问:
Relation between dimensions of two subspaces of $\ell^2.$
Fix some some double sequence $(y_{n,m})_{n,m\in \mathbb{N}}\subset \mathbb{C}$ with $\sum_{n,m}|y_{n,m}|^2<\infty.$ Let
$$A = \{x\in\ell^2: \sum_{n=1}^\infty y_{n,1}x_n = 0 \,\,\wedge\,\, \sum_{n=1}^\infty y_{n,m+1}x_n = -x_m\,\,\forall m\geq 1 \},$$
$$B = \{x\in\ell^2: \sum_{n=1}^\infty y_{m,n}x_n = -x_{m+1}\,\,\forall m\geq 1\}.$$
I want to show that $\text{dim } A = \text{dim } B - 1.$ I plan to do it in two steps:

Show that $A'= \{x\in\ell^2: \sum_{n=1}^\infty y_{n,m+1}x_n = -x_m\,\,\forall m\geq 1\}$ has the same dimension as $B$ via some clever isomorphism.

Show that intersecting $A'$ with $\{x\in\ell^2: \sum_{n=1}^\infty y_{n,1}x_n = 0\}$ reduces the dimension by $1.$

I am stuck on both steps. I think I'm missing some insight on what $A'$ and $B$ look like and how they depend on the choice of $y.$ I think it might be better to replace $y$ in the definitions of $A$, $A'$ and $B$ by the complex conjugate $\bar{y},$ to use tools from Hilbert spaces, since we can then write these sets as orthogonal complements, but even then I am still stuck. Any help with either step would be much appreciated.

答: This is false. If $y_{n,m}=0$ for all $n,m$ then $A =\ell^{2}$ and $B$ is one dimensional.


问:
Find a numerical solution for an ODE with an initial condition which is not at the beginning
This may sound really weird...

Consider the following Cauchy problem:

Find $y(t)$ for all $t \geq 0$ such that:
$$\begin{cases} \dot{y} & = f(t, y)\\ y(T) &= Y \end{cases},$$
for some $T > 0$.

I know that, under suitable conditions, this Cauchy problem has exactly one solution.

Anyway, all numerical tools that I know are meant (or I believe so!) to solve problems like:

Find $y(t)$ for all $t \geq 0$ such that:
$$\begin{cases} \dot{y} & = f(t, y)\\ y(0) &= Y \end{cases},$$

How can I solve the first problem?

答:
Ancient trade secret: You can apply the numerical methods also with a negative step size.

Most numerical tools also support this, the sequence of evaluation points, if given, just has to be monotonous. This makes two calls to the ODE solver routine, one for the interval $[0,T]$ and one for $[T,T_{end}]$

Test of concept: Plot solutions of $y'(x)=\frac{x-y}{x+y}$ with value $y(2)$ around $2$ for $x\in[-1,5]$.

```<code>f=lambda y,x: (x-y)/(x+y)
x1 = np.linspace(2,-1,150)
x2 = np.linspace(2,5,150)
for y0 in 2+np.linspace(-0.5,0.5,6):
    y1,y2 = (odeint(f,y0,x) for x in (x1,x2))
    l, = plt.plot(x1,y1)
    c = l.get_color()
    plt.plot(x2,y2,c=c,label=f"y(2)={y0}")
plt.grid(); plt.legend()
plt.show()
</code>```

ContentLicense: CC BY-SA 4.0

Body:
Ancient trade secret: You can apply the numerical methods also with a negative step size.

Most numerical tools also support this, the sequence of evaluation points, if given, just has to be monotonous. This makes two calls to the ODE solver routine, one for the interval $[0,T]$ and one for $[T,T_{end}]$

Test of concept: Plot solutions of $y'(x)=\frac{x-y}{x+y}$ with value $y(2)$ around $2$ for $x\in[-1,5]$.

f=lambda y,x: (x-y)/(x+y)
x1 = np.linspace(2,-1,150)
x2 = np.linspace(2,5,150)
for y0 in 2+np.linspace(-0.5,0.5,6):
    y1,y2 = (odeint(f,y0,x) for x in (x1,x2))
    l, = plt.plot(x1,y1)
    c = l.get_color()
    plt.plot(x2,y2,c=c,label=f"y(2)={y0}")
plt.grid(); plt.legend()
plt.show()


问:
What is the (parametric) intersection of a plane and a sphere?
Can someone please show me how to prove that the intersection of the plane

$$x+y+z=0$$  and the sphere

$$x^2+y^2+z^2=1$$

can be expressed as

$$x(t)=\frac{\cos t-\sqrt3 \cdot\sin t}{\sqrt6}$$

$$y(t)=\frac{\cos t+\sqrt3 \cdot\sin t}{\sqrt6}$$

$$z(t)=\frac{-2\cos t}{\sqrt6}$$

Ps: Also, why am I not getting the correct notation? I am using a macbook pro (safari) if that is a concern?

答:
(Originally I had $u,v=x+y,x-y$, but after going through half the problem, I considered that $2u,2v=x+y,x-y$ would be a little more convenient.)

amanda, after your substitution, we consider $x^2+xy+y^2={1\over2}$.

Here are is a nice choice for the next substitution,

\begin{align}
2u&=x+y,\\
2v&=x-y.
\end{align}

We next consider

\begin{align}
4u^2&=x^2+y^2+2xy,\\
4v^2&=x^2+y^2-2xy.
\end{align}

To get $x^2+y^2+xy$, we need $2u^2+2v^2+(u^2-v^2)$. Thus our problem is transformed to

$$6u^2+2v^2=1.$$

Then we have parametric solutions

$$u=\sqrt{1\over6}\cos t,\ v=\sqrt{1\over2}\sin t.$$

From there, we find for instance

$$x=u+v={\cos t+\sqrt3\sin t\over\sqrt6},$$

and the rest should be straightforward.

To get the sign to match what is shown in your equations, I can think of several ways to go.

You could simply note the symmetry in $x$ and $y$ and switch them (which suggests that if we had made the substitution $2v=y-x$ instead of $x-y$, that would have worked also).

From $6u^2+2v^2=1$, we could have chosen $v=-\sqrt{1\over2}\sin t$.

If we make the substitution $t=-s$, then we have $x={\cos (-s)+\sqrt3\sin (-s)\over\sqrt6}={\cos s-\sqrt3\sin s\over\sqrt6}$. Now, call it $t$ again.

ContentLicense: CC BY-SA 3.0


问:
Geometric distribution Abel's partial summation expectation
It is $0 < q \le 1$ and P=Geom(q) the geometric distribution on $\Omega=\mathbb{N}$ and $X(\omega)=\omega$.
Determine EX and Var(X) with  Abel's partial summation:

$\sum_{k=1}^{n}a_kb_k = \sum_{k=1}^{n}a_kb_n +\sum_{k=1}^{n-1}(\sum_{j=1}^{k}a_j)(b_k-b_{k+1})$
$\forall n \in \mathbb{N},a_j,b_j \in \mathbb{R},j \in \{1,...,n\}$.

Consider the special case $q=1$ and don't take the limit $n \to \infty$ at the start.

答:
It is perhaps more convenient to first let $p = 1-q$.  Then if we use your choices $a_k = (1-q)^{k-1} = p^{k-1}$, $b_k = k$, the formula
$$\sum_{k=1}^n a_k b_k = b_n \sum_{k=1}^n a_k + \sum_{k=1}^{n-1} \sum_{j=1}^k a_j (b_k - b_{k+1}) \tag{1}$$ becomes
$$\begin{align}
\sum_{k=1}^n k p^{k-1}
&= n \sum_{k=1}^n p^{k-1} - \sum_{k=1}^{n-1} \sum_{j=1}^k p^{j-1} \\
&= n \frac{1 - p^n}{1-p} - \sum_{k=1}^{n-1} \frac{1 - p^k}{1-p} \\
&= \frac{1}{1-p} \left( n(1-p^n) - (n-1) + \sum_{k=1}^{n-1} p^k \right) \\
&= \frac{1}{1-p} \left(1 - np^n + \frac{p-p^n}{1-p} \right) \\
&= \frac{1}{(1-p)^2} ( 1 - np^n - p + np^{n+1} + p - p^n ) \\
&= \frac{1 - (n+1)p^n + np^{n+1}}{(1-p)^2}.
\end{align}$$
Then as $n \to \infty$, for $0 < p < 1$, the numerator tends to $1$ and the expectation is $$\operatorname{E}[X] = q \sum_{k=1}^n a_k b_k = \frac{q}{(1-p)^2} = \frac{1}{q}.$$

To perform the variance calculation using the same method, it is natural to choose $b_k = k^2$, but I would recommend choosing $b_k = k(k-1)$ instead.  This will give $\operatorname{E}[X(X-1)] = \operatorname{E}[X^2] - \operatorname{E}[X]$, from which we can still recover the variance.


问:
Minimizer of a convex function.
I found this result on some lectures about optimization:
.

Let $f : X \rightarrow \mathbb{R}$ a differentiable convex function. Then $x$ is a minimizer of $f$ if and only if

$$\langle x'-x,\nabla f(x)\rangle \ge 0 \qquad \forall x'$$

Shouldn't be $\langle x'-x,\nabla f(x)\rangle = 0 \quad \forall x'$, if $x$ is a minimizer of $f$?

答:
I'm assuming here that $\ X\ $ is an open convex subset of some finite dimensional Euclidean space $\ \mathbb{R}^n\ $ (although the result will also hold in more general cases too).

If $\ \left\langle x^\prime -x, \nabla f(x)\right\rangle\ge0\ $ for all $\ x^\prime\ $, then we get $\ \left\langle y, \nabla f(x)\right\rangle\ge0\ $ for any $\ y\ $ by taking $\ x^\prime=y-x\ $, and $\ \left\langle y, \nabla f(x)\right\rangle\le0\ $ by taking $\ x^\prime=-y-x\ $. Hence, the condition $\ \left\langle x^\prime -x, \nabla f(x)\right\rangle\ge0\ $ for all $\ x^\prime\ $ is equivalent to the condition $\ \left\langle y, \nabla f(x)\right\rangle=0\ $ for all $\ y\ $, or, equivalently, $\ \nabla f(x)=0\ $.  For a differentiable convex function, this is a necessary and sufficient for $\ x\ $ to be a minimiser of $\ f\ $.

Title: Minimizer of a convex function.


问:
Solving systems of equations using matrices
I'm teaching myself how to solve systems of equations using matrices and row operations. I get the basic mechanics of it (legal operations, etc.), but it seems like it's kind-of a crapshoot deciding where to begin, and choosing the "wrong" operation to start with can lead to a really difficult problem.

My question is: Are there any rules of thumb for deciding which operations to begin with and how to proceed?

For instance, given the system:

```<code>2x - 9y - z = -23
 x + 9y - 5z= 2
3x + y + z = 24
</code>```

What row operation would you begin with, and where would you go from there? I'm not asking you to do it for me, but for insight into your thought process regarding how to proceed.

答:
There is an algorithm for this process called Gauss-Jordan elimination.

Applying Gauss-Jordan to the example given:

1) The first step is to get a leading one in the first row of your matrix.  The easiest way to do this is to switch rows 1 and 2.  This gives the matrix

$$
\left[\begin{array}{ccc|c}
1 & 9 & -5 & 2
\\
2 & -9 & -1 & -23
\\
3 & 1 & 1 & 24
\end{array}\right]
$$

2) Next, since you have a leading 1 in the first row, add multiples of the first row to all other rows so that ll other entries in the first column are zero.

3) Then apply the same process on column 2.  That is, get a leading one in the second row if possible...


问:
Converting to cylindrical coordinates
The given integral is: $$\int_\frac{-\sqrt3}2^\frac{\sqrt3}2\int_\frac{1}2^\sqrt{1-x^2}\int_\sqrt{x^2+y^2}^{2-x^2-y^2} \frac{x}y dzdydx$$

I converted it as: $$\int_0^\pi \int_0^1\int_r^{2-r} \cot\theta r dzdrd\theta$$

We were only asked to set up the integral and not solve it but I tried to check what the answer to this would be and it will be undefined. What did I do wrong?

答: You are wrong about the values that $\theta$ can take: it can take any value in $\left[\frac\pi6,\frac{5\pi}6\right]$. For each such $\theta$, the values that $r$ can take go from $\frac{\cos\theta}2$ to $1$. Finally, the values that $z$ can take go from $r$ to $2-r^2$. So, you should have got$$\int_{\pi/6}^{5\pi/6}\int_{\csc(\theta)/2}^1\int_r^{2-r^2}\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$

Title: Converting to cylindrical coordinates


问:
Is it true that $w\wedge v=0\implies w=v\wedge\phi$ for some $\phi\in\bigwedge^{k-1}V$?
In particular, what if $v=0$ and $w\ne0$? Then $w\wedge v=0$, but $w$ cannot equal $v\wedge\phi$ for any $\phi$, because $v\wedge\phi=0$ for all $\phi$.

I'm considering $v\in V$, $w\in\bigwedge^kV$, and $\phi\in\bigwedge^{k-1}V$.

答:
As you have shown, the statement in the title is not true.

However, it is true if $v \neq 0$ (at least if $V$ is finite-dimensional). To see this, first complete $\{v\}$ to a basis, i.e. choose a basis $\{v_1, \dots, v_n\}$ for $V$ such that $v_1 = v$. There is an induced basis for $\bigwedge^kV$ given by $\{v_{i_1}\wedge\dots\wedge v_{i_k} \mid 1 \leq i_1 < \dots < i_k \leq n\}$ so

$$w = \sum_{1 \leq i_1 < \dots < i_k \leq n}a_{i_1,\dots,i_k}v_{i_1}\wedge\dots\wedge v_{i_k}.$$

If $i_1 = 1$ then $v\wedge v_{i_1}\wedge\dots\wedge v_{i_k} = v\wedge v\wedge\dots\wedge v_{i_k} = 0$, so

$$v\wedge w = \sum_{1 < i_1 < \dots < i_k \leq n}a_{i_1,\dots,i_k}v\wedge v_{i_1}\wedge\dots\wedge v_{i_k} = \sum_{1 < i_1 < \dots < i_k \leq n}a_{i_1,\dots,i_k}v_1\wedge v_{i_1}\wedge\dots\wedge v_{i_k}.$$

As $\{v_{j_1}\wedge\dots\wedge v_{j_{k+1}} \mid 1 \leq j_1 < \dots < j_{k+1} \leq n\}$ is a basis for $\bigwedge^{k+1}V$ and $v\wedge w = 0$, we see that $a_{i_1,\dots,i_k} = 0$ for $i_1 > 1$. Therefore

\begin{align*}
w &= \sum_{1 < i_2 < \dots < i_k \leq n}a_{1,i_2,\dots,i_k}v_1\wedge v_{i_2}\wedge\dots\wedge v_{i_k}\\
&= v_1\wedge\left(\sum_{1 < i_2 < \dots < i_k \leq n}a_{1,i_2,\dots,i_k}v_{i_2}\wedge\dots\wedge v_{i_k}\right)\\
&= v\wedge\phi
\end{align*}

where

$$\phi = \sum_{1 < i_2 < \dots < i_k \leq n}a_{1,i_2,\dots,i_k}v_{i_2}\wedge\dots\wedge v_{i_k} \in \bigwedge\nolimits^{k-1}V.$$


问:
How to show C_e is closed and not dense in C.
Let $C_{e}([-1,1],\mathbb{R})$ denote the set of even functions in $C([-1,1],\mathbb{R})$

(a) Show $C_e$ is closed and not dense in $C$.

(b) show the even polynomials are dense in $C_e$, but not in $C$.

I can't start on it... I can't catch any clue..

答: Assume there is a sequence of even functions $f_{i}$ converge to $x$ under $\sup$ norm such that $$\sup |f_{i}(x)-x|<\epsilon$$ for large enough $i$. Then this implies $$\int f_{i}=\int x$$ on $[-R,R]$. But this is impossible unless $\int f_{i}\rightarrow 0$ on $[0,R]$ for any $R$. However this itself is absurd. $(2)$ is similar. </p>


问:
Maximum Angular Velocity Object falling disk
We have a small problem. We designed a drum with a small rope around it, at the end of the rope a small weight is attached. We want to calculate the maximum angular velocity of the drum. The rope is 1m long. The weight attached are of a mass of 200 gram. The drum is hollow with an outer radius of R1 and inner radius of R2. And the drum has a mass of m kg.
How can we calculate the maximum angular or rotational velocity of the drum?</p>

答:
A diagram would do wonders in this case :D.
Since you omitted the fact that there is air resistance on the mass and roll friction on the drum, the only limiting factor is the length of the rope. Lets assume for simplicity that the mass (denoted with $M$) is located at the side of the drum initially:
$$Mg = Ma = F_t$$
Where $g$ is the gravitational acceleration, $a$ the acceleration of the mass and $F_t$ is the tangential force on the disk. Furthermore:
$$F_tR_1 = I\alpha$$
where $I$ is the moment of inertia of the disk (this can be computed using standard formulas found in your book or online) and $\alpha$ is the rotational acceleration. From here up to the moment the rope is fully unwound, these relations remain constant. This means that you have to compute when the mass has traveled $1m$ (through integrating for instance). Compute the rotational velocity using the resulting time from the angular acceleration.

EDIT: I am completely forgetting one fact, this only works when the rope is actually way longer and only travels 1m. If the endpoint of the rope has passed the startpoint of the mass (eg the side of the disk), the gravitional acceleration of the mass will not be transferred directly to the tangential force on the disk, but a component of it is transferred to the normal force $F_n$ on the disk (this points towards the center):
$$F_t = \text{cos}(\theta)Ma$$
$$F_n = \text{sin}(\theta)Ma$$
where $\theta = 0$ at the startpoint, so at the bottom point of the disk, $\theta = 0.5\pi$ all the force from the mass is countered by the normal force on the disk. Since the tangential force will be negative for any angle afterwards, the maximum angular velocity will be reached here.


问:
Can irregular polygons with unique vertices be guaranteed of a unique center point? Formula?
To visualize this, I've created the following diagram.

I have sets of 80 numbers each. Here, just to draw it easily, I'm showing them as sets of 12 numbers each. The sets S1 and S2. Each set can contain 12 numbers ranging from 1 to 10, and the order of the number in the set matters. i.e. {1,2,3} is not the same as {1,3,2}.

In 2D space, there are 10 concentric circles as shown by the dotted lines in the image. Each circle represents the possible values that can be contained in a set. Each of the straight lines jutting out of the center of the image, represents the position of each number in the set, starting from the "start" line. So since S1 has 3 as the first number, the first red dot is placed on the third concentric circle at the start line. Moving counter-clockwise, the next number is 7, which gets placed on the next line at circle 7 and so on.

Circles placed as such, form the red and green polygons shown in the image.

Question:

I want to find a unique center point for each such polygon formed by any such set containing 12 numbers where the numbers may be generated randomly. I've placed a red and green plus point for what I'd assume might be the approximate centers of the polygons.

I'm not particular about whether only the vertices of the polygon are
considered for finding the center or whether the sides of the polygon
are also considered.

I'm also not particular about whether the center point lies within

the polygon or not.

I just want to ensure that each unique polygon should definitely have a unique center point on the given 2D space. Is there a formula to calculate it and is it possible to ensure a unique point? Do remember that I'll eventually be using sets of 80 numbers each, containing numbers ranging from 1 to 10.

I've seen the centroid formulae and this, but I'm not sure if they'd mathematically ensure unique center points.

答:
For simplicity, I would use the centroid of a finite number of points, which is simply calculating the arithmetic mean of each coordinate. This is as if you have an equal weight at each of the points, and then find the balancing point. This is simple to implement without errors.

The other formulations you link to assume the weight is uniformly distributed across the whole area of the polygon. That will (if implemented correctly, and if given the vertices in order around the polygon) also give a uniquely determined centroid, but is slightly different to the above, and I see no reason why it would be preferred over the simpler one.

Note that as pointed out in the comments, there will always be distinct polygons that have the same centroid. There will be many that have the centroid in the origin (e.g. any polygon with rotational symmetry). There are also ones with matching centroids that are not in the origin, but those will be much rarer. However, given that you are using 80-gons, each vertex with 10 possible values, the probability of getting matching centroids is very small.


问:
Preimage of Prime Divisors is finite union of Prime divisors
I am reading the book ‘An Introduction to Birational Geometry of Algebraic Varieties’ by Shigeru Iitaka. In section 2.11 we attempt to define the pullback of a prime divisor. In particular, we have the following:

Suppose $f:V\rightarrow Z$ is a morphism of varieties where $V$ is a normal variety and $Z$ is locally factorial. (Here, locally factorial means all local rings are UFDs).

Suppose $\Gamma$ is a prime divisor on $Z$ such that $f(V) \not\subset \Gamma$ (so that $f^{-1}(\Gamma)$ is the not the whole of $V$) and also $f(V) \cap \Gamma \neq \emptyset$ (so that $f^{-1}(\Gamma)$ is not empty.)

He then proceed to claim the following: “By theorem 2.7, $f^{-1}(\Gamma)$ is union of (finitely many) prime divisors $W_1,\cdots, W_r$ of $V$. For context, here is theorem 2.7 of the book:

Theorem 2.7: Let $R$ be a Noetherian integral domain and $f$ be a nonzero and nonunit element in $R$. Then every minimal prime ideal of $fR$ is of height 1; hence, $\operatorname{ht}(fR)=1$.

My question is: How do we use theorem 2.7 to conclude that the preimage $f^{-1}(\Gamma)$ of a prime divisor must be union of finitely many prime divisors?

Any help given would be greatly appreciated! Thanks!

答:
I gave this question some thought and I think I may have found out why. (Although there are still some parts which I am not entirely sure) Anyway here goes:

Given a morphism $f:V\rightarrow Z$ satisfying hypothesis as above, since $Z$ is locally factorial, any prime divisor $\Gamma$ on $Z$ is locally principal, i.e. we have an affine open cover $\{U_\lambda:= \operatorname{Spec}A_\lambda\}_\lambda$  such that $V_\lambda := \Gamma\cap U_\lambda$ can be written as $V(\varphi_\lambda)$ for some element $\varphi_\lambda \in A_\lambda$.

We choose an affine open subset $\operatorname{Spec} B_\lambda$ of $V$ mapping into $U_\lambda$ by $f$, i.e. we consider the restriction $f: \operatorname{Spec}B_\lambda \rightarrow \operatorname{Spec}A_\lambda$. This in turn gives a ring homomorphism $\theta: A_\lambda \rightarrow B_\lambda$.

We know that by the properties of $f$ and $\theta$, one has $f^{-1}(V(\varphi_\lambda)) = V(\theta(\varphi_\lambda) B_\lambda)$.

Now, by the theorem 2.7, we have every minimal prime ideal of $\theta(\varphi_\lambda) B_\lambda$ is of height 1. Hence coupled with the fact that $B_\lambda$ is Noetherian, we have $f^{-1}(V(\varphi_\lambda))$ is a finite union of $V(\mathfrak{q}_\lambda)$’s where each $\mathfrak{q}_\lambda$ are height 1 prime ideals in $B_\lambda$.

Hence locally, the statement holds. How can I translate statement to the global case?


问:
If $Ax$ and $Ay$ are linearly independent, then are $x$ and $y$ linearly independent?
Suppose matrix $A$ and vectors $x$ and $y$ are conformable for the products $Ax$ and $Ay$.
If $Ax$ and $Ay$ are linearly independent, then are $x$ and $y$ linearly independent?

答:
Hint: proof by contradiction.

Suppose that $\alpha x+\beta y=0$, $\alpha$ and $\beta$ are not equal to 0 simultaneously. Then multiplying  both sides by $A$ from the left, we get:
$$
A(\alpha x+ \beta y)=0 \\
A \cdot \alpha x + A \cdot \beta y=0\\
$$

I bet you can finish it from here.


问:
Definite integration with natural logarithm
$$\int_0^a \ln(x)\ln(a-x)\,dx$$

How to do this? I couldn't proceed at all.

($\ln$ is natural logarithm.)

Body:
$a>0$
\begin{align}J&=\int_0^a \ln(x)\ln(a-x)\,dx\\
&=\frac{1}{2}\int_0^a \ln^2(x)dx+\frac{1}{2}\int_0^a \ln^2(a-t)dt-\frac{1}{2}\int_0^a \ln^2\left(\frac{u}{a-u}\right)du\\
&\overset{x=a-t,x=\frac{u}{2-u}}=\int_0^a \ln^2(x)dx-\frac{a}{2}\int_0^\infty \frac{\ln^2 x}{(1+x)^2}dx\\
&=\int_0^a \ln^2(x)dx-\frac{a}{2}\int_0^1 \frac{\ln^2 x}{(1+x)^2}dx-\frac{a}{2}\int_1^\infty \frac{\ln^2 u}{(1+u)^2}du\\
&\overset{x=\frac{1}{y}}=\int_0^a \ln^2(x)dx-a\int_0^1 \frac{\ln^2 x}{(1+x)^2}dx\\
&\overset{\text{IBP}}=\int_0^a \ln^2(x)dx+a\left[\left(\frac{1}{1+x}-1\right)\ln^2 x\right]_0^1-2a\int_0^1 \left(\frac{1}{1+x}-1\right)\frac{\ln x}{x}dx\\
&=\int_0^a \ln^2(x)dx+2a\int_0^1 \frac{\ln x}{1+x}dx\\
&=\int_0^a \ln^2(x)dx+2a\int_0^1 \frac{\ln x}{1-x}dx-2a\int_0^1 \frac{2u\ln u}{1-u^2}du\\
&\overset{x=u^2}=\int_0^a \ln^2(x)dx+2a\int_0^1 \frac{\ln x}{1-x}dx-a\int_0^1 \frac{\ln x}{1-x}dx\\
&=\int_0^a \ln^2(x)dx+a\int_0^1 \frac{\ln x}{1-x}dx\\
&=\Big[x\left(\ln^2 x-2\ln x+2\right)\Big]_0^a+a\times -\frac{\pi^2}{6}\\
&=\boxed{a\left(\ln^2 a-2\ln a+2\right)-\frac{a\pi^2}{6}}
\end{align}

NB: i assume that:  $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\zeta(2)=-\frac{\pi^2}{6}$


问:
Why $\int_{0}^{\pi} \tan(\theta +ia) \text{d}\theta = \frac{1}{2} \int_{0}^{2\pi} \tan(\theta + ia) \text{d}\theta$
I need to prove that equality in order to solve the integral using the remainder theorem, but I don't really know how to do it. Any help is appreciated.</p>

答:
$\begin{align}&\int_0^{2\pi}\tan(\theta+i\alpha) d\theta=\int_0^{\pi}\tan(\theta+i\alpha)d\theta+\int_{\pi}^{2\pi}\tan(\theta+i\alpha)d\theta \end{align}$

In the second integral change $\theta \to \theta+\pi$. Thus the limits in the second integral become $0$ to $\pi$

So we have,

$\begin{align}&\int_0^{2\pi}\tan(\theta+i\alpha) d\theta=\int_0^{\pi}\tan(\theta+i\alpha)d\theta+\int_{0}^{\pi}\tan(\pi+\theta+i\alpha)d\theta \end{align}$

Using $\tan(\pi+x) = \tan(x)$,

$\begin{align}&\int_0^{2\pi}\tan(\theta+i\alpha) d\theta=\int_0^{\pi}\tan(\theta+i\alpha)d\theta+\int_{0}^{\pi}\tan(\theta+i\alpha)d\theta
 = 2\int_0^{\pi}\tan(\theta+i\alpha)d\theta\end{align}$

or

$\begin{align} \int_0^{\pi}\tan(\theta+i\alpha)d\theta = \frac12\int_0^{2\pi}\tan(\theta+i\alpha) d\theta\end{align}$


问:
Reference request: about consistency of adding any amount Quine atoms to ZF-Reg.?
It is known that we can add to ZF-Reg. any amount of Quine atoms (singletons that are elements of themselves), we can even add a proper class of them.

When that result was first proved?

Does it depend on forcing?

答:
I think I can reach to an answer to this question of mine. It can be easily seen in a labeled mereological interpretation of set theory.

Take any transitive model $M$ of ZFC. Define a new membership relation $``\epsilon"$ on elements of $M$ as:

$x \epsilon y \iff \exists z \in y : z=\{x\} $

Now we define class as a set of singletons.

Now $x$ is said to be a pure class if and only if $x$ is $\epsilon$-hereditarily "class", that is $x$ is a class and every element of the $\epsilon$ transitive closure of $x$, denoted by $T^\epsilon(x)$, is a class also

Define: $T^\epsilon(x)=\{m| \forall t [\forall y \epsilon x ( y \epsilon t) \land \forall z \forall k (z \epsilon t \land k \epsilon z \to k \epsilon t) \Rightarrow m \epsilon t]\}$

Let $S$ be the set of all pure classes in $M$; then clearly $\langle S, \epsilon \rangle $ is a model of ZFC.

If we want for example to have ZF-Ext.+ two distinct empty objects exist, then we define $S^*$ as the set of all sets in $M$ that are $\epsilon$-hereditarily $``class \ or \ \{\emptyset\}"$. Now clearly $\langle S^*, \epsilon \rangle$ would be a model of ZF-Ext. with only two distinct empty objects.

Now via a similar approach its easy to prove the existence of any amount of Quine atoms in ZF-Reg.

To do that, it's essential to recognize that sets that are not sets of singletons are regarded here as not classes and those would constitute the Ur-elements of our theory. And we can control how much we have of those in our universe of discourse.

For example we can allow all non classes but only up to some stage $V_\alpha$ to be in the $\epsilon$ transitive closures of our sets, that is we take a subset $S_\alpha$ of our model $M$, that is the set of all sets that are  $\epsilon$-hereditarily $``class \ \lor \in V_\alpha"$, [that is, non-classes that are not elements of $V_\alpha$ are shunned from being elements of the $\epsilon$-transitive closure of elements of $S_\alpha$]; define a new membership relation $\epsilon^*$ as:

$x \epsilon^* y \iff [class(y) \land x \epsilon y] \lor [\neg class(y) \land x=y]$

Then $\langle S_\alpha, \epsilon^* \rangle$ would be a model of ZF-Reg. with $|V_\alpha|$ many Quine atoms.

If we want a proper class of Quine atoms, then we define the above membership relation over the whole of our original model $M$ of ZFC.


问:
Localisation of an integral domain
My question started as follows:
An intermediate ring $\Bbb Z\subseteq R\subseteq\Bbb Q$ arises as a localisation.
In particular it holds that $R=S_{P}^{-1}\Bbb Z$, where $S_{P}$ is the multiplicative
subset of $\Bbb Z$ generated by the primes which are invertible in $R$, i.e. the
set of primes $\{p:1/p\in R\}$. The crucial part to prove this is the property

$$\forall a/b\in\Bbb Q \ \ \ \ (a/b\in R \iff 1/b\in R) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$

which follows easily from the Bézout identity.

Now, I want to find an intermediate ring $\Bbb Q[x,y]\subseteq R\subseteq\Bbb Q(x,y)$ which
is not a localisation of $\Bbb Q[x,y]$. So I came up with the ring $\Bbb Q[x,y,x/y]$, in which
while it holds that $x/y\in R$, I managed to prove that $1/y\not\in R$.
So if I assume that $R$ is a localisation of $\Bbb Q[x,y]$ at a multiplicative closed subset $S\subseteq\Bbb Q[x,y]$ then I am pretty sure that
taking into account the fact that $\Bbb Q[x,y]$ is a domain we have the property $(*)$, which will end the proof. However, I cannot prove it until now. So I made up the conjecture below:

Conjecture: Let $A$ be an integral domain and $S\subseteq A$ a multiplicative subset of $A$. Then for every $a/b\in \mathrm{Frac}(A)$ it holds that $a/b\in S^{-1}A\iff 1/b\in S^{-1}A$.

答:
As written, the conjecture is false.

Let $A=\mathbb{Z}$ and $S$ the set of odd numbers.  Observe that $S$ is a multiplicatively closed.  We also observe that $2/2\in\operatorname{Frac}(A)$ is in $S^{-1}A$ since $2/2=1/1$ and $1/1\in S^{-1}A$.  Note, however, that $1/2\not\in S^{-1}A$ since if $1/2=a/s$ with $a\in A$ and $s\in S$, then $2a=s$, but the LHS is even while the RHS is odd.  A contradiction.

Now, one might argue that this example is unfair since $2/2$ is not in lowest terms.  However, if $A$ is not a unique factorization domain (or worse), lowest terms might not make sense.

Title: Localisation of an integral domain


问:
$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$
I'm having trouble understanding how to apply the $\frac{d}{dx}$when taking the anti-derivative.

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$$
In class it was mentioned we'll end up taking the derivative of $e^{x^{2}}$ which is $2xe^{x^{2}}$

My guess of solving this is

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt = [2t^{\frac{1}{2}}]_0^{e^{x^2}}$$

Then
$$[2(e^{x^{2}})^{\frac{1}{2}}]-[2(0)^{\frac{1}{2}}]=2(e^{x^{2}})^{\frac{1}{2}}$$
Am i missing something?

Body:
$$f(t)=t^{-\frac{1}{2}} \quad ,\int_{t=0}^{t=e^{x^2}} f(t)\, \mathrm dt =F(e^{x^2})-F(0)$$
$$\frac {\mathrm d} {\mathrm dx}\left[F(e^{x^2})-F(0) \right]=\frac {\mathrm d} {\mathrm dx}\left[F(e^{x^2}) \right]=\frac {\mathrm d F(e^{x^2})} {\mathrm dx}=\frac {\mathrm d \, F(e^{x^2})} {\mathrm d \, e^{x^2}}\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x}=f(e^{x^2})\,\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x}$$
$$f(e^{x^2})\,\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x}=f(e^{x^2})\,\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x^2}\,\frac {\mathrm d \, x^2 } {\mathrm d \,x}=(e^{x^2})^{-\frac{1}{2}} \, e^{x^2} \, 2x= e^{-\frac{x^2}{2}+{x^2}} \, 2x = e^{\frac{x^2}{2}} \, 2x $$


问:
Probability of at least $k$ consecutive heads after $n$ fair coin tosses
I've seen some discussions on similar problems but they seem to be overly complex, so I'm wondering if I'm missing something. I obtained the following: $$P(n, k) = 1 - \left(\frac{2^k-1}{2^k}\right)^{n-k}$$ Since for any consecutive $k$ tosses the probability that at least one tails show up is $\frac{2^k-1}{2^k}$ and the total number of these $k$ - sequences is $n-k$, the formula should hold.

答:
You're missing the fact that these events are not independent. (Also, but less important, there are $n-k+1$ possible places it could occur, not $n-k$; if $n=k$ there is still one possible place, etc.)

For example, take $k=2$ and $n=3$. Your formula (after correcting $n-k$ to $n-k+1$) gives $1-(3/4)^2=7/16$. However the actual probability is $3/8$, since you need the middle toss and at least one other to be heads.

What's going on here is that if the first consecutive pair aren't all heads, that increases the chance that the second toss was a tail, so increases the chance that the second pair also aren't all heads.


问:
Atan2 identities
I was reading through the definition of $atan2$ function from this page

For convenience

$$
\text{atan2}(y,x) =
\begin{cases}
\text{atan}(y/x) & x >0 \\
\pi + \text{atan}(y/x) & x < 0, y \geq 0 \\
-\pi + \text{atan}(y/x) & x < 0, y < 0 \\
\pi/2 & x = 0, y > 0 \\
-\pi/2 & x = 0, y < 0
\end{cases}
$$

I was trying to find a relationship between $\text{atan2}(y,x)$ and $\text{atan2}(-y,x)$
and I've derived

$$
\text{atan2}(y,x) + \text{atan2}(-y,x) = 2\pi \chi_{(-\infty,0)\times(-\infty,0]}(x,y)
$$

Where $\chi_A(x,y)$ is $1$ if $(x,y) \in A$ and $0$ otherwise.
My question is is the identity above valid? What I did to derive was literally computing based $\text{atan2}(-y,x)$ for all the cases above when $y$ flips sign and adding the original expression and the one with the $y$'s sign flipped together. It mostly makes sense to me except that corner case $2\pi$.

答:
That is not correct. For $(x, y) \ne (0, 0)$ is $\operatorname{atan2}(y, x)$ the angle between the positive x-axis and the ray from the origin to the point $(x, y)$ in the Euclidean plane, with $-\pi < \operatorname{atan2}(y, x) \le \pi$.

The points $(x, -y)$ and $(x, y)$ are symmetric with respect to the x-axis. If the points lie on the negative real axis ($x<0$ and $y=0)$ then
$$
 \operatorname{atan2}(y, x) + \operatorname{atan2}(-y, x) = \pi + \pi = 2 \pi \, .
$$
Otherwise the angles between the x-axis and the rays to these points have the same magnitude, but opposite sign in the range $(-\pi, \pi)$, so that
$$
\operatorname{atan2}(y, x) + \operatorname{atan2}(-y, x) = 0\, .
$$

In your notation that would be
$$
\operatorname{atan2}(y, x) + \operatorname{atan2}(-y, x) = 2\pi \chi_{(-\infty,0)\times \{ 0 \}}(x,y) \, .
$$

Title: Atan2 identities


问:
Determining if a sequence of functions is not uniformly continuous.
We let   $f_n(x) = x^{n}\sin(nx)$ and let $S=(-1,1)$.

The limit function seems to be $0$ on $S$.

I want to show that $f_n(x)$ is uniformly convergent on any compact subset of $S$ but not on $S$.

One strategy is to find a sequence $x_k$ converging to $1$, such that $|f_n(x_k) - f(x_k)| \geq \epsilon$.

But I am not sure quite sure how to go about it.

答:
You wrote "uniformly continuous ", but I think it should read "uniformly convergent".

If $C$ is a compact subset of $S$, then there is $0 \le q <1$ such that $|x| \le q$ for all $x \in C.$

Then: $|f_n(x)| \le |x|^n \le q^n$ for all $n \in \mathbb N$ and all $x \in C.$

This shows that $(f_n)$ converges uniformly on $C$.

Let $x_n:=1- \frac{1}{n}.$ It is your turn to show that the sequence $(f_n(x_n))$ does not converge to $0$. This shows that $(f_n)$ does not converges uniformly on $S$.


问:
What does it mean for a matrix to be orthogonally diagonalizable?
I'm a little confused as to when a matrix is orthogonally diagonalizable.

I understand that if symmetric, it's always orthogonally diagonalizable, but in what other cases can you orthogonally diagonalize a matrix?

答:
A square matrix is said to be orthogonally diagonalizable if there exist an orhtogonal matrix $P$ such that $P^{-1}AP$ is a diagonal matrix.

A square  matrix $A$ is orthogonally diagonalizable $\Leftrightarrow$ $A$ is symmetric.


问:
Solving an equation that involves both ROUND and ROUND-DOWN
$\left [ x,2 \right ]$ means to round $x$ to $2$ decimal places. For examples, $\left [ \pi,2 \right ]=3.14, \left [ \sqrt{9},2 \right ]=3.00,\left [ \sqrt{5},2 \right ]=2.24$

$\left \lfloor x,2 \right \rfloor$ means to round $x$ down to $2$ decimal places. For examples, $\left \lfloor 1.978,2 \right \rfloor=1.97,\left \lfloor 5.019,2 \right \rfloor=5.01,\left \lfloor 8.999,2 \right \rfloor=8.99$

How to find the minimum value of $\left [ s,2 \right ]$ such that

$$n\left [ b,2 \right ]+\left [ 0.00155n\left [ b,2 \right ],2 \right ]+\left \lfloor 0.15 \left [ 0.00155n\left [ b,2 \right ],2 \right ],2 \right \rfloor$$
$$\le n\left [ s,2 \right ]-\left [ 0.00155n\left [ s,2 \right ],2 \right ]-\left \lfloor 0.15 \left [ 0.00155n\left [ s,2 \right ],2 \right ],2 \right \rfloor$$

Where $n$ is a natural number, $b,s$ are positive real numbers.

Clarification with an example: (This example is taken from a real scenario):

$n$ is the number of units to be bought, say $350$

$\left [ b,2 \right ]$ is the buying (without commission and tax) price of one unit, say $28.62$

Commission is calculated as $0.00155 \times 350 \times 28.62 = 15.52635$ but rounding that to the nears $2$ decimal places, that is $15.53$

Tax is $15 \text{%}$ of that commission $=0.15 \times 15.53 = 2.3295$ but rounding that down to $2$ decimal places, that is $2.32$

So, this costs $350 \times 28.62 + 15.53 +2.32=10034.85$

So, the question is about finding the selling price $\left [ s,2 \right ]$ so that we have no losses.

By trial and error, $\left [ s,2 \right ]=28.73$

I think, it only looks complicated but actually easy for some of you. Please help me, your help would be appreciated. Thanks!

答:
With $B=100b$ and $S=100s$ (which are already integers without rounding!), multiply your inequality with $100$ to arrive at
$$nB+\left [ 0.00155nB \right ]+\left \lfloor 0.15 \left [ 0.00155nB\right ] \right \rfloor\\
\le nS-\left [ 0.00155nS \right ]-\left \lfloor 0.15 \left [ 0.00155nS \right ] \right \rfloor$$
In math, we are happier with the floor function, so rewrite as
$$nB+\lfloor 0.00155nB +0.5\rfloor+\bigl \lfloor 0.15 \lfloor 0.00155nB+0.5\rfloor \bigr \rfloor\\
\le nS-\lfloor  0.00155nS +0.5\rfloor -\bigl \lfloor 0.15 \lfloor  0.00155nS +0.5\rfloor \bigr \rfloor$$
Now note that $\lfloor x\rfloor = x-\theta$ for some $\theta$ (depending on $x$) with $0\le \theta<1$.
This allows us to write
$$nB+ 0.00155nB +0.5-\theta_1+ 0.15 ( 0.00155nB+0.5) -0.15\theta_2 -\theta_3\\
\le nS-0.00155nS -0.5+\theta_4 - 0.15(  0.00155nS +0.5)+0.15\theta_5+\theta_6$$
and rearrange as
$$ 1.0017825nB +0.575-2.15\theta_7\le 0.9982175 nS-0.575+2.15\theta_8,$$
or
$$S\ge B+\frac{1426}{399287} B +\frac{460000}{399287n}-\frac{1720000\theta_9}{399287n}\approx B+\frac{1426}{399287} B + (1.152-4.3\theta_9)\frac1n$$
It seems natural to let $S$ be the right hand side rounded up, but we don't know $\theta_9$.  Hence with

$$ S_0=B+\left\lceil\frac{1426 B-1260000}{399287} \right\rceil$$
we can simply try if $S=S_0$ fulfills the original inequality. If it doesn't, try $S=S_0+1$ and after that $S=S_0+2$ and so on.
Note that for $n>4$, we will never have to check $S_0+2$, for $n>2$, we will never have to check $S_0+3$, and even for $n=1$, we need not check $S_0+5$ or above.


问:
Evaluate $\int_{0}^{2\pi} \frac{\cos 2\theta}{2+\cos \theta} d\theta$.
I'm trying to solve evaluate $\int_{0}^{2\pi} \frac{\cos 2\theta}{2+\cos \theta} d\theta$.  What I did is the Cauchy-Residue Theorem approach.

I can have $z = e^{i\theta}$ where $\theta \in [0,2\pi]$.  So $dz/(iz) = d\theta$.  Since $\cos \theta = \frac{z+z^{-1}}{2}$ and $\sin \theta = \frac{z-z^{-1}}{2i}$, it follows that $\cos 2\theta = \frac{z^2+z^{-2}}{2}$.

So substituting, we have $\int_{|z| = 1}  \frac{z^4+1}{z^2+4z+1}dz$.  But I'm stuck in here.  Can you help me? Please!!! I will truly appreciate all the help.

Body:
Let $z_1=-2+\sqrt{3}$ and $z_2=-2-\sqrt{3}$. Then $z_1,z_2$ are the zeros of $z^2+4z+1,$ we have $|z_2| >1$ and $|z_2|<1.$ Thus

$$\int_{|z| = 1}  \frac{z^4+1}{z^2+4z+1}dz= 2 \pi i Res(f;z_1),$$

where $f(z)=\frac{z^4+1}{z^2+4z+1}.$ $f$ has at $z_1$ a simple pole, thus

$$Res(f;z_1)= \lim_{z \to z_1}(z-z_1)f(z).$$

$\lim_{z \to z_1}(z-z_1)f(z)$ is easy to calculate.


问:
Find the probability (Conditional and total probability needed).

In the first box you have 8 white balls and 4 black balls, in the second box you have 5 white balls and 7 black balls, in the third box you have 6 white and 6 black balls and in the fourth box you have 3 white balls and 9 black balls. Without looking at we chose a box and take 2 balls from the chosen box. Find the probability:

a) Chosen balls are white,

b)chosen balls are different colors ,

c) chosen balls are black.

I started (a) by writing the hypotheses H1: balls are chosen from the first box, H2: From the second box, H3: From the third box and H4: From the fourth box, A and the Event A: Chosen balls are white. Now my question for (a) is do I have to use Total Probability Formula if yes then P(hypotheses) is it going to be the same for all hypotheses 1/4,  what is going to be conditional probability :PH1(A).
If someone can answer is it enough to do the (a) cuz (c) is similar, and please help with (b) cuz I don't even know how to start, actually I'm not very sure. Thank you!

答:
In your questions a), b) and c) there is not any conditional probability requested.

To solve the posted questions, as you mentioned, Total probability theorem is enough

example

a)

Two white balls (without replacement, I suppose) are drawn

$$\mathbb{P}[WW]=\frac{1}{4}\Bigg[\frac{8\cdot7}{12\cdot11}+\frac{5\cdot4}{12\cdot11}+\frac{6\cdot5}{12\cdot11}+\frac{3\cdot2}{12\cdot11}\Bigg]$$

et cetera et cetera


问:
A problem about convergence in distribution to $N(0,1)$.

Let $(Y_{n})_{n\geq 1}$ be a sequence of  independents random variables such that $$\mathbb{P}[Y_{n}=1]=\mathbb{P}[Y_{n}=-1]=\frac{1}{2n}, \quad \mathbb{P}[Y_{n}=0]=1-\frac{1}{n}, \quad n \geq 1$$
Let $S_{n}=\sum_{i=1}^{n}Y_{i}$ and prove that $$\frac{S_{n}}{n}\overset{a.s}{\to}0 \quad \text{and} \quad \frac{S_{n}}{\sqrt{\log(n)}}\overset{D}{\to}N(0,1)$$

My approach:
For the first part, it's to say prove that $\frac{S_{n}}{n} \overset{a.s}{\to} 0$ I think that since that for $n\geq 1$, we know that
\begin{eqnarray*}
Y_{n}=1 \implies \mathbb{P}[Y_{n}=1]=\frac{1}{2n}\\
Y_{n}=-1 \implies \mathbb{P}[Y_{n}=-1]=\frac{1}{2n}\\
Y_{n}=0 \implies \mathbb{P}[Y_{n}=0]=1-\frac{1}{n}\\
\end{eqnarray*}
other values is not possible. So, the support for $Y_{n}$ is $S_{Y_{n}}={-1,0,1}$. So, by definition we have $$\mathbb{E}Y_{n}=\sum_{y_{n}\in S_{Y_{n}}}y_{n}\mathbb{P}[Y_{n}=y]=\sum_{y_{n}\in \{-1,0,1\}}y_{n}\mathbb{P}[Y_{n}=y]=0$$
similarly, we have $$\mathbb{V}Y_{n}=\mathbb{E}Y_{n}^{2}-(\mathbb{E}Y_{n})^{2}=\frac{1}{n}$$
and since that $\displaystyle \sum_{n=1}^{\infty}\frac{\mathbb{V}Y_{n}}{n^{2}}<\infty$, so by SLLN-Kolmogorov with have $$\frac{S_{n}-\mathbb{E}S_{n}}{n}=\frac{S_{n}}{n} \overset{a.s}{\to} 0$$
For the second part, to prove that $\frac{S_{n}}{\sqrt{\log(n)}}\overset{D}{\to} N(0,1)$ but don't have the hypothesis identical distribution, so I can't use the central limit theorem. How can I solve this part?

Note: I know that if $Y_{1},Y_{2},\ldots$ are independent random variables and let $\displaystyle S_{n}=\sum_{i=1}^{n}Y_{n}$. For each $i$ let $\mu_{i}=\mathbb{E}Y_{i}$ and $\sigma^{2}_{i}=\mathbb{V}Y_{i}$ and let $m_{n}=\sum_{i=1}^{\infty}\mu_{i}$ and $s_{n}^{2}=\sum_{i=1}^{n}\sigma_{i}^{2}$ mean and variance respectively. Suppose that a) $s_{n}^{2}\to \infty \quad \text{as} \quad n \to \infty$ and b) there exists a constant $M$ sucha that $\mathbb{P}[|X_{i}|\leq M]=1, \forall i$. So we have $$\frac{S_{n}-m_{n}}{s_{n}}\overset{D}{\to}N(0,1)$$

答:
First, the characteristic function of $S_n/\sqrt{\ln n}$ is given by
$$
\varphi_{S_n/\sqrt{\ln n}}(t)=\prod_{i=1}^n\left[\frac{\cos(t/\sqrt{\ln n})-1}{i}+1\right].
$$
Taking logs and noticing that $\ln(1+x)\approx x$ for small $x$, one gets,
\begin{align}
\ln\!\left(\varphi_{S_n/\sqrt{\ln n}}(t)\right)&\approx \left(\cos\!\left(t/\sqrt{\ln n}\right)-1\right)\sum_{i=1}^n \frac{1}{i}\\
&=\left(\cos\!\left(t/\sqrt{\ln n}\right)-1\right)\ln n \times R_n\to -\frac{t^2}{2},
\end{align}
where $R_n\to 1$ as $n\to\infty$.

Alternatively, setting $g_n(t)=\cos(t/\sqrt{\ln n})$,
$$
\varphi_{S_n/\sqrt{\ln n}}(t)=\frac{\Gamma(g_n(t)+n)}{\Gamma(n+1)\Gamma(g_n(t))}\approx \frac{n^{g_n(t)-1}}{\Gamma(g_n(t))}\to e^{-t^2/2},
$$
where we used the following approximation: $\Gamma(n+a)/\Gamma(n+b)\approx n^{a-b}$ for large $n$.


问:
Proof regarding unitary self-adjoin linear operators
I'm suck on how to do the following Linear Algebra proof:

Let $T$ be a self-adjoint operator on a finite-dimensional inner product space $V$.

Prove that for all $x \in V$, $||T(x)\pm ix||^2=||T(x)||^2+||x||^2.$

Deduce that $T-iI$ is invertible and that $[(T-iI)^{-1}]^{*}=(T+iI)^{-1}.$

Furthermore, show that $(T+iI)(T-iI)^{-1}$ is unitary.

My attempt at a solution (to the first part):

$||T(x)\pm ix||^2=\left< T(x)\pm ix, T(x)\pm ix\right>$

$=\left< T(x), T(x) \pm ix\right>\pm \left<ix,  T(x)\pm ix \right>$

...

$=\left<T(x), T(x) \right>+ \left<x,x \right>$

$=||T(x)||^2+||x||^2$

The ... is the part I'm stuck on (I know, it's the bulk of the first part).
I have yet to consider the next parts since I'm still stuck on this one.

Any help would be appreciated! Thanks.

答:
we have $$(Tx+ix,Tx+ix)=(Tx,Tx)+(ix, Tx)+(Tx, ix)+(ix,ix)=|Tx|^{2}+i(x,Tx)-i(x,Tx)+|x|^{2}$$where I assume you define the inner product to be Hermitian. I think for $Tx-ix$ it should be similar. The rest should be leave as an exercise for you; they are not that difficult.

To solve the last one, notice we have $(T-iI)^{-1}$'s adjoint to be $(T+iI)^{-1}$, and $(T+iI)$'a adjoint to be $T-iI$. The first one can be proved by expanding $(T-iI)^{-1}$ as $(I-iT^{-1})^{-1}T^{-1}$, then use geometric series.

The second one follows by $(Tx+ix,y)=(Tx,y)+(x,-iy)=(x,Ty)+(x-iy)=(x,(T-I)y)$.

If we want $(T+iI)(T-iI)^{-1}$ to be unitary, then we want
$$((T+iI)(T-iI)^{-1}x,(T+iI)(T-iI)^{-1}x)=(x,x)$$ for all $x$. Moving around this is the same as
$$((T^{2}+I)(T-iI)^{-1}x,(T-iI)^{-1}x)=(x,x)$$ but this is the same as
$$((T+iI)x, (T-iI)^{-1}x)=(x,x)$$
and the result follows because we know $(T-iI)^{*}=(T+iI)^{-1}$.


问:
Expected number of distinct elements when drawing from a multiset with replacement
Suppose I have a multiset $M$ with elements $1,\dots,m$ and respective multiplicities $n_1, \dots, n_m$:
$$
(1, n_1) \\
(2, n_2) \\
\dots \\
(m, n_m)
$$

What is the expected number of distinct elements in a sample of size $k$ (with replacement) from $M$?

答:
Hint:

What is the probability that element $1$ was selected at least once in your $k$ draws?

What is the probability that element $2$ was selected at least once in your $k$ draws?

Letting $X_1 = \begin{cases}0&\text{if }1\text{ was not selected}\\1&\text{if }1\text{ was selected at least once}\end{cases}$ and similarly defining $X_2,\dots, X_m$, how can you describe the random variable counting the number of distinct elements selected using these random variables?

 $X = X_1+X_2+\dots+X_m$

Apply linearity of expected value.

 $E[X]=E[X_1+X_2+\dots+X_m]=E[X_1]+E[X_2]+\dots+E[X_m]$


问:
Is this function Riemann integrable??
Let $f[0,1]\to [0,1]$ given by
$$f(x):=\begin{cases}
0& x\notin Q\\
0& x=0\\
\dfrac{1}{q+1}& x=\dfrac{q}{p}, x\in \mathbb{Q}\setminus \{0\}, \gcd(p,q)=1
\end{cases}$$
Is $f$ Riemann integrable?

I know Thomae function is Riemann integrable but
This fuction is some different. I ve'spent a lot of time
but I can't solve this function is Riemann integrable.

答:
It is. Let's proof that it's continuous in any irrational point.

For simplicity, choose any irrational $x > 0$ (negative case is similar).

If $q < n$ and $p > \frac{2n}{\varepsilon}$, then $\frac{q}{p} < \frac{\varepsilon}{2}$. So for any $n$ there are only finitely many points $\frac{q}{p}$ with $q < n$ in $\frac{\varepsilon}{2}$-neighborhood of $x$. None of this points coincide with $x$ (as it's irrational), so some neighborhood of $x$ contains no such points. And $f$ is at most $\frac{1}{n}$ in this neighborhood.

Thus, for any $n$, $x$ has neighborhood $U_n(x)$ s.t. $|f(U_n(x))| < \frac{1}{n}$. It means $\lim\limits_{y \to x} f(y) = 0$. And as $f(x) = 0$ too, it means $f$ is continuous in $x$.

As $f$ is continuous in all irrational points, it's continuous almost everywhere, and thus Riemann integrable by Lebesgue criteria (as it's also bounded).

Title: Is this function Riemann integrable??


答:
Let $X_i \in \{0,1\}$ be the indicator random variable that the sample contains element $i$.

The expected value $E(X_i)$ is the probability that the sample contains element $i$:

$$\begin{align*}
E(X_i) &= \Pr(\text{the sample contains element }i)\\
&= 1-\left(1-\frac{n_i}{s}\right)^k
\end{align*}$$

where $s$ is the size of the multiset including duplicates:

$$s = n_1 + n_2 + \cdots + n_m$$

Then the required expected number of distinct elements in a sample of size $k$ (with replacement) is

$$\begin{align*}
E &= E\left(\sum_{i=1}^m X_i\right)\\
&= \sum_{i=1}^m E(X_i)\\
&= \sum_{i=1}^{m}\left[1-\left(1-\frac{n_i}{s}\right)^k\right]\\
&= m - \sum_{i=1}^{m}\left(1-\frac{n_i}{s}\right)^k
\end{align*}$$

Title: Expected number of distinct elements when drawing from a multiset with replacement

Body:
Suppose I have a multiset $M$ with elements $1,\dots,m$ and respective multiplicities $n_1, \dots, n_m$:
$$
(1, n_1) \\
(2, n_2) \\
\dots \\
(m, n_m)
$$

What is the expected number of distinct elements in a sample of size $k$ (with replacement) from $M$?


问:
Definition of $T_pM$ where $p \in \partial M$
The following excerpt is from Tu's textbook:

A germ of smooth functions at $p$ is an equivalence class of such functions. With the usual addition, multiplication, and scalar multiplication of germs, the set $C^\infty_p(M)$ of germs of $C^\infty$ functions at $p$ is an $\mathbb R$-algebra. The tangent space $T_pM$ at $p$ is then defined to be the vector space of all point-derivations on the algebra $C^\infty_p(M)$.

For example, for $p$ in the boundary of the upper half-plane $\mathcal H^2$, $\partial/\partial x|_p$ and $\partial/\partial y|_p$ are both derivations on $C^\infty_p(\mathcal H^2)$...

I am unsure about the last bolded statement. I agree that $\partial/\partial y|_p$ exists in the positive $y$-direction, but it is not well-defined for the negative $y$-direction (i.e. $-\partial/\partial y|_p$ is not well-defined.) Can anyone help me understand this concept?

Thank you

答:
Tu defines $T_pM$ as the vector space of all point-derivations on the algebra $C_p^\infty(M)$. A derivation is a more abstract concept than an ordinary directional derivative for smooth functions $f  : U \to \mathbb R$ where $U$ is open in $\mathbb R^n$ or open in $\mathcal H^n$.

In your example the ordinary directional derivatives $\partial/\partial x|_p$ and $\partial/\partial y|_p$ are derivations on $C_p^\infty(\mathcal H^2)$ and they form a basis of $\mathcal H^2$. We do not need to regard $-\partial/\partial x|_p$ and $-\partial/\partial y|_p$ as directional derivates, they are certainly derivations in the formal sense.

Nevertheless note that all directional derivatives $\partial/\partial v|_p$ with $v \in \mathbb R^2$ are defined. In fact
$$\partial/\partial v|_p = \lim_{h \to 0} \frac{f(p+hv) -f(p)}{h}$$
where we of course only take $h$ such that $p + hv \in \mathcal H^2$. If $v \in \mathcal H^2$, then this requires $h \ge 0$, if $v \notin \mathcal H^2$, then it requires $h \le 0$. This shows that as expected $\partial/\partial{(-v)}|_p = -\partial/\partial v|_p$.


问:
Motorcycle Velocity Word Problem using Intergrals
A motorcycle, starting from rest, speeds up with a constant acceleration of 2.6m/s2. After it has travelled 120m, it slows down with a constant acceleration of −1.5m/s2 until it attains a speed of 12m/s. What is the distance travelled by the motorcycle at that point?

I have set up the info as follows:

$
t=0\\
a(0)=2.6m/s\\
v(0)=0 \\
s(0)=0\\
$

$
t=t_1\\
a(t_1)=-1.5 m/s\\
v(t_1)=?\\
s(t_1)=120m\\
$

$
t=t_f\\
v(t_f)=12m/s\\
s(t_f)=n
$

So,

$
a(t)=2.6m/s\\
s(t) \rightarrow \frac{2.6t^2}{2}=120 \rightarrow t=\sqrt \frac{240}{2.6} \\
v(t) \rightarrow 2.6t \rightarrow 4\sqrt{39} \approx 24.97 \\
$

At this point I am stuck.
I am guessing my next step is to use $v(t_f)-v(t_1) \int a dt$.

But don't understand how to apply it to find $s(t_f)=n$.

答:
$v(t_2) = v(t_1) + a t_2$

i.e $12 = 24.98 - 1.5t_2 \implies t_2 \approx 8.65$

(please note the total time is $t_f = t_1 + t_2)$.

Distance traveled in time $\displaystyle t_2, s_2 = \int_0^{t_2} (24.98 - 1.5t)dt = 24.98t_2 - 0.75{t_2}^2$

Now calculate $s_2$. Total distance traveled $ = s_1 + s_2 = 120 + s_2\,$.


问:
Lexicographic Product of two Graphs
How can I prove that the lexicographic product of bipartite graphs is not bipartite? I would appreciate any help on this!</p>

答:
As per Brandon du Preez's comment, you should include definitions in your question for completeness (especially given the incredible number of graph product definitions.) It would also pay to edit your question with what you've attempted so far.

If you want to show the lexicographic product (see also graph composition) of bipartite graphs is not necessarily bipartite, it would be easy to show via counter-example.

Hint: Take the lexicographic product of the simplest bipartite graph you can think of with itself.

If you want to be more adventurous, you can prove the following:

The lexicographic product of two graphs is bipartite if and only if one factor is $\overline{K_t}$ and the other is bipartite.

You can probably use contradiction for the forward implication. Use the definition for the converse.

Note $\overline{K_t}$ is the complement of the complete graph, i.e. the graph with $t$ independent vertices as its vertex set.

Title: Lexicographic Product of two Graphs


问:
Solving systems of equations using matrices
I'm teaching myself how to solve systems of equations using matrices and row operations. I get the basic mechanics of it (legal operations, etc.), but it seems like it's kind-of a crapshoot deciding where to begin, and choosing the "wrong" operation to start with can lead to a really difficult problem.

My question is: Are there any rules of thumb for deciding which operations to begin with and how to proceed?

For instance, given the system:

```<code>2x - 9y - z = -23
 x + 9y - 5z= 2
3x + y + z = 24
</code>```

What row operation would you begin with, and where would you go from there? I'm not asking you to do it for me, but for insight into your thought process regarding how to proceed.

答:
JavaMan is right that the algorithm works.

But if you're just looking for rules of thumb: the $-z$ and $+z$ in the first and third rows jump out at me -- adding one row to the other will create a zero.  Ditto for $-9y$ and $+9y$.  To leverage both of those, you need to add the first row to the second row, then to the third row.  That should get you well on your way.


问:
Iteration with limit (eigenvalues/convergence)
Let $x_0 \in \mathbb{R^2}$. Let $A$ be a matrix with eigenvalues $\lambda_+$ and $\lambda_-$ and associated eigenvectors $p_+$ and $p_-$. Let $\eta >0$.

It's $x_0=\alpha_+p_++\alpha_-p_-$ for $\alpha_+,\alpha_- \in \mathbb{R}$ and $x_{k+1}=\alpha_+(1-\eta \lambda_+)^{k+1}p_++\alpha_-(1-\eta \lambda_-)^{k+1}p_-$

My question is:

Why $x_0 \in \mathrm{span} \lbrace p_+ \rbrace $ implies that $\lim\limits_{k\to\infty}x_k=\textbf{0}$ for $\eta >0$ chosen sufficiently small such that $|1-\eta \lambda_+|<1$ is satisfied?

My idea was:

Since $x_0 \in \mathrm{span} \lbrace p_+ \rbrace $ it follows that $\alpha_-=0$.

So $x_0=\alpha_+p_+ \Rightarrow \lim\limits_{k\to\infty}x_k \in \mathrm{span} \lbrace p_+ \rbrace $

Now I don't know how to continue. Is this step true or how can it be shown?

答:
As you said $\alpha_-=0$, so:
$$x_{k}=\alpha_+(1-\eta \lambda_+)^{k}p_+$$
Since:
$$\lim_{k\to \infty} ||x_{k}||=\lim_{k\to \infty} ||\alpha_+(1-\eta \lambda_+)^{k}p_+||$$
We have by the properties of norm:
$$\lim_{k\to \infty} ||x_{k}||=\lim_{k\to \infty} |\alpha_+||1-\eta \lambda_+|^{k}||p_+||$$
Then:
$$\lim_{k\to \infty} ||x_{k}||=|\alpha_+| ||p_+|| \lim_{k\to \infty} |1-\eta \lambda_+|^{k}$$
So if $ |1-\eta \lambda_+|<1$, we have:
$$\lim_{k\to \infty} ||x_{k}||=0$$
Which by definition of limit in a normed vector space is equivalent to:
$$\lim_{k\to \infty} x_{k}=0$$


问:
Positive definiteness of $\frac{1}{2}(A+A^T)$ and positive eigenvalues of $A$
Given a (non-symmetric) matrix $A \in \mathbb{R}^{n\times n}$, is it possible to relate the sign of its eigenvalues (only the real part) to the definiteness of the corresponding symmetric matrix

$$\frac{1}{2}(A+A^T)?$$

I am very confused due to the non-consistent definitions of positive definiteness of matrices across the literature:

in engineering or applied mathematics textbooks, a general matrix $M$ is positive definite $\iff$ all eigenvalues have positive real parts.

in classical matrix analysis textbooks, however, the matrix $M$ needs to be Hermitian.

I was wondering if this is kind of justified by the statement above relating the signs of eigenvalues.

答:
Counterexample (if I understand the question correctly):

$$A:=\begin{pmatrix}2&2\\-2&-1\end{pmatrix},\qquad \tfrac{1}{2}(A+A^\top)=\begin{pmatrix}2&0\\0&-1\end{pmatrix}$$

The eigenvalues of $A$ are $\frac{1}{2}\pm \frac{1}{2}i\sqrt7$, yet its symmetric part $\tfrac{1}{2}(A+A^\top)$ is not positive definite.


问:
closed intervals and limit of a sequence
Closed intervals are gotten by " Closing up" intervals so that sequences must limit to points in the intervals itself.

Does this notion generalize?
How are closed sets related to limits of sequence?

答:
Let $A$ be a subset of $ \mathbb R^n$ (or more general a subset of a metric space).

Then we have:

$A$ is closed $ \iff$ for each convergent sequence $(x_n)$ in $A$ it holds that $ \lim_{n \to \infty}x_n \in A.$


问:
The kernel of the symmetrizing map.
Define the symmetrizing map:
$S: \mathbb{Q}[x_1,...x_n]\rightarrow \mathbb{Q}[f_1,...f_n], f\mapsto \frac{1}{n!}\sum_{\sigma\in S_n}\sigma\cdot f$
where $f_i$ are the elementary symmetric polynomials and the action $\cdot$ is the permutation of variables.
We have that if there exists a $\sigma\in S_n$ such that $\sigma\cdot f = -f$, then $S(f)=0.$ Does the converse hold?

答: Take the polynomial $f(x_1,x_2,x_3)=x_1^2+a_2x_2^2+a_3x_3^2$, ($a_2$,$a_3$ are determined later). You can check that if $S(f)=0$ then $a_2,a_3$ must satisfy $2+a_2+a_3=0$. Taking $a_2=\frac{3}{2}$ forces $a_3$ to be $\frac{-7}{2}$. Hence for this polynomial there is no $\sigma$ such that $\sigma (f) = -f$

Title: The kernel of the symmetrizing map.


问:
Does this particular solution to a linear ode with varying coefficient vanish for all t and is it transferable to matrices?
I want to show that a particular initial value solution to the following differential equation is zero for all $t$.
The differential equation is
$$
\frac{d}{dt}f(t) = g(t)f(t)
$$
and the initial condition is $f(t_0) = 0$. I conclude that in this case the first derivative and thus  all higher order derivatives are zero at $t_0$. Since all time derivatives of higher order vanish and the initial value is zero, the particular solution to this problem is $f(t)=0 \ \forall \ t$.

All higher order derivatives can be factorized into a form where $f(t)$ appears as multiplicative factor by plugging in the one order lower derivative.

$$\begin{aligned}
f^{(n)} &= g^{(n-1)}f^{(0)} + g^{(n-2)}f^{(n-1)} \ \forall \ n>1 \\
f^{(1)} &= g^{(0)}f^{(0)} \\
f^{(0)}(t_0) = 0  &\Rightarrow  f^{(n)}(t_0) = 0 \\
f_{particular}(t) &= 0
\end{aligned}$$
with $$f^{(n)} = \frac{d^n f}{d t^n} $$

Is this correct or am i missing something ?

Assuming that this is correct, is it transferable to matrix equations where $f$ and $g$ are matrices ?

EDIT:

In 1 dimension the solution can be given as
$$
f(t) = f(t_0)\exp\left(\int^t_{t_0}d\tau g(\tau)\right)
$$
which shows that $f(t)$ vanishes for $f(t_0)=0$. Is there a solution of similar  form, when going to a matrix differential equation $f\rightarrow F, g\rightarrow G$ of the form,
$$
\frac{d}{dt}F(t) = G(t) F(t)
$$
where $G(t)$ does not commute with its integral $\int d\tau G(\tau)$. Can the solution still be brought into something of the form
$$
F(t) = F(t_0) \cdot (\dots)
$$
?

I didn't manage to do derive such a form for a non commuting matrix $G(t)$ but i think my argument for $f$ should also hold for matrices. I guess i need to include the assumption that the derivatives of $G$ exist.

Is my argument then valid for matrices ?

答:
Your arguments use the fact that $g$ has derivatives of all orders. But this is not neccesary.

If we only assume that $g$ is continuous, then $g$ has an anti-derivative $G$. The general solution of the linear first order differential equation

$$\frac{d}{dt}f(t) = g(t)f(t)$$

is given by

$$f(t)=C e^{G(t)}, $$

where $C \in \mathbb R.$ If $f(t_0)=0$, then $Ce^{G(t_0)}=0,$, hence $C=0.$

This gives $f(t)=0 \ \forall \ t.$


问:
$AB-AA^TABB^TA^Tvv^TAB$ is not full rank
My question is very simple.

$A \in \mathbb R^{2 \times 3}, B \in \mathbb R^{3 \times 3}, v \in \mathbb R^{2 \times 1}$

I'm looking for matrices $A,B,v$ of the dimensions above, such that:

a) $AB$ has full row rank.

b) The matrix $AB-AA^TABB^TA^Tvv^TAB$ does not have full row rank.

I've been looking for an example like that for a while, beginning to think maybe one doesn't exist.

Anyone care to try find an example?

答:
Notice that $AB-AA^TABB^TA^Tvv^TAB = (I-AA^TABB^TA^Tvv^T)AB$ and the eigenvalues of $AA^TABB^TA^Tvv^T$ are $0$ and $v^TAA^TABB^TA^Tv$, so it is enough that
$v^TAA^TABB^TA^Tv = 1$

In order to find the simplest example, take $B=I$ and
$$
A =
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
$$
so in this case, $v^TAA^TABB^TA^Tv = v^Tv$ and any norm one $v$ is sufficient.


问:
Converting Between Non-Decimal Bases?
How can a number in an arbitrary base be converted to another random base - for example from 9.2E9B03 base 17 to base 5? I would prefer a generalized procedure avoiding an intermediary base 10 conversion - which I already know how to do.</p>

答:
If you can do arithmetic in either base, it is possible to convert.

For example, one can do repeated divisions by 5 in base 17, to get the base 5 number from the units.  Or, one can feed the digits of base 17 into x'= 17x+d, to get the number in base 5.

Invoice arithmetic works for converting any columned system to any other.  It relies on short division and multiplication.

For example, 1 acre costs £1. The first column then consists of the acre being divided into roods, 10 perches and perches.  The price is likewise divided to 5/-, 1/3, and 1.5 d.  One then buys so many acres, roods and perches at the listed price, so the conversion is done without recall to any base.

```<code>   1 acre    costs £1 0s 0d   3 acres cost      £3 0 0
   1 rood   costs     5s 0d
  10 perches costs    1s 3d   20 perches costs  £  2s 6d
   1 perch   costs      1.5d   7 perches costs       10.5d
                                                 --------
                                                £ 3 3s 4.5d
</code>```

For the base-17 to base 5 example, one prices the cost of each 17th in base 5' and each second 17th, etc.  You can run an alternating 4 as well, so tje expression in base 17, is a sum of 0.4, 0.1, 0.04, 0.01, etc.  You then buy these like 9 = 2*4+1 etc.

```<code>    base 17      base 5                    base 5
   9.2E9B03
   4.000000     4.000000000      * 2 =   13.000000000
   1.000000     1.000000000      * 1 =    1.000000000
   0.400000     0.104201213      * 0 =    0.000000000
   0.100000     0.014133413      * 2 =    0.033321332
   0.040000     0.001313302      * 3 =    0.010000411
   0.010000     0.000101034      * 3 =    0.000303212
   0.004000
                 -                      ==============
                                         14.04412410
</code>```

Title: Converting Between Non-Decimal Bases?

Body:
If you can do arithmetic in either base, it is possible to convert.

For example, one can do repeated divisions by 5 in base 17, to get the base 5 number from the units.  Or, one can feed the digits of base 17 into x'= 17x+d, to get the number in base 5.

Invoice arithmetic works for converting any columned system to any other.  It relies on short division and multiplication.

For example, 1 acre costs £1. The first column then consists of the acre being divided into roods, 10 perches and perches.  The price is likewise divided to 5/-, 1/3, and 1.5 d.  One then buys so many acres, roods and perches at the listed price, so the conversion is done without recall to any base.

   1 acre    costs £1 0s 0d   3 acres cost      £3 0 0
   1 rood   costs     5s 0d
  10 perches costs    1s 3d   20 perches costs  £  2s 6d
   1 perch   costs      1.5d   7 perches costs       10.5d
                                                 --------
                                                £ 3 3s 4.5d

For the base-17 to base 5 example, one prices the cost of each 17th in base 5' and each second 17th, etc.  You can run an alternating 4 as well, so tje expression in base 17, is a sum of 0.4, 0.1, 0.04, 0.01, etc.  You then buy these like 9 = 2*4+1 etc.

    base 17      base 5                    base 5
   9.2E9B03
   4.000000     4.000000000      * 2 =   13.000000000
   1.000000     1.000000000      * 1 =    1.000000000
   0.400000     0.104201213      * 0 =    0.000000000
   0.100000     0.014133413      * 2 =    0.033321332
   0.040000     0.001313302      * 3 =    0.010000411
   0.010000     0.000101034      * 3 =    0.000303212
   0.004000
                 -                      ==============
                                         14.04412410


问:
A question on symmetric states of a Markov chain
Suppose two distinct states $i,j$ satisfy
\begin{align*}
\mathbb {P}\left(\tau_{j}<\tau_{i}\vert X_{0}=i\right)=\mathbb {P}\left(\tau_{i}<\tau_{j}\vert X_{0}=j\right),
\end{align*}
where $\tau_{k}:=\inf \left\{n\geq 1:X_{n}=k\right\}$. Show that, if $X_{0}=i$, the expected number of visits to $j$ prior to re-visiting $i$ is one. Suppose $N$ is the number of visits to $j$ before visiting to $i$ and $\theta=\mathbb {P}\left(\tau_{j}<\tau_{i}\vert X_{0}=i\right)=\mathbb {P}\left(\tau_{i}<\tau_{j}\vert X_{0}=j\right)$. When I looked at the solution, I don't quite understand why we have the following derivation:
\begin{align*}
\mathbb {P}(N\geq k\vert X_{0}=i)=\theta (1-\theta)^{k-1} \quad \text{for $k\geq 1$}\Longrightarrow \mathbb {E}(N\vert X_{0}=i)=\sum_{k=1}^{\infty}\theta (1-\theta)^{k-1}=1
\end{align*}
Since I know that in fact
\begin{align*}
\mathbb {P}(N= k\vert X_{0}=i)=\theta (1-\theta)^{k-1}\theta \quad \text{for $k\geq 1$}
\end{align*}
so what's wrong with the following formula? Thanks a lot!
\begin{align*}
\mathbb {E}(N\vert X_{0}=i)=\sum_{k=1}^{\infty}k\cdot\theta (1-\theta)^{k-1}\theta
\end{align*}

答:
Nothing is wrong actually: note that
$$
\mathbb {P}(N\geq k\vert X_{0}=i)=\sum_{j=k}^\infty \mathbb {P}(N=j\vert X_{0}=i)=\theta^2\sum_{j=k}^\infty (1-\theta)^{j-1}=\theta^2(1-\theta)^{k-1}\sum_{j=0}^\infty (1-\theta)^{j}=\theta (1-\theta)^{k-1}.
$$
You can recover the result from the last equality in your post by writing
$$
\mathbb {E}(N\vert X_{0}=i)=\sum_{k=1}^{\infty}k\cdot\theta (1-\theta)^{k-1}\theta=\sum_{k=1}^{\infty}\sum_{j=1}^k  \theta (1-\theta)^{k-1}\theta$$
and switch the sums over $j$ and $k$. This gives
$$
\mathbb {E}(N\vert X_{0}=i)=\theta^2\sum_{j=1}^\infty\sum_{k=j}^\infty (1-\theta)^{k-1}=\theta^2\sum_{j=1}^\infty\sum_{k=0}^\infty (1-\theta)^{k+j-1}=\theta^2\sum_{j=1}^\infty (1-\theta)^{j-1}\sum_{k=0}^\infty(1-\theta)^k
$$
and the last two series are both equal to $1/\theta$.


答:
Example:
$$
A = \pmatrix{1&0&0\\0&1&0}, \quad B = \pmatrix{1&0&0\\0&1&0\\0&0&1}, \quad v = \pmatrix{1\\0}.
$$
We find that $AB$ has full rank, but
$$
AB-AA^TABB^TA^Tvv^TAB = A - AA^Tv(A^Tv)^T = \pmatrix{0&0&0\\0&1&0}
$$
does not.

Title: $AB-AA^TABB^TA^Tvv^TAB$ is not full rank

Body:
My question is very simple.

$A \in \mathbb R^{2 \times 3}, B \in \mathbb R^{3 \times 3}, v \in \mathbb R^{2 \times 1}$

I'm looking for matrices $A,B,v$ of the dimensions above, such that:

a) $AB$ has full row rank.

b) The matrix $AB-AA^TABB^TA^Tvv^TAB$ does not have full row rank.

I've been looking for an example like that for a while, beginning to think maybe one doesn't exist.

Anyone care to try find an example?


问:
Sequence and Floor fucntion $a_n = a_{n-1} + a_{\lfloor \frac{n}{3} \rfloor}$
$a_0=5, a_1 =2$
$$a_n = a_{n-1} + a_{\lfloor \frac{n}{3} \rfloor}$$
Prove that there are infinitely many $i \in \mathbb{N}$ such that $13 \mid a_i$

I have no idea for this problem, please help me.
You can see that $i = 5, 10, 22, ...$

$a_2=7, a_3=9, a_4=11, a_5=13, a_6=20, a_7=27, a_8=34, a_9=43, a_{10}=52, ..., a_{22}=247$

答:
Assume the contrary, i.e., the set $I=\{5,10,22,\ldots\}$ of indices where $a_n\equiv 0\pmod{13}$ is finite (but non-empty!).
Let $N=\max I$ (so certainly $I\ge 22$).
Then $$a_{3N-1}\equiv a_{3N}\equiv a_{3N+1}\equiv a_{3N+2}\pmod {13}.$$ Next, the thirteen terms $a_{9N-4},\ldots, a_{9N+7}$ are in an arithmetic progression modulo $13$, i.e.,
$$ a_{9N-4+k}\equiv a_{9N-4}+k\cdot a_{3N}\pmod {13}$$
for $k=0,1,\ldots, 12$. As $a_{9N-4+k}\equiv 0$ would contradict $9N-4+k>N$, we see from the pigeon-hole principle that two of these terms must be congruent, i.e.,
$$ a_{9N-4}+k_1\cdot a_{3N}\equiv a_{9N-4}+k_2\cdot a_{3N}\pmod {13}$$
with $0\le k_1<k_2\le 12$, and from this
$$ (k_2-k_1)a_{3N}\equiv 0\pmod{13}.$$
As $13$ is prime, we must have $k_2-k_1\equiv 0$ or $a_{3N}\equiv 0$, both of which are absurd.

Apparently, the same argument works with any prime $p$ and sequence with recursion
$$ a_n=a_{n-1}+a_{\lfloor n/d\rfloor},$$
provided $p\le d^2+d+1$ and there exists at least on (small) $a_n$ that is a multiple of $p$.


问:
Are the two spans equal?
Let $x$,$y$ and $z$ be three linearly independent vectors. Explain, with justification, whether or not
$\operatorname{span}\{x,\,y,\,z\}=\operatorname{span}\{x+y,\,y+z,\,z+x\}$.

答:
Assume W := v ∈ span{x+y, y+z, z+x} then

v = a(x+y) + b(y+z) + c(z+x)

= (a+c)x + (a+b)y + (b+c)z

so that if v ∈ W then v ∈ U, i.e. W ⊆ U. Assume v ∈ U then

v = ax + by + cz

= ((a+b−c)/2)(x+y) + ((b+c−a)/2)(y+z) + ((a−b+c)/2)(z+x) ∈ W

i.e. U ⊆ W.

Title: Are the two spans equal?


问:
Confusion on the definition of Simple connectedness
The problem I am working on asks me to prove that

$X$ is simply connected iff any two paths in $X$ is path-homotopic.

If I'm not mistaken, I need two conditions for simple connectedness. (i) $X$ is path-connected (ii) fundamental group is trivial.

For the reverse direction of this proof, I'm stuck on proving (i) $X$ is path-connected. In the extreme case, if there are just two discrete points in $X$ in the metric space, it wouldn't contradict the condition that any two paths in $X$ is path-homotopic (vacuously holds) while failing the (i) condition.
I just want to know if the question lacks some additional condition or there is some misunderstanding about the definition of simple-connectedness. I would appreciate any of your comments.

答:
It is wrong.

First, the condition "any two paths in $X$ is path-homotopic" cannot be satisfied unless $X$ has a single point. As pointed by Gae. S. in his answer, path homotopy requires that paths have the same endpoints. Thus the author was imprecise. We can study two precisions:

Any two paths with the same endpoints are path-homotopic.

Any two paths are freely homotopic.

But 2. does not characterize simple connectedness, it only characterizes path connectedness. This comes from the fact that each path is freely homotopic to a constant path, and constant paths are freely homotopic iff their images are contained in the same path component of $X$.

Variant 1. performs better, it characterizes spaces $X$ in which all path components are simply connected. However, if $X$ is not path connected, then it is not simply connected.

Thus it is correct to state

$X$ is simply connected iff it is path connected and satisfies 1.


答:
Try following these steps (practice a bunch of them and you start seeing approaches, but there are no hard and fast rules on what is optimal).

Swap R1 and R3

Subtract (1/3) R1 from R2

Multiply R2 by (3/2)

Multiply R3 by -1

Add (2/3) R1 to R3

Multiply R3 by 3

Swap R2 with R3

Subtract (13/29) R2 from R3

Multiply R3 by (-29/297)

Subtract 5*R3 from R2

Subtract R3 from R1

Subtract (1/29) R2 from R1

Divide R1 by 3

Divide R2 by 29

You will end up with:

$$
\left[\begin{array}{ccc|c}
1 & 0 & 0 & 5\\0 & 1 & 0 & 3\\0 & 0 & 1 & 6
\end{array}\right]
$$

Title: Solving systems of equations using matrices

Body:
I'm teaching myself how to solve systems of equations using matrices and row operations. I get the basic mechanics of it (legal operations, etc.), but it seems like it's kind-of a crapshoot deciding where to begin, and choosing the "wrong" operation to start with can lead to a really difficult problem.

My question is: Are there any rules of thumb for deciding which operations to begin with and how to proceed?

For instance, given the system:

```<code>2x - 9y - z = -23
 x + 9y - 5z= 2
3x + y + z = 24
</code>```

What row operation would you begin with, and where would you go from there? I'm not asking you to do it for me, but for insight into your thought process regarding how to proceed.


问:
Find the extrema values of a multivariable function with one inequality constraint
Question: Find the absolute maximum and minimum values of $f(x, y) = xy^2$ on the set $D=\{{(x, y), x^2+y^2 <= 3}\}$.

My attempted Solution: First I computed the following gradients of the functions:

Gradient of $f(x,y)=<y^2, 2xy>$

Gradient of $g(x,y)=<2x, 2y>$

Critical point 1: Since $<y^2, 2xy> = <0, 0>$, $(x, y) = (0, 0)$

By using the Lagrange multipliers, I obtained the following three functions:

$$y^2=2x\lambda$$
$$2xy=2y\lambda$$
$$x^2+y^2=3$$

However, I'm not sure how to continue on from this point to compute the critical points. Since $x$ and $y$ can be zero in both these cases, I'd assume that we can't divide either from both sides of the equation, thus making it hard to solve.

I'd be much appreciated if someone could show me how this is done. Thanks in advance!

答:
$f(x,y) = xy^2$

With condition $x^2 + y^2 \leq 3$

Based on our function $(f(x,y) = xy^2)$, one thing to note is that our extrema would likely occur on the boundary $x^2 + y^2 = 3$.

Continuing from the point that you got to in your question,

$y^2=2x\lambda \implies \lambda = \frac{y^2}{2x} \, $ if $x \ne 0$

$2xy=2y\lambda \implies \lambda = x \,$ or $ \,x = 0$

For $x = 0, y = 0$

For $x \ne 0$, equating $\lambda, \, y = \pm \sqrt2 \, x$

Based on the condition, $-1 \leq x \leq 1$

So potential points to check for min and max are $(-1, - \sqrt2), (-1, \sqrt 2), (1, -\sqrt2), (1, \sqrt 2), (0, 0)$.


问:
Gaussian convolution
$X$ and $Y$ are standard normal independent random variables, let $Z=XY$. Is the distribution of $Z$ equal to $N(0,\frac{1}{2})$?

I tried using convolution to get the distribution of $Z$. Is it correct?

答:
The sum Z=X+Y is Gaussian distributed, with the variance being the sum of variances of X and Y.
It can "easily" be shown using the characteristic function (Fourier of the probability density function).

I don't think the product is Gaussian.

Title: Gaussian convolution


问:
The Limit of a Wallis-like Integral
Let

$\displaystyle I_n=\int^{\pi}_{0} e^x \sin^n x dx$,

then, what number does

$\displaystyle \sqrt{n} I_n$

tend to?

According to wolfram alpha it tends to 12.05... but I want to know what is this number.

I posted similarly titled question about two months ago: The limit and Asymptotic Behavior Wallis-like Integral but the theme here is not that.

答:
There is a reduction formula for the antiderivative. Have a look at equation $2.662.1$ in Table of Integrals, Series, and Products (Seventh Edition)  by I.S. Gradshteyn and I.M. Ryzhik.

If you enjoy hypergeometric functions, there is a formal solution. Using the bounds, if I am not mistaken
$$I_n=-\frac{i  e^{\frac{i \pi  n}{2}} \left(e^{\pi(1+ni)}-1\right) \Gamma
   \left(-\frac{n+i}{2}\right) \Gamma (n+1)}{2^{n+1}\Gamma \left(\frac{n-i}{2}+1\right)}$$

How to take the limit or the asymptotics of the monster ???

Computing it for $n=10^k$, some results (waiting for an answer !)
$$\left(
\begin{array}{cc}
k & \text{result} \\
 1 & 12.30548841 \\
 2 & 12.08766068 \\
 3 & 12.06108749 \\
 4 & 12.05838001 \\
 5 & 12.05810876 \\
 6 & 12.05808163
\end{array}
\right)$$

Title: The Limit of a Wallis-like Integral


问:
For a symmetric matrix, the geometric and algebraic multiplicities are equal
Could anyone tell me how to prove the following proposition?

Let $\lambda$ be an eigenvalue of a symmetric matrix. Then, its geometric multiplicity equals its algebraic multiplicity.

答: If $A$ is real symmetric then it is diagonalizable.  That is, there is some orthogonal $P$ such that $AP=PD$, where $D$ is diagonal.  The columns of $P$ are each eigenvectors, and form a basis.  Hence all geometric multiplicities equal algebraic multiplicities.</p>


问:
When do we use Jacobian matrix or some other tests for stability of linear systems
I am studying linear systems of differential equations and I am very new to the subject. Now, given a linear system $\dot{x}=Ax$ for some matrix $A$ with real entries, when we want to find stability of the origin - the equilibrium point- we may use

The test that uses $\det{A}=\delta$ and $Tr(A)=\tau$ so that we check the conditions on the sign of $\delta,\tau$ or $\tau^2-4\delta$

We first find the equilibrium points of the system and then evaluate them using the Jacobian matrix $Df(x)$. Then, depending on the signs of the real part of the eigenvalues of $Df(x^*)$, we say that for example the equilibrium point $x^*$ is a saddle.

However, I am not sure about when to use the first one or the second one. Can someone clarify the uses for  these tests?

答:
Both tests are meant to be applied to a differential equation of the form
$$
\dot {\mathbf x} = f(\mathbf x),
$$
where $\mathbf x(t) \in \Bbb R^n$ for all $t \in \Bbb R$. The second test can be applied to any differential equation of this form as long as the function $f:\Bbb R^n \to \Bbb R^n$ is differentiable. The first test applies only to the specific case that $n = 2$ (so that $\mathbf x(t) = (x_1(t),x_2(t))$) and
$$
f(\mathbf x) = A \mathbf x
$$
for some ($2 \times 2$) matrix $A$.

If your problem is of the form $\dot {\mathbf x} = A \mathbf x$ and $A$ has size $2 \times 2$, then apply method 1 (since it is easier). In all other cases, method 1 does not work but method 2 does.


问:
Show that $H(U)$ the space of holomorphic functions on $U$ is not normable.
I'm trying to show the following result:

Let $U \subset \mathbb{C}$ be an open set. Show that the vector space $H(U)=\{f:U \to \mathbb{C} \; \text{analytic}\}$ is not normable.

I know that $H(U)$ is a Frechet space. If we take $A_{k}:=\{z \in U\ : |z| \leq k \; , \; \text{dist}(\partial U,z) \geq \frac{1}{k} \}$ and $p_{k}(f)=\sup_{z \in A_{k}}|f(z)|$ for each $k \in \mathbb{N}$, we have
$\cup_{k}A_{k}=U$ and metric

$$d(f,g)=\sum_{k=1}^{\infty}\frac{1}{2^{k}} \frac{p_{k}(f-g)}{1+p_{k}(f-g)}.$$

I'm not sure how to work with this. Any help would be greatly appreciated.

答:
Supposing  by contradiction that $H(U)$ is normable via some norm $\|{\cdot}\|$, observe that  the set
$$
  B:= \{f\in H(U): \|f\|<1\}
  $$
is open, so there exists some $\varepsilon >0$, such that
$$
  \Omega :=\{f\in H(U): d(f, 0)≤\varepsilon \}\subseteq  B.
  $$

Notice moreover  that, for every $z$ in $U$, the point evaluation
$$
  \delta _z: f\in  H(U) \mapsto  f(z)\in  \mathbb C
  $$
is a continuous linear functional.  So,
given any $f$ in $\Omega $, and any $z\in U$, we  have that
$$
  |f(z)|  \leq  \|\delta _z\| \|f\| \leq   \|\delta _z\|.
  \tag 1
  $$

This is the first indication that something is wrong, since the functions in $H(U)$ are not necessarily bounded.
Choose $n$ such that $\sum_{k=n+1}^\infty 2^{-k}≤\varepsilon /2$ and notice that, since
$$
  A_1\subseteq A_2\subseteq \cdots \subseteq A_n
  $$
we have that
$p_1\leq p_2\leq \cdots \leq p_n$, whence for every $f$ in $H(U)$ we have
$$
  d(f,0) =
  \sum_{k=1}^\infty \frac1{2^k} \frac{p_k(f)}{1+p_k(f)} \leq  $$$$ \leq
  \sum_{k=1}^n\frac1{2^k} \frac{p_k(f)}{1+p_k(f)} +   \sum_{k=n+1}^n\frac1{2^k} \leq  $$$$ \leq
  \Big(\sum_{k=1}^n\frac1{2^k}\Big) \frac{p_n(f)}{1+p_n(f)} +  \frac\varepsilon 2 \leq
  \frac{p_n(f)}{1+p_n(f)} +  \frac\varepsilon 2.
  $$
Chosing $\delta >0$ such that
$$
  0\leq x\leq \delta \Rightarrow \frac{x}{1+x}\leq \varepsilon /2,
  $$
we then deduce that
$$
  p_n(f)\leq \delta  \Rightarrow d(f,0)\leq \varepsilon.
  $$

Summing up  we have by (1) that
$$
  \sup_{z\in A_n}|f(z)| \leq  \delta  \Rightarrow  f\in \Omega   \Rightarrow  |f(z)| \leq  \|\delta _z\|,\quad \forall z\in  U.
  \tag 2
  $$

Picking any point $z_0$ in $U\setminus A_n$ it is now easy to choose some holomorphic function $f$ on $U$ that is
bounded by $\delta $ on $A_n$, and takes on an arbitrarily large value on $z_0$, e.g. bigger that $ \|\delta _{z_0}\|$, thus
contradicting (2).


问:
Proving that the limit is greater than any $f(x)$
Let $f:\mathbb R\to \mathbb R$ and $t\in \mathbb R$, $A=\{x\in [b,c) \mid f(x) > t\}$  so that $\inf(A) \notin A$.

Let $M\in \mathbb R$, so that $\lim_{x\to \inf(A)}f(x)=M$.

Prove that $M\geq t$.

I showed that, since $\inf(A)\notin A$, $f(\inf(A))\leq t$.

Then I tried showing that from the definition of limit and that $\lim_{x\to \inf(A)}f(x)=M$,
it means that "around the point" $\inf(A)$, the function is $M$.
What brought me to the opposite of what I needed to prove. I got to the conclusion that:
$f(\inf(A))\leq t \Rightarrow M \leq t$

What did I get wrong?

答: I don't understand how you got to the opposite conclusion.  Write $a=\inf(A)$.  There is a sequence $(x_n)$ such that $x_n\in A,\ x_n>a,\ x_n\to a$ by the definition of $\inf$.  Now $$f(x_n)>t,$$ since $x_n\in A$.  Taking the limit as $x\to a$ on both sides gives $$M\geq t$$.


问:
Transform line equation with barycentric coordinates
From some testing I have done with Desmos, it appears that straight lines, when every point on that line is transformed using barycentric coordinates from one triangle to another, will stay straight. Not only that, but parallel lines appear to stay parallel.

This leads me to suspect that it is possible to get an equation for the gradient and y intercept of a line, after it has been transformed using barycentric coordinates from one set of three points to another. However, I am not sure where to even start on creating such an equation.
I suppose you could transform two independent points on the line and then calculate the equation from both of those, but surely there is a neater method?

The problem I really need to solve is to find the minimum (and therefore perpendicular) distance between two parallel lines when transformed from one triangle to another (with barycentric coordinates) and so I feel getting the transformed lines' equations is a good place to start, but if there is some alternate method, that would also be great to know.

答:
First, a bit of background so that others also understand what this is about.

We have a non-degenerate triangle (one that has nonzero area, so is not a point or a line) with vertices $\vec{p}_1$, $\vec{p}_2$, and $\vec{p}_3$.  Any point $\vec{p}$ within this triangle can be specified via barycentric coordinates $u$, $v$, $w$:
$$\begin{aligned}
\vec{p} &= u \vec{p}_1 + v \vec{p}_2 + w \vec{p}_3 \\
u + v + w &= 1 \\
0 &\le u \le 1 \\
0 &\le v \le 1 \\
0 &\le w \le 1 \\
\end{aligned}$$
This also implies that $w = 1 - u - v$, i.e.
$$\vec{p} = u \vec{p}_1 + v \vec{p}_2 + (1 - u - v) \vec{p}_3$$
Reordering the terms above we can also write it equivalently as
$$\vec{p} = \vec{p}_3 + u (\vec{p}_1 - \vec{p}_3) + v (\vec{p}_2 - \vec{p}_3)$$
which is equivalent to a linear (but not necessarily orthogonal) coordinate system in $u$,$v$ with origin at $\vec{p}_3$, $u$ axis $\vec{p}_1-\vec{p}_3$, and $v$ axis $\vec{p}_2-\vec{p}_3$.
This, in turn, means that the transformation from barycentric coordinate system $\overline{b} = (u, v)$ (with $0 \le u, v, w \le 1$ and $w = 1 - u - v$) to Euclidean 2D coordinate system can be described as
$$\begin{aligned}
\vec{p} &= \vec{p}_3 + \mathbf{M} \overline{b} \\
\mathbf{M} &= \left[ \begin{matrix} \vec{p}_1 - \vec{p}_3 & \vec{p}_2 - \vec{p}_3 \end{matrix} \right ] \\
\end{aligned}$$
(Note, $\mathbf{M}$ is a matrix with 2 columns, and as many rows as there are dimensions in the Euclidean coordinate system.)

With this, we have accidentally proven that the transform between Euclidean coordinates and barycentric coordinates is an affine transformation, because it is the composition of two functions: a translation and a linear map.

By definition, affine transformations preserve lines and parallelism.

In two dimensions, the inverse mapping is
$$\overline{b} = \mathbf{M}^{-1} \left(\vec{p} - \vec{p}_3\right)$$
where
$$\mathbf{M} = \left[\begin{matrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{matrix}\right] \quad \iff \quad \mathbf{M}^{-1} = \left[\begin{matrix}
\frac{m_{22}}{m_{11} m_{22} - m_{12} m_{21}} &
\frac{- m_{12}}{m_{11} m_{22} - m_{12} m_{21}} \\
\frac{- m_{21}}{m_{11} m_{22} - m_{12} m_{21}} &
\frac{m_{11}}{m_{11} m_{22} - m_{12} m_{21}} \end{matrix}\right]$$
In three dimensions, the easiest is to check which component in $(\vec{p}_1 - \vec{p}_3)\times(\vec{p}_2 - \vec{p}_3)$ is the largest in magnitude, and drop that Euclidean coordinate component, and do the above transform using only the two coordinate axes left.  It is guaranteed to provide exact correct results, if the triangle is nondegenerate.

If the triangle is degenerate, then the common denominator is zero, $m_{11} m_{22} - m_{12} m_{21} = 0$, and no inverse exists.

There are many different ways to describe lines in barycentric coordinates.  Remember, that although there are three coordinates, one of them can always be trivially eliminated ($w = 1 - u - v$), and is usually omitted, especially in computer programs and simulations using barycentric coordinates.  (It can be confusing to see barycentric coordinates with only two components, but one only needs to remember that the third is always related to the two via $w = 1 - u - v$, because full barycentric coordinates sum to 1.)

(If we allow full real range in $u$ and $v$ and $w$, with $w = 1 - u - v$, the barycentric coordinates can cover the full plane the non-degenerate triangle is in.  This means that if we know the triangle has nonzero area – is not a point or a line –, we can safely use $u$, $v$ coordinates anywhere on the plane, not only inside the triangle.  It is then useful to remember that the triangle edges are at $u = 0$, $v = 0$, and $w = 0$ i.e. $u + v = 1$.)

However, affine transforms do not necessarily conserve distance.  Even though the distance between the two lines were to stay constant in Euclidean coordinates, just rotating them can change their distance in barycentric coordinates.  This is because barycentric coordinates are not orthonormal.

Therefore, I would recommend describing each line in the barycentric coordinates using two points. (This means that if you actually have a point and a direction, you can use the point+direction as the second point, even though it may be outside the triangle.)
Then, transform the four points back to Euclidean coordinates.  Convert back to point ($\vec{a}_0$ for one line, $\vec{b}_0$ for the other) and direction (difference between the two points, $\vec{a}_\Delta$ for one line, and $\vec{b}_\Delta$ for the other).

The cosine of the angle $\varphi$ between the two lines is
$$\cos\varphi = \frac{\left\lvert \vec{a}_\Delta \cdot \vec{b}_\Delta \right\rvert}{\left\lVert\vec{a}_\Delta\right\rVert ~ \left\lVert\vec{b}_\Delta\right\rVert}$$
We take the absolute value of the dot product, because we consider line $\vec{a}_0$,$\vec{a}_\Delta$ the same as $\vec{a}_0$,$\vec{a}_\Delta$.
If the cosine is close enough to 1, you can consider them parallel.

When the two lines are parallel, we can solve their distance $d$ by using a temporary variable $t$ so that $\vec{a}_0 + t \vec{a}_\Delta$ is at minimum distance to $\vec{b}_0$.  This occurs when their distance reaches an extremum, and the solution is
$$t = \frac{\vec{a}_\Delta \cdot \left( \vec{b}_0 - \vec{a}_0 \right) }{\left\lVert \vec{a}_\Delta \right\rVert^2}$$
Then,
$$d = \left\lVert \vec{a}_0 + t \vec{a}_\Delta - \vec{b}_0 \right\rVert$$

Depending on how you describe your lines, it is usually possible to combine the above steps, so that fewer actual arithmetic operations are needed as plain reading of the above would indicate.  That is, I didn't write the above to be "optimum" in any way; I tried to make it as easy to understand as possible.

I recommend you use a Computer Algebra System such as free Maxima (wxMaxima is the graphical user interfaced one) available for all operating systems to explore the details.  In particular, if the direction vectors have unit lengths in Euclidean coordinates, you'll find a lot of the math related to lines becomes much, much simpler.  (You'll be able to omit many $\lVert\vec{a}_\Delta\rVert$ and $\lVert\vec{b}_\Delta\rVert$ multipliers and divisors, in particular.)


答:
Let $a=x+y, b=y+z, c=z+x.$

Then $$\pmatrix{a\\b\\c} = \pmatrix{1&1&0\\0&1&1\\1&0&1}\pmatrix{x\\y\\z}$$
and
$$ \pmatrix{x\\y\\z}= \frac{1}{2}\pmatrix{1&-1&1\\1&1&-1\\-1&1&1}\pmatrix{a\\b\\c}$$

Since there is an invertible transformation from one set of vectors to the other, the spans are equivalent.

Title: Are the two spans equal?

Body:
Let $x$,$y$ and $z$ be three linearly independent vectors. Explain, with justification, whether or not
$\operatorname{span}\{x,\,y,\,z\}=\operatorname{span}\{x+y,\,y+z,\,z+x\}$.


问:
Tangent of Ellipse
Given the equation of an ellipse $x^2+4y^2=36$, I am asked to find the equation of the 2 tangent lines that intersect at the point $(12,3)$. I am not sure of what to do after doing the implicit differentiation, I tried forming an equation for the gradient, $(y_1-3)/(x_1-12)$ then equate it to $\mathrm{d}y/\mathrm{d}x$. I'm not sure if I'm approaching it the right way but I can't find an answer.

答:
If $ax^2+by^2=c$ then
$2ax dx+2by dy=0$ so
$y'=dy/dx=-ax/(by)$.

Therefore the equation of the tangent at $(u,v)$ is
$(y-u)/(x-v)=-au/(bv)$.

If this line passes through $(x,y)$,
this gives one equation for $(u,v)$.
The second equation is given by
$(u,v)$ on the ellipse, so
$au^2+bv^2=c$.

Solving these gives $u$ and $v$ in terms of
$x, y, a, b,$ and $c$.

Title: Tangent of Ellipse


问:
Nonlinear system for origin translation
the nonlinear system

$$x'=a+bx+cx^2-x^3$$

by translating the origin, we can eliminate any second-degree terms, such that

$$x'=\mu_1+\mu_2x-x^3$$

how to do this?

答: You set $x=\frac{c}3+y$, then the quadratic terms in the equation for $y$ cancel. This is exactly the same as when you transform a cubic equation into normal form.

Title: Nonlinear system for origin translation


答:
An easy way to understand this (if the required topology is uniform convergence on compacts) is to consider the "derivative" operator $d:H(U) \rightarrow H(U)$. Suppose $H(U)$ is normable. Then $d$ is sequentially continuous so has a bounded operator norm.
In particular, its eigenvalues are bounded.

But, for each $\lambda \in \mathbb{C}$, $\lambda$ is an eigenvalue of $d$ with eigenvector $z \in U \longmapsto e^{\lambda z}$, so we get a contradiction.

Title: Show that $H(U)$ the space of holomorphic functions on $U$ is not normable.

Body:
I'm trying to show the following result:

Let $U \subset \mathbb{C}$ be an open set. Show that the vector space $H(U)=\{f:U \to \mathbb{C} \; \text{analytic}\}$ is not normable.

I know that $H(U)$ is a Frechet space. If we take $A_{k}:=\{z \in U\ : |z| \leq k \; , \; \text{dist}(\partial U,z) \geq \frac{1}{k} \}$ and $p_{k}(f)=\sup_{z \in A_{k}}|f(z)|$ for each $k \in \mathbb{N}$, we have
$\cup_{k}A_{k}=U$ and metric

$$d(f,g)=\sum_{k=1}^{\infty}\frac{1}{2^{k}} \frac{p_{k}(f-g)}{1+p_{k}(f-g)}.$$

I'm not sure how to work with this. Any help would be greatly appreciated.


问:
square complex number sum
Is it possible to demonstrate that

$\sum_{j=0}^{n}x_{j}y_{j}^{*}\sum_{j=0}^{n}x_{j}^{*}y_{j}=|\sum_{j=0}^{n}x_{j}y_{j}^{*}|^{2}$

where * denotes the complex conjugate.

Thanks in advance

答: If you write $z =\sum_{j=0}^{n}x_{j}y_{j}^{*}$, then the assertion is simply that $$zz^*=|z|^2$$ which is, of course, true.

Title: square complex number sum


问:
Composition of two matrix transformations
Consider two matrices acting on points in $(0,1)^2$ in the real plane, $$h_s=\begin{pmatrix}   e^{-e^{s}} & 0 \\   0 & e^{-e^{-s}}   \end{pmatrix}.$$

and $$g_s=\begin{pmatrix}   1-e^{-e^{s}} & 0 \\   0 & e^{-e^{-s}}   \end{pmatrix}.$$

How do you compose these transformations?

I tried to do $h_s \circ g_s$ but got stuck because I haven't dealt with composition of matrices before.

I have done function composition, so I am familiar with that concept.

答:
Treat a point $(x,y)$ as a $2 \times 1$ vector $\begin{bmatrix} x \\ y \end{bmatrix}$.  You may want to start with usual calculations for function compositions

\begin{align}
g_s \left( \begin{bmatrix} x \\ y \end{bmatrix} \right)
&= \begin{pmatrix} 1-e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (1-e^{-e^{s}})x \\ e^{-e^{-s}}y \end{pmatrix}. \tag{1} \label{my1} \\
(h_s \circ g_s) \left( \begin{bmatrix} x \\ y \end{bmatrix} \right)
&= h_s\left( g_s\left( \begin{bmatrix} x \\ y \end{bmatrix} \right) \right) \\
&= h_s\left( \begin{pmatrix} (1-e^{-e^{s}})x \\ e^{-e^{-s}}y \end{pmatrix} \right) \\
&= \begin{pmatrix} e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix} \begin{pmatrix} (1-e^{-e^{s}})x \\ e^{-e^{-s}}y \end{pmatrix} \\
&= \begin{pmatrix} e^{-e^{s}} (1-e^{-e^{s}})x \\ e^{-2e^{-s}}y \end{pmatrix}.
\end{align}

From this, it's clear that $h_s \circ g_s = \begin{pmatrix} e^{-e^{s}}(1-e^{-e^{s}}) & 0 \\ 0 & e^{-2e^{-s}} \end{pmatrix}$.

However, it's preferable to describe $g_s$ and $h_s$ in words.  In this way, you see this directly without writing the vector.  Note that in a matrix product, the row $\begin{pmatrix} 1 & 0 \end{pmatrix}$ on the left means to "pick the first row in the matrix/vector on its right-hand side".  It's easy to see why the first row of the right-hand side in \eqref{my1} is $(1-e^{-e^s})x$.  The second row follows similarly.

With this verbal description of matrix product in mind, you can find the matrix product
$$ \begin{pmatrix} e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix} \begin{pmatrix} 1-e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix} $$
using the matrices associated with $h_s$ and $g_s$.  Note that the question asks for $h_s \circ g_s$, so you have to put
$$\text{matrix for } h_s \cdot \text{matrix for } g_s.$$
Note that the matrices in this question are diagonal matrices, so you can simply do an element-by-element multiplication on the diagonal.


问:
Isomorphism of algebras vs isomorphism of categories of representations
Let $A, B$ be finite-dimensional algebras over a field, so that their categories of modules $A\text{-mod}$ and $B\text{-mod}$ are finite abelian in the sense of EGNO Tensor categories.

Given an isomorphism $f:A \to B$ of algebras, the pullback $f^* : B\text{-mod} \to A\text{-mod}$ is an isomorphism of finite abelian categories. Furthermore, it commutes with the forgetful functors, in the sense that
\begin{align}
    \hom_A(A, f^*V) \cong \hom_B(B, V),
    \quad g \mapsto (b \mapsto b.g(1))
\end{align}
natural in $B$-modules $V$. Here, $A$ and $B$ are the regular modules. (actually, this isomorphism is

I was wondering if we can reverse this statement.
However, I don't see it, and I don't understand where isomorphism vs equivalence plays a role.

So far I did

take $F: B\text{-mod} \to A\text{-mod}$ an equivalence of finite abelian categories

such that there exists $\nu : \hom_A(A, F-) \overset{\sim}{\Rightarrow} \hom_B(B,-)$

We have the vector space isomorphism $FB \cong \hom_A(A,FB) \xrightarrow{\nu_B} \hom_B(B,B) \cong B$.
Note also that $F$ is an equivalence so $A \cong FGA$, where $G$ is a quasi-inverse of $F$.
In particular, then,
\begin{align}
    \hom_B(B,V) \cong \hom_A(A,FV) \cong \hom_A(FGA,FV) \cong \hom_B(GA,V)
\end{align}
whence by Yoneda $A \cong FB$ in $A\text{-mod}$.
In particular, $A \cong B$ as vector spaces.

I feel I'm overlooking something very obvious here. There at least should be some condition on $\nu$; and I guess $F$ should actually do nothing on morphisms?

Any hints would be appreciated.

答:
$\newcommand{\id}{\operatorname{id}}
\newcommand{\tensor}{\otimes}$

This follows  from Lemma 1.6 in Hopf Monads (Bruguieres, Virelizier).
Indeed, if $A$ is an algebra, then $A\otimes -$ is a monad on $\text{Vect}$, and the category of modules over the monad is precisely the category of modules over $A$.

EDIT:
I will now be more explicit.
Without loss we can actually assume that the forgetful functor sends a module to its underlying vector space (this functor is isomorphic to the one I wrote above).
Let's at first not restrict to equivalences/isomorphisms of categories.
So our initial set-up is the following:

$F : B\text{-mod} \to A\text{-mod}$ a functor (+ appropriate adjectives) which commutes with the forgetful functors to $\text{Vect}$.

This already implies two things:

$Ff = f$, i.e. $F$ doesn't do anything on intertwiners.

$F(V,\rho) = (V,\hat{\rho})$ with $\hat{\rho} : A\otimes V\to V$, for a $B$-module $(V,\rho)$. [Note that we denote a module by a pair consisting of a vector space and an action]

Define for any vector space the linear map
\begin{align}
    f_{V} = \widehat{m_B \tensor \id_V}
    \circ \id_A \tensor 1_B \tensor \id_V
    : A \tensor V \to B \tensor V \ ,
\end{align}
where $1_B$ is the unit of $B$.
Note that this is natural in $V$ and thus in fact determined by its value $f=f_k$ on the field, i.e. $f_V = f \tensor \id_V.$

Let $(V,\rho) \in B\text{-mod}$.
Consider the free $B$-module $(B \tensor V, m_B \tensor \id_V)$, where $m_B$ is the multiplication of $B$.
We can consider $\rho$ as a morphism in $B\text{-mod}$, from the free module above to $(V,\rho)$.
Then in particular $F\rho = \rho$ is an $A$-module map, which means
\begin{align}
    F\rho \circ \widehat{m_B \tensor \id_V}
    = \widehat{\rho} \circ \id_A \tensor F\rho \ .
\end{align}
Thus we find for the interplay between $\rho$ and $f_{V}$
\begin{align}
    F\rho \circ f_{V}
    &= F\rho \circ
    \widehat{m_B \tensor \id_V}
    \circ \id_A \tensor 1_B \tensor \id_V
    \\
    &= \widehat{\rho}
    \circ \id_A \tensor F\rho
    \circ \id_A \tensor 1_B \tensor \id_V
    \\
    &=
    \widehat{\rho}
    \ .
\end{align}
That is, we have that $\widehat{\rho} = \rho \circ f_{V} = \rho \circ f \tensor \id_V$ as linear maps.

Finally, the claim that $F = f^*$ follows once one has proved that the following are equivalent for a linear map $f:A \to B$ between algebras:

$f$ is an algebra map

If $(V,\rho)$ is a $B$-module, then $(V, \rho \circ f \tensor \id_V)$ is an $A$-module.

I will leave this as a good exercise (if done correctly, the proof takes less than one A6-sized page)


问:
If $\lim_n(x_{n+1}-x_n)=4$, can the sequence $\{\sin(x_n)\}$ be convergent?
If $(x_n)$ is a sequence such that $\lim_{n\to\infty}(x_{n+1}-x_n)=4$, is it possible that the sequence $(\sin(x_n))$ can be convergent or does it have to be divergent?

The reason why I asked this is because if $0<\lim_{n\to\infty}(x_{n+1}-x_n)<\pi$, then $(\sin(x_n))$ has to be divergent as we can find two subsequences of it such that one is a positive sequence and the other one is a negative sequence. However, if it is $4$ then we are not able to find such two subsequences. That's why I wonder what the answer is to this question.

答:
Let's suppose that $(\sin(x_n))$ is convergent to a limit $l \in \mathbb{R}$. Then you have
$$\sin(x_{n+1})=\sin(x_{n+1}-x_n + x_n)=\sin(x_{n+1}-x_n)\cos(x_n) + \cos(x_{n+1}-x_n)\sin(x_n) \quad \quad (1)$$

$$\sin(x_{n-1})=\sin(x_n - (x_n - x_{n-1}))=\sin(x_n)\cos(x_n-x_{n-1}) - \cos(x_n)\sin(x_n-x_{n-1}) \quad (2)$$

Let $n \rightarrow +\infty$ in $(1)$ and $(2)$ : you get that $(\cos(x_n))$ converges to a limit $l'$ satisfying
$$l=\sin(4)l'+\cos(4)l \quad \quad (1')$$
$$l = l \cos(4) - l'\sin(4) \quad \quad (2')$$

If you sum $(1')$ and $(2')$, you get
$$2l = 2l \cos(4)$$

so $$l=0$$

and $(1')$ gives then $l'=0$, which contradicts the fact that $l^2 + l'^2 = 1$ (by letting $n \rightarrow +\infty$ in $\cos^2(x_n) + \sin^2(x_n)=1$)


问:
proof of $\lim\limits_{x \to a}\sin(x) = \sin(a)$
Prove with definition that $\lim\limits_{x \to a} \sin(x) = \sin(a)$. Is it always true?

答:
Since $\left|\sin x- \sin a\right|=\left|2\sin\left(\frac{x-a}{2}\right)\cos\left(\frac{x+a}{2}\right)\right|\rightarrow 0$ as $\left|x-a\right|\rightarrow 0, $ the desired result follows.

Further, since sine function is continuous, the above limits is always valid.