问:
Understanding resolution derivates of a clause $C$ from a set of clauses $\mathcal{C}$.
We know that a resolution derivative of a clause $C$ from a set of clauses $\mathcal{C}$ is a sequence $(C_1, ..., C_n)$, so that:

$C_n = C$ and

for all $1 \leq i \leq n$ the following holds: $C_i \in \mathcal{C}$ or there is $j,k < i$ with $C_i \in \text{Res}(C_j,C_k)$.

If this holds, then $C$ is a resolution proof from $\mathcal{C}$ and is written as $\mathcal{C} \vdash_R C$. However, a resolution rebuttal of a set of clauses $\mathcal{C}$ is a resolution derivative of the empty clause $\square$.

I understand the part with the $\square$ because there is an example to illustrate it. However, I am honestly lost about the entire first section. What is happening here? I need to understand that section so that I can understand resolution calculus, so any help is appreciated.

PS - this is translated from German and I'm not sure if some of the terms translate well in English. So if there is anything wrong, please mention it in the comments.

答:
Let $a, b, d, a', \cdots$ be propositional variables. Let $A, D$ be disjuncts of literals. Let $W1$ and $W2$ be conjuncts of disjuncts of literals within a complete wff.

In some resolution proof settings, each clause is a disjunction of literals. A literal is either a propositional variable by itself or a negated propositional variable.

I'm going to answer a slightly different question than the one you posed ... at least initially. I will assume that there is some prefix of clauses $C_1, \cdots , C_w$ that are independent of each other and constitute the premises of the proof $\mathcal{C}$. In the question that you posed, only $C_1$ is the only clause that is not a product of a resolution step. I'll address this problem a little bit later.

If we have two clauses such that the left clause has a positive literal and the right clause has the corresponding negative literal, then we can infer a new clause that is formed by joining the remaining literals from both the left and right clauses.

```<code>a \/ a' \/ b       !b \/ d \/ d'
--------------------------------
        a \/ a' \/ d \/ d'
</code>```

Let $A$ and $D$ be metavariables ... standing in for arbitrary disjunctions of literals.

```<code>A \/ b     !b \/ D
------------------
      A \/ D
</code>```

This inference rule is valid if only if $(A \lor b) \land (\lnot b \lor D) \implies A \lor D$ is an unconditional tautology. We can convince ourselves that this is true using truth tables.

The turnstile symbol $\vdash$, possibly with subscripts, is often used to denote that the thing on the right hand side is a syntactic consequence of the things on the left hand side. It can also be used to denote that a proof of some sort exists given some premises.

In the cases I've seen, $\vdash$ is frequently used to represent an inductively defined predicate on two sets of well-formed formulas. That seems to be the case here, $\mathcal{C}$ is both an ordered list of premises and a proof.

The fact that $C_1$ is the only unconstrained clause in the list makes me think that in the presentation in your book, $C_1$ is not restricted to being a disjunction of literals.

We can deal with this problem by insisting that $C_1$ is in conjunctive normal form, i.e. it is a conjunction of disjunctions of literals.

In this case, we can modify our inference rule slightly by ignoring all but one conjunct in each of the input clauses. Let $W1$ and $W2$ each stand for an arbitrary number of ignored conjuncts in each of our premises

```<code>W1 /\ (A \/ b)     (!b \/ D) /\ W2
----------------------------------
            (A \/ D)
</code>```

Because of the semantics of and, we can treat $W1$ and $W2$ the same way we treat the other clauses that aren't being used as premises.

Also, we can do the following with both conjuncts being taken from the same clause. We're forced to do this for $C_2$.

```<code>   (A \/ b) /\ W1 /\ (!b \/ D)
----------------------------------
            (A \/ D)
</code>```

If we set up the inference rule this way, there are very strong limits on what $\mathcal{C}$ can end in (it must end with a disjunction of literals). I don't know whether this is acceptable.

Also, note that $\vdash_R$ is an inductive relation.

$ J \vdash_R K $ is true if and only if there's a resolution step sending two conjuncts of $J$ to $K$

$ \Gamma, J \vdash_R K $ is true if and only if there's a resolution step sending some conjuncts in $ \Gamma \cup \{J\}$ to $K$ and $\Gamma \vdash_R J$.

Body:
Let $a, b, d, a', \cdots$ be propositional variables. Let $A, D$ be disjuncts of literals. Let $W1$ and $W2$ be conjuncts of disjuncts of literals within a complete wff.

In some resolution proof settings, each clause is a disjunction of literals. A literal is either a propositional variable by itself or a negated propositional variable.

I'm going to answer a slightly different question than the one you posed ... at least initially. I will assume that there is some prefix of clauses $C_1, \cdots , C_w$ that are independent of each other and constitute the premises of the proof $\mathcal{C}$. In the question that you posed, only $C_1$ is the only clause that is not a product of a resolution step. I'll address this problem a little bit later.

If we have two clauses such that the left clause has a positive literal and the right clause has the corresponding negative literal, then we can infer a new clause that is formed by joining the remaining literals from both the left and right clauses.

a \/ a' \/ b       !b \/ d \/ d'
--------------------------------
        a \/ a' \/ d \/ d'

Let $A$ and $D$ be metavariables ... standing in for arbitrary disjunctions of literals.

A \/ b     !b \/ D
------------------
      A \/ D

This inference rule is valid if only if $(A \lor b) \land (\lnot b \lor D) \implies A \lor D$ is an unconditional tautology. We can convince ourselves that this is true using truth tables.

The turnstile symbol $\vdash$, possibly with subscripts, is often used to denote that the thing on the right hand side is a syntactic consequence of the things on the left hand side. It can also be used to denote that a proof of some sort exists given some premises.

In the cases I've seen, $\vdash$ is frequently used to represent an inductively defined predicate on two sets of well-formed formulas. That seems to be the case here, $\mathcal{C}$ is both an ordered list of premises and a proof.

The fact that $C_1$ is the only unconstrained clause in the list makes me think that in the presentation in your book, $C_1$ is not restricted to being a disjunction of literals.

We can deal with this problem by insisting that $C_1$ is in conjunctive normal form, i.e. it is a conjunction of disjunctions of literals.

In this case, we can modify our inference rule slightly by ignoring all but one conjunct in each of the input clauses. Let $W1$ and $W2$ each stand for an arbitrary number of ignored conjuncts in each of our premises

W1 /\ (A \/ b)     (!b \/ D) /\ W2
----------------------------------
            (A \/ D)

Because of the semantics of and, we can treat $W1$ and $W2$ the same way we treat the other clauses that aren't being used as premises.

Also, we can do the following with both conjuncts being taken from the same clause. We're forced to do this for $C_2$.

   (A \/ b) /\ W1 /\ (!b \/ D)
----------------------------------
            (A \/ D)

If we set up the inference rule this way, there are very strong limits on what $\mathcal{C}$ can end in (it must end with a disjunction of literals). I don't know whether this is acceptable.

Also, note that $\vdash_R$ is an inductive relation.

$ J \vdash_R K $ is true if and only if there's a resolution step sending two conjuncts of $J$ to $K$

$ \Gamma, J \vdash_R K $ is true if and only if there's a resolution step sending some conjuncts in $ \Gamma \cup \{J\}$ to $K$ and $\Gamma \vdash_R J$.


问:
How do I calculate compounding interest with a continously changing interest rate?
If the interest rate itself is transformed after each time interval (for example following a parabola, or some other easily calculated trajectory) what would be the equation for this? As an example, imagine an interest rate that is multiplied by .95 each time period. How do I work this into the central equation?

Thanks

答:
Let $x(t)$ be the wealth at time $t$.
The differential equation describing the evolution of wealth under continuous compounding is $x^{\prime}(t)=r(t)x(t)$ where $r(t)$ is the interest rate at time $t$.

Remark.
It is important to note that the interest rate is specified in the same units of time as $t$.
For example, if $t$ is measured in years, then $r(t)$ is a per-annum interest rate.

You can obtain the case of discrete compounding by choosing some fixed times $t_{0}<t_{1}<\cdots<t_{N}$ at which payments occur and discretizing the ODE to obtain a recurrence.
That is,
$$
\frac{x_{n+1}-x_{n}}{t_{n+1}-t_{n}}=r(t_{n})x_{n}.
$$
Given an initial wealth $x_0$ at time $t_0$, you can use the above to obtain the wealth at subsequent times.

ContentLicense: CC BY-SA 4.0


问:
Same eigenvalues, different eigenvectors but orthogonal
I am using a two different computational libraries to calculate the eigenvectors and eigenvalues of a symmetric matrix. The results show that the eigenvalues calculated with both libraries are exactly the same, however, the eigenvectors differ. Nevertheless, both seem to be correct since their eigenvectors are orthogonal and the factorization is also correct. How can that be possible? I asked the developers and although they seemed confused, they said that "probably" has to do with the directions of the vectors.

If it is possible, is there a kind of margin error between those two different range of eigenvectors so I could compare them and make sure they are in range (they are right)?

答:
There is no promise of uniqueness for eigenvectors, (at least, not without a lot of constraints decided upon ahead of time.)

If $v$ is an eigenvector for $\lambda$, then so is $\alpha v$ for every $\alpha$ in your field $F$. It is not alarming to have different eigenvectors for a single eigenvalue.

It is also not alarming to have linearly independent eigenvectors with the same eigenvalue. In fact, since your matrix is symmetric, it is diagonalizable, and so an eigenvalue of multiplicity $n$ must have $n$ linearly independent eigenvectors for that eigenvalue.

ContentLicense: CC BY-SA 3.0


问:
Proof of the fact that $[0,2]$ is not a Jordan null set.
The means of Jordan null set is as following.

$A \subset \mathbb{R}^1$ is a Jordan null set

$\iff$
$\forall \epsilon >0, \ \exists U_1, U_2, \cdots , U_m (\text{each } U_i(=[a_i, b_i]) \text{ is bounded closed-interval});$

\begin{equation}
A \subset \cup_{i=1}^m U_i \text{  and } \sum_{i=1}^m (b_i-a_i) < \epsilon.
\end{equation}

Then, I have to consider whether $A:=[0,2]$ is a Jordan null set or not.
Intuitively, I think $[0,2]$ is not a Jordan null set, but I cannot prove this.

In order to prove this, I have to prepare $\exists \epsilon$ and $\forall U_1, \cdots U_m$ such that $[0,2] \subset \cup_{i=1}^m U_i \text{  and } \sum_{i=1}^m (b_i-a_i) \geqq \epsilon $.

But I don't come up with how I can prove.
Please tell me some ideas.

答:
If $A$ is Jordan-null then for any $\epsilon$ there is a finite family $F=\{(c_i,d_i):i\in I\}$ of bounded $open$ intervals with $c_i<d_i$ for each $i,$  such that $A\subseteq \cup F$ and $\sum_{i\in I}(d_i-c_i)<\epsilon.$

Proof: Let $U=\{[a_i,b_i]:i\in I\}$ have $m$ members, with $m\in \Bbb N,$ and with $a_i\le b_i$ for each $i,$ such that $A\subseteq \cup U$ and $$\sum_{i\in I}(b_i-a_i)<\epsilon /2.$$ The idea is to take $c_i<a_i$ and $b_i<d_i$ where each $a_i-c_i$ and each $d_i-b_i$ is small enough that $$\sum_{i\in I} (d_i-b_i)+(a_i-c_i)<\epsilon /2$$ so that $$\sum_{i\in I}(d_i-c_i)= \sum_{i\in I}(b_i-a_i) + \sum_{i\in I} (d_i-b_i)+(a_i-c_i)<\epsilon /2 +\epsilon /2.$$ For example let $d_i-b_i=a_i-c_i=(\epsilon /42)(1/m).$

So if $A $ is Jordan-null then $A$ is Lebesgue-null.


问:
Proving a graph with less than $\frac{k(k+1)}{2}$ edges is $k$-colorable
We are to assume the graph has no loop edges.

I don't know where to begin this problem. I have tested small values of $k$ and it seems to hold up, but that isn't a proof. I have in my notes that $e \leq 3v-6$ where $v$ and $e$ are vertices and edges, respectively. So I tried this:

So I know $e < \frac{k(k+1)}{2}$ and $e < 3v-6$, but I can't really do anything with this I don't think. The problem has a restriction on $e$ but not on $v$. Apparently that is important because my professor stressed that. Would it help if I wrote $v$ in terms of $k$? If there are less than $\frac{k(k+1)}{2}$ edges, then what can I say about the number of vertices?

I am hoping that I am on the right track here using this formula to solve this problem. Any advice is greatly appreciated.

答:
We shall prove the contrapositive -  if a graph is not $k$-colorable, then it has at least $\frac{k(k+1)}{2}$ edges. Suppose that a graph $G$ has chromatic number $\chi(G)>k$.

Then for any pair of colors (under a $\chi(G)$-coloring), there must be an edge between two vertices of those colors (otherwise, you could merge both colors).

Since there are $\binom{\chi(G)}{2}$ such pairs, this implies that there must be at least $\binom{\chi(G)}{2} \geq \binom{k+1}{2}$ edges.

Thus, if there are less than $\binom{k+1}{2}$ edges, then $\chi(G)\leq k$ and the graph is $k$-colorable.


问:
Full-Binary Tree with leaf nodes
Prove that if $T$ is a full binary tree with $i$ internal vertices (i.e. non-terminal nodes), then $T$ has $i+1$ terminal vertices (i.e. leaf nodes) and $2i+1$ total vertices.

From the definition of full binary tree, we know that $i_{th}$ node has either left child (i.e. $2i$), or right child (i.e. $2i+1$), and it seems to me that this problem will utilize such property, but I could not exactly relate to it. What are possible approaches for this question?

Update: Here is my solution without handshake lemma, and would be happy yo know whether it is correct or not:

The vertices of $T$ consist of the vertices that are children (of some parent) and the vertices that are not children (of any parent). There is one nonchild—the root. Since there are $i$ internal vertices, each having two children, there are $2i$ children. Thus the total number of vertices of $T$ is $2i + 1$, and the number of terminal vertices is
$(2i+1)−i = i+1$.

Note: Just upvote if it is right solution, no need to further comment on this question.

答:
Hint

Let $x$ be number of nodes and $y$ number of edges.

In any tree $y = x-1$

The sum of degrees of nodes $= 2y$

Suppose $x>1$. Root has degree $2$. A leaf has degree $1$. Any non-terminal node that is not root has degree $3$.

Title: Full-Binary Tree with leaf nodes


问:
Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$
When I'm trying to find the limit of  $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :

$\dfrac{\sqrt{1-\cos x}}{\sin x} =  \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)}  =   \dfrac{\sqrt{x^2(1+2\epsilon_1(x))}}{\sqrt{2}x(1+\epsilon_2(x))} = \dfrac{|x|}{\sqrt{2}x}\dfrac{\sqrt{1+2\epsilon_1(x)}}{1+\epsilon_2(x)} $

But I can't seem to do it  properly  using Landau notation

I wrote :

$ \dfrac{\sqrt{\frac{x^2}{2}+o(x^2)}}{x+o(x)} $

and I'm stuck... I don't know how to carry these o(x) to the end

Could anyone please show me what the step-by-step solution using Landau notation looks like when written properly ?

答:
It is the same as in the "$\epsilon$" notation. For numerator, we want $\sqrt{x^2\left(\frac{1}{2}+o(1)\right)}$, which is $|x|\sqrt{\frac{1}{2}+o(1)}$. In the denominator, we have $x(1+o(1))$.

Remark: Note that the limit as $x\to 0$ does not exist, though the limit as $x$ approaches $0$ from the left does, and the limit as $x$ approaches $0$ from the right does.


问:
Determine the supremum and infimum of sets
We consider the sets \begin{equation*}S_1:=\left \{\frac{2^n}{n}\mid n\in \mathbb{N}\right \} \ \text{ and } \ S_2:=\left \{\frac{1}{n}-2^{-n}\mid n\in \mathbb{N}\right \}\end{equation*} Determine the supremum and the infimum (if possible) and check if these are maximum and minimum respectively.

For the first set, it is not bounded from above, is it?

If we consider the second set do we consider it as a sum of sets $\left \{\frac{1}{n}\right \}$ and $\left \{-2^{-n}\right \}$ ?

答:
The sequence $\frac{2^n}{n}$ is increasing an unbounded, so the infimum (and minimum) is $\frac{2^1}{1}=2$ and there is no supremum/maximum. Regarding the second questions you dot not sum sets...

$S_2$ is a set of numbers obtained by the formula $\frac 1n -2^{-n}$, it has no common elements with the sets you mention. The sequence $\frac 1n -2^{-n}$ is decreasing and tends to zero, so de supremum/maximum is $\frac 11 -2^{-1}= \frac 12$, the infimum is zero, but it is not the minimum.


问:
If a periodic function has discontinuity then the series of fourier coefficients diverge
Let $ f:\mathbb{R} \to \mathbb{C} $ be periodic function and continuous in the interval $ [-\pi,\pi) $. (The period is $2\pi $ ).

We are given that $ lim_{x \to \pi^-} f(x) \neq f(-\pi) $, prove that $ \sum_{k \in \mathbb{Z}} | \hat{f}(k) | = \infty $
.

I tried to assume for contradiction that the sum actually converge, then by the M-test we'll get that Fourier series of $ f $ converge, and thus converge to $f$ in any point $ x_0 $ where $ f $ is continuois at. But Im not sure  how to get the contradiction.

Ive been thinking about it for a while, any ideas would help.
Thanks in advance

答: If $\sum |\hat {f} (k)| <\infty$ then $g(x)=\sum \hat {f} (k) e^{ikx}$ defines a continuous function $g$ with $g(x+2\pi)=g(x)$ for all $x$. Note that the series is uniformly convergent. You an calculate $\int g(x)e^{-ikx} dx$ by integrating term by term. You will find that $f$ and $g$ have the same Fourier coefficients. This implies that $f=g$ almost everywhere. By continuity we get $f(x)=g(x)$ for all $x <\pi$.  But $g(\pi)=g(-\pi)$ so $\lim_{x \to \pi -} f(x)=\lim_{x \to \pi -} g(x)=g(\pi)=g(-\pi)=f(-\pi)$.


问:
Function of a Markov chain is a Markov chain
Given a Markov chain $X_n$ with it's states taking values in the set $S$. $f$ is a function, $f:S\rightarrow\mathbb{R}$. If a $f(X_n)$ is also a Markov chain, prove that either $f$ is injective or $f$ is constant.  I'm totally clueless as to how I can proceed.

答:
The statement you're trying to prove is false.

For example, let $S =\{1,2,3\}$. Let $X_n$ be a Markov chain with transition matrix
$$
\left[
    \begin{array}{ccc}
      1 & 0 & 0 \\
      0 & 1 & 0 \\
      0 & 0 & 1
    \end{array}
  \right].
$$
Let $f(x) = 1$ if $x=1$, and $2$ if $x \in \{2,3\}$. Then $f$ is neither injective nor constant, but $f(X_{n})$ is a Markov chain with transition matrix
$$
\left[
    \begin{array}{cc}
      1 & 0\\
      0 & 1\\
    \end{array}
  \right].
$$

The above example was not irreducible. Here's an irreducible example:
$$
\left[
    \begin{array}{ccc}
      1/3 & 1/3 & 1/3 \\
      1 & 0 & 0 \\
1 & 0 & 0
    \end{array}
  \right].
$$
We keep the same $f$. Then $f(X_{n})$ has transition matrix
$$
\left[
    \begin{array}{cc}
      1/3 & 2/3\\
      1 & 0\\
    \end{array}
  \right].
$$

Here's a more complicated example:
$$
\left[
    \begin{array}{ccc}
      2/6 & 1/6 & 3/6 \\
      1/4 & 2/4 & 1/4 \\
      1/4 & 1/4 & 2/4
    \end{array}
  \right].
$$
Then $f(X_{n})$ has transition matrix
$$
\left[
    \begin{array}{cc}
      1/3 & 2/3\\
      1/4 & 3/4\\
    \end{array}
  \right].
$$

As suggested in the second comment on the question, it's possible to merge states with the same outgoing probabilities, but in the last example, states 2 and 3 do not have the same outgoing probabilities. They only have the same outgoing probability to state 1.


问:
Finding cosets and coset leaders of [6,3] code
I was given a generator matrix:

$$ G =\begin{bmatrix}
1 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1
\end{bmatrix} $$

From there I was able to find that the code C is:

$$ C = \{\{0 0 0 0 0 0\}, \{1 0 0 1 1 1\}, \{0 1 0 1 0 1\}, \{0 0 1 0 1 1\}, \{1 1 0 0 1 0\}, \{1 0 1 1 0 0\}, \{0 1 1 1 1 0\}, \{1 1 1 0 0 0\}\} $$

I am trying to find the cosets and coset leaders. I know that the following will be coset leaders:

$$ \{000000\}, \{000001\}, \{000010\}, \{000100\}, \{001000\}, \{010000\}, \{100000\} $$

However, based on the [n, k] code, I know I should have 8 coset leaders. How do I go about finding the final one that is missing? Is there an algorithm for it?

答:
After the automatic start of listing all the vectors of weight $<d_{min}/2$ this task often involves a bit of case-by-case analysis. I'm not sure if there is an efficient general algorithm. Just go about it systematically starting from the low weight end. For my research I once needed to classify and enumerate the coset leaders of the Reed-Muller code $R(5,2)$, and I recall it being a fun exercise.

The next possible weight of a coset leader is two. There are $\binom62=15$ vectors of weight two. Let's denote by $x_i$ the vector with five zeros and a single $1$ at position $i$, $i=1,2,\ldots,6$. No vector of weight two is in the coset $C$, so we check those in $x_i+C, i=1,\ldots,6$. We get them by adding a codeword of weight three to $x_i$, so we go through the four weight three codewords $010101$,
$001011$, $110010$, $101100$ ($111000$ is not a codeword, the sum of all three rows of the generator matrix is $111001$).

$x_1+C$ has $010010$ and $001100$,

$x_2+C$ has $000101$ and $100010$,

$x_3+C$ has $000011$ and $100100$,

$x_4+C$ has $010001$ and $101000$,

$x_5+C$ has $001001$ and $110000$,

$x_6+C$ has $010100$ and $001010$.

We see that the list is missing the weight two vectors
$$y_1=100001,y_2=011000,y_3=000110.$$
They must thus all be the leaders of the remaining coset. In particular, they must belong to the same coset. Indeed,
$$
\begin{aligned}
y_1+y_2&=111001\in C\quad \text{(the correction in my comment)}\\
y_1+y_3&=100111\in C
\end{aligned}
$$
proving that $y_2,y_3\in y_1+C$.

Whenever the weight of a coset $\ge d_{min}/2$ we cannot expect a coset to have a unique leader. This remaining coset has three leaders.

An afterthought. This code is gotten by shortening a perfect $[7,4,3]$ binary Hamming code $C^+$ with generator matrix
$$ G^+ =\begin{bmatrix}
1 & 0 & 0 & 1 & 1 & 1 & 0\\
0 & 1 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 1 & 1 & 0\\
0 & 0 & 0 & 1 & 1 & 0 & 1
\end{bmatrix}$$
in the last bit position. Because $C^+$ is a perfect single-error-correcting code, the entire space $\{0,1\}^7$ is covered by $C^+$ itself together with the seven weight one cosets $x_i+C^+, i=1,2,\ldots,7$. The three weight two words that are leaders of the remaining coset of $C$ are gotten by shortening the seventh bit from $x+x_7$ where $x$ ranges over those weight three words of $C^+$ that have the seventh bit on.


问:
Doubt in Hoffman and Kunze Linear Algebra Section 8.5
Here is the proof from the book I was reading

Now, I know about linear transformations and how in its matrix representation , the columns are coordinates of the image of basis vectors in the transformation. I don't know how and why but today while reading this proof I thought of working this proof out and it turned out that matrix multiplication doesn't make sense to me anymore.It is written that $T \alpha_1 = A_{11} \alpha_1$ . Now this makes sense  since first column is the coordinate of image of first basis vector. But if I go by explicit matrix multiplication, it turns out be something else and I got confused .Basically how to prove that this is correct by explicit matrix-multiplication?? This might too stupid to ask but I am badly stuck.

Here is what I mean by explicit multiplication:

答:
You are making several category errors, all based on confusing abstract vectors with their coordinate representations given a certain basis.

First, $T\alpha_1$ is a vector in the abstract vector space $V$. It is not necessarily a column vector in $\mathbb{R}^n$. Thus your first equation $T\alpha_1=A\alpha_1$ is already an error. You cannot multiply the $n\times n$ matrix $A$ and the abstract vector $\alpha_1$; the right side of your equation is simply nonsense. You are confusing the vector itself with its coordinate representation.

You can, however, multiply $A$ and the coordinate representation of $\alpha_1$ in the given basis, which is just the column vector $(1, 0, \ldots, 0)^T$. This is your second error: you represented $\alpha_1$ using arbitrary coordinates $x_i$ when in fact the coordinates should be $x_1=1$ and $x_i=0$ otherwise, because $\alpha_1$ is part of the basis. When you represent $\alpha_1$ correctly in coordinates, the product is the vector $(A_{11}, 0, \ldots, 0)^T$, which is indeed $A_{11}$ multiplied by the (correct) coordinate representation of $\alpha_1$.

But all of this is overkill. What you need to recall is that the numbers in the first column of $A$ tell you the coordinates of the abstract vector $T\alpha_1$ with respect to the given basis. (There is a theorem in the book that says this.) Since $A$ is upper triangular, the only nonzero coordinate is $A_{11}$, which tells you that the abstract vector $T\alpha_1$ is just $A_{11}\alpha_1$, as desired.


问:
Homogeneous first and second order difference equations
I don't really get what I am meant to be doing here, ALSO (...)- ANYTHING inside brackets are subscripts since I don't know how to do that here.
Solve the following first order difference eqn :
$$2\,x_{n+1} + 3\,x_n = 0$$
I know the answer is X(n)=A[-3/2]^n (sorry gave you the wrong answer must have confused you my bad!

I have worked it out
$$2\,x+3=0\quad X_n=A\left(-\frac 32\right)^n$$
$$2\,x=-3$$
$$x=-\frac 32$$
and the formula $x_n=C.[a^n]$ but I actually don't really know or understand what I am doing :(

this will probably seem very easy to most of you mathematicians out there but could you please explain every single step?

答:
The basic idea is to assume that any solution has the form $$
  x_n = n^kz^n
$$
for some $z$. For a particular choice of $k$, that turns the (homogenous) equation into a polynomial in $z$, and the roots of the polynomial thus correspond to the solutions of your difference equation.

You always start with $k=0$, i.e. $x_n = z^n$, and only needs to consider higher values of $k$ if that doesn't yield "enough" solutions.


问:
The Universal Property of Tangent Spaces
There are many different ways to define the tangent space $T_pM$ at some point $p\in M$, just like there are different ways to define the tensor product $V_1\otimes\cdots\otimes V_n $ or the cartesian product $V_1\times\cdots\times V_n$ of vector spaces. In the latter cases, all definitions have a universal property that guaranturees the existence of exactly one natural isomomorphism from one set to the other.

Does $T_pM$ have such a universal property, too?

By the way, I searched for this on the Internet and I found this post on Mathoverflow, but I understand neither the question, nor the answers, as they are very advanced (I am an undergraduate). I'm hoping that there is an easier answer like in the case of the tensor product. (For comparison, here is an explanation for the case of the cartesian product that I like.)

答:
$\newcommand{\dif}{\mathrm{d}}$
The following is inspired by the pages 362-3 of Tammo tom Dieck's book "Algebraic Topology". The definition of tangent spaces and the differential exploits the fact that we have already defined the differential of a differentiable function $f\colon U\subset\mathbf{R}^n\to\mathbf{R^n}$: The differential
$$Df\colon U\to L(\mathbf{R}^n,\mathbf{R^n})$$
is given by the Jacobian:
$$
Df(a)(x)=\partial_if^j(a)x^ie_j=\begin{pmatrix}\partial_if^1(a)x^i\\
\vdots\\
\partial_if^n(a)x^i
\end{pmatrix}=\begin{pmatrix}\partial_1f^1(a)&\cdots&\partial_nf^1(a)\\
\vdots&&\vdots\\
\partial_1f^n(a)&\cdots&\partial_nf^n(a)
\end{pmatrix}\cdot\begin{pmatrix}
x^1\\
\vdots\\
x^n
\end{pmatrix}$$

The universal property

Let $M$ be a differentiable, $n$-dimensional manifold - a manifold with an atlas such that all transition charts are differentiable. Consider the following subset of $A$ for some $p\in M$:
\begin{equation*}
    A_p:=\left\{x\in A:p\in\text{Dom}(x)\right\}
\end{equation*}
(According to the definition of an atlas, $A_p\neq\emptyset$ for all $p\in M$.)

A tangent space at $p\in M$ is a pair $(V,\phi)$, consisting of a $n$-dimensional real vector space $V=:T_pM$ and a map
\begin{align*}
\phi\colon A_p&\to L\left(T_pM,\mathbf R^n\right)\\
x&\mapsto\phi(x)=:\dif x_p
\end{align*}
with the following properties:

$\dif x_p$ is an isomorphism for all $x\in A_p$. Thus,
\begin{equation*}
\left(\frac{\partial}{\partial x^1}|_p,\ldots,\frac{\partial}{\partial x^n}|_p\right):=(\mathrm{d}x_p^{-1}(e_1),\ldots,\mathrm{d}x_p^{-1}(e_n))
\end{equation*}
is a basis of $T_pM$ - the basis induced by $x$.

We have
\begin{equation}\tag{1}
D(y\circ x^{-1})_{x(p)}=\dif y_p\circ\dif x_p{}^{-1}
\end{equation}
for all $x,y\in A_p$.

Lemma. The tangent space is defined up to a natural isomorphism:

Let $(V,i)$ and $(V',i')$ be tangent spaces at $p$. An isomorphism $f\in L(V',V)$ is called natural, if there is an $x\in A_p$ such that $f=\mathrm{d}x_p^{-1}\circ\mathrm{d}x_p'$. You can prove that there is exactly one natural isomorphism using the chain rule.

Definition - The Differential

Let $(M,A)$ and $(M',A')$ be two differentiable manifolds.

A function $f\colon M\to M'$ is called differentiable at $p\in M$ if $y\circ f\circ x^{-1}$ is differentiable at $x(p)$ for some pair $(x,y)\in A_p\times A'{}_{f(p)}$. In this case, $y\circ f\circ x^{-1}$ is differentiable at $x(p)$ for all $(x,y)\in A_p\times A'{}_{f(p)}$.

The differential of a differentiable function $f\colon M\to M'$ is defined as
\begin{equation*}
    \mathrm{d} f_p:=\mathrm{d}y_{f(p)}^{-1}\circ D(y\circ f\circ x^{-1})_{x(p)}\circ\mathrm{d}x_p\in L(T_pM,T_{f(p)}M')
\end{equation*}
You can prove that $\mathrm{d} f_p$ doesn't depend on the choice of $(x,y)\in A_p\times A'_{f(p)}$ by using the chain rule.

Let me conclude with a useful equation:
\begin{equation*}
    \mathrm{d}f_p\frac{\partial}{\partial x^i}|_p=\underbrace{(\mathrm{d}y_{f(p)}^{-1}\circ\underbrace{D(y\circ f\circ x^{-1})_{x(p)}\circ\underbrace{\mathrm{d}x_p)\frac{\partial}{\partial x^i}|_p}_{=e_i}}_{=\partial_i(y^j\circ f\circ x^{-1})_{x(p)}e_j}}_{=\partial_i(y^j\circ f\circ x^{-1})_{x(p)}\mathrm{d}y_{f(p)}^{-1}e_j}=\partial_i(y^j\circ f\circ x^{-1})_{x(p)}\frac{\partial}{\partial y^j}|_{f(p)}
\end{equation*}


问:
Dual of 7 dimensional regular simplex
I have the vertices of a 7-simplex that can be inscribed in a 7-cube. Given this information, how can I find the vertices of its dual simplex? (Note that the vertices of this dual should be the vertices of the cube)</p>

答:
A dual of a simplex is produced by central symmetry. Just invert all coordinates, and that would be it.

This is not necessarily the case with all other self-dual polytopes (certainly not with $F_4$, for example), but that's another story.

How did I know that? Well, let's say our n-hypercube has all coordinates +1 or -1, with center at 0. A dual of an n-simplex must be another simplex, which sits in such a position that its vertices are at the centers of (n-1)-hyperfaces of our simplex, scaled accordingly. What are those hyperfaces? Why, that's simple: each of them contains all but one vertices of the simplex. What is the sum of coordinates of all but one vertices? Again, that's simple: just the coordinates of the remaining vertex, taken with the opposite sign.

So it goes.

Title: Dual of 7 dimensional regular simplex


问:
Find the condition number and show its instability
If I have $f(a,b)=a+b$ and first I have to find the condition number:

I got to $(cond f)(a,b)=\dfrac{a}{a+b}+ \dfrac{b}{a+b}=1$
But the rest of the question says to show the subtraction of close numbers will cause instability in this question?
How can I do it?

答:
Let $\text{fl}(x+y)$ the floating point representation of the sum of $x$ and $y$, where actually it holds $$\text{fl}(x+y)=\text{fl}(\text{fl}(x) + \text{fl}(y))$$ Compute the relative error:

$$\epsilon = \frac{|x+y - \text{fl}(x+y)|}{|x+y|} \leq \frac{|x|}{|x+y|}\epsilon_x + \frac{|y|}{|x+y|}\epsilon_y $$

where $\epsilon_x =\frac{|x - \text{fl}(x)|}{|x|}$ is the relative error in $x$.

The computation above shows that, whenever $ x +y \approx 0 $ you have large relative error as the denominator becomes larger and larger. This is exactly what your question is asking.


问:
Raise a number to the "y" power without using exponents.
This is kind of a spinoff on my question Divide by a number without dividing.

Can anyone think of some clever ways to raise any given number to any given power without using an exponent anywhere in your equation/formula?

$$x^{y}=z$$

答:
You can always use the Taylor series for $f(u) = e^u$.

$$
x^y = 1 + y \ln x + \frac{(y \ln x)(y \ln x)}{2!} + \frac{(y \ln x)(y \ln x)(y \ln x)}{3!} + \cdots
$$


问:
Question regarding notation of definition of operations on set families
Are those two notations saying the same? If not, how can the text in the first equation be avoided?

$$\bigcap_{i \in I} M := \{ x \in G | x \in M_i \text{ for all } i \in I \}$$

$$\bigcap_{i \in I} M := \{x \in G | \forall i \in I : x \in M_i\}$$

答: Yes, they are equivalent. $\forall$ does exactly mean "for all". If you replace the symbol there are hardly any differences at all, besides order.


问:
Finding Integer Solutions to $x^2-y^2-n=0$
I am trying to find the integer solutions to the equation $x^2-y^2-n=0$.

Effectively, I am trying to find when the difference of two perfect squares is $n$

I have tried using a modified Newton's method to find the nearest point where the equation reaches parity; however, it ends up finding non-integer solutions. I am wondering if there is a way to factor/find all the integer solutions to this equation.

For my algorithm, taking the square root is prohibitively costly and I am trying to avoid using it.

Thanks for the help!

答:
Suppose $(x, 3)=1$, also $(y, 3)=1$, the due to FLT we have:

$x^2-y^2 \equiv (x-y) \ mod (3)$

So a family of solution can be found if $x-y =3k$ such that:

$x^2-y^2= 3m$

Examples:

$x=5$, $y=2$ $\rightarrow 25-4=21=3\times 7$

Another family of solution can be found when $x\equiv ± 1$ also $y \equiv ± 1 \mod (3)$ in this case we have:

$x^2-y^2\equiv (1-1=0)mod (3)$

For example $x=7\equiv 1 \mod (3)$, $y=2 \equiv -1 \mod (3)$ then:

$49-4=45= 3\times 15$


答:
If you are willing to skip Landau notation, you can just note (when $0<x<\pi$), $\sin x = \sqrt{\sin^2 x}=\sqrt{1-\cos^2 x}$. Then:

$$\frac{\sqrt{1-\cos x}}{\sin x} = \sqrt{\frac{1-\cos x}{1-\cos^2 x}}=\frac{1}{\sqrt{1+\cos x}}$$

Note this is why there is no general limit as $x\to 0$. When $-\pi<x<0$, the value of the expression is $\frac{-1}{\sqrt{1+\cos x}}$.  You can only get a limit as $x\to 0+$ or as $x\to 0-$, not a general limit.

Title: Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$

Body:
When I'm trying to find the limit of  $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :

$\dfrac{\sqrt{1-\cos x}}{\sin x} =  \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)}  =   \dfrac{\sqrt{x^2(1+2\epsilon_1(x))}}{\sqrt{2}x(1+\epsilon_2(x))} = \dfrac{|x|}{\sqrt{2}x}\dfrac{\sqrt{1+2\epsilon_1(x)}}{1+\epsilon_2(x)} $

But I can't seem to do it  properly  using Landau notation

I wrote :

$ \dfrac{\sqrt{\frac{x^2}{2}+o(x^2)}}{x+o(x)} $

and I'm stuck... I don't know how to carry these o(x) to the end

Could anyone please show me what the step-by-step solution using Landau notation looks like when written properly ?


问:
To show that gcd (a,b) ,gcd (b,c),gcd (c,d),gcd (c,a)=1
Suppose 91 distinct positive integers greater than 1 are given such that there are at least 456 pairs, among them which are relatively prime. Show that one can find four integers a, b, c, d among
them such that gcd (a, b) = gcd (b, c) = gcd (c, d) = gcd (d, a) = 1.

My approach was to use number theory ,initually I used the theorems on HCF to prove,Eulers theorem and all,but failed.
A rigorous proof is needed

答:
Take one pair $(a,b)$ from the $456$ mentioned pairs. Then $a$ and $b$ are relatively prime, which means that $\gcd(a,b)=1$. Now choose $c=a$ and $d=b$. The resulting four intergers $a,b,c,d$ fullfill the requirements as they are all among the $91$ integers and:

$$\begin{eqnarray}
\gcd(a,b)&=&1 \\
\gcd(b,c)&=&\gcd(b,a)=1 \\
\gcd(c,d)&=&\gcd(a,b)=1 \\
\gcd(d,a)&=&\gcd(b,a)=1
\end{eqnarray} $$

qed


问:
How would I prove the identity $(x^2-y^2)=(x-y)(x+y)$, if a true identity at all?
In going through old Monterey Physics Flash Games, I approached this current step in some derivation in connection to an elastic collision, where an identity was referenced.

$(x^2-y^2)=(x-y)(x+y)$

However, I have never even heard of this identity; I can not find it online, and in plugging in $x=3$ and $y=5$ (arbitrarily) results in different values. Is this even a real identity? Why is it used if it is not true? If there is some truth in it, how would I prove it? I am in general confused as to why this constitutes a logical step in derivation if it is not even true. (As another point of consideration, is this identity true only for certain values?)

答:
You should know that for any numbers

$$a\times (b+c)=a\times b+a\times c$$

Let’s apply that with $a=x-y$ and $b=x$, $c=y$

We then have

$$(x-y)\times (x+y)=(x-y)\times x+(x-y)\times y$$

Doing the same with $(x-y)\times x=x^2-y\times x$ and $(x-y)\times y=x\times y-y^2$ and adding all that together we get

$$(x-y)\times (x+y)=x^2+x\times y-y\times x-y^2=x^2-y^2$$


问:
Question on finding more than one group homomorphisms from one group to another
Question:

How many group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$ are there?

My attempts:

For the group of $\mathbb Z/3$ with addition as the operation, I know that it refers to the group of modulo 3.

And for $S_4$, I think it refers to the symmetry group of the all the permutations of 4 elements, which is a group consisting of 24 elements.

And I know that for the group $(\mathbb Z/3, +)$ to be a homomorphism to $S_4$, I need to check if $f(ab) = f(a)f(b)$, for f being the mapping from group $(\mathbb Z/3, +)$ to $S_4$ and $a, b $ are any elements in $(\mathbb Z/3, +). $

My Question 1:
As I am not sure on the operation in $S_4$, I am quite stuck at the checking here.

My Question 2:
If $f(ab) = f(a)f(b)$ is true, it stands for there is only ONE homomorphism exist from $(\mathbb Z/3, +)$ to $S_4$, right? But I think the question gave hints that there is more than ONE homomorphism from $(\mathbb Z/3, +)$ to $S_4$. So how to find out the other group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$?

Please assist, thank you so much!

答:
This is based on group actions.

A homomorphism $\varphi\colon \Bbb Z_3 \to S_4$ is equivalent to an action $\mathcal{A}\colon \Bbb Z_3\times X\to X$, where $X:=\{1,2,3,4\}$. Now, the stabilizers are subgroups of the group which acts, so for every $i\in X$, either $\operatorname{Stab}(i)=\{0\}$ or $\operatorname{Stab}(i)=\Bbb Z_3$; accordingly, an orbit's size is either $3$ or $1$, respectively (Orbit-Stabilizer theorem). But the set of the orbits is a partition of $X$, and hence the only allowed orbit equations are $4=1+1+1+1$ and $4=1+3$. The former corresponds to the trivial homomorphism (if all the stabilizers are equal to $\Bbb Z_3$, then the kernel of the homomorphism is $\bigcap_{i=1}^4\operatorname{Stab}(i)=\Bbb Z_3$, and all the group elements are mapped to $()_{S_4}$); the latter corresponds to four embeddings, in fact: a) three out of the four stabilizers are trivial; b) the singleton orbit can be any among $O(i), i=1,2,3,4$.


问:
Why $\sqrt {xy} = \sqrt x \sqrt y$ and same for division
Why $\sqrt {xy} = \sqrt x \sqrt y$ and same for division?
I found question like these on website but I don't anything about precalculus and not even whole algebra so I want to know the proof with basic concepts that can explain it.

答:
It follows from laws of exponents.

$$(ab)^m=a^mb^m$$

Since $\sqrt x=x^{1/2}$, it follows that $$\sqrt{xy}=(xy)^{1/2}=x^{1/2}y^{1/2}=\sqrt x\sqrt y$$ for non-negative $x,y$.

As for division, simply by understanding that $$\frac{x}{y}=x\cdot\frac1y$$ and following the above reasoning should be sufficient in showing that the same applies.


问:
Extension of isometry
The original question is Here(In polar decomposition(A=RU) decomposition, How to extend Isometry into whole complex Euclidean space)

But I make brief conditions that we need.

Let $X$ is complex Euclidean space and $A$ is linear mapping of $X$

Conditions

i) $A$ is not invertible,

ii) $AA^*=R^2$ where $AA^*$, $R$ is both nonnegative self-adjoint mapping

iii) $\parallel Rx\parallel = \parallel A^*x\parallel$ for any $x\in X$

iv) If $Rx=Ry$, $A^*x=A^*y$

For any $u$ in the range of $R$, $u=Rx$, we can define $Vu$ as $A^*x$

By upper condition, $V$ is isometry mapping of $Range(R)$ to $Range(A^*)$

---Here, I can't extend $V$ to whole space, $X$

I tried with following steps.

For any $x\in X$, $x$ can be decomposed by $x=u+v$ where $u\in Range(R), (u,v)=0$

Let $u=Ra$ for some $a$.

Then, $\parallel Vx\parallel = \parallel A^*a+Vv\parallel$

To make $V$ as unitary $\parallel Vv\parallel$ should be $\parallel v\parallel$

Also, $Vv$ must orthogonal with $Range(A^*)$

But I can't how to define this $Vv$ so that we can make sure existence of mapping $V$ which is unitary.

答:
One has $\ker(R) = \ker(A^*)$, hence $ran(R) = ran(R^*) = \ker(R)^\perp = \ker(A^*)^\perp$. It follows that the space $E$ is the orthogonal direct sum
\begin{equation}
E = \ker(A^*)^\perp \perp \ker(A^*)=ran(R)\perp \ker(A^*)
\end{equation}
You already defined $V$ on $ran(R)$, it remains to define it on $\ker(A^*)$. As $V$ sends $ran(R)$ onto $ran(A^*)=\ker(A)^\perp$ and you want a unitary operator, you need to send $\ker(A^*)$ to $\ker(A)$. Take any linear isometry
\begin{equation}
\varphi: \ker(A^*)\to\ker(A)
\end{equation}
and define $V x = \varphi(x)$ for all $x\in\ker(A^*)$. This extension of $V$ should solve the problem.

Title: Extension of isometry


问:
Non-vanishing 2-form on quartic surface.
Let $S\subset \mathbb P^3$ be a quartic surface defined by a homogeneous degree 4 polynomial $F\in k[x_0,x_1,x_2,x_3]$. $S$ is a K3 surface, so it has a unique non-vanishing $(2,0)$-form $\omega$ up to scalar.

How can this $\omega$ be computed?

答:
If $S$ is smooth, $\omega$ can be calculated as a Poincaré residue (see Griffiths and Harris, pg. 147, for details).

It boils down to the following:  in an affine patch with coordinates $z_1,z_2,z_3$, where $F$ is represented by $f(z_1,z_2,z_3)$,
$$
\omega = \int_{f=0} \frac{dz_1\wedge dz_2\wedge dz_3}{f}
$$
where the integral is a higher-dimensional contour integral -- it is taken over the boundary of some arbitrarily smooth tubular neighbourhood of $S$.


问:
Infimum proof, with function
I have this question here:

Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ and let $m \in \mathbb{R}$. Consider the set:

$A = \{x \in [a,b) \mid f(x) > m\}$,  $A$ is not empty.

Prove that $s=\inf(A)$ exists and that $a\leq s<b$.

Can I say that $A$ is bounded from below by $a$? What is the explanation for that? Is this because $a\leq x<b$? But what about $f(x)$?

From here, should I use $\varepsilon$ somehow?

Thank you for your help and for your time!

答:
Your set $A$ is, by definition, a subset of $[a,b)$. So, $a$ is a lower bound of $A$ and therefore, since $\inf A$ is the greatest lower bound of $A$, $a\leqslant\inf A$.

And there is some $c\in A$. Since $c\in[a,b)$, $c<b$. So, $\inf A\leqslant c<b$.

Title: Infimum proof, with function


问:
The number of roots for this equation
To find the number of roots of $2\cos^{2} \theta=1+3^{\sin \theta}+3^{\csc \theta}$ in the interval [0,$2\pi$]

I was only able to simplify it to

$\cos 2\theta=3^{\sin \theta}+3^{\csc \theta}$

I'm uncertain on how to proceed after this .

Thanks in advance guys.

答:
Partial:
You have that
$$3^{\sin(\theta)}+3^{\csc(\theta)}=\cos(2\theta)\leq 1$$
So:

$$3^{\sin(\theta)}\leq 1 \Rightarrow \sin(\theta)\leq 0$$
But $\sin(\theta)\neq 0$(otherwise cosecant is not defined), so $\sin(\theta)<0$

So a first restriction is $\theta \in (\pi,2\pi)$
Notice now that the LHS function, and the RHS function are symmetrical respect to $\theta=\frac 32 \pi$. So you can restrict even more to $(\pi,\frac 32 \pi)$(I verified and $\frac 32 \pi$ is not a solution).
Moreover $cos(2\theta)$ has to be $>0$(it's sum of two exponentials in base 3). So our interval reduces even more to $(\pi,\frac 54 \pi)$.

Title: The number of roots for this equation


问:
Is there a suitable name for spaces where quasi-components are singletons?
I am familiar with totally disconnected spaces. All components are singletons there.

Is there some terminology for spaces that satisfy the (stronger) condition of having only quasi-components that are singletons?

I thought of a name myself (totally separated) but maybe this is in use already for something else.

Also I am willing to subject to what is already there.

Thank you.

答: I believe "totally separated" actually is the standard term already - see e.g. here or here. Note that the latter source defines "totally separated" as "for all distinct points $p,q$, there is a partition of the space into open sets $U,V$ with $p\in U$ and $q\in V$," but that's clearly equivalent to all quasicomponents being singletons.


问:
If $f$ is a mapping from $\overline{B_1(0)}$ to itself that is continuous and analytic, prove that $f$ has a fixed point.
I am new to this but I kept on looking answer about this problem

Suppose $\displaystyle{f: \overline{B_1(0)} \rightarrow \overline{B_1(0)}}$ is continuous and $f$ is analytic in $\displaystyle{B_1(0)}$.  Prove that $f(z)$ has a fixed point in $\displaystyle{\overline{B_1(0)}}$.

I don't even know how to start with this to be honest.  Can you help me guide through it? I would appreciate it.  Thanks

答:
Edit: When $f$ is a self-mapping of the boundary circle, then you may also want to know that $f$ is a finite Blaschke product (see https://en.m.wikipedia.org/wiki/Blaschke_product); that is, $f$ can be written as
$$f(z)=e^{i\theta}\prod_{k=1}^n\frac{z-a_k}{1-\overline{a_k}z}\,$$
where $\{a_k\}_k$ is a sequence of complex numbers (with possible repetitions) lying entirely in $B_1(0,1)$, and $\theta$ is a real number. Note that $f$ is defined for all $z\in\mathbb{C}\setminus\{1/\overline{a_k}\}_k$, and that we have the identity
$$\overline{f(z)}f(\frac{1}{\overline{z}})=1\,,$$
for all $z\in\mathbb{C}\setminus\{1/\overline{a_k}\}_k$ . It follows that if $z\ne 0$ and $f(z)=z$, then we also have that $f(\frac{1}{\overline{z}})= \frac{1}{\overline{z}}$. Thus it suffices to show that $f$ has at least one fixed point $z$ in $\mathbb{C}\setminus\{1/\overline{a_k}\}_k$ then it would follow that either $z$ or $\frac{1}{\overline{z}}$ lies in $\overline{B_1(0,1)}$.
But observe that the fixed point equation $f(z)=z$ in $\mathbb{C}\setminus\{1/\overline{a_k}\}_k$ is equivalent to solving for the zeroes of the polynomial
$$g(z):=e^{i\theta}\prod_{k=1}^n(z-a_k)-z\prod_{k=1}^n(1-\overline{a_k}z)\,,$$
which exist thanks to the Fundamental Theorem of Algebra.


问:
Non differentiability of self-similar processes
Proposition.(Non differentiability of self-similar processes) Suppose $(X_t)$ is H-self-similar with stationary increments for some $H\in (0,1)$. Then for every fixed $t_0$,
$$\limsup_{t\downarrow t_0}\frac{\lvert X_t-X_{t_0}\rvert}{t-t_0}=\infty,$$
i.e sample paths of H-self-similar processes are nowhere differentiable with probability $1$.

Proof. Without loss of generality we choose $t_0=0$. Let $\{t_n\}$ be a sequence such that $t_n\downarrow 0$. Then, by H-self-similarity, $X_0=0$ a.s. Let $x>0$ be a real number. Observe that the event
$$\bigg\{\sup_{0\le s\le t_{n+1}}\frac{\lvert X_s\rvert}{s}>x\bigg\}\subseteq\bigg\{\sup_{0\le s\le t_{n}}\frac{\lvert X_s\rvert}{s}>x\bigg\},$$ then

\begin{equation}
\begin{split}
\mathbb{P}\bigg\{\lim_n\sup_{0\le s \le t_n}\frac{\lvert X_s\rvert}{s}>x\bigg\} = &\lim_n\mathbb{P}\bigg\{\sup_{0\le s \le t_n}\frac{\lvert X_s\rvert}{s}>x\bigg\}\\
= &\limsup_n\mathbb{P}\bigg\{\sup_{0\le s \le t_n}\frac{\lvert X_s\rvert}{s}>x\bigg\}\\
\ge & \limsup_n\mathbb{P}\bigg\{\frac{\lvert X_{t_n}\rvert}{t_n}>x\bigg\}\\
= & \limsup_n\mathbb{P}\bigg\{\lvert X_1\rvert >xt_n^{1-H}\bigg\}\\
\color{red}{=} & \mathbb{P}\big\{\lvert X_1\rvert>0\big\}
\end{split}
\end{equation}

Question I did not understand the inequality in red. Could someone explain it to me? Thanks.

I observed that the events sequence
$$\bigg\{\lvert X_1\rvert> xt_n^{1-H}\bigg\}\subseteq \bigg\{\lvert X_1\rvert> xt_{n+1}^{1-H}\bigg\}$$

答:
Note that since $H < 1$ and $t_n \downarrow 0$ one has that
$$\{|X_1| > 0\} = \bigcap_{n \geq 1} \{|X_1| > xt_n^{1-H}\}$$
so that (since the sets in the intersection on the right hand side are decreasing in $n$) by elementary properties of probability measures
$$\mathbb{P}(\{|X_1| > 0\}) = \mathbb{P}\left(\bigcap_{n \geq 1} \{|X_1| > xt_n^{1-H}\}\right) = \lim_{n \to \infty} \mathbb{P}(\{|X_1| > xt_n^{1-H}\}).$$


答:
Since $a$ is a lower bound of $A$ (i.e. $a\leq x$ for all $x\in A$) we need to show that it is indeed the greatest lower bound of $A$  (this is the definition of an infimum). Suppose for contradiction there exists an element $c>a$ that is also a lower bound of $A$. But $a\in A$ and by assumption $a<c$, hence $c$ can’t be a lower bound on $A$. Therefore $\inf(A)$ exists, namely $a$.

Next we need to show that $a\leq a < b$. Clearly this boils down to showing that $a<b$. Note that $b$ is an upper bound of $A$ and also $b\notin A$. This implies $x<b$ for all $x\in A$. In particular, $a<b$, as required.

Title: Infimum proof, with function

Body:
I have this question here:

Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ and let $m \in \mathbb{R}$. Consider the set:

$A = \{x \in [a,b) \mid f(x) > m\}$,  $A$ is not empty.

Prove that $s=\inf(A)$ exists and that $a\leq s<b$.

Can I say that $A$ is bounded from below by $a$? What is the explanation for that? Is this because $a\leq x<b$? But what about $f(x)$?

From here, should I use $\varepsilon$ somehow?

Thank you for your help and for your time!


问:
Simplifying fractions - Ending up with wrong sign
I've been trying to simplify this
$$
1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)}
$$
to get it to that
$$
1-\frac{(n+3)-1}{(n+2)(n+3)}
$$
but I always end up with this
$$
1-\frac{(n+3)+1}{(n+2)(n+3)}
$$
Any ideas of where I'm going wrong?
Wolfram Alpha gets it to correct form but it doesn't show me the steps (even in pro version)

Thanks

Body:
Just in case you want to see a full simplification:

\begin{align*}
1-\frac{1}{n+2}+\frac{1}{(n+2)(n+3)} &= \frac{(n+2)(n+3)}{(n+2)(n+3)} - \frac{(n+3)}{(n+2)(n+3)}+\frac{1}{(n+2)(n+3)} \\
&= \frac{(n+2)(n+3)-(n+3)+1}{(n+2)(n+3)}\\
&= \frac{(n^2+5n+6) -n-2 }{(n+2)(n+3)} \\
&= \frac{n^2+4n+4}{(n+2)(n+3)} \\
&= \frac{(n+2)^2}{(n+2)(n+3)} \\
&= \frac{n+2}{n+3} \\
\end{align*}
Provided $n\neq -2$.


问:
First order condition of Lagrange and Integral.
I'm study Economics and I have a math question about the first order condition on an integral. Let's me explain:

First, the Lagrangian for the problem (Optimal Allocation of Consumption Expenditures Dixit-Stiglitz Model) is:

$$\mathcal{L} = \left[\int_0^1 C_t(i)^{1-\frac{1}{\theta}}  di\right]^{\theta /(\theta-1)} - \lambda \left(\int_0^1 P_t(i) C_t(i) di \right)$$

The first order condition in the book is something like this:

$$\left[\int_0^1 C_t(i)^{1-\frac{1}{\theta}}  di\right]^{1 /(\theta-1)}  C_t(i)^{-1/\theta} = \lambda P_t(i)$$

Do you know why? I tried leibniz formula, fundamental theorem of calculus, but nothing works.

Thank you !

答:
I'm reading the question now and I think the FOC could be obtained with derivation rules, even if integrals make the expression apparently difficoult.
We have to find the optimal quantity consumed of every good in our model so we need to derive the lagrangian respect to $C_t(i)$ so:

$\displaystyle\frac {d\mathcal{L}}{dC_t(i)}=0$ and it is equal to $\displaystyle \frac{\theta}{\theta-1}\Big[\int_{0}^{1}C_t(i)^{1-\frac{1}{\theta}}di\Big]^{\frac{\theta-\theta+1}{\theta-1}}\cdot \Big(1-\frac{1}{\theta}\Big)C_t(i)^{1-\frac{1}{\theta}-1} -\lambda P_t(i)=0$

I think the reason because the summation is not present in the derivative is that we have to derive respect all $C_t(i)$ and here, we have an infinite quantity of them, because we have an integral, so we write the result for one of them, I think. When we derive the integral I think we can deal with it as we are dealing with a normal function because here we're supposing that the quantity consumed for each variety increases a little bit, so we have to do a integral over a quantity consumed a little bit higher of each variety $C_t(i)$. Leibniz's rule I think it can't be used because is for integrand functions with two variables. Maybe you forgot the share of income for the manufacturing good into the parenthesis multiplied by $\lambda$ from which you've to subtract $\displaystyle \int_{0}^{1} P_t(i)C_t(i) di$. Please someone correct me if I made mistakes because I'm not 100% sure of everything I've written.

What's the name of the book you're using?


问:
Explanation for some parts of this proof.
We want to prove that in a UFD the gcd exists.

In the proof we let $R$ be a UFD and $a,b \in R$
So we suppose that $a,b$ are not units because if one of them were the unit, the proof is trivial.

The proof now let $a,b$ be both non units.
Well we know that we can write $$a=T_1×...×T_m$$ and $$b=R_1×....×R_m$$ were $T_i$ and $R_j$ are irreducible elements.
Now we can rewrite them as $$a=p_1^{α_1}.....p_n^{α_n}$$
And $$b=p_1^{β_1}....p_n^{β_n}$$ were $p_i$ are prime obviously.
Here's my 1st question: how did we let the prime elements in both $a$ and $b$ be equal and with same index $n$?? Shouldn't they be different? Shouldn't we say that $b=q_1^{β_1}...q_m^{β_m}$??

Also:
it defines $s_i=min(α_i,β_i)$ and let $d=p_1^{s_1}...p_n^{s_n}$. again how are the primes equals to the case of $a$ and $b$?
The proof proceeded to show that $d$ is the gcd.
I can see why $d$ is the common divisor.
To show that d is the gcd, we let $x|a$ and $x|b$
Then if x is a unit then $x|d$
But what if x is non unit??

The proof needs details that the author didn't include or they're trivial but I don't see them.
Please help! :(

答:
The trick is that we allow the exponents $\alpha_i,\beta_i$ to be zero, and the list $p_1,\dots,p_n$ contains all primes that divide either $a$ or $b$.

The other important thing is that in an UFD, $a\,|\,b$ iff for every prime $p$ and exponent $\gamma\in\Bbb N$ we have $p^\gamma\,|\,a\implies p^\gamma\,|\,b$, that is, for every prime $p$, if the exponent of $p$ in $a$ is $\alpha$ and in $b$ is $\beta$ (and a priori they can be zero as well), then $\alpha\le\beta$.


答:
$\sqrt x$ is defined to be the unique non-negative number such that $\sqrt x^2=x$. The same goes for $\sqrt y$ and $\sqrt{xy}$. They, too, are the unique non-negative numbers whose square is $y$ or $xy$, respectively.
Now since $\sqrt x$ and $\sqrt y$ are non-negative, their product $\sqrt x\sqrt y$ is also non-negative. In addition, we have that
$$\left(\sqrt x\sqrt y\right)^2=\sqrt x^2\sqrt y^2=xy.$$
So $\sqrt x\sqrt y$ is a non-negative number whose square is $xy$. So by definition, $\sqrt{xy}$ must be this number. Or more concisely:
$$\sqrt x\sqrt y=\sqrt{xy}.$$
For division, we must first show that $\sqrt{\frac{1}{x}}=\frac{1}{\sqrt x}$. This works exactly the same: $\frac{1}{\sqrt x}$ is non-negative and its square is $\frac{1}{x}$, so by definition it is the square root of $\frac{1}{x}$. Now apply both rules to get
$$\sqrt{\frac{x}{y}}=\sqrt{x\frac{1}{y}}=\sqrt x\sqrt{\frac1y}=\sqrt x\frac{1}{\sqrt y}=\frac{\sqrt x}{\sqrt y}.$$

Body:
Why $\sqrt {xy} = \sqrt x \sqrt y$ and same for division?
I found question like these on website but I don't anything about precalculus and not even whole algebra so I want to know the proof with basic concepts that can explain it.


问:
Maximum/minimum values for two-dimensional type-II discrete cosine transform matrix
When encoding a JPEG image, the pixels are encoded as an 8x8 matrix of values in the range [-128...127]. A two dimensional type-II DCT is applied to the matrix and the result is compressed further.

Is there a way to calculate the maximum and minimum output values for any given element of the the matrix? I need to analyse a matrix and determine if it is possibly the result of a DCT transform. If a value of a given element is 'out of bounds' then I can discard the data as it cannot possibly be output of a valid DCT transform.

Probably the best solution would be to perform a reverse DCT and see if any values are outside the range [-128...127] but most of the time I will not have the value of element at [0,0] available.

Any ideas?

答:
A classical DCT-II transformation writes:

$$
D[u,v] = \frac{1}{4} C_u C_v \sum_{y=0}^7 \sum_{x=0}^7 I[x,y] \cos\frac{(2x+1)u\pi}{16} \cos\frac{(2y+1)v\pi}{16}
$$

I will leave the outer constant $\frac{1}{4} C_u C_v$ aside for the reasoning, and write the inner kernel:

$$C[u,v]= \cos\frac{(2x+1)u\pi}{16} \cos\frac{(2y+1)v\pi}{16}.$$

As a consequence, the summation rewrites:

$$\sum_{y=0}^7 \sum_{x=0}^7 I[x,y]C[u,v].$$

You can find an upper bound by maximizing each product, choosing $I[x,y]$ carefully: if $C[u,v]>0$, pick $I[x,y]=127$, otherwise set $I[x,y]=-128$ (hence each $I[x,y]C[u,v]$ product is maximally positive). Similarly, a lower bound is given for $I[x,y]=-128$ when $C[u,v]>0$ and $I[x,y]=127$ when $C[u,v]\le 0$.

Thus, you just have to produce a matrix of the signs of $C[u,v]$, and turn the +1 into 127, and the -1 into -128, and compute both sums of products.

Plugin in the outer constant $\frac{1}{4} C_u C_v$ will give you possible lower and upper bounds.

Two last thoughts: first, your input matrix might be integer-valued, hence you can add some rounding to the above values. Second, the input values are signed integers in the $[-128,127]$ range, which is not the most usual. Pixel values have often in  $[0,255]$. I thus suspect that an offset could be consider as well.


问:
Show why the proof for the following is false $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm1$
Let $u,m,n$ be three integers.

If $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm 1.$

If $\gcd(u,m) = 1$, then $1 = us + mt$ for some integers $s, t$.

If $u \mid mn$, then
$us = mn$ for some integer s.

Hence, $1 = mn + mt = m(n + t)$, which
implies that $m|1$, and therefore $m = 1$.

First dips into discrete math and having trouble with proofs

I've been playing around with various values and found $u=2, m=3, n=4,$ to be a counterexample to the proof but have yet to find the flash of inspiration needed to understand why.

Any help is appreciated.

答: There exist integers $s ,t$ with $1=us+mt.$ There exists integer $s'$ with $us'=mn.$ IF $s=s'$ THEN $m=\pm 1.$ But if $s\ne s'$ then ... ?


问:
Finding remainder using Fermat's Little Theorm

Use Fermat's Little Theorem to find the remainder of $5^{15}$ divided by $1337$.

I know Fermat's Little Theorem, but failed to understand how to use it in this.

output.944

答:
We'll forego using any 'heavy' theory and see what we can come up with to manually crank out the answer.

Anticipating what is to come we write out,

$\tag 1 \;(13 + k)10^2 \equiv k10^2 - 37 \pmod{1337}$

Getting to work,

$\tag 2 5^5 = 3125 = 3100 +  25$

Since $31 = 13\cdot2 + 5$,

$\tag 3 5^5 = (13*2 + 5) 10^2 + 25 = 2(13 + 3) 10^2 - 75$

and so by $\text{(1)}$,

$\tag 4 5^5 \equiv 2(3\cdot10^2-37)-75 \equiv 2 (263) -75\equiv 451 \pmod{1331} $

With the $\text{(4)}$ work behind us, and observing that

$\quad 451 = 450+1 \land 450 = 2 \cdot 3^2 \cdot 5^2$

we apply the binomial theorem,

$\tag 5 (2 \cdot 3^2 \cdot 5^2 + 1)^3= 2^3 \cdot 3^6 \cdot 5^6 + 3\cdot 2^2 \cdot 3^4 \cdot 5^4 + 3\cdot 2 \cdot 3^2 \cdot 5^2+ 1$

Lets us modulo reduce the first (and largest) term on the rhs of  $\text{(5)}$ using mental arithmetic and $\text{(1)}$ and, at the start, $\text{(4)}$:

$\quad 2^3 \cdot 3^6 \cdot 5^6 \equiv 8 \cdot 9 \cdot 9 \cdot 9 \cdot 5 \cdot 451 =$

$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot 9 \cdot 2255 =$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot 9 \cdot (2300 -45) \equiv$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot 9 \cdot (1000-37 -45) \equiv$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot 9 \cdot 419 \cdot (-1) =$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot (3600 + 90 + 81) \cdot (-1) \equiv$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot (2300 -37 + 90 + 81) \cdot (-1) \equiv$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot (1000 - 37 -37 + 90 + 81) \cdot (-1) \equiv$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot 9 \cdot 240 =$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot (2100 + 60) \equiv$

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot (800-37 +60) \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 9 \cdot (514) \cdot (-1) = $

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot (4500 + 90 + 36) \cdot (-1) \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot (3200 - 37 + 90 + 36) \cdot (-1) \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot (1900 -37 - 37 + 90 + 36) \cdot (-1) \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot (600 - 37 -37 - 37 + 90 + 36) \cdot (-1) = $

$\quad \quad \quad \quad \quad \quad \; \;  8 \cdot 615 \cdot (-1) = $

$\quad \quad \quad \quad \quad \quad \; \;  2 \cdot 4 \cdot 615 \cdot (-1) = $

$\quad \quad \quad \quad \quad \quad \; \;  2 \cdot (2400 +60) \cdot (-1) \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  2 \cdot (1160 - 37) \cdot (-1) \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  2 \cdot 214 \equiv $

$\quad \quad \quad \quad \quad \quad \; \;  428 \pmod{1337}$

Next we modulo reduce the second term on the rhs of  $\text{(5)}$:

$\quad 3\cdot 2^2 \cdot 3^4 \cdot 5^4 =$

$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot 50^2 =$

$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot 2500 \equiv$

$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot (1200 -37)  \equiv$

$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot 174 \cdot (-1) =$

$\quad \quad \quad \quad \quad \quad \; \; 3^3 \cdot (1500 + 30 + 36) \cdot (-1)\equiv$

$\quad \quad \quad \quad \quad \quad \; \; 3^3 \cdot (200 - 37 + 30 + 36) \cdot (-1)=$

$\quad \quad \quad \quad \quad \quad \; \; 3^3 \cdot 229 \cdot (-1)=$

$\quad \quad \quad \quad \quad \quad \; \; 3^1 \cdot (1800 + 180 + 81) \cdot (-1)=$

$\quad \quad \quad \quad \quad \quad \; \; 3^1 \cdot (2000 + 61) \cdot (-1) \equiv$

$\quad \quad \quad \quad \quad \quad \; \; 3^1 \cdot 724 \cdot (-1) \equiv$

$\quad \quad \quad \quad \quad \quad \; \; 502 \pmod{1337}$

After modulo reducing the final two terms $\text{(5)}$ we can write

$\tag 6 5^{15} \equiv 451^3 \equiv 428 + 502 + 13 + 1 \equiv 944 \pmod{1337}$


问:
Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$
When I'm trying to find the limit of  $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :

$\dfrac{\sqrt{1-\cos x}}{\sin x} =  \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)}  =   \dfrac{\sqrt{x^2(1+2\epsilon_1(x))}}{\sqrt{2}x(1+\epsilon_2(x))} = \dfrac{|x|}{\sqrt{2}x}\dfrac{\sqrt{1+2\epsilon_1(x)}}{1+\epsilon_2(x)} $

But I can't seem to do it  properly  using Landau notation

I wrote :

$ \dfrac{\sqrt{\frac{x^2}{2}+o(x^2)}}{x+o(x)} $

and I'm stuck... I don't know how to carry these o(x) to the end

Could anyone please show me what the step-by-step solution using Landau notation looks like when written properly ?

Body:
Now
$$\frac{\sqrt{1 - \cos(x)}}{\sin(x)} = \frac{\sqrt{2 \sin^2(x/2)}}{\sin(x)}
= \sqrt{2} \frac{ |\sin(x/2)|}{\sin(x)} = \sqrt{2} \frac{|x/2| + O(x^3)}{x + O(x^3)}\\
= \sqrt{2} \frac{\text{sgn}(x)/2 + O(x^2)}{1 + O(x^2)} = \text{sgn}(x)/\sqrt{2} + O(x^2)$$


问:
Does limit x tends to 1 sin inverse x exist?
One thought says that since the limit x tends to 1+ does not exist therefore the limit does not exist. But other thought says that since the function is not defined at 1+, i.e. it is not in the domain of the function so we must not discuss limit over there and the limit is by default the limit x tends to 1-. Which thought is correct?</p>

答:
let's have an example :
$f(x) = \frac{x² - 25}{x-5}.$ Find $f(5)$.
As we know, if we find $f(5)$, the answer will be not defined.

But now let's find limit $x$ tends to $5$ for the same function.
When we will have an answer, we will come to know that the limiting value of the given function for $x=5$ approaches $10$.

So we can conclude that, AT A POINT, LIMITING VALUE OF FUNCTION MAY OR MAY NOT BE EXIST IRRESPECTIVE OF FUNCTION BEING DEFINED OR NOT DEFINED AT THAT POINT.

So limit value of $\sin^{-1} x$ exist and also limit exists, too if and only if we choose the value of $x$ from its domain.


问:
What is the probability that three babies are boys, given that at least one is a boy?
I know this is a simple problem but I am arguing with a friend about its solution, so I want to show him an "official" proof!
Suppose that in any birth, the probability to have a boy is $48.5\%$.

If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?

If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

For the first question, we calculate the probability of one NOT having a boy, which is $1-0.485 = 0.515$ and then the required probability of all three not having a boy is $0.515^3 = 0.1365$ so the probability that at least one will have a boy is $1-0.1365 = 0.8634 = 86.34\%$.

For the second question, since the three events are independent, the probability that all three will have a boy given that at least one will have a boy is equal to the probability that the other two will have a boy. Is it $0.485^2$? I am not sure about the second one.

答:
In the second problem, please note your set excludes the case where none of the fetus is a baby boy as you know at least one of them is.

So the probability in the second case that all three fetuses are baby boys $ \displaystyle = \frac{0.485^3}{1-0.515^3}$


答:
now, I am not sure about an "official" sol but here is mine.

now in these cases, I would use sets and a Venn diagram. let's name the persons 1,2,3
now let A = {events|1 has a boy} and similarly for B and C so the Venn diagram would look like-Venn diagram

so now the if at least one of them has a boy then we are in A$\cup$B$\cup$C and so the required probability is n(A$\cap$B$\cap$C)/n(A$\cup$B$\cup$C) = $0.485^3/(1-0.515^3)$

Title: What is the probability that three babies are boys, given that at least one is a boy?

Body:
I know this is a simple problem but I am arguing with a friend about its solution, so I want to show him an "official" proof!
Suppose that in any birth, the probability to have a boy is $48.5\%$.

If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?

If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

For the first question, we calculate the probability of one NOT having a boy, which is $1-0.485 = 0.515$ and then the required probability of all three not having a boy is $0.515^3 = 0.1365$ so the probability that at least one will have a boy is $1-0.1365 = 0.8634 = 86.34\%$.

For the second question, since the three events are independent, the probability that all three will have a boy given that at least one will have a boy is equal to the probability that the other two will have a boy. Is it $0.485^2$? I am not sure about the second one.


答: Let $X\in\{0,1,2,3\}$ be the number of boys. Let $A$ be the event that $X=3$ and $B$ be the event that $X\ge1$. By definition, we have  ${\rm P}(A\cap B)={\rm P}(A\mid B){\rm P}(B)$ so $${\rm P}(X=3\mid X\ge1)=\frac{{\rm P}(X=3\cap X\ge1)}{{\rm P}(X\ge1)}=\frac{{\rm P}(X=3)}{{\rm P}(X\ge1)}.$$ You have calculated ${\rm P}(X\ge1)$ in the first part and evaluating ${\rm P}(X=3)$ is straightforward.


问:
How to infer average per unit time in a time series data?
This may be a very basic math but I am  stuck in a type of inference. Below is an illustration of a function say $y = f(t)$. This is shown continuous, but we have a sample data with a 10 sec interval. The unit is $W/m^2$. We are interested in average value of this per unit time, say per minute or per hour.

When I take naively the average between $(t_1,t_2)$, I would only get the avearage $\bar{y}$ within the time period $(t_1,t_2)$. Extending this inference further, at any point of time $t$ within $(t_1,t_2)$, the expected $y$ would be $\bar{y}$.

To convert to kWh/m2, we need the average of $y$ per unit time, say per hour. How do we get to that from here? As per my understanding, $\bar{y}$ is instantaneous expected value of y at time t. But then what would be next step here for average $y$ per hour, say $\bar{y_1}$

答:
The area under the curve is $$\int\limits_{t_1}^{t_2} f(t)\, dt =  \bar y (t_2 - t_1)$$

If $t$ is measured in hours $(h)$, and $f(t)$ and $\bar y$ are measured in $W/m^2$, then the integral has units $Wh/m^2$.

If $t$ is measured in seconds $(s)$, and $f(t)$ and $\bar y$ are measured in $W/m^2$, then the integral has units $Ws/m^2$.

To get from a numerical answer with units $Ws/m^2$ to $Wh/m^2$, divide by $60^2=3600$.

To get from a numerical answer with units $Wh/m^2$ to $kWh/m^2$, divide by $1000$.


问:
Distribution of the sum of n independent geometric random variables
I already tried to search an answer for this question, but none of the other posts were satisfactory.

If we have n independent random variables $X_1,\dots,X_n$ where each $X_i$ is distributed according to $q_i(1 - q_i)^k, k \in \mathbb{Z}_+$, is the sum $S_n = \sum_{i = 1}^{n}X_i$ a geometric random variable?

I tried to see how this works in the case of $n = 2$ by computing the probability generating functions for $X_1$ and $X_2$ and then comparing whether or not $\phi_{S_2}(s) = \phi_{X_1}(s)\phi_{X_2}(s)$ has the form of $\frac{p}{1 - s(1 - p)}$ for some $p$. Opening the expression for $\phi_{S_2}(s)$ gives $\phi_{X_1}(s)\phi_{X_2}(s) = \frac{q_1}{1 - s(1 - q_1)}\frac{q_2}{1 - s(1 - q_2)} = \frac{q_1q_2}{1 - s((1 - q_1) + (1 - q_2) + s(1-q_1)(1-q_2))}$. To me this does not have the desired shape, but I do not really know how to perform more deeper analysis. So what gives?

答:
The MGF of your geometric distribution is the following

$$\frac{p}{1-(1-p)e^t}$$

Being $X_i$ iid, MGF of $Y=\Sigma_i X_i$ is

$$\frac{p^n}{[1-(1-p)e^t]^n}$$

Which is the MGF of a Negative Binomial


问:
A simple question about a bounded function
Let $f$ be a function defined on $[0,\infty)$. If $|f(x)| \leq M$ for all $ x \in [0, \infty)$, then can I say   $$ \exists C,R >0 : |f(x)| \leq \frac{C}{(1+x)^2}\;\;(x \geq R) $$
is equivalent to $$ \exists C >0 : |f(x)| \leq \frac{C}{(1+x)^2} \;\; (\forall x \in [0,\infty))\;?$$
I think this is true because for $ x \in [0, R)$, $$ |f(x)| \leq M = \frac{M(1+x)^2}{(1+x)^2} \le \frac{M(1+R)^2}{(1+x)^2} = \frac{C'}{(1+x)^2}.$$
I want to know that my statements are true or not. Thank you. </p>

答: Your proof is correct.  In particular, you have shown that if there exist $R$ and $C$ as specified in the first equation, you can take $C$ in the second equation to be the maximum of $C$ and $C'$.  Thus the first equation implies the second.  And the second equation also implies the first (pick $R$ to be anything at all).  Thus, the two equations are equivalent.</p>


问:
Is $y(x)=0$ a solution to the differential equation, $y=y'$?
I think I read or was told that the natural exponential function, $e^x$ is the only solution to $y=y'$, and that it originally was defined by that property.

But isn't $y(x)=0$ one too?

If so, $e^x$ isn't unique with that property, though $y(x)=0$ obviously isn't a very interesting function.

Have I been misinformed or have I misunderstood something?

答: What you read was probably that $e^x$ is the only solution of $y'=y$ such that $y(0)=1$. Actually the set  of all solutions of that differential equation is $\{ce^x\mid c\in\Bbb R\}$. Take $c=0$, and you'll get the null function.


答: actually, you are right but misinformed also, actual complete set of functions with the property $y = y'$ is $y = ke^x$ for all $k \in R$

Title: Is $y(x)=0$ a solution to the differential equation, $y=y'$?

Body:
I think I read or was told that the natural exponential function, $e^x$ is the only solution to $y=y'$, and that it originally was defined by that property.

But isn't $y(x)=0$ one too?

If so, $e^x$ isn't unique with that property, though $y(x)=0$ obviously isn't a very interesting function.

Have I been misinformed or have I misunderstood something?


问:
Poisson distribution and Probability question
I am trying to solve a question but I don't have any idea about how to solve it, could you please help me?</p>
```<code>The log-ons to a network follow a Poisson process with an average of 180 counts per hour.
a) Probability that 5 counts occur in a minute?
b) Normal approximation (with correction) of the probability that
log-on counts to the network in a minute is between 1 and 5 (end-points are included).
</code>```

答:
For part (a) we have the following:

The mean ($\lambda$) is given as $180$ counts per HOUR, however, the question asks about the probability that $5$ counts in a MINUTE, in order to work with that we need both of them to have the same unit, the easiest conversion would be to convert the given mean to minutes; $180$ counts per hour is equal to $3$ counts per minute since we have to divide by $60$.
Now since we fixed this we can apply the Poisson distribution:

$P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}$ where $\lambda$ is our mean .

Here we have $\lambda=3$ and $k=5$, applying our formula we get:

$P(X=k)=\frac{3^5 e^{-3}}{5!}=0.10081881...$

For part (b):

When approximating a Poisson distribution by the normal distribution we have the following:  Poisson($\lambda$)~Normal($\mu=\lambda$, $\sigma^2=\lambda$), note that the larger $\lambda$ is the better the approximation. Here we will keep using $\lambda=3$ since they are still asking about the average in minutes.

For the sake of understanding and to avoid confusion the question in part (b) can be rephrased as:

The log-on per minute count to a network follow a normal distribution with $\mu=3$ and $\sigma^2=3$, find the probability that we get between $1$ and $5$ log-ons per minute.

This means we need $P(1<=X<=5)$, as mentioned in the question we need to do "correction", the more technical word here is continuity correction which yields our probability range to be $P(0.5<=X<=5.5)$ can transform this to the standard normal distribution to use the table, note that if $X$ is a normal distribution then $\frac{X-\mu}{\sigma}$ is a standard normal distribution $Z$ we here we need $P(\frac{0.5-3}{\sqrt 3}<=\frac{X-3}{\sqrt3}<=\frac{5.5-3}{\sqrt3}$) that is $P(\frac{-2.5}{\sqrt 3}<=Z<=\frac{2.5}{\sqrt 3}$), converting these to decimal values in order to use the table we get $P(-1.4437<=Z<=0.14437)$, since the standard distribution table only gives us $(P(X<=x)$ we need one last step and that is:

$P(1.4437<=Z<=1.4437)=P(Z<=1.4437)-P(Z<=-1.4437)$
Using the table that is equal to: $0.92507-0.07493=0.85014$

Note that the exact values aren't in the table (this is pretty normal and happens in most exercises), I used $P(Z<=-1.44)-P(Z<=-1.44)$ which is pretty close to the numbers given and the result doesn't change much.

I don't know what you have learned and haven't learned yet so if anything I used is unclear feel free to comment.

Some good sources for further reading and examples:
http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Distrib_Poisson

http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Limits_Norm2Poisson


问:
Quadratic surfaces and intersection curves
You are given a second order surface $Ax^2+By^2+Cz^2=α$ and a plane $x+z=β$. Determine the type of the surface and the type of the curve at the intersection of the surface and the plane.

Basically, I need to write a program, that will do task, getting as parameters $A, B, C, \alpha,\beta$.
So, the best way I found to make this, is to just check every equation by coefficients signs to determine quadratic surface. But what about curves of intersection? Is there a simple way to determine their type only by given parameters?

答:
Comment:

When you intersect the surface with plane $p(x, z)=x+z=\beta$ that means you eliminate y in the surface equation. Now you have curve $Ax^2+Cz^2=\alpha$  in (x,z) plane; we may rewrite this equation as:

$\frac {x^2}{C}+\frac{z^2}{A}=\frac {\alpha}{AC}$

The coordinate of  points of intersection are the solution of this system of equatn:

$\begin{cases}\frac {x^2}{C}+\frac{z^2}{A}=\frac {\alpha}{AC}\\x+z=\beta \end{cases}$

If $A, C. \alpha>0$ then you have an ellipse. If [$A<0$ and $C>0$]or [$C<0 $ and $A>0$], then Hyperbola is the intersection.


问:
Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$.

Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$. If $DE = 15$ cm, $EF = 9$ cm, find $FG$.

What I Tried: Here is a picture :-

I have given variables for each of the sides, and there are a lot of things which I can do here.

I can use Pythagoras theorem on a lot of triangles, and more importantly, there are a whole lot of similar triangles.

For example, $\Delta DFC \sim \Delta GFB \sim \Delta GDA$ . From here :-
$$\Rightarrow \frac{(z-a)}{z} = \frac{z}{(z + y)} = \frac{24}{(24 + x)}$$
And :-
$$\Rightarrow \frac{z}{a} = \frac{(24 + x)}{x} = \frac{(z + y)}{y}$$
And :-
$$\Rightarrow \frac{(z - a)}{a} = \frac{z}{y} = \frac{24}{x}$$
And using Pythagoras Theorem on multiple triangles gives even more equations. But I am confused here how to correctly use them and finally get the value of $x$, particularly because it is difficult for me to solve $4$ variables from here.

Can anyone help me? Thank You.

答:
Triangles $AED$ and $CEF$ are similar, and the ratio of their sides is $15:9=5:3$.

Hence this is the ratio of $AD:CF$.

Therefore $BF:CF=2:3$.

Triangles $ACF$, $GBF$ are also similar, with ratio $3:2$. Since $DF=24$, $FG=16$.


答:
$CE$ is angle bisector. So use $$\dfrac{FE}{ED}=\dfrac{CF}{CD}=\dfrac{3}{5}$$

Thus $$\dfrac{FB}{CF}=\dfrac{2}{3}=\dfrac{FG}{FD}$$

Body:
Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$. If $DE = 15$ cm, $EF = 9$ cm, find $FG$.

What I Tried: Here is a picture :-

I have given variables for each of the sides, and there are a lot of things which I can do here.

I can use Pythagoras theorem on a lot of triangles, and more importantly, there are a whole lot of similar triangles.

For example, $\Delta DFC \sim \Delta GFB \sim \Delta GDA$ . From here :-
$$\Rightarrow \frac{(z-a)}{z} = \frac{z}{(z + y)} = \frac{24}{(24 + x)}$$
And :-
$$\Rightarrow \frac{z}{a} = \frac{(24 + x)}{x} = \frac{(z + y)}{y}$$
And :-
$$\Rightarrow \frac{(z - a)}{a} = \frac{z}{y} = \frac{24}{x}$$
And using Pythagoras Theorem on multiple triangles gives even more equations. But I am confused here how to correctly use them and finally get the value of $x$, particularly because it is difficult for me to solve $4$ variables from here.

Can anyone help me? Thank You.


答:
observe $\Delta DAE \sim \Delta FCE$

$ 15/9 = z/(z-a) $ so we got 'a' interms of 'z'

observe $\Delta GAD \sim \Delta GBF$

$ z/a = 24+x/x $ so done as we had gotten a in terms of z
and we'll get $ x = 16 $

Title: Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$.


问:
How to sketch $f(z) = \frac{1}{64}i \times z$ on the complex plane?
How do I sketch a plane of points on the complex plane that is multiplied by $\frac{1}{64}i$? I mean $f(z) = \frac{1}{64}i\times z$. Do I just rotate my existing plane by $\frac{1}{64}\times\frac{\pi}{2}$ to the left?

答: No. You rotate (clockwise) by an angle $\frac\pi2$, and then you apply a contraction centered at the origin in which everything is reduced by a factor of $64$.


问:
What is the probability of 4th player to make it to the final?
There are 8 players in a tennis tournament. At the beginning of the tournament, the players are matched at random. For each subsequent round, the losers are eliminated and the remaining players are matched at random.
What is the probability that the 4th make it to the finals?

The 4th one only reach the final when it is not allowed to play with first 3 players and the two second best players get eliminated in the second round, and it will only happen when in the first round the three best players play with each other so the number of favourable cases are
$$\frac{4C1}{7C1} \cdot \frac{3C2}{3C2+3C1 \cdot 3C1} \cdot \frac{1}{4C2}$$

答:
It is not clearly stated, but I take it that the assumption is that a higher ranked player will always defeat a lower ranked player

Since player ranked $4$ has to clear two rounds to reach the final, it should be obvious that $3$ lower ranked players must in effect be in the same half despite random matchings later. Place "our" player anywhere, and place $3$ lower ranked players in the same half,

Then we can solve very simply as $Pr=\frac 4 7\cdot\frac 3 6\cdot\frac2 5=\frac 4{35}$


问:
simply connected open as a union of convex open subsets
Let $U$ be a simply connected and connected open subset of $\mathbb R^2$. Is it always possible to write
$U$ as a finite (non-disjoint) union of convex open subsets of $\mathbb R^2$?

答:
No.

Here is an example. Define
$$U = \mathbb R \times (0,2) \setminus \mathbb Z \times [1,2) .$$
This set is contractible. In fact, $\mathbb R \times \{\frac{1}{2}\}$ is a strong deformation retract of $U$.

Now assume that $U = \bigcup_{i=1}^m C_i$ with convex $C_i$ (it is irrelevant whether they are open or not). One of the $C_i$ must contain more than one of the points $z_n = (n - \frac{1}{2},\frac{3}{2}) \in U$, say $z_{n_1}, z_{n_2} \in C_j$ with $n_1 < n_2$. But then $C_j$ must contain the line segment $S$ connecting $z_{n_1}$ an $z_{n_2}$. But then $(n_1,\frac{3}{2}) \in S \subset C_j \subset U$ which contradicts the definition of $U$.


问:
A problem to calculate antisymmetric relations and the recurring equations

This problem was asked in an entrance examination.To admit I could not solve it.How do we calculate those antisymmetric relations,can someone help.
And how to proceed about the recurring equation for non crossing relations.
I need some help.

This problem is lying unanswered...could anyone solve it

Any hint or something to move the prove ahead is needed

Has anyone even seen this problem

答:
For (i), each pair $(a,b)\in S\times S$ with $a<b$ gives three choices (of including at most one of $(a,b)$ and $(b,a)$ into $R$). And all these choices are independent. Thus, the number of antisymmetric relations on $S=\{1,\ldots,k\}$ is equal to $3^{k(k-1)/2}$.

As for (ii), let's call a relation good if it is "non-crossing" and "without isolated elements". Observe that if $R$ is a good relation from $\{1,\ldots,k\}$ to $\{1,\ldots,n\}$, then we must have $(k,n)\in R$ (indeed, because of the no-isolated-elements requirement, we have $(a,n)\in R$ and $(k,b)\in R$ for some $a$ and $b$; now, if both $a<k$ and $b<n$, the non-crossing requirement is violated). Now, if we remove $(k,n)$ from $R$, either $k$ or $n$ (or both) become isolated (otherwise again a contradiction with non-crossing). This gives a recurrence $$f(k,n)=f(k-1,n)+f(k,n-1)+f(k-1,n-1)$$ for $k,n>1$, with initial conditions $f(k,1)=f(1,n)=f(1,1)=1$.

For (iii), we can do even more, and prove that for $k,n\geqslant0$ $$f(k+1,n+1)=\sum_{j=0}^{\min\{k,n\}}2^j\binom{k}{j}\binom{n}{j}.$$ This can be done just by checking the recurrence and the initial conditions above. For those who don't have this result at hand, a possible approach is to find the generating function $$\sum_{k,n>0}f(k,n)x^k y^n=\frac{xy}{1-x-y-xy}$$ (out of the recurrence and the initial conditions) and expand it back into a double power series (say, w.r.t. $y$ using the geometric series, then w.r.t. $x$ using the "negative binomial" series).


问:
If $f$ is a mapping from $\overline{B_1(0)}$ to itself that is continuous and analytic, prove that $f$ has a fixed point.
I am new to this but I kept on looking answer about this problem

Suppose $\displaystyle{f: \overline{B_1(0)} \rightarrow \overline{B_1(0)}}$ is continuous and $f$ is analytic in $\displaystyle{B_1(0)}$.  Prove that $f(z)$ has a fixed point in $\displaystyle{\overline{B_1(0)}}$.

I don't even know how to start with this to be honest.  Can you help me guide through it? I would appreciate it.  Thanks

答:
If we consider the Poincare distance on the unit disk
$$d(z_1, z_2) = 2\sinh^{-1} \frac{|z_1-z_2|}{\sqrt{(1-|z_1|^2)(1-|z_2|^2)}}$$
then any holomorphic map from $D(0,1)$ to itself does not increase the distance between points. This is called the Schwarz-Pick Theorem

Now, if $0<\rho < 1$, the one checks right away that
$$d(\rho w_1, \rho w_2) < d(w_1, w_2)$$
for every $z_1\ne z_2$ in $D(0,1)$.

We conclude that $g =\rho f\colon D[0,\rho]\to D[0,\rho]$, continuous, and decreasing the distance between points. It follows that $g$ has a fixed point. Indeed, otherwise consider a point $z$ such that $d(z, g(z))$ is smallest and $>0$. But then $d(g(z), g^2(z)) < d(z, g(z))$, contradiction.

Now take a sequence $\rho_n \to 1$. There exists $z_n \in D[0, \rho_n]$ such that $\rho_n f(z_n) = z_n$. The sequence $z_n$ has a subsequence convergent to a point $z\in B[0,1]$. We have $f(z) = z$.

Note that it is possible that $f$ has a fixed point only on $S^1$. For example, consider
$$f(z) = \frac{3+4 i}{5}\cdot \frac{z -\frac{1+ 2 i}{5}}{ 1- \frac{1-2i}{5} z}$$

with unique fixed point $z=1$.


问:
Maximum and Minimum possible Ratios between number of nodes in a path graph and number of nodes in its full independent set
Suppose, we consider a path graph (straight line, no branching) with an arbitrary number of nodes n. Suppose we have two cases where independent set (IS) is always full: 1. IS is maximum possible size 2. IS in full but not maximum possible size.

Case 1:

Suppose, we consider a maximum possible independent set for that graph where i is the number of nodes in that set.
Is there a constant that denotes what the minimum possible ratio i/n is such that there is no such path graph where that ratio would be less than this constant?

If so, what's the name and value of this constant? My math background is very limited: I tried searching for it, but didn't find anything.

Case 2:

Similarly to Case 1, what's the minimum possible ratio between a number of nodes in the minimal possible full independent set (i.e., such a configuration of an independent set where it's not a maximum independent set, and yet no more nodes can be added to it: E.g., for a path graph of length 5, [1, 0, 1, 0, 1] is maximum possible (3 nodes) VS. [0, 1, 0, 1, 0] is a minimum possible full (2 nodes) independent set) and a number of nodes in a path graph i/n?

Basically, I am trying to find what the ratio bounds are for when we cannot add any more nodes to an independent set.

Thank you.

答:
The size of a maximum independent set is $\lceil n/2 \rceil$ (take every other node), so the ratio is $n/\lceil n/2 \rceil$, which is $2$ if $n$ is even and in the interval $[1,2)$ if $n$ is odd.

Update after you added Case 2 to the question: The minimum size of a maximal independent set is called the independent domination number of the graph.
For a path on $n$ nodes, this number is $\lceil n/3 \rceil$ by taking every third vertex.  The ratio $n/\lceil n/3 \rceil$ is $3$ if $n$ is a multiple of $3$ and in the interval $[1,3)$ otherwise.

In summary, for a path on $n$ nodes the ratio of the size of a maximal independent set and the number of nodes is always in the interval $[1,3]$.


问:
Proving/Disproving a limit with absolute value
Assume f: R -> R, if |f(x) − f(y)| = |x − y|,
Then limx→∞ |f(x)| = ∞

I'm new to limits and was wondering how would I prove/disprove such a claim?

答:
If $ |a| = |b| \Rightarrow a = \pm b$

so $ f(x) - f(y) = x-y \Rightarrow f(x) = x+c , c\in R$ (we get this by putting y = 0)

or

similarly $f(x) = -x +c$

and now we can easily verify that the limit is true...


问:
Is $L=\{\,a^ib^jc^k \mid i=2j=3k\, \}$ a Turing language?
Is $L= \{a^ib^jc^k \mid i=2j=3k\}$ a Turing language?

At first I thought it was possible to scan the tape and for each $c$ delete the amount of $a$ by the amount of $b$, it works when the input is $aaaaaabbbcc$ but if the input is $aaaaaaaaaaaabbbbbbcccc$ it is not.

How can I figure it out the algorithm, which technique can I use to solve this?

答:
A simple way to solve your question is to start with the context-sensitive language $C= \{a^nb^nc^n \mid n \geqslant 0\}$ and use the fact that the class of context-sensitive languages is closed under non-erasing morphisms. Now, consider the non-erasing morphism $f$ defined by $f(a) = a^6$, $f(b) = b^3$ and $f(c) = c^2$. Then
$$f(C)  = \{a^{6n}b^{3n}c^{2n} \mid n \geqslant 0 \} = L.$$ Thus $L$ is context-sensitive.


问:
There are $5$ different apples, $4$ different oranges, and $3$ different bananas. Number of ways to select at least one of each kind?

There are $5$ different apples, $4$ different oranges, and $3$ different bananas.  Number of ways to select at least one of each kind?

My first approach: I selected one of each kind first and then I have $9$ fruits remaining, of which I can choose in $2$ ways [either include or exclude].

$${5\choose1} \cdot {4\choose1} \cdot {3\choose1} \cdot (2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot  2 \cdot 2 \cdot 2 \cdot 2)$$

My second approach:
$$\bigg({5\choose1}+{5\choose2}+{5\choose3}+{5\choose4}+{5\choose5}\biggl)\bigg({4\choose1}+{4\choose2}+{4\choose3}+{4\choose4}\bigg)\bigg({3\choose1}+{3\choose2}+{3\choose3}\bigg)$$

but the two methods give different answers.

I know my second approach is correct, but what is going wrong in my first approach?

You can help me by giving small example.

答:
There are a total of $5 + 4 + 3 = 12$ fruits.  You can either select or not select each fruit.  Thus, if we did not have the restriction that at least one fruit of each type must be selected, there would be $2^{12} = 4096$ possible ways to select the fruits (including the choice of not selecting any of them).  Your first approach gives
$$\binom{5}{1}\binom{4}{1}\binom{3}{1} \cdot 2^9 = 30,720$$
an indication you are over counting.

Let's focus on the apples.  According to the count in your first approach, there are
$$\binom{5}{1} \cdot 2^4 = 80$$
ways to select at least one apple rather than
$$\binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 31$$
ways to select at least one apple.  This is because you count each selection with exactly $k$ apples $k$ times, once for each way you could have designated one of those $k$ apples as the apple you selected.  To illustrate this, consider the following two examples.

Suppose the five apples are Ambrosia, Fuji, Gala, Granny Smith, and Honeycrisp. You count the selection of an Ambrosia apple and a Fuji apple twice, depending on which of those apples you designate as the apple you have selected.

$$
\begin{array}{l l}
\text{apple} & \text{additional apples}\\ \hline
\text{Ambrosia} & \text{Fuji}\\
\text{Fuji} & \text{Ambrosia}
\end{array}
$$

You count the selection of an Ambrosia apple, a Fuji apple, and a Gala apple three times, depending on which of those apples you designate as the apple you have selected.

$$
\begin{array}{l l}
\text{apple} & \text{additional apples}\\ \hline
\text{Ambrosia} & \text{Fuji, Gala}\\
\text{Fuji} & \text{Ambrosia, Gala}\\
\text{Gala} & \text{Ambrosia, Fuji}
\end{array}
$$

Since your first approach counts each selection of exactly $k$ apples $k$ times, your first approach yields
$$1 \cdot \binom{5}{1} + \color{red}{2} \cdot \binom{5}{2} + \color{red}{3} \cdot \binom{5}{3} + \color{red}{4} \cdot \binom{5}{4} + \color{red}{5} \cdot \binom{5}{5} = \color{red}{5 \cdot 2^4} = \color{red}{80}$$
ways to select at least one apple,
$$1 \cdot \binom{4}{1} +  \color{red}{2} \cdot \binom{4}{2} + \color{red}{3} \cdot \binom{4}{3} + \color{red}{4} \cdot \binom{4}{4} = \color{red}{4 \cdot 2^3} = \color{red}{32}$$
ways to pick at least one orange, and
$$1 \cdot \binom{3}{1} +  \color{red}{2} \cdot \binom{3}{2} + \color{red}{3} \cdot \binom{3}{3} = \color{red}{3 \cdot 2^2} = \color{red}{12}$$
ways to pick at least one banana, giving your total of
$$\color{red}{80 \cdot 32 \cdot 12} = \color{red}{30,720}$$
ways to select at least one fruit of each type.

A final note:  There are $2^n$ subsets of a set with $n$ elements since we can either choose to include or not include each of the $n$ elements in a subset.  Of these $2^n$ subsets, $2^n - 1$ are nonempty.  Hence, there are $2^5 - 1$ ways to select at least one of the five apples, $2^4 - 1$ ways to select at least one of the four oranges, and $2^3 - 1$ ways to select at least one of the three bananas.  Therefore, there are
$$(2^5 - 1)(2^4 - 1)(2^3 - 1) = 31 \cdot 15 \cdot 7 = 3255$$
ways to select at least one fruit of each type.  Notice that
$$\left[\binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5}\right]\left[\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}\right]\left[\binom{3}{1} + \binom{3}{2} + \binom{3}{3}\right] = 31 \cdot 15 \cdot 7 = 3255$$


问:
Compute, up to isomorphism, all groups of order $21$
I'm trying to solve this problem from my group theory course:

Compute, up to isomorphism, all groups of order $21$

I've first considered the abelian case, whih gives me as Elemental Divisors $3$, $7$, so one group is
$$\mathbb{Z}_3\oplus \mathbb{Z}_7=\boxed{\mathbb{Z}_{21}}$$
After this, I don't know what to do. I have trouble with this kind of problems in general, where I'm asked to find all groups of certain order (The abelian case is easy to handle, I have trouble when the group is non-abelian). What is a general tip for solving this kind of problems? What is the solution in my example? Any help will be appreciated, thanks in advance.

答:
I just found my answer:

Given $|G|=21=3\cdot 7$ (product of two primes), and given that $3|(7-1)$, from my textbook I know that the only possible groups (up to isomorphism) are:

$G\cong\mathbb{Z}_3\oplus\mathbb{Z}_7\cong\mathbb{Z}_{21}$

$G\cong\mathbb{Z}_7\rtimes\mathbb{Z}_3$


问:
If matrix $B$ is skew-symmetric with eigenvalue $r$, then $-r$ is also an eigenvalue of $B$
I've considered the determinants of both $B$ and $-B^T$, and found that

$\det(B) = (-1)^n \det(B)$, where $B$ is a $n \times n$ matrix.

I've also tried the approach where $r$ is an eigenvalue of $B$ iff $\det(B-rI) = 0$, but as determinants are not preserved over addition/subtraction,
I'm not really sure what to do from there.

答:
Consider an arbitrary commutative ring $\mathbf{A}$, a natural number $n \in \mathbb{N}$ and an antisymmetric matrix $M \in \mathscr{M}_n(\mathbf{A})$, square of order $n$ over $\mathbf{A}$. I use the term antisymmetry as a synonym of skew-symmetry, meaning that the relation $M^{\mathrm{t}}=-M$ takes place.

Let $\chi_T \in \mathbf{A}[X]$ denote the characteristic polynomial of arbitrary matrix $T$. We then have by definition that:
$$\begin{align*}
\chi_{M^{\mathrm{t}}}&=\det\left(X\mathrm{I}_n-M^{\mathrm{t}}\right)\\
&=\det(X\mathrm{I}_n+M)\\
&=\det\left(-(-X\mathrm{I}_n-M)\right)\\
&=(-1)^n\det((-X)\mathrm{I}_n-M)\\
&=(-1)^n\chi_M(-X).
\end{align*}$$
On the other hand, it is also generally the case that:
$$\begin{align*}
\chi_{M^{\mathrm{t}}}&=\det\left(X\mathrm{I}_n-M^{\mathrm{t}}\right)\\
&=\det(X\mathrm{I}_n-M)^{\mathrm{t}}\\
&=\det(X\mathrm{I}_n-M)\\
&=\chi_M.
\end{align*}$$
We thus infer the relation $\chi_M=(-1)^n\chi_M(-X)$. If $\mathbf{A}$ is a field of characteristic different from $2$, we gather that the coefficients of $\chi_M$ at degrees $k$ of opposite parity to $n$ are null (and thus in particular there exists no invertible antisymmetric matrix square of odd order over a commutative field of characteristic distinct from $2$).

This means that given the decomposition $\chi_M=\displaystyle\prod_{k=1}^n(X-\lambda_k)$ for the characteristic polynomial of $M$ we also have the decomposition $\chi_M=\displaystyle\prod_{k=1}^n(X+\lambda_k)$. Therefore, if $\mathbf{A}$ is a field and $\lambda$ an eigenvalue of $M$ the opposite $-\lambda$ must also be an eigenvalue of $M$ of the same multiplicity.

In particular, when $\mathbf{A}$ is an algebraically closed field such that $\mathrm{char}\mathbf{A}\neq 2$, the characteristic polynomial of $M$ can be written as $\chi_M=X^h\displaystyle\prod_{\lambda \in R}\left(X-\lambda^2\right)^{k_{\lambda}}$, where $R \subseteq \mathbf{A}^{\times}$ is a finite subset such that $|R \cap \{\pm \mu\}| \leqslant 1$ for each scalar $\mu \in \mathbf{A}$ and $k \in \left(\mathbb{N}^{\times}\right)^R$ is the family of multiplicities, $k_{\lambda}$ being the common multiplicity that both $\lambda$ and $-\lambda$ have in $\chi_M$.