问:
A question regarding existence of Cyclic extension ( pg 296 Question 6)
Consider the following problem:  from algebra by Thomas Hungerford Section 5.

If char $K = p \neq $0, let $K_p = ${ $u^p - u | u\in K$ }
then show that  a cyclic extension field F of K of degree p exists if and only if $K \neq K_p$ .

Let K $\neq K_p$ , but I couldn't make any progress on this side.

Now, for the opposite, it is given that  Cyclic extension exists and let K =$K_p$ , so $u^p - u=u$ and I obtained a contradiction taht the polynomial is neither irreducible nor separable.

But I need help for the converse part.

Ty!

答: Let $K\ne K_p$ and $a\in K\setminus K_p$, i.e. the polynomial $f=x^p-x-a$ has no roots in $K$. By this answer $f$ is irreducible. Let $L=K(u)$ where $u$ is a zero of $f$ in some extension field. A straightforward computation shows that $u+r$ for $r\in\Bbb F_p$ (the prime field) are also zeros of $f$ (or just see the linked answer again). It follows that $L/K$ is Galois and that the Galois group is generated by $u\mapsto u+1$, i.e. it is cyclic.


问:
For $0 \leq a_n, b_n$: Is $\varliminf a_n \varlimsup b_n \leq \varlimsup a_n b_n $ true?
The sequences are assumed to be bounded!

I suspect it is, since $\varliminf a_n + \varlimsup b_n \leq \varlimsup (a_n+b_n)$ is. But I have not managed to prove it.

I thought it possible to mirror the following trick when proving the inequality for sums:

$\varlimsup b_n = \varlimsup (a_n + b_n - a_n) \leq \varlimsup (a_n + b_n ) + \varlimsup(-a_n) = \varlimsup (a_n + b_n )  -\varliminf(a_n)$

But the incipient idea fell short rather fast:

$0 = \varlimsup (a_n b_n - a_n b_n) \leq  \varlimsup (a_n b_n) + \varlimsup (-a_n b_n)$

I was hoping to manipulate $\varlimsup (-a_n b_n)$ in some ingenious way, but have not succeded.

Would love to read any ideas or hints.

答: The fact that the exponential function is continuous and increasing implies that liminfs and limsups pass through the exponential function. Setting $a_n=e^{A_n} ,\ b_n= e^{B_n}$ this turns the inequality into $$\exp(\liminf A_n + \limsup B_n) \le \exp(\limsup (A_n + B_n))$$ Since the exponential function is increasing we can cancel it on both sides which reduces it to the additive known result. Since this result is true for bounded sequences (including negative values), we are done.

ContentLicense: CC BY-SA 4.0


问:
Cauchy's Integral
Can we use Cauchy's integral formula to evaluate $\oint_{|z|=1}\frac{zdz}{\sin z}$? If so, how?

答: As @saulspatz noted, $z=0$ isn't a true singularity of the integrand, nor is any with $|z|<\pi$. The integrand has a Maclaurin series $1+\tfrac16z^2+O(z^4)$; in particular, there's no $\tfrac1z$ term, so the integral is $0$.

Title: Cauchy's Integral


问:
Tricks to interpret $\mathbb{R}^{\mathbb{R}}$
So I have the following question of how to interpret $\mathbb{R}^{\mathbb{R}}$. Is this just an "infinite" tuple of real numbers?

Since for example $ (x, y) \in \mathbb{R}^2$, $(x, y, z) \in \mathbb{R}^3$, etc. Is $(x_1, x_2, ... , x_n) \in \mathbb{R}^n$? Or is there another way that I should interpret this?

答: Yes: it is the set of all functions from $\Bbb R$ into $\Bbb R$. In fact, a function $f\colon\Bbb R\longrightarrow\Bbb R$ can be seen as the element $\bigl(f(x)\bigr)_{x\in\Bbb R}$ of $\Bbb R^{\Bbb R}$. And an element $(a_\lambda)_{\lambda\in\Bbb R}$ of $\Bbb R^{\Bbb R}$ can be seen as the function$$\begin{array}{ccc}\Bbb R&\longrightarrow&\Bbb R\\\lambda&\mapsto&a_\lambda.\end{array}$$


答: @s.harp is right. For sets $A,\,B$, the set of functions from $A$ to $B$ is denoted $B^A$ because its cardinality is $|B|^{|A|}$ (this is trivial when both cardinalities are finite, but is more general the definition of exponentiation of cardinalities). So $\Bbb R^{\Bbb R}$ is the set of functions from $\Bbb R$ to $\Bbb R$.

Title: Tricks to interpret $\mathbb{R}^{\mathbb{R}}$

Body:
So I have the following question of how to interpret $\mathbb{R}^{\mathbb{R}}$. Is this just an "infinite" tuple of real numbers?

Since for example $ (x, y) \in \mathbb{R}^2$, $(x, y, z) \in \mathbb{R}^3$, etc. Is $(x_1, x_2, ... , x_n) \in \mathbb{R}^n$? Or is there another way that I should interpret this?


问:
2D convolution of two circles
Imagine if two circles exist with definitions of $f_1(r) = circ(\frac{r}{R_1})$ and $f_2(r) = circ(\frac{r}{R_2})$ where circ is defined in a 2d dimension as:

$  circ(\frac{r}{R}) =
\begin{cases}
1,  & r \leq R  \\
0, & oth.
\end{cases}$

Where $r = \sqrt{x^2+y^2}$ in the 2d dimension. How is the convolution $ f_1(x,y) ** f_2(x,y)$ or $f_1(r) ** f_2(r)$ calculated? Notably, I'm very curious about the visual representation of the convolution.

答:
For circularly symmetric functions $f(r)$ and $g(r)$ we have

$$\begin{align*}(f**g)(x,y) &= \int_{-\infty}^\infty\int_{-\infty}^\infty f(x',y')g(x-x',y-y')dx' dy'\\
\\
&= \int_{-\infty}^\infty\int_{-\infty}^\infty f\left(\sqrt{x'^2+y'^2}\right)g\left(\sqrt{\left[x-x'\right]^2+\left[y-y'\right]^2}\right)dx' dy'\\
\\
&= \int_0^{2\pi}\int_0^\infty f\left(r'\right)g\left(\sqrt{\left[r\cos\theta-r'\cos\theta'\right]^2+\left[r\sin\theta-r'\sin\theta'\right]^2}\right)r'dr' d\theta'\\
\\
&= \int_0^{2\pi}\int_0^\infty f\left(r'\right)g\left(\sqrt{r^2+r'^2-2rr'\cos\theta\cos\theta'-2rr'\sin\theta\sin\theta'}\right)r'dr' d\theta'\\
\\
& (\text{pick }\theta=0 \text{, since the functions are circularly symmetric})\\
\\
&= \int_0^{2\pi}\int_0^\infty f\left(r'\right)g\left(\sqrt{r^2+r'^2-2rr'\cos\theta'}\right)r'dr' d\theta'\\
\\
&= (f**g)(r)
\end{align*}$$

which hopefully makes some sense, in that the convolution is multiplying and summing contributions from $g(r)$ with a law of cosines hypotenuse type of displacement.

I cannot provide you with any alternative visual representation for the 2D polar convolution.  I can say the operation is exactly equivalent to the 2D cartesian convolution of the two functions.

So let's assign $f(r) = \mathrm{circ}(r/a)$ and $g(r)=\mathrm{circ}(r/b)$, with $a<b$ without loss of generality since convolution is commutative.

In the $(r',\theta')$ plane, $f(r')$ has boundary $r'=a$, which is a circle of radius $a$, centered at $r'=0.$

In the $(r',\theta')$ plane, $g\left(\sqrt{r'^2+r^2-2rr'\cos\theta'}\right)$ has a boundary given by $b=\sqrt{r'^2+r^2-2rr'\cos\theta'}$, which is a circle of radius $b$, centered at $(r'=r, \theta' =0)$.

This convolution now depends on $4$ separate intervals for $r$, since we're looking at how two disks of different size overlap in the $(r',\theta')$ plane, as the center, $(r'=r, \theta'=0)$, of one disk is varied in $r'$ along the $\theta'=0$ line.

The four intervals are

$$\begin{align*}0 \le &r \le b-a\\
b-a \le &r \le b\\
b \le &r \le b+a \\
b+a \le &r < \infty\\
\end{align*}$$

I. Interval $0 \le r \le b-a$

This one is easy. The larger disk completely overlaps the smaller disk, so the smaller disk dictates the bounds of the integrals.  Hence, convolution in this interval is

$$(f**g)(r) = 2\int_0^\pi \int_0^a r'dr'd\theta' = a^2\pi$$

II. Interval $b+a \le r < \infty$

This one is easy. The larger disk has no overlap with the smaller disk.  Hence the convolution in this interval is

$$(f**g)(r) = 0$$

III. Interval $b-a \le r \le b$

This one gets messy, but something that makes it slightly simpler is that the origin, $r'=0$, is still included in the region of overlap of the disks.

We have to break this convolution up into at least two double integrals, to capture the area of the overlap of the disk.  (It's probably smarter/cleaner to break it up into 3 simpler integrals, but I'll go with the following 2 for now.)

To find the two regions, we need to find the angles $\pm \theta'_c$ where the boundaries of the 2 disks intersect.  So substituting the equation for the one boundary, $r'=a$, into the equation for the other boundary:

$$b = \sqrt{a^2 +r^2 -2ar\cos\theta'_c}$$

we end up with

$$\theta'_c = \cos^{-1}\left[\dfrac{r}{2a}\left(1-\dfrac{(b+a)(b-a)}{r^2}\right)\right]$$

The first region, bounded by $f(r')$ on the right, is simple to set integral bounds for and compute the integral.

The second region, bounded by $g\left(\sqrt{r'^2+r^2-2rr'\cos\theta'}\right)$ on the left, requires us to use the quadratic formula to find the upper bound of integration on $r'$.

Starting from

$$b= \sqrt{r'^2 +r^2 -2rr'\cos\theta'}$$

followed by some algebra and the quadratic formula we get

$$r' =r\left[\cos\theta' \pm \sqrt{\left(\dfrac{b}{r}\right)^2-\sin^2\theta'}\right]$$

of which we take the $+$ branch to get radii $r' > 0$.

So the convolution in this interval is

$$\begin{align*}(f**g)(r) &= 2\int_0^{\theta'_c}\int_0^a r'dr'd\theta'\\
&\quad + 2\int_{\theta'_c}^\pi \int_0^{r\left[\cos\theta' + \sqrt{\left(\frac{b}{r}\right)^2-\sin^2\theta'}\right]} r'dr'd\theta'\\
\\
&= a^2 \cos^{-1}\left[\dfrac{r}{2a}\left(1-\dfrac{(b+a)(b-a)}{r^2}\right)\right]\\
&\quad + r^2\int_{\theta'_c}^\pi \left[\cos\theta' + \sqrt{\left(\frac{b}{r}\right)^2-\sin^2\theta'}\right]^2 d\theta'\\
\\
&= a^2 \cos^{-1}\left[\dfrac{r}{2a}\left(1-\dfrac{(b+a)(b-a)}{r^2}\right)\right]\\
&\quad + r^2\left.\left[\sin\theta'\left(\cos\theta' + \sqrt{\left(\frac{b}{r}\right)^2-\sin^2\theta'}\right) \\
\quad + \left(\dfrac{b}{r}\right)^2\left(\theta'+\tan^{-1}\left[\dfrac{\sin\theta'}{\sqrt{\left(\frac{b}{r}\right)^2-\sin^2\theta'}}\right]\right)\right] \right\rvert_{\theta'_c}^\pi \\
\\
&= \left(a^2-b^2\right) \cos^{-1}\left[\dfrac{r}{2a}\left(1-\dfrac{(b+a)(b-a)}{r^2}\right)\right]\\
&\quad+b^2\left[\pi-\tan^{-1}\left(\dfrac{r\sin\theta'_c}{\sqrt{b^2-r^2\sin^2\theta'_c}}\right)\right] \\
&\quad - r\sin\theta'_c\left(r\cos\theta_c' + \sqrt{b^2-r^2\sin^2\theta'_c}\right) \\
\end{align*}$$

which is pretty messy.

There is a way to compute the convolution in this interval using two circular sectors and then subtracting off the area of a triangle, which should yield a nicer expression.

IV. Interval $b \le r \le b+a$

This one is also messy.  What makes it worse then the previous is that the origin is not included in the overlap of the disks, and the equation for the arc on the left hand side of the the region of overlap is not defined for all $\theta'$, since $\dfrac{b}{r} < 1$.  It's still computable, but care must be taken to use intervals where $\sin\theta' \le \dfrac{b}{r}$.

I'll leave this interval as an exercise for the reader. :)

Title: 2D convolution of two circles


答:
In general, the notation $X^Y$ means the set of all functions from $Y$ to $X$. In particular, $\mathbb{R}^{\mathbb{R}}$ means all the real-valued functions on $\mathbb{R}$.

The set $\mathbb{R}^n$ can be thought as $\mathbb{R}^{\{1,2,\cdots,n\}}$, i.e., the set of all functions $f:\{1,2,\cdots,n\}\to\mathbb{R}$.

The set $\mathbb{R}^{\mathbb{R}}$ can be interpreted as an infinite Cartesian product.


问:
What would a 3-way convolution be?
Convolution is defined as
$$
(f*g)(t) = \int^\infty_{-\infty} f(\tau) g(t-\tau)d\tau
$$

If I want to do a 3-way convolution like
$$
3way(f,g,h)(t) = \int^\infty_{-\infty}\int^\infty_{-\infty} f(\tau) g(t-\tau)h(t-\upsilon)d\tau
d\upsilon
$$

What is it normally known as?

Is it different from a normal convolution of 3 functions which is defined as

$$
(f*g*h)(t) = \int^\infty_{-\infty}\left(\int^\infty_{-\infty} f(\tau) g(\upsilon-\tau)d\tau\right)
h(t-\upsilon)d\upsilon
$$

答:
The usual way of writing a convolution is conducive to explicit calculations but it obscures the underlying symmetry of the definition. If you write it as

$$(f \ast g)(t) = \int_{\tau_1 + \tau_2 = t} f(\tau_1) g(\tau_2) \, d \mu$$

where $\mu$ refers to Lebesgue measure then it's clear how to generalize to any number of functions: the $n$-fold convolution is

$$(f_1 \ast \dots \ast f_n)(t) = \int_{\tau_1 + \dots + \tau_n = t} f_1(\tau_1) \dots \dots f_n(\tau_n) \, d \mu.$$

This definition preserves the full symmetry of convolution: you can arbitrarily permute the functions $f_i$ and you get the same result. Writing it down as an ordinary repeated integral requires breaking this symmetry, e.g. by writing $\tau_n = t - \tau_1 - \dots - \tau_{n-1}$ and writing it as an integral over $\tau_1, \dots \tau_{n-1}$. For $n = 3$, for example, this gives

$$(f \ast g \ast h)(t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\tau_1) g(\tau_2) h(t - \tau_1 - \tau_2) \, d \tau_1 \, d \tau_2.$$

What you describe as "a normal convolution" is the same as this, just with a different parameterization, namely $\tau_1 = \tau, \tau_2 = v - \tau$.

Title: What would a 3-way convolution be?


问:
Volume of a cone given constraints on $z$
This question is pretty straightforward, however, it's been a couple years since I've had to solve an integral like this!

Compute the volume bounded by $x^2 + y^2 \le z^2$ such that $c_1 \le z \le c_2$ where $c_1,c_2 \in \Bbb{R}_+$

I believe the first equation is a cone so I think the solution is something like
$$\int_0^1\int_{0}^{2\pi}\int_{c_1}^{c_2}dzrd\theta dr$$
however, I'm not confident in the bounds I set.

答: No; that's not the volume of that cone; instead, it's the volume of the cylinder$$\left\{(x,y,z)\in\Bbb R^3\,\middle|\,x^2+y^2\leqslant1\wedge z\in[c_1,c_2]\right\}.$$Note that\begin{align}x^2+y^2\leqslant z^2&\iff r^2\leqslant z^2\\&\iff r\leqslant z,\end{align}since $z,r\geqslant0$. So, it should be$$\int_0^{2\pi}\int_{c_1}^{c_2}\int_0^zr\,\mathrm dr\,\mathrm dz\,\mathrm d\theta$$


问:
Surjective homomorphism of Lie algebras implies surjective homomorphism of the Lie Algebra of the maximal compact subgroup
I tried it before, here is another try.
We have commutative Lie-groups $H$ and $G$ and a Lie-group hom. $f: G \rightarrow H$ such that the corresponding map of Lie-algebras $df: L(G) \rightarrow L(H)$ is surjective. Furthermore $H$ is compact. Let $G^1$ be the maximal compact subgroup of $G$. I want to show that the map of Lie-algebras $df: L(G^1) \rightarrow L(H)$ is also surjective. Can this be true? I am not so familiar with that kind of arguments, sorry.

Thanks for your patience.

答: This is false. Take $G = \mathbb{R}, H = S^1$ and $f : \mathbb{R} \to S^1$ to be the exponential map. $df$ is an isomorphism and $H$ is compact. But $G$ has trivial maximal compact subgroup.


答:
You don't need the spherical coordinates in the particular case.

The volume is given by
$$
V=\int_{c_1}^{c_2}\left(\iint_{x^2+y^2\le z^2} 1 dxdy\right)dz
=\int_{c_1}^{c_2}\pi z^2dz=\frac{\pi}{3}z^3\big|_{c_1}^{c_2}
$$

Title: Volume of a cone given constraints on $z$

Body:
This question is pretty straightforward, however, it's been a couple years since I've had to solve an integral like this!

Compute the volume bounded by $x^2 + y^2 \le z^2$ such that $c_1 \le z \le c_2$ where $c_1,c_2 \in \Bbb{R}_+$

I believe the first equation is a cone so I think the solution is something like
$$\int_0^1\int_{0}^{2\pi}\int_{c_1}^{c_2}dzrd\theta dr$$
however, I'm not confident in the bounds I set.


问:
Random mapping and entropy ordering
Let $X_1,X_2$ be discrete random variables such that $H(X_1)<H(X_2)$ where $H()$ is the entropy. We know that for any random mapping $T$ which is invertible ($T$ is a function of $\omega$ and $X$, by invertible we mean invertible for every fixed $\omega$) and independent of a random variable $X$ the following is true $$H(T(X))\geq H(T(X)|T)=H(T^{-1}T(X)|T)=H(X),$$ and we have $$H(T(X_1))\geq H(X_1)\text{ and }H(T(X_2))\geq H(X_2).$$ If $X_1$, $X_2$, $T(X_1 )$ and $T(X_2 )$ have the same support, is it true that $$H(T(X_1))\leq H(T(X_2))? $$

The reason for assuming the same support is the following: If the cardinality of the support of $X_2$ is greater than the cardinality of the support of $X_1$ and  $P(T(X_1))$, $P(T(X_2))$ are uniform then the above does not hold.

Please correct me if I made any mistakes. Thank you for your time.

答:
The following constructs a counterexample on three letters.

Take an $X$ that is distributed on $\{1,2,3\}$ under the law $(p_1, p_2, p_3)$, and let $T$ be picked uniformly at random from $\{T_a, T_b\}$ independently of $X$ for $T_a = (1,2,3), T_b = (2,1,3),$ where the vectors should be read as $(T_*(1), T_*(2), T_*(3))$. Notice that $T$ is a supported on invertible functions as desired. Further, the action of $T$ is to smooth out the mass on $\{1,2\}$ and to leave the mass on $3$ unchanged. Concretely, $$ P(T(X) = 1) = P(T(X) = 2) = \frac{p_1 + p_2}{2}, P(T(X) = 3) = p_3.$$

Now suppose we took an $X_1 \sim p$ with $p_3 = 1/3$ but $p_1$ and $p_2$ very skewed, and we took an $X_2 \sim q$ with $q_3 > 1/3$ but $q_1 = q_2$. Then $T(X_1)$ will be uniform but $T(X_2)$ has the same law as $X_2$. So as long as $H(X_2) > H(X_1)$ this yields a counterexample.

Concretely, take $X_1 \sim (p, 2/3 - p, 1/3)$ and $X_2 \sim (1/4, 1/4,1/2)$. By the argument above, $H(T(X_2)) = H(X_2) < \log_2(3) = H(T(X_1))$. So all we have to ensure is the existence of $\alpha$ such that $H(\alpha, 2/3 - \alpha, 1/3) < H(1/4, 1/4, 1/2)$. This exists by a continuity argument near $p = 0$, but we can be very explicit - take $\alpha = 1/6$. Then $H(X_1) = 2/3 + \log_2(3)/2 < 3/2 = H(X_2).$

Title: Random mapping and entropy ordering


问:
point of intersection of two lines in barycentric coordinate system
I am looking for an efficient way to determine the intersection point of two lines which go through a triangle (face) of a 3D triangular surface mesh.

For both lines I know the two points at which they intersect with the edges of a triangle (face).
Denoted $P_A^1, P_B^1$ for the first line and $P_A^2, P_B^2$ for the second line (See illustrative example figure).

Based on my investigations my preferred approach would be to:

Transform the the points of intersection for both lines ($P_A^1, P_B^1$ and $P_A^2, P_B^2$) into barycentric coordinates resulting in $B_A^1, B_B^1$ and $B_A^2, B_B^2$.

Define two lines: $L_1$ which goes through $B_A^1, B_B^1$ and $L_2$ going through $B_A^2, B_B^2$ based on the two-point form defined in section 4.1.1. of this document

Determine the point $B_I$ as the barycentric coordinates where $L_1$ and $L_2$ intersect based on the equations in section 4.3 of this document.

Transform $B_I$ back into the cartesian coordinate system to have its 3D coordinates, denoted as $P_I$.

Unfortunately steps 2 and 3 do not lead to meaningful results (i.e., the intersection points fall outside the triangle) and I have doubts that I am applying equations in section 4.1.1. and section 4.3 properly.

Even when using a simple example based on the upper illustration in this figure here by defining $B^1_A=(1,0,0), B^1_B=(0,1/2,1/2)$ and $B^2_A=(0,1,0), B^2_B=(1/2,0,1/2)$ I cannot determine $B_I$ correctly as $B_I=(1/3,1/3,1/3)$ with the above procedure.

答:
For clarity, let's use $(u, v, w)$ for barycentric coordinates.

Let's say the first line passes through points $(u_1 , v_1 , w_1)$ and $(u_2 , v_2 , w_2)$ in barycentric coordinates; and the second line through points $(u_3, v_3, w_3)$ and $(u_4, v_4, w_4)$.  If we parametrise these using $t_1$ and $t_2$, we have lines
$$\left\lbrace ~ \begin{aligned}
L_1(t_1) &= \Bigr(u_1 + t_1 (u_2 - u_1) ,~ v_1 + t_1 (v_2 - v_1) ,~ w_1 + t_1 ( w_2 - w_1) \Bigr) \\
L_2(t_2) &= \Bigr(u_3 + t_2 (u_4 - u_3) ,~ v_3 + t_2 (v_4 - v_3) ,~ w_3 + t_2 ( w_4 - w_3) \Bigr) \\
\end{aligned} \right.$$
At the intersection, $L_1(t_1) = L_2(t_2)$.

By definition, $u + v + w = 1$ for all $u, v, w \in \mathbb{R}$.  Therefore, we can pick any pair ($u, v$ or $u, w$ or $v, w$) of barycentric coordinates.  If we pick $u$ and $v$ (and therefore $w = 1 - u - v$), we have a system of two linear  equations and two unknowns $t_1$ and $t_2$:
$$\left\lbrace ~ \begin{aligned}
u_1 + t_1 ( u_2 - u_1 ) &= u_3 + t_2 ( u_4 - u_3 ) \\
v_1 + t_1 ( v_2 - v_1 ) &= v_3 + t_2 ( v_4 - v_3 ) \\
\end{aligned}\right.$$
which has exactly one solution:
$$\left\lbrace ~ \begin{aligned}
t_1 &= \frac{ u_1 (v_4 - v_3) + u_3 (v_1 - v_4) + u_4 (v_3 - v_1) }{ (u_1 - u_2)(v_4 - v_3) - (u_4 - u_3)(v_1 - v_2) } \\
t_2 &= \frac{ u_1 (v_2 - v_3) + u_2 (v_3 - v_1) + u_3 (v_1 - v_2) }{ (u_1 - u_2)(v_4 - v_3) - (u_4 - u_3)(v_1 - v_2) } \\
\end{aligned}\right.$$
where the denominator is the same for both, and is zero if and only if the lines are parallel.

If $0 \le t_1 \le 1$, then the intersection is on the line segment between the specified points on the first line.

If $0 \le t_2 \le 1$, then the intersection is on the line segment between the specified points on the second line.

The intersection is at $(u, v, w)$,
$$\left\lbrace ~ \begin{aligned}
u &= u_1 + t_1 ( u_2 - u_1 ) = u_3 + t_2 ( u_4 - u_3 ) \\
v &= v_1 + t_1 ( v_2 - v_1 ) = v_3 + t_2 ( v_4 - v_3 ) \\
w &= w_1 + t_1 ( w_2 - w_1 ) = w_3 + t_2 ( w_4 - w_3 ) \\
\end{aligned} \right.$$
where you can use either of the right sides.

The point is inside the triangle if and only if $0 \le u \le 1$, $0 \le v \le 1$, and $0 \le w \le 1 \iff 0 \le u + v \le 1$.

If you have the Cartesian coordinates $(x_1, y_1, z_1)$ corresponding to $(u_1, v_1, w_1)$, and so on, you can calculate the 3D coordinates of the intersection $(x, y, z)$ using
$$\left\lbrace ~ \begin{aligned}
x &= x_1 + t_1 ( x_2 - x_1 ) = x_3 + t_2 ( x_4 - x_3 ) \\
y &= y_1 + t_1 ( y_2 - y_1 ) = y_3 + t_2 ( y_4 - y_3 ) \\
z &= z_1 + t_1 ( z_2 - z_1 ) = z_3 + t_2 ( z_4 - z_3 ) \\
\end{aligned} \right.$$
again using either one of the right sides, as they are equal at the intersection by definition.


问:
Understanding the definition of exterior derivative
What shown below is a reference form Analysis on Manifolds by James Munkres

So I saw that some authors define the differential operetor $d$ as a function from $\bigcup_{k=0}^\infty\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)$ to $\bigcup_{k=1}^\infty\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)$ such that
$$
d\omega\in\Omega^{k+1}(A)
$$
for any $\omega\in\Omega^k(A)$ but anyway to give any definition I think that it is importat to prove  that if $h$ and $k$ are two different naturals then
$$
\Omega^k(A)\cap\Omega^h(A)=\emptyset
$$
because otherwise we do not how calculate the quantity $d\omega$ if $\omega$ is a $h$-form and $k$-form. Well it is not hard to prove that if $h,k\ge 1$ and $h\neq k$ then the above identity holds. Indeed if $x$ and $y$ are two point of $\Bbb R^n$ then
$$
\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)\cap\mathcal{L}^h\big(\mathcal T_y(\Bbb R^n)\big)=\emptyset
$$
when they are different and when they are not. However unfortunately I did not be able to prove that
$$
\Omega^0(A)\cap\Omega^k(A)=\emptyset
$$
for any $k>0$ so I ask to prove it. In particualr I know that if there exist a scalar function $f$ in $\Omega^0(A)\cap\Omega^k(A)$ for some $k>0$ then
$$
f(x)\in\Bbb R\cap\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)
$$
for any $x\in A$ and thus the statement follows proving that
$$
\Bbb R\cap\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)=\emptyset
$$
but unfortunately I did not to prove it. Anyway if we define a $0$-tensor as an element of the set $\Bbb R^{\{0\}}$ where $0$ is the null vector of $\mathcal T_x(\Bbb R^n)$ then the above identity becames
$$
\Bbb R^{\{0\}}\cap\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)=\emptyset
$$
and it trivially holds because $\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)\subseteq\Bbb R^{\big(\mathcal T_x(\Bbb R^n)\big)^k}$ and
$$
\Bbb R^{\{0\}}\cap \Bbb R^{\big(\mathcal T_x(\Bbb R^n)\big)^k}=\emptyset
$$
so that the above identity follows. So could someone help me, please?

答:
In practice, we do not actually care whether $\Omega^k(\mathbb{R}^n)$ and $\Omega^h(\mathbb{R}^n)$ are literally disjoint sets or not when $k \ne h$.  Instead, in a completely formal development, we would make it explicit that we are talking about a family of maps $d^{(k)} : \Omega^k(\mathbb{R}^n) \to \Omega^{k+1}(\mathbb{R}^n)$.  However, in most situations it will be clear from context which map $d^{(k)}$ we must be thinking of applying so that putting in the explicit index would be unnecessary noise.

For a somewhat more complex example of what I mean by that last sentence, we will also have a wedge product $\wedge$ which formally is actually a family of maps ${\wedge}^{(k,\ell)} : \Omega^k(\mathbb{R}^n) \times \Omega^\ell(\mathbb{R}^n) \to \Omega^{k+\ell}(\mathbb{R}^n)$.  Then one result you can state is: for $\omega \in \Omega^k(\mathbb{R}^n)$ and $\tau \in \Omega^\ell(\mathbb{R}^n)$, we have $$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^k \omega \wedge d\tau.$$  Now, from this context, it is straightforward to go through and assign what indices would have to be filled in for a more formal system to be able to accept the statement:
$$d^{(k+\ell)}(\omega \wedge^{(k,\ell)} \tau) = (d^{(k)} \omega) \wedge^{(k+1,\ell)} \tau +^{(k+\ell+1)} [(-1)^k \cdot^{(k)} \omega] \wedge^{(k,\ell+1)} (d^{(\ell)} \tau).$$
(Note that here, I have also made it explicit that vector addition and scalar multiplication of differential forms are also formally indexed families, i.e. we will never try to add $\omega \in \Omega^k(\mathbb{R}^n)$ to $\tau \in \Omega^h(\mathbb{R}^n)$ unless $k = h$.)

As a side note, this point of view, that we don't really care in practice whether different sets might actually overlap or not, is part of the idea behind "type theory" as a language for mathematics.  In most type theories, the question of whether elements of unrelated types are equal becomes something that it is not even meaningful to ask, in the sense that the corresponding formulas would not qualify as "well-formed formulas".  And I'm personally convinced that outside of axiomatic set theory, a vast majority of mathematics is actually done and thought about in some version of type theory rather than in ZFC.  Certainly, ZFC is useful from a theoretical point of view because of its restricted language and few axioms, and type theory is usually not too difficult to build "on top of" ZFC.  A useful analogy to a computer science situation is to think of ZFC as a form of "machine code", and think of type theory as a higher-level language which can be translated to machine code.


问:
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$
I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?

Thanks.

答:
Just to throw in another way to do this, since the equation has no $x^2$ term, the sum of its roots must be zero.  If it has multiple real roots, then they are all real and (to add up to zero) at least one root must be negative.

Complete a square for $a$ and rearrange:
$$(a+x/2)^2 = x^3 + \frac{x^2}{4} - 1$$

Basic analysis for RHS shows that it's negative when $x\le0$, so this equation cannot have a negative root.

Therefore the equation can only have one real root.


问:
Summable function that is not infinitesimal
This comes out from this exercise:

Let $u$ Harmonic on $\mathbb{R}^n$ and $\int_{_{\mathbb{R}^n}}u(x)^2 dx<+\infty$ prove that $u(x)=0$.

I want to use that $\mathbb{R}^n$ and $\int_{_{\mathbb{R}^n}}u(x)^2 dx<+\infty$ implies $\lim_{|x|\to+\infty}|u(x)|=0$. I am not sure of this property, is it real?

I mean, one can prove the exercise and than it shows that is real, but to prove the exercise one use Cauchy Shwartz on $\langle 1,u\rangle$ (i can give more datails).

So the question is when does $\mathbb{R}^n$ and $\int_{_{\mathbb{R}^n}}u(x)^2 dx<+\infty$ implies $\lim_{|x|\to+\infty}|u(x)|=0$ in general?

答:
I don't think you can have what what you ask. You really need to use specific properties of harmonic functions. Indeed, consider an arbitrary nonnegative function $\varphi \in C_c^{\infty} (\mathbb{R})$ such
that $\int_{\mathbb{R}} \varphi (x) d x = 1$. Define the infinitely differentiable function (no more with compact support)
$$ \Psi (x) := \sum_{n \in \mathbb{N}} \varphi \left( \frac{x - n}{r_n}
   \right) $$
where $(r_n)\to 0$ is a sequence of positive real numbers that shrinks the support
of $\varphi$. Further conditions on $(r_n)$ to be found later. We have
$$ \int_{\mathbb{R}} \Psi (x) d x = \sum_{n \in \mathbb{N}}
   \int_{\mathbb{R}} \varphi \left( \frac{x - n}{r_n} \right) d x =\sum_{n \in \mathbb{N}} r_n \int_{\mathbb{R}} \varphi
   (x) d x = \sum_{n \in \mathbb{N}} r_n . $$
Therefore if you choose $(r_n)\to 0$ so that the series converges to some $c > 0$, then
$$ \int_{\mathbb{R}} \Psi (x) d x = c. $$
However
$$\limsup_{x \rightarrow + \infty} \Psi (x)= \sup_{x \in \mathbb{R}} \varphi
(x) \, .$$


问:
Let $B(n)$ be the number of bit strings that contain “$01$” derive a recurrence relation for $B(n)$ and solve iteratively
I believe I have found the recurrence relation to be
$$B\left(n\right)=B\left(n-1\right)+2^{n-1}-1$$
with Initial Condition $B(0)=0$ (I am a bit unsure about the initial condition though but I think it is correct)

Now I am trying to solve $B(n)$ using iteration, this is what I have so far:
$$B\left(n\right)=B\left(n-1\right)+2^{n-1}-1$$
$$B\left(n\right)=B\left(n-2\right)+2^{n-2}+2^{n-1}-\left(2\right)1$$
$$B\left(n\right)=B\left(n-3\right)+2^{n-3}+2^{n-2}+2^{n-1}-\left(3\right)1$$
$$=B\left(n-k\right)+2^{n-k}+2^{n-\left(k-1\right)}+2^{n-\left(k-2\right)}-\left(k\right)1$$
and then I let n=k since the intial condition is $B(0)$ but this is where I get confused I am not sure what to do from here?
$$=B\left(0\right)+2^0+2^1+2^2+...+2^{\left(?\right)}-n$$

Body:
You’re on the right track:

$$\begin{align*}
B(n)&=B(n-1)+2^{n-1}-1\\
&=B(n-2)+2^{n-2}+2^{n-1}-2\\
&=B(n-3)+2^{n-3}+2^{n-2}+2^{n-1}-3\\
&\;\;\vdots\\
&=B(n-k)-k+\sum_{i=1}^k2^{n-i}\\
&\;\;\vdots\\
&=B(0)-n+\sum_{i=1}^n2^{n-i}\\
&=-n+\sum_{i=0}^{n-1}2^i\\
&=-n+2^n-1\\
&=2^n-n-1\,.
\end{align*}$$


答:
Yes, $B_0=B_1=0$.  Let $A_n = 2^n - B_n$ be the number of bit strings that do not contain $01$, so $A_0=1$.  For $n>1$, condition on whether the first bit is $0$ or $1$ to obtain recurrence relation
$$A_n = 1 + A_{n-1},$$
which implies that
$$A_n = n+1.\tag1$$
Hence
$$B_n = 2^n - A_n = 2^n - (1 + A_{n-1}) = 2^n - 1 - (2^{n-1} - B_{n-1}) = 2^{n-1} - 1 + B_{n-1},$$
or just use $(1)$ to conclude that $B_n = 2^n - n - 1$.

Title: Let $B(n)$ be the number of bit strings that contain “$01$” derive a recurrence relation for $B(n)$ and solve iteratively

Body:
I believe I have found the recurrence relation to be
$$B\left(n\right)=B\left(n-1\right)+2^{n-1}-1$$
with Initial Condition $B(0)=0$ (I am a bit unsure about the initial condition though but I think it is correct)

Now I am trying to solve $B(n)$ using iteration, this is what I have so far:
$$B\left(n\right)=B\left(n-1\right)+2^{n-1}-1$$
$$B\left(n\right)=B\left(n-2\right)+2^{n-2}+2^{n-1}-\left(2\right)1$$
$$B\left(n\right)=B\left(n-3\right)+2^{n-3}+2^{n-2}+2^{n-1}-\left(3\right)1$$
$$=B\left(n-k\right)+2^{n-k}+2^{n-\left(k-1\right)}+2^{n-\left(k-2\right)}-\left(k\right)1$$
and then I let n=k since the intial condition is $B(0)$ but this is where I get confused I am not sure what to do from here?
$$=B\left(0\right)+2^0+2^1+2^2+...+2^{\left(?\right)}-n$$


问:
Weak convergence of sums of inner products
My problem is the following; suppose $H$ an infinite dimensional separable Hilbert space, with orthonormal basis $(e_n)_{n\in\mathbb{N}}$. Define $f_N:=\frac{1}{N}\sum_{n=1}^{N^2}{e_n},$ for all $N\geq1.$ Show $f_N\rightarrow0$ weakly.

My previous attempts have gone through use of the triangle inequality and Cauchy-Schwarz inequality, yielding a non-useful bound tending to infinity with $N,$ through something along the lines of taking arbitrary $x\in H$ and saying

$\sum_{n=1}^{N^2}{\langle e_n,x\rangle}\leq|\sum_{n=1}^{N^2}{\langle e_n,x\rangle}|\leq\sum_{n=1}^{N^2}{|\langle e_n,x\rangle|}\leq\sum_{n=1}^{N^2}{||e_n||||x||}=N^2||x||,$

such that we may conclude $\langle f_N, x\rangle=\frac{1}{N}\sum_{n=1}^{N^2}{\langle e_n,x\rangle}\leq\frac{1}{N}N^2||x||=N||x||\rightarrow\infty,$ as $N\rightarrow\infty.$

What am I missing? - and would the use of $e_n\overset{n\rightarrow\infty}{\rightarrow}0$ weakly together with $\frac{1}{N}\overset{N\rightarrow\infty}{\rightarrow}0,$ in any way be sufficient?

答: Let $x \in H$ and $\epsilon >0$. Since $\sum_j |\langle x, e_j \rangle|^{2} <\infty$ there exist $k$ such that $\sum\limits_{j=k}^{\infty} |\langle x, e_j \rangle|^{2} <\epsilon$.  Now $\frac 1 N \sum\limits_{j=1}^{N^{2}} |\langle x, e_j \rangle| \leq \frac 1 N\sum\limits_{j=1}^{k-1} |\langle x, e_j \rangle|+\frac 1 N\sum\limits_{j=k}^{N^{2}} |\langle x, e_j \rangle|$ and $\sum\limits_{j=k}^{N^{2}} |\langle x, e_j \rangle| \leq (\sum\limits_{j=k}^{N^{2}} |\langle x, e_j \rangle|^{2})^{1/2} (N^{2}-k+1)^{1/2}$ by C-S inequality. Can you finish?


问:
Prove $-a\cdot e\cdot \ln(x)\le x^{-a}$
Today during a Calc I exam we(me and my classmates) have been ask to prove $$-a\cdot e\cdot \ln(x)\le x^{-a}$$, $\forall x>0$ and $\forall a\in \mathbb{R}$. But, noneone in the room has known how to do it. Anyone knows how to?

Thanks in advance

答:
The inequality is obviously true if $a=0$. If $a\not=0$, let $x=e^{-u/a}$. The inequality to prove becomes $eu\le e^u$ for all $u\in\mathbb{R}$.  Now $f(u)=e^u-eu$ has a unique critical point at $u=1$, that being where $f'(u)=e^u-e=0$.  Since $f''(u)=e^u\gt0$ for all $u$, the critical point at $u=1$ is a global minimum, and thus

$$e^u-eu=f(u)\ge f(1)=e^1-e\cdot1=0$$

for all $u$, which proves the desired inequality.

Title: Prove $-a\cdot e\cdot \ln(x)\le x^{-a}$


问:
Confusion about Hartshorne 5.16.1
Let $X$ be a scheme over a Noetherian scheme $Y$. This remark claims:

$X$ is projective over $Y$ if and and only if $X$ is proper over $Y$ and there exists an invertible very ample sheaf $\mathscr{L}$ on $X$.

Projectivity implies these two condistions seems to be clear. For the other direction, it cites an exercise before:

$f:X\to Z$ separated and of finite type over Noetherian $Y$, then any closed subscheme $C$  of $X$ which is proper over $Y$ has image $f(C)$ (with induced image subscheme structure) proper over $Y$.

In this case, $Z=\mathbb{P}^n_Y$ and $i:X\hookrightarrow Z$ an given immersion. Then $X,Z$ are both separated and of finite type over $Y$. Take $C=X$. Then we see the image closed. Then $i$ is a closed immersion, so $f$ is projective. Somehow, I do not need the existence of this invertible sheaf $\mathscr{L}$. Did I miss anything?

答: The very ample invertible sheaf $\mathcal{L}$ is how you get the immersion $X\to \Bbb P^n_Y$: Hartshorne's definition of a very ample sheaf $\mathcal{L}$ relative to a morphism $X\to Y$ is one so that $\mathcal{L}$ is $i^*\mathcal{O}_{\Bbb P^n_Y}(1)$ for some immersion $i:X\to \Bbb P^n_Y$. In general, there is no reason for a scheme $X$ over $Y$ to have an immersion to $\Bbb P^n_Y$: see for instance exercise II.7.13, where a proper scheme over $k$ with no immersion in to $\Bbb P^n_k$ is constructed.

Title: Confusion about Hartshorne 5.16.1


问:
Minimizing the volume of a cylinder using fixed surface area
Considering that I want my cylinder to have a fixed surface area of 0.3m^2, how can I minimize the volume of the cylinder. I have already tried to optimize it but I am only able to maximize the volume.</p>

答:
In the limits, the volume approaches 0, there technically isn't a minimum in the sense of 'local extrema'. Consider the following:

$$A=2\pi( r^2 + rh)=0.3$$

$$h =\frac{0.3}{2\pi r}-r$$

$$V=\pi r^2 h = \pi r^2 (\frac{0.3}{2\pi r}-r)=\frac{0.3r}{2} - \pi r^3$$

Notice as $r \rightarrow 0$, $V \rightarrow 0$, and of course $h \rightarrow \infty$.


问:
Why are there problems when interpolating $f(x)=\arctan(10x)$?
Given $f(x)=\arctan(10x)$, there would be a problem when we interpolate it by using Lagrange's method. This would have something to do with the derivatives of $f(x)$. I plotted some derivatives of $f$ but I did not come up with an answer. Can anybody tell me what goes wrong if we interpolate $arctan(10x)$ and why this comes from the derivatives of $f(x)$. thanks :-)

Oh.. second question: I also would like to know why this problem does not occur when using cubic Hermite interpolation... In addition, I want to know why this problem does not occur when using cubic spline interpolation ? I can not figure this out

答:
If what you got is something like this:

where dashed lines are the interpolant.

Then this is Runge's phenomenon. What you said is a typical example of this phenomenon that using high degree polynomial to approximate a continuous function on evenly spaced sample points. This happens when we approximate some smooth function on a given interval, given $n$ evenly distributed sample points, and using a single globally defined polynomial.

If you use cubic Hermite spline, we would not have this phenomenon because of this interpolation's piecewise nature, it is piecewisely defined on each small interval. We have multiple locally defined polynomials, we only glue these different polynomials together using continuity (namely $f$ and $f'$ are continuous across the sample points).

ContentLicense: CC BY-SA 3.0


问:
Pointwise and uniform convergence of $f_n(x)=\cos \frac{nx}{1+n^2}$
Study the pointwise and uniform convergence of $f_n(x)=\cos \frac{nx}{1+n^2}$ for $x \in \mathbb{R}$.

If $x=0$ it is $f_n(0)=1$, for $x \ne 0$ it is
$$\lim_{n \to \infty} \cos\frac{nx}{1+n^2}=\cos 0=1$$
So $f_n(x)$ converges pointwise to $f(x)=1$ for all $x \in \mathbb{R}$. To study the uniform convergence I must evaluate
$$\lim_{n \to \infty} \sup_{x \in \mathbb{R}}\left|\cos \frac{nx}{1+n^2}-1\right|=\lim_{n \to \infty} \sup_{x \in \mathbb{R}}\left(1-\cos \frac{nx}{1+n^2}\right)$$
Since the derivative of $1-\cos \frac{nx}{1+n^2}$ exists for all $x\in\mathbb{R}$, it is
$$\frac{d}{dx} \left(1-\cos \frac{nx}{1+n^2}\right)=\frac{n}{1+n^2} \sin \frac{nx}{1+n^2} \geq 0$$
$$\iff 2k\pi \leq \frac{nx}{1+n^2} \leq (2k+1)\pi \iff \frac{1+n^2}{n}2k\pi \leq x \leq \frac{1+n^2}{n}(2k+1)\pi$$
For $k\in\mathbb{Z}$, so
$$\lim_{n \to \infty} \sup_{x \in \mathbb{R}}\left(1-\cos \frac{nx}{1+n^2}\right)=\lim_{n \to \infty} \left(1-\cos \frac{n \left(\frac{1+n^2}{n}(2k+1)\pi\right)}{1+n^2}\right)=\lim_{n\to\infty}(1-\cos\pi)=2$$
So $f_n$ is not uniformly convergent in $\mathbb{R}$.
Is this correct? Some more questions:

I've noticed that $f_n(\frac{1+n^2}{n})=\cos 1$, can I conclude that since there exists at least one $\bar{x}=\frac{1+n^2}{n}$ such that $|f_n(\bar{x})-f(\bar{x})|=1-\cos 1 \ne 0$ so $f_n$ cannot be uniform convergent in whole $\mathbb{R}$ because it it will never be $|f_n(x)-f(x)|<\varepsilon$ for all $x \in \mathbb{R}$ because of $\bar{x}$. Is this correct?

It is $\cos \frac{nx}{1+n^2} \approx 1-\frac{n^2 x^2}{2(1+n^2)^2}$ as $n \to \infty$, can this be useful to evaluate the limit of the supremum? Or is it useless because the asymptotic behaviour can't give informations about a supremum?

Is there a way to identify if there are some subsets of $\mathbb{R}$ where $f_n$ can converge uniformly to $f$?

答:
For $(1)$, the answer is yes.
Another simple way to see this is that $\text{Im}(f_n) = \text{Im}(\cos) = [-1,1]$ for all $n$, and since $\text{Im}(f) = \{1\}$ convergence cannot be uniform.

For $(2)$, I don't really see how that helps.
If $x$ is fixed and you're letting $n$ grow, all you're getting help with is calculating the pointwise limit.

For $(3)$, notice that as $n\to\infty$, $n/(1+n^2)$ approaches $0$.
I guess the natural approach here would be to consider what happens if $x$ is bounded.
And then you'd have to ask yourself what happens if $x$ is not bounded?


问:
What is the remainder when 4 to the power 1000 is divided by 7
What is the remainder when $4^{1000}$ is divided by 7?
In my book the problem is solved, but I am unable to understand the approach. Please help me understand -

Solution  -

To find the Cyclicity, we keep finding the remainders until any
  remainder repeats itself. It can be understood with the following
  example:

No./7    ->    $4^1$  $4^2$  $4^3$  $4^4$  $4^5$  $4^6$  $4^7$  $4^8$

Remainder -> 4 2 1 4 2 1 4 2

Now $4^4$ gives us the same remainder as $4^1$, so the Cyclicity is of
  3 (Because remainders start repeating themselves after $4^3$

So any power of 3 or multiple of 3 will give the remainder of 1. So,
  $4^{999}$ will give remainder 1.

Final remainder is 4.

Now I don't understand the last line. Please explain, how the remainder comes down to 4?

答: We have $\varphi(7)=6$, therefore, $4^6\equiv 1\  (\mathrm{mod}\ 7)$ by Euler's theorem. Now, $ (4^6)^{166}=4^{996}\equiv 1\ (\mathrm{mod}\ 7)$. Since $4^4\equiv 4\ (\mathrm{mod}\ 7)$, it follows that $4^{1000}=4^{996}\times 4^4\equiv 4\ (\mathrm{mod}\ 7)$.


问:
Find the sum of the series $2^{-\frac{1}{n}} - 2^{-\frac{1}{n+1}}$
Find the sum of the series $\sum_{n=1}^\infty (\frac{1}{2^\frac{1}{n}}-\frac{1}{2^\frac{1}{n+1}})$.  Expanding we get

$$(\frac{1}{2}-\frac{1}{2^\frac{1}{2}})+(\frac{1}{2^\frac{1}{2}}-\frac{1}{2^\frac{1}{3}})+\cdots+(\frac{1}{2^\frac{1}{n}}-\frac{1}{2^\frac{1}{n+1}}).
$$
Cancelling equal and opposite terms we get $(\frac{1}{2}-\frac{1}{2^\frac{1}{n+1}})$.
If we write it as $\frac{1}{2}-(\frac{1}{2})^\frac{1}{n+1}$, and use $\lim_{n\to\infty} x^\frac{1}{n}=0$, for $|x|<1$, then the answer that we get is $\frac{1}{2}$, but if we take the common factor $\frac{1}{2}$ out,
then it becomes $\frac{1}{2}(1-(\frac{1}{2})^\frac{-1}{1+1/n})=\frac{1}{2}(1-2^\frac{1}{1+1/n}).$ As $n\to\infty$ it simplifies to $\frac{1}{2}(1-2)=-\frac{1}{2}$ OR as the author has done, $$S_n=\frac{1}{2}-\frac{1}{2^\frac{1}{n+1}} \implies \lim_{n\to\infty} S_n=\frac{1}{2}-\frac{1}{1}=-\frac{1}{2}.$$  What is my mistake?

答:
A CW answer to remove this question from the Unanswered queue.

As Gerry Myerson remarks, we have for $0 < x < 1$:

$$\lim_{n\to\infty} x^{1/n} = 1$$

Other than that, your computation is correct.


答:
You want to solve the problem:

$$\tag{P} \begin{cases} \text{minimize:} & V = \pi r^2 h \\ \text{u. c.:} & 2\pi (r^2 + r h) = 0.3 \\ & r,h > 0 \end{cases}$$

From the equality constraint you get:

$$\pi rh = 0.15 - \pi r^2$$

therefore your two-variables problem rewrites as one-variable optimization problem, i.e.:

$$\tag{P'} \begin{cases} \text{minimize:} & V = r (0.15 - \pi r^2) \\ \text{u. c.:} & 0 < r < \sqrt{\frac{0.15}{\pi}} \end{cases}\; .$$

Function $V = r (0.15 - \pi r^2)$ is of class $C^\infty$ and differentiating you find:

$$\begin{split} V^\prime &= 0.15 - 3\pi r^2 \\ V^{\prime \prime} &= -6\pi r \end{split}$$

hence $V$ is concave in $]0,\sqrt{\frac{0.15}{\pi}}[$ and takes its infimum in one of the extrema, i.e. either for $r \nearrow \sqrt{\frac{0.15}{\pi}}$ or for $r\searrow 0$; in either case you find $V \to 0$, which is not the mimimum because $0$ is not a value taken by function $V$ on $]0,\sqrt{\frac{0.15}{\pi}}[$.

Thus the minimum problem (P) has no solution at all.

Title: Minimizing the volume of a cylinder using fixed surface area

Body: Considering that I want my cylinder to have a fixed surface area of 0.3m^2, how can I minimize the volume of the cylinder. I have already tried to optimize it but I am only able to maximize the volume.</p>


问:
Measurable group homomorphisms are continuous
I am trying to solve the following exercise:

Let $f:\mathbb{R}^d\to\mathbb{C}$ be a group homomorphism between $(\mathbb{R}^d,+)$ and $(\mathbb{C},+)$, i.e., $f(x+y) = f(x)+f(y)$. Show that if $f$ is Lebesgue measurable, then $f$ is continuous.

The hint to the exercise says to demonstrate that for any open ball $B_{\mathbb{C}}(0,r)$ centred at the origin of the complex plane, there is at least one $z\in\mathbb{C}$ such that the set $f^{-1}(z+B_{\mathbb{C}}(0,r))$ has positive measure. One then has to use Steinhaus theorem, which states that for any $E\subseteq \mathbb{R}^d: \mu(E)>0$ the set $E-E$ contains a neighbourhood of the origin.

Unfortunately, I do not understand how the hint relates to the exercise (i.e., how this leads to "an inverse image of any open ball/set is an open set). What I have done so far is:

If $f^{-1}(z+B(0,r))$ is null for every $z$, then all of $\mathbb{R}^d$ is null by expressing the latter type as a countable union of sets of former type. Thus, there is at least one such $z$.

$f(0) = 0$, so we use the origin as a starting point.

答:
It is straightforward to check that $f(rx) = r f(x)$ for any rational $r$.

Suppose that in some neighbourhood $B(0, \delta)$, the function $f$ is bounded by $K$. Then we see that if $x \in B(0, {1 \over n} \delta)$ we have $f(x) \in B(0, {1 \over n} K)$. In particular, $f$ is continuous at the origin.

Since $f(x+y) = f(x)+f(y)$, we see that if $f$ is continuous at the origin, it is continuous at any $y \in \mathbb{R}^n$.

As an aside, if $r_n \to \lambda$ where $r_n$ are rationals, we see that $f(\lambda x) = \lim_n f(r_nx) = \lim_n r_n f(x) = \lambda f(x)$, hence $f$ is
linear.

To see where the neighbourhood of the origin comes from, suppose
$E= f^{-1}(z+B(0,r))$ has positive measure. Then
$E-E$ contains some neighbourhood of the origin, say $B(0,\delta)$.

Hence if $x \in B(0,\delta) \subset E-E$, we have $x = e_1 -e_2$ for some $e_k \in E$. Then $f(e_k) = z+z_k$ where $z_k \in B(0,r)$, and so
$f(x) = z_1-z_2 \in B(0,2r)$.


问:
Show that $Y$ is a closed subspace of $\ell^2$
This might be a straight forward problem but I wouldn't ask if I knew how to continue.
Apologies in advance, I am not sure how to use the mathematical formatting.

We are currently busy with inner product spaces and I'm struggling to prove that:
$$Y = \{x\mid x=(\xi_j) \in \ell^2, \xi_{2n}=0, n \in \mathbb{N}\}$$
is a closed subspace of $\ell^2$.

From this we need to find orthogonal complement of $Y$ which I should manage to do.
I just need to show the subspace is closed.

workings:

$\ell^2 = \{\xi_1,\xi_2,\xi_3,\xi_4,\ldots\}$,
$\sum_{i=1}^\infty |\xi_i|^2$ finite therefore convergent series.

$$Y = \{\xi_1,0,\xi_3,0,\xi_5,\ldots\}$$

$Y$ is a proper subspace of $\ell^2$ the separable infinite dimensional Hilbert space.
Clearly the sum |of the elements of Y|^2 converges however it will be slower convergence than $\ell^2$'s elements. Am I missing something elementary? I dont know how to put it together.

Also if $x_n \in Y$ and $x_n \to x$ implies $x \in Y$ then $Y$ will be closed.
How do I write out the formal proof?

Thank you!

答:
If $x_n\to x$ (in the $\ell^2$ norm) strongly, in particular it converges weakly, i.e.
$$
x_n\to x \text{ (weak) iff }\langle x_n, y \rangle \to \langle x, y \rangle, \;\forall \; y\in \ell^2.
$$
Then you have
$$
\langle x_n, e_{2k} \rangle=0 \to \langle x, e_{2k} \rangle=0,
$$
where $e_{2k}=(0,\dots,1,\dots,0)$ and the $1$ is the $2k$th entry.
Thus, every $2k$th entry of $x$ vanishes. This implies that $x\in Y$</p>


问:
Suppose X and Y have joint density $f(x, y) = 1512x^2y^5$ for $0 \leq x, 0 \leq y, x+y < 1$ and $f(x,y) = 0$ otherwise
I wish to double check my answers so that I know how to handle bounds when the region of $x$ and $y$ is a triangle.

a. Are $X$ and $Y$ independent?

No because the region of x and y. $0 \leq x, 0 \leq y, x+y < 1$ forms a triangle so we can automatically tell that it's dependent as the region must be rectangular to even be considered independent.

b. Find $P(X < 1/2)$.

I found that the density is $f(x) = \int_0^{1-x}1512x^2y^5dy = 252x^2(1-x)^6$ for $0 \leq x <1$
so $P(X < 1/2) = \int_0^{1/2}252x^2[1-x]^6 dx = 233/256$

c. Find $P(Y < 1/2)$.

I found that the density is $f(y) = \int_0^{1-y}1512x^2y^5dx = 504y^5(1-y)^3$ for $0\leq y < 1$
So $P(Y< 1/2) = \int_0^{1/2} 504y^5(1-y)^3 dy = 65/256 $

d. Find $P(X + Y < 1/2)$.

$P(X+Y < 1/2) = \int_0^{1/2}\int_0^{\cfrac{1}{2}-y}1512x^2y^5dxdy = 1/512$

I just wish to double check if my methodology and bounds of integration are correct. When I draw a picture I think I got the correct answers but I just want to make sure.

Thank you for any help!

答:
Your integrals look correct. Just a few comments.

a. Alternatively, you can find the marginal density functions $f_X$ and $f_Y$ and compare the product $f_X(x)f_Y(y)$ with $f(x,y)$.

bc. You should not use the same symbol $f$ for both the density functions $X$ and $Y$ (and the joint density). To avoid writing the number $1512$ over and over again, you can introduce $A=1512$, and write the joint density as $f(x,y)=Ax^2y^5$. You can then substitute the number in the last step.

The density of $X$ is
$$
f_X(x)=\int_{0}^{1-x}f(x,y)dy=A\int_{0}^{1-x}x^2y^5,\quad 0\le x\le 1
$$
and the density of $Y$ is
$$
f_Y(y)=\int_0^{1-y}f(x,y)dx=A\int_0^{1-y}x^2y^5,\quad 0\le y\le 1
$$

Once you find $f_X$ and $f_Y$, you then have
$$
P(X<1/2)=\int_{-\infty}^\infty f_X(x)dx=\int_0^{1/2}f_X(x)dx
$$
$$
P(Y<1/2)=\int_0^{1/2}f_Y(y)dy
$$

d. For the last question
$$
P(X+Y<1/2)=\int_{0}^{1/2}\left(\int_0^{1/2-y}Ax^2y^5dx\right)dy
$$


问:
R-orientations of Manifolds
Let $M$ be an $n$-manifold and $R$ a commutative ring with $1$.  Then for every $x\in M$ the relative homology group $H_{n}(M,M-{x};R)$ is isomorphic to $R$ and an $R$-orientation of $M$ is defined to be a function that assigns to each $x\in M$ a unit in $H_{n}(M,M-{x};R)$ subject to a local consistency condition.  This definition is where I am struggling.  This isomorphism is merely a group isomorphism, so how does it make sense to talk about a unit in $H_{n}(M,M-{x};R)$?

There is no canonical group isomorphism $H_{n}(M,M-{x};R)\rightarrow R$, and even if there was this doesn't seem helpful since homology groups aren't rings.  It seems like we are being invited to use what ever ismorphism $H_{n}(M,M-{x};R)\rightarrow R$ in order to associate the homology groups $H_{n}(M,M-{x};R)$ with $R$.  This would be fine if it's the case that if $\alpha$ and $\beta$ are any two isomorphisms $H_{n}(M,M-{x};R)\rightarrow R$ that there exists a ring isomorphism $\phi:R\rightarrow R$ such that $\phi\circ \beta=\alpha\circ id_{H_n}$.  And also if $B$ is a ball in $\mathbb{R}^{n}$ containging $x$ that the isomorphism $H_{n}(M,M-{B};R)\rightarrow H_{n}(M,M-{x};R)$ must send units to units.  But are these things true?

答:
No, it is not merely a group isomorphism. It is a module isomorphism.

Let $R$ be a ring, and consider $R$ as a module over itself. Suppose $\varphi: R \to R$ is a module homomorphism. Then $\varphi(r) = \varphi(1 \cdot r) = \varphi(1) \cdot r$ for all $r \in R$, by the fact that this is a module homomorphism.

So every module endomorphism of $R$ is of the form "multiplication by an element of $R$". Write $\varphi_a: R \to R$ for the map $\varphi_a(r) = ar$.

Then $\varphi_a \varphi_b = \varphi_{ab}$. It follows that $\varphi_a$ is invertible as a module homomorphism if and only if $a$ is invertible as an element of $R$. As products of units are units, it follows that any module isomorphism $\varphi: R \to R$ sends units to units.

Note that "units in $R$" are definable module-theoretically, which provides an alternative proof.

An element $r \in R$ is a unit if and only if the submodule generated by $r$ --- also known as the ideal $(r)$ --- is the entire module $R$. A unit of a ring is the same as a generator of $R$, considered as a module over itself (that is, we say $m \in M$ is a generator of a module if $mR = M$; notice that if $M = R$ considered as a module over itself, the condition $rR = R$ is precisely the condition that $r$ is a unit). As isomorphisms send module generators to module generators, it follows that a module isomorphism $R \to R$ sends units to units.

Title: R-orientations of Manifolds


问:
Fundamental group of theAlexandroff compactification of a subspace of $\mathbb R^2$

Consider $\mathbb R^2$ with the euclidean topology and let $X=\{(x,y)\in \mathbb R^2:x^3-y=0\}$ be a subspace of the real plane.

(a) Is $X$ connected, locally compact and closed?

(b) Calculate the fundamental group of the compactification $\hat X=X\cup \{\infty\}$ with base point in $(0,0)$.

For the request (a) I had no problems, but I'm not sure about how approaching the second point of the exercise.

I'm used to computing fundamental groups (using van Kampen's theorems or via homotopic equivalence) of topological spaces like the torus , the Klein bottle, particular surfaces in $\mathbb R^2$ or in $\mathbb R^3$ and so on. I always worked with Alexandroff compactification showing, for example, certain topological properties of generic $\hat X$ spaces.

In this problem we have to compute the fundamental group of the compactification of $\Gamma_{y=x^3}$ and I'm not able (maybe it's easy) to visualize it, so I have difficulties in identifying the classes of homotopy of the loop that generate the topological space.

Thank for you for your help.

答:
This community wiki solution is intended to clear the question from the unanswered queue.

Your question has been answered in the comments. The space $X$ is the graph of the function  $f : \mathbb R \to \mathbb R, f(x) = x^3$. Thus $X \approx \mathbb R$.

(a) Yes.

(b) $\hat X  \approx S^1$, thus $\pi_1(\hat X) \approx \mathbb Z$.


问:
Knot being a subset of another knot.
Can you show an example of a nontrivial knot $ K $ in $ \mathbb{R}^{4} $ such that $ K \cap \mathbb{R}^{3} $ is a nontrivial knot in $ \mathbb{R}^{3} $ ? It would be nice if you used parametric formulas.

答:
Any slice knot gives an example. Consider the three sphere $S^3$ as the boundary of the four ball $B^4$. A knot $K\subset S^3$ is (smoothly) slice if it bounds a smooth disk in $B^4$ (called a slice disk).

The slice disk can be doubled to obtain a knotted $S^2$ in $S^4$ as follows. Consider the four sphere $S^4$ as two $4$-balls glued along their common $S^3$ boundary: $S^4 = B^4_1 \cup_{S^3} B^4_2$. If $K\subset S^3$ is slice, then $K$ bounds smooth disks $D_1\subset B^4_1$ and $D_2\subset B^4_2$. The union $D_1\cup D_2$ is a knotted $S^2$ in $S^4$ that contains the knot $K$ as a subset.

Just to have some visuals, the picture below depicts a knot and a slice disk. The disk looks like it intersects itself, but the colors are supposed to indicate position in the fourth dimension. Because the arc of intersection of the disk with itself is colored differently on each intersecting page, the disk is embedded in $B^4$.

I don't know enough to write down parametric formulas for the disks, but perhaps someone can improve on my answer.

Title: Knot being a subset of another knot.


问:
Expected value of a Poisson process problem
The problem is as follows:

A pizza parlor serves n different types of pizza, and is visited by
a number $K$ of customers in a given period of time, where $K$ is a Poisson random
variable with mean $\lambda$. Each customer orders a single pizza, with all types of pizza
being equally likely, independent of the number of other customers and the types of
pizza they order. Find the expected number of different types of pizzas ordered.

Here is how I did it:

Let $X$ be the random variable for the number of different types of pizzas ordered.

Let $X_i$ be the indicator variable for when the $i^{th}$ pizza is ordered at least once.

Thus we have:
$$X = X_1 + ... + X_n$$
And:
$$E[X] = nE[X_i]$$

$E[X_i]$ is simply the probability that at least one person orders the $i^{th}$ pizza which is also $1-P(no\:one\:orders)$. The probability that a customer arrives and orders the $i^{th}$ pizza at any infinitesimal period $\delta$ is $\lambda\delta\frac{1}{n}$, thus, we can think of the arrival and ordering process as a Poisson random variable with parameter $\frac{\lambda}{n}$.

Thus, the probability there are no orders in some given time interval $t$ is:
$$P_P(0,t) = e^{\frac{\lambda}{n}t}$$

Here, $P_P$ is the probability distribution for the number of arrivals $k$ in some interval $\tau$:
$$P_P(k, \tau) = \frac{(\lambda\tau)^ke^{-\lambda\tau}}{k!}$$

Finally, we have
$$E[X] = n(1-e^{\frac{\lambda}{n}t})$$

However, the solution seems to not use any time interval, instead the answer is:

$$E[X] = n(1-e^{\frac{\lambda}{n}})$$

Does anyone know why this is the case? Don't we have to specify a time interval for such a problem? The solution seems to directly use the Poisson distribution which does not depend on an interval. If so, what is $n$ and $p$ here?

答:
Implicitly you're assuming $t=1$ unit of time, so that the number of customers arriving is actually Poisson $\lambda$.

Here's another approach that might be a bit clearer. Suppose $K=k$. Then easily
$$\mathbb E[X_i\mid K=k]=1-\mathbb P(\text{no one gets }i)=1-\left(1-\frac{1}{n}\right)^k.$$
Hence $\mathbb E[X\mid K=k]=n\left(1-(1-1/n)^k\right)$. So by the law of total expectation,
$$\mathbb E[X]=\mathbb E[\mathbb E[X\mid K]]=\mathbb E\left[n\left(1-(1-1/n)^K\right)\right]=n\left(1-\mathbb E\left[(1-1/n)^K\right]\right).$$
Now use PGF of the Poisson distribution $\mathbb E\left[t^K\right]=e^{\lambda(t-1)}$ to get the result $$\mathbb E[X]=n\left(1-e^{-\lambda/n}\right).$$
It looks like you're missing a minus sign in the exponent -- the expression that you've written could be negative!


问:
Prove that $(Q\leftrightarrow Q')\rightarrow (P\leftrightarrow P') $ is a tautology
Let $P,Q'$ be propositions and $Q$ a subformula of $P$ ($Q\in sub(P)$). Prove that if we get the formula $P'$ from $P$ by replacing the occurrence of the formula $Q$ with the formula $Q'$, then
$$(Q\leftrightarrow Q')\rightarrow (P\leftrightarrow P') $$
is a tautology.

I am having a hard time understanding the problem so I can't really get started on it. I just tried to do some truth table business, but it didn't really go anywhere. Maybe because I didn't know how to factor in the assumption that $Q\in sub(P)$. Help would be appreciated.

答:
What does the implication $(Q\leftrightarrow Q')\rightarrow (P\leftrightarrow P')$ say: if Q and Q' have same truth value then P and P' have same truth value. Now, it should be obvious why this implication holds.

If you are still not convinced, lets try to take it slowly. How we should prove an implication? Well, we should conjecture antecedent and then derive consequent. So, lets assume that $Q\leftrightarrow Q'$ holds. Then, if we replace every occurrence of $Q$ with $Q'$ in $P$ we will get formula $P'$. But this formula has same truth value as $P'$ because $Q$ has same truth value as $Q'$. So, it is true that $P\leftrightarrow P'$. Thus, we proved the implication $(Q\leftrightarrow Q')\rightarrow (P\leftrightarrow P')$.

In other words, formulas $P$ and $P'$ are semantically equivalent even if they are  potentially different formulas (takan as sequences of symbols). For example, formulas $\neg(p\land q)$ and $\neg p \lor \neg q$ are different but they have same truth tables (check this). So we can exchange them in other formulas without altering truth values of those formulas.

Formally, this statement is proved by induction on subformulas of $P$.


问:
interpolation of i.i.d. random variables
Let $\{X_k\}_{k\in \mathbb{Z}}$ be a sequence of iid, zero-mean and unit-variance Gaussian random variables. Let $Y(t), t\in\mathbb{R}$ be a continuous-time random process built by linear interpolation of $\{X_k\}_{k\in \mathbb{Z}}$. What is the first order pdf of $Y(t)$, i.e., $f(y;t)$?

Amendment:

What if the $\{X_k\}_{k\in \mathbb{Z}}$ is an iid sequence of random variables uniformly distributed over $(0,1)$?

答:
I think I got it:

Let $\tau=t-\lfloor{t}\rfloor$. Then $Y(t)=(1-\tau)X(\lfloor t\rfloor)+\tau X(\lceil t\rceil)$. At any instance $t$, $Y(t)$ is a random variable with the characteristic function:

$$\Phi_{Y(t)}(\omega) = E\{ e^{j\omega Y(t)} \} = E\{ e^{j\omega [(1-\tau)X(\lfloor t\rfloor)+\tau X(\lceil t\rceil)]} \}$$

Since $X(\lfloor t\rfloor)$ and $X(\lceil t\rceil)$ are independent:

$$\Phi_{Y(t)}(\omega) = E\{ e^{j\omega [(1-\tau)X(\lfloor t\rfloor)}\} E\{e^{j\omega\tau X(\lceil t\rceil)]} \}$$

Let $f_X(x)$ and $\Phi_X(\nu)$ denote the pdf and the characteristic function of $X$, respectively. Then

$$ \Phi_{Y(t)}(\omega) = \Phi_X((1-\tau)\omega)) \Phi_X(\tau\omega))$$

Taking the inverse Fourier transform:

$$ f_Y(y;t) = \dfrac{1}{1-\tau}f_x\left(\dfrac{y}{1-\tau}\right) * \dfrac{1}{\tau}f_x\left(\dfrac{y}{\tau}\right)$$

where $*$ denotes the convolution operator.

If $X_k$'s are independent and zero-mean Gaussian, any $Y(t)$ is also a zero-mean Gaussian and it suffices to evaluate the variance at $t$. However, this approach does not work if they are not Gaussian. As for the amended part, where $X$'s are uniformly distributed as

$$f_X(x) = \begin{cases}
1 & 0<x<1\\
0 & \text{o.w.}\\
\end{cases}$$

for $\tau=t-\lfloor{t}\rfloor$, let $\alpha = \max\{\tau, 1-\tau\}$. So,

$$ f_Y(y;t) = \begin{cases}
\frac{\tau}{\alpha(1-\alpha)} & 0<y<1-\alpha\\
\frac{1}{\alpha} & 1-\alpha<y<\alpha \\
\frac{1-\tau}{\alpha(1-\alpha)} & \alpha<y<1 \\
0 & \text{o.w.}
\end{cases}$$


问:
Find the inverse of $f(x)=\sqrt{3x^2 +1}$
Find the inverse of $f(x)=\sqrt{3x^2 +1}$.

We don't know if $f$ is invertible so we have to prove it, but how?

$f$ bijective $\Leftrightarrow $ $f$ invertible

Is $f$ injective? $ x \neq y \Rightarrow f(x) \neq f(y) $

$f(x) = f(y) $

$\sqrt{3x^2 +1} = \sqrt{3y^2 +1}$

$ x^2 = y^2 $

$x = y$

No injective $\Rightarrow$ No bijective $\Rightarrow$ No invertible.

Is that right?

答:
If you are looking at the "natural domain" of $f$, then it is not invertible since
$f(1)=f(-1)$.

If you look at a subset of the natural domain, say $(0,\infty)$, then
setting $y=\sqrt{3x^2+1}$ and solve for $x$, you get
$$
x=\sqrt{\frac{y^2-1}{3}},\quad y>1
$$
which gives you the inverse of $f:(0,\infty)\to\mathbb{R}$ where
$$
f^{-1}(y)=\sqrt{\frac{y^2-1}{3}},\quad y>1
$$


答: When we talk about functions and inverses, we should be careful about domains and codomains. Function $f$ can be defined on whole $\mathbb R$, but then it will not be injective, because $f(x)=f(-x)$ for all $x\in\mathbb R$. If we restrict function to $[0,\infty)$ then $f$ will be injective and its image will be $[1,\infty)$. Similar thing happens if we restrict $f$ to $(-\infty,0]$.

Body:
Find the inverse of $f(x)=\sqrt{3x^2 +1}$.

We don't know if $f$ is invertible so we have to prove it, but how?

$f$ bijective $\Leftrightarrow $ $f$ invertible

Is $f$ injective? $ x \neq y \Rightarrow f(x) \neq f(y) $

$f(x) = f(y) $

$\sqrt{3x^2 +1} = \sqrt{3y^2 +1}$

$ x^2 = y^2 $

$x = y$

No injective $\Rightarrow$ No bijective $\Rightarrow$ No invertible.

Is that right?


问:
What is the convex hull of $\text{conv}(u_1,u_2,\cdots,u_p)+\text{conv}(v_1,v_2,\cdots,v_s)$?
Let $u_i, i= 1,\cdots,p$ and $v_j, j= 1,\cdots,s$ be finitely many vectors in $\mathbb{R}^n$. Show that

$$
\text{conv}(u_1,u_2,\cdots,u_p)+\text{conv}(v_1,v_2,\cdots,v_s)=\text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}
$$

We need to show

$$
x+y \in \text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}
$$

where $x \in \text{conv}(u_1,u_2,\cdots,u_p)$ and $y \in \text{conv}(v_1,v_2,\cdots,v_s)$. Also, we need to show

$$
z \in \text{conv}(u_1,u_2,\cdots,u_p)+\text{conv}(v_1,v_2,\cdots,v_s)
$$

where $z \in \text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}$.

I have tried the following for the first one:

Let $x \in \text{conv}(u_1,u_2,\cdots,u_p)$ so $x=\sum_{i=1}^p\lambda_iu_i$ where $\sum_{i=1}^p\lambda_i=1$. Also, Let $y \in \text{conv}(v_1,v_2,\cdots,v_s)$ so $x=\sum_{j=1}^s\mu_jv_j$ where $\sum_{j=1}^s\mu_j=1$.

Summing them

$$x+y=\lambda_1u_1+\lambda_2u_2+\cdots+\lambda_pu_p+\mu_1v_1+\mu_2v_2+\cdots+\mu_sv_s.$$

Now the question is how we can get something in the form of $\text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}$?

答:
Here is an less verbose and more general way. Suppose you have two sets $S_1,S_2$ and we let
$C_1= \operatorname{co} S_1$, $C_2= \operatorname{co} S_2$.

Then we have
$S_1+S_2 \subset C_1+C_2$ and so $\operatorname{co}  (S_1+S_2) \subset C_1+C_2$.

We have $S_1+S_2 \subset\operatorname{co}  (S_1+S_2) $. Suppose
$c \in C_1$, then $c= \sum_k \lambda(k) s_1(k)$ where the $s_1(k) \in S_1$
and the $\lambda(k)$ are convex multipliers. Since $\operatorname{co}  (S_1+S_2)$ is convex, it follows that $c+s_2 = \sum_k \lambda(k) (s_1(k) +s_2) \in \operatorname{co}  (S_1+S_2)$ for any $s_2 \in S_2$, and so $C_1 +S_2 \subset \operatorname{co}  (S_1+S_2)$. A similar argument shows
that $C_1+C_2 \subset \operatorname{co}  (S_1+S_2)$.


问:
RSA encryption theory - modulo theory
I'm a bit mathematically challenged and have been working on the RSA cipher (good start). I can find the public and private keys and know how to work do modulo operations on a calculator. The problem is that I can't do them when the numbers get to high. For example say I have:

$$10^{541} \bmod{2923} = C$$
The numbers involved here become very large and don't display fully on a calculator, if it can even handle the numbers (mine is crap). What I am wondering is if there is a better method to work out the ciphertext or plaintext that will work for largish numbers.

N.B. I'm not a mathematician, I'm in computing but was told my question would be better posed here.

答:
You should familiarize yourself with the square and multiply algorithm. That way you never need to multiply together numbers larger than the modulus (here 2923).

The ideas is crudely that you first calculate remainders $x_k$ of $10^{2^k}$ modulo $2923$ for as high $k$ as needede by repeated squaring:
$$
10^2\equiv 100\pmod{2923},
$$
$$
10^4=(10^2)^2=10000\equiv1231\pmod{2923},
$$
(because $10000=8769+1231=3\cdot2923+1231)$,
$$
10^8=(10^4)^2\equiv1231^2\equiv x_3 \pmod{2923},
$$
(can't be bothered to compute $x_3$ now, sorry)
$$
10^{16}=(10^8)^2\equiv x_3^2\equiv x_4\pmod{2923},
$$
and so on all the way up to
$$
10^{512}=(10^{256})^2\equiv x_8^2\equiv x_9\pmod{2923}.
$$
At this point you know integers $x_k,k=0,1,2,\ldots,9$ such that
$0\le x_k<2923$ and $10^{2^k}\equiv x_k\pmod{2923}$. You had to calculate the squares of nine integers (all smaller than $2923$) and the remainders of those squares. Then you can finish off by multiplying a chosen selection of those numbers and the remainde of that product:
$$
10^{541}=10^{512+16+8+4+1}=10^{512}\cdot10^{16}\cdot10^8\cdot10^4\cdot10
\equiv x_9x_4x_3x_2x_0.
$$

The usual description of square and multiply organizes the calculations a bit more efficiently (so that you don't have to store so many intermediate results), but the general idea is as above.

Title: RSA encryption theory - modulo theory


问:
Operation precedence in coordinate transformation
Consider two orthogonal, right-handed frames whose orientations and positions are as depicted in the below picture.

In order to bring the axes of frame {A} onto those of frame {B}, we can first rotate {A} about axis $Z_A$ $\alpha$ degrees counterclockwise and then translate an amount of s in the $X_A$ direction. We could do the opposite. The two operations do not yield the same result, what is more, first translating then rotating produces a false result.

Rotate and translate:

$$
    \begin{bmatrix}
    1 & 0 & s \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}
    \begin{bmatrix}
    \cos\alpha & -\sin\alpha & 0 \\
    \sin\alpha & \cos\alpha & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}=\begin{bmatrix}
    \cos\alpha & -\sin\alpha & s \\
    \sin\alpha & \cos\alpha & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}
$$
which is true; translate, then rotate:
$$ \begin{bmatrix}
    \cos\alpha & -\sin\alpha & 0 \\
    \sin\alpha & \cos\alpha & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}
    \begin{bmatrix}
    1 & 0 & s \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}
   =\begin{bmatrix}
    \cos\alpha & -\sin\alpha & s\cdot\cos\alpha \\
    \sin\alpha & \cos\alpha & s\cdot\sin\alpha \\
    0 & 0 & 1 \\
    \end{bmatrix}
$$
which is false.

I know that the order of operations should be maintained when there are at least two rotations, but here there are just one translation and one rotation. My questions are,

Why don't one translation and one rotation commute?

Why did I get a false matrix when I first translated?

答:
Matrix $$\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\\sin\alpha & \cos\alpha & 0 \\0 & 0 & 1 \end{bmatrix}$$ is matrix of rotation around coordinate origin. In second case you want to rotate vectors around point $Z_B$, and matrix of that rotation is more complicated.

In general, when you want to rotate vectors about some point $A$, then you should translate $A$ to origin, do rotation of vector, and then translate rotated vector back to $A$.


问:
How to find density of one random variable if joint pdf contains $3$ random variables?
My book asks me the following question:

Let $X, Y, Z$ have joint density
$f(x, y, z) = 6$, for $0 < x < y < z < 1$
and $f(x, y, z) = 0$ otherwise.

a. Are $X$ and $Y$ and $Z$ independent?

I wrote no because it doesn't seem like the region of $0 < x < y < z < 1$ is rectangular so they aren't independent.

b. Find the density $f(x)$ of X.

c. Find the density $f (y)$ of Y .

d. Find the density $f(z)$ of Z

I need help finding the density of one random variable for example $f(x)$ if the joint pdf of three random variables are given.

Can someone give me the formula or methodology to do so? If I can figure out how to do one, I can definitely do the rest.

Thank you for any hints!

答:
The power density function of $X$ $f_X$ is given by :

$$ f_X(x) = \int_{(y,z)\in \mathbb{R}^2} f(x,y,z)dydz$$

By Fubini-Tonelli, you can integrate either w.r.t. $y$ or $z$ first, so you get

$$  f_X(x) = \int_{y\in \mathbb{R}} \int_{z\in \mathbb{R}} f(x,y,z)dzdy = \int_{y\in \mathbb{R}} \int_{z\in \mathbb{R}} 6 \times \textbf{1}_{\{0 < x < y < z < 1\}}dzdy $$

Using the definition of $f$, only taking into account the regions of the plane where it is non-zero :

$$ f_X(x) = \int_{y = x}^1 \int_{z = y}^1 6dzdy = \int_{y = x}^1 6 \times (1-y)dy$$

And finally you get

$$ f_X(x) = 3 \times (1 - x)^2 $$ (you can check that this is indeed a density function on $[0,1]$)

Now you have to proceed in a similar way to compute $f_Y$ and $f_Z$. Hope that helps.


问:
Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.

Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.

We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?

答:
(A comment rather than an answer
but much easier to enter)

This is an improvement
of my answer
Proving or disproving: $a^b<b^a$?
where I showed that
if $x^{1/x} = y^{1/y}$
and
$1 < x < y$
then
$y > e^2/x$.

This is an improvement since
$(e-1)^2/(x-1)+1- e^2/x
=(e - x)^2/((x - 1) x)
\gt 0$.


问:
$(t\mapsto t^ae^{bt} )_{a\in \mathbb{R}_+,b\in\mathbb{R}}$ is linearly independent of $\mathcal{F}(\mathbb{R},\mathbb{R})$ over $\mathbb{R}$.
We consider the family  $F=(t\mapsto t^ae^{bt} )_{a\in \mathbb{R}_+,b\in\mathbb{R}}$ of functions of space $\mathcal{F}(\mathbb{R},\mathbb{R})$.
Show that $F$  is linearly independent of $\mathcal{F}(\mathbb{R},\mathbb{R})$ over $\mathbb{R}$.

An idea please.

答:
Some details omitted:

Suppose $\sum_k \alpha_k t^{a_k}e^{t b_k} = 0$ for all $t$.

Order the $(a_k,b_k)$ such that $(a,b) < (c,d)$ if $b<d$ or $b=d$ and $a<c$.

By reordering we can suppose that the $(a_k,b_k)$ are in increasing order.

Consider ${1 \over t^{a_n}e^{t b_n}} \sum_k \alpha_k t^{a_k}e^{t b_k} $ and let $t \to \infty$, this shows that $\alpha_n = 0$.

Rinse & repeat.


问:
Proving two subgroups of the same order are both normal
Let $X$ be the set of all subgroups of $G$, and let the group action be defined by conjugation, that is for $a\in X$:
$$g.A = gAg^{-1}$$
Now let $A$ and $B$ be the only subgroups of $G$ of order $n$ (N.B. that $G$ may have other subgroups, but they will all have order $\neq n$). Suppose $|G|$ is odd, prove that $A$ and $B$ are both normal in $G$. We are given the hint to use the second form of the orbit-stabiliser theorem given to us by: $$|\rm{Orb}_G(x)| = {|G|\over{|\rm{Stab}_G(x)|}}$$
In a previous part of the question we have proven that if we have $K\in \rm{Orb}_G(A)$ that $|K| = |A|$.

My immediate thoughts for this is that if $A,B$ are normal then $\rm{Stab}_G(A)=G$ but have no idea how to use the fact that $|G|$ is odd or how that could factor in, is there something really obvious that I'm missing?

答:
Suppose $A$ is not normal in $G$. Since conjugate subgroups have the same order, $B$ must be in the orbit of $A$ when $G$ acts on it by conjugation. So $

As there are no other subgroups of that order, $2=|Orb_G(A)|$.

But then, $2$ divides $|G|$ but this is odd!.


问:
If the pair of random variables X and Y have $f(x, y) = 180x^2y^2$ , for $0 < x, 0 < y$, and $x + y < 1$ and $f(x,y)=0$ otherwise
Find: $P(X + Y > 1/2)$

I know the answer is $63/64$. But how do you find that answer? I'm really struggling with what the bounds of integration would be for a question like this. Can someone show me?
I tried $\int_{1/2}^1\int_0^{1-x}180x^2y^2dydx$ but this gives me 11/32 which isn't the correct answer?

What do I need to do to derive $63/64$?

Thank you for any help

答:
Watch lectures on calculating area, volume using integrals there are tons of them on youtube.
Bounds are easy to figure out if you draw graphs. I will show you how it is so easy to solve these types of questions using graphs.

$P(X+Y\ge\frac{1}{2})=1-P(X+Y<\frac{1}{2})=\int\int f(x,y)dy dx$

The region inside the red outlined triangle is your required area where you have to figure out bounds.

Draw a vertical strip like shown below and go bottom to up(You can also start drawing a horizontal strip and go left to right).

From the bottom, we see $y$ touches the $x$-axis so $y=0$ and strip goes up and touches $y=\frac{1}{2}-x$ and drag thiis vertical strip left to right we see $x=0$ initially and goes till $x=\frac{1}{2}$

so we have just found our bounds.

$1-\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-x} f(x,y)dy dx$ is required probability.


问:
Prove $g(x)$ is irreducible over $F$.
Let $F$ be a field and let $f(x)$ be a polynomial of degree $n \geq 1$ in $F[x]$. Let $g(x) = f\big(\frac{1}{x}\big)x^n$.

Assume $f(0) \neq 0$. Prove that $f(x)$ is irreducible over $F$ if and only if $g(x)$ is irreducible over $F$.

$\mathbf{My~Attempt:}$

Prove of if:

Assume that $f(x)$ is irreducible over $F$, WTP: $g(x)$ is irreducible over $F$.

First, notice that $F[x] = \{ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 ~|~ a_i \in F \text{ and for some } n \in \mathbb{N} \cup \{ 0 \} \}$.

Since, we let $f(x)$ be a polynomial of degree $n \geq 1$ in $F[x]$.

Then, $f(x) = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0$, for each constant coefficient $b_i \in F$.

Since, we assumed that $f(x)$ is irreducible over $F$ and we know that $f(0) \neq 0$.

Then, by defintion of irreducible, if $f(x) = h(x)j(x)$ with $h(x), j(x) \in F[x]$, then $h(x)$ or $j(x)$  is a unit in $F[x]$.

Now I don't know how to prove that $g(x) = f\big(\frac{1}{x}\big)x^n = b_n + b_{n-1} x + \cdots + b_1 x^{n-1} + b_0 x^n$ whhere $g(x) \in F[x]$ is irreducible.

Also, I have no idea only how to prove the only if which assume $g(x)$ is irreducible over $F$ and prove $f(x)$ is irreducible over $F$.

答:
Suppose $g$ is reducible, let $g=h_1\cdot h_2$, with $n_1=\deg h_1$, $n_2=\deg h_2$. Then
$$\left(x^{n_1}h_1(1/x)\right)\cdot\left(x^{n_2}h_2(1/x)\right)=x^ng(1/x)=f(x),$$
so $f$ is reducible. Hence $f$ irreducible $\implies$ $g$ irreducible.

You can then basically run this argument in reverse.

Title: Prove $g(x)$ is irreducible over $F$.


问:
Expected value of mean of N Chi squared random variables
I understand that here they have used law of large numbers. But according to the law the result should equal the expected value of the random variable. But how is the expected value of the r.v only equal to the expectation of the first chi squared r.v. shouldn't it include all the r.v. till n. X1,X2,...,Xn are iid and normally distributed. Then as n approaches infinity what value does summation of Xi squared/n (where i=1,2,..,n) converge to</p>

答:
We are given random variables are iid.Let $Y=X^2$

we know $Y \sim \chi_{1}^2$

$E(Y)=1$

$E\left(\frac{Y_1+Y_2.....+Y_n}{n}\right)=\frac{1}{n}(E(Y_1)+E(Y_2)..+E(Y_{n})))=\frac{n}{n}=1$

Also if you don't know about chisquare result you can use $E(X^2)=V(X)+E(X)^2=1+0$

There you can even use $\frac{1}{n}E(X_{1}^2+X_{2}^2...X_{n}^2)=\frac{n}{n}$

So as $n$ goes large we get $1$


答:
If $f$ is reducible, then $f(x)=q_1(x)q_2(x)$ and so $g(x)=q_1(\frac{1}{x})q_2(\frac{1}{x})x^n$. But $n=\deg(q_1)+\deg(q_2)$, and so $$g(x)=x^{\deg(q_1)}q_1(\frac{1}{x})x^{\deg(q_2)}q_2(\frac{1}{x})$$

where $x^{\deg(q_1)}q_1(\frac{1}{x})$ is a polynomial. This tells us that $g(x)$ is reducible. This should show you the "only if" part.

Title: Prove $g(x)$ is irreducible over $F$.

Body:
Let $F$ be a field and let $f(x)$ be a polynomial of degree $n \geq 1$ in $F[x]$. Let $g(x) = f\big(\frac{1}{x}\big)x^n$.

Assume $f(0) \neq 0$. Prove that $f(x)$ is irreducible over $F$ if and only if $g(x)$ is irreducible over $F$.

$\mathbf{My~Attempt:}$

Prove of if:

Assume that $f(x)$ is irreducible over $F$, WTP: $g(x)$ is irreducible over $F$.

First, notice that $F[x] = \{ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 ~|~ a_i \in F \text{ and for some } n \in \mathbb{N} \cup \{ 0 \} \}$.

Since, we let $f(x)$ be a polynomial of degree $n \geq 1$ in $F[x]$.

Then, $f(x) = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0$, for each constant coefficient $b_i \in F$.

Since, we assumed that $f(x)$ is irreducible over $F$ and we know that $f(0) \neq 0$.

Then, by defintion of irreducible, if $f(x) = h(x)j(x)$ with $h(x), j(x) \in F[x]$, then $h(x)$ or $j(x)$  is a unit in $F[x]$.

Now I don't know how to prove that $g(x) = f\big(\frac{1}{x}\big)x^n = b_n + b_{n-1} x + \cdots + b_1 x^{n-1} + b_0 x^n$ whhere $g(x) \in F[x]$ is irreducible.

Also, I have no idea only how to prove the only if which assume $g(x)$ is irreducible over $F$ and prove $f(x)$ is irreducible over $F$.


问:
Help solving a PDE similar to the heat equation
In my course on PDEs, I was given this PDE similar to the heat equation, which we solve for $t \geq 0$ and $0 \leq x \leq 1$

$u_t=\alpha u_{xx} + u$ with $\alpha$ a positive constant. The conditions provided are $u_x(0,t)=u_x(1,t)=0$ and $u(x,0)=\sin^2(\pi x)$.

I was not told how to solve this equation. This looks similar to the heat equation which I know can be solved using separation of variables. I am not very familiar with separation of variables or Fourier series, so I was wondering, can this be solved with separation of variables? I found myself unable to solve this equation using separation of variables, probably due to lack of familiarity. Can anyone please show me how to solve this? I thank all helpers.

答:
Make the substitution

$$v = e^{-t}u$$

then

$$v_t = -e^{-t}u + e^{-t}u_t$$
$$v_{xx} = e^{-t}u_{xx}$$

Now your original equation was:

$$u_t = \alpha u_{xx} + u$$

$$u_t-u-\alpha u_{xx} = 0$$

Multiply across by $e^{-t}$

$$e^{-t}u_t-e^{-t}u-e^{-t}\alpha u_{xx} = 0$$

Plugging in what you derived

$$v_t - \alpha v_{xx} = 0$$

For initial conditions:

$$v_x(0,t) = e^{-t}u_x(0,t) = e^{-t}(0)= 0$$

$$v_x(1,t) = e^{-t}u_x(1,t)= e^{-t}(0) = 0$$

$$v(x,0) = e^{0}u(x,0) = \sin^2(\pi x)$$

Hence solve the system (Heat Equation):

$$v_t - \alpha v_{xx} = 0$$

$$v_x(0,t) = 0$$

$$v_x(1,t)  = 0$$

$$v(x,0) = \sin^2(\pi x)$$

To get your final answer. remember

$$v = e^{-t}u$$

$$u = e^{t}v$$


问:
Decomposition of a meromorphic function into a singular and holomorphic part
Let $X$ be a normed space, $\Omega\subseteq\mathbb C$ be open and $f:\Omega\to X$ be meromorphic in the sense that for all $z_0\in\Omega$, there is a $k_{z_0}\in\mathbb N_0$, $\left(a^{(z_0)}_k\right)_{k\ge-k_{z_0}}\subseteq X$ and a neighborhood $N_{z_0}\subseteq\Omega$ of $z_0$ with $$f(z)=\underbrace{\sum_{k=-k_{z_0}}^{-1}a_k^{(z_0)}(z-z_0)^k}_{=:\:m_{z_0}(z)}+\underbrace{\sum_{k=0}^\infty a_k^{(z_0)}(z-z_0)^k}_{=:\:h_{z_0}(z)}\tag1$$ for all $z\in N_{z_0}\setminus\{z_0\}$. Let $P$ denote the subset of $\Omega$ for which $k_{z_0}$ cannot be chosen to be $0$.

EDIT: The basic idea of this question is to consider a region $B$ of $f$, where $f$ has at most $|F|$ poles, and then to consider the Laurent expansion at the first pole, substract the principal part to obtain a function which has ony $|F|-1$ poles and continue this process until we arrive at a holomorphic function. This should somehow be similar to what is done in this answer, but I don't get why we arrive at a holomorphic function on all of $B$ at the end, since the "holomorphic parts" of each expansion are only defined on a small neighborhood of the corresponding pole ...

Let $B\subseteq\mathbb C$ be bounded and open with $\overline B\subseteq\Omega$. Are we able to show that for $F:=B\cap P$, we can decompose $f$ on $B$ into $$f(z)=\sum_{z_0\in F}\tilde m_{z_0}(z)+h_F(z)\;\;\;\text{for all }z\in B\tag2$$ for some holomorphic $h_F:B\to X$ and $$\tilde m_{z_0}(z)=\sum_{k=-k_{z_0}}^{-1}b_k^{(z_0)}(z-z_0)^k\;\;\;\text{for all }z\in\Omega$$ for some suitable $\left(b^{(z_0)}_k\right)_{k\ge-k_{z_0}}\subseteq X$?

Since $\overline B$ is compact and $P$ is discrete, $F$ must be finite and hence the sum over $F$ in $(2)$ is finite.

We may clearly assume that $N_{z_0}$ is open for all $z_0\in F$, if that would be an issue.

My idea is the following: Consider $z_i\in P$ with $z_1\ne z_2$. We know that $P$ is discrete and hence there is a $\varepsilon>0$ with $$|z_1-z_2|\ge\varepsilon\tag3.$$ Shrinking $N_{z_i}$ if necessary, we may assume that $$N_{z_i}\subseteq B_{\varepsilon/2}(z_i)\tag4$$ and hence $$N_{z_1}\cap N_{z_2}=\emptyset.\tag5$$ Let $$N_{\{z_1,\:z_2\}}:=N_{z_1}\cup N_{z_2}$$ and $$m_{\{z_1,\:z_2\}}:=\left.\begin{cases}m_{z_1}(z)&\text{, if }z\in N_{z_1}\\m_{z_2}(z)&\text{, if }z\in N_{z_2}\end{cases}\right\}\;\;\;\text{for }z\in N_{\{z_1,\:z_2\}}$$ (since the sum in the definition of $m_{z_i}$ is finite, $m_{z_i}$ is well-defined on all of $\Omega$). Shrinking $N_{z_i}$ if necessary, we may assume that $N_{z_1}$ and hence $N_{\{z_1,\:z_2\}}$ is open. If I'm not missing something, $$h_{\{z_1,\:z_2\}}(z):=\left.\begin{cases}h_{z_1}(z)&\text{, if }z\in N_{z_1}\\h_{z_2}(z)&\text{, if }z\in N_{z_2}\end{cases}\right\}\;\;\;\text{for }z\in N_{\{z_1,\:z_2\}}$$ should be holomorphic.

In the same way we should be able to construct a holomorphic function $h_F$ on $$N_F:=\bigcup_{z_0\in F}N_{z_0},$$ shrinking the $N_{z_0}$, if necessary, to ensure that this union is disjoint, with $$h_F(z)=h_{z_0}(z)\;\;\;\text{for all }z\in N_{z_0}\text{ and }z_0\in F\tag6.$$ On the other hand, by definition of $P$, the function $f$ should be holomorphic on the open set $B\setminus N_F$. Thus, the extension of $h_F$ to $B$, by setting it to $f$ on $B\setminus N_F$, should be holomorphic as well.

Now we may define $$m_F(z):=m_{z_0}(z)\;\;\;\text{for }z\in N_{z_0}\text{ and }z_0\in F.$$ However, while each individual $m_{z_0}$ can clearly be extended to all of $\Omega$, we don't arrive at the desired decomposition yet. How do we need to the choose $\left(b^{(z_0)}_k\right)_{k\ge-k_{z_0}}$?

答:
$\frac{\pi^2}{\sin^2(\pi z)}$ is meromorphic,

$$\frac{\pi^2}{\sin^2(\pi z)}-\sum_{n=-N}^N \frac1{(z-n)^2}$$ is holomorphic on $|z|<N+1$ (complex differentiable at every point, a local property). The Cauchy integral formula implies that it is analytic and equal to its Taylor series at $0$ for $|z|<N+1$.

It works the same way when substracting finitely many poles to any meromorphic function.

When substracting infinitely many poles it gets more complicated, for example with  $\frac{2i\pi z}{e^{2i\pi z}-1}$ then subtracting $\sum_n \frac{n}{z-n}$ won't work as it doesn't converge, instead we'll look at $\sum_n \frac{n}{z-n}+1+z/n$, generalizing this is the purpose of https://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem


问:
Equivalent definitions of convexity for $f\in\mathcal C^1(\mathbb R^n)$
I noticed many identical questions on this site relating to equivalent definitions of convexity, see for example 668679, 740938, 318692, 1717542, 1761801, 3019331, 3047518 and there might be more.

So I wanted to create a "big list" of equivalent definitions of convex functions, I will start with some equivalent definitions and post my proofs in my answer below. Feel free to add any other equivalent definitions. This question can then serve as an abstract duplicate.

Claim. Let $n\in\mathbb N$ and $f\in\mathcal C^1(\mathbb R^n)$. Then the following are equivalent:

$f$ is convex, i.e. $f(\lambda x_1 + (1-\lambda x_2))\le\lambda f(x_1)+(1-\lambda)f(x_2)$ for all $\lambda\in[0,1]$ and $x_1,x_2\in\mathbb R^n$.

For all $x,y\in\mathbb R^n$, $\langle\nabla f(x)-\nabla f(y),x-y\rangle\geq0$.

For every $x,y\in\mathbb R^n$, $f(y)\geq f(x)+\langle y-x,\nabla f(x)\rangle$.

答:
Proof of equivalence of 1., 2. and 3.

1.$\implies$3. Let $f$ be convex. Then, by definition of the gradient, for every $x,y\in\mathbb R^n$,
\begin{align}
\langle y-x, \nabla f(x)\rangle &= \lim_{h\to0}\frac{f(x+h(y-x))-f(x)}{h}\\&=\lim_{h\to0}\frac{f((1-h)x+hy)-f(x)}{h}\\&=\lim_{h\to0, h>0}\frac{f((1-h)x+hy)-f(x)}{h}
\\&\overset{\text{convexity}}\le \lim_{h\to0, h>0}\frac{hf(y)-hf(x)}h\\&=f(y)-f(x).
\end{align}

3.$\implies$2. By assumption 3., we have, for every $x,y\in\mathbb R^n$, \begin{equation}\label1\tag1f(y)\geq f(x)+\langle y-x,\nabla f(x)\rangle\end{equation} but also \begin{equation}\label2\tag2f(x)\geq f(y)+\langle x-y,\nabla f(y)\rangle.\end{equation}

\eqref{2} is equivalent to $$-f(y)+\langle y-x,\nabla f(y)\rangle\geq -f(x),$$ which added to \eqref{1} yields $$\langle y-x,\nabla f(y)\rangle\geq\langle y-x,\nabla f(x)\rangle$$ and thus the desired result.

2.$\implies$1. Suppose that 2. is true and define $g:[0,1]\to\mathbb R$, $g(t)\overset{\text{Def.}}=f(x+t (y-x))$. Then $g$ is convex as, for $t\in[0,1]$,
\begin{equation}
g'(t)=\lim_{h\to0}\frac{f(t y +(1-t)x + h(y-x))- f(t y+(1-t)x)}{h}
= \langle y-x, \nabla f(t y +(1-t)x)\rangle,
\end{equation}
so that, for $0\le t_1< t_2\le 1$,
\begin{align}
g'(t_2)-g'(t_1)&=\langle y-x, \nabla f(x+t_2 (y-x)) -\nabla f(x+t_1(y-x))\rangle
\\
&=\frac1{t_2-t_1}\langle (x+t_2 (y-x)) - (x+t_1(y-x)), \nabla f(x+t_2 (y-x)) -\nabla f(x+t_1(y-x))\rangle
\\
&\geq 0,
\end{align}
by assumption 3.

Hence $g'$ is increasing and thus $g$ is indeed convex (proof).

So we have $$g(\lambda)\le\lambda g(1)+(1-\lambda) g(0)$$ for every $\lambda\in[0,1]$, which, by definition of $g$, is exactly $$f((1-\lambda)x+\lambda y)\le(1-\lambda) f(x)+\lambda f(y).$$

$\square$


问:
Derivative of matrix inverse using definition of Frechet derivative
Let $F : Gl_n(\mathbb{R}) \rightarrow Gl_n(\mathbb{R})$ is given by $a \mapsto a^{-1}$ Show that the Frechet derivative is given by $D_A F(h) = -a^{-1} h a^{-1}$.

Work done so far:

\begin{align*}
||F(u_0 + h) - f(u_0) + u_0^{-1}hu_0^{-1}|| &= || \Sigma_{k = 0}^{\infty}(-1)^k(u_0^{-1}h)^ku_0^{-1} - u_0^{-1} + u_0^{-1}hu_0^{-1}|| \\
&= ||\Sigma_{k \geq 2} (-1)^k(u_0^{-1}h)^ku_0^{-1}||
\end{align*}

I am not sure the last element approaches zero as h goes to zero. If someone could explain that part.

答:
If you know the product rule -- derivative of $fg$ at $A$ is $H\mapsto f'(A)(H)\cdot g(A) + f(A)\cdot g'(A)(H)$ -- then you can apply it to the function $G=\mathrm{id}\cdot F$, i.e. $G(A)=AA^{-1}=I_n$ to get
\begin{align*}
0=G'(A)&=\mathrm{id}'(A)(H)\cdot F(A)+\mathrm{id}(A)\cdot F'(A)(H)\\
&= H\cdot F(A)+A\cdot F'(A)(H),
\end{align*}
which rearranges to $F'(A)(H)=-A^{-1}HA^{-1}$.


问:
Meager set union Bernstein set with empty interior
Fix a meager set $M\subset \Bbb R$. of course  $\text{int} M=\emptyset$. I want to construct a Bernstein set $B\subset \Bbb R$ such that $$\text{int}(M\cup B)=\emptyset \tag{1}$$

We know it is true for any Bernstein set that $\text{int}B=\emptyset.$ I am not sure such $B$ can be found to make (1) true.

Any help will be appreciated greatly

答:
In fact, we have a stronger result:

Suppose $B$ is Bernstein and $M$ is meager. Then $int(M\cup B)=\emptyset$.

In the interest of brevity, I'll use an oversized hammer for the problem: the perfect set property for $G_\delta$ sets, that every uncountable $G_\delta$ set has a perfect subset.

Suppose $int(M\cup B)\not=\emptyset$ and $M$ is meager. Let $(a,b)\subseteq M$. Then $B\cap (a,b)$ is comeager in $(a,b)$.  We then observe:

If $X$ is nowhere dense, then so is its closure $\overline{X}$; so every nowhere dense set is contained in a closed nowhere dense set.

Consequently, every meager set is contained in a meager $F_\sigma$ set.

And so dually every comeager set has a comeager $G_\delta$ subset.

And all of this relativizes to a nondegenerate open interval like $(a,b)$. So $B$ contains an uncountable $G_\delta$ set, and hence is not Bernstein by the perfect set property.


问:
What is the bias (as a function of number of points) of the average of data as an estimator of the mean of the Weibull distribution?
The Weibull distribution's cumulative distribution function  is:

$
F=1-\exp\left[-(\frac{x}{b})^c\right]
$

By differentiating to obtain the pdf, and integrating it with respect to $x$ to obtain the first moment, i.e. the mean, we get ($\Gamma$ is the euler gamma function):

$
\mu=b\Gamma(1/c+1)
$

If we have n experimental data thought to be Weibull-distributed, we can calculate its average, i.e:

$
\bar x=\frac{1}{n}\sum_{i=1}^{n}x_i
$

What is the bias of $\bar x$ with respect to $\mu$ as a function of $n$? I have found papers about the bias of estimators for the distribution parameters i.e. $b$ and $c$. I can also do Montecarlo with some reasonable values of $b$, $c$, and $m$ to get a feel, but I am wonder if there is a theoretical expression that is known, but that I am missing because maybe I am not searching for the right keywords. Anybody knows the answer or a paper that could point me in the right direction?

答: If $(X_1, X_2, \ldots, X_n)$ is an iid sample with $X_i \sim \operatorname{Weibull}(b, c)$ with parametrization $F_X(x) = 1 - \exp(-(x/b)^c)$, then as you have already calculated, $$\operatorname{E}[X_i] = \mu = b \Gamma(1/c + 1).$$  Therefore, by linearity of expectation, $$\operatorname{E}[\bar X] = \operatorname{E}\left[\frac{1}{n} \sum_{i=1}^n X_i\right] = \frac{1}{n} \sum_{i=1}^n \operatorname{E}[X_i] = \frac{1}{n} \sum_{i=1}^n \mu = \mu.$$  Therefore, the sample mean $\bar X$ is unbiased.


问:
How can we derive the Interpolation formula for higher order?
Let, $$y = mx + c \tag 1$$ is the equation of a straight line.

Let, it pass through the point $$(x_0, y_0).$$

So, from (1) we find,
$$c = y_0 + mx_0 \tag 2$$
On the other hand, from the formula of slope we find
$$m=\frac {y_1-y_0}{x_1-x_0} \tag 3$$
Putting (3) in (2) we find
$$c = y_0 - {\frac {y_1-y_0}{x_1-x_0}} x_0 \tag 4$$
Putting (3) and (4) into (1) we find
$$y = y_1 \frac {x-x_0}{x_1-x_0} + y_0 \frac {x-x_1}{x_0-x_1}$$
$$\Longrightarrow P(x) = y_1 \frac {x-x_0}{x_1-x_0} + y_0 \frac {x-x_1}{x_0-x_1} \tag 5$$

Putting $x  = x_0$ into (5) we find,
$$P(x_0) = y_0 \tag 6$$
Putting $x  = x_1$ into (5) we find
$$P(x_1) = y_1 \tag 7$$
Putting (6) and (7) into (5) we find
$$\Longrightarrow P(x) = P(x_1) \frac {x-x_0}{x_1-x_0} + P(x_0) \frac {x-x_1}{x_0-x_1} \tag 8 $$
Ok. Now, how to find this:

$P(x)=$

<img src="https://i.stack.imgur.com/3jDcM.gif" alt="$$=> P(x) = P(x_1) \frac {x-x_0}{x_1-x_0} + P(x_0) \frac {x-x_1}{x_0-x_1}$$" />
$\tag 9$

答:
The idea is:

$y(x)=a+bx=c_0(x-x_0)+c_1(x-x_1)$

for some choice of $c_0$ and $c_1$ s.t. $y(x_i)=y_i, (i=0,1)$.

$y(x)=a+bx+cx^2=c_0(x-x_1)(x-x_2)+c_1(x-x_0)(x-x_1)+c_2(x-x_0)(x-x_1)$ for some choice of $c_0,c_1$ and $c_2$ s.t. $y(x_i)=y_i, (i=0,1,2)$.

Now generalize this for $n$-degree polynomial.


问:
Let $G$ a tree, $M$ maximum matching of $G$, then some leaf (vertex of degree 1) of $G$ is $M-$saturated.
Let $G$ a tree, $M$ maximum matching of $G$, then some leaf (vertex of degree 1) of $G$ is $M-$saturated.

I am trying to do this problem by contradiction. Thus, since there are no M-saturated leaves, then if we take some vertex $x$ that is leaf, since it has degree 1, then there exists a vertex $y$ such that $xy\notin M$.

But I don't know if that result is useful to get the contradiction

答: Hint: if no leaf is saturated, pick two leaves and consider the unique path between them.  Can you see how to create a larger matching?</p>


问:
Prove or disprove: If $v_1,v_2,...,v_n$ are vectors in $\mathbb{R}^n$...
and for every $v_i$ , $i=1,..,n$ the sum of coordinates of $v_i$ is $0$, then $v_1,...,v_n$ and linearly dependent.

What I tried to do: This questions was given after I learnt about matrix similarity and eigen vectors and eigen values, so I tried to think in that direction, I couldn't really get the question saying "the sum of coordinates of $v_i$ is $0$), because I always saw coordinates come with basis so I can use them? (would be appreciated if someone explains in what way should I understand that sentence).

But I proceed to think in that direction and reached something: I chose matrix $A$ to be all ones, ($a_{ij}=1\space \forall i,j) $, and let $T(v)=Av$ a linear transformation, and here I got stuck but my idea was to try to say by contradiction that they're linearly  independent and reach contradiction if I find that $A$, is similiar to the zero matrix (if I use the sum of coordinates), and then explain that they can't be similar with different ranks, and prove the statement. But  I have no idea how to continue and my weak understanding of the question scares me if I'm in the right direction so I came here to ask.

Appreciate all the help, Thanks in advance to everyone.

答:
HINT-A system of $n$ vectors $v_1,v_2,.....,v_n$ is linearly dependent if
$\sum_{i=1}^nc_iv_i=0$ holds for some non zero vector $(c_1,c_2,.....,c_n)$.

In your case, indeed this non zero vector is $(1,1,.....,1)$.


问:
Problem on hyperbolic hyperboloid generated by a rotation
This is the problem:

In $\mathbb{E}^3$ we consider the conic $\gamma$ of equations $x=yz-2=0$ , the line $a$ of equations $x=y+z=0$ and the surface $Q$, that is generated by the rotation of $\gamma$ around $a$.

(a) After having explained the equation, show that $Q$ is a hyperbolic hyperboloid.

(b) Describe the cartesian equations of the straight lines contained in Q, and the minimum circle of Q.

(c) Let $Q'$ projective closure of $Q$ in $\mathbb{P} ^ 3$, determine a plane $\pi$ in  $\mathbb{P} ^ 3$, through P = (2:0:0:1), such that the trace of Q' in the space affine $\mathbb{A}: = \mathbb{P} ^ 3/\pi$ is a paraboloid

I try to complete (a): I found the plane perpendicular to $ a $, but I do not know how to impose conditions to generate Q

答:
Let's see about (a). $x=0$ tells you that both your curves live in the $y$-$z$-plane. So you might want to draw or plot an image of the situation in that plane.

The equation $y+z=0$ for $a$ can be re-interpreted as the line through the origin and in direction $(0,1,-1)$. One possible approach is rotating your coordinate system in such a way that this axis coincides with one of the coordinate axes. To map that line onto the $y$ axis, you can use

$$ M = \begin{pmatrix}
1 & 0 & 0 \\
0 & \frac12\sqrt2 & -\frac12\sqrt2 \\
0 & \frac12\sqrt2 & \frac12\sqrt2
\end{pmatrix} $$

A point $p=(x,y,z)$ lies on the rotated curve $\gamma'$ iff its preimage lies on $\gamma$, i.e. if $\gamma(M^{-1}\cdot p)=0$.

$$\gamma': x=\tfrac12z^2-\tfrac12y^2-2=0$$

Now you can rotate that around the $y$ axis. The term $z^2$ in the above equation describes the squared distance from the $y$ axis in the $y$-$z$-plane, so you rotate by writing $y^2+z^2$ instead of $z^2$.

$$Q': \tfrac12x^2+\tfrac12z^2-\tfrac12y^2-2=0$$

Now rotate everything back. A point $p$ lies on $Q$ if its transformed version $M\cdot p$ lies on $Q'$, i.e. $Q'(M\cdot p)=0$.

$$Q: \tfrac12x^2+yz-2=0$$

Now you have the equation of the hyperboloid. To see that it is hyperbolic, you can intersect the hyperboloid with the line. Simply insert a generic point on the line, i.e. $\lambda\cdot(0,1,-1)$, into the equation of the hyperboloid.

$$Q\left(\lambda\cdot\begin{pmatrix}0\\1\\-1\end{pmatrix}\right) =
-\lambda^2-2=0$$

This equation has no real solution, which means that the hyperboloid does not intersect the line. Therefore your hyperboloid has only a single sheet. I assume that this is what you mean by “hyperbolic” in this context. If not, please clarify since I can't find a good reference of this use of the term. Note that this decision of the line not intersecting the hyperboloid could have been deduced from $\gamma$ and $a$ alone, without the equation for $Q$.


答:
We may assume $u_0 = I$ wlog, as multiplication is continuous and
$$ (u_0 + h)^{-1} - u_0^{-1} + u_0^{-1} h u_0^{-1} = u_0^{-1}\left( (I + hu_0^{-1})^{-1} - I + hu_0^{-1} \right). $$

Using the Neumann series, we have
$$ (I+h)^{-1} - I + h = \sum_{k\ge 2} (-h)^k = (-h)^2 \sum_{k\ge 0} (-h)^k = h^2 (I+h)^{-1} = o(\|h\|), $$
as
$$ \frac{\| h^2 (I+h)^{-1} \|}{\|h\|} \le \frac{\|h\|^2 \|(I+h)^{-1}\|}{\|h\|} = \|h\|\|(I+h)^{-1}\| \to 0 \cdot 1. $$

Notes:
If you already know that $F$ is differentiable, then you can simply apply the product rule as in @jlammy's answer. $F$ is in fact smooth, as it is a rational function by Cramer's rule.

Title: Derivative of matrix inverse using definition of Frechet derivative

Body:
Let $F : Gl_n(\mathbb{R}) \rightarrow Gl_n(\mathbb{R})$ is given by $a \mapsto a^{-1}$ Show that the Frechet derivative is given by $D_A F(h) = -a^{-1} h a^{-1}$.

Work done so far:

\begin{align*}
||F(u_0 + h) - f(u_0) + u_0^{-1}hu_0^{-1}|| &= || \Sigma_{k = 0}^{\infty}(-1)^k(u_0^{-1}h)^ku_0^{-1} - u_0^{-1} + u_0^{-1}hu_0^{-1}|| \\
&= ||\Sigma_{k \geq 2} (-1)^k(u_0^{-1}h)^ku_0^{-1}||
\end{align*}

I am not sure the last element approaches zero as h goes to zero. If someone could explain that part.


问:
Approximating measures by open sets and compact sets.
I'm having trouble starting two similar proofs:

Let $\epsilon > 0$. And let $E$ be a measurable set of finite measure.

Prove that there is an open set $U$ containing $E$ such that $m(U \setminus E) < \epsilon$.

Similarly, prove there is a compact set $K$ contained in $E$ such that $m(E \setminus K) < \epsilon$.

Any hints are much appreciated.

NOTE: $m$ is the Lebesgue outer measure. And $E \subseteq \mathbb{R}$ is measurable if $m(A) \geq m(A \cap E) + m(A \setminus E)$.

答:
Hossien proved that there is an open set $O$ that contains $E$ such that $m(O\backslash E)<\epsilon$. Let me try to prove that there is a compact set $K$ contained in $E$ such that $m(E\backslash K)<\epsilon$.

Since $E$ is measurable and measurable sets are an algebra, the complement of $E$, $E^c$, is also measurable. Then, because Hossien proved it, there exists an open set $O$ such that $m(O\backslash E^c)<\epsilon$. But the complement of $O$, $O^c$, is a closed set contained in $E$. Therefore it is compact because it is closed and bounded and we are in $\mathbb{R}$. Moreover,
$$ m(E\backslash O^c) = m(E\cap O) = m(O\cap E) = m(O\backslash E^c) < \epsilon $$

And then we proved that $K=O^c$ is that set.