kw_L | I+J=K+L)\subseteq Ker\theta$.

I'm having some trouble with the other inclusion:

let $f\in Ker\theta$ and let $M=w_{I_1}\dots w_{I_n}$ a monomial of degree $n$ appearing in $f$ with some non zero constant coefficient. Since $\theta(f)=0$, there must be another monomial $M'$ of degree $n$ appearing in $f$ with the same coefficient multiplied by $-1$ such that $M'=w_{J_1}\dots w_{J_n}$ and $I_1+\dots+ I_n=J_1+\dots+J_n$. I would like to show that $M-M'\in (w_Iw_J-w_Kw_L | I+J=K+L) $.

By induction on $n$:

If n=2 then $M-M'$ is exactly a generator of $(w_Iw_J-w_Kw_L | I+J=K+L)$.

Assume $n>2$ I would like to write $M-M'$ as a combination of $(N-N')$, monomials of degree $n-1$, and generators of $(w_Iw_J-w_Kw_L | I+J=K+L)$ but I don't know how to do this.

I don't know if it is the right way...

答:
If you view the "Veronese embedding" $v$ as a closed immersion of varieties/schemes

$$v_d: \mathbb{P}^n \rightarrow \mathbb{P}^{\binom{n+1}{d}-1}$$

you may construct an inverse map $v_d^{-1}:v_d(\mathbb{P}^n) \rightarrow \mathbb{P}^n$ with the property that

$$v_d \circ v_d^{-1}=v_d^{-1}\circ v_d = Id$$

hence $v_d$ is an isomorphism onto its image. In the below link this is done in the simplest case and it should be possible to generalize the calculation. Maybe this calculation can be used to prove that the ideal above is a prime ideal since projective space is irreducible hence the ideal of $v_d(\mathbb{P}^n)$ in $k[y_1,..,y_N]$ is a prime ideal.

Veronese embedding


问:
Link between two gradient definitions
Let $M$ a $n$ dimensional embedded Riemannian manifold of $\mathbb{R}^{n+1}$. There are two definitions of $\nabla_M f$ that seem different, but I think they should be related:

Def 1: $\nabla_M f = \nabla f - \langle\nabla f, n\rangle n$, where $n$ is the outer unit normal.

Def 2: Definition through parametrization.

By Def 1, one sees that $\nabla_M f \in \mathbb{R}^{n+1}$, while using Def 2 (in practice), we get $\nabla_M f \in \mathbb{R}^{n}$.

How to relate this two definitions? (Maybe, behind my confusion there is some hidden identification).

Edit 1:

Take, e.g., $f(x,y,z)=\frac{1}{2}x^2$ on $\mathbb{R}^3$. Then
by Def 1: $\nabla_{S^2} f(x,y,z)=(x-x^3, -x^2y,-x^2z)$ and by Def 2: $\nabla_{S^2} f(\phi(\theta,\varphi))=(-\frac{1}{2} \sin(2\theta),\frac{1}{2} \sin(2\varphi) \cos^2 \theta)$.

But in Riemannian geometry, they said that they are the same, they even prove one from the other.

答:
Mind that $Def. 1$ is like all derivatives and gradients only define at a given point on $M$ where $f$ is sufficiently steady otherwise there is need to derive from directions towards the selected points. If there is set in $M$ where all the gradients exist then the gradient can be written in the abstracted form of Your question set. This inherent in the definition of gradients.

There very same is true is the definition of $f$ is some kind of path on $M$. The point matters not any paths along which the gradient is taken. The gradient exists if it is independent of the path of differentiation.

Your example is not so suitable since $f:\Bbb R^{3} \rightarrow \Bbb R$. This is not like in Your definition $\Bbb R^{3}$ with $n=2$ and Your $n$ is $1$.

Your $S^{2}$ is meant the surface of the unit sphere in $\Bbb R^{3}$ which has the dimension 2 but You forgot that $n$ from the $Def.1 $s is a unit vector. It has to have the form
$$\vec n=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$

So
$\vec \nabla f(x,y,z)=(x,0,0)$
without any restriction to the sphere $S^{2}$.
An example normal on the $S^{2}$ is:
$\vec n=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$

$$\vec \nabla_{S^{2}} f(x,y,z)=(x,0,0)-<\vec \nabla f,\vec n>\vec n$$
$$=(x,0,0)-<(x,0,0),\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}>\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$
$$=(x,0,0)-\frac{x^2}{\sqrt{x^2+y^2+z^2}}\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$
$$=\frac{(x(x^2+y^2+z^2)-x^2,-x^{2}y,-x^2z)}{x^2+y^2+z^2}$$
Your results is rather different!
$\vec n$ is directed outwards. The same is true is $-\vec n$ is used but oriented inwards. There are other normals on $S^{2}$.

This shows up that the normal has to be specified in the definition before calculating the gradient. And the normal has to be assigned to a point on the unid sphere centered around zero for example the point (0,0,1) where the gradient becomes (0,0,-1).

I in the second example the same $f$ is used why are there only two components of the gradient? $r=1$ is only the definition of the unit sphere in spherical coordinates is not allowed to ignore the $r$ derivatives.

As a check transform the function $f$ properly. $x=r sin(\theta) cos(\phi)$ so
$$f(r,\phi,\theta)=\frac{r^2 sin(\theta)^2 cos(\phi)^2}{2}$$

The gradient is now with the projection on the normal:

Form this point with use the unit sphere restriction for our example normal. The example normal is the $\vec e_{r}$ unit vector of the coordinate system. $r=1$.
So the second term is
$$-(cos(\phi)cos(\theta))^2\vec e_{r}$$
The $\vec e_{r}$ is simply $cos(\theta)^2$. The others remain unchanged at $r=1$.

ContentLicense: CC BY-SA 4.0

Title: Link between two gradient definitions


问:
Does this graph already have a name?
I am studying the Cartesian product of such a graph with other graphs. I want to ask if it has a known name?

By the way, for some special graphs, is there a good website to check?

答:
There is no well established name for this particular graph. However, there is a well known graph which is similar to some extend.

It is called the Banner graph

It is obtained by joining a single vertex of the cycle graph $C_4$ to a pendant vertex. Our graph is obtained by joining two non-adjacent vertices of $C_4$ with two pendant vertices.

Some comments to the website given by Greg Martin:

The string "ElD?" is the graph6 representation of your graph. This format was specified by Brendan McKay here. It is used in most computational frameworks, such as Mathematica, SageMath, etc. Using sage, we can parse this format directly into sage graph objects as follows

```<code>g=Graph("ElD?")
g.show()
</code>```

yielding to

The string "X84" is the internal name in the database ISGCI. This can be verified using

```<code>t = graph_classes.smallgraphs()
</code>```

which returns a set in which "X84" is used for internal reference only.

Title: Does this graph already have a name?

Body:
There is no well established name for this particular graph. However, there is a well known graph which is similar to some extend.

It is called the Banner graph

It is obtained by joining a single vertex of the cycle graph $C_4$ to a pendant vertex. Our graph is obtained by joining two non-adjacent vertices of $C_4$ with two pendant vertices.

Some comments to the website given by Greg Martin:

The string "ElD?" is the graph6 representation of your graph. This format was specified by Brendan McKay here. It is used in most computational frameworks, such as Mathematica, SageMath, etc. Using sage, we can parse this format directly into sage graph objects as follows

g=Graph("ElD?")
g.show()

yielding to

The string "X84" is the internal name in the database ISGCI. This can be verified using

t = graph_classes.smallgraphs()

which returns a set in which "X84" is used for internal reference only.


问:
Why is the integral $C^\infty$
I am reading Differentiable manifolds from Warner.

In order to prove that the dimension of the tangent space is the same as the dimension of the manifold, they use the following calculus lemma -

If $g$ is of class $C^k$ ($k \geq 2$) on a convex open subset $U$ about $p$ in $\mathbb{R}^d$, then for each $q \in U$,
$g(q)\ =\ g(p) + \sum_{i=1}^d \frac{\partial g}{\partial r_i}|_p (r_i(q)-r_i(p))+\sum_{i,j}(r_i(q)-r_i(p))(r_j(q)-r_j(p))\int_0^1(1-t)\frac{\partial^2g}{\partial r_i\partial r_j}|_{(p+t(q-p))} dt.$

This is the Taylor expansion. It further says, if $g\in C^\infty$, then the integral as a function of $q$ is of class $C^\infty$. How is this? Do we have to use fundamental theorem of calculus or something like that?

答: The expression $\int_0^1(1-t)\frac{\partial^2g}{\partial r_i\partial r_j}|_{(p+t(q-p))} dt$ can be differentiated under the integral sign to obtain all of its partial derivatives.  Since $g$ is $C^\infty$, so is the integral.</p>

ContentLicense: CC BY-SA 3.0

Title: Why is the integral $C^\infty$


问:
Binomial expansion of $(1-x)^n$
We've been given with following binomial expansion

$$(x+1)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3\cdots$$

How can I get the formula of $(1-x)^n$

答:
I'm not sure how appropriate it is to answer questions this old, but compared to the methods above, I feel the easiest way to see the answer to this question is to take

```<code>a = -x
</code>```

And substitute that into the binomial expansion:

```<code>(1+a)^n
</code>```

This yields exactly the ordinary expansion. Then, by substituting -x for a, we see that the solution is simply the ordinary binomial expansion with alternating signs, just as everyone else has suggested.

In my opinion, this substitution is the best way to see "how" to get the binomial expansion, as the OP originally asked, because it demonstrates a method which reduces the problem to the expression OP already has, but shows how one can eliminate the added complexity of the minus sign, and explicitly justifies the treatment of -x used in the other answers.

Title: Binomial expansion of $(1-x)^n$

Body:
I'm not sure how appropriate it is to answer questions this old, but compared to the methods above, I feel the easiest way to see the answer to this question is to take

a = -x

And substitute that into the binomial expansion:

(1+a)^n

This yields exactly the ordinary expansion. Then, by substituting -x for a, we see that the solution is simply the ordinary binomial expansion with alternating signs, just as everyone else has suggested.

In my opinion, this substitution is the best way to see "how" to get the binomial expansion, as the OP originally asked, because it demonstrates a method which reduces the problem to the expression OP already has, but shows how one can eliminate the added complexity of the minus sign, and explicitly justifies the treatment of -x used in the other answers.


问:
Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$
Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.

Body:
With $n:=2m+1$, the polynomial is

$$(2m)^2(2m+2)^2(4m^2+4m+2)^2(16m^4+32m^3+24m^2+8m+2)
\\=2^7m^2(m+1)^2(2m^2+2m+1)(8m^4+16m^3+12m^2+4m+1).$$

The last two factors are always odd and the remaining ones form at least a multiple of $2^9$.

More precisely, if the multiplicity of the factor $2$ in the prime factorization of $\dfrac{n-1}2$ is $p>0$, then $k=2p+7$, otherwise $k=2q+7$ where $q$ is the multiplicity of $2$ in $\dfrac{n+1}2$.

E.g., with $n=15$, $p=0,q=3$ and $k=13$. Indeed, $15^{12}-15^8-15^4+1=2^{13}\,15837863153$, and with $n=17$, $p=3$ and $17^{12}-17^8-17^4+1=2^{13}\,71120027025$.


问:
Integral of $\int_1^N\frac{\{x\}}xdx$?
For $N\in\mathbb{N}$ how can the following integral be computed?

$$
\int_1^N\frac{\{x\}}{x}dx
$$

The notation $\{x\}$ is the fractional part of $x$, so $\{x\}=x-\lfloor x\rfloor$.

Apparently, the integral evaluates to $N-1-N\ln{N}+\sum_{n=1}^N\ln{n}$. How to show this fact?

After graphing $\{x\}/x$, I think I must find the area under each 'slice':

$$
\sum_{n=1}^{N-1}\int_n^{n+1}\frac{\{x\}}{x}dx
$$
But I cannot figure out an expression for the area under each slice. Is this the correct approach? Thanks!

答:
Try using this:

$$\int_{n}^{n+1} {\lfloor x\rfloor\, dx \over x}  = \int_{n}^{n+1} {n\, dx \over x}  =
n(\log(n+1) - \log(n)).$$

Title: Integral of $\int_1^N\frac{\{x\}}xdx$?


问:
The derivation of the product in Sobolev space
Let $f \in W^{1,\infty}(0,T)$
and $g \in W^{1,1}(0,T)$ $(0<T\leq\infty)$.

Is $(fg)'= f'g+fg'$ true?

Thanks

答:
Yes.

Such a statement will usually be proven by smooth approximation. In the reflexive range, so the corresponding statement for $f \in W^{1,p}(0,T)$ and $g \in W^{1,p'}(0,T)$ with $1 < p < \infty$, this is fully straight forward since smooth functions on $[0,T]$ are dense. This is however not the case for $W^{1,\infty}(0,T)$, so a bit more must be done.

One way could be like this. Extend $f$ and $g$ to functions $F \in W^{1,1}(\mathbb{R})$ and $G \in W^{1,\infty}(\mathbb{R})$ by your favorite reflection or so. Then pick a standard mollifier family $(\varphi_n)$ and set $f_n := (\varphi_n \star F)$, and $g_n$ analogously. Then (the restrictions to $(0,T)$ of) $f_n,g_n$ are smooth on $[0,T]$, with $f_n \to f$ and $g_n \to g$ in $W^{1,1}(0,T)$ and also each (by embedding) uniformly on $[0,T]$. Moreover,
$$\|g_n'\|_{L^{\infty}(0,T)} \leq \|g_n'\|_{L^{\infty}(\mathbb{R})} \leq \|G'\|_{L^{\infty}(\mathbb{R})}$$
(this is a standard mollifier property), so there is a weakly-$\star$ convergent subsequence w.r.t. the pair $L^1(0,T)^\star = L^\infty(0,T)$ of $(g_n')$ (Banach-Alaoglu theorem) which we do not relabel. Since $g$ is the uniform limit of $(g_n)$, the weak-$\star$ limit of $(g_n')$ must also be $g'$. But then \begin{multline*}(fg)(t) - (fg)(0) \quad \longleftarrow \quad (f_ng_n)(t) - (f_ng_n)(0) \\ = \int_0^t \bigl[f_n'g_n + f_ng_n'\bigr] \quad \longrightarrow  \quad \int_0^t \bigl[f'g + fg'\bigr]\end{multline*}
for every $t \in (0,T)$. In particular, $fg \in W^{1,1}(0,T)$ and the weak derivative is $$(fg)' = f'g + fg'.$$

OwnerDisplayName: Hannes


问:
A question on inverse functions
Let $f:\mathbb R \to \mathbb R$ is a strictly increasing function and $f^{-1}$ is its inverse function. It satisfies:

$f(x_1)+x_1=a$; $f^{-1}(x_2)+x_2=a$.

What is the value of $x_1+x_2$? Thanks for your help.

答: It seems the following. Since the function $f$ is strictly increasing, then  the functions $f^{-1}$ and $g(x)=f(x)+x$ are strictly increasing too. We have $a=g(x_1)=g(f^{-1}(x_2))$. Then $x_1=f^{-1}(x_2)$ and $x_1+x_2=f^{-1}(x_2)+x_2=a$.</p>

Title: A question on inverse functions


Title: Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$

Body:
Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.

Body:
$$(2r+1)^2+1\equiv8\cdot\dfrac{r(r+1)}2+2=8p+1\text{(say)}\equiv2\pmod8$$

$$(2r+1)^4+1=(8p+1)^2+1\equiv2\pmod{16}$$

Now $n^2-1=(2r+1)^2-1=8\cdot\dfrac{r(r+1)}2$

$\implies k\ge2\cdot3+2\cdot1+1$


答:
First, we start by factorising the expression just like you did:
$$(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$$
And we know, since $n$ is odd, each of the above terms is even: $(n-1)$, $(n+1)$ are even. $n^2$ and $n^4$ are also odd, so $(n^2 +1)$ and $(n^4+1)$ are even too. This means that we have 7 even factors.

BUT, there's one more important thing here, the two even numbers before and after an odd number, one of them has to be a multiple of 4. And this means that either $(n+1)$ or $(n-1)$ is a multiple of 4, not just 2, and because both terms are squared, we have to add 2 to our original result, 7, which makes our final result 9.


问:
Question about the standard orthonormal basis of nomalized eigenvectors of a positive definite matrix
Given a symmetric positive definite matrix $A$. May someone please explain the following statement for me?

Let $v_1, ..., v_n$ be the standard orthonormal basis of $A$. We define a vector, that is not orthogonal to $v_1$ by $\sum_{i=1}^n c_iv_i$ Hence, we obtain $\sqrt{\sum_{i=1}^n c_i^2}=1$.
Unfortunately in my lecture there were no details on $c_i$. Does somebody know why this is true?

答:
What is a basis of a matrix ??

Let $w=\sum_{i=1}^n c_iv_i.$ Let us denote the usual inner product on $ \mathbb R^n$ (or $ \mathbb C^n$) by $(\cdot, \cdot)$ and the induced norm $|| \cdot||= \sqrt{(\cdot, \cdot)}.$

By Pythagoras we get

$$||w||^2= \sum_{i=1}^n||c_i v_i||^2= \sum_{i=1}^nc_i^2.$$

If $||w||=1$, then we have  $\sum_{i=1}^nc_i^2=1$.

Do we have $||w||=1 ?$


答: The answer $k=9$ is correct but I don't think your reasoning is good. For example for expression $(n+1)^2(n^2+1)^2(n^4+1)$ this doesn't work because for $n=3$ the result is divided by $2^7$ but for $n=5$ it is divided only by $2^5$. I think the way to solve this is too look at expressions $n-1$, $n+1$, $n^2+1$, $n^4+1$ and try to find what's the least amount of 2's that could be dividing each of them.


问:
Confusion about equation of a plane
This is a confusion that I've always had, that I've never quite gotten rid of. (I'm definitely making a foolish mistake somewhere, I know... Sorry) Suppose $x,v \in \mathbb{R}^3$, and $c \in \mathbb{R}$. The equation of a plane in $\mathbb{R}^3$ is given: $\{x: \langle x, v\rangle=c\}$, where the inner product denotes the dot product. The reason why this is a plane is, for an $x_0$ such that $\langle x_0, v\rangle=c$, we have $\langle x-x_0, v\rangle=0$, so $x-x_0$ is a plane that is perpendicular to the vector $v$.

But alternately, using $x \cdot v = |x| |v| \cos (\theta)$ where $\theta$ is the angle between $x,v$, shouldn't $\{x: \langle x, v\rangle=c\}$ look like a circle around the vector $v$ in mid air (i.e. away from the origin of $\mathbb{R}^3$), where the vector $x_0$ (defined as above) has been rotated around the vector $v$? Because rotating the vector $x_0$ around the vector $v$, the angle between $x_0$ and $v$ stays equal, while there is no change in the length of the two vectors. But clearly this is wrong, as it must be a plane, not a circle.

答:
In the situation you ask about where $\,P := \{x: \langle x, v\rangle=c\}\,$
is the equation of a plane in $\,\mathbb{R}^3,\,$ there exists a unique point
$\,O\in P\,$ where the angle $\,\theta\,$ between $\,O\,$ and $\,v\,$ is $\,0.\,$
This point $\,O\,$ is the origin in a polar coordinate system for the plane
$\,P.\,$ The points $\,x\,$ with a constant angle between them and $\,v\,$ are a
constant distance from $\,O\,$ in the plane and hence they form a circle.
Conversely, all circles in the plane $\,P\,$ with a fixed distance from the origin
$\,O\,$ are described by $\,\{x\in P: |x| = d\}\,$ for some $\,d\,$ or alternatively  $\,\{x\in P: |x-O| = r\}\,$ for some $\,r.\,$

Title: Confusion about equation of a plane


问:
How to find shortest distance from triangle to point in 2D space?
I have a triangle with three corners located at $(-6,7), (2,5)$ and $(-4,8)$. I need to calculate shortest distance from nearest edge of triangle to the point $(-4,7)$. I can't figure out how to do that, could I have some advise?

答:
You find the equations of the 3 lines that make the triangle:
$$m = \frac {y_2-y_1} {x_2-x_1} \\ y - y_1 = m(x-x_1)$$
Then you can simplify, and use this formula to find distance between $(-4,7)$ and each line, for a line of equation $ax+by+c=0$:
$$\operatorname{distance}= \frac {ax+by+c} {\sqrt {a^2 + b^2}}$$


问:
Maximum Likelihood Estimation for Markov Chains
According to calculations in this paper
https://www.stat.cmu.edu/~cshalizi/462/lectures/06/markov-mle.pdf

The first line just uses the definition of conditional probability, but the second
line actually uses the Markov property — that the future is independent of the
past, given the present.

Then the equation (4) becomes equation (5). I did not understand the logic of using $N_{ij}$ here. If I assume that $n=3$ ,then the equation (4) is

$$Pr(X_1,X_2,X_3) =  Pr(X_1=x_1)Pr(x_2|x_1)Pr(x_3|x_2) $$

Even if we use the $N_{ij}$ , it will always be equal to the $1$.So why do we need to use $N_{ij}$?

Thank you

答:
You can see that by taking an example, say a MC with just $k=2$ states $\{1,2\}$. Then you run it for $n=10$-steps. Then the formula (4) is still valid but note that each $p_{ij}$ can not just occur once. The total number of transitions will be $9$ (assuming the initial time point is $1$). Therefore some of those $p_{ij}$'s must repeat and how many times? Precisely the number of times the MC jumps from $i$ to $j$.

More precisely, equation (3) or (4) are probabilities of a path visited by the MC. Take for example (in the $2$-states case) a path $(x_1=1,x_2=1,x_3=2,x_4=1,x_5=1,x_6=2,x_7=2,x_8=1,x_9=1,x_{10}=2)$. Then using equation (3) or (4) gives probability of this path is equal to
$$
Pr(X_1=1)p_{11}p_{12}p_{12}p_{21}p_{11}\ldots p_{12}= Pr(X_1=1)p_{11}^{3}p_{12}^{3}p_{21}^{2}p_{22}.
$$


问:
$\lim_{n \to \infty} f_{n}(x_{n})=f(a)$
$(f_{n})_{n}$ a row of continuous function in $[0,1]$ who converges pointswise to a continuous function $f:[0,1]\to \mathbb{R}$.

Can you conclude dat for every $a \in [0,1]$ and every row $(x_{n})_{n} \in [0,1]$ who converges to $a$ :

$\lim_{n \to \infty} f_{n}(x_{n})=f(a)$ ?

I had a prove for if $f_{n}$ converges uniformly to $f$, but I can't see where it goes wrong when $f_{n}$ converges pointswise to $f$. Here my prove for $f_n$ converges uniformly.

take $a \in [0,1]$ and $(x_n)_n \in [0,1]$ randomly with $(x_n)_n \to a$. take $\epsilon_1$>0 random but fixed. Because $(x_n)_n \to a$ there exist an $n_1 \in \mathbb{N}$ so that for every $n>n_1$ , $|x_n -a|<\epsilon_1$.

Now take $\epsilon_2$>0 random.  Because $(f_n)_n$ is a row of continuous functions and  $|x_n -a|<\epsilon_1$ , $|f_n(x_n)-f_n(a)|<\epsilon_2/2$.
Because $(f_n)_n$ converges uniformly to $f$ there exist an $n_2 \in \mathbb{N}$ so that for $x \in [0,1]$ and $n>n_2$, $|f_n(x)-f(x)|<\epsilon_2/2$.

Take now $n_0=\min\{n_1,n_2\}$. Take $n>n_0$ randomly then $|f_n(x_n)-f(a)|=|f_n(x_n)+f_n(a)-f_n(a)-f(a)|<|f_n(x_n)-f_n(a)|+|f_n(a)-f(a)|<\epsilon_2$

EDIT: as an counter example for pointwise I have

$f_n:[0,1]\to \mathbb{R}: x\to \left\{ \begin{matrix}\mbox{2nx} & 0<=x<1/2n\\ \mbox{2-2nx } & 1/2n <=x<=1/n\\ \mbox{0} & 1/n<x<=1 \end{matrix}\right.$

with $(x_n)_n=1/2n$

Now I just want to know why my prove wouldn't work if I assume pointswise convergence.

答:
Let $f_1(x):=2x$ for $0\leqslant x\leqslant\frac{1}{2}$, and let $f_1(x):=f_1(1-x)$ for $\frac{1}{2}\leqslant x\leqslant 1$, and let $f_1(x):=0$ for $x>1$.

Now define $f_n$ on $[0,1]$ by $f_n(x):=f_1(nx)$.

Then $f_n(x)\to 0$ on $[0,1]$, so the limit function $f$ is identically $0$

However with $x_n=\frac{1}{2n}$ we have that $x_n\to 0$ but $f_n(x_n)\to 1\not=f(0)$.

Title: $\lim_{n \to \infty} f_{n}(x_{n})=f(a)$


问:
Prove $P(A\cap B\cap C)=P(B\cap C)$.
If $A, B, C$ are events, and $P(A\vert B)=1$, prove $P(A\cap B\cap C)=P(B\cap C)$.

I have tried to prove it.

$P(A\vert B)=1\iff \dfrac{P(A\cap B)}{P(B)}=1\iff P(A\cap B)=P(B)$.

Is it right, if
$$P(A\cap B)=P(B)$$
then
$$A\cap B=B,$$
so
$$A\cap B\cap C=B\cap C,$$
and we have
$$P(A\cap B\cap C)=P(B\cap C)?$$

答:
Let $X_A, X_B$, and $X_C$ be the indicator variables corresponding to the events $A, B,$ and $C$. From the given relation, we have
$\mathbb{E}(X_B(1-X_A))=0$.
However, due to the non-negativity of the term within expectation, we have almost surely, $X_B(1-X_A)=0$.
The result now follows by multiplying both sides by $X_C$ and taking expectation.

Title: Prove $P(A\cap B\cap C)=P(B\cap C)$.


问:
Image of a group equal to his kernel
If we have a function $f: G\rightarrow G$ , is there an example of a group $G$ that $imf=kerf$?

I try to think about the trivial group but there is another one?

Note: $f$ is Homomorphism.

答:
Sure, there's many. Here's a family:

Take any group $H$, let $G=H\times H$ and consider
\begin{align}
H \times H &\longrightarrow H\times H, \\
(x, y) &\longmapsto (y, e_H).
\end{align}
Kernel and image are equal to $H\times\{e_H\}$.

Title: Image of a group equal to his kernel


答:
Take $f\colon C_2\times C_2\rightarrow C_2\times C_2$, given by $(x,y)\mapsto (y,1)$.

Then $\ker(f)=C_2\times 1$ and $C_2\times 1=f(G)={\rm im}(G)$.

Title: Image of a group equal to his kernel

Body:
If we have a function $f: G\rightarrow G$ , is there an example of a group $G$ that $imf=kerf$?

I try to think about the trivial group but there is another one?

Note: $f$ is Homomorphism.


问:
$L^1(\mathbb{P}) \neq L^1(\mathbb{Q})$ in general?
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $L^1(\mathbb{P})$ denote the collection of all random variables that are $\mathbb{P}$-integrable.

Now let $\mathbb{Q}$ be an absolutely continuous probability measure w.r.t $\mathbb{P}$ and denote its Radon-Nikodym derivative by $Z$.  Denote the collection of random variables that are $\mathbb{Q}$-integrable by $L^1(\mathbb{Q})$.  We know that $X \in L^1(\mathbb{Q})$ if and only if $ZX \in L^1(\mathbb{P})$.

Is it true that if $Z \notin L^\infty$, then there exists $X \in L^1(\mathbb{P})$ such that $\mathbb{E}[ZX] = \infty$?

I am fairly sure this is true, but I am not sure how to prove it.

答:
By assumption, as $Z \notin L^{\infty}$, we have $$\mathbb{P}(Z > n^2) > 0, \quad n \in \mathbb{N}.$$ Therefore, we may define $$X = \sum_{n=1}^{\infty}\frac{1}{n^2} \frac{\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)} \in L^1(\mathbb{P}).$$ However, it follows that

\begin{align*}\mathbb{E}[ZX] = \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{Z\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)}\right] \geq \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{n^2\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)}\right] = \sum_{n=1}^{\infty}1 = \infty.\end{align*}


问:
Elliptic curve, number of elements of finite order
I've been looking at this problem for some time now and just can't seem to get the right idea. The problem is:

Consider the elliptic curve $C:y^2=x^3+bx$ defined over the rational numbers with $b$ a non-zero integer.
Show that there are either 2 or 4 points of finite order.

Here's what I know:
If $p$ is a prime with $p\equiv 3\mod{4}$ and $p\not| 2b$ then there are $p+1$ points of finite order when considering $C$ over $F_p$. So by the reduction modulo $p$ theorem the number of points with finite order must divide $p+1$.

From this I can deduce that 2 and 4 are options since $p+1\equiv 0\mod{4}$ and that 1 is not an option since I have the point at infinity and $(0,0)$.

But how do I argue that these are the only possibilities i.e. for all $p$ satisfying the above 4 is the only factor that all the $p+1$'s have in common.

答:
You are almost done! Let $q$ be an odd prime. We have to construct a prime $p \equiv 3 \mod 4$ such that $q \nmid p+1$, and such that $p \nmid b$.

First consider the case $q \neq 3$. Consider the arithmetic progression $3+4qk$. By Dirichlet's theorem, it contains infinitely many primes; take one sufficiently large so that it doesn't divide $b$, say $p=3+4qk_0$, which is $\equiv 3 \mod 4$. Then $p+1 = 4(1+qk_0)$ is not divisible by $q$.

For $q=3$, consider instead the progression $7+12k$. Any prime $p$ in this progression is $\equiv 3 \mod 4$ and satisfies $p+1 \equiv -1 \mod 3$. Once again, taking it to be sufficiently large we can make sure that it doesn't divide $b$.

Now, can you show that there exist infinitely many primes $p \equiv 3 \mod 4$ such that $8 \nmid p+1$?


问:
Finding complex roots of quadratic equation to the power of 3
Here's the problem:

Find the 6 complex numbers solutions of the equation $(x^2+x+1)^3=8i$, presenting your solutions in exponential form

I'm acquainted with solving complex roots of quadratic equations, but not one with exponents.

Here is my working so far: I'm not sure if I am approaching this question right

$(x^2+x+1)^3 = 8i$

$(x^2+x+1) = 2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}$

$(x+\frac{1}{2})^2+\frac{3}{4}=2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}$

$(x+\frac{1}{2})^2=2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}-\frac{3}{4}$

$x+\frac{1}{2}= \sqrt{2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}-\frac{3}{4}}$

答: The number $8i$ is equal to $2^3e^{\pi i/2}$. Therefore, its cubic roots are $2e^{\pi i/6}$, $2e^{5\pi i/6}$, and $2e^{3\pi i/2}(=-2i)$. So, solve the equations $x^2+x+1=\omega$, for each $\omega\in\left\{2e^{\pi i/6},2e^{5\pi i/6},-i\right\}$.


答:
As $8i=(-2i)^3$

$$\left(\dfrac{x^2+x+1}{-2i}\right)^3=1$$

$\implies \dfrac{x^2+x+1}{-2i}=w$  where $w$ is a cube root of unity

$\iff x^2+x+1+2iw=0$

Title: Finding complex roots of quadratic equation to the power of 3

Body:
Here's the problem:

Find the 6 complex numbers solutions of the equation $(x^2+x+1)^3=8i$, presenting your solutions in exponential form

I'm acquainted with solving complex roots of quadratic equations, but not one with exponents.

Here is my working so far: I'm not sure if I am approaching this question right

$(x^2+x+1)^3 = 8i$

$(x^2+x+1) = 2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}$

$(x+\frac{1}{2})^2+\frac{3}{4}=2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}$

$(x+\frac{1}{2})^2=2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}-\frac{3}{4}$

$x+\frac{1}{2}= \sqrt{2e^{\frac{i\pi}{6}+\frac{2ki\pi}{3}}-\frac{3}{4}}$


问:
I need to define the expression for the function f(x).
The function f: R → R is differentiable and satisfies the conditions

f(2) = 1

f(x) + xf´(x) = 2 with all x ∈ R.

Define the expression for the function f.

答:
$$xf'+x=2 \implies \frac{d(xf)}{dx}=2~\text{(integration)}\implies xf =2x+C \implies f(x)=2+\frac{C}{x}$$ Next $f(2)=1 \implies C=-2.$
So $f(x)=2(1-1/x)$ but it is not continuous and differentiable at $x=0$.


问:
how to calculate the curvature
The Question is: Let $\gamma(t)$ be a regular curve lies on a sphere $S^2$ with center $(0, 0, 0)$ and radius $r$. Show that the curvature of $\gamma$ is non-zero, i.e., $κ \ne 0$.

My question:

We define $k$ as $k=\frac{(||\gamma'' \times \gamma'||)}{|| \gamma' || ^3} $

The equation of the sphere is: $x^2 + y^2 + z^2 = r$

So, we have $\gamma = (\sqrt{r} \cos u \sin v,\sqrt{r} \sin u \sin v, \sqrt{r} \cos v$)

How can I calculate $\gamma'$ and $\gamma''$ ? because $\gamma$ depends on two variables $u,v$

Thank you.

答: The quickest way of seeing this is by using the formula $k^2=k_g^2 + k_n^2$.  Here $k_g$ is the geodesic curvature and $k_n$ is the normal curvature.  In the case of the sphere the normal curvature is the inverse of the radius of the sphere.  Hence curvature is positive.</p>

Title: how to calculate the curvature


问:
What is the sum of the coefficients?
If $h(x) = x^4+ax^3+bx^2+cx+d$ then what is $a+b+c+d$?

I try:
\begin{align}
x=2: 2^4+2^3a+2^2b+2c+d = 3 &\implies 8a+4b+2c+d = -13 \label{I} \tag{I}\\
x=-2:-2^4-2^3a-2^2b-2c+d=3 &\implies  -8a+4b-2c+d = -13 \label{II} \tag{II}\\
\eqref{I} + \eqref{II} \colon 8b +2d = -26 &\implies \boxed{4b+d = -13} \ em \tag{I}\\
8a-13+2c=-13 &\implies 8a+2c = 0 \implies\boxed{4a+c = 0}
\end{align}

I stop here..don't find another equation

答:
By its graph $\;h\;$ is even and thus $\;a=c=0\;$. we're also given $\;h(2)=3\;$ and thus

$$3=h(2)=2^4+b\cdot2^3+d=16+8b+d\implies 8b+d=-13$$

...and I think some data is missing since it isn't possible to know from the given graph what is, for example, $\;h(0)=d\;$ ...unless you're allowed to use calculus and evaluate the first derivative, where it vanishes and etc. But you didn't tag this and didn't say anything about this...

Assuming you can use calculus: the first derivative vanishes at $\;x=2\;$ , thus

$$h'(2)=2\cdot2\cdot(2\cdot2^2+b)=0\implies b=-8$$

and then you the value of $\;d\;$ and we're done

Title: What is the sum of the coefficients?


问:
Distribution of the cosine of an angle of a random oriented vector in a sphere
I can't wrap my head around a problem, I have a methane molecule, which is basically a tetrahedron, and I orient one of its bonds (let's say from atom 1 and atom 2) along the $z$ axis of a  cubic box.

Now I know that the probability distribution for the angle $A$ between the 1-2 vector and the $z$ axis would be shaped as a sine function, but i think I understood that $\cos(A)$ instead should have a flat probability $P[\cos(A)]$ distribution (from https://www.lockhaven.edu/~dsimanek/puzzles/hatbox.htm#:~:text=While%20puzzling%20how%20to%20find,radius%20R%20and%20height%202R.%22) .

Now If I make my molecule rotate around the  $y$ axis, so that I have 360 structures all equispaced of 1 degree I expected to see a flat $P[\cos(A)]$ distribution, and instead i get a distribution shaped like this:

Then I tried to get random numbers between 0 and 360 and use them to rotate the molecule around $x, y$ and $z$ of that number of degrees, and I still get the same distribution.

Then I tried to get three random numbers between $-1$ and $1$ and using arccos of those numbers to rotate the molecule around $x,y$ and $z$. And I still get the same shape of the distribution.

Can someone help me wrap my head around this?

答:
What you are looking for is the distribution of the random variable $Y=\cos (X)$, where the angle $X$ is a random variable with the probability distribution function $$f_X(x)=\frac1\pi\quad\text {for}\quad 0 \le x \le \pi. $$
[Note that we need not to consider the range $[\pi,2\pi]$ in view of $\cos(2\pi-x)=\cos x$.]

Let us solve the problem using the distribution function technique. First we find the cumulative probability distribution function:

$$F_Y (y)=P (Y\le y)=P (\cos (X)\le y)=P (X\ge \arccos (y))\\
=\int\limits_{\arccos(y)}^\pi f_X (x)dx=1-\frac1\pi\arccos (y) .$$
The probability density function is then:
$$f_Y (y)=\frac d {dy}F_Y(y)=\frac1 {\pi\sqrt {1-y^2}},
$$
in (at least qualitative) agreement with the result of your numerical simulation.


问:
regular value of homotopy F is regular value for F_t
Let $M,N$ be smooth manifolds of the same dimension, let $I=[0,1]$ and say we have a smooth map $F:M \times I \to N$ with regular value $y \in N$. Is $y$ then also a regular value of $F_t := F(\cdot,t)$ for all $t \in I$?

答:
No. Consider $M = N = S^1$, parameterized by an angle $\theta$. Define

$$
F(\theta, t) = t
$$
Then the derivative of $F$ at $(\theta, t)$ is
$$
\pmatrix{0\\1}
$$
which is maximal rank, hence $F$ is regular at $(\theta, t)$.

On the other hand, $F_0(\theta, t) = 0$ for all $\theta$, hence its derivative is the $0$-matrix, which does not have maximal rank.


问:
Try to show the marginal distribution of a normal random variable with a normal mean is indeed normal?
Let $X_1,...,X_n$ be a random sample from a $n(\theta,\sigma^2)$ population, and suppose that the prior distribution on $\theta$ is $n(\mu,\tau^2)$. Here we assume that $\sigma^2,\mu,\tau^2$ are all known. Show that $m(\bar{x}|\sigma^2,\mu,\tau^2)$, the marginal distribution of $\bar{X}$, is $n(\mu, (\sigma^2/n)+\tau^2)$.

I am trying to solve this question by MGF. The general mgf of normal distribution is $M_X(t)=exp(t\mu+\frac{1}{2}t^2 \sigma^2)$.

Hence, the mgf of $\bar{X}$ is $exp(\bar{x}\theta+\frac{1}{2}\bar{x}^2 \sigma^2/n)$.

The mgf of $\theta$ is   $exp(\theta \mu+\frac{1}{2}\theta^2 \tau^2)$.

I assume the mgf of the marginal distribution of $\bar{X}$ is $\int exp(\bar{x}\theta+\frac{1}{2}\bar{x}^2 \sigma^2/n) exp(\theta \mu+\frac{1}{2}\theta^2 \tau^2) d\theta$? Am I correct? If yes, how to simplify this integral?

答:
Interesting problem.

First observe that $(\overline{X}_n|\theta)\sim N\left(\theta;\frac{\sigma^2}{n}\right)$

To show that, marginally, $\overline{X}_n\sim N\left(\mu;\frac{\sigma^2}{n}+\tau^2\right)$, let's set

$$Y=\overline{X}_n-\theta$$

and observe that, $\forall \theta$, the conditional rv $(Y|\theta)\sim N\left(0;\frac{\sigma^2}{n}\right)$ thus $Y\perp\!\!\!\perp\theta$ ($Y$ is independent from $\theta$) and thus it results that

$$\overline{X}_n=Y+\theta$$

is the sum of 2 independent Gaussians and for this reason its distribution is still normal with mean $0+\mu=\mu$ and variance $\frac{\sigma^2}{n}+\tau^2$, that is:

$$\overline{X}_n\sim N\left(\mu;\frac{\sigma^2}{n}+\tau^2\right)$$


答:
Hint

I believe there is a better approach for this problem.

Considering the graph, the polinomial $h(x)-3$ has a double root at $-2$ and $2$. So, it can be writen as

$$h(x)-3=k(x-2)^2(x+2)^2=k(x^2-4)^2.$$

In addition, $h(1)$ give you the sum of the coefficients of $h(x)$.

Can you finish?

Title: What is the sum of the coefficients?

Body:
If $h(x) = x^4+ax^3+bx^2+cx+d$ then what is $a+b+c+d$?

I try:
\begin{align}
x=2: 2^4+2^3a+2^2b+2c+d = 3 &\implies 8a+4b+2c+d = -13 \label{I} \tag{I}\\
x=-2:-2^4-2^3a-2^2b-2c+d=3 &\implies  -8a+4b-2c+d = -13 \label{II} \tag{II}\\
\eqref{I} + \eqref{II} \colon 8b +2d = -26 &\implies \boxed{4b+d = -13} \ em \tag{I}\\
8a-13+2c=-13 &\implies 8a+2c = 0 \implies\boxed{4a+c = 0}
\end{align}

I stop here..don't find another equation


问:
Density function of given Interval
The Density function of a random variable x is given by

```<code>f(x) = x^2 + 3x + 2 on the interval [1, 4].
</code>```

The functions is zero if x > 4 or x < 1

What is the average value of a sample of a million random numbers generated from this function?

答:
There are a couple conceptual hurdles I will address. In regards to problems involving density functions I have never seen before, one of the first things I will do is first verify that the function given to me is indeed a density function.

By definition, a density function must be nonnegative and integrate to one so we verify this first. Now, $f$ is clearly nonnegative, i.e. $f\geq 0$, everywhere outside the interval $[1,4]$. Further investigation into the function on $[1,4]$ also reveal $f$ satisfies the nonnegativity criteria. Next, we check to see if $f$ integrates to one. Let $\mathbf 1_{A}$ denote the indicator function which, by definition, is one when $A$ is true and zero otherwise. Then, integrating our function $f$ we have
$$
M
=\int_{-\infty}^\infty f(x)\,\mathrm dx
=\int_{-\infty}^\infty (x^2 + 3x + 2) \mathbf 1_{x\in[1,4]}\,\mathrm dx
=\int_1^4 (x^2 + 3x + 2)\,\mathrm dx
=\frac{99}{2}.
$$
So we see that $f$, as given, is not a density function. As such, we will define the density function by normalizing $f$ via
$$
g_X(x)=\frac{f(x)}{M}=\frac{2}{99}(x^2 + 3x + 2) \mathbf 1_{x\in[1,4]}.
$$
Now that we have a density function our next question is:

What is the average value of a sample of a million random numbers generated from this function?

Using inverse transform sampling I generated one-million observations from $g_X$ and took the mean to get: $\bar x=2.8629$. Generating another million and taking the mean gives: $\bar x=2.86388$. Indeed, if we keep repeating this procedure we will get a different answer every time which is due to the fact that the data generating the averages is random (or at least pseudo-random)! As such, our question cannot be answered.

Now, we can assign an expected value to our density $g_X$ which can loosely be thought of as the "average value" obtained from an infinite number of samples. Using the definition of expected value, if $X\sim g_X$, then
$$
\mathsf E X=\int_{-\infty}^\infty xg_X(x)\,\mathrm dx=\int_1^4 x\frac{2}{99}(x^2 + 3x + 2)\,\mathrm dx=63/22=2.86364.
$$

Title: Density function of given Interval


问:
Integral of $\int_1^N\frac{\{x\}}xdx$?
For $N\in\mathbb{N}$ how can the following integral be computed?

$$
\int_1^N\frac{\{x\}}{x}dx
$$

The notation $\{x\}$ is the fractional part of $x$, so $\{x\}=x-\lfloor x\rfloor$.

Apparently, the integral evaluates to $N-1-N\ln{N}+\sum_{n=1}^N\ln{n}$. How to show this fact?

After graphing $\{x\}/x$, I think I must find the area under each 'slice':

$$
\sum_{n=1}^{N-1}\int_n^{n+1}\frac{\{x\}}{x}dx
$$
But I cannot figure out an expression for the area under each slice. Is this the correct approach? Thanks!

Title: Integral of $\int_1^N\frac{\{x\}}xdx$?

Body:
As you know the integral is $$ \begin{align} \sum_{n=1}^{N-1} \int_{n}^{n+1} \dfrac{\{x \}}{x}dx &=  \sum_{n=1}^{N-1} \int_{n}^{n+1} \dfrac{x- \lfloor x \rfloor}{x}dx \\
&=  \sum_{n=1}^{N-1} \int_{n}^{n+1} 1- n \dfrac{1}{x}dx \\
&= N-1 - \sum_{n=1}^{N-1} n \ln(n+1) - n\ln(n) \\
&= N-1 - \sum_{n=1}^{N-1} (n+1) \ln(n+1) - n\ln(n) - \ln(n+1) \end{align} $$

And now you can continue


问:
Find all solutions $z\in \mathbb C$ for the following equation: $z^2 +3\bar{z} -2=0$
The question is

Find all solutions $z\in \mathbb C$ for the following equation: $z^2 +3\bar{z} -2=0$

I have attempted numerous methods of approaching this question, from trying to substitute $x+iy$ and $x-iy$ respectively, in addition to substituting $z^2$ for $z\bar z$, but with no luck. I would really appreciate if you were able to provide some direction so I know where to start. Thank you!

答:
No need for real and imaginary parts here.

From $z^2 +3\bar z = 2$ we have  $\overline{z^2 +3\bar z} = \bar 2$.  Now $\overline{z^2 +3\bar z}= \bar z^2+3z$.
Hence we have
$$\begin{align}
z^2 +3\bar z &= 2\\
\bar z^2+3 z &=2
\end{align}
$$

Subtracting gives
$$(z-\bar z)(z+\bar z-3)=0,$$
that is $z=\bar z$ (hence $z$ is real in this case) or $\bar z =3-z$.  Both lead to quadratic equation: the first has two real solutions, the latter two complex solutions.


问:
Still not understanding a detail in a posted proof of an algebraic inequality
While working through a proof I am struggling with one section in it where the following estimate is established:

For real numbers $\,a,b,c\,$ which satisfy $\,0<a,b\,$ and $\,0\leqslant c\leqslant a,b\,$ one has
$$\frac ab + \frac ba \;\leqslant\; \frac{a^2+bc}{b^2+c^2} + \frac{b^2+ac}{a^2+c^2}\tag{1}\,.$$

The line of argumentation starts from $\,ac+bc\leqslant a^2+b^2\,$
which is clear from the givens. $\checkmark$

Then
$$ab(a-b)+c(bc-a^2)\leqslant ab(a-b)+c(b^2-ac)\tag{2}$$
follows. $\checkmark$

Despite the WLOG-assumption $b\leqslant a$, I cannot grasp the subsequent implication
$$\frac{ab(a-b) +bc^2-a^2c}{a(a^2+c^2)}\;\leqslant\;\frac{ab(a-b)+b^2c-ac^2}{b(b^2+c^2)}\,,\tag{007}$$

which goes along the guideline "Increase the numerator & Decrease the denominator", because

both numerators in $(007)$ alias $(2)$ may be negative!

I hope you see my difficulty in understanding this one.

Thus may I ask if $(1)$ is correct & how it can be proven eventually?

The said proof belongs to an inequality on AoPS, and the solution which I'd like to understand was posted on 24/04/2016 by luofangxiang.
A screen-shot of it is also included, to avoid missunderstanding which one is meant.

Edit to summarise:

The RHS of $(2)$ cannot get negative. This rectifies my above-noted statement in italics.

From mathlove's answer one deduces
$$ab(a-b) +b^2c -ac^2 \:=\: (a-b)(b-c)(a-c) + c(a-c)^2 + c^2(b-c)\:\geqslant\: 0\,.$$
Thus overall, the estimate $(1)$ is a correct one.

答:
If $0\leqslant c\leqslant b\leqslant a$, then RHS of $(2)$ is non-negative.

Letting $b=c+p$ and $a=c+p+q$ where $p,q\geqslant 0$, we get
$$\text{RHS of $(2)$}=c^2 p + c p^2 + 2 c p q + c q^2 + p^2 q + p q^2\geqslant 0$$

So, $(007)$ is correct.


问:
Complement of a Disconnected Graph is Connected
Let $G$ be a disconnected graph. Prove that its complement $\bar{G}$ is connected.

I would like to check if my proof of the above (rather famous) problem is valid. It is as follows:

Since $G$ is disconnected, its vertex set can be partitioned into $2$ disjoint vertex sets, $V_1$ and $V_2,$  such that each vertex is only adjacent to vertices in the same set as it. Hence, in $\bar{G}$, each vertex is adjacent to all the vertices in the other set, and will not be adjacent to any vertex in the same set as it. in other words, $\bar{G}$ is a complete bipartite graph, which is clearly connected, so we are done.

答:
This is not correct.  It is in general not true that each vertex in $V_1$ will be adjacent to all other vertices in $V_1$. That would only be the case with graphs that are a union of several complete graphs.

So the complementary graph is certainly not always a complete bipartite graph, it could have more edges as well.  But the important thing here is that this complete bipartite graphs is always a subgraph of $\bar{G}$ containing all vertices of $\bar{G}$. Therefore it is indeed connected.


问:
If $x\le y$, prove $\lfloor{x}\rfloor \le \lfloor{y}\rfloor$
I've just started my infinitesimal course, and I would like to see if I got the essence of floor right.

Let $x, y \in \mathbb R$ and $x\le y$ , prove that $\left\lfloor x \right\rfloor\le \left\lfloor y \right\rfloor$.

So, we know from the definition of floor that:

A. $\left\lfloor x \right\rfloor\le x < \left\lfloor x \right\rfloor +1$

B. $\left\lfloor y \right\rfloor \le y < \left\lfloor y \right\rfloor +1$

Now, since $x\le y$:

$\left\lfloor x \right\rfloor \le x \le y \rightarrow \left\lfloor x \right\rfloor \le y$

Also from the definition, we understand:

A. $\left\lfloor x \right\rfloor = max\{n\in \mathbb Z | n\le x\} $

B. $\left\lfloor y \right\rfloor = max\{n\in \mathbb Z | n \le y\}$

Is it okay to derive from here since $x\le y$ and $\left\lfloor x \right\rfloor \le y$

That also $\left\lfloor x \right\rfloor \le \left\lfloor y \right\rfloor$?

Thanks!

答:
Since floor is defined as a maximum of a bounded above set of numbers, you want to use the following property of maxima of bounded above subsets of $\mathbb R$:

When $A\subseteq B$ then $\max A \le \max B$.

Now note that when $x\le y$ you have
$$
\{\,n\in\mathbb Z\mid n\le x\,\} \subseteq \{\,n\in\mathbb Z\mid n\le y\,\}
$$
by transitivity of $\le$ and thus the above identity yields $\lfloor x\rfloor \le \lfloor y\rfloor$.


问:
Proving the intersection of an open set by induction
I am trying to make sense of this question:

$S_n$ = $(-\frac{1}{n},\frac{1}{n})$,  $n\in \mathbb{N}$

Find $\cup_{n\geq1}S_n$,  $\cap_{n\geq1}S_n$, then use induction to show that $\cap_{1\leq k \leq n} = (-\frac{1}{n},\frac{1}{n})$

I've already found $\cup_{n\geq1}S_n = S_1 = (-1,1)$ and  $\cap_{n\geq1}S_n=\emptyset$ .
However, I am unsure on how to go about doing the induction step for  $\cap_{1\leq k \leq n} = S_n =(-\frac{1}{n},\frac{1}{n})$

答:
$\bigcap_{n\geq1}S_n=\emptyset$ is not true. We have $\bigcap_{n\geq1}S_n=\{0\}.$

$$\bigcap_{1\leq k \leq n+1}S_k = (\bigcap_{1\leq k \leq n}S_k) \cap S_{n+1} =S_n \cap S_{n+1}=S_{n+1}.$$


答:
Another (perhaps more direct) way too see this is to split the problem into two cases:
if $y \ge \lfloor x \rfloor + 1$ then $\lfloor y \rfloor \ge \lfloor x \rfloor + 1$ since this quantity is in $\mathbb{Z}$, hence $\lfloor y \rfloor \ge \lfloor x \rfloor $.

Otherwise, $x \le y < \lfloor x \rfloor + 1$ in which case $\lfloor y \rfloor = \lfloor x \rfloor$ since this is the biggest integer smaller than $y$.

In both case we conclude that $\lfloor x \rfloor \le \lfloor y \rfloor$.

Title: If $x\le y$, prove $\lfloor{x}\rfloor \le \lfloor{y}\rfloor$

Body:
I've just started my infinitesimal course, and I would like to see if I got the essence of floor right.

Let $x, y \in \mathbb R$ and $x\le y$ , prove that $\left\lfloor x \right\rfloor\le \left\lfloor y \right\rfloor$.

So, we know from the definition of floor that:

A. $\left\lfloor x \right\rfloor\le x < \left\lfloor x \right\rfloor +1$

B. $\left\lfloor y \right\rfloor \le y < \left\lfloor y \right\rfloor +1$

Now, since $x\le y$:

$\left\lfloor x \right\rfloor \le x \le y \rightarrow \left\lfloor x \right\rfloor \le y$

Also from the definition, we understand:

A. $\left\lfloor x \right\rfloor = max\{n\in \mathbb Z | n\le x\} $

B. $\left\lfloor y \right\rfloor = max\{n\in \mathbb Z | n \le y\}$

Is it okay to derive from here since $x\le y$ and $\left\lfloor x \right\rfloor \le y$

That also $\left\lfloor x \right\rfloor \le \left\lfloor y \right\rfloor$?

Thanks!


问:
Optimal strategy in three-player number-picking game
Consider a game in which there are three players. Call them Player $1$, Player $2$, and Player $3$. Here are the rules:

Each player is supposed to select an integer between $1$ and $100$.

Player $1$'s number is randomly generated. Player $2$ and Player $3$ both know that Player $1$'s number is randomly generated.

The person with the largest number has to pay the other two people the number that each one of them said (i.e. say Player $1$ picks $5$, Player $2$ picks $70$ and Player $3$ picks $90$. In this case, Player $3$ pays $5$ to Player $1$ and $70$ to Player $2$).

Let's suppose you are Player $3$. Furthermore, suppose that Player $2$ plays optimally. What's the best strategy if you want to maximize profit?

I solved the $n = 2$ (two-player game) case here: Optimal strategy in probability-based game

I want to extend it to $n = 3$, but I can't figure it out. I would appreciate any help.

答:
Actually, there are only 2 players in this game. Player 1 does not choose a strategy between a set of strategies.

This game could be represented as $\mathcal{G}(X_1,X_2,H_1,H_2)$, where $X_i=[0,100]\cap\Bbb{Z}$, for i={1,2} and:

$$H_i(x_1,x_2)=
\begin{cases}
E(Z_i),  & \text{if $x_i> x_j$} \\
\frac{E(Z_i)}{2},  & \text{if $x_i= x_j$} \\\\
x_i, & \text{$x_i<x_j$}
\end{cases}$$

Now suposse $x_j<x_i$:  $ E(Z_i\mid x_i > Y)=-(\frac{x_i+1}{2}+x_j)$

$E(Z_i) = \left(\frac{1 - x_i}{100}\right)\left(\frac{x_i + 1}{2}+x_j\right) + \left(\frac{100x_i - x_i^2}{100}\right)=\frac{-3x_i^2-2x_ix_j+200x_i+2x_j+1}{200}$

The new optimum is $x_i=33-2x_j$, for  $ i,j=\left\lbrace1,2\right\rbrace$, $j\neq i$.

The solution to that system is $(11,11)$, which is the Nash equilibrium of $\mathcal{G}$.

Edit: Now assuming that if two of them tie, they will pay half. Notice that if they tie, the payment function had the same optimum as the original.
This solution is consistent with the previous game, if both of them tie, they will have to pay $\frac{x+x}{2}=x$ to each other, so the benefits will be $0$.


答:
$\bigcap_{n\geq 1} S_n$ is actually not the empty set. There is a single number that is an element of $S_n$ for all $n$.

For the induction, take your time and go slowly. Remember, for induction, you have the following steps:

Take a statement $P(n)$.

Prove the statement $P(1)$.

Assume that the statement $P(n)$ is true *for some $n$.

Prove, from 3., that $P(n+1)$ is true.

Steps 3 and 4 can be rewritten into one step, i.e. "Prove that $P(n)\implies P(n+1)$ is true.

In your case, clearly, the statement $P(n)$ should be $P(n)="\bigcap_{1\leq k\leq n}S_k=\left(-\frac1n, \frac1n\right)$".

Now, it's time for the induction proof. Start slow:

Write down what $P(1)$ is. Can you prove this statement?

Write down what $P(n)$ is. Yes, again. Just do it. Trust me.

Directly below $P(n)$, write down what $P(n+1)$ is. Observe the two statements. Stare at them for a while. Try and figure out how you could prove $P(n+1)$ from $P(n)$.

Hint (for step 3 above):  $$\bigcap_{1\leq k\leq n+1} S_k= \bigcap_{1\leq k \leq n} S_k \cap S_{n+1}.$$

Title: Proving the intersection of an open set by induction

Body:
I am trying to make sense of this question:

$S_n$ = $(-\frac{1}{n},\frac{1}{n})$,  $n\in \mathbb{N}$

Find $\cup_{n\geq1}S_n$,  $\cap_{n\geq1}S_n$, then use induction to show that $\cap_{1\leq k \leq n} = (-\frac{1}{n},\frac{1}{n})$

I've already found $\cup_{n\geq1}S_n = S_1 = (-1,1)$ and  $\cap_{n\geq1}S_n=\emptyset$ .
However, I am unsure on how to go about doing the induction step for  $\cap_{1\leq k \leq n} = S_n =(-\frac{1}{n},\frac{1}{n})$


问:
Is the logarithm function injective (one-to-one)?
Is the logarithm function injective (or, one-to-one)?

In other words, does $\log_2(x) = \log_2(y) \implies x = y$?

I.e., as $x$ and $y$ are in the same log base, can I just drop the logs?

Thanks!

答:
What does it mean for a number $a$ to be equal to $\log_2(x)$? It means that $2^a=x$.

Can you use this to answer the question?


问:
Find the base of the linear function $T(p(x))$
Let $T:P_1(\mathbb{R})\rightarrow \mathbb{R}^{2}$ where $P_1$ are the polynomial with the most degree 1.

And $T(P(x))=(P(0),P(2))$ with the standard basis $b=\left \{ 1,x \right \},g=\left \{ e_1,e_2 \right \}$

I want to find $\left [ T \right ]^{g}_{b}$

so I take from the first element of the $b$ basis $p(1)=1+1=2$ $(p(x)=1+x)$ so $T(P(1))=(2,2)$
then $T(P(x))=(P(0),P(2))=(1+0,1+2)=(1,3)$
so  $\left [ T \right ]^{g}_{b}=\begin{pmatrix}
2 & 1\\
2 & 3
\end{pmatrix}$
the answer I should get is $\begin{pmatrix}
1 & 0\\
1 & 2
\end{pmatrix}$  I am assuming I just do row operations to take that matrix or am i doing something wrong?

答:
Let $p_1(x)=1$ and $p_2(x)=x.$ Then

$$T(p_1)=(1,1)=1 \cdot e_1+1 \cdot e_2.$$

Hence the first column of $\left [ T \right ]^{g}_{b}$ is given by $(1,1)^T.$

We have

$$T(p_2)=(0,2)=0 \cdot e_1+2 \cdot e_2.$$

Hence the second column of $\left [ T \right ]^{g}_{b}$ is given by $(0,2)^T.$


问:
Sign of Cycle Notation
I know that the sign of cycle notation permutation of length $\displaystyle k$, will be: $\displaystyle ( -1)^{k-1}$.

The question is, what happens when in a cycle there are multiple orbits?

From Wikipedia: "The length of a cycle is the number of elements of its largest orbit.".

https://en.wikipedia.org/wiki/Cyclic_permutation

Some examples:

$\displaystyle ( 123)( 31) ,$ max $\displaystyle k=3$, $\displaystyle ( -1)^{3-1} =1$. On the other hand, I know this cycle has sign $\displaystyle -1$.

$\displaystyle ( 12)( 34)$, max $\displaystyle k=2$, $\displaystyle ( -1)^{2-1} =-1$. On the other hand, I know this cycle has sign $\displaystyle 1$.

Also, I know that each cycle can be written as an aggregiation of transformation (ie: $\displaystyle ( p_{0} p_{1} \dotsc p_{k}) =( p_{0} p_{1}) \cdotp ( p_{1} p_{2}) \cdotp \dotsc \cdotp ( p_{k-1} p_{k})$.

So under their assumption, I might argue each cycle has max lenght $\displaystyle k=2$, and therefor have sign $\displaystyle -1$.

A possible solution to this question, is that they referred to ONLY DISJOINT cycles. But here two questions arise:

In example 2 they were disjoint and still didn't work

And if so, then how can you quickly calculate the sign of not disjoint cycles? (without having to unroll them).

Bottom line, could anyway clarify how to calculate the sign of cycles with multiple orbits? Both for joints and disjoint ones.

Thanks

答:
The context of this question is permutations of a finite set.
Every such permutation is uniquely decomposed into the product of disjoint
cycles including cycles of length $1$ which correspond to fixed points.
The order of the cycles in the product does not
matter since disjoint cycles commute with each other. The sign (or signature)
of any cycle of length $\,k\,$ is defined as $\,(-1)^{k-1}.\,$ The sign of the
product of disjoint cycles is the product of the signs of each cycle. Note that
the sign of a cycle of length $1$ is $1$. Thus, for any permutation in the
disjoint cycles representation, the fixed point cycles can be removed without
changing the sign of the permutation.

Your question

The question is, what happens when in a cycle there are multiple orbits?

is answered by the decomposition of any permutation into disjoint cycles (orbits).

Your Wikipedia quote

The length of a cycle is the number of elements of its largest orbit.

is missing the context of the definition of a cycle stated previously as

[...] a bijective function $\,\sigma : X \to X ,\,$ is called a cycle if the action on $\,X\,$ of the subgroup generated by $\,\sigma\,$ has at most one orbit with more than a single element.

and, in this context only, the length of a cycle is the length of
its largest orbit because all of the other orbits have length $1$.

Your question

[...] how can you quickly calculate the sign of not disjoint cycles? (without having to unroll them).

can be answered by stating that I don't think there is any quick way
to calculate the sign of not disjoint cycles unless you first find the
disjoint cycle decomposition of the permutation.

Title: Sign of Cycle Notation


答:
I would go

$$\lfloor x\rfloor\le x\le y\lt\lfloor y\rfloor+1\implies-1\lt\lfloor y\rfloor-\lfloor x\rfloor\implies0\le\lfloor y\rfloor-\lfloor x\rfloor$$

where the final implication uses the fact that $\lfloor y\rfloor-\lfloor x\rfloor$ is an integer greater than $-1$.


问:
Complex numbers: With conjugate
I've just started calculating complex numbers (last time I calculated with complex numbers was in high school) and I've already got stuck at this exercise:

$$3z-i\bar z = 7-5i$$

where $\bar z$ is the conjugate of z.

What I've tested so far is to set $z=x+yi$

and with further calculations I've reached this equation

$$3(x+yi)-i(x-yi)=7-5i \implies 3x+3yi-xi+yi^2=7-5i$$

The result should be $z=2-i$.

答:
A slightly different approach:

Start with the given equation

$3z-i\overline z = 7-5i$.

Any root of this equation must also solve the conjugate equation

$3\overline z-iz = 7+5i$.

The second equation forces $\overline z=[7+(z+5)i]/3$, so the original equation becomes by substitution

$3z-i[7+(z+5)i]/3 = 7-5i$.

This contains only $z$ and can be solved by the usual methods for linear equations to give $\color{blue}{z=2-i}$.

As a final step, check by direct evaluation that this root satisfies the original equation.

Title: Complex numbers: With conjugate


问:
Putting colored balls into colored urns

There are six balls and urns each. They are equally distributed in three colors: R, W, and B. Place the balls randomly in the six urns, arranged linearly. What is the probability that no ball is in an urn of the same color?

We can think of this as a matching problem, consider $X = \{r_1,r_2,b_1,b_2,w_1,w_2\}$. We want to find the number of bijections $f:X\to X$ such that:

$f(r_i) \ne r_j$ for $1\le i,j\le 2$

$f(b_i) \ne b_j$ for $1\le i,j\le 2$

$f(w_i) \ne w_j$ for $1\le i,j\le 2$

The total number of bijections is $6! = 720$. Define a few sets:

$R = \{f: X\to X \text{ is a bijection }: f(r_i)\ne r_j, 1\le i,j\le 2\}$

$B = \{f: X\to X \text{ is a bijection }: f(b_i)\ne b_j, 1\le i,j\le 2\}$

$W = \{f: X\to X \text{ is a bijection }: f(w_i)\ne w_j, 1\le i,j\le 2\}$

The answer we require then is $1 - \frac{N}{720}$, where $N = |(R\cap B\cap W)^c| = |R^c\cup B^c\cup W^c|$. Using inclusion-exclusion principle, we know that, $$|R^c\cup B^c\cup W^c| = |R^c| + |B^c| + |W^c| - |R^c\cap B^c| - |R^c\cap W^c| - |W^c\cap B^c| + |R^c \cap B^c\cap W^c|$$

Any other ways to approach the problem? Thanks a lot!

答:
My solution counts the permissible arrangements directly.

We have six urns - two each of colors R, W and B. We call them R, W and B. Same for six balls and we call them r, w and b.

We first pick two balls of a color - say r. There are

i) $4$ ways to put both r in urns of one color ($2$ ways to go into W and $2$ ways to go into B).

ii) $8$ ways to put them in urns of different colors ($4$ choices for the first ball and then $2$ choices for the second ball)

In case of $(i)$, if both r have gone into W, both b have to be in R and both w must go into B. This can be done in $2 \cdot 2$ ways.

In case of $(ii)$, we have two R and one each of W and B empty. One of the w must be in B ($2$ ways) and one of the b must be in W (again $2$ ways). Then remaining two balls can go into R in $2$ ways.

So total number of permissible arrangements $ = \small \displaystyle 4 \cdot 2 \cdot 2 + 8 \cdot 2 \cdot 2 \cdot 2 = 80$

So the desired probability $ = \small \displaystyle \frac{80}{6!} = \frac{1}{9}$

Title: Putting colored balls into colored urns


问:
Simplifying Factorials to solve for a variable
I was solving a problem in discrete math and i got the answer down to:

$$26^8 = (n-7)(n-7)!$$

I was wondering how you would solve this for n, if that's even possible. Or do you think I made a mistake getting to this point?

答:
Making the problem more general, the problem is to find $n$ such that
$$(n-a)\,(n-a)!=k$$ where $k$ is a large number.

Making the approximation $(n-a)\,(n-a)!\sim (n-a+1)!$, we just need the inverse of the factorial.

On the site, @gary proposed a superb approximation of the inverse factorial (have a look here). Applied to the problem, this will give
$$n \sim a+ \frac{{\log \left( {\frac{k}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{k}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2}$$ where $W(.)$ is Lambert function.

Using $a=6$ and $k=26^8$ this would give,as an approximation
$n \sim 20.3242$ while the "exact" solution of your problem is $n=20.3521$


问:
Reason behind the condition for multiplication of matrices
Why is the product of two matrices defined only when the number of columns in $A$ is equal to the number of rows in $B$?

答:
The product of matrices $BA$ is only well defined when the number of columns of $A$ equals the number of rows of $B$ because that is how matrix multiplication is defined. The real question is then why is matrix multiplication defined this way?

If we have a $n \times m$ matrix $A$ (with entries in $\mathbb{R}$) then we can define a linear transformation $T: \mathbb{R}^m \rightarrow \mathbb{R}^n$ by $T(\textbf{x}) = A\textbf{x}$.

Likewise, if we have a $m \times l$ matrix $B$ then we can define a linear transformation $S: \mathbb{R}^m \rightarrow \mathbb{R}^l$ by $S(\textbf{x}) = B\textbf{x}$.

Of course, we can compose these linear transformations to get another linear transformation $S \circ T: \mathbb{R}^m \rightarrow \mathbb{R}^l$. This map is only well defined because the codomain of $T$ equals the domain of $S$.

But $S \circ T(\textbf{x})=S(A\textbf{x})=B(A\textbf{x})=(BA)\textbf{x}$.

We can now see that $S \circ T$ is only a well defined linear transformation when the number of columns of $A$ (the codomain of $T$) equals the number of rows of $B$ (the domain of $S$).

Matrix multiplication was defined in the way it is specifically so that composition of the corresponding linear transformations corresponds to matrix multiplication.


问:
integral of $\sin(\sqrt(ax^4+bx^3+cx^2+dx+e))$
How does one approach integrating:

$f(x)=\sin(\sqrt{ax^4+bx^3+cx^2+dx+e})$?

I have tried with substitution, which yields the following. I am quite new to this complexity of integration, so please tell me if there are any mistakes.

$\displaystyle\int f(x)dx=\displaystyle\int \sin(\sqrt{ax^4+bx^3+cx^2+dx+e})dx$

$u=\sqrt{ax^4+bx^3+cx^2+dx+e}$

$\displaystyle\int f(x)dx=\displaystyle\int \sin(u)dx$

Next, try to write $dx$ as a function $du$ and $u$.

$du=\dfrac{4ax^3+3bx^2+2cx+d}{\sqrt{ax^4+bx^3+cx^2+dx+e}}dx$

$du=\dfrac{4ax^3+3bx^2+2cx+d}{u}dx$

Here I get stuck, because I don't know what to do with the top part of the fraction to get rid of the remaining x's. I see that:

$4ax^3+3bx^2+2cx+d=\dfrac{d}{dx}u^2$,

but I am not sure what I can do with that.

Guidance is really appreciated.

How I got to this integral

For context, I am creating a model of a quad rotor drone. In this case simplified to a single continuously running motor. I will leave most of the physics formulas out of this and try to make things concise.

Angle of the drone under a single running motor in axis-angle representation.

$\underrightarrow{\theta}=\theta \underrightarrow{e},\lvert e\rvert=1$

$\underrightarrow{\theta}(t)=\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}+t\dot{\underrightarrow{\theta}}(0)+\underrightarrow{\theta}(0)$, with $\ddot{\underrightarrow{\theta}}$ constant.

$\theta(t)=\sqrt{(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_x+t\dot{\underrightarrow{\theta}}_x(0)+\underrightarrow{\theta}_x(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_y+t\dot{\underrightarrow{\theta}}_y(0)+\underrightarrow{\theta}_y(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_z+t\dot{\underrightarrow{\theta}}_z(0)+\underrightarrow{\theta}_z(0))^2}$

$\underrightarrow{e}(t)=\dfrac{\underrightarrow{\theta}(t)}{\theta(t)}$

Next, velocity of the drone, and where the complicated integral comes in.

Acceleration under the (constant) force of one rotor

$\underrightarrow{\alpha}=\dfrac{1}{m}\underrightarrow{F^{'}}=\dfrac{1}{m}(\cos(\theta)\underrightarrow{F}+\sin(\theta)(\underrightarrow{e}\times\underrightarrow{F})+\underrightarrow{e}(1-\cos(\theta))(\underrightarrow{e}\cdot\underrightarrow{F}))$

with $F$ the force applied by the rotor, and $F^{'}$ that force after it is rotated by $\underrightarrow{\theta}$.

Integrating this to obtain velocity

$\underrightarrow{v}=\dfrac{1}{m}\displaystyle\int\underrightarrow{a}(t)\underrightarrow{F}+\underrightarrow{b}(t)+\underrightarrow{c}(t)dt$, with

$\underrightarrow{a}(t)=\cos(\theta(t))$

$\underrightarrow{b}(t)=\sin(\theta(t))(\underrightarrow{e}(t)\times\underrightarrow{F})$

$\underrightarrow{c}(t)=\underrightarrow{e}(t)(1-\cos(\theta(t)))(\underrightarrow{e}(t)\cdot\underrightarrow{F})$

I approached the integral by splitting into the above three parts. This question is about integrating part a. In all its large form it looks like

$\displaystyle\int\underrightarrow{a}(t)dt=\displaystyle\int\cos(\sqrt{(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_x+t\dot{\underrightarrow{\theta}}_x(0)+\underrightarrow{\theta}_x(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_y+t\dot{\underrightarrow{\theta}}_y(0)+\underrightarrow{\theta}_y(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_z+t\dot{\underrightarrow{\theta}}_z(0)+\underrightarrow{\theta}_z(0))^2})dt$

Which I simplified to the original equation given in this question. (I now see that it's actually $\cos$ and not $\sin$, but it's all the same)

答:
comment

For example, consider $\int\sin\sqrt{x^4+1}\;dx$.  Change variables using $y^2 = x^4+1$ to get
$$
\int\frac{y\sin y}{2(y^2-1)^{3/4}}\;dy
$$
This integral probably is not an elementary function.  It is not known to Maple.  I did not find it in Gradshteyn & Ryzhik.


问:
consider a semicircle with a diameter AB where the length is 4, and a point C on the circular arc.
if the area of triangle ABC is  a half of the  maximum  and point C  is nearer to point A than point B  then the angle CAB  is? I have already found out that the  maximum area of triangle ABC is 4.But I have no idea on how to continue.</p>

答:
Drop a perpendicular from point $C$ on $AB$ and call the feet of this perpendicular $D$. Let the centre of the semicircle be $O$.

Observe that, in right $\triangle CDO$, $CD=1=\frac {1}{2} 2=\frac {1}{2} CO$ and hence it is a $30-60-90$ triangle with $\angle COD=30^{\circ}$. Now find $\angle CAB$ taking advantage of the fact that $\triangle COA$ is isosceles.


答:
Up to the cited algorithm, the solutions $a,b$ (provided $a\le b$) are
those and only those pairs of numbers $(a,b)$ that for
each $n \in \mathbb N $ are calculated by the formulas:$a_n=b_{n-1},\, b_n=k^2b_{n-1}-a_{n-1},\,a_0=0,\,b_0=k$, here $k=29$.

The RSolve command of Mathematica cracks it:

```<code>RSolve[{a[n] == b[n - 1], b[n] == 29^2*b[n - 1] - a[n - 1], a[0] == 0,
b[0] == 29}, {a, b}, n]
</code>```

$$a(n)=-\frac{29 \left(2 \left(\sqrt{707277}+841\right)\right)^{-n} \left(4^n-\left(\sqrt{707277}+841\right)^{2 n}\right)}{\sqrt{707277}},$$ $$b(n)=29 U_n\left(\frac{841}{2}\right). $$
in the above $U_n$ stands for Chebyshev polynomials. It remains to compare this solution with the solution found by Reduce.

Title: Are all the solutions produced by Mathematica?

Body:
Here is a very strong and impressive result of the Reduce command.

```<code>Reduce[a^2 + b^2 == 841*(a*b + 1), {a, b}, PositiveIntegers]
</code>```

performs (a == 24389 && b == 29) and 6 infinite series of the solutions, one of these is (The results are too big for $\LaTeX$.)

(C[1] \[Element] Integers && C[1] >= 0 &&  a == (1/1414554)(-(24389/ 2) (-707277 (297410399 - 353640 Sqrt[707277])^C[1] + 841 Sqrt[707277] (297410399 - 353640 Sqrt[707277])^C[1] -  707277 (297410399 + 353640 Sqrt[707277])^C[1] - 841 Sqrt[707277] (297410399 + 353640 Sqrt[707277])^C[1]) -  20511033/ 2 (-841 (297410399 - 353640 Sqrt[707277])^C[1] + Sqrt[707277] (297410399 - 353640 Sqrt[707277])^C[1] -  841 (297410399 + 353640 Sqrt[707277])^C[1] -  Sqrt[707277] (297410399 + 353640 Sqrt[707277])^C[1])) && b == -(1/1414554) 29 (-707277 (297410399 - 353640 Sqrt[707277])^C[1] + 841 Sqrt[707277] (297410399 - 353640 Sqrt[707277])^C[1] -  707277 (297410399 + 353640 Sqrt[707277])^C[1] - 841 Sqrt[707277] (297410399 + 353640 Sqrt[707277])^C[1]))

This is problem 16 from Kvant 2020, # 10, p. 42 (in Russian). Its solution is given in p. 60 ibid .
However, no explicit formulas are presented there, but only an algorithm to obtain all the solutions in PositiveIntegers.

My questions are: are all the solutions produced by Reduce? how to establish it? I hope the answer to the first question is yes.

Body:
Up to the cited algorithm, the solutions $a,b$ (provided $a\le b$) are
those and only those pairs of numbers $(a,b)$ that for
each $n \in \mathbb N $ are calculated by the formulas:$a_n=b_{n-1},\, b_n=k^2b_{n-1}-a_{n-1},\,a_0=0,\,b_0=k$, here $k=29$.

The RSolve command of Mathematica cracks it:

RSolve[{a[n] == b[n - 1], b[n] == 29^2*b[n - 1] - a[n - 1], a[0] == 0,
b[0] == 29}, {a, b}, n]

$$a(n)=-\frac{29 \left(2 \left(\sqrt{707277}+841\right)\right)^{-n} \left(4^n-\left(\sqrt{707277}+841\right)^{2 n}\right)}{\sqrt{707277}},$$ $$b(n)=29 U_n\left(\frac{841}{2}\right). $$
in the above $U_n$ stands for Chebyshev polynomials. It remains to compare this solution with the solution found by Reduce.


问:
$\prod_{k=1}^{\infty}(1-\alpha_k)>0\iff\sum_{k=1}^{\infty}\alpha_k<\infty\;\;\;\ \alpha_k\in(0,1)$ Proof verification
I would like to ensure that my logic works here for proving the following:
$$\prod_{k=1}^{\infty}(1-\alpha_k)>0\iff\sum_{k=1}^{\infty}\alpha_k<\infty\;\;\;\;\;\;, \alpha_k\in(0,1)$$
So, for
$\Rightarrow $:
$$\sum_{k=1}^{\infty}\alpha_k=\infty \Rightarrow \alpha_k \not\to0
\Rightarrow (1-\alpha_k)\not\to1\Longrightarrow \\
1-\alpha_k\leq1 \land (1-\alpha_k<1 \;\text{almost always}) \Longrightarrow\\
\prod_{k=1}^{\infty}(1-\alpha_k)=0
$$
Where the before last statement expresses fact that this is infinitely many times strictly decreasing sequence, that is never increasing that is bounded by only by zero

And for $\Leftarrow$:

if $\sum_{k=1}^{\infty}\alpha_k<\infty$ then $\alpha_k\to0$. By using continuity, lets speak of

$$\exp(\lim_{k\to\infty}\log(\prod_{j=1}^{k}(1-\alpha_j)))
$$
and it is enough to prove that
$$\lim_{k\to\infty}\log(\prod_{j=1}^{k}(1-\alpha_j)) \\
=\lim_{k\to\infty}\sum_{j=1}^{k}\log(1-\alpha_j)$$ exists and is not $\pm\infty$, because this means that the argument of the $\exp(*)$ is not $-\infty$ which means that the product is greater than $0$.

And this happens because $-\log(1-\alpha_j)>0$ and $\frac{-\log(1-\alpha_j)}{\alpha_j}\to_{\alpha_j\to0}1$ implies that our serie converges as we wished.

答:
The $\implies$ direction is wrong, the divergence of the series does not imply that $a_n \not\to 0$. You have the right idea in the $\impliedby$ direction, and that can be used for both directions.

We have the general inequality
$$
  \frac{x}{1+x} \le \log(1+x) \le x
$$
for $x > -1$, so that
$$
 - \sum_{k=1}^n \frac{\alpha_k}{1-\alpha_k} \le \log \prod_{k=1}^n (1-\alpha_k) \le -\sum_{k=1}^n \alpha_k \, .
$$

If $\sum_{k=1}^\infty \alpha_k = \infty$ then the right-hand side diverges to $-\infty$, so that $\prod_{k=1}^n (1-\alpha_k) \to 0$.

If $\sum_{k=1}^\infty \alpha_k < \infty$ then $\alpha_k < 1/2$ for all sufficiently large $k$, so that the left-hand side converges to a finite value $c$. It follows that $\prod_{k=1}^n (1-\alpha_k)$ is bounded below by the positive value $e^c$, and therefore $\prod_{k=1}^\infty (1-\alpha_k) \ge e^c > 0$.

Remark: Here we have used $\prod_{k=1}^{\infty}(1-\alpha_k) = \lim_{n\to \infty}\prod_{k=1}^{n}(1-\alpha_k)$ as definition of the infinite product, the limit exists because the partial products are positive and decreasing. Some sources do not allow zero as the limit of the partial product, or call such a product “diverging to zero.”


问:
Minimize a linear function over the unit ball

Minimize a linear function over the unit ball. (without using Lagrangian)

We have a problem: minimize $c^Ty$ subject to $y^Ty \le 1$.

By the Cauchy-Schwarz inequality we have $c^Ty \le \|c\|_2\|y\|_2 \le \|c\|_2$. So, if the problem was to maximize we would get $c^Ty = \|c\|_2, y=  c/\|c\|_2$.

The answer to the problem is $y^{*} = -c/\|c\|_2$. However, I don't see how it can be derived. One thought is:

The problem is equivalent to: maximize $-c^Ty = (-c)^Ty$ subject to $y^Ty \le 1$.

$(-c)^Ty \le \|c\|_2, y = -c/\|c\|_2$

答:
You can write the Lagrangian of the problem:

$\mathcal{L}(y, \lambda) = c^T y - \lambda(y^Ty - 1)$ with $\lambda$ the dual variable associated with the constraint.

The optimality conditions are:

$\frac{\partial \mathcal{L}}{\partial y} = c - 2 \lambda y = 0$

$y^Ty \le 1$

$\lambda(y^Ty - 1) = 0$

$\lambda \le 0$

We must have $\lambda < 0$ (constraint is active at the solution: $y^T y = 1$), so $y = \frac{c}{2 \lambda}$, which we inject in the 3rd equation, solve for $\lambda$, substitute in $y$ and we get the result.


问:
What does "Group" generalize?
In linear algebra, I've learned that the concept of "vector space" is nothing more than the linear space's abstract generalization.
And thanks to this, I've never forgotten the axioms for the vector spaces.

Although the axioms (or the definition) of "group" in abstract algebra are much more straightforward, I can't figure out what a concept of group generalizes.
It seems the concept of group generalizes the coordinate transformation (in an active sense, or just operation), but I am not sure of it.
Could somebody please elaborate on this?

答: Groups don't "generalise" but "abstract" the notion of a symmetry group. One of the simplest example of one, given your understanding of vector spaces, is the additive group of a vector space which is the group of translations.</p>

OwnerDisplayName: 윤종인

Title: What does "Group" generalize?

OwnerDisplayName: Mozibur Ullah


答:
Groups generalize a lot of things actually, because their definition is really easy and only imposes a few "loose" conditions:

existence of an identity element (in an additive group, it would be $0$, in a multiplicative group, $1$, ...)

every element has an inverse (for example, for $\mathbb{R}$ endowed with addition, this means that for every $x \in \mathbb{R}$, we can find some $-x$ element)

the operation is associative (which is a very natural property for most operations so it doesn't hurt)

The reason groups are so useful is because we noticed many spaces, when endowed with some operation, behave in much the same way. Here are a few examples of groups:

$\mathbb{R}$ with addition

$\mathbb{R}\backslash\{0\}$ with multiplication

$O_n$, the group of orthogonal matrices of size $n$. It gets really useful in physics, when you want to talk about angular momentum for example (because orthogonal matrices are basically rotations + reflections). And the closely related $SU_n$, used in quantum mechanics.

$GL_n$ (group of invertible matrices), $SL_n$ (group of matrices with determinant $1$),...

All of those are infinite groups, and they look very different from each other at first, but since they all have a group structure, we can treat them using group theory tools and theorems (which are often very powerful, and there is a lot of material on the subject).

Intuitively, it's not as straightforward to see it as a generalization of something "concrete" (like a vector space is the generalization of our usual space), because groups can come in so many different forms. The way I see it is that at some point, we started thinking "ok, this structure comes up so often in maths that we'll give it a name, and start treating all of those spaces the same".

And while all the groups mentioned above are infinite, there are plenty of finite groups, which are just as useful:

$D_{2n}$, which you can see as the symmetry group of an $n-$gon (consisting of rotations and reflections, much in the same way as $O_n$ does. But $D_{2n}$ only contains symmetries of the $n-$gon, instead of symmetries of the whole space)

$S_n$, the symmetry group of $n$ elements (ie the set of permutations of the numbers $1,...n$, endowed with composition)

etc...

The reason it's so useful is, say you come up with a whole new space which you've never seen in your life. Well, if you manage to find an operation on it which makes it a group, then that tells you a great deal of things about your space without you having to do anything.

Title: What does "Group" generalize?

Body:
In linear algebra, I've learned that the concept of "vector space" is nothing more than the linear space's abstract generalization.
And thanks to this, I've never forgotten the axioms for the vector spaces.

Although the axioms (or the definition) of "group" in abstract algebra are much more straightforward, I can't figure out what a concept of group generalizes.
It seems the concept of group generalizes the coordinate transformation (in an active sense, or just operation), but I am not sure of it.
Could somebody please elaborate on this?

OwnerDisplayName: Azur


答:
Before going to groups, it would be easier, actually, to start with semigroups. A semigroup is like a group, but you have only one axiom: associativity.

A semigroup generalizes the idea of a compositionally-closed set of functions - that is, functions mapping some domain set into itself, such that if we compose any two functions from the set, that composite is also in the set. Function composition is associative, so any such collection of functions is a semigroup under composition. It is also possible to prove the converse: that there exists a domain set and set of self-functions on that set which have the same structure as any given semigroup.

Now if we require the functions to be bijective, i.e. invertible self-maps, then that generalizes to what is known as a quasigroup.

Finally, a group is what happens when we include all inverses of the invertible maps into the quasigroup. Every group corresponds to a set of bijective self-maps, closed under both composition and inversion, on some domain. For example, the cyclic groups $\mathbb{Z}_n$ correspond to cyclic permutations of a set of $n$ elements (think of the "rotate" operations that, say, turn "hello" into "lohel" or "llohe".), or if you want a different domain, plane rotations by $\frac{\tau}{n}\ \mathrm{rad}$.

Title: What does "Group" generalize?


问:
stochastic integral equation
For $0 \leq t \leq T$, define

$$Z_t:=\exp {\left\lbrace \int_0^t X_sdW_s - \frac 12 \int_0^t X_s^2ds \right\rbrace }$$

Show that this process satisfies the stochastic integral equation

$$Z_t=1+\int_0^tZ_s X_s dW_s , \qquad 0≤t≤T$$

答:
You have $Z_t = f(Y_t)$ where $f(y) = \mathrm e^y$ and
$$
  \mathrm dY_t = -\frac12 X^2_t\mathrm dt + X_t\mathrm dW_t.
$$
Just apply Ito's lemma to $f(Y_t).$</p>

Title: stochastic integral equation


问:
What is $\frac{\partial }{\partial x_1}$ equal to if we have $y_1=\frac{x^2_1}{x_2},\:y_2=log\left(x_1\right)$?
We got:

$$\frac{\partial }{\partial x_1}=\frac{\partial y_1\:}{\partial \:x_1}\frac{\partial \:}{\partial \:y_1}+\frac{\partial \:y_2}{\partial \:x_1}\frac{\partial \:}{\partial \:y_2}$$

But isn't this wrong? Because it would mean that:

$$\frac{\partial }{\partial x_1}=\frac{\partial \:}{\partial \:x_1}+\frac{\partial \:}{\partial \:x_1}=2\frac{\partial \:}{\partial \:x_1}$$

答:
$\frac{\partial }{\partial x_1}$ is an operator that works the same way as the derivative in one variable but for functions of several variables. For example if $f(x_1,x_2)=x_1^2+x_1x_2+x_2^3$ then $\frac{\partial }{\partial x_1}f(x_1,x_2)=2x_1+x_2$ ,  $x_2$ is considered a constant when you compute the derivative in order to $x_1$. Note now that if $f(x_1,x_2)=1$ then $\frac{\partial }{\partial x_1}f(x_1,x_2)=0$ because it does not depend on $x_1$.

Writing $\frac{\partial }{\partial x_1}f(x_1,x_2)$ where $f(x_1,x_2)=1$ differently with the help of $\frac{\partial }{\partial x_1}y_1(x_1,x_2)$ and $\frac{\partial }{\partial x_1}y_2(x_1,x_2)$:

$\frac{\partial }{\partial x_1}y_1(x_1,x_2)=2\frac{x_1}{x_2}  \text{ and } \frac{\partial }{\partial x_1}y_2(x_1,x_2)=\frac{1}{x_1}$
$$\text{Therefore } 0=\frac{\partial }{\partial x_1}f(x_1,x_2)=0\times\frac{\partial }{\partial x_1}y_1(x_1,x_2)+0\times\frac{\partial }{\partial x_1}y_2(x_1,x_2)$$

Now imagine that $x_1=x_1(t) $ and $x_2=x_2(t)$ then

$\frac{d }{dt}y_1(x_1,x_2)=\frac{\partial }{\partial x_1}y_1(x_1,x_2)\frac{d}{dt}x_1(t)+\frac{\partial }{\partial x_2}y_1(x_1,x_2)\frac{d}{dt}x_2(t)$


答:
The answer provided by Winged Reptile is well done ($+1$). But allow me to say something more.

Sometimes, in a Linear Algebra course, ones starts by studying the algebra of matrices, before studying what a linear transformation is (as it was in my case).

Of course that when one starts to study linear transformations, all becomes clearer. But if you haven’t yet study linear transformations, you can think as follows.

Given two vectors of dimension $n$, you can compute the inner product of those two vectors. For example, if $u = (u_1,u_2,\dots,u_n)$ and $v=(v_1,v_2,\dots,v_n)$, with the usual inner product, you get

\begin{align*}
u \cdot v = u_1 v_1 + u_2 v_2 + \dots + u_n v_n = \sum_{i=1}^n u_iv_i.
\end{align*}

Hence, you can think of the product of two matrices as a generalization of this inner product. The ideia is that one should compute the inner product of the rows vectors of the first matrix with the column vectors of the second matrix.

Since the inner product is only defined for vector with the same dimensions, we require that the number of columns of the first matrix to be equal to the number of rows of the second matrix.

Once again, I point that the explanation is given by the fact that in this way, we can compute the composition of two linear transformations by multiplying the two matrices associated to those linear transformations. But if you haven’t yet hear about linear transformations, this is a nice way to get a sense of the definition of product of matrices.

Title: Reason behind the condition for multiplication of matrices

Body: Why is the product of two matrices defined only when the number of columns in $A$ is equal to the number of rows in $B$?


问:
Limit point of a sequence with pairwise different elements
Let $(a_n)_{n\in \mathbb{N}}$ be a convergent sequence in $\emptyset\neq D\subseteq \mathbb{C}$, where $D$ is an open set, with $a_i\neq a_j$ for all $i\neq j$ and limit $a\in D$. Can we conclude that $a$ is a limit point of $D$?

My attempt: Taking a limit $j\to\infty$ in $a_i\neq a_j$ for all $i\neq j$ yields $a_i\neq a$ for all $i$. Then $(a_n)$ is a sequence in $D\setminus \{a\}$ with $a_n\to a$ and $a$ is a limit point of $D$.

However, I guess we can not take the limit $j\to\infty$ as the sequence $a_1=0,\,\, a_n=\frac{1}{n}$ for all $n\geq 2$ satisfies $a_i\neq a_j$ for all $i\neq j$, but $\lim_{n\to\infty} a_n = 0 = a_1$.

答: I guess I have found the answer to my question. Let us assume that $a$ is an isolated point of $D$. Then, according to this post, we have that $a_n = a$ eventually. Thus, we find $n_0\in\mathbb{N}$ such that $a_n = a$ for all $n\geq n_0$. Choosing $n=n_0$ and $n=n_0+1$ yields $a_{n_0} = a = a_{n_0+1}$, a contradiction. But $a$ is an isolated point if and only if it is not a limit point by the post above, so $a$ is indeed a limit point.


问:
Do you know this function?
I'm looking for a real-valued function with the following properties:

$\lim \limits_{x \to 0} f(x) = 1$

$\lim \limits_{x \to 0^+} f'(x) = -\infty$

$\lim \limits_{x \to \infty} f(x) = 0$ and $0<f(x)\leq 1$ always

$1-\sqrt{x}$ fits the first two criteria and may just need the right factor. I would also appreciate ideas on how to approach problems like this generally. Thanks in advance.

答:
the simplest solutions are $\frac{1}{x^\alpha+1}$ and $(1+x)^\alpha-x^\alpha$ with $\alpha \in (0,1)$

every function of the form $f(x) = 1 - \sqrt{1-g(x)}$ with $g(0) = 1$ and  $\lim \limits_{x \to \infty} g(x) = 0$ works, some examples are $g(x)=e^{-x}$ or $\frac{1}{x+1}$.

some trig solutions are: $f(x) = \frac{\pi}{2} \operatorname{arccsc}(x+1)$ and $\exp(x^\alpha)$

Title: Do you know this function?


问:
Bennett's Inequality to Bernstein's Inequality
Bennett's Inequality is stated with a rather unintuitive function,

$$
h(u) = (1+u) \log(1+u) - u
$$

See here.  I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,

$$
h(u) \ge \frac{ u^2 }{ 2 + \frac{2}{3} u}
$$

and plugging it back into Bennett's Inequality.  However, I can't see where this expression comes from.  Could someone point me in the right direction?

答:
While the answer by @Arash outlines the general strategy of deriving the inequality,
the explicit computation is missing.
I am providing it here in the hope that it will be helpful for future readers.

The actual computation of the derivative is highly simplified by first simplifying
the last term in the definition of
$$
  f(x)
  = (1 + x) \ln(1 + x) - x - \frac{x^2}{2 + \frac{2}{3} x}
$$
as follows:
$$
  \frac{x^2}{2 + \frac{2}{3} x}
  = \frac{3}{2} \frac{x^2}{3 + x}
  = \frac{3}{2} \frac{x \cdot (3 + x) - 3 x}{3 + x}
  = \frac{3}{2} x - \frac{9}{2} \frac{x}{x + 3}.
$$
Therefore, we see
\begin{align*}
  \require{cancel}
  f(x)
  & = (1 + x) \ln (1 + x) - \frac{5}{2} x + \frac{9}{2} \frac{x}{x + 3} , \\
  f'(x)
  & = \ln(1 + x) + 1 - \frac{5}{2} + \frac{9}{2} \frac{x+3 - x}{(x+3)^2} \\
  & = \ln(1 + x) - \frac{3}{2} + \frac{3^3}{2} (x+3)^{-2}, \\
  f''(x)
  & = \frac{1}{1 + x} - 3^3 \cdot (x+3)^{-3} \\
  & = \frac{(x+3)^3 - 27 \cdot (1 + x)}{(1 + x) (3+x)^3} \\
  & = \frac{x^{3} + 9 x^{2} + \bcancel{27 x} + \cancel{27} - \cancel{27} - \bcancel{27 x}}
           {(1 + x) (3+x)^3} \\
  & = \frac{x^{3} + 9 x^{2}}
           {(1 + x) (3+x)^3}
    \geq 0
\end{align*}
for $x \geq 0$, which is the only case we are interested in.

Since $f(0) = 0$ and $f'(0) = 0$, this implies $f' \geq 0$ and then $f \geq 0$
on $[0,\infty)$.


问:
What is the partial derivative for z = f(η), where η=g(x,y).
For solving the partial derivative, I can find the example for $z = f(x,y),\; x = g(t)$, and $y=h(t)$. Also the case for $z = f(x,y),\; x = g(s,t)$, and $y=h(s,t)$ from internet.

However I cannot find the example case for  $z = f(\eta)$, where $\eta=g(x,y).$

For this case, does the partial derivative respect to $x$ or $y$ is just

$$\frac{\partial z}{\partial x} = \frac{d f}{d \xi}\frac{\partial \xi}{\partial x} $$

Also, can I solve $\frac{dz}{dx}$ for this cases? or they are essentially same?

答:
If $z$ is a function of $x$ and $y$, then there is no $\frac{dz}{dx}$, only $\frac{\partial z}{\partial x}$.  Your formula for the partial derivative is correct.
There is also a general form of the chain rule for a function of n variables each of which is a function of m variables.


问:
Find covariance of estimator and derivative of the log-likelihood function
Problem:

Given an estimator $\hat k$. The estimation method is either max likelihood or other method. We know that it's unbiased.
Let $L$ be the likelihood function and $\ell = ln L$.

Find $\Bbb Cov( \frac{d \ell}{d k},\hat k)$

My attempt:

$$\Bbb Cov( \frac{d \ell}{d k},\hat k) = E( \frac{d \ell}{d k} \cdot \hat k) - E( \frac{d \ell}{d k}) \cdot E( \hat k)$$

As unbiased $E( \hat k) = k$. Also,  $\frac{d \ell}{d k}$ is score function. We know that $E(s(k)) = 0$
Hence, we have

$$E( \frac{d \ell}{d k} \cdot \hat k) - E( \frac{d \ell}{d k}) \cdot E( \hat k) =E( \frac{d \ell}{d k} \cdot \hat k) $$

I decided to take $\hat k$ into the differential.

$$=E( \frac{d \ell  \cdot \hat k}{d k})$$

Since the expectation is linear, I can take the derivative out of it.

$$=  \frac{d E(\ell  \cdot \hat k)}{d k}$$

However, I am still stuck. Am I doing something wrong?

Note. The friend of mine did manage to solve this, so the task has enough information to be solved.

答:
As all the commentators noted, the likelihood function makes a little sense if your estimator is obtained by, say, the method of moments. I would assume that, in this case, they don't correlate, so the covariance is zero.

Let $\hat k$ be MLE.

As unbiased, let replace the true parameter $k$ in log-likelihood function with $E[\hat k]$. Then, if you swap the derivative and expectation, you will have something like $\ell (\hat k)$. The MLE maximizes the log-likelihood function, consequently, the derivative is zero. The further steps are trivial.


答:
I just suggest a "quick" proof (I agree it's not an answer, but there is not enough place in the comment to write it as a comment).

If : You have that $\alpha _k\to 0$ when $k\to \infty $, and thus, $-\log(1-\alpha _k)\sim \alpha _k$ when $k\to \infty $. Therefore $\log\left(\prod_{k\geq 1}(1-\alpha _k)\right)>-\infty $, and thus $\prod_{k\geq 1}(1-\alpha _k)>0$.

Only if: You have that $-\sum_{k\geq 1}\log(1-\alpha _k)<\infty $ and thus $\alpha _k\to 0$ when $n\to \infty $. Therefore $\alpha _k\sim -\log(1-\alpha _k)$ when $k\to \infty $, and thus $\sum_{k\geq 1}\alpha _k<\infty $.

Title: $\prod_{k=1}^{\infty}(1-\alpha_k)>0\iff\sum_{k=1}^{\infty}\alpha_k<\infty\;\;\;\ \alpha_k\in(0,1)$ Proof verification

Body:
I would like to ensure that my logic works here for proving the following:
$$\prod_{k=1}^{\infty}(1-\alpha_k)>0\iff\sum_{k=1}^{\infty}\alpha_k<\infty\;\;\;\;\;\;, \alpha_k\in(0,1)$$
So, for
$\Rightarrow $:
$$\sum_{k=1}^{\infty}\alpha_k=\infty \Rightarrow \alpha_k \not\to0
\Rightarrow (1-\alpha_k)\not\to1\Longrightarrow \\
1-\alpha_k\leq1 \land (1-\alpha_k<1 \;\text{almost always}) \Longrightarrow\\
\prod_{k=1}^{\infty}(1-\alpha_k)=0
$$
Where the before last statement expresses fact that this is infinitely many times strictly decreasing sequence, that is never increasing that is bounded by only by zero

And for $\Leftarrow$:

if $\sum_{k=1}^{\infty}\alpha_k<\infty$ then $\alpha_k\to0$. By using continuity, lets speak of

$$\exp(\lim_{k\to\infty}\log(\prod_{j=1}^{k}(1-\alpha_j)))
$$
and it is enough to prove that
$$\lim_{k\to\infty}\log(\prod_{j=1}^{k}(1-\alpha_j)) \\
=\lim_{k\to\infty}\sum_{j=1}^{k}\log(1-\alpha_j)$$ exists and is not $\pm\infty$, because this means that the argument of the $\exp(*)$ is not $-\infty$ which means that the product is greater than $0$.

And this happens because $-\log(1-\alpha_j)>0$ and $\frac{-\log(1-\alpha_j)}{\alpha_j}\to_{\alpha_j\to0}1$ implies that our serie converges as we wished.


问:
Let $2Z:= \left \{ 2k | k\in Z \right \}$ be the collection of all even integers. Show that $N \cup2Z$ is denumerable.

Let $2Z:= \left \{ 2k | k\in Z \right \}$ be the collection of all
even integers.

Show that $N \cup2Z$ is denumerable.

I first try to build a map $f:N\rightarrow N \cup2Z$ to be bijective, to show $N \cup2Z$ is denumberable.

However, since all even integers $2Z$ contains negative integers, and $N$ doesn't, I do not think they would have the same cardinality.

Hence there is no such bijective map.

I am wondering if I've missed something here.

Some hint or correction would be very helpful.

答:
To emphasize, "our set having a bijection with $\Bbb N$" can be rephrased as "our set can be listed in a sequence, every element eventually appearing in the sequence exactly once"

Our set in question here is $2\Bbb Z\cup \Bbb N$.

An example of a sequence which touches each element exactly once might be something like:

$$0,1,2,-2,3,4,-4,5,6,-6,7,8,-8,9,10,-10,11,12,-12,\dots$$

If you give me your favorite number like $42$ that you know to be in the set, I could tell you where in the sequence it would be.  (Counting starting from zero, it would be the 62'nd term in the sequence)

Now... you might ask, "why does a sequence like this imply there is a bijection?"  To that, you have to understand what a sequence is in the first place.  The sequence is the bijection.  A sequence, formally, is a function $f$ from $\Bbb N$ to a set.  The "first element of the sequence" is simply $f(1)$ and the "second element of the sequence" is simply $f(2)$ and so on...  We just choose to short-hand and write the sequence as $0,1,2,-2,3,4,-4,\dots$ rather than $f(x)=\begin{cases}0&\text{if}~x=0\\1&\text{if}~x=1\\2&\text{if}~x=2\\-2&\text{if}~x=3\\3&\text{if}~x=4\\\vdots\end{cases}$.  The sequence will be bijective per our selection of the sequence being one in which every element eventually appears in our sequence exactly once.

The only challenge remaining should be to formalize the sequence to write it in a more compact form.

 Further hint, try writing it in the form $f(x)=\begin{cases}\dots&\text{if }x\equiv 0\pmod{3}\\\dots&\text{if }x\equiv 1\pmod{3}\\\dots &\text{if }x\equiv 2\pmod{3}\end{cases}$

Alternatively, if you merely wish to show they are countable and are willing to skip finding an explicit bijection, you can see that your set is a superset of $\Bbb N$ and a subset of $\Bbb Z$, both of which you should know by example are countable.  Your statement though "However, since all even integers 2Z contains negative integers, and N doesn't, I do not think they would have the same cardinality" implies to me that you are unaware of the fact that $\Bbb Z$ is also countable.


问:
Prove that $(n-1)!\le n^n e^{-n}e\le n!$, where n is a positive integer.
Theorem:

If $n$ is a positive integer Then $$(n-1)!\le n^n e^{-n}e\le n!$$

It’s from the book Short Calculus By Serge Lang unfortunately he didn’t make a full explanation of this theorem, he said evaluate this integral:
$$\int_1^{n} \ln (x) dx$$ and then exponentiate the result so I did this:
$$\int_1^{n} \ln (x) dx= x\ln (x)-x \Big|_1^n=n\ln(n)-n+1$$
Now exponentiating:
$$e^{n\ln(n)-n+1}=n^n e^{-n}e$$
so far everything clear but how can i find the bounds $(n-1)!$ and $n!$

Lang said that you can find them using Upper and Lower Sums, and using the partition $(1,2,3...n)$ but I don't know how to do this although I know what is the upper and the lower sums:
$$L_1^{n}(P,f)=\sum_{i=0}^{n-1}f(s_i)(x_{i+1}-x_i)=\sum_{i=0}^{n-1}f(s_i)$$
where $s_i$ is the minima of function ($\ln$) in the interval $[x_{i+1}, x_i]$ so :$$L_1^{n}(P,f)=\sum_{i=0}^{n-1}\ln(s_i)=\ln(\prod_{i=0}^{n-1}s_i)$$
Here I don't know how to exponentiating.

答:
If you do a left hand sum, in this case a lower sum, of $\int_{1}^{n} \ln x dx$, you are finding $1\cdot \ln 1 + 1\cdot \ln 2 + ... + 1\cdot \ln (n-1)$ which by the laws of logarithms equals $\ln (1\cdot2\cdot...\cdot(n-1)) = \ln ((n-1)!)$.  Similarly for the right hand sum to get $\ln(n!)$.

EDIT: $\ln x$ is an increasing function, so the smallest value in an interval is always at the left edge, the lower limit, in this case the left hand sum = lower sum.

Likewise the largest value in any interval is the rightmost value for an increasing function, so the upper sum is the same as the right hand sum.


问:
Verifying the continuous operation on a topological semigroup
If we are given a topological semigroup $G$ (i.e. it just has an associative operation "$\cdot$", without other properties). Now we want to verify this operation $"\cdot"$ is continuous. We denote a function $(x,\cdot)$ by $f_x(\cdot)$(i.e. $f_x(y)=x\cdot y$.)

I want to prove a claim: $(\cdot,\cdot):X\times X\rightarrow X$ is continuous if and only if $f_x:X\rightarrow X$ is continuous for all $x\in X$.

Is it right? How can we prove it? Or could you give some counterexample?  Thank you very much.

答:
This may fail even for groups.

Consider the set $B(H)$  of all bounded,  linear transformations on the Hilbert
space $H=\ell ^2({\mathbb Z})$.  The so called weak operator topology (WOT)  on $B(H)$  is defined to be the coarsest topology relative to which
the maps
$$
  T\mapsto \langle T(x), y\rangle
  $$
are continuous,  for every $x$ and $y$ in $H$.

Define $\text{GL}(H)$ to be the group of all invertible elements of $B(H)$ (the multiplication operation being the composition
of operators),  equipped with the induced WOT topology.

It is
then an easy exercise  to prove that multiplication is separately continuous.
However it is  not jointly continuous, and here is why.  Consider the bilateral shift operator $B$, defined on the
standard orthonormal basis $\{e_n\}_{n\in {\mathbb Z}}$ of $H=\ell ^2({\mathbb Z})$ by
$$
  B(e_n)=e_{n+1}, \quad \forall n\in  {\mathbb Z}.
  $$

It may be easily proved that the powers $B^n$ converge to zero in the WOT as $n\to \pm \infty $.  Consequently  the invertible
operators
$$
  T_n = B^n+2I
  $$
converge to $2I$ and hence
$$
  (\lim_{n\to \infty }T_n)  (\lim_{n\to \infty }T_{-n}) = (2I)(2I) = 4I.
  $$
However
$$
  T_nT_{-n} = (B^n+2I)(B^{-n}+2I) = I + 2B_n + 2 B^{-n} + 4I \to 5I,
  $$
so multiplication is not jointly continuous.