问:
Prove that if $n\geq\text{lcm}(a,b)$ and $\gcd(a,b)|n$ then $n=xa+yb$ for some integers $x,y\geq 0$
I thought I had it, but then I realized I didn't. Even just a hint—am I going in the right direction or should I try something completely different?

We know that $\gcd(a,b)=wa+zb$ for some integers $w,z$. Then since $\gcd(a,b)|n$ we have that $k \gcd(a,b) = n$ for some integer $k$. Then $n=kwa+kzb$, so letting $x=kw$ and $b=kz$ gives the result.

Except that it doesn't because I realized that there's no promises about $kw$ and $kz$ being nonnegative and in fact I think that in general they are not.

I think there must be some way to reconcile this using $n\geq\text{lcm}(a,b)$ but I am not sure how to. Or maybe this argument was doomed from the start. Anyone have a hint?

答:
Hint: If $n=xa + yb$, then $n=(x-b)a + (y+a)b$.

If you start with $x>0$ and $y<0$, you can keep applying this hint repeatedly, until $y$ becomes positive. Of course, $x$ will keep decreasing at the same time... but can it become negative before $y$ jumps above zero?


问:
Hint: vector space $V=\{f:S\rightarrow F\}$ is a finite dimensional vector space over F if and only if $S$ is finite.
No answers, please! Just a hint.

Let $S$ be a set and let $F$ be a field. Prove that the vector space $V=\{f:S\rightarrow F\}$ of all maps from $S$ to $F$ is a finite-dimensional vector space over $F$ if and only if $S$ is finite.

EDIT: Thank you all for the hints. This is my proof. How is it?

Suppose $S$ is finite. Let's define $\delta_s$ as follows:
$$\delta_s(x)=\begin{cases}
1&x=s\\
0&x\neq s
\end{cases}$$
We can express any function $f$ in terms of $\{\delta_s \mid s \in S\}$: $f = \sum_{s \in S} a_i \delta_s$, so $\{\delta_s \mid s \in S\}$ is a finite spanning set. Thus, $V$ is a finite-dimensional vector space. Conversely, suppose to the contrary that $S$ is infinite. Then  $\{\delta_s \mid s \in S\}$ is an infinite spanning set. However, this contradicts the fact that $V$ is a finite-dimensional vector space, and so $S$ must be finite.

答:
Hints:

Define $f+g$ and $af$ for $a\in F$ and $f,g\in V$.

Find a basis of $V$.

ContentLicense: CC BY-SA 4.0


答: Hint: think about the functions $(\delta_s:s\in S)$ defined by $\delta_s(x)=1$ if $x=s$ and $\delta_s(x)=0$ if $x\ne s$.

Title: Hint: vector space $V=\{f:S\rightarrow F\}$ is a finite dimensional vector space over F if and only if $S$ is finite.

Body:
No answers, please! Just a hint.

Let $S$ be a set and let $F$ be a field. Prove that the vector space $V=\{f:S\rightarrow F\}$ of all maps from $S$ to $F$ is a finite-dimensional vector space over $F$ if and only if $S$ is finite.

EDIT: Thank you all for the hints. This is my proof. How is it?

Suppose $S$ is finite. Let's define $\delta_s$ as follows:
$$\delta_s(x)=\begin{cases}
1&x=s\\
0&x\neq s
\end{cases}$$
We can express any function $f$ in terms of $\{\delta_s \mid s \in S\}$: $f = \sum_{s \in S} a_i \delta_s$, so $\{\delta_s \mid s \in S\}$ is a finite spanning set. Thus, $V$ is a finite-dimensional vector space. Conversely, suppose to the contrary that $S$ is infinite. Then  $\{\delta_s \mid s \in S\}$ is an infinite spanning set. However, this contradicts the fact that $V$ is a finite-dimensional vector space, and so $S$ must be finite.


问:
Has the set of all ideals of a ring a ring-like structure?
After learning about ideals of a ring and the basic operations that can be made with them, I started wondering about the following situation.

Let $R$ be a commutative ring with unity (for simplicity), and denote by $I(R)$ the set of all ideals of $R$. The sum of two ideals, $I$ and $J$, is given by: $I+J=\{i+j:i \in I,j \in J\}$, and their product is: $IJ=\{\sum_{i=1}^n x_iy_i:x_i \in I, y_i \in J, n \in \mathbb{N}^\ast \}$. Then:

$i)$ $(I(R),+)$ is a commutative monoid, with identity element $(0)$;

$ii)$ $(I(R),\cdot)$ is also a commutative monoid, with identity element $R$;

$iii)$ We have distributivity: $I(J+K)=IJ+IK$ and $(I+J)K=IK+JK$, for $I,J,K \in I(R)$.

Now my questions:

$1)$ Are all of the above, $i),ii),iii)$ correct? I am quite unsure about the product being associative (I failed to prove for myself, so any proof/indication would be appreciated). The other axioms seem to hold.

$2)$ If $1)$ is true, then what type of structure is $(I(R),+,\cdot)$?

Thank you in advance!

答:
Yes, your claims are all true. To see the associativity $(IJ)K = I(JK)$ (as you asked), take an element
$x \in (IJ)K$.
It can be written as
$$x = \sum_i a_ib_i$$
with $a_i \in IJ$ and $b_i \in K$. We can further write $a_i = \sum_{j=1}^{n_i}  x_j^i y_j^i$ with $x_j^ i \in I$ and $y_j^i \in J$. Thus
$$x= \sum_i \sum_{j=1}^{n_i} (x_j^iy_j^i)b_i = \sum_i \sum_{j=1}^{n_i} x_j^i (y_j^i b_i) \in I(JK)$$
and the other inclusion is similar. We thus see that your associativity follows from associativity of ring multiplication, as could be expected.


问:
Show that this ODE with boundary conditions has a unique solution
We have the ODE $$-\ddot{x}+q(t)x=g(t),\quad x(a)=x_a,\, x(b)=x_b$$
with $g\in C([a,b],\mathbb{R}$ and $q\in C([a,b],\mathbb{R}_0^+)$. Show that there is a unique solution.

First of all, I transformed this problem in the form
$$\dot{X}(t)=U(t)X(t)+b(t),
$$
where $X=(x,\dot{x})^T$, $U(t)=\begin{pmatrix}0 & 1 \\ q(t) & 0 \end{pmatrix}$, $b=-(0, g(t))^T$. Also we can write the boundary condition as
$$AX(a)+BX(b)=r(X(a),X(b)),\quad A=\begin{pmatrix}1 & 0 \\0 & 0\end{pmatrix},\, B=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}.
$$
I know that there is a theorem that states that this problem has a unique solution if
$$E=A+BW(b,a)
$$
is invertible, where
$$ \frac{\partial}{\partial t}W(t,a)=f_x(t,X(t))W(t,a)=U(t)W(t,a).
$$
How would I show that? Or is there some other way to show it?

EDIT: Forgot the "x" in the equation.

答:
You already saw that you can apply linear algebra to the related initial value problem. If $x_0$ is the solution to the ODE with IC $x_0(a)=x_a$ and $\dot x_0(a)=0$, and $y$ the solution to the homogeneous equation $-\ddot y+qy=0$ with IC $y(a)=0$, $\dot y(a)=1$, then $x=x_0+v_ay$ is a solution of the first ODE with $x(a)=x_a$ and $\dot x_a=v_a$.

Now the question can be answered when the second boundary condition can be satisfied.

If $y(b)\ne 0$, then $x_b=x(b)=x_0(b)+v_ay(b)$ can be solved for $v_a$.

If $y(b)= 0$, and accidentally $x_0(b)=x_b$, then any initial speed will land at the second boundary condition.

In the remaining case there are no solutions. Up to here this is true for any second order linear DE. Now if $y(a)=y(b)=0$ for this specific case, and as $\dot y(a)=1$, there has to be a maximum $y(c)>0$ inside the interval. But then $0>\ddot y(c)=q(c)y(c)>0$, which is impossible.


问:
Does this IVP have a unique solution for all $x \in \mathbb R$
Is $\displaystyle {dy\over dx}=\sin(y)$ with initial conditions $y(X)=Y$ guaranteed to have a unique solution for all $x\in\mathbb R$? </p>

答:
Yes, there is a unique solution to this problem on all $\mathbf R$. The local existence and uniqueness follow from the standard theorems, since the right hand side is $C^\infty(-\infty,\infty)$ in both variables. To show that the solution exists globally note that $|\sin y|\leq 1$ for any $y$, hence the solution to the problem with the initial conditions, e.g., $y(0)=0$ is between $y_{-}=-x$ and $y_{+}=x$. Since for any $x$ $y_{-}\leq y(x)\leq y_{+}$, the solution cannot blow up and can be always extended on the whole real line.

Added: However, a better argument might be that for any initial conditions $y(x_0)=y_0$ there are always two equlibria $y_{-}$ and $y_{+}$ such that $\sin y_{-}=\sin y_{+}=0$ and $y_{-}<y_0<y_{+}$, hence the solution is bounded, and hence can be extended for any interval, including $\mathbf R$.

ContentLicense: CC BY-SA 3.0


问:
Dimension of the space of cusp forms
I guess this is rather basic, and as much a reality check as a question.

We have the classical equations for the dimension of the space of modular forms of weight $k$ for $\Gamma(1)$:

$$\dim M_{12j+r} = \left\{
  \begin{array}{lr}
    j+1 &  r=0,4,6,8,10\\
    j & otherwise
  \end{array}
\right.
$$

Now, in Daniel Bump's "Automorphic Forms and Representations", it says:

$$\dim M_{k}=\dim S_{k}+1$$ for $k \geq 4$. I guess that is a mistake, since the leading discussion makes clear that the extra modular form is an Eisenstein series, and those only exist for even $k \geq 4$. Or am I perhaps missing something?

Computing $\dim M_{k}$ for $k \leq 16$ gives me:

$0,0,0,1,0,1,0,1,0,1,0,2,1,1,1,2$

By the preceding reasoning, this means that the first cusp shows up in $k=12$, the very famous $\Delta$. The second would be $k=16$ (which one is that?).

Or to put it another way, $\dim S_{k}$ would be:

$0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1$

* My question is, is this correct?

答: The second one would be the product $E_4 \; \Delta$ which is an eigenform (multiplicative coefficients, completely non-obvious at first) because the cuspidal subspace is 1-dimensional.

Title: Dimension of the space of cusp forms


问:
Conditional Markov Chain
I am interested in a method of how to compute the expected number of steps for absorpion in a Markov Chain with only one absorption node and given the starting and pre-final node (before absorption).

So, if the transition probability matrix is given, for example by:

$P = \begin{bmatrix}
0 & 1/5 & 2/5 & 1/5 & 1/5\\
1/5 & 0 & 1/5 & 1/5 & 2/5\\
2/5 & 0 & 1/5 & 1/5 & 1/5\\
1/5 & 1/5 & 0 & 1/5 & 2/5\\
0 & 0 & 0 &0 & 1
\end{bmatrix}$

what will be the expected number of steps for being absorbed given that we start in node 1 (first line) and the last state before absorption is node 2 (second line) ?

So what I tried so far... I know the very known technique of using the fundamental matrix $N$ and then computing $N = (I-Q)^{-1}$ where $Q$ is the recurrent probabilities matrix (from the canonical form of $P$) and the expected number of steps would be the first entry of the vector $(I-Q)^{-1}1$, where $1$ is the five dimensional vector with all entries equal to $1$, in the case where we dont know the information about where we are absorbed.

Then, as we know that we are absorbed from node two, I tried to replace all the probabilties of the last columns by $0$ (execept for line $2$ and for the aborption node that I kept $2/5$ and $1$, respectively), then I am not pretty sure of how re-normalizing all the others...

Thank you very much!

答:
I'm slightly correcting my comments, and turning it into an answer. Let $P$ be the Markov chain you're interested in. Write $j_i$ for the initial state, $j_f$ for the last state before absorption, and $j_a$ for the absorption state. Write $P'$ for the sub-Markov chain consisting of all states except $j_a$. If $i$ and $j$ are states in a Markov chain, write $\mu(i,j)$ for the expected amount of steps needed to reach $j$ given that you start at $i$.

To find the expected amount of time to reach absorption given that you're in state $j_f$ right before that, what you'll want to do is make a summation over the amount of times that you've passed through $j_f$ before landing in $j_a$. With probability $P_{j_f,j_a}$, you'll need to pass through $j_f$ only once. With probability $(1 - P_{j_f,j_a}) P_{j_f,j_a}$, you'll need to pass through $j_f$ twice. And so on. Given that you pass through $j_f$ precisely $n+1$ times, the expected amount of time that you'll need to reach $j_a$ is
$$\mu(j_i,j_f) + n \cdot\mu(j_f,j_f) + 1,$$
where $\mu$ is computed in $P'$ (not in $P$; this is because we assume that on our path from $j_f$ back to itself, we don't get absorbed along the way). Now it's a matter of appropriate summation. Modulo errors I think the answer will be
$$E(T) = \sum_{n = 0}^\infty P_{j_f,j_a} (1 - P_{j_f,j_a})^{n} \big(\mu(j_i,j_f) + n \cdot \mu(j_f,j_f) + 1\big)$$
which can be simplified to
$$E(T) = \mu(j_i,j_f) + 1 + \sum_{n = 0}^\infty n P_{j_f,j_a} (1-P_{j_f,j_a})^{n} \mu(j_f,j_f),$$
and yet further to
$$E(T) = \mu(j_i,j_f) + \frac{1-P_{j_f,j_a} }{P_{j_f,j_a}}\mu(j_f,j_f) + 1.$$

Here's an approach for finding $\mu$. In a general Markov chain $Q$, the expected amount of time needed to pass from state $i$ to state $j$ satisfies the recurrence relation
$$\mu(i,j) = 1 + \sum_{k \neq j} Q_{i,k} \mu(k,j),$$
where the sum is taken over all states $k$ not equal to $j$. This formula should be enough to find $\mu(j_i,j_f)$ and $\mu(j_f,j_f)$ in $P'$ numerically.

Title: Conditional Markov Chain


问:
Why $\mathbb Q^n$ is not a lattice of $\mathbb R^n$?
We can show that $\mathbb Z^n$, additive subgroup of $\mathbb R^n$, is a lattice and intuitively see that it might not be possible to generate $\mathbb Q^n$ as integral multiple of $m$, $m\leq n$, linearly independent vectors in $\mathbb R^n$. But can you give a proof, I am having difficulty grasping the definition of lattices itself.

答:
$\Bbb Q^n$ cannot be finitely generated as an abelian group (for $n\geq 1$). For simplicity assume that $n=1$, the general case is similar. Let $r_1,\dots,r_k$ be rational numbers and $d\in\Bbb Z$ a common denominator of them. Then it is clear that the (additive) subgroup of $\Bbb Q$ generated by the $r_1,\dots,r_k$ consists only of numbers which can be written as a fraction with $d$ in the denominator. So $r_1,\dots,r_k$ cannot generate $\Bbb Q$.

So $\Bbb Q^n$ is a non-finitely generated abelian group, hence it is in particular not possible to write $\Bbb Q^n$ as the $\Bbb Z$-span of finitely many linearly independent vectors in $\Bbb R^n$.


问:
Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.
Determine all three-digit numbers N having the property that N is divisible
by 11, and N/11 is equal to the sum of the squares of the digits of N.</p>

答:
To start, it's important to define some variables for the 3-digit number set let's define the set A = {100,101,102...,997,998,999}.

Make A the domain of a function f(x), such that f(x) = x mod 11, and now for all value of x which f(x)=0, we have the set S defined as {10,11,12,13...87,88,89,90}

(realize that neither 9 nor 91 is not contained in the set because and 9 x 11 =99 and 91 x 11 = 1001, are not 3-digit numbers)

Now, in order to define which are the square digits of n let's define our 3-digit number in function of our new set S.

S = {10,11,12...87,88,89,90} where S is set set of numbers in the format "xy", where x and y are natural numbers

eg.S = {xy} = {10} => x=1,y=0 and so on.

11 S = {110,121,132,...,968,979,990}

Multiply the "literal" set by 11
xy
x 11

and separate between the decimal units

if (x+y)<10

C
D
U

x
(x+y)
y

or

C
D
U

x+1
(x+y)mod 10
y

and the sum the square of those digits must be equal to xy itself

and we'll have 2 cases

#1st case
for this case, we'll have to choose 2 digits 0-9, such that there's no zero on the first digit and the sum of those digits is less than 10 , which gives us a total of 45 comparisons. but now let's find a way to make those faster

C
D
U

x
(x+y)
y

x²+(x+y)²+y² = "xy" (the quotation marks are to represent that xy is a number not a product)

Expanding...
x²+x²+2xy+y²+y²= "xy" => 2(x²+xy+y²)= "xy" =>

x²+xy+y² = $\frac{"xy"}{2}$

as we defined earlier x and y are natural numbers and this means that y must be divisible by 2 which reduces ours comparisons to 25.

we can also make this number of comparisons smaller once we start iterating setting a value to x, starting at 1

for x=1

1²+y+y²= $\frac{"1y"}{2}$ (where "1y" could be {10,12,14,16,18})

but before the comparison you could test if the number is in the desired range by doing y(y+1) $\le$ $\frac{(x+1)"0"}{2}$ -x²

y(y+1) $\le$ $\frac{(1+1)"0"}{2}$-1 =>

y(y+1) $\le$ 10-1 => y(y+1) $\le$ 9, and with this affirmation we can reduce the number of possibilities for 1, from 5 to only 2 which are:

for y=0

y(y+1)= $\frac{"1y"}{2}$-1 =>
0(0+1)= $\frac{"10"}{2}$-1 => 0 = 4 false

for y=2

y(y+1)= $\frac{"1y"}{2}$-1 =>
2(2+1)= $\frac{"12"}{2}$-1 => 6 = 5 false

for x=2

y(y+1) $\le$ $\frac{(2+1)"0"}{2}$ -2² and with this affirmation we can reduce the number of possibilities for 2, from 4 to 3(from now on this approach is not effective anymore to use) which are:

for y=0

y(y+1)= $\frac{"2y"}{2}$-4 =>
0(0+1)= $\frac{"20"}{2}$-4 => 0 = 6 false

for y=2

y(y+1)= $\frac{"2y"}{2}$-4 =>
2(2+1)= $\frac{"22"}{2}$-4 => 6 = 7 false

for y=4

y(y+1)= $\frac{"2y"}{2}$-4 =>
4(4+1)= $\frac{"24"}{2}$-4 => 20 = 8 false

for x=3

for y=0

y(y+1)= $\frac{"3y"}{2}$-9 =>
0(0+1)= $\frac{"30"}{2}$-9 => 0 = 6 false

for y=2

y(y+1)= $\frac{"3y"}{2}$-9 =>
2(2+1)= $\frac{"32"}{2}$-9 => 6 = 7 false

for y=4

y(y+1)= $\frac{"3y"}{2}$-9 =>
4(4+1)= $\frac{"34"}{2}$-9 => 20 = 8 false

for x=4

for y=0

y(y+1)= $\frac{"4y"}{2}$-16 =>
0(0+1)= $\frac{"40"}{2}$-16 => 0 = 4 false

for y=2

y(y+1)= $\frac{"4y"}{2}$-16 =>
2(2+1)= $\frac{"42"}{2}$-16 => 6 = 5 false

for y=4

y(y+1)= $\frac{"4y"}{2}$-16 =>
4(4+1)= $\frac{"44"}{2}$-16 => 20 = 4 false

for x=5

for y=0

y(y+1)= $\frac{"5y"}{2}$-25 =>
0(0+1)= $\frac{"50"}{2}$-25 => 0 = 0 TRUE

for y=2

y(y+1)= $\frac{"5y"}{2}$-25 =>
2(2+1)= $\frac{"52"}{2}$-25 => 6 = 1 false

for y=4

y(y+1)= $\frac{"5y"}{2}$-25 =>
4(4+1)= $\frac{"54"}{2}$-25 => 20 = 2 false

for x=6

for y=0

y(y+1)= $\frac{"6y"}{2}$-36 =>
0(0+1)= $\frac{"60"}{2}$-36 => 0 = -6 false

for y=2

y(y+1)= $\frac{"6y"}{2}$-36 =>
2(2+1)= $\frac{"62"}{2}$-36 => 6 = -5 false

Since we're receiving negative values as output now, let's assume that no other value greater than 6 will reach our goal so, let's move on to the final step

as we have tested for values smaller than 6 let's take a guess and start the second case at 6

#2nd case
(x+y)$\ge$ 10

we will have the following equation:

(x+1)² + [(x+y) mod 10]²+ y² = "xy"

for x=6
[(6+y) mod 10]²+ y² = "6y" - (6+1)²

for y=4

[(6+y) mod 10]²+ y² = "6y" - 49

[(6+4) mod 10]²+ 4² = "64" - 49 => 16 = 15 false

for y=5

[(6+y) mod 10]²+ y² = "6y" - 49

[(6+5) mod 10]²+ 5² = "65" - 49 => 26 = 16 false

for y=6

[(6+y) mod 10]²+ y² = "6y" - 49

[(6+6) mod 10]²+ 6² = "66" - 49 => 40 = 17 false

as we've also seen by this sequence the left side is growing must faster than the right side, therefore, it will never be equal, let's move on for the next value of x

for x=7
[(7+y) mod 10]²+ y² = "7y" - (7+1)²

for y = 3

[(7+y) mod 10]²+ y² = "7y" - 64

[(7+3) mod 10]²+ 3² = "73" - 64² => 9 = 9 TRUE

for y = 4

[(7+y) mod 10]²+ y² = "7y" - 64

[(7+4) mod 10]²+ 4² = "74" - 64² => 17 = 10 false

for the same reason than the last one the left side now is growing faster than the right side so they will never meet again

for x=8

[(8+y) mod 10]²+ y² = "8y" - (8+1)²

for y = 2

[(8+y) mod 10]²+ y² = "8y" - 9²

[(8+2) mod 10]²+ 2² = "82" - 81 => 4 = 1 false

from now on the left side side of the equation will always be greater than the right one so, we have found every number that multiplied by eleven, the sum of their digits squared equals itself.

The number we found out were = {50,73}

To set them in the format of the question, all you gotta do is multiply them by 11

and there you go

50x11 = 550
73x11 = 803

Solution = {550, 803}

I hope this answer is useful to you, the resolution got more extensive than I imagined, and the great challenge of questions like this is to answer them fast, you can learn new methods to go faster, but the most important part is training. Thank you for the question. Please, if there's any new question or suggestion in order to make my resolutions better please let me know.


问:
Proof $(A\oplus B) \times C = C \times (A-B)$
$(A\oplus B) \times C = C \times (A-B)$

For starters, $((A\cap B)^c\cap (A\cup B)) \times C  = C \times (A \cap B^c)$

I don't know how to continue from here since there's Cartesian product

Body:
It's false. Let $A=\{1\}$, $B=\{2\}$ and $C=\{3\}$ so that
$$
(A\oplus B)\times C= \{1,2\}\times\{3\} = \{(1,3),(2,3)\}
$$
while
$$
C\times (A-B) = \{3\}\times\{1\} = \{(3,1)\}.
$$


问:
If $d(x,A)=0\forall x\in X$ for some subset $A$ of $X$, does it follow that $A$ is dense?
If $d(x,A)=0 \:\:\forall x\in X$ for some subset $A$ of $X$ then $A$ is dense in $X$, right?

Once I did one problem which says $d(x,A)=0\Leftrightarrow x\in \bar{A}$ so by the condition here we get $\bar{A}=X$.

Other options were

$2$. $A$ is closed

$3$. $A=X$

$4$. $A$ is compact

A counterexample for $2, 3$ and $4$ is $A=\mathbb{Q}\subseteq\mathbb{R}=X$

答: Look at $B(x,\epsilon)$. Since $d(x,A)=0 \forall x \in X$, $\exists a \in A$ such that $a\in B(x,\epsilon)$. Since these open balls form a base of the topology of $X$, every open set $U \in X$ contains an element of $A$. Hence $A$ is dense in $X$.</p>


问:
Why the probability of single element event is "zero" on continuous interval?
Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.

cool thanks!

答:
Suppose otherwise, so that the chance of picking any element is positive, say $p$. Assuming uniform distribution, this probability is equal for all values in the interval by definition. Then for each of the infinitely many values in the interval, it is chosen with probability $p$, so the total probability would be infinite, which is absurd.

Of course, you don't need a uniform distribution either. You can easily convince yourself that summing uncountably many positive values is going to give you an infinite result.


问:
Flow doesn't change volume
Given an autonomous equation $y'=f(y) \in \mathbb{R} ^m$, where $\mathrm{div} f=0$ for whole domain of equation. Show that flow $\phi^t$ of this equation doesn't change volume, ie. whenever $\phi^t$ is defined on a measurable set $E$ then $\mathrm{Vol} \left(\phi^t\left(E\right) \right) = \mathrm{Vol} \left(E\right) $.

I have tried doing it by some kind of change of variables in integration, knowing that $\phi^t$ is a diffeomorphism, but I couldn't finish it off.

答:
As the point dynamic is approximately $y(t+dt)=y(t)+f(y(t))dt$, the tangent space is propagated over the time step $dt$ via the linear map $I+f'(y)dt$. The volume change is the determinant of that matrix, and
$$
\det(I+f'(y)dt)=1+{\rm tr}(f'(y))dt+O(dt^2)=1+{\rm div}(f)_{y(t)}dt+O(dt^2)
$$

Title: Flow doesn't change volume


问:
How to find the third root of a complex number by transforming the complex number into a root
I'd like to find all 3rd roots of this number z = i - 1. Now I've found formulas on how to do it; First we transform the complex number into this form

$$ \sqrt[n]{r} * e^{i\frac {\phi + 2k\pi}{n}} $$ Where n should be 3 (because of 3rd root) and k should be k = n - 1 (Including 0)

Now what is not really clear to me is what to do with this -1? If I transform this I'd pressume I get this.

$$\sqrt[3]{-1} * e^{i\frac {\pi/2 + 2k\pi}{3}}$$

But than what? If it was a 2nd root I could say that it is simply i (the root of a negative number) but since it is a 3rd root I'm not quite sure what to do next?

Now for the phi (angle) I just assumed this;  $$i = e^{i \frac {\pi}{2}}$$ And I'm not too sure about this

Thanks!

EDIT: With great help from the people in the comments; I think I've solved it.So as already stated in the comments; $$ r = \sqrt 2, \phi = \frac{3\pi}{4} $$ Now If i would to input that into my formula we should  get this;

$$ \sqrt[6]{2} * e^{i\frac{3pi/4 + 2k\pi}{3}} $$

Now as stated before k = 0... n-1;

k = 0 --> $$ \sqrt[6]{2} * e^{i(\pi/4)} $$
k = 1 --> $$ \sqrt[6]{2} * e^{i(11\pi/12)} $$
k = 2 --> $$ \sqrt[6]{2} * e^{i(19\pi/12)} $$

You reckon I did this correctly?

答: Note that\begin{align}i-1&=-1+i\\&=\sqrt2\left(-\frac1{\sqrt2}+\frac i{\sqrt2}\right)\\&=\sqrt2e^{3\pi i/4}.\end{align}Therefore, the cubic roots of $i-1$ are$$\sqrt[6]2e^{\pi i/4},\ \sqrt[6]2e^{\pi i/4+2\pi i/3}(=\sqrt[6]2e^{11\pi i/12})\text{, and }\sqrt[6]2e^{\pi i/4+4\pi i/3}(=\sqrt[6]2e^{19\pi i/12}).$$


问:
How to use central difference formula if step size is not constant
In the problem I am solving, table values are given for function and it says to find $f''(0.5)$ using second order central difference formula. I know the formula which is
$\frac{f(x+\triangle x)-2f(x)+f(x-\triangle x)}{{\triangle x}^2}$. But the problem is, table values are as follows(I am listing values around $0.5$ only).

$f(0.48)=1.336, f(0.5)=1.405,f(0.51)=1.481$

I am confused what value should I use as $\triangle x$.

答:
If, as I commented, you build the quadratic, working with whole numbers
$$f(x)=\frac{822}{25}-\frac{7927}{60}x+\frac{415}{3}x^2$$

So $$f''(x)=2\times\frac{415}{3}=276.66667$$


问:
Company A has 40% of German cars and company B has 10% of German cars. All other cars are Korean.Help needed solving Bayesian Probability
There are 2 taxi companies. Company A has 40% of German cars and company B has 10% of German cars. All other cars are Korean. Taxi aggregator assigns orders to A or B with probabilities 1/3 and 2/3 respectively. If German car arrives, what's the probability that it's from company B?

So my solution for this problem would be using Bayes Theorem

According to me the question is asking to find P(car being from company B|Being German)
So using Bayes theorem it would be equal to

$P(B|G)=(P(B) * P(G|B))/P(G)$  and that would be

$P(B|G)=((2/3)*0.1)/0.25= 4/15$ which is wrong. I would like to know where am i making mistake

Here P(B) is probability of car being from company B
P(G) is probability of car being German

I found P(G) considering there are 100 cars in  both company A and B so $50/200=1/4=0.25$

答:
Law of total probability:

$P(G) = P(G|A)P(A) + P(G|B) P(B) = 0.4*1/3 + 0.1*2/3 = 0.2$.

By the way, the cars should be Mercedes Benz please.


答:
I find these much easier to think about visually. The number you start with is $i-1$, or in the more standard notation $-1+i$. That number is located in the second quadrant of the complex plane. Its modulus (i.e., its absolute value, or distance from $0$) is $\sqrt2$, and its argument (the angle between the positive real axis and a line from $0$ to our point) is $\frac{3\pi}{4}$.

You'll find one cube root by taking the cube root of the modulus, and dividing the argument by $3$. This comes from DeMoivre's theorem on powers of complex numbers. Thus, the cube root of $2^{1/2}$ is $2^{1/6}$, and one third of $\frac{3\pi}{4}$ is $\frac{\pi}{4}$.

Therefore, the principal cube root will be $2^{1/6}$ units from $0$, in the direction of $1+i$. In polar form, it's $2^{1/6}\cdot e^{i\pi/4}$ Considering that the real and imaginary parts are equal, and their squares sum to the square of $2^{1/6}$, we obtain the number $a+bi$, where $a=b=2^{-1/3}$. Writing it with radicals, that's $$\frac{1}{\sqrt[3]{2}}+ \frac{1}{\sqrt[3]{2}}i$$

The other two cube roots will have the same modulus, but their arguments are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ greater than the argument of the principal root.

Does this answer your question?

Title: How to find the third root of a complex number by transforming the complex number into a root

Body:
I'd like to find all 3rd roots of this number z = i - 1. Now I've found formulas on how to do it; First we transform the complex number into this form

$$ \sqrt[n]{r} * e^{i\frac {\phi + 2k\pi}{n}} $$ Where n should be 3 (because of 3rd root) and k should be k = n - 1 (Including 0)

Now what is not really clear to me is what to do with this -1? If I transform this I'd pressume I get this.

$$\sqrt[3]{-1} * e^{i\frac {\pi/2 + 2k\pi}{3}}$$

But than what? If it was a 2nd root I could say that it is simply i (the root of a negative number) but since it is a 3rd root I'm not quite sure what to do next?

Now for the phi (angle) I just assumed this;  $$i = e^{i \frac {\pi}{2}}$$ And I'm not too sure about this

Thanks!

EDIT: With great help from the people in the comments; I think I've solved it.So as already stated in the comments; $$ r = \sqrt 2, \phi = \frac{3\pi}{4} $$ Now If i would to input that into my formula we should  get this;

$$ \sqrt[6]{2} * e^{i\frac{3pi/4 + 2k\pi}{3}} $$

Now as stated before k = 0... n-1;

k = 0 --> $$ \sqrt[6]{2} * e^{i(\pi/4)} $$
k = 1 --> $$ \sqrt[6]{2} * e^{i(11\pi/12)} $$
k = 2 --> $$ \sqrt[6]{2} * e^{i(19\pi/12)} $$

You reckon I did this correctly?


问:
Conic equation: ${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$ Converting this conic to "standard form"
This is a conic equation:

${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$

I tried a way like this:

1.)${B^{2}-4AC <0} --> ((-4)^2-4x4x7 = 16-112 = -96 < 0)$ "So this is an ellipse"

2.)${Tan2θ = \frac{B}{A-C} = \frac{4}{3} ->θ = 27}$ "degree"

3.) (I saw them using these formulas on a site) There are formulas for ${A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}$ Taking B = 0 to get rid of the xy expression, and the others are:

${A^{'} = Acos^{2}θ+Bcosθsinθ+Csin^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 2,96.

${B^{'} = 0 }$

${C^{'} = Asin^{2}θ+Bsinθcosθ+Ccos^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 4,73

${D^{'} = Dcosθ+Esinθ}$ Plug in the angle "θ" 27 degrees and A 'is about 13,38

${E^{'} = -Dsinθ+Ecosθ}$ Plug in the angle "θ" 27 degrees and A 'is about -0,06

${F^{'} = F = -9}$

4.)Since this is an ellipse according to 1, the standard form of the general equation given is:

${\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}$

it must be.

Finally,
I've got this:

${2,96x^{2}+4,73y^{2}+13,38x-0,06y=9}$

Well, what should I do now? As a result of my various researches, there are completely different solutions and I am confused.

I request your help

答:
The conic can be written as $P(x,y)=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
4 &-2 \\
-2 & 7
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}+2\begin{bmatrix}
6 & 3
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}-9=0$

We want to find a matrix $P \in SO(2)$ such that $P^\top AP$, where $A=\begin{bmatrix}
4  &-2 \\
-2 & 7
\end{bmatrix}$. So you calculate the characteristic polynomial of $A$, which is $c_A(t)=t^2-11t+24$. The eigenvalues are $t_1=8$ and $t_2=3$. With some calculations you will find that the eigenspaces are: $V_8=\mathcal{L}\begin{Bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{5}}\\
\frac{-2}{\sqrt{5}}
\end{bmatrix}\\
\end{Bmatrix}$ and $V_3=\mathcal{L}\begin{Bmatrix}
\begin{bmatrix}
\frac{2}{\sqrt{5}}\\
\frac{1}{\sqrt{5}}
\end{bmatrix}\\
\end{Bmatrix}$. The matrix we were looking for is $P=\begin{bmatrix}
\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\
\frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{bmatrix}$. We can change coordinates as following $\begin{bmatrix}
x\\
y\end{bmatrix}=P\begin{bmatrix}
x'\\
y'\end{bmatrix}$. In this way the ellipse will have its axis parallel to the $x$ and $y$ axis of the coordinate plane. Doing some calculations, you will find that the rotated ellipse has equation: $P'(x,y)=8(x')^2+3(y')^2+6\sqrt5y'-9=0$. Its major axis lies on the $y-$axis. If you want it to lie on the $x-$axis you have to rotate the conic we found of an angle of $\pi/2$. Eventually, by completing the square, you can translate the ellipse such that its center corresponds with the origin $(0,0)$. This last two passages should not be too difficult.


问:
find the equation of the tangents to a quadratic curve through the point $(-7,1)$
Given a quadratic curve $\mathcal{C}: X^2+2XY+4Y^2-2X+6Y+6=0$, determine its type and find the equation of the tangents to $\mathcal{C}$ through the point $(-7,1)$.

So far I've written the matrix of $\mathcal{C}$ as
$$A = \begin{bmatrix}6 & -1 & 3 \\ -1 & 1 & 1 \\ 3 &1 &4\end{bmatrix}.$$

I used the fact that $\det A\neq 0$ and $\begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix}>0$ to say that this is an ellipse, by some theorems we did in class. Now I know that if a point $P(x,y)$ is on the curve, then the unique tangent at this point is the polar line which would be calculated by taking the homogeneous vector of $P$ and multiplying it by A as follows $(1:x:y)A$. However $(-7,1)$ is not on the curve although I calculated $x=\frac{16}{7}$ to be its tangent if it was. so I am stuck here. What could I use to find the tangents since the point is not on the curve? If we write $t_P: y=ax+b$ then $t_P$ must contain $P$ and intersect $\mathcal{C}$ at a point, but I am not sure what point that would be.

答:
An approach, which contains some elements of calculus, is considered here to find the tangents to the quadadratic curve, the equation of which is given as
$$f \left(x,y\right) =x^2-xy-y^2-2x+2y+1=0. \tag{1}$$

We start by differentiating the equation as shown below to obtain the slope of the tangent to the curve (1) at an arbitrary point $Q\left(x,\space y\right)$ lying on it
$$2xdx-ydx-xdy-2ydy-2dx+2dy=0\quad\rightarrow\quad \frac{dy}{dx}= \frac{2x-y-2}{ x+2y-2}. \tag{2}$$

If this tangent passes through a point $P \left(4, -2\right)$, which, as we know, does not lie on the given curve, we can obtain a second expression to describe its slope as,
$$\frac{dy}{dx}=\frac{y+2}{x-4}. \tag{3}$$

We can equate (3) to (2) to establish the following quadratic relationship between $x$ and $y$.
$$\frac{2x-y-2}{ x+2y-2}= \frac{y+2}{x-4}\quad\rightarrow\quad x^2-xy-y^2-6x+y+6=0 \tag{4}$$

Now, we strive to express $y$ in terms of $x$ as simple as possible. In order to obtain a linear relationship between $x$ and $y$, we need to subtract (4) from (1). The result is shown below.
$$4x-y-5=0 \quad\rightarrow\quad y=5-4x \tag{5}$$

Next, we seek for an equation with one variable, so that we can solve for that variable to determine either $x-$ or $y-$coordinate of the point $Q$. If we proceed by replacing $y$ in (1) using its value given by (5), we end up with the following quadratic equation of a single variable.
$$11x^2 - 25x+14=0 \quad\rightarrow\quad \left(x-1\right) \left(11x-14\right) = 0 \tag{6}$$

This equation  has two real roots – meaning that there are two tangents that go through the point $P$. The roots of (6) are
$$x_1=1 \quad\mathrm{and}\quad x_2=\frac{14}{11}. \tag{7}$$

Now, we can use (3) to find the slope of each tangent. But, first we have to substitute the value given by (5) in (3) to obtain
$$\frac{dy}{dx}=\frac{y+2}{x-4}=\frac{7-4x}{x-4}. \tag{8} $$

According to  (7) and (8), the slopes are
$$ \left(\frac{dy}{dx} \right)_1 = -1 \quad\mathrm{and}\quad \left(\frac{dy}{dx} \right)_2 =-\frac{7}{10}.$$

Therefore, the equations of the two tangents can be written as
$$y+x+c_1=0 \quad\mathrm{and}\quad y+\frac{7}{10}x+c_2=0.$$

Since the point $P \left(4, -2\right)$ sits on both these tangents, we can find the $y$-intercepts $c_1$ and $c_2$ as
$$c_1=-2 \quad\mathrm{and}\quad c_2=-\frac{4}{5}.$$


问:
Does $P(A\cap B) + P(A\cap B^c) = P(A)$?
Based purely on intuition, it would seem that the following statement is true, when thinking of the events as sets:

$$P(A\cap B) + P(A\cap B^c) = P(A)$$

However, I am not sure if this is true, and cannot find out how to prove it, or describe a straight forward intuition of it.

Is this equation true? And if not, is there another way to link $P(A\cap B)$ and $P(A\cap B^c)$ with $P(A)$?

答:
As $A\cap B$ and $A\cap B^c$ are disjoint, we have
$$P(A\cap B)+P(A\cap B^c)=P((A\cap B)\cup(A\cap B^c))=B(A\cap(B\cup B^c))=P(A).  $$</p>


问:
Time-invariance of a MIMO system
According to my book, given a MIMO system with n inputs and m outputs (corresponding to the n inputs), the model itself is said to be time-invariant if a time shifting of inputs causes a time shifting of outputs. Anyway I am not sure if each output is a function of only one input or not because of "corresponding to the n inputs".</p>

答: A system is (regardless of the amount of inputs / outputs) time-invariant if applying the same input to the same initial state at time $t$ yields the same response if it is applied at time $t+\tau$. This means that if I push a cart right now, its response should be the same as if I push this cart an hour from now, and the same a day from now and so on. Nothing changes if the same input is applied at a different time, as such its a time-invariant system. an example of a time-variant system could be a leaf spring that looses tension over time due to wear and corrosion. As the response at time $t=0$ (when it has just been made) and $t = \tau$ (a large time of doing nothing) alters.

Title: Time-invariance of a MIMO system


答:
Hint: In the formula you used $r$ and $\phi$   are the absolute value and the argument of the complex number $z$.

The absolute value and argument of a complex number $z=x+iy$ can be computed as:
$$
r=\sqrt{x^2+y^2}, \quad \phi=\begin {cases}\arccos\frac xr,&y\ge0\\
2\pi-\arccos\frac xr,&y<0\\
\end {cases}.
$$

Particularly for $z=-1+i$ :
$$r=\sqrt2,\quad \phi=\frac{3\pi}4.
$$


问:
Topological groups and cardinality of their commutator group (example request)
What would be examples of non abelian, non discrete, second countable compact topological groups with a finite or at most countable commutator (derived) subgroup ?</p>

答:
Proposition. Let $G$ be a compact group with in which $[G,G]$ is countable. Then $[G,G]$ is finite, and the center of $G$ is open of finite index.

[So $G$ is extension of a finite group (as normal subgroup) by an abelian compact group.]

Note that a priori, it is not clear that the closure of $[G,G]$ is countable, so this is not immediate.

Here's how to prove this. Suppose that $[G,G]$ is countable. For fixed $y$, consider $x\mapsto xyx^{-1}y^{-1}$. It has a countable image. Right-translating, we deduce that the continuous $x\mapsto xyx^{-1}$ has countable image. But this is an orbit map (for $G$ acting on itself by conjugacy), so the image is both compact and homogeneous, and countable. So it is finite. That is, $y$ has a finite conjugacy class. Since this holds for every $y$, each conjugacy class in $G$ is finite. A group with this property is called FC-group.

So the main result of Hofmann and Russo, which describe FC-groups that admit a compact group topology, applies. (Ref: K. Hofmann, F. Russo. The probability that x and y commute in a compact group. Math. Proc. Cambridge Philos. Soc. 153 (2012), no. 3, 557–571. ArXiv link.


问:
$|\sum\limits_{k\in \mathbb{Z}}P(X=k)e^{ikt}|=1$ implies $X$ is almost surely constant
Question :

Let's suppose $P$ is a probability, $X$ a random variable defined on $\mathbb{Z}$.

Let's suppose we have $|\sum\limits_{k\in \mathbb{Z}}P(X=k)e^{ikt}|=1$ for all $t\in \mathbb{R}$.

Prove that $X$ is almost surely constant.

My attempt :

If we consider the case where $X$ is defined on $\mathbb{N}$, we have $1=|\sum\limits_{k\in \mathbb{Z}}P(X=k)e^{ikt}|\leqslant \sum\limits_{k\in \mathbb{Z}}P(X=k) = 1$ so we have equality in the triangular inequality, but it's a infinite inequality so I'm not sure I can apply the result for the equality case for infinite sum.

Could someone help me ? (there's surely an easiest way than what I tried)

答:
Write $p_m:=P(X=m)$.
The correct conclusion is: There must exist some integer $m$ such that $p_m=1$ and $p_k=0$ for all $k \ne m$. Indeed, otherwise there exist distinct integers $m,k$ such that $p_m p_k>0$. Fix $t$ so that $(m-k)t$ is not a multiple of $2\pi$.  In that case $|p_me^{imt}+p_k e^{ikt}|=p_m+p_k-\epsilon$ for some positive $\epsilon$  by the condition for equality in the triangle inequality [1]. Therefore for $N>|m|,|k|$ we have
$$|\sum_{n =-N}^N p_ne^{int}|  \le p_m+p_k-\epsilon+ \sum_{n \in [-N,N]: \, \,  n \ne m,k} p_n=-\epsilon+\sum_{n \in [-N,N]} p_n .$$
Now take a limit as $N \to \infty$ to obtain a contradiction.

[1] Equality of triangle inequality in complex numbers


答:
That sum you wrote down is called the characteristic function $\phi(t)=\mathbb E[e^{itX}]$.

Let $Y$ be an iid copy of $X$, then the characteristic function of $X-Y$ is $$\mathbb E[e^{itX}]\cdot\mathbb E[e^{-itY}]=\lvert\phi(t)\rvert^2=1.$$
The unique distribution with characteristic function identically $1$ is the rv that is almost surely $0$, so $X=Y$ almost surely. Then since $X$, $Y$ are iid,
$$\mathbb P(X\leq x)=\mathbb P(X\leq x, Y\leq x)=\mathbb P(X\leq x)^2\implies\mathbb P(X\leq x)\in\{0,1\}.$$
This is clearly enough to imply that there is some $a$ in the image of $X$ (which is $\mathbb Z$ in this case) such that $\mathbb P(X=a)=1$.

Title: $|\sum\limits_{k\in \mathbb{Z}}P(X=k)e^{ikt}|=1$ implies $X$ is almost surely constant

Body:
Question :

Let's suppose $P$ is a probability, $X$ a random variable defined on $\mathbb{Z}$.

Let's suppose we have $|\sum\limits_{k\in \mathbb{Z}}P(X=k)e^{ikt}|=1$ for all $t\in \mathbb{R}$.

Prove that $X$ is almost surely constant.

My attempt :

If we consider the case where $X$ is defined on $\mathbb{N}$, we have $1=|\sum\limits_{k\in \mathbb{Z}}P(X=k)e^{ikt}|\leqslant \sum\limits_{k\in \mathbb{Z}}P(X=k) = 1$ so we have equality in the triangular inequality, but it's a infinite inequality so I'm not sure I can apply the result for the equality case for infinite sum.

Could someone help me ? (there's surely an easiest way than what I tried)


问:
Bilinear Form and Rank Proof
I found a task in the internet about bilinear forms that I would like to unterstand. I changed a few variables but it goes like this. Suppose that $\psi: U \times V \rightarrow K$ is a bilinear form of rank $r$ on finite dimensional vector spaces $U$ and $V$
over $K$. I want to show that there exist bases $e_1, . . . , e_m$ for $U$ and $f_1, . . . , f_n$ for $V$ such that
$$\psi \big{(}\sum_{i=1}^m x_ie_i, \sum_{j=1}^n y_jf_j \big{)}=\sum_{k=1}^r x_ky_k$$ for all $x_1, . . . , x_m, y_1, . . . , y_n \in K$. Then how to find the dimensions of the left and right kernels of $\psi$? Thanks in advance for advice of any kind!

答:
Let

$$\begin{array}{l|rcl}
\psi_1 : & U & \longrightarrow & V^* \\
    & u & \longmapsto & (v \mapsto \psi(u,v)) \end{array}$$

The rank of $\psi_1$ is equal to the one of $\psi$, i.e. equal to $r$. Let $\{f_1^*, \dots, f_r^*\}$ be a basis of $\text{im}\ \psi_1$ that we complement into a basis
$\{f_1^*, \dots, f_r^*, f_{r+1}^*, \dots, f_n^*\}$ of $V^*$. We can find $e_1, \dots , e_r \in U$ such that $\psi_1(e_i) = f_i^*$ for $1 \le i \le r$. $\{e_1, \dots , e_r\}$ is linearly independent as $\{f_1^*, \dots, f_r^*\}$ is. We again complement $\{e_1, \dots , e_r\}$ into a basis $\{e_1, \dots , e_r, e_{r+1}, \dots, e_m\}$ of $U$. $\{e_{r+1}, \dots, e_m\}$ is a basis of $\ker \psi_1$.

Now let $\{f_1, \dots, f_n\} \subseteq V$ be the dual basis of $\{f_1^*, \dots, f_r^*, f_{r+1}^*, \dots, f_n^*\}$.

For $1 \le i \le r$  and $j \in \{1,\dots, n\}$ we have

$$\psi(e_i,f_j) = (\psi_1(e_i))(f_j)=f_i^*(f_j)=\delta_{ij}$$ where $\delta_{ij}$ stands for Kronecker delta.

While for $i \gt r$  and $j \in \{1,\dots, n\}$

$$\psi(e_i,f_j) = (\psi_1(e_i))(f_j)=0(f_j)=0.$$

Which allows to conclude to the desired result by bilinearity of $\psi$.

Note: I implicitly proved the result mentioned in jlammy response that is called the rank normal form of a matrix.

Title: Bilinear Form and Rank Proof


问:
Bijection between a set of injections and a union of bijection
Consider $A$ a set with $n$ elements and $k\leq n$.

Show that
$$
\phi:\left\{  f:\left\{  1,2,\ldots,k\right\}  \rightarrow A\mid f\text{
injective}\right\}  \rightarrow
{\bigsqcup\limits_{B\subset A,card(B)=k}}
\left\{  g:\left\{  1,2,\ldots,k\right\}  \rightarrow B\mid g\text{
bijective}\right\}
$$
by constructing an explicit bijection.

By generalizing the construction above, show that
$$
\Phi:\left\{  f:A\rightarrow C\mid f\text{ injective}\right\}  \rightarrow
{\bigsqcup\limits_{B\subset C.card(C)=n}}
\left\{  g:B\rightarrow C\mid g\text{ bijective}\right\}  .
$$

I attempted to start from the observation that every injective function $f$
has an image $\operatorname{Im}f$ as subset of $A$ with exactly $k$ elements.
Reciprocally every subset of $A$ with $k$ elements is the image of some
injection. But there are $k!$ injections that maps a fixed set $C$ into a
subset of $A$ with $k$ elements. Now, every
$$
f:\left\{  1,\ldots,k\right\}  \rightarrow A
$$
gives us a bijection $g:\left\{ 1,\ldots,k\right\}  \rightarrow B,B\subset A$
and $card(B)=k$. From this point, I don't know how to move forward. How can I
find an explicit bijection?

答:
$\require{HTML}$You basically answered your own question already.

Given sets $|X|$ and $|Y|$ with $|X|\le |Y|$ we have a bijection
\begin{align}
\{\text{injections $X\to Y$}\} &\longleftrightarrow \bigsqcup_{\substack{Z\subset Y,\\|Z|=|X|}} \{\text{bijections $X\to Z$}\}, \\
(f\colon X\to Y) &\longmapsto (X\to \operatorname{Im}(f), x\mapsto f(x)),\\
(X\to Y, x\mapsto g(x))&\,\style{display: inline-block; transform: scaleX(-1)}{\longmapsto} (g\colon X\to Z).
\end{align}


问:
compute usig cauchy integral formula, $ \int_\Gamma \frac{dz}{z^2+1}$
The question is:
$$ \int_\Gamma \frac{dz}{z^2+1}$$

Here is how i thought, we can divide it into two contour and then deform continuously the inner curve to a circle around $i$ but my problem is here how should i deform the bigger contour around $-i$ i couldn't do that, because we can't pass through singularity, what do you think? any suggestion would be great,

Thanks

答: Read the fine print viz. $\oint_\Gamma f(z)dz=2\pi i\sum_{w\in S}\color{blue}{I(\Gamma,\,w)}\color{red}{\operatorname{Res}(f,\,w)}$, with $S$ the set of poles of a holomorphic $f$ the closed rectifiable curve $\Gamma$ encloses, so $w\in S$ has winding number $I(\Gamma,\,w)$. Your contour integral is$$2\pi i\left(\color{blue}{2}\color{red}{\lim_{z\to i}\frac{1}{z+i}}+\color{blue}{1}\color{red}{\lim_{z\to-i}\frac{1}{z-i}}\right)=2\pi i(2/(2i)+1/(-2i))=\pi.$$


问:
Given two sets $A$ and $B$, is it true that $|B^A|=\max\{|B|,|\mathcal{P}(A)|\}$?
My intuition is that this is true, because it makes sense for little sets (with cardinalities $|\mathbb{N}|,|2^\mathbb{N}|$). But i'm not sure at all that this is actually true.

If it's not, what's the real relation between $|B^A|$ and $|B|,|A|$.

(Obviously talking about infinite sets)

答:
Surely not true for finite sets. Counterexample

$$A=\{1,2\} \text{ and } B = \{1,2,3\}.$$

And if Singular cardinals hypothesis is assumed, the result is again wrong as

$$\text{if } \vert \mathcal P(A) \vert \lt \vert B \vert \text{ and } \text{cf} \vert B \vert \le \vert A \vert \text{ then } \vert B^A \vert = \vert B \vert^+$$


问:
Multiobjective optimization
I need some clarification on multi objective optimization. I would like to know if a problem has three objectives with completely different  variables, should such a problem be solved as three  independent single objective optimization problem  or could the problem be solved using a multiobjective optimization.  I would appreciate your feedback.  Thank you.

eg. Min
f1=a1*x1 +a2*x2;
f2= a3*x3 +a4*x4;
f3=a5*x5 + a6*x6

答:
At a time, you have asked too many different questions.
Whether a single objective or multiobjective optimization is required, that depends on your aim of study.
Whether variables can be separated or not is completely different problem, it defines the nature of objective functions and their interrelationship.

In order to get clear grasp on the multiobjective optimizatiom, you can refer to the following research article.

https://www.researchgate.net/publication/341752284_Pareto_dominance_based_Multiobjective_Cohort_Intelligence_algorithm

OR

http://dx.doi.org/10.1016/j.ins.2020.05.019

From the article, you can see that when objectives are conflicting, multiobjective optimization is most suitable, otherwise you can use single objective optimization.

Title: Multiobjective optimization


答:
Yes it is true. In fact is one of the basic properties.

$$
P(A)=P(A\cap\Omega)=P(A\cap\left(B\cup B^c\right))=P((A\cap B)\cup (A \cap B^c))=P(A\cap B)+P(A\cap B^c)
$$

Title: Does $P(A\cap B) + P(A\cap B^c) = P(A)$?

Body:
Based purely on intuition, it would seem that the following statement is true, when thinking of the events as sets:

$$P(A\cap B) + P(A\cap B^c) = P(A)$$

However, I am not sure if this is true, and cannot find out how to prove it, or describe a straight forward intuition of it.

Is this equation true? And if not, is there another way to link $P(A\cap B)$ and $P(A\cap B^c)$ with $P(A)$?


答:
Hint.

The result follows from the residue theorem for closed contours.

If you are only allowed to use the Cauchy integral formula with simple closed contours, first rewrite the integral as
$$
\int_\Gamma\frac{1}{z^2+1}\,dz=\frac{1}{2i}\left(\int_\Gamma\frac{1}{z-i}dz-\int_\Gamma\frac{1}{z+i}dz\right)
$$

Then after you split the contour $\Gamma$, handle $\frac{1}{z-i}$ and $\frac{1}{z+i}$ separately with the Cauchy integral formula.

Title: compute usig cauchy integral formula, $ \int_\Gamma \frac{dz}{z^2+1}$

Body:
The question is:
$$ \int_\Gamma \frac{dz}{z^2+1}$$

Here is how i thought, we can divide it into two contour and then deform continuously the inner curve to a circle around $i$ but my problem is here how should i deform the bigger contour around $-i$ i couldn't do that, because we can't pass through singularity, what do you think? any suggestion would be great,

Thanks


问:
How do we compute this specific $2$ variables integral?
Let $X=\mathcal{C}_{0}(\mathbb{R}^{2})\subset \overline{\mathcal{C_{c}}(\mathbb{R}^{2})}$ the closure in norm $\|\cdot\|_{+\infty}$ of continous functions with compact support in $\mathbb{R}^{2}$. Given a sequence $(a_{n})\subset\mathbb{R}^{+}$, for every $n\in\mathbb{N}$ and $u\in X$ we set:
$$T_{n}u=\int_{-a_{n}}^{a_{n}}u(x,nx)dx$$
once shown $T_{n}$ is a linear operator, how do we compute its norm? Initially one may observe that
$$|x|\leq a_{n}\qquad\text{and}\qquad|nx|\leq a_{n}\implies |x|\leq \frac{a_{n}}{n}$$
But I'm not sure how to change the first integral:
$$\int^{\frac{a_{n}}{n}}_{-\frac{a_{n}}{n}}\int_{-a_{n}}^{a_{n}}u(x,y)dxdy\qquad\text{or}\qquad\frac{1}{n}\int^{\frac{a_{n}}{n}}_{-\frac{a_{n}}{n}}\int_{-a_{n}}^{a_{n}}u(x,y)dxdy$$
the question is actually stupid but it got me confused. This is just the first part of an exercise, and giving the wrong answer would change the rest quite drastically.

答:
The goal is to compute $||T_n|| := \sup_{||u||_\infty=1}\{ |T(u)|\}$ where  $||u||_\infty$ is the sup norm on $C_0(\mathbb{R}^2)$

By definition of the sup norm we have $|u(x,nx)| \leq ||u||_\infty$ so we have $$|T_n(u)| = \left| \int_{-a_n}^{a_n} u(x, nx) dx \right| \leq  \int_{-a_n}^{a_n} |u(x, nx)| dx \leq  \int_{-a_n}^{a_n} ||u||_\infty dx = 2a_n ||u||_\infty$$

So $||T_n||\leq 2a_n$

The next step  is to produce a function $u\in C_0(\mathbb{R}^2)$ that makes the above inequalities an equality. For this take any function which is 1 on the rectangle $K = [-a_n, a_n]\times [-na_n, na_n]\subset \mathbb{R}^2$, zero outside an open $O$ containing $K$ and bounded by $1$ everywhere.

Now $$|T_n(u_n)| = \left| \int_{-a_n}^{a_n} u_n(x, nx) dx \right| =  \left| \int_{-a_n}^{a_n} 1 dx \right| =  2a_n $$


问:
Solve $\int_{0}^{1}\int_{0}^{x}\int_{z}^{x}x\cos (y^{2})dydzdx$
I want to solve this triple integral: $$\int_{0}^{1}\int_{0}^{x}\int_{z}^{x}x\cos (y^{2})dydzdx$$

I tried to change the variables but I still couldn't make it. Is there any clue how to start?

答: The variable $x$ can take any value from $0$ to $1$ and, for each value of $x$, $z$ can take any value from $0$ to $x$ and $y$ can take any value from $z$ to $x$. So, if you fix $x\in[0,1]$, $y$ can take any value from $0$ to $x$ and $z$ can take any value from $0$ to $y$. Therefore\begin{align}\int_0^1\int_0^x\int_z^xx\cos(y^2)\,\mathrm dy\,\mathrm dz\,\mathrm dx&=\int_0^1\int_0^x\int_0^yx\cos(y^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx\\&=\int_0^1\int_0^xxy\cos(y^2)\,\mathrm dy\,\mathrm dx\\&=\int_0^1\frac12x\sin(x^2)\,\mathrm dx\\&=\frac14-\frac{\cos(1)}4.\end{align}


问:
In a Renewal Process, show that sum of convolutions is convergent
I have been doing a course on coursera on stochastic processes. The first process that was introduced was Renewal process, defined as follows;

$$S_n = S_{n-1} + \xi_n$$

where $\xi_n$ are IID.

We define $F^{n*}$ as convolution of $n$ IID variables, C.D.F, $F(x) = P\{\xi <= x\}$, and $F(0) = 0$. The claim is,

$$\sum_{n=1}^\infty F^{n *} < \infty.$$

I would have thought that this has something to do with CLT. However, I cannot seem to take the next step. I cannot seem to make any kind of intuitive reasoning why this should be so. The fact that these are positive random variables does not seem to be much of a clue.

Any thoughts, Any clues ideas would be of great help. I much appreciate it and thank you folks in advance.

答:
The proof is really not that difficult, It is a bit more elaborate than the accepted solution. I think there are a couple of interesting things going on here. I thought it will be useful for other people that are learning this.

Given $S_n = S_{n-1} + \xi_n$ where $\xi_i$ are i.i.d. distribution $\mathbf{F}$, and $\mathbf{F}(0)$, we have (1) $\mathcal{U}(t) = \sum_1^\infty \mathbf{F}^*_n(t) < \infty$, (2) $\mathbf{E} \mathbf{N}_t = \mathcal{U}(t)$. Where $\mathbf{F}^*_n(t)$ is the convolution of $n$ i.i.d. variables.

The proof goes as follows.

To prove the identity (1) we do the following, $\sum_1^\infty \mathbf{F}^*_n(t) = \lim_{n \rightarrow \infty } \sum_1^n \mathbf{F}_i^*$, note that $\mathbf{F} : \mathbb{R^+} \rightarrow [0,1]$, $\mathbf{F}^*_n(t) \ge \mathbf{F}^*_{n+1}(t), \ \forall t$ and $ \mathbf{F}^*_n(t): \mathbb{R^+} \rightarrow [0,1]$. Clearly the sequence of functions $\mathbf{F}^*_n$ is monotone decreasing, and $\sum_1^\infty \mathbf{F}^*_n(t) \le \sum_1^\infty \mathbf{F}^n(t)$.
Notice that, $\sum_1^\infty \mathbf{F}^n(t) = \mathbf{F}(t) + \mathbf{F}^2(t) + \mathbf{F}^3(t) + \cdots$ (independence of $\xi_i$, and $\mathbf{F}(\alpha) = \mathbf{P}(\xi_i < \alpha)$) and that $0 \le \mathbf{F} (t) \le 1$. This is a geometric series that converges for $0 \le \mathbf{F}(t) < 1$. Consider now a point $t_0$ such that $\mathbf{F}(t_0) = 1$. The interesting aspect of the convolution is that, there is an $r$ such that $\mathbf{F}^*_r(t_0) < 1$. This follows from the fact that $\mathbf{F}(t_0) = 1$ implies random variable is concentrated in a region $[0, t_0]$. The sum of $n$ such i.i.d. random variables is concentrated in the region $[0, nt_0]$. This implies that there is a $r < n$ for some $n$ such that $\mathbf{F}^*_r(t_0) < 1$. Using the fact that $\mathbf{F}^*_n(t) \ge \mathbf{F}^*_{n+1}(t), \ \forall t$ we can write the following,
$\mathcal{U}(t_0) = \sum_1^\infty \mathbf{F}^*_n(t_0) = \sum_1^r \mathbf{F}^*_n(t_0) + \sum_{r+1}^\infty \mathbf{F}^*_k(t_0) \le \sum_1^r \mathbf{F}^*_n(t_0) + \sum_{r+1}^\infty (\mathbf{F}^*_r(t_0))^k$, which converges.

The identity (2) is fairly trivial to prove. Let us look at $\{\mathbf{N}_t = n\} = \{\mathbf{N}_t \ge n\} - \{ \mathbf{N}_t > n\} = \{\mathbf{N}_t \ge n\} - \{ \mathbf{N}_t >= n + 1\} = \{\mathbf{S}_n \le t\} - \{\mathbf{S}_{n+1} \le t\}$. Taking probabilities we get,
$\mathbf{P}(\{\mathbf{N}_t = n\}) = \mathbf{P}(\{\mathbf{S}_n \le t\} - \{\mathbf{S}_{n+1} \le t\}) = \mathbf{P}(\{\mathbf{S}_n \le t\}) - \mathbf{P}(\{\mathbf{S}_{n+1} \le t\}) = \mathbf{F}^*_n(t) - \mathbf{F}^*_{n+1}(t)$

We use the fact that, $\{S_n = t\} \bigcup \{S_{n+1} \le t\} = \{ S_n \le t\}$, and $\{S_n = t\} \bigcap \{S_{n+1} \le t\} = \emptyset$  implies sets are mutually exclusive.

$\mathbf{E} \mathbf{N}_t$ = $\sum_1^\infty k \mathbf{P}(\{\mathbf{N}_t = k\}) = \sum_1^\infty k (\mathbf{F}^*_k(t) - \mathbf{F}^*_{k+1}(t)) = \mathbf{F}^*_1 - \mathbf{F}^*_2 + 2 \mathbf{F}^*_2 - 2 \mathbf{F}^*_3 + 3 \mathbf{F}^*_3 + \cdots = \sum_1^\infty \mathbf{F}^*_k (t) = \mathcal{U}(t) $


问:
Express $\frac{\ln\left(\frac{1}{2}\right)}{80000000}$ as a negative exponent
I would like to express $\dfrac{\ln\left(\frac{1}{2}\right)}{80000000}$ as a negative exponent. Using Desmos online calculator I tried this variant:

Why are the results different? I expected/goal was to be the same result.

How can I express $\dfrac{\ln\left(\frac{1}{2}\right)}{80000000}$ not as a fraction?

Body:
$$a\ln b=\ln\left(b^a\right)$$

Using this fact, we can write

$$\frac{\ln\left(\frac12\right)}{80000000}=\frac1{80000000}\ln\left(\frac12\right)=\ln\left(\left(\frac12\right)^{\frac1{80000000}}\right)=\ln\left(\frac{1^{\frac1{80000000}}}{2^{\frac1{80000000}}}\right)=\ln\left(2^{-\frac1{80000000}}\right)$$


问:
Whether every measurable function is Radon–Nikodym derivative?
The definition of  Radon Nikodym derivative https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem.
Each measurable function from a measure space to another measurable space induces a
measure on its range space see Lemma 17. http://www.stat.cmu.edu/~arinaldo/Teaching/36752/S18/Notes/lec_notes_2.pdf

My question is whether every measurable function is Radon–Nikodym derivative or not?
If not which kind of measurable function is?

答:
First let us recall some key concepts. Note that for  measurable function to be Radon Nikodym derivative  it must take values in $\bar{\Bbb R}$, that is $[-\infty, +\infty]$ or, in the case of complex measure, take values in $\Bbb C$. Let us first see de case where $f$ take value in  $\bar{\Bbb R}$.

Recall that  $f^+$ is the positive part of $f$, that is, $f^+ = \max \{ f, 0\}$  and $f^-$ is the negative part of $f$, that is, $f^- = \max \{ -f, 0\}$. So $f^+ , f^- \geqslant 0$ and $f=f^+ - f^-$. Clearly, $f$ is measurable if and only if $f^+$ and $f^-$ are measurable. Noe let us prove:

Let $(X, \Sigma, \mu)$ be a measure space  and  $f: X \rightarrow \bar{\Bbb R}$ be a measurable function. Then  $f$ is a Radon Nikodym derivative  if an only if $\int_X f^- d\mu < +\infty$ or  $\int_X f^+ d\mu < +\infty$.

Proof: ($\Rightarrow$) Suppose $f$ is a Radon Nikodym derivative. Then $\nu$ defined for each $E \in \Sigma$ by
$$ \nu(E) = \int_E f d\mu $$
is a measure. Note that
$$ \nu(E) = \int_E f d\mu = \int_E f^+ d\mu - \int_E f^- d\mu $$
In particular $\nu(X)$ must be defined, so $\int_X f^- d\mu < +\infty$ or  $\int_X f^+ d\mu < +\infty$.

($\Leftarrow$) Suppose $\int_X f^- d\mu < +\infty$ or  $\int_X f^+ d\mu < +\infty$. So, if $\int_X f^- d\mu < +\infty$, the, for all $E \in \Sigma$,
$$ 0 \leqslant  \int_E f^- d\mu \leqslant \int_X f^- d\mu < +\infty$$
So $ \nu(E) = \int_E f d\mu = \int_E f^+ d\mu - \int_E f^- d\mu $ is well-defined. It is immediate consequence of the properties of the integral that $\nu$ is measure. So $f$ is is a Radon Nikodym derivative.

In a completely similar way, if $\int_X f^+ d\mu < +\infty$, we can conclude that $f$ is is a Radon Nikodym derivative.

Remark: For the case of complex valued measures, leu us write $\Re f$ to indicate the real part of $f$ and $\Im f$ to indicate the imaginary part of $f$. the result is

Let $(X, \Sigma, \mu)$ be a measure space  and  $f: X \rightarrow \Bbb C$ be a measurable function. Then  $f$ is a Radon Nikodym derivative  if an only if two conditions hold:

$\int_X (\Re f)^- d\mu < +\infty$ or  $\int_X (\Re f)^+ d\mu < +\infty$;

$\int_X (\Im f)^- d\mu < +\infty$ or  $\int_X (\Im f)^+ d\mu < +\infty$.

The proof is completely analogous to proof of the case of $f$ taking values in $\bar{ \Bbb R}$.


问:
Efficient techniques for multiobjective optimization for a beginner
I want to minimize a function $f(x)$ at multiple positions of $x$, e.g., from $x_1$ to $x_4$, $x_{17}$ to $x_{20}$, $x_{51}$ to $x_{54}$ etc. Each value of $x$ is an independent sinusoidal function. The value of $f(x)$ depends on amplitudes of $x_1, x_2, \ldots$. Since there is an upper constraint on $x_i^2$, I want to find the optimal values of the $x_i$'s to minimize $f(x)$. I know little about optimization and have solved this problem to get minimum value at single range of values ie. $x_{1}$ to $x_{4}$. Being new to optimization, I am not sure whether I can call it multiobjective optimization? To get minimum value, which optimization technique will be feasible for the problem? </p>

答:
The actual functions must be known beforehand in order to assess whether conflicting objectives exist.
As a simple test, you can check whether increasing f1 reduces f2. If it is so, then you can apply multiobjective optimization.

In order to get further details, you can refer to the following article.

http://dx.doi.org/10.1016/j.ins.2020.05.019

OR
https://www.researchgate.net/publication/341752284_Pareto_dominance_based_Multiobjective_Cohort_Intelligence_algorithm


答:
You need to determine which order of integration will make the calculations possible. Notice that the region of integration is defined by $0 \leq z \leq y \leq x \leq 1$.

As written, you need to first integrate with respect to $y$, but that will not work since you have $\cos(y^2)$ and no $y$ outside that.

What if you integrate first with respect to $x$? That is possible, and the integral becomes:
$$\int_0^1 \int_z^1 \int_y^1 x\cos(y^2) \ dx \ dy \ dz \quad \text{or} \quad \int_0^1\int_0^y\int_y^1 x\cos(y^2) \ dx \ dz \ dy.$$
In either case the innermost integral is:
$$\int_y^1 x \cos(y^2) \ dx = \frac{1}{2}x^2\cos(y^2)\Big|_{x=y}^{x=1}=\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2).$$
We can't integrate either of these with respect to $y$, so we would want to integrate with respect to $z$ next:
\begin{align*}\int_0^y \left(\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2)\right) \ dz &=\left(\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2)\right)z\Big|_{x=0}^{x=y}\\
&= y\left(\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2)\right)\\
&=\frac{1}{2}\cos(y^2)(y-y^2).
\end{align*}
Again we can't integrate with respect to $y$.

So now we backtrack and instead try integrating with respect to $z$ first. The innermost integral is:
$$\int_0^y x\cos(y^2) \ dz = xz\cos(y^2)\Big|_{z=0}^{z=y}=xy\cos(y^2).$$
We could integrate with respect to $x$ or $y$ next. But notice that integrating with respect to $x$ will introduce a $\cos(y^2)$ and a $y^2\cos(y^2)$ term (since the bounds on $x$ are $y \leq x \leq 1$, and the antiderivative of $x$ is $\frac{1}{2}x^2$). So we should integrate with respect to $y$ next, which we can do using a $u$-substitution. After that, it turns out you can also do the final integral with respect to $x$. So we have determined that the best order of integration is:
$$\int_0^1 \int_0^x \int_0^y x \cos(y^2) \ dz \ dy \ dx.$$

Altogether there are $3!=6$ possible orders of integration to consider, and you need to be systematic about checking the possibilities. We could immediately rule out two of them ($dy \ dx \ dz$ and $dy \ dz \ dx$) because we cannot do the innermost integral. For the other four, we had to do some trial and error, but it can be sped up by thinking ahead about the next step, as I did above when deciding whether to integrate $xy\cos(y^2)$ with respect to $x$ or $y$ first.

Title: Solve $\int_{0}^{1}\int_{0}^{x}\int_{z}^{x}x\cos (y^{2})dydzdx$

Body:
I want to solve this triple integral: $$\int_{0}^{1}\int_{0}^{x}\int_{z}^{x}x\cos (y^{2})dydzdx$$

I tried to change the variables but I still couldn't make it. Is there any clue how to start?


Title: Express $\frac{\ln\left(\frac{1}{2}\right)}{80000000}$ as a negative exponent

Body:
I would like to express $\dfrac{\ln\left(\frac{1}{2}\right)}{80000000}$ as a negative exponent. Using Desmos online calculator I tried this variant:

Why are the results different? I expected/goal was to be the same result.

How can I express $\dfrac{\ln\left(\frac{1}{2}\right)}{80000000}$ not as a fraction?

Body:
\begin{align}
\frac{\ln(\frac12)}{8\cdot 10^7}
&=
\frac{1}{8\cdot 10^7}\cdot \ln(\frac12)\\
&=
\frac{1}8{\cdot 10^{-7}}\cdot \ln(\frac12)\\
&=\ln\left((\frac12)^{\frac{1}8{\cdot 10^{-7}}}\right)\\
&\ne\ln\left((\frac12)^{-8{\cdot 10^{7}}}\right)
\end{align}


问:
A is a matrix of integers , prove that A+I is invertible
Question:

$2 \le d \in \Bbb Z$

Let $A \in M_n(\Bbb Q) s.t$ All of it's elements are integers divisible by d.

Prove that $I+A$ is invertible.

What I thought:

I thought of using the determinant of A and extracting $d^2$ from it, but I can't use that on det(A+I)... tried also to calculate the determinant for n=2, and got a polynomial of 2nd degree (With no idea on how to prove that it doesn't equal 0).

答:
Look at the laplace expansion of the determinant of $A+I$ (http://en.wikipedia.org/wiki/Laplace_expansion)

Your matrix has the form $$
  A = d\left(
  \begin{matrix}
    k_{1,1} &\ldots &k_{1,n} \\
    \vdots  &\ddots &\vdots \\
    k_{n,1} &\ldots &k_{n,n}
  \end{matrix}\right)
$$
and by laplace expanding the determinant using the first row you get $$
  \det(A+I) = (dk_{1,1} + 1)C_{1,1} + dk_{1,2}C_{1,2} + \ldots
$$
Note that $C_{1,1} = \det(A_1 + I)$ where $$
  A_1 = d\left(
  \begin{matrix}
    k_{2,2} &\ldots &k_{2,n} \\
    \vdots  &\ddots &\vdots \\
    k_{n,2} &\ldots &k_{n,n}
  \end{matrix}\right) \text{.}
$$

Edit:: (I previously suggested to decude from $\det(A+I)=0$ that $C_{1,1}$ is zero, but that's not as obvious as I initially thought, since you only get "zero or a multiple of $d$". You'd need an extra induction step to show that the multiple of $d$ case doesn't happen. One can avoid that by looking at everything modulo $d$)

From the above, you get by induction that  $$
  \det(A+I) \equiv C_{1,1} \equiv \det(A_1+I) \equiv \ldots \equiv \det(A_{n-1}+I) \equiv 1\mod d \text{.}
$$


问:
Compact-open topology for the space of continuous paths
My question is probably silly... There is a cryptic claim in Hatcher's book page 530 that I do not understand about the compact-open topology on the set of continuous maps from the segment $[0,1]$ to a topological space $X$. I guess that it is a misunderstanding with the English words.

It is claimed that a basis of open subsets in this particular case consists of the open sets $\bigcap_i M(K_i,U_i)$ where the $K_i$'s are a partition of $[0,1]$ into nonoverlapping closed intervals. Consider the open subset $M(\mathbb{K},U)$ where $\mathbb{K}$ is the Cantor set. I do not see how to find an open of the form $\bigcap_i M(K_i,U_i)$ contained in $M(\mathbb{K},U)$, unless the partition is not necessarily finite: in that case, $M(\mathbb{K},U)=\bigcap_{b\in \mathbb{K}} M(\{b\},U) \cap \bigcap_{n\geq 0} M([a_n,b_n],X)$ where the $[a_n,b_n]$ are the closure of all connected components of the complement of $\mathbb{K}$.

What does Hatcher mean exactly ?

答:
It seems like Hatcher is claiming something different (emphasis mine):

In [the case of the compact open topology on $X^I$] one can check that a system of basic neighborhoods of a path $f: I \to X$ consists of the open sets $\bigcap_i M(K_i, U_i)$ where the $K_i$'s are a partition of $I$ into nonoverlapping closed intervals and...

On the other hand, the subbasis of the topology definitionally consists of the sets $M(K, U)$. A basis of the topology then consists of finite intersections of such sets. (Note that this means there is no "overlapping" condition on the basic open sets.) This means that your open subset $M(\mathbb K, U)$ is already a basic open set.


答:
Let $M_{ij}=\psi(e_i,f_j)$ be the matrix representing the bilinear form, in the bases $\{e_i\}$ and $\{f_i\}$. It is a theorem that we can choose bases so that the form of $M$ is
$$M=\begin{pmatrix}I_r&0\\0&0\end{pmatrix}.$$
Then in these bases it is clear that $\psi(x,y)=x^\top My$ acts as you want.

The left kernel is then the span of $e_{r+1},\dots,e_m$ (corresponding to those bottom rows of zeros in the matrix), so the dimension is $m-r$. Similarly the right kernel is the span of $f_{r+1},\dots,f_n$ which has dimension $n-r$.

Title: Bilinear Form and Rank Proof

Body:
I found a task in the internet about bilinear forms that I would like to unterstand. I changed a few variables but it goes like this. Suppose that $\psi: U \times V \rightarrow K$ is a bilinear form of rank $r$ on finite dimensional vector spaces $U$ and $V$
over $K$. I want to show that there exist bases $e_1, . . . , e_m$ for $U$ and $f_1, . . . , f_n$ for $V$ such that
$$\psi \big{(}\sum_{i=1}^m x_ie_i, \sum_{j=1}^n y_jf_j \big{)}=\sum_{k=1}^r x_ky_k$$ for all $x_1, . . . , x_m, y_1, . . . , y_n \in K$. Then how to find the dimensions of the left and right kernels of $\psi$? Thanks in advance for advice of any kind!


问:
How can i calculate the contour integral of this trigonometric function?
Here's my integral $\int_0^{2\pi} \frac{cos(n\theta)}{cosha+acos\theta}d\theta$ where $|a|<1$

I tried $\int_0^{2\pi} \frac{e^{in\theta}}{cosha+acos\theta}d\theta$ but stuck here because of denominator.

Then i used $cos\theta=\frac{1}{2}(z+\frac{1}{z})=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $cos(n\theta)=\frac{1}{2}(e^{in\theta}+e^{-in\theta})$

So my integral turns into $\oint \frac{1/2(e^{in\theta}+e^{-in\theta})}{cosha+\frac{a}{2}(e^{i\theta}+e^{-i\theta})}\frac{dz}{iz}$
But nothing useful comes from here since $cosh (a)$ is so restrictive.

How can I calculate this?

答:
HINTS.

First of all rewrite the integral in a more "pleasing" way:

$$\int_0^{2\pi} \frac{\cos(n\theta)}{A + B\cos(\theta)}\ \text{d}\theta$$

Where of course $A = \cosh(a)$, $B = a$.

Then apply the Complex Step as you "almost" did:

$$z = e^{i\theta} ~~~~~~~~~~~\text{d}\theta = \frac{\text{d}z}{iz} ~~~~~~~~~~~ \cos(\theta) = \frac{1}{2}\left(z^2 + 1\over z\right) ~~~~~~~~~~~ \cos(n\theta) = \Re\left(z^n\right)$$

This leads to

$$-2i\int_{|z| = 1} \frac{\Re(z^n)\ \text{d}z}{ z^2 + bz + 1}$$

Where $b = B/A$.

Now it's all about to understand what $b^2 - 4$ is to get the hypothetical poles.

Poles are at

$$z_{1,\ 2} = \frac{-b \pm\sqrt{b^2-4}}{2}$$

It's not easy to determine which one lies inside the unit circle due to the unknown parameter $a$, but suppose we "know" that only

$$z = \frac{-b + \sqrt{b^2-4}}{2}$$

lies inside the unit circle (remember that $b = B/A = \cosh(a)/a$, so it would be immediate with numbers).

Then by residues theorem we have

$$\text{Integral} = -2i\left(2\pi i \times \sum_k \text{Residues}[f(z), z_k]\right)$$

In our case

$$f(z) = \frac{\Re(z^n)}{\left(z - \frac{-b + \sqrt{b^2-4}}{2}\right)\left(z - \frac{-b-\sqrt{b^2-4}}{2}\right)}$$

Whence

$$\text{Residue} = \lim_\left\{z\to \frac{-b + \sqrt{b^2-4}}{2}\right\} \ \frac{-b + \sqrt{b^2-4}}{2}\cdot f(z) = \dfrac{(\sqrt{b^2-4}-b)^n}{2^n \sqrt{b^2-4}}$$

And finally:

$$\text{Integral} = -2i\left(2\pi i \times \text{Res}\right) = 4\pi \dfrac{(\sqrt{b^2-4}-b)^n}{2^n \sqrt{b^2-4}}$$

Where again $b = \left(\dfrac{\cosh(a)}{a}\right)^{-1}$


问:
Cramer's rule doesn't work in this case. Why?
Here I have these equations,
x + y + z = 2,
2x + 3y + 2z = 5,
2x + 3y + 15z = 5
When solving normally, we get (x,y,z) = (1,1,0) but when used cramer's rule, we get
Delta = 0 , Delta x = 13, Deltta y = 13, Delta z = 0. It means that the given system of equation is inconsistent. How can this be possible?

EDIT : Determinant is not 0, it's 13. I just wrote the terms wrong when calculating :p

答:
$$\Delta=\det\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 2 & 3 & 2 \\
 2 & 3 & 15 \\
\end{array}
\right)=13$$
and
$$\Delta_x=\det\left(
\begin{array}{ccc}
 2 & 1 & 1 \\
 5 & 3 & 2 \\
 5 & 3 & 15 \\
\end{array}
\right)=13$$
To solve with Cramer's rule you must do
$$x=\frac{\Delta_x}{\Delta}=\frac{13}{13}=1$$
And so on for the other unknowns


问:
Stuck in a case of proving that $L_1$ is complete
First of all, in order to prove that $L_1$ is complete, I have to prove a couple of lemmas, where $L_1$ is like below:

Description of $L_1 :$

Primary connectives: $\lnot$ , $\lor$

$A\rightarrow B$ is $\lnot A \lor B$

Axioms:

(A1) $B \lor B \rightarrow B$

(A2) $B \rightarrow B \lor C$

(A3) $B \lor C \rightarrow C \lor B$

(A4) $(C \rightarrow D ) \rightarrow (B \lor C \rightarrow B \lor D)$

The only rule of inference is $MP$

But I am stuck in a case of proving one of the lemmas. I am stuck in proving that:

$\lnot B ,\lnot C \vdash_{L_1} \lnot(B \lor C)$

The problem is, none of the formulas that I have proved so far has anything like $\lnot (A \lor B)$ in it. So I am clueless. Do I have to prove some more lemmas or is it something trivial?

答:
I found a proof using prover9. Here is an outline of the proof it finds:

$\lnot B \vdash B \to X$ for any $X$ from A2 using definition of $\to$.

$\lnot B \vdash (X \lor B) \to (X \lor X)$ for any $X$ from 1 and A4.

$\vdash Y \lor (X \lor X) \to (Y \lor X)$ for any $X$ and $Y$ from A1 and A4.

$\lnot B \vdash (X \lor B) \to X$ for any $X$ from 2 and 3 with $Y =\lnot (X \lor B)$ using definition of $\to$.

$\lnot B \vdash (\lnot(B \lor C) \lor B) \to \lnot (B \lor C)$ from 4 with $X =\lnot (B \lor C)$.

$\lnot C \vdash (Y \lor C) \to Y$ for any $Y$ proved just like 4 with $C$ in place of $B$.

$\lnot C \vdash \lnot(B \lor C) \lor B$ from 6 with $Y = B$ and definition of $\to$.

$\lnot B, \lnot C \vdash \lnot(B \lor C)$ from 5 and 7.

Of course, a proof on more structured lines as suggested in the comments would be more informative.


问:
Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule?

I wonder if there is another way to do this.
I did this question by using Cosine Rule.
I tried to use similar triangle, but it seems that there is not enough information to support the property. The best I could is Side and Angle + unknown..

Edited

COSINE RULE

$$102^{2} = 126^{2} + 60^{2} - [ 2*{60}*{126}*Cos{A}] $$
Where A = angle (RPQ)

After finding the angle, I then use trigonometry to find

$$Cos{A} = \frac{PS}{60}$$

答:
Yep. Here's the clever way:

Simplify the lengths:

PR=10

RQ=17

PQ=21

Consider the largest side as the base (PQ in this case)

Think of two right angles with hypotenuses 10 and 17.

 8-15-17 and 6-8-10. Note that 8 is common in both and 6+15=21. Hence the scaled-down RS = 8

Now, fill the rest in the diagram, and then, scale up.


答: Hint:  Write the Pythagorean formula for each of the right triangles.  You have two equations in two unknowns.</p>

Title: Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule?

Body:
I wonder if there is another way to do this.
I did this question by using Cosine Rule.
I tried to use similar triangle, but it seems that there is not enough information to support the property. The best I could is Side and Angle + unknown..

Edited

COSINE RULE

$$102^{2} = 126^{2} + 60^{2} - [ 2*{60}*{126}*Cos{A}] $$
Where A = angle (RPQ)

After finding the angle, I then use trigonometry to find

$$Cos{A} = \frac{PS}{60}$$


答: Writing $h$ for $RS$ and $a$ for $PS$, we have $a^2 + h^2 = 60^2$ (angle $PSR$ is right because angle $QSR$ is) and $(126-a)^2 + h^2 = 102^2$.  If I subtract the second equation from the first I learn that $a^2 - (126-a)^2 = 60^2 - 102^2$.  If I do some algebra I see that the $a^2$ terms on the left hand side of this equation drop out, and I can solve directly for $a$. Knowing $a$ makes the rest easy.


问:
Determining whether a vector belongs in $V=\text{span}(v_1,v_2,v_3).$

Let $v_1=(1,1,0,3), v_2=(1,2,1,1)$ and $v_3=(0,2,1,0)$ be vectors of $\Bbb{R^4}$ and $V=\text{span}(v_1,v_2,v_3).$ Determine if $v=(5,7,8,0)$ belongs in $V$.

So I would need to reduce the following matrix $$\begin{bmatrix}v_1 && v_2&&v_3 \mid v \end{bmatrix}.$$

However this matrix seems to be inconsistent so in conclusion $v \notin V$?

答: Yes, that will work. Or you can compute the determinant of the matrix$$\begin{bmatrix}1 & 1 & 0 & 5 \\ 1 & 2 & 2 & 7 \\ 0 & 1 & 1 & 8 \\ 3 & 1 & 0 & 0\end{bmatrix}.$$Since it is not $0$ (actually, it is $-13$), the four vectors are linearly independent and therefore $v$ is not a linear combination of the other three.


问:
Finding eigenvalues of block type matrix
If I know the eigenvalues of matrix $A$, is there any way to calculate eigenvalues of $B$ using some tricks? Here $A$ is an $m\times m$, $O$ is the zero matrix and $I$ denotes $m \times m$ identity matrix. I want to calculate the eigenvalues of the matrix $B$.

$B = \left( \begin{array}{ccc}
A & I-kI \\
I & O \end{array} \right) $.

Here $k$ is any positive integer. I tried to calculate the eigenvalues of the matrix $B$ by taking $k=1$. But I could not anything relevant. Is it possible to find any relation between eigenvalues of $B$  and eigenvalues of $A$? What can be said if $A$ is a symmetric matrix?
Thanks in advance

答:
The eigenvalues of B can be found by the expression:

$$ \det \left(
\left( \begin{array}{cc}
A & I-kI \\
I & O \end{array} \right)
-
\left( \begin{array}{cc}
\lambda I & O \\
O & \lambda I \end{array} \right)
\right)
=
\det \left(
 \begin{array}{cc}
A-\lambda I & I-kI \\
I & -\lambda I \end{array}
\right)
=
0$$

There are relations for calculating the determinant of a block matrix:

$$\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(A)\times  \det\left(D - C A^{-1} B\right) = \det(D)\times  \det\left(A - B D^{-1} C\right)$$

$$
\det \left(
 \begin{array}{cc}
A-\lambda I & I-kI \\
I & -\lambda I \end{array}
\right)
=
\det(A-\lambda I)\times  \det\left(-\lambda I -  (A-\lambda I)^{-1} (I-kI)\right)
=
\det(-\lambda I)\times  \det\left( (A-\lambda I) - (I-kI) (-\lambda I)^{-1} \right)
=
0$$

Looking into the first equation, when $\lambda$ is also an eigenvalue of $A$, $\det(A-\lambda I)$ is equal to zero which means that $(A-\lambda I)$ is not invertible and this equation can't be used.

On the other hand, the second equation is only valid if $\lambda$ is different from zero. So if all the eigenvalues of $A$ are different from zero and $k=1$, the eigenvalues of $A$ are also eigenvalues of $B$.


问:
Simplification in the curvature tensor identity
In Milnor's Morse Theory, he defines the curvature $R$ of an affine connection $\vdash$ as $$R(X,Y)Z=-X\vdash(Y\vdash Z)+Y\vdash(X\vdash Z)+[X,Y]\vdash Z.$$ In his proof of Lemma 9.3 part (2), which states that $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$, he writes that, "Since all three terms of (2) are tensors, it is sufficient to prove (2) when the bracket products $[X,Y]$, $[X,Z]$, and $[Y,Z]$ are all zero."

Now, I understand how to prove this lemma without using this simplification, so I don't really need a proof of this lemma. But I don't really understand how this "wlog" follows at all from tensor-ness.

I did see the following: Bianchi identity proof : why we can consider $[X,Y]=[Y,Z]=[X,Z]=0$?
However, as I commented on that post a couple days ago, it seems that the answer implies that all vectors fields $X,Y$ have $[X,Y]=0$, which is obviously false.

Thanks!

答:
The lemma i know is:

For $v_1,\dots,v_n\in T_pM$ there are vectorfields  $ X_1,\dots, X_n\in\Gamma TM$ with $v_i= (X_i)_p$ and $[ X_i, X_j]_q=0$  for all $q$ in a neighbourhood of $p$.

This can be achieved by writing the $v_i$ as linear combinations of some coordinate vectorfields and then multiplying the linear combinations of coordinate vectorfields by a bump function which is $1$ in a neighborhood of $p$.

Now tensoriality of a map $\beta:(\Gamma TM)^n\to\Gamma E$ implies that $\beta(X_1,\dots, X_n)_p$ only depends on the values of the $X_i$ at $p$, so to show for example $\beta(X_1,\dots, X_n)_p=0$, by the above lemma one can assume $[X_i,X_j]=0$ in a neighboorhod of $p$.

This of course is weaker than assuming $[X_i,X_j]=0$ everywhere, but in most cases sufficient, since if $\nabla$ is a connection then $(\nabla_XY)_p$ only depends on the value of $X$ at $p$ and the values of $Y$ in a neighborhood of $p$.


答:
You are right.

$v\in\textrm{span}(v_1,v_2,v_3)$ means there exist $x,y,z$ such that
$$
xv_1+yv_2+zv_3=v\tag{1}
$$
In other words, the linear system $AX=v^T$, where $A=[v_1^T v_2^T v_3^T]$ has a solution $X=(x,y,z)^T$.

The augmented matrix $[A\mid v^T]$ being inconsistent means that (1) has no solution and thus $$
v\not\in\textrm{span}(v_1,v_2,v_3).$$

Title: Determining whether a vector belongs in $V=\text{span}(v_1,v_2,v_3).$

Body:
Let $v_1=(1,1,0,3), v_2=(1,2,1,1)$ and $v_3=(0,2,1,0)$ be vectors of $\Bbb{R^4}$ and $V=\text{span}(v_1,v_2,v_3).$ Determine if $v=(5,7,8,0)$ belongs in $V$.

So I would need to reduce the following matrix $$\begin{bmatrix}v_1 && v_2&&v_3 \mid v \end{bmatrix}.$$

However this matrix seems to be inconsistent so in conclusion $v \notin V$?


问:
Relation between Stirling numbers of first kind and harmonic numbers
We have the following nice relations for Striling numbers of the first kind

$${n\brack 2} = \Gamma(n) H_{n-1}$$
$${n\brack 3} = \frac{\Gamma(n)}{2} \big((H_{n-1})^2-H_{n-1}^{(2)}\big)$$

Where

$$H^{(p)}_n = \sum^n_{k=1} \frac{1}{k^p}, \,\,\,H^{(1)}_n \equiv H_n$$

Questions

I want an algebraic proof (not combinatorial) for the previous relations.

Is there a "simple" general formula in terms of the harmonic numbers for

$${n\brack k} = {?}$$

答: An idea to get a general formula is given in this Wikipedia article: we write $$\frac{1}{n!}\sum_{k=0}^n{n+1\brack k+1}x^k=\prod_{k=1}^n\left(1+\frac xk\right)=\exp\sum_{k=1}^n\log\left(1+\frac xk\right)\\=\exp\sum_{k=1}^n\sum_{j=1}^\infty\frac{(-1)^{j-1}}{j}\left(\frac xk\right)^j=\exp\sum_{j=1}^\infty\frac{(-1)^{j-1}}{j}H_n^{(j)}x^j$$ and apply exponentiation of a power series. The result can be written as $${n+1\brack k+1}=n!(-1)^k\sum_{\substack{a_1,\,a_2,\,\ldots,\,a_k\,\geqslant\,0\\a_1+2a_2+\ldots+ka_k=k}}\prod_{j=1}^k\frac{1}{a_j!}\left(-\frac{H_n^{(j)}}{j}\right)^{a_j}.$$


问:
Does $\int_{I}f(t)g(t)dt = 0$ for every $g|_{\partial I} = 0$ implies $f\equiv 0$?
Let $I=[t_{0},t_{1}]$ be a bounded interval of $\mathbb{R}$ and suppose $f: I \to \mathbb{R}$ is $C^{k}(I)$ for some $k \ge 2$. Suppose that:
$$\int_{t_{0}}^{t_{1}}f(t)g(t) dt = 0$$
for every $C^{k}(I)$ function $g:I\to \mathbb{R}$ such that $g(t_{0})=g(t_{1}) = 0$. Does it follow that $f(t) \equiv 0$?

答:
Elaborating on leoli1's comment, if your $f$ is such that $f(t) = r > 0$ for some $t\in (t_0, t_1)$ then there exists a open $O\subset (t_0, t_1)$, and a closed set $K\subset O$ such that $f(O) \in (0, \infty)$ and $t$ is an interior point of $K$. From known theory we can construct a smooth function (hence also $C^k$) $\phi: [t_0, t_1]\to [0,1]$ such that  $\phi(K) = 1$ and $\phi$ is zero outside of $O$.

For the record, these kind of "bump" functions can be constructed out of the (smooth) partition of unity subordinate to the cover $$\{O, [t_0, t_1]\backslash K\}$$ of $[t_0, t_1]$, explicitly, just take $\phi$ to be function in the partition of unity subordinate to the open $O$ in the above cover.

Clearly $$\int_{t_0}^{t_1}f(t) \phi(t) > 0$$
so we must have  $f(t) = 0$ for all $t\in (t_0, t_1)$ and by continuity of $f$ also on the boundary points. $t_0$ and $t_1$.


问:
Given $z^{z^x}=x$, estimate $\frac{dx}{dz}$
I'm reading Knoebel's Exponentials Reiterated, page 242 and have very hard time understanding the following:

If $z^{z^x}=x$, then $\frac{dx}{dz}\gt 0$ on the open line
$$L=\{<z,x>|0\lt z\lt e^{-e}\,\text{and}\, x=e^{-1}\}.$$

This doesn't make sense to me. Because if $z^{z^x}=x$, then $x=e^{-1}$ implies $z=e^{-e}$, which is not in the open interval $\left(0,e^{-e}\right)$.

I know that, if $j(z,x)=z^{z^x}-x$, then
$$dj(z,x)=z^{z^x}z^{x-1}(x\ln z+1)\,dz+\left(z^{z^x}z^x\ln ^2z-1\right)\,dx,$$
and there is a unique trajectory satisfying $j(z,x)=0$ when it is not true that $z=e^{-e}$ and $x=e^{-1}$ simultaneously by the Implicit function theorem,
but how could that be used to prove $\frac{dx}{dz}\gt 0$ on the line?

答:
I think Knoebel meant $dx/dz>0$ where $z\in(0,e^{-e})$ and $x\in(0,e^{-1})$. Otherwise we would just be evaluating the derivative at one point, namely $(x,z)=(e^{-1},e^{-e})$.

Note that $z^{z^x}=x$ is equivalent to $\ln\ln x=\ln\ln z+x\ln z$. Implicit differentiation gives $$\frac{dx}{dz}=\frac{x\ln x(1+x\ln z)}{z\ln z(1-x\ln x\ln z)}$$ after collecting terms. The derivative is positive if $\frac{1+x\ln z}{1-x\ln x\ln z}>0$ which is true following a case-by-case analysis of $1-x\ln x\ln z\in(-\infty,0)$, $(0,1/e)$, $(1/e,1)$ and $(1,\infty)$.


问:
Clique-Width - Confusion over Proof
I am working through some Graph Theory notes, and do not understand a line from the proof of a Theorem. Here is the statement of the Theorem.

$cw(G)=max\{cw(H)$ : $H$ is a prime induced subgraph of $G$$\}$.

The proof uses induction on $n=|V(G)|$ and starts by dealing with the case that $G$ is prime itself, which is easy to follow. It then continues as follows:

Let $G$ be non-prime, and let $M_{1},...M_{p}$ be maximal non-trivial modules of $G$. By contracting each $M_{i}$ into a single vertex $m_{i}$ we obtain the characteristic graph $G_{0}$ of $G$. We separately  construct expressions representing the graphs $G_{0}$ and $G[M_{i}]$ for each i, assuming by the induction that each expression uses at most $max\{cw(H)$ : $H$ is a prime induced subgraph of $G$$\}$ different labels. If in the expression defining $G_{0}$ the vertex $m_{i}$ is created with label $j$, we finish the construction of the graph $G[M_{i}]$ by renaming all labels to $j$. Then in the tree describing $G_{0}$ we replace the node creating $m_{i}$ with the root of the tree creating $G[M_{i}]$. The resulting tree represents $G$ and uses at most $max\{cw(H)$ : $H$ is a prime induced subgraph of $G$$\}$ labels, as required. $\square$

Ok so here are my thoughts. It agree that the construction of each $G[M_{i}]$ requires at most $max\{cw(H) \space : \space H$ is a prime induced subgraph of $G\}$ labels, by the induction hypothesis and the fact that any prime induced subgraph of $G[M_{i}]$ is a prime induced subgraph of $G$. But when constructing $G$ by taking the construction of $G_{0}$ and 'editing' in the sub-constructions of the $G[M_{i}]$, we have to make sure that the labels used in construction of each $G[M_{i}]$ have not already been used in the construction of $G$. For instance, when we come to the creation of the last vertex $m_{p}$ of $G_{0}$, although we may only need less than or equal to the required number of labels for the construction $G[M_{p}]$, these labels have to be chosen to not 'clash' with the labels occupied by the previous $m_{i}$, so we cannot be sure we are not using too many labels?

Would really appreciate some help on this.

Thanks

Misha:

答:
From the point of view of the formula:

If $X$ is an expression for creating $G[M_i]$, then $\rho_{1 \to j}(\rho_{2 \to j}(\dots \rho_{k \to j}(X)\dots))$ is an expression for creating a $G[M_i]$ with all vertices labeled $j$.

In the description for $G_0$, if we replace the vertex creation $j (m_i)$ with the expression $\rho_{1 \to j}(\rho_{2 \to j}(\dots \rho_{k \to j}(X)\dots))$, then instead of creating a single vertex $m_i$ with label $j$, we'll get all of $G[M_i]$. Then, because the labels are the same, the edges between $G[M_i]$ and the rest of $G$ will all be the same as the edges between $m_i$ and the rest of $G_0$, exactly what we want from a module.

Order of operations ensures that whatever happens in $\rho_{1 \to j}(\rho_{2 \to j}(\dots \rho_{k \to j}(X)\dots))$ does not affect the rest of the formula.

From the point of view of the tree:

The creation of $G[M_1], G[M_2], \dots$ are happening in separate subtrees. When we do a relabeling $\rho_{i \to j}$, a edge join $\eta_{i,j}$, or anything else, that node only affects the things in the subtree below it. So we don't have to worry about the same labels being used for a different purpose in another subtree.

And no matter if we're looking at a single $j(m_i)$ node, or a complicated subtree creating a uniformly $j$-labeled $G[M_i]$, the next thing we see above it is a disjoint union node connecting it to the rest of the graph. That is what separates labels in the subtree from labels used anywhere else.

Title: Clique-Width - Confusion over Proof


问:
Banach spaces admitting a coarser topology for which the closed unit ball is compact.
Let $X$ be a Banach space and let $B$ be its closed unit ball.

It is well known that $B$  is compact in the weak topology provided $B$ is reflexive.   Otherwise,   if $X$ is at least
a  dual space, then $B$ is compact in the weak* topology.

These examples show that, in many situations, there exists a Hausdorff locally convex topology on $X$, coarser than the
norm topology,  relative to which
$B$ is compact.

For $X=c_0$, I cannot think of such a topology and I doubt it exists.  The reason is I feel the sequence $\{x_n\}_n$ given
by
$$
  x_n= (1,1,\ldots, 1,0,0, 0\ldots)
  $$
($n$ ones) cannot possibly have a cluster point in any sensible topology.

Question.  Given a Banach space $X$,  is there always  a Hausdorff locally convex topology on $X$, coarser than the
norm topology, relative to which the
closed unit ball is compact? Can one characterize the spaces for which this is true?
What if $X=c_0$?

答:
Krein–Milman theorem — A compact convex subset of a Hausdorff locally convex topological vector space is equal to the closed convex hull of its extreme points.

Since the unit ball of $c_0$ has no extreme points, we are done.


问:
Kolmogorov Differential Equations governing random dynamical system
I am reading the original paper (Translation) by Kolmogorov "On Analytical Methods in Probability Theory". I do not quite understand some of the conclusions in the paper.

This is also on the wiki page,

https://en.wikipedia.org/wiki/Kolmogorov_equations_(Markov_jump_process)

I do not understand the condition $\sum_k A_{jk} (t) = 0$. This is supposed to follow from the fact that $\sum_j P_{ij}(t,u) = 1$.

答:
Renaming indices you get to

$$
A_{jk} = \left. \frac{\partial P_{jk}}{\partial u} \right|_{u=t}
$$

so that

\begin{eqnarray}
\sum_k A_{jk} &=& \sum_k\left. \frac{\partial P_{jk}}{\partial u} \right|_{u=t} \\
&=& \left.\frac{\partial}{\partial u}\left(\color{blue}{\sum_k P_{jk} }\right)\right|_{u=t} \\
&=& \left.\frac{\partial}{\partial u}\left(\color{blue}{1 }\right)\right|_{u=t} \\
&=& 0
\end{eqnarray}

so that


问:
What are all the integral solutions of $n!=m(m^2-1)$?
Observe that:

$3!=2(2^2-1)$

$4!=3(3^2-1)$

$5!=5(5^2-1)$

$6!=9(9^2-1)$

Question: What are all the integral solutions of $n!=m(m^2-1)$?

I guess it is just $(n,m) = (3,2),(4,3),(5,5),(6,9)$, but how to prove that there is no other one?

I checked that there is no other one for $n<20$.

Let reformulate the problem using Cardano's formula:    consider the cubic equation $$x^3+px+q=0,$$
its discriminant is $\Delta = -(4p^3+27p^2)$. Our case corresponds to $(p,q) = (-1,-n!)$, so $\Delta = 4-27(n!)^2 <0$. Thus the cubic has one real root and two non-real complex conjugate roots, and by Cardano's formula, the real root is $$ \left(-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}} \right)^{1/3} +\left(-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}\right)^{1/3} $$
So the problem "reduces" to ask whether there exists an integer $n>6$ such that

$$ \left(\frac{n!}{2} +\sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3} +\left(\frac{n!}{2} - \sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3}  $$
is also an integer. It is not clear that it helps...

答:
In Richard Guy's third edition of Unsolved Problems in Number Theory, he writes on p.301 in D25 Equations involving factorial $n$. that

Simmons notes that $n!=(m-1)m(m+1)$ for $(m,n)=(2,3),(3,4),(5,5),$ and $(9,6)$ and asks if there are other solutions. More generally he asks if there are any other solutions of $n!+x=x^k.$

This Simmons is in reference to:

Gustavus J. Simmons, A factorial conjecture, J. Recreational Math., 1(1968) 38.

So it would seem that this problem was open in 2004 when Guy wrote the third edition of his book, but I don't know if it has been solved in the last 17 years.


答:
Hint:

$$\begin{cases}PS^2+SR^2&=60^2,\\QS^2+SR^2&=102^2,\\QS+PS&=126.\end{cases}$$

From this,

$$QS-PS=\frac{QS^2-PS^2}{QS+PS}=54$$ and the rest easily follows.

$$PS=36,QS=90,SR=48.$$

Alternatively, express that the heights of the two right triangles are the same:

$$60^2-PS^2=102^2-(126-PS)^2,$$

or after simplification

$$252\,PS=9072.$$


问:
Finding Area of a shape
I'm doing some revision and I'm a bit stuck. How do you find out the area of this shape?

I know you have to do 10cm x 2cm = 20cm devided by 2 = 10cm - 2cm x ? (something) = area.

I'm not sure what the cm of the line next to 10cm is (the question mark bit). I mean would you do 13cm - 10cm = 3cm to find the missing length. Or would it be 2cm like the other side?

Would it be 6cm squared or 4cm squared.

Btw that corner bit is not part of the shape, it's only the 2 sides traingle without the right angle... that's just added on so I can work it out.

P.S. I know this is an easy question but it's bugging me.

答: Use Heron's formula to find the area of a triangle given its sides.</p>

Title: Finding Area of a shape


问:
compute usig cauchy integral formula, $ \int_\Gamma \frac{dz}{z^2+1}$
The question is:
$$ \int_\Gamma \frac{dz}{z^2+1}$$

Here is how i thought, we can divide it into two contour and then deform continuously the inner curve to a circle around $i$ but my problem is here how should i deform the bigger contour around $-i$ i couldn't do that, because we can't pass through singularity, what do you think? any suggestion would be great,

Thanks

答:
The "three curves" solution.  We use three curves.

The first curve surrounds $-i$ but not $i$.  Notice the direction was reversed on the inner loop.

The second and thrid curves surround $i$ but not $-i$.

Add the three together to get the original curve.


问:
If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where f is a bijection}\}$ mean?
If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where $f$ is a bijection}\}$ mean.
From my understanding it means the Set containing the function $f$ that maps elements of $X$ onto elements of $X$, where $f$ is a bijection.
However, in the proof to see whether $S(X)$ is a group under composition, they said that composition is a closed binary operation for $S(X)$ because
if $f$ is a bijection, then $f(g)$ is also a bijection so $f(g)$ is in $S(X)$

But where did they get $g$ from? Why can you randomly talk about other functions if $S(X)$ only contains the function $f$ that is bijective.

Thanks in advance

答: In a sense, $S(X)$ "comes prior" to (abstract) groups: are precisely the four properties fulfilled by $S(X)$ endowed with map composition as operation (closure, associativity, identity, inverses), that the definition of (abstract) group just mimics and axiomatizes for any set endowed with a binary operation.


问:
Should the distance between two disjoint sets always equal the distance between their infimum and supremum?
Let $A$ and $B$ be two subsets of a metric space $X \subset \mathbb{R}$, where the metric is the usual notion of distance. Clearly if $A \cap B \neq \varnothing$, then $\mathrm{dist}(A, B) = \inf\{d(a, b)\mid a \in A, b \in B\} = 0$. So suppose that $A \cap B = \varnothing$ and $\forall b \in B, \forall a \in A: b > a$.  If we are talking only about metric spaces like $X$, it feels  intuitive that the distance between $A$ and $B$ should equal the distance between their respective supremum and infimum. That is, $\mathrm{dist}(A, B) = d(\sup(A), \inf(B))$. (I think this could be proven by showing a contradiction with respect to the triangle inequality property of a metric.)

So my question is that how is this intuitive feeling of distance between two sets generalized, if at all, and what properties does the surrounding space $X$ need to satisfy? Is it necessary and sufficient for $X$ to obey the order axioms and density of real numbers, or whatever their general variants are called?

答: Your assumption that, if $A,B\subset\Bbb R$, then$$\operatorname{dist}(A,B)=|\sup(A)-\inf(B)|\tag1$$is wrong. Note that if you exchange $A$ and $B$, then the number $\operatorname{dist}(A,B)$ reamins the same, whereas $|\sup(A)-\inf(B)|$ changes (in general). Anyway, if$$A=\left\{\frac1n\,\middle|\,n\in\Bbb N\right\}\quad\text{and}\quad B=\left\{\frac{\sqrt2}n\,\middle|\,n\in\Bbb N\right\}$$then $\operatorname{dist}(A,B)=0$, but $\inf(A)=\inf(B)=0$, $\sup(A)=1$ and $\sup(B)=\sqrt2$. So you neither have $(1)$ nor $\operatorname{dist}(A,B)=|\sup(B)-\inf(A)|$.


问:
Gradience wrt a Position Vector
I found the following equations in a dynamics textbook,

The gravitational potential energy for any two particles in a n-partice system is given by,

$V_{ij} = - \frac {G  m_i  m_j}{r_{ij}}$

where $r_{ij}$ is the distance between $m_i$ and $m_j$. The total potential energy of the system is,

$V = \frac{1}{2} \sum_{i = 1}^{n} \sum_{j = 1}^{n}V_{ij}$ (i $\neq$ j)

If $R_i$ is the postion vector of the $i^{th}$ particle, then

$\frac{\partial{V}} {\partial {\vec{R_{i}}}} = - \frac{\partial{V}}{\partial{\vec{r_{ji}}}} + \frac{\partial{V}}{\partial{\vec{r_{ij}}}} = -2 \frac{\partial{V}}{\partial{\vec{r_{ij}}}}$

What does it mean to take the derivative of a Scalar Function$(V)$ with respect to a vector$(\vec{R_1})$? Is it directional derivative?

I've been trying all day to get the last equation. I would be very grateful if somebody could help me(or mention some reference perhaps). I don't really know which part of math is used to get the last equation.

Thanks in advance.

答:
Let’s do it for two particles – generalization will be straightforward.
The potential energy of two particles is $V_{ij} = - \frac {Gm_i  m_j}{r_{ij}}=-\frac {Gm_i  m_j}{|\vec{r_{ij}}|}=- \frac {Gm_i  m_j}{|\vec{R_{i}}-\vec{R_{j}}|}$,  because $r_{ij} =|\vec{R_{i}}-\vec{R_{j}}|$.

By definition $\frac{\partial}{\partial\vec{R}}V({R_{i}})=\frac{\partial} {\partial{R_x}}\vec{e_{x}}+\frac{\partial}{\partial{R_y}}\vec{e_y}+\frac{\partial} {\partial{R_z}}\vec{e_z}$, where vectors $\vec{e_x}, \vec{e_y}, \vec{e_z}$ are directed along axis X, Y and Z correspondingly and each has a unit length.

Next,
$\frac{\partial}{\partial\vec{R_1}}\frac{1}{|\vec{R_1}-\vec{R_2}|}=(\frac{\partial} {\partial{R_{1x}}}\vec{e_x}+\frac{\partial}{\partial{R_{1y}}}\vec{e_y}$$+\frac{\partial} {\partial{R_{1z}}}\vec{e_z})\frac{1}{\sqrt{(R_{1x}- R_{2x})^2+(R_{1y}- R_{2y})^2+(R_{1z}- R_{2z})^2}}$.

Taking derivatives we get explicitly:
$\frac{\partial}{\partial\vec{R_1}}\frac{1}{|\vec{R_1}-\vec{R_2}|}=-\left((R_{1x}- R_{2x})\vec{e_x}+(R_{1y}-R_{2y})\vec{e_y}+(R_{1z}- R_{2z})\vec{e_z}\right)(\frac{1}{\sqrt{(R_{1x}- R_{2x})^2+(R_{1y}- R_{2y})^2+(R_{1z}- R_{2z})^2)}})^{3}$$=-\vec{r_{12}}\left((R_{1x}- R_{2x})^2+(R_{1y}- R_{2y})^2+(R_{1z}- R_{2z})^2\right)^{-\frac{3}{2}}$

If you take the derivatives with respect to $\vec{R_2}$ you will get the same formula, but with the negative sing.

Title: Gradience wrt a Position Vector