问:
Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$

I have tried to fiddle with it as follows:

$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividing both sides by $6$ gives us
$$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$

Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.

Body:
$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$

Divide by$12^x$

$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$

$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$

We put :$t=(\frac{3}{2}) ^x$

$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$

Multiply by $t$

$\Rightarrow $ $ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$

We can see $-1$ is a solution of the equation

So:after division by $t+1$ we see

$ t^2 - \frac{13}{6}x+1=0$

$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $

So :

$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $

$\Rightarrow $ $x_1=1$

And

$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $

$\Rightarrow $ $x_2=-1$

Finally

$x=-1$ or $ x= 1$


问:
Component-wise averaging of similar quaternions while handling quaternion "double cover issue"
To average together quaternions in a well-defined way, the eigendecomposition method of Markley et al. may be used, from Averaging Quaternions, Journal of Guidance, Control, and Dynamics, 30(4):1193-1196, June 2007, Eqs. (12) and (13).

However, if a set of all quaternions are close to each other (meaning that they represent very similar rotations), then element-wise averaging of the quaternions followed by normalization may produce a sufficiently "central" quaternion. (Elementwise averaging is much faster than the eigendecomposition, which is important for some applications.)

However, the quaternions $\bf{q}$ and $\bf-{q}$ represent the same rotation (sometimes called the "double cover issue" of quaternions), so element-wise averaging cannot be applied without first somehow making sure that any quaternions that are to be averaged lie within the same "half" of the rotation group SO(3).

There are several possible methods for "standardizing" each quaternion in a set of quaternions so that the double-cover issue is not a problem, and I wrote about these in this answer, but I am not sure which of these methods is correct (or optimal, and under what assumptions). Some possible methods for standardizing all quaternions ${\bf q}_i \in Q$ (while ensuring that each quaternion still represents the same rotation) include the following:

If the $w$ component is negative, negate the quaternion (i.e. replace ${\bf q}_i$ with $-{\bf q}_i$), so that the $w$ component is positive for all quaternions in the set $Q$.

Take the dot product of ${\bf q}_1$ with all subsequent quaternions ${\bf q}_i$, for $2 \le i \le N$, and negate any of the subsequent quaternions whose dot product with ${\bf q}_i$ is negative.

For each quaternion, measure the angle of rotation about the rotation axis of the quaternion, and normalize it so it always rotates the "short way around", i.e. such that $-\pi \le \theta \le \pi$. If it rotates the "long way around", i.e. $\theta \lt -\pi$ or $\theta \gt \pi$, then negate the quaternion.

These sometimes produce the same result, but they all produce different results in some cases (i.e. they all can negate different quaternions in a set of quaternions) -- therefore they are not equivalent.

What is the best way to deal with quaternions in a standardized way in order to overcome the double cover issue in situations like this?

Note that it is not just element-wise averaging of quaternions that can cause the double cover issue to affect the results. Another example is the swing-twist decomposition: in a naive implementation, the recovered rotation component around a given axis can represent either a rotation "the short way around" or a rotation "the long way around", which can lead to some unexpected or unstable results if you care only about the rotation about the axis, not the full quaternion.

答:
As in this answer,
let's define
$d(\mathbf p, \mathbf q) \triangleq  1 - (\mathbf p \cdot \mathbf q)^2$
to represent the dissimilarity (or "distance") between two quaternions,
where $\mathbf p \cdot \mathbf q$ is the usual componentwise inner product of the quaternions treated as four-dimensional vectors.

On the assumption that we are only going to average together quaternions that represent similar orientations,
let's suppose that we have a set $Q$ containing some finite positive number of unit quaternions and that there exists some unit quaternion $\mathbf q_0$ (not necessarily a member of $Q$) such that for every $\mathbf q \in Q,$

$$ d(\mathbf q_0, \mathbf q) < \frac12. \tag1 $$

For component-wise averaging to be a good method, I think we would actually want the dissimilarity to be much smaller than this bound.
I chose $\frac12$ merely because it is small enough to establish a property I want.
If a set $Q$ admits a tighter bound, that's fine; what follows will be just as true, but the final result may be even better.

In particular, $d(\mathbf q_0, \mathbf q) < \frac12$ implies that
$\lvert \mathbf q_0 \cdot \mathbf q\rvert > \frac{\sqrt2}2$,
which implies that either $\mathbf q_0 \cdot \mathbf q > \frac{\sqrt2}2$ and
the angle between $\mathbf q_0$ and $\mathbf q$ is less than $\frac\pi4$,
or $-\mathbf q_0 \cdot \mathbf q > \frac{\sqrt2}2$ and
the angle between $-\mathbf q_0$ and $\mathbf q$ is less than $\frac\pi4$.

This also implies for any two quaternions $\mathbf p,\mathbf q \in Q,$
that $\mathbf q_0 \cdot \mathbf p$ and $\mathbf q_0 \cdot \mathbf q$
both have signs (positive or negative), that if these signs are the same then
the angle between $\mathbf p$ and $\mathbf q$
is less than than $\frac\pi2$ and therefore $\mathbf p \cdot \mathbf q > 0,$
and that if the signs are opposite then
the angle between $\mathbf p$ and $\mathbf q$
is greater than than $\frac\pi2$ and therefore $\mathbf p \cdot \mathbf q < 0.$

So we can partition $Q$ into two subsets:
the subset $Q_+ = \{\mathbf q\in Q \mid \mathbf q_0 \cdot \mathbf q > 0\}$
and $Q_- = \{\mathbf q\in Q \mid \mathbf q_0 \cdot \mathbf q < 0\}$.
Any two quaternions from one subset will have a positive dot product,
whereas any two quaternions from different subsets will have a negative dot product.

Now consider method 2. If the quaternion $\mathbf q_1$ is in $Q_+$, then after replacing $\mathbf q_i$ with $-\mathbf q_i$ whenever $\mathbf q_1\cdot\mathbf q_i<0,$
all the quaternions will be in $Q_+$
and the final result of averaging these quaternions and normalizing the result
will be some quaternion $\bar{\mathbf q}.$
On the other hand, $\mathbf q_1$ is in $Q_-$, then after replacing $\mathbf q_i$ with $-\mathbf q_i$ whenever $\mathbf q_1\cdot\mathbf q_i<0,$
all the quaternions will be in $Q_-$
and the final result will be $-\bar{\mathbf q},$
that is, the exact opposite of the quaternion we would have gotten if
$\mathbf q_1$ were in $Q_+$,
representing the exact same rotation.

Hence, given a finite set of orientations that are sufficiently similar, the final result is completely independent of which of the two possible quaternions is selected to represent each orientation.
Moreover, the quaternions that are figured into the final average are all relatively close together on the $3$-sphere;
whereas if you take any method that is not equivalent to this one,
the difference between the method must manifest in the fact that the alternative method averages one or more quaternions from $Q_+$ with one or more quaternions from $Q_-$, which will certainly introduce worse undesired cancellation effects than using quaternions from only one subset.

I would therefore choose method 2.

ContentLicense: CC BY-SA 4.0


问:
$\sigma(\sigma(p^2)) \neq 2p^2$ for all odd primes $p$.
How to prove that $\sigma(\sigma(p^2)) \neq 2p^2$ for all odd primes $p$?

I know that $\sigma(p^2)=1+p+p^2$ but I can't progress anymore.

答:
First consider $p \geq 11$, and assume on the contrary that $\sigma(\sigma(p^2))=\sigma(1+p+p^2)=2p^2$.

If $1+p+p^2$ has at least 3 distinct prime factors $q, r, s$, then write $1+p+p^2=q^ar^bs^cm$, then $2p^2=\sigma(1+p+p^2)=\sigma(q^a)\sigma(r^b)\sigma(s^c)\sigma(m)$. If $p \nmid \sigma(q^a)$, then since $\sigma(q^a) \mid 2p^2$, we have $\sigma(q^a) \mid 2$, so $2 \geq \sigma(q^a)=1+q+ \ldots +q^a \geq 3$, a contradiction. Thus $p \mid \sigma(q^a)$. Similarly $p \mid \sigma(q^b), \sigma(s^c)$, so $p^3 \mid \sigma(q^a)\sigma(r^b)\sigma(s^c)\sigma(m)=2p^2$, a contradiction.

Therefore we have $1+p+p^2=q^ar^b, q \not =r, a, b \in \mathbb{Z}^+$ or $1+p+p^2=q^a, a \in \mathbb{Z}^+$, where $q, r$ are primes. Note that $1+p+p^2$ is odd, so $q, r$ must be odd.

Let us first consider $1+p+p^2=q^ar^b$. Then $2p^2=\sigma(1+p+p^2)=\sigma(q^a)\sigma(r^b)$. As above, if $p \nmid \sigma(q^a)$, we have $\sigma(q^a) \mid 2$, so $2 \geq \sigma(q^a) \geq 3$, a contradiction. Therefore $p \mid \sigma(q^a)$, and similarly $p \mid \sigma(r^b)$. WLOG assume $\sigma(q^a) \leq \sigma(r^b)$, so that we have $$p=\sigma(q^a)=1+q+ \ldots +q^a, 2p=\sigma(r^b)=1+r+ \ldots +r^b$$

If $r \not =3$, then $r \geq 5, q \geq 3$, so

$$p=\frac{q^a-1}{q-1}+q^a \leq \frac{q^a-1}{2}+q^a, 2p=\frac{r^b-1}{r-1}+r^b \leq \frac{r^b-1}{4}+r^b$$

$$2p+1 \leq 3q^a, 8p+1 \leq 5r^b$$

$$(2p+1)(8p+1) \leq 15q^ar^b=15(1+p+p^2)$$

$$p^2-5p-14 \leq 0$$

However $p^2-5p-14 \geq 11p-5p-14=6p-14>0$, a contradiction.

Therefore $r=3$, so $q \not =3$, so $q \geq 5$. We have

$$1+p+p^2=q^a3^b, p=\frac{q^a-1}{q-1}+q^a \leq \frac{q^a-1}{4}+q^a, 2p=1+3+ \ldots +3^b=\frac{3^b-1}{2}+3^b$$

$$4p+1 \leq 5q^a, 4p+1 \leq 3(3^b)$$

$$(4p+1)^2 \leq 15(q^a3^b)=15(1+p+p^2)$$

$$p^2-7p-14 \leq 0$$

However $p^2-7p-14 \geq 11p-7p-14=4p-14>0$, a contradiction.

Therefore we are left with the case $1+p+p^2=q^a$, where $q \geq 3$ is an odd prime. Then $$2p^2=\sigma(q^a)=1+q+ \ldots +q^a=\frac{q^a-1}{q-1}+q^a \leq \frac{q^a-1}{2}+q^a$$

$$4p^2+1 \leq 3q^a=3(1+p+p^2)$$

$$p^2-3p-2 \leq 0$$

However $p^2-3p-2 \geq 11p-3p-2=8p-2>0$, a contradiction.

Therefore $\sigma(\sigma(p^2)) \not =2p^2$ for $p \geq 11$.

It remains to check $p=3, 5$, or $7$. We have:

$$\sigma(\sigma(3^2))=\sigma(13)=14 \not =2(3^2)$$
$$\sigma(\sigma(5^2))=\sigma(31)=32 \not =2(5^2)$$
$$\sigma(\sigma(7^2))=\sigma(57)=\sigma(3)\sigma(19)=80 \not =2(7^2)$$

Therefore $\sigma(\sigma(p^2)) \not =2p^2$ for all odd primes $p$.

ContentLicense: CC BY-SA 3.0


问:
Showing if the following can be a perfect square
I'm trying to see if $$\frac{n(n^2+1) }{2}$$ can be a perfect square for $n$ a positive integer, but I have no idea how to...

答:
Observe that $n$ and $n^2+1$ are coprime. Also if $ab$ is square with coprime $a,b$, both $a,b$ must be square.

Now there are two cases:

Case 1: $n$ is even. Then both $\dfrac n2$ and $n^2+1$ are squares. In particular the latter condition forces $n=0$.

Case 2: $n$ is odd. Then both $n$ and $\dfrac {n^2+1}2$ are squares. Writing $n = x^2$ we have the equation $x^4+1 = 2y^2$ for some integer $y$. As shown here (Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$.), this equation only has trivial solutions $x = \pm1$, $y= \pm 1$, giving the unique solution $n=1$.

The solution there observes that $(x^2\pm1)^2 = 2y^2\pm2x^2$, and thus $(x^4-1)^2 = 4y^4-4x^4$.

From $x^4+1=2y^2$, $x$ is odd, so the equation can be rewritten as $y^4-x^4=\left(\dfrac {x^4-1}2\right)^2$, and by Fermat's Right Triangle Theorem there are no nontrivial solutions.


问:
Is it true that any 3-regular graph containing K_{3,2} as subgraph must be non-planar?
It might seem intuitively clear once we try to draw down such a graph. But I cannot figure out a formal proof by applying Kuratowski's theorem yet. I am aware that the construction of 3-regular bipartite graph is a mysterious hard problem. See https://hal.inria.fr/hal-00990486/document.</p>

答:
No.
The following graph is $3$-regular, with a $K_{2,3}$ subgraph, and planar. I found it starting with $K_{2,3}$ and making it $3$-regular.

Even better, the following graph is $3$-regular, planar, and with $K_{2,3}$ as an induced subgraph.


问:
Functional and relational inductive set
Before we get into the actual problem we would like to first look at some background of the problem:

First we are given a set $U$ and an initial set $B\subseteq U$,

and a class $\mathcal{F}$ of functions containing just two members $f:U\times U\to U$ and $g:U\to U$.

Say a subset $S$ of $U$ is closed under $f$ and $g$ iff whenever elements $x$ and $y$ belongs to $S$, then so also do $f(x,y)$ and $g(x)$.

Say that $S$ is inductive iff $B\subseteq S$ and $S$ is closed under $f$ and $g$.

Let $C^*$ be the intersection of all the inductive subsets of $U$.

Define a construction sequence to be a finite sequence $(x_1,...,x_n)$ of elements of $U$ such that for each $i\leq n$ we have at least one of $$x_i\in B$$ $$x_i=f(x_j,x_k)\text{ for some }j<i,k<i$$ $$x_i=g(x_j)\text{ for some }j<i.$$

Let $C_*$ be the set of all points $x$ such that some construction sequence ends with $x$.

Then one can show that $C^*=C_*$.

Now, the problem:

We can generalize the discussion by requiring of $\mathcal{F}$ only that it be a class of relations on $U$. $C_*$ is defined as before except that $(x_0,x_1,...,x_n)$ is now a construction sequence provided that for each $i\leq n$ we have either $x_i\in B$ or $(x_{j_1},...,x_{j_k},x_i)\in R$ for some $R\in\mathcal{F}$ and some $j_1,...,j_k$ all less than $i$. Give the correct definition of $C^*$ and show that $C^*=C_*$.

My approach:

Say a relation $R$ is $i+1$-valued if $R\subseteq U^{i+1}$.

Say a subset $S$ of $U$ is closed under relations $R_i$'s in $\mathcal{F}$ iff for any $x_1,...,x_j\in S$ and $j+1$-valued $R_j\in \mathcal{F}$, if $x\in U$ such that $(x_1,...,x_j,x)\in R_j$, then $x\in S$.

Say that $S$ is $R$-inductive iff $B\subset S$ and $S$ is closed under $R_i$'s.

Let $C^*$ be the intersection of all the $R$-inductive subsets of $U$.

Now, I wonder whether my definition is correct or not... Also I want to show that $C_*$ is $R$-inductive so that $C^*\subseteq C_*$. Can any one help me on this please?

答:
I think my definition of $C^*$ is correct because I can show $C^*=C_*$ as follows:

First, to show $C^*\subseteq C_*$, we just need to show that $C_*$ is $R$-inductive. First, since $(x)$ with $x\in B$ is a construction sequence that ends with $x$, $B\subseteq C_*$. And suppose that $x_1,...,x_i\in C_*$, that $R_i\in\mathcal{F}$ is an $i+1$-valued relation, and that $x\in U$ such that $(x_1,...,x_i,x)\in R_i$. we can concatenate the construction sequences of $x_j$'s and $(x_1,...,x_i,x)$ to obtain the construction sequence that ends with $x$. So $x\in C_*$ showing $C_*$ is closed under $R_i$'s and hence is $R$-inductive.

To show $C_*\subseteq C^*$, take $x\in C_*$ and let $(x_1,...,x_n,x)$ be a construction sequence that ends with $x$. Note that $x_1\in B\subseteq C^*$. And by ordinary induction on $j$, we can see that $x_j\in C^*$ because of the fact that $C^*$ is closed under the $R_i$'s.

Title: Functional and relational inductive set


问:
Why do we name equations more in applied math?
In pure math, equations aren't often named. For instance, we might define that $f$ is a (real) polynomial function iff there exists a finite sequence of real numbers $a_i$ such that $$f(x)=a_0 + a_1 x + \cdots a_n x^n.$$

Notice, though, that we haven't defined an equation, but rather a kind of function.

Furthermore, we might define that a solution multiset of a polynomial function $f$ is a multiset of real numbers $x$ such that $f(x)=0$, with the additional requirement that the multiplicities are correct. Again, note that we've defined a way of getting a multiset from a function; technically, this is another function! Its certainly not an equation.

However, in applied math, I've noticed a lot more naming of equations / syntactical objects. Like, a text might define that a polynomial equation is an equation of the form $$a_0 + a_1 x + \cdots a_n x^n = 0.$$

Why is this the case?

答:
Applied mathematicians need more help remembering things
than pure mathematicians do.</p>


问:
Unsolved arithmetic exercise for printing numbers in book
Here is the exercise from the "Arithmetic" book (ISBN: 5-02-013764-2).

There are 1989 numbers had printed to enumerate book's pages. How many
  pages in book?

Honestly I decided to refresh my arithmetic knowledge at all but stuck with that problem.
Here's my decision:

there are 9 numbers to print for pages 1-9 (9 * 1);

there are 200 numbers to print for pages 10-99 (100 * 2);

1989 - 209 = 1780;

1780 /3 = 593,3(3) - that is the problem

I clearly understand how simple it must be and maybe something is wrong with my brain. But please, tell me where I made mistake.

答:
Can this formula be used to determine the following challenge

I have a book with xyz pages. The pages start at 1. The numbers are made from little stickon labels. Of which had only 1 digit on it. Eg page 30 would be 2 labels 3,0
The number of labels required was ZYX  how many pages in the book ?


问:
Finding an Idempotent Operator
Taken from Conway's A course in functional analysis chapter 2, section 3, problem 1:

Let $\mathcal{H}$ be the two-dimensional real Hilbert space $\mathbb{R}^2$, let $\mathcal{M} = \{(x, 0)\in \mathbb{R}^2: x \in \mathbb{R}\}$ and let $\mathcal{N} = \{(x, x\tan\theta): x \in \mathbb{R}\}$, where $0<\theta<\frac{1}{2}\pi$. find a formula for the idempotnent $E_{\theta}$ with Ran$E_\theta = \mathcal{M}$ and Ker$E_\theta = \mathcal{N}$. Show that $||E_\theta|| = (\sin(\theta)||^{-1}$

My attempt at a solution is below:

An idempotent operator $E_\theta$ is one such that $E_\theta^2 = E_\theta$. I was hoping to let $E(x, y) = (\sqrt{x^2+y^2}, 0)$. It is idempotent and  clearly satisfies the condition on the range, but not the kernel. The norm of this map is also $\frac{1}{\cos(\theta)}$ which is a little bit off the norm the operator is supposed to be.

Any tips would be greatly appreciated.

答:
That chapter in Conway is about linear operators. Obviously the question expects $E_\theta$ to be linear; otherwise the operator is not uniquely defined.

The set of  linear operators on $\mathbb R^2$ is usually identified with the $2\times 2$ matrices. So you want
$$
E_\theta=\begin{bmatrix} a&b\\ c&d\end{bmatrix}.
$$
To be in $\mathcal M$, we want the second component of $E_\theta \begin{bmatrix} x\\ y\end{bmatrix} $ to be zero. That gives us
$$
cx+dy=0,\qquad x,y\in\mathbb R.
$$
As this should hold for all $x,y$ we are forced to take $c=d=0$. We also want $E_\theta$ to be an idempotent, so we need
$$
\begin{bmatrix} a&b\\0&0\end{bmatrix}=E_\theta=E_\theta^2=\begin{bmatrix} a^2&ab\\0&0\end{bmatrix}.
$$
If $a=0$ we get $E_\theta=0$, which wouldn't have the right kernel nor image. So $a=1$.

With the range of $E_\theta$ and idempotency guaranteed, let's go for the kernel. We want
$$
\begin{bmatrix} 0\\0\end{bmatrix}=\begin{bmatrix} 1&b\\0&0\end{bmatrix}\begin{bmatrix} x\\ x\tan\theta\end{bmatrix}=\begin{bmatrix} x(1+b\tan\theta)\\0\end{bmatrix}.
$$
As this should hold for all $x$, this forces $b=-\frac1{\tan\theta}$. The norm of $E_\theta$ can be calculated by definition, but it is easier to do
$$
\|E_\theta\|^2=\|E_\theta E_\theta^*\|=\biggl\|\begin{bmatrix} 1&-\cot\theta\\0&0\end{bmatrix}\begin{bmatrix} 1&0\\-\cot\theta&0\end{bmatrix}\biggr\|
=\biggl\|\begin{bmatrix} 1+\cot^2\theta&0\\0&0\end{bmatrix}\biggr\|=\frac1{\sin^2\theta}.
$$

Title: Finding an Idempotent Operator


问:
Some questions related to the limit $\lim_{x \rightarrow \infty}(\lim_{y \rightarrow \infty} (\frac{x}{y}) )$
This is a relatively strange question, title says the original question. Intuition says this limit is 1, and graphing the function $f(x,y)=\frac{x}{y}$ this confirms it.
$$\lim_{x \rightarrow \infty}(\lim_{y \rightarrow \infty} (\frac{x}{y}) ) = 1$$

I do not know how, but online calculators disagree. Wolfram alpha widgets seems to think that this limit doesn't exist and symbolab somehow ended up with 0.

However this is where I don't understand certain things. I encountered this issue in the context of converting an infinite sum to an integral.

It is known that an infinite sum can be often rewritten as a Reimann sum in the following way:

$$\int_{a}^{b}f(x)dx=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(a+\frac{b-a}{n}k))\frac{b-a}{n}$$

Various numerical examples suggest that this is true. The next step is where I do not understand where I went wrong. It makes sense to think that if we are taking an integral from $0$ to $\infty$ we can write it in the following way:

$$\lim_{b \rightarrow \infty}\int_{0}^{b}f(x)dx=\lim_{b \rightarrow \infty}\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(\frac{b}{n}k))\frac{b}{n}$$

Since both limits go to infinity, the quantity of $\frac{b}{n}$ should go to one, as previously established. So I expected the following to hold:

$$\lim_{b \rightarrow \infty}\int_{0}^{b}f(x)dx=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(k))$$

However, this apparently isn't true, according to numerical calculations. Consider the following infinite sum, stolen shamelessly from someone else's question:

$$\sum_{k=1}^{\infty}(\frac{1}{5^k+2})$$

Letting $f(x)=\frac{1}{5^k+2}$ we have:

$$\sum_{k=1}^{\infty}(\frac{1}{5^k+2})=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(k))=\lim_{b \rightarrow \infty}\int_{0}^{b}f(x)dx=\lim_{b \rightarrow \infty}\int_{0}^{b}\frac{1}{5^x+2}dx$$
$$\sum_{k=1}^{\infty}(\frac{1}{5^k+2})=\int_{0}^{\infty}\frac{1}{5^x+2}dx$$

According to desmos, the RHS evaluates to ~0.3413 and the LHS evaluates to ~0.1898. So this begs the question of what I did wrong. For anyone interested, here are the images of $\int_{0}^{x}\frac{1}{5^{t}+2}dt$ and $\sum_{n=1}^{x}\left(\frac{1}{5^{n}+2}\right)$ in red and blue respectively, drawn by desmos.

答:
The problem you're running into is that limits are much more slippery in multiple variables. You have this two variable function $f(x,y)=x/y$. Somehow you want to ask "what does this function do as $x$ and $y$ both go to infinity?" The problem is that there are many ways to go off to infinity. For instance, we could march off along the line $y=x$, which seems to be what you've done. Restricting to this line, we have
$$\lim_{x \to \infty} f(x,x) = \lim_{x \to \infty} 1 = 1.$$
But you could've gone off along the line $y=2x$ and gotten a limit
$$\lim_{x \to \infty} f(x,2x) = \lim_{x \to \infty} 1/2 = 1/2.$$
Even worse, you could've gone off along a parabola $x=y^2$ and seen
$$\lim_{y \to \infty} f(y^2,y) = \lim_{y \to \infty} y = \infty.$$

Another approach is that the limit $\lim_{x \to \infty, y \to \infty} f(x,y)$ should be the same as $\lim_{(x,y) \to (0,0)} f(1/x,1/y)$. But in your case $f(1/x,1/y)= y/x$, and this function clearly doesn't have a limit at the origin (nor is it defined there).


问:
Question about orientations and loops of the $n-1$ faces of an n-simplex $\Delta^n$
$S$ is the $\Delta$-complex formed by considering $\Delta^n$ along with its faces. Let the vertices be $v_0, v_1, ... v_n$.

Then $C_n(S) = \langle[v_0 v_1..v_n]\rangle$
and $C_{n-1}(S) = \langle\{[v_0....\hat{v_i} ..v_n] : i \leq n \} \rangle$.

Now obviously $\partial_{n-1}(\Sigma_{i\leq n} (-1)^i[v_0..\hat{v_i}..v_n]) = 0$

Suppose I have $X \subseteq \partial\Delta^n$ a $\Delta$ complex. I want to show that if $H_{n-1}(X) \not = 0$, then $X = \partial \Delta^n$, but this amounts to showing that there can't be a cycle consisting of $< n+1$ many faces of the generators of $C_{n-1}(S)$. That is, I have to show that if $\partial_{n-1}(\Sigma_{i\leq m}N_i\times[v_0..\hat{v_i}..v_n]) = 0$, where $N_{i}$ are non-zero integers,then $m = n$. Thus $X$ contains all $n+1$ faces and then $X = \partial\Delta^n$.

This is intuitively pretty obvious, but I'm not sure how to rigorously show this. Hopefully this isn't too dumb of a question.

Is it obvious as saying that then $\Sigma_{i\leq m}N_i\times[v_0..\hat{v_i}..v_n]$ would then be in $\partial\Delta^n$'s $ker(\partial_{n-1})$, but then that woud be pushing back the question to showing that for $\partial\Delta^n$, $ker(\partial_{n-1}) = \langle\Sigma_{i\leq n}(-1)^i[v_0..\hat{v_i}..v_n]\rangle$, which, again is very obvious, but I'm blanking how to rigorously prove it.

Perhaps I'm missing something incredibly obvious however.

答:
The $(n-1)$-st homology of your set $X \subset \partial \Delta^n$ is the same as the $(n-1)$-st homology of the union of the (closed) $(n-1)$-dimensional faces of $X$. So we can assume that $X$ is just a union of $(n-1)$-dimensional faces. If $X \not= \partial \Delta^n$, then roughly $X \subset \partial \Delta^n$ looks like the complement of an $(n-1)$-ball in the $(n-1)$-sphere $S^{n-1}$, which is contractible.

Even if it's hard to see why $X$ looks like the complement of a ball, it's at least the complement of a finite set of open balls in $S^{n-1}$. And such a thing deformation retracts to an $(n-2)$-complex.


问:
Something is wrong with my proof by induction solution
I am trying to prove a formula by induction but I cant get it right. Something is wrong somewhere and I dont know where. Any help will be greatly appreciated.

We want to use induction to prove that:
$$\sum_{k=0}^n {n \choose k} = 2^n$$
Base Case 1: When n = 0
$$ {0 \choose 0} = 2^0 $$
Base Case 2: When n = 1
$${1 \choose 0} + { 1 \choose 1} = 2 ^ 1$$
Lets assume that the equation is true for $p < n$ where p$ >= 1$
Then the equation is true for $p - 1$
$$\sum_{k=0}^{p-1}{p-1 \choose k} = 2^{p-1}$$
$${p-1 \choose 0} + \sum_{k=1}^{p-1}{p-1 \choose k} = 2^{p-1}$$
$$\sum_{k=1}^{p-1}{p-1 \choose k} = 2^{p-1} - 1$$
Also
$$\sum_{{k - 1=0}}^{p-1}{p-1 \choose k-1} = 2^{p-1}$$
$$\sum_{k = 1}^{p-1}{p-1 \choose k-1} = 2^{p-1}$$
When we add these two equations we get:
$$\sum_{k=1}^{p-1}{p-1 \choose k} + \sum_{k = 1}^{p-1}{p-1 \choose k-1} = 2^{p-1} - 1 + 2^{p-1}$$
$$\sum_{k=1}^{p-1}{p-1 \choose k} + \sum_{k = 1}^{p-1}{p-1 \choose k-1} = 2^p - 1$$
$$\sum_{k=1}^{p-1}{({p-1 \choose k} + {p-1 \choose k-1})} = 2^p - 1$$
$$\sum_{k=1}^{p-1}{p \choose k} = 2^p - 1$$
This is where I get stuck. I can't convert it to
$$\sum_{k=0}^{p}{p \choose k} = 2^p $$

答:
$$\sum_{k = 1}^{p-1}{p-1 \choose k-1} = 2^{p-1}$$ is not correct. It should be $$\sum_{k = 1}^{p}{p-1 \choose k-1} = 2^{p-1}$$.

$ \sum\limits_{k=1}^{p-1} \binom {p} {k}=\binom {p} {1}+\binom {p} {2}+...+\binom {p} {p-1}$ and $ \sum\limits_{k=0}^{p} \binom {p} {k}=\binom {p} {0}+\binom {p} {1}+...+\binom {p} {p}$. What differences do you find between these two sums? If you remember that $\binom {p} {0}=\binom {p} {p } (=1)$ then you can complete your argument.


问:
How to find point of perpendicular bisector given equation of line and point?
It would be really awesome if you guys could give me a hand on this question!

Question: Given that the equation of a line is $y=-\sqrt{3}x$. This line can be also expressed in $|z-1|=|z-k|$, where $k \in\Bbb C$. Find $k$ in cartesian form.

What I've done so far:

Since the perpendicular gradient is $\frac{1}{\sqrt3}$, I made the equation $\frac{y-0}{x+1}=\frac{1}{\sqrt3}$. This lead me to no-where

I've also considered using $cis\frac{-2\pi}{3}$, since it's perpendicular, but I'm confused on what to do from here

The answer is $-\frac{1}{2} - \frac{\sqrt3}{2}i$

Please correct me if I am wrong.

答:
I think drawing out your problem will help you understand the equation you seek

Now, the line given is $y = -\sqrt3 x$. This line passes through the origin, and has a slope of $\tan \theta = -\sqrt3$

Now, the equation $|z-1| = |z-k|$ describes the line that is the perpendicular bisector of the points $(1,0)$ and $(k_x, k_y)$

What is the slope of the line segment joining these two points?

$$m  = \frac{y_2 - y_1}{x_2 - x_1} = \frac{k_y - 0}{k_x - 1}$$

Now, another equation we have for $m$ comes from the perpendicular

$$m\times(- \sqrt3) = -1$$

$$\implies m = \frac{1}{\sqrt3}$$

So now you have two equations to help find $k_x, k_y$ -

$$\sqrt3 (k_y) = k_x - 1$$

$$\frac{k_y}{2} = -\sqrt3 \left(\frac{k_x+1}{2}\right)$$


答:
The line is $y=-\sqrt3 x$,

so a line that passes through $(1,0)$ and is perpendicular to the line is $y=\dfrac x{\sqrt3}-\dfrac1{\sqrt3}.$

These lines intersect when $-\sqrt3 x=\dfrac x{\sqrt3}-\dfrac1{\sqrt3}$ ; i.e., $x=\dfrac14$ and $y=\dfrac{-\sqrt3}4$.

Now double the segment from $(1,0) $ to $(\frac14,\frac{-\sqrt3}4)$ to get $(\frac{-1}2,\frac{-\sqrt3}2).$

Title: How to find point of perpendicular bisector given equation of line and point?

Body:
It would be really awesome if you guys could give me a hand on this question!

Question: Given that the equation of a line is $y=-\sqrt{3}x$. This line can be also expressed in $|z-1|=|z-k|$, where $k \in\Bbb C$. Find $k$ in cartesian form.

What I've done so far:

Since the perpendicular gradient is $\frac{1}{\sqrt3}$, I made the equation $\frac{y-0}{x+1}=\frac{1}{\sqrt3}$. This lead me to no-where

I've also considered using $cis\frac{-2\pi}{3}$, since it's perpendicular, but I'm confused on what to do from here

The answer is $-\frac{1}{2} - \frac{\sqrt3}{2}i$

Please correct me if I am wrong.


问:
Prove that for any integer $n>1$, there exists a set of $n$ positive integers such that, for any two numbers among them , $a-b$ divides $a+b$

Prove that for any integer $n>1$, there exists a set of $n$ positive integers such that, for any two numbers among them (say $a$ and $b$), $a-b$ divides $a+b$

I have come up with three strategies to tackle this problem:

(i) Try to construct a set satisfying the conditions

(ii) Induction

(iii) Try to prove it by contradiction. (Which is, I think very difficult to do so)

I have tried smaller examples hoping to find a pattern. Tried arithmetic, geometric series, but no luck. It is very hard to  even come up with an example for $n=5$. We can make some simple observations like $(n,n+1)$ and $(n,n+2)$ always work. But the thing with this problem, that is making it difficult, is the rule should be followed by every two numbers in the set.

Induction definitely fails, fix any number $a$, then condition $a-x|a+x$ can also be written as $a-x|2a$. Which means there are only finitely many values of $x$ which satisfies the condition. So, we can't rely on induction

I am not sure, how can we go about using (ii)? Or is there any other method?

答:
It's too early to say that induction fails just because the simplest thing to try doesn't work: adding a new element to an existing set.

The next thing to try is: given an $n$-element set $S$ for which this property holds, can we turn it into a different $n$-element set for which it will be easier to add a new element?

I thought of two things that we can do to change $S$:

We can multiply every element of $S$ by a fixed factor $k$. This doesn't break divisibility, but also doesn't seem super useful.

We can add $k$ to every element of $S$, provided that $a-b \mid 2k$ for every pair $a,b \in S$. There are infinitely many such $k$; for instance, the LCM of the pairwise differences.

Okay, now if we can shift $S$ over, is there a convenient element we can try adding to $S$?

Adding the element $0$ always works: the divisibility condition is that $a - 0 \mid a+0$. Of course, $0$ is not a positive integer, but we can shift the elements over to make everything positive.

So now we have a method to increase the size of any set $S$ with this property by $1$:

Let $S' = S \cup \{0\}$.

Let $k$ be the LCM of all pairwise differences in $S'$, and let $S'' = S'+k$.

The result $S''$ after the second step is the bigger set we wanted.

In principle, once we have an inductive proof, we can try to understand the construction it gives, and simplify it to a direct argument. But in this case, starting with the $2$-element set $\{1,2\}$, I don't see a nice pattern for what happens later:

$\{1,2\}$ becomes $\{1,2,3\}$

$\{1,2,3\}$ becomes $\{6,7,8,9\}$

$\{6,7,8,9\}$ becomes $\{504, 510, 511, 512, 513\}$

$\{504, 510, 511, 512, 513\}$ becomes $\{11408463360, 11408463864, 11408463870, 11408463871, 11408463872, 11408463873\}$

I decided not to continue.


问:
Explanation for these transformations of integrals
I've recently found the following transformations:

$$\int _{a} ^{\infty} \frac{\ln x}{x^2 + a^2}\,dx = \int _{a} ^{0} \frac{\ln x}{x^2 + a^2}\,dx$$

$$\int _{0} ^{\pi/2} \log(\sin(\tfrac{\pi}{2}-u))\,du = \int _{0} ^{\pi/2}\log(\sin u)\, du = \int _{\pi/2} ^{\pi}\log(\sin u)\, du$$

Could you explain to me why that is?

答:
$(1)$

$$ - \int _a ^{0} \frac{\ln x}{x^2 + a^2}\,dx+\int _a ^{\infty} \frac{\ln x}{x^2 + a^2}\,dx$$

$$= \int _0^a \frac{\ln x}{x^2 + a^2}\,dx+\int _a ^{\infty} \frac{\ln x}{x^2 + a^2}\,dx$$

$$= \int _0^\infty \frac{\ln x}{x^2 + a^2}\,dx=I\text{(say,)}$$

Putting $x=a\tan \theta\implies dx=a\sec^2\theta d\theta$

When $x=0, \theta=0, x=\infty\implies \theta=\frac\pi2$

$$\text{So,}I=\int_0^\frac\pi2\frac{\ln(a\tan\theta)}{a^2 \sec^2\theta}a\sec^2\theta d\theta$$

$$\implies a\cdot I=\ln a\int_0^\frac\pi2d\theta+\int_0^\frac\pi2 \ln \tan\theta d\theta$$

Now, $\ln a\int_0^\frac\pi2d\theta=\ln a\left(\frac\pi2-0\right)$

$I_1=\int_0^\frac\pi2 \ln \tan\theta d\theta=\int_0^\frac\pi2 \ln \tan \left(\frac\pi2+0-\theta\right) d\theta $ as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

$I_1=\int_0^\frac\pi2 \ln \cot\theta d\theta$
$=\int_0^\frac\pi2 \ln (\tan\theta)^{-1} d\theta=-\int_0^\frac\pi2 \ln \tan\theta d\theta=-I_1\implies I_1=0$

$\implies a\cdot I=\frac\pi2\ln a\ne0$ unless $a=1$

$(2)$

$$\int _{0} ^{\pi/2} \log(\sin(\tfrac{\pi}{2}-u))\,du =\int _{0} ^{\pi/2} \log(\cos u)\,du $$

$$=\int _{0} ^{\pi/2} \log \cos \left(\frac\pi2+0-u\right)\,du  \text{ as } \int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

$$=\int _{0} ^{\pi/2} \log(\sin u)\,du $$

Put $v=\pi-u\implies \sin u=\sin(\pi-v)=\sin v$


问:
Question of the minimal sufficient statistics of beta-distribution

The beta distribution with parameters $\alpha>0$ and $\beta>0$ has
density$$f(x|\alpha,\beta)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},0<x<1$$
Suppose $X_1,...,X_n$ are independently and identically distributed
from a beta distribution.

Determine a minimal sufficient statistic (for the family of joint    distributions) if $\alpha$ and $\beta$ very freely

Determine a minimal sufficient statistic if $\alpha=2\beta$

Determine a minimal sufficient statistic if $\alpha=\beta^2$

The above is the setup of our homework question for the statistical inference course. For (a),

$$\prod f(y_i|\alpha,\beta)=\frac{1}{B^n(\alpha,\beta)}\exp\{(\alpha-1)\sum\log y_i+(\beta-1)\sum\log(1-y_i)\}$$

$$\frac{\prod f(x_i|\alpha,\beta)}{\prod f(y_i|\alpha,\beta)}=\exp\{(\alpha-1)\sum(\log x_i-\log y_i)+(\beta-1)\sum(\log(1-x_i)-\log(1-y_i))\}$$

If $\sum\log x_i=\sum\log y_i, \sum \log(1-x_i)=\sum\log(1-y_i)$, the ratio is constant, therefore by Lehmann-Scheffe theorem, $$\{T_1=\sum\log x_i,T_2=\sum\log(1-x_i)\}$$ is a minimal sufficient statistic for $(\alpha,\beta)$.

The above is how I approached this question. I first used the factorization criterion to find the sufficient statistics and prove it is also minimal sufficient. However, coming to the second part, I tried to approach the question the same way:

$$f(x|\beta)=\frac{\Gamma(3\beta)}{\Gamma(2\beta)\Gamma(\beta)}x^{2\beta-1}(1-x)^{\beta-1}$$

$$=\frac{\Gamma(3\beta)}{\Gamma(2\beta)\Gamma(\beta)}\frac{1}{x(1-x)}\exp\{2\beta\log x+\beta\log (1-x)\}$$

$$=\frac{\Gamma(3\beta)}{\Gamma(2\beta)\Gamma(\beta)}\frac{1}{x(1-x)}\exp\{\beta(2\log x+\log(1-x))\}$$

$$=\frac{\Gamma(3\beta)}{\Gamma(2\beta)\Gamma(\beta)}\frac{1}{x(1-x)}\exp\{\beta\log x^2(1-x)\}$$

$$T(X)=\sum\log x_i^2(1-x_i)=2T_1+T_2$$

However, what I have done so far, in my view, is merely showing that the statistic is sufficient but how to show it is also the minimal sufficient statistic I do not know. I tried to find the answer online and found This material from Purdue's website. The question looks exactly the same, and the procedure for the first two questions seem not so different from what I have done. I am perplexed about how come the answer directly jump to that the statistic is also minimal sufficient? Did I miss any theorem? Also does $T(X)=\sum\log x_i^2(1-x_i)=2T_1+T_2$ actually carry with some implications?

Also, when approaching this question I have a more general and vague question that I want to ask. It appears that other than making a exponential family in an exponential form, Fisher-Neyman Factorization Theorem is enough to establish the sufficient statistics, an it is usually in the product form: here is just an example for your reference . It always appear that the product can be transformed to summation through exponential and logarithm somehow. But when I studied the concepts in statistics I always struggled to make distinctions between the distribution of a statistic and the joint distribution of a set of random variables. I guess it is because most of the time I am dealing with normal distribution; therefore the summation of x and x-square just naturally appear in the joint distribution and I always thought when we take the product we obtain something connects with mean and variance. But these concepts turn out very confusing when I am dealing with other distributions.

For example, the beta distribution. So far from what I have learnt online, both $\prod x_i$,$\prod (1-x_i)$ and $\sum\log x_n , \sum \log (1-x_n)$ seem both valid. However, when considering them as statistics I feel very hard to understand what we are trying to estimate from these statistics. Also it appear to me they are estimating totally different thing, how come they are both efficient? If they are both efficient there must be a one-to-one correspondence if I am not mistaken. How should I link the concept of sufficient statistics with through estimates I obtained from LME and other estimation techniques?

I hope my question make sense. Thank you so much for your time and patience!

答:
$a)$ We have that $T(\mathbf{X})=\left(\sum_{i=1}^n\log(X_i),\sum_{i=1}^n\log(1-X_i)\right)$ is sufficient for the family $\{B(\alpha,\beta),\alpha>0,\beta>0\}$ since for all $\mathbf{x}\in\mathbb R^n$, $\alpha>0$, and $\beta>0$,

$$\begin{align*}
f(\mathbf{x}\mid\alpha,\beta)
&=\left[\frac{\Gamma\left(\alpha+\beta\right)}{\Gamma\left(\alpha\right)\left(\beta\right)}\right]^n\left(\prod_{i=1}^n x_i\right)^{\alpha-1}\left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}\prod_{i=1}^n\mathbb{1}_{[0,1]}(x_i)\\\\
&=\underbrace{\exp\left[\left(\alpha-1\right)\sum_{i=1}^n\log(x_i)+(\beta-1)\sum_{i=1}^n\log(1-x_i)+n\log\left(\frac{\Gamma\left(\alpha+\beta\right)}{\Gamma\left(\alpha\right)\left(\beta\right)}\right)\right]}_{g(T(\mathbf{x})\mid\alpha,\beta)}\underbrace{\mathbb{1}_{[0,1]}(x_{(1)})\mathbb{1}_{[0,1]}(x_{(n)})}_{h(\mathbf{x})}
\end{align*}$$

We also have that $T(\mathbf{X})=\left(\sum_{i=1}^n\log(X_i),\sum_{i=1}^n\log(1-X_i)\right)$ is a minimal sufficient statistic since for any two possible sample points $\mathbf{x}$ and $\mathbf{y}$

$$\begin{align*}
\frac{f\left(\mathbf{x}\mid\alpha,\beta\right)}{f\left(\mathbf{y}\mid\alpha,\beta\right)}
&=\frac{\exp\left[\left(\alpha-1\right)\sum_{i=1}^n\log(x_i)+(\beta-1)\sum_{i=1}^n\log(1-x_i)+n\log\left(\frac{\Gamma\left(\alpha+\beta\right)}{\Gamma\left(\alpha\right)\left(\beta\right)}\right)\right]}{\exp\left[\left(\alpha-1\right)\sum_{i=1}^n\log(y_i)+(\beta-1)\sum_{i=1}^n\log(1-y_i)+n\log\left(\frac{\Gamma\left(\alpha+\beta\right)}{\Gamma\left(\alpha\right)\left(\beta\right)}\right)\right]}\\\\
&=\exp\left[\left(\alpha-1\right)\left[\sum_{i=1}^n\log(x_i)-\sum_{i=1}^n\log(y_i)\right]+(\beta-1)\left[\sum_{i=1}^n\log(1-x_i)-\sum_{i=1}^n\log(1-y_i)\right]\right]
\end{align*}$$

does not depend on $\alpha$ or $\beta$ if and only if $T(\mathbf{x})=\left(\sum_{i=1}^n\log(x_i),\sum_{i=1}^n\log(1-x_i)\right)$ equals $T(\mathbf{y})=\left(\sum_{i=1}^n\log(y_i),\sum_{i=1}^n\log(1-y_i)\right)$

$b)$ We have that $T(\mathbf{X})=\sum_{i=1}^n\log\left(X_i^2(1-X_i)\right)$ is sufficient for the family $\{B(\alpha,\beta),\alpha=2\beta,\beta>0\}$ since for all $\mathbf{x}\in\mathbb R^n$ and $\beta>0$,

$$\begin{align*}
f(\mathbf{x}\mid\alpha=2\beta,\beta)
&=\left[\frac{\Gamma\left(2\beta+\beta\right)}{\Gamma\left(2\beta\right)\left(\beta\right)}\right]^n\left(\prod_{i=1}^n x_i\right)^{2\beta-1}\left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}\prod_{i=1}^n\mathbb{1}_{[0,1]}(x_i)\\\\
&=\exp\left[\left(2\beta-1\right)\sum_{i=1}^n\log(x_i)+(\beta-1)\sum_{i=1}^n\log(1-x_i)+n\log\left(\frac{\Gamma\left(3\beta\right)}{\Gamma\left(2\beta\right)\left(\beta\right)}\right)\right]\mathbb{1}_{[0,1]}(x_{(1)})\mathbb{1}_{[0,1]}(x_{(n)})\\\\
&=\exp\left[(\beta-1)\sum_{i=1}^n\log\left(x_i^2(1-x_i)\right)+\sum_{i=1}^n\log(x_i)+n\log\left(\frac{\Gamma\left(3\beta\right)}{\Gamma\left(2\beta\right)\left(\beta\right)}\right)\right]\mathbb{1}_{[0,1]}(x_{(1)})\mathbb{1}_{[0,1]}(x_{(n)})\\\\
&=\underbrace{\exp\left[(\beta-1)\sum_{i=1}^n\log\left(x_i^2(1-x_i)\right)+n\log\left(\frac{\Gamma\left(3\beta\right)}{\Gamma\left(2\beta\right)\left(\beta\right)}\right)\right]}_{g(T(\mathbf{x})\mid\alpha=2\beta,\beta)}\underbrace{\exp\left(\sum_{i=1}^n\log(x_i)\right)\mathbb{1}_{[0,1]}(x_{(1)})\mathbb{1}_{[0,1]}(x_{(n)})}_{h(\mathbf{x})}\\\\
\end{align*}$$

We also have that $T(\mathbf{X})=\sum_{i=1}^n\log\left(X_i^2(1-X_i)\right)$ is a minimal sufficient statistic since for any two possible sample points $\mathbf{x}$ and $\mathbf{y}$

$$\begin{align*}
\frac{f\left(\mathbf{x}\mid\alpha=2\beta,\beta\right)}{f\left(\mathbf{y}\mid\alpha=2\beta,\beta\right)}
&=\frac{\exp\left[\left(\beta-1\right)\sum_{i=1}^n\log\left(x_i^2(1-x_i)\right)+n\log\left(\frac{\Gamma\left(3\beta\right)}{\Gamma\left(2\beta\right)\left(\beta\right)}\right)\right]\exp\left(\sum_{i=1}^n\log(x_i)\right)}{\exp\left[\left(\beta-1\right)\sum_{i=1}^n\log\left(y_i^2(1-y_i)\right)+n\log\left(\frac{\Gamma\left(3\beta\right)}{\Gamma\left(2\beta\right)\left(\beta\right)}\right)\right]\exp\left(\sum_{i=1}^n\log(y_i)\right)}\\\\
&=\exp\left[\left(\beta-1\right)\left[\sum_{i=1}^n\log\left(x_i^2(1-x_i)\right)-\sum_{i=1}^n\log\left(y_i^2(1-y_i)\right)\right]+\left[\sum_{i=1}^n\log(x_i)-\sum_{i=1}^n\log(y_i)\right]\right]
\end{align*}$$

does not depend on $\beta$ if and only if $T(\mathbf{x})=\sum_{i=1}^n\log\left(x_i^2(1-x_i)\right)$ equals $T(\mathbf{y})=\sum_{i=1}^n\log\left(y_i^2(1-y_i)\right)$

$c)$ We have that $T(\mathbf{X})=\left(\sum_{i=1}^n\log(X_i),\sum_{i=1}^n\log(1-X_i)\right)$ is sufficient for the family $\{B(\alpha,\beta),\alpha=\beta^2,\beta>0\}$ since for all $\mathbf{x}\in\mathbb R^n$ and $\beta>0$,

$$\begin{align*}
f(\mathbf{x}\mid\alpha=\beta^2,\beta)
&=\left[\frac{\Gamma\left(\beta^2+\beta\right)}{\Gamma\left(\beta^2\right)\left(\beta\right)}\right]^n\left(\prod_{i=1}^n x_i\right)^{\beta^2-1}\left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}\prod_{i=1}^n\mathbb{1}_{[0,1]}(x_i)\\\\
&=\underbrace{\exp\left[\left(\beta^2-1\right)\sum_{i=1}^n\log(x_i)+(\beta-1)\sum_{i=1}^n\log(1-x_i)+n\log\left(\frac{\Gamma\left(\beta^2+\beta\right)}{\Gamma\left(\beta^2\right)\left(\beta\right)}\right)\right]}_{g(T(\mathbf{x})\mid\alpha=\beta^2,\beta)}\underbrace{\mathbb{1}_{[0,1]}(x_{(1)})\mathbb{1}_{[0,1]}(x_{(n)})}_{h(\mathbf{x})}
\end{align*}$$

We also have that $T(\mathbf{X})=\left(\sum_{i=1}^n\log(X_i),\sum_{i=1}^n\log(1-X_i)\right)$ is a minimal sufficient statistic since for any two possible sample points $\mathbf{x}$ and $\mathbf{y}$

$$\begin{align*}
\frac{f\left(\mathbf{x}\mid\alpha=\beta^2,\beta\right)}{f\left(\mathbf{y}\mid\alpha=\beta^2,\beta\right)}
&=\frac{\exp\left[\left(\beta^2-1\right)\sum_{i=1}^n\log(x_i)+(\beta-1)\sum_{i=1}^n\log(1-x_i)+n\log\left(\frac{\Gamma\left(\beta^2+\beta\right)}{\Gamma\left(\beta^2\right)\left(\beta\right)}\right)\right]}{\exp\left[\left(\beta^2-1\right)\sum_{i=1}^n\log(y_i)+(\beta-1)\sum_{i=1}^n\log(1-y_i)+n\log\left(\frac{\Gamma\left(\beta^2+\beta\right)}{\Gamma\left(\beta^2\right)\left(\beta\right)}\right)\right]}\\\\
&=\exp\left[\left(\beta^2-1\right)\left[\sum_{i=1}^n\log(x_i)-\sum_{i=1}^n\log(y_i)\right]+(\beta-1)\left[\sum_{i=1}^n\log(1-x_i)-\sum_{i=1}^n\log(1-y_i)\right]\right]
\end{align*}$$

does not depend on $\beta$ if and only if $T(\mathbf{x})=\left(\sum_{i=1}^n\log(x_i),\sum_{i=1}^n\log(1-x_i)\right)$ equals $T(\mathbf{y})=\left(\sum_{i=1}^n\log(y_i),\sum_{i=1}^n\log(1-y_i)\right)$


问:
There are 'n' candies and 't' boxes. Find the number of ways to place the candies in the boxes for each of the conditions (given in the problem).
There are 'n' candies and 't' boxes. Find the number of ways to place the candies in the boxes for each of the conditions (all the candies must be spread out) :

(a) candies and boxes are different;

(b) candies of the same diiferent boxes should not be empty cartons:

(c) candies equally new, boxes are different;

Edit :(d) candies are different, the boxes are the same, there should be no empty boxes;

(e) candies of different boxes alike.(Edit : candies are different, boxes are equal)

Specify the display type that matches the placement, if possible.

My Answers :

(a) Each layout is encoded with a word of $n$ letters from the alphabet of 't' letters $\implies$ possible $n^t$ variants.

(b) Write, $n$ candies in the form of balls ina line, we need to put $(t-1)$ partition in $(n-1)$ place, but we can't put two partitions on one place, so we get : $^{t-1}C_{n-1}$.

(c) First we need to choose a candy in a box (in the first box $n$ ways, in the second box $(n-1)$, $\cdots$ in the $t-th$ : $(n-t+1)$ ways $\implies$ total $n!/(n-t)!$ methods,

and then we distribute the remaining $(n-t)$ candies into $t$ boxes, this is encoded with owrds from the alphabets (for each candy $t$ variants) $\implies$ $t^{n-t}$.

The Answer is : $\frac {n!}{(n-t)!}*t^(n-t)$

Are my answers (a), (b) & (c) are correct?

For (d) & (e) I don't know how to proceed? Please help me.

答:
I'm re-writing your conditions slightly to be clear about each case.

(a) $n$ different candies into $t$ different boxes:

Since each candy can be placed into one of $t$ different boxes, the number of ways will be

$$t \cdot t \cdot t\cdot t \cdots t = t^n$$

(b) $n$ identical candies into $t$ different boxes, no box to remain empty:

Your solution is correct.

(c) $n$ identical candies into $t$ different boxes, some boxes may remain empty:

[I'm taking "candies equally new" to mean they're identical; if that's not the case please specify what "equally new" means.]

Since the candies are identical, there's no need to choose a candy for each box as you're doing in the early part of your solution. The simplest way to do this is similar to your solution for (b), except now some boxes may remain empty; thus we need the number of arrangements of $(t-1)$ partitions and $n$ candies, i.e.

$$\frac{(n+t-1)!}{n!(t-1)!} = \binom{n+t-1}{n} \quad \text{or} \quad \binom{n+t-1}{t-1} $$

(d) & (e) $n$ different candies into $t$ identical boxes

The solution(s) here are known as Stirling Numbers of the Second Kind which have no closed formula but do satisfy a recurrence relationship.

For (d) if all the boxes are to be unempty, then the number of ways is $$S(n,t)$$ For (e) if some boxes may remain empty, then the number of ways is $$S(n,1) + S(n,2) + S(n,3) + \cdot + S(n,t) = \sum_{k=1}^t S(n,k)$$.


答:
Here is an intuitive way to solve the problem: $|z-1|=|z-k|$ is the set of all points equidistant from $(1,0)$ and $k$. This is the perpendicular bisector of the line segment joining $(1,0)$ and $k$. Thus the point $k$ is the reflection of $(1,0)$ with respect to the perpedicular bisector $y=-x\sqrt3$.

Say $k=(a,b)$, then mid-point of $k$ and $(1,0)$ lies on bisector, i.e. $\frac b2=-\sqrt3(1+a)/2\implies b=-(1+a)\sqrt3$  .

Further the slope of segment joining $k$ and $(1,0)$ is $1/\sqrt3$, so $\frac{b}{a-1}=1/\sqrt 3$. Solving these two equations gives $a=-1/2,b=-\sqrt3/2$.


问:
Calculate $\sum_{n\geq 1} \frac{1}{5^{n}+2}$

Calculate $$\sum_{n\geq 1} \frac{1}{5^{n}+2}$$

My approach: I know by the ratio test that the series converges.

However, how can I calculate the sum?

Body:
Without special functions, we can make approximations. Let us write the infinite summation as
$$S_p=\sum_{n=1}^p \frac 1{5^n+2}+\sum_{n=p+1}^\infty \frac 1{5^n}\frac1{1+2\times 5^{-n}}$$ and write
$$\frac1{1+2\times 5^{-n}}=\sum_{k=0}^\infty (-1)^k \,2^k \times 5^{-nk}$$ Summing from $k=0$ to $k=6$ for example
$$\left(
\begin{array}{cc}
p & S_p \\
 1 & \color{red}{0.18976287}954938663595977935271628162746328531817419 \\
 2 & \color{red}{0.18976287877266}367299681638975331866450032235521123 \\
 3 & \color{red}{0.1897628787726615593}3180851573757063300425936308525 \\
 4 & \color{red}{0.1897628787726615593263284}9366930907000585425941698 \\
 5 & \color{red}{0.1897628787726615593263284796045}6167280726135887333 \\
 6 & \color{red}{0.189762878772661559326328479604525648}62133819307056 \\
 7 & \color{red}{0.189762878772661559326328479604525648529106}83382480 \\
 8 & \color{red}{0.189762878772661559326328479604525648529106597707}69 \\
 9 & \color{red}{0.18976287877266155932632847960452564852910659770708}
\end{array}
\right)$$


问:
Determine the $n$-th power of matrix
Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1  \\ }$$ May you help me with the answer since the book states :  $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n &  3\cdot4^n & 2+4^n  \\ }$$ while mine misses the ${1\over 6}$. So who is right?

$$$$Thanks in advance for any help you are able to provide.

Title: Determine the $n$-th power of matrix

Body: I think about  $$\pmatrix{\left(2^{2n-1}-\frac{2}{3}(4^n-1)\right) & 2^{2n-1} & \frac{2}{3}(4^n-1) \\ \left(2^{2n-1}-\frac{2}{3}(4^n-1)+1\right) & 2^{2n-1} & \frac{2}{3}(4^n-1)+1 \\\left(2^{2n-1}-\frac{2}{3}(4^n-1)+1\right) & 2^{2n-1} & \frac{2}{3}(4^n-1)+1  \\ }$$


问:
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$
Question

Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.

My work

Let $k=x^2+y^2$

Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ that I can maximize?

答:
$x^2 + y^2$ is essentially the square of the distance of a point $(x, y)$ from the origin. Now, the question is asking us to find two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ which lie on the circle and are at a minimum and maximum distance from the origin.

It makes sense that these two points will lie on the line joining the center of the circle and the origin. I won't be proving this here, but it should be pretty obvious upon drawing a rough diagram.

For the given circle

$$
x^2 - 4x + y^2 + 3 = 0
$$

the center is $(2,0)$. Hence, we draw a line passing through this point and the origin (the red line in the below diagram).

The red line is

$$
y = 0
$$

Now we find the point of intersection between this line and the circle.

$$
x^2 - 4x + 3 = 0 \\
(x - 3)(x - 1) = 0 \\
x = 1, 3
$$

Since $y = 0$, the maximum and minimum value of $x^2 + y^2$ is $1$ and $9$.


问:
How to show $f(x) \leq (1-t)f(x)+tg(x) \leq g(x)$
Given $f(x) < g(x)$ and $f(x), g(x), t \in[0,1]$, how can I show $f(x) \leq (1-t)f(x)+tg(x) \leq g(x)$?

It is easy to show $f(x) \leq (1-t)f(x)+tg(x)$. Indeed, $(1-t)f(x)+tg(x) = f(x)-tf(x)+tg(x) = f(x)+t(g(x)-f(x))$. Given that $f(x) \leq g(x)$ it follows $f(x) \leq (1-t)f(x) +g(x).$

However, I am having difficulties to show $(1-t)f(x)+tg(x) \leq g(x)$. One argument would be to show at $t=0$ we have $(1-t)f(x) \leq g(x)$ and $t=1$ we will get $g(x) \leq g(x)$. But I am not sure if this is good enough since we need to check for $t \in (0,1)$.

I would like an answer that does not depend on the specific choices of $t$. When I showed $f(x) \leq (1-t)f(x)+tg(x)$, the argument did not depend on the choice of $t$ and some sort of squeeze theorem.

答:
You could also just let $k=1-t$. Then $k \in [0,1]$ and

$$\begin{align}
(1-t)f(x)+tg(x) &= k f(x) + (1-k) g(x) \\
&= g(x) + k \left[ f(x) - g(x) \right] \\
&\le g(x)
\end{align}$$


问:
How do you break two absolute values inside double integral?
How would I split the two absolute values of this double integral?

$\int_{-1}^1\int_{-1}^1(|x|+|y|)dxdy$

The answer key shows the integral = 2 with work:

$\int_{-1}^1(|x|+|y|)dx$

split into

$\int_{-1}^0((-x)-y)dx$ + $\int_{0}^1((x)-y)dx$

but why does y become -y instead of staying |y| inside the dx integral? Could I not carry |y| over and split it inside the dy integral?

答:
Note that
\begin{eqnarray*}
\int_{-1}^{1}\int_{-1}^{1} |x|+|y|dxdy&=&\int_{-1}^{1} \left(\int_{-1}^{1}|x|dx+\int_{-1}^{1}|y|dx \right)dy\\
&=&\int_{-1}^{1} \left(2\int_{0}^{1}|x|dx+\int_{-1}^{1}|y|dx \right)dy\\
&=&\int_{-1}^{1}1+2|y|dy\\
&=&\boxed{4}
\end{eqnarray*}

Explication:

We need to calculate $$\int_{-1}^{1}1+2|y|dy$$
so we can see by linearity,
$$\int_{-1}^{1}1+2|y|dy=\int_{-1}^{1}1dy+\int_{-1}^{1}|y|dy$$
Now, we can see that $|y|$ is an even function and interval $[-1,1]$ ($f$ is even par function if $f(-x)=f(x)$ for all $x$) and is symmetric about $0$, so $$\int_{-1}^{1}|y|dy=2\int_{0}^{1}|y|dy$$ and since that $y$ is always positive when $0<x<1$, so we can to make $|y|=y$, and also $$2\int_{0}^{1}|y|dy=2\int_{0}^{1}ydy=2\left.\left(\frac{y^{2}}{2} \right)\right|_{0}^{1}=2$$so, we can see that $$\int_{-1}^{1}1+2|y|dy=2+\int_{-1}^{1}dy=4$$

Similarly we can see that $$\int_{-1}^{1}|x|+|y|dx=1+2|y|$$


问:
What is the meaning of an "irreducible representation"?
What does it mean to talk about the "irreducible representatives of SO(3)"?  I'm struggling to understand the concept of irreducible representations.  Could someone give a concrete example for someone unfamiliar with group theory or representation theory?</p>

答:
if $\rho:G\to{\rm GL}_n(\mathbb{C})$ is a finite dimensional complex representation of the group $G$, then $\rho$ is irreducible if whenever $\rho(G)(V)\subseteq V$ then $V=\mathbb{C}^n \text{ or } 0$ (there are no non-trivial subspaces of $\mathbb{C}^n$ left invariant by $\rho(G)$).

here's a "visual" example.  the real representation of ${\rm SO}_3(\mathbb{R})$ (the elements acting on $\mathbb{R}^3$ by rotations about the origin) is irreducible (there is no line or plane in $\mathbb{R}^3$ that is invariant under all rotations).


答:
The area of integration is a square of side length $2$ centred at the origin and with sides parallel to the axes. $$|x|+|y|=\begin{cases}x+y,&0\le x,y\le1\text{ (Ist quadrant)}\\-x+y,&0\le -x,y\le1\text{ (IInd quadrant)}\\-x-y,&0\le-x,-y\le1\text{ (IIIrd quadrant)}\\x-y,&0\le x,-y\le1\text{ (IVth quadrant)}\end{cases}$$and hence the integral is split into four integrals, one over each quadrant.

Note that what you propose is also correct. You can chose to retain $|y|$ as is and split it up in the $dy$ integral. $$\int_{-1}^1(|x|+|y|)dx=2|y|+\int_{-1}^1|x|dx=2|y|+1$$so$$I=\int_{-1}^1(2|y|+1)dy=4$$

Title: How do you break two absolute values inside double integral?

Body:
How would I split the two absolute values of this double integral?

$\int_{-1}^1\int_{-1}^1(|x|+|y|)dxdy$

The answer key shows the integral = 2 with work:

$\int_{-1}^1(|x|+|y|)dx$

split into

$\int_{-1}^0((-x)-y)dx$ + $\int_{0}^1((x)-y)dx$

but why does y become -y instead of staying |y| inside the dx integral? Could I not carry |y| over and split it inside the dy integral?


Body:
You could use the fact that
$$ \int |x|\,dx=\frac{1}{2}x|x|+c $$

Then

\begin{eqnarray}
\int_{-1}^{1}\int_{-1}^{1} |x|+|y|\,dx\,dy&=&\int_{-1}^{1}\left[\frac{1}{2}x|x|+x|y|\right]_{-1}^1\,dx\,dy\\
&=&\int_{-1}^{1}1+2|y|\,dy\\
&=&\left[\,y+y|y|\,\right]_{-1}^1\\
&=&4
\end{eqnarray}


答:
You can solve this in a purely algebraic way as follows:

$z=0$ lies on the line, hence

$$|0-1| = |0-k| \Rightarrow |k|=1$$

$z = 1-\sqrt 3 i$ lies also on the line. Hence, plugging this into the equation $|z-1| = |z-k|$ gives

$$\sqrt 3 = |1-\sqrt 3 i - k|$$

Squaring and setting $k = \cos t + i \sin t$ gives

$$3 = (1-\cos t)^2 + (-\sqrt 3 - \sin t)^2$$

Expanding and rearranging gives

$$-1 = -\cos t + \sqrt 3 \sin t = 2\cos \left(t-\frac{2\pi}{3}\right)$$

The solutions for $t \in [0,2\pi)$ are $t=0$ and $t=\frac{4\pi}{3}$. Hence, the looked for $k$ is

$$k = \cos\left(\frac{4\pi}{3}\right) + i \sin \left(\frac{4\pi}{3}\right) = -\frac 12 - \frac{\sqrt 3}{2}i$$


问:
Probability of 5 of them having the same birthday while the other 10 having different birthday.
There are 15 students. Assume that there are 365 days in a year, what is the probability that 5 of them have the same birthday on any day of a year (the other 10 students have different birthday)?

$$
P\left(1^{st\ }in\ 5\ to\ have\ same\ birthday\right)=\frac{365}{365}=1
$$
The rest of the 4 students then need to have same birthday as the first student, making probability for each of them, $\frac{1}{365}$.
$$
P=\frac{365}{365}\times\frac{1}{365}\times\frac{1}{365}\times \frac{1}{365}\times\frac{1}{365}\times\frac{364}{365}\times \frac{363}{365}\times\frac{362}{365}\times\frac{361}{365}\times\frac{360}{365}\times\frac{359}{365}\times\frac{358}{365}\times\frac{357}{365}\times\frac{356}{365}\times\frac{355}{365} = 4.839 \times 10^{-11}
$$

Initially, I am using this way to figure out the probability and I am exploring another way.

$$356^{15} = 2.7189 \times 10^{38} $$

This shows the 365 possibilities for their birthday.
There are 11 ways for the birthdays to be selected which decreases after each birthday being selected.

$$
N_{1} = 365 \times 364 \times363\times362\times361\times360\times359\times358\times357\times356\times355 = 1.3157 \times10^{28}
$$
$$
N_2 = {{15}\choose{11}} = 1365
$$
$N_2$ is that there are 11 possible combinations in the 15 students.

$$
P = \frac{{N_1}{N_2}}{D} = 6.605 \times 10^{-8}
$$

Can someone please explain why is the right way of doing it and why one of them is wrong?

答:
Your first solution is almost correct; you just have to include the choice of which 5 of the 15 students would share the same birthday; i.e. the answer should be

$$\left[ 4.389 \times 10^{-11} \right] \cdot \binom{15}{5} \approx 1.453 \times 10^{-7}$$

The second solution doesn't work because in your computation $365 \times 364 \times  \cdots \times 355$, you're assigning an unnecessary ordering to the 10 dates chosen for the people with different birthdays. Also, the second step of calculating $\binom{15}{11}$ doesn't work - which 11 people are being chosen?

Note a simple method to count the number of ways is to break the process into smaller steps:

Step (1): Choose the birth date that five people will share - this can be done in $365$ ways.

Step (2): Choose which five people will share the above date - this can be done in $\binom{15}{5}$ ways.

Step (3): Choose the birth dates for the 10 remaining people - this can be done in $\binom{364}{10}$ ways.

Step (4): Assign the 10 birth dates to the 10 remaining people - this can be done in $10!$ ways.

Thus the required probability is:

$$\frac{365 \binom{15}{5} \binom{364}{10} 10!}{365^{15}} \approx 1.453 \times 10^{-7}$$


答:
In most cases with some function which changes definition as we vary the domain(*), we integrate it by splitting the domain into fragments where a certain definition of the function pertains and integrate over those partitions.

You could actually carry the |y| and split it after that, I demonstrate as follows:

$$ \int _{-1}^1 \int_{-1}^1 |x| dx dy + \int _{-1}^1 \int_{-1}^1 |y| dx dy$$

Now actually, there is a trick, both the integrals are exactly the same but with different variables, so we actually only have to think of one and twice the answer to it. Let's take twice the one on the left,

$$ 2 \int _{-1}^1 \int_{-1}^1 |x| dx dy $$

Now, The easy part is to integrate with $y$ first:

$$ 2 \int_{-1}^1 \int_{-1}^1 |x| dy dx$$

Since the bounds aren't dependent on $x$ , i.e: we are integrating over a rectangle, $|x|$ is simply a constant and hence:

$$ 2 \int_{-1}^1 [y]_{-1}^1 |x| dx$$

Or,

$$ 4 \int_{-1}^1 |x| dx$$

Now this is integral is where the first part of my answer comes in:

$$ 4 [ \int_0^1 |x| dx +  \int_{-1}^0 |x| dx]$$

Now, for the first integral, we can simply remove that modulus, and the second we negate because the domain is negative:

$$ 4 [ \int_0^1 x dx + \int_{-1}^0 -x dx]$$

Or,

$$ 4 [ [\frac{x^2}{2}]_{0}^1  - [\frac{x^2}{2}]_{-1}^0]$$

This evaluates to :

$$ 4$$

*:Usually, functions are consistent rules like $x^3$ where we cube the quantity given or more generally do some algorithm of calculations on the number to get our answer.

However, there is certain function who has their rule/ algorithm to evaluate changing as we move across domains. For example, the modulus function has a rule that we should take negative for $x<0$ and positive for $x>0$. Ultimately these two rules form one rule which has two specific cases.

Anyways the point I'm trying to convey is that split the domain such that we can easily evaluate the quantity for each subrule


问:
Order type of strings under lexicographical order
Let $A$ be a finite, totally-ordered alphabet (set of characters), and let $S$ denote the set of all strings (of any length, possibly infinite) over this alphabet, ordered lexicographically. Lexicographical ordering is defined as follows: for any two strings $s_1,s_2 \in S$, $s_1 < s_2$ iff the first character in $s_1$ that differs from the corresponding character in $s_2$ is the smaller of the two. (The edge case in which one of the strings is a proper prefix of the other is handled by thinking of strings as being terminated by a special character that is smaller than all characters in $A$).

Let $\alpha$ denote the order type of $[|A|] \stackrel{def}{=} \{1, \ldots, |A|\}$, so that $\alpha^\mathbb{N}$ denotes the product $\prod_{i = 1}^{\infty} \alpha$ (i.e. the order type of $[|A|]^{\mathbb{N}}$, the set of all infinite sequences in $[|A|]$, where for any $(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}} \in [|A|]^{\mathbb{N}}$ we have $(x_n)_{n \in \mathbb{N}} < (y_n)_{n \in \mathbb{N}}$ iff they're not equal and the elements at the last differing position $i$ satisfy $x_i < y_i$).

Would I be right in saying that the order type of $S$ is $\alpha^\mathbb{N}$? Is there any standard notation for this kind of order type? Also, what would happen if $A$ were not finite, but say infinitely countable, would the order be $\omega^\mathbb{N}$ by extension?

答:
Start with $[0,1]$, and split each dyadic rational in $(0,1)$ into two adjacent points; call the resulting linear order $X$.

More technically, let $D$ be the set of dyadic rationals in $(0,1)$. Let $Y=[0,1]\times\{0,1\}$, and let $X=\{\langle x,i\rangle\in Y:i=0\text{ or }x\in D\}$, ordered lexicographically.

When $|A|=2$, the set of infinite sequences over $A$ ordered lexicographically is order-isomorphic to $X$: essentially we’re just looking at the binary expansions of numbers in $[0,1]$ but not identifying the two expansions for dyadic rationals. Thus, $\langle 0,1,1,1,\ldots\rangle$ immediately precedes $\langle 1,0,0,0,\ldots\rangle$. As long as $2\le|A|<\omega$, the set of infinite sequences over $A$ ordered lexicographically will be order-isomorphic to a set like $X$, but built from a countable dense subset of $(0,1)$ different from $D$.

The lexicographic order on $S$ is rather more complicated. In the example above, for instance, the immediate successor of $\langle 0,1,1,1,\ldots\rangle$ is now $\langle 1\rangle$, which is immediately followed by $\langle 1,0\rangle$, and that by $\langle 1,0,0\rangle$, and so on, so that a whole sequence converging monotonically to $\langle 1,0,0,0,\ldots\rangle$ is inserted between $\langle 0,1,1,1,\ldots\rangle$ and $\langle 1,0,0,0,\ldots\rangle$.

More generally, let $\langle d_0,d_1,\ldots,d_n\rangle$ be a finite sequence such that $d_n\ne 0$; then $\langle d_0,d_1,\ldots,d_n\rangle$ is the immediate successor of $\langle d_0,d_1,\ldots,d_n-1,a,a,a,\ldots\rangle$, where $a=\max A$. It is immediately followed by $\langle d_0,d_1,\ldots,d_n,0\rangle$, which is immediately followed by $\langle d_0,d_1,\ldots,d_n,0,0\rangle$, and so on. Finally, the order has an initial sequence of order type $\omega$, namely, $\langle 0\rangle$, $\langle 0,0\rangle$, $\langle 0,0,0\rangle$, and so on, converging up to the infinite zero sequence.


问:
Smooth path definition
I keep getting confused with the definition of a smooth path. Here is a definition from William T. Shaw's Complex Analysis with Mathematica:

A path $\phi$ is a continuous mapping from a segment of the real axis into the complex numbers; i.e. $\phi:[a,b]\rightarrow C$.

A path $\phi$ is smooth if it is a differentiable path, and furthermore, the derivative map $\phi':[a,b]\rightarrow C$ is continuous.

OK, now here is my path: $\phi:[-2,2]\rightarrow C$ by $\phi(t) = t^2 + i t^3$. Now, I believe it is differentiable:

$$\phi'(t)=2t+i 3t^2$$

And I believe that $\phi'$ is continuous on $[-2,2]$. However, here is the image of the path:

See the sharp cusp at (0,0)? This is a smooth path?

I am obviously missing a subtle point.

D.

答:
In short, a smooth curve $\vec{c}: [a,b]\rightarrow C$ must:

map distinct points on $(a,b)$ to distinct values

have a $\vec{c}'(t)$ that exists, is continuous, and $\neq\vec{0}$ at any point in $(a,b)$

So no, the example you have given is not smooth, because at $(0,0)$, $\vec{c}'(t)=\vec{0}$ so that it can turn around, which violates condition 2.

(I was going to leave this as a comment for posterity's sake, but unfortunately do not have the reputation to :/)

Title: Smooth path definition


问:
Symmetry group 3 and writing elements in terms of each other
In the group $S_3$ and let $a,b$ be distinct elements of order two. Express each non-identity element in $S_3$ as a word in $a,b$.

I am simply confused by what this means. I know $S_3 = \{e, (1\; 2), (1\; 3), (2\; 3), (1\; 2\; 3), (1 \;3 \;2)\}$, but I am struggling with how to express the other terms as "words."

答: By word, they just mean a string (ie a product) of $a$ and $b$. As $a$ and $b$ are the elements of order 2, you can just arbitrarily say that $a$ is (12) and $b$ is (23). Now write the other elements as products of $a$ and $b$. For example what element is $ab$?


问:
The correspondence of topologies of different definitions of local fields?
I'm recently looking about the classification theorem of local fields, starting from the definitions.
One of the common definition is locally compact Hausdorff non-discrete topological field. The other is locally compact field completed with a non-trivial absolute value.

Given a field in the first definition, it's common to define a Haar measure on it which turns out inducing an absolute value in the second definition. Reversely any aboslute value induces a metric topology on the field.

My question is, are they inverse to each other? If so, any such topological field has an unique topology.

答:
I dive into Weil's book ''Basic Number Theory'' and get the answer, which is yes.
The following is an outline of the proof.

We will show one direction here ''the metric topology induced by the Haar measure agrees with the original topology''.  The other direction can be shown by first classifying all the local fields and checking them explicitly.

Given a locally compact Hausdorff non-discrete topological field $(K,\tau)$, there exists a Haar measure $\mu$ which has value $1$ on some compact neighborhood $X$ of $0$ containing a open neighborhood $U$ of $0$, and induces an absolute value $\alpha:K\to \mathbb{R}_{\geq 0}$ s.t. $\alpha (s):=\frac{\mu(sX)}{\mu(X)}$ (true for any non-null measurable set $X$) and $\alpha(0):=0$.

It's easy to see $\alpha(ab)=\alpha(a)\alpha(b)$ from the left translation invariant property of Haar measure. The topology $\tau'$ on $K$ induced by $\alpha$ is generated by the open balls $B(x,r)$ with center $x$ and radius $r$. So it suffices to show that the open balls generates $\tau$ as well. By homogeneous property of topological field, it suffices to show $B(0,r)$s form a neighborhood basis of $0$ in $\tau$, i.e. $\forall r>0,B(0,r)\in \tau$ and $\forall U\in \tau$ s.t. $0\in U$,$\exists r>0$ s.t. $B(0,r)\subset U$.

Part I: $\alpha$ is continuous.

So we want to show $\forall \epsilon>0,\forall a\in K,\exists U\in \tau\text{ s.t. }a\in U,\forall x\in U,-\epsilon\leq \alpha(x)-\alpha(a)\leq \epsilon$. If so then $B(0,r)=\alpha^{-1}([0,r))$ is automatically open.

Note that $\alpha$ is continous if and only if $\alpha$ is upper semicontinuous and lower semicontinuous at all $a\in K$.  We will show that $\alpha$ is continuous at $0$ and upper semicontinuous on the open subset $K^\times$ (every singleton is closed by Hausdorff).

Claim: If $\alpha$ is upper semicontinous on $K^\times$, then it is continuous on $K^\times$.

Proof: As $K^\times$ is open, $\alpha$ is uppser semicontinuous on $K^\times$ iff $\alpha|_{K^\times}$ is upper semicontinuous iff $\alpha^{-1}:K^\times \to \mathbb{R}_{>0}$ is lower semicontinous. Fix $\epsilon>0$ and $a\in K^\times$, $\exists U,V\in \tau$ s.t. $a\in U,a^{-1}\in V$, $\forall x\in U$ we have
$$\alpha(x)\leq \alpha(a)+\epsilon$$
and $\forall x\in V^{-1}$ we have
$$\alpha(a^{-1})^{-1}-\epsilon\leq \alpha(x^{-1})^{-1}$$
so $\alpha(a)-\epsilon\leq \alpha(x)\leq \alpha(a)+\epsilon$ for all $x\in U\cap V^{-1}\ni a$.$\square$

Now we deal with the equation $\alpha(x)\leq \alpha(a)+\epsilon$. We don't know much about $\alpha$ so we multiply $\mu(X)$ for some non-null measurable set $X$ which can be chosen later. So we just need $\mu(xX)\leq \mu(aX)+\epsilon$.

Note that Haar measure is by definition outer regular, i.e. it can be approximated from above by open sets. This is how we get $\epsilon$. So $\exists V\in \tau$ s.t. $aX\subset V$ and $\mu(V)\leq \mu(aX)+\epsilon$.

Hence we need to find the open subset $U\ni a$ s.t. $\forall x\in U,xX\subset V$ so that $\mu(xX)\leq \mu (V)$. Equivalently, enlarge from $aX\subset V$ to $UX\subset V$.

As the mutliplication map $\pi:K\times K\to K,(a,b)\to ab$ is continuous, $\forall x\in X,\exists U_x,X_x\in \tau$ s.t. $a\in U_x,x\in X_x$ and $U_x X_x\subset V$. Clearly $(\bigcap_{x\in X} U_x)X\subset V$ so $U:=\bigcap_{x\in X} U_x$ is our candidate.

Using locally compactness, pick compact $X$ containing a open neighborhood of $0$ which has measure $1$. There exists a finite subset $I\subset X$ s.t. $X\subset \bigcup_{x\in I}X_x$ hence $U:=\bigcap_{x\in I}U_x\ni a$ satisfies $UX\subset V$. Thus $\alpha$ is upper semicontinuous on $K$ (the proof works with $a=0$).  As $\alpha(0)=0$ and $\alpha$ is non-negative, $\alpha$ is continuous at $0$. So $\alpha$ is continuous on $K$.

Part II: $B(0,r)$s form a neighborhood basis of $0$.

Fix a open $U\ni 0$, we want to find a radius $r$ s.t. $B(0,r)\subset U$. $B(0,r)$ is something checked by $\alpha$. Equivalently we need to show $\alpha^{-1}([0,r))\subset U$, equivalently $\alpha(U^c)\subset [r,\infty)$, i.e. bounded below by a positive number.

Recall that continuous image of compact set is compact, so bounded and closed in $\mathbb{R}_{\geq 0}$, moreover $0\notin U^c$ so $0\notin \alpha(U^c)$, if $U^c$ is compact then $\alpha(U^c)$ is bounded below by a positive element so we are done.

Now we use locally compactness, pick $X\ni 0$ compact which contains a open $V\ni 0$. And consider $U\cap V \subset X$. Rename $U\cap V$ as $U$. So $\alpha(U)\subset \alpha(X)$ is bounded above by $m$ for some $m>0$. Hence $U\subset \overline{B}(0,m):=\alpha^{-1}([0,m])$.

So $U^c=(\overline{B}(0,m)-U)\coprod \overline{B}(0,m)^c$ where $\overline{B}(0,m)-U$ is a closed subset of compact (to be shown in Part III) so compact. Both $\alpha(\overline{B}(0,m)-U)$ and $\alpha(\overline{B}(0,m)^c)$ is bounded below, the result follows.

Part III: $B_m:=\overline{B}(0,m)$ is compact.

Again we utilize locally compactness. $B_m$ is already closed so it suffices to show it lies in a compact set. There exists a compact neighborhood $X$ of $0$ which contains a open neighborhood $U$ of $0$, and we can deform it using translation and multiplication so that it maintains compact and covers $B_m$. As $\alpha$ is multiplicative, it is natural to consider multiplication by a big enough element.

To find an arbitrarily large element, it suffices to find an arbitrarily small non-zero element and then take inverse. As $K$ is not discrete, $\{0\}$ can not be open, $B(0,\epsilon)$ is always non-trivial. Hence $\alpha$ is unbounded.

We would like to show $B_m\subset tX$ for some large $t$ or equivalently $t B_m\subset X$ for some small $t$. Fix any non-zero $t\in B(0,1)$, $\{t^n\}_{n\geq 1}$ can be arbitrarily small. For every element $a$ in $B_m$ or actually any element, we expect $t^n a\to 0$ so the open neighborhood $U$ of $0$ must contain some $t^n a$.

Clearly any limit point of $\{t^n a\}$ is still a limit point under $\alpha$, which is $0$. As $\alpha$ is non-zero except at $0$, $0$ is the only possible limit point. If $a=0$ we are done, if $a\neq 0$ it's equivalent to consider $\{t^n\}$ using the homeomorphism $s\mapsto sa^{-1}$. And we could pick non-zero $t\in B(0,1)\cap U\cap W$ where $W$ is an open neighborhood of $0$ s.t. $WX\subset U$ (see above), which must be non-trivial since $\{0\}$ is not open. Therefore $\{t^n\}$ lies in $X$ which is compact so limit point compact (every infinite subset has a limit point), and it has a limit point which must be $0$.

For each $a\in B_m$ denote $n(a)$ the smallest integer s.t. $t^{n(a)}a\in X$, if $n(a)$ is bounded above by some $N$ then $B_m$ will be covered by $\bigcup_{i=0}^N t^{-i}X$ so compact. It suffices to consider $n(a)$ s.t. $n(a)\geq 1$, so $t^{n(a)-1}a\notin X$ and hence $t^{n(a)}a\notin tX$. So $t^{n(a)}a\in X\cap (tX)^c$. As $\alpha(a)\leq m$, to show $n(a)$ is bounded above, it suffices to show $\alpha(X\cap (tX)^c)$ is bounded below by a positive number.

Let $C=\overline{X\cap (tX)^c}$, then $C$ is a closed subset of $X$ hence compact. Also $X\cap (tX)^c\subset (tX)^c \subset (tU)^c$, $C$ is a closed subset of $(tU)^c$, so $0\notin C$. Therefore $\alpha(C)$ is bounded below by a positive integer. The result follows.


问:
Textbook advice- Dynamical Systems and Differential Equations
I am currently an undergrad math major taking a gap year (because my university is entirely online this semester).  For this year, I have signed up for a "Directed Reading Program" with a graduate student whose specialty involves dynamical systems.  For this program, I am supposed to read through a textbook that we can discuss.  Two of her suggestions were Nonlinear Dynamics and Chaos by Steven H. Strogatz and Differential Equations, Dynamical Systems, and an Introduction to Chaos by Hirsch, Smale, and Devaney.

As I took a look at those books, I realized an additional reason why reading a book such as those could be useful: although I took a differential equations course at my local community college when I was in high school, I don't remember them all that well.  My university's math department is very theory-oriented, so I may never have the opportunity to take a DiffEQ course as an undergrad, though, as a math major who may want to go into something more applied, I feel as though a high level of comfort with differential equations would be nice to have.  Looking through the two textbooks online, neither appears to cover Laplace transforms, which I remember to have been an entire unit in my community-college course.  Because of this, I am having doubts about the efficacy of the two books with respect to giving me said comfort.  However, the books seem to be fantastic with respect to gaining a deeper understanding of the material, so I am not trying to criticize.

Two questions:

Between Strogatz and Hirsch/Smale/Devaney, which would you recommend?

In light of the above (the lack of coverage of topics such as Laplace transforms), do you think I ought to, in addition to one of those two books, spend time with Ordinary Differential Equations by Tenenbaum and Pollard (which I got for Christmas or something awhile back but haven't spent much time with)?

答:
Let me preface by saying that analysis of ODEs runs along two different directions. The first is to solve ODEs, in the sense that you end up with exact or approximate solutions of the ODE. This is taught in a first course, in terms of manipulation of simpler ODEs such as first and second order with constant coefficients or numerical approximations of solutions. Application of Laplace transforms is also limited to linear ODEs afaik.

The second direction takes an abstract setting and fits into a larger study called dynamical systems. One no longer looks for solutions of the ODEs in the sense of a closed form description of a function. Instead the focus is on the qualitative behavior of solutions. The first two books you listed mostly take this approach.
The reason for the abstract setting is two fold: It gives you a simple way of dealing with a large class of equations. But more importantly, once you start looking at nonlinear ODEs, the "mode" of analysis changes drastically.
For example, instead of solutions, one looks to qualify solutions based on their long term behavior. This leads to notions of stable and unstable solutions.

Although I cannot answer your first question, as I haven't read these books, I can say that I've heard good things about both. In any case, spending time with basic books like Tenenbaum and Pollard will be beneficial, especially if you do not remember much from your first course.


问:
VC dimension of union and intersection
Let $\mathbb{A}$ and $\mathbb{B}$ be sets of subsets of $X$ of finite VC dimension. Show that:

(a) If $\mathbb{C}=\{A\cap B : A \in \mathbb{A},B \in \mathbb{B}\}$, then $\prod_{\mathbb{C}}(n)\leq\prod_{\mathbb{A}}(n) \cdot \prod_{\mathbb{B}}(n) $

(b) If $\mathbb{C}= \mathbb{A}\cup \mathbb{B}$, then $\prod_{\mathbb{C}}(n)\leq\prod_{\mathbb{A}}(n) + \prod_{\mathbb{B}}(n) $

The definition is of Vapnik-Chernovenkis (VC) dimension of a set of classifiers $\mathbb{H}$ is $$\text{VC}(\mathbb{H})=\text{max}\{n\in\mathbb{N}:\prod_{\mathbb{H}}(n)=2^n\}$$

The growth function $\prod_{\mathbb{H}}$ is defined as:

$$\prod_{\mathbb{H}}(n)=\text{max}_{\mathbb{x}\in X^n}|\{h(\mathbb{x}):h\in \mathbb{H}\}|.$$

Attempt for (a):

Fix a set $X$ of $n$ points. Let $Y_1,...,Y_k$ be the set of intersections of $\mathbb{A}$ with $X$. By defintion of $\prod_{\mathbb{A}}(X),k\leq\prod_{\mathbb{A}}(X)\leq \prod_{\mathbb{A}}(n)$. By definition $\prod_{\mathbb{B}}(Y_i)$, the intersection of the $\mathbb{B}$ with $Y_i$ is at most $\prod_{\mathbb{B}}(Y_i)\leq \prod_{\mathbb{B}}(n)$. Thus, the number of sets intersections of $\mathbb{C}$ with $X$ is at most

$$k \prod_{\mathbb{B}}(Y_i)\leq \prod_{\mathbb{A}}(n) \prod_{\mathbb{B}}(n)$$

I'm not sure how to go further for (a)

答:
$\newcommand{\m}[1]{\mathbb{#1}}$
$\newcommand{\r}[1]{\mathrm{#1}}$Yes, I think the answer sort of follows from definition, as I mentioned. But because this is bountied I'd like to put just the little more effort into this one.

What is this VC dimension of a set of "classifiers"? Well, basically a classifier on a set is like a rule : if an element is part of the classifying set, then it satisfies that rule, otherwise it does not. Now, a set of classifiers shatter a set, if every subset of that set can be shown as exactly those following a certain set of rules. Basically, for this set, if you wanted to isolate any fixed subset , you could do it using the given classifiers.

For example, suppose you had $\{1,2,3,4,5\}$ and you want to get the subset $\{2,3,5\}$ by applying some rules. If one of the rules was "prime numbers only" then you can apply this rule to the set and get $\{2,3,5\}$. But suppose there were only the rules : "only even numbers" and "only odd numbers" then you can't actually get $\{2,3,5\}$ from these classifiers alone , and would need other rules to help.

The "strength" of a classification is obviously based on how many elements it can show apart from one another. Hence, the largest possible set that can be shattered by a set of classifiers, is its VC dimension. The exercise that you have been assigned, tells you that if you take the intersection of two sets of classifiers, and similarly if you take their union, these sets of classifiers have limited strength (a statement which can be translated into VC dimension bounds).

Rigorously, in your language now, starting on part $b$. We want to show that $\Pi_{\m{C}}(n) \leq \Pi_{\m{A}}(n) + \Pi_{\m{B}}(n)$. So let $\r{x} \in X^n$, and consider the set $|\{h(\r{x}) : h \in \m{C}\}|$. We know that $\m{C} = \m{A} \cup \m{B}$, so in particular we have $$\{h(\r{x}) : h \in \m{C}\} = \{h(\r{x}) : h \in \m{A}\} \cup \{h(\r{x}) : h \in \m{B}\} \\
\implies |\{h(\r{x}) : h \in \m{C}\}| \leq |\{h(\r{x}) : h \in \m{A}\}| +|\{h(\r{x}) : h \in \m{B}\}|$$

and now we can take the maximum over $\r{x} \in X^n$  on both sides, and the maximum is sub-additive over sets so it splits on the left hand side, leading to :
$$
\max_{\r{x} \in X^n} |\{h(\r{x}) : h \in \m{C}\}| \leq \max_{\r{x} \in X^n}\left\{|\{h(\r{x}) : h \in \m{A}\}| +|\{h(\r{x}) : h \in \m{B}\}|\right\} \\
\leq \max_{\r{x} \in X^n}\left\{|\{h(\r{x}) : h \in \m{A}\}|\right\} +\max_{\r{x} \in X^n}\left\{|\{h(\r{x}) : h \in \m{B}\}|\right\}
$$

which gives the result.

Your logic for $a$ is actually quite good, but just has a few writing errors.

Suppose $\m{C} = \{A \cap B : A \in \m{A} , B \in \m{B}\}$, and let $\r{x} \in X^n$. Consider the set $\{h(\r{x}) : h \in \m{C}\}$, we would like to describe this set. To do that, note that $h = p \cap q$ for some $p \in \m{A}$ and $q \in \m{B}$, so in particular we have $\{h(x) : h \in \m{C}\} = \{q(p(\r{x})) : p \in \m{A} , q \in \m{B}\}$. Why is this true? Well, $h(\r{x})$ is applying rule $h$ on $\r{x}$, but rule $h$ is basically two rules put together, one of $\m{A}$ and one of $\m{B}$, so applying $h$ is like applying both these rules (the order of rule application does not matter, because at the heart of it , rule application is intersection of sets which is commutative and associative. So I could have written $p(q(\r{x}))$ above as well).

From here we get :
$$
\{h(x) : h \in \m{C}\} = \{q(p(\r{x})) : p \in \m{A} , q \in \m{B}\} = \cup_{p \in \m{A}}\{q(p(\r{x})) : q \in \m{B}\}
$$

and now taking the sizes and using the union bound along with the definitions :
$$
|\{h(x) : h \in \m{C}\}| \leq \sum_{p \in \m{A}} |\{q(p(\r{x})) : q \in \m{B}\}| = \sum_{a \in T} |\{q(a) : q \in \m{B}\}|
$$

where $T = \{p(\r{x}) : p \in \m{A}\}$. But then $|T| \leq \Pi_{\m{A}}$ and each summand above is $\leq \Pi_{\m{B}}$, so for each $\r{x}$ the LHS is less than or equal to $\Pi_{\m{A}}\Pi_{\m{B}}$ and the result follows by taking a maximum over $\r{x}$ on the LHS.

To give both proofs in words :

$b$ : If you take the union of  two sets of classifiers, then any set classified by the union must already have been previously classified by at least one of the two original sets, therefore the union bound may be used.

$a$ : If you form a "refined" classifying set by combining classifiers from both kinds one after the other and taking all such pairs, then any set which is classified by this "refinement" must go through a classifier of one kind, followed by a classifier of the other kind. These may be treated independently : the classifier of the first kind can only produce so many different sets regardless of input, and the same goes for the second. Finally, we can just multiply these together, as we do for independent events in combinatorics.

Title: VC dimension of union and intersection


问:
Canonical homomorphism between $K/\mathcal O_K$ and $S^1$
Is well know that there a isomorphism

1. $$\mathbb Q_p/\mathbb Z_p\approx \mathbb Z[1/p]/\mathbb
    Z\hookrightarrow \mathbb R/\mathbb Z\approx \mathbb S^1.$$

Now, let $K/\mathbb Q_p$ be a finite extension and let $\mathcal O_K$ be the ring of integers of $K$ i.e. $\mathcal O_K:=\{x\in K:\vert x\vert_p\leq 1\}$. And consider the additive quotient $K/\mathcal O_K$.

My question is if there a natural homomorphism as in the preceding case i.e.
a   homomorphism $K/\mathcal O_K\to \mathbb S^1$ such that if $K=\mathbb Q_p$ then the homomorphism $K/\mathcal O_K\to \mathbb S^1$ coincides with previous homomorphism $\mathbb Q_p/\mathbb Z_p\approx \mathbb Z[1/p]/\mathbb
    Z\hookrightarrow \mathbb R/\mathbb Z\approx \mathbb S^1$ ?

thanks you all.

I would appreciate any answer or reference.

答:
$$O_K = \sum_{j=1}^n b_j \Bbb{Z}_p, \qquad K = \sum_{j=1}^n b_j \Bbb{Q}_p$$ where the $b_j$ have valuation $\in [0,1)$, $b_1=1$, $n=[O_K:\Bbb{Z}_p]$.

$$K/O_K=
\sum_{j=1}^n (b_j \Bbb{Q}_p/b_j \Bbb{Z}_p)\cong (\Bbb{Q}_p/\Bbb{Z}_p)^n$$

Let $$f \in Hom(\Bbb{Q}_p/\Bbb{Z}_p,\Bbb{R/Z}),\qquad f(\frac{a}{p^k}+\Bbb{Z}_p)=\frac{a}{p^k}+\Bbb{Z},\qquad a\in \Bbb{Z}$$
Then $$c \to f(c.)$$ is an isomorphism $$\Bbb{Z}_p\to Hom(\Bbb{Q}_p/\Bbb{Z}_p,\Bbb{R/Z})$$
Whence $$Hom(K/O_K,\Bbb{R/Z})= \{ (\sum_{j=1}^n b_j A_j+O_K \to \sum_{j=1} f(c_j A_j)+\Bbb{Z}),c\in \Bbb{Z}_p^n\}$$
(where $A\in \Bbb{Q}_p^n$)

Once it is clear to you then for $p\nmid n$ there is some kind of canonical map, which is $$f \circ \frac1n  Tr_{K/\Bbb{Q}}$$


问:
Find the limit of the function without L'Hôpital's rule
I have a problem, such as:

$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$

How do I solve this problem without using L'Hôpital's rule or small-o?

Thanks!

答:
Composing Taylor series
$$y= \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}\implies\log(y)=\frac{1}{x^2}\big[\log(\cos(ax))-\log(\cos(bx))\big]$$

Using
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$
$$\cos(cx)=1-\frac{c^2 x^2}{2}+\frac{c^4 x^4}{24}+O\left(x^6\right)$$
$$\log(\cos(ax))=-\frac{c^2 x^2}{2}-\frac{c^4 x^4}{12}+O\left(x^6\right)$$

Use it twice (for $c=a$ and $c=b$ and subtract to get
$$\log(\cos(ax))-\log(\cos(bx))=\frac{1}{2} x^2 \left(b^2-a^2\right)+\frac{1}{12} x^4
   \left(b^4-a^4\right)+O\left(x^6\right)$$
$$\log(y)=\frac{1}{2} \left(b^2-a^2\right)+\frac{1}{12} x^2
   \left(b^4-a^4\right)+O\left(x^4\right)$$
$$y=e^{\log(y)}=e^{\frac {b^2-a^2}2}\left(1+\frac{1}{12} x^2 \left(b^4-a^4\right)+O\left(x^4\right)\right)$$ which shows the limit and also how it is approached.


问:
Proof an Equation is Wrong

“For all numbers $j$ and $k$, $(j + k)^2 = j^2 + k^2$.”

How would you prove this is wrong??

When I times out the bracket with the number it makes $j^2 + k^2$ but that's wrong.

Should I put a number for $j$ and $k$ and then times it out?

Title: Proof an Equation is Wrong

Body:
$(j + k)^2$

If $j = 3$ and $k = 2$

= $(3+2)^2$

= $(3+2)(3+2)$

= $9+6+6+4$

= $25$

However, if we were to say that

$(j+k)^2=j^2+k^2$

then:

= $3^2+2^2$

= $9+4$

= 13

Which means that $(j + k)^2 \neq 3^2+2^2$


问:
Irreducibility of iterations of a polynomial
I am look for some nice examples or conditions that force $f^{(n)}(x)$ to be irreducible (this is composition). I looked at things online and it seemed that topic is quite complex and involves calculating discriminants. I would like to do my own little exposition/project on perhaps an example where I could prove $f^{(n)}(x)$ is irreducible using elementary methods. I know a little bit of dynamical systems and field theory up until basics of Galois.

Does such an example exist? Perhaps something akin to that? In general I am looking for something I could write extensively about that relates to iterations of polynomials, irreducibility is the first thing that came to mind.

Edit: I ran some tests with $x^2+1$ and it seems the first 10 compositions are irreducible so I think considering this polynomial could be a good start. I am just not sure what to do as I never actually considered reducibility of compositions.

答:
Let $f(x) = (x-1)^2 + 1$.  Then $f^{(n)}(x)$ is irreducible over $\mathbb Q$ for all $n\ge 1$, and we can show this using just Eisenstein's criterion.

First we deduce the closed form:

$$ f^{(n)}(x) = (x-1)^{2^n} + 1 $$

This can be shown easily by induction, but it can be seen essentially by inspection if we write:

$$ f(x) = \tau^{-1} \circ g \circ \tau(x) $$

where $g(x) = x^2$ and $\tau(x) = x-1$ (so that $\tau^{-1}(x) = x+1$).

Thus $f^{(n)}(x) = \tau^{-1} \circ g^{(n)} \circ \tau(x)$, in agreement with the closed form above, since $g^{(n)}(x) = x^{2^n}$.

Now apply Eisenstein's criterion with $p=2$:

$$ f^{(n)}(x) = (x-1)^{2^n} + 1 = x^{2^n} + \left[ \sum_{k=1}^{2^n -1} \binom{2^n}{k} (-x)^k \right] + 2 $$

For proofs that $p=2$ divides each binomial coefficient $\binom{2^n}{k}$ for $1\le k \le 2^n -1$ see the previous Math.SE Question Prime dividing the binomial coefficients.


问:
How to derive: $\left(\forall x\right)\left[Lx\rightarrow\left(Mx\land N x\right)\right]\vdash\left(\forall x\right)\left(Lx\rightarrow M x\right)$
I'm trying to derive:
$\left(\forall x\right)\left[Lx\rightarrow\left(Mx\land N x\right)\right]\vdash\left(\forall x\right)\left(Lx\rightarrow M x\right)$

I thought I was on the right track with:

$1 \hspace{5 mm}(1) \left(\forall x\right)\left[Lx\rightarrow\left(Mx\land N x\right)\right]\hspace{10 mm}A\\
1\hspace{5 mm}(2)\hspace{10 mm}   La\rightarrow\left(Ma\land N a\right) \ \hspace{8 mm}1,\forall E\\
2\hspace{5 mm}(3)\hspace{10 mm}   La \hspace{42 mm}A\\
2\hspace{5 mm}(4)\hspace{10 mm}   Ma\ \land Na \ \hspace{20 mm}2,3\rightarrow E\\
2\hspace{5 mm}(5)\hspace{10 mm}   Ma \hspace{40 mm}4, \land E4\\
1,2\hspace{1 mm}(6)\hspace{10 mm}   La\rightarrow Ma \hspace{23 mm}2,4\rightarrow I\\
1,2\hspace{5 mm}(7)\hspace{10 mm}   \left(\forall x\right)\left(Lx\rightarrow M x\right) \hspace{10 mm}6, \forall I\\
$

However, I think by assuming the antecedent in step (3) that I've done something wrong.
Is that step allowed or is there a more efficient way to derive the conclusion?
Or can anyone provide any insights that I may have missed?

答:
Here is another approach (convenient if you doubt about the introduction of arbitrary assumptions):

$$
\begin{array}{lll}
(1)&(\forall x)[Lx\to(Mx\land Nx)]&\text{Premise}\\
(2)&La\to(Ma\land Na)&\text{\(\forall\) elimination (1)}\\
(3)&\neg La\lor(Ma\land Na)&\text{\(\to\) equivalence (2)}\\
(4)&(\neg La\lor Ma)\land(\neg La\lor Na)&\text{\(\lor\) distributive (3)}\\
(5)&\neg La\lor Ma&\text{\(\land\) elimination (4)}\\
(6)&La\to Ma&\text{\(\to\) equivalence (5)}\\
(7)&(\forall x)(Lx\to Mx)&\text{\(\forall\) introduction (6)}\\
\end{array}
$$


问:
If $f(1/z)$ has a pole at z=0 then $f(z)$ is polynomial.
The condition is that $f$ is entire and injective, but the fact that it is injective might be an overkill. How can I prove this statement? I am having trouble understanding what the sufficient condition for the polynomial is.

答:
I moved my comment here since it was closer to an answer than a comment:

Since f is entire it has a Taylor series at the origin which converges everywhere:

$$f(z)=\sum_{n=0}^\infty a_nz^n, |z|<\infty$$

Plug in 1/z into this expression. This will become a Laurent series, which you know properties about since you know $z=0$ is a pole of $f(\frac{1}{z})$. In particular, you know that the Laurent series has finitely many terms.


答:
So, as it turns out, any two of $(12),(13)$ and $(23)$ generate $S_3$.

That just means that any element of $S_3$ can be written as a finite product of, say $(12)$ and $(13)$ 's.  (We should consider the inverses of these elements as well.  But of course they are their own inverses.)

So let's see: $e=(12)(12),(12)=(12),(13)=(13),(23)=(12)(13)(12),(123)=(13)(12)$.

And I leave $(132)$ for you.

Title: Symmetry group 3 and writing elements in terms of each other

Body:
In the group $S_3$ and let $a,b$ be distinct elements of order two. Express each non-identity element in $S_3$ as a word in $a,b$.

I am simply confused by what this means. I know $S_3 = \{e, (1\; 2), (1\; 3), (2\; 3), (1\; 2\; 3), (1 \;3 \;2)\}$, but I am struggling with how to express the other terms as "words."


问:
How to find a positive semidefinite matrix satisfying the following condition?
Consider such a problem:
$$\mathbf A\mathbf x=\mathbf b$$
where the vectors $\mathbf x$ and $\mathbf b$ are known.

I want to find a positive semidefinite matrix $\mathbf A$ satisfying the aforementioned equation.

What conditions do $\mathbf x$ and $\mathbf b$ need to satisfy?

Thanks a lot.

答:
I have chosen to provide the general ($n$ dimensional case) as a separate answer. I will use the letter $Y$ instead of $B$, i.e., being given $X$ and $Y$, find a symmetric positive semi-definite matrix $A$ such that

$$AX=Y$$

Let $s=X^TY$ be the dot product of $X$ and $Y$.

As remarked by @Fenris, a necessary condition for the existence of $A$ is that $s \ge 0$.

It is in fact sufficient. Here is why.

Let $\{U_1, U_2, \cdots U_{n-1}\}$ be any basis of the orthogonal subspace to $X$; then a solution is

$$A=\sum_{k=1}^{k=n-1} \alpha_k U_kU_k^T+\dfrac1s YY^T$$

for any sequence $\alpha_1,\cdots \alpha_{n-1}$ of positive numbers (s is assumed $\ne 0$). Indeed,

$$AX=\sum_{k=1}^{k=n-1}  \alpha_k U_k\underbrace{U_k^TX}_0+\dfrac1s Y\underbrace{Y^TX}_s=Y$$

and for any vector $V$:

$$V^TAV=\left(\sum_{k=1}^{k=n-1}  \alpha_k V^T U_k U_k^T V\right)+\dfrac1s V^TYY^TV = \sum_{k=1}^{k=n-1} \alpha_k\left(U_k^T V\right)^2+\dfrac1s(Y^TV)^2 \ge 0$$

establishing that $A$ is positive semi-definite.

Remark: a very particular case is obtained by taking all the $\alpha_k=0$...


问:
Number theory question including concepts of primes and divisibility.
prove that $(n!)^{(n-1)!}$ divides (n!)!

I know this question already exists but i'm looking for a purely number theory proof, no combinatorics.

My attempt: I tried to go about the concept of largest prime power that divides n! , which is given by [n/p]+[n/$p^2$] +...... upto infinity (where [.] is the greatest integer function). So i tried to prove that largest power of prime p that divides (n!)! ≥ largest power of p that divides $(n!)^{(n-1)!}$, but i ended up with an ugly inequality with no idea how to proceed further.

Any help would be appreciated, cheers!

答:
Pick a prime number $p$. For every $m\in  \mathbb{Z}$ let $\alpha_p(m)$ be the exponent of $p$ in factorization of $m$. Then
$$\alpha_p(n!) = \sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]$$
and
$$\alpha_p((n!)!) = \sum_{k=1}^{+\infty}\bigg[\frac{n!}{p^k}\bigg]$$
In particular, we have
$$\alpha_p\big((n!)^{(n-1)!}\big) = (n-1)!\cdot \alpha_p(n!) = (n-1)!\cdot \bigg(\sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]\bigg)$$
Since
$$x\cdot [y] \leq [x\cdot y]$$
for $x,y>0$ and $x\in \mathbb{Z}$, we deduce that
$$\alpha_p\big((n!)^{(n-1)!}\big) = (n-1)!\cdot \bigg(\sum_{k=1}^{+\infty}\bigg[\frac{n}{p^k}\bigg]\bigg) =$$
$$= \sum_{k=1}^{+\infty}(n-1)!\cdot \bigg[\frac{n}{p^k}\bigg] \leq \sum_{k=1}^{+\infty}\bigg[(n-1)!\cdot \frac{n}{p^k}\bigg] = $$
$$= \sum_{k=1}^{+\infty}\bigg[\frac{n!}{p^k}\bigg] = \alpha_p\big((n!)!\big)$$


问:
List the members and find the cardinality of the sets
Could anyone help me understanding this question ?

List the members and find the cardinality of the following sets :
$$A=\{x\mid x \text{ is a real number such that } x^4=4\}$$
$$B=\{x\mid x = 2n - 5, x \text{ and }n \text{ are natural numbers}\}$$
$$C=\{x\mid x \text{ is a prime number less than } 100\}$$
$$D=\{x\mid x \text{ is the square of an integer and } x<200\}$$

For $C$ and $D$, I was able to solve it. For $A$, I'm not sure but I think it's equal to the empty set (there is no number that fit the formula).

I found an answer that say $B=\{1, 3, 5, 7, \dots\}$ but I didn't get it!!
Could you help me understand how to solve $B$ please!!

答:
Welcome.

Cardinality means just the size of the set.
How many numbers are there in the set of natural numbers $\mathbb{N}$?

For $A$, just power both sides by $\frac{1}{4}$, is this a real number? and since $x$ is even powered, you have both positive and negative solutions.

For $B$, first you know that $n$ is a natural number, meaning $n = 1, 2, 3, 4, 5, \dots$. Just plug in $n$ into the equation of $x$, what $x$ do you get? From this $x$, which of them are also natural numbers?

Hope it helps,


问:
How to show that $\sum_{i\ge i_0} \frac{i^2}{4^i}\to 0$
How to show that: for all $\varepsilon>0$, take a $i_0=i_0(\varepsilon)$ such that

$$\sum_{i\ge i_0} \frac{i^2}{4^i}\le \varepsilon.$$

I know that is convergence to zero by the ratio test but how to choose $i_0$?

答: Hint: Use induction to prove that $\frac {i^{2}} {4^{i}} \leq \frac  1{(1.5)^{i}}$ for all $i \geq 1$. Since $\sum_{i \geq i_0} \frac  1{(1.5)^{i}}$ is a geometric sum you can compute it explicitly and find $i_0$ such that the sum is less than $\epsilon$.


问:
Algebra exercise
If $a^4+a^3+a^2+a+1=0$ find the value of $a^{2000}+a^{2010}+1$

I got this problem in a book and tried to solve it.I multiplied with suitable powers of a and added and subtracted alternatively to get $a^{2010}+a^{2008}+a^{2006}+a^{2005}+a^{2004}+a^{2002}+a^{2000}=0$ but i can't figure what to do.I tried to replace $a^{2005}$ by multiplying the parent equation by $a^{2003}$ but nothing useful came.

Any help would be appreciated.Thanks in advance.

Title: Algebra exercise

Body:
Hint

$$(a-1)(a^4+a^3+a^2+a+1)=?$$

Now $a^{2000}=(a^5)^{400}=?$


问:
Probability exercise: Two individuals agree to meet
Two individuals agree to meet within the limits of the agreed period $\ l\ $. The first individual to arrive waits for the time $\ a <l$, and then leaves. What is the probability that they will meet?

I tried to use geometric probability.

let Ω = {$(x, y) | 0≤x, y≤ l$}

Using coordinate axes,

Let $(a_1) <(a_2)$ such that $(a_2) - (a_1) = a$
then the area where the coordinate pairs are found should be $[(a_1) ^ 2 - (a_2) ^ 2]$
and the area of Ω = $(l ^ 2)$

So the probability is P (A) = $([(a_1) ^ 2 - (a_2) ^ 2]) / (l ^ 2)$

coordinate axes

答:
This is the correct drawing

The probability that they meet is the area of the purlple region on the total area, say 1 minus the area of the two white triangles on the total area

$$1-\frac{(l-a)^2}{l^2}$$