问:
Expected value of visits state in finite markov chain
I have 4 states {human, dog, cat, floor). Flea can move between each state (always with same probability). In state human flea is dying (end game), in state {cat, dog} flea eats, and on floor flea is starving. Count expected values of meals{dog, cat}, starting from floor and before dying in human.

This is my transition matrix (3 states - Hungry, Meal, Dead)
$$ A = \left( \begin{matrix} 0 & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & 1 \end{matrix} \right) $$

I think about counting expected values of visit of state Meal (starting from Hungry and before die in Dead), but I don't have any idea, how to start

答:
The answer is either $1.5$ or $2$, depending on the interpretation of the problem.

Let $e_f$ be the expected number of future meals if the flea is currently on the floor, $e_c$ the number of future meals if the flea is on the cat, and $e_d$ the number of future meals if it is on the dog.  Clearly $e_c=e_d$.

Under the assumption that the flea always changes state, we have $$\begin{align}
e_f&=\frac23(e_c+1)\\
e_c&=\frac13e_f+\frac13(e_d+1)
\end{align}$$
Solving we get $e_c=\frac54,\ e_f=\frac32$.

Under the assumption that the flea may stay in the same state, we have $$\begin{align}
e_f&=\frac14e_f+\frac12(e_c+1)\\
e_c&=\frac14+\frac12(e_c+1)
\end{align}$$
and we get $e_c=e_f=2$

I've done two simulations, one for each case  In the first case, and the results agree with these.  In an earlier version of this post, I said that the simulation in the first case gave $1.7$, but there must have been some error in the code.  I've rewritten it, and it gives $1.5$.

Here's my python script for the first simulation:

```<code>from random import random

floor, cat, dog, human = range(4)

def test(trials):
    meals  = 0
    for _ in range(trials):
        state = floor
        while state != human:
            r = random()
            if state == floor:
                if r <= 1/3:
                    state = cat
                elif r <= 2/3:
                    state = dog
                else:
                    state = human
            elif state == cat:
                if r <= 1/3:
                    state = floor
                elif r <= 2/3:
                    state = dog
                else:
                    state = human
            elif state == dog:
                if r <= 1/3:
                    state = floor
                elif r <=2/3:
                    state = cat
                else:
                    state = human
            if state in (cat,dog):
                meals += 1
    return meals/trials
</code>```

Here's my python script for the second case:

```<code>from random import random

floor, cat, dog, human = range(4)

def test(trials):
    meals  = 0
    for _ in range(trials):
        state = floor
        while state != human:
            r = random()
            if state == floor:
                if r <= 1/4:
                    state = cat
                elif r <= 1/2:
                    state = dog
                elif r <= 3/4:
                    state = floor
                else:
                    state = human
            elif state == cat:
                if r <= 1/4:
                    state = floor
                elif r <= 1/2:
                    state = dog
                elif r <= 3/4:
                    state = cat
                else:
                    state = human
            elif state == dog:
                if r <= 1/4:
                    state = floor
                elif r <=1/2:
                    state = cat
                elif r <= 3/4:
                    state = dog
                else:
                    state = human
            if state in (cat,dog):
                meals += 1
    return meals/trials

</code>```

Surely we should be able to state this in term of the transition matrix.

EDIT

I haven't had a chance to read it yet, but section 9.2 of these lecture notes addresses the question.

Body:
The answer is either $1.5$ or $2$, depending on the interpretation of the problem.

Let $e_f$ be the expected number of future meals if the flea is currently on the floor, $e_c$ the number of future meals if the flea is on the cat, and $e_d$ the number of future meals if it is on the dog.  Clearly $e_c=e_d$.

Under the assumption that the flea always changes state, we have $$\begin{align}
e_f&=\frac23(e_c+1)\\
e_c&=\frac13e_f+\frac13(e_d+1)
\end{align}$$
Solving we get $e_c=\frac54,\ e_f=\frac32$.

Under the assumption that the flea may stay in the same state, we have $$\begin{align}
e_f&=\frac14e_f+\frac12(e_c+1)\\
e_c&=\frac14+\frac12(e_c+1)
\end{align}$$
and we get $e_c=e_f=2$

I've done two simulations, one for each case  In the first case, and the results agree with these.  In an earlier version of this post, I said that the simulation in the first case gave $1.7$, but there must have been some error in the code.  I've rewritten it, and it gives $1.5$.

Here's my python script for the first simulation:

from random import random

floor, cat, dog, human = range(4)

def test(trials):
    meals  = 0
    for _ in range(trials):
        state = floor
        while state != human:
            r = random()
            if state == floor:
                if r <= 1/3:
                    state = cat
                elif r <= 2/3:
                    state = dog
                else:
                    state = human
            elif state == cat:
                if r <= 1/3:
                    state = floor
                elif r <= 2/3:
                    state = dog
                else:
                    state = human
            elif state == dog:
                if r <= 1/3:
                    state = floor
                elif r <=2/3:
                    state = cat
                else:
                    state = human
            if state in (cat,dog):
                meals += 1
    return meals/trials

Here's my python script for the second case:

from random import random

floor, cat, dog, human = range(4)

def test(trials):
    meals  = 0
    for _ in range(trials):
        state = floor
        while state != human:
            r = random()
            if state == floor:
                if r <= 1/4:
                    state = cat
                elif r <= 1/2:
                    state = dog
                elif r <= 3/4:
                    state = floor
                else:
                    state = human
            elif state == cat:
                if r <= 1/4:
                    state = floor
                elif r <= 1/2:
                    state = dog
                elif r <= 3/4:
                    state = cat
                else:
                    state = human
            elif state == dog:
                if r <= 1/4:
                    state = floor
                elif r <=1/2:
                    state = cat
                elif r <= 3/4:
                    state = dog
                else:
                    state = human
            if state in (cat,dog):
                meals += 1
    return meals/trials

Surely we should be able to state this in term of the transition matrix.

EDIT

I haven't had a chance to read it yet, but section 9.2 of these lecture notes addresses the question.


问:
How can I calculate using parallel axis theorem the moment of inertia of a cylinder relative to a main body?
The problem is the following. There is large body A to which we know the center of gravity coordinates relative to the center of gravity of a shape (that is part of A), specifically a uniform density cylinder, which we call B. Below is a picture of this statement, this is not to scale, and I apologize if the axis look out of place.

The new axis $x',y',z'$ are parallel in their respective planes to the original $x,y,z$ axis. In other words, the new center of gravity merely follows a translational change, and not a rotational one (no yaw, pitch, or roll angle). Now, my job is to calculate the Inertia Tensor of the body A, knowing the inertia tensor without the addition of this body B, but also knowing the center of gravity of A including body B, in the reference frame $x'y'z'$.

This seems fairly straightforward, but I am having some doubts. Here is my current work.

$$I_{xx}=I_{xx_A}+I_{xx_B}+m_B(d_x)^2$$

$$I_{yy}=I_{yy_A}+I_{yy_B}+m_B(d_y)^2$$

$$I_{zz}=I_{zz_A}+I_{zz_B}+m_B(d_z)^2$$

To simplify things, let us assume that the distance $d_y$ is zero. This means the cylinder is offset from the center of gravity of the entire larger body A in only one plane. This also means our inertia tensor will be axisymmetric but will also be a diagonal (we also assume that the body A has a diagonal tensor with and without B).

The result is then:

$$I_{xx}=I_{xx_A}+I_{xx_B}+m_B(d_x)^2$$

$$I_{yy}=I_{yy_A}+I_{yy_B}$$

$$I_{zz}=I_{zz_A}+I_{zz_B}+m_B(d_z)^2$$

And our inertia tensor is $[I_{xx}\ 0\ 0;0\ I_{yy}\ 0;0\  0\ I_{zz}]$

Background:

The purpose of doing this is for determining the moment of inertia of a launch vehicle. This cylinder approximates the shapes of one of the engines. To me, the problem seems fairly straightforward. However, when I look at an example in my book, they seem to be doing a completely different calculation. Speficially, the $md^2$ part is written as $m*(l-lx-x_{cg})$ where $m$ is the mass of the entire launch vehicle ($m_A$ in this example), $lx$ is the height of the center of gravity location relative to the bottom most part of the launcher (the part that touches the ground, in this case the bottom of our cylinder), and $x_{cg}$ is the center of gravity of the entire launch vehicle (which we know).

答:
The cylinder $c$ has a moment of inertia w.r.t. its centre of gravity $G$
$$I_c^{(G)} = \frac{m}{12} \begin{bmatrix}
    6r^2 & 0 & 0\\
    0 & 3r^2+h^2 & 0 \\
    0 & 0 & 3r^3+h^2
  \end{bmatrix}.$$
(Notice, however, that textbooks usually define x and z in reverse.)

Apply the parallel axis theorem to find the moment of inertia of the cylinder w.r.t the axes of entity $A$
$$I_c^{(A)} = I_c^{(G)} + m \operatorname{veck}^{-T}\!\!\left(r^{GA}\right) \operatorname{veck}^{-1}\!\!\left(r^{GA}\right)$$
with the inverse vectorise skew operator
$$\operatorname{veck}^{-1} X = \operatorname{veck}^{-1}\left(\left[x\quad y\quad z\right]^T\right) =   \begin{bmatrix}
    0 & -z & y \\
    z & 0 & -x \\
    -y & x & 0
  \end{bmatrix}.$$

Inserting the distance between $G$ and $A$
$$r^{GA} = \begin{bmatrix}
    r_x \\
    0 \\
    r_z
  \end{bmatrix},$$
you obtain
$$I_c^{(A)} = I_c^{(G)} + m \begin{bmatrix}
    r^2_z & 0 &-r_x r_z\\
    0 & r^2_x + r^2_z & 0 \\
    -r_x r_z & 0 & r^2_x
  \end{bmatrix}.$$

ContentLicense: CC BY-SA 4.0


问:
Why the group $\langle x,y\mid x^3, y^3, yxyxy\rangle$ is not trivial?
This comes from Artin Second Edition, page 219. Artin defined  $G=\langle x,y\mid x^3, y^3, yxyxy\rangle$ , and uses the Todd-Coxeter Algorithm to show that the subgroup $h=\langle y\rangle$  has index 1, and therefore he gets $G=H$ is the cyclic group of order 3.

My question is: why $\langle y\rangle$ is not trivial? The Todd-Coxeter Algorithm gives no extra information about $y$  except it acts trivially on $\langle y\rangle$ as any other elements in $G$ do. So how do we know that $y\neq 1$?

Many thanks!

答:
The group $G$ is cyclic of order $3$, generated by the element $y$.

Here are two proofs. Both are pretty similar, the main difference is how you conclude that the group is non-trivial (implicit in the first, explicit in the second).

First proof: Presentations. Add in the new generator $z:=yx$, and then remove $x$ via the relevant Tietze transformation to get:
$$\langle y, z\mid (y^{-1}z)^3, y^3, z^2y\rangle.$$
We then see that the group is cyclic, generated by $z$ (as $y=z^{-2}$). So lets find the relevant presentation:
$$\langle z\mid (z^3)^3, (z^2)^3\rangle=\langle z\mid z^9, z^6\rangle.$$
This simplifies to:
$$\langle z\mid z^3\rangle.$$
Then it is generated by $y$ as $z=yx$ is a generator, and therefore so is $y=(xy)^{-2}$.

Second proof: Use the abelianisation. By the relation $yxyxy$, we see that $y^{-1}\in\langle yx\rangle$, and so $x, y\in\langle yx\rangle$. Therefore, $yx$ generates our group $G=\langle x, y\rangle$. This means that $G$ is abelian, so $(yx)^3=y^3x^3=1$. Therefore, $G$ is either cyclic of order three or trivial. As the comments to the question point out, the map which kills $x$ has non-trivial image, and so $G$ is non-trivial. Therefore, $G$ is cyclic of order $3$, and generated by $yx$. Conclude that $G=\langle y\rangle$ as in the first proof (or by noting that the map which kills $x$ much be an isomorphism, so $x$ is trivial and so $yx=y$).


问:
Practical Uses for the Tensor Product
From reading a bit of chapter section 18 in Differential Geometry, by Loring W. Tu, my understanding of tensor products is that they allow one to represent multilinear maps with linear maps.

I was simply curious whether there is any practical use for the tensor product (perhaps in computer science or physics, etc.). Are there other interpretations of tensor products?

答:
If $\varepsilon^1,\varepsilon^2$ are the projectors on $\mathbb R^2$, that is for any $v=(v^1,v^2)^{\top}$ then $\varepsilon^i(v)=v^i$, so the tensor product
$$\varepsilon^1\wedge\varepsilon^2=\varepsilon^1\otimes\varepsilon^2-\varepsilon^2\otimes\varepsilon^1,$$
satisfies, for a pair of vectors
$$\varepsilon^1\wedge\varepsilon^2(v,w)=v^1w^2-w^1v^2,$$
which is the determinant of the data given by $v,w$ and at the same time the area of the parallelogram that vector $v,w$ spawn.

Title: Practical Uses for the Tensor Product


问:
Number of combinations
You are given K prime numbers, bigger than 6, find the number of different number that can be made of those prime numbers(using 1 number, 2 numbers ..., k numbers).

Obviously you need to get the number of combiantions that you'll get using only 1 digit, 2 digits and so on and at the end add them all. It's easy, but when one or more digits appear more than once I encounter a problem.

Here's example:

You are given:

2 2 3 5 7 11 13

All the combinations are of two digits are: (2,2), (2,3), (2,5), (2,7), (2,11), (2,13), (3,5), (3,7), (3,11), (3,13), (5,7), (5,11), (5,13), (7,11), (7,13), (11,13).

There are 16 combinations, but using the formula for combinations i get 21, because it counts the (2,3), using the first 2 and (2,3) using the second 2 as different combinations.

I see a pattern there, that I need to find the number of combiantions and then subtract the number of combinations with the number that appear more than once and the prime number that are different of that number.. In this case there are 5 such combinations and 21-5=16, gives the right result.

But what when the number of same prime numbers is bigger?. For ex:

2, 2, 2, 3, 3, 5, 7, 7, 11, 13

答:
Your problem is in that you are trying to count the number that can be made with 1 prime, 2 primes, 3 primes, etc. separately.  The computation becomes simple when you stop making this distinction.

In your example $2,2,2,3,3,5,7,7,11,13$, the number can have $0,1,2,$ or $3$ twos, $0,1,$ or $2$ threes, etc., so the total number of possible products is
$$
4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 2 - 1 = 287
$$
The $-1$ is to subtract the case of multiplying no primes together, i.e. the $1$ case.  Each possibility is unique by the fundamental theorem of arithmetic.  And as long as you don't actually need the specific values for 2 primes, 3 primes, etc. this method is sufficient.

ContentLicense: CC BY-SA 3.0

Title: Number of combinations


问:
Show that convex hull of vector states is weak$^*$-dense
Let $A\subseteq B(H)$ be a $C^*$-algebra. If necessary, we may assume $A$ is a von Neumann algebra or/and that $A$ acts non-degenerately on $H$. Consider the vector state
$$\omega_{\xi}: A \to \mathbb{C}: a \mapsto \langle a \xi, \xi\rangle$$
on $A$. Let $C$ be the weak$^*$-closure of the convex hull of $\{\omega_\xi: \|\xi\|=1\}$. Is it true that $C = A^*?$

Attempt: Suppose to the contrary that $C\subsetneq A^*$. Choose $\omega \in A^*\setminus C$. By the geometric Hahn-Banach theorem, there is $\lambda \in \mathbb{R}$ and $m \in M$ with
$$\Re \omega_{\xi}(m) < \lambda < \Re \omega(m)$$
for all $\xi \in H$. Is it possible to obtain a contradiction from here?

答:
The result is not true in general. Take for example an infinite dimensional, separable Hilbert space $\mathcal{H}$ and let $A=\mathcal{K(H)}$, the compact operators on $\mathcal{H}$. Note that $\mathcal{K(H)}\subset\mathcal{B(H)}$ is a non-degenerate representation. It is true that the pure states of $\mathcal{K(H)}$ are precisely the vector states $\omega_\xi$, where $\xi$ runs over the unit vectors on $\mathcal{H}$. A proof of this can be found in Murphy's book, example 5.1.1.

Now by the Krein-Milman theorem, we know that the weak-* closed convex hull of $\text{PS}(A)\cup\{0\}$ (where $\text{PS}(A)$ denotes the pure states on $A$) is equal to the set of positive linear functionals on $A$ of norm at most $1$. This is of course a proper subset of the dual space $A^*$.

But, if we go as far away as we can from the compacts, Glimm's lemma can give something relevant to what you guessed, (but not exactly $C=A^*$). Check out Brown-Ozawa, Chapter 1 for a statement and a very brief (and difficult imo!) proof of Glimm's lemma:

Under the hypothesis that $A\cap\mathcal{K(H)}=0$, each state on $A$ can be approximated in the weak-* topology by a net of vector states. Moreover, if $A$ is separable, then each state of $A$ can be approximated in the weak-* topology by a sequence of vector states $\omega_{\xi_n}$ so that $(\xi_n)$ is an orthonormal sequence. Now combining this with the fact that $A^*$ is the linear span of the states $S(A)$, gives something close to what you are describing here.

Edit

As (definitely rightfully) requested in the comments by QuantumSpace, I am showing that the weak-star closure of $\{\omega_\xi\}_\xi$ is equal to the state space, when the algebra is unital and non-degenerately represented. This is based on the answer of Ruy on this post. The "standard exercise" I am referring to can be proved easily by using a beautiful result of linear algebra: see this post.

Let $A\subset B(H)$ be a unital $C^*$-algebra that is non-degenerately represented, i.e. $1_A=\text{id}_H$. Then the set $F=\{\omega_\xi:\xi\in H, \|\xi\|=1\}$ is weak-star dense in the state space $S(A)$.

Of course $F\subset S(A)$ and let's take its weak-star closed convex hull, which I will call $E$, which is again a subset of $S(A)$, since the weak-star limit of states is a state, since $A$ is unital. Now assume that $\tau\in S(A)\setminus E$. By the Hahn-Banach separation theorem we can find a weak-star continuous functional $\phi:A^*\to\mathbb{C}$ and a real number $t$ so that
$$\text{Re}(\phi(\rho))<t<\text{Re}(\phi(\tau))$$
for all $\rho\in E$. It is a standard exercise that every weak-star continuous functional on $A^*$ is an evaluation functional, so $\phi=\text{ev}_a$ for some $a\in A$. By writing $a=x+iy$ for $x,y\in A_{sa}$ we have that $\phi(\sigma)=\text{ev}_a(\sigma)=\sigma(a)=\sigma(x)+i\sigma(y)$, so the real part of $\phi(\sigma)$ is equal to $\sigma(x)$ for all $\sigma\geq0$ in $A^*$. The inequality thus becomes
$$\rho(x)<t<\tau(x) $$
for all $\rho\in E$. In particular we have that
$$\langle x\xi,\xi\rangle_H<t=\langle t\cdot\text{id}_{H}\xi,\xi\rangle$$
for all unit vectors $\xi$, so, if $\eta\in H$ is arbitrary, then
$$\frac{1}{\|\eta\|^2}\langle x\eta,\eta\rangle<\frac{1}{\|\eta\|^2}\langle t\cdot\text{id}_H\eta,\eta\rangle$$
so we can wipe out the $1/\|\eta\|^2$ factor and thus $t\cdot\text{id}_{H}\geq x$ (as operators, i.e. this inequality takes place in $A$). But now $\tau$ is a state, so $\tau(x)\leq\tau(t\cdot\text{id}_H)=t\cdot\tau(1_A)=t$, a contradiction.


问:
Prove that $T$ is compact $\iff$ dim ran $T < \infty$
If we have the linear operator $T : X \rightarrow X$ is bounded and satisfies $T^2 = T$.

Then I have to prove that $T$ is compact $\iff$ dim ran $T < \infty$.

For $\leftarrow$ I was thinking of:

We have that $T$ has finite rank, then Im$T$ is a finite-dimensional normed space. Furthermore, for any bounded sequence $\{ x_n \}$ in $X$, the sequence $\{ T x_n \}$ is bounded in Im$T$, and since Bolzano-Weierstrass theorem stated that each bounded sequenc has a convergent subsequence. I think we can conclude that $T$ is compact.

But I could use some help for $\rightarrow$.

Thanks in advance.

答:
If $T^2=T$, then $\left. T\right\rvert_{\operatorname{ran}T}:(\operatorname{ran}T,\lVert \cdot\rVert)\to (\operatorname{ran}T,\lVert\cdot\rVert)$ is a bijective isometry and therefore, by Riesz lemma, the set $T\left(B_X(0,1)\cap\operatorname{ran}T\right)=T\left(B_{\operatorname{ran}T}(0,1)\right)=B_{\operatorname{ran}T}(0,1)$ won't be totally bounded unless $\dim\operatorname{ran}T<\aleph_0$.

"Continuous of finite rank implies compact" is a standard result that you may find in any book: namely, you may use your favourite formulation of the fact that in $\Bbb R^n$ totally bounded subsets and bounded subsets are the same, and the image of the ball by a continuous operator is bounded.


问:
A different way to calculate $\int_0^\infty \tfrac{1}{1+x^n} dx$.
Most of you might have stumbled upon this integral, when studying complex analysis, namely via contour integration, but today I asked myself if there is a different way to calculate it. I'm almost sure that I'm not the first one to come up with this technique, but I wanted to share it with you and would like to know if anyone of you knows another approach.

EDIT
I made a mistake regarding the format of this post, so I posted my own approach as an answer now.

答:
Here are two approaches.

Beta Function and Euler's Reflection Formula
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{1+x^n}
&=\frac1n\int_0^\infty\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x\tag{1a}\\
&=\frac1n\,\Gamma\!\left(\frac1n\right)\Gamma\!\left(1-\frac1n\right)\tag{1b}\\[3pt]
&=\frac\pi n\csc\left(\frac\pi n\right)\tag{1c}
\end{align}
$$
Explanation:

$\text{(1a)}$: substitute $x\mapsto x^{1/n}$

$\text{(1b)}$: Beta Integral

$\text{(1c)}$: Euler's Reflection Formula

Applying Elementary (Completely Real) Tools

$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{1+x^n}
&=\frac1n\int_0^\infty\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x\tag{2a}\\
&=\frac1n\int_0^1\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x+\frac1n\int_0^1\frac{x^{-\frac1n}}{1+x}\,\mathrm{d}x\tag{2b}\\
&=\frac1n\int_0^1\sum_{k=0}^\infty(-1)^kx^{k+\frac1n-1}\,\mathrm{d}x+\frac1n\int_0^1\sum_{k=0}^\infty(-1)^kx^{k-\frac1n}\,\mathrm{d}x\tag{2c}\\
&=\frac1n\sum_{k=0}^\infty\frac{(-1)^k}{k+\frac1n}+\frac1n\sum_{k=0}^\infty\frac{(-1)^k}{k-\frac1n+1}\tag{2d}\\
&=\frac1n\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\frac1n}\tag{2e}\\[3pt]
&=\frac\pi{n}\csc\left(\frac\pi{n}\right)\tag{2f}
\end{align}
$$
Explanation:

$\text{(2a)}$: substitute $x\mapsto x^{1/n}$

$\text{(2b)}$: break the domain of integration into $[0,1]$ and $[1,\infty)$

$\phantom{\text{(2b):}}$ substitute $x\mapsto1/x$ on $[1,\infty)$

$\text{(2c)}$: Taylor series for $\frac1{1+x}$

$\text{(2d)}$: integrate

$\text{(2e)}$: substitute $k\mapsto-k-1$ in the right-hand sum

$\text{(2f)}$: apply $(8)$ from this answer


问:
How does $m_2g = \frac{3gm_2}{4} + \frac{9gm}{2}$ lead to $m_2 = \frac{9/2m } {1 - 3/4} = 18m$?
$$m_2g = \frac{3gm_2}{4} + \frac{9gm}{2}$$

Then, my teacher wrote that

$$m_2 = \frac{9/2m } {1 - 3/4} = 18m$$

I want to know how did my teacher derive this expression . I am getting it at all.

答:
$$m_{2}g=\frac{3gm_{2}}{4}+\frac{9gm}{2}$$

Subtract $\frac{3gm_{2}}{4}$ from both sides of equation

$$m_{2}g-\frac{3gm_{2}}{4}=\frac{9gm}{2}$$

$$\implies m_{2}g\left(1-\frac{3}{4}\right)=\frac{9gm}{2}$$

$$\implies m_{2}g\left(\frac{4-3}{4}\right)=\frac{9gm}{2}$$

$$\implies \frac{m_{2}g}{4}=\frac{9gm}{2}$$

Multiply by $4$ on both sides of equation

$$m_{2}g=\frac{9gm}{2}×4$$

$$\implies m_{2}g=18gm$$


问:
First Course in Linear Algebra book suggestions?
I'm starting maths degree this September and hoping to do some reading on Linear Algebra before I start. What are some good introductory books?

Is there something that starts from the very beginning but also covers stuff in depth? (Is there a book that will be useful for both 1st year and 2nd year linear algebra modules?)

It would also be great if the book has lots of exercises (proof questions!) and solutions given as well.

Thanks in advance.

答: I really liked Linear Algebra Done Right by Sheldon Axler. It starts from the very beginning and contains mostly proof questions; but unfortunately there are no solutions.</p>


问:
On the centres of the dihedral groups
In an proof that I recently read, the following 'fact' is used, where $D_{2n}$ denotes the dihedral group of order $2n$:

If $n$ is even, then $D_{2n} \cong C_2 \times D_n$.

The (short) given justification is that the centre $Z(D_{2n}) \cong C_2$, whenever $n$ is even, and is trivial provided that $n$ is odd.

However, here is a result that a friend of mine found in the litterature which contradicts the previous one.

Suppose that 4 divides $n$. Then $D_{2n}$ is not isomorphic to $C_2 \times D_n$.

Proof: Suppose otherwise. We know that $Z(D_{2n}) \cong C_2$. By assumption, $D_{2n} \cong C_2 \times D_n$, therefore $Z(D_{2n}) \cong Z(C_2 \times D_{n})$. Since in a direct product $A \times B$, the groups $A$ and $B$ commute, we obtain that $Z(C_2 \times D_{n}) = Z(C_2) \times Z(D_n) \cong C_2 \times C_2$, a contradiction. QED.

Now comes my first question : Is the above proof correct?

Second question : Given a finite group $G$ whose centre is not trivial, does it always exist a group $H$ such that $G \cong Z(G) \times H$?

If so, then given $n$ a multiple of 4, the dihedral group $D_{2n}$ could be written as a direct product $C_2 \times H$. Here comes my third question : What should be $H$ since we know that it cannot have nontrivial centre (and, in particular, since $H$ cannot be $D_n$)?

答:
Question 1. The proof is a correct proof that the original assertion (that for $n$ even you have $D_{2n}\cong C_2\times D_n$) is false. Indeed, you can simply do it with $n=4$: $D_4$ is abelian, the Klein $4$-group; $D_8$ is not abelian, but $C_2 \times D_4$ is abelian, so $D_8$ cannot be isomorphic to $C_2\times D_4$. The assertion about $D_{2n}$ is false.

Question 2. No, it's not true in general. Take a nonabelian group of order $p^3$; then $Z(G)\cong C_p$. If you had $G\cong Z(G)\times H$, then $H$ would be of order $p^2$, hence abelian, so $G$ would be abelian (a product of two abelian groups).

Question 3. Moot.

Title: On the centres of the dihedral groups


问:
What is the difference between an implicit ordinary differential equation and a differential algebraic equation?
I'm rather confused on this particular point. What is the difference between an implicit ordinary differential equation of the form:

```<code>x' = f(x',x,t);
</code>```

and a differential algebraic equation of the form:

```<code>f(x',x,t) = 0;
</code>```

I've read some documentation on them online, and while some places mention that implicit ODEs are a special class of DAEs, they're not very clear on the point. Also, is there any specific consideration that needs to be made in regards to numerical solutions of such problems?

答: See the book Computer Methods for Ordinary Differential Equations and Differential-Algebraic Equations by Ascher and Petzold.</p>


问:
Do quotienting and contraction commute?
$\DeclareMathOperator{\Spec}{Spec}$

Let $A \subset B$ be an integral extension of rings. Consider the map $f:\Spec B \to \Spec A$ by $f(Q) = Q^c = Q \cap A$. Show that $f$ is a closed map.

My idea was to consider a typical closed set $V(I)$ of $\Spec B$ and show that $f[V(I)] = V(I^c)$.

The inclusion $(\subset)$ was easy to show. It is the reverse direction that I am struggling with.

Here's what I have done: Consider $P \in V(I^c)$. Thus, $P \in \Spec A$ and $P \supset I^c$.

Note that the extension $A/I^c \hookrightarrow B/I$ is integral and $P/I^c$ is a prime in $A/I^c$.

Then, by the lying over theorem, it follows that there exists a prime $Q/I \in \Spec(B/I)$ contracting to $P/I^c$.

Thus, we have gotten hold of a prime ideal $Q \in V(I).$ I am now not sure how to prove that $Q \cap A = P$. (Is this even true?)

An alternate approach could have been to directly use the lying over theorem for $P$ to conclude that there exists a prime $Q \in \Spec B$ lying over $P$. But then it wasn't clear why $Q \supset I$.

答:
Recall that the inclusion $i : R/I^c \hookrightarrow S/I$ is given as $i(r + I^c) = r + I$. Regarding $R/I^c$ as a subset of $S/I$ under this map, we show that $$(Q/I) \cap (R/I^c) = (Q \cap R)/I^c.$$
(You already know that the left set is $P/I^c$.) Note that $Q \cap R$ indeed contains $I^c = I \cap R.$

$(\subset)$ Let $x \in (Q/I) \cap (R/I^c)$. Then $x = r + I^c = q + I$ for some $r \in R$ and $q \in Q$. However, we identify $r + I^c$ with $r + I$ in $S/I$. Thus, we see that $r + I = q + I$ in $S/I$ and thus, $r - q \in I \subset Q$.

Thus, we see that $r \in q + Q = Q$. Since $r \in R$ to begin with, we conclude that $r \in Q \cap R$ and thus, $x = r + I^c \in (Q \cap R)/I^c$.

$(\supset)$ Let $x \in (Q \cap R)/I^c$.

Then, $x = y + I^c$ for some $y \in Q \cap R$. Under the identification, we have $$x = y + I^c = y + I.$$
The second expression shows that $x \in R/I^c$ and the third that $x \in Q/I$.

Conclusion. The main issue here is getting around the (abuse of) notation and then concluding that $$P/I^c = (Q/I) \cap (R/I^c) = (Q \cap R)/I^c.$$

Now, $P$ and $Q \cap R$ are ideals containing $I^c$. By the ideal correspondence, we see that $P = Q \cap R$.

Title: Do quotienting and contraction commute?


答:
For $\rightarrow:$ let $(x_n)$ be a bounded sequence in $ker(I-T).$ Hence $Tx_n=x_n.$ Since $T$ is compact, $(Tx_n)=(x_n)$ contains a convergent subsequence. Therefore $ker(I-T)$ has the Bolzano-Weierstrass property. Thus $ \dim ker(I-T) < \infty.$

Since $T^2=T$, we have $ran( T)=ker(I-T).$ This gives $ \dim ran(T) < \infty.$

Title: Prove that $T$ is compact $\iff$ dim ran $T < \infty$

Body:
If we have the linear operator $T : X \rightarrow X$ is bounded and satisfies $T^2 = T$.

Then I have to prove that $T$ is compact $\iff$ dim ran $T < \infty$.

For $\leftarrow$ I was thinking of:

We have that $T$ has finite rank, then Im$T$ is a finite-dimensional normed space. Furthermore, for any bounded sequence $\{ x_n \}$ in $X$, the sequence $\{ T x_n \}$ is bounded in Im$T$, and since Bolzano-Weierstrass theorem stated that each bounded sequenc has a convergent subsequence. I think we can conclude that $T$ is compact.

But I could use some help for $\rightarrow$.

Thanks in advance.


问:
Definition of Complex Convolution?
The convolution of two real value functions f and g is defined to be

$$(f \ast g)(t) = \int_{-\infty}^\infty f(\tau)g(t-\tau)\,d\tau$$
If I extend this idea into higher dimension (Let's say in a unit disc or maybe $
\mathbb{C}$, a set of complex number), can we defined a convolution of two complex value function?

My guess is going to be a double integral in some way. Could you please precisely state it?

答:
I believe the result of Fourier convolution defined in formula (1) below is only valid for $t\in\mathbb{R}$, whereas the result of a Mellin convolution such as the two defined in formulas (2) and (3) below can also be valid for at least a subset of the complexes such as $\Re(s)>0$.

(1) $\quad (f \ast g)(t)=\int\limits_{-\infty}^\infty f(\tau)\,g(t-\tau)\,d\tau$

(2) $\quad (f \ast_{M_1} g)(t)=\int\limits_0^\infty f(x)\,g\left(\frac{y}{x}\right) \frac{dx}{x}$

(3) $\quad (f \ast_{M_2} g)(t)=\int\limits_0^\infty f(x)\,g(y\,x)\,dx$

For examples, see formulas (52) - (70) in this answer I posted to one of my own questions on Math StackExchange.

Title: Definition of Complex Convolution?


答: Lang's Linear Algebra starts at the very beginning and covers stuff in depth.

Title: First Course in Linear Algebra book suggestions?

Body:
I'm starting maths degree this September and hoping to do some reading on Linear Algebra before I start. What are some good introductory books?

Is there something that starts from the very beginning but also covers stuff in depth? (Is there a book that will be useful for both 1st year and 2nd year linear algebra modules?)

It would also be great if the book has lots of exercises (proof questions!) and solutions given as well.

Thanks in advance.


问:
Stability of the modified predator-prey equations?
So I'm facing a modified predator-prey model:

$$ \dfrac{dU}{dt} = aU\left(1 - \dfrac{U}{K_U}\right) - cUV,$$

$$ \dfrac{dV}{dt} = ceUV - bV,$$ where $a, b, c, e > 0$.

To determine the stability, I found out the equilibria at $(0, 0)$, $(K_U, 0)$ and $(b/(ce), a/c - (ab)/(ce^2 K_U))$. The 1st, trivial one should be unstable like the original (without the $1 - U/K_U$ factor), so my concerns lie on the other 2.

If I understand correctly, the stability of the 2nd equilibrium should depend on whether $ceK_U$ or $b$ is larger. If the former $\leq$ latter, it is still stable. Did I misunderstand anything?

For the last one, I'm very confused about how to handle the eigenvalue equation $$\lambda(ab + ceK_U \lambda) + (b/e)(ae - (ab)/(cK_U)) = 0.$$ The usual quadratic formula gives another quadratic equation about $ceK_U$ within the square root (let's denote it as $\sigma$). It is still okay when $\sigma$ is real -- when $\sigma \leq a^2 b^2$, the node is stable. Since the square root looks complicated, (a) should I not use quadratic formula to determine the eigenvalues? (b) When $\sigma$ is complex, what happens to the equation system? For example, does the solution attract?

Thanks in advance.

答:
No.  When $ceK_U=b$, the equation gives $\dot V=(U-K_U)V$ so for a small time there will be exponential growth in number of predators if $U>K_U$.  So the equilibrium is not stable.

Only the real part matters.  The imaginary part of eigenvalue measures the rate of trajectories rotating as it spirals in (or out) near the equilibrium so doesn't affect stability.  If $ceK_U>b$ this equilibrium is stable since it is either a pair of complex conjugate eigenvalues with negative real part $-ab/(2ceK_U)$, or two negative reals.  If $ceK_U<b$ then there is a negative and a positive eigenvalue, so unstable.  Equality means you will need to analyse what happens along the degenerate direction.


问:
Isn't it partial derivative?
Now I'm studying financial math.

This material gets me in trouble What I learned in partial derivative is like red written memo in the picture.

However I couldn't derive the result using the way I learned.
So I guess It little bit different with partial derivative. (Is it outcome from stochastic derivative..?)

To derive those equation in the picture, ( under the line : 'Changing variables from S_1, S_2 to ξ we must use the following for the derivatives.' )
What study-material I should study?

Please give me some hint for those derivation.
Thanks.

答:
I'll give a quick example to illustrate what the author is talking about. This isn't very formal, but let's suppose we have some expression
$$f(S_1,S_2)=\frac{S_1}{S_2}+\frac{S_2}{S_1}$$
Let's say we use a change of variable $\xi=S_1/S_2$. Then
$$f(S_1,S_2)=\frac{S_1}{S_2}+\frac{S_2}{S_1}=g(\xi)=\xi+\frac{1}{\xi}$$
Now notice that
$$(\partial_{S_1}f)(S_1,S_2)=\frac{1}{S_2}+\frac{-S_2}{{S_1}^2}$$
Whereas
$$(\partial_\xi g)(\xi)=1+\frac{-1}{\xi^2}$$
$$\partial_\xi g\left(\frac{S_1}{S_2}\right)=1-\frac{{S_2}^2}{{S_1}^2}=S_2\cdot\left(\frac{1}{S_2}+\frac{-S_2}{{S_1}^2}\right)=S_2\cdot\partial_{S_1}f(S_1,S_2)$$
Now since $$g\left(\frac{S_1}{S_2}\right)=f(S_1,S_2)$$
We have
$$\partial_\xi=S_2\partial_{S_1}$$
$$\partial_{S_1}=\frac{1}{S_2}\partial_{\xi}$$
This works regardless of the functions $f,g$ that we use as long as they are differentiable. I just think that having a concrete example helps in understanding.

Title: Isn't it partial derivative?


答:
Let $\mu$ be a complex eigenvalue of $B$. Because $B$ is antisymmetric, then $\mu \in i\mathbb{R}$, so $\mu^2$ is a nonpositive real number. But $\mu^2$ is eigenvalue of $B^2$, so of $A^2$, but the eigenvalues of $A^2$ are all nonnegative since $A$ is symmetric. So $\mu=0$.
You get $B=0$, so $A^2=0$, so $A$ is real symmetric and nilpotent, so $A=0$.

Title: Let $A$ is a real symmetric matrix, $B$ is a real antisymmetric matrix, $A^2 = B^2$, prove $A = B = 0$.

Body:
Let $\boldsymbol{A}$ is a real symmetric matrix, $\boldsymbol{B}$ is a real antisymmetric matrix, $\boldsymbol{A}^2 = \boldsymbol{B}^2$, prove $\boldsymbol{A} = \boldsymbol{B} = \boldsymbol{0}$.

I tried the second-order matrix.
Let $\boldsymbol{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{bmatrix}$,
$\boldsymbol{B} = \begin{bmatrix} b_{11} & b_{12} \\ {-b_{12}} & b_{22} \end{bmatrix}$, $a_{ij}, b_{ij} \in \mathbb{R}$.

\begin{align}
\boldsymbol{A}^2 &= \begin{bmatrix} {a_{11}^2 + a_{12}^2} & {a_{11}a_{12} + a_{12}a_{22}} \\ {a_{11}a_{12} + a_{12}a_{22}} & {a_{12}^2 + a_{22}^2} \end{bmatrix} \\
\boldsymbol{B}^2 &= \begin{bmatrix} {b_{11}^2 - b_{12}^2} & {b_{11}b_{12} + b_{12}b_{22}} \\ {-b_{11}b_{12} - b_{12}b_{22}} & {-b_{12}^2 + b_{22}^2} \end{bmatrix}
\end{align}

Because $\boldsymbol{A}^2 = \boldsymbol{B}^2$, I got
\begin{align}
a_{11}^2 + a_{12}^2 &= b_{11}^2 - b_{12}^2 \\
a_{11}a_{12} + a_{12}a_{22} &= b_{11}b_{12} + b_{12}b_{22} \\
a_{11}a_{12} + a_{12}a_{22} &= -b_{11}b_{12} - b_{12}b_{22} \\
a_{12}^2 + b_{22}^2 &= -b_{12}^2 + b_{22}^2
\end{align}

Hence,
\begin{align}
a_{11}a_{12}+a_{12}a_{22}=b_{11}b_{12}+b_{12}b_{22}=0
\end{align}

I lost my momentum here.


问:
find the coordinates of the vector $(a,b,c)?$
Suppose $a_1 = (1,0,-i) $ , $a_2 =(1+i , 1-i,1)$ , $a_3 =(i,i,i)$ in $\mathbb{C}^3$ . What are the coordinates  of the vector $(a,b,c)$ in this basis ?

My attempt : Here obviously $a_1,a_2$ and $a_3$  will form  basis since  det $A \neq 0$

take $A=\begin{bmatrix}
    1 &   0   & -i \\
    1+i & 1-i & 1  \\
    i &   i & i
\end{bmatrix}$

and  we have  $$A^{-1}=\begin{bmatrix}
    \frac{1-2i}{5} &   \frac{1-2i}{5}   &  \frac{3-i}{5}\\
     \frac{1-2i}{5}& \frac{1+3i}{5} & \frac{-2-i}{5}  \\
    \frac{-2+4i}{5} & \frac{-2-i}{5}   & \frac{-1-3i}{5}
\end{bmatrix}$$

The coordinates  of the vector $(a,b,c)$ in this basis =$A^{-1}  \begin{bmatrix}  a \\ b\\c\end{bmatrix}=\begin{bmatrix}
    \frac{1-2i}{5}a +   \frac{1-2i}{5}b   +  \frac{3-i}{5}c\\
     \frac{1-2i}{5}a+\frac{1+3i}{5}b + \frac{-2-i}{5} c \\
    \frac{-2+4i}{5}a + \frac{-2-i}{5} b + \frac{-1-3i}{5}c
\end{bmatrix}$

Is my solution  is correct or not ?

答:
No, it isn't. Take the vector $(1,0,-i)$ for example. Then you get

$$A^{-1}\begin{bmatrix}
1\\
0\\
-i
\end{bmatrix}=\begin{bmatrix}
-i\\
0\\
-1+i
\end{bmatrix}$$

But $(1,0,-i)$ is the first element of your basis so you should be getting $(1,0,0)$.

To see where you went wrong, look at how you constructed your change of basis matrix $A$.

If $A^{-1}$ takes a vector written in the canonical basis and gives you the vector written in the $\{a_1,a_2,a_3\}$ basis, $A$ should do the opposite. Then what should you get for
$$A \begin{bmatrix}
1\\
0\\
0
\end{bmatrix}?$$
Notice that this tells you what $A$'s first column should be.


问:
Prove that a polynomial of degree $d$ has at most $d$ roots (without induction)

Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$.

Is it possible to prove this without using induction on degree? If so, how can I do this?

答:
If $p(x)$ were to have more than $d$ distinct roots in $F$, then it would have at least $d+1$ linear factors $(x - r_1), (x - r_2), \cdots$.  This is impossible.

(Edit: see also Inceptio's comment.)


答:
Let us denote the usual inner product on $ \mathbb R^n$ by $(\cdot|\cdot).$ Then

$$||Ax||^2=(Ax|Ax)=(A^TAx|x)=(A^2x|x)=(B^2x|x)=(Bx|B^Tx)=(Bx|-Bx)=-(Bx|Bx)=-||Bx||^2$$

for all $x \in \mathbb R^n.$

Hence $-||Bx||^2 \ge 0$ for all $x \in \mathbb R^n.$ This gives

$||Bx||^2 = 0$ for all $x \in \mathbb R^n$ and thus $||Ax||^2 = 0$ for all $x \in \mathbb R^n$ .

Cosequence: $A=B=0.$


问:
Are $C^1$, $BV$ functions dense in set of all continuous $BV$ functions with respect to $BV$ norm?
I was wondering whether any $BV$ function can be approximated arbitrarily well with a piecewise linear function in sense of $BV$ norm.

Let:

$A$ - be the set of all continuous functions $[0,1]\rightarrow\mathbb{R}^n$ of bounded variation which map $0$ to $0$,
$B$ - be the set of all continuous piecewise linear functions $[0,1]\rightarrow\mathbb{R}^n$ of bounded variation which map $0$ to $0$.

Is $B$ dense in $A$ with respect to the $BV$ norm: $\left\|f-g\right\|=\text{var}(f-g)$ ($\text{var}$ is like the variation...)?

More generally, is something like the opposite of the next statement true?

The statement that I know is true is:

$f_n\xrightarrow[BV]{}f\Rightarrow f_n\xrightarrow[\left\|\cdot\right\|_\infty]{}f$, when $n\rightarrow\infty$,

but can it be that the opposite is also true? Something like:

$f_n\xrightarrow[\left\|\cdot\right\|_\infty]{}f\wedge\text{var}(f_n)\rightarrow\text{var}(f)\Rightarrow f_n\xrightarrow[BV]{}f$, when $n\rightarrow\infty$.

I ask the second question because if it was true, then I would be able to answer the first question with a construction.

答:
No, the type of convergence you're considering is too strong.  (In the answer below, I'm assuming $B$ is the set of continuous, piecewise linear functions on $[0,1]$, hence a subset of $A$.)

On the definition of $A$:  The first thing to check is that $A$ is actually a Banach space as defined.  This isn't completely obvious, not only because of the requirement that the functions vanish at zero but also because of the continuity requirement.

It is worth recalling that if $f : [0,1] \to \mathbb{R}$ has bounded variation, then there is a finite (signed) Borel measure $\mu_{f}$ on $[0,1]$ acting as the weak derivative of $f$:
\begin{equation*}
\int_{0}^{1} f(x) g'(x) \, dx = f(1^{-}) g(1) - f(0^{+}) g(0) -\int_{0}^{1} g(\xi) \, \mu_{f}(d\xi) \quad \text{if} \, \, g \in C^{\infty}((0,1)).
\end{equation*}
It is possible to show that if $[a,b] \subseteq [0,1]$, then
\begin{equation*}
TV(f; [a,b]) = \|\mu_{f}\|([a,b]),
\end{equation*}
where $\|\mu_{f}\|$ is the total variation measure of $\mu$ (i.e. if $\mu_{f} = \mu^{+}_{f} - \mu^{-}_{f}$ is the Hahn decomposition of $\mu$, then $\|\mu_{f}\| = \mu^{+}_{f} + \mu^{-}_{f}$).  In particular,
\begin{equation*}
\text{var}(f) = \|\mu_{f}\|([0,1]).
\end{equation*}

After recalling the previous preliminaries, it becomes clear that $A$ is a Banach space.  Indeed, if $(f_{m})_{m \in \mathbb{N}}$ is a Cauchy sequence in $A$ as defined above, then $(\mu_{f_{m}})_{m \in \mathbb{N}}$ is Cauchy with respect to the total variation norm $\|\cdot\|_{TV}$ defined by $\|\nu\|_{TV} = \|\nu\|([0,1])$.  In particular, it converges to some measure $\mu$.  Further, since the continuity of the functions $(f_{m})_{m \in \mathbb{N}}$ implies $\|\mu_{f_{m}}\|(\{x\}) = 0$ independently of the point $x \in [0,1]$ or the index $m \in \mathbb{N}$, the total variation convergence implies $\|\mu(\{x\})\| = 0$ independently of $x \in [0,1]$.  Finally, using the following identity (which follows by choosing an appropriate sequence of $g$ in the definition of weak derivative above)
\begin{equation*}
f_{m}(x) = f_{m}(0) + \mu_{f_{m}}([0,x]) = \mu_{f_{m}}([0,x]) \quad \text{if} \, \, x \in [0,1], \, \, m \in \mathbb{N},
\end{equation*}
we see that we can define a function $f : [0,1] \to \mathbb{R}$ by
\begin{equation*}
f(x) = \mu([0,x]) = \lim_{m \to \infty} \mu_{f_{m}}([0,x]) = \lim_{m \to \infty} f_{m}(x).
\end{equation*}
In fact, the convergence $f_{m} \to f$ is uniform and $\mu(\{0\}) = 0$ so $f(0) = 0$.  This proves $A$ is a Banach space.

$B$ is not dense:  In the previous discussion, we actually saw that the norm $\|\cdot\| = \text{var}(\cdot)$ is slightly stronger than the supremum norm.  It's not enough to make $B$ (or even the $C^{1}$ functions) dense in $A$.

The key here is that $B$ is a subset of $W^{1,1}([0,1])$, but many functions in $A$ are not in $W^{1,1}([0,1])$.

The Sobolev space $W^{1,1}([0,1])$ is the space of (absolutely) continuous functions $f : [0,1] \to \mathbb{R}$ with a weak derivative $\mu_{f}$ satisfying $\mu_{f} \ll dx$.  (Here $dx$ is the Lebesgue measure on $[0,1]$.)  Usually we define the measurable function $f'$ by $f' = \frac{d \mu}{dx}$.  It is possible to show that $f$ is differentiable at almost every point in $[0,1]$ and its derivative equals $f'$ at those points.

$W^{1,1}([0,1])$ is made a Banach space with the following norm:
\begin{equation*}
\|f\|_{W^{1,1}([0,1])} = \int_{0}^{1} \left( |f(x)| + |f'(x)| \right) \, dx.
\end{equation*}

If $f \in W^{1,1}([0,1])$, then an exercise shows that $TV(f; \cdot)$ can be computed as
\begin{equation*}
TV(f; [a,b]) = \int_{a}^{b} |f'(x)| \, dx.
\end{equation*}
Hence the norm $\|\cdot\|$ defined above in $A$ induces the same topology on $B$ as the norm $\|\cdot\|_{W^{1,1}([0,1])}$.  Thus, $\overline{B} \subseteq A \cap W^{1,1}([0,1])$.  (In fact, equality holds.)

On the other hand, $A \setminus W^{1,1}([0,1])$ is not empty.  If $\mu$ is any singularly continuous measure in $[0,1]$ (that is, a measure that is singular with respect to $dx$, but has no atoms), then the function $f(x) = \mu([0,x])$ is in $A$, but not in $W^{1,1}([0,1])$.  (An example of such a function is the "Devil's Staircase.")


问:
Why is this set from Thomae's function finite?
Let $f(a):[0, 1] \to \mathbb{R}$ be Thomae's function, which is restated below:
$$f(a)=\begin{cases}
0 &a\text{ is irrational}\\
\frac{1}{n} &a\text{ is rational and } n \text{ is the smallest positive integer s.t. } a=\frac{m}{n} \text{ for some } m\in \mathbb{Z}^+
\end{cases}$$

Now consider the set $\{x \in[0, 1]:f(x) \ge \varepsilon/2\}$. Could someone please explain why this set is finite?

答:
Let $\varepsilon$ be arbitrary. By the Archimedian principle, there exists an $n\in\mathbb N$ such that $\varepsilon > \frac1n$.

Thus, if $f(a)>\varepsilon$, then $a=\frac cd$ where $c\leq d<n$. Thus, there are at most $n^2$ choices of $a$ that map to a value $>\varepsilon$.


问:
Practical Uses for the Tensor Product
From reading a bit of chapter section 18 in Differential Geometry, by Loring W. Tu, my understanding of tensor products is that they allow one to represent multilinear maps with linear maps.

I was simply curious whether there is any practical use for the tensor product (perhaps in computer science or physics, etc.). Are there other interpretations of tensor products?

答:
In quantum mechanics, the state of a physical system (say, a particle) is an element of a Hilbert space $\mathcal H$. If we take two such systems (say, two particles) with Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, and consider them together, then the state of the physical system will be an element of the tensor product $\mathcal H_1\otimes\mathcal H_2$ of Hilbert spaces.

Also, in relativity, we  work on a pseudo-Riemannian manifold $M$, on which we have a metric which gives us the spacetime interval between two events. A metric on a pseudo-Riemannian manifold is just a non-degenerate, symmetric bilinear form on every tangent space $T_pM$ of the manifold, which we can interpret as an element from the tensor product $T_pM^\ast\otimes T_pM^\ast$ of the duals of the tangent spaces.

Title: Practical Uses for the Tensor Product


问:
Continuity of a definite integral
Let $f : \mathbb{R} \to \mathbb{C}$ be an absolutely integrable function, and let $F: \mathbb{R} \to \mathbb{C}$ be the definite integral $$F(x):=\int_{-\infty}^x f(t)\,\mathrm{d}t.$$ I need to show that $F$ is continuous, i.e., that for every $x_0 \in \mathbb{R}$ and $\epsilon >0$ there exists $\delta > 0$ such that $|F(x)-F(x_0)|=|\int_{[x,x_0]} f(t) dt|<\epsilon$ whenever $|x-x_o|< \delta$. Is there a simple way to do this?

答:
We can show that it is sequentially continous. Consider a sequence $(x_n)$ convergning to $x\in \mathbb{R}$ and  the associed sequence of functions $f_n(x) = \chi_{[-\infty,x_n]}f(t)$. We have $|f_n|(t) \leq |f|(t)$ which is integrable. By the Lebesgue's dominated convegrence theorem, we get $$\lim_{n\to \infty}F(x_n) = \lim_{n\to \infty}\int_{\mathbb{R}}f_n(t) \mathrm{d}t =  \int_{\mathbb{R}}\chi_{[-\infty,x]}f(t)\mathrm{d}t = F(x).$$

A function is sequentially continous iff it is continous in «$\epsilon-\delta$»

Title: Continuity of a definite integral


问:
In the following figure: $AE = ED$ , $BD = DC$. Prove that $BE = \frac{EF}{3}$
The question is : $AE = ED$ , $BD = DC$. Prove that $BE = \cfrac{EF}{3}$

Approach:

Realisations:

$AD$ is a median of triangle $ABC$.

$BE$ is a median of triangle $DBA$.

$E$ is a centroid.

I tried to prove these triangles similar, but I was unsuccessful as one criteria is always missing.

Please help me with this problem.

How to do this using Ceva's Theorem?

答:
Apply Menelaus' Theorem following B-C-A-D to see:

$$\frac{BD}{BC} \cdot \frac{FC}{AF}\cdot \frac{AE}{ED} =1 \implies \frac{AF}{FC}=\frac{1}{2}$$

Apply Menelaus' Theorem again, this time following A-C-B-F to see:

$$\frac{AF}{AC} \cdot \frac{DC}{BD}\cdot \frac{BE}{EF}=1 \implies \frac{BE}{EF}=3$$


问:
Prove: $(A_1 \wedge A_2 \wedge \dots \wedge A_n) \to B\iff A_1 \to (A_2 \to (\dots \to (A_n \to B)\dots))$
I want to prove the following, using induction for every $n \geq 0$:
$$(A_1 \wedge A_2 \wedge \dots \wedge A_n) \to B\iff A_1 \to (A_2 \to (\dots \to (A_n \to B)\dots)).$$
It is an exercise regarding propositional logic. $\iff$ stands for "(tauto)logically equivalent".

My base cases for induction are (maybe I do not need all of them):

$n=0: B \iff B$ (always holds)

$n=1: A_1 \to B \iff A_1 \to B$ (always holds)

$n=2: (A_1 \wedge A_2) \to  B \iff A_1 \to (A_2 \to B) $ (proven here, using a boolean table)

I am getting stuck at the induction step. Assuming that it holds for n:
$$((A_1 \wedge A_2 \wedge \dots \wedge A_n) \to B) \to(A_1 \to (A_2 \to (\dots \to (A_n \to B)\dots))) \text{ is always true}\\
((A_1 \wedge A_2 \wedge \dots \wedge A_n) \to B) \leftarrow (A_1 \to (A_2 \to (\dots \to (A_n \to B)\dots))) \text{ is always true}$$

I do not know why it should hold for n+1:
$$(A_1 \wedge A_2 \wedge \dots \wedge A_n \wedge A_{n+1}) \to B \iff A_1 \to (A_2 \to (\dots \to (A_n \to (A_{n+1} \to B))\dots)).$$

Any tips on how I can use the assumptions? I was also thinking about strong induction.
Tipps would be greatly appreciated before giving me a full solution. But I am glad for any input.

答:
Ansatz:

$(A_1\wedge A_2\wedge\ldots \wedge A_n \wedge A_{n+1})\rightarrow B$

equivalent to

$(A_1\wedge A_2\wedge\ldots \wedge (A_n \wedge A_{n+1}))\rightarrow B$

equivalent by induction to

$A_1\rightarrow (A_2\rightarrow(\ldots\rightarrow (A_{n-1}\rightarrow ((A_n\wedge A_{n+1}\rightarrow B)...))$

equivalent by $n=2$:

$(A_n\wedge A_{n+1})\rightarrow B$ equivalent to $A_n\rightarrow (A_{n+1}\rightarrow B)$

to

$A_1\rightarrow (A_2\rightarrow(\ldots\rightarrow (A_{n-1}\rightarrow (A_n\rightarrow A_{n+1}\rightarrow B)...))$


答:
Denote areas as [.] and establish the ratio
$$1=\frac{AE}{ED}= \frac{[ABF]}{[DBF]} = \frac{\frac{AF}{AC}[ABC]}{\frac12 (1-\frac{AF}{AC})[ABC]}\implies \frac{AF}{AC}=\frac13
$$
Then
$$
\frac{BE}{EF}= \frac{[ABD]}{[AFD]}
=  \frac{\frac12[ABC]}{\frac12 \frac{AF}{AC}[ABC]}=\frac{AC}{AF}=3
$$

Title: In the following figure: $AE = ED$ , $BD = DC$. Prove that $BE = \frac{EF}{3}$

Body:
The question is : $AE = ED$ , $BD = DC$. Prove that $BE = \cfrac{EF}{3}$

Approach:

Realisations:

$AD$ is a median of triangle $ABC$.

$BE$ is a median of triangle $DBA$.

$E$ is a centroid.

I tried to prove these triangles similar, but I was unsuccessful as one criteria is always missing.

Please help me with this problem.

How to do this using Ceva's Theorem?


问:
Understanding differences in Geodetic (WGS84) to ECEF equations?
Methods suggested in this, that and there all recommend the following:

$$x = R \cos(\theta) \cos(\phi)$$
$$y = R \cos(\theta) \sin(\phi)$$
$$z = R \sin(\theta),$$

where latitude is $\theta$, longitude is $\phi$ and the Earth's approximate radius is $R$ (6371km).

However, I am not sure why these equations are using $R$ rather than their documented form:

$$x = (N(\theta) + h) \cos(\theta) \cos(\phi)$$
$$y = (N(\theta) + h) \cos(\theta) \sin(\phi)$$
$$z = \left(\frac{b^2}{a^2} N(\theta) + h\right) \sin(\theta),$$

where altitude is $h$, semi-major axis is $a$, semi-minor axis is $b$ and the prime vertical radius of curvature is $N(\theta)$. Further, we can say:

$$N(\theta) = \frac{a}{\sqrt{1-{e^2}\sin^{2}(\theta)}}$$
and...
$$e^2 = 1 - \frac{b^2}{a^2}$$
thus...
$$z = (1 - e^2)  (N(\theta) + h)  \sin(\theta).$$

Now for my purposes, I am treating all points with an altitude of 1, so the $h$ can be eliminated. What I am not comprehending is how they are arriving at:

$$N(\theta) = R$$

and...

$$(1-e^2) * N(\theta) = R$$

Is it safe to use the radius of the Earth as an approximation or should I use the WGS84 constraints (semi-major and first eccentricity) in my computations?

答:
Comments

The earth is not perfectly spherical. There is a flattening at the poles.

The WGS84 modifies spherical coordinates in two ways.

First in order to take into account the equatorial bulge through eccentricity factor $e$.

Secondly if distances are measured from an aircraft or spacecraft at an altitude $h$ above the earth this altitude should be added on to earth's radius.

Parallel lines unlike meridians are not geodesics even in the ideal spherical earth model $(e=0)$. At higher latitudes distance measured along latitudes and shortest path geodesic is more pronounced.


答:
From Paulo answer

i notice  that  to find the coordinate

$$A^{-1}\begin{bmatrix}
1\\
0\\
-i
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$

must satisfied

Therefore the  coordinates  of the vector $(a,b,c)$ in this basis =$(A^{-1})^T  \begin{bmatrix}  a \\ b\\c\end{bmatrix}=\begin{bmatrix}
    \frac{1-2i}{5}a +   \frac{1-2i}{5}b   +  \frac{-2+4i}{5}c\\
     \frac{1-2i}{5}a+\frac{1+3i}{5}b + \frac{-2-i}{5} c \\
    \frac{3-i}{5}a + \frac{-2-i}{5} b + \frac{-1-3i}{5}c
\end{bmatrix}$

Title: find the coordinates of the vector $(a,b,c)?$

Body:
Suppose $a_1 = (1,0,-i) $ , $a_2 =(1+i , 1-i,1)$ , $a_3 =(i,i,i)$ in $\mathbb{C}^3$ . What are the coordinates  of the vector $(a,b,c)$ in this basis ?

My attempt : Here obviously $a_1,a_2$ and $a_3$  will form  basis since  det $A \neq 0$

take $A=\begin{bmatrix}
    1 &   0   & -i \\
    1+i & 1-i & 1  \\
    i &   i & i
\end{bmatrix}$

and  we have  $$A^{-1}=\begin{bmatrix}
    \frac{1-2i}{5} &   \frac{1-2i}{5}   &  \frac{3-i}{5}\\
     \frac{1-2i}{5}& \frac{1+3i}{5} & \frac{-2-i}{5}  \\
    \frac{-2+4i}{5} & \frac{-2-i}{5}   & \frac{-1-3i}{5}
\end{bmatrix}$$

The coordinates  of the vector $(a,b,c)$ in this basis =$A^{-1}  \begin{bmatrix}  a \\ b\\c\end{bmatrix}=\begin{bmatrix}
    \frac{1-2i}{5}a +   \frac{1-2i}{5}b   +  \frac{3-i}{5}c\\
     \frac{1-2i}{5}a+\frac{1+3i}{5}b + \frac{-2-i}{5} c \\
    \frac{-2+4i}{5}a + \frac{-2-i}{5} b + \frac{-1-3i}{5}c
\end{bmatrix}$

Is my solution  is correct or not ?


问:
Derivation of the least square estimator for multiple linear regression
I find a derivation of the least square estimator for multiple linear regression, but there some part I am not fully understand some part in the. The derivation is following:

Starting from $y= Xb +\epsilon $, which really is just the same as

$\begin{bmatrix}
y_{1} \\
y_{2} \\
\vdots \\
y_{N}
\end{bmatrix}
=
\begin{bmatrix}
x_{11} & x_{12} & \cdots & x_{1K} \\
x_{21} & x_{22} & \cdots & x_{2K} \\
\vdots & \ddots & \ddots & \vdots \\
x_{N1} & x_{N2} & \cdots & x_{NK}
\end{bmatrix}
*
\begin{bmatrix}
b_{1} \\
b_{2} \\
\vdots \\
b_{K}
\end{bmatrix}
+
\begin{bmatrix}
\epsilon_{1} \\
\epsilon_{2} \\
\vdots \\
\epsilon_{N}
\end{bmatrix} $

let $e = y - Xb$, then it all comes down to minimzing $e'e$:

$e'e = \begin{bmatrix}
e_{1} & e_{2} & \cdots & e_{N} \\
\end{bmatrix}
\begin{bmatrix}
e_{1} \\
e_{2} \\
\vdots \\
e_{N}
\end{bmatrix} = \sum_{i=1}^{N}e_{i}^{2}
$

So minimizing $e'e$ gives us:

$min_{b}$ $e'e = (y-Xb)'(y-Xb)$

$min_{b}$ $e'e = y'y - 2b'X'y + b'X'Xb$

$\frac{\partial(e'e)}{\partial b} = -2X'y + 2X'Xb \stackrel{!}{=} 0$

$X'Xb=X'y$

$b=(X'X)^{-1}X'y$

my problem is about this part "$min_{b}$ $e'e = (y-Xb)'(y-Xb)$", I believe the ' notation at here is the transpose, but I could not get the result same as above, if I just open the bracket as usual
\begin{align*}
min_{b}e'e &= (y-Xb)'(y-Xb)\\
& = y'y - y'Xb - b'X'y + b'X'Xb
\end{align*}which is not same as what the derivation got "$min_{b}$ $e'e = y'y - 2b'X'y + b'X'Xb$", could anyone explain to me how could the derivation get this matrix equation?

答:
This is because
$$b'X'y = (y'Xb)'$$
combined with the fact that they are both scalar.
A transposed scalar equals itself.


问:
Given a probability density function $f(x) = \begin{cases} 0 & x < 100 \\ \frac{a}{x^2} & x \geq 100 \end{cases}$. Probability distribution function?
Given a probability density function:

$$ f(x) = \begin{cases} 0 & x < 100 \\ \\ \frac{a}{x^2} & x \geq 100 \end{cases} $$

What is the probability distribution function F(x)?

We know that: $F(y) = \int_{-\infty}^{y}f(x)dx$

When $x < 100$:
$$
\begin{equation}
\begin{split}
F(y) &= \int_{-\infty}^{y}0dx \\\\
     &= 0 \Biggr|_{-\infty}^{y} \\\\
     &= 0
\end{split}
\end{equation}
$$

When $x \geq 100$:
$$
\begin{equation}
\begin{split}
F(y) &= \int_{-\infty}^{100}0dx + \int_{100}^{y}\frac{a}{x^2}dx \\\\
     &= 0 \Biggr|_{-\infty}^{100} + \left( -\frac{a}{x} \right) \Biggr|_{100}^{y} \\\\
     &= -\frac{a}{y} + \frac{a}{100} \\\\
\end{split}
\end{equation}
$$

So:
$$ F(y) = \begin{cases} 0 & y < 100 \\ \\ -\frac{a}{y} + \frac{a}{100} & y \geq 100 \end{cases}$$

When $x >= 100$, $F(x)$ should be equal to 1 but it contains $a$.

I think that $F(x) = 1$ and $a = 100$ but do not know how to prove it.

答:
You know $\lim\limits_{y \to \infty} F(y) =1$ since this is a probability distribution on the real numbers

You have $\lim\limits_{y \to \infty} \left(-\frac{a}{y} + \frac{a}{100}\right) = \frac{a}{100} $

So you need $\frac{a}{100} =1 $ and thus $a=100$, and that is a proof.


答:
Physics: Good examples are provided on the wikipedia page for tensors.

CS: Tensors are a fundamental part of machine learning. It's not for nothing that Google's machine learning software is called TensorFlow

Math: Tensors are used widely, but one notable area is representation theory

In general, tensors are essentially higher dimensional versions of matrices, so it's not completely surprising that they would arise in certain settings.

Title: Practical Uses for the Tensor Product

Body:
From reading a bit of chapter section 18 in Differential Geometry, by Loring W. Tu, my understanding of tensor products is that they allow one to represent multilinear maps with linear maps.

I was simply curious whether there is any practical use for the tensor product (perhaps in computer science or physics, etc.). Are there other interpretations of tensor products?


问:
Periodic function's integral
Let $b > 0 \in \mathbb R$, a continuous function $f : [0, b] \to \mathbb R$, and a periodic and continuous function $p : \mathbb R \to \mathbb R$ of period $T = 1$.

If $p \geq 0$ on $\mathbb R$ et $\int_0^1 p(t) dt = 1$, prove that
$$\lim_{n \to \infty} \int_0^b f(t)p(nt) dt = \int_0^bf(t) dt. $$

I was told that we should use Riemann/Darboux sums, is there any another way to do it ? If not how to do it with Riemann sums ?

答:
This is what I would do:

$1)$ Prove that the statement is true for $p(x)=e^{2\pi i x}$.

$2)$ Prove that it is true for trigonometric polynomials $$p(x)=\sum\limits_{j=-m}^ma_je^{2\pi i jx} \quad with \quad\sum\limits_{j=-m}^ma_j=1,$$ ie, for any element belonging to the span of $\{e^{2\pi inx}:n\in\mathbb{Z}\}$ with norm equal to 1.

$3)$ Use that $\{e^{2\pi inx}:n\in\mathbb{Z}\}$ is an orthonormal complete system, which implies that $\{e^{2\pi inx}:n\in\mathbb{Z}\}$ is dense in $L^2(\mathbb{T})$. In other words, any function $p$ fulfilling the hypothesis can be approximated by a sequence of trigonometric polynomials that satisfy the original property.

Note: I solved the problem assuming that $f$´s domain was $[0,1]$ instead of $[0,b]$, but this should be easy to fix.

Title: Periodic function's integral


问:
Probability question on union sets
Consider $P(H/E)>P(H)$. Is it true that $P(H\cup¬E/E)>P(H\cup
¬E)$.

I indicate with the negation sign the complement of a set
.

答:
Throw a fair die and consider $H=\{1,3,5\}$ being an odd value and $E=\{2,3,5\}$ a prime value.

Then $\mathbb P(H \mid E) = \frac23 > \frac12 = \mathbb P(H )$:

this is $\{3,5\}$ out of $\{2,3,5\}$, compared to $\{1,3,5\}$ out of $\{1,2,3,4,5,6\}$

but $\mathbb P(H \cup \lnot E \mid E) = \frac23 < \frac56 = \mathbb P(H \cup \lnot E)$:

this is $\{3,5\}$ out of $\{2,3,5\}$, compared to $\{1,3,4,5,6\}$ out of $\{1,2,3,4,5,6\}$

Title: Probability question on union sets


问:
Billy and Bob start with two rectangular grid which can be modeled as a chocolate

Billy and Bob start with two rectangular grid which can be modeled as
a chocolate. They take turns cutting and eating the chocolate: on
their turn, they can cut the chocolate along an edge parallel to the
sides (thus resulting in three rectangles) and consume one of the
three pieces, so that his partner has two chocolates, and therefore
can make an analogous move. A player loses if they cannot make a cut.
Assuming optimal play, who wins if:

(a) the starting rectangles are $1\times 2020$ and $2\times 4040$,

(b) the starting rectangles are $100\times100$ and $100\times 500$, and

(c) generalize.

My friend and I tested the first case (a) extensively, and ended up with somewhat of a mirroring thing if there are two $1\times n$ rectangles resultant. However, if $n$ is even one player could theoretically cut the $1\times n$ into two $1\times\frac n2$ rectangles, and discord the other $1\times n$.

I don't know how to continue.

答:
I will write each game position as an ordered quadruple $(a,b,c,d)$, meaning an $a\times b$ bar and a $c\times d$ bar.

Call a number vile if it has an even number of twos in its prime factorization, and dopey if it has an odd number of twos. This terminology was originally conceived by Fraenkel.

Let $(e_0,e_1,e_2,\dots)=(1,3,4,5,\dots)$ denote the sorted, zero-indexed list of vile numbers, and let $(d_0,d_1,d_2,\dots)=(2,6,8,10,\dots)$ be the sorted list of dopey numbers. Finally, let $f(n)$ denote the index of $n$ in its list, regardless if that list is $e$ or $d$. That is, $f(3)=1$, $f(10)=3$, $f(4)=f(8)=2$.

Let $S$ be the of positions $(a,b,c,d)$  where all of these conditions are satisfied:

$a$ and $b$ have the same type (both vile or both dopey).

$c$ and $d$ have the same type.

$f(a)\oplus f(b)\oplus f(c)\oplus f(d)=0$, where $\oplus$ is the nim-sum.

A position is losing if and only if it is in $S$.

I found this pattern with help of Vepir and a computer program, while Jaap Scherphuis came up with the proof of its correctness.

We need to show that any move from $S$ results in a position outside of $S$, and every position not in $S$ has a legal move to $S$. First, here is the winning strategy to take any position outside $S$ to one in $S$:

Suppose $a$ and $b$ have different types, WLOG $a$ is vile and $b$ is dopey. A winning move is to cut the $a\times b$ bar in half perpendicularly to the $b$ side, and eat the other bar, leaving the position $(a,b/2,a,b/2)$. Since $b$ is dopey, it follows that $b/2$ is vile, so it is easy to check the nim-sum condition and verify this position is in $S$.
 The same logic applies when $c$ and $d$ have different types.

Otherwise, you will have $f(a)\oplus f(b)\oplus f(c) \oplus f(d)\neq 0$. The idea is to make a move which changes a single index of one of the coordinates, without changing its type, and which reduces the index corresponding to the winning move on the nim game with heap sizes $f(a),f(b),f(c),f(d)$.

Now, suppose $(a,b,c,d)\in S$. Any move where you eat one of the pieces from the bar you broke will change exactly one of the coordinates. This necessarily leaves a position outside of $S$, because for any choice of three coordinates, there is a unique choice of a fourth coordinate for which the position is in $S$. Therefore, we restrict out attention to moves where you eat the bar you did not break, of the form
$$
(a,b,c,d)\implies (a',b,a-a',b)
$$
where $1\le a' \le a-1$. In order for this new position to be in $S$, $a'$ and $a-a'$ must both have the same type as $b$, and it must be true that $f(a')=f(a-a')$. If $a'$ and $a-a'$ have the same type and index, they must be equal, so the move would have to be $(a/2, b,a/2,b)$. But $a/2$ has the opposite type of $a$, so this is not in $S$.


问:
easy homework: Equivalence classes, how do they look?
Let's say that I got a

```<code> set = { Arnold, Harrison }
</code>```

and I want to display the equivalence class of [ Harrison ]

The actual condition for the relation doesn't matter in this case so let's just say that {Arnold} is the only relation to Harrison.

This would be displayed as:

```<code>[ Harrison ] = { Arnold }
</code>```

However, if we have this set instead:

```<code>set = { Arnold, Harrison, {Arnold, Harrison}, { Arnold, Arnold, Harrison} }
</code>```

And we say that this time, Harrison has a relation to Arnold but also {Arnold, Harrison} and even {Arnold, Arnold, Harrison}.

How is this expressed as an equivalence class?

```<code>[ Harrison ] = {Arnold, Harrison}
</code>```

or...

```<code>[ Harrison ] = { Arnold, {Arnold, Harrison}, {Arnold, Arnold, Harrison}  }
</code>```

or... what?!

答:
Note that the equivalence class of Harrison must always contain Harrison. Otherwise, if I understand your relations correctly the second answer is closer to being correct.

[Harrison] = {Harrison, Arnold, {Arnold, Harrison}, {Arnold, Arnold, Harrison}}

Note that usually we do not repeat elements of a set, so that this can be more concisely written as

[Harrison] = {Harrison, Arnold, {Arnold, Harrison}}

Since {Arnold, Harrison} = {Arnold, Arnold, Harrison}


问:
Constructing a singular simplex from other two singular simplexes.
Suppose $\sigma_1:\Delta^k \rightarrow X$ is a singular $k$-simplex and $\sigma_2:\Delta^l \rightarrow X$ is a singular $l$-simplex. Is there a singular $(k+l)$-simplex, $\sigma: \Delta^{k+l} \rightarrow X$, such that $\sigma|_{[v_1,\dots,v_k]} = \sigma_1$ and $\sigma|_{[v_k,\dots,v_{k+l}]}=\sigma_2$?

答:
Well, since you want a singular simplex you only require the mapping $\sigma: \Delta^{k+l}\to X$ to be continuous. So you can certainly, given $\sigma_1$ and $\sigma_2$, identify $k$ and $l$ faces of $\Delta^{k+l}$ corresponding to the simplices above, and just arbitrarily define how $\sigma$ maps the other faces (in a continuous way, of course).

For example, suppose that you have two singular $1$-simplices $\alpha$ and $\beta$. Then you can identify two of the sides of a singular $2$-simplex with $\alpha$ and $\beta$, but there is a third face which you may define arbitrarily.


问:
A question about $\det(A^TA)$
Assume that $A\in\mathbb{R}^{m\times n}$, where $m\leq n$. The aim is to find a lower and an upper bound for $\mathrm{det}(A^TA)$. What can be said about this issue?

答:
As described in the other answer, $\det(A^TA)=\sigma^2_1\cdots\sigma^2_n$.

Using the GM-AM inequality $$\det{A^TA}=\sigma^2_1\cdots\sigma^2_n\le\left(\frac{\sigma_1^2+\cdots+\sigma_n^2}{n}\right)^{n}=\left(\frac{\|A\|^2_2}{n}\right)^{n}=\left(\tfrac{1}{n}\sum_{ij}a_{ij}^2\right)^{n}$$

Title: A question about $\det(A^TA)$


问:
Practical significance of $\frac{17}{21}$=$\frac{1}{2}$+$\frac{1}{6}$+$\frac{1}{7}$
In an article of ancient maths I found that Babylonians also had a great idea of maths.
For practicle purposes, if I want to divide 17 grain bushels among 21 workers, the equation would be $\frac{17}{21}$≈0.8. It means I have to give 0.8 grain bushels to each worker.
Since $\frac{17}{21}$ can be written as $\frac{1}{2}$+$\frac{1}{6}$+$\frac{1}{7}$, what does this equation implies to the real world?
There is also written that it has practical meaning then I am wondered what's that practical meaning of this equation.

答: The point is just that everyone can get the right amount of grain if I give everyone half a bushel, a sixth of a bushel, and a seventh of a bushel. The article you originally linked goes on to note that the obvious greedy algorithm outputs  1/2 + 1/4 + 1/17 + 1/1428, which is an extremely unhelpful division (who has the ability to divide a bushel into 1428ths?).</p>


问:
Graph optimization problem linked with maximal spanning tree.
Let $G$ be a non-oriented connexe graph made of $N$ nodes $B_i$. We suppose that they are not any self-connected node.

For every $(i,j) \in \{1, \dots, N \}$, we note $\nu_{ij} \geq 0$ a positive weigth on the associated edge $ij$.  If $\nu_{ij}=0$, that simply means that the nodes $B_i$ and $B_j$ are not really connected by an edge.

For a collection of real $(b_{ij})_{1 \leq i<j \leq N}$, consider the following problem of optimization :

\begin{equation}
\min_{(u_1,\dots,u_N) \in \mathbb{R}^N} \sum_{i=1}^N \sum_{j=1}^{i-1}
 |b_{ij}+u_i-u_j| \nu_{ij}
\tag{1}
\end{equation}
Solving this problem is equivalent to cut some edges out of the graph $G$, the idea is to chose well the coefficient $u_i$ so as to erase the biggest contributions $\nu_{ij}$. I would like to prove that the edges which are going to be 'erased' (i.e the edges such as $|b_{ij}+u_i-u_j|=0$) are exactly the edges of a maximal spanning tree $T$ of $G$.

I recall here the defitinition of a maximal spanning tree :

A graph $T=((B_i)_{1 \leq N},A_T)$ is said to be a maximal spanning tree of $G$ if for every $(i,j) \in (1, \dots, N)$, the edge $ij$ has a weight of $x_{ij} \nu_{ij}$, with the $(x_{ij})_{1 \leq i < j \leq N}$ be a solution of :

\begin{align}
\max_{x_{ij} \in (0,1)^{\frac{N(N-1)}{2}}}  \sum_{i=1}^N \sum_{j=1}^{i-1} x_{ij} \nu_{ij}
\tag{2}
\end{align}
with the following constraints :

the graphe formed with coefficient $x_{ij} h_{ij}$ has no cycle in it

$$\sum_{j=i+1}^N x_{ij} + \sum_{j=1}^{i-1} x_{ji}  \geq 1, \quad  \forall i \in \{1,\dots,N \}$$

These two conditions insures that the graph obtained with coefficient $x_{ij} \nu_{ij}$ is indeed a tree.

Remark that I'm not sure that the result I want to prove is true. I'have tried this construction on several little graphs but I found no counter-example, but I'm not very used to graph theory. Any remarks or advices are welcomed.

答:
Your objective function $f : u \mapsto \sum_i \sum_j |b_{ij} + u_i - u_j| v_{ij}$ has upper- and lower-partial derivatives for each $u_k$. For a specific $u_i$, the upper-partial derivative will be $\frac {\partial f} {\partial u_i^+} = \sum_{j \in P_i} v_{ij} - \sum_{j \notin P_i} v_{ij}$, where $j \in P_i$ iff $b_{ij} + u_i - u_j \ge 0$ ; and the lower-partial $\frac {\partial f} {\partial u_i^-}$ is the same with a strict inequality.

The global minimum has to be a local minimum, so at the optimal solution (which exists : if $||u|| \to \infty$ then $f(u) \to \infty$) the upper partial derivative needs to be $\le 0$ and the lower partial derivative needs to be $\ge 0$. Moreover, since your objective function is convex, this condition is sufficient.

A cool property is that the curvature is null (second partial derivatives are zero) when defined, so as your intuition suggests there exists a solution where each $u_i$ is at point where its partial derivative isn't defined (the upper- and lower- are different), which means that for each $i$ there is a $j$ such that $b_{ij} = u_i - u_j$.

But that point isn't necessarily the one with the biggest $v_{ij}$. It's just a point where there's a compensation between the positive $v_{ij}$ and the negative $v_{ij}$ in the partials... It's more probable that it occurs around a "large" $v_{ij}$, but not necessary.

To solve your problem, you should rather look towards convex optimization methods. I'm a little rusty on those so I won't give you direct references, but I'm sure someone else will be happy to do it. :)


问:
Why is $\det (A-\lambda I)=0$?
I'm not sure I understand the logic behind why $\det (A-\lambda I)=0$ for any non-trivial solution to $(A-\lambda I)x=0$.</p>

答:
If there exists a non-trivial $x$ s.t. $(A - \lambda I)x=0$, that implies that the mapping $(A-\lambda I)$ must project the vector $x$ onto $0$.

This could only be the case if $x$ lies in the null space of $(A-\lambda I)$. If $(A-\lambda I)$ has a null space, that implies that its rows are not linearly independent.

We know for matrices with non linearly independent rows that $det(A-\lambda I) = 0$.
(Consider that with linearly dependent rows, we could use matrix row operations to change an entire row of the matrix to zeroes.)

Title: Why is $\det (A-\lambda I)=0$?


问:
solving $\frac{\pi}{2\sqrt{pq}}$=$\frac{\partial g}{\partial q}$ + $\frac{\partial g}{\partial p}$
I was thinking about an integral and got it into the form:

$$\frac{\pi}{2\sqrt{pq}}=\frac{\partial g}{\partial q} + \frac{\partial g}{\partial p}$$

where $g$ is the integral as a function of $q$ and $p$.

I've not really done much on solving PDEs so was wondering if anyone could outline how you'd go about solving this.

I was thinking the symmetry between $q$ and $p$ is important, but really not sure how to approach the problem.

答:
$$\frac{\partial g(p,q)}{\partial q} + \frac{\partial g(p,q)}{\partial p}=\frac{\pi}{2\sqrt{pq}}$$
The Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dp}{1}=\frac{dq}{1}=\frac{dg}{\frac{\pi}{2\sqrt{pq}}}$$
A first characteristic equation comes from solving $\frac{dp}{1}=\frac{dq}{1}$ :
$$p-q=c_1$$
A second charactristic equation comes from solving $\frac{dp}{1}=\frac{dg}{\frac{\pi}{2\sqrt{pq}}}$ with $q=p-c_1$ :

$dg=\frac{\pi}{2\sqrt{p(p-c_1)}}dp$

$g=\int \frac{\pi}{2\sqrt{p(p-c_1)}}dp=\frac{1}{2}\ln\left|p-\frac12 c_1+\sqrt{p(p-c_1)}\right|+c_2$
$$g-\frac{1}{2}\ln\left|p-\frac12 c_1+\sqrt{p(p-c_1)}\right|=c_2$$
The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ leads to :
$$\boxed{g(p,q)=\frac{1}{2}\ln\left|\frac{p+q}{2}+\sqrt{p\,q}\right|+F\left(p-q\right)}$$
$F$ is an arbitrary function.


问:
First Course in Linear Algebra book suggestions?
I'm starting maths degree this September and hoping to do some reading on Linear Algebra before I start. What are some good introductory books?

Is there something that starts from the very beginning but also covers stuff in depth? (Is there a book that will be useful for both 1st year and 2nd year linear algebra modules?)

It would also be great if the book has lots of exercises (proof questions!) and solutions given as well.

Thanks in advance.

答: A book that is absolutely terrific for self-study is Halmos's Linear Algebra Problem Book. It's a bit different from most other introductory linear algebra books and it's definitely worth a look. Basically it is about 300 pages of carefully crafted problems with hints and complete solutions. All theory is presented as problems and is then further explained in the solutions. I think it's a very nice approach.</p>


问:
A pointwise proposition for $L^2(\mathbb{R}^3)$ functions
Suppose $f\in L^2(\mathbb{R}^3)$, do we have that

$f(x)\rightarrow 0$ as $|x|\rightarrow \infty$?

If this proposition holds, how to prove it? If not, pls give a counter example.

答:
No. Consider
$$
f(x) = \left\{
    \begin{array}{ll}
        1 & \mbox{if } x \in \mathbb{Q}^3 \\
        0 & \mbox{otherwise.}
    \end{array}
\right.
$$


问:
Writing Back-Propogation out in terms of Matrix Multiplication
I am trying to write my own backpropogation algorithm for neural networks for a class. For any specific weight of my network I could easily take the derivative, but for computational speed I want to write the derivatives out in terms of matrix expressions that could be simply coded and executed. Let us use the following notation: let $x\in\mathbb{R}^p$ be our set of $p$ inputs to our network. Let us have $m$ hidden layers of nodes each consisting of $n$ nodes. We will write the values at the $i^{th}$ hidden layer as $z_i$, which are retieved from multipling the $i^{th}$ weight matrix $W_i$ by the previous layer a chosen function $f$ (sigmoid, RELU, etc). Finally, we consider only $1$ output node (since we could loop through the output nodes and do this for all of them, although it'd be nice to generalize).

As such we can say that our network is given as
$$
z_1 = f(W_1 x)
$$
$$
z_2 = f(W_2 z_1)
$$
$$
\vdots
$$
$$
z_n = f(W_n z_{n-1})
$$
$$
\hat{y} = f(W_{n+1} z_n)
$$

Using Mean Squared Error we get that our objective function for a given point is
$$
MSE = (\hat{y} - y)^2
$$

The first update matrix (which is actually a vector) was fairly easy to find the update for. I found that
$$
\frac{\partial(MSE)}{\partial W_{n+1}} = 2(\hat{y}-y)f'(W_{n+1}z_n) z_n = k_n z_n
$$

where $k_n=2(\hat{y}-y)f'(W_{n+1}z_n)$. From here the next layer is also special I found that if we let
$$
\frac{\partial(MSE)}{\partial W_n} = k_n (W_{n+1}\circ f'(W_n z_{n-1}))\otimes z_{n-1} = k_{n-1}\otimes z_{n-1}
$$

where $\circ$ represents the element wise multiplication and $\otimes$ represents the outer product. The final formula I get is the one that should (in theory) work for every layer following, however I am getting a code error on the final step that the matrix dimensions I have do not properly line up, which means I must have a problem with my formula. I find that if
$$
k_i = (k_{i+1}^T W_i)\circ f'(W_i z_{i-1})
$$
then we can say that
$$
\frac{\partial (MSE)}{\partial W_i} = k_i \otimes z_{i-1}
$$

If anyone can help me find an error that'd be much appreicated!

答:
First, note that your $W_{n+1}$ should be a vector to get 1d output. So we need to treat the cases where an index $=n+1$ anywhere differently from cases where all indices $<n+1$.

Using differentials we have
$$dz_i=f'(W_iz_{i-1})\circ dW_iz_{i-1}+f'(W_iz_{i-1})\circ W_idz_{i-1}$$
Using the chain rule for differentials we have
$$dMSE[z_{i+1}(z_i),dz_i]=dMSE[z_{i+1}, dz_{i+1}(z_i,dz_i)]=\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\cdot dz_{i+1}(z_i,dz_i)=$$
$$=\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\cdot f'(W_{i+1}z_i)\circ W_{i+1}dz_i=W_{i+1}^T\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\circ f'(W_{i+1}z_i)\cdot dz_i$$
$\cdot$ in the formulas is the dot product.

Note that in the formula above it is assumed that $W_{i+1}dz$ is a matrix-vector product, so works only for $i+1<n+1$.
So the derivative is
$$\frac{\partial MSE}{\partial z_i}=W_{i+1}^T\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\circ f'(W_{i+1}z_i)$$
Also from the chain rule and the formulas above we have
$$dMSE[z_i(W_i),dW_i]=dMSE[z_i,dz_i(W_i,dW_i)]=$$
$$=W_{i+1}^T\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\circ f'(W_{i+1}z_i)\cdot dz_i(W_i,dW_i)=$$
$$=W_{i+1}^T\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\circ f'(W_{i+1}z_i)\circ f'(W_iz_{i-1})z_{i-1}^T\cdot dW_i$$
So the derivative is
$$\frac{\partial MSE}{\partial W_i}=W_{i+1}^T\frac{\partial MSE(z_{i+1})}{\partial z_{i+1}}\circ f'(W_{i+1}z_i)\circ f'(W_iz_{i-1})z_{i-1}^T$$
To get the closed form formula we have
$$dMSE(z_n, dz_n)=2(\hat{y}-y)f'(W_{n+1}z_n)W_{n+1}\cdot dz_n$$
$$\frac{\partial MSE(z_n)}{\partial z_n}=2(\hat{y}-y)f'(W_{n+1}z_n)W_{n+1}=k$$
So for $i<n$ we have
$$\frac{\partial MSE}{\partial z_i}=W_{i+1}^T\dots W_n^Tk\circ f'(W_nz_{n-1})\circ\dots\circ f'(W_{i+1}z_i)$$
You can plug it into the formula for $\frac{\partial MSE}{\partial W_i}$ to get the closed form of the derivative.

However if you really want to implement backpropogation, this would be an incorrect approach.
The correct approach is to implement backward pass for each of your module with the input of the derivative from above of your computational tree.
For example, taking linear layer $l(W,h)=Wh$ and derivative from above $\frac{\partial MSE}{\partial l}=A$ you have
$$dMSE(l(W,h);dW,dh)=A\cdot dl(W,h;dW,dh)=A\cdot dWh+A\cdot Wdh=$$
$$=h^TA\cdot dW+AW^T\cdot dh$$
So that on backpropagation step this layer gets $A$ and values from forward pass as an input, passes $AW^T$ further (or rather down the chain) and also gives you $\frac{\partial MSE}{\partial W}=h^TA$ for the gradient descent on $W$. The value of $h$ you know from the forward pass.


问:
Cross product and determinants
The determinant of a $3 \times 3$ matrix gives the volume of the parallepiped formed by 3 vectors.  With the cross product, say to find the torque, we find the $3 \times 3$ determinant value of displacement vector $\mathbf{r}$ and force $\mathbf{F}$,
\begin{align}
\boldsymbol{\tau} &= \mathbf{r} \times \mathbf{F}
\\
&=
\begin{vmatrix}
  \mathbf{i}&\mathbf{j}&\mathbf{k}\\
  r_1&r_2&r_3\\
  F_1&F_2&F_3\\
\end{vmatrix},
\end{align}
and the definition of the cross product yields the area vector.

But the determinant by definition is volume. How is the definition of the determinant changed to find the cross product ?

答:
When we write a cross-product as a determinant, we are really abusing notation.  If you look at the expression,
\begin{align}
\mathbf{a\times b}
&= \begin{vmatrix}
  \mathbf{i}&\mathbf{j}&\mathbf{k}\\
  a_1&a_2&a_3\\
  b_1&b_2&b_3\\
\end{vmatrix}
\\
&=
(a_2b_3 - a_3b_2)\mathbf{i} -(a_1b_3 - a_3b_1)\mathbf{j} +(a_1b_2 - a_2b_1)\mathbf{k}
\\
&=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix},
\end{align}
you can see that in each term we have a product of two dimensionful `length' coordinates, giving an interpretation as area.

When we talk about it being numerically the area of a parallogram, we are talking about the norm of this vector,
\begin{align}
\left\lVert
\mathbf{a\times b}
\right\rVert
&=
\sqrt{(a_2b_3 - a_3b_2)^2 + (a_1b_3 - a_3b_1)^2 + (a_1b_2 - a_2b_1)^2}.
\end{align}
Counting the dimensions is easiest if we just consider
\begin{align}
&\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix}
&\mathbf{b} = \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix},
\end{align}
then
\begin{align}
\left\lVert
\mathbf{a\times b}
\right\rVert
&=
\sqrt{(a_1b_2)^2}
\\
&=
a_1 b_2,
\end{align}
ie an area.

Why do we understand a determinant as corresponding to volume?  Consider instead the scalar triple product,
\begin{align}
\left(\mathbf{a\times b}\right) \cdot \mathbf{c}
&=
\left( a_2b_3-a_3b_2 \right) c_1 +
\left( a_3b_1-a_1b_3 \right) c_2 +
\left( a_1b_2-a_2b_1 \right) c_3
\\
&=
\begin{vmatrix}
  c_1&c_2&c_3\\
  a_1&a_2&a_3\\
  b_1&b_2&b_3\\
\end{vmatrix}.
\end{align}

Then if we count the dimensions, eg by choosing
\begin{align}
&\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix}
&\mathbf{b} &= \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix}
&\mathbf{c} = \begin{pmatrix} 0 \\ 0 \\ c_3\end{pmatrix},
\end{align}
to simplify the algebra,
we get
$$
\left(\mathbf{a\times b}\right) \cdot \mathbf{c}
=
a_1 b_2 c_3,
$$
a product of three lengths and hence a volume.

Final comment. Why do we abuse notation to write a cross-product as a determinant?
Really, we are interested in the only totally-antisymmetric tensor in three dimensions, called the Levi-Civita, or alternating, symbol $\varepsilon_{ijk}$, defined as
$$
 \varepsilon_{ijk} =
  \begin{cases}
         +1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1), \text{ or } (3,1,2), \\
         -1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2), \text{ or } (2,1,3), \\
    \;\;\,0 & \text{if } i = j, \text{ or } j = k, \text{ or } k = i
\end{cases}.
$$

Both the cross-product and determinant are properly defined using this, with
$$
\mathbf{a}\times\mathbf{b} = \sum_{i,j,k=1}^3 \mathbf{e}_i \;\varepsilon_{ijk} \, a_j b_k
$$
and
$$
\det A
=
\begin{vmatrix}
  A_{11}&A_{12}&A_{13}\\
  A_{21}&A_{22}&A_{23}\\
  A_{31}&A_{32}&A_{33}\\
\end{vmatrix}
=
\sum_{i,j,k=1}^3
\varepsilon_{ijk} \,
A_{1i}
A_{2j}
A_{3k}
$$
which allows us to use our mnemonic for calculating the latter (an expansion in the minors, the $2\times 2$ submatrices) to remember how to calculate the former.

Title: Cross product and determinants

OwnerDisplayName: JCW


问:
Prove that a polynomial of degree $d$ has at most $d$ roots (without induction)

Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$.

Is it possible to prove this without using induction on degree? If so, how can I do this?

答:
I'm not sure if using the residue theorem counts, and this only works over $\Bbb C$, but :

Look at $\int P'(z)/P(z) \;dz$ where you integrate on a circle $|z| = r > |x_i|$.

As $r \to \infty$, this is equivalent to $\int d/z \;dz = 2id\pi$.

Meanwhile, the residue theorem says that the integral is $\sum 2i\pi \mu_i$ where $\mu_i$ is the multiplicity of the root $x_i$. Hence the number of roots, counted with their multiplicity, is equal to $d$, the degree of the polynomial.


问:
Manifolds with boundaries: Why map boundary points onto boundary points?
The notion of a manifold with boundary has just been introduced in my script. If $\mathbb{R}^n_+$ is defined as $\mathbb{R}^n_+ := \{ (x^1, \ldots, x^n) \in \mathbb{R}^n \; | \; x^n \geq 0 \}$, then a manifold with boundary $M$ is an atlas whos charts $\phi : M \longrightarrow V \subseteq \mathbb{R}^n_+$.

Now, as far as I understand, it is required that the preimage of each point in $\mathbb{R}^n$ with $x^n = 0$ is a boundary point on $M$. In other words each boundary point of $M$ gets mapped onto the hyperplane defined by $x^n = 0$.

I don't really understand this requirement. I think my intuition of either this definition or of a manifold/coordinate chart is wrong. I always understood the coordinate charts as local coordinates of the manifold in Euclidean space. A ball in 3 dimensions for example can be described by $x,y,z$. But describing its boundary with $z = 0$ is pretty uncommon when parametrising a ball, isn't it?

How is that definition or the coordinate charts of a manifold to be interpreted with respect to this example?

答:
Get rid of that image of a manifold given by an embedding into some $\Bbb R^n$. An atlas describes the manifold without reference of any surrounding space (if such a space exists at all). Think of a classical atlas of the (surface of the) Earth: Nowhere do you work with three-dimensional coordinates, but rather on each page of the atlas is a flatt part of $\Bbb R^2$ (unlike when working with a globe instead of an atlas), you can reference a point on Earth (provided it is in the region depicted in that map) by two coordinates only (e.g., distance from left and from top edge of the page). Thanks to the various maps overlapping (though with some distortions), you can manage to navigate across all of the world.

Now for a manifold with boundary, imagine we cut a hole into the globe, e.g., we remove the continent on Antarctica and deny its existence, or we imagine that its boundary is some unpassable boundary (perhaps an ice wall guarded by armed penguins?). If you look at your standard atlas from your shelf and imagine that you simply erase all parts of Antarctica as terra incognita or more precisely as terra non existens, you will get some maps that extend only to parts of the pages, and only for some directions there are maps that show some extension. For convenience, you may assume that all those maps showing some boundary are distorted to make the a priori wildly curved boundary straight and transform the map further to bring that straight boundary to the $y=0$ line and perhaps write "Here be Dragons" on the other side of that line. (It may be necessary to split up some maps into several, namely when you originally had a map that showed the complete closed boundary of Antarctica). This doesn't change the fact that the atlas can be used to navigate the modified world, it just simplifies our notion of a maps with boundary and thereby simplifies working with such an atlas.


问:
Probability of a coin dropping in a box
suppose I am going to toss a coin of diameter 1 inch in a square box that has length of 2 inch .Whats the probability of the coin dropping in the box but will not touch the border?</p>

答:
With a simple drawing you realize that the probability is

$$\frac{1^2}{2^2}=\frac{1}{4}$$

that is the area of the red square divided by the total area

Title: Probability of a coin dropping in a box


问:
There is a point on the road from where the car with initially empty fuel tank can complete a loop.
I need help proving the following statement:

There are $n$ gas stations along a circular road. The total amount of fuel at these stations is exactly the amount needed for a car to complete one loop. Show that there is a point on the road from where the car with initially empty fuel tank can complete a loop.

Here is my progress:

Note that a successful starting point must always be at one of the gas stations since; otherwise, the car will not be able to move due to the empty fuel tank. Before tackling the proof, consider the following example with four gas stations. Label the gas stations

$g_1, g_2, g_3, g_4$

so that the amount of fuel at gas station $g_i$ is $x_i$ for $1 \leq i \leq 4$. Define the distance between consecutive gas stations to be $d(g_i, g_{i+1})$. Then $x_1 + x_2 + x_3 + x_4 = 1$ and $d(g_1, g_{2}) + d(g_2, g_{3}) + d(g_3, g_4) + d(g_4, g_1) = 1$. We want to make $d(g_i, g_{i+1})$ and $x_1, ..., x_4$ arbitrary so; for example, consider $x_1 =0.2,x_2 = 0.3 ,x_3 = 0.4, x_4= 0.1$ and $d(g_1, g_{2}) = 0.3 , d(g_2, g_{3}) = 0.1 , d(g_3, g_4)= 0.2, d(g_4, g_1) = 0.4$. We want to find a starting gas station so that we can get back to it. If the starting point is at $g_2$ and fill the car with fuel, call it $X$, we have $X=0.3$. We can then travel to the next gas station $g_3$ but will be left with $X = 0.3 - d(g_2, g_{3}) = 0.2$. After fueling at $g_3$, $X = 0.7$ and traveling to $g_4$ we have $X= 0.7 - d(g_3, g_4) = 0.5$. After fueling at $g_4$, $X = 0.6$ and traveling to $g_1$ we have $X= 0.6 - d(g_4, g_1) = 0.2$. Therefore, we have completed a loop so at least one starting point exists and that is at $g_2$.

claim:

If $\sum_{i =1}^k x_i = 1$ and $\sum_{i=1}^k d(g_i, g_{i+1}) = 1$

then there exists an $x_i$ where $i = 1, ..., k$ such that $x_i \geq d(g_i, g_{i+1})$.

proof of claim:
By way of contradiction suppose $x_i < d(g_i, g_{i+1})$

for some $1 \leq i \leq k$. Then, $\sum_{i=1}^k d(g_i, g_{i+1}) = 1$ implies $\sum_{i = 1}^k x_i < 1$ which is a contradiction.

proof of problem:
We induct on the number of gas stations. Suppose there is only one gas station along the circular road such that the total amount of fuel at this station is exactly the amount needed for a car to complete one loop. Therefore, there is only one possible starting point for the car which is at the single gas station. Now, suppose there is always a successful starting point when there are $k$ gas stations. That is, for gas stations

$g_1, ..., g_k$ with fuel amounts of $x_1, ..., x_k$, respectively, such that $d(g_1, g_2) + ...+ d(g_k, g_1) = 1$ and $x_1 + ... + x_k = 1$.
The total amount of fuel at the
$k$ stations is exactly the amount needed for a car to complete one loop. We want to show that there is always a successful starting point when there are $k+1$ gas stations so consider a gas station with $k+1$ gas stations. By the claim above, there is at least one gas station that contains enough gas to reach the next gas station, call this gas station $h_1$ and consider the gas station that comes
after
$h_1$; call it, $h_2$.

I am not sure where to go from here.

An idea:

Combining the fuel from $h_1$, $h_2$ and proceeding up to $h_k$, we have enough fuel for a car with an initial empty fuel tank to make a loop so that a starting point exists, namely at $h_1$.

But how do we know that the fuel from $h_2$ will get us to the next gas station?

答:
Let me take a somewhat different approach. Starting from any given gas station, you can get some distance around the loop before running out of gas. So some starting point will get you the maximum possible distance around the loop. We might as well number things so that gas station $1$ is a distance-maximal starting point.

Now if the maximum distance isn't the complete loop, then we must have run out of gas with at least one station unvisited. Let's say we fell short of gas station $m\le n$. Now a bit of backward induction takes over: there couldn't be enough gas in station $n$ to get to station $1$, or we would have done better by starting at station $n$ instead of station $1$. Likewise, there couldn't be enough gas in stations $n-1$ and $n$ combined to get from station $n-1$ to station $1$. And so forth, down to station $m$. But this is a contradiction: the combined gas in stations $m$ to $n$ is enough to drive the entire rest of the way, from where you ran out of gas all the way to station $1$, so it must be more than enough to get from station $m$ to station $1$.


问:
Continuous coloring of a Mandelbrot fractal
I've recently started making a small fractal app in Javascript using the famous Mandelbrot bulb $(z = z^2 + c)$. I've been trying to find the best method of coloring the points on the complex plane, and I've come across some very interesting ideas:

http://linas.org/art-gallery/escape/escape.html

http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Real_Escape_Time

So I've been focusing on this "normalized iteration count algorithm" and I've run into a small mental hurdle sorting out what needs to be calculated. Basically, I'm confused about the first equation that shows up in that second link. Assuming we have a complex number $Z$ which is the end result of $n$ iterations. I can't quite figure out what "Zx2" or "zy2" is and how how one might add them together or take the square root of their sum. Is that just shorthand for some basic complex operators, or is there something else going on? Basically, my problem is that I'm not particularly good at reading mathematical notation in this area.

Anyway...any directional guidance you can offer would be extremely helpful. Thanks in advance!

答:
Based on the C code given there, Zx is just the real part $\Re z$ and Zy is just the imaginary part $\Im z$.

Remember that the complex iteration formula for the Mandelbrot set

$z=z^2+c$

when expanded to real variables looks like

$x=x^2-y^2+u$

$y=2xy+v$

where $z=x+iy$ and $c=u+iv$.

The variables with a 2 appended are the squares, as can be seen from how they were defined, so for instance $|z|=\sqrt{x^2+y^2}$ is sqrt(Zx2+Zy2) which is used for determining how much the iteration diverges (escapes to infinity), which is used by the coloring functions.

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