问:
How to solve this problem without using the binomial expansion?
Consider the following problem:

On a train route of 18 stations, in how many ways can a train stop at 4 stations so that there are at least 2 and at most 6 stations between any two of them?

The above is the problem I have a question about. Meanwhile, consider another similar problem:

A box contains 10 red, 12 yellow and 8 green balls. In how many ways can 14 balls be selected if at least one ball of each colour must be selected?

The above is a sticks-and-stars problem with an upper bound on the number of stars between two sticks. We can reduce the above problem to permuting 2 sticks (3 colours) and 11 stars (taking 1 red, 1 yellow and 1 green ball of the 14 required). The total number of permutations is $13 \choose 2$. From this we subtract the illegal cases like the one below:

$$**********|*|$$

where the space to the left of the first stick represents red balls. Note that there are 10 stars here, but we clearly have only 9 red balls in our budget. There are two ways this particular illegal case could arise, and we can find them by permuting everything to the right of the first stick (there are $2 \choose 1$ permutations). Another illegal case:

$$***********\space |\space |$$

Total number of permutations of this illegal case = $1 \choose 1$

We can repeat the above process for green balls, of which we can take only 7. (Everything to the left of the first stick is green balls now). The total number of permutations where there are more than 7 green balls is ${4 \choose 1} + {3 \choose 1} + {2 \choose1} + {1\choose 1} $.

Finally, we subtract the number of illegal permutations from the number of total permutations to get: $${13 \choose 2} - {2 \choose 1} - {1 \choose 1} - {4 \choose 1} - {3 \choose 1} - {2 \choose 1} - {1 \choose 1} = 65 \text{ legal cases}$$

My teacher uses the binomial expansion to solve most combinatorial problems, including this one (he finds the coefficient of a particular power of $t$ in a specific $p(t)$). I do not like his method as it is not at all intuitive. However, my own approach works only because the order of the sticks wasn't important (we deemed everything to the left of the first stick to be red balls in the first case and green balls in the other). It will not work when counting stations.

To illustrate what I mean, consider the below:

Take away 6 stations from 18, since there must at least be 2 stations between each of the 4 selected stations. Now, our problem simplifies to: let the four stations be represented by sticks and the stations between them be stars; find the number of permutations where there are between zero and four stars between any two sticks. For example, these are allowed:

$$*|**|**|**|*$$ $$*****|*|**||$$

But this is not allowed: $$|******|**||$$

How do I modify my method to solve this problem? Other combinatorial approaches are also appreciated.

答:
If $x_1,\ldots,x_5$ are the numbers of stars before the first bar, between consecutive bars, and after the last bar, we want the number of solutions in non-negative integers to the equation

$$x_1+x_2+x_3+x_4+x_5=8$$

subject to the conditions that $x_2,x_3,x_4\le 4$. This can be solved as an inclusion-exclusion problem. Without the limits on $x_2,x_3$, and $x_4$ there are $\binom{12}4$ solutions. How many solutions exceed the limit on $x_2$?

To count those, we can assume that $x_2\ge 5$, replace it by $y_2=x_2-5$, and count solutions in non-negative integers to

$$x_1+y_2+x_3+x_4+x_5=3\,,$$

of which there are $\binom74$. The same goes for $x_3$ and $x_4$, so a better approximation to the desired result is $\binom{12}4-3\binom74$. And in fact this is the desired result, since it is impossible for more than one of $x_2,x_3$, and $x_4$ to exceed its limit: there are

$$\binom{12}4-3\binom74=495-3\cdot 35=390$$

solutions meeting the stated conditions.


问:
Proof Möbius inversion theorem without using Dirichlet convolutions
I’m reading a textbook on number theory and i just finished reading möbius inversion theorem

I also read the proof and fully understand it
But I have a question though

How can I Prove möbius inversion theorem directly and without using Dirichlet convolutions.

Can someone drop a hint or give an answer?

Thanks in advance

答:
I suppose the meaning of "without using Dirichlet convolutions" is to avoid using its associative and commutative, or multiplicative properties.

Let $f(n)$ and $g(n)$ be arbitrary arithmetic functions such that

$$
g(n)=\sum_{d|n}f(d)
$$

Then we have

$$
\begin{aligned}
\sum_{d|n}g(d)\mu\left(\frac nd\right)
&=\sum_{ab=n}\mu(a)g(b) \\
&=\sum_{ab=n}\mu(a)\sum_{cd=b}f(c) \\
&=\sum_{acd=n}\mu(a)f(c) \\
&=\sum_{rc=n}f(c)\color{blue}{\sum_{ad=r}\mu(a)}
\end{aligned}
$$

Now, the core part of our problem is to handle the blue part. Since $\mu(a)=0$ whenever $a$ is divisible by some square number $>1$, we may run the summation only on factors $p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$ of $r$ with $a_m\in\{0,1\}$:

$$
\sum_{ad=r}\mu(a)=\sum_{a_m\in(0,1)}\mu(p_1^{a_1}p_2^{a_2}\dots p_k^{a_k})
$$

Since $\mu(p)=-1$ for all primes, and for $a_m\in\{0,1\}$, there are exactly $\binom nt$ ways to make $a_1+a_2+\cdots+a_k=t$, so we arrive at this after interchanging the summation symbol

$$
\sum_{ad=r}\mu(a)=\sum_{t=0}^k(-1)^t\binom kt
$$

When $r=1$, we have $k=0$, so this evaluates to one. However when $r>1$ this becomes zero due to binomial theorem. Finally, we obtain

$$
\sum_{ad=r}\mu(a)=\begin{cases}
1 & r=1 \\
0 & r\ne1
\end{cases}
$$

Plugging this result into the blue formula, we deduce

$$
\sum_{d|n}g(d)\mu\left(\frac nd\right)=\sum_{c=n}f(c)=f(n)
$$

Thus completing the derivation.

ContentLicense: CC BY-SA 4.0


问:
Simple proof showing a subset has smaller cardinal
In Shoenfield, page 254, I read a proof of the following fact: $x \subseteq y \implies Card(x) \le Card(y)$.

Recall

$Card(x)$ is the least ordinal $\alpha$ that is bijective to $x$.

Thus the above inequality just means that $Card(x) = Card(y) \lor Card(x) \in Card(y)$.

Simple proof

The proof uses a remark issued from the proof that the axiom of choice implies Zermelo's theorem. I'm looking for a direct proof which assumes already Zermelo's theorem to be equivalent to the axiom of choice and simple facts about ordinals such as the facts gathered here.

答:
Here is my work on the way Shoenfield proves it. It does not provide an alternative proof. As it can be seen, the procedure doesn't use Zermelo's theorem. As one can see, simplicity comes when understanding that each proof is essentially counting the elements of a particular object using the ordinals.

Count of a set

Every set is equipotent to an ordinal.

Proof

Let $A$ be a set. We are going to count the elements of $A$ and put them in bijection with the elements of some ordinal $\alpha$. First thing to realize is that if we count an element, then we shouldn't count its parts (if they also belong to $A$) because their parts come for free to the ordinal $\alpha$.

So set $\delta = \mathcal{P}(A) \setminus \{\emptyset\}$ and $g$ its choice function. Since $A$ is a set, it cannot be equal to $V$ and there exists $b \notin A$. We will use transfinite recursion to define such an iteration. Here is the function $F: V \to V$: [
F x =
\begin{cases}
 g(A \setminus range(x)) & range(x) \subset A \\
 b                       & otherwise
 \end{cases}
] here is the corresponding transfinite recursive function $G: On \to V$: [
F \xi =
\begin{cases}
  g(A \setminus range(F \upharpoonright \xi)) & range(F \upharpoonright \xi) \subset A \\
  b                                           & otherwise
  \end{cases}
] Some calculation will show $F 0 = g A \in A$, $F 1 = g(A \setminus g A) \in A \setminus g A$ when $gA \subset A, \ldots$. So we already see that at each iteration we choose a fresh element and that iteration stops when the previously chosen elements exhaust $A$. Iteration has to stop since otherwise that would give an injection $On \to A$ contradicting that $A$ is a set. By transfinite induction, there will be a least stopping ordinal $\alpha$. At that point, $F\upharpoonright_{\alpha}: \alpha \to A$ is an injection and $range(F \upharpoonright_{\alpha}) = \{ F x. x \in \alpha \} \subseteq A$ cannot be a proper subset of $A$, thus $range(F \upharpoonright_{\alpha}) = A$. Thus, $F\upharpoonright_{\alpha}$ is a bijection with $\alpha$.

Count of a set of ordinals

If $x \subset \sigma \in On$ then $\exists \tau \in On. \tau \le \sigma$ and $\tau$ is bijective to $x$.

Proof

The counting theorem on sets gives as an ordinal $\tau$ that is equipotent to $x$ via a counting function $F: \tau \to x$ such that $F(\rho) = g ` (x - range(F \upharpoonright \rho))$. By contradiction, if $\sigma < \tau$ then it is mapped by $F$ to $F `\sigma \in x \subset \sigma \implies F `\sigma < \sigma$. But this doesn't match our intuition that we are counting our set $x$. So let $\rho$ be the least ordinal such that this unordering happens, i.e. $F `\rho < \rho$</span>. Then necessarily <span class="math-container">$F ` F ` \rho = F ` \rho$. This violates the injectivity of F.

What we wanted

Note that $x$ will be bijective with a subset of $Card(y)$. So by the count of a set of ordinals. There is $\tau \le Card(y)$ bijective to $x$. So $Card(x) \le \tau \le Card(y)$.


问:
Can power set of any set equals to $\phi$
In my topology quiz this question was asked and I couldn't do it right. So, I am posting it here.

Let A be any set . Let P(A) be the power set of A, that is the set of all subsets of A, Then how is the statement" P(A)=$\phi$ for some A" a false statement?

I think if A =$\phi$ then P(A) is also $\phi$ but in answer key it is marked as false ?

Why?

答:
The powerset of a set $X$ has $2^{|X|}$ elements.
So the powerset of any set has at least one element (as $2^0=1$), and therefore can not be empty.


问:
In $\Delta ABC$ $\angle BAC = 40^\circ, AB = 10$ and $AC = 6$. Points $D$ and $E$ lie on $AB$ and $AC$ respectively.

In $\Delta ABC$, $\angle BAC = 40^\circ, AB = 10$ and $AC = 6$. Points $D$ and $E$ lie on $AB$ and $AC$ respectively. What is the minimum possible value of $BE + DE + CD?$

What I Tried: Here is a picture :-

I have already seen the solution in the Art of Problem Solving Website, because this is one of the $2014$ AMC $12$A Problems.
Check the solutions here :- https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20

The problem, however, is that all the solutions use Law of Cosines as Trigonometry, and I prefer using simple geometry techniques to solve this problem.

So is it possible to solve this without Trigonometry? Thank You.

答:
In this construction we have $|AB_1|=|AB|=10$, $|AC_1|=|AC|=6$,
$|ED_1|=|ED|$, $|D_1C_1|=|DC|$.

Since $\angle C_1GA=90^\circ$ and
$\angle GAC_1=180^\circ-3\cdot40^\circ=60^\circ$,
$\triangle C_1GA$ is half of the equilateral triangle
with the altitude $C_1G$, hence
\begin{align}
|AG|&=\tfrac12\,|AC_1|=3
,\\
|C_1G|&=\sqrt{|AG|^2+|AC_1|^2}=\tfrac{\sqrt3}2\,|AC_1|=3\sqrt3
,\\
|BG|&=|AG|+|AB|=13
,\\
|BC_1|&=|BE|+|ED_1|+|D_1C_1|
\\
&=|BE|+|ED|+|DC|
\\
&=\sqrt{|BG|^2+|C_1G|^2}=\sqrt{196}=14
.
\end{align}


问:
Finding a kind of matrix
Given $h > 0$, let

$$B := \begin{pmatrix} 0 & -2 \\ \frac{1}{2h} & 0 \end{pmatrix}$$

or, more generally, let

$$B := \begin{pmatrix} 0^{n\times n} & -2 I^{n\times n}  \\ \frac{1}{2h}I^{n\times n}  & 0^{n\times n} \end{pmatrix}$$

Is there an explicit form for $e^{B}$?

答:
If  $\bar{\textbf{x}}(t)' = B\bar{\textbf{x}}(t) $ then $ \bar{\textbf{x}}(t) = e^{tB}\begin{pmatrix} x_0\\ y_0 \end{pmatrix}$

It will be fairly easy to solve the ODEs $x' = -2y$ and $y' = \frac{1}{h}x$
Then from this you will have a solution

Title: Finding a kind of matrix


问:
Question regarding syntax of $\bigcap$ definition
I know what this definition is supposed to do. However just looking at the syntax, i cannot explain it to myself.

$$
\bigcap M := \{x \mid \forall y \in M : x \in y\}
$$

Could somebody please explain me how to read this definition (not what it does, but rather how to read it close to the definition's syntax so that it is easier to understand)?

答:
It would be easier to see it this way: Let $A$ and $B$ be sets. Let $M=\{A,B\}.$ Define $$\bigcap M = \bigcap_{y\in M} y.$$

Then $\bigcap M = A \cap B,$ and if you expand the definition above you can see that it fits the one you provided.


问:
Let $\alpha$ be a solution of a monic quadratic polynomial with integer entries and $|\alpha|=1$.Then prove that $\alpha^{12}=1$.
($a$) Let $\alpha$ be a solution of a monic quadratic polynomial with integer entries and $|\alpha|=1$.Then prove that $\alpha^{12}=1$.

($b$)Let $A \in M_2(\mathbb{Z})$ such that $A^n=I$ for some $n$ then show that $A^{12}=I$

completely stuck on it.can I get some help?

答:
Since $\alpha$ is a zero of the polynomial we ca write the polynomial as

$$P(x)=(x-\alpha)(x-\beta)$$
for some $\beta\in\mathbb C$. Now either $\alpha$ is real, then $\alpha=\pm 1$ and we are done, or it is complex $\alpha=e^{i\phi}$ for some $\phi$. Then $\beta=e^{-i\phi}$, because the complex zeros of a real polynomial come in conjugate pairs. Now the coefficients of $P$ are integers, thus $\alpha+\beta=2\cos(\phi)=p$ for some $p\in \mathbb Z$.
Thus $$\phi=\arccos(\frac p2)\in\{\frac \pi3,\frac{2\pi}{3},\pi,0,\frac\pi2\}+2\pi\mathbb Z$$
In any case $$\alpha^{12}=e^{12i\phi}=1$$

For the second question: The characteristic polynomial of $A$ satisfies the conditions on the polynomial in part a). The (complex) eigenvalues are solutions of this polynomial and since $A^n=I$, the matrix $A$ is diagonalisable over $\mathbb C$  (This is clear if the eigenvalues are distinct and easy to see if they coincide). In particular the $n^{th}$ power of the eigenvalues is 1, which implies $|\alpha|=1$. Finally $A$ is (up to base-change) a diagonal matrix where the entries saitsfy $\alpha^{12}=\beta^{12}=1$.

ContentLicense: CC BY-SA 3.0


问:
Prove $\sum_{d\mid n}\mu(d)=[\frac{1}{n}]$
I'm reading a direct proof for mobius inversion theorem but I don’t know how the writer make the assumption that $$\sum_{d\mid n}\mu(d)=\left[\frac{1}{n}\right]$$

Can someone please explain and prove this?

答:
Observe that

$$\left\lfloor\frac1n\right\rfloor=\begin{cases}
1,&\text{if }n=1\\
0,&\text{if }n>0\,,
\end{cases}$$

so this is just a fancy way of saying that

$$\sum_{d\mid n}\mu(d)=\begin{cases}
1,&\text{if }n=1\\
0,&\text{if }n>0\,.
\end{cases}\tag{1}$$

By definition $\mu(1)=1$, so $(1)$ holds for $n=1$. If $p$ is any prime,

$$\begin{align*}
\sum_{d\mid p^k}\mu(d)&=\sum_{i=0}^k\mu(p^i)\\
&=\mu(1)+\mu(p)+\sum_{i=2}^k\mu(p^i)\\
&=1+(-1)+0\\
&=0\,.
\end{align*}$$

The result now follows from the fact that $\mu$ is multiplicative.


答:
$$
\begin{array}{cccccc}
\bigcap M
 & :=
 & \{
 & x
 & \mid
 & \forall
 &  y  \in M :
 &  x \in y
 & \}  \\
\text{This “$\bigcap M$” thing}
 & \text{means}
 & \text{the set of}
 & \text{all $x$}
 & \text{such that}
 & \text{for every}
 & \text{ set $y$ in $M$,}
 & \text{$x$ is in set $y$}
 & \text{.}
\end{array}
$$

Title: Question regarding syntax of $\bigcap$ definition

Body:
I know what this definition is supposed to do. However just looking at the syntax, i cannot explain it to myself.

$$
\bigcap M := \{x \mid \forall y \in M : x \in y\}
$$

Could somebody please explain me how to read this definition (not what it does, but rather how to read it close to the definition's syntax so that it is easier to understand)?


问:
Finding $x$ with $\langle\, Ax, x\rangle > 0$ when matrix $A$ has at least one eigenvalue with positive real part.
Given a real matrix $A_{n\times n}$ with at least one eigenvalue having positive real part, how can I find $x\in\mathbb{R}^n$ such that $\langle\, Ax, x\rangle > 0$?

I tried to use the real part of the eigenvector, but it didn't quite work.
With $x=u+vi$ eigenvector of the eigenvalue $\lambda = a+bi$, $a>0$, we have $Au = au-bv$. Then,

$$\langle\, Au, u\rangle = a|u|^2-b \langle\, u, v\rangle.$$

But I don't know if the RHS is always greater than 0.

答:
We have $Ax = \lambda x$. This means that
$$
Au + i Av = A(u + vi) = (a + bi)(u + vi) = (au - bv) + (bu + av)i.
$$
Now, note that
$$
\langle Au,u \rangle + \langle Av,v \rangle = \langle au - bv,u\rangle + \langle bu + av,v \rangle\\
= a \|u\|^2 + a \|v\|^2 - b \langle v,u \rangle + b \langle u,v \rangle\\
= a (\|u\|^2 + \|v\|^2) > 0.
$$
It follows that we must have $\langle Au,u \rangle > 0$ or $\langle Av,v \rangle > 0$.


问:
Why is $f(x,y,z)$ equal to $1$ when trying to find the length of a curve
Can someone explain to me why the integrand of line integral is $f(x,y,z)=1$ when trying to find the length of the curve? Click the link below to access the image/ example for reference purposes

Flight of an eagle An eagle soars on the ascending spiral path
$$C\colon \mathbf r(t) = \langle x(t),y(t),z(t)\rangle = \left\langle 2400cos\frac t2, 2400\sin\frac t2, 500t\right\rangle,$$
where $x$, $y$ and $z$ are measured in feet and $t$ is measured in minutes. How far does the eagle travel over  $0\le t\le 10$?

Because we are finding the length of a curve, the integrand in this integral is $f(x,y,z)=1$.

答: This is for essentially the same reason that for $a < b$ we can find the length of interval $[a,b]$ via $$\int_{a}^{b} \,dx = b - a.$$  Let's consider the case of a curve $C$.  If we want to compute it's length $L$, one way we can do that is to think of approximating the length of the curve by breaking the curve into many small line segments.  We can first approximate the length in this manner by sampling $n+1$ points $(x_{i},y_{i},z_{i})$ along the curve, and finding the distance between two consecutive points $(x_{i},y_{i},z_{i})$ and $(x_{i+1},y_{i+1},z_{i+1})$ via $$\Delta s_{i} = \sqrt{(x_{i+1} - x_{i})^{2} + (y_{i+1} - y_{i})^{2} + (y_{i+1} - y_{i})^{2}}.$$  Then, we can say that $$L \approx \sum_{i = 1}^{n}\Delta s_{i}.$$  If we take the limit as the size of the greatest distance between consecutive points goes to zero we get the actual length: $$L = \lim_{\|\Delta s\| \to 0}\sum_{i = 1}^{n}\Delta s_{i}.$$  Note, however, that we can rewrite this as $$L = \lim_{\|\Delta s\| \to 0}\sum_{i = 1}^{n}\Delta s_{i} = \lim_{\|\Delta s\| \to 0}\sum_{i = 1}^{n}1\cdot \Delta s_{i} = \int_{C}\,ds.$$

OwnerDisplayName: lilkarti


问:
Inductive proof $f(n) = n$ with function composition.
I need help with a inductive proof. Currently got no idea how to start with this assignment.

Assignment:

Let $f : \mathbb{N}_0 \to \mathbb{N}_0$ be a function, so that the inequality:

$$f \circ f(n) < f(n+1)$$ exists. $\forall n \in \mathbb{N}_0$

Show that $f(n) = n$. $\forall n \in \mathbb{N}_0$

I am currently struggling to understand where to start with this one. Thanks for any help in advance.

答:
Being an inductive proof, we start with the smallest number we can find.
In $\mathbb{N}_0$, the smallest number is $0$.
Hence the  first thing we want to show is that $f(0)=0$

You can start by thinking what would happen if $f(0) \neq 0$, considering that $f \circ f (n) < f(n+1) $ for all $n$.

(I will edit if I reach to go further, but the start of the proof should be in this direction)


问:
How do you call a partition of a totally ordered set that....
A partition of a set S is a collection P={B1,B2,...,Bk}  of subset of S such that for each i, j, ∈ {1,2,...,k}:

Bi ≠ ∅

Bi ⋂ Bj ≠ ∅

B1 ⋃ B2 ⋃...⋃ Bk = S

now if S is a totally ordered set and P is also equipped with a partial order ≼ how do you call a partition that meets the following additional condition?

for x ∈ Bi, y x ∈ Bj, if x<y then i≤j (or B i ≼ Bj)

For example if S= {1, 2, 3, 4} we can have the following partitions of S into two subsets:

1/234, 12/34 and 123/4 (but not 13/24 nor 14/23)

I've come accross the following definitions: monotone partition, consecutive partition...but I don't seem to find a commonly accepted terminology nor any body of litterature that would study these ...  (I know it is subset of partitions and non-crossing partitions)

Any directions, hints for me?

Many thanks

P.S: to give a bit of context, I wanted to study this type of partitions in details (their lattice structure, total number given binomials as suggested by Donald S. etc...)

答:
$13/2$ avoiding partitions. This means that we cannot have $a_1 < a_2 < a_3$ & $ a_1,a_3 \in B_i, a_2 \in B_j$ ($i \neq j$)

With this restriction we get binomials and without this we get stirling numbers of the second kind.


问:
Elliptic polar coordinate
Does anybody know why the elliptic polar coordinate of the ellipse is :
$$ \begin{cases}
      x=a \ r\cos\theta \\
      \\
      y=b \ r\sin\theta
   \end{cases}
$$

答:
The general equation of an ellipse centerd at the origin and focus on the same Axis is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$
If you make the substitutio $X=x/a$, $Y=y/b$, you'll get $$X^2+Y^2=1$$ that is, the unit circumpherence. Using standard polar coordinates here, $$X=r\cos\theta,\ Y=r\sin\theta.$$ Finally, undo the substitution and you'll get your Elliptic Polar Coordinates.

Title: Elliptic polar coordinate


答:
When $n=1$ we have $$\sum_{ d|n} \mu(d)=\mu(1)=1$$.

When $n>1$, then write $n=p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$, we get
$$
\begin{align*}
   \sum_{ d|n} \mu(d)&=\sum_{ 0 \leq b_i \leq a_i} \mu(p_1^{b_1}p_2^{b_2} \dots p_k^{b_k})\\
 &=\mu(1)+\mu(p_1)+\dots \mu(p_n)+\mu(p_1p_2)\dots \mu(p_{n-1}p_n)+\dots\mu(p_1 \dots p_n)\\
 &=1(-1)^0+{n \choose 1}(-1)^1+{n \choose 2}(-1)^2+{n \choose 3}(-1)^3+ \dots {n \choose n}(-1)^n \\
 &=(1-1)^n \\
 &=0.
 \end{align*}$$
Hence $$\sum_{ d|n} \mu(d)=\begin{cases}
1,&\text{if }n=1\\
0,&\text{if }n>0\
\end{cases}=\left\lfloor\frac1n\right\rfloor$$

This proves your result. Note that this proof doesn't assume the multiplicative property.

Title: Prove $\sum_{d\mid n}\mu(d)=[\frac{1}{n}]$

Body:
I'm reading a direct proof for mobius inversion theorem but I don’t know how the writer make the assumption that $$\sum_{d\mid n}\mu(d)=\left[\frac{1}{n}\right]$$

Can someone please explain and prove this?


问:
Vector Limit Proof
If $x_n,y_n,x,y$ are vectors in $\mathbb{R}^m$ and $x_n\rightarrow x$, $y_n\rightarrow y$ prove that $\langle x_n,y_n\rangle\rightarrow\langle x,y\rangle$

So far I have tried to use Cauchy-Schwartz inequality to get $\lvert\langle x_n,y_n\rangle\rvert \leq\lvert\lvert x_n-x\rvert\rvert \cdot   \lvert\lvert y_n-y\rvert\rvert \rightarrow 0$. From here I am stuck. Can I simply say by definition of limit we then have that $\langle x_n,y_n\rangle\rightarrow\langle x,y\rangle$?

答:
I think you meant $|\langle x_n-x,y_n-y\rangle|\leq ||x_n-x||\cdot ||y_n-y||$. From there, if you want an $N$-$\varepsilon$ proof, let $\varepsilon>0$ and take $N=\max\{N_1,N_2\}$, where $N_1,N_2$ are the numbers corresponding to the convergence of $x_n$ to $x$ and $y_n$ to $y$ respectively, for $\sqrt{\varepsilon}$. Hence, if $n\geq N$, then $$|\langle x_n-x,y_n-y\rangle|\leq ||x_n-x||\cdot ||y_n-y||<\sqrt{\varepsilon}\cdot\sqrt{\varepsilon}=\varepsilon.$$

If you didn't get there (so I was wrong interpreting you result), you can do the following $$|\langle x_n,y_n\rangle-\langle x,y\rangle|=|\langle x_n,y_n\rangle-\langle x_n,y\rangle+\langle x_n,y\rangle-\langle x,y\rangle|=|\langle x_n,y_n-y\rangle+\langle x_n-x,y\rangle|.$$ From there, use triangle inequality, C-S for each inner product, the fact that $||x_n||$ is bounded (it converges) and an argument similar to the one above with $N$-$\varepsilon$.

Title: Vector Limit Proof


问:
How to find the variance of the estimator, $\frac{1}{\bar{X}}$, for $X_i \sim \operatorname{Expo}(\lambda)$?
So I used the method of moments to find an estimator for $\lambda$: $\frac{1}{\bar{X}}$.

However, I'm having trouble deriving the variance of it:

$$\operatorname{Var}\left(\frac{1}{\bar{X}}\right) = \operatorname{Var} \left(\frac{n}{\sum_{i=1}^n X_i}\right)$$
$$= n^2 \operatorname{Var}\left(\frac{1}{\sum_{i=1}^n X_i}\right)$$

I'm not sure how to proceed from here given the denominator is a sum of $X_i$'s.

答:
The distribution of $\sum_{i=1}^n X_i$ is a gamma distribution:
$$
\frac 1 {\Gamma(n)} (\lambda x)^{n-1} e^{-\lambda x} (\lambda \, dx) \quad \text{for } x>0.
$$
Consequently
\begin{align}
\operatorname E\left( \frac 1 {\sum_{i=1}^n X_i} \right) & = \int_0^\infty \frac 1 x\cdot \frac 1 {\Gamma(n)} (\lambda x)^{n-1} e^{-\lambda x} (\lambda \, dx) \\[8pt]
& = \lambda \int_0^\infty \frac 1 {\lambda x} \cdot \frac 1 {\Gamma(n)} (\lambda x)^{n-1} e^{-\lambda x} (\lambda \, dx) \\[8pt]
& = \lambda \int_0^\infty \frac 1 u \cdot \frac 1 {\Gamma(n)} u^{n-1} e^{-u} \, du \\[8pt]
& = \frac \lambda {\Gamma(n)} \int_0^\infty u^{n-2} e^{-u} \, du \\[8pt]
& = \frac{\lambda \Gamma(n-1)}{\Gamma(n)} = \frac \lambda {n-1}.
\end{align}
Therefore
$$
\operatorname E\left( \frac 1 {\overline X} \right) = \frac{n\lambda}{n-1}.
$$
Similarly
\begin{align}
\operatorname E\left( \frac 1 {\left( \sum_{i=1}^n X_i \right)^2} \right) & = \frac {\lambda^2} {\Gamma(n)} \int_0^\infty \frac 1 {u^2} \cdot u^{n-1} e^{-u} \, du \\[8pt]
& = \frac{\lambda^2\Gamma(n-2)}{\Gamma(n)} = \frac{\lambda^2}{(n-1)(n-2)}
\end{align}
so
$$
\operatorname E\left( \frac 1 {\overline X^2} \right) = \frac{n^2\lambda^2}{(n-1)(n-2)}.
$$
Finally
\begin{align}
\operatorname{var}\left(\frac 1 {\overline X}\right) & = \operatorname E\left(\frac 1 {\overline X^2} \right) - \left( \operatorname E\left(\frac 1 {\overline X} \right) \right)^2 \\[8pt]
& = \lambda^2 \left( \frac{n^2}{(n-1)(n-2)} - \frac{n^2}{(n-1)^2} \right) \\[8pt]
& = \frac{n^2\lambda^2}{(n-1)^2(n-2)}.
\end{align}


问:
Function with given properties in range $[0,1]$
I'm looking for a function with the following properties:

$$f : [0,1] \to [0,1]$$
$$f(0) = 1$$
$$f(1) = 0$$
$$f'(0) = -\infty$$
$$f'(1) = 0$$

It should look like the $\frac{1}{x}$ but with the given properties. Thanks for your help.

答: Consider $$f(x) =1- \sqrt{x(2-x)}$$

Body: Consider $$f(x) =1- \sqrt{x(2-x)}$$


问:
Best program for creating educational math animations?
I'm looking for recommendations on what program to use for creating mathematical animations. These animations will be used in creating educational videos for high school math -- Trigonometry first, but others will come later. I am particularly focusing on teaching the material from a conceptual basis, and to that end, I want to create high-quality animations that can help students understand the ideas they're working with.

What program would be particularly well suited for making math animations aimed at teaching high school level material? I'm willing to learn a scripting/programming language (or at least enough of one) if that's the best choice. Upfront time investment is okay: I'll likely be making quite a few of these animations over the next few years.

Requirements

Plotting (2d): Needs to be able to plot single-variable functions and then animate through manipulation of one or more parameters.

Export Animations: It must be possible to export the animation to some form of movie file format or animated GIF.

Not Required, but Nice

Graphics Primitives: The program should have various primitives (line segments, circles, etc.) that are also capable of animation. This can be worked around by doing enough plotting, but it's much more elegant to have access to primitives.

Runs on Mac: I do most of my work on my Mac laptop, so that's the preferred OS.

Free (preferably open-source): Not much budget at the moment. Additionally, I plan to open-source all of the materials I use in creating the videos, so that's less useful if the work is done on a paid program.

Here's my current list of possibilities.

Free Options

GeoGebra

GIFsmos (Takes Desmos animations as inputs and spits out GIFs)

Paid Options

Mathematica

MATLAB

Maple

Non-Math-Focused Option

Adobe Flash (Can't [I think?] handle plotting, but might be best choice for working with simple graphics.)

If you think one of the above candidates is a great choice, please let me know why you think so in your answer. If you think that there's another option better than the above, please tell me about it.

Summary: What is the best program for making math animations, knowing that the intent is to use them in videos teaching high school math?

答:
There's this open-source tool that's gotten pretty popular recently. It's called Manim - the math animation engine, originally developed by Grant Sanderson (who is the man behind the popular math YouTube channel 3b1b) while he was an undergrad.

His videos feature extremely pleasing animations which served as an ultimate motivation for me to study  quantum mechanics. He made his tool open-source and from then the manim engine has gained huge popularity among many math YouTubers. The fact that it's Python-based makes it even more appealing and easy-to-use.

Though the official documentation is a little lagging behind (I feel) there are many forks on GitHub, some of which have better documentation like this one and this one. There are fairly good number of tutorials that teach animation in manim. It's evident that there's a growing community adopting this fairly new tool.


问:
Finding Vectors in cartesian form
I am stuck on this question could you please help me.

Find,in Cartesian form, the equations of the straight line through the point with position vector (-1,2,-3) parallel to the direction given by (2,1,-2)

Any help is appreciated.
Thanks

OwnerDisplayName: Matt

Title: Finding Vectors in cartesian form

Body: $$x=-1+2t\\y=2+t\\z=-3-2t$$ or simply $$(x,y,z)=(-1,2,-3)+t(2,1,-2)$$


答:
A line $L$ is uniquely specified by giving any point $P_0=(x_0,y_0,z_0)$ on $L$ and any vector $u=(a,b,c)$ parallel to $L$ as follows:
$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{y-y_0}{c}$$</p>

Title: Finding Vectors in cartesian form

Body:
I am stuck on this question could you please help me.

Find,in Cartesian form, the equations of the straight line through the point with position vector (-1,2,-3) parallel to the direction given by (2,1,-2)

Any help is appreciated.
Thanks


问:
Holomorphic function taking all values near a singularity
I recall a theorem in complex analysis about a holomorphic function having to take all different possible values in the neighbourhood of a singularity. Can someone point me to the name of this theorem please?

Thanks in advance.

答:
You are probably thinking of (Great) Picard's Theorem:

Great Picard's Theorem: If an analytic function $f$ has an essential
singularity at a point $w$, then on any punctured neighborhood of $w$, $f(z)$ takes on all possible complex values, with at most a single
exception, infinitely often.


问:
If $k[X_1,.., X_n]/I$ is a field non algebraic extension of $k$, how to show there exists an algebraically independent subset of $\{X_1+I,..,X_n+I\}$?
Let $A = k[X_1,.., X_n]/I$ where $k$ is a field and $I$ is some ideal.
Suppose $A$ is a field and $A/k$ is not an algebraic extension. Let $x_i = X_i + I$. I want to show that we can find (relabeling the variables if necessary) there exist $\{ x_1,.., x_r \}$ that is algebraically independent over $k$ and each $x_{r+1},.., x_n$ is algebraic over $k(x_1, .., x_n)$?

答:
"Zariski's lemma" states that, for any field $k$, if a field is finitely generated as a $k$-algebra then it is a finite field extension of $k$. Proof of this is below.

However, this means that the circumstances of your problem cannot arise; $A$ is finitely generated as a $k$-algebra (by the $x_i$), and hence – since it is a field – must be a finite and thus algebraic extension of $k$.

Lemma 1: If $E$ and $F=E(x)$ are fields, and $F$ is finitely generated as an $E$-algebra, then $F$ is a finite field extension of $E$.

Proof: Say $F=E[f_1,\ldots,f_n]$, and suppose for contradiction that $F$ is not finite over $E$. Then $x$ must be transcendental over $E$, and so $F$ is the field of rational functions in $x$ over $E$. This means that each $f_i=p_i/q_i$ for some $p_i,q_i\in E[x]$. Consider the element $r:=\frac{1}{x\prod_{i=1}^nq_i+1}\in F$; there must be some polynomial $g\in E[t_1,\ldots,t_n]$ such that $r=g(f_1,\ldots,f_n)$. We can clear denominators in $g(f_1,\ldots,f_n)$ by multiplying through with a suitable power of $\prod_{i=1}^nq_i$, giving $s:=(\prod_{i=1}^nq_i)^kg(f_1,\ldots,f_n)\in E[x]$ for some $k\in\mathbb{N}$, and hence $\frac{1}{x\prod_{i=1}^nq_i+1}=\frac{1}{(\prod_{i=1}^nq_i)^k}s$. But this means $(\prod_{i=1}^nq_i)^k=(x\prod_{i=1}^nq_i+1)s$, and this is a contradiction, since $x\prod_{i=1}^nq_i+1$ and $\prod_{i=1}^nq_i$ (and hence $x\prod_{i=1}^nq_i+1$ and $(\prod_{i=1}^nq_i)^k$) are coprime in $E[x]$.

Lemma 2: Let $A\subseteq B\subseteq C$ rings with $A$ Noetherian. If $C$ is finitely generated as an $A$-algebra and $C$ is finitely generated as a $B$-module, then $B$ is finitely generated as an $A$-algebra.

Proof: See eg here.

Theorem (Zariski's lemma): Suppose $F:E$ is a field extension and that $F$ is finitely generated as an $E$-algebra. Then $F$ is finite over $E$.

Proof: Say $F=E[x_1,\ldots,x_n]$; we prove by induction on $n$. The case $n=0$ is clear. Now suppose the result holds for $n-1$, and let $E'=E(x_n)\subseteq F$. By the tower law, $[F:E]=[F:E'][E':E]$, and so it suffices to show (i) that $F$ is a finite extension of $E'$ and (ii) that $E'$ is a finite extension of $E$. For (i), note that $F=E'[x_1,\ldots,x_{n-1}]$, and so – since $E'$ is a field – by the induction hypothesis we have that $F$ is a finite field extension of $E'$. For (ii), note that (i) allows us to apply lemma 2 and show that $E'$ is finitely generated as an $E$-algebra. By lemma 1, this means that $E'=E(x_n)$ is in fact a finite extension of $E$, so (ii) holds and we are done.


问:
Totally disconnected space is 0-dimensional
Let $X\neq \emptyset$ be a compact, perfect metrizable space. (This is a segment of an exercise, I do not know if every property will be involved in the proof)

I want to proof that $X$ is totally disconnected implies that $X$ is $0$-dimensional.

For that I have to give a basis $\mathcal{B}$ where every element is closed and open.

I tried to study the sets in $X$ which are closed and open, to construct a basis. But until now, I was not able to give such a set, besides the trivial $\emptyset$ and $X$.

$X$ is totally disconnected. So $\{x\}$ is closed. But $\{x\}$ is not open as $X$ is perfect.

I know that the clopen sets can not be finite, as this would contradict that every singleton is not open.

I do not know if this "strategy" will eventually lead to a proof.

Do you have a hint?
Thanks in advance.

答:
You need only that $X$ is totally disconnected, locally compact, and Hausdorff. Let $x\in X$, and let $U$ be an open nbhd of $x$; we want to show that $x$ has a clopen nbhd contained in $U$. $X$ is locally compact, so $x$ has an open nbhd $V$ such that $\operatorname{cl}V\subseteq U$ and $\operatorname{cl}V$ is compact. If $V$ is clopen, we’re done.

Otherwise, the total disconnectedness of $X$ implies that for each $y\in\operatorname{bdry}V$ there is set $H_y$ such that $x\in H_y$, $y\notin H_y$, and $H_y$ is clopen in $\operatorname{cl}V$. Then each $H_y$ is closed in $X$, so $\{X\setminus H_y:y\in\operatorname{bdry}V\}$ is an open cover of $\operatorname{bdry}V$. And $\operatorname{bdry}V$ is compact, so there is a finite $F\subseteq\operatorname{bdry}V$ such that

$$\operatorname{bdry}V\subseteq\bigcup_{y\in F}(X\setminus H_y)\,.$$

Let $H=\bigcap_{y\in F}H_y$; clearly $x\in H$, and $H$ is clopen in $\operatorname{cl}V$. Moreover, $H\cap\operatorname{bdry}V=\varnothing$, so $H\subseteq V\subseteq U$. Thus, $H$ is closed in the closed set $\operatorname{cl}V$ and open in the open set $V$, so $H$ is clopen in $X$ and in particular is a clopen nbhd of $x$ contained in $U$.


答:
Supposing you want $f(0) = 1, f(1) = 0, f'(0) = +\infty, f'(1) = 0$ and $f:[0,1]\to \mathbb R$

$f(x)= \sqrt x$ will give you $f'(0)=\infty$ but $f(0)=0$

$f(x)= 1+\sqrt x$ will give you $f'(0)=\infty$ and $f(0)=1$ but $f(1)=2$

$f(x)=(x-1)^2$ will give you $f(0)=1$ and $f(1)=0$ and $f'(1)=0$ but $f'(0)=-2$

$f(x)= (1+\sqrt x)(x-1)^2$ will give you $f(0)=1$ and $f(1)=0$ and $f'(1)=0$ and $f'(0)=\infty$ and looks like

As Joshua Wang has commented, this cannot avoid exceeding $1$ near $x=0$ as you start at $f(0)=1$ with a positive derivative

Title: Function with given properties in range $[0,1]$

Body:
I'm looking for a function with the following properties:

$$f : [0,1] \to [0,1]$$
$$f(0) = 1$$
$$f(1) = 0$$
$$f'(0) = -\infty$$
$$f'(1) = 0$$

It should look like the $\frac{1}{x}$ but with the given properties. Thanks for your help.


问:
How do I show the sequence converge to zero in measure?
Let $X=[0,1]$ with a Lebesgue measure. Let $f_n(x)=x^n$ for $x \in [0,1]$. Prove that $f_{n}$ converge to zero in measure.

My idea: Clearly, $f(x)= 0 $ for $0 \leq x <1$, and $1$ for $x=1$, then by definition

$\mu(\{x: |x^{n}|\geq 1\}) \rightarrow$ $0$ as $n \rightarrow \infty$.

Is this solution correct?

答:
In my opinion the simplest solution is the following:

$f_n$ converges a.e. to $f = 0$, moreover X has finite Lebesgue measure, so a.e. convergence implies the one in measure.


问:
Verify $||T||_{L(X,Y)}$ norm is given by $||T||_{L(X,Y)} = \sup_{x \in X, ||x||_X = 1} ||T(x)||_Y$.
Can somebody help me verify the problem

If $L(X,Y)$ is the set of all bounded and linear operator, then for every $T$ which is linear and bounded, $||T||_{L(X,Y)}$ norm is given by $||T||_{L(X,Y)} = \sup_{x \in X - \{0\}, ||x||_X \leq 1} \frac{||T(x)||_Y}{||x||_X} = \sup_{x \in X, ||x||_X = 1} ||T(x)||_Y$.

I don't know where to start with that.

答:
If $x \in X, x \ne 0$, then $u := \frac1{\|x\|_X}x \in X$ and $\|u\|_X = \frac1{\|x\|_X}\|x\|_X = 1$. And
$$\|T(u)\|_Y = \left\|T\left(\frac x{\|x\|_X}\right)\right\|_Y = \frac{\left\|T\left(x\right)\right\|_Y}{\|x\|_X}$$ since $T$ is linear and norms commute with positive multipliers.

So every value in $$S_1 := \left\{\frac{\|T(x)\|_Y}{\|x\|_X}\ \middle|\ x \in X, \|x\|_X \le 1, x\ne 0\right\}$$ is also in $$S_2 := \left\{\|T(u)\|_Y\mid u \in X, \|u\|_X = 1\right\}$$ I.e., $S_1 \subset S_2$. And manifestly $S_2 \subset S_1$, so in fact $S_1 = S_2$. And being the same set, they have the same supremum.

Note that this same argument also works on $$S_3 = \left\{\frac{\|T(x)\|_Y}{\|x\|_X}\ \middle|\ x \in X, x\ne 0\right\}$$ So the $\|x\|_X \le 1$ part of the first supremum was pointless.

As for why $\|T\|_{L(X, Y)}$ is the same value, that is because this is the definition of $\|T\|_{L(X, Y)}$.


问:
Solve a cubic polynomial given that one root is four times a second root?
How exactly would you solve the equation:

"Solve the equation $10x^3+23x^2+5x−2=0$ given that one root is four times a second root."

How would you go about solving this? Any help would be greatly appreciated.

答:
We know that a cubic polynomial has three zeroes over the complex numbers (including the case where we have double or triple zeroes). So, we can write any cubic polynomial as $a (x-z_1)(x-z_2)(x-z_3)$, where $z_1, z_2, z_3$ are its zeroes.

In this case, we can assume that $z_2 = 4z_1$ by the side condition. We get $$10x^3+23x^2+5x-2 = a (x-z_1)(x-4z_1)(x-z_3).$$
By equating coefficients, we find:
$$\begin{align}
 a &= 10\\
 -a(5z_1 + z_3) &= 23\\
 a(4z_1^2+5z_1z_3) &= 5\\
 a(-4z_1^2z_3) &= -2
 \end{align}$$
Solving this system should give you a quadratic equation for $z_1$, from which the result follows.


问:
Finding the number of real zeros for the equation $11^x + 13^x + 17^x-19^x=0 $
This is a question from a mathematics contest where we are asked about the zeros of the equation $ 11^x + 13^x + 17^x -19^x=0.$
The options are:

1: No real root

2:Only one real root

3: Two real roots

4: More than two real roots.

I want to know a general method to solve these type of questions. I know that if we were given some other function say polynomial, or trigonometric function, etc then we can look into derivative of the function and then by analysing the graph or the maxima and minima we can say about the zeros.

However in this case I am unable to use such things. Any help is highly appreciated.

答:
We can rewrite the equation: $$11^x+13^x+17^x-19^x=0\implies 19^x\left[\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1\right]=0$$

and note that $$f(x)=\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1$$ is a continuous, strictly decreasing function.
With

$\lim_{x\to-\infty}f(x)=\infty$

$\lim_{x\to\infty}f(x)=-1$

$19^x>0$ for all $x\in\mathbb{R}$

we can say that the equation has exactly one real root.


答: For any $\epsilon\in(0,1)$, the set $\{x\in[0,1];x^n\geqslant\epsilon\}$ is the interval $[\epsilon^{1/n},1]$ and is (Lebesgue) measure is $1-\epsilon^{1/n}$ which tends to $0$ as $n\to\infty$.

Title: How do I show the sequence converge to zero in measure?

Body:
Let $X=[0,1]$ with a Lebesgue measure. Let $f_n(x)=x^n$ for $x \in [0,1]$. Prove that $f_{n}$ converge to zero in measure.

My idea: Clearly, $f(x)= 0 $ for $0 \leq x <1$, and $1$ for $x=1$, then by definition

$\mu(\{x: |x^{n}|\geq 1\}) \rightarrow$ $0$ as $n \rightarrow \infty$.

Is this solution correct?


答:
THe derivative method is, probably the best one. But here, you may begin with some algebraic trick

REcall that $$11^x + 13^x + 17^x -19^x=0 \iff  \left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1=0.$$
Define now $F(x):=\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1$ wich is continuous and differentiable on $\Bbb R$.

$F(0)=2>0$ and $\lim_{x\to+\infty}F(x)=-1<0$ (so, there is $M\in\Bbb R$ such that $F(M)<0$). An application of Bolzano's Theorem leads to the existence of, at least, one real (positive, in fact) solution.

But, $F'(x)=\left(\frac{11}{19}\right)^x\ln(11/19)+\left(\frac{13}{19}\right)^x\ln(13/19)+\left(\frac{17}{19}\right)^x\ln(17/19)<0$ for all $x\in\Bbb R$, so the solution is unique.

Title: Finding the number of real zeros for the equation $11^x + 13^x + 17^x-19^x=0 $

Body:
This is a question from a mathematics contest where we are asked about the zeros of the equation $ 11^x + 13^x + 17^x -19^x=0.$
The options are:

1: No real root

2:Only one real root

3: Two real roots

4: More than two real roots.

I want to know a general method to solve these type of questions. I know that if we were given some other function say polynomial, or trigonometric function, etc then we can look into derivative of the function and then by analysing the graph or the maxima and minima we can say about the zeros.

However in this case I am unable to use such things. Any help is highly appreciated.


答:
Homogenize the inequality by writing (replace "1"s):
$$
(2a+b)(2b+c)(2c+a)+(a+b+c+a+2b)(a+b+c+b+2c)(a+b+c+c+2a) \leq 9(a+b+c)^3
$$
Evaluating the terms gives the following equivalent inequality to be shown
$$
a^3 + b^3 + c^3 - 3 abc  \ge 0
$$
Now since the variables are not restricted to nonnegative ones, this must be proved "by hand" (not by AM-GM). Observe the identity, which can be shown by multiplying everything out:
$$
a^3 + b^3 + c^3 - 3 abc  = \frac12 (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)
$$
Since $a+b+c=1$, this gives the required
$$
a^3 + b^3 + c^3 - 3 abc  = \frac12 ((a-b)^2 + (b-c)^2 + (c-a)^2) \ge 0
$$
and equality only holds for $a=b=c=\frac13$.

$\qquad \Box$

Title: If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that $(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$


问:
The commutator subgroup of a quotient in terms of the commutator subgroup and the kernel
Let $H$ be the normal subgroup of $G$.

Is it true that $(H[G,G])/H$ is isomorphic to $[G/H,G/H]$?

If so, I want to make a surjective homomorphism $\phi\colon H[G,G]\to [G/H,G/H]$ with Kernel $H$ to prove it.

But, if I define $\phi(h[g_1,g_2])=[g_1/H,g_2/H]$ then I have some trouble.

答:
An easy argument certainly follows user8268's comment.

A few things should be noticed (as they might not seem totally obvious). At first since $H$ is normal in $G$,
$$ [g_1H,g_2H] = [g_1,g_2]H \in G/H.$$
Now restricting the canonical epimorphism $G\to G/H$ to $H[G,G]$, we may call it $\phi$, gives us a mapping, that satisfies
$$ \phi(h[g_1,g_2])= [g_1,g_2]H = [g_1H,g_2H] \in [G/H,G/H]$$
for $h\in H$ and $g_1,g_2\in G$. Since $H[G,G]$ is generated by the elements $h[g_1,g_2]$, $\phi(H[G,G])$ is certainly generated by the elements $[g_1H,g_2H]$, that generate $ [G/H,G/H]$. Hence $\phi$ is surjective and its kernel is still $H$ as it is included in $H[G,G]$.


问:
Prove $\sqrt x$ is uniformly continuous on $[1,+\infty[$
I have done the following:

For any given $\;\varepsilon>0\;,\;$ let $\;\delta=2\varepsilon>0\;,$

$\forall\;x,y\in[1,+\infty[\;$ with $\;|x−y|<\delta\;,$

since $\;\sqrt{x}+\sqrt{y}\ge2\;,\;$ we get that

$\left|\sqrt{x}-\sqrt{y}\right|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}\le\dfrac{|x-y|}{2}<\dfrac{\delta}{2}=\varepsilon\;.$

Hence, $\;\sqrt{x}\;$ is uniformly continuous on  $\;[1,+\infty[\;.$

Am I missing something? Is this proof correct?

答:
The function $\sqrt{x}$ is uniformly continuous on $[0,\infty)$. It is uniformly continuous on $[0,1]$ because it is continuous, and because $[0,1]$ is compact. To prove uniform continuity on $[1,\infty)$, you can do this directly, as you suggest
$$
      \sqrt{x}-\sqrt{y} = \frac{x-y}{\sqrt{x}+\sqrt{y}}, \\
         |\sqrt{x}-\sqrt{y}| \le \frac{1}{2}|x-y|.
$$
Combining the results on $[0,1]$ and $[1,\infty)$ gives you what you want.


问:
How to identify when representations of solutions to trigonometric equations match (eg, $(2n\pm1)\frac\pi2$ and $(2n+1)\frac\pi2$)
I'm not sure if "identifying" questions are allowed or not, if not a link or a point in the right direction would be great!

I am not able to tell if the two trigonometric equations are the same or not just by looking at them, unless I substitute values, which is not viable in competitive examinations.

For example,
$$(2n\pm1)\frac{\pi}{2}\quad\text{and}\quad (2n+1)\frac{\pi}{2}$$

Let's say I get the first one, and the second one is one of four options. Is it possible to see them both and tell they are the same?

Another example:

$$(4n\pm1)\frac{\pi}{4}\quad\text{and}\quad(2n+1)\frac{\pi}{4}$$

答:
Plugging points is an unreliable method to show that two functions are equal.  It is too easy to choose points that coincidentally align, with the functions not being the same everywhere.

What you have to do is use identities to show that these functions are the same.

The basic ones are:

The definitions:

$\tan x = \frac {\sin x}{\cos x}\\
\sec x = \frac {1}{\cos x}\\
\csc x = \frac {1}{\sec x}$

The Pythagorean Theorem:

$\cos^2 x + \sin^2 x = 1$

The complementary angles:

$\sin x = \cos (\frac {\pi}{2} - x)\\
\tan x = \cot (\frac {\pi}{2} - x)\\
\sec x = \csc (\frac {\pi}{2} - x)$

The even / odd symmetry:

$\sin (-x) = -\sin x\\
\cos (-x) = \cos x$

Angle addition:

$\sin (A+B) = \sin A\cos B + \cos A\sin B\\
\cos (A+B) = \cos A\cos B - \sin A\sin B$

And rest follow from there.

e.g.

$\cos^2 x = \frac 12 \cos 2x + \frac 12\\
\frac 12 \cos (x+x) + \frac 12 \\
\frac 12 (\cos x\cos x - \sin x\sin x) + \frac 12 \\
\frac 12 \cos^2 x + \frac 12 (1- \sin^2 x) \\
\frac 12 \cos^2 x + \frac 12 (\cos^2 x) \\
\cos^2 x$


答:
Let $$f(x)=(11/19)^x+(13/19)^x+(17/19)^x-1$$
$$\implies f'(x)=(11/19)^x \ln (11/19)+ (13/19)^x \ln (13/19)+(17/19)^x \ln (17/19)<0$$ Hence $f(x)$ is monotonically decreasing so $f(x)=0$ will have atmost one real root and $f(0)=2 >0$ and $f(\infty)=-1<0.$ So by IVT, $f(x)=0$ has one real root.
Combining both statements $f(x)=0$, has exactly one real root.


答:
A faster  way to get the same result:

$Y=\Sigma_i X_i\sim Gamma(n;\lambda)$

Thus

$\frac{1}{Y}\sim\text{Inverse Gamma}$ and thus $E\Big[\frac{1}{Y}\Big]=\frac{\lambda}{n-1}$

Concluding,

$$E\Bigg[\frac{1}{\overline{X}}_n\Bigg]=\frac{n}{n-1}\lambda$$

Similar reasoning to get  immediately variance, without any calculation

Title: How to find the variance of the estimator, $\frac{1}{\bar{X}}$, for $X_i \sim \operatorname{Expo}(\lambda)$?

Body:
So I used the method of moments to find an estimator for $\lambda$: $\frac{1}{\bar{X}}$.

However, I'm having trouble deriving the variance of it:

$$\operatorname{Var}\left(\frac{1}{\bar{X}}\right) = \operatorname{Var} \left(\frac{n}{\sum_{i=1}^n X_i}\right)$$
$$= n^2 \operatorname{Var}\left(\frac{1}{\sum_{i=1}^n X_i}\right)$$

I'm not sure how to proceed from here given the denominator is a sum of $X_i$'s.


问:
Evaluate $I = \iint_{R}\frac{x}{y^{4}}\,dx\,dy$ over a given region

Evaluate the double integral $$I = \iint_{R}\frac{x}{y^{4}}\,dx\,dy$$ over the region $R = \{(x,y)\mid 0\le x \le3, 3\le y \le5\}$.

Hi I was wondering if anyone could help me on this question which involves Taylor's Expansion.

This is what I have done so far:

From using the given boundaries I have created the integral:

$$\int_3^5\int_0^3 \frac{x}{y^4}\, dx\,dy$$

Then I worked out the first integral:

$$\int_0^3\frac{x}{y^4}\, dx$$

which gives me $\frac{9}{2y^4}$ and then I put this into the next integral that is with respect to $y$:

$$\int_3^5\frac{9}{2y^4} \,dy$$

Which gives me $3.415703704\times 10^{-3}$, which is evidently wrong as the correct answer is: $49/1125$

Can someone please help me on where I am going wrong? Thanks in advance!

Body:
Note that:

$$\int\int_{R}\frac{x}{y^{4}}\ dx\ dy = \int_{3}^{5}\int_{0}^{3}\frac{x}{y^{4}}\ dx\ dy = \int_{3}^{5}\bigg(\frac{x^{2}}{2y^{4}}\bigg)\bigg\vert_{0}^{3}\ dy = \int_{3}^{5}\frac{9}{2y^{4}}\ dy$$

$$= -\frac{3}{2y^{3}}\bigg\vert_{3}^{5} = -\frac{3}{250} + \frac{1}{18} = \boxed{\frac{49}{1125}}$$


问:
Differentiable scalar fields question.
Let $f,g:\mathbb{R}^3\to\mathbb{R}$ be twice continuously differentiable scalar fields.
Which of the following statements is false?

A: If $S$ is any simple piecewise-smooth surface with unit normal vector $\mathbf{n}$, then $\iint_S (\nabla\times\nabla f)\cdot\mathbf{n} \,dS=0$

B: If $S$ is the unit sphere with the outward pointing normal vector $\mathbf{n}$, then $\iint_S \nabla f\cdot\mathbf{n} \,dS=4\pi$

C: If the directional derivative of $f$ along the unit tangent vector of a simple piecewise-smooth curve $C$ is zero at every point on $\space\space\space\space\space C$, then $\int_C \nabla f\cdot d\mathbf{r}=0$

D: The vector field $\nabla f \times \nabla g$ is solenoidal

What I know so far...

So, I believe statement B to be false since I know that $\iint_S dS=4\pi$ where $S$ is the unit sphere. Also, I know D to be true if $f$ and $g$ are twice continuously differentiable scalar fields - so that rules that out. My gut tells me C to be true since $C$ is zero at every point. It's from here I'm not sure which is false... any help would be great!

答:
For A: compute $\nabla\times \nabla f=rot(\nabla f)$ using the definitions.

For B: take a constant field and compute.

For C: use that $\frac{d f}{d \vec{v}}=\nabla f\cdot \vec{v}$ and that $\int_C \vec{F} \cdot d\vec{r}=\int_C(\vec{F}\cdot \vec{T}) dr$, for vector fields $\vec{F}$.

Title: Differentiable scalar fields question.


问:
Consider a stick, cut twice, probability the smallest is 1/5
Consider a stick of unit length, you take two points (uniform independent), and then break the stick on these two points. You now got 3 segment. What is the probability the smallest segment  is less than 1/5?

$X, Y$ be the two cuts.

$P(X < 1/5  \text{ and } Y-X < 1/5 \text{ and } (1-Y-X)<1/5). $

答:
Let $P(S)$ be the probability that the smallest segment is less than $\frac{1}{5}$; let $X$ be the location of the first cut, and $Y$ be the location of the second cut.

If $X < \frac{1}{5}$ or $X > \frac{4}{5}$, then you already know that one of the segments has length less than $\frac{1}{5}$, so the conditional probability $P\big(S \vert X \in [0,\frac{1}{5}) \cup (\frac{4}{5},1]\big)=1$.

If $\frac{2}{5} < X < \frac{3}{5}$, then the possible locations where the second cut could facilitate $S$ are the regions of length $\frac{1}{5}$ on either side of $X$, and at either end of the stick, all of which are non-overlapping.
Therefore, the conditional probability $P(S \vert X \in (\frac{2}{5},\frac{3}{5}]) = \frac{4}{5}$.

Moving $X$ left of $\frac{2}{5}$, you linearly lose area in which $Y$ can land to facilitate $S$, until right before $X=\frac{1}{5}$, the good region is only length $\frac{3}{5}$. The same argument follows on the other side.

Combining all of these results, I end up with a piecewise function for the conditional probability of $P(S|X)$:
$$
P(S \vert X) = \begin{cases}
1 & 0 \leq X < \frac{1}{5} \\
\frac{2}{5} + X & \frac{1}{5} \leq X < \frac{2}{5} \\
\frac{4}{5} & \frac{2}{5} \leq X < \frac{3}{5} \\
\frac{7}{5} - X & \frac{3}{5} \leq X < \frac{4}{5} \\
1 & \frac{3}{5} \leq X \leq 1
\end{cases}
$$
and by integrating that function over $X$, I find that $P(S)=\frac{21}{25}=84\%$.


答:
Hint:
Every well-bred young people should know the formula
$$\int\frac{\mathrm d t}{t^n}=-\frac 1{(n-1)t^{n-1}} \qquad ({}+C)$$

Title: Evaluate $I = \iint_{R}\frac{x}{y^{4}}\,dx\,dy$ over a given region

Body:
Evaluate the double integral $$I = \iint_{R}\frac{x}{y^{4}}\,dx\,dy$$ over the region $R = \{(x,y)\mid 0\le x \le3, 3\le y \le5\}$.

Hi I was wondering if anyone could help me on this question which involves Taylor's Expansion.

This is what I have done so far:

From using the given boundaries I have created the integral:

$$\int_3^5\int_0^3 \frac{x}{y^4}\, dx\,dy$$

Then I worked out the first integral:

$$\int_0^3\frac{x}{y^4}\, dx$$

which gives me $\frac{9}{2y^4}$ and then I put this into the next integral that is with respect to $y$:

$$\int_3^5\frac{9}{2y^4} \,dy$$

Which gives me $3.415703704\times 10^{-3}$, which is evidently wrong as the correct answer is: $49/1125$

Can someone please help me on where I am going wrong? Thanks in advance!


问:
How did the author take the gradient to get $\overline{W} \Leftarrow \overline{W} - \alpha \nabla_{W} L_i$?
I am currently studying the textbook Neural Networks and Deep Learning by Charu C. Aggarwal. Chapter 1.2.1.1 What Objective Function Is the Perceptron Optimizing? says the following:

Can we find a smooth loss function, whose gradient turns out to be the perceptron update? The number of classification errors in a binary classification problem can be written in the form of a $0/1$ loss function for training data point $(\overline{X_i}, y_i)$ as follows:

$$L_i^{(0/1)} = \dfrac{1}{2} (y_i - \text{sign}\{ \overline{W} \cdot \overline{X_i} \})^2 = 1 - y_i \cdot \text{sign} \{ \overline{W} \cdot \overline{X_i} \} \tag{1.7}$$

The simplification to the right-hand side of the above objective function is obtained by setting both $y_i^2$ and $\text{sign}\{ \overline{W} \cdot \overline{X_i} \}^2$ to $1$, since they are obtained by squaring a value drawn from $\{ -1, +1 \}$. However, this objective function is not differentiable, because it has a staircase-like shape, especially when it is added over multiple points. Note that the $0/1$ loss above is dominated by the term $- y_i \cdot \text{sign} \{ \overline{W} \cdot \overline{X_i} \}$, in which the sign function causes most of the problems associated with non-differentiability. Since neural networks are defined by gradient-based optimization, we need to define a smooth objective function that is responsible for the perceptron updates. It can be shown that the updates of the perceptron implicitly optimize the perceptron criterion. This objective function is defined by dropping the sign function in the above $0/1$ loss and setting negative values to $0$ in order to treat all correct predictions in a uniform and lossless way:
$$L_i = \max\{ - y_i ( \overline{W} \cdot \overline{X_i} ), 0 \} \tag{1.8}$$
The reader is encouraged to use calculus to verify that the gradient of this smoothed objective function leads to the perceptron update, and the update of the perceptron is essentially $\overline{W} \Leftarrow \overline{W} - \alpha \nabla_{W} L_i$.

How did the author take the gradient to get $\overline{W} \Leftarrow \overline{W} - \alpha \nabla_{W} L_i$?

答:
Some strategies I have found useful when self-studying ML (before I found out about stackexchange), is that if you are struggling with a particular author's presentation, consult other reputable sources on the same topic, as another author may present it in a slightly different way that might be more amenable to how you are thinking about the problem. Furthermore, I've found that studying this way allows you to get a more well-rounded perspective, as different authors, depending on their field of specialisation, will often emphasise different aspects of the same topic.

I haven't seen this particular presentation before, but much of the part on how 0/1 loss  is incompatible with gradient-based methods, due to discontinuity/non-differentiability coheres with what I know from the sources I used to learn this.

Loss function - intuition.

Where there is a minor difference, but one which has consequences for intuition of what is going on, is the specification of the perceptron update criterion in terms of the loss function per training point outlined here:

$$L_i = \max\{-y_i (\bar{W} \cdot \bar{X}_i), 0 \}$$

Crucially, the author hasn't been entirely explicit about emphasising the fact that contributions to gradient updates only occur for misclassified points (although that can be deduced from the above). As that might be unclear at this stage, here is how it works. Let's rewrite the loss per training point piece-wise:

$$L_i =\max\{-y_i (\bar{W} \cdot \bar{X}_i), 0 \} =
\begin{cases}
-y_i(\bar{W} \cdot \bar{X}_i) &\text{if} \space -y_i(\bar{W} \cdot \bar{X}_i) > 0 \\
0 &\text{otherwise} \\
\end{cases}$$

Now for correctly classified points, $L_i = 0$. Why? Because your label $y_i \in \{-1, +1\}$ will have the same sign as your prediction $\bar{W} \cdot \bar{X}_i$, due to the classification being correct. This means that $y_i(\bar{W} \cdot \bar{X}_i) > 0$ and hence $-y_i(\bar{W} \cdot \bar{X}_i) < 0$, and $L_i = 0$.

Now for incorrectly classified points, $L_i = -y_i(\bar{W} \cdot \bar{X}_i)$. Why? The sign of the label $y_i$ will be different to your prediction $\bar{W} \cdot \bar{X}_i$, and so $y_i(\bar{W} \cdot \bar{X}_i) < 0$, meaning that $-y_i(\bar{W} \cdot \bar{X}_i) > 0$, and hence $L_i = -y_i(\bar{W} \cdot \bar{X}_i)$.

Hopefully, this clarifies the fact that the loss function $L_i$ is assigning a loss of $0$ to correctly classified points, and a loss of $-y_i(\bar{W} \cdot \bar{X}_i)$ to incorrectly classified points.

Stochastic gradient descent and the gradient of the loss function.

Recall we want to set $\bar{W}$ to minimise the sum of individual losses over the entire data set, that is we want to minimise $\sum^n_{i=1} L_i$. In the case of the perceptron, because we cannot just minimise $\sum^n_{i=1} L_i$ analytically by solving $\nabla_{W}\sum^n_{i=1} L_i = 0$, we use stochastic gradient descent.

So the expression $\bar{W} \leftarrow \bar{W} - \alpha \nabla_{W} L_i$ just refers to iteratively updating the parameter vector using stochastic gradient descent. It is stochastic because you are using only one training point $(\bar{X}_i, y_i)$ at a time to update the parameter vector $\bar{W}$. $\alpha$ is just a learning rate parameter.

Where gradients have been taken is in $\nabla_{W} L_i$, which is the gradient of the loss on an individual training point.

Now, to see that contributions to gradient updates of the parameter only occur for misclassified points, notice that when a point is correctly classified $L_i = 0  \implies \nabla_{W} L_i = 0$, and that when a point is incorrectly classified $L_i = -y_i (\bar{W} \cdot \bar{X}_i) \implies \nabla_{W} L_i = -y_i \bar{X}_i$.

Yielding, for misclassified points only:

$$\bar{W}^{(t+1)} = \bar{W}^{(t)} - \alpha \nabla_{W}L_i = \bar{W}^{(t)} + \alpha y_i \bar{X}_i$$

Where $t$ refers to iteration number. How does this yield the perceptron algorithm? Start with an initial guess $\bar{W}^{(0)}$. Out of all misclassified points, i.e. those points where the sign of the label $y_i$ is not the same as the sign of your prediction $\bar{W}^{(0)} \cdot \bar{X}_i$, randomly select one and update $\bar{W}$ using the above formula to get $\bar{W}^{(1)}$, and repeat...

References.

There are absolutely no guarantees that the algorithm converges. For further info on convergence and a clear exposition on the perceptron to supplement your readings, see p192-196 of "Pattern Recognition and Machine Learning" by Christopher Bishop. Lecture 12 of this series here is also excellent. In particular, the latter contains a neat result that can be proved using elementary tools - a lower bound on the total number of misclassifications the perceptron can make on a dataset under certain conditions, due to Novikoff and Block (1962).


问:
Solve a linear equation with 3 unknowns and 1 parameter
$$(a+1)x+y+3z=1$$$$8x+2y+(a+3)z=2$$$$3x+y+2z=-1$$

This question can be calculated with Gauss-elimination and I want to take away the y by taking $-2$ from the middle and $-1$ from the top and keep the third one as it is, my question is how do I calculate it with a parameter?

答:
$$
\left \{ \begin{array}{lcr}
(a+1)x+y+3z & = & 1 \\
8x+2y+(a+3)z & = & 2 \\
3x+y+2z & = & -1
\end{array}\right . \\
\left \{ \begin{array}{lcr}
2(a+1)x+2y+6z & = & 2 \\
8x+2y+(a+3)z & = & 2 \\
3x+y+2z & = & -1
\end{array}\right . \\
\left \{ \begin{array}{lcr}
2(a-3)x+(3-a)z & = & 0 \\
8x+2y+(a+3)z & = & 2 \\
3x+y+2z & = & -1
\end{array}\right . \\
\left \{ \begin{array}{lcr}
(a-3)(2x-z) & = & 0 \\
8x+2y+(a+3)z & = & 2 \\
3x+y+2z & = & -1
\end{array}\right .
$$
So, from here, you can conclude that if $a = 3$, the system gets overdefined (number of variables exceeds number of equations), and since first equation will give you nothing, second and third equations can be restated taking into account $a = 3$
$$
\left \{ \begin{array}{lcr}
8x+2y+6z & = & 2 \\
3x+y+2z & = & -1
\end{array}\right .
$$
Subtract, doubled second from first
$$
2x+2z =4
$$
or $z = 2 - x$, substitute it to third equation back, so you'll get
$$
y = -1 - 3x - 4+2x = -x-5
$$
As you may see, due to the overdefinition, you observed one possible scenario – parametric solution, i.e. when one of the variables can take any value, uniquely defining other two.

Now, let's assume that $a \ne 3$. In this case initial system can be restated as
$$
\left \{ \begin{array}{lcr}
2x-z & = & 0 \\
8x+2y+(a+3)z & = & 2 \\
3x+y+2z & = & -1
\end{array}\right .
$$
First equation yields $z = 2x$. Consider latter two
$$
\left \{ \begin{array}{lcr}
8x+2y+2(a+3)x & = & 2 \\
3x+y+4x & = & -1
\end{array}\right . \\
\left \{ \begin{array}{lcr}
2(a+7)x+2y & = & 2 \\
7x+y & = & -1
\end{array}\right . \\
\left \{ \begin{array}{lcr}
(a+7)x+y & = & 1 \\
7x+y & = & -1
\end{array}\right .
$$
Consider first equation minus second
$$
ax+7x+y-7x-y = ax = 2
$$
Again, you can see that $a \ne 0$. If it is, first and second equations are inconsistent, so you won't get any solution from there.

And finally, you can find that
$$
x= \frac 2a
$$
Substitute it to the equations
$$
y = -1 - 7x = -1 - \frac {14}a = -\frac {a+14}a
$$
and recalling initial equation $z = 2x$ you can finally find that
$$
\left [ \begin{array}{lc}
a \notin \{0,3\} & \left \{ \begin{array}{ccr}
x & = & \frac 2a \\
y & = & -\frac {a+14}a \\
z & = & \frac 4a
\end{array} \right . \\
a = 3 & \left \{
\begin{array}{ccr}
y & = & -x-5 \\
z & = & 2-x
\end{array}
\right .
\end{array} \right .
$$

Disclaimer

Even though this solution is overly filled with elementary things that could and even should have been easily omitted, I've solely decided that it might be useful for you. You can check answers provided here with Wolfram alpha by this link.


问:
Distribution of the Sum of Uniform Random Variable and Another Random Variable
This could have been asked somewhere else, but I couldn't find it anywhere. Assume $X$ and $Y$ are random variables which take values in $\mathbb{Z}_n$ - i.e., $\{0, 1, 2, ..., n-1\}$. Also assume that $X$ is uniformly distributed. What is the distribution $X+Y$?

My approach: Let $Z$ be a random variable defined by $X+Y$. Then, since the maximum value that $X$ and $Y$ can take is $n-1$, $Z$ can take a maximum number of $2(n-1) = 2n -2$, which means that $Z$ takes values in $\mathbb{Z}_{2n-1}$. Now to get the probability mass function of $Z$, we have that
\begin{align*}
P(Z = z) &= P(X+Y = z)\\
&= P(X= x, Y = z -x) \\
&= P(X=x) P(Y = z-x) \\
&= \frac{1}{n} P(Y = z - x)
\end{align*}
which shows that $Z$ has the same distribution as $Y$ and it takes values in $\mathbb{Z}_{2n-1}$. Is my logic correct? If not, how else can I find the distribution of $X+Y$? I would appreciate any help on this!

答:
Taken almost verbatim from here:

If $X$ and $Y$ are two independent integer-valued random variables with distribution functions $m_{1}(x)$ and $m_{2}(x)$, respectively.
The convolution of $m_{1}(x)$ and $m_{2}(x)$ is the distribution $m_{3}(x) = m_{1}(x) * m_{2}(x)$ given by
$$
m_{3}(j) = \sum_{k} m_{1}(k) \cdot m_{2}(j-k).
$$
The function $m_{3}(x)$ is the distribution function of the random variable $Z = X + Y$.

I think your issue comes from your third equality, since you have to remember that
$$
P(A,B) = P(A|B) P(B),
$$
where $P(A|B)$ will equal $P(A)$ only if $A$ and $B$ are independent.
In your case, the probability that $Y=z-x$ is not going to be independent from the probability that $X=x$.

The linked chapter from which I copied the quote about convolution has many more details and examples which may be of interest to you.


问:
How can I find all $x$ such that $1+(x+1)2^x$ is a perfect square?
How can I find all $x$ such that $1+(x+1)2^x$ is a perfect square?

I only find that when $x = 4$, the value is $81$ is the square of $9$, but I cannot get a general solution for this. Could anyone help me with that?

答:
Write $(x+1)2^x=n^2-1$ for integers $n,x$.

For $x \leq -1$, notice that $2^{-x}(n^2-1)=x+1\leq 0$, which implies $n^2-1 \leq 0$ (since $2^{-x}>0$) and so $n=0,1$, from which only $n=1,x=-1$ works in the original problem.

For $0 \leq x \leq 4$, we can see that only $x=4$, $n=9$ works (simply inspect the five cases by hand).

For $x \geq 5$, notice that $4 \mid (n-1)(n+1)$, and also $\gcd(n-1,n+1) \mid 2$, so exactly one of terms $n-1$ or $n+1$ is divisible by $2^{x-1}$ (and possibly higher power, but that is not important here) and the other term is divisible by $2$ (and no by higher power of $2$). So either $n-1=d2^{x-1}$ or $n+1=d2^{x-1}$ where $d \geq 1$ is an integer. Either way, $$(x+1)2^x = n^2-1 \geq (d2^{x-1}-2)(d2^{x-1}),$$
and after division by $2^x$ and using $d \geq 1$ we get
$$x+1 \geq (d2^{x-2}-1)d \geq 2^{x-2}-1,$$
impossible for $x \geq 5$ (you can prove the opposite inequality by induction).


问:
Given $p \in \mathbb R^N$ and $X$ a symmetric matrix, find $\varphi \in C^2$ such that $D\varphi = p, D^2\varphi = X$ at a given point
In their paper User's guide to viscosity solutions of second order differential partial equations, Crandall, Ishii and Lions define the superjet of an upper semicontinuous function $u$ at $\hat x \in \mathcal O$, $\mathcal O$ a locally compact subset of $\mathbb R^N$, to be the set of $(p, X) \in \mathbb R^N \times \mathcal S_N$ ($\mathcal S_N$ the set of symmetric matrices) such that
\begin{equation}
u(x) \leq u(\hat x) + \langle p, x - \hat x \rangle + \frac 12 \langle X(x - \hat x), x - \hat x \rangle + o(|x - \hat x|^2) \quad \quad (*)
\end{equation}
holds. The superjet is denoted $J_{\mathcal O}^{2, +}u(\hat x)$. It is left as an exercise that
$$
J_{\mathcal O}^{2, +}u(\hat x) = \{(D\varphi(\hat x), D^2 \varphi(\hat x)) \ : \ \varphi \in C^2 \text{ and } u - \varphi \text{ has a local maximum at } \hat x\}.
$$

My question is: given $(p, X) \in \mathbb R^N \times \mathcal S_N$ such that $(*)$ holds, how to find such a $\varphi$?

Thanks in advance.

答:
Given a $(p, X)$ in this set, we can construct a $\phi$ as follows, by using these as data to construct $\phi$ with a specified first/second derivative. Some extra work has to be done to ensure this local maxima property. First, set
$$
\tilde\phi(x) = u(\hat x) + \langle p, x - \hat x\rangle + \frac{1}{2}\langle X(x - \hat x), x - \hat x\rangle.
$$
From the definition of $\tilde\phi$ we see that $\tilde\phi$ is $C^2$ with $D\tilde\phi(\hat x) = p$ and $D^2 \tilde\phi(\hat x) = X$. By the fact that $(p, X)$ was in the superjet, we have
$$
u(x) \leq \tilde\phi(x)  + o(|x - \hat x|^2).
$$
This isn't quite what we want, since we want $u - \tilde\phi$ to have a local max at $\hat x$, i.e $u - \tilde \phi \leq 0$ (since $u(\hat x) = \tilde\phi(\hat x)$). To fix this, we use the definition of little-o notation. There exists a $\delta > 0$ such that for $|x- \hat x| < \delta$, $\frac{1}{|x - \hat x|^2}o(|x - \hat x|^2) \leq 1$, hence the little-o error term is $\leq |x - \hat x|^2$ on $B(\hat x, \delta)$. Define now
$$
\phi(x) = \tilde\phi(x) + 2|x - \hat x|^2.
$$
Then $\phi$ has $D\phi(\hat x) = D\tilde\phi(\hat x)$, $u(\hat x) = \phi(\hat x)$, and for all $x \in B(\hat x, \delta)$, we have
$$
u(x) - \phi(x) = u(x) - \tilde\phi(x) - 2|x - \hat x|^2  \leq |x - \hat x|^2 - 2|x - \hat x|^2 = - |x - \hat x|^2 \leq 0.
$$
Thus $u - \phi$ has a local maximum at $x = \hat x$. Note also that
$$
D^2\phi(\hat x) = D^2\tilde\phi(\hat x) + 2I
$$
and thus
$$
u(x) \leq u(\hat x) + \langle D\phi(\hat x), (x - \hat x)\rangle + \langle D^2\phi(\hat x)(x - \hat x), x - \hat x \rangle + o(|x - \hat x|^2),
$$
since $D^2\phi \geq D^2 \tilde \phi$ (the identity matrix is positive definite). Thus $(D\phi(\hat x), D^2\phi(\hat x))$ is in the superjet.


问:
Let $\alpha$ be a solution of a monic quadratic polynomial with integer entries and $|\alpha|=1$.Then prove that $\alpha^{12}=1$.
($a$) Let $\alpha$ be a solution of a monic quadratic polynomial with integer entries and $|\alpha|=1$.Then prove that $\alpha^{12}=1$.

($b$)Let $A \in M_2(\mathbb{Z})$ such that $A^n=I$ for some $n$ then show that $A^{12}=I$

completely stuck on it.can I get some help?

答:
1) If $\,\alpha\in\Bbb R\;$ then $\,\alpha=\pm1\implies \alpha^{12}=1\,$ and we're done, otherwise:

2) $\;\alpha\in\Bbb C-\Bbb R\implies\;$ also $\,\bar\alpha\,$ is a root of the same monic integer polynomial $\,x^2+bx+c\in\Bbb Z[x]\,$ .

But then

$$\alpha+\bar\alpha=2\text{Re}(\alpha)=-b\;,\;\;\alpha\bar\alpha=c=|\alpha|^2=1\implies$$

Writing $\,\alpha=e^{ix}\;,\;x\in\Bbb R\,$ , we get:

$$-1\le\cos x=-\frac b2\le1\implies -2\le -b\le 2\le \implies-2\le b\le 2\implies$$

$$\implies b\in\{\,-2,\,-1,\,0,\,1,\,2\,\}\implies\cos x\in\left\{\;0\,,\frac12\,,\,1\;\right\} \implies $$

$$\implies x\in\pm\left\{\;0\,,\,\frac\pi3\,,\,\frac\pi2\,,\,\frac{2\pi}3\;\right\}\ldots\ldots$$

Well, now finish the proof.


问:
Is the convolution of projected probability measures equal to the original measure?
Let $V$ be a real vector space, $\mathcal{A}$ a $\sigma$-algebra over $V$ and $\mu$ a probability measure on $\mathcal{A}$. Suppose there are real vector spaces $V_1$, $V_2$ such that $V = V_1 \oplus V_2$. We then have canonical projections $\pi^1 : V \to V_1$ and $\pi^2 : V \to V_2$ as well as canonical inclusion maps $\iota^1 : V_1 \to V, v_1 \mapsto \left( v_1, 0 \right) \in V_1 \oplus V_2 = V$ and $\iota^2$ analogously.

We may now construct the convolution of the iterated pushforward measures as

\begin{equation}
\nu = \left( \iota^1_\ast \pi^1_\ast \mu \right) \ast \left( \iota^2_\ast \pi^2_\ast \mu \right) \, .
\end{equation}

Do we have $\mu = \nu$? Intuitively, this should be the case but I seem to be unable to prove it. If this is false, please provide a counterexample and/or some criteria when one might expect it to be true.

答: This is false in general.  Let $V = \mathbb{R}^2$ and $\mu$ be (normalized) $1$-dimensional Lebesgue measure on the line segment $\{(t,t) : 0 \leq t \leq 1 \}$.  Then $\pi^1_* \mu$ is Lebesgue measure on the $x$-axis restricted to $[0,1]$ and $\pi^2_*\mu$ is Lebesgue measure  on the $y$-axis restricted to $[0,1]$.  But the convolution of these two measures is $2$-dimensional Lebesgue measure on the unit square in $\mathbb{R}^2$.


问:
Is subtracting covariances legal in the case?
I want to find the Covariance of $cW$ and $Y - cV$. Which of the following is correct?

$$\mathrm{Cov}(cW, y - cV) = c \mathrm{Cov}(W, Y) + c^2 \mathrm{Cov}(W, V)$$

$$\mathrm{or}$$

$$\mathrm{Cov}(cW, y - cV) = c \mathrm{Cov}(W, Y) - c^2 \mathrm{Cov}(W, V)$$

答:
(I'm going to assume that $V$, $W$, and $Y$ are all random variables, and that where you wrote $Z$ you meant $Y$.)

We know that $\mathrm{cov}(X,Y) = E[XY] - E[X]E[Y]$, so in your case,
$$
\begin{aligned}
\mathrm{cov}(cW,Y-cV) &= E[(cW)(Y-cV)] - E[cW]E[Y-cV] \\
&= E[cWY - c^{2}WV] - E[cW]E[Y-cV] \\
&= cE[WY] - c^{2}E[WV] - cE[W]E[Y] + c^{2}E[W]E[V] \\
&= c\Big\{E[WY] - E[W]E[Y]\Big\} - c^{2} \Big\{ E[WV] - E[W]E[V] \Big\} \\
&= c \cdot \mathrm{cov}(W,Y) - c^{2} \cdot \mathrm{cov}(W,V).
\end{aligned}
$$


问:
Bivariate Normal Transformation
I have $X, Y$ governed by a Bivariate Normal($\mu_{X} = 0, \mu_{y} = 0, \sigma^2_{x} = 1, \sigma^2_{y} = 1, \rho=\rho$). $H, J$ are from the same joint distribution as $X,Y$ and are independent. If $G = cX + (H - aJ)$, what is the joint distribution $f_{X,G}(x,g)$? I am experiencing troubles with the transformation aspect of this.

答:
The random vector $U:=(X,G)^T$ is a linear transformation of the random vector $V:=(X,Y,H,J)^T$, specifically
$$
U:=\begin{pmatrix}X\\G\end{pmatrix}=
\begin{pmatrix}1&0&0&0\\c&0&1&-a\end{pmatrix}
\begin{pmatrix}X\\Y\\H\\J\end{pmatrix}=BV,
$$
where we write $B$ for the $2\times 4$ matrix of constants. Since $V$ has multivariate Gaussian distribution, it follows that $U$ also has Gaussian distribution. It remains to determine the mean vector and covariance matrix for $U$. Clearly $U$ has mean vector zero, so the covariance matrix for $U$ has the simple form
$$
\operatorname{Var}(U)=E(UU^T)=E[(BV)(BV)^T]=E(BVV^TB^T)=BE(VV^T) B^T=B\Sigma B^T
$$
where $\Sigma$ is the covariance matrix for $V$. Now plug in.

Title: Bivariate Normal Transformation


问:
A cylinder is circumscribed about a sphere. If their volumes are denoted by $C$ and $S$, find $C$ as a function of $S$
Here is the problem.

A cylinder is circumscribed about a sphere. If their volumes are denoted by $C$ and $S$, find $C$ as a function of $S$

My (Amended) Attempt:

[Based on the correction suggested by herbSteinberg]

Let $r$ be the radius of the sphere.

Then the height of the cylinder is $2r$, and the radius of the base  is $r$. So the volume $C$ of the cylinder is given by
$$
C = \pi r^2 (2r) = 2 \pi r^3. \tag{1}
$$

And, the volume $S$ of the sphere is given by
$$
S = \frac43 \pi r^3. \tag{2}
$$

From (2), we obtain
$$
r^3 = \frac{3}{4 \pi} S = \frac{ 3S }{4 \pi},
$$
and hence
$$
r = \sqrt[3]{ \frac{3S}{4 \pi} }. \tag{3}
$$

Finally, putting the value of $r$ from (3) into (1), we get
$$
C = 2 \pi \left( \sqrt[3]{ \frac{3S}{4 \pi} } \right)^3 = 2 \pi \left( \frac{3S}{4 \pi} \right) = \frac32 S.
$$

Is my solution correct in each and every detail? Or, are there any errors of approach or answer?

答:
As noted in the comments, the radius of the cylinder is just $r$. Otherwise your work appears to be correct, but you've made things harder on yourself than necessary. Note that $r^3$ appears in both formulas, so once you have solved for $r^3 = \frac{3S}{4\pi}$, you can immediately use this expression in place of $r^3$ in the formula for $C$, going straight to the last equality.

In other words:

$$S = \frac{4}{3}\pi r^3 \Rightarrow r^3 = \frac{3S}{4\pi}$$
$$C = \pi r^2 \cdot (2r) = 2\pi r^3 = 2\pi\cdot\frac{3S}{4\pi}=\frac{3S}{2} $$


问:
Help me solve a trigonometric equation
I am doing some work in RF circuit design. I need to solve an equation for my design:

$$\frac 1{\cos(t_1)}+\frac 1{\sin(t_1)} =\frac 1{\cos(t_2)}+\frac 1{\sin(t_2)}$$

(I created a nicely typed image of this equation but setting don't allow me to upload unless I get at least 10 !]

Kindly help me find a general solution of the above equation (i.e. to find relation between $t_2$ and $t_1$). I may solve easier equations (like $\,\cos t=0.5$ !), but seems complex to me.
I observe that at $t_1=t_2=\dfrac {\pi}4$, this equation holds true. But I am unable to proceed further.

Thanks.
Mohammad A Maktoomi.

答:
Let's rewrite your equation :
$$\frac {\sin(t_1)+\cos(t_1)}{\sin(t_1)\cos(t_1)} =\frac {\sin(t_2)+\cos(t_2)}{\sin(t_2)\cos(t_2)}$$
that you may multiply by $\frac 12\sin\left(\frac{\pi}4\right)$ and rewrite as :
$$\frac {\sin\left(t_1+\frac {\pi}4\right)}{\sin(2\;t_1)} =\frac {\sin\left(t_2+\frac {\pi}4\right)}{\sin(2\;t_2)}$$
We could 'linearize' this to obtain :
$$\sin\left(t_1+\frac {\pi}4\right)\sin(2\;t_2)=\sin\left(t_2+\frac {\pi}4\right)\sin(2\;t_1)$$

But it will be more convenient to set $\;x_1:=t_1+\frac {\pi}4,\ x_2:=t_2+\frac {\pi}4\;$ and get (since $\sin\bigl(2x-\frac {\pi}2\bigr)=-\cos(2x)$ and reverting the fractions) :
$$\frac{\cos(2\;x_2)}{\sin(x_2)}=\frac{\cos(2\;x_1)}{\sin(x_1)}$$

and rewrite :

$$\frac{\cos(2\;x)}{\sin(x)}=\frac {1-2\sin(x)^2}{\sin(x)}=\frac 1{\sin(x)}-2\;\sin(x)$$
Since the solutions of $\frac 1s-2s=r$ are obtained by resolving $2s^2+rs-1=0$ with solutions $\;s=\frac{-r\pm\sqrt{8+r^2}}4$ we may deduce, for $s=\sin(x_2)$ and $r=\frac{\cos(2\;x_1)}{\sin(x_1)}$, that :

\begin{align}
\sin(x_2)&=\frac{-\cos(2\;x_1)\pm\sqrt{8\;\sin(x_1)^2+\cos(2\;x_1)^2}}{4\;\sin(x_1)}\\
&=\frac{-\cos(2\;x_1)\pm\left(2-\cos(2\;x_1)\right)}{4\;\sin(x_1)}\\
\end{align}
With the two different solutions :
\begin{align}
\sin(x_2)&=\frac{1-\cos(2\;x_1)}{2\;\sin(x_1)}=\sin(x_1)\\
\sin(x_2)&=-\frac{1}{2\;\sin(x_1)}\\
\end{align}

From this we deduce the four different solutions (modulo $2\pi$) :

\begin{align}
x_2&=x_1\\
x_2&=-\arcsin\left(\frac{1}{2\;\sin(x_1)}\right)\\
x_2&=\pi-x_1\\
x_2&=\pi+\arcsin\left(\frac{1}{2\;\sin(x_1)}\right)\\
\end{align}

The substitution $t_1=x_1-\frac {\pi}4$ and $t_2=x_2-\frac {\pi}4$ should give you the wished solutions.

Title: Help me solve a trigonometric equation


答: The radius of the sphere should be equal to the radius of the cylinder face. Though you are correct about the height of the cylinder being twice the radius of the sphere.</p>

Body:
Here is the problem.

A cylinder is circumscribed about a sphere. If their volumes are denoted by $C$ and $S$, find $C$ as a function of $S$

My (Amended) Attempt:

[Based on the correction suggested by herbSteinberg]

Let $r$ be the radius of the sphere.

Then the height of the cylinder is $2r$, and the radius of the base  is $r$. So the volume $C$ of the cylinder is given by
$$
C = \pi r^2 (2r) = 2 \pi r^3. \tag{1}
$$

And, the volume $S$ of the sphere is given by
$$
S = \frac43 \pi r^3. \tag{2}
$$

From (2), we obtain
$$
r^3 = \frac{3}{4 \pi} S = \frac{ 3S }{4 \pi},
$$
and hence
$$
r = \sqrt[3]{ \frac{3S}{4 \pi} }. \tag{3}
$$

Finally, putting the value of $r$ from (3) into (1), we get
$$
C = 2 \pi \left( \sqrt[3]{ \frac{3S}{4 \pi} } \right)^3 = 2 \pi \left( \frac{3S}{4 \pi} \right) = \frac32 S.
$$

Is my solution correct in each and every detail? Or, are there any errors of approach or answer?


问:
Construct an elementary chain of order type $\mathbb Q$
Let $T$ be a complete first-order theory over a countable language $L$, such that $T$ has an infinite model.

It is well-known that we can construct an elementary chain of size-$\aleph_0$ models of order type the reverse ordering of any countable ordinal, see e.g. here. Simply, a possible proof uses Gödel's compactness theorem as well as augmenting the language with suitable Skolem functions.

It is even more well-known that we can construct an elementary chain of size-$\aleph_0$ models of order type any countable ordinal. Moreover, the ordinary or reversed ordering of any countable ordinal continuously embeds into the ordering of $\mathbb Q$.

Question 1: Is it always possible to construct an elementary chain $\{M_q:q\in\mathbb Q\}$ of models of $T$, such that every $\|M_q\|=\aleph_0$, every $M_p\precneqq M_q$ whenever $p<q$, and moreover for all $q\in\mathbb Q$, it holds that
\begin{equation}\tag{$\dagger$}\bigcup_{i<q}M_i=M_q=\bigcap_{j>q}M_j
\end{equation}

Question 2: If the answer to Question 1 is NO, then under what conditions for $T$ can the answer become YES?

Thanks in advance!

Simple observations:

(1) Every discontinuity of the chain must happen at Dedekind cuts of irrarional numbers. For example, if $T$ is the theory of infinite sets, then let $S$ be the set of all irrational algebraics, and the models $M_q=\{s\in S:s<q\}$ work as desired.

(2) By repeating the argument mentioned above, it is clear that the answer is YES when $(\dagger)$ is weakened to
$$\bigcup_{i<q}M_i\subseteq M_q\subseteq \bigcap_{j>q} M_j
$$
and it's easy to find examples such that both subset relations modified as above are cofinitely strict. Thus by replacing each $M_q$ with their upper/lower bounds, we see that the answer is also YES when either of the two equalities in $(\dagger)$ is omitted.

(3) It suffices for $(\dagger)$ to hold only for a dense subset of an interval in $\mathbb Q$, since $\mathbb Q$ would continuously embed into it.

答:
We can do much better! I'll explain how to construct an $\mathbb{R}\setminus \mathbb{Q}$-indexed chain of countable models with the continuity properties you want. This is more general, since you can get a $\mathbb{Q}$-indexed chain by taking the subchain indexed by $(\pi + \mathbb{Q})\subseteq \mathbb{R}\setminus \mathbb{Q}$.

$T$ is a complete first-order theory over a countable language $L$, with an infinite model. Let $T'$ be a Skolemization of $T$ (and note that the Skolemized language $L'$ is still countable). We'll work in $T'$, and then you can take the reduct of everything back to $L$ at the end of the day.

Let $I = (a_q)_{q\in \mathbb{Q}}$ be any non-constant indiscernible sequence in a model of $T'$. For $r\in \mathbb{R}\setminus \mathbb{Q}$, let $I_{<r} = \{a_q\mid q\in \mathbb{Q}, q<r\}$, and define $M_r = \text{dcl}(I_{<r})$. Now let's check your conditions.

For any $r\in \mathbb{R}\setminus \mathbb{Q}$, $|M_r| = \aleph_0$, since $I_{<r}$ is countable and the definable closure (in a countable language) of a countable set is countable.

When $r<s$ in $\mathbb{R}\setminus \mathbb{Q}$, $I_{<r} \subseteq I_{<s}$, so $M_r\subseteq M_s$, and since we have Skolem functions, $M_r\preceq M_s$. Now pick some $q\in\mathbb{Q}$ with $r<q<s$. Then $a_q\in I_{<s}\subseteq M_s$. To show that $M_r\precneqq M_s$, I claim that $a_q\notin M_r$. So suppose for contradiction that $a_q\in M_r = \text{dcl}(I_{<r})$. Then there are $a_{p_1},\dots,a_{p_k}\in I_{<r}$ and a formula $\varphi(a_{p_1},\dots,a_{p_k},y)$ defining $a_q$. Pick some $q'>q$ in $\mathbb{Q}$. By indiscernibility, we also have  $\varphi(a_{p_1},\dots,a_{p_k},a_{q'})$, so $a_q = a_{q'}$. But then by indiscernibility, the sequence $I$ is constant, which is a contradiction.

Fix $r\in \mathbb{R}\setminus \mathbb{Q}$. We have $M_s\subseteq M_r$ for all  $s<r$, so $\bigcup_{s<r}M_s \subseteq M_r$. For the other inclusion, let $b\in M_r = \text{dcl}(I_{<r})$. Then there are $a_{p_1},\dots,a_{p_k}\in I_{<r}$ with $p_1<\dots<p_k<r$ and a formula $\varphi(a_{p_1},\dots,a_{p_k},y)$ defining $b$. Pick some $s\in \mathbb{R}\setminus \mathbb{Q}$ with $p_k < s < r$. Then $a_{p_1},\dots,a_{p_k}\in I_{<s}$, so $b\in \text{dcl}(I_{<s}) = M_s$. Thus $M_r\subseteq \bigcup_{s<r} M_s$.

Again, fix $r\in \mathbb{R}\setminus \mathbb{Q}$. We have $M_r\subseteq M_s$ for all $s>r$, so $M_r\subseteq \bigcap_{s>r} M_s$.  For the other inclusion,  let $b\in \bigcap_{s>r} M_s$. Pick some $s>r$ in $\mathbb{R}\setminus \mathbb{Q}$, so $b\in M_s = \text{dcl}(I_{<s})$. Then there are $a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell}\in I_{<s}$ with $p_1<\dots<p_k<r<q_1<\dots<q_\ell<s$ and a formula $\varphi(a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell},y)$ defining $b$. Note that we use the fact that $r$ is irrational here: it is important that none of the $p_i$ or $q_j$ are equal to $r$. Pick some  $s'\in \mathbb{R}\setminus \mathbb{Q}$ with $r < s' < q_1$. Then also $b\in M_{s'} = \text{dcl}(I_{<s'})$, so there are $a_{p'_1},\dots,a_{p'_{k'}},a_{q'_1},\dots,a_{q'_{\ell'}}\in I_{<s'}$ with $p'_1<\dots<p'_{k'}<r<q'_1<\dots<q'_{\ell'}<s'$ and a formula $\psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q'_1},\dots,a_{q'_{\ell'}},y)$ defining $b$. Then we have: $$\forall x\, (\varphi(a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell},y) \leftrightarrow \psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q'_1},\dots,a_{q'_{\ell'}},y)).$$
Now we will slide the $a_{q'_i}$ into $M_r$. Let $p^* = \max(p_k,p'_{k'})<r$, and pick $q''_1,\dots,q''_{\ell'}\in \mathbb{Q}$ with $p^*<q''_1<\dots<q''_{\ell'} < r$. By indiscernibility, we have: $$\forall x\, (\varphi(a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell},y) \leftrightarrow \psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q''_1},\dots,a_{q''_{\ell'}},y)).$$
In other words, $\psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q''_1},\dots,a_{q''_{\ell'}},y)$ defines $b$. But $a_{p'_1},\dots,a_{p'_{k'}},a_{q''_1},\dots,a_{q''_{\ell'}}\in I_{<r}$, so  $b\in \text{dcl}(I_{<r}) = M_r$. Thus $\bigcap_{s>r} M_s\subseteq M_r$.

Note that the phrase "coheir sequence" didn't appear in this construction. In fact, the only thing special about our indiscernible sequence $I$ is that it was non-constant. The point is that the type of  any indiscernible sequence is finitely satisfiable in any initial segment without an upper bound,  just by indiscernibility. So the tail segment $I_{> r}$ is automatically a coheir sequence over $M_r = \text{dcl}(I_{<r})$.

The reason we had to pick our indiscernible sequence more carefully in the question you linked to is that there  we wanted to make an arbitrary model $M$ the intersection of an elementary chain of models. So we had to take special care to ensure that we built the elementary chain from a coheir sequence - the finite satisfiability in $M$ was not automatic the way it is for models that come from the indiscernible sequence itself.


问:
Which of the following statements regarding quadratic forms in $3$ variables

Please give a hint for this problem.

答:
The following theorems will be helpful:

Theorem(Sylvester's law of inertia): Let $\phi$ be a symmetric bilinear form on a finite-dimensional real vector space $V$. Then there exists unique non-negative integers $p$, $q$ such that $\phi$ is represented by the matrix
$$\begin{pmatrix}I_p & 0 & 0\\
0 & -I_q & 0\\
0 & 0 & 0\end{pmatrix}$$
with respect to some basis. Then $p+q=r(\phi)$ is the rank of this bilinear form.

Theorem: Let $\phi$ be a symmetric bilinear form over a complex vector space $V$. Then there exists a basis $(\mathbf v_1,\cdots, \mathbf v_m)$ for $V$ such that $\phi$ is represented by
$$\begin{pmatrix}I_p & 0 \\
0 & 0\end{pmatrix}.$$
with respect to this basis, where $p = r(\phi)$ is the rank of this bilinear form.

Also note that the rank of a bilinear form is invariant under isomorphisms. Hope this helps!

Body: Please give a hint for this problem.


问:
Prove $\sqrt x$ is uniformly continuous on $[1,+\infty[$
I have done the following:

For any given $\;\varepsilon>0\;,\;$ let $\;\delta=2\varepsilon>0\;,$

$\forall\;x,y\in[1,+\infty[\;$ with $\;|x−y|<\delta\;,$

since $\;\sqrt{x}+\sqrt{y}\ge2\;,\;$ we get that

$\left|\sqrt{x}-\sqrt{y}\right|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}\le\dfrac{|x-y|}{2}<\dfrac{\delta}{2}=\varepsilon\;.$

Hence, $\;\sqrt{x}\;$ is uniformly continuous on  $\;[1,+\infty[\;.$

Am I missing something? Is this proof correct?

答: The simplest way I see is to note that $\sqrt{x}$ is Lipschitz on $[1, +\infty[$.  (To show that it is Lipschitz, compute the upper bound on the absolute value of the derivative of $\sqrt{x}$.)  Lipschitz continuity implies uniform continuity.


问:
Solve the given system of quadratic equations

$$x^2 + xy + xz = 2$$
$$y^2 + yz + xy = 3$$
$$z^2 + zx + yz = 4$$

I tried solving it but i just ended up with
$$(x-y)(x+y+z) = -1$$
$$(y-z)(x+y+z) = -1$$
$$(x-z)(x+y+z) = -2$$

I'm not sure what to do. Any hints?

答: I solved it by solving for (x+y+z) in all three original equations. Then you can compare the results to each other.</p>


答:
Take
$$ x = 2t \; , \; \;  y = 3 t \; , \; \;  z = 4t \; . \;$$
We know this is valid because $x+y+z$ must be nonzero, so too the individual letters, and
$$ \frac{y}{x} = \frac{3}{2} \; , \; \; \frac{z}{x} = \frac{4}{2} \; . $$Then $$ x+y+z = 9t  $$

Title: Solve the given system of quadratic equations

Body:
$$x^2 + xy + xz = 2$$
$$y^2 + yz + xy = 3$$
$$z^2 + zx + yz = 4$$

I tried solving it but i just ended up with
$$(x-y)(x+y+z) = -1$$
$$(y-z)(x+y+z) = -1$$
$$(x-z)(x+y+z) = -2$$

I'm not sure what to do. Any hints?


问:
Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions.
Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions, both solutions are negative. I want to prove this by contradiction.

Proof: If we let $a,b,c = 1$, we have $x^{2} + x + 1 = 0$

or $x^{2} + x = -1$. If $x$ is positive the sum of two positive terms can not be $< 0$

hence $x$ must be negative. But no matter the $x$ we pick there will be no solution to $x^{2} + x = -1$.

So $a,b,c$ must also be $ < 0$. If $a,b,c < 0$ we have $-ax^{2} -bx -c = 0$

or  $-ax^{2} -bx = c$. If $x$ is positive then resulting equation is $negative - negative = positive$, which is not possible. So $x$ must be negative.

Have I shown by contradiction, both solutions are negative? How can I improve my answer?

答:
Let $r_1,r_2$ be the real roots. Then $r_1+r_2=-\frac{b}{a}$ and $r_1r_2=\frac{c}{a}$.
Since $c/a>0$, both roots have the same sign, but $-b/a<0$ so they both must be negative.


答:
That is not correct. A proof by contradiction would be as follows: if $r$ is a root of $ax^2+bx+c$ and $r\geqslant0$, then $ar^2+br+c=0$. But this is not true, because $ar^2+br+c\geqslant c>0$. Therefore $r<0$.

You should keep in mind that proving it form a single polynomial ($x^2+x+1$, in your case) doesn't prove it in general.

Title: Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions.

Body:
Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions, both solutions are negative. I want to prove this by contradiction.

Proof: If we let $a,b,c = 1$, we have $x^{2} + x + 1 = 0$

or $x^{2} + x = -1$. If $x$ is positive the sum of two positive terms can not be $< 0$

hence $x$ must be negative. But no matter the $x$ we pick there will be no solution to $x^{2} + x = -1$.

So $a,b,c$ must also be $ < 0$. If $a,b,c < 0$ we have $-ax^{2} -bx -c = 0$

or  $-ax^{2} -bx = c$. If $x$ is positive then resulting equation is $negative - negative = positive$, which is not possible. So $x$ must be negative.

Have I shown by contradiction, both solutions are negative? How can I improve my answer?


问:
The meaning of E[X|Y] = E[X]
In class today it was mentioned that when E[X|Y] = E[X] , the two random variables X and Y are uncorrelated, i.e. cov(X, Y) = 0.

I'm curious as to why that's the case.

Why are the two variables X and Y uncorrelated when E[X|Y] = E[X]? The professor just mentioned it as passing as if it was a natural thing, but I can't really understand how this property is proved.

Could anyone provide an explanation regarding why "E[X|Y] = E[X]"  means that  "covariance of X and Y = 0 ( cov(X,Y)=0 )"? Thanks.

答:
By the tower rule ($E[E[Z |X]]=E[Z]$) we can write

$$ E[ E[X Y | Y ] ] = E[X Y]$$

But $E[XY |Y] = Y E[X|Y]$ and $E[X|Y]=E[X]$ by hypothesis. Then the result follows.

Notice that the reverse implication is not true.

Title: The meaning of E[X|Y] = E[X]


答: Solve first equation for $y$, substitute in the others, simplify...


问:
Increasing the tail of exponential function
Suppose I generate a number of random number according to the exponential distribution and convert them into probability values in R.

```<code>prob <-  data.frame(round(rexp(2000 ,0.5) , digits = 4))
colnames(prob) <- "V"
prob$V <- prob$V / max(prob$V)
hist(prob$V , prob= TRUE, breaks =30, main=" ", xlab= "Probability Vals", col="red")
lines(density(prob$V, adjust=2), lty="dotted", col="darkblue", lwd=2)
</code>```

When I plot the distribution of the probability values, here how it looks;

figure

I would like to obtain something like

second figure

I have found something called exponentially modified Gaussian distribution and select the parameters as $\mu= -3, \sigma =1, \lambda =0.25$ as shown at this wiki page. However, I am not getting the desired output. The R code is

```<code>prob <-  data.frame(round(rexp(2000 ,0.25)+ dnorm(2000, mean = -3, sd = 1, log = FALSE)   , digits = 4))
</code>```

I was wondering if someone help me out.

答:
No matter what distribution you choose it will go to zero as $x$ goes to $\infty$. You could potentially ask that it goes to zero much slower than an exponential distribution, and there are many ways to achieve this. For instance you could choose the cdf to be $1-1/f$, where $f$ is a very slow growing function like $\log x$ or $\log \log x$. (To do this you might have to shift also to make it zero at $x=0$).

Putting this all together for $f = \log \log x$, say, a pdf with qualitatively very large tail is
$$
\frac{1}{(x+e^e)\log (x+e^e) \log^2\log (x+e^e)}
$$


Title: The meaning of E[X|Y] = E[X]

Body:
In class today it was mentioned that when E[X|Y] = E[X] , the two random variables X and Y are uncorrelated, i.e. cov(X, Y) = 0.

I'm curious as to why that's the case.

Why are the two variables X and Y uncorrelated when E[X|Y] = E[X]? The professor just mentioned it as passing as if it was a natural thing, but I can't really understand how this property is proved.

Could anyone provide an explanation regarding why "E[X|Y] = E[X]"  means that  "covariance of X and Y = 0 ( cov(X,Y)=0 )"? Thanks.

Body: $$E[XY]=E[E[XY|Y]]=E[Y\underbrace{E[X|Y]}_{E[X]}]=E[X]E[Y]$$


问:
Different versions of the entropy term in the entropy-regularized Wasserstein distance
\begin{equation}
\mathcal{W}_\epsilon(\alpha, \beta) = \min_{\pi\in \Pi(\alpha\beta)} \int c(x,y) \mathrm{d}\pi(x,y) + \epsilon H(\pi \| \alpha \otimes \beta)
\end{equation}
Cuturi (2013) introduced the entropy-regularized Wasserstein distance, or Sinkhorn distance, shown above, where $\epsilon $ is the regularization parameter and $H(\pi \| \alpha \otimes \beta)$ is the relative entropy, or KL-divergence, between the transport plan and the marginal probabilities.

But I have seen the $H(\cdot)$ term shown in two different ways, one with entropy and the other with relative entropy:

\begin{align}
H(\pi) &= \int \pi(x,y) \ln \pi(x,y) \\
H(\pi \| \alpha \otimes \beta)  &= \int \ln \left(\frac{\mathrm{d}\pi (x,y)}{\mathrm{d}\alpha(x)  \mathrm{d}\beta(y)  } \right) \mathrm{d}\pi (x,y)
\end{align}

How are the last two lines equal or connected to each other? Obviously they're not the same, so why are there two different versions running around?

答:
These two are actually equivalent up to a constant when $\pi$ is a coupling of $\alpha$ and $\beta$. I'll assume that $\pi,\alpha, \beta$ all have densities.
We can then write:

$$
H(\pi||\alpha\otimes \beta) = \int\ln\left(\frac{d\pi}{d\alpha d\beta} \right)d\pi = \int \pi(x,y) \ln\left(\frac{\pi(x,y)}{\alpha(x)\beta(y)} \right) dx dy
$$

Note that $\pi(x,y)$ is the density with respect to the Lebesgue measure, and the same can be said for $\alpha(x)$ and $\beta(y)$. Therefore:

$$
H(\pi||\alpha\otimes \beta) = \int\pi(x,y)\ln \pi(x,y) dx dy
- \int\pi(x,y)\ln(\alpha(x))dxdy
- \int\pi(x,y)\ln(\beta(y))dxdy =\\
= \int \pi(x,y) \ln\pi(x,y) dx dy - \int\alpha(x)\ln\alpha(x) dx
-\int \beta(y) \ln \beta(y) dy = H(\pi) - H(\alpha) - H(\beta)
$$

Since $\alpha$ and $\beta$ are fixed, we get $H(\pi) + C$, where $C$ is a constant.


问:
Finding Gaussian function with specific level curves
I’d like to find a Gaussian function with specific level curves as attached in the image, that is; the first quadrant with $0 \leq x,y \leq 1$, symmetric around $y=x$ and somewhat egg shaped (preferably converging to a ridge by manipulating any of the constants below). I’ve tried the following function: $$f(x,y)=e^{-a(x-0.5)^2-b(x-0.5)(y-0.5)-c(y-0.5)^2}$$ See this Geogebra illustration. What are the optimal values of the constants $a,b,c$ to get the level curves as shown in the image?

答: Try $a=c=30$, and $b=-50$. To be a Gaussian, you must have $a,c>0$. Since you want it to be symmetric around $x=y$, you get $a=c$. Start by setting $b=0$. You will get a Gaussian with a circular symmetry. If you increase $b$, you will deform the Gaussian such that the long axis is parallel to $y=-x$. So in order to get it to have the long axis along $y=x$ you need to have a negative $b$. Also note that it can't be too negative or too positive. (in the limit $b=\pm 2a$ you get a long ridge)


问:
Largest possible radius of a circle lying in a parabola

Circle of radius $r$ touches the parabola $y^2+12x=0$ at its vertex. Centre of circle lies left of the vertex and circle lies entirely within the parabola. What is the largest possible value of $r$?

So my book has given the solution as follows:

The equation of the circle can be taken as: $(x+r)^2+y^2=r^2$

and when we solve the equation of the circle and the parabola, we get $x=0$ or $x=12-2r$.

Then, $12-2r≥0$ and finally, the largest possible value of $r$ is $6$.

This is where I got stuck as I'm not able to understand why that condition must be true. I get that the circle must lie within the parabola...

Can someone please explain this condition to me?

答:
Under given constraints, the maximal value of the radius
$r_{\max}=\tfrac1\kappa$,
where $\kappa$
is the curvature of the parabola at its vertex.

For the given parabola $y^2=-12x$
we can use a convenient
parametric representation

\begin{align}
y(t)&=t
,\quad
x(t)=-\tfrac1{12}\,t^2
,\\
y'(t)&=1
,\quad
x'(t)=-\tfrac1{6}\,t
,\\
y''(t)&=0
,\quad
x''(t)=-\tfrac1{6}
,
\end{align}

so we can use a known expression for
the curvature of parametric curve

\begin{align}
\kappa(t)&=
\frac{x'y''-y'x''}{\sqrt{(x'^2+y'^2)^3}}
\\
&=\frac{-\tfrac1{6}\,t\cdot0-1\cdot(-\tfrac16)}
{\sqrt{\Big(\tfrac1{36}\,t^2+1\Big)^3}}
=\frac{36}{\sqrt{(t^2+36)^3}}
.
\end{align}

The vertex is located at the point $(x(t),y(t))|_{t=0}$,
so the answer is

\begin{align}
r_{\max}&=\frac1{\kappa(0)}
=\frac{\sqrt{(0^2+36)^3}}{36}
=6
.
\end{align}