问:
Does the summation symbol imply precedence over multiplication?
Does

$$a\sum_{n=1}^N x_n$$ mean $$a(x_1 + x_2 + x_3 ... + x_N)$$ or $$a * x_1 + x_2 + x_3 ... + x_N$$

Edit: So apparently the first option is the correct one, but why? Multiplication has precedence over summation, right?

答:
It means the former.

Let's think that we have some numbers $x_1, x_2, x_3 \dots x_n$

Now, we multiply each by a constant $a$ and add them. We get : $$ax_1 + ax_2 + \dots ax_n\\ = \sum^n_{i=1}ax_i$$

Using distributive property of addition, we can rewrite this as : $$a(x_1 + x_2 + \dots +x_n)\\= a\sum^n_{i=1}x_i$$


问:
How prove this matrix $rank (A_{11}A_{12})=rank(A_{12})$
let $A_{n\times n}$ is a positive semidefinite matrix,
and
$$A=\begin{bmatrix}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{bmatrix}$$
where $A_{11}$ is $m\times m,m<n $ matrix
prove or disprove
$$rank(A_{11}A_{12})=rank(A_{12})$$</p>

答: Hint: If $A$ is Hermitian, then $A_{11}$ must have full rank by Sylvester's criterion.</p>

ContentLicense: CC BY-SA 3.0


问:
Sum with two indices, how to handle the condition $i\neq j$?
I'm given
$$
\sum_{i,j=1, i \neq j}^{N} ij \tag 1
$$
I assume this is a short hand for

$$
\sum_{i=1}^{N}
\Bigg (
\sum_{j=1}^{N} ij
\Bigg )
\qquad? \tag 2
$$
But where does $i\neq j$ belong, to the inner or outer sum?

Example with $N=2$ and $i\neq j$ in the inner sum:
\begin{align}
\sum_{i,j=1 \\i\neq j}^{2} ij
&=
\sum_{i=1}^{2}
\Bigg (
\sum_{j=1, i \neq j}^{2} ij
\Bigg )
\tag 3
\\
&=\sum_{i=1}^{2}
\Bigg (
i1+i2
\Bigg )
\tag 4
\\
&=
(11+12 + 21 + 22)
\tag 5
\\
&\text{\{Now discard 11 och 22\}}\\
&=23
\end{align}

I guess the end result is correct, but I don't think my approach is correct ("Now discard"). How should I handle the condition $i\neq j$?

答:
The sum $\sum_{i, j}^N ij$ could be thought as the iterative product of a column of natural numbers from $1$ to $N$ with $N$ lines of a list of numbers from $1$ to $N$. That makes $N^2$ computations:
$$\sum_{i, j}^N ij = (1\times 1 + 1\times2 +...+1\times N) + (2\times 1 + 2\times2 +...+2\times N) + (...) + (N \times 1 + N \times 2 +... + N \times N)$$

Now with the condition $i \neq j$ you exclude all products whose indices are the same (meaning every $1\times1, 2\times2 ... N\times N$). Now you have $N \times (N-1)$ computations.
$$\sum_{i, j : i \neq j}^N ij = (1\times2 +...+1\times N) + (...) + (N \times 1 + N \times 2 +... + N \times N-1)$$

You could also think of it as:
$$\sum_{i, j : i \neq j}^N ij =\sum_{i, j}^N ij - \sum_{i}^N ii $$

ContentLicense: CC BY-SA 4.0


答:
You can't get it wrong with a multiplication on the left, the sum is taken as a whole and clearly

$$a\sum b_i=a\left(\sum b_i\right).$$

Malicious guy could tell you that there is ambiguity in

$$\sum a_i\ b$$ which could be interpreted as

$$\left(\sum a_i\right)b$$

or
$$\sum \left(a_i\ b\right)$$

but by the distributivity law, these are equivalent.

Different is

$$\sum a_i+b$$ which could be understood as

$$\left(\sum a_i\right)+b$$ or $$\sum\left(a_i+b\right).$$

Without parenthesis, the first interpretation holds.

Title: Does the summation symbol imply precedence over multiplication?

Body:
Does

$$a\sum_{n=1}^N x_n$$ mean $$a(x_1 + x_2 + x_3 ... + x_N)$$ or $$a * x_1 + x_2 + x_3 ... + x_N$$

Edit: So apparently the first option is the correct one, but why? Multiplication has precedence over summation, right?


问:
12 two digit numbers. Show that at least two of them has a difference of the form aa
Given 12 different two digit numbers. Show that at least two of them has a difference of the form aa.

Answer: Two digits equal each other in modulo 11.

I don't understand. Why modulo 11? How does that translate to a number of the form aa?
An example is 33 and 22. 33-22=11=aa

答:
This is an application of The Pigeon Hole Principle.

Note that in $12$ different numbers, at least two of them will have the same remainder when divided by $11$. In other words, at least two of them equal each other in modulo $11$.

This implies that their difference is divisible by $11$. And since their difference is a two-digit number, it must be in the form $aa$.


问:
Is it possible to "massage" (via shear transformations) a parallelogram with integer-coordinate vertices into an axis-aligned rectangle?
(The problem is my original, unless there's prior art I'm unaware of.)

Given a parallelogram whose vertices have all integer coordinates, you can give it a "massage". Each "move" of the massage is a shear transformation where one side remains fixed and the opposite side is shifted by a vector parallel to that side and also with integer coordinates. That vector will necessarily be a multiple of the vector equal to the side itself (understood as a vector) divided by GCD(x,y) where x and y are the coordinates of the side-as-vector.

Question:

Is it possible to massage any such parallelogram into an axis-aligned rectangle? If so, how hard is the problem of finding the right sequence of moves? I understand that it is at least as hard as integer factorization, because the solution explicitly gives a factorization of the (necessarily integer) area of the parallelogram, which is preserved by moves. But is it harder? Is the sequence of moves easily computed from either a bifactorization or full factorization of the area?

EDIT: Revisiting this with a fresh mind and after Jean Marie's comments:

We can fix a vertex to the origin and always move a "free" side so the fixed vertex stays at origin; this simplifies the model.

If we manage to bring either side to an axis, it's one more trivial move (a shear along that axis) and we're done.

If we could massage the parallellogram so that one of the side-vectors have non-coprime coordinates, then the GCD of the coordinates would necessarily divide the area. This is because that side would contain lattice points along its length, allowing us to slice the parallelogram evenly into smaller ones with integer area, which is clearly only possible if the number of the smaller ones divides the area. We could then keep massaging the small one, and the big one would notionally follow. So a necessary condition for general possibility is to always be able to massage a prime area $p$ into a $1 \times p$ rectangle. Always nice to reduce (part of) a problem to primes!

答:
A provisional answer whose merit is to give a simplified version of your problem.

Do we agree that any translation can be decomposed into (at most) four successive "massages"? See figure below: going from Red parallelogram to Blue parallelogram, is done by transiting through the Black parallelogram. One can move from R to B by two "massages", from B to R as well by two "massages".

Then we can assume WLOG that our parallelogram (P) has one of its vertices at the origin, the 3 other ones being

$$u=\binom{a}{b}, \ v=\binom{c}{d}, \ u+v=\binom{a+c}{b+d}$$

Let us assume that the area of (P) is equal to the determinant of $u,v$, i.e., $ad-bc$ assumed WLOG positive is factorizable as $L \times W$. Then, the problem boils down to being able to find a sequence of "massages" sending parallelogram (P) onto rectangle with vertices:

$$\binom{0}{0}, \ u'=\binom{L}{0}, \ v'=\binom{0}{W}, \ u'+v'=\binom{L}{W}$$

which is possible (working backwards) if matrix $M$ defined by:

$$M\begin{pmatrix}L&0\\0&W\end{pmatrix}=\begin{pmatrix}a&c\\b&d\end{pmatrix} \iff M=\begin{pmatrix}a/L&c/W\\b/L&d/W\end{pmatrix}$$

has integer entries.

Remark: one can check that $\det(M)=1$ which is necessary for area preservation.

I stop here because I want to see first if you agree with what I have said.


问:
Arrangements in a circle with a fixed gap
Hello could anyone tell me if I'm on the right track? I'm a little confused with circular arrangements

This is the question : How many ways can 8 persons sit in a circle with 10 seats such that there is always exactly one person between the empty two seats?

is the answer $\frac{9!}{2}$•8 ?

My reasoning is to treat the person in the middle and the two empty chairs as one group, so then there are $\frac{9!}{2}$ possible permutations for that arrangement since the two empty chairs are identical in each arrangement. Then the individual in the middle could be any of the 8 people, resulting in $\frac{9!}{2}$•8

答:
By convention, in a circular permutation, only the relative order of the objects matters.  Therefore, we can take the person who sits between the two empty seats as our reference point.

There are eight ways to choose the person who sits between two empty seats.  As we proceed clockwise around the table from that person, there are $7!$ ways to seat the remaining seven people in the remaining seven seats, that is, the seats we have not reserved for the person who sits between two empty seats and the two empty seats.  Hence, there are
$$8 \cdot 7! = 8!$$
such seating arrangements.

The problem with your approach is that the two empty seats must be separated by exactly one seat.  Therefore, you do not have $10$ objects to permute since the location of the solitary person also determines the positions of the empty seats.


答:
This is a notational convention: things inside a summation ($\Sigma$) or product ($\Pi$) sign are considered to be a single entity for terms of other operations applied to them.  So while multiplication of single elements takes precedence over addition, the summation or product sign binds "more tightly".

So,
$$ax_1 +x_2$$ applies the $a$ only to $x_1$, but
$$a\sum_{i=1}^2 x_i$$ expands out to $ax_1 +ax_2$ always.

When you start dealing with infinite sums and products this is important.  In an infinite-dimensional Banach space the (formal) sum
$$\sum_{j\in {\mathbb N}} \alpha_j x_j$$
may or may not be an element of the space.  (For example, if the Banach space is $c_0$ and each $\alpha_j =1$ and the $\{x_j\}_{j\in {\mathbb N}}$ are the standard basis then the resulting sum is not an element of $c_0$ but rather of its bidual, $l_\infty$.)  Multiplying the formal sum by something can affect which space it lives in then, so you need to be accurate in recording what you're doing.


问:
Pick the highest of two (or $n$) independent uniformly distributed random numbers - average value?
With "random number" I mean an independent uniformly distributed random number.

One

Picking one random number is easy: When I pick a random number from $x$ to $y$, the average value is $(x+y)/2$.

Two

I'm no maths expert, nevertheless I was able to work out a solution for whole numbers: When I pick the highest of two random whole numbers from $x$ to $y$ ...

Let $n$ be the count of possible results $n = y - x + 1$.

I looked at the probability of every single possible outcome and noticed an arithmetic sequence. I knew the sequence had to start with the rarest possibility: Rolling two times in a row the lowest value.

$$p_1 = \frac{1}{n^2}$$

And I knew the sequence had to have a sum of 100 %. This made it possible to calculate the last, the $n$th, element:

$$p_n = \frac{2}{n} - p_1$$

Based on that it was easy to calculate the difference between elements and subsequently the formula for the sequence.

$$p_i = \frac{2in-2i-n+1}{n^3-n^2}$$

All I had to do now, was multiplying the probabilities with their respective values and sum this.

$$
\sum_{i=1}^{n} \frac{2in-2i-n+1}{n^3-n^2}(x+i-1)
$$

Questions

What sort of distribution is this, how is it called? I can't name it and I can hardly search for it.

What is the solution for picking the highest of two random real numbers from $x$ to $y$? I feel a little bit lost, because my approach fails, because there are infinite real numbers.

What is the solution for picking the highest of $c$ random real numbers from $x$ to $y$?

答:
Here is exactly what you are looking for for all types of random variables.

For $n$ iid discrete uniform random variables the PDF is

$$P(Y_{max}=y)=\left(\frac{\lfloor y\rfloor-a+1}{b-a+1}\right)^n-\left(\frac{\lfloor y\rfloor-a}{b-a+1}\right)^n$$

For $n$ iid continuous uniform random variables the PDF is

$$f_{Y_{max}}(y)=\begin{cases}n\left(\frac{1}{b-a}\right)^{n-1}\left(\frac{y-a}{b-a}\right)^{n} &,y\in[a,b]\\
0&,\text{otherwise}\\
\end{cases}$$


问:
Prove or disprove: $(A-B)-C=(A-C)- B$ for sets $A$, $B$, $C$

Prove or disprove: For sets $A$, $B$, $C$,
$$(A-B)-C=(A-C)- B$$

Now my attempt is

$$(x\in A \;\text{and}\; x\notin B) \;\text{and}\; x\notin C \tag1$$

We can write it as

$$(x\in A  \;\text{and}\; x\notin C)  \;\text{and}\; x\notin B \tag2$$

which proves the statement.

Is it right?

答:
Let $X$ be a set and $A,B,C \subset X.$

$A$ \ $B=A \cap B^c $(where $B^c$ is the complement of $B$ in $X$).

Then

LHS:

$(A \cap B^c)\cap C^c;$

RHS:

$(A \cap C^c) \cap B^c.$

Using associativity and commutative we are done.


问:
Inexistence of a finite $\frac{1}{2}$-net

Let $C\left( [0;1]  \right)$ be the set of all continuous functions
$f \colon [0;1] \to \mathbb{R}$, and $\| f \|_\infty := \sup\limits_{0 \leq x \leq 1} |f(x)| $
for all $f \in C\left( [0;1] \right)$. Show
that in the normed space $\left( C\left( [0;1] \right), \| \cdot \|_\infty \right)$
, the closed ball $\overline{B}_1(0)$ (with center
$g \equiv 0$, radius $1$) has no finite $\dfrac{1}{2}$-net!

Let $(X,d)$ be a metric space, $\varepsilon > 0$ and $A \subset X$. Then $A$ is said to be an $\varepsilon$-net if $ \bigcup\limits_{a \in A} B_{\varepsilon}(a) = X $.

I don't know how to prove this, can anyone help, please?

答:
It's worth considering carefully what the ball you're trying to cover looks like.
This ball $\overline{B_1} (0)$ is all the continuous functions $[0,1]\rightarrow \mathbb{R}$ that only take values within distance $1$ of $0$. In other words, your ball is all continuous functions $[0,1]\rightarrow [-1,1]$.

To show it has no finite $\frac{1}{2}$ net, you need to show that given any finite set $A = \{f_1, f_2, \dots, f_n\}$ of continuous functions $[0,1]\rightarrow [-1,1]$, there exists some continuous function $g: [0,1]\rightarrow [-1,1]$ such that $||g-f_i||_\infty > \frac{1}{2}$ for all $f_i$ (i.e., the function $g$ only has to take one single value far away from each $f_i$, of which there are finitely many).

For one possible way of doing this, click the spoiler below:

 Pick $n$ distinct points $0 < x_1 < x_2 < \dots < x_n < 1$ in $(0,1)$, and let $y_i = f_i(x_i)$. We construct a function $g$ such that $|g(x_i)-f_i(x_i)| > \frac{1}{2}$ as follows: if $y_i \geq 0$, let $g(x_i) = -1$, and if $y_i < 0$, let $g(x_i) = 1$. We can extend this to a continuous function by just making $g$ a straight line from $g(x_i)$ to $g(x_{i+1})$ on the intervals $[x_i, x_{i+1}]$. The function $g$ does not lie in any $\frac{1}{2}$ net of $A$, and we are done. For a picture example, click the next spoiler.


问:
Basic logic: modus ponens and left-nested conditionals
Consider (1) and (2):

(1) (a ⊃ b) ⊃ a

(2) a

do (1) and (2) together with modus ponens entail (3) b? My guess is no, but am I right? Could someone explain why?

答:
You are right. To appply modus ponens with $a$ as one of the premises to conclude $b$, you would need as a second premise an implication of the form $(a ⊃ b)$. But there is no such premise among the formulas you gave, so modus ponens is not applicable.

The inference is invalid because with $a \mapsto True, b \mapsto False$ there exists a counter model which makes both premises true but the conclusion false.


问:
how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded?
I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$

答:
Claim. If $a_n$ is increasing, then $a_n<3$.

Otherwise, if $a_n\ge 3$, for some $n\in\mathbb N$, then

Case A. $a_n=3$, then $a_{n+1}=\sqrt{a_n+6}=\sqrt{3+6}=\sqrt{9}=3$. Hence $a_n$ is not increasing.

Case B. $a_n>3$. Then $a_{n+1}>a_n>3$ and
$$
a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{\sqrt{(a_n+6}-a_n)(\sqrt{a_n+6}-a_n)}{\sqrt{a_n+6}+a_n}\\=\frac{a_n+6 -a_n^2}{\sqrt{a_n+6}+a_n}=-\frac{(a_n+2)(a_n-3)}{\sqrt{a_n+6}+a_n}<0.
$$
Contradiction.

Hence, $a_n$ is upper bounded by $3$ and lower bounded by $a_1$.


问:
Proof verification: $f:S\to S$ is bijective $\iff\exists ! g,h:S\to S$ s. t.$g\circ f=f\quad\&\quad f\circ h=f.$

Prove that a function $f:S\to S$ is bijective iff there exist unique functions $g,h:S\to S$ s. t.

$$g\circ f=f\quad\&\quad f\circ h=f.$$

My attempt:

Direction $\boxed{\Leftarrow}$

If the functions $f,g:S\to S$ are unique, then $f$ is bijective. I tried to use the contraposition:

If $f$ isn't bijective, then $f$ or $g$ isn't unique.

If $f$ isn't bijective, then it isn't injective or surjective.

If $f$ isn't injective, then there are some $x_1,x_2\in S,x_1\ne x_2$ and $f(x_1)=f(x_2)$.

Let $S_i:=\{x\in S\mid f(x)=y_i,\},\space i\in\mathcal I$ and let $\space T=S\setminus\bigcup\limits_{i\in\mathcal I} S_i$ where $f$ is injective. We can define $h$ in the following way:

$$h(x):=\begin{cases} x'_i\in S_i,&\color{red}{x\in S_i,  x\ne x_i'}\\ x,& x\in T\end{cases}$$

So, my idea is to permute the $x$' es which give the same output. I first thought we can start with $h(x_1)=x_2\space\&\space h(x_2)=x_1$ whenever $f(x_1)=f(x_2)$, but I tried to inductively generalize it. As we can construct $h$ different from identity, which also satisfies the condition, $h$ isn't unique.

If $f$ isn't surjective, then $\operatorname{Ran}(f)\subset S$. We can define $g$ in the following way:
$$g(x)=\begin{cases}x,&x\in\operatorname{Ran}(f)\\k(x), &x\notin\operatorname{Ran}(f)\end{cases}$$

where $k:S\to S$ is an arbitrary function. Therefore, $g$ isn't unique either.

Direction $\boxed{\Rightarrow}$

If $f$ is bijective, then $g$ and $h$ are unique.

As $f$ is bijective and $g\circ f=f,\space g\circ f$ has to be bijective as well. If so, $g$ has to be surjective, but, since $f$ is surjective, $g$ also has to be injective. Then, I concluded that $g$ is the identity on $S$.

On the other hand, since $f$ is injective, $f(h(x))=f(x)\implies h(x)=x=g(x)$, which is unique.

May I ask if my arguments are valid and how I could improve what I've done so far?

Thank you in advance!

答:
The first part looks good. The only thing I would point out is this:

If so, $g$ has to be surjective, but, since $f$ is surjective, $g$ also has to be injective.

This statement is only true when $S$ is finite (I suggest thinking of counterexamples for functions on $\mathbb R$). The correct way to approach it is like this: first, existence is clear as the identity function on $S$ is a function satisfying $\operatorname{id_S}\circ f = f$ and $f \circ \operatorname{id_S} = f$. Now, let $g : S \to S$ be some other function such that $g \circ f = f$. Then $$g \circ f = f = \operatorname{id_S} \circ f$$ and since $f$ is surjective, we can conclude that $g = \operatorname{id_S}$. Similarly, if there is a function $h : S \to S$ such that $f \circ h = f$, then $$f \circ h = f = f \circ \operatorname{id_S} $$ and since $f$ is injective, we conclude that $h = \operatorname{id_S}$. This completes the proof of uniqueness.

Alternatively, if we have $g \circ f = \operatorname{id_S} \circ f$ (resp. $f \circ h = f \circ \operatorname{id_S}$), we can arrive at the same result by composing on the right (resp. left) with the inverse of $f$ which exists if and only if $f$ is bijective. However, you only need injectivity for left cancellation, and surjectivity for right cancellation.


问:
How to prove this problem about 'the limit under the integral sign'?

problem:let ${f_n}$ be a sequence of nonnegative measurable function on $\Bbb R$ that converges pointwise on $\Bbb R$ to $f$ and $f$ be integrable over $\Bbb R$,show that:
If $\int_ \Bbb R f=\lim_{n\to \infty}\int_ \Bbb R f_n $,then $\int_E f=\lim_{n\to \infty}\int_E f_n $ for any measurable set $E$.

I want to use formula :$\int_E f=\int_ \Bbb R f \mathcal X_E$,but i don't know how to use the condition "$\int_ \Bbb R f=\lim_{n\to \infty}\int_ \Bbb R f_n $ "

答:
lemma:let${x_n}$ and ${y_n}$ be sequence,${x_n}$ is converges,then we have:$\lim inf (x_n+y_n)$=$\lim x_n$ +$\lim inf y_n$.

Proof:we can see $f_n\mathcal X_E $ converges to $f\mathcal X_E$.by fatou's lemma :we have $\int_\Bbb R (f-f\mathcal X_E) ≤ \lim inf\int_\Bbb R (f_n-f_n\mathcal X_E)$
By lemma: we have$\lim inf\int_\Bbb R (f_n-f_n\mathcal X_E)$=$\int_\Bbb R f -\lim supf_n\mathcal X_E$,then we have
$\int_\Bbb R f\mathcal X_E≥\lim sup\int _\Bbb Rf_n\mathcal X_E$.
By fatou's lemma,we have
$\int_E f≤\lim inf \int_E f_n$.
As we desired.


问:
Power series representation of $(1+x^2)\arctan x$
so i've been looking for a cleaner way of solving this question, finding a Power series representation of the following:
$$
\left(1+x^2\right)\arctan\left(x\right)
$$
knowing the power series of arctan i got:
$$
\sum_{n=0}^{\infty \:}\frac{\left(-1\right)^n\cdot x^{2n+1}}{2n+1}\:+\:\sum_{n=0}^{\infty}\frac{\left(-1\right)^n\cdot \:x^{2n+3}}{2n+1}
$$
which after opening up the sum expressions amounts to :
$$
x+\:\:\sum_{n=1}^{\infty}\frac{\left(-1\right)^n\cdot \:2\cdot x^{2n+1}}{\left(2n\right)^2-1}
$$

I am sure there is a better way of doing this, maybe differentiation or integration? I've tried but i can't figure it out.
Any help would be greatly appreciated :]

Body:
If we have a polynomial of the form:

$$ P(x) = \sum a_i x^i$$

Another one of form:

$$ Q(x)  = \sum b_j x^j$$

$$ P(x) \cdot Q(x) = \sum a_i \cdot b_j x^{i+j}$$

Sub :

$$ i+j = u$$

Leads to:

$$ i= u-j$$

$$ P(x) \cdot Q(x)  = \sum_{u=0}^{\infty} \sum_{j=0}^u a_{u-j} b_j x^u$$

Or, r.h.s:

$$ \sum_{u=0}^{\infty} x^u \sum_{j=0}^u a_{u-j} b_j$$

Now call:

$$  \sum_{j=0}^u a_{u-j} b_j =  c_u$$

Or,

$$ \sum_{j=0}^u a_j b_{u-j} = c_u$$

For this particular case:

$$ a= \{ a_0 , a_1, a_2 ,0  ,0 , 0 ,0 ,0... \} =  \{ 1 , 0 , 1 , 0 , 0 , 0 ....\}$$

Hence:

$$  (1) b_{u} + (1)b_{u-2} = c_u$$

Now:

$$ u \to 2k+1$$

Or,

$$ b_{2k+1}  + b_{2k-1} = c_{2k+1}$$

For the $ \tan^{-1} $ series:

$$ b_{2k+1} = \frac{(-1)^k}{2k+1}$$

And,

$$ b_{2k-1} = \frac{ (-1)^{k-1} }{ 2k-1}$$

Hence:

$$ c_{2k+1} = \frac{(-1)^{k-1}}{2k-1} + \frac{ (-1)^k}{2k+1}$$

Hence our series is:

$$ \sum_{k=0}^{\infty} c_{2k+1} x^{2k+1}$$
DONE!


问:
Some alternative examples to the question "Are there two irrational numbers $x$ and $y$ such that $x^y$ is rational?"
This question is a classic and is on Stack Exchange several times, but I am looking for some atypical answers. The basic question, as you all already know is, "Find two irrational numbers $a$ and $b$ such that $a^b$ is rational."

There are two very common answers. The first being the classic $(\sqrt{2}^{\sqrt{2}})^\sqrt{2} = 2$ argument (in which the irrationality of $\sqrt{2}^{\sqrt{2}}$ happens to be irrelevant) and the second being the $\sqrt{2}^{2\log_2(3)} = 2$ example. These are both trivial and traditional proofs, but are there any other examples not usually given? A bunch more examples would be nice.  It would also be helpful to show that the two numbers $a$ and $b$ are irrational, as some of these proofs, like $\pi$ and $e$ are not elementary.

Cheers.

答:
It is easy to establish that $x^x$ is irrational when $x$ is a positive non-integer rational number (see for instance, https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1031393807). Consider the equation $$x^{x+n}=q\,,$$
where $n\ge 0$ is an integer and $q>0$ is a non-integer rational number (Note: A solution to this exists by the Intermediate Value Theorem).

Claim: $x$ is an irrational number

Proof: Suppose $x$ is a rational number, then we have
$$x^x=x^{-n}q\,.$$
Thus, $x^x$ is a rational number and thus $x$ must be a positive integer; but $x^{-n}q$ cannot be an integer, which is absurd.


问:
CW Structure and connectedness.
Suppose $X$ be a $k$-connected CW-complex. I want to know Is it possible to find the value of $k$ from the CW-structure? In other words, Suppose $X$ has $n_i$, $i$-cell. What is the maximum value of $k$ such that $X$ be a $k$-connected topological space?

答:
If $X$ has exactly one $0$-cell and no other cells until dimension $n$, then $X$ is $(n-1)$-connected. This is a consequence of the cellular approximation theorem.

Unfortunately this is the most you can say about $X$ without more information. For instance it is possible that $X$ may be more highly connected. As an example consider the contractible space $D^{n+1}=S^n\cup e^{n+1}$. To see now why it is impossible to read off the connectivity of an arbitrary CW complex from just its list of cells compare $D^{n+1}$ with $S^n\vee S^{n+1}$.

Title: CW Structure and connectedness.


答:
From the definition for the summation notation:
$$\sum_{n=1}^N x_n = x_1 + x_2 +...+x_{N-1} +x_N$$

Now you multiply $a$ to both sides and apply the distributive property:
$$a\sum_{n=1}^N x_n = a(x_1 + x_2 +...+x_{N-1} +x_N) = ax_1 + ax_2 +...+ax_{N-1} +ax_N$$


问:
$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$, $a_n\in \mathbb Q$ and not $a_n$ not monotonic
How can I construct a sequence $\{a_n\}$ of positive rational numbers, which is not monotonic such that $$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$$
I thought a lot on this question but I always come up with a monotonic sequence. Any help will be appreciated.</p>

答:
if $\sum_{n=0}^{\infty}(-1)^n a_n $ is absolutely convergent then all rearrangement will still converge to the same value.

So take any sum $\sum_{n=0}^{\infty}(-1)^n a_n = \pi$ that is absolutely convergent and rearrange them in a way that are not monotonic.


问:
Deriving AM-GM in a way that reflects my intuition.
One of my core intuitions with inequalities (especially the olympiad kind in which this is very explicit, but this has more profound influences on how I understand the relationship between inequalities) is that in a homogenous inequality, on average, terms that are more mixed ($x^2yz$) are smaller than their less-mixed counterparts ($x^4$).

With this bit of intuition, many inequalities seem trivially true, like am-gm:

$$x^3+y^3+z^3\geq xyz + xyz +xyz$$

Or this one:

$$x^2y + y^2z + z^2x\geq xyz+xyz+xyz$$

And a generalisation of part of this idea: Murihead's Inequality.

I want to find a proof (by which I mean not particularly formal derivation) of AM-GM, that adequately captures 'mixed terms being smaller on average' as being the reason for the inequality. I think this would satisfy some innate and inexplicable urge.

There may not exist a satisfying answer to this question (though I would certainly no one if I saw it), but hopefully, I have communicated the kind of thing I'm looking for.

答:
For cyclic inequalities the Muirhead's idea does not work.

For example $$(4,3,0)\succ(4,2,1),$$ but the inequality
$$\sum_{cyc}x^4y^3\geq\sum_{cyc}x^4y^2z$$ is wrong for positives $x$, $y$ and $z$ (take $y<z$ and $x\rightarrow+\infty$).

Actually.

If for non-negatives $\alpha$, $\beta$, $\gamma$, $\delta$ and $\epsilon$ $$(\alpha,\beta,0)\succ(\gamma,\delta,\epsilon)$$ and the inequality
$$\sum_{cyc}x^{\alpha}y^{\beta}\geq\sum_{cyc}x^{\gamma}y^{\delta}z^{\epsilon}$$ is true for any positives $x$, $y$ and $z$, so we can prove it by AM-GM.

This fact was proved by Vasc more than $40$ years ago.


问:
$1$, $z$, and $z^{2019}$ are vertices of an equilateral triangle

Find all complex numbers $z$, such that the points, in the complex plane, which correspond to $1$, $z$, and $z^{2019}$ form an equilateral triangle.

Thanks in advance.

So far I got $$\frac{z^{2019}-1}{z-1}-\cos(\pi/3)-i\sin(\pi/3)=0$$ and $$\frac{z^{2019}-1}{z-1}-\cos(\pi/3)+i\sin(\pi/3)=0$$ and my task will be completed if I knew how many different roots these two polynomials have?

答:
Let $z$ be any root of the polynomial
$$ \begin{align}f(X)&=\frac{(X^{2019}-1)^3+(X-1)^3}{(X^{2019}-1)+(X-1)}\\
&=(X^{2019}-1)^2-(X^{2019}-1)(X-1)+(X-1)^2\\
&=X^{4038}-X^{2020}-X^{2019}+X^2-X+1.\end{align}$$

Then  $a=1$, $b=z$, $c=z^{2019}$ form an equilateral triangle.
Indeed, if $z=1$, we have a degenerate equilateral triangle, which we may or may not accept as one of the possible solutoins.
In all other cases, let $w=\frac{a-c}{b-a}$ and observe that
$$ 0=(a-c)^2+(a-c)(b-a)+(b-a)^2=(w^2+w+1)(b-a)^2,$$
i.e., $w$ is a primitive third root of unity. It follows that $|a-c|=|b-a|$, but also
$$ c-b=-(a-c)-(b-a)=(-w-1)(b-a)=\overline w(b-a),$$
i.e., also $|c-b|=|b-a|$.
Hence for each of the roots of $f$ (with $z=1$ counted or not), we obtain an equilateral triangle, and there are no other solutions.
Note that $$f'(X)=4038X^{4037}-2020X^{2019}-2019X^{2019}+2X-1$$
and by a straightforward but laborious calculation, we find that
$$ \gcd(f(X),f'(X))=X-1,$$
i.e., $z=1$ is the only root of $f$ that has higher (namely, double) multiplicity. The remaining $4036$ simple roots of $f$ are proper solutions to the problem.


问:
To find the x and y-intercepts of the line $ax+by+c=0$
Please check if I've solved the problem in the correct way:

The problem is as follows:

Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.

My solution:

We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.

Now we solve the equation $ax+by+c=0$ for $y$:

$ax + by + c=0$

$ax + by = -c$

$by = -ax -c$

$y = \frac {-ax-c}{b}$

$\because x = 0$ at y-intercept,

$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$

$y = -\frac cb$.

The point at which the line crosses the y-axis is $(0,-\frac cb)$

Now we solve the equation $ax+by+c=0$ for $x$:

$ax+by+c=0$

$ax+by=-c$

$ax = -by-c$

$x = \frac {-by-c}{a}$

$\because y = 0$ at x-intercept

$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$

$x = -\frac ca$

The point at which the line crosses the x-axis is $(-\frac ca,0)$

答:
Your work is exemplary, Samama. You know your definitions well, and your answers are entirely correct.

A nice "shortcut" is to take advantage of what you already know:

the $x$ intercept is the value of $x$ when $y = 0$, and

the $y$ intercept is the value of $y$ when $x = 0$.

$ax + by + c = 0 \iff ax + by = -c$

x intercept

We can use  $\;ax + by = -c\;$ to solve for $x$ when $y = 0$, by plugging in $0$ for $y,\;$ right at the start:

$$ax + \underbrace{by}_{y = 0} = -c \iff ax = - c \iff x = \frac {-c}{a}$$

y intercept

And we can do the same to solve for $y$ when $x = 0$, by plugging in $0$ for $x,\,$ right at the start:

$$\underbrace{ax}_{x = 0} + by = -c \;\;\iff\;\; by = -c \;\;\iff \;\; y = \frac{-c}{b}$$


问:
A question regarding inequalities and some binomial theorem
Let $m, n \in \mathbb N$ with $m<n$ show that $ \frac {1}{m^k} \binom{m}{k} <\frac {1}{n^k} \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}}$ for all $k = 2, ... , m$

Now in order to start this question off I've decided to expand the binomial for the $ \frac {1}{m^k} \binom{m}{k} <\frac {1}{n^k} \binom{n}{k}$ following parts and it's left me with $\frac {1}{m^k}(\frac {m(m-1)*...*(m-k+1)}{k!}) $ and the same for the one with n in place of the m. Now I'm not sure where to go from here or what I should do.

答:
Idea: Induction over $k$ could help here

Verify the statement for $k=2$

Assume that for a certain $k$ you have  $ \frac {1}{m^k} \binom{m}{k} <\frac {1}{n^k} \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}}$

and prove that

$ \frac {1}{m^{k+1}} \binom{m}{k+1} <\frac {1}{n^{k+1}} \binom{n}{k+1} \leq \frac{1}{(k+1)!} \leq \frac{1}{2^{k}}$


问:
Prove that certain $\alpha_m:C_n\rightarrow C_n$ is automorphism if $\gcd(m,n)=1$
I need help with the following abstract algebra problem:

Given $C_n$ the cyclic group of order $n$, we define the map
\begin{align*}   \alpha_m : C_n &\to C_n\\   x &\mapsto x^m \end{align*} Prove that $\alpha_m$ is automorfism if and only if
$gcd(m,n)=1$.

The first part is proving that $\alpha_m$ is homomorfism, which is easy seen by doing this:
$$\alpha_m(xy)=(xy)^m=x^my^m=\alpha_m(x)\alpha_m(y).$$
Now I need to prove it's injective and surjective. For injectivity, I assume there are some $x$ and $y$ that verify $x^m=y^m$. And here's where I'm stucked. Maybe considering $m=kn+r$? Any help will be appreciated.

答:
Hint : $\gcd(n,m)=1$ implies that there exists $u,v$ such that $un+vm=1$. Thus

$$x=x^{1}=x^{un+vm} = (x^n)^u (x^m)^v = (x^m)^v = (y^m)^v = (y^m)^v (y^n)^u = y^{mv+un} = y$$

Notice also that proving injectivity is enough, since the map is a morphism from a finite group into itself.

Edit : For the converse, you can prove that if $\alpha_m$ is surjective, then $\gcd(n,m)=1$. Let's write $(C_n, \times)$ as $(\mathbb{Z}/n\mathbb{Z}, +)$. Then if $\alpha_m$ is surjective, there exists $x \in \mathbb{Z}/n\mathbb{Z}$ such that
$$mx=1 \in \mathbb{Z}/n\mathbb{Z}$$
which means that there exists $k$ such that $mx + kn = 1$ in $\mathbb{Z}$, so $\gcd(n,m)=1$.


问:
proof for that exactly one normal can be drawn to a parabola from an exterior point.
It is well known that a maximum of three normals can be drawn from the interior of a parabola. But I am interested in the proof for the fact that exactly one normal can be drawn to a parabola from an exterior point wrt to the parabola.

So far I have done this.

WLOG we consider $y^2=4ax$.

Let the exterior point be $(h,k)$.

We must have $$k^2-4ah> 0\tag1$$

Also the equation of a normal to a parabola is $$y=mx-2am-am^3$$
so $(h,k)$ satisfies $$am^3+2am-mh+k=0$$

All I have to do is show that there is only one real root possible for this cubic using $(1)$.
Can anyone complete the proof?

Alternative methods are welcome!

NoteBy a normal i mean that a segment that is external to parabola

Thank you!

答:
First of all, if by normal you mean any line that intersects the parabola at 90 degrees then this is not true: take any point with positive $x$ on $y=x^2$ and draw a normal line $l$ at $(x, x^2)$. It will  intersect the part of the parabola with negative $x$s somewhere, say at $(x_n, x_n^2)$. Then any point to the left of that  $(x_n, x_n^2)$ will have a normal which intersects $l$ at a point $p$ external to the parabola. This $p$ has two "normals" to the parabola. (This is why purely algebraic arguments that find such normal lines will not give uniqueness.) However, if by normal you mean a segment that is entirely external to the parabola, then this is true for the exterior of any closed convex set in $\mathbb{R}^n$ - hence for the exterior of the graph of any convex function (more precisely, for the "overgraph" $S=\{(\vec{x}, y)| y\geq f(\vec{x})\}$ for any convex function $f$), including of course the parabola  $f(x)=ax^2+bx+c$.

Let's show this. Firstly, for any closed convex set $S$ and for any point $p$ not in $S$ there exists unique point $n$ in $S$ which is closest to $p$. This $n$ will lie on the boundary and, when the boundary is differentiable $np$ will be normal to it (see below for a proof of all of this). Moreover, since there are no points of $S$ closer to $p$ than $n$, the whole segment $np$ lies outside of $S$.

Now, why doesn't there exist a second normal segment? Suppose there is one, say $pn'$. Then the normal $T_{n'}$ to $pn'$ through $n'$ is tangent  to the boundary of $S$ at $n'$.  By convexity of $S$, $S$ lies on one side of this tangent $T_{n'}$. Since the segment $pn'$ is outside $S$, we must have $S$ on the opposite side of $T_{n'}$ from $p$. But then $n$ is on the opposite side of $T_{n'}$ from $p$, and in particular is further from $p$ than $n'$, which is a contradiction. This completes the proof.

Now, for completeness, the existence of the closest point and it's normality:

Existence of the closest point: Pick a point $s\in S$ let $S_b=S\cap B(p, |p, s|)$ be the intersection of $S$ with closed ball around $p$. Now $S_b$ is closed and bounded, so the distance function to $p$ being continuous has a minimum on $S_b$. Suppose this minimum is $m$ and is achieved at $n\in S_b$. Now any other point in $S$ is either not in $S_b$, and so has distance to $p$ bigger than $|p-s|$ and so also bigger than $|p-n|$, or is in $S_b$ and so has distance to $p$ at least $|p-n|$. We conclude that $n$ is a point at which the minimum distance from $p$ to $S$ is achieved.

The closest point is on the boundary: Draw the segment $pn$. If $n$ is in the interior of $S$ there would be other points of the segment that lie in $S$; but then they would be closer to $p$ than $n$, which is impossible. So $n$ is on the boundary.

Uniqueness of the closest point: Draw the tangent  $T_n$ to $S$ at $n$ (in general, "a support hyperplane"). Then all of $S$ is on one side of $T_n$ by convexity, and it is the opposite side from $p$ (otherwise some point $s' \in S$ is on the same side of $T_n$ as $p$ then, by convexity os $S$, the segment $s'n$ is in $S$, but for points on that segment close to $n$ the distance to $p$ is smaller than that of $n$, which is a contradiction). But all points on opposite side of $T_n$ from $p$ have higher distance to $p$ than $n$ -- and, in particular, all points of $S$ do. Thus $n$ is the unique point of $S$ closest to $p$.

Normality of the closest point: Now assuming  the boundary   $\partial S$ is differentiable, let's see why $pn$ is normal to $\partial S$.  If $np$ is not normal to $\gamma$ at $n$, then the tangent $T_n$ of $\partial S$ at $n$ has points very near $n$ that are closer to $p$ than $n$; but then since near $n$ the $\partial S$ itself follows the tangent closely, there will be points of $\partial S \subset S$ that are closer to $p$ than $n$, which is not true. So $pn$ is orthogonal to $\partial S$ as wanted.


问:
In $\triangle ABC, AB = 28, BC = 21$ and $CA = 14$. Points $D$ and $E$ are on $AB$ with $AD = 7$ and $\angle ACD = \angle BCE$

In $\triangle ABC, AB = 28, BC = 21$ and $CA = 14$. Points $D$ and $E$ are on $AB$ with $AD = 7$ and $\angle ACD = \angle BCE$. Find $BE$.

What I Tried: Here is a picture :-

I know the side-lengths of the triangle so I can find out their altitudes too using Heron's Formula, but that didn't give me any useful information. The fact is that I can't use the side-lengths in any way, neither the triangles, because there are no similar triangles here. Angle-chasing, I don't think it is going to help. I did not try Trigonometry cause I am a little weak at it.

Can anyone give me any ideas for this problem? Thank You.

答:
We have $\triangle DAC \sim \triangle CAB$ by SAS ($\angle A$ common)

$$\therefore \angle B = \angle ACD = \angle BCE$$

So $\triangle BEC$ is isosceles with $BE=CE$. Drop $EF \perp BC$.

$\triangle BEF \cong \triangle CEF$

$BF = 21/2$

By cosine rule in $\triangle ABC$,

$$ \cos B = \dfrac{28^2 + 21^2 - 14^2 }{2\cdot 28 \cdot 21} = \dfrac{7}{8} $$

Easy enough,

in right $\triangle BEF$,
$$ BE \cos B = BF $$
$$ \Rightarrow \boxed{BE = 12}$$


问:
Showing two versions of the central limit theorem are equivalent (are they?)
I am trying to read Norris' book on Markov chains. In particular, at page 160 he states the central limit theorem :

Theorem 4.4.1 (Central limit theorem). Let $X_1, X_2,...$ be a sequence of independent and identically distributed real-valued random variables with mean 0 and variance $t ∈ (0,∞)$. Then, for all bounded continuous functions $f$, as $n → ∞$ we have
$\mathbb{E}[f(X_1+\ldots+X_n)/\sqrt{n}]\to \int_{\mathbb{R}}f(x)\phi_t(x)dx$

I thought until now that the central limit theorem was

Let $X_1, X_2,...$ be a sequence of independent and identically distributed real-valued random variables with mean 0 and variance $t ∈ (0,∞)$. Then, for all bounded continuous functions $f$, as $n → ∞$ we have
$(X_1+\ldots X_n)/\sqrt{n}\xrightarrow{d}\mathcal{N}(0,t)$, where "$\xrightarrow{d}$" denotes convergence in distribution.

I don't really understand if these are equivalent. In the first theorem, $f$ has to be bounded so we cannot take the identity $f(x)=x$ which is not bounded right?

And I don't understand how to see that these formulation are equivalent...

答:
$Y_n \to Y$ in distribution if and only if $Ef(Y_n) \to Ef(Y)$ for every bounded continuous function $f$. [ For more information see https://en.wikipedia.org/wiki/Convergence_of_random_variables ]

If $Y$ has a density $\phi$ then we can write $Ef(Y)$ as $\int f(x)\phi (x) dx$. In our case the density is $\phi_t$ defined by $\phi_t(x)=\frac 1 {t\sqrt {2 \pi}}e^{-x^{2}/2t}$.


答:
Let $x=e^{i\pi/\sqrt2}=\cos(\pi/\sqrt2)+i\sin(\pi/\sqrt2)$ and $y=\sqrt2$. We have that $x$ is irrational since $\sin(\pi/\sqrt2)\not=0$ (so $x$ isn't even real, much less rational), and $y$ is irrational by for the usual elementary reasons. And $x^y=e^{i\pi}=-1$ is rational.

Added later: I am tacitly using a definition of irrational number as a number that is not rational. I see however that the set of irrational numbers is often explicitly defined to be the subset of real numbers that are not rational. So this may or may not be a suitable example, depending on what you mean by irrational.

Title: Some alternative examples to the question "Are there two irrational numbers $x$ and $y$ such that $x^y$ is rational?"

Body:
This question is a classic and is on Stack Exchange several times, but I am looking for some atypical answers. The basic question, as you all already know is, "Find two irrational numbers $a$ and $b$ such that $a^b$ is rational."

There are two very common answers. The first being the classic $(\sqrt{2}^{\sqrt{2}})^\sqrt{2} = 2$ argument (in which the irrationality of $\sqrt{2}^{\sqrt{2}}$ happens to be irrelevant) and the second being the $\sqrt{2}^{2\log_2(3)} = 2$ example. These are both trivial and traditional proofs, but are there any other examples not usually given? A bunch more examples would be nice.  It would also be helpful to show that the two numbers $a$ and $b$ are irrational, as some of these proofs, like $\pi$ and $e$ are not elementary.

Cheers.


答:
Here's a non-formal trick which can help you on these recurrence problems. But one must be careful and prove that $a_n > 0$ and monotonic first so that $b \geq 0$. You can prove $a_n > 0$ easily with induction and since you've already proven it is increasing let's get straight to the point.
$$\lim a_{n+1} = \lim \sqrt{6+a_n} = b = \lim a_n$$

So $b = \sqrt{6+b} \Leftrightarrow b^2-b-6 = 0 \Leftrightarrow (x+2)(x-3) = 0 \Leftrightarrow b = 3 \lor b = -2$. Since $b \geq 0$ then $b = 3$.

We conclude that the limit of this monotonic and positive recurrence sequence is $3$, so it becomes closer and closer to 3 from below. Meaning $a_n \rightarrow 3^-$.

This should not be your proof; just a side calculation that helps you get the supremum for $\{a_n\}_{n=1}^{\infty}$. We will use it to show that $a_n$ is bounded above.

Now we must demonstrate it with a formal proof (by induction) that for all $n \in \mathbb{N}, a_n < 3$. Since we are saying "for all $n$" it must be also true for $n+1 = n'$.

Our base case ($n=1$) is trivial since $a_1 =
\sqrt{6} < 3$.

Now assuming $p(n):= a_n < 3$ is true, we must show that $p(n+1) = a_{n+1} < 3$ is also true.

In fact,
$$a_n < 3 \Leftrightarrow a_n + 6 < 3 + 6 \Leftrightarrow \sqrt{6 + a_n} < \sqrt{3 + 6} = 3$$

But $\sqrt{6+a_n} = a_{n+1}$ for which $a_{n+1} < 3$ and the condition holds true for all $n$.

So we proved by induction that $0 < a_n < 3$ for all $n \in \mathbb{N}$.

Title: how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded?

Body: I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$


问:
How to properly get log or exp parameters given a sample table of a exp/log curve?
What is the correct process to get the correct formula of conversion between X and Y (in both directions) where

```<code>        Y       X
        1       0
        0.5     -6
        0.25    -12
        0.125   -18
</code>```

I would say, intuitively, that

X is 6 times a log2 of Y, but I learnt to avoid using intuition, if I can't get something more formal, I would not trust my intuition.

Also Y will be necessarily > 0

答:
You are correct. Note that

$$\begin{array}{c|c|}
 & \text{Y} & \text{X} \\
 & \big(\frac{1}{2}\big)^{0} & 0\cdot(-6) \\
 & \big(\frac{1}{2}\big)^{1} & 1\cdot(-6) \\
 & \big(\frac{1}{2}\big)^{2} & 2\cdot(-6) \\
&  \big(\frac{1}{2}\big)^{3} & 3\cdot(-6)
\end{array}$$

So we have  $$Y=\big(\frac{1}{2})^{-\frac{X}{6}}=2^{\frac{X}{6}}$$  and it follows that $X=6\log_2{Y}$ for $Y>0$.

You can try a solution of the form $Y=a^{bX}$, substitute $(Y,X)=(1,0)$, $(Y,X)=(0.5,-6)$ and solve for $a$ and $b$. Then check it satisfies the remaining coordinates. Generally it is not easy to find a "nice" solution.


问:
In simplex method should right hand sides be positive or non-negative?
let's say we have an optimal basic feasible solution for an LP problem solved by simplex method. In this solution, should the RHSs (right hand side) be positive or non-negative?

Because I read that if a RHS is zero (which means a basic variable takes on the value zero) then there is degeneracy. Thus the zero causes degeneracy on the RHS. But in the definition of Simplex Method it says that RHSs sohuld be non-negative. So which one? I am confused.

答: You cannot choose to avoid degeneracy. Degeneracy is not caused by a particular assignment of the variables. It is part of the given problem (e.g. linear dependency of the given constraints, vertices that are determined by more equations than necessary, linear dependency between constraints and objective function etc.) If you have an optimal basic feasible solution with active variables that take the value zero, this assignment of variables indicates that there is degeneracy, it does not cause it. There is nothing wrong with active variables taking the value zero. (You might have to introduce special handling for those, if you actually implement the simplex method.)


问:
In how many ways you can form $6$-digit numbers using the digits $0, 1, 2, 3,$ and $4$ with exactly two digit $4$ in each number?
In how many ways you can form $6$-digit numbers using the digits $0, 1, 2, 3,$ and $4$ with exactly two digit $4$ in each number?

I am trying to solve this problem using combination.

I think (but not that sure) we can calculate the total number of 6-digit numbers as follows

$_6C_2 × 4 × 4 × 4 × 4$ = $8,256$

I am confused on how I can consider in the calculation that the digit 0 is not allowed to be in the hundred thousands digit.

Any comment and suggestion will be much appreciated.
Thank you in advance.

答:
First, forget about zero. You have to pick two of the six places for two digits 4. You can do that in $\binom62$ different ways. The rest can be filled with 0, 1, 2, 3 in $4^4$ different ways so the total number of possiblities is:

$$\binom 62 4^4$$

However we have to eliminate all numbers starting with zero. If you fix 0 in the first place, it's like asking how many 5-digit numbers we can create with two digits 4, plus digits 0,1,2,3. By using similar reasoning, you can pick $\binom52$ positions for $4$ and you can fill the rest in $4^3$ different ways:

$$\binom52 4^3$$

So the result is:

$$\binom 62 4^4-\binom52 4^3$$


问:
Continuity of the homeomorphic map between $S^n$ and $R^n\cup \infty$
In topology, the n-dimension Euclidean space with an added point is described as being homeomorphic to the n-sphere.

One possible map that defines the "similarity" would be $f:S^n \mapsto R^n \cup \infty$ given by:
$$ f(x_0,...,x_n) = \begin{cases}  (\frac{x_1}{1-x_0},\frac{x_2}{1-x_0}...,\frac{x_n}{1-x_0})    &\text{ if } x_0 \ne 1\\
          \infty &\text{ if } x_0 = 1\end{cases}$$

The topology on $X^+ = R \cup \{ \infty \}$ is given by the following closure operator $\tau^+ : P(X^+) \to P(X^+)$ on the power set of $X^+$:

$$ \tau^+(S) = \begin{cases}
\tau (S)    &\text{ if } S \subseteq R^n \text{and bounded}\\
          \tau(S) \cup \{\infty \} &\text{ if S is not bounded and} \subseteq R^n\\
\tau(S - \infty) \cup \{\infty \} &\text{ if } \infty \in S\\
\end{cases}$$
for any set $S \subseteq R^n \cup \{\infty \}$, where $\tau : P(R^n) \to P(R^n)$ is the standard closure operator on $R^n$.
Now I can see that if $x_0 \ne 1$ then this function is clearly continuous as it would be defined by rational functions, but how does one go about proving that the function is continuous when its image is infinity? Similarly what is the process of showing that the inverse function is continuous (I can see that this function is surjective and injective).

答:
To see the continuity, you have to show that if you have a sequence $x_k=(x_{k,0},x_{k,1},\ldots,x_{k,n})$ with $x_{k,0}\to1$ then there is some $j\in\{1,\dotsc,n\}$ such that $(x_{k,j}/(1-{x_{k,0}}))_k$ is unbounded.

To see that the inverse $g=f^{-1}$ is continuous, it suffices to show that preimages of open sets are open or (passing to the complements) that preimages of closed sets are closed. But preimages of closed subsets under $g$ are images of compact sets under $f$. So it suffices to show that compact subsets of the image space are closed. Depending on what you know already about topology, this might be trivial (e.g. it follows from the fact that the image space is Hausdorff).


问:
Ordering balls in a line. probability of the event $\{\text{ no ball in the first 4 balls is blue }\}$
given $8$ red balls, $5$ blue balls, $9$ green balls. assume all the ordering options have the equal probability. what is the probability of the event $$\{\text{ no ball in the first 4 balls is blue }\}$$

My attempt:

My probability space is $$\Omega = \binom{17}{4}$$ this is because I have to choose 4 balls without the $5$ blues. But I am not sure if this is true because it can be treated "choose 4 balls without 5 reds".

In addition to that, I need to determine the power of my event group wich is pretty hard for me to understand how should I do it.

I hope some one can help

答:
Your space of events $\Omega$ is the set of all essentially unique orders. By "essentially unique" I mean that, if two balls of the same colour are switched, the element of the set $\Omega$ which identifies the new sequence doesn't change. The size of this set is
\begin{equation}
\#\Omega=\frac{22!}{8!\cdot 5!\cdot 9!}.
\end{equation}
This is because, once a permutation of the $22$ balls is fixed, you're allowed to re-arrange them as long as you don't switch two balls of two different colours. The event $E$ you're interested in is that none of the first four balls is blue, therefore the $5$ blue balls must appear in the last $18$ ones. How many are the ways in which I can choose a subset of $5$ elements in a set of $18$ elements? This number is given by
\begin{equation}
a=\binom{18}{5}.
\end{equation}
Once you have fixed the position of the five blue balls, every disposition of the remaining $17$ balls is allowed, and there are exactly
\begin{equation}
b=\frac{17!}{8!\cdot 9!}
\end{equation}
of these dispositions. Therefore $\#E=a\cdot b$ and computing $\#E/\#\Omega$ gives the solution to your problem.


问:
What is the group $\operatorname{Aut}_{\mathbb{Q}} { \mathbb{Q}{\bigl(\sqrt2, \sqrt3\,\bigr)}} $
This question was given to us by instructor in class notes of Course on Field theory and there are no questions worked out in his notes which are posted online . So, I am looking for an explanation of his question which is solved( but explanation was not good).

Find group  $\operatorname{Aut}_{\mathbb{Q}} { \mathbb{Q}{\bigl(\sqrt2, \sqrt3\,\bigr)}} $ ?

Answer posted : let $\sigma \in \operatorname{Aut}_{\mathbb{Q}} \bigl(\mathbb{Q}\bigl(\sqrt2, \sqrt3\,\bigr)\bigr) $ , then $\sigma$ is completely determined by $\sigma(\sqrt{2}\,)$ and $\sigma(\sqrt{3}\,)$.

Why so? Which result is used?

(solution continued) Only possibility is $\sigma(\sqrt2\,) = \sqrt2, -\sqrt2 \:$ and $\sigma(\sqrt3\,) =\sqrt3, -\sqrt3$.

Again no reasoning is given!!

So, there are 4-distinct $\Bbb Q$-homomorphisms (understood).

And $\operatorname{Aut}_{\mathbb{Q}}\bigl(\mathbb{Q}\bigl(\sqrt2, \sqrt3)\bigr) $ is isomorphic to $\mathbb{Z}_2 × \mathbb{Z}_2$.

Why not isomorphic to $\mathbb{Z}_4$?

I have asked 3 questions but I could really not understand the problem after trying to think a lot and I would be grateful for your help.

答:
First question: $\sigma$ is completely determined by the images of $\sqrt 2$ and $\sqrt 3$ simply because any element in $\mathbf Q\bigl(\sqrt2,\sqrt 3\,\bigr)$ is a polynomial in $\sqrt 2$ and $\sqrt 3$, and $\sigma$ is a ring  homomorphism.

Second question: the image of $\sqrt 2$ satisfies the same minimal equation, since $\sigma$ is a $\bf Q$-homomorphism. Same  reason for $\sqrt 3$.

Last question: The group of automorphisms can't be $\mathbf Z/4\mathbf Z$ because  no automorphism of this extension has order $4$, as the $\sigma^i(\sqrt 2\,)$ and the $\sigma^i(\sqrt 3)$ are roots of the quadratic polynomials $X^2-2 $ and $X^2-3$ respectively, and a quadratic polynomial has at most two roots.


问:
a question about poles of meromorphic functions

Justify true or false : If $f$ is meromorphic and has an infinite number of poles, then every closed disk $|z|\leq R$ $(0<R<\infty)$
contains only a finite number of poles inside it.

Considering $f(z)=\tan z$ , we can see that its poles are $\displaystyle z=\frac{(2n+1)\pi}{2}$ , which are contained finitely in some of the closed disks . I wonder if there is a counterexample by which we can show that it is not true for all meromorphic functions as such. Am I on the right track, or is this statement indeed true? Any help is appreciated.

N.B. - There is also a statement I remember that if $f$ has an infinite number of poles inside a simple closed contour $C$, then some subsequence of them would converge to non-isolated singularity of $f$. But I'm not  quite sure whether that would help in this case.

答: Meromorphic functions are analytic except for isolated singularities (which are poles). Isolated singularities form a set with no limit points. Since closed disks are compact there can only be finite number of isolated singularities  inside them.</p>


答: If $f$ is meromorphic, then $f=\frac gh$, where $g$ and $h$ ara anlytic functions, and $h$ is not the null function. Every pole of $f$ is a zero of $h$. And a closed disk is compact, and therefore it can only contain finitely many zeros of $h$; otherwise, it would follow from the identity theorem that $h$ is the null function, because then the set of zeros of $h$ would have an accumulation point.

Title: a question about poles of meromorphic functions

Body:
Justify true or false : If $f$ is meromorphic and has an infinite number of poles, then every closed disk $|z|\leq R$ $(0<R<\infty)$
contains only a finite number of poles inside it.

Considering $f(z)=\tan z$ , we can see that its poles are $\displaystyle z=\frac{(2n+1)\pi}{2}$ , which are contained finitely in some of the closed disks . I wonder if there is a counterexample by which we can show that it is not true for all meromorphic functions as such. Am I on the right track, or is this statement indeed true? Any help is appreciated.

N.B. - There is also a statement I remember that if $f$ has an infinite number of poles inside a simple closed contour $C$, then some subsequence of them would converge to non-isolated singularity of $f$. But I'm not  quite sure whether that would help in this case.


问:
Completely factoring $a^3 + b^3 + c^3 - 3abc$?
How do I completely factor $a^3 + b^3 + c^3 - 3abc$ over the complex numbers? The first thing I did is factor this into $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$. But I am not sure how to factor $a^2 + b^2 + c^2 - ab - bc - ac$ over the complex numbers. Can anyone help me?

答: $a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$, where $\omega$ is the root of $x^2+x+1=0$.


答: Meromorphic functions are defined to those which have possible singularities at pole only. Now, if there are infinite number of poles, say, $(z_n)$inside a closed disk, Heine-Borel will tell you that there must be a limit point inside that disk of the poles, say, $z_0$. But then this $z_0$ is a non-isolated singularity, contrary to the definition of meromorphic functions.


问:
Showing ${\frak c}:=|\Bbb{R}|=2^{\aleph_0}$ using only the axioms of the reals
It is a well-known fact that the cardinality of the continuum, ${\frak c}:=|\Bbb{R}|$, is the same as the cardinality of the powerset of the natural numbers, but I've tasked myself with proving the result from scratch, and naturally a whole bevy of little issues come up that I hadn't thought of to begin with. The kicker here is that I want to prove this using only the axioms of the real numbers, not any construction-dependent facts.

It is sufficient by Schroeder-Bernstein to prove ${\frak c}\le2^{\aleph_0}$ and ${\frak c}\ge2^{\aleph_0}$, and one direction is clear: the function $f(A)=\sum_{i\in A}3^{-i}$ for $A\subseteq\Bbb{N}$ is an injective map from ${\cal P}(\Bbb{N})\approx2^{\aleph_0}$ to the Cantor set, which is a subset of the reals (note that by using $3^{-i}$ instead of $2^{-i}$, I avoid that silly technicality about non-unique decimal expansions), so ${\frak c}\ge2^{\aleph_0}$.

To prove the converse, I don't know whether it can be done using only the axioms (stated here). Given a construction of the reals by Dedekind cuts, we note that each real number is a set of rationals, so $\Bbb{R}\subseteq{\cal P}(\Bbb{Q})\Rightarrow{\frak c}\le2^\Bbb{Q}=2^{\aleph_0}$. Alternatively, if we use Cauchy sequences, we know that each real number is a an equivalence class of sequences of rational numbers, so ${\frak c}\le_*\Bbb{Q^N}=2^{\aleph_0}$ (where the $\le_*$ indicates the existence of a surjection), and after an application of AC, we get ${\frak c}\le2^{\aleph_0}$.

But if I had no construction, can it even be done? I've heard some stuff about $\Bbb{R}$ being "the unique order-complete field up to isomorphism" or something like that, so it should be possible to use a pure axiomatic approach to get the result, but the principal counterexample in my mind is the surreal numbers $\sf{No}$, which (I believe) also satisfy all the listed axioms (except the existence one, since $\sf{No}\supseteq\sf{On}$), and obviously there are a lot more surreal numbers than real numbers. Are there missing axioms that I need to prove this?

答:
I do not know whether this meets your requirements, but:

The Cantor function $f:[0,1]\to[0,1]$ is surjective, continuous, and non-decreasing. This function $f$ has the property that the restriction $f\upharpoonright C$ of $f$ to the Cantor set $C$ is also surjective. I have included a graph showing what the function looks like, which should also make its definition clear. (Image found here.)

For a good survey of the Cantor function and its properties, see

O. Dovgoshey, O. Martio, V. Ryazanov, and M. Vuorinen. The Cantor function, Expo. Math. 24 (1), (2006), 1–37. MR2195181 (2006k:26005).

We can define an injection $g:(0,1)\to C$ by letting $g(x)$ be the first $t\in C$ with $f(t)=x$ (using that $C$ is closed). Since $|C|=2^{\aleph_0}$, we have a proof that $\mathfrak c\le 2^{\aleph_0}$, at least if we accept that $\mathbb R$ and $(0,1)$ have the same size. If we need to argue about this, there are even continuous bijections (such as some appropriate translation/dilation of arc-tangent).


问:
What distribution do sum of sub-Gaussian variable with different variance follow?
Suppose that X = $\{X_1, ..., X_n\}$ and each $X_i$ is sub-Gaussian with variance parameter $\sigma_i^2$. What distribution does $\sum^n_{i=1} X_i$ follow? I believe it is still sub Gaussian but is the variance parameter $\sum^n_{i=1} \sigma_i^2$?

答:
This could be helpful for you:
https://ocw.mit.edu/courses/mathematics/18-s997-high-dimensional-statistics-spring-2015/lecture-notes/MIT18_S997S15_Chapter1.pdf

I starts in the link above with page 14. Read the pages from 19 to 21:
Sums of independent sub-Gaussian random variables


问:
How to calculate the distance from a point on ellipse arc to the chord
Now I have an ellipse-arc starting at $A$ and ending at $B$, so $AB$ is the chord of the ellipse-arc. I want to know how to calculate the maximum distance from the points on the ellipise-arc to the chord.

答:
Not a full answer, but too long to be a comment:

Let $A,B\in\mathbb{R}^2$ be (given) points which lie in the ellipse, and the ellipse to be represented by all points $x\in\mathbb{R}^2$ which comply $x^TMx=1$ for some positive definite matrix $M$. Moreover, the chord from $A$ to $B$ is parametrized by all points $y$ which comply $y=\lambda A + (1-\lambda)B$ with $\lambda\in[0,1]$. Thus, if I understood correctly, you want to solve the optimization program:
$$
\begin{aligned}
D=\max_{x,\lambda}\ &\|x-(\lambda A + (1-\lambda)B)\|\\
s.t.\ \ & x^TMx=1\\
& 0\leq \lambda\leq 1
\end{aligned}
$$
Or equivalently
$$
\begin{aligned}
D=\max_{x,\lambda}\ &(x-(\lambda A + (1-\lambda)B))^T(x-(\lambda A + (1-\lambda)B))\\
s.t.\ \ & x^TMx=1\\
& 0\leq \lambda\leq 1
\end{aligned}
$$
Note that if you ignore the constraint $0\leq \lambda\leq 1$ thus, the solution would be $D=\infty$ since points in the line $\lambda A + (1-\lambda)B$ can go all the way to infinity. So we can't ignore $0\leq \lambda\leq 1$. I'm saying this since if you were interested in the minimum distance instead of the maximum, you could ignore $0\leq \lambda\leq 1$, and thus use Lagrange multipliers to solve the previous problem and maybe find an explicit formula.

Now, in your case, at this point I see no other way but to solve this problem numerically. Due to the geometry of the problem, I think one may be able to show that there is a unique maximizer (but I don't have this clear right now). So using a numerical solver such as $\texttt{fmincon}$ from MATLAB should be able to do the work.

Now, note that my approach could be an overkill. Another contributor, more skilled in geometry than me, may find a clever argument to solve this problem more easily.


问:
Length of composition series and injective homomorphisms

Let $f:M\rightarrow N$ be an injective $R$-module homomorphism. Show that $l(M)\leq l(N)$ where $l(M)$ denote the number of nonzero submodules in a composition series of $M$.

My solution:
Since $f$ is injective, we have  Ker($f$)=$\{0\}$. Define $g:M \rightarrow \frac{M}{\text{Ker}(f)}$ by $g(x)=x+{0}$. Hence,$g$ is isomorphism. Therefore, $M$ is isomorphic to $\frac{M}{\text{Ker}(f)}$. Thus, $l(M)=l(\frac{M}{\text{Ker}(f)})$. By isomorphism theorem, we have $\frac{M}{\text{Ker}(f)}$ isomophic to Im$(f) \leq N$. Thus, $l(M)=l(\text{Im}(f))\leq l(N)$.

I'm not really sure about my answer. Is there any elementary and better solution? Thanks a lot.

答: Let $0=M_0\subsetneq \cdots \subsetneq M_n=M$ be a composition series for $M$. Since $f$ is an injective homomorphism, $f(M_i)$ is a submodule for each $M_i$ and strict inclusions are preserved, so $f(M_0)\subsetneq \cdots\subsetneq f(M_n)\subseteq N$, thus $l(N)\geq n=l(M)$.</p>


问:
Little Laurent Polynomial Identity
I am trying to prove the following statement. I don't have any assumptions on the the base field $\Bbbk$ yet, but I am happy to restrict this if required.

Statement: Let $p(h)\in\Bbbk[h^{\pm 1}]$. If $p(h)+p(h^{-1})=m(h)(h^{-1}-h),$ then $p(h)=g(h)(h^{-1}-h)$, where $g(h)\in\Bbbk[h^{\pm 1}]$.

In other words, if $h^{-1}-h$ is a factor of $p(h)+p(h^{-1})$, then $h^{-1}-h$ is a factor of $p(h)$.

For example, let $m(h)=(h^{-1}-h)$ then $m(h)(h^{-1}-h)=h^2-2+h^{-2}$. Rearranging this we get $h^2-1+h^{-2}-1$. So $p(h)=h^2-1=-h(h^{-1}-h).$

Notes

The equation $p(h)+p(h^{-1})=m(h)(h^{-1}-h)$ implies that $m(h)=-m(h^{-1})$. To see this, apply the map $h\mapsto h^{-1}$.

If $p(h)=g(h)(h^{-1}-h)$, then
$$g(h)(h^{-1}-h)+g(h^{-1})(h-h^{-1})=\left[g(h)-g(h^{-1})\right](h^{-1}-h).$$

答:
The statement is true (there might be some issues with fields of characteristic $2$, but the statement will be true for all other fields). Let $\mathbb{k}$ be a field of characteristic not equal to $2$. Let $p \in \mathbb{k}[X,X^{-1}]$ be an arbitrary Laurent polynomial, say
$$p(X) = \sum_{k=-d}^d a_kX^k.$$
Let us write
$$P(X) = p(X) + p(X^{-1}) = \sum_{k=-d}^d (a_k + a_{-k})X^k.$$
The condition
$$P(X) = p(X) + p(X^{-1}) = m(X)(X-X^{-1})$$
implies that $P(1) = 0$ and $P(-1) = 0$. Equivalently
$$\sum_{k=-d}^d(a_k + a_{-k}) = 2\sum_{k=-d}^da_k= 0,\qquad \text{and} \qquad \sum_{k=-d}^d(-1)^k(a_k + a_{-k}) = 2\sum_{k=-d}^d(-1)^ka_k= 0.$$
This is true if and only if $p(1) = p(-1) = 0$, which is true if and only if $p(X) = g(X)(X-X^{-1})$ for some $g \in \mathbb{k}[X,X^{-1}]$. To see why the latter statement is true, it's easier to pass to (regular) polynomials. Let $\tilde{p}$ be the associated polynomial given by $\tilde{p}(X) = X^dp(X)$. Then $p(1) = p(-1) = 0$ also implies that $\tilde{p}(1) = \tilde{p}(-1) = 0$. The polynomial remainder theorem then implies that $(X^2 -1)\mid \tilde{p}(X)$, i.e., there exists some polynomial $\tilde{g}(X)$ such that
$$\tilde{p}(X) = \tilde{g}(X)(X^2-1).$$
Dividing through by $X^d$ gives us
$$p(X) = g(X)(X-X^{-1}),$$
where $g(X) = X^{-(d-1)}\tilde{g}(X)$.

Title: Little Laurent Polynomial Identity


问:
Can I directly apply the theorem 1=(a/d,b/d) for this? Let a and b be two positive even integers. Prove that (a, b) = 2(a/2, b/2)
Let d=(a,b)

a and b are even integers

$→$ $2|a$ $and$ $2|b$

$→$ $2|ax + by$ ($x,y$ are some integers)

$→$ $2|d$

Then we can express,

$a =2p,$ $b=2q$, $d= 2k$ ($k,p,q$ are some integers)

then $d=(a,b)=(2p,2q)$

We know that $(a/d,b/d)=1$

Then,

$1 = (2p/2k,2q/2k)$

$1 = (p/k,q/k)$

$k = (p,q)$

$d/2 = (a/2,b/2)$ (since $k=d/2,p=a/2,q=b/2$)

$d = 2(a/2,b/2)$

My problem is, Can I directly apply the theorem $(a/d,b/d)=1$ to prove this?

Please, Can anyone offer some assistance, please?

答:
Yes, you can directly apply the theorem, but you don't need to. See this.

Here's a proof for the theorem:

Let $d=(a,b)$, then there exists $x,y\in\mathbb Z$ with $a=dx$ and $b=dy$.

We want to show that $(x,y)=1$. Suppose that $(x,y)=d'>1$. Then there exist integers $m,n$ with $x=d'm$ and $y=d'n$, but then we have $a=dx=dd'm$ and $b=dy=dd'n$, so $(a,b)>d$, which is a contradiction since $d=(a,b)$. Thus we have $(x,y)=1$.

In other words, since the greatest common divisor, $d$, is the largest positive integer that divides $a$ and $b$, if $(\frac{a}{d},\frac{b}{d})>1,$ then $d$ will not be the greatest common divisor of $a$ and $b$ because $\frac{a}{d}$ and $\frac{b}{d}$ have a common divisor bigger than $1$.


问:
proof of existence of partition
Let  be a partition of a set . Show that there exist a set  and a surjection :→ such that ={{∈∶()=}∶∈}.

How can I attempt this question?

答: Define $B$ to be the image of the bijection $g\colon \mathcal{C} \to B$ that maps each separate subset of $A$ to a distinct element of $B$. This is clearly well-defined, if by simple enumeration. Then define the function $f\colon A \to B$ as follows: $$\forall x\in A, f(x)=y \iff x\in g^{-1}(y).$$ Note that the inverse of $g$ is well-defined too, since it is a bijection. The proof that  $f$ is a surjection and the given condition is satisfied follows immediately.

Title: proof of existence of partition


问:
Applications of the Fibonacci sequence
The Fibonacci sequence is very well known, and is often explained with a story about how many rabbits there are after $n$ generations if they each produce a new pair every generation. Is there any other reason you would care about the Fibonacci sequence?

答: it pops up every now and then. For example, positions (F2n, F2n+1) are winning ones in Wythoff's Game.

Title: Applications of the Fibonacci sequence

ContentLicense: CC BY-SA 2.5


问:
Real life usage of Benford's Law
I recently discovered Benford's Law. I find it very fascinating. I'm wondering what are some of the real life uses of Benford's law. Specific examples would be great.</p>

答:
I recently attended a lecture on this. So I'll list a few interesting points that I remember.

The original discovery was by Newcomb who noticed that tables of logarithms were more worn out in the front than the back. It was rediscovered by Benford in 1938.

Areas of countries, areas of rivers and sizes of populations obey this rule. A more surprising example claims the numbers that appear in newspapers also follow this rule. In spite of the many observances of Benford's law, not a lot of instances are fully explained. I believe there have been proposals to provide simple models to associate an 'exponential-like' behavior of the areas of countries, but I don't really know anything about them.

Another note on the forensic accounting example mentioned by workmad3, I remember the lecturer pointed out that evidence pertaining to Benford's law is considered admissible in court.

There are things that don't really obey Benford's law, for example: the number of pages in books.

and that's more or less what I can recall from the talk, hope that helps :)

Title: Real life usage of Benford's Law


问:
Spectral sequences: equivalence of exact couples and classic (?) method
By the 'classic' method I mean the construction of the spectral sequence associated to a filtration as found in Weibel's book p. 133-134. There is also the method of construction through exact couples due to Massey. I would like to find a proof that these two methods are equivalent. To elaborate, in the classic method we define certain objects $A^{r}_{p,p+q}$ and then objects $Z^{r}_{p,p+q}$ and $B^{r}_{p,p+q}$ from $A^{r}_{p,p+q}$ to eventually construct the spectral sequence $E^r$. With exact couples, we avoid this step but we can then define $Z^{r}_{p,p+q}$ as those cycles that 'never die' and $B^{r}_{p,p+q}$ as those elements that eventually bound. How do we show that these two definitions of $Z^{r}_{p,p+q}$ and $B^{r}_{p,p+q}$ are equivalent? I tried to do it but got bogged down in an inductive nightmare. A reference would be great.</p>

答:
You can find this in McClearly's "A user's guide to spectral sequences". </p>

<p>It is in Chapter 2 (page 42 in my edition). It's not as bad as you would think! </p>

Body:
You can find this in McClearly's "A user's guide to spectral sequences".

It is in Chapter 2 (page 42 in my edition). It's not as bad as you would think!


答:
Sketch of proof:

Let $C=A_1 \cup A_2 \cup \dots \cup A_n$ be this partition (one doesn't need to have a finite partition, just to simplify)

Then you can construct $f$ such that $\forall x,y\in A, \exists i \in \{1,2,\dots,n\}, x,y \in A_i \iff f(x)=f(y)$

This means that if $x,y$ are from the same part (of the partition), you construct $f$ such that their images are equal.

Then $B$ is some set including these images... (here in the finite case $B=\{1,2,\dots,n\}$.

Title: proof of existence of partition

Body:
Let  be a partition of a set . Show that there exist a set  and a surjection :→ such that ={{∈∶()=}∶∈}.

How can I attempt this question?


问:
If $A\subseteq B\cup C$ then $A\subseteq B$ or $A\subseteq C$.
The question is to prove or disprove

If $A\subseteq B\cup C$ then $A\subseteq B$ or $A\subseteq C$.

I know that this is wrong and can easily be disproved with an example, but I tried to prove it and I actually came up with proof that I know is wrong but I don't know why it's wrong, I started by converting it to this:
$$x\in A\implies x\in B\cup C$$
$$x\notin A\lor x\in B\lor x\in C$$
$$(P\lor P\equiv A,A\lor B\equiv B\lor A)\text{ so}$$
$$x\notin A\lor x\in B\lor x\notin A\lor x\in C$$
$$(x\in A\implies x\in B)\lor(x\in A\implies x\in C)$$
$$A\subseteq B\lor A\subseteq C$$
and I want to know why this proof is wrong and where exactly is my mistake.

答:
Everything down to your fifth line is essentially correct (modulo the typo in line 4).

However, you're missing quantifiers. $A \subseteq B \cup C$ is not equivalent to $x \in A \implies x \in B \cup C$: the latter has an unbound $x$ in it, whereas the former doesn't. It is instead equivalent to $\forall x (x \in A \implies x \in B \cup C)$. Continuing that through to the end, we get, in line five, $\forall x ((x \in A \implies x \in B) \vee (x \in A \implies x \in C))$. That is not the same as $(\forall x (x \in A \implies x \in B)) \vee (\forall x (x \in A \implies x \in C)),$ which is what is equivalent to $A \subseteq B$ or $A \subseteq C$.

And, indeed, the result is false: let $B$ and $C$ be such that neither $B \subseteq C$ nor $C \subseteq B$. Then with $A := B \cup C$, we obviously have $A \subseteq B \cup C$, but we cannot have $A \subseteq B$ (else we'd have $C \subseteq B$) or $A \subseteq C$ (else $B \subseteq C$).


答:
Line 4 contains a typo ($B$ should be $C$). Your mistake is that you neglect quantifiers. Essentially, your mistake is that you used

$$\forall x: (P(x)  \lor Q(x))\implies (\forall x: P(x)) \lor (\forall x: Q(x))$$

But this is definitely not true.

Body:
The question is to prove or disprove

If $A\subseteq B\cup C$ then $A\subseteq B$ or $A\subseteq C$.

I know that this is wrong and can easily be disproved with an example, but I tried to prove it and I actually came up with proof that I know is wrong but I don't know why it's wrong, I started by converting it to this:
$$x\in A\implies x\in B\cup C$$
$$x\notin A\lor x\in B\lor x\in C$$
$$(P\lor P\equiv A,A\lor B\equiv B\lor A)\text{ so}$$
$$x\notin A\lor x\in B\lor x\notin A\lor x\in C$$
$$(x\in A\implies x\in B)\lor(x\in A\implies x\in C)$$
$$A\subseteq B\lor A\subseteq C$$
and I want to know why this proof is wrong and where exactly is my mistake.


问:
Show that $ \sqrt[3]{(z-1)(z-2)(z-3)} $ can be defined on the entire plane minus [1, 3].
My professor gave me the above problem as test preparation, and I am feeling a bit lost at how to solve it. I looked up the answer in the back of my book, and I didn't really understand the logic behind it. I was hoping someone could shed some insight into the problem solving process since I will have a similar, but not identical, question on my test. Without further ado, here is the solution:

Let $$ f(z) = \sqrt[3]{(z-1)(z-2)(z-3)} = \text{exp}(\frac{1}{3}\text{log}[(z-1)(z-2)(z-3)]) $$

where

$$ \text{log}((z-1)(z-2)(z-3)) = \int_4^z{\frac{[(z-1)(z-2)(z-3)]'}{(z-1)(z-2)(z-3)}dz} + \text{log} (6) \text{, for } z \in \mathbb{C} \backslash (-\infty, 3]$$

Then, it can be shown that $ f(z) $ defines a function that is continuous at all points $ x \in \mathbb{R} \text{ such that } x \lt 1 $.

There are many things I don't understand about this, mostly related to the choice of logarithm. Why is the integral from 4 to z? Why do you add log(6)? Why is it defined on $ \mathbb{C} \backslash (-\infty, 3] $? How can I figure out how to solve problems like this in similar cases?

Furthermore, what is the easiest way to show continuity on the real axis where x < 1? Thanks for the help.

答: Note that, the function has three branch points, namely $z=1,2,3$. So, in order to make the function a single-valued function you need to take a branch cut in the complex plane. In this case, we can take the interval $[1,3]$ as branch cut. This will make the function defined on $\mathbb{C}\backslash[1,3]$.  </p>


问:
Hamming cube has two minimal cuts $S_1, S_2$ such that $E = S_1 \cup S_2$
Consider the d-dimensional Hamming cube $H_d$.

Show: There are two minimal cuts $S_1, S_2$ of $H_d$ such that $E = S_1 \cup S_2$.

Hint: Consider the case where $d=2$ and construct two minimal cuts for $H_3$. Then generalize the construction for all $d \ge 2$.

Thanks for your ideas/thoughts..

答:
Every Hammingcube $H_d$ is obviously consisting in two Hammingcubes $H_{d-1}$ (copies, as it were), where the copied vertices are connected by an edge...

A cut is by definition given, when the set of vertices can be partitioned in two sets in such a way that the edges lie in between the two sets.

So, suppose one can partition a given d-dimensional Hammingcube in the described fashion. Then appearently one can construct a $d+1$-dimensional Hammingcube $H_{d+1}$ like so:

copy $H_d = H_d'$

invert the vertex colouring in $H_d'$

connect every vertex in $H_d$ to its copy in $H_d'$ by an edge


问:
primitive root of 1 in algebraically closed set
Let $K$ be a field, and character is $p$.
If $p$ does not divide $n$, primitive $n$-th root of unity exists in the algebraically closed extension of $K$.

I want to find example, where $p$ devides $n$,there is no unity in the algebraically closed extension of K, thank you.

答:
If $K$ is any field of characteristic $p$ then the following equality holds in $\overline{K}[x]$:

$x^p-1=(x-1)^p$

And so $1$ is the only root of unity of order $p$. So there are no primitive roots of order $p$ in $\overline{K}$.


问:
Random Shuffle of Groups
Let's suppose that we have 54 peoples and we arrange them into 9 groups of equal size, so this means that each group will have 6 persons in it. I want to find a procedure, such that the groups are shuffled deterministic in order that people will not meet several times in a group. I couldn't come of with an idea so far.

Since each person will meet then 5 new people in each group, this means that we can shuffle the groups up to 10 times. So I will decrease the complexity of this question to come up with a mix such that we will build 10 times new groups. Adding a random element in the shuffling would be an extra!

答:
Note that $54 = 2 \times 3^3$. This can be visualized as stacking two Rubik's cubes on top of each other. Just imagine that the cubes can be rearranged anywhere within a layer including the central position.

Let us say the top layer of cubes represents 9 individuals who are the group leaders and they remain at their assigned positions throughout the shuffles. Let other 5 layers of cubes represent other members who can move around to other groups under other leaders. Note that there is no need for them to move across layers, because doing so will pair them up with the same leader (who is situated vertically on the top layer). It is easy to see this holds for all layers. A group at any time instant can be visualized as a single column of vertically stacked cubes in this formation.

We can assign positions $0$-$7$ for each of the boundary (non-central) cubes in a layer and the position $9$ to the central cube in the layer. In each shuffle, the cubes are rearranged by incrementing their position modulo $9$ by a certain number $c_k$ using the relation:

$$pos_k(n) = pos_k(n-1) + c_k \mod 9$$

where, $pos_k(n)$ is the new position for a cell previously in position $n-1$ in a layer $k$.

$c_k \in \{ 1, 2, 5, 7\}$ only because if we choose $3, 6, 9$, there will be shorter periods modulo $9$ and if we choose $4, 8$, there will be a duplication with $2$. Any number greater than $9$ reduces to one of these cases modulo $9$.

But we have 5 layers under the top layer and we cannot choose a value for $c_k$ for one of the layers. Therefore, a duplicate pairing will occur in the first shuffle itself. Moving the leaders to different positions doesn't solve this problem.

So, to answer the question: 10 shuffles are not possible with 54 people without duplicate pairings. After the initial arrangement, not even 1 shuffle is possible.

On the question of replacement: Replacing an entire layer of 9 people in each shuffle with new people will allow 4 layers to take on different positions with the last layer filled with all new people in each shuffle that permits grouping without any duplicate pairing.

Title: Random Shuffle of Groups


问:
Show that the Boolean algebra of finite & cofinite subsets of $U$ is not complete

If $U$ is infinite, show that the Boolean algebra of subsets of $U$ that are finite or cofinite (i.e. their complement is finite) is not complete. A Boolean algebra $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is said to be complete if every non-empty subset of $B$ has a greatest lower bound (g.l.b).

Let $U'=\{X\subseteq U: X\text{ is finite, or}X^c=U\backslash X\text{ is finite}\}$. I think the Boolean algebra of concern here is $\mathcal{B}:= (U',\subseteq, \cup,\cap,^c,\phi, U)$, right?

From what I understand, the g.l.b operation here is just $\bigcap$, i.e. if $\phi\neq U''\subseteq U'$, then g.l.b($U''$) = $\bigcap_{z\in U''}z$, is that correct?

If that is the case, then the Boolean algebra looks complete* to me - which is contrary to what is required to be proved. Where am I going wrong?

*Looks complete because two sets will always have some intersection, empty or not!

答:
It is closed under finite intersections, but completeness requires closure under arbitrary intersections. Let $X_i = U \setminus \{i\}$, where $i$ ranges over some infinite subset $I$ of $U$ such that $U \setminus I$ is also infinite (e.g., if $U = \Bbb{R}$, you could take $I = \Bbb{Z}$). Then $\bigcap_{i \in I} X_i = U \setminus I$ which is not in $U'$.

More formally, the family $\{X_i \mid i \in I\}$ has no g.l.b. because any finite (resp. cofinite) set $A$ such that $A \subseteq X_i$ for all $i$ must be a proper subset of $\bigcap_{i \in I} X_i = U \setminus I$ and hence can be extended to give a finite (resp. cofinite) set $B \supset A$ such that $B \subseteq X_i$ for all $i$.


问:
Line intersecting three (or four) given lines
How can I find a fourth line $L$ that intersects three given lines $L_1$, $L_2$, $L_3$ in 3D space?

We can assume that $L_1$, $L_2$, $L_3$ are in "general position", so no two of them are coplanar, etc.

I'm not even sure that three lines is enough to uniquely define $L$, actually. If three lines is not enough, how many do I need?

The question is related to this one. Specifically, see idea #4 in my list of suggested approaches. It requires finding a line that intersects with a few given ones.

Edit:

Apparently, I need four lines, not three, to uniquely define $L$. So, how can I construct a fifth line that interesects four given ones?

I found this paper, and this one, but they are both difficult for me to read. Surely there must have been solutions before 2008, and, if so, I'm hoping that these are easier to understand.

答:
To build up on Ted Shifrin's comment, assuming you have four skew lines $L_1$, $L_2$, $L_3$ and $L_4$, a costructive and an easy way to find a fifth line that intersects all the four lines would be as follows:

Construct two ruled surfaces (hyperboloids or hyperbolic paraboloids) uniquely defined by $L_1$, $L_2$, $L_3$ and $L_1$, $L_2$, $L_4$ that form the reguli of the quadrics $H_{123}$ and $H_{124}$. Find out the intersection between those two quadrics which definitely contains $L_1$ and $L_2$. Along with these two lines, the intersection must also contain two other lines (from the complementary reguli of the quadrics), say $M$ and $N$ that intersect all four given lines. Here's an example:
$$L_1=(-3,-5,-2)+ t(1,0,0)$$
$$L_2=(-3,-5,2)+ t(0,1,0)$$
$$L_3=(3,5,0)+ t(0,0,1)$$
$$L_4=(3,10,0)+ t(1,0,1)$$
$$H_{123}=4\,xy-10\,xz+6\,yz-60$$
$$H_{124}=4\,xy-15\,xz+4\,yz+15\,{z}^{2}-10\,x+4\,y-25\,z-130$$
The intersection of these two quadrics yields 4 lines (I did it by finding the Gröbner basis of the ideal $\langle H_{123}, H_{124} \rangle$ using lexicographic ordering and factoring the first polynomial, there are other ways to do it and almost all computational environments have a Gröbner basis command if you want to check it out):
$$L_1: z+2=y+5=0 $$
$$L_2: -z+2=x+3=0 $$
$$M: x+ \left( -\frac {1}{10}\,\sqrt {73}-{\frac {7}{10}} \right) y+\frac {1}{2}\,\sqrt {73}+
\frac {1}{2}= 5\,z\sqrt {73}+10\,\sqrt {73}+8\,y-55\,Z-70=0 $$
$$N: x+ \left( \frac {1}{10}\,\sqrt {73}-{\frac {7}{10}} \right) y-\frac {1}{2}\,\sqrt {73}+\frac {1}{2}= 5\,z\sqrt {73}+10\,\sqrt {73}-8\,y+55\,Z+70=0 $$
where $M$ and $N$ intersect $L_i, i=1,2,3,4$.


问:
Let $f : {\bf R}^2 \to {\bf R}^2$ and $f(x,y) = (x^2+2xy+y^2, 2x + 2y).$ Determine if $f$ is locally injective.

Let $f : {\bf R}^2  \to {\bf R}^2$ and $f(x,y) = (x^2+2xy+y^2, 2x + 2y).$ Determine if $f$ is locally injective.

Noting first that $f$ is $C^1$ and computing the Jacobi one has that $$J_f(x,y) = \begin{bmatrix} 2x+2y && 2y+ 2x \\ 2 && 2\end{bmatrix}.$$

Thus the determinant $\det(J_f(x)) = 0$. However, this only means that the Jacobian matrix isn't invertible. The function might still be injective.

Any hints on how should I go about this and determine if the function is actually injective or not?

答: It is not an injective function. If you assume $f(a,b) = f(c,d)$ for some real numbers $a,b,c,d$ then you only get the equality $a+b=c+d$, which is not guaranteed to be $a = c$ and $b = d$. As mentioned above, taking $(x,y)$ and $(x-\epsilon, y + \epsilon)$ is one of the simplest example.


问:
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$
I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$

Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.

答:
I have one different idea, variation of constant.

If one has a recurrence relation of the form
$$ a_{n+1} = \alpha_n a_n + \beta_n,\quad n\geq 0, $$
then we can make the ansatz
$$ a_n^{(p)} = C_n \prod_{j=}^{n-1} \alpha_j $$
and get
$$ C_{n+1} \prod_{j=0}^n \alpha_j = \alpha_n C_n \prod_{j=0}^{n-1} \alpha_j + \beta_n $$
$$ \Longrightarrow C_{n+1} = C_n + \frac{\beta_n}{\prod_{j=0}^n \alpha_j} $$
$$ \Longrightarrow C_n = \sum_{l=0}^{n-1} \frac{\beta_l}{\prod_{j=0}^l \alpha_j} + C_0$$
and therefore
$$ a_n = \sum_{l=0}^{n-1} \frac{\beta_l}{\prod_{j=0}^l \alpha_j} \prod_{j=0}^{n-1} \alpha_j + C \prod_{j=0}^{n-1} \alpha_j = \sum_{l=0}^{n-1} \frac{12 H_{l+1}}{\prod_{j=0}^l \frac{j-3}{j+1}} \prod_{j=0}^{n-1} \frac{j-3}{j+1} + C \prod_{j=0}^{n-1} \frac{j-3}{j+1} \\= \sum_{l=0}^{n-1} 12 H_{l+1} \prod_{j=l+1}^{n-1} \frac{j-3}{j+1} + C \prod_{j=0}^{n-1} \frac{j-3}{j+1},$$
if we define
$$ \alpha_n = \frac{n-3}{n+1},\qquad \beta_n = 12 H_{n+1}.$$
I do not know how to continue from here, maybe someone knows how to simplify this terms to get to the solution?


问:
Ordering balls in a line. probability of the event $\{\text{ no ball in the first 4 balls is blue }\}$
given $8$ red balls, $5$ blue balls, $9$ green balls. assume all the ordering options have the equal probability. what is the probability of the event $$\{\text{ no ball in the first 4 balls is blue }\}$$

My attempt:

My probability space is $$\Omega = \binom{17}{4}$$ this is because I have to choose 4 balls without the $5$ blues. But I am not sure if this is true because it can be treated "choose 4 balls without 5 reds".

In addition to that, I need to determine the power of my event group wich is pretty hard for me to understand how should I do it.

I hope some one can help

答:
Method 1: We find the probability that none of the first four balls selected are blue.

There are $8 + 5 + 9 = 22$ balls.  You can select four of them in
$$\binom{22}{4}$$
ways.  Of these $22$ balls, $22 - 5 = 17$ are not blue.  Thus, the number of ways you can select four balls so that none of them are blue is
$$\binom{17}{4}$$
Therefore, the probability that no ball in the first four balls is blue is
$$\frac{\dbinom{17}{4}}{\dbinom{22}{4}}$$

Method 2:  We calculate the probability that there are no blue balls in the first four positions of an arrangement of the balls in a line.

If we instead took our sample space to be all possible arrangements of $8$ red, $5$ blue, and $9$ green balls, we would have to choose $8$ positions for the red balls, $5$ of the remaining $14$ positions for the blue balls, and all $9$ of the remaining positions for the green balls, giving
$$\binom{22}{8}\binom{14}{5} = \frac{22!}{8!14!} \cdot \frac{14!}{5!9!} = \frac{22!}{8!5!9!}$$
possible arrangements, where the factors in the bottom represent, respectively, the number of ways we could permute the eight red balls among themselves, the five blue balls among themselves, and nine green balls among themselves since permuting balls of the same color among does not produce an arrangement distinguishable from the given arrangement.

As for the favorable cases, we must choose five of the last $22 - 4 = 18$ positions for the blue balls since none of them can appear in the first four positions.  Once we have done so, we must choose which eight of the remaining $222 - 5 = 17$ positions will be filled with red balls.  All of the remaining nine positions must be filled with green balls.  Hence, the number of favorable cases is
$$\binom{18}{5}\binom{17}{8}$$
Thus, the desired probability is
$$\frac{\dbinom{18}{5}\dbinom{17}{8}}{\dbinom{22}{8}\dbinom{14}{5}}$$

As you can verify, the two methods yield the same answer.


问:
How many numbers from $1$ to $99999$ have a digit-sum of $8$?
How many numbers from $1$ to $99999$ have a digit-sum of $8$?

Why is the answer ${8+4\choose 4}$?

Does the following method work?

Answer is the number of ways to split 8 into 5 digits,

i.e. number of ways to insert 4 lines among a row of 8 objects.

答:
Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5} \geq 0$ be the 5 digits of the number. Then the number of numbers from 1 to 99999 that have a digit sum of 8 are the integer solutions to the equation $x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 8$. This problem is combinatorially identical to the number of distributions of $8$ indistinguishable objects to $5$ distinguishable people, hence, ${8 + 4 \choose 4}$. For more on why that is, see my answer here:

Combinatorics Distribution - Number of integer solutions Concept Explanation


问:
How to multiply a vector and a square matrix with Kronecker product, and know the answer's shape?

$\mathbf{1}_n \in \mathbb{I}^{n\times 1}$ is a vector of ones with shape $n\times 1$

$\mathbf{I}_m \in \mathbb{I}^{m\times m}$ is an identity matrix with shape $m\times m$

What is the answer to, and shape of, $\mathbf{1}_n^\top \otimes  \mathbf{I}_m$ ?

$$\mathbf{1}_n^\top \otimes  \mathbf{I}_m  = \overset{(1\times n)}{\begin{pmatrix}
1 & 1 \dots 1
\end{pmatrix}}
\otimes
\overset{(m\times m)}{\begin{pmatrix}
1 & 0 & \dots & 0 \\
0 & 1 & \dots & 0 \\
\vdots &\vdots &  \ddots & \vdots \\
0 & 0 & \dots & 1
\end{pmatrix}}
=?
$$

答:
It's $$\underbrace{[I_m|I_m|\cdots|I_m]}_{n \ \text{times}}$$

because is $A$ is $p \times q$ and $B$ is $m \times m$:

$$A\otimes B=\begin{pmatrix}a_{11}B&\cdots&a_{1q}B\\ \vdots&&\vdots\\
a_{p1}B&\cdots&a_{pq}B\end{pmatrix}\tag{1}$$

Therefore the shape of $\mathbf{1}_n^\top \otimes  \mathbf{I}_m$ is $m \times mn$.


问:
How to find distance between numbers in a finite set that loops?
If we take a finite set of numbers and make them cycle at a specific threshold. For example with 5 as the threshold:

$1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \rightarrow etc...$

How can we find the distance between any 2 elements, considering that they form a cycle? So on that example:

$distance(1, 1) = 0$

$distance(1, 2) = 1$

$distance(1, 3) = 2$

$distance(1, 4) = 2$

$distance(1, 5) = 1$

Sorry for the messy explanation. I lack formal math education, and I'm just trying to figure out an code algorithm for a side project I'm developing. Thanks.

答: Let d  = $|a-b|$, the absolute value of the difference between them. Then the distance is the smaller of $d$ and $5-d$.


答:
Yes, the reasoning below the line in your question is correct, though it can be expanded for greater clarity. Lay out a row of $8$ stars, say:

$$********$$

Now insert $4$ dividers to break them up into $5$ groups, e.g.,

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

From left to right read off the number of stars in each of the $5$ groups: $$10304$$

The result is clearly a number between $1$ and $99999$ whose digits sum to $8$. And the procedure is clearly reversible, so the number of ways of inserting the $4$ dividers really is the number of integers in which we’re interested. For example, starting with $352=00352$, we get

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

The string of stars and dividers is a string of $8+4=12$ objects, and the $4$ dividers can go anywhere in this string, so there are $\binom{12}4$ ways to place them and therefore $\binom{12}4$ numbers of the desired kind.

Body:
How many numbers from $1$ to $99999$ have a digit-sum of $8$?

Why is the answer ${8+4\choose 4}$?

Does the following method work?

Answer is the number of ways to split 8 into 5 digits,

i.e. number of ways to insert 4 lines among a row of 8 objects.


答:
I think what you want is this: if the threshold is $n$, then for $a \le b$:
$$
d(a, b) = \left\{
\begin{array}{l{\quad}l}
b - a &\mbox{if $b - a\le n/2$}\\
a + n - b &\mbox{if $b - a > n/2$}
\end{array} \right.
$$

and for $a > b$, $d(a, b) = d(b, a)$.

I.e., in the case when $a \le b$, you go up from $a$ to $b$ unless going up from $b$ to $a + n$ would be shorter.

Title: How to find distance between numbers in a finite set that loops?

Body:
If we take a finite set of numbers and make them cycle at a specific threshold. For example with 5 as the threshold:

$1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 1 \rightarrow etc...$

How can we find the distance between any 2 elements, considering that they form a cycle? So on that example:

$distance(1, 1) = 0$

$distance(1, 2) = 1$

$distance(1, 3) = 2$

$distance(1, 4) = 2$

$distance(1, 5) = 1$

Sorry for the messy explanation. I lack formal math education, and I'm just trying to figure out an code algorithm for a side project I'm developing. Thanks.


问:
Lagrange Interpolation Polynomials and functions of Matrices
Say we are given a matrix $A$ over the field $\mathbb{C}$ with minimal polynomial $\mu_A(\lambda)=\lambda^2(\lambda-1)^2$ so we know $A \in \mathbb{C}^{m \times m}$ where $m \in \mathbb{N}_{\geq 4}$. Then if we were to find a formula for $A^n$, we let $h(z)=z^n$, we want to find a degree 3 polynomial

$$
g(z) = \alpha_0 + \alpha_1\cdot z + \alpha_2\cdot z^2 + \alpha_3\cdot z^3.
$$
We known that $g(0) = h(0), \;g'(0) = h'(0),\; g(1) = h(1),\; g'(1) = h'(1)$ (from our minimal polynomial). This gives $g(z) = (3-m)z^2 + (m-2)z^3$. Does this mean that
$$
A^m = (3-m)\cdot A^2 + (m-2)\cdot A^3, \; m\in \mathbb{N}_{\geq 4}.
$$
Does this formula work, I'm unsure of the logic behind Lagrange Interpolations.

Just to make sure my computation is correct we get 4 equations
$$
1=\alpha_0 + \alpha_1 + \alpha_2 + \alpha_3 \\
m= \alpha_1 + 2\alpha_2 + 3\alpha_3 \\
0 = \alpha_0\\
0 = \alpha_1
$$
thus $a_3 = m-2, \; a_2 = 3-m$.

答:
You should have made an error in your computations as $\mu_A(z) = z^4-2z^3+z^2$ and therefore

$$z^4= 1 \cdot \mu(z) + 2 z^3 - z^2$$ which doesn't match

$$(4-m)\cdot z^2 + (m-3)\cdot z^3$$ for $m=4$.

Your error is that $h_m^\prime(z) = \left(z^m\right)^\prime = m z^{m-1}$. So $m= \alpha_1 + 2\alpha_2 + 3\alpha_3$ and not $m-1= \alpha_1 + 2\alpha_2 + 3\alpha_3$.

And your overall logic is correct. You are doing the Euclidean division of $h^m$ by $\mu_a$, i.e.

$$h_m = q_m \mu_A + r_m$$ with $\deg r_m <\deg \mu_A$.

As $\mu_A(0)=\mu_A(1)=\mu_A^\prime(0)=\mu_A^\prime(1)=0$, the equations you used are correct to find $r_m(z) = (m-2) z^3 + (3-m) z^2$.

Note: as $h$ ang $g$ depend on $m$, better to use indices to name them.


问:
Induced homomorphism in homology between Moore spaces.
I am searching for the homomorphism induced $x_{*}: H_{2}(M) \rightarrow H_{2}(N)$ by the maps given in the following proposition:

My questions are:

1-I have found this question Moore space, induced map in homology here on this site, does that means that the induced homomorphism I am searching for does not exist?

2- Also, is it correct to describe $x$ as a map(the author of the paper did this)?If not, how can I describe $x$ correctly? one of my professors told me that $x$ is not a map it is a universal quantifier.

EDIT:

**I want to understand in general what is the case in which we can say there is no induced homomorphisms between more spaces, could anyone help me answer this question please? **

答:
If $X$ is a co-H-group and $\alpha,\beta:X\rightarrow Y$ are (homotopy classes of) maps then the sum $\alpha+\beta:X\rightarrow Y$ is defined with the co-H-structure on $X$ and the induced map in homology satisfies
$$(\alpha+\beta)_*=\alpha_*+\beta_*.$$
The element $r\cdot \alpha$ for $r\in\mathbb{Z}$ is obtained as the composite
$$r\cdot\alpha=\alpha\circ\underline r$$
where $\underline r\in[X,X]$ is $r$-times $id_X$. Since the map $\underline r$ induces multiplication by $r$ on $H_*X$ we get
$$(r\cdot \alpha)_*=r\cdot \alpha_*.$$

Having covered the relevant background lets address your question. I'm going to have to guess at what the notation means since you haven't included any details for me to go by. Thus I'm assuming that $M=M(\mathbb{Z}_m,2)$ is the fixed space $S^2\cup_{\underline m}e^3$ where $\underline m:S^2\rightarrow S^2$ is as above, that $i':S^2\hookrightarrow M$ is the inclusion of the bottom cell, that $q:M\rightarrow S^3$ pinches to the top cell, and that $\phi\in\pi_3S^2$ is the Hopf map.

Then from the discussion above we get
$$(r\lambda+k(i'\phi q))_*=(r\lambda)_*+(k(i'\phi q))_*=r\cdot \lambda_*+k\cdot(i'\phi q)_*=r\cdot \lambda_*$$
since $\phi_*$ is zero for degree reasons.


问:
Let $X$ be a compact Kaehler manifold. Is the group of holomorphic isometries of $X$ compact?
Let $X$ be a compact Kaehler manifold. Then it was shown by Bochner and Montgomery that, with respect to the compact open topology, $Aut_{\mathbb{C}}(X)$, the group of complex automorphisms of $X$, can be endowed with a structure of complex Lie group. Is it true that the group of holomorphic isometries $Iso_{\mathbb{C}}(X)$ is a compact subgroup of $Aut_{\mathbb{C}}X$?

答:
A holomorphic map is an isometry iff it preserves the geodesic distance (the latter implying, as $X$ is compact – that’s not entirely trivial – bijectivity).

Now, the set of geodesic distance-preserving maps $X \rightarrow X$ is compact by Ascoli, and the set of holomorphic maps $X \rightarrow X$ is closed in $C(X,X)$ (for the compact open topology).

So their intersection is compact, and it is $Iso_{\mathbb{C}}(X)$.


问:
Order of convergence of Fixed Point Iteration
I have to determine the order of convergence of this iteration specification but I don't know how to do this in $\mathbb{R}^n$. (So far, I've only done this in $\mathbb{R}$).
Any help is highly appreciated (and sorry for my bad english).

$x^{(k+1)} = \begin{pmatrix}
0 & -1/2 & -1/4\\
-1/2 & 0 & 0\\
0 & -1/2 & 0
\end {pmatrix}x^{(k)}+\begin{pmatrix} 3/4 \\ -2 \\ -1/2 \end{pmatrix}$
for $k \geq 0$

$x^{(k)} = \begin{pmatrix} x_1^{(k)} \\ x_2^{(k)} \\ x_3^{(k)} \end{pmatrix}$

答: It works the same in $\Bbb R^n$ as in $\Bbb R$.  Find the limit, which will be a vector.  Add a different epsilon to each component of the vector, apply the iteration, and see how much closer to the limit the new value is.  You will have three orders of convergence, one for each component.  Report the poorest as the overall value.


问:
Let $\vec{AA_1} + \vec{BB_1} + \vec{CC_1} = 0$. Prove that the triangle is equilateral.
Let $A A_1, BB_1, CC_1$ be the angle bisectors   of triangle $ABC$ and $\vec{AA_1}+\vec{BB_1}+\vec{CC_1}=0.$ Prove that $ABC$ is equilateral.

I know  that  the vectors  $\vec{AA_1}$ and $\frac{\vec{AB}}{|AB|}+\frac{\vec{AC}}{|AC|}$ ( and so on) are collinear  but not find how it can help.

Any ideas?

答: If we write this equation with position vectors we get $${1\over 3}(\vec{A}+\vec{B}+\vec{C}) = {1\over 3}(\vec{A}_1+\vec{B}_1+\vec{C}_1)$$ which means that triangles $ABC$ and $A_1B_1C_1$ share the same gravity center. But this means that every angle bisector is also a median which means $\triangle AA_1B \cong \triangle AA_1C$. But this means that angles at $B$ and $C$ are the same and further as angle at $A$ and we are done.